Re: QFT, divergent power series, and all that... [Archive]

Charles Francis

Apr7-04, 08:56 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <e7f834be.0404020428.14843d78@posting.google.co m>, Frank\nHellmann <C.i.m@gmx.net> writes\n>\n>How should a poor student make sense of this? We start with a\n>lagrangian, apply some more or less arbitrary prescriptions to get a\n>qunatum theory, which we can not solve, and then when we develop it\n>the predictions are divergent, but the lowest order terms are in\n>excellent agreement with experiment. WTF?\n>\n>Can anybody shed some light on this?\n\nScharf sheds, Finite QED, some light on quantisation. The arbitrary\nprescriptions for producing operators are not so bad if you work through\nFock space and construct them.\n>\n>To ask some direct questions:\n>Are those powerseries divergent in the sense that they run of to\n>infinity, or non convergent but finite (like summing over an infininte\n>series of -2 and +2, evidently bounded yet not convergent)?\n\nThey are divergent.\n\n>What is one to make of this divergence physically?\n\nThere isn\'t a generally accepted answer. My own view is that qed breaks\ndown on very small time/distance scales, and as a result the series\nbecomes less and less accurate for higher order terms.\n\n\n\n\nRegards\n\n--\nCharles Francis\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <e7f834be.0404020428.14843d78@posting.google.com>, Frank
Hellmann <C.i.m@gmx.net> writes
>
>How should a poor student make sense of this? We start with a
>lagrangian, apply some more or less arbitrary prescriptions to get a
>qunatum theory, which we can not solve, and then when we develop it
>the predictions are divergent, but the lowest order terms are in
>excellent agreement with experiment. WTF?
>
>Can anybody shed some light on this?

Scharf sheds, Finite QED, some light on quantisation. The arbitrary
prescriptions for producing operators are not so bad if you work through
Fock space and construct them.
>
>To ask some direct questions:
>Are those powerseries divergent in the sense that they run of to
>infinity, or non convergent but finite (like summing over an infininte
>series of -2 and +2, evidently bounded yet not convergent)?

They are divergent.

>What is one to make of this divergence physically?

There isn't a generally accepted answer. My own view is that qed breaks
down on very small time/distance scales, and as a result the series
becomes less and less accurate for higher order terms.

Regards

--
Charles Francis

Italo Vecchi

Apr7-04, 09:27 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4sapf\\$c99\\$1@lfa222122.richmond.edu>...\ n> In article <e7f834be.0404020428.14843d78@posting.google.com >, Frank\n> Hellmann <C.i.m@gmx.net> writes\n......\n>\n> >What is one to make of this divergence physically?\n>\n> There isn\'t a generally accepted answer. My own view is that qed breaks\n> down on very small time/distance scales, and as a result the series\n> becomes less and less accurate for higher order terms.\n>\n>\nIn his "Dialog on quantum gravity" Rovelli states that "the weak field\napproximation fails for GR" because it "is based on Feynman integrals\nthat sum over infinite momenta, namely over regions of arbitrarily\nsmall volume". A result of QLG is that "there is literally no volume\nsmaller than the Planck volume" so that "it makes no sense to\nintegrate over degrees of freedom far smaller than the Planck length"\nsince such regions "literally do not exist".\n\nNow, according to your view above, if "there is literally no volume\nsmaller than the Planck volume" that would impact divergences in QFT\ntoo. Right?\n\nRegards,\n\nIV\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4sapf$c99$1@lfa222122.richmond.edu>...
> In article <e7f834be.0404020428.14843d78@posting.google.com>, Frank
> Hellmann <C.i.m@gmx.net> writes
......
>
> >What is one to make of this divergence physically?
>
> There isn't a generally accepted answer. My own view is that qed breaks
> down on very small time/distance scales, and as a result the series
> becomes less and less accurate for higher order terms.
>
>
In his "Dialog on quantum gravity" Rovelli states that "the weak field
approximation fails for GR" because it "is based on Feynman integrals
that sum over infinite momenta, namely over regions of arbitrarily
small volume". A result of QLG is that "there is literally no volume
smaller than the Planck volume" so that "it makes no sense to
integrate over degrees of freedom far smaller than the Planck length"
since such regions "literally do not exist".

Now, according to your view above, if "there is literally no volume
smaller than the Planck volume" that would impact divergences in QFT
too. Right?

Regards,

IV

Arnold Neumaier

Apr8-04, 02:27 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Frank Hellmann wrote:\n> Ok, here is a poor undergrad student struggling with his first courses\n> on QFT... Overall an frustrating experience. After the beautifully\n> concise and mathematically (or at least formulistically) precise\n> treatment of mechanics and quantum mechanics (particularly in the\n> books of Landau Lifschitz and Dirac respectively) this appears\n> horribly complicated and obscure (I\'m mostly working with Weinbergs\n> book, any other recommendations?).\n>\n> One particular aspect is really driving me mad though, particularly\n> because none of the books I\'m reading appears to have much to say on\n> it.\n>\n> It appears that the pertubation series does generally not converge.\n> Now my prof in advanced stat phys had to say something on this, but it\n> was very hand wavy and included various incantations of a mystic\n> concept called Borel summation.\n>\n> How should a poor student make sense of this? We start with a\n> lagrangian, apply some more or less arbitrary prescriptions to get a\n> qunatum theory, which we can not solve, and then when we develop it\n> the predictions are divergent, but the lowest order terms are in\n> excellent agreement with experiment. WTF?\n>\n> Can anybody shed some light on this?\n\n\n> What is one to make of this divergence physically?\n> What is Borel summation? (I have found applications and simple\n> definitions on the web, and in earlier threads but it remains\n> mysterious to me, particularly it is said that the original value of\n> the sum can be extracted from the Borel sum, how would one even define\n> the "original value of the sum" if this sum is divergent?)\n\nSee\nhttp://www.lns.cornell.edu/spr/2003-05/msg0051062.html\nand context.\n\n\n> How is it\'s use justified in QFT? (in particular, is it just a\n> mathematical operation to get something out or is there physics\n> involved at that step)\n\nIn certain cases, where nonperturbative QM applies, one can show that\nthe nonperturbative result satisfies the properties needed to show\nthat Borel summation of the perturbative expansion reproduces the\nnonperturbative result.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann wrote:
> Ok, here is a poor undergrad student struggling with his first courses
> on QFT... Overall an frustrating experience. After the beautifully
> concise and mathematically (or at least formulistically) precise
> treatment of mechanics and quantum mechanics (particularly in the
> books of Landau Lifschitz and Dirac respectively) this appears
> horribly complicated and obscure (I'm mostly working with Weinbergs
> book, any other recommendations?).
>
> One particular aspect is really driving me mad though, particularly
> because none of the books I'm reading appears to have much to say on
> it.
>
> It appears that the pertubation series does generally not converge.
> Now my prof in advanced stat phys had to say something on this, but it
> was very hand wavy and included various incantations of a mystic
> concept called Borel summation.
>
> How should a poor student make sense of this? We start with a
> lagrangian, apply some more or less arbitrary prescriptions to get a
> qunatum theory, which we can not solve, and then when we develop it
> the predictions are divergent, but the lowest order terms are in
> excellent agreement with experiment. WTF?
>
> Can anybody shed some light on this?

> What is one to make of this divergence physically?
> What is Borel summation? (I have found applications and simple
> definitions on the web, and in earlier threads but it remains
> mysterious to me, particularly it is said that the original value of
> the sum can be extracted from the Borel sum, how would one even define
> the "original value of the sum" if this sum is divergent?)

See
http://www.lns.cornell.edu/spr/2003-05/msg0051062.html
and context.

> How is it's use justified in QFT? (in particular, is it just a
> mathematical operation to get something out or is there physics
> involved at that step)

In certain cases, where nonperturbative QM applies, one can show that
the nonperturbative result satisfies the properties needed to show
that Borel summation of the perturbative expansion reproduces the
nonperturbative result.

Arnold Neumaier

Charles Francis

Apr8-04, 06:32 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <61789046.0404061053.6a1a619d@posting.google.com >, Italo\nVecchi <vecchi@weirdtech.com> writes\n>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4sap\n>f\\$c99\\$1@lfa222122.richmond.edu>. ..\n>> In article <e7f834be.0404020428.14843d78@posting.google.com >, Frank\n>> Hellmann <C.i.m@gmx.net> writes\n>.....\n>>\n>> >What is one to make of this divergence physically?\n>>\n>> There isn\'t a generally accepted answer. My own view is that qed breaks\n>> down on very small time/distance scales, and as a result the series\n>> becomes less and less accurate for higher order terms.\n>>\n>>\n>In his "Dialog on quantum gravity" Rovelli states that "the weak field\n>approximation fails for GR" because it "is based on Feynman integrals\n>that sum over infinite momenta, namely over regions of arbitrarily\n>small volume". A result of QLG is that "there is literally no volume\n>smaller than the Planck volume" so that "it makes no sense to\n>integrate over degrees of freedom far smaller than the Planck length"\n>since such regions "literally do not exist".\n\nFor my own reasons I think the limiting distance is much smaller, the\nSchwarzschild radius of an electron. But I agree with the principle.\n>\n>Now, according to your view above, if "there is literally no volume\n>smaller than the Planck volume" that would impact divergences in QFT\n>too. Right?\n\nSpecifically in would impact the Landau Pole, which is the last\nremaining divergence which does not already have a resolution. If there\nis a lower bound on time/distance there will be also be a bound on the\nnumber of vertices possible in a Feynman diagram, and the series will\nterminate.\n\n\n\n\nRegards\n\n--\nCharles Francis\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <61789046.0404061053.6a1a619d@posting.google.com>, Italo
Vecchi <vecchi@weirdtech.com> writes
>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4sap
>f$c99$1@lfa222122.richmond.edu>...
>> In article <e7f834be.0404020428.14843d78@posting.google.com>, Frank
>> Hellmann <C.i.m@gmx.net> writes
>.....
>>
>> >What is one to make of this divergence physically?
>>
>> There isn't a generally accepted answer. My own view is that qed breaks
>> down on very small time/distance scales, and as a result the series
>> becomes less and less accurate for higher order terms.
>>
>>
>In his "Dialog on quantum gravity" Rovelli states that "the weak field
>approximation fails for GR" because it "is based on Feynman integrals
>that sum over infinite momenta, namely over regions of arbitrarily
>small volume". A result of QLG is that "there is literally no volume
>smaller than the Planck volume" so that "it makes no sense to
>integrate over degrees of freedom far smaller than the Planck length"
>since such regions "literally do not exist".

For my own reasons I think the limiting distance is much smaller, the
Schwarzschild radius of an electron. But I agree with the principle.
>
>Now, according to your view above, if "there is literally no volume
>smaller than the Planck volume" that would impact divergences in QFT
>too. Right?

Specifically in would impact the Landau Pole, which is the last
remaining divergence which does not already have a resolution. If there
is a lower bound on time/distance there will be also be a bound on the
number of vertices possible in a Feynman diagram, and the series will
terminate.

Regards

--
Charles Francis

Italo Vecchi

Apr11-04, 11:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nCharles Francis <charles@clef.demon.co.uk> wrote in message news:<c54jt4\\$gt4\\$1@lfa222122.richmond.edu>...\ n> In article <61789046.0404061053.6a1a619d@posting.google.com >, Italo\n> Vecchi <vecchi@weirdtech.com> writes\n....\n> >In his "Dialog on quantum gravity" Rovelli states that "the weak field\n> >approximation fails for GR" because it "is based on Feynman integrals\n> >that sum over infinite momenta, namely over regions of arbitrarily\n> >small volume". A result of QLG is that "there is literally no volume\n> >smaller than the Planck volume" so that "it makes no sense to\n> >integrate over degrees of freedom far smaller than the Planck length"\n> >since such regions "literally do not exist".\n>\n> For my own reasons I think the limiting distance is much smaller, the\n> Schwarzschild radius of an electron. But I agree with the principle.\n> >\n> >Now, according to your view above, if "there is literally no volume\n> >smaller than the Planck volume" that would impact divergences in QFT\n> >too. Right?\n>\n> Specifically in would impact the Landau Pole, which is the last\n> remaining divergence which does not already have a resolution. If there\n> is a lower bound on time/distance there will be also be a bound on the\n> number of vertices possible in a Feynman diagram, and the series will\n> terminate.\n>\nThanks, very interesting.\nI wonder what\'s so special about the Schwarzschild radius of (of all\nthings) the electron.\nI also wonder at which scales QFT series start to diverge from\nexperimental results ,i.e. which terms do you actually have to throw\naway since they ruin your predictions instead of enhancing them. I\nguess that could provide an answer to the Schwarzschild electron\nradius vs Planck length issue.\n\nRegards,\n\nIV\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c54jt4$gt4$1@lfa222122.richmond.edu>...
> In article <61789046.0404061053.6a1a619d@posting.google.com>, Italo
> Vecchi <vecchi@weirdtech.com> writes
....
> >In his "Dialog on quantum gravity" Rovelli states that "the weak field
> >approximation fails for GR" because it "is based on Feynman integrals
> >that sum over infinite momenta, namely over regions of arbitrarily
> >small volume". A result of QLG is that "there is literally no volume
> >smaller than the Planck volume" so that "it makes no sense to
> >integrate over degrees of freedom far smaller than the Planck length"
> >since such regions "literally do not exist".
>
> For my own reasons I think the limiting distance is much smaller, the
> Schwarzschild radius of an electron. But I agree with the principle.
> >
> >Now, according to your view above, if "there is literally no volume
> >smaller than the Planck volume" that would impact divergences in QFT
> >too. Right?
>
> Specifically in would impact the Landau Pole, which is the last
> remaining divergence which does not already have a resolution. If there
> is a lower bound on time/distance there will be also be a bound on the
> number of vertices possible in a Feynman diagram, and the series will
> terminate.
>
Thanks, very interesting.
I wonder what's so special about the Schwarzschild radius of (of all
things) the electron.
I also wonder at which scales QFT series start to diverge from
experimental results ,i.e. which terms do you actually have to throw
away since they ruin your predictions instead of enhancing them. I
guess that could provide an answer to the Schwarzschild electron
radius vs Planck length issue.

Regards,

IV

Matthew A. Nobes

Apr11-04, 12:06 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\nIn article <61789046.0404090027.3c4943c2@posting.google.com>, \nItalo Vecchi wrote:\n\n> I also wonder at which scales QFT series start to diverge from\n> experimental results ,i.e. which terms do you actually have to throw\n> away since they ruin your predictions instead of enhancing them.\n\nThe rough guide for asymtotic series is that they get better until\nyou exceed the inverse of the expansion parameter number of terms.\nSo in QED, the expansion parameter is 1/137, so you might\nexpect the approximation to get better for 100 terms or so.\n\nIn QCD it\'s more like 1/10, and there are examples of series\nwhich show very poor convergence (though they still provide\nreasonable apporximations). I do lattice QCD perturbation\ntheory, and here the coupling is more like 0.2-0.3 so we\ndon\'t expect much beyond 3-4 terms would be any use.\n\nMatthew Nobes\n--\nmanobes@sdf.lonestar.org\nSDF Public Access UNIX System - http://sdf.lonestar.org\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <61789046.0404090027.3c4943c2@posting.google.com>,
Italo Vecchi wrote:

> I also wonder at which scales QFT series start to diverge from
> experimental results ,i.e. which terms do you actually have to throw
> away since they ruin your predictions instead of enhancing them.

The rough guide for asymtotic series is that they get better until
you exceed the inverse of the expansion parameter number of terms.
So in QED, the expansion parameter is 1/137, so you might
expect the approximation to get better for 100 terms or so.

In QCD it's more like 1/10, and there are examples of series
which show very poor convergence (though they still provide
reasonable apporximations). I do lattice QCD perturbation
theory, and here the coupling is more like .2-0.3 so we
don't expect much beyond 3-4 terms would be any use.

Matthew Nobes
--
manobes@sdf.lonestar.org
SDF Public Access UNIX System - http://sdf.lonestar.org

Frank Hellmann

Apr13-04, 11:29 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nThank you a lot for the pointers. I particularly like Weinbergs book\nso far because it appears to keep the physics in the foreground, while\nlinking them extensively to the mathematics (at least in the first few\nchapters which are all I have studied yet...), but for the time being\nI don\'t need to understand just pass exams, so I\'m gonna tackle this\nagain later, including the examples by Mr Baez which shed quite some\nlight on these issues.\n\nI have one additional mathematical question though, every now and then\nthe term operator valued distribution crops up. Now for ordinary QM we\nhave distributions, where we identify the Hilbert/Schwartz Space used\nto construct them in the first place naturally through the integral\nrepresentation of the distributions:\nf_d(g) = int dx (f(x)*g(x))\nwith f element of the Hilbert Space (which contains the Schwartz\nspace) and g element of the Schwartz space and f_d the distribution\nnaturally assosciated with f. This assosciation of course preserves\nthe vector space structure of the function space, and allows us to\ndefine a large number of nice tools on the Hilbert space in a very\nnaturall fashion.\n\nNow for operator valued distributions, how does this construction\nproceed? How does one identify operator valued functions of a\nparticular space with the distributions? Is there a similarily\nnaturall way to do it?\n\ncheers,\nFrank Hellmann\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Thank you a lot for the pointers. I particularly like Weinbergs book
so far because it appears to keep the physics in the foreground, while
linking them extensively to the mathematics (at least in the first few
chapters which are all I have studied yet...), but for the time being
I don't need to understand just pass exams, so I'm gonna tackle this
again later, including the examples by Mr Baez which shed quite some
light on these issues.

I have one additional mathematical question though, every now and then
the term operator valued distribution crops up. Now for ordinary QM we
have distributions, where we identify the Hilbert/Schwartz Space used
to construct them in the first place naturally through the integral
representation of the distributions:
f_d(g) = \int dx (f(x)*g(x))
with f element of the Hilbert Space (which contains the Schwartz
space) and g element of the Schwartz space and f_d the distribution
naturally assosciated with f. This assosciation of course preserves
the vector space structure of the function space, and allows us to
define a large number of nice tools on the Hilbert space in a very
naturall fashion.

Now for operator valued distributions, how does this construction
proceed? How does one identify operator valued functions of a
particular space with the distributions? Is there a similarily
naturall way to do it?

cheers,
Frank Hellmann

Arnold Neumaier

Apr14-04, 03:17 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Frank Hellmann wrote:\n\n> I have one additional mathematical question though, every now and then\n> the term operator valued distribution crops up. Now for ordinary QM we\n> have distributions, where we identify the Hilbert/Schwartz Space used\n> to construct them in the first place naturally through the integral\n> representation of the distributions:\n> f_d(g) = int dx (f(x)*g(x))\n> with f element of the Hilbert Space (which contains the Schwartz\n> space) and g element of the Schwartz space and f_d the distribution\n> naturally assosciated with f. This assosciation of course preserves\n> the vector space structure of the function space, and allows us to\n> define a large number of nice tools on the Hilbert space in a very\n> naturall fashion.\n>\n> Now for operator valued distributions, how does this construction\n> proceed? How does one identify operator valued functions of a\n> particular space with the distributions? Is there a similarily\n> natural way to do it?\n\nIf phi(x) is the operator valued distribution, and g a Schwartz test\nfunction then\nphi(g) = int dx phi(x) g(x)\nis a good operator.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann wrote:

> I have one additional mathematical question though, every now and then
> the term operator valued distribution crops up. Now for ordinary QM we
> have distributions, where we identify the Hilbert/Schwartz Space used
> to construct them in the first place naturally through the integral
> representation of the distributions:
> f_d(g) = \int dx (f(x)*g(x))
> with f element of the Hilbert Space (which contains the Schwartz
> space) and g element of the Schwartz space and f_d the distribution
> naturally assosciated with f. This assosciation of course preserves
> the vector space structure of the function space, and allows us to
> define a large number of nice tools on the Hilbert space in a very
> naturall fashion.
>
> Now for operator valued distributions, how does this construction
> proceed? How does one identify operator valued functions of a
> particular space with the distributions? Is there a similarily
> natural way to do it?

If \phi(x) is the operator valued distribution, and g a Schwartz test
function then
\phi(g) = \int dx \phi(x) g(x)
is a good operator.

Arnold Neumaier

Arnold Neumaier

Apr14-04, 08:36 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\n\n\nMatthew A. Nobes wrote:\n> In article <61789046.0404090027.3c4943c2@posting.google.com>, \n> Italo Vecchi wrote:\n>\n>\n>>I also wonder at which scales QFT series start to diverge from\n>>experimental results ,i.e. which terms do you actually have to throw\n>>away since they ruin your predictions instead of enhancing them.\n>\n>\n> The rough guide for asymtotic series is that they get better until\n> you exceed the inverse of the expansion parameter number of terms.\n\nThis is a very poor guide.\nIt very much depends on the magnitude of the terms in the\nseries. Generally, an asymptotic series can be trusted as long as the\nfirst discarded term is smaller than the last used term, though even\nthis rule may be fallacious.\n\nOne can study these things quite well with functions which have\nknown asymptotic expansions (Watson\'s lemma, etc.)\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Matthew A. Nobes wrote:
> In article <61789046.0404090027.3c4943c2@posting.google.com>,
> Italo Vecchi wrote:
>
>
>>I also wonder at which scales QFT series start to diverge from
>>experimental results ,i.e. which terms do you actually have to throw
>>away since they ruin your predictions instead of enhancing them.
>
>
> The rough guide for asymtotic series is that they get better until
> you exceed the inverse of the expansion parameter number of terms.

This is a very poor guide.
It very much depends on the magnitude of the terms in the
series. Generally, an asymptotic series can be trusted as long as the
first discarded term is smaller than the last used term, though even
this rule may be fallacious.

One can study these things quite well with functions which have
known asymptotic expansions (Watson's lemma, etc.)

Arnold Neumaier

Italo Vecchi

Apr15-04, 11:29 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<407BE7B2.6070206@univie.ac.at>...\n> Matthew A. Nobes wrote:\n....\n> > The rough guide for asymtotic series is that they get better until\n> > you exceed the inverse of the expansion parameter number of terms.\n>\n> This is a very poor guide.\n> It very much depends on the magnitude of the terms in the\n> series. Generally, an asymptotic series can be trusted as long as the\n> first discarded term is smaller than the last used term, though even\n> this rule may be fallacious.\n>\n> One can study these things quite well with functions which have\n> known asymptotic expansions (Watson\'s lemma, etc.)\n>\n\nThe problem here is that we do not have a mathematical function, only\nexperimental results. I was suggesting that the onset of divergence\nfrom measured data may correspond to the lower bound on physically\nmeaningful time/distance.\n\nIV\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<407BE7B2.6070206@univie.ac.at>...
> Matthew A. Nobes wrote:
....
> > The rough guide for asymtotic series is that they get better until
> > you exceed the inverse of the expansion parameter number of terms.
>
> This is a very poor guide.
> It very much depends on the magnitude of the terms in the
> series. Generally, an asymptotic series can be trusted as long as the
> first discarded term is smaller than the last used term, though even
> this rule may be fallacious.
>
> One can study these things quite well with functions which have
> known asymptotic expansions (Watson's lemma, etc.)
>

The problem here is that we do not have a mathematical function, only
experimental results. I was suggesting that the onset of divergence
from measured data may correspond to the lower bound on physically
meaningful time/distance.

IV

Frank Hellmann

Apr16-04, 02:28 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> If phi(x) is the operator valued distribution, and g a Schwartz test\n> function then\n> phi(g) = int dx phi(x) g(x)\n> is a good operator.\n>\n\nHu?\nI don\'t see what you are doing there...\nIs this to imply that the dual space is taken as the space of mapings\nfrom the operaor valued functions to the operators instead of to the\nreals?\n\nAnd if not:\n\nAssuming phi(x) is an operator valued function, and ovd (phi) is a\ndistribution maping from the space of operator valued functions into\nthe reals, then for any test function g (x) we could get the function\nphi(x)g(x), that is phi operating on g, which if we integrate gives a\nfunction in the reals that depends on linearly on phi and in some way\non g. Thus at least naivly it appears that there is no naturall way to\nidentify the distribution ovd (phi,g) := int dx phi(x) g(x) with the\noperator valued function phi (x), as there is an additional dependency\non g.\n\nDoes every possible function g then furnish an equivalent\nidentification for the function space in the dual space? Is the g\ndependency therefore irrelevant?\n\ncheers,\nFrank.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>> If \phi(x) is the operator valued distribution, and g a Schwartz test
> function then
> \phi(g) = \int dx \phi(x) g(x)
> is a good operator.
>

Hu?
I don't see what you are doing there...
Is this to imply that the dual space is taken as the space of mapings
from the operaor valued functions to the operators instead of to the
reals?

And if not:

Assuming \phi(x) is an operator valued function, and ovd (\phi) is a
distribution maping from the space of operator valued functions into
the reals, then for any test function g (x) we could get the function
\phi(x)g(x), that is \phi operating on g, which if we integrate gives a
function in the reals that depends on linearly on \phi and in some way
on g. Thus at least naivly it appears that there is no naturall way to
identify the distribution ovd (\phi,g) := \int dx \phi(x) g(x) with the
operator valued function \phi (x), as there is an additional dependency
on g.

Does every possible function g then furnish an equivalent
identification for the function space in the dual space? Is the g
dependency therefore irrelevant?

cheers,
Frank.

Arnold Neumaier

Apr16-04, 02:28 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Italo Vecchi wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<407BE7B2.6070206@univie.ac.at>...\n>\n>>Matt hew A. Nobes wrote:\n>\n> ...\n>\n>>>The rough guide for asymtotic series is that they get better until\n>>>you exceed the inverse of the expansion parameter number of terms.\n>>\n>>This is a very poor guide.\n>>It very much depends on the magnitude of the terms in the\n>>series. Generally, an asymptotic series can be trusted as long as the\n>>first discarded term is smaller than the last used term, though even\n>>this rule may be fallacious.\n>>\n>>One can study these things quite well with functions which have\n>>known asymptotic expansions (Watson\'s lemma, etc.)\n>>\n>\n>\n> The problem here is that we do not have a mathematical function, only\n> experimental results. I was suggesting that the onset of divergence\n> from measured data may correspond to the lower bound on physically\n> meaningful time/distance.\n\nWith experimental results you just have numbers, and not infinite\nseries, so questions of convergence do not occur.\n\n\nOn the other hand, if you know of an infinite series a finite number of\nterms only, the result can be, strictly speaking, anything.\nBut usually one applies some extrapolation algorithm\n(e.g., the epsilon or eta algorithm) to get a meaningful guess for the\nlimit, and estimates the error by doing the same several times,\nkeeping a variable number of terms. The difference between\nconsecutive results can count as a reasonable (though not foolproof)\nerror estimate of these results.\n\nBut to have reliable bounds you need to know an exact definition of\nwhat you are approximating, and work from there.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Italo Vecchi wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<407BE7B2.6070206@univie.ac.at>...
>
>>Matthew A. Nobes wrote:
>
> ...
>
>>>The rough guide for asymtotic series is that they get better until
>>>you exceed the inverse of the expansion parameter number of terms.
>>
>>This is a very poor guide.
>>It very much depends on the magnitude of the terms in the
>>series. Generally, an asymptotic series can be trusted as long as the
>>first discarded term is smaller than the last used term, though even
>>this rule may be fallacious.
>>
>>One can study these things quite well with functions which have
>>known asymptotic expansions (Watson's lemma, etc.)
>>
>
>
> The problem here is that we do not have a mathematical function, only
> experimental results. I was suggesting that the onset of divergence
> from measured data may correspond to the lower bound on physically
> meaningful time/distance.

With experimental results you just have numbers, and not infinite
series, so questions of convergence do not occur.

On the other hand, if you know of an infinite series a finite number of
terms only, the result can be, strictly speaking, anything.
But usually one applies some extrapolation algorithm
(e.g., the \epsilon or \eta algorithm) to get a meaningful guess for the
limit, and estimates the error by doing the same several times,
keeping a variable number of terms. The difference between
consecutive results can count as a reasonable (though not foolproof)
error estimate of these results.

But to have reliable bounds you need to know an exact definition of
what you are approximating, and work from there.

Arnold Neumaier

MM

Apr17-04, 05:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>John Baez wrote:\n\n> Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable\n> and compute its Abel sum.\n>\n> Exercise: Show that Abel summation dominates (C,k) summation for\n> all k.\n>\n> Exercise: Find a series that is Abel summable but not (C,k) summable\n> for any k.\n\nOK, I eventually blundered my way through the Cesaro exercises, but I\ndon\'t know how to do the first Abel exercise.\n\nPlease take pity and give some more hints.\n\n- MM.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>John Baez wrote:

> Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable
> and compute its Abel sum.
>
> Exercise: Show that Abel summation dominates (C,k) summation for
> all k.
>
> Exercise: Find a series that is Abel summable but not (C,k) summable
> for any k.

OK, I eventually blundered my way through the Cesaro exercises, but I
don't know how to do the first Abel exercise.

Please take pity and give some more hints.

- MM.

Arnold Neumaier

Apr17-04, 05:08 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Frank Hellmann wrote:\n>>If phi(x) is the operator valued distribution, and g a Schwartz test\n>>function then\n>> phi(g) = int dx phi(x) g(x)\n>>is a good operator.\n>>\n>\n>\n> Hu?\n> I don\'t see what you are doing there...\n> Is this to imply that the dual space is taken as the space of mapings\n> from the operaor valued functions to the operators instead of to the\n> reals?\n\nNo. Probably I\'d have said that\nphihat(g) = int dx phi(x) g(x)\nis a good operator. If one works rigorously, one assumes the\nexistence of the left hand side as a mapping from test functions\nto operators, and regards the right hand side just as a different\nnotation for it.\n\n\n> And if not:\n>\n> Assuming phi(x) is an operator valued function, and ovd (phi) is a\n> distribution maping from the space of operator valued functions into\n> the reals,\n\nI don\'t understand how you read that into what I wrote.\nIn my statement, phi(x) is not a function but a distribution,\nand g is a mollifier that smears the singularity so that one\ngets something nice.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann wrote:
>>If \phi(x) is the operator valued distribution, and g a Schwartz test
>>function then
>> \phi(g) = \int dx \phi(x) g(x)
>>is a good operator.
>>
>
>
> Hu?
> I don't see what you are doing there...
> Is this to imply that the dual space is taken as the space of mapings
> from the operaor valued functions to the operators instead of to the
> reals?

No. Probably I'd have said that
phihat(g) = \int dx \phi(x) g(x)
is a good operator. If one works rigorously, one assumes the
existence of the left hand side as a mapping from test functions
to operators, and regards the right hand side just as a different
notation for it.

> And if not:
>
> Assuming \phi(x) is an operator valued function, and ovd (\phi) is a
> distribution maping from the space of operator valued functions into
> the reals,

I don't understand how you read that into what I wrote.
In my statement, \phi(x) is not a function but a distribution,
and g is a mollifier that smears the singularity so that one
gets something nice.

Arnold Neumaier

Frank Hellmann

Apr19-04, 02:14 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>> > And if not:\n> >\n> > Assuming phi(x) is an operator valued function, and ovd (phi) is a\n> > distribution maping from the space of operator valued functions into\n> > the reals,\n>\n> I don\'t understand how you read that into what I wrote.\n> In my statement, phi(x) is not a function but a distribution,\n> and g is a mollifier that smears the singularity so that one\n> gets something nice.\n>\n>\n> Arnold Neumaier\n\n\nWell a distribution is a linear mapping (aka operator) from the test\nfunctions into the reals so phihat(g) is certainly a distribution, an\nphi(x) it\'s integral core, as phihat(g) is real valued if phi(x) and\ng(x) are both real valued.\nI don\'t see where operator valued distributions enter the picture with\nwhat you wrote, this seems to be just standard distributions to me.\n\nIf I\'m way of the mark do you know a good textbook that contains an\nintroduction?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>> > And if not:
> >
> > Assuming \phi(x) is an operator valued function, and ovd (\phi) is a
> > distribution maping from the space of operator valued functions into
> > the reals,
>
> I don't understand how you read that into what I wrote.
> In my statement, \phi(x) is not a function but a distribution,
> and g is a mollifier that smears the singularity so that one
> gets something nice.
>
>
> Arnold Neumaier

Well a distribution is a linear mapping (aka operator) from the test
functions into the reals so phihat(g) is certainly a distribution, an
\phi(x) it's integral core, as phihat(g) is real valued if \phi(x) and
g(x) are both real valued.
I don't see where operator valued distributions enter the picture with
what you wrote, this seems to be just standard distributions to me.

If I'm way of the mark do you know a good textbook that contains an
introduction?

ebunn@lfa221051.richmond.edu

Apr20-04, 02:34 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <38b55e8c.0404152345.219b8d58@posting.google.com>, \nMM <mikem@despammed.com> wrote:\n>John Baez wrote:\n>\n> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable\n> > and compute its Abel sum.\n> >\n> > Exercise: Show that Abel summation dominates (C,k) summation for\n> > all k.\n> >\n> > Exercise: Find a series that is Abel summable but not (C,k) summable\n> > for any k.\n>\n>OK, I eventually blundered my way through the Cesaro exercises, but I\n>don\'t know how to do the first Abel exercise.\n\nI think I\'ve got this one!\n\nLet me add a 0th term and change the signs (so that my solution will be the\nnegative of the original): we want to evaluate\n\n0^k - 1^k + 2^k - ... = \\sum_j (-1)^j j^k.\n\nTo do Abel summation, we\'re supposed to consider the function f\ndefined by\n\nf(x) = \\sum_j (-1)^j j^k x^j.\n\nLet D be the operator defined by\n\nDg = x (dg/dx).\n\nThen D(x^j) = j x^j, so\n\nf(x) = \\sum_j (-1)^j D^k(x^j)\n= D^k(\\sum_j (-x^j))\n= D^k(1/(1+x)).\n\nLike any good physicist, I\'m being cavalier about issues of uniform\nconvergence and the like, switching orders of derivatives and sums\nwith wild abandon. But I think this is all OK for |x| < 1.\n\nNow we have to evaluate this thing at x = 1. If my kung fu were\nstronger, I\'d have been able to do this in some clever way. What\nI actually did was give it to Maple, evaluate it explicitly\nfor k from 1 to 20, and feed the results into the Encyclopedia of\nInteger Sequences.\n\n[Incidentally, anyone who\'s read this far into this post would\nprobably like the Encyclopedia of Integer Sequences. If you don\'t\nknow about it, check out http://www.research.att.com/~njas/sequences/ ]\n\nAfter playing around a bit, I found that the sum is\n\n-(2^(k+1) - 1) B_{k+1} / (k+1),\n\nwhere B_n is the nth Bernoulli number.\n\nRemember that I flipped the sign at the beginning, so the solution to\nthe original problem the negative of this.\n\nI haven\'t actually proved this last expression, but it works for k=1\nthrough 20, and I also spot-checked it at k=100, so I\'m morally sure\nit\'s right. I suppose that it should be easy enough to prove by\nmessing around a bit with the definitions of the Bernoulli numbers.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <38b55e8c.0404152345.219b8d58@posting.google.com>,
MM <mikem@despammed.com> wrote:
>John Baez wrote:
>
> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable
> > and compute its Abel sum.
> >
> > Exercise: Show that Abel summation dominates (C,k) summation for
> > all k.
> >
> > Exercise: Find a series that is Abel summable but not (C,k) summable
> > for any k.
>
>OK, I eventually blundered my way through the Cesaro exercises, but I
>don't know how to do the first Abel exercise.

I think I've got this one!

Let me add a 0th term and change the signs (so that my solution will be the
negative of the original): we want to evaluate

0^k - 1^k + 2^k - .[/itex].. = \sum_j (-1)^j j^k.

To do Abel summation, we're supposed to consider the function f
defined by

f(x) = \sum_j (-1)^j j^k x^j.

Let D be the operator defined by

Dg = x (dg/dx).

Then D(x^j) = j x^j, so

f(x) = \sum_j (-1)^j D^k(x^j)= D^k(\sum_j (-x^j))= D^k(1/(1+x)).

Like any good physicist, I'm being cavalier about issues of uniform
convergence and the like, switching orders of derivatives and sums
with wild abandon. But I think this is all OK for |x| < 1.

Now we have to evaluate this thing at x = 1. If my kung fu were
stronger, I'd have been able to do this in some clever way. What
I actually did was give it to Maple, evaluate it explicitly
for k from 1 to 20, and feed the results into the Encyclopedia of
Integer Sequences.

[Incidentally, anyone who's read this far into this post would
probably like the Encyclopedia of Integer Sequences. If you don't
know about it, check out http://www.research.att.com/~njas/sequences/ ]

After playing around a bit, I found that the sum is

[itex]-(2^(k+1) - 1) B_{k+1} / (k+1),

where B_n is the nth Bernoulli number.

Remember that I flipped the sign at the beginning, so the solution to
the original problem the negative of this.

I haven't actually proved this last expression, but it works for k=1
through 20, and I also spot-checked it at k=100, so I'm morally sure
it's right. I suppose that it should be easy enough to prove by
messing around a bit with the definitions of the Bernoulli numbers.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

John Baez

Apr20-04, 02:35 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <38b55e8c.0404152345.219b8d58@posting.google.com>, \nMM <mikem@despammed.com> wrote:\n\n>John Baez wrote:\n\n> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable\n> > and compute its Abel sum.\n> >\n> > Exercise: Show that Abel summation dominates (C,k) summation for\n> > all k.\n\n>OK, I eventually blundered my way through the Cesaro exercises, but I\n>don\'t know how to do the first Abel exercise.\n>\n>Please take pity and give some more hints.\n\nI apologize for giving such a hard one to start with!\n\nHow about this:\n\n1 - 2 + 3 - 4 + ....\n\nTo Abel sum this, you need to compute the limit of\n\nsum (-1)^{n+1} n x^n\n\nas x approaches 1 from below. The trick is to find an explicit\nclosed form for this sum, and then take the limit.\n\nIf you have trouble with the "trick" part, maybe it\'s good to\nthink a bit more generally. If you know\n\nsum a_n x^n = f(x)\n\nthen how is\n\nsum n a_n x^n\n\nrelated to f(x)? This is a nice thing to know. If you know it,\nyou can reduce the Abel sum\n\n1 - 2 + 3 - 4 + .... := lim_{x -> 1} sum (-1)^{n+1} n x^n\n\nto this easier one:\n\n1 - 1 + 1 - 1 + .... := lim_{x -> 1} sum (-1)^{n+1} x^n\n\nYou can do the latter one with a geometric series! Alternatively, you can\ndo it using Cesaro summation, together with the theorem that whenever\nCesaro summation works, Abel summation does too - and gives the same answer.\nIf you did the Cesaro summation problems, presumably you know already\nthat Cesaro summation gives\n\n1 - 1 + 1 - 1 + .... = 1/2\n\nIf you get stuck, I can tell you about some webpages of mine that\ngive the answer to the Abel sum\n\n1 - 2 + 3 - 4 + ...\n\nand more generally\n\n1^k - 2^k + 3^k - 4^k + ...\n\nThis stuff is fun! It\'s like black magic.\n\n........................................ ....................................\n\nPuzzle 19: As of February 2004, five of the ten richest people in the\nworld had the same last name. What is it?\n\nIf you get stuck, go here:\n\nhttp://www.math.ucr.edu/home/baez/puzzles/19.html\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <38b55e8c.0404152345.219b8d58@posting.google.com>,
MM <mikem@despammed.com> wrote:

>John Baez wrote:

> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable
> > and compute its Abel sum.
> >
> > Exercise: Show that Abel summation dominates (C,k) summation for
> > all k.

>OK, I eventually blundered my way through the Cesaro exercises, but I
>don't know how to do the first Abel exercise.
>
>Please take pity and give some more hints.

I apologize for giving such a hard one to start with!

How about this:

1 - 2 + 3 - 4 + .[/itex]...

To Abel sum this, you need to compute the limit of

sum (-1)^{n+1} n x^n

as x approaches 1 from below. The trick is to find an explicit
closed form for this sum, and then take the limit.

If you have trouble with the "trick" part, maybe it's good to
think a bit more generally. If you know

sum a_n x^n = f(x)

then how is

sum n a_n x^n

related to f(x)? This is a nice thing to know. If you know it,
you can reduce the Abel sum

1 - 2 + 3 - 4 + .... := lim_{x -> 1} sum (-1)^{n+1} n x^n

to this easier one:

1 - 1 + 1 - 1 + .... := lim_{x -> 1} sum (-1)^{n+1} x^n

You can do the latter one with a geometric series! Alternatively, you can
do it using Cesaro summation, together with the theorem that whenever
Cesaro summation works, Abel summation does too - and gives the same answer.
If you did the Cesaro summation problems, presumably you know already
that Cesaro summation gives

1 - 1 + 1 - 1 + ..[itex].. = 1/2

If you get stuck, I can tell you about some webpages of mine that
give the answer to the Abel sum

1 - 2 + 3 - 4 + ...

and more generally

1^k - 2^k + 3^k - 4^k + ...

This stuff is fun! It's like black magic.

.................................................. ..........................

Puzzle 19: As of February 2004, five of the ten richest people in the
world had the same last name. What is it?

If you get stuck, go here:

http://www.math.ucr.edu/home/baez/puzzles/19.html

ebunn@lfa221051.richmond.edu

Apr21-04, 04:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <c617t2\\$d3g\\$1@lfa222122.richmond.edu>,\n<ebunn @lfa221051.richmond.edu> wrote:\n>In article <38b55e8c.0404152345.219b8d58@posting.google.com>, \n>MM <mikem@despammed.com> wrote:\n>>John Baez wrote:\n>>\n>> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable\n>> > and compute its Abel sum.\n>> >\n>> > Exercise: Show that Abel summation dominates (C,k) summation for\n>> > all k.\n>> >\n>> > Exercise: Find a series that is Abel summable but not (C,k) summable\n>> > for any k.\n>>\n>>OK, I eventually blundered my way through the Cesaro exercises, but I\n>>don\'t know how to do the first Abel exercise.\n>\n>I think I\'ve got this one!\n\nI think I probably know the answer to the last one too.\n\n(In case anyone doesn\'t want to see it, here\'s some blank space\nbefore I write my answer.)\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n \n\n\nWe should look at alternating sums whose terms blow up faster than any\npower, since we already know that the alternating series of kth powers\nis (C,k) summable. That suggests we try\n\n\\sum_{j=0}^\\infty (-1)^j exp(j a)\n= 1 - exp(a) + exp(2a) - exp(3a) + ...\n\nwith a some positive constant.\n\nThe nth partial sum of this guy grows exponentially with n as\nn->infinity, as does the average of the first n partial sums. So it\'s\ncertainly not Cesaro summable. I\'m pretty sure the iterated averages\nneeded to compute the (C,k) sum also blow up exponentially, although I\nhaven\'t proved that.\n\nBut it is Abel summable. The function to consider is\n\nf(x) = \\sum_{j=0}^\\infty (-1)^j exp(ja - jt)\n\nwhere t = -ln x. That\'s just a geometric series:\n\nf(x) = 1/(1+exp(a-t)).\n\nTake the limit as x->1, i.e. as t->0, and you\'re done. The Abel sum\nis\n\n1/(1+exp(a)).\n\nDid I get it right?\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <c617t2$d3g$1@lfa222122.richmond.edu>,
<ebunn@lfa221051.richmond.edu> wrote:
>In article <38b55e8c.0404152345.219b8d58@posting.google.com>,
>MM <mikem@despammed.com> wrote:
>>John Baez wrote:
>>
>> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable
>> > and compute its Abel sum.
>> >
>> > Exercise: Show that Abel summation dominates (C,k) summation for
>> > all k.
>> >
>> > Exercise: Find a series that is Abel summable but not (C,k) summable
>> > for any k.
>>
>>OK, I eventually blundered my way through the Cesaro exercises, but I
>>don't know how to do the first Abel exercise.
>
>I think I've got this one!

I think I probably know the answer to the last one too.

(In case anyone doesn't want to see it, here's some blank space
before I write my answer.)

We should look at alternating sums whose terms blow up faster than any
power, since we already know that the alternating series of kth powers
is (C,k) summable. That suggests we try

\sum_{j=0}^\infty (-1)^j \exp(j a)= 1 - \exp(a) + \exp(2a) - \exp(3a) + .[/itex]..

with a some positive constant.

The nth partial sum of this guy grows exponentially with n as
n->infinity, as does the average of the first n partial sums. So it's
certainly not Cesaro summable. I'm pretty sure the iterated averages
needed to compute the (C,k) sum also blow up exponentially, although I
haven't proved that.

But it is Abel summable. The function to consider is

f(x) = \sum_{j=0}^\infty (-1)^j \exp(ja - jt)

where t = -ln x. That's just a geometric series:

f(x) = 1/(1+\exp(a-t)).

Take the limit as x->1, i.e. as t->0, and you're done. The Abel sum
is

1/(1+\exp(a)).

Did I get it right?

[itex]-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

Haelfix

Apr21-04, 04:24 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"It very much depends on the magnitude of the terms in the\nseries. Generally, an asymptotic series can be trusted as long as the\nfirst discarded term is smaller than the last used term, though even\nthis rule may be fallacious."\n\nActually, this is precisely the condition that leads to the rule that\n1/alpha is the place where to truncate.\n\nRoughly speaking QFT leads to series that look like U ~ A^m m! m^alpha\n\nUsing your condition... U(m+1) alpha ^ (m+1) / U(m) alpha^(m) ~ 1\n\n=> m alpha ~ 1 => m ~ 1/alpha\n\nWhen you are more careful you end up with factors of pi, and things\nlike the number of colors in QCD.. (but they tend to cancel roughly )\netc etc\n\nThis is actually a problem for some QCD calculations, since people\nhave calculated to 4 loops, and they are right on the onset of\ndivergences.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"It very much depends on the magnitude of the terms in the
series. Generally, an asymptotic series can be trusted as long as the
first discarded term is smaller than the last used term, though even
this rule may be fallacious."

Actually, this is precisely the condition that leads to the rule that
1/\alpha is the place where to truncate.

Roughly speaking QFT leads to series that look like U ~ A^m m! m^\alpha

Using your condition... U(m+1) \alpha ^ (m+1) / U(m) \alpha^(m) ~ 1=> m \alpha ~ 1 => m ~ 1/\alpha

When you are more careful you end up with factors of \pi, and things
like the number of colors in QCD.. (but they tend to cancel roughly )
etc etc

This is actually a problem for some QCD calculations, since people
have calculated to 4 loops, and they are right on the onset of
divergences.

Alejandro

Apr21-04, 04:25 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<407BE7B2.6070206@univie.ac.at>...\n> Matthew A. Nobes wrote:\n\n> > The rough guide for asymtotic series is that they get better until\n> > you exceed the inverse of the expansion parameter number of terms.\n>\n> This is a very poor guide.\n\nOn the contrary, it seems it has some physical meaning. It tells that\nan N-legs diagram for an interaction V(x) could be assimilated as\na one-leg diagram for an interaction N V(x). It has sense.\n\nI wonder, Matthew, where did you got your argument from? I\'d like\nto understand when this rule can be applied (and when can it be\nfallacious).\n\n> It very much depends on the magnitude of the terms in the\n> series. Generally, an asymptotic series can be trusted as long as the\n> first discarded term is smaller than the last used term, though even\n> this rule may be fallacious.\n>\n> One can study these things quite well with functions which have\n> known asymptotic expansions (Watson\'s lemma, etc.)\n>\n> Arnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<407BE7B2.6070206@univie.ac.at>...
> Matthew A. Nobes wrote:

> > The rough guide for asymtotic series is that they get better until
> > you exceed the inverse of the expansion parameter number of terms.
>
> This is a very poor guide.

On the contrary, it seems it has some physical meaning. It tells that
an N-legs diagram for an interaction V(x) could be assimilated as
a one-leg diagram for an interaction N V(x). It has sense.

I wonder, Matthew, where did you got your argument from? I'd like
to understand when this rule can be applied (and when can it be
fallacious).

> It very much depends on the magnitude of the terms in the
> series. Generally, an asymptotic series can be trusted as long as the
> first discarded term is smaller than the last used term, though even
> this rule may be fallacious.
>
> One can study these things quite well with functions which have
> known asymptotic expansions (Watson's lemma, etc.)
>
> Arnold Neumaier

Arnold Neumaier

Apr22-04, 03:51 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Frank Hellmann wrote:\n>>>And if not:\n>>>\n>>>Assuming phi(x) is an operator valued function, and ovd (phi) is a\n>>>distribution maping from the space of operator valued functions into\n>>>the reals,\n>>\n>>I don\'t understand how you read that into what I wrote.\n>>In my statement, phi(x) is not a function but a distribution,\n>>and g is a mollifier that smears the singularity so that one\n>>gets something nice.\n>\n> Well a distribution is a linear mapping (aka operator) from the test\n> functions into the reals so phihat(g) is certainly a distribution,\n\nSince you didn\'t seem to understand the first version,\nphi(g) = int dx phi(x) g(x),\nI used the second time the physicists way of talking about\ndistributions, distinguishing between the object with x as argument\nand the object with g as argument:\nphihat(g) = int dx phi(x) g(x)\nhere phihat is the mathematician\'s distribution, a mapping which\ntakes a test function g and turns it [since it is an operator-valued\ndistribuiton] into an operator phihat(g),\nwhile\nphi(x) = phihat(delta(x))\nis the physicists distribution, obtained by applying phihat to\na singular object which, strictly speaking, is forbidden.\nBut with a little care, one encounters no problems unless one starts\nto multiply such singular objects. The intuitive interpretation is\nthat one replaces the delta by a sequence of smooth hat functions g_l\nconverging weakly to the delta function, and takes phi(x) to be the\nlimit of the phihat(g_l(x)), which are ordinary operators.\nOf course, this limit also makes sense only weakly,\nand after multiplying with a terst function and integration\nyou are back to phihat.\n\n\n> If I\'m way off the mark do you know a good textbook that contains an\n> introduction?\n\nThe book on QED by Scharf (not a textbook) works directly with the\nphi(g). Usual textbooks are usually loose with regard to rigor.\nPerhaps Streater/Wightman (Spin, Statistics and all that) has a careful\nexposition.\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Frank Hellmann wrote:
>>>And if not:
>>>
>>>Assuming \phi(x) is an operator valued function, and ovd (\phi) is a
>>>distribution maping from the space of operator valued functions into
>>>the reals,
>>
>>I don't understand how you read that into what I wrote.
>>In my statement, \phi(x) is not a function but a distribution,
>>and g is a mollifier that smears the singularity so that one
>>gets something nice.
>
> Well a distribution is a linear mapping (aka operator) from the test
> functions into the reals so phihat(g) is certainly a distribution,

Since you didn't seem to understand the first version,
\phi(g) = \int dx \phi(x) g(x),
I used the second time the physicists way of talking about
distributions, distinguishing between the object with x as argument
and the object with g as argument:
phihat(g) = \int dx \phi(x) g(x)
here phihat is the mathematician's distribution, a mapping which
takes a test function g and turns it [since it is an operator-valued
distribuiton] into an operator phihat(g),
while
\phi(x) = phihat(\delta(x))
is the physicists distribution, obtained by applying phihat to
a singular object which, strictly speaking, is forbidden.
But with a little care, one encounters no problems unless one starts
to multiply such singular objects. The intuitive interpretation is
that one replaces the \delta by a sequence of smooth hat functions g_l
converging weakly to the \delta function, and takes \phi(x) to be the
limit of the phihat(g_l(x)), which are ordinary operators.
Of course, this limit also makes sense only weakly,
and after multiplying with a terst function and integration
you are back to phihat.

> If I'm way off the mark do you know a good textbook that contains an
> introduction?

The book on QED by Scharf (not a textbook) works directly with the
\phi(g). Usual textbooks are usually loose with regard to rigor.
Perhaps Streater/Wightman (Spin, Statistics and all that) has a careful
exposition.

Arnold Neumaier

Alejandro

Apr22-04, 04:07 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c54jt4\\$gt4\\$1@lfa222122.richmond.edu>...\ n\n> Specifically in would impact the Landau Pole, which is the last\n> remaining divergence which does not already have a resolution. If there\n> is a lower bound on time/distance there will be also be a bound on the\n> number of vertices possible in a Feynman diagram, and the series will\n> terminate.\n\nBut in some sense, Matthew\'s bound (above in this thread) is already\na bound on the number of vertices, isn\'t it?\n\nHmm hmm I wonder if we could even relate both limitations.\n\nAlejandro\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c54jt4$gt4$1@lfa222122.richmond.edu>...

> Specifically in would impact the Landau Pole, which is the last
> remaining divergence which does not already have a resolution. If there
> is a lower bound on time/distance there will be also be a bound on the
> number of vertices possible in a Feynman diagram, and the series will
> terminate.

But in some sense, Matthew's bound (above in this thread) is already
a bound on the number of vertices, isn't it?

Hmm hmm I wonder if we could even relate both limitations.

Alejandro

Arnold Neumaier

Apr24-04, 12:14 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Haelfix wrote:\n> "It very much depends on the magnitude of the terms in the\n> series. Generally, an asymptotic series can be trusted as long as the\n> first discarded term is smaller than the last used term, though even\n> this rule may be fallacious."\n>\n> Actually, this is precisely the condition that leads to the rule that\n> 1/alpha is the place where to truncate.\n>\n> Roughly speaking QFT leads to series that look like U ~ A^m m! m^alpha\n>\n> Using your condition... U(m+1) alpha ^ (m+1) / U(m) alpha^(m) ~ 1\n>\n> => m alpha ~ 1 => m ~ 1/alpha\n\nNo, it depends very much on the relative size of the U(m).\nIn practice, these are unknown. Nobody has the fiantest idea of how large\nthe 137th coefficient of an expansion in the fine structure constant is.\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Haelfix wrote:
> "It very much depends on the magnitude of the terms in the
> series. Generally, an asymptotic series can be trusted as long as the
> first discarded term is smaller than the last used term, though even
> this rule may be fallacious."
>
> Actually, this is precisely the condition that leads to the rule that
> 1/\alpha is the place where to truncate.
>
> Roughly speaking QFT leads to series that look like U ~ A^m m! m^\alpha
>
> Using your condition... U(m+1) \alpha ^ (m+1) / U(m) \alpha^(m) ~ 1
>
> => m \alpha ~ 1 => m ~ 1/\alpha

No, it depends very much on the relative size of the U(m).
In practice, these are unknown. Nobody has the fiantest idea of how large
the 137th coefficient of an expansion in the fine structure constant is.

Arnold Neumaier

MM

Apr27-04, 03:00 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>John Baez wrote:\n\n>>> Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel\n>>> summable and compute its Abel sum.\n\n[...]\n\n> 1 - 2 + 3 - 4 + ....\n>\n> To Abel sum this, you need to compute the limit of\n>\n> sum (-1)^{n+1} n x^n\n>\n> as x approaches 1 from below. The trick is to find an explicit closed\n> closed form for this sum, and then take the limit.\n>\n> If you have trouble with the "trick" part, maybe it\'s good to think a\n> bit more generally. If you know\n>\n> sum a_n x^n = f(x)\n>\n> then how is\n>\n> sum n a_n x^n\n>\n> related to f(x)? [...]\n\nOK, I think get the idea. One expresses things in terms of various\nderivatives of f(x). So the 1^k... series requires the k\'th derivative of\nf(x) and various lower derivatives, with coefficients that become ever more\npainful to calculate. Doing this by brute force for the first few values of\nk, I arrived at the same answers that Ted obtained with his expression\ninvolving Bernoulli numbers. But I have no idea how to derive the general\nexpression directly. Are other tricks involved?\n\n\n>>> Exercise: Show that Abel summation dominates (C,k) summation\n>>> for all k.\n\nIn an old Schaum Outline "advanced calculus" book I found out how to prove\nthat (C,k+1) dominates (C,k). But although it mentions Abel summation it\ndoesn\'t give the proof of dominance over Cesaro. How does one approach\nthat?\n\n\n> [...] I can tell you about some webpages of mine that\n> give the answer to the Abel sum.\n\nYes please. I\'m sure I\'ll learn something more by reading them.\n\n\n> This stuff is fun! It\'s like black magic.\n\nYes and yes. And this is all before you even get to Borel summation.\nI hope you\'ll move on to that soon.\n\n\n- MikeM.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>John Baez wrote:

>>> Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel
>>> summable and compute its Abel sum.

[...]

> 1 - 2 + 3 - 4 + ....
>
> To Abel sum this, you need to compute the limit of
>
> sum (-1)^{n+1} n x^n
>
> as x approaches 1 from below. The trick is to find an explicit closed
> closed form for this sum, and then take the limit.
>
> If you have trouble with the "trick" part, maybe it's good to think a
> bit more generally. If you know
>
> sum a_n x^n = f(x)
>
> then how is
>
> sum n a_n x^n
>
> related to f(x)? [...]

OK, I think get the idea. One expresses things in terms of various
derivatives of f(x). So the 1^k... series requires the k'th derivative of
f(x) and various lower derivatives, with coefficients that become ever more
painful to calculate. Doing this by brute force for the first few values of
k, I arrived at the same answers that Ted obtained with his expression
involving Bernoulli numbers. But I have no idea how to derive the general
expression directly. Are other tricks involved?

>>> Exercise: Show that Abel summation dominates (C,k) summation
>>> for all k.

In an old Schaum Outline "advanced calculus" book I found out how to prove
that (C,k+1) dominates (C,k). But although it mentions Abel summation it
doesn't give the proof of dominance over Cesaro. How does one approach
that?

> [...] I can tell you about some webpages of mine that
> give the answer to the Abel sum.

Yes please. I'm sure I'll learn something more by reading them.

> This stuff is fun! It's like black magic.

Yes and yes. And this is all before you even get to Borel summation.
I hope you'll move on to that soon.

- MikeM.

John Baez

Apr28-04, 02:48 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <c617t2\\$d3g\\$1@lfa222122.richmond.edu>,\n<ebunn @lfa221051.richmond.edu> wrote:\n\n>>John Baez wrote:\n\n>> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable\n>> > and compute its Abel sum.\n\n> I think I\'ve got this one!\n\nHey, I thought you were too busy for this sort of game!\nPerhaps you just couldn\'t resist the urge to know what this\ndivergent series sums to? It\'s a very pretty series....\n\n> Let me add a 0th term and change the signs (so that my solution will be the\n> negative of the original): we want to evaluate\n>\n> 0^k - 1^k + 2^k - ... = \\sum_j (-1)^j j^k.\n\nOkay.\n\n> To do Abel summation, we\'re supposed to consider the function f\n> defined by\n>\n> f(x) = \\sum_j (-1)^j j^k x^j.\n\nRight.\n\n> Let D be the operator defined by\n>\n> Dg = x (dg/dx).\n\nAha! This is the "trick" I spoke of, which lets you express any\nseries of the form\n\nG(x) = \\sum_j j a_j x^j\n\nin terms of\n\ng(x) = \\sum_j a_j x^j\n\nNamely:\n\nG = Dg\n\nThis "D" is sometimes called the "Euler operator", since Euler\nnoticed that its eigenfunctions are the homogeneous functions,\nas you note:\n\n> D(x^j) = j x^j\n\nIn the Fock representation of the harmonic oscillator, where\nx^j represents the jth energy eigenstate, D is also called the\n"harmonic oscillator Hamiltonian" - or the "number operator".\nThis is how I motivate this game here:\n\nhttp://math.ucr.edu/home/baez/qg-winter2004/zeta.pdf\n\neventually leading to a proof that\n\n1 - 2 + 3 - 4 + ... = 1/4\n\nwhen we use Abel summation.\n\nBut the same trick works for anything like\n\n> 0^k - 1^k + 2^k - ... = \\sum_j (-1)^j j^k.\n\nLet\'s see how you do it!\n\n> f(x) = \\sum_j (-1)^j D^k(x^j)\n> = D^k(\\sum_j (-x^j))\n> = D^k(1/(1+x)).\n>\n> Like any good physicist, I\'m being cavalier about issues of uniform\n> convergence and the like, switching orders of derivatives and sums\n> with wild abandon.\n\nGood!\n\n> But I think this is all OK for |x| < 1.\n\nYes, you can always pass a derivative through the sum when you\nhave a Taylor series and you\'re working *within* the radius of\nconvergence.\n\n>Now we have to evaluate this thing at x = 1. If my kung fu were\n>stronger, I\'d have been able to do this in some clever way. What\n>I actually did was give it to Maple, evaluate it explicitly\n>for k from 1 to 20, and feed the results into the Encyclopedia of\n>Integer Sequences.\n\nThat\'s pretty clever! Ya gotta crack eggs to make omelettes.\nOnce you know the name of the sequence you\'re dealing with, you\ncan usually read up on it and *prove* what the Encyclopedia helped\nyou *guess*... assuming you need a proof.\n\n> [Incidentally, anyone who\'s read this far into this post would\n> probably like the Encyclopedia of Integer Sequences. If you don\'t\n> know about it, check out http://www.research.att.com/~njas/sequences/ ]\n\nJim Dolan loves to crack problems with the help of this marvelous\nwebsite, which was originated by Sloane of "Conway & Sloane" fame.\nHe\'s used it to solve problems in algebraic topology, n-category theory...\nThe point is that an encyclopedia of integer sequences is\npowerful general-purpose pattern recognition tool!\n\nBtw, have you heard about the sequence whose ith term is 1 more than\nthe ith term in the ith sequence in the Encyclopedia of Integer Sequences?\n\nAn easy diagonal argument proves that this sequence cannot be *any*\nof the sequences in the Encyclopedia!\n\nBut in fact, it *is* in the Encyclopedia!\n\nHow can this be?\n\nAnyway:\n\n>After playing around a bit, I found that the sum is\n>\n>-(2^(k+1) - 1) B_{k+1} / (k+1),\n>\n>where B_n is the nth Bernoulli number.\n\nGREAT! This looks right!\n\nIf you look at\n\nhttp://math.ucr.edu/home/baez/qg-winter2004/zeta.pdf\n\nyou\'ll see that Euler took the zeta function\n\nzeta(-k) = 1^k + 2^k + 3^k + ...\n\nmultiplied it by 2^k to get\n\n2^k zeta(-k) = 2^k + 4^k + 6^k + ...\n\nand subtracted twice the latter from the former to get\n\n(1 - 2^{k+1}) zeta(-k) = 1^k - 2^k + 3^k - 4^k + ...\n\nThen he worked out the Abel sum of the right-hand side and,\nlike any good physicist, used this to evaluate zeta(-k) for k = 1,2,3,...\nwithout worrying that his argument was invalid for those values of k.\nFor example, he got\n\nzeta(-1) = 1 + 2 + 3 + 4 = -1/12.\n\nThis sort of thing was justified, later, by Riemann and others.\n\nBut we can turn around Euler\'s dirty trick to figure out the Abel sum\n\n1^k - 2^k + 3^k - 4^k + ...\n\nin terms of the Riemann zeta function! Namely,\n\n1^k - 2^k + 3^k - 4^k + ... = (1 - 2^{k+1}) zeta(-k)\n\n(That was easy.) Then, with some more work, we can show that\n\nzeta(-k) = -B_k/(k+1)\n\nwhere B_k is the kth Bernoulli number. We then get your result:\n\n1^k - 2^k + 3^k - 4^k + ... = (2^{k+1} - 1) B_k / (k+1)\n\nThis is a fairly sneaky and nonrigorous approach to the problem;\nI mention it just because it\'s famous. There has to be a simpler\nand more rigorous one that starts with your observation\n\n1^k - 2^k + 3^k - 4^k + ... = lim_{x -> 1} D^k(1/(1+x)).\n\ntogether with the definition of the Bernoulli numbers:\n\nx/(e^x - 1) = sum_k B_k x^k / k!\n\nNotice: the latter implies that\n\nlim_{x -> 0} D^k(x/(e^x - 1)) = B_k\n\nIt should just require some change of variables to work out\n\nlim_{x -> 1} D^k(1/(1+x))\n\nIf anyone wants more fun with Bernoulli numbers, try this:\n\nhttp://math.ucr.edu/home/baez/qg-winter2004/bernoulli.pdf\n\nIf anyone gets stuck, they can also get answers to the\nhomework problems I\'ve referred to above:\n\nhttp://math.ucr.edu/home/baez/qg-winter2004/zeta.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/zeta_jeffrey.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/zeta_miguel.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/zeta_erin.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/zeta_derek.pdf\n\nhttp://math.ucr.edu/home/baez/qg-winter2004/bernoulli.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/bernoulli_jeffrey.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/bernoulli_miguel.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/bernoulli_erin.pdf\nhttp://math.ucr.edu/home/baez/qg-winter2004/bernoulli_derek.pdf\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <c617t2$d3g$1@lfa222122.richmond.edu>,
<ebunn@lfa221051.richmond.edu> wrote:

>>John Baez wrote:

>> > Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel summable
>> > and compute its Abel sum.

> I think I've got this one!

Hey, I thought you were too busy for this sort of game!
Perhaps you just couldn't resist the urge to know what this
divergent series sums to? It's a very pretty series....

> Let me add a 0th term and change the signs (so that my solution will be the
> negative of the original): we want to evaluate
>
> 0^k - 1^k + 2^k - ... = \sum_j (-1)^j j^k.

Okay.

> To do Abel summation, we're supposed to consider the function f
> defined by
>
> f(x) = \sum_j (-1)^j j^k x^j.

Right.

> Let D be the operator defined by
>
> Dg = x (dg/dx).

Aha! This is the "trick" I spoke of, which lets you express any
series of the form

G(x) = \sum_j j a_j x^j

in terms of

g(x) = \sum_j a_j x^j

Namely:

G = Dg

This "D" is sometimes called the "Euler operator", since Euler
noticed that its eigenfunctions are the homogeneous functions,
as you note:

> D(x^j) = j x^j

In the Fock representation of the harmonic oscillator, where
x^j represents the jth energy eigenstate, D is also called the
"harmonic oscillator Hamiltonian" - or the "number operator".
This is how I motivate this game here:

http://math.ucr.edu/home/baez/qg-winter2004/\zeta.pdf

eventually leading to a proof that

1 - 2 + 3 - 4 + .[/itex].. = 1/4

when we use Abel summation.

But the same trick works for anything like

> 0^k - 1^k + 2^k - ... = \sum_j (-1)^j j^k.

Let's see how you do it!

> f(x) = \sum_j (-1)^j D^k(x^j)
> = D^k(\sum_j (-x^j))
> = D^k(1/(1+x)).
>
> Like any good physicist, I'm being cavalier about issues of uniform
> convergence and the like, switching orders of derivatives and sums
> with wild abandon.

Good!

> But I think this is all OK for |x| < 1.

Yes, you can always pass a derivative through the sum when you
have a Taylor series and you're working *within* the radius of
convergence.

>Now we have to evaluate this thing at x = 1. If my kung fu were
>stronger, I'd have been able to do this in some clever way. What
>I actually did was give it to Maple, evaluate it explicitly
>for k from 1 to 20, and feed the results into the Encyclopedia of
>Integer Sequences.

That's pretty clever! Ya gotta crack eggs to make omelettes.
Once you know the name of the sequence you're dealing with, you
can usually read up on it and *prove* what the Encyclopedia helped
you *guess*... assuming you need a proof.

> [Incidentally, anyone who's read this far into this post would
> probably like the Encyclopedia of Integer Sequences. If you don't
> know about it, check out http://www.research.att.com/~njas/sequences/ ]

Jim Dolan loves to crack problems with the help of this marvelous
website, which was originated by Sloane of "Conway & Sloane" fame.
He's used it to solve problems in algebraic topology, n-category theory...
The point is that an encyclopedia of integer sequences is
powerful general-purpose pattern recognition tool!

Btw, have you heard about the sequence whose ith term is 1 more than
the ith term in the ith sequence in the Encyclopedia of Integer Sequences?

An easy diagonal argument proves that this sequence cannot be *any*
of the sequences in the Encyclopedia!

But in fact, it *is* in the Encyclopedia!

How can this be?

Anyway:

>After playing around a bit, I found that the sum is
>
>-(2^(k+1) - 1) B_{k+1} / (k+1),
>
>where B_n is the nth Bernoulli number.

GREAT! This looks right!

If you look at

http://math.ucr.edu/home/baez/qg-winter2004/\zeta.pdf

you'll see that Euler took the \zeta function

\zeta(-k) = 1^k + 2^k + 3^k + ...

multiplied it by 2^k to get

2^k \zeta(-k) = 2^k + 4^k + 6^k + ...

and subtracted twice the latter from the former to get

(1 - 2^{k+1}) \zeta(-k) = 1^k - 2^k + 3^k - 4^k + ...

Then he worked out the Abel sum of the right-hand side and,
like any good physicist, used this to evaluate \zeta(-k) for k = 1,2,3,...
without worrying that his argument was invalid for those values of k.
For example, he got

\zeta(-1) = 1 + 2 + 3 + 4 = -1/12.

This sort of thing was justified, later, by Riemann and others.

But we can turn around Euler's dirty trick to figure out the Abel sum

1^k - 2^k + 3^k - 4^k + ...

in terms of the Riemann \zeta function! Namely,

1^k - 2^k + 3^k - 4^k + ... = (1 - 2^{k+1}) \zeta(-k)

(That was easy.) Then, with some more work, we can show that

\zeta(-k) = -B_k/(k+1)

where B_k is the kth Bernoulli number. We then get your result:

1^k - 2^k + 3^k - 4^k + ... = (2^{k+1} - 1) B_k / (k+1)

This is a fairly sneaky and nonrigorous approach to the problem;
I mention it just because it's famous. There has to be a simpler
and more rigorous one that starts with your observation

1^k - 2^k + 3^k - 4^k + ... = lim_{x -> 1} D^k(1/(1+x)).

together with the definition of the Bernoulli numbers:

[itex]x/(e^x - 1) = sum_k B_k x^k / k!

Notice: the latter implies that

lim_{x -> 0} D^k(x/(e^x - 1)) = B_k

It should just require some change of variables to work out

lim_{x -> 1} D^k(1/(1+x))

If anyone wants more fun with Bernoulli numbers, try this:

http://math.ucr.edu/home/baez/qg-winter2004/bernoulli.pdf

If anyone gets stuck, they can also get answers to the
homework problems I've referred to above:

http://math.ucr.edu/home/baez/qg-winter2004/\zeta.pdf
http://math.ucr.edu/home/baez/qg-winter2004/\zeta_jeffrey.pdf
http://math.ucr.edu/home/baez/qg-winter2004/\zeta_miguel.pdf
http://math.ucr.edu/home/baez/qg-winter2004/\zeta_erin.pdf
http://math.ucr.edu/home/baez/qg-winter2004/\zeta_derek.pdf

http://math.ucr.edu/home/baez/qg-winter2004/bernoulli.pdf
http://math.ucr.edu/home/baez/qg-winter2004/bernoulli_jeffrey.pdf
http://math.ucr.edu/home/baez/qg-winter2004/bernoulli_miguel.pdf
http://math.ucr.edu/home/baez/qg-winter2004/bernoulli_erin.pdf
http://math.ucr.edu/home/baez/qg-winter2004/bernoulli_derek.pdf

MM

Apr28-04, 02:30 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>John Baez wrote:\n\n>> Exercise: Find a series that is Abel summable but not (C,k) summable\n>> for any k.\n\nTed wrote:\n\n> I think I probably know the answer to the last one too.\n\nI don\'t think I understand your solution properly...\n\n> We should look at alternating sums whose terms blow up faster than any\n> power, since we already know that the alternating series of kth powers\n> is (C,k) summable. That suggests we try\n>\n> \\sum_{j=0}^\\infty (-1)^j exp(j a)\n> = 1 - exp(a) + exp(2a) - exp(3a) + ...\n>\n> with a some positive constant.\n\nHmmm, this looks like a geometric series 1 + r + r^2 + r^3 + ...\nwith r = (-exp(a)). It converges if |r|<1 and diverges otherwise.\nIn your case, |r| > 1 so we would normally expect divergence unless\nsome tricky miracle comes to our rescue.\n\n> But it is Abel summable. The function to consider is\n>\n> f(x) = \\sum_{j=0}^\\infty (-1)^j exp(ja - jt)\n>\n> where t = -ln x. That\'s just a geometric series:\n>\n> f(x) = 1/(1+exp(a-t)).\n\nBut doesn\'t the usual formula for summing a geometric series only work\nhere if |exp(a-t)| < 1? I.e: (a-t) < 0, i.e: a < t.\n\n> Take the limit as x->1, i.e. as t->0, and you\'re done.\n\nTaking the limit as x->1 from below means taking t->0 from above.\nThat means that your use of the geometric sum formula is only valid\nif a->0 also. I.e: only if your series is merely 1 - 1 + 1 - 1 + ....\n\n\n> The Abel sum is\n>\n> 1/(1+exp(a)).\n>\n> Did I get it right?\n\nIf that were correct, then couldn\'t you sum *any* alternating\ngeometric series easily regardless of whether |r|<1 ?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>John Baez wrote:

>> Exercise: Find a series that is Abel summable but not (C,k) summable
>> for any k.

Ted wrote:

> I think I probably know the answer to the last one too.

I don't think I understand your solution properly...

> We should look at alternating sums whose terms blow up faster than any
> power, since we already know that the alternating series of kth powers
> is (C,k) summable. That suggests we try
>
> \sum_{j=0}^\infty (-1)^j \exp(j a)
> = 1 - \exp(a) + \exp(2a) - \exp(3a) + ...
>
> with a some positive constant.

Hmmm, this looks like a geometric series 1 + r + r^2 + r^3 + ...
with r = (-\exp(a)). It converges if |r|<1 and diverges otherwise.
In your case, |r| > 1 so we would normally expect divergence unless
some tricky miracle comes to our rescue.

> But it is Abel summable. The function to consider is
>
> f(x) = \sum_{j=0}^\infty (-1)^j \exp(ja - jt)
>
> where t = -ln x. That's just a geometric series:
>
> f(x) = 1/(1+\exp(a-t)).

But doesn't the usual formula for summing a geometric series only work
here if |\exp(a-t)| < 1? I.e: (a-t) < 0, i.e: a < t.

> Take the limit as x->1, i.e. as t->0, and you're done.

Taking the limit as x->1 from below means taking t->0 from above.
That means that your use of the geometric sum formula is only valid
if a->0 also. I.e: only if your series is merely 1 - 1 + 1 - 1 + ....

> The Abel sum is
>
> 1/(1+\exp(a)).
>
> Did I get it right?

If that were correct, then couldn't you sum *any* alternating
geometric series easily regardless of whether |r|<1 ?

John Baez

Apr28-04, 02:30 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <c63d1e\\$dvu\\$1@lfa222122.richmond.edu>,\n<ebunn @lfa221051.richmond.edu> wrote:\n\n>>> John Baez wrote:\n\n>>> > Exercise: Find a series that is Abel summable but not (C,k) summable\n>>> > for any k.\n\n>We should look at alternating sums whose terms blow up faster than any\n>power, since we already know that the alternating series of kth powers\n>is (C,k) summable. That suggests we try\n>\n>\\sum_{j=0}^\\infty (-1)^j exp(j a)\n>= 1 - exp(a) + exp(2a) - exp(3a) + ...\n>\n>with a some positive constant.\n\nI don\'t think this is Abel summable unless you play a sneaky trick -\nwhich you do. Abel summation tames divergent sums by inserting an\nexponential damping factor and then removing it. This exponential\ndamping factor is good enough to cure sums where the terms grow like\na polynomial. But, it shouldn\'t be good enough to cure sums where the\nterms grow like an exponential, like yours!\n\nIn case this doesn\'t make sense, let\'s just see if your sum is Abel\nsummable - without playing any sneaky tricks. To do this, we look at\n\nsum_{j=0}^infty (-1)^j exp(j a) x^j\n\nand see if we can take the limit as x -> 1 from below. If we write\n\nx = exp(-t)\n\nthis is the same as the limit where b -> 0 from above.\nThis is nice because\n\nx^j = exp(-j t)\n\nso our sum becomes\n\nsum_{j=0}^infty (-1)^j exp(j (a - t))\n\nSo, let\'s try to take the limit as t -> 0.\n\nWhoops! As soon as t gets below a, this sum diverges!\n\nSo, the original sum was NOT Abel summable.\n\nUnless you play a sneaky trick....\n\n>The nth partial sum of this guy grows exponentially with n as\n>n->infinity, as does the average of the first n partial sums. So it\'s\n>certainly not Cesaro summable. I\'m pretty sure the iterated averages\n>needed to compute the (C,k) sum also blow up exponentially, although I\n>haven\'t proved that.\n\nThat\'s true. (C,k) summation is basically good enough to handle\nalternating sums where terms grow polynomially. Abel summation can\nhandle these too. But as I\'ve shown, Abel summation CANNOT handle\nalternating sums where the terms grow exponentially! So, we really\nneed an example where the rate of growth is intermediate between\npolynomial and exponential!\n\nBut now I\'m giving away the answer to my own puzzle - tsk tsk.\nLet\'s instead see your sneaky trick.\n\n>But it is Abel summable.\n\nYeah, right.\n\n> The function to consider is\n>\n>f(x) = \\sum_{j=0}^\\infty (-1)^j exp(ja - jt)\n>\n>where t = -ln x. That\'s just a geometric series:\n>\n>f(x) = 1/(1+exp(a-t)).\n\nNote this blow up when t = a.\n\n>Take the limit as x->1, i.e. as t->0, and you\'re done.\n\nAha! Even though the sum diverges when t gets below a,\nyou don\'t care! You instead work with your closed-form\n*answer* to the sum, and analytically continue around the\npole at t = a. Very sneaky!\n\n>The Abel sum is\n>\n>1/(1+exp(a)).\n>\n>Did I get it right?\n\nWell, you didn\'t answer *my* puzzle, but you came up with\nsomething nice. In the usual definition of Abel summation,\nwe demand that\n\nsum a_n x^n\n\nconverge for all x < 1, and demand that the limit of this sum\nas x -> 1 from below exists.\n\nBUT, you\'re right that we can often analytically continue around\nsome value of x < 1 where the sum diverges, and evaluate this\nanalytic continuation at x = 1.\n\nSo, we\'ve got "Bunn summation". I believe it dominates\nall the (C,k) summations and also Abel summation, meaning it\nagrees with them when they\'re both defined, but can sum more\nseries. I wonder if it agrees with Borel summation when both\nare defined, and whether it\'s dominated by Borel summation!\n\nBut my original puzzle remains...\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <c63d1e$dvu$1@lfa222122.richmond.edu>,
<ebunn@lfa221051.richmond.edu> wrote:

>>> John Baez wrote:

>>> > Exercise: Find a series that is Abel summable but not (C,k) summable
>>> > for any k.

>We should look at alternating sums whose terms blow up faster than any
>power, since we already know that the alternating series of kth powers
>is (C,k) summable. That suggests we try
>
>\sum_{j=0}^\infty (-1)^j \exp(j a)>= 1 - \exp(a) + \exp(2a) - \exp(3a) + ...
>
>with a some positive constant.

I don't think this is Abel summable unless you play a sneaky trick -
which you do. Abel summation tames divergent sums by inserting an
exponential damping factor and then removing it. This exponential
damping factor is good enough to cure sums where the terms grow like
a polynomial. But, it shouldn't be good enough to cure sums where the
terms grow like an exponential, like yours!

In case this doesn't make sense, let's just see if your sum is Abel
summable - without playing any sneaky tricks. To do this, we look at

sum_{j=0}^\infty (-1)^j \exp(j a) x^j

and see if we can take the limit as x -> 1 from below. If we write

x = \exp(-t)

this is the same as the limit where b -> from above.
This is nice because

x^j = \exp(-j t)

so our sum becomes

sum_{j=0}^\infty (-1)^j \exp(j (a - t))

So, let's try to take the limit as t -> .

Whoops! As soon as t gets below a, this sum diverges!

So, the original sum was NOT Abel summable.

Unless you play a sneaky trick....

>The nth partial sum of this guy grows exponentially with n as
>n->infinity, as does the average of the first n partial sums. So it's
>certainly not Cesaro summable. I'm pretty sure the iterated averages
>needed to compute the (C,k) sum also blow up exponentially, although I
>haven't proved that.

That's true. (C,k) summation is basically good enough to handle
alternating sums where terms grow polynomially. Abel summation can
handle these too. But as I've shown, Abel summation CANNOT handle
alternating sums where the terms grow exponentially! So, we really
need an example where the rate of growth is intermediate between
polynomial and exponential!

But now I'm giving away the answer to my own puzzle - tsk tsk.
Let's instead see your sneaky trick.

>But it is Abel summable.

Yeah, right.

> The function to consider is
>
>f(x) = \sum_{j=0}^\infty (-1)^j \exp(ja - jt)
>
>where t = -ln x. That's just a geometric series:
>
>f(x) = 1/(1+\exp(a-t)).

Note this blow up when t = a.

>Take the limit as x->1, i.e. as t->0, and you're done.

Aha! Even though the sum diverges when t gets below a,
you don't care! You instead work with your closed-form
*answer* to the sum, and analytically continue around the
pole at t = a. Very sneaky!

>The Abel sum is
>
>1/(1+\exp(a)).
>
>Did I get it right?

Well, you didn't answer *my* puzzle, but you came up with
something nice. In the usual definition of Abel summation,
we demand that

sum a_n x^n

converge for all x < 1, and demand that the limit of this sum
as x -> 1 from below exists.

BUT, you're right that we can often analytically continue around
some value of x < 1 where the sum diverges, and evaluate this
analytic continuation at x = 1.

So, we've got "Bunn summation". I believe it dominates
all the (C,k) summations and also Abel summation, meaning it
agrees with them when they're both defined, but can sum more
series. I wonder if it agrees with Borel summation when both
are defined, and whether it's dominated by Borel summation!

But my original puzzle remains...

John Baez

Apr28-04, 03:26 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <38b55e8c.0404260058.5a849fdc@posting.google.com>, \nMM <mikem@despammed.com> wrote:\n\n>John Baez wrote:\n\n> >>> Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel\n> >>> summable and compute its Abel sum.\n\nFor example:\n\n> > 1 - 2 + 3 - 4 + ...\n> >\n> > To Abel sum this, you need to compute the limit of\n> >\n> > sum (-1)^{n+1} n x^n\n> >\n> > as x approaches 1 from below. The trick is to find an explicit closed\n> > closed form for this sum, and then take the limit.\n\nIn case anyone didn\'t get it yet, here\'s how it goes:\n\nsum (-1)^{n+1} n x^n = x d/dx [sum (-1)^{n+1} x^n]\n\n= x d/dx [-1/(1 + x)]\n\n= x / (1 + x)^2\n\nand then taking the limit as x approaches 1 from below, we get 1/4.\n\nBut what about in general, for\n\n1^k - 2^k + 3^k - 4^k + ... ?\n\nWell, this is the limit as x approaches 1 from below of\n\nsum (-1)^{n+1} n^k x^n = (x d/dx)^k [-1/(1 + x)]\n\nbut to actually work this out is rather messy.\n\n>Doing this by brute force for the first few values of\n>k, I arrived at the same answers that Ted obtained with his expression\n>involving Bernoulli numbers. But I have no idea how to derive the general\n>expression directly. Are other tricks involved?\n\nIn another post on this thread, I describe an ultra-clever nonrigorous\nmethod due to Euler. But that\'s really overkill; I believe the\nonly "trick" you really need is the definition of Bernoulli numbers!\n\nIn other words: the pattern is sufficiently complicated, but sufficiently\nsimilar to the pattern that shows up here:\n\nx/(e^x - 1) = sum_k B_k x^k / k!\n\nthat it\'s best to express your answer in terms of the Bernoulli numbers\nB_k.\n\nIt\'s good to know a bit about Bernoulli numbers - at least enough\nto guess when they\'re showing up, and where to learn more about\nthem. Probably all you *need* to remember is that they show up\nwhen you do this sum:\n\n1^k + 2^k + ... + n^k\n\nas well as this Abel sum:\n\n1^k - 2^k + 3^k - 4^k + ...\n\nand also this sum, the Riemann zeta function:\n\nzeta(k) = 1^{-k} + 2^{-k} + 3^{-k} + ...\n\nwhen k is either even and positive, or a negative even integer.\n\n(In the latter case the sum diverges, but becomes well-defined\nvia analytic continuation; this is a special case of "zeta function\nregularization", which shows up a lot in quantum field theory.)\n\nYou can learn a bit more about Bernoulli numbers here:\n\nHomework for "week 6" here: http://math.ucr.edu/home/baez/qg-winter2004/\n\nHomework for "week 7" here: http://math.ucr.edu/home/baez/qg-winter2004/\n\n> >>> Exercise: Show that Abel summation dominates (C,k) summation\n> >>> for all k.\n>\n>In an old Schaum Outline "advanced calculus" book I found out how to prove\n>that (C,k+1) dominates (C,k). But although it mentions Abel summation it\n>doesn\'t give the proof of dominance over Cesaro. How does one approach\n>that?\n\nUmm, are you trying to show the part about them giving the same\nanswer when they agree, or the part about Abel handling cases that\nCesaro can\'t handle?\n\n> > [...] I can tell you about some webpages of mine that\n> > give the answer to the Abel sum.\n>\n>Yes please. I\'m sure I\'ll learn something more by reading them.\n\nThis is the one I was talking about:\n\nHomework for "week 5" here: http://math.ucr.edu/home/baez/qg-winter2004/\n\nIn all these homeworks, there\'ll be some concepts that will\nseem mysterious unless you read the course notes also available at\nthis webpage... but you can try to just ignore those concepts,\nsince they\'re somewhat tangential.\n\n>Yes and yes. And this is all before you even get to Borel summation.\n>I hope you\'ll move on to that soon.\n\nOh yeah? Hmm. Maybe I\'ll start by giving a cute example of\na series we can handle with Borel summation, but not with the\nmethods we\'ve been discussing so far:\n\n0! + 1!/x + 2!/x^2 + 3!/x^3 + ...\n\nDoes this look familiar?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <38b55e8c.0404260058.5a849fdc@posting.google.com>,
MM <mikem@despammed.com> wrote:

>John Baez wrote:

> >>> Exercise: Show that 1^k - 2^k + 3^k - 4^k + ... is Abel
> >>> summable and compute its Abel sum.

For example:

> > 1 - 2 + 3 - 4 + ...
> >
> > To Abel sum this, you need to compute the limit of
> >
> > sum (-1)^{n+1} n x^n
> >
> > as x approaches 1 from below. The trick is to find an explicit closed
> > closed form for this sum, and then take the limit.

In case anyone didn't get it yet, here's how it goes:

sum (-1)^{n+1} n x^n = x d/dx [sum (-1)^{n+1} x^n]= x d/dx [-1/(1 + x)]= x / (1 + x)^2

and then taking the limit as x approaches 1 from below, we get 1/4.

But what about in general, for

1^k - 2^k + 3^k - 4^k + .[/itex].. ?

Well, this is the limit as x approaches 1 from below of

sum (-1)^{n+1} n^k x^n = (x d/dx)^k [-1/(1 + x)]

but to actually work this out is rather messy.

>Doing this by brute force for the first few values of
>k, I arrived at the same answers that Ted obtained with his expression
>involving Bernoulli numbers. But I have no idea how to derive the general
>expression directly. Are other tricks involved?

In another post on this thread, I describe an ultra-clever nonrigorous
method due to Euler. But that's really overkill; I believe the
only "trick" you really need is the definition of Bernoulli numbers!

In other words: the pattern is sufficiently complicated, but sufficiently
similar to the pattern that shows up here:

x/(e^x - 1) = sum_k B_k x^k / k!

that it's best to express your answer in terms of the Bernoulli numbers
B_k.

It's good to know a bit about Bernoulli numbers - at least enough
to guess when they're showing up, and where to learn more about
them. Probably all you *need* to remember is that they show up
when you do this sum:

1^k + 2^k + ..[itex]. + n^k

as well as this Abel sum:

1^k - 2^k + 3^k - 4^k + ...

and also this sum, the Riemann \zeta function:

\zeta(k) = 1^{-k} + 2^{-k} + 3^{-k} + ...

when k is either even and positive, or a negative even integer.

(In the latter case the sum diverges, but becomes well-defined
via analytic continuation; this is a special case of "\zeta function
regularization", which shows up a lot in quantum field theory.)

You can learn a bit more about Bernoulli numbers here:

Homework for "week 6" here: http://math.ucr.edu/home/baez/qg-winter2004/

Homework for "week 7" here: http://math.ucr.edu/home/baez/qg-winter2004/

> >>> Exercise: Show that Abel summation dominates (C,k) summation
> >>> for all k.
>
>In an old Schaum Outline "advanced calculus" book I found out how to prove
>that (C,k+1) dominates (C,k). But although it mentions Abel summation it
>doesn't give the proof of dominance over Cesaro. How does one approach
>that?

Umm, are you trying to show the part about them giving the same
answer when they agree, or the part about Abel handling cases that
Cesaro can't handle?

> > [...] I can tell you about some webpages of mine that
> > give the answer to the Abel sum.
>
>Yes please. I'm sure I'll learn something more by reading them.

This is the one I was talking about:

Homework for "week 5" here: http://math.ucr.edu/home/baez/qg-winter2004/

In all these homeworks, there'll be some concepts that will
seem mysterious unless you read the course notes also available at
this webpage... but you can try to just ignore those concepts,
since they're somewhat tangential.

>Yes and yes. And this is all before you even get to Borel summation.
>I hope you'll move on to that soon.

Oh yeah? Hmm. Maybe I'll start by giving a cute example of
a series we can handle with Borel summation, but not with the
methods we've been discussing so far:

0! + 1!/x + 2!/x^2 + 3!/x^3 + ...

Does this look familiar?

ebunn@lfa221051.richmond.edu

Apr30-04, 03:00 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <38b55e8c.0404271746.2e48a87a@posting.google.com>, \nMM <mikem@despammed.com> wrote:\n\n>Ted wrote:\n>\n> > I think I probably know the answer to the last one too.\n>\n>I don\'t think I understand your solution properly...\n\nYou\'re being too kind to me. I think you understand perfectly why my\nanswer was completely wrong! John Baez has just posted a response to\nmy post explaining in detail why I was wrong.\n\nThe bottom line seems to be that a series that\'s Abel summable but not\n(C,k) summable for any k must have terms that grow faster than any\npower but slower than exponentially. So I guess maybe something like\n\n\\sum_j (-1)^j exp(j^a)\n\nfor 0<a<1 might fit the bill. Clearly the function you need to consider\nfor the Abel sum,\n\n\\sum_j (-1)^j exp(j^a - j t)\n\nconverges for all positive t (since the decaying exponential dominates\nas j->infinity), but I have no idea how to evaluate it explicitly or\nto show that its limit exists as t->0.\n\n[...]\n\n>If that were correct, then couldn\'t you sum *any* alternating\n>geometric series easily regardless of whether |r|<1 ?\n\nIf my solution were correct, then it would mean that any alternating\ngeometric series was Abel summable, regardless of whether |r|<1. Of\ncourse, that doesn\'t mean such a series would be convergent in the\nusual sense. What we all learned way back when is still true, after\nall: a geometric series with |r| >= 1 doesn\'t converge.\n\nI\'ll conjecture that that statement is in fact true for Borel\nsummation. That is, any geometric series, even with |r| >= 1, is\nBorel summable. My evidence for this: in his post, John wonders\nwhether that method that I used for summing is dominated by\nBorel summation. He no doubt knows whether geometric series are\nBorel summable. If they weren\'t, he would know that what he calls\n"Bunn summation" is not dominated by Borel summation.\n\n(Since I don\'t even know the definition of Borel summation, I can\'t\ncome up with a better argument than that!)\n\n-Ted\n\n\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <38b55e8c.0404271746.2e48a87a@posting.google.com>,
MM <mikem@despammed.com> wrote:

>Ted wrote:
>
> > I think I probably know the answer to the last one too.
>
>I don't think I understand your solution properly...

You're being too kind to me. I think you understand perfectly why my
answer was completely wrong! John Baez has just posted a response to
my post explaining in detail why I was wrong.

The bottom line seems to be that a series that's Abel summable but not
(C,k) summable for any k must have terms that grow faster than any
power but slower than exponentially. So I guess maybe something like

\sum_j (-1)^j \exp(j^a)

for 0<a<1 might fit the bill. Clearly the function you need to consider
for the Abel sum,

\sum_j (-1)^j \exp(j^a - j t)

converges for all positive t (since the decaying exponential dominates
as j->infinity), but I have no idea how to evaluate it explicitly or
to show that its limit exists as t->0.

[...]

>If that were correct, then couldn't you sum *any* alternating
>geometric series easily regardless of whether |r|<1 ?

If my solution were correct, then it would mean that any alternating
geometric series was Abel summable, regardless of whether |r|<1. Of
course, that doesn't mean such a series would be convergent in the
usual sense. What we all learned way back when is still true, after
all: a geometric series with |r| >= 1 doesn't converge.

I'll conjecture that that statement is in fact true for Borel
summation. That is, any geometric series, even with |r| >= 1, is
Borel summable. My evidence for this: in his post, John wonders
whether that method that I used for summing is dominated by
Borel summation. He no doubt knows whether geometric series are
Borel summable. If they weren't, he would know that what he calls
"Bunn summation" is not dominated by Borel summation.

(Since I don't even know the definition of Borel summation, I can't
come up with a better argument than that!)

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

MM

Apr30-04, 03:01 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>JB:\n>>> Exercise: Show that Abel summation dominates (C,k) summation\n>>> for all k.\n\nMe:\n>> [...] I found out how to prove\n>> that (C,k+1) dominates (C,k). But although it mentions Abel\n>> summation it doesn\'t give the proof of dominance over Cesaro.\n>> How does one approach that?\n\nJB:\n> Umm, are you trying to show the part about them giving the same\n> answer when they agree, or the part about Abel handling cases\n> that Cesaro can\'t handle?\n\nI meant the former.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>JB:
>>> Exercise: Show that Abel summation dominates (C,k) summation
>>> for all k.

Me:
>> [...] I found out how to prove
>> that (C,k+1) dominates (C,k). But although it mentions Abel
>> summation it doesn't give the proof of dominance over Cesaro.
>> How does one approach that?

JB:
> Umm, are you trying to show the part about them giving the same
> answer when they agree, or the part about Abel handling cases
> that Cesaro can't handle?

I meant the former.

Haelfix

May3-04, 05:52 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n> No, it depends very much on the relative size of the U(m).\n> In practice, these are unknown. Nobody has the fiantest idea of how large\n> the 137th coefficient of an expansion in the fine structure constant is.\n>\n> Arnold Neumaier\n\nI\'m sorry, I don\'t agree. If we are indeed dealing with an asymptotic\nseries (and that isn\'t rigorous but thats what most people believe)\nthen the above bound is valid.\n\nYou don\'t need to know what the 137th entry is.. You just need to know\nthat the division between the 137th and 138th is large.\n\nIf you take the form I gave above for the U(m) you won\'t get strange\noscillatory contributions for the series. You get precisely one\nturning point, and thats the place where we like to call it quits.\n\nIncidentally, the problem doesn\'t appear in practise for the EM fine\nstructure constant series, since you would have to calculate 100-200\nor so loops before any problem becomes manifest.\n\nThis is a known result, that matches the behaviour for lower\ndimensional field theories that have been analytically solved. In 4d,\nyou get the renormalon behaviour, and the eqn is different. But we\nexpect the general argument to hold true nonetheless.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>No, it depends very much on the relative size of the U(m).
> In practice, these are unknown. Nobody has the fiantest idea of how large
> the 137th coefficient of an expansion in the fine structure constant is.
>
> Arnold Neumaier

I'm sorry, I don't agree. If we are indeed dealing with an asymptotic
series (and that isn't rigorous but thats what most people believe)
then the above bound is valid.

You don't need to know what the 137th entry is.. You just need to know
that the division between the 137th and 138th is large.

If you take the form I gave above for the U(m) you won't get strange
oscillatory contributions for the series. You get precisely one
turning point, and thats the place where we like to call it quits.

Incidentally, the problem doesn't appear in practise for the EM fine
structure constant series, since you would have to calculate 100-200
or so loops before any problem becomes manifest.

This is a known result, that matches the behaviour for lower
dimensional field theories that have been analytically solved. In 4d,
you get the renormalon behaviour, and the eqn is different. But we
expect the general argument to hold true nonetheless.

Arnold Neumaier

May3-04, 09:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nHaelfix wrote:\n>>No, it depends very much on the relative size of the U(m).\n>>In practice, these are unknown. Nobody has the fiantest idea of how large\n>>the 137th coefficient of an expansion in the fine structure constant is.\n>>\n>>Arnold Neumaier\n>\n>\n> I\'m sorry, I don\'t agree. If we are indeed dealing with an asymptotic\n> series (and that isn\'t rigorous but thats what most people believe)\n> then the above bound is valid.\n\nNo. You need much stronger assumptions than that, since everything depends\non the growth behavior of the coefficients.\n\n\n> You don\'t need to know what the 137th entry is.. You just need to know\n> that the division between the 137th and 138th is large.\n\nBut how do you know that? Consider the asymptotic series\nsum_n (n!)^s alpha^n\nFor fixed alpha, the index of the smallest contribution (which is\nabout the place where to stop) is at n=[alpha^{-1/s}], which\nstrongly depends on s. Only for s=1 we get n=137 [if alpha=1/137],\nfor s=2 we have n=11, for s=1/2 about n=18769.\n\nSo to prove your recipe correct you\'d have to prove that the\nQED contributions grow approximately like n!, not much slower and\nnot much larger. Could you please refer me to a paper where this is\nproved?\n\n\nArnold Neumaier\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Haelfix wrote:
>>No, it depends very much on the relative size of the U(m).
>>In practice, these are unknown. Nobody has the fiantest idea of how large
>>the 137th coefficient of an expansion in the fine structure constant is.
>>
>>Arnold Neumaier
>
>
> I'm sorry, I don't agree. If we are indeed dealing with an asymptotic
> series (and that isn't rigorous but thats what most people believe)
> then the above bound is valid.

No. You need much stronger assumptions than that, since everything depends
on the growth behavior of the coefficients.

> You don't need to know what the 137th entry is.. You just need to know
> that the division between the 137th and 138th is large.

But how do you know that? Consider the asymptotic series
sum_n (n!)^s \alpha^n
For fixed \alpha, the index of the smallest contribution (which is
about the place where to stop) is at n=[\alpha^{-1/s}], which
strongly depends on s. Only for s=1 we get n=137 [if \alpha=1/137],
for s=2 we have n=11, for s=1/2 about n=18769.

So to prove your recipe correct you'd have to prove that the
QED contributions grow approximately like n!, not much slower and
not much larger. Could you please refer me to a paper where this is
proved?

Arnold Neumaier

Esa A E Peuha

May6-04, 08:52 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>baez@galaxy.ucr.edu (John Baez) writes:\n\n> Btw, have you heard about the sequence whose ith term is 1 more than\n> the ith term in the ith sequence in the Encyclopedia of Integer Sequences?\n>\n> An easy diagonal argument proves that this sequence cannot be *any*\n> of the sequences in the Encyclopedia!\n>\n> But in fact, it *is* in the Encyclopedia!\n>\n> How can this be?\n\nThat\'s a nice paradox. :-)\n\nHere\'s something else that may be of interest. If given a recurrence\nrelation x_n = a * x_{n-1} + b, anyone can easily express x_n as\nA_n * X_0 + B_n * b. However, if the relation is instead\nx_n = a * x_{n-1} + b * x_{n-2} + c, then it\'s not so easy to find the\ncoefficients in x_n = A_n * x_1 + B_n * x_0 + C_n * c (although once you\nknow A_n, B_n and C_n follow easily). If the relation is\nx_n = a * x_{n-1} + b * x_{n-2} + c * x_{n-3} + d, it\'s even more\ndifficult, not to mention longer relations. The challenge here is to\nsolve this problem for _all_ such relations at once.\n\n--\nEsa Peuha\nstudent of mathematics at the University of Helsinki\nhttp://www.helsinki.fi/~peuha/\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>baez@galaxy.ucr.edu (John Baez) writes:

> Btw, have you heard about the sequence whose ith term is 1 more than
> the ith term in the ith sequence in the Encyclopedia of Integer Sequences?
>
> An easy diagonal argument proves that this sequence cannot be *any*
> of the sequences in the Encyclopedia!
>
> But in fact, it *is* in the Encyclopedia!
>
> How can this be?

That's a nice paradox. :-)

Here's something else that may be of interest. If given a recurrence
relation x_n = a * x_{n-1} + b, anyone can easily express x_n as
A_n * X_0 + B_n * b. However, if the relation is instead
x_n = a * x_{n-1} + b * x_{n-2} + c, then it's not so easy to find the
coefficients in x_n = A_n * x_1 + B_n * x_0 + C_n * c (although once you
know A_n, B_n and C_n follow easily). If the relation is
x_n = a * x_{n-1} + b * x_{n-2} + c * x_{n-3} + d, it's even more
difficult, not to mention longer relations. The challenge here is to
solve this problem for _all_ such relations at once.

--
Esa Peuha
student of mathematics at the University of Helsinki
http://www.helsinki.fi/~peuha/

Aaron Bergman

May7-04, 07:39 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <86p1xlzjiti.fsf@sirppi.helsinki.fi>,\nEsa A E Peuha <esa.peuha@helsinki.fi> wrote:\n\n>\n> Here\'s something else that may be of interest. If given a recurrence\n> relation x_n = a * x_{n-1} + b, anyone can easily express x_n as\n> A_n * X_0 + B_n * b. However, if the relation is instead\n> x_n = a * x_{n-1} + b * x_{n-2} + c, then it\'s not so easy to find the\n> coefficients in x_n = A_n * x_1 + B_n * x_0 + C_n * c (although once you\n> know A_n, B_n and C_n follow easily). If the relation is\n> x_n = a * x_{n-1} + b * x_{n-2} + c * x_{n-3} + d, it\'s even more\n> difficult, not to mention longer relations. The challenge here is to\n> solve this problem for _all_ such relations at once.\n\nJust use generating functions and partial fractions.\n\nAaron\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <86p1xlzjiti.fsf@sirppi.helsinki.fi>,
Esa A E Peuha <esa.peuha@helsinki.fi> wrote:

>
> Here's something else that may be of interest. If given a recurrence
> relation x_n = a * x_{n-1} + b, anyone can easily express x_n as
> A_n * X_0 + B_n * b. However, if the relation is instead
> x_n = a * x_{n-1} + b * x_{n-2} + c, then it's not so easy to find the
> coefficients in x_n = A_n * x_1 + B_n * x_0 + C_n * c (although once you
> know A_n, B_n and C_n follow easily). If the relation is
> x_n = a * x_{n-1} + b * x_{n-2} + c * x_{n-3} + d, it's even more
> difficult, not to mention longer relations. The challenge here is to
> solve this problem for _all_ such relations at once.

Just use generating functions and partial fractions.

Aaron

ebunn@lfa221051.richmond.edu

May10-04, 06:02 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn article <c6kec0\\$n5\\$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu> wrote:\n\n>Hey, I thought you were too busy for this sort of game!\n>Perhaps you just couldn\'t resist the urge to know what this\n>divergent series sums to? It\'s a very pretty series....\n\nYes, I have a bad habit of letting myself get distracted by\ninteresting puzzles with no direct application to what I\'m supposed to\nbe doing. Actually, whether this is a bad habit or not depends\nentirely on whether I end up getting tenure: as long as I do, it will\nhave been a good habit. After all, I do learn things this way, and as\nEmil Faber said, "Knowledge is Good."\n\n>Btw, have you heard about the sequence whose ith term is 1 more than\n>the ith term in the ith sequence in the Encyclopedia of Integer Sequences?\n>\n>An easy diagonal argument proves that this sequence cannot be *any*\n>of the sequences in the Encyclopedia!\n>\n>But in fact, it *is* in the Encyclopedia!\n>\n>How can this be?\n\nI think the Encylopedia does include sequences that terminate after\nfinitely many terms. If this is the nth sequence in the Encyclopedia,\npresumably it has only n-1 terms.\n\n-Ted\n\n--\n[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <c6kec0$n5$1@glue.ucr.edu>, John Baez <baez@galaxy.ucr.edu> wrote:

>Hey, I thought you were too busy for this sort of game!
>Perhaps you just couldn't resist the urge to know what this
>divergent series sums to? It's a very pretty series....

Yes, I have a bad habit of letting myself get distracted by
interesting puzzles with no direct application to what I'm supposed to
be doing. Actually, whether this is a bad habit or not depends
entirely on whether I end up getting tenure: as long as I do, it will
have been a good habit. After all, I do learn things this way, and as
Emil Faber said, "Knowledge is Good."

>Btw, have you heard about the sequence whose ith term is 1 more than
>the ith term in the ith sequence in the Encyclopedia of Integer Sequences?
>
>An easy diagonal argument proves that this sequence cannot be *any*
>of the sequences in the Encyclopedia!
>
>But in fact, it *is* in the Encyclopedia!
>
>How can this be?

I think the Encylopedia does include sequences that terminate after
finitely many terms. If this is the nth sequence in the Encyclopedia,
presumably it has only n-1 terms.

-Ted

--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.]

Esa A E Peuha

May11-04, 06:11 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Aaron Bergman <abergman@physics.utexas.edu> writes:\n\n> Just use generating functions and partial fractions.\n\nWell, if you can figure out the generating functions... (I couldn\'t,\nso I just recognizes the pattern in the coefficients and managed to\nconvince myself that it holds for all possible relations.)\n\n--\nEsa Peuha\nstudent of mathematics at the University of Helsinki\nhttp://www.helsinki.fi/~peuha/\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Aaron Bergman <abergman@physics.utexas.edu> writes:

> Just use generating functions and partial fractions.

Well, if you can figure out the generating functions... (I couldn't,
so I just recognizes the pattern in the coefficients and managed to
convince myself that it holds for all possible relations.)

--
Esa Peuha
student of mathematics at the University of Helsinki
http://www.helsinki.fi/~peuha/

Aaron Bergman

May14-04, 05:05 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <86pisf3tmly.fsf@sirppi.helsinki.fi>,\nEsa A E Peuha <esa.peuha@helsinki.fi> wrote:\n\n> Aaron Bergman <abergman@physics.utexas.edu> writes:\n>\n> > Just use generating functions and partial fractions.\n>\n> Well, if you can figure out the generating functions... (I couldn\'t,\n> so I just recognizes the pattern in the coefficients and managed to\n> convince myself that it holds for all possible relations.)\n\nYou have\n\nx_{n+3} = a x_{n+2} + b x_{n+1} + c x_{n}\n\nlet g(z) = \\sum x_n z^n. Now, multiply both sides of the above by z^n\nand sum:\n\n\\sum x_{n+3} z^n = a \\sum x_{n+2} z^n + b \\sum x_{n+1} z^n + c g(z)\n\n\\sum_{n=0}^\\infty x_{n+k} z^n = \\sum_{n=k}^\\infty x_n z^{n-k} =\nz^{-k} (g(z) - \\sum_{n=0}^(k-1) x_n z^n)\n\nGiving\n\nz^-3 (g(z) - x_0 - z x_1 - z^2 x_2) =\na z^-2 (g(z) - x_0 - z x_1) + b z^-1 (g(z) - x_0) + c g(z)\n\nimplies\n\ng(z) - a z g(z) - b z^2 g(z) - c z^3 g(z) =\n(1 - a z - b z^2) x_0 + (z - a z^2) x_1 + z^2 x_2\n\nso\n(1 - a z - b z^2) x_0 + (z - a z^2) x_1 + z^2 x_2\ng(z) = -------------------------------------------------\n1 - a z - b z^2 - c z^3\n\nFun, huh? I hope all my algebra\'s right....\n\nAnyways, factor the cubic and you can do partial fractions.\n\nA standard case is a = b = 1, c = 0, x_0 = 0, x_1 = 0, x_2 = 1 which are\njust the Fibonacci numbers:\n\ng(z) = z^2 / (1 - z - z^2)\n\nThe roots of 1 - z - z^2 are (-1/2)(1 -+ sqrt(5)) = 2/(1 +- sqrt(5)).\nCall them 1/r and 1/s. Then we have\n\ng(z) = z^2 (1 / sqrt(5))( r / (1 - r z) - s / (1 - s z))\n= z^2 / sqrt(5) \\sum (r^{n+1} z^n - s^{n+1} z^n)\n\nSo, x_{n+2} = (1/sqrt(5)) ( r^{n+1} - s^{n+1} )\n\nwhich is just the Binet formula for the Fibonacci numbers.\n\nAaron\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <86pisf3tmly.fsf@sirppi.helsinki.fi>,
Esa A E Peuha <esa.peuha@helsinki.fi> wrote:

> Aaron Bergman <abergman@physics.utexas.edu> writes:
>
> > Just use generating functions and partial fractions.
>
> Well, if you can figure out the generating functions... (I couldn't,
> so I just recognizes the pattern in the coefficients and managed to
> convince myself that it holds for all possible relations.)

You have

x_{n+3} = a x_{n+2} + b x_{n+1} + c x_{n}

let g(z) = \sum x_n z^n. Now, multiply both sides of the above by z^n
and sum:

\sum x_{n+3} z^n = a \sum x_{n+2} z^n + b \sum x_{n+1} z^n + c g(z)\sum_{n=0}^\infty x_{n+k} z^n = \sum_{n=k}^\infty x_n z^{n-k} =z^{-k} (g(z) - \sum_{n=0}^(k-1) x_n z^n)

Giving

z^-3 (g(z) - x_0 - z x_1 - z^2 x_2) =a z^-2 (g(z) - x_0 - z x_1) + b z^-1 (g(z) - x_0) + c g(z)

implies

g(z) - a z g(z) - b z^2 g(z) - c z^3 g(z) =(1 - a z - b z^2) x_0 + (z - a z^2) x_1 + z^2 x_2

so
(1 - a z - b z^2) x_0 + (z - a z^2) x_1 + z^2 x_2g(z) = -------------------------------------------------
1 - a z - b z^2 - c z^3

Fun, huh? I hope all my algebra's right....

Anyways, factor the cubic and you can do partial fractions.

A standard case is a = b = 1, c = 0, x_0 = 0, x_1 = 0, x_2 = 1 which are
just the Fibonacci numbers:

g(z) = z^2 / (1 - z - z^2)

The roots of 1 - z - z^2 are (-1/2)(1 -+ \sqrt(5)) = 2/(1 +- \sqrt(5)).
Call them 1/r and 1/s. Then we have

g(z) = z^2 (1 / \sqrt(5))( r / (1 - r z) - s / (1 - s z))= z^2 / \sqrt(5) \sum (r^{n+1} z^n - s^{n+1} z^n)

So, x_{n+2} = (1/\sqrt(5)) ( r^{n+1} - s^{n+1} )

which is just the Binet formula for the Fibonacci numbers.

Aaron

MM

May20-04, 02:24 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nI wrote:\n\n>> [...]\n>> I hope you\'ll move on to [Borel summation] soon.\n\nJohn Baez wrote:\n\n> Oh yeah? Hmm. Maybe I\'ll start by giving a cute example\n> of a series we can handle with Borel summation, but not\n> with [Cesaro or Abel summation].\n>\n> 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....\n>\n> Does this look familiar?\n\nFeeling a bit intimidated by your "oh yeah?", I went away\nand searched the web for stuff about Borel summation.\n\nIFUC, one can sum series like the above if one knows\nabout this trick:\n\nn! = Int{0,inf} z^n exp(-z) dz\n\nOne uses it like this:\n\nS = Sum_n n!/x^n\n\n= Int{0,inf} Sum_n (z^n/x^n) exp(-z) dz\n\n= Int{0,inf} (1/(1 - z/x)) exp(-z) dz\n\n= -x Int{0,inf} exp(-xy)/(y-1) dz\n\nI.e: if we can sum the series without the factorials to a function f\nwhose Laplace transform exists, we\'re done.\n\nBut consulting my table of Laplace transforms, it looks as if the\ntransform above can\'t be done unless we start with a different sum S\',\nwhere x has been replaced by -x to make it an alternating series.\nRedoing the above then gives:\n\nS\' = x Int{0,inf} exp(-xy)/(y+1) dz\n\n= x Laplace{1/(1+y)}\n\n= x exp(x) E_1(x) // According to my LT tables.\n\nI\'m not entirely sure what the E_1(x) is. An Euler function perhaps?\n\nAnother question: I read that Borel summability is also called\n"1-summability". Presumably there\'s also 2-summability, etc. Is that\njust applying the Borel technique twice, or something trickier?\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>I wrote:

>> [...]
>> I hope you'll move on to [Borel summation] soon.

John Baez wrote:

> Oh yeah? Hmm. Maybe I'll start by giving a cute example
> of a series we can handle with Borel summation, but not
> with [Cesaro or Abel summation].
>
> 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....
>
> Does this look familiar?

Feeling a bit intimidated by your "oh yeah?", I went away
and searched the web for stuff about Borel summation.

IFUC, one can sum series like the above if one knows
about this trick:

n! = \Int{0,inf} z^n \exp(-z) dz

One uses it like this:

S = Sum_n n!/x^n= \Int{0,inf} Sum_n (z^n/x^n) \exp(-z) dz= \Int{0,inf} (1/(1 - z/x)) \exp(-z) dz= -x \Int{0,inf} \exp(-xy)/(y-1) dz

I.e: if we can sum the series without the factorials to a function f
whose Laplace transform exists, we're done.

But consulting my table of Laplace transforms, it looks as if the
transform above can't be done unless we start with a different sum S',
where x has been replaced by -x to make it an alternating series.
Redoing the above then gives:

S' = x \Int{0,inf} \exp(-xy)/(y+1) dz

= x Laplace{1/(1+y)}

= x \exp(x) E_1(x) // According to my LT tables.

I'm not entirely sure what the E_1(x) is. An Euler function perhaps?

Another question: I read that Borel summability is also called
"1-summability". Presumably there's also 2-summability, etc. Is that
just applying the Borel technique twice, or something trickier?

Ulmo

May24-04, 05:31 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nmikem@despammed.com (MM) wrote in message news:<38b55e8c.0405200000.78c30c5b@posting.google. com>...\n> I wrote:\n>\n> >> [...]\n> >> I hope you\'ll move on to [Borel summation] soon.\n>\n> John Baez wrote:\n>\n> > Oh yeah? Hmm. Maybe I\'ll start by giving a cute example\n> > of a series we can handle with Borel summation, but not\n> > with [Cesaro or Abel summation].\n> >\n> > 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....\n> >\n> > Does this look familiar?\n>\n> Feeling a bit intimidated by your "oh yeah?", I went away\n> and searched the web for stuff about Borel summation.\n>\n> IFUC, one can sum series like the above if one knows\n> about this trick:\n>\n> n! = Int{0,inf} z^n exp(-z) dz\n>\n> One uses it like this:\n>\n> S = Sum_n n!/x^n\n>\n> = Int{0,inf} Sum_n (z^n/x^n) exp(-z) dz\n>\n> = Int{0,inf} (1/(1 - z/x)) exp(-z) dz\n>\n> = -x Int{0,inf} exp(-xy)/(y-1) dz\n>\n> I.e: if we can sum the series without the factorials to a function f\n> whose Laplace transform exists, we\'re done.\n>\n> But consulting my table of Laplace transforms, it looks as if the\n> transform above can\'t be done unless we start with a different sum S\',\n> where x has been replaced by -x to make it an alternating series.\n> Redoing the above then gives:\n>\n> S\' = x Int{0,inf} exp(-xy)/(y+1) dz\n>\n> = x Laplace{1/(1+y)}\n>\n> = x exp(x) E_1(x) // According to my LT tables.\n>\n> I\'m not entirely sure what the E_1(x) is. An Euler function perhaps?\n>\n> Another question: I read that Borel summability is also called\n> "1-summability". Presumably there\'s also 2-summability, etc. Is that\n> just applying the Borel technique twice, or something trickier?\n\nThere is another numerical technique called Mellin-Barnes\nregularization. Can you speak about that?\n\nDavid\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>mikem@despammed.com (MM) wrote in message news:<38b55e8c.0405200000.78c30c5b@posting.google.com>...
> I wrote:
>
> >> [...]
> >> I hope you'll move on to [Borel summation] soon.
>
> John Baez wrote:
>
> > Oh yeah? Hmm. Maybe I'll start by giving a cute example
> > of a series we can handle with Borel summation, but not
> > with [Cesaro or Abel summation].
> >
> > 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....
> >
> > Does this look familiar?
>
> Feeling a bit intimidated by your "oh yeah?", I went away
> and searched the web for stuff about Borel summation.
>
> IFUC, one can sum series like the above if one knows
> about this trick:
>
> n! = \Int{0,inf} z^n \exp(-z) dz
>
> One uses it like this:
>
> S = Sum_n n!/x^n
>
> = \Int{0,inf} Sum_n (z^n/x^n) \exp(-z) dz
>
> = \Int{0,inf} (1/(1 - z/x)) \exp(-z) dz
>
> = -x \Int{0,inf} \exp(-xy)/(y-1) dz
>
> I.e: if we can sum the series without the factorials to a function f
> whose Laplace transform exists, we're done.
>
> But consulting my table of Laplace transforms, it looks as if the
> transform above can't be done unless we start with a different sum S',
> where x has been replaced by -x to make it an alternating series.
> Redoing the above then gives:
>
> S' = x \Int{0,inf} \exp(-xy)/(y+1) dz
>
> = x Laplace{1/(1+y)}
>
> = x \exp(x) E_1(x) // According to my LT tables.
>
> I'm not entirely sure what the E_1(x) is. An Euler function perhaps?
>
> Another question: I read that Borel summability is also called
> "1-summability". Presumably there's also 2-summability, etc. Is that
> just applying the Borel technique twice, or something trickier?

There is another numerical technique called Mellin-Barnes
regularization. Can you speak about that?

David

John Baez

May24-04, 05:31 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nIn article <38b55e8c.0405200000.78c30c5b@posting.google.com>, \nMM wrote:\n\n>John Baez wrote:\n\n>> MM wrote:\n\n>>> I hope you\'ll move on to [Borel summation] soon.\n\n> > Oh yeah? Hmm. Maybe I\'ll start by giving a cute example\n> > of a series we can handle with Borel summation, but not\n> > with [Cesaro or Abel summation].\n> >\n> > 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....\n> >\n> > Does this look familiar?\n\nYou were supposed to say:\n\nYeah! It looks like an insane upside-down version of the\nfamiliar Taylor series:\n\nexp(x) = 1/0! + x/1! + x^2/2! + x^3/3! + ....\n\n.... but of course its radius of convergence is zero!\n\n>Feeling a bit intimidated by your "oh yeah?", I went away\n>and searched the web for stuff about Borel summation.\n\nI was just saying "oh yeah?" because I don\'t understand\nBorel summation very well, so I can\'t explain it very well,\nso I had to figure out what to do...\n\n.... so I tossed out a really cute example of a series\none can tackle using this method.\n\nBut I\'m glad this remark got you to figure out Borel summation\nfor yourself!\n\n>IFUC, one can sum series like the above if one knows\n>about this trick:\n>\n> n! = Int{0,inf} z^n exp(-z) dz\n\nRight.\n\n>One uses it like this:\n>\n> S = Sum_n n!/x^n\n>\n> = Int{0,inf} Sum_n (z^n/x^n) exp(-z) dz\n>\n> = Int{0,inf} (1/(1 - z/x)) exp(-z) dz\n>\n> = -x Int{0,inf} exp(-xy)/(y-1) dz\n>\n>I.e: if we can sum the series without the factorials to a function f\n>whose Laplace transform exists, we\'re done.\n\nRight.\n\n>But consulting my table of Laplace transforms, it looks as if the\n>transform above can\'t be done unless we start with a different sum S\',\n>where x has been replaced by -x to make it an alternating series.\n\nThat sounds right. I just preferred the look of this:\n\n0! + 1!/x + 2!/x^2 + 3!/x^3 + ....\n\nto this\n\n0! - 1!/x + 2!/x^2 - 3!/x^3 + ....\n\n>Redoing the above then gives:\n>\n> S\' = x Int{0,inf} exp(-xy)/(y+1) dz\n>\n> = x Laplace{1/(1+y)}\n>\n> = x exp(x) E_1(x) // According to my LT tables.\n>\n>I\'m not entirely sure what the E_1(x) is. An Euler function perhaps?\n\nIt\'s an "exponential integral" function - the simplest one, actually.\nFor x with positive real part it\'s defined by\n\nE_1(x) = int_x^infinity (exp(-t)/t) dt\n\nand it\'s an interesting puzzle to start with this and derive the\nasymptotic series\n\nE_1(x) = (exp(-x)/x) (0! - 1!/x + 2!/x^2 - 3!/x^3 + ....)\n\nYou\'ve done most of the work involve in going the reverse\ndirection, starting with\n\n0! - 1!/x + 2!/x^2 - 3!/x^3 + ....\n\nand Borel summing it to get\n\nx exp(x) E_1(x)\n\nThe exponential integral E_1(x) is just a slight variant of\n\nEi(x) = integral_{-infinity}^x (exp(t)/t) dt\n\nand it\'s a close relative of the more famous logarithmic integral\n\nli(x) = integral_0^x (dt / ln t)\n= Ei(ln(x))\n\nwhich is a good estimate for the number of primes less than x.\nThese are good functions to be aware of, since they come from\nsome of the simplest integrals that can\'t be done using elementary\nfunctions.\n\nAbramowitz and Stegun is a good place to read about these...\nand now you can get it online! For E_1, try this:\n\nhttp://jove.prohosting.com/~skripty/page_228.htm\n\n>Another question: I read that Borel summability is also called\n>"1-summability". Presumably there\'s also 2-summability, etc. Is that\n>just applying the Borel technique twice, or something trickier?\n\nBorel summation is about where my knowledge of summation methods\nfizzles out. Presumably there\'s a hierarchy of even more powerful\nsummation methods, of which it\'s the first, but I don\'t know what\nthese are called. Your guess sounds good to me!\n\nExtra credit: which rock band has a song called "Cesaro\nSummability", and what are the lyrics to this song?\n\nDoes anyone know why they called the song this??\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <38b55e8c.0405200000.78c30c5b@posting.google.com>,
MM wrote:

>John Baez wrote:

>> MM wrote:

>>> I hope you'll move on to [Borel summation] soon.

> > Oh yeah? Hmm. Maybe I'll start by giving a cute example
> > of a series we can handle with Borel summation, but not
> > with [Cesaro or Abel summation].
> >
> > 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....
> >
> > Does this look familiar?

You were supposed to say:

Yeah! It looks like an insane upside-down version of the
familiar Taylor series:

\exp(x) = 1/0! + x/1! + x^2/2! + x^3/3! + .[/itex]...

.... but of course its radius of convergence is zero!

>Feeling a bit intimidated by your "oh yeah?", I went away
>and searched the web for stuff about Borel summation.

I was just saying "oh yeah?" because I don't understand
Borel summation very well, so I can't explain it very well,
so I had to figure out what to do...

.... so I tossed out a really cute example of a series
one can tackle using this method.

But I'm glad this remark got you to figure out Borel summation
for yourself!

>IFUC, one can sum series like the above if one knows
>about this trick:
>
> n! = \Int{0,inf} z^n \exp(-z) dz

Right.

>One uses it like this:
>
> S = Sum_n n!/x^n
>
> = \Int{0,inf} Sum_n (z^n/x^n) \exp(-z) dz
>
> = \Int{0,inf} (1/(1 - z/x)) \exp(-z) dz
>
> = -x \Int{0,inf} \exp(-xy)/(y-1) dz
>
>I.e: if we can sum the series without the factorials to a function f
>whose Laplace transform exists, we're done.

Right.

>But consulting my table of Laplace transforms, it looks as if the
>transform above can't be done unless we start with a different sum S',
>where x has been replaced by -x to make it an alternating series.

That sounds right. I just preferred the look of this:

0! + 1!/x + 2!/x^2 + 3!/x^3 + ....

to this

0! - 1!/x + 2!/x^2 - 3!/x^3 + ....

>Redoing the above then gives:
>
> S' = x \Int{0,inf} \exp(-xy)/(y+1) dz
>
> = x Laplace{1/(1+y)}
>
> = x \exp(x) E_1(x) // According to my LT tables.
>
>I'm not entirely sure what the E_1(x) is. An Euler function perhaps?

It's an "exponential integral" function - the simplest one, actually.
For x with positive real part it's defined by

E_1(x) = \int_x^infinity (\exp(-t)/t) dt

and it's an interesting puzzle to start with this and derive the
asymptotic series

E_1(x) = (\exp(-x)/x) (0! - 1!/x + 2!/x^2 - 3!/x^3 + ....)

You've done most of the work involve in going the reverse
direction, starting with

0! - 1!/x + 2!/x^2 - 3!/x^3 + ....

and Borel summing it to get

[itex]x \exp(x) E_1(x)

The exponential integral E_1(x) is just a slight variant of

Ei(x) = integral_{-infinity}^x (\exp(t)/t) dt

and it's a close relative of the more famous logarithmic integral

li(x) = integral_0^x (dt / ln t)= Ei(ln(x))

which is a good estimate for the number of primes less than x.
These are good functions to be aware of, since they come from
some of the simplest integrals that can't be done using elementary
functions.

Abramowitz and Stegun is a good place to read about these...
and now you can get it online! For E_1, try this:

http://jove.prohosting.com/~skripty/page_228.htm

>Another question: I read that Borel summability is also called
>"1-summability". Presumably there's also 2-summability, etc. Is that
>just applying the Borel technique twice, or something trickier?

Borel summation is about where my knowledge of summation methods
fizzles out. Presumably there's a hierarchy of even more powerful
summation methods, of which it's the first, but I don't know what
these are called. Your guess sounds good to me!

Extra credit: which rock band has a song called "Cesaro
Summability", and what are the lyrics to this song?

Does anyone know why they called the song this??

Daniel Doro Ferrante

May25-04, 02:32 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 24 May 2004 05:31:07 -0400, John Baez <baez@galaxy.ucr.edu> wrote:\n\n[snipping unnecessary stuff...]\n> Extra credit: which rock band has a song called "Cesaro\n> Summability", and what are the lyrics to this song?\n>\n\nIf i\'m not mistaken, the band name is "Tool" and this song is\ninstrumental.\n\nLinks include:\n<http://www.moron.nl/lyrics.php?id=10629&artist=Tool>\n<http://www.songmeanings.net/lyric.php?lid=3458764513820543198>\n<http://www.eecs.berkeley.edu/~tmatthew/tool.htm>\n\n\n> Does anyone know why they called the song this??\n>\n\nAs far as Google is concerned, i couldn\'t find any explanation for this\nsong!\n\n\n--\nUsing M2, Opera\'s revolutionary e-mail client: http://www.opera.com/m2/\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 24 May 2004 05:31:07 -0400, John Baez <baez@galaxy.ucr.edu> wrote:

[snipping unnecessary stuff...]
> Extra credit: which rock band has a song called "Cesaro
> Summability", and what are the lyrics to this song?
>

If i'm not mistaken, the band name is "Tool" and this song is
instrumental.

Links include:
<http://www.moron.nl/lyrics.php?id=10629&artist=Tool>
<http://www.songmeanings.net/lyric.php?lid=3458764513820543198>
<http://www.eecs.berkeley.edu/~tmatthew/tool.htm>

> Does anyone know why they called the song this??
>

As far as Google is concerned, i couldn't find any explanation for this
song!

--
Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/

MM

May26-04, 06:12 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nJohn Baez wrote:\n\n>>> 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....\n>>>\n>>> Does this look familiar?\n>\n> You were supposed to say:\n>\n> Yeah! It looks like an insane upside-down version of the\n> familiar Taylor series:\n>\n> exp(x) = 1/0! + x/1! + x^2/2! + x^3/3! + ....\n>\n> ... but of course its radius of convergence is zero!\n\nGrr. And there was I, racking my brains trying to match it\nup with some series in QED.\n\n\n> [E_1(x)] is an "exponential integral" function [...]\n\nOK. Thanks, I can make sense of that now when I work out\nthe Laplace transform explicitly.\n\n\n> Abramowitz and Stegun is a good place to read about\n> [exponential integral functions] ...\n> and now you can get it online! For E_1, try this:\n>\n> http://jove.prohosting.com/~skripty/page_228.htm\n\nWow. I\'ll definitely bookmark that URL. Thanks.\n\n> [...]\n> Borel summation is about where my knowledge of summation methods\n> fizzles out. Presumably there\'s a hierarchy of even more powerful\n> summation methods [...]\n\nYes, it seems like there\'s a whole industry devoted just to "QFT\nas a Resummation Problem". While searching the web I also came\nacross a paper where the author has a summation technique involving\na super-duper transform with an exp(-exp(-t)) as the kernel. It\'s\nclaimed to have a much larger radius of convergence than Borel.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>John Baez wrote:

>>> 0! + 1!/x + 2!/x^2 + 3!/x^3 + ....
>>>
>>> Does this look familiar?
>
> You were supposed to say:
>
> Yeah! It looks like an insane upside-down version of the
> familiar Taylor series:
>
> \exp(x) = 1/0! + x/1! + x^2/2! + x^3/3! + ....
>
> ... but of course its radius of convergence is zero!

Grr. And there was I, racking my brains trying to match it
up with some series in QED.

> [E_1(x)] is an "exponential integral" function [...]

OK. Thanks, I can make sense of that now when I work out
the Laplace transform explicitly.

> Abramowitz and Stegun is a good place to read about
> [exponential integral functions] ...
> and now you can get it online! For E_1, try this:
>
> http://jove.prohosting.com/~skripty/page_228.htm

Wow. I'll definitely bookmark that URL. Thanks.

> [...]
> Borel summation is about where my knowledge of summation methods
> fizzles out. Presumably there's a hierarchy of even more powerful
> summation methods [...]

Yes, it seems like there's a whole industry devoted just to "QFT
as a Resummation Problem". While searching the web I also came
across a paper where the author has a summation technique involving
a super-duper transform with an \exp(-\exp(-t)) as the kernel. It's
claimed to have a much larger radius of convergence than Borel.

MM

May29-04, 12:54 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Ulmo wrote:\n\n> There is another numerical technique called\n> Mellin-Barnes regularization. Can you speak\n> about that?\n\nI don\'t know much more than that, and what\'s\nin the paper by Kowalenko & Rawlinson:\nhttp://www.iop.org/EJ/article/0305-4470/31/38/002/a838l2.pdf\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Ulmo wrote:

> There is another numerical technique called
> Mellin-Barnes regularization. Can you speak
> about that?

I don't know much more than that, and what's
in the paper by Kowalenko & Rawlinson:
http://www.iop.org/EJ/article/0305-4470/31/38/002/a838l2.pdf