To PI or not to PI challenge. Stern-Brocot

arithmetic · May 7, 2010

Dear readers,

At my webpage:
http://domingogomez.web.officelive.com/brocotfraction.aspx

I have posed a challenge entitled "To PI or not to PI", it deals with
a generalized continued fraction (bifurcated fraction) whose
coeficients are all the Stern-Brocot fractions, that is, all the positive
rational numbers in their reduced forms.
This very special fractal fraction converges to a very specific value
close to PI.
I hope you find of some interest this challenge.

Domingo Gomez Morin

Xitami · May 7, 2010

Code:

/* PARI/GP */
sbf(a,b,c,d,h)={
	h--;
	if(h<=0, return((a+c)/(b+d)));
	return( (a+c)/(b+d)+sbf(a+c,b+d,c,d,h)/sbf(a,b,a+c,b+d,h));
}

/* faster */
sbF(a,b,c,d,h)={
	h--;
	if(h<=0, return( 0.0 + (a+c)/(b+d)));
	return( (a+c)/(b+d)+sbF(a+c,b+d,c,d,h)/sbF(a,b,a+c,b+d,h));
}

sbf(0,1,1,0,3) = 13/5 = 2.6

for(i=0,18,print(i" "sbf(0,1,1,0,i)+.))
0 1.0000000000000000000000000000000000000000000000000
1 1.0000000000000000000000000000000000000000000000000
2 5.0000000000000000000000000000000000000000000000000
3 2.6000000000000000000000000000000000000000000000000
4 3.4626379400945874934314240672622175512348922753547
5 2.9388088765046610642144686035478929406022355808575
6 3.2735185568492236282348448216538099507168494439501
7 3.0347992453939067319279281866337440790873147702144
8 3.2099684860688422980323591345066881944750995102337
9 3.0738506312040716737872747524887840798894383527984
10 3.1814697094275065992958010264534251536404002098575
11 3.0934945502880832477021949766849855361715888355938
12 3.1663219592217772133098797759764429309295906541622
13 3.1047408775416900905987339929906019611661463991880
14 3.1573142239629423684606802771536687757026460862718
15 3.1117718703279799014367140110065699129312803115134
16 3.1515201645394594475594374322542040402712357473677
17 3.1164582339359882762699683957496012711553724723406
18 3.1475716978017248990506977242760373644809795171547
number of terms: 2^(i+1)-1
2^19-1=524287

C program, doubles (34 359 738 367 terms)

Code:

  0   1.00000000000000000
  1   1.00000000000000000
  2   5.00000000000000000
  3   2.60000000000000010
  4   3.46263794009458700
  5   2.93880887650466120
  6   3.27351855684922330
  7   3.03479924539390660
  8   3.20996848606884240
  9   3.07385063120407190
 10   3.18146970942750640
 11   3.09349455028808280
 12   3.16632195922177710
 13   3.10474087754169000
 14   3.15731422396294280
 15   3.11177187032798040
 16   3.15152016453945950
 17   3.11645823393598760
 18   3.14757169780172540
 19   3.11973774954432280
 20   3.14475925418342990
 21   3.12212230271433850
 22   3.14268465110109930
 23   3.12391046202905590
 24   3.14111024668793840
 25   3.12528588642403580
 26   3.13988701664436620
 27   3.12636658593696290
 28   3.13891762963651200
 29   3.12723119902446500
 30   3.13813631476721520
 31   3.12793376938022140
 32   3.13749732276145020
 33   3.12851243406977100
 34   3.13696802810121510

Code:

sumalt_partial(p, n=#p)={ local(b,c,s);
	b=2^(2*n-1);
	c=b;
	s=0;
	forstep(k=n-1,0,-1,
		s+= b*p[k+1];
		b*= ((2*k+1)*(k+1))/(2*(n+k)*(n-k));
		c+= b;
	);
	s/c
}

t[] computed in C

Code:

PARI/GP>for(i=1,#t,print(i"   "altsum_partial(t,i)))
1   0.66666666666
2   0.94117647058
3   2.28282828282
4   3.12790294627
5   3.31575100366
6   3.27005398107
7   3.20530124624
8   3.16657805849
9   3.14828078568
10   3.1403458694
11   3.1368920474
12   3.1352752842
13   3.1344294778
14   3.1339363396
15   3.1336247926
16   3.1334175662
17   3.1332751260
18   3.1331749115
19   3.1331030537
20   3.1330506392
21   3.1330117843
22   3.1329825333
23   3.1329601875
24   3.1329428795
25   3.1329293002
26   3.1329185187
27   3.1329098646
28   3.1329028481
29   3.1328971071
30   3.1328923702
31   3.1328884319
32   3.1328851345
33   3.1328823558
34   3.1328800237

New, fundamental constant 3.13288...

arithmetic · May 7, 2010

Xitami says:
>New, fundamental constant 3.13288...

Well that seems to be the answer to Shakespeare's question: Not to PI, just close.

That was fast Xitami ¡¡¡, thanks a lot, let's see what others have to say about
your result.
Please let me know your name (by email) so I can include it for the records in my webpage,
or let me know if you prefer to use "Xitami".

I think this new constant might be called: "Universal Rational-Mean Constant"

Cheers,

Domingo Gomez

arithmetic · May 7, 2010

When using Farey fractions instead of Stern-Brocot's then the new constant is approx.:

1.226...

which is not "Universal"

because it does not embrace all the positive rational numbers in their reduced forms but just those from 0 to 1.

So this one could be called: "Farey Rational Constant"

Could the reader find a relation between both of them?

Xitami · May 7, 2010

alternting variant

Code:

        2/1 + ---
1/2 -  ----------
        1/2 + ---

  0      1.00000000000000000
  1      1.00000000000000000
  2     -3.00000000000000000
  3      1.00000000000000000
  4     -6.91232876712328980
  5      0.14328134217554114
  6     -3.87884017589414440
  7      0.69975974137673203
  8     -8.44206995530308470
  9     -0.76694421386509692
 10     -3.65503802208626280
 11      2.35578956476438380
 12      6.23413307523992440
 13    -19.28387468639876800
 14      2.89341005066847280
 15      0.32136221024384604
 16      1.16624268747289440
 17      0.80821242069640231
 18      1.04301471087375220
 19      0.95481756005451746
 20      1.01034312761793440
 21      0.98854958616575817
 22      1.00233394417308630
 23      0.99836436971475051
 24      1.00006518857518570
 25      0.99998636872615365
 26      1.00000017834156240
 27      0.99999838543925079
 28      1.00000000642554520
 29      0.98723877363144263
 30      1.18727245568190480
 31     15.64720874911006400
 32      0.99427527762970824
strange

Regards, Xitami.

Xitami · May 7, 2010

arithmetic said:

When using Farey fractions instead of Stern-Brocot's then the new constant is approx.:
1.226...
...

1.226... ?
pari/gp> for(i=1,20,print(sbf(0,1,1,1.,i)))
0.500000000000000000
2.500000000000000000
1.491379310344827586
1.913773073178070399
1.627260322338834205
1.819867311604092935
1.675122681756296973
1.785265731369331449
1.697124869319332277
1.768470830315575050
1.709007940366553484
1.759014903604435405
1.716146298615272456
1.753157353718325420
1.720768312861207621
1.749275249557767505
1.723932145152701193
1.746568856199571096
1.726192767930516595
1.744606292952420387

arithmetic · May 7, 2010

Yes Xitami, it depends on the colocation of the initial fractions (0/1, 1/1)
or (1/1, 0/1) (for the fractal fraction it means: above or below respectively)

That is, starting with: 1/2+ (2/3)/(1/3)...
or
starting with 1/2+ (1/3)/(2/3) ...

got it? So both constants are right: 1.226... and 1.7...

The same applies for the

Shakespeare's constant 3.13288...
please try by using (1/0, 0/1) instead of (0/1, 1/0) for that case, and you will see...

All of them are new constants (My conjecture is that they are Transcendental Numbers)

Interesting, isn't it?

arithmetic · May 7, 2010

just a typo: in my last mesasge I said "colocation" and I meant "LOCATION"

Xitami · May 8, 2010

arithmetic said:

Yes Xitami, it depends on the location of the initial fractions (0/1, 1/1)
or (1/1, 0/1)...

Yes
for i = 1..32

Code:

      0  oo              oo  0             0  1              1  0              1  oo             oo  1
------------------------------------------------------------------------------------------------------------
    0       1           1      0         0      1          1      0          1       1         1       1
    -       -           -      -         -      -          -      -          -       -         -       -
    1       0           0      1         1      1          1      1          1       0         0       1
------------------------------------------------------------------------------------------------------------[
[1.00000000000000, 1.00000000000000, 0.50000000000000, 0.50000000000000, 2.00000000000000, 2.00000000000000],
[5.00000000000000, 1.25000000000000, 2.50000000000000, 1.00000000000000, 4.00000000000000, 2.50000000000000],
[2.60000000000000, 1.40000000000000, 1.49137931034483, 1.15340909090909, 3.67272727272727, 2.63448275862069],
[3.46263794009459, 1.43781235126130, 1.91377307317807, 1.17893945899528, 3.71044022189325, 2.63542013102534],
[2.93880887650466, 1.44734402880067, 1.62726032233883, 1.19791711626815, 3.69960653966179, 2.63610535624420],
[3.27351855684922, 1.45442687388447, 1.81986731160409, 1.20683409722323, 3.70306463236905, 2.63646628086641],
[3.03479924539391, 1.45774683559641, 1.67512268175630, 1.21264226915310, 3.70196833698054, 2.63648082287349],
[3.20996848606884, 1.45994731258142, 1.78526573136933, 1.21650578913913, 3.70237446386729, 2.63650648379447],
[3.07385063120407, 1.46140822964650, 1.69712486931933, 1.21912169269838, 3.70222649553624, 2.63650960649757],
[3.18146970942751, 1.46239986749675, 1.76847083031558, 1.22107287639257, 3.70228404560910, 2.63651175275555],
[3.09349455028808, 1.46313955366077, 1.70900794036655, 1.22249580904769, 3.70226142970045, 2.63651227368401],
[3.16632195922178, 1.46367916480036, 1.75901490360444, 1.22360568989495, 3.70227057182131, 2.63651250401425],
[3.10474087754169, 1.46410008980877, 1.71614629861527, 1.22446640266835, 3.70226682040408, 2.63651258450038],
[3.15731422396294, 1.46442653445570, 1.75315735371832, 1.22515692996179, 3.70226838384100, 2.63651261462383],
[3.11177187032798, 1.46468843849495, 1.72076831286121, 1.22571570267877, 3.70226772362143, 2.63651262714428],
[3.15152016453946, 1.46490037258285, 1.74927524955777, 1.22617487185787, 3.70226800534697, 2.63651263166333],
[3.11645823393599, 1.46507452956305, 1.72393214515270, 1.22655727672556, 3.70226788386056, 2.63651263367106],
[3.14757169780173, 1.46521957113390, 1.74656885619957, 1.22687824862140, 3.70226793666468, 2.63651263441218],
[3.11973774954432, 1.46534131208324, 1.72619276793052, 1.22715108699814, 3.70226791352349, 2.63651263474630],
[3.14475925418343, 1.46544479659443, 1.74460629295242, 1.22738428973423, 3.70226792373012, 2.63651263487502],
[3.12212230271434, 1.46553324778298, 1.72786411537339, 1.22758564599825, 3.70226791919895, 2.63651263493274],
[3.14268465110110, 1.46560961997042, 1.74313747561861, 1.22776037989534, 3.70226792122129, 2.63651263495598],
[3.12391046202906, 1.46567589459545, 1.72913465153928, 1.22791317933125, 3.70226792031393, 2.63651263496631],
[3.14111024668794, 1.46573384972492, 1.74200936653856, 1.22804745180117, 3.70226792072287, 2.63651263497062],
[3.12528588642404, 1.46578477778274, 1.73012308207908, 1.22816613189901, 3.70226792053776, 2.63651263497252],
[3.13988701664437, 1.46582979182719, 1.74112401178017, 1.22827151794382, 3.70226792062188, 2.63651263497334],
[3.12636658593696, 1.46586976358497, 1.73090719777399, 1.22836552708201, 3.70226792058352, 2.63651263497370],
[3.13891762963651, 1.46590542020834, 1.74041633194309, 1.22844974450798, 3.70226792060107, 2.63651263497386],
[3.12723119902447, 1.46593736294390, 1.73153970353667, 1.22852547356149, 3.70226792059301, 2.63651263497393],
[3.13813631476722, 1.46596608613394, 1.73984170648085, 1.22859382814243, 3.70226792059672, 2.63651263497396],
[3.12793376938022, 1.46599201226834, 1.73205733694769, 1.22865572486824, 3.70226792059501, 2.63651263497398],
[3.13749732276145, 1.46601548900992, 1.73936870714773, 1.22871196056974, 3.70226792059580, 2.63651263497398]];
----------------------------------------------------------------------------------------------------------------
 3.13286308929902, 1.46633409868672, 1.73582085736194, 1.22950308242259, 3.70226792059555, 2.63651263497399 Wynn
 3.13288122714498, 1.46562502661973, 1.73583922196094, 1.22779888812588, 3.70226792059623, 2.63651263487680 SumAlt
 3.13294527734985, 1.46483896033168, 1.73589528981583, 1.22606704915966, 3.70226791233679, 2.63651256289041 Aver

arithmetic · May 8, 2010

Great job Xitami ¡¡¡

arithmetic · May 8, 2010

Well I made a lookup at Plouffe's inverter:
http://pi.lacim.uqam.ca/

and, of course, we need better approximations to grasp a bit of the essence
of all this.

Please Xitami take a lookup of all those values at Plouffe's inverter.

Xitami · May 10, 2010

[URL]http://fotoo.pl/zdjecia/files/2010-05/a512eda9.png[/URL]

arithmetic · May 15, 2010

Indeed, many thanks Xitami.
It would be great if you could make a brief explanation on the amazing graph you
have posted, thanks again. :approve:

On the other hand, I would like to make the following question:

Does anyone here know about any book or paper describing the ARITHMETIC MEAN
of ALL RATIONAL NUMBERS?
It seems to me that its value converges to 1.5, but I would like to know about any previous analysis on this in the math-literature.

Based on such value 1.5, it is quite interesting to realize that the "EQUILIBRIUM POINT"
of all rational numbers (between 0/1 and 1/0) is not 1, but 1.5.

Isn't 1.5 interesting?

¿?¿?¿?¿?¿?¿?

Next task for me: HARMONIC MEAN OF ALL RATIONAL NUMBERS

arithmetic · May 15, 2010

>Next task for me: HARMONIC MEAN OF ALL RATIONAL NUMBERS

Answer:
Harmonic mean = 1/(Arithmetic Mean)

Xitami · May 16, 2010

about graph

Code:

sbf(a,b,c,d,h){
	h--;
	
	plot( a+c , b+d );
	
	if(h<=0) return(a+c)/(b+d);
	return (a+c)/(b+d) + sbf(a+c,b+d,c,d,h) / sbf(a,b,a+c,b+d,h);
}

A(0,oo) A= 1+B[i-1] / C[i-1]
B (1,oo)
C (0,1)

Code:

h | A | B | C ---+----------------------+-----------------------+---------------------- 01 | 1.00 | 2.000000000000000000 | 0.500000000000000000 02 | 5.000000000000000000 | 4.000000000000000000 | 2.500000000000000000 03 | 2.600000000000000000 | 3.672727272727272400 | 1.491379310344827600 04 | 3.462637940094587251 | 3.710440221893247100 | 1.913773073178070450 05 | 2.938808876504660973 | 3.699606539661788800 | 1.627260322338834450 06 | 3.273518556849223286 | 3.703064632369045900 | 1.819867311604092850 07 | 3.034799245393906756 | 3.701968336980543000 | 1.675122681756296950 08 | 3.209968486068842485 | 3.702374463867286600 | 1.785265731369331200 09 | 3.073850631204071915 | 3.702226495536238700 | 1.697124869319332350 10 | 3.181469709427506392 | 3.702284045609098000 | 1.768470830315575400 11 | 3.093494550288082851 | 3.702261429700445900 | 1.709007940366553250 12 | 3.166321959221777252 | 3.702270571821310700 | 1.759014903604435600 13 | 3.104740877541689813 | 3.702266820404082700 | 1.716146298615272250 14 | 3.157314223962942769 | 3.702268383840999500 | 1.753157353718324950 15 | 3.111771870327980238 | 3.702267723621433400 | 1.720768312861207450 16 | 3.151520164539459630 | 3.702268005346967900 | 1.749275249557767900 17 | 3.116458233935987872 | 3.702267883860555800 | 1.723932145152700750 18 | 3.147571697801725247 | 3.702267936664682700 | 1.746568856199571800 19 | 3.119737749544323044 | 3.702267913523485300 | 1.726192767930516700 20 | 3.144759254183429791 | 3.702267923730116400 | 1.744606292952420400 21 | 3.122122302714338615 | 3.702267919198950300 | 1.727864115373394000 22 | 3.142684651101099294 | 3.702267921221285500 | 1.743137475618610650 23 | 3.123910462029055894 | 3.702267920313926800 | 1.729134651539278500 24 | 3.141110246687938288 | 3.702267920722872600 | 1.742009366538558100 25 | 3.125285886424035834 | 3.702267920537761500 | 1.730123082079081600 26 | 3.139887016644366005 | 3.702267920621877400 | 1.741124011780174050 27 | 3.126366585936963090 | 3.702267920583515600 | 1.730907197773988050 28 | 3.138917629636511884 | 3.702267920601069600 | 1.740416331943094300 29 | 3.127231199024464900 | 3.702267920593012000 | 1.731539703536669950 30 | 3.138136314767215297 | 3.702267920596721100 | 1.739841706480854450 31 | 3.127933769380221235 | 3.702267920595009100 | 1.732057336947687500 32 | 3.137497322761450205 | 3.702267920595801300 | 1.739368707147728200 33 | 3.128512434069770803 | 3.702267920595433600 | 1.732486341353947300 34 | 3.136968028101215221 | 3.702267920595604600 | 1.738974675742910300 35 | 3.128994730192923877 | 3.702267920595525100 | 1.732845868834206550 36 | 3.136524654143805385 | 3.702267920595562000 | 1.738642940698007300 37 | 3.129400944801941102 | 3.702267920595545100 | 1.733150161872439850 38 | 3.136149539746592745 | 3.702267920595553100 | 1.738361013035048200 39 | 3.129746291382634296 | - | -

arithmetic · May 16, 2010

Doskonały Xitami. Wielkie dzięki.

Now I can see that 3.702267920595553 does not appear at Plouffe's Inverter database
http://pi.lacim.uqam.ca/
Now, it seems to me that it is very dificult to get enough decimal digits for the other cases.
I wonder if it would be necesary to use a supercomputer, not just CUDA.
Am I wrong?

Now to the question on the arithmetic, harmonic and geometric mean of all rational numbers,
I would really appreciate if someone could bring me any reference on papers or books relating this topic.

The fact that the arithmetic mean of all rational numbers is 1.5 means (at least, from my point of view) that if all rational numbers were "entities" all of them having the same weight, then :
The center of gravity of all Rational numbers is 1.5

Indeed, it is very important to me to find any previous works on this matter.

Arithmetic Mean of all rational numbers = 3/2
Harmonic Mean of all rational numbers = 2/3
Geometric Mean of all rational numbers = 1

Many thanks,
Domingo Gomez Morin

arithmetic · May 16, 2010

Has anyone checked that the arithmetic mean of all rational numbers is 3/2 ?

Am I wrong?

If I am not wrong then try to imagine this: An infinite number of point-masses located
in the real line at the same position of all the Rational-Numbers, then, when computing
their Center of Mass it yields: 3/2.

¿?
got it?

How can this be physically posible?
Are there the same number of point-masses within (0, 3/2) than within (3/2, Infinite)?

How can this be mathematically posible?
Are there the same number of rational numbers within (0, 3/2) than within (3/2, Infinite)?

Why is 3/2 the equilibrium point of all rational numbers?

Regards,
Domingo Gomez Morin

CRGreathouse · May 16, 2010

arithmetic said:

Has anyone checked that the arithmetic mean of all rational numbers is 3/2 ?

Am I wrong?

It's ill-defined. Depending on what order you choose, you can get a divergent series or convergence to any value desired.

arithmetic · May 16, 2010

CRGreathouse said:

It's ill-defined. Depending on what order you choose, you can get a divergent series or convergence to any value desired.

Yes, it might be ill-defined but what exactly is ill-defined here, let's see,
The Stern-Brocot Tree is supposed to bring out all rational numbers in their reduced forms, So at an infinite stage I will have all rational numbers, isn't it?
Now, when computing the arithmetic mean of all the Stern-Brocot fractions at any stage it
always yields approximations closer and closer to 3/2, indeed, no matter what stage of the
Stern-Brocot Tree you are considering.

If it is ill-defined to say that the arithmetic mean of all rational numbers is 3/2, then
it is ill-defined to say that the Stern-Brocot Tree brings out ALL the rational numbers.

I understand that I have chosen the "Stern-Brocot Order", however this also means
that it is also ill-defined to say that the Stern-Brocot Tree yields all the rational numbers.

not exactly but kind of Zeno's paradoxe...

CRGreathouse · May 16, 2010

arithmetic said:

Yes, it might be ill-defined but what exactly is ill-defined here, let's see,
The Stern-Brocot Tree is supposed to bring out all rational numbers in their reduced forms, So at an infinite stage I will have all rational numbers, isn't it?

It does generate all the rationals, yes.

arithmetic said:

Now, when computing the arithmetic mean of all the Stern-Brocot fractions at any stage it
always yields approximations closer and closer to 3/2, indeed, no matter what stage of the
Stern-Brocot Tree you are considering.

I trust you on this point, I'm not going to verify it tonight.

arithmetic said:

If it is ill-defined to say that the arithmetic mean of all rational numbers is 3/2, then
it is ill-defined to say that the Stern-Brocot Tree brings out ALL the rational numbers.

False.

arithmetic said:

I understand that I have chosen the "Stern-Brocot Order", however this also means
that it is also ill-defined to say that the Stern-Brocot Tree yields all the rational numbers.

False.

arithmetic said:

not exactly but kind of Zeno's paradoxe...

How so?

arithmetic · May 17, 2010

CRGreathouse said:

It does generate all the rationals, yes.

>>Now, when computing the arithmetic mean of all the Stern-Brocot fractions at any stage >>it always yields approximations closer and closer to 3/2, indeed, no matter what stage of >>the Stern-Brocot Tree you are considering.

>I trust you on this point, I'm not going to verify it tonight.

O.K., I agree with you that the rest of the statements are false,
also I hope you have the time to verify that the value I am talking about is certainly: 3/2,
by then, it would be great if you could be so kind to explain the reason that makes ill-defined the value of the arithmetic mean (3/2) of all the Stern-Brocot rational numbers which
at an infinite stage are suposed to be all the rational numbers .
Many thanks for your help, sincerely, I find all this so confusing.

Domingo Gomez Morin

Xitami · May 17, 2010

Calcin-Wilf tree

Shakespeare's like constant = 1.36787803140495...
"Rational average" = 3/2

-------------------------------------------------------------------------------------
othervise
[URL]http://fotoo.pl/zdjecia/files/2010-05/c0ebcfc5.gif[/URL]
GCD(n, d)=1
"rational average" computed in triangles or rectangles diverge
-------------------------------------------------------------------------------------
large black regions on my graph...

arithmetic · May 18, 2010

Wonderful graphs Xitami, indeed.

Shakespeare's like constant = 1.36787803140495...

Xitami, Where does Shakespeare's value come from?

"Rational average" = 3/2
-------------------------------------------------------------------------------------
"rational average" computed in triangles or rectangles diverge
-------------------------------------------------------------------------------------
large black regions on my graph...

Yes, I understand all the problem related to divergent series.
However, conceptually the value 3/2 comes from computing the average of all the Stern-Brocot fractions at once, thus If one assures that the Stern-Brocot Tree yields all the rational numbers at once at an infinite stage and If the average at each stage is closer and closer to 3/2, then it is hard to admit that the Stern-Brocot tree yield all the rational numbers, because at each stage (no matter what stage) the average of all its fractions converges to 3/2.

In the case of using other "orders" (triangles, rectangles) for computing the average is another very different thing, because you are not assuring that your "order" yields all the rational numbers at each stage, you are not using the whole Stern-Brocot Tree as in the previous case but just choosing arbitrary sequences extracted from the Stern-Brocot tree.

So I find all this intriguing even when I am acquainted with the old problem (since Euler's days) of handling divergent series.
That's it.
I understand that 3/2 can't be but I think it has a very important meaning.

Xitami · May 18, 2010

Code:

Calkin-Wilf-Hamlet :-)
     1    1/3 + ...
     - + -------
1    2    3/2 + ...
- + --------------- = 1.367878...
1    2    2/3 + ...
     - + -------
     1    3/1 + ...

--------------------------------------------------

on infinite stage table contain ALL rational numbers

by the way

\lim_{n\to\infty}\frac{1+\sum_{i=2}^{n}2\varphi(i)}{n^2}=\frac{6}{\pi^2}

arithmetic · May 19, 2010

Angel's Falls are fantastic, yes, my country

I can see that 1.36787803140495 does not appear at Plouffe's inverter database, cheers ¡

I came up with the average's stuff because I was wondering about the meaning of all
these fractal values, their validity and their significance.
I think that might be the main point to deal with all these very special fractal-Stern-Brocot fractions.
I have reasons to think they all are transcendental numbers.

Now, I am also eager to know about your formula for 6/Pi^2.
Was your formula derived from the standard continued fraction expression of 6/Pi^2?In reference to those fractal values (original Shakespeare's constant) which
cannot be checked at Plouffe's database because of lack of enough exact digits,
Do you think it would be necesary to use a supercomputer instead of just CUDA?

CRGreathouse · May 19, 2010

arithmetic said:

it would be great if you could be so kind to explain the reason that makes ill-defined the value of the arithmetic mean (3/2) of all the Stern-Brocot rational numbers which
at an infinite stage are suposed to be all the rational numbers .

OK, here's a concrete example. Let R_n be the Stern-Brocot sequence. Let's define a new sequence T_n(x) for some arbitrary rational number x >= 3/2.

Also, let S_1(x) = T_1(x) and S_{n+1}(x) = S_n(x) + T_{n+1}(x) so that S is the sum of the values of T through the nth term.

Let T_1(x) = R_1. Then let T_{2n}(x) = 2nx - S_{2n-1}(x) if that value is not equal to T_1, T_2, ..., T_{2n-1}. Otherwise, let T_{2n}(x) = 2nx - S_{2n-1}(x) + 1/k, where k >= n is the smallest integer k such that 2nx - S_{2n-1}(x) + 1/k is not equal to any of T_1, T_2, ..., T_{2n-1}. (Clearly, n <= k <= 3n.)

For n > 1, let T_{2n-1}(x) = R_n if that value is not equal to T_1, T_2, ..., T_{2n}. Otherwise, let T_{2n}(x) = R_k for the smallest k > n such that R_k is not equal to any of T_1, T_2, ..., T_{2n-1}. (Clearly, n < k <= 3n.)

It's easy to see that for any sequence R_n enumerating the rationals there is a subsequence with average converging to our chosen x. In the case of Stern-Brocot (or any other sequence of rationals which does not grow too quickly or jump around too much) it's clear that the average of the whole sequence converges to x.

So we can pick (almost) any value we want and find an enumeration of the rationals with an average of that value. There is no unique value for it to cluster around.

arithmetic · May 20, 2010

CRGreathouse said:

OK, here's a concrete example. Let R_n be the Stern-Brocot sequence. Let's define a new sequence T_n(x) for some arbitrary rational number x >= 3/2.

Also, let S_1(x) = T_1(x) and S_{n+1}(x) = S_n(x) + T_{n+1}(x) so that S is the sum of the values of T through the nth term.

Let T_1(x) = R_1. Then let T_{2n}(x) = 2nx - S_{2n-1}(x) if that value is not equal to T_1, T_2, ..., T_{2n-1}. Otherwise, let T_{2n}(x) = 2nx - S_{2n-1}(x) + 1/k, where k >= n is the smallest integer k such that 2nx - S_{2n-1}(x) + 1/k is not equal to any of T_1, T_2, ..., T_{2n-1}. (Clearly, n <= k <= 3n.)

For n > 1, let T_{2n-1}(x) = R_n if that value is not equal to T_1, T_2, ..., T_{2n}. Otherwise, let T_{2n}(x) = R_k for the smallest k > n such that R_k is not equal to any of T_1, T_2, ..., T_{2n-1}. (Clearly, n < k <= 3n.)

It's easy to see that for any sequence R_n enumerating the rationals there is a subsequence with average converging to our chosen x. In the case of Stern-Brocot (or any other sequence of rationals which does not grow too quickly or jump around too much) it's clear that the average of the whole sequence converges to x.

So we can pick (almost) any value we want and find an enumeration of the rationals with an average of that value. There is no unique value for it to cluster around.

thanks,
Yes, there is a good book: "Mathematical fallacies and paradoxes" author: Bryan Bunch, which I have enjoyed so much, it is full of examples on this subject.

arithmetic · May 25, 2010

I have added two new entries to the Generalized Continued Fractions menu at:
http://domingogomez.web.officelive.com/gcf.aspx

Of course the issue on the arithmetic mean of all stern-brocot fractions is not related to the convergence of these fractal fractions, but it is interesting to clearly understand
the behavior of Stern-Brocot sequences.
The fractal fractions always yield roots of polynomial equations of any degree.

arithmetic · Feb 13, 2011

Hi Xitami,

My Fractal fraction for the Golden Mean appeared in the January-2011 issue of the American Mathematical Monthly, No. 118, page 65.

I hope you could take a look at such very special continued fraction.
There is much more on this, but I prefer to let time talk.

Warmest regards,

Domingo Gomez Morin
http://domingogomez.web.officelive.com/

To PI or not to PI challenge. Stern-Brocot

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