Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Strange sums

  1. Apr 14, 2006 #1
    "strange" sums...

    Let be the next 2 sums in the form:

    [tex] f(x)+f(x+1)+f(x+2)+.......... [/tex] (1) and

    [tex] f(x)+f(2x)+f(3x)+............ [/tex] (2)

    how would you calculate them?..well i used a "non-rigorous" but i think correct method to derive their sums..let be the infinitesimal generator

    [tex] D=d/dx [/tex] (traslation) [tex] xD=x(d/dx) [/tex] (dilatation)

    the 2 series above can be "summed" as:

    [tex] (1+e^{D}+e^{2D}+e^{3D}+..........)f(x) [/tex]

    [tex](1+e^{xD}+e^{2xD}+e^{3xD}+..........)f(x) [/tex]

    now we put [tex] (1/D)f(x)=F(x) [/tex] and [tex](1/xD)f(x)=F(x)/x [/tex]

    where F(x) is the "primitive" of f(x) then we would have..

    [tex] \sum_{n=0}^{\infty}\frac{B_{n}}{n!}F^{n}(x) [/tex]

    [tex] \sum_{n=0}^{\infty}\frac{B_{n}}{n!}x^{n-1}F^{n}(x) [/tex]

    these 2 expressions above would be the sum for (1) and (2) respectively where the B are the Bernoulli,s number and (n) means n-th derivative of F.
  2. jcsd
  3. Apr 15, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What on earth allows you to assume that the f are differentiable? And even then what does raising e to the power d/dx do? Let what be an infinitesimal generator of what? What's the/a primitive of f? Why invent a name for it that you don't ever use again? Not that the primitive makes much sense (inverting d/dx?).

    You really need to focus on explaining what you mean properly (not rigorously; you have a completely unnecessary issue with rigour).

    If by F^n you mean F raised to the power n (F(x)^n), then it cannot be a correct formula, for obvious reasons which I'll let you figure out (hint what if F(x)=0 for some x).

    If it means F applied n times, then my gut reaction is that it still will be false, not that it is clear what F actually is.

    Your second sum is also not defined at x=0.
    Last edited: Apr 15, 2006
  4. Apr 15, 2006 #3
    -No.... [tex] F(x)=\int{dx}f(x) [/tex] and [tex] \frac{d^{n}F}{dx^{n}}=F^(n)(x) [/tex] by the way you say is not correct Euler himself used (1) in the same manner i do to prove that:

    [tex] 1+2+3+4+5+6+...........=(-1/12) [/tex] just put f(x)=x and apply the differential operator..by the way is a known result that:

    [tex] e^{aD}f(x)=f(x+a) [/tex] or [tex] e^{xD}=f(2x) [/tex] just take a look at "Wikipedia" (english version)..chao.
  5. Apr 15, 2006 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    There you go again. I didn't say it was not a correct method I said that what you wrote doesn't make sense. You have a very odd sense of what is 'universally known and accepted' terminology, and a poor way of presenting things; partly this is a language problem, partly it is not.

    If you ask a question, why not explain the terms properly? However, I'll take a guess that e^D is short hand for the operator

    [tex] \sum \frac{1}{n!}\frac{d^n}{dx^n}[/tex]

    is that about right?

    And you assert Euler proved that the sum of all the positive integers is -1/12, do you? If indeed he did demonstrate that then it hardly implies that the method is correct, does it?

    Note that F is only defined up to an additive constant. Which choice of constant should it be?

    Anyway, you're still assuming f to be a smooth function, yet you didn't say that did you? It's things like that that you need to be careful about.

    Was that bit about F^(n) meant to be an answer to the question I asked about whether F^n meant F applied n times or F(x) raised to the power n? Because if it was you've just changed notation from your first post, and your latex is incorrect. You need to enclose all of the things you want in the superscript inside curly braces.
    Last edited: Apr 15, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook