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Series of functions help.

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the series,

    [tex]\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}[/tex]

    defines a continuous function f on the domain of convergence. What is this domain? In addition, write a series representation of ( f ' ) and determine the domain of convergence of this series to ( f ' ).

    2. Relevant equations



    3. The attempt at a solution

    I need abit of help with this problem. If somebody could point me in the direction I would be very happy.

    It looks to me that the series in question might be smaller than the series 1/ (k^2) and therefore converges on a domain of all real numbers.

    I had a question about the wording, "the series defines a continuous function f on the domain of convergence". Does this mean that I am looking for the function that this series uniformly or pointwise converges to?
     
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  3. Feb 24, 2009 #2

    Office_Shredder

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    In a manner of speaking, but you can consider it more directly. If, given a value of x, the sum converges, then we can define f(x) = the value of the summation for x, and this is the function they're referring to.
     
  4. Feb 25, 2009 #3
    I have that for,

    [tex]x=0 , \sum_{n=1}^{\infty}\frac{1}{n^2}[/tex]

    [tex]x=1 , \sum_{n=1}^{\infty}\frac{1}{1+n^2}[/tex]

    [tex]x=2 , \sum_{n=1}^{\infty}\frac{1}{4+n^2}[/tex]

    I know that these converge to 0. So no matter what are x values, this series converges. I would say that the domain of convergence is all of R. Am I right with that?

    So did the series define the function f(x)=0. I am having trouble understanding what this means.
     
  5. Feb 26, 2009 #4
    Ok, so this series converges to some function f and I must find what the interval of x is in which this happens.

    Second, if we take the derivative of f, say ( f ' ), the same series also converges to ( f ' ) but on a different (or same) interval ?
     
  6. Feb 26, 2009 #5

    lanedance

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    why not have a look at the term by term differentiation of f? i think it asks for this is the question
     
    Last edited: Feb 26, 2009
  7. Feb 26, 2009 #6

    HallsofIvy

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    Surelly you know that showing that a function converges for 3 values of x does NOT show that it converges for ALL x. Also, it is clear that those DO NOT "convert to 0". The very first one starts 1+ 1/4+ 1/9 and has only positive terms.
     
  8. Feb 26, 2009 #7
    I have a theorem that says that the radius of convergence of a power series is the same as the radius of convergence of the term by term differentiation of the power series.

    I didn't know if this was a power series or not. I cannot see how to separate the a_n and the x^n if it is, and if it isn't does the result apply also to series like the one I have brought up?

    I am confused about this though. I though that I could show that the summation for 1/k^2 converges by the ratio test by proving that for n large enough the terms of the ratio are all less than 1. Then for any other x in R, x^2 is positive and would be adding more to the denominator, thus making it even smaller and thus would converge also.

    What am I confusing here? I understand that we are adding positive terms now though, and the series all converge to something that is not necessarily 0.
     
  9. Feb 26, 2009 #8
    I understand the problem better now. I don't think I need to find the equation of the function that it converges to. (It would look like a bell curve though, I'm guessing)

    I used the ratio test (with the x's in there) and determined that the series converges for all x in R. (R = infinity). I then did a term by term differentiation and performed the ratio test on that to get the same radius of convergence (all x in R).

    I've learned a couple things here about the function. Is there anything else I am missing? (Assuming that doing the 2 ratio tests were valid?)
     
  10. Feb 26, 2009 #9

    lanedance

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    not sure what your lecturer wants.... but you have shown both f & f' converge on a radius of covergence, and determined this radius

    Reading the question I think you might still need to show continuity on the domain of convergence... maybe a quick epsilon delta?
     
  11. Feb 26, 2009 #10
    I was wondering if I could do that. Do you mean something like this?

    [tex]| \sum_{n=1}^{\infty}\frac{1}{x^2+n^2}- \sum_{n=1}^{\infty}\frac{1}{a^2+n^2}|[/tex]

    I tried this to no avail after I combined? the series into one. I found one |x-a| in there but it seemed like nothing else.
     
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