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Expressing a series as a function

  1. Aug 29, 2010 #1
    1. The problem statement, all variables and given/known data

    I seem to have forgotten how to express a series as a function. I took calc 2 over the summer so we rushed through everything and i don't think all that information got a chance to settle. So here is the question. I'm going to try using syntax but i will write it out as well as this is the first time i use syntax.

    If
    [tex]\sum \frac{x^{2n}}{n!}[/tex] = f(x) , then express f in terms of standard functions.

    sum (n=0 to infinity) of (x^(2n)) / (n!)


    2. Relevant equations




    3. The attempt at a solution

    No attempt, i don't even know where to start.
     
  2. jcsd
  3. Aug 29, 2010 #2
  4. Aug 29, 2010 #3
    Thank you, i now know which chapter to re-read. Greatly appreciated.
     
  5. Aug 29, 2010 #4
    yup, looks like the Taylor series of exp(t) where t = x^2 :)
     
  6. Aug 29, 2010 #5
    I can't seem to be able to come to the conclusion that you have gomunkul51 unless i follow the method used from the following site

    http://www.cliffsnotes.com/study_guide/Taylor-Series.topicArticleId-19736,articleId-19727.html" [Broken]

    I know i should probably just stay quiet and deal with it, but is there any other way other than the one provided. Simply substituting the [tex]x^{2}[/tex] in for x is easy but requires the memorization of the taylor series expansion of [tex]e^{x}[/tex] I would prefer to be able to break it down using only the general taylor series formula. For example i tried the following:

    f(x) = [tex]e^{x^{2}}[/tex]
    f ' (x) = 2x[tex]e^{x^{2}}[/tex]
    ...
    [tex]f^{(n)}[/tex](x) = [tex]2^{n}[/tex][tex]x^{n}[/tex][tex]e^{x^{2}}[/tex]

    f(0) = 1
    f ' (0) = 0
    ...
    [tex]f^{(n)}[/tex](0) = ???

    So using this method would be one way but it calls for f(x) to be chosen and becomes a guess and check deal if you don't have the experience with taylor series to be able to recognize the modified taylor series expansion of exp(x). My question: is it possible to go backwards from a taylor series to a function without the memorization that comes from experience dealing with common taylor series?

    If not I appreciate your help.

    P.S. Sorry for the sloppy syntax, im not used to it.
     
    Last edited by a moderator: May 4, 2017
  7. Aug 30, 2010 #6
    I advice you to memorize exp(x), sin(x), cos(x) and 1/(1-x) Taylor expansions they are everywhere! calculus, Complex Analysis...

    You have to be able to recognize what the series looks like! this is the human approach :)
    when you did recognized it look like some Taylor expansion you know, you can try to modify the original Taylor expansion like this: substitute x=f(t) or use differentiation and integration of the Taylor series.

    What you tried to do with:
    f(x) = [tex]
    e^{x^{2}}
    [/tex]

    is trying to check if it is really equals to the series? because to do so you had to propose that you series equals to that function.

    also you didn't differentiate correctly:
    f(x) = exp(x^2)
    f'(x) = 2x*exp(x^2)
    f''(x) = 2exp(x^2) + 4x^2*exp(x^2)
    ...
     
  8. Aug 30, 2010 #7
    Ok, thank you for your advice. I guess i will do a bunch of questions to drill it into my brain. I wrote down those taylor series before your reply because i kept seeing them online. Well, at least i only have to memorize 4. Again thank you.
     
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