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Looking for a function with derivatives of

  1. Dec 5, 2009 #1
    I'm searching for a function expressed explicitly as a non-infinite series where
    f(0)=1/1
    f'(0)=1/3
    f''(0)=1/5
    ...
    fn(0)=1/(2n+1)
    ...

    Does it exist? What is it?
     
  2. jcsd
  3. Dec 5, 2009 #2

    Mute

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    If it's a strictly finite series then it's just an (n+1)-th degree polynomial with the coefficient of each x^k being f^(k)(0)/k!.

    [tex]f(x) = 1 + f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f^{(3)}(0)x^3 + \dots + \frac{1}{n!}f^{(n)}(0)x^n.[/tex]

    Are you really just looking for a finite series?
     
    Last edited: Dec 5, 2009
  4. Dec 5, 2009 #3

    mathman

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    Slight error on your part: kth term has k! as divisor.
     
  5. Dec 5, 2009 #4

    Mute

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    Whoops! You're right, I've corrected the post now.
     
  6. Dec 6, 2009 #5
    Nope, this is what I'm looking to avoid. I can make a maclaurin series out of it, but that doesn't help me. What I mean is that I'm hoping it can be expressed without an implied infinite pattern. For instance, I don't want
    [tex]
    f(x) = 1 + f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f^{(3)}(0)x^3 + \dots + \frac{1}{n!}f^{(n)}(0)x^n + \dots
    [/tex]

    I want
    [tex]
    f(x) = e^x
    [/tex]

    For the function I'm thinking of, [tex]f^n(0)=\frac{1}{2n+1}[/tex] for all n from 0 to infinity, so it wouldn't terminate as a Maclaurin series.
     
  7. Dec 6, 2009 #6

    HallsofIvy

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    I don't understand what you mean by "an implied infinite pattern". Yes, [itex]f(x)= 1+ x+ \frac{1}{2}x^2+ \dots+ \frac{1}{n!} x^n+ \dots[/itex] is an infinite series and is equal to ex.

    But [itex]f(x)= 1+ x+ \frac{1}{2}x^2+ \dots+ \frac{1}{n!} x^n[/itex], for fixed n, is not. It is a finite series and is a polynomial. If n= 3, that is [itex]1+ x+ (1/2)x^2+ (1/6)x^3[/itex].

    So that is NOT a "non-infinite" series as you said in your original post. I have no idea what you are asking for. Nor do I know what you mean by "wouldn't terminate as a MacLaurin series".

    Any "power series" that sums to a given analytic function is the Taylor's series of that function about some point.
     
  8. Dec 6, 2009 #7

    arildno

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    Let us set up the relevant power series here:
    [tex]F(x)=\sum_{n=0}^{\infty}\frac{1}{n!(2n+1)}x^{n}[/tex]

    Now, assume x>=0, and set [tex]x=y^{2}[/tex]
    and we have:
    [tex]F(x)=G(y)=\sum_{n=0}^{\infty}\frac{1}{n!(2n+1)}y^{2n}[/tex]
    Then, we have:
    [tex]yG(y)=\sum_{n=0}^{\infty}\frac{1}{n!(2n+1)}y^{2n+1}[/tex], and:
    [tex]\frac{d}{dy}(yG(y))=\sum_{n=0}^{\infty}\frac{1}{n!}(y^{2})^{n}=e^{y^{2}}[/tex]

    We therefore have:
    [tex]yG(y)=\int_{0}^{y}e^{t^{2}}dt\to{F}(x)=\frac{1}{\sqrt{x}}\int_{0}^{\sqrt{x}}e^{t^{2}}dt[/tex]
    Thus, your desired function can be described in terms of a definite integral.

    Hmm..something has occurred at x=0 that I don't like, it seems F(0) equals zero.
     
    Last edited: Dec 6, 2009
  9. Dec 6, 2009 #8
    To HallsofIvy:
    Sorry for the confusion.. I was trying to express that I wanted a function that is finitely expressible. It's true that I wasn't interested in a series.

    A maclaurin series expansion will terminate if the nth derivative of the function is always zero after some n.

    To arildno:
    I can assure you your work is correct, because this is the inverse of what I'm trying to do. I was looking for a solution to [tex]\int e^{(x^2)}dx[/tex], and figured that if I could find a nice closed form for f(x), then [tex]\int e^{(x^2)}dx=xf(x^2)[/tex].
    At this point I'm going to assume what I'm looking for doesn't exist or it would have been found already.

    Don't worry about f(0)=0; the +c can take care of that. Actually, would x=0 even work with the 1/sqrt(x) at the start?
     
    Last edited: Dec 6, 2009
  10. Dec 6, 2009 #9

    Gib Z

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    It's quite well known that integral has no elementary solution, and it can be proven there is no solution, it's not just because no one has found it yet. Doing so however requires the Risch Algorithm, something I have only a vague idea how it works.
     
  11. Dec 7, 2009 #10

    arildno

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    Dearly Missed

    Correction:

    F(0) can be set at 1, and still be a continuous function.
     
  12. Dec 8, 2009 #11

    for this case i do not know, but for the case [tex] \frac{1}{(2n+1)2n!} [/tex]

    there is a generating function [tex] 2x^{-1}\int_{0}^{x}dtsin(t)/t =g(x) [/tex]

    taking the inverse laplace transfomr of this function you will get a function f(x) so its taylor series includes 1/(2n+1) in its derivatives
     
    Last edited by a moderator: Dec 9, 2009
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