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Math Help - Finding Taylor series for f

  1. #1
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    Finding Taylor series for f

    Hello there, I was having some trouble with this problem:

    Find the Taylor series for f centered at 4 if:

    \[f^n(0) = \frac{(-1)^nn!}{3^n(n+1)}\]

    What is the radius of convergence of the Taylor series?

    Help appreciates.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Finding Taylor series for f

    Quote Originally Posted by Intrusion View Post
    Hello there, I was having some trouble with this problem:

    Find the Taylor series for f centered at 4 if:

    \[f^n(0) = \frac{(-1)^nn!}{3^n(n+1)}\]

    What is the radius of convergence of the Taylor series?

    Help appreciates.
    To find the Taylor series for f centered at 4, you need the values of f^n(4).
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Finding Taylor series for f

    Quote Originally Posted by Intrusion View Post
    Hello there, I was having some trouble with this problem:

    Find the Taylor series for f centered at 4 if:

    \[f^n(0) = \frac{(-1)^nn!}{3^n(n+1)}\]

    What is the radius of convergence of the Taylor series?

    Help appreciates.
    The Taylor expansion of f(*) around z=0 is...

    f(z)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ z^{n}}{3^{n}\ (n+1)} (1)

    ... and the series (1) has radius of convergence 3 because the singularity of the function in z=-3. Therefore the series of f(*) centered in z=4 has radius of convergence 7...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 18th 2011 at 05:06 AM.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Finding Taylor series for f

    The problem is ill defined, we have no sufficient information about f . For example

    f(x)=\begin{Bmatrix} \displaystyle \sum_{n=0}^{\infty}\dfrac{(-1)^n}{3^n(n+1)}x^n& \mbox{if}& |x|<3\\x-4 & \mbox{if}& |x|\geq 3\end{matrix}\quad g(x)=\begin{Bmatrix} \displaystyle \sum_{n=0}^{\infty}\dfrac{(-1)^n}{3^n(n+1)}x^n& \mbox{if}& |x|<3\\|x-4| & \mbox{if}& |x|\geq 3\end{matrix}

    We have f^{(n)}(0)=g^{(n)}(0)=\dfrac{(-1)^nn!}{3^n(n+1)} . However: a) The Taylor series of f centered at 4 has radius of convergence +\infty . b) The Taylor series of g centered at 4 does not exist .
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: Finding Taylor series for f

    Quote Originally Posted by chisigma View Post
    The Taylor expansion of f(*) around z=0 is...

    f(z)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ z^{n}}{3^{n}\ (n+1)} (1)

    ... and the series (1) has radius of convergence 3 because the singularity of the function in z=-3. Therefore the series of f(*) centered in z=4 has radius of convergence 7...
    May be that a further explanation from me is necessary... if we consider the series expansion...

    g(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ x^{n}}{3^{n}}= \frac{1}{1+\frac{x}{3}} (1)

    ...it is not too difficult to obtain...

    f(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n}\ x^{n}}{3^{n}\ (n+1)}= \frac{1}{x}\ \int_{0}^{x} g(\xi)\ \d \xi = \frac{3\ \ln (1+\frac{x}{3})}{x} (2)

    Now the f(*) defined in (2) is analytic for x>-3 and that means that its series expansion centered in x_{0}>-3 has radius of convergence x_{0}+3 ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 19th 2011 at 12:47 PM. Reason: a trivial typing error...
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Finding Taylor series for f

    Quote Originally Posted by chisigma View Post
    May be that a further explanation from me is necessary...
    Well, independently of your explanation I would insist that the only solution to the problem is: The radius of convergence of the Taylor series for f centered at x_0=4 (if exists) can be any \rho\in[0,+\infty] depending on the expression of f^{(n)}(4) .
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