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Math Help - Taylor Series Help

  1. #1
    Newbie vortex's Avatar
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    Lightbulb Taylor Series Help

    Hi everybody, I hope anyone could help


    1. The problem statement, all variables and given/known data

    Find the first three terms of the Taylor series for f(x) at c.




    2. Relevant equations

    f(x)= f(c) + f'(c).(x-c)/1! + f"(c).(x-c)^2/2! + f'''(c).(x-c)^3/3! + ... + fn(c).(x-c)^n/n! + ...

    3. The attempt at a solution

    what I understand is that I have to find the followings:
    f'(x), f''(x), f'''(c)
    and
    f'(c), f''(c), f'''(c)

    is that right?

    well, to find f'(x) I used the product rule

    d/dx (uv) = u'v + uv'

    u = x
    u' = 1

    v = e^x
    v' = e^x

    d/dx (xe^x) = e^x + xe^x
    = e^x(1+x)

    but unfortunately i couldn't go forward!!


    the answer in the back of the book is
    Last edited by vortex; December 6th 2010 at 09:21 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    As alternative approach You can consider that y(x)= x\ e^{x} is solution of the DE...

    \displaystyle y^{'} = y + e^{x} , y(-1)= -\frac{1}{e} (1)

    From (1) You derive that...

    \displaystyle y(-1)= - \frac{1}{e}

    \displaystyle y^{'} = y + e^{x} \implies y^{'}(-1)= 0

    \displaystyle y^{''} = y^{'} + e^{x} \implies y^{''}(-1)= \frac{1}{e}

    \displaystyle y^{(3)} = y^{''} + e^{x} \implies y^{(3)}(-1) = \frac{2}{e}

    ...

    \displaystyle y^{(n)} = y^{(n-1)} + e^{x} \implies y^{(n)} (-1)= \frac{n-1}{e}

    ... so that is...

    \displaystyle x\ e^{x} = \frac{1}{e} \sum_{n=0}^{\infty} \frac{n-1}{n!}\ (x+1)^{n} (2)



    Merry Christmas from Italy

    \chi \sigma
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  3. #3
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    Suppose g(x) = x, and h(x) = e^{x} , so that f(x) = g(x)\cdot h(x).

    Observe that g^{n}(x) = \begin{cases}<br />
1\mbox{ if } n = 1\\<br />
0\mbox{ if } n\ge 2 \end{cases} and h^n(x) = e^x, for all n \in\mathbb{N}.

    Using Leibniz's theorem, we get:

    [tex]\displaystyle f^{n}(x)\bigg|_{x=-1} = \sum_{0 \le k \le n}\binom{n}{k}g^{n-k}(x)h^{k}(x)\bigg|_{x=-1} = \sum_{0 \le k \le n}\binom{n}{k}x^{n-k}{e}^{kx}\bigg|_{x=-1}[/Math]

    . . . . . . . . . . \displaystyle  = \sum_{0 \le k \le  n}\binom{n}{k}e^{x(n-k)}x^{k}\bigg|_{x=-1} = e^{x}\left(x+n\right)\bigg|_{x=-1} = \frac{1}{e}\left(n-1}\right).

    Thus, as Chisigma found, by Taylor's theorem we have:

    \displaystyle f(x) = \sum_{n=0} ^{\infty} \frac {f^{(n)}(c)}{n!} \, (x-c)^{n} = \sum_{n=0} ^{\infty} \frac {\frac{1}{e}\left(n-1\right)}{n!} \, (x+1)^{n} = \frac{1}{e}\sum_{n=0} ^{\infty} \frac {\left(n-1\right)}{n!} \, (x+1)^{n}.
    Last edited by TheCoffeeMachine; December 6th 2010 at 01:40 PM.
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  4. #4
    Newbie vortex's Avatar
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    Thumbs up

    Many, many thanks for all your help

    Thank you (chisigma), and Merry Christmas from Saudi Arabia : )

    Tkank you (
    TheCoffeeMachine), your cooperation is highly appreciated



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