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Math Help - Taylor Series Expansion Function

  1. #1
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    Unhappy Taylor Series Expansion Function

    What is the nth order term in the Taylor series expansion of the function

    f(x)=exp x about x=-3

    Can someone please help me out with the full workings and some explanation-I have an Aptitude test to write tomorrow i have been out of school for the past 8 years.i really need help!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pringles View Post
    What is the nth order term in the Taylor series expansion of the function

    f(x)=exp x about x=-3

    Can someone please help me out with the full workings and some explanation-I have an Aptitude test to write tomorrow i have been out of school for the past 8 years.i really need help!
    The Taylor series about a point x=x_0 is defined as:

    \sum_{n=0}^{\infty}\frac{f^{\left(n\right)}\left(x  _0\right)}{n!}(x-x_0)^n

    Since f(x)=e^{x}, we see that a derivative of any order would always yield of e^{x}. Evaluated at x=3, this would give us e^{3}.

    Thus, the Taylor series for this function at the given point is \sum_{n=0}^{\infty}\frac{e^3}{n!}(x-3)^n. Now, we see that the nth term of this series is \color{red}\boxed{\frac{e^3}{n!}\left(x-3\right)^n}.

    Does this make sense?
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  3. #3
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    x=-3

    x=-3

    So I guess the modified answer would be like this:

    e^-3/n!(x+3)^n ??

    thanks
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pringles View Post
    x=-3

    So I guess the modified answer would be like this:

    e^-3/n!(x+3)^n ??

    thanks
    woops! I didn't see that negative in front of the three.

    Yes, your modification is correct.
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  5. #5
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    Thumbs up Thanks Very much

    Thanks very much
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