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Math Help - taylor series at b=1

  1. #1
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    taylor series at b=1

    hi,
    how would i find the taylor series for xe^x at b=1
    i know the known taylor series for e^x but what do I do if there's an x in front and b=1

    thanks
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  2. #2
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    Just write it as x(e^x) = x(1 + x + ...)

    EDIT: I used the Maclaurin series, but you get the idea.
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  3. #3
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    how

    how would this look in the sigma form? I check the answer and it had some 'e' there. how about b=1
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  4. #4
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    If f(x)=xe^x, and g(x)=e^x:

    g(1+t)=f(1) + f'(1)t + \frac{f''(1)t^2}{2!} + O(t^3) = e^1 + {e^1}t + {e^1}t^2/2 + O(t^3)

    Since x = 1+t, we get e(1+t)(1+ t + t^2/2 + O(t^3))

    Or in Sigma form:

    \sum_{n=0}^{\infty}e(t+1) \frac{t^{n}}{(n)!}
    Last edited by Spec; October 20th 2008 at 01:08 PM. Reason: b=1 and not 2 :(
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  5. #5
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    ?

    where did u get 2+t?
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  6. #6
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    Quote Originally Posted by khuezy View Post
    where did u get 2+t?
    Sorry, I read your original post as b=2 for some reason.
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  7. #7
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    <br /> <br />
\sum_{n=0}^{\infty}e(t+1) \frac{t^{n}}{(n)!}<br />

    this isn't in the taylor series formation right?
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  8. #8
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    Quote Originally Posted by khuezy View Post
    <br /> <br />
\sum_{n=0}^{\infty}e(t+1) \frac{t^{n}}{(n)!}<br />

    this isn't in the taylor series formation right?
    It's the same as <br /> <br />
\sum_{n=0}^{\infty}xe \frac{(x-1)^{n}}{(n)!}<br />
since x=1+t \implies t=x-1
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  9. #9
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    thanks

    if b=2 then
    <br /> <br />
\sum_{n=0}^{\infty}xe^{2} \frac{(x-2)^{n}}{(n)!}<br />
    ?
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  10. #10
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    Quote Originally Posted by khuezy View Post
    if b=2 then
    <br /> <br />
\sum_{n=0}^{\infty}xe^{2} \frac{(x-2)^{n}}{(n)!}<br />
    ?
    Correct.
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  11. #11
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    <br /> <br />
e + e\sum_{n=0}^{\infty} \frac{(n+1)}{(n)!} (x-1)^{2}<br />

    that's the answer key to the problem. How come they look so different?
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