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Math Help - nth derivative taylor series

  1. #1
    Senior Member polymerase's Avatar
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    nth derivative taylor series

    Let f^{(n)}(a) denote the n-th derivative of f at a. if f(x)=xe^{2x}, then f^{(64)}(0)=

    I can get the answer using the old fashion way, but when I lookeed at the solution using power series, I don't understand one of the steps. heres the soultion my prof has:

    e^x=\sum_{n=0}^\infty \frac{x^n}{n!}
    f(x)=xe^{2x}=\sum_{n=0}^\infty \frac{2^nx^{n+1}}{n!}=\sum_{m=1}^\infty \frac{2^{m-1}x^{m}}{(m-1)!}= \boxed{\sum_{m=1}^\infty \frac{f^{(m)}(0)}{m!}x^m}

    hence, f^{(m)}(0)=\frac{2^{m-1}m!}{(m-1)!}

    I understand every step except the step to the result in the box. How are they equal?
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  2. #2
    Moo
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    Hello,

    The thing in the box is the result of the Taylor series : Taylor series - Wikipedia, the free encyclopedia
    It is said that the sum begins at m=1, but here it doesn't matter as f(0)=0

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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    The thing in the box is the result of the Taylor series : Taylor series - Wikipedia, the free encyclopedia
    It is said that the sum begins at m=1, but here it doesn't matter as f(0)=0

    nvm i figured it out...
    What I was really trying to get at was how does \sum_{m=1}^\infty \frac{f^{(m)}(0)}{m!}x^m=\sum_{n=0}^\infty \frac{2^nx^{n+1}}{n!}
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  4. #4
    Moo
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    nvm ?

    The left hand part comes from the Taylor series
    The right hand part comes from the power series :-)

    e^x=\sum_{n \geq 0} \frac{x^n}{n!}

    Hence e^{2x}=\sum_{n \geq 0} \frac{(2x)^n}{n!}=\sum_{n \geq 0} \frac{2^n \ x^n}{n!}

    -> xe^{2x}=\sum_{n \geq 0} \frac{2^n \ x^{n+1}}{n!}
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Moo View Post
    nvm ?
    never mind
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