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I've been struggling with this problem:

Find the Taylor series representation for $xe^{2x}$

I was able to find the Taylor series for $e^{2x}$ (centered at a=k) in a previous exercise which I found to be $\sum_{n=0}^\infty \frac{2^ne^{2k}(x-k)^n}{n!}$

I've looked on wolfram alpha and the answer seems to be $x{\sum_{n=0}^\infty \frac{2^ne^{2k}(x-k)^n}{n!}}$

I just don't know if that's right and if so, how to arrive at that conclusion. There is no grade associated with this. It's a practice exercise and it's been a while since I've had to use Taylor series so it's infuriating me that I can't figure this out.

Any help would be appreciated!

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  • $\begingroup$ Multiply $x$ to the series you've already found, no? Seems like that's the answer WA is suggesting. $\endgroup$ – Kaster Aug 27 '15 at 23:27
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Hint. You may just write $$ xe^{2x}=(x-k)\times\color{blue}{e^{2x}}+k\times\color{blue}{e^{2x}}\tag1 $$ then insert $$ \color{blue}{e^{2x}}=\sum_{n=0}^\infty \frac{2^ne^{2k}(x-k)^n}{n!}\tag2 $$ in $(1)$.

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  • $\begingroup$ What's the point? Couldn't you just substitute expansion for $e^{2x}$ directly to the $x e^{2x}$? $\endgroup$ – Kaster Aug 27 '15 at 23:29
  • $\begingroup$ @Kaster The Taylor series is at $x=k$ not at $x=0$... $\endgroup$ – Olivier Oloa Aug 27 '15 at 23:30
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    $\begingroup$ @Kaster Because a Taylor series at $x=k$ has to be of the form $$ \sum \limits_{n = 0}^\infty \frac{a_n}{n!}(x - k)^n$$ $\endgroup$ – Olivier Oloa Aug 27 '15 at 23:39
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    $\begingroup$ @OlivierOloa Thank you very much for your help. I'll see if my algebra will allow me to arrive at that answer. Thanks again! $\endgroup$ – WhatsAGuitar Aug 27 '15 at 23:53
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    $\begingroup$ @OlivierOloa thank you very much! $\endgroup$ – WhatsAGuitar Aug 28 '15 at 0:17

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