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Find the series expansion of $\left(e^{(x-1)}\right)^2$. I thought maybe I could use binomial expansion but that is only for $(1+x)^n$, so now I am unsure how to proceed. I could set $(x-1)^2=n$ and use the series expansion for $e^n$.

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  • $\begingroup$ do you mean $e^{((x-1)^2)}$ or $(e^{(x-1)})^2$ $\endgroup$ – Jimmy R. Oct 22 '14 at 16:27
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Since it is unclear whether you mean $e^{\left((x-1)^2\right)}$ or $\left(e^{(x-1)}\right)^2$, I consider both cases: $$e^{\left((x-1)^2\right)}=\sum_{n=0}^{\infty}\dfrac{\left((x-1)^2\right)^n}{n!}=\sum_{n=0}^{\infty}\dfrac{(x-1)^{2n}}{n!}$$ but $$\left(e^{(x-1)}\right)^2=e^{(x-1)+(x-1)}=e^{2(x-1)}=\sum_{n=0}^{\infty}\dfrac{\left(2(x-1)\right)^n}{n!}$$


The second term can be also written as $$\left(e^{(x-1)}\right)^2=e^{2(x-1)}=e^{-2}\cdot e^{2x}=e^{-2}\cdot\left(1+\dfrac{2}{1!}x^1+\dfrac{2^2}{2!}x^2+\dfrac{2^3}{3!}x^3+\dfrac{2^4}{4!}x^4+\ldots\right)$$ which is the Taylor expansion of the function $f(x)=e^{2x}$ at $a=0$.

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  • $\begingroup$ I meant the second one, (e^(x−1))^2, do you have any idea how to find that from basic series expansions, not just by using the formula? $\endgroup$ – user183782 Oct 22 '14 at 16:37
  • $\begingroup$ See my edit above $\endgroup$ – Jimmy R. Oct 22 '14 at 16:45
  • $\begingroup$ @user183782 Use the binomial theorem on $\lim_{n\to\infty}(1+\frac xn)^n$ to get the Taylor expansion of $e^x$. $\endgroup$ – Akiva Weinberger Oct 22 '14 at 18:20

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