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I tried creating the Taylor series of $e^{\frac{x^2}{2}}$ around 0. With The general case (around a): $$e^{\frac{a^2}{2}}(x)=\sum_{i=0}^n\frac{a^ie^{\frac{a^2}{2}}}{i!}(x-a)^i+\frac{\xi^{n+1}e^{\frac{\xi^2}{2}}}{(n+1)!}(x-a)^{n+1}$$

So when $a=0$ I get that $f(x) = 1$ but $e^{\frac{\xi^2}{2}}$ is bound so I don't think the remainder blows up, so what is happening here?

Thanks.

Eloo
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  • You have to determine the derivates of the function at $x_0=a$ – Peter May 01 '18 at 14:19
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    Should be just the $e^x$ maclaurin series but with $x^2/2$ in place of $x$ – Andrew Li May 01 '18 at 14:23
  • I fixed it, but my problem still persists. $(e^{\frac{x^2}{2}})' = xe^{\frac{x^2}{2}}$ no? – Eloo May 01 '18 at 14:23
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    Yes but $(e^{\frac{x^2}{2}})'' = (1+x^2) e^{\frac{x^2}{2}}$ so not all terms are zero. Your expression for the $n$th derivative is not correct. The $n$th derivative is $f^{(n)}(x) = p_n(x) e^{\frac{x^2}{2}}$ where $p_n(x)$ is some $n$th order polynomial with $p_n(0) = 0$ when $n$ is odd and non-zero when $n$ is even. The Taylor series for $a=0$ is most easily found from that of $e^{x}$ by taking $x\to x^2/2$ so $e^{\frac{x^2}{2}} = \sum_{k=0}^\infty \frac{x^{2n}}{2^nn!}$ – Winther May 01 '18 at 14:24
  • Related questions: How to get nth derivative of $e^{x^2/2}$ – Winther May 01 '18 at 16:07

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