I'm learning basic things about power series. The author of the text I'm reading writes that it is rather straight forward to see that $\sum_{k=0}^{\infty} \dfrac{(-1)^kx^{2k}}{2^kk!} = \exp \left(\dfrac{-x^2}{2} \right)$

I don't understand how I'm supposed to see this. I don't know if one could do it using Taylor series but I don't think that's what the author is suggesting since they have not been covered in relation to power series yet, so I would like an answer that does not use Taylor series.

The only way I've learnt to evaluate an actual value of a power series is to rewrite it like the derivative or integral of a geometric series, but I can't see how I would do that here.

EDIT: I'm not sure about using the power series definition of $e$ since $e$ hasn't really been defined in the text. I'm going to leave the question open for a while in case anyone could come up with another somewhat simple solution.

  • $\begingroup$ It is immediate once you know that $\exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}$. $\endgroup$
    – reuns
    Jul 29 '19 at 17:34

I don't know how you are defining $\exp{(z)}$ but many definitions of the function would be using the power series representation. That is to say $$\exp{(z)}=\sum_{k=0}^\infty \frac{z^k}{k!},\,\forall z\in\mathbb{C}$$ Thus your series is clearly equivalent to $$\sum_{k=0}^\infty \frac{(-x^2/2)^k}{k!}=\exp{\left(-\frac{x^2}2\right)}$$ by definition.

  • $\begingroup$ Thanks for your answer. I'm not sure that's what the author had in mind since $e$ hasn't realy been defined in the text. I'll accept this answer in a while in case no other answer has appeared that might seem more reasonable for my situation. $\endgroup$ Jul 29 '19 at 17:40
  • $\begingroup$ @DancingIceCream If you are unsure, take the derivative of the given summation and you should get the original sum multiplied by $-x$ which implies that $f'(x)=-xf(x)$. This differential equation has the unique solution $f(x)=\exp{(-x^2/2)}$ with the obvious boundary condition $f(0)=0$. $\endgroup$ Jul 29 '19 at 17:42
  • $\begingroup$ Ok, well that's cool! That might have been what the author had in mind. Thanks! $\endgroup$ Jul 29 '19 at 17:53

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