Show $ \sum_{n=0}^\infty \frac{(-1)^n \cdot 2^{-n} \cdot x^{2n+1}}{n!} = x \cdot \text{e}^{-\frac{x^2}{2}} $

Question

I have to show that the following are equivalent: $$ \sum_{n=0}^\infty \frac{(-1)^n \cdot 2^{-n} \cdot x^{2n+1}}{n!} = x \cdot \text{e}^{-\frac{x^2}{2}} $$ and $$ \sum_{n=0}^\infty \frac{(4n^2+2n) \cdot (-1)^n \cdot 2^{-n} \cdot x^{2n-1}}{n!} = x \cdot (x^2-3) \cdot \text{e}^{-\frac{x^2}{2}} $$ I know that the following is true: $$ \sum_{n=0}^\infty \frac{x^n}{n!}=\text{e}^x $$ But I don't know how to work with it. I hope i can get some help to solve this question.

Thanks in advance

Answer 1

You can plug in values for the power series. For example, if you have

$$e^x = \displaystyle\sum_{n=0}^\infty{\frac{x^n}{n!}}$$

plug in $(-x^2)/2$ in place of $x$ to get

$$\begin{align} e^{-(x^2/2)} &= \displaystyle\sum_{n=0}^\infty{\frac{(-(x^2/2))^n}{n!}} \\ &=\displaystyle\sum_{n=0}^\infty{\frac{(-1)^n(x^2/2)^n}{ n!}} \\ &=\displaystyle\sum_{n=0}^\infty{\frac{(-1)^n 2^{-n} x^{2n}}{ n!}} \end{align}$$

Now the original series asked for $x \cdot e^{(-x^2)/2}$, and you now have $e^{(-x^2)/2}$. I'm sure you can figure out how to multiply the series on the right and by $x$.

Can you do the second problem? It's actually just more arithmetic.