How do I write $e^{(-x/2)}$ as a summation?
I am new to power series. I know how to write $e^x$ as a summation, but i do not know how that helps me.
$\endgroup$ 33 Answers
$\begingroup$You know how to write $e^x$ as a power series, meaning that you know how to write
$$\exp(y) = e^y = \sum_{n = 0}^\infty \frac{y^n}{n!}$$
You might know that this is valid for all $y$. What you want to do is write out the series for $\exp(-x/2)$. Fortunately, you can get this from the series that you know.
All you need to do is substitute $-x/2$ in for $y$ in the expansion that you know, which I wrote above.
$\endgroup$ 2 $\begingroup$let $t=\frac{-x}{2}$
$$e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + ...$$
$\endgroup$ $\begingroup$This problem is easy if you don't confuse yourself by using the same variable for multiple things. Here's an expression for $e^x$:
$$e^x = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$$
And here's an expression for $e^u$:
$$e^u = \sum_{n=0}^{\infty} \frac{u^{n}}{n!}$$
The takeaway here is that that the exponent of $e$ is simply a "dummy variable" - this means you can plug in anything for the variable and it will give you the correct equation! So in your example, you want to find $e^{x^{-2}}$. Just let $u = -x^2$, and plug in that value for $u$ wherever you see a $u$ in the series:
$$ e^{x^{-2}} = \sum_{n=0}^{\infty} \frac{{(x^{-2})}^{n}}{n!} = \sum_{n=0}^{\infty} \frac{x^{-2n}}{n!} \\ = \frac{x^0}{0!} + \frac{x^{-2}}{1!} + \frac{x^{-4}}{2!} + \frac{x^{-6}}{3!} + \frac{x^{-8}}{4!} \cdots \\ = 1 + \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{6x^6} + \frac{1}{24x^8} + \cdots $$
$\endgroup$ 2
