# Find a closed form from the given power series

I have the power series $\sum_{n=0}^{\infty} {z^{2n}\over{n!}}$, how do I find the closed form for this power series.

I am aware that $e^z=\sum_{n=0}^{\infty} {z^{n}\over{n!}}$, so I tried to manipulate this in order to be able to use $e^z$, but it did not work out.

-
Then I just have $e^{z^2}$ as an answer, right? –  Akaichan Mar 18 '14 at 10:05
I am so silly! :( –  Akaichan Mar 18 '14 at 10:06

You know the series for $e^w$. Let $w=z^2$.

-

It helps to rename the variables so you don't have two copies of $z$ that mean different things.

You want to find the series

$$\sum \frac{z^{2n}}{n!}$$

and you know

$$e^x = \sum \frac{x^n}{n!}$$

and you're hoping the first series matches the pattern of the second series. Sometimes, you can match the pattern simply by setting things equal: you were hoping

$$\frac{z^{2n}}{n!} = \frac{x^n}{n!}$$

and we see that we can easily cancel out all of the occurances of $n$, leaving gives $x = z^2$.

-

As you said, $$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$

Suppose we change $z$ to $z^2$, then we get

$$e^{(z^2)} = \sum_{n=0}^{\infty} \frac{(z^2)^n}{n!}$$

Which by the rules of exponents simplifies to

$$\sum_{n=0}^{\infty} \frac{z^{2n}}{n!}$$

-