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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.

I am very new to this, so please help

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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
up vote 6 down vote accepted

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

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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 $x = z^2$.

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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

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

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