## How do you find the Maclaurin series for \(e^{x^{2}}\)?

Answer 1:

The Maclaurin series is simply the Taylor series centered around a = 0.

The Taylor series formula is:

Now, we need to take some derivatives. Let’s go to n = 4.

Now we can construct the Maclaurin series:

See the pattern? It requires a bit of manipulation to figure out, but the constant happens to be \(\frac{n !}{\left(\frac{n}{2}\right) !}\) while the n! in the actual formula remains there. That cancels out to give \(\left(\frac{n}{2}\right) !\) in the denominator of the simplified answer.

n = 0,2,4,6……………

Answer 2:

Alternatively, just substitute \(x^{2}\) for \(x\) in the well known series for \(e^{x}\) to find: