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: