Taylor Polynomial of $e^{-x^2}$

Question

What is $P_4$, the 4th degree Taylor Polynomial of $e^{-x^2}$ around $x=0$?

I thought I try: $u=-x^2$

So for $e^u$: $P_4(u)=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}$

Therefore I thought for $e^{-x^2}$: $P_4(x)=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}$

But... apparently the right answer is for $e^{-x^2}$: $P_4(x)=1-x^2+\frac{x^4}{2!}$

What is my mistake? Why is the last answer the right answer?

Answer 1

Why is $P_4(x) = 1 - x^2 + \frac{x^4}{2!}$?

Because that is the degree 4 Taylor polynomial. You gave the degree 8 Taylor polynomial.

Answer 2

Simply because 4th degree does not mean with 4 powers of x.