## Taylor series question

The Taylor Series of f(x) = exp(-x^2) at x = 0 is 1-x^2...

Why is this?

The formula for Taylor Series is f(x) = f(0) + (x/1!)(f'(0)) + (x^2/2!)(f''(0)) + ...

and f'(x) = -2x(exp(-x^2)) therefore f'(0) = 0???

Can someone please explain why it is 1-x^2???

 Although f'(0) = 0, yet f''(0) will NOT be zero.
 The formula is (x^n)/n! (with the sigma crap in front) 0! is 1 and x^0 is 1 Just substitute x=-t^2 and you get the series for e^-t^2

## Taylor series question

 Quote by HomogenousCow The formula is (x^n)/n! (with the sigma crap in front) 0! is 1 and x^0 is 1 Just substitute x=-t^2 and you get the series for e^-t^2
Then why doesn't the second derivative term have an x^4?

 Here's what HomogenousCow is trying to say. Consider the expansion of ##\exp(t)## where ##t=-x^2##. The Taylor expansion of ##\exp(t)## is $$\sum_{n=0}^\infty \frac{t^n}{n!},$$ so substituting gives you $$e^{-x^2}=\sum_{n=0}^\infty \frac{(-x^2)^n}{n!}=\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}.$$ The second derivative term does not have an ##x^4## by its definition: Each term is given by ##\frac{f^{(n)}(0)x^n}{n!}## (for MacLaurin; for Taylor about ##c##, it's ##\frac{f^{(n)}(c)(x-c)^n}{n!}##). When you plug in 2, you get $$\frac{f''(0)x^2}{2!}.$$ In this case, ##f''(0)=-2##. This cancels with the ##2!## in the denominator, so you end up with ##1-x^2## as the first two nontrivial terms of the expansion.

Recognitions:
Gold Member
It looks like you are trying to combine two different ways of finding the Taylor series of $e^{-x^2}$
1) Using the basic definition: the "$x^2$" term has coefficient f''(0)/2 so that the second derivative term, by definition, involves $x^2$, not $x^4$.
2) Replacing x in the Taylor's series for $e^x$ with $-x^2$. In that case, the term that you get from the second derivative of $e^x$ has $x^4$ but that has nothing to do with the second derivative of $e^{-x^2}$. That term comes from replacing x with $-x^2$ in the $f'(0) x$ term of the Taylor series for $e^x$.