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## Homework Statement

Find the exact value for the 100th Derivative of [tex]f(x) = e^{-x^{2}}[/tex] evaluated at x = 0

## Homework Equations

[tex]e^{x} = \sum^{\infty}_{0} \frac{x^{n}}{n!}[/tex]

[tex]e^{-x^{2}} = \sum^{\infty}_{0} \frac{(-1)^{n}x^{2n}}{n!}[/tex]

## The Attempt at a Solution

The only thing I can think of to do is to get the 100th term of the series which is

[tex]\frac{x^{200}}{100!}[/tex]

factor out an [tex]x^{n}[/tex] to make it look like the general term of a MacLaurin Series

[tex]\frac{x^{100}}{100!}x^{100}[/tex]

and I thought that since

[tex]C_{n} = \frac{f^{(n)}(0)}{n!}[/tex]

I would just solve for the derivative

[tex]C_{100} = \frac{x^{100}}{100!} = \frac{f^{(100)}(0)}{100!}[/tex]

But..I can't do anything with that.

I even attempted to use n = 50, so the coefficient would not have an X in it, but that just gives me the derivative is 1...in fact if I do it like that it says the derivative is always 1.

I know that to be not the case because if you evaluate the 2nd derivative by hand you get -2, and the 4th is 12