# Find the 100th Derivative of e^(-x^2)

1. Dec 3, 2008

### iceman713

1. The problem statement, all variables and given/known data

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

2. Relevant equations
$$e^{x} = \sum^{\infty}_{0} \frac{x^{n}}{n!}$$

$$e^{-x^{2}} = \sum^{\infty}_{0} \frac{(-1)^{n}x^{2n}}{n!}$$

3. 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
$$\frac{x^{200}}{100!}$$
factor out an $$x^{n}$$ to make it look like the general term of a MacLaurin Series
$$\frac{x^{100}}{100!}x^{100}$$

and I thought that since
$$C_{n} = \frac{f^{(n)}(0)}{n!}$$

I would just solve for the derivative
$$C_{100} = \frac{x^{100}}{100!} = \frac{f^{(100)}(0)}{100!}$$

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

2. Dec 3, 2008

### neo86

you are on the right track...
the only term which gives you something nonzero at the 100th derivative it n=50
so you need to evaluate the 100th derivative of x^100/50!, but this is simply 100!/50!, if you want to get the exact number you would need to plug that into some CAS. You could also just use Stirling's formula to get a very good approximation.

3. Dec 3, 2008

### iceman713

Thanks very much. It's a little awkward to think that the 100th term in the series is not the 100th term in the original Taylor/MacLaurin series but I suppose that's what I didn't realize.