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

  • Thread starter iceman713
  • Start date
  • #1
7
0

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
 

Answers and Replies

  • #2
21
1
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
7
0
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.

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.
 

Related Threads on Find the 100th Derivative of e^(-x^2)

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
8K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
5
Views
36K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
8
Views
3K
Replies
10
Views
3K
Top