nth derivative of e^(x^2)

[Hawkwings]

My friend has this math problem... find an equation for the nth derivative of e^(x^2). Here's what I told him.

http://wims.unice.fr/wims/wims.cgi?modu ... unction.en

enter the function, check all 5 derivative checkboxes, watch stuff happen. You can then enter the last equation again and take even more derivatives.

9th derivative of e^(x^2)=
512 x^9 e^(x^2) + 9216 x^7 e^(x^2) + 48384 x^5 e^(x^2) + 80640 x^3 e^(x^2) + 30240 x e^(x^2)

The pattern is definitely there, but I'm not sure how to write it out. For example, the number of terms is n/2+1 for even n, and n/2+0.5 for odd n. The coefficient for the first term is easy, 2^n, but for the others it's trickier. The last one also depends on whether n is even or odd. For every odd n, the previous coefficient is multiplied by 2+4((n-1)/2). I haven't tried working out the middle ones yet.

So yeah, I can see a pattern, I just don't know how to write it out.

So I'm curious now... how would a general equation for this be represented? How would a general equation deal with the changing number of terms in the answer? Better yet, what is the general equation?

(I feel like I should probably know the answer to my first two questions, so I reserve the right to idiocy)

[Kuroneko]

They satisfy the recurrence relation Pn+1 = 2xPn + 2nPn-1. This does have a closed form, albeit as an ugly summation involving a double factorial and the falling factorial: P2n+1 = 2n+1(2n+1)!!x{ 1 + Sum0<k<n+1[ (-4)k(-n)kx2k/(2k+1)! ] }. So for n = 4, we have 30240x[ 1 + (8/3)x² + (8/5)x4 + (32/105)x6 + (16/945)x8 ]. An even-termed solution is very similar; in general, look up the Hermite polynomials.

[Feil]

Do the derivative of the first few terms without adding like terms after every step and the pattern should become clear. d/dx(ke^[x2]*x^m) kicks off one term at 2ke^[x2]*x^[m+1] and one term at mke^[x2]*x^[m-1].

The highest term is multiplied by 2^n where n=m+1 times the original constant, which in your case is just 2^n.

2^n*x^n*e^[x2]

The ratio of the m+1 term to the m term is always 2, but the m term has gone away. So the m term was 2^n-1

The ratio of the m-1 term term to the m+1 term is always m/2 which is 2^n-2, but remember that the m-1 term includes k, which is one power of 2 less than the m+1 term, so

2^n*x^n*e^[x2] + 2^n-2*2^n-1*x^n-2*e^[x2]

but that term is added to a term of the same order that was kicked off by the m-2 term, which is the other half of the m term

that term was 2^n-3*2^n-2*x^n-2*e^[x2]

add those together to get

2^n*x^n*e^[x2] + [2^n-2]([2^n-3]+[2^n-1])[x^n-2]*e^[x2]

from there it is easy because every next term follows the same pattern until you hit n=1 where the pattern terminates because the associated m=n-1 term is 0 and 0 times anything is 0 and the derivative of 0 is 0.

I'm sure there is a more elegant way of writing it but I'll leave that to someone else

[Hawkwings]

Ahh, I knew this would draw Kuroneko out of hiding... Thanks for the quick response! I can safely say that I certainly did not know the answers to my first two questions, at least not in the way you two are describing it.

[Surlethe]

If you want an exercise, prove the closed form via induction.

[Feil]

I made a mistake in my post. I wrote

The ratio of the m-1 term term to the m+1 term is always m/2 which is 2^n-2[THIS PART IS WRONG], but remember that the m-1 term includes k, which is one power of 2 less than the m+1 term, so

2^n*x^n*e^[x2] + 2^n-2*2^n-1*x^n-2*e^[x2]

This should be

The ratio of the m-1 term term to the m+1 term is always m/2 which is n-1/2, but remember that the m-1 term includes k, which is one power of 2 less than the m+1 term, so we can put the 1/2 into that and make it 3 powers of 2 less, or:

2^n*x^n*e^[x2] + 2^n-2*1/2 *n-1*x^n-2*e^[x2]

Also, the same-order term from the derivative of the m-2 term is, of course, NOT what I said it was but rather it is just 2^n-2*x^n-2*e^[x2] because it is the counterpart of the highest-order term (the same but with m-2 increased to n-2 = m-1).

So the combined term is
2^n*x^n*e^[x2] + 2^n-2 *(1/2*n-1*+ 1)*x^n-2*e^[x2]

unless I made another mistake that I didn't notice.