- #1

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[tex]\frac{d^n}{dx^n}e^{a x^2}[/tex]

I am going to find the result of above derivitative when x -> 0.

- Thread starter KFC
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- #1

- 488

- 4

[tex]\frac{d^n}{dx^n}e^{a x^2}[/tex]

I am going to find the result of above derivitative when x -> 0.

- #2

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P.S. I would have tried to take some derivatives myself, but it is bed time now...lol...

- #3

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The answer to this question would most probably be (2xa)^n*(e)^a(x)^2

- #4

Mute

Homework Helper

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Most certainly not! You're going to have to use the chain rule every time, which is going to result in some polynomail times [itex]e^{ax^2}[/itex]. In fact, the result is going to be related to the Hermite polynomials, defined byThe answer to this question would most probably be (2xa)^n*(e)^a(x)^2

[tex]H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex].

Getting the form desired in the OP amounts to setting [itex]x \rightarrow iax[/itex].

http://en.wikipedia.org/wiki/Hermite_polynomials

(Note I used the "physicist definition" of the Hermite polynomials defined on the wikipedia page).

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- #5

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We define a new variable u = i sqrt(a) x:

[tex]\frac{d^n}{dx^n}e^{a x^2}[/tex]

I am going to find the result of above derivitative when x -> 0.

[tex]\frac{d^n}{dx^n}e^{a x^2} |_{x = 0} = (i \sqrt(a))^n \frac{d^n}{du^n}e^{-u^2}|_{u = 0} =

(i \sqrt(a))^n (-1)^n e^{-u^2}H_n (u) |_{u = 0}

=(-2 i \sqrt(a))^n \frac{\sqrt{\pi }}{\Gamma \left(\frac{1-n}{2}\right)} [/tex]

where the last step has used the form of the Hermite numbers:

http://mathworld.wolfram.com/HermiteNumber.html

Note also that when n is odd the Gamma function blows up and so the value we seek is zero, when n is even the imaginary unit disappears and we obtain a real result.

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