Timeline for $n$-th derivative of $\exp\left(-\frac{\lambda(x-\mu)^2}{2\mu^2x}\right)$
Current License: CC BY-SA 4.0
6 events
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| Apr 24 at 23:34 | comment | added | qifeng618 | @StevenClark I am sorry that I misunderstood your words “ Your derivation down to and including your second to last formula”. Thank you very much for your upvoting this answer. I don’t know why I mistakenly verified the boxed formula several times, perhaps my software Mathematica has bugs or some computer virus intruded into my computer. This time is a lesson for me. Anyway, to the end, we now confirm the boxed formula correct | |
| Apr 24 at 13:26 | comment | added | Steven Clark | I didn't down-vote your answer, I actually like it because it eliminates the recursion in my answer. I didn't actually try your final formula before since you originally indicated it wasn't correct, but I see now it evaluates correctly. I upvoted to cancel the downvote. | |
| Apr 24 at 9:06 | comment | added | qifeng618 | @StevenClark I don't find any error in my proof. I numerically verified the boxed derivative formula by the software Mathematica 12 and didn't find it wrong again. I don't know what happened before. Consequently, the boxed derivative formula is correct. By the way, I want to know why you voted down this answer. | |
| Apr 24 at 8:55 | history | edited | qifeng618 | CC BY-SA 4.0 |
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| Apr 24 at 0:41 | comment | added | Steven Clark | Your derivation down to and including your second to last formula $$\frac{\partial^n e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}}}{\partial x^n}=e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}} \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}$$ seems to validate, so there must be a mistake in the final step where you moved some of the terms out of the two sums. | |
| Apr 23 at 9:29 | history | answered | qifeng618 | CC BY-SA 4.0 |