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I have a simple double sum as function:

u[m_, n_] = Sum[m^M/M! * n^N/N!, {M, 0, Infinity}, {N, 0, M - 1}]

and would like to compute the partial derivatives.

D[u[m, n], m]

I can't get the output neatly here, but it basically gives me (with correct indices on the summations)

$$ \left(\sum \sum \frac{m^{M-1}M n^N}{M! N!}\right)[m, n] + \left(\sum \sum \frac{m^{M} n^N}{M! N!}\right)^{(1, 0)}[m, n]$$

I would have thought that the first term would be the correct answer. I looked up the $(1, 0)$ notation and it apparently means "partial derivative w.r.t. the first input" - but isn't that exactly what I've put in here?

How am I supposed to understand this result?

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    $\begingroup$ Might be better to differentiate the summand instead of the sum. Also, avoid using N since it is a built-in function. $\endgroup$ – J. M. will be back soon Jul 27 '17 at 11:52
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    $\begingroup$ I only seem to get the first term (11.0.1.0 OSX) $\endgroup$ – Ruud3.1415 Jul 27 '17 at 11:55
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I think you got the wrong result because you evaluated some code that is different from what included in the question. I think you used:

u = Sum[m^M/M! * n^N/N!, {M, 0, Infinity}, {N, 0, M - 1}]

and forgot to Clear[u].

To differentiate appropriately you need Inactive[Sum] and replace the built in N by e.g. esc N esc.

Clear[u]
u[m_, n_] = Sum[m^M/M!*n^Ν/Ν!, {M, 0, Infinity}, {Ν, 0, M - 1}]
uSym[m_, n_] = Inactive[Sum][m^M/M!*n^Ν/Ν!, {M, 0, Infinity}, {Ν, 0, M - 1}]
uDm[m_, n_] = Activate[D[uSym[m, n], m]];

$\sum _{M=0}^{\infty } \sum _{Ν=0}^{M-1} \frac{M m^{M-1} n^Ν}{M! Ν!}$

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