Maple derivative with sum() and product()

Question

I am new to maple and have run into an issue optimizing wrt a term within sum() and or product(). Suppose I want to optimize a profit function function given by

\begin{align} Q : = p \cdot A \prod_{l=1}^{L-1} z_l^{a_l} - \sum_{l=1}^{L-1}w_lz_l \end{align}

wrt to input $z_l$. diff(Q,z[l]) yields nonsense and diff(Q,z[1]) zero. I am having trouble finding alternative methods generating the FOC. Any help is appreciated.

Differentiating by $z_l$ does not make sense as there is nothing like that in your expression (it exists only inside the products/sums, but not in the expression itself). Also differentiating by $z_1$ probably does not work because Maple does not know the $L$, it is just too generic ($L$ could be $1$ for what it knows, and there would be no $z_1$ in that case, but still this behavior looks a little bit strange). I would try defining $Q$ as a function of $L$, and then check what Maple gives for specific values of $L$ when differentiating by $z_1$ etc... — Sil, Sep 02 '18 at 09:06

score 0 · Answer 1 · answered Oct 14 '19 at 15:52

As @Sil already has mentioned in a comment below your question, your problem is that you have a undetermined sum and product. For example assume you want to take derivative with respect to z2, how can Maple check if there is a z2 in your expression? L is not defined so it's not clear what indices exist in your expression. What you can do is as following. If you need the Q expression for different values of L, then define a function of L or a procedure with L as its input. So when you want to compute something related to Q of a specific L, you just put Q equal to the output of that procedure for your choice of L, then do the computations.

Qproc := proc(L) 
local l; 
return(p*A*product(z[l]^a[l],l=1..L-1)-add(w[l]*z[l],l=1..L-1); 
end proc:

This procedure returns you your Q expression for a given choice of L. Let's say I want $L=3$. Then I do as following.

Q:=Qproc(3);

The Maple output will be;

Then you can ask derivatives with respect to z1 and z2, because now they are in your expression (not like when you had an undetermined sum and product!).

for l to 3 - 1 do
print(expand(z[l]*diff(Q,z[l])));
end do;

The Maple output is;

Maple derivative with sum() and product()

1 Answers1