@sakhan To get all the nonlinear terms at once, first get the linear terms and the constant terms and subtract from the original. What's left is the nonlinear terms.

Nice. So your degree 9 polynomial is of the form f(x^3) where f is degree 3, hence it is solvable.

@User7843 Markiyan's answer shows that it is not pixelation. And, yes, I noticed the absence of any improvemnet with varying the thickness. I wonder what is going on that causes this. Markiyan, you probably have an idea.

Whoever gave me a vote up for this answer should take it away.

@User7843 Markiyan's answer shows that it is not pixelation. And, yes, I noticed the absence of any improvemnet with varying the thickness. I wonder what is going on that causes this. Markiyan, you probably have an idea.

Whoever gave me a vote up for this answer should take it away.

@sakhan

You wrote:

I may be wrong but I think getting all the integrals is easy, we just need to do
seq( op(i, lhs(eq10_1), i=1..nops(lhs(eq10_1));

You're missing a right parenthesis after the first 10_1. But, putting that in, that command will only give you the terms, not the integrals within the terms. Indeed that command is equivalent to simply op(lhs(eq10_1));

If you want the terms that have a[j] to the 1st power, and do not have a[j+1] or a[j+1/2], then do

coeff(coeff(coeff(lhs(eq10_1), a[j], 1), a[j+1], 0), a[j+1/2], 0);

Likewise, for any exponents m, n, k, the terms for a[j]^m*a[j+1]^n*a[j+1/2]^k can be extracted by

coeff(coeff(coeff(lhs(eq10_1), a[j], m), a[j+1], n), a[j+1/2], k);

Does that do what you want?

@sakhan

You wrote:

I may be wrong but I think getting all the integrals is easy, we just need to do
seq( op(i, lhs(eq10_1), i=1..nops(lhs(eq10_1));

You're missing a right parenthesis after the first 10_1. But, putting that in, that command will only give you the terms, not the integrals within the terms. Indeed that command is equivalent to simply op(lhs(eq10_1));

If you want the terms that have a[j] to the 1st power, and do not have a[j+1] or a[j+1/2], then do

coeff(coeff(coeff(lhs(eq10_1), a[j], 1), a[j+1], 0), a[j+1/2], 0);

Likewise, for any exponents m, n, k, the terms for a[j]^m*a[j+1]^n*a[j+1/2]^k can be extracted by

coeff(coeff(coeff(lhs(eq10_1), a[j], m), a[j+1], n), a[j+1/2], k);

Does that do what you want?

@sakhan Yes, the indices do change on different runs. That's one reason that I suggest using indets.

@sakhan Yes, the indices do change on different runs. That's one reason that I suggest using indets.

To proceed, we need numeric values for a, b, and c.

Do you mean a situation like this?

for x in L do
   if cond(x) then break end if;
   ...
end do;

Or like this?

for k to nops(L) while not cond(L[k]) do
   ...
end do;

Or somethings else, perhaps with seq? And, do you want to make sure that cond is applied to every element of the list, perhaps because cond has a side effect?

You'll need to show the rest of the code, especially the point at which you get the error message.  Also, I don't see any derivatives, so how are they ODEs?

This might be a change in newer versions of Maple, but currently a with command performed inside a procedure has no effect on other commands in the procedure. But the execution of such a procedure causes the with to be performed globally. The procedures that are called within a procedure depend on what name-bindings are in effect at the time the procedure was defined, not the time that it is executed.

This might be a change in newer versions of Maple, but currently a with command performed inside a procedure has no effect on other commands in the procedure. But the execution of such a procedure causes the with to be performed globally. The procedures that are called within a procedure depend on what name-bindings are in effect at the time the procedure was defined, not the time that it is executed.

@Markiyan Hirnyk At the bottom of Markelov's LiveJournal post to which you gave a link there a link to a 1988 article in Russian: http://kvant.mccme.ru/1988/06/tri_formuly_ramanudzhana.htm. I don't read Russian, but it looks like this article attributes the formula to a Ramanujan. I am not sure. Could you check that please?

@Markiyan Hirnyk Thank you Markiyan for that stunningly beautiful identity. Where did you get that? It is reminiscent of Ramanujan.

The identity can indeed be proved in Maple. I feel that the proof warrants a separate post, so I am branching off. Put any replies there.