@brian bovril I got Joe's code to work after a bunch of fiddling with my output settings. I am running in a Standard worksheet with 1D input (Maple Input). I load the code out of a Code Edit Region, but the good old bold red Maple Input works as well. In order for it do anything interesting, however, the output display needs to be set to "Typeset Notation" and the Typesetting Level needs to be set to "Maple Standard". These settings are accessed via the Tools => Options => Display menu.

Joe, where can I find documentation for these MathML codes? Particularly, I am interested in the identifier names the begin `#mrow. There doesn't seem to be anything about it in packages TypeSetting, MathML, DocumentTools, Worksheet, or XMLtools.

Minor point, easily correctable: I think you missed the point about the searches needing to be case insensitive. Perhaps you haven't seen the opening credits of the show. For example, the actor name Anna Gunn is shown as AnNa Gunn, with the Na highlighted as the sodium block from the periodic table.

@Markiyan Hirnyk Well, I did say that the questioner's ranges defined a nonagon, not a hexagon. It makes a nice looking plot though. Could be a new airplane design. :-). I have verified that the provided corrected ranges do define a right hexagon, the plot of which is, of course, boring.

@Markiyan Hirnyk Well, I did say that the questioner's ranges defined a nonagon, not a hexagon. It makes a nice looking plot though. Could be a new airplane design. :-). I have verified that the provided corrected ranges do define a right hexagon, the plot of which is, of course, boring.

@Kitonum Here's your recursion written in more standard Maple. In particular, note the remember, thisproc, and the "seeding" of the remember table in the last line. Also note that this technique avoids the need to give an initial value for a(2).

a:= proc(n::posint)
   option remember;
   local a:= thisproc, k:= irem(n,3);
   a(n-1) + `if`(k=2, -1, a(n-2) + `if`(k=1, 2, 0))
end proc:
a(1):= 2:  

Please post the questions on the list so that the questions and the answers become searchable by all users until the end of time. Also, please start a new thread for the questions.

That being said, I have a few comments on your worksheet:

1. What do you think evala does? Can you check whether it is actually doing something in those places where you use it? As far I know, there's no reason to think that it would do anything for your expressions.
2. About the thing with assuming h > 0: Yes I do think it's a bug, but I'm not sure. I've tried to make a simple case where the same thing happens, but I haven't been able to do it yet.
3. You should take out all uses of assuming h > 0 and assuming t[i] > 0. Use assume(t[i] > 0) instead, and let the original assumption on h handle the h. I've tested this on your code, and I get much better simplifications this way.
4. Regarding your main question about extracting the coefficients for the system of equations: I'm sure that there's an easier way, but I don't yet know exactly what that way is. I'll need to look at it more. If you post that question in a new thread, it is much more likely that someone else will read it and answer it. It seems like it's just you and me in this current thread now.

Please post the questions on the list so that the questions and the answers become searchable by all users until the end of time. Also, please start a new thread for the questions.

That being said, I have a few comments on your worksheet:

1. What do you think evala does? Can you check whether it is actually doing something in those places where you use it? As far I know, there's no reason to think that it would do anything for your expressions.
2. About the thing with assuming h > 0: Yes I do think it's a bug, but I'm not sure. I've tried to make a simple case where the same thing happens, but I haven't been able to do it yet.
3. You should take out all uses of assuming h > 0 and assuming t[i] > 0. Use assume(t[i] > 0) instead, and let the original assumption on h handle the h. I've tested this on your code, and I get much better simplifications this way.
4. Regarding your main question about extracting the coefficients for the system of equations: I'm sure that there's an easier way, but I don't yet know exactly what that way is. I'll need to look at it more. If you post that question in a new thread, it is much more likely that someone else will read it and answer it. It seems like it's just you and me in this current thread now.

Wow, that's a tour-de-force.

We were working on essentially the same thing at the same time. I at first considered using an Array of Records, but settled on a listlist to reduce the overkill factor (see my procedure below).

An aside, not related to the Asker's problem: I've noticed that you code a lot of ModuleApplys. Is there any benefit to making a procedure into a module whose only export is the ModuleApply? or is it just habit and style at this point? When do module locals get garbage collected?

Wow, that's a tour-de-force.

We were working on essentially the same thing at the same time. I at first considered using an Array of Records, but settled on a listlist to reduce the overkill factor (see my procedure below).

An aside, not related to the Asker's problem: I've noticed that you code a lot of ModuleApplys. Is there any benefit to making a procedure into a module whose only export is the ModuleApply? or is it just habit and style at this point? When do module locals get garbage collected?

Please rewrite your question. Perhaps there are some missing characters that were dropped by the editor? You can use the Preview option (next to Submit) in the editor to check how the post will look after it's posted.

Your last three equations are

phi[i]:= piecewise(...);

and likewise for phi[i+1/2] and phi[i+1]. I think that those should be

phi[j]:= r-> piecewise(...);

and likewise for phi[j+1/2] and phi[j+1]. Is that right? If I do that, then all the integrals in eq10 are evaluated. The resulting expression is large, but I can apply simplify(%) and it simplifies nicely all at once. The final expression fits on one screen. How much free memory do you have?

Your last three equations are

phi[i]:= piecewise(...);

and likewise for phi[i+1/2] and phi[i+1]. I think that those should be

phi[j]:= r-> piecewise(...);

and likewise for phi[j+1/2] and phi[j+1]. Is that right? If I do that, then all the integrals in eq10 are evaluated. The resulting expression is large, but I can apply simplify(%) and it simplifies nicely all at once. The final expression fits on one screen. How much free memory do you have?

Are you saying that you want to take the tickmark positions that are chosen by default and make their labels shorter? Or are you saying that you want to choose both the positions of the tickmarks (as in your prior question) and the way their labels are printed? The latter is easier than the former.

Are you saying that you want to take the tickmark positions that are chosen by default and make their labels shorter? Or are you saying that you want to choose both the positions of the tickmarks (as in your prior question) and the way their labels are printed? The latter is easier than the former.

@sakhan I think that I may have missed part of the computation; maybe I got i's and j's mixed up. For whatever reason, my eq10 is only a half screen in size---not that complicated. Please upload the worksheet with all computations starting from the restart. Yes, I am sure that I can find a way to simplify each integral as it is done.

@sakhan I think that I may have missed part of the computation; maybe I got i's and j's mixed up. For whatever reason, my eq10 is only a half screen in size---not that complicated. Please upload the worksheet with all computations starting from the restart. Yes, I am sure that I can find a way to simplify each integral as it is done.