@Mac Dude 

Good answer. Would you kindly please provide an example of what you mean by "series will not work on functions"? It's not that I doubt what you say; I just don't understand it.

@casperyc Sorry that I didn't see your comment earlier.

You asked:

I am a bit confused why (how) would you  think of reducing the accuracy?

Why: Because the error message that one gets when using the default accuracy suggests that an answer may be possible at a reduced accuracy, and it even suggests what to reduce the accuracy to.

Sol:= dsolve({ODE, BCs}, numeric, method= bvp[middefer], maxmesh= 2^15);

Error, (in dsolve/numeric/bvp) unable to achieve requested accuracy of 0.1e-5 with maximum 32768 point mesh (was able to get 0.14e-2), consider increasing `maxmesh` or using larger `abserr`

How: By adjusting the parameters relerr and/or abserr.

And what does the Reduced accuracy answer tell us?

Nothing in this case. In some other case it may be the only way to get an answer, and the answer produced will be accurate to the given accuracy.

In my experience, sometimes when I got underflow (tooo small) issue, I would increase accuracy, to get something out of it.

Numeric solution of BVPs is much trickier than IVPs. It is difficult to increase the accuracy without increasing the mesh size, and the maximum mesh size is 2^15. Larger mesh sizes mean a larger system of linear equations to solve, which introduces its own accuracy problems.

@Kitonum "Runs" refers to contiguous groups of the same character. So the run lengths in your random example would be 2, 1, 1, 2, 1, 3, 2, 2, 1, 2, 1, 1, 1, 1, 2, 3, 1, 3.

I have trouble following your worksheet because of the use of context-menu commands. Can you redo the worksheet with typed commands rather than context-menu commands?

@litun Yes, it's clear now. But it seems to me that fnormal handles that. Can you show me an example where fnormal does not do what you want?

How can you have a term with a factor of 0?

Doesn't fnormal do exactly what you want?

@georgepan 

There is something missing from your Reply. I still don't have the complete equation. Please attach a worksheet or enter a formula in plain text.

Is the equation correct as I entered it? 

@Kitonum Your example doesn't quite work the same way in Maple 18. It produces this plot:

The problem can be corrected by changing the order in the display command.

plots[display](B, A, scaling= constrained);

@oldstudent The simpler solution only works if the shaded region is under the entire x-range of the curve. It can be shortened to

plot(sqrt(x), x= 0..1, filled);

@oldstudent I highlighted.

While "Numerically Solve" will usually give the same results, the process used to obtain the results is vastly different.

Preben: Thanks for the information. I edited the Answer to reflect it.

@Markiyan Hirnyk I did read the help for GlobalOptima and I still can't tell why it returned such an answer for the OP's command. It doesn't matter if the constraints are changed to {x >= 0, y >= 0, z >= 0, x+y+z = 3}; it still doesn't work.

I immediately foresee numerous problems with your plan. My advice is to give up and use strings. Note that there are a few syntaxes that have been developed for the unambigous one-dimensional input of chemical formulae. Look up SMILES and SYBYL in Wikipedia.

@acer At the risk of going more tangential....

The concept of gc real time and how to measure it does not quite make sense to me. The only way that I can make sense of it is if the processing of all the other Maple processes is suspended while gc is being done. Do you know if that is true?

Please give your equation in normal characters (Maple 1D input) or attach a worksheet. What you have posted as your equation is meaningless to Maple.

@brian bovril Okay, here's a procedure to do it.

Display:= proc(res::list(indexed=nonnegint))
local S,e;
     S:= `%+`(seq(`if`(rhs(e)=0, [][],
          `if`(rhs(e)=1, ithprime(op(lhs(e))), `%*`(ithprime(op(lhs(e))), rhs(e)))
          ), e= res
     ));
     InertForm:-Display(InertForm:-Value(S) = S)
end proc:

Now just make the last line PF

Display(res[2])