@pepegna90 I can't execute your worksheet because I do not have the original 136 x 117 Array. Please upload the complete executable worksheet and any necessary data files.

@Mac Dude Change the procedure `print/delta` to

`print/delta`:= proc(x) local delta; delta*convert(x,`local`) end proc:

and likewise for `print/Delta`. By converting the variables to locals, they won't be able to cancel. Nor will the deltas cancel in an expression such as delta(y)/delta(x).

Your xpr2 was in fact a hidden thingy that just printed as delta. You can reveal the hidden nature of any expression with lprint:

lprint(xpr2);
delta(x)/x

@pepegna90 There is a bug in MaplePrimes that prevents file uploads from working about half of the time. Unfortunately, this has happened to your file. Please try uploading it again, or use a file hosting service.

@pepegna90 In order to diagnose your problem further, I need the actual worksheet instead of just a screenshot. Try uploading it here. Or post it on one of those file hosting services.

@Mac Dude I think that you should ask this in a separate Question thread. What do you want to do with a+b*(c+d)? Do you want to extract the coefficient of (c+d)?

@pepegna90 For now, use ArrayTools:-AddAlongDimension and don't try to do it with with(ArrayTools).

@Mac Dude Since delta is not a variable, there can be no meaningful response to limit(delta(x), delta= 0). To differentiate with respect to delta(x), I believe that you can use Physics:-Fundiff, but I can't tell you much more than that.

@Axel Vogt Sorry, I don't understand your Answer. Have you used alias?

dsolve({ode1, bcs1}, T(z)) returns a fairly compact solution in terms of erf. Presumably you already know that. So, what's wrong with that solution?

@brian bovril Thank you for your interest in the Logic Problem package, Brian. The problem needs to be "massaged" somewhat before it can be coded into the package. In order to use the package, the "variables" EN, Dept, and Sport each need to have the same number of possible values---eight in this case. But there are only three values for Dept. If we knew how many people worked in each department, then we could add extra values to Dept. For example, if we knew that there were three in personnel, three in marketing, and two in administration then we could make the values person||(1..3), market||(1..3), and admin||(1..2). But we don't know the breakdown between the departments.

However, we do know that there are only three possible breakdowns because there are "not more than three of them in any department". By the pigeonhole principle, the breakdown must be 3-3-2, 3-2-3, or 2-3-3. Furthermore, we know "Fay works in Personnel with only Alex". So, there are two in personnel and three in the other two. So we set

Dept:= [person||(1..2), market||(1..3), admin||(1..3)];

I'll work on coding the rest of problem later.

@pepegna90 I don't see the screenshot.

@pepegna90 To get the sum of the ith row of Matrix M, use add(x, x in M[i, ..]);

Note that style=line is only appropriate if the points are sorted. It is better to vary the symbol (as noted by Christopher).

plot(
     [[seq([n, sin((1/10)*n)], n=0..30)], [seq([n, cos((1/10)*n)], n=0..30)]],
     style= point, symbol= [cross, diamond]
);

@jerbertz In your implementation of RLopez's method, you have print~([TestList])[]. The TestList should be TestSeq, the same TestSeq that you used with the other methods. Then it will work as advertised.

@Andriy Sorry, the file is not attached.