An alternative to the explicit option is to apply allvalues to a result returned by solve which contains RootOfs.

solve({x=k/4,y=-k/3,z=3*k/8,2*x^2+3*y^2+4*z^2=9},{x,y,z,k});
allvalues(%);
 

You wrote:

I will have to draw some tick marks on for r=0 and r=1, in Inkscape. I don't know why there isn't an (obvious) option to have tickmarks but not lines, or to move the lines so that they look nice.

No need to use another program. The tickmarks and the gridlines can be specified independently. So you can get your r=0 and r=1 tickmarks like this:

Download polarplot.mw

The plot options are not obvious because there are so many of them. I found this solution on page ?plot,axis.

You need to make it so that there is no tickmark on the radial axis that is equal to the full length of the axis. In this case, we need to remove the tickmark at r = 1. We do this with the axis[radial]= [tickmarks= ...] option. The other options below are to make the plot otherwise exactly like in your post, which I assume you did with the menus.

Download polarplot.mw

Download trigrad1314.mw

I made a minor change to deal with some of the long display labels. Here's a repost of the worksheet with the new code and some new examples. As for colored labels, I think something may be possible along those lines: colors (and other formatting information) can be encoded into names. I'll think about it.

Download FunctionTree.mw

Your expression for the integrand has unbalanced parentheses. I think you're missing a left parenthesis "(" at the very beginning. Once you repost a correction, I might be able to help you more. But I think that it's unlikely that you will get a symbolic answer.

The command is plots:-listcontplot(M, contours= [$1..20]). But you need to lay out your points in a grid form, which involves some sorting and approximating. I vaguely recall seeing a tool for this a many years ago. I will look for it.

Does the error occur in CalculateMatrix or after? Put a print statement after the CalculateMatrix call to find out.

It is a system of linear equations. It would be much more efficient to solve it with matrix operations (i.e. LinearAlgebra:-LinearSolve). Would a floating-point solution be acceptable? That would use much less memory.

You can convert the arrows, or anything else, to Cartesian coordinates, and put them on a polar plot like this:

Download polarplot.mw

Use Sum (with a capital S), instead of sum, and you will get the answer that you expect.

I have no idea what that escaped variable _O and the rest of that huge mess represents. Surely the result of differentiating the lowercase sum is a bug. However, I will note that the fault lies with sum, not diff, because you get a nearly identical mess if you just do the derivative in your head and enter

sum(sin(theta[j](t))*cos(theta[j](t))*diff(theta[j](t), t), j= 1..i-1);

It looks like your Alphabet is shifted by one. I get "Y"=58, "e"=70, "s"=84. Regardless of that, the encryption steps are the same. The essential command is the inert modular exponentiation operator x &^ e mod n. I also did the decryption, partly for to verify the encryption and partly for my own amusement.

with(StringTools):
Alphabet:=cat(Select(IsPrintable,convert([seq(i,i=1..255)],bytes))):
AlphaToNum:= table([seq](Alphabet[k] = k, k= 1..length(Alphabet))):
plaintext:= "Yes":
NumText:= [seq](AlphaToNum[c], c in plaintext);
                          [58, 70, 84]
n, e:= 119, 7:
Crypted:= (c-> c &^ e mod n) ~ (NumText);
                         [114, 77, 84]
To decrypt:
f:= ifactor(n);
                            (7) (17)
phi:= expand(map(x-> x-1, f));
                               96
d:= 1/e mod phi;
                               55
Decrypted:= (c-> c &^ d mod n) ~ (Crypted);
                          [58, 70, 84]
cat(seq(Alphabet[k], k in Decrypted));
                             "Yes"

Save your file in Excel. I think as a CSV file would be best (Comma-Separated Values). Then in Maple, give the command ImportData(); and follow the dialog. Once you get that done, post a followup to this, and we can proceed with the plotting. Also post a followup if you have any trouble with the dialog.

mrmathexpert wrote:

How can I control the specific regions that I want to highlight? For example, from pi/2 to 3pi/2: outside or the left side of the y-axis bounded; and from 3pi/2 to pi/2, outside or the right side of the y-axis bounded.

You can do the left side and the right side separately, each side consisting of two polarplot commands, and paste all four together with display. I'm not sure if this is what you were trying to describe:

Download polarplot.mw

Note that the order of the plots in the display command matters: They are printed in the reverse of the order that they appear in the command. I don't think this is documented.

GraphTheory:-Graph vertices need to be integers, symbols, or strings; arbitrary expressions are not allowed. So 2, `*`, function, cos, y, and z are valid; but 2*y, 3*z, and cos(x+2*y) are not.

The command that you want is subsindets (or possibly evalindets). Here's an appropriate analogy: subsindets is to indets as applyop is to op.

I think you should collect like terms with respect to the integrals, like this

collect(eq2, Int);

so that each specific integrand only occurs once.

Tell me exactly which integrands you are trying to select. You said "like" phi[i](y)*diff(w(y), y). Do you mean exactly that integrand? Or would phi[i](y)*diff(w(y), y)^2 also count as "like"?

Once you answer that, I can give you a type specification that you can use with subsindets.