Carl Love

## 26109 Reputation

11 years, 60 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

## I get it...

@Joe Riel Finally, I get it! Thank you for your patient explanation. Put more succintly (I think), in all cases I was missing the choice of whether the element(s) of (A xor B) were or were not in C. In the |AB|=4 case, there is one such element, for a factor of 2; in the |AB|=3 case, there are two such elements, for a factor of 2^2.

I editted my solution at the top of this subthread to reflect these corrections.

## PolynomialIdeals...

My experience with this, admittedly very limited, is that the ?PolynomialIdeals package gives you access the Groebner package with a more ideal-oriented syntax.

## You're right, I was using the wrong vers...

@Markiyan Hirnyk You're right, I was using the wrong version of Maple. I just sent you an email with more info.

## You're right, I was using the wrong vers...

@Markiyan Hirnyk You're right, I was using the wrong version of Maple. I just sent you an email with more info.

## A simulation in Maple...

@Markiyan Hirnyk By doing a simulation in Maple, I know that my answer is wrong and that Joe's edit of his original answer is correct. But I either don't understand or don't agree with Joe's analysis of what was wrong with my answer. My simulation showed that my count of the |AB|=4 case was short by a factor of 2, and my counts of both subcases of the |AB|=3 case were short by a factor of 4. Here's the simulation code:

S:= combinat[powerset](5):
n:= 0: ab3:= 0: ab4:= 0: ab5:= 0: b4:= 0:
for A in S do  for B in S do  for C in S do
if nops(A union B) = 5 and nops(A intersect B intersect C) = 3 then
n:= n+1;
abn:= nops(A intersect B);
if abn=3 then
ab3:= ab3+1;
if nops(B)=4 then b4:= b4+1 end if
elif abn=4 then ab4:= ab4+1
end if
end if
end do  end do  end do:
n, ab3, ab4, b4;

So, the final answer is 250*binomial(30,5)*2^25, which is the same as what Joe got, although he expressed it differently.

## A simulation in Maple...

@Markiyan Hirnyk By doing a simulation in Maple, I know that my answer is wrong and that Joe's edit of his original answer is correct. But I either don't understand or don't agree with Joe's analysis of what was wrong with my answer. My simulation showed that my count of the |AB|=4 case was short by a factor of 2, and my counts of both subcases of the |AB|=3 case were short by a factor of 4. Here's the simulation code:

S:= combinat[powerset](5):
n:= 0: ab3:= 0: ab4:= 0: ab5:= 0: b4:= 0:
for A in S do  for B in S do  for C in S do
if nops(A union B) = 5 and nops(A intersect B intersect C) = 3 then
n:= n+1;
abn:= nops(A intersect B);
if abn=3 then
ab3:= ab3+1;
if nops(B)=4 then b4:= b4+1 end if
elif abn=4 then ab4:= ab4+1
end if
end if
end do  end do  end do:
n, ab3, ab4, b4;

So, the final answer is 250*binomial(30,5)*2^25, which is the same as what Joe got, although he expressed it differently.

## Code Edit Region...

The best way to enter multi-line code into a Maple worksheet is to use a Code Edit Region. Place your cursor where you want the window, then select Code Edit Region from the Insert menu. It gives you a Notepad-like editor with tabs and hanging indents. Also, you can use ordinary Enter (or Return) instead of Shift-Enter (Shift-Return).

Please upload your worksheet, not just a screen shot of it, so that we can work with it. Use the green up arrow that is the last thing on the tool bar of the MaplePrimes editor. And please upload a photo of a hand-drawn sketch of what you think the plot should look like. My first impression is that we can use the RootOf comand with the selector option.

## I doubt it...

I doubt that it could be expressed in radical form without the I, because it comes from solving a cubic. Attempting to use evalc and simplify puts it back into a trigonometric form.

## It could be any integer number of degree...

"... if the number of degrees is divisible by three" seems like a rather weak statement! Here I prove that the trigonometric functions for any integer number of degrees can be expressed in radicals. Below, I derive cos(20°) in radicals (because the expression is relatively simple) and cos(1°). Clearly, once you have cos(1°) in radicals you can get any trigonometric function of any integer number of degrees in radicals by applying secondary-school-level identities.

I also used my font chart from my previous post to find that the degree symbol is character number 176, so I use that below.

It is surprising that convert(..., radical) does not apply these techniques. It seems limited to multiple-of-three degrees.

 > restart;
 > °:= Pi/180:
 > cos(3*`20°`) = cos(60*°);

 > expand(%);

 > solve({%, cos(`20°`) > 0}, cos(`20°`), Explicit);

 > solve({%, cos(`1°`) > 0}, cos(`1°`), Explicit);