Carl Love

Carl Love

27656 Reputation

25 Badges

12 years, 124 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

MaplePrimes Activity


These are replies submitted by Carl Love

@Axel Vogt 

Yes, I recall writing something like that, but I don't recall posting it. If you saw it, then I did. Someday soon, I'll have to go download all that Yahoo Maple stuff. I haven't looked at it in many years.

The gist of code in question, IIRC, is that if

theta=p*Pi/q/2^m/3^n

for integers p, q, m, n with q = 1, 5, 7, or 11, then the trig functions of θ can be expressed in (complex) radicals. I can't recall if my code handled every case.

The case q = 11 is quite interesting because it involves solving a quintic. I wonder if there are higher values of q for which the polynomial is solvable (even though Maple can't solve it). Does anyone here know? For q odd, the polynomial to solve for cos(Pi/2/q) is of the form x*p(x^2) where degree(p) = (q-1)/2 and all the roots are real and in (0,1).

What if q is a Fermat prime? For q=17, that polynomial is degree 8. Maple should be able to compute the galois group, but I don't know how to interpret the results. Seems like there should be a connection between this and Gauss's proof of the compass-and-straight-edge constructability of a regular n-gon when n is a Fermat prime. (Gauss's wanted his tombstone to be engraved with a regular 17-gon.)

It's not so simple. You used 12 instead of 16. Try it with 16.

It's not so simple. You used 12 instead of 16. Try it with 16.

@Markiyan Hirnyk The Maple approach is a sledgehammer that I don't like because it involves checking 2^15 cases to do something that should be (at least in principle) doable with secondary-school combinatorics. But I had to find out where my analysis went wrong.

@Markiyan Hirnyk The Maple approach is a sledgehammer that I don't like because it involves checking 2^15 cases to do something that should be (at least in principle) doable with secondary-school combinatorics. But I had to find out where my analysis went wrong.

I could show you how to do it with evalhf, but it is a little complicated. So, first I ask Did you mean evalf? I don't see any benefit to using evalhf in this case. So let me know if you really want evalhf, and why.

@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.

@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.

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

@PlpPlp Upload the long expression.

@PlpPlp Upload the long expression.

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

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

@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.

@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.

First 678 679 680 681 682 683 684 Last Page 680 of 704