Comments on: Product of polygon diagonals

By: Eway

Eway — Fri, 15 Mar 2013 15:57:09 +0000

I have discussed your proof with a Maths professor and unfortunately him and me both agree that it is not valid, because of the impossibility of the substitution of z=1 in the last part of your proof.

The part:

(z^(n-1) + z^(n-2) + … 1) = (z – ω)(z – ω^2) … (z – ω^(n-1)).

Is only valid if z is not equal to 1, making the subsitution of z=1 into the equation impossible.

By: Michael Lee

Michael Lee — Wed, 13 Mar 2013 15:50:52 +0000

@ Ralph: Because (z-1)(z-w)(z-w^2)...(z-w^(n-1)) is an alternate factorization of z^(n-1), since all the different values of w^k are roots of z^(n-1) - 1 = 0. Dividing by (z - 1) on both sides results in the last equation in the blog post. Essentially, this proof is setting two different (but equally valid) factorizations of z^(n - 1) against each other, and plugging in z = 1.

By: Ralph

Ralph — Sat, 23 Feb 2013 23:12:11 +0000

I have a question: how do you get from
z^n-1 /(z–1)
to
(z-w)(z-w^2 )(z-w^3 )…(z-w^(n-1))

By: Amadeus

Amadeus — Sun, 03 Feb 2013 13:40:50 +0000

Hey people i just wanted to ask that if i would put z=1 and prove that 0=0 , is it a valid proof? Cause usually if you prove something you should not put values, but rather do it in general … so is there any other way of proving this conjecture? Would it be helpfull to further simplify the left hand side and then the right hand side or wouldn’t i get a solution??
Thanks for the answer in advance!

By: John

John — Sat, 29 Dec 2012 13:29:47 +0000

If by pure mathematical you mean pure geometrical, yes, you can prove the result geometrically, though the proof is much longer. I believe the book I cite gives a geometrical proof in a special case.

By: Elize

Elize — Sat, 29 Dec 2012 13:27:02 +0000

The proof by complex numbers is elegant, but it does not help me to understand the result. Does anyone know a pure mathematical explanation of the beautyfull theses?

By: John

John — Fri, 30 Nov 2012 20:52:11 +0000

You and I are substituting z=1 into different equations. Stick z=1 into
(z^n-1 + z^n-2 + … 1) = (z – ω)(z – ω^2) … (z – ωn-1) and you get 1 + 1 + … + 1 = n on the left side and the product of the diagonal lenghts on the right.

By: Peter Fry

Peter Fry — Fri, 30 Nov 2012 19:36:22 +0000

John, surely if you say
“Since
(z^n – 1) = (z – 1)(z^n-1 + z^n-2 + … 1)
it follows that
(z^n-1 + z^n-2 + … 1) = (z – ω)(z – ω2) … (z – ωn-1).
The result follows from evaluating the expression above at z = 1.”

if you put z=1 into the expression you just prove that 0 = 0
since (z^n – 1) = 1 – 1 = 0
and (z – 1)(z^n-1 + z^n-2 + … 1) = (0)(z^n-1 + z^n-2 + … 1) = 0

Maybe my maths is a bit rusty after 50 years!

By: darvish kamalia

darvish kamalia — Wed, 28 Nov 2012 14:27:16 +0000

Nevermind figured it out

By: darvish kamalia

darvish kamalia — Fri, 23 Nov 2012 19:30:26 +0000

Also, could you please elaborate on the second step, specifically why we can remove the modulus/absolute value signs and also the last step?

Thanks

By: darvish kamalia

darvish kamalia — Fri, 23 Nov 2012 16:59:32 +0000

Thanks for the quick reply

Just to clarify, |a-b| means the modulus of the complex number formed by subtracting complex number b from another complex number a?

By: John

John — Fri, 23 Nov 2012 16:42:36 +0000

Darvish: The distance between two complex numbers a and b is |a - b|.

By: darvish kamalia

darvish kamalia — Fri, 23 Nov 2012 16:40:00 +0000

Can you tell me why the length of the diagonal is given by |1-ω|?

By: g

Tue, 20 Nov 2012 23:28:48 +0000

No need for any consideration of conjugate pairs. The product of the absolute values = the absolute value of the product, so just work out the product. It happens that it’s positive and real and hence is its own absolute value, but you don’t need to know ahead of time that it’s going to turn out that way.

By: Gabriel Verret

Gabriel Verret — Tue, 20 Nov 2012 20:46:46 +0000

Just a slight clarification : for n even, -1 is a root and does not come in a conjugate pair, but you can still remove the absolute value.

By: Scott

Scott — Tue, 20 Nov 2012 20:00:10 +0000

Yes, I stumbled for a moment over that second to last step too. What a beautiful little gem of a proof!

By: John

John — Tue, 20 Nov 2012 16:38:23 +0000

GlennF: You’re right. That part could use more explanation.

By: GlennF

GlennF — Tue, 20 Nov 2012 16:22:29 +0000

Amusing result. Maybe my morning coffee hadn’t kicked in… but I found the second the last step was a bit murky until I realized that what you were saying was that (z^n-1) must vanish at all the powers of omega, and thus given (z^n-1)=(z-1)(z^n-1 + … + 1), it follows that the sum must factor as stated. (If one worries about it, a few moments thought will also show that the constant factor must be one.)

By: Marshall Thompson

Marshall Thompson — Tue, 20 Nov 2012 15:15:39 +0000

Reminds me very much of the IB HL portfolio task for 2012-2013 (page 6). Cool stuff!