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Your Wife

Wait, you're a math geek? #282?

Le Bohemian

Hmmm wha? Oh... oh yeah, yeah, I totally understood all of that....... Yeah.


dave glasser

OK, let's finish up that last little issue. You had z^n - 1 = (z-1)(z-r_2)...(z-r_n). Divide through by z-1:

(z^n-1)/(z-1) = (z-r_2)...(z-r_n). But it's a basic theorem about partial sums of geometric series that the left side is 1 + z + z^2 + ... + z^{n-1}. Now plug in 1 on both sides: you get 1 + 1 + ... + 1 = n on the left side, and (1-r_2)...(1-r_n) on the right side, which as you mentioned earlier must have a magnitude equal to the product of the chords. QED!


Am I studpid or doesn't CMFE work for n=2?

It only works if the two marks are respectively on the exact opposite side of the circle.


~~~ Me stupid ~~~~

I missed the words "equally spaced"...

~~~ Me stupid ~~~~



I like your use of the formula for a partial geometric sum...it simplifies the proof a lot. But I think it still has the same problem: dividing by z-1 is only valid if z is never 1, and then later you use z=1. Same issue in a different outfit, I think.

dave glasser

Hmm. So, I guess, you're happy to believe that 1 + z + ... + z^{n-1} = (z-r_2)...(z-r_n) for all z other than 1, right? (Because the only thing I would have invalidated by the division would have been z=1.) And both sides are degree n-1 polynomials, so like you said in a different part of the proof they have to be the *identical*, because two polynomials are just not going to disagree in precisely one point.

Another way to look at it is that we're just manipulating objects in the ring of polynomials C[z], so that the statement that (z^n - 1)/(z-1) = 1 + z + ... z^{n-1} is not a statement about values that functions take at various values of z, but of two polynomials being the same object.

(I guess in retrospect you can apply that exact argument to your original proof too!)


Now, I always thought the coolest math fact ever was e^(pi*i) + 1 = 0.

This is cool, too, though.


"...they have to be the *identical*, because two polynomials are just not going to disagree in precisely one point."

An excellent point. I'm convinced!

Kristjan Kannike

It does work for n = 2. Remember that the unit circle is of radius 1. The length of its diametre, the chord between two equally spaced marks, is 2, of course.


The problem reduced, in geometric analysis, to one of the following forms.

I'm using Prod[f(i),a, b] to represent the product of f(a)*f(a+1)*f(a+2)..*f(b)

If n is even,
productOfSegments = (1/2)*Prod[ 2*Sin[i*pi/n], 1, n/2]^2

If n is odd,
productOfSegments = Prod[2*Sin[i*pi/n], 1, (n-2)/2]^2

I think I typed that in correctly. Hmm. I've not done much with products and the like before, so I'm not sure where to go from there. Interesting, though.

Cat lover

Are you sure the diagonal lengths of an n-gon can be *transcendental*? If so, can I have a reference? Certainly they aren't necessarily constructible (in the Euclidean sense) or even expressable in radicals, but my intuition says that they should at least be algebraic. Oh, and, very cool math fact, btw :-) Slick proof, too!


Thanks for the correction, cat lover. I agree that those numbers are probably algebraic. My HEAD knows that there are algebraic numbers that can't be written with roots, but I think I don't know it yet in my heart of hearts.

kevin cline

The diagonal lengths are definitely algebraic, since the end points are algebraic numbers, and the distance function is an algebraic function.


Ran across your site by looking for interesting math problems. For the above problem you can avoid dividing by zero if you first move RHS to LHS and then factor out the Z. Then equate other factor to zero and solve. QED


Generalize and prove results for z^n=a+bi, where |a+bi|=1.
What happens when |a+bi|≠1


Can you explain why you substituted z with Z+1 at that one part in your proof? Why can you do so, and why do you do so?

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