If
¥
å
n=0
a_{k n}=0

then all a_m=0?

By Brad Rodgers on Saturday, September 07, 2002 - 11:11 pm:

Is it true that, if for all k,

¥
å
n=0

a_{k n}=0

,
then all a_m = 0? [I can prove the converse ;)]

This would be a generalization of the theorem:

Under the identification of the place with the set of complex numbers, let f:R²®R be a function such that for all regular n-gon A₁ A₂ ...A_n in the complex plane,

f(A₁ ) + f(A₂ ) + ... + f(A_n ) = 0,

then f(z) = 0 for all z.

I know how to prove this. Any ideas on how to prove the first statement (or disprove it)?

Brad

By David Loeffler on Monday, September 09, 2002 - 01:56 pm:

I think we can build such a series up by induction, using finite blocks of terms.

For any k, I suggest we build a block of terms of length (k+1)!, a₀ ... a_(k+1)!-1. Letting a₀=1 this is obviously possible for k=0.

Suppose we can do such a block for a particular k. Then let a_r(k+1)!+s=-1/(k+1)a_s for s=0...(k+1)!-1, r=1...k+1 - so we are appending k+1 new blocks, to attain a_n for k=0 ... (k+2)!-1. We now see that k+1 will cancel:
((k+2)!-1)/(k+1))
å
i=0
a_i(k+1)=0

. Since the block length (k+2)! is a multiple of k+1, the infinite sum is 0 (since if we continue the construction, the blocks are multiplied by ever smaller factors, so the sequence of partial sums oscillates between bounds that are tending to 0 from above and below)

That's rather confusing (it confuses me and I wrote it ), so here's an example of how we build it up in blocks:

1....

1 -1 .... (dealing with k=1)

(1 -1) (-1/2 1/2) (-1/2 1/2) (k=1 and 2)

[(1 -1) (-1/2 1/2) (-1/2 1/2)][(1/3 -1/3) (1/6 -1/6) (1/6 -1/6)][(1/3 -1/3) (1/6 -1/6) (1/6 -1/6)] (k=1, 2, 3)

The key is that k+1 divides the number of terms obtained at the k^th step, so when going along the sequence in steps of k+1 we land on the first term in each new block; so we pick out terms in the same positions in corresponding blocks.

David

By David Loeffler on Monday, September 09, 2002 - 02:03 pm:

PS. Pretty result, that one about polygons, isn't it? I came across it in IMO training in spring last year. I can't see how you're relating it to this one though.

David

By Michael Doré on Monday, September 09, 2002 - 02:33 pm:

Well if you assume f (as a function from C to C) has a power series, then you can substitute in complex exponentials (representing vertices of polygons centred on the origin) and add these together, and you get the sums Brad is referring to. But obviously if f has a power series the result is trivial anyway - it is trivial for any continuous f.