# Strange resummation

1. Sep 16, 2006

### lokofer

According to Hardy's book "divergent series" the sum:

$$g(x)=1/2+cos(x)+cos(2x)+cos(3x)+.......$$ has sum equal to 0 (¡¡¡¡¡¡) then if we integrate in the sense:

$$\int_{0}^{b}f(x)\sum_{n=0}^{\infty}cos(nx) dx \rightarrow (-1/2)\int_{0}^{b}f(x)dx$$ since the sum "regularized" has the value 0..but is this true?...can you manipulate divergent series giving them a "sum" although they diverge and even in this case that is clearly 0?..

2. Sep 16, 2006

### Hurkyl

Staff Emeritus
You're omitting a lot of details; if you paid attention to them you might have your answer already.

I'll assume Hardy knows what he's talking about and that you have faithfully represented him... and I'll try and guess at what the missing details are.

The sum of functions

$$\sum_{n = 0}^{+\infty} \cos nx$$

certainly doesn't exist. But that doesn't mean you can't do some other operation to this series to get a number. I'll select a certain class of "test functions", and define a "regularizaion" operation (which I'll call $\mathcal{R}\sum$) on a sequence $f_n$ of functions to be:

$$\mathcal{R}\sum_{n = 0}^{+\infty} f_n(x) = g(x)$$

if and only if

$$\sum_{n = 0}^{+\infty} \int_0^b k(x) f_n(x) \, dx = \int_0^b k(x) g(x) \, dx$$

for every "test function" $k$.

The other way I could interpret this is due to an annoying notation often used in physics that is rarely explicitly explained. If a physicist wrote what you had written, I would expect that they meant $\cos nx$ does not refer to the cosine function, but instead to the operator that maps:

$$k(x) \rightarrow \int_0^b k(x) \cos nx \, dx$$

Then, I imagine for a certain class of "test functions", the infinite sum of these operators does, in fact, converge to the operator $-1/2$ that maps

$$k(x) \rightarrow \int_0^b k(x) (-1/2) \, dx$$

Last edited: Sep 16, 2006
3. Sep 17, 2006

### lokofer

I was refering to a sum over "cosines"...what i have done comes from the fact that "Poisson sum formula"..is:

$$\sum_{n=-\infty}^{\infty}f(n)= \int_{-\infty}^{\infty}f(x)(1+ \sum_{n=0}^{\infty}Cos(nx))$$

So from this you could "Suppose"(¿?) that the sum of cosines tends to -1/2 .. in fact i think you can deduce this identity by taking the "Laplace discrete trasnform:

$$\sum_{n=0}^{\infty}cos(na)e^{-sn}$$ and take the limit s-->0+ (tends to 0 by the right side) to get this identity...my objective in proposing that is to get a "resummation" method for divergent integrals that appear in QFT theory (Quantum mehcanics) of the form:

$$\int_{0}^{\infty}dx x^{m}$$ m>0 or m=0 so you can use "Poisson sum formula" to get a relation between these integrals and the sum:

$$1+2^{m}+3^{m}+................. =\zeta (-m)$$ m>0