Re: Sinc sum


"Kaba" <none@xxxxxxxx> wrote in message
news:MPG.223028766b9a7f159899a4@xxxxxxxxxxxxxxxxx
Let
sinc(x) = sin(pi * x) / (pi * x)

Does it hold that
for all x in [0, 1[ in R: sum[i in Z] sinc(x + i) = 1
?

I'd expect yes. How to prove it?

--
http://kaba.hilvi.org

you have

S = sum(n=-oo,+oo) sin(pi*(x+n)) / (pi*(x+n))

consider then the following S_p = sin(pi*x) / pi * sum(n=-p,+p) (-1)^n /
(x+n). Now

S_p = (-1)^p * { 1/(x-p) - 1/(x+1-p) + .....-1/(x-1) + 1/x - 1/(x+1) +
1/(x+2) -....+1/(x+p) ) +1/x

and so we have

S_p = 1/x + (2*x*pi^2)*sum(k=1,p) (-1)^(k+p) / ((pi*x)^2 - pi^2*k^2)

We may write this as

S_p = pi/(pi*x) + (2*x*pi^2)*sum(k=1,oo) (-1)^(k) / ((pi*x)^2 -
i^2*k^2) - (2*x*pi^2)*sum(k=p+1,oo) (-1)^(k+p) / ((pi*x)^2 - pi^2*k^2)

since the series are absolutely convergent. Note that the first series and
the first term is simply pi*csc(pi*x), and so as p->oo we have therefore

S _oo= pi*csc(pi*x)

Your series is simply

S = sin(pi*x)/pi * S_oo = 1.

The result here hinges on a particular series representation of csc which is
only valid for x not equal to an integer. I'm not sure how to extend this
derivation to x=1...but you possibly don't need to because i think you can
appeal to Dirichlet's test (?) which should extend the derivation to all
real x.

cheers
moth


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