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Topic: sum of csc
Replies: 0

 stainburg Posts: 1 From: china Registered: 10/18/10
sum of csc
Posted: Oct 18, 2010 10:42 PM

sum{csc[pi*(n+.5)/N]} with n
where
k+1<=n<=N-k-2
0<=k<=N-1

n,k,N are natural numbers

[Difficulty]
I doubt if it's concerned about the gamma function.
gamma(x)*gamma(1-x)=pi*csc(pi*x)

[Thoughts]
This question is from the problem X(k+1/2)=DFT[cot(pi*(n+.5)/N)*exp(-1j*pi*n/N)],where DFT denotes the discrete fourier transform.

while x(n) denotes a time sequence,X(k) denotes its DFT sequence and we have
X(k)=DFT[x(n)]=sum[x(n)*exp(-1j*2*pi*n*k/N)],(0<=n<=N-1),where N is the length of x(n),k is the sampling frequency,n is the sampling time.
X(k+1/2)=DFT[x(n)*exp(-1j*pi*n/N)]
=(1/N)*{sum[X(g)]+sum[X(g)*cot[pi*(k-g+1/2)/N]]},with 0<=g<=N-1.
and sum[X(g)*cot[pi*(k-g+1/2)/N]] denotes a circle convolution with X(k)and cot[pi*(k+1/2)/N].
I have known that
X(k)=DFT[cot(pi*(n+.5)/N)]=T(k)-N*exp(1j*pi*k/N),
where T(k) is a sampling function, which is equal to
[N,0,0,0,0...].
so
X(k+1/2)=DFT[cot(pi*(n+1/2)/N)*exp(-1j*pi*n/N)]
can be expressed by
X(k+1/2)=cot(pi*(k+1/2)/N)+1j
-exp(1j*pi*(k+1/2)/N)*{sum{csc[pi*(g1+1/2)/N]}-sum{csc[pi*(g2+1/2)/N]}}
with, 0<=g1<=k,k+1<=g2<=N-1,0<=k<=N-1
and there comes my question.