Prove csc⁡(x)=∑k=−∞∞(−1)kx+kπ

Question

Prove  csc⁡(x)=∑k=−∞∞(−1)kx+kπ

Wilson Mitchell · Accepted Answer

We can use also this well known summation formula (a consequence of the residue theorem):

\sum_{k \in Z} {(- 1)}^{k} f (k) = - \sum [residues of π \csc (π z) f (z) a t {f (z)}^{'} s poles]

so if we take

f (z) = \frac{1}{x + z π}

we get

\sum_{k \in z} \frac{{(- 1)}^{k}}{x + k π} = - R e s_{z = - \frac{x}{π}} (\frac{π \csc (π z)}{x + z π}) = \csc (x)

as wanted.

lorugb · Accepted Answer

In this answer (using complex methods) and in this answer (using real methods), it is shown in detail that

\sum_{k = - \infty}^{\infty} \frac{1}{z + k} = π \cot (π z)

(1)
(1) is the sum for even and odd k. The sum for even k would be

\sum_{k = - \infty}^{\infty} \frac{1}{z + 2 k} = \frac{π}{2} \cot (\frac{π z}{2})

(2)
The sum for even minus the sum for odd would be twice (2) minus (1)

\sum_{k = - \infty}^{\infty} \frac{{(- 1)}^{k}}{z + k} = π \cot (\frac{π z}{2}) - π \cot (π z)

= π \frac{1 + \cos (π z)}{\sin (π z)} - π \frac{\cos (π z)}{\sin (π z)}

= π \csc (π z)

Therefore,

\sum_{k = - \infty}^{\infty} \frac{{(- 1)}^{k}}{z + k π} = \csc (z)

RizerMix · Accepted Answer

We begin by expanding the function cos⁡(xy) in a Fourier series, cos⁡(xy)=a0/2+∑k=1∞akcos⁡(ky) (1) for x∈[−π/π]. The Fourier coefficients (1) are given by ak=2π∫0πcos⁡(xy)cos⁡(ky)dy =1π(−1)ksin⁡(πx)(1x+k+1x−k) (2) Substituting (2) into (1), setting y=0, and dividing by sin⁡(πx) reveals πcsc⁡(πx)=1y+∑n=1∞(−1)k(1x−k+1x+k) =∑k=−∞∞(−1)kx−k =∑k=−∞∞(−1)kx+k (3) Finally, enforcing the substitution x→x/π and dividing by π in (3) yields the coveted result csc⁡(x)=∑k=−∞∞(−1)kx+kπ

Prove \csc(x)=\sum_{k=-\infty}^\infty\frac{(-1)^k}{x+k\pi}

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