nLab cotangent function

Contents

Contents

Idea

In trigonometry the cotangent function is one of the basic trigonometric functions.

Definition

The contangent is the ratio of the cosine and the sine:

cot(x)=cos(x)sin(x) \cot(x) = \frac{\cos(x)}{\sin(x)}

Properties

Relations with tangent function

  • The cotangent and tangent are reciprocal to each other: cotx=1tanx\cot x = \frac1{\tan x}.

  • The cotangent and tangent are complementary to each other: cotx=tan(π2x)\cot x = \tan (\frac{\pi}{2} - x).

  • Double angle formula: tanx=cotx2cot2x\tan x = \cot x - 2\cot 2x.

Relation to Bernoulli numbers

The hyperbolic analog cothx=e x+e xe xe x\coth x = \frac{e^x + e^{-x}}{e^x - e^{-x}} is related to cotx\cot x via the formula

cotx=icothix.\cot x = i\coth i x.

Meanwhile cothx\coth x is related to the Bernoulli numbers B nB_n, defined by the exponential generating function

xe x1= n0B nx nn!,\frac{x}{e^x - 1} = \sum_{n \geq 0} \frac{B_n x^n}{n!},

through a series of equations

xe x1+x2=x(2+e x1)2(e x1)=x2e x+1e x1=x2e x/2+e x/2e x/2e x/2=x2cothx2.\frac{x}{e^x - 1} + \frac{x}{2} = \frac{x(2 + e^x - 1)}{2(e^x - 1)} = \frac{x}{2} \frac{e^x + 1}{e^x - 1} = \frac{x}{2} \frac{e^{x/2} + e^{-x/2}}{e^{x/2} - e^{-x/2}} = \frac{x}{2}\coth \frac{x}{2}.

Notice the right side defines an even function. Therefore

x2cothx2= n0B 2nx 2n(2n)!\frac{x}{2}\coth \frac{x}{2} = \sum_{n \geq 0} \frac{B_{2n} x^{2n}}{(2n)!}

and so

xcotx=ixcothix= n0B 2n(2ix) 2n(2n)!= n0(1) nB 2n2 2nx 2n(2n)!.x \cot x = i x \coth i x = \sum_{n \geq 0} \frac{B_{2n} (2i x)^{2n}}{(2n)!} = \sum_{n \geq 0} (-1)^n \frac{B_{2n} 2^{2n} x^{2n}}{(2n)!}.

“Eisenstein” series expansion

The cotangent is the logarithmic derivative? of the sine function:

cotx=(log(sinx)).\cot x = (\log (\sin x))'.

Applying this observation to the Euler-Weierstrass product formula for the sine function (see there for a proof):

sin(πx)=πx n=1 (1x 2n 2)\sin (\pi x) = \pi x \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2}\right)

one obtains the following summation formula for the cotangent:

πcot(πx)=1x+ n=1 (1x+n+1xn)\pi\, \cot (\pi x) = \frac1{x} + \sum_{n=1}^\infty \left(\frac1{x + n} + \frac1{x - n}\right)

This expansion was used by Eisenstein as a starting point for developing the theory of trigonometric functions; Eisenstein’s account of elliptic functions (cf. the eponymous Eisenstein series), developed further by Weierstrass, Kronecker, and others, runs parallel to his trigonometric theory, as explained later by Weil. For some more details, see these notes by Varadarajan.

Relation to zeta function values

Proposition

The power series identity

πxcotπx=12 k1ζ(2k)x 2k\pi x \cot \pi x = 1 - 2\sum_{k \geq 1} \zeta(2k) x^{2k}

holds over an open domain where the series converges, |x|<1{|x|} \lt 1.

Proof

From the Eisenstein expansion, we have

πxcot(πx)=x(1x+ n=1 (1x+n+1xn))=1+ n02x 2x 2n 2=12 n1x 2/n 21x 2/n 2.\pi x\, \cot (\pi x) = x\, \left(\frac1{x} + \sum_{n=1}^\infty \left(\frac1{x + n} + \frac1{x - n}\right)\right) = 1 + \sum_{n \geq 0} \frac{2x^2}{x^2 - n^2} = 1 - 2\sum_{n \geq 1} \frac{x^2/n^2}{1 - x^2/n^2}.

By a geometric series expansion, the last expression is

12 n1 k1(x 2n 2) k=12 k1x 2k n11n 2k1 - 2\sum_{n \geq 1} \sum_{k \geq 1} \left(\frac{x^2}{n^2}\right)^k = 1 - 2\sum_{k \geq 1} x^{2k} \sum_{n \geq 1} \frac1{n^{2k}}

which is the same as 12 k1ζ(2k)x 2k1 - 2\sum_{k \geq 1} \zeta(2k)x^{2k}.

References

Last revised on June 7, 2023 at 10:39:03. See the history of this page for a list of all contributions to it.