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: cot x = 1 tan x \cot x = \frac1{\tan x}
The cotangent and tangent are complementary to each other: cot x = tan ( π 2 − x ) \cot x = \tan (\frac{\pi}{2} - x)
Double angle formula: tan x = cot x − 2 cot 2 x \tan x = \cot x - 2\cot 2x
Relation to Bernoulli numbers
The hyperbolic analog coth x = e x + e − x e x − e − x \coth x = \frac{e^x + e^{-x}}{e^x - e^{-x}} cot x \cot x
cot x = i coth i x . \cot x = i\coth i x. Meanwhile coth x \coth x Bernoulli numbers B n B_n exponential generating function
x e x − 1 = ∑ n ≥ 0 B n x n n ! , \frac{x}{e^x - 1} = \sum_{n \geq 0} \frac{B_n x^n}{n!}, through a series of equations
x e x − 1 + x 2 = x ( 2 + e x − 1 ) 2 ( e x − 1 ) = x 2 e x + 1 e x − 1 = x 2 e x / 2 + e − x / 2 e x / 2 − e − x / 2 = x 2 coth x 2 . \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
x 2 coth x 2 = ∑ n ≥ 0 B 2 n x 2 n ( 2 n ) ! \frac{x}{2}\coth \frac{x}{2} = \sum_{n \geq 0} \frac{B_{2n} x^{2n}}{(2n)!} and so
x cot x = i x coth i x = ∑ n ≥ 0 B 2 n ( 2 i x ) 2 n ( 2 n ) ! = ∑ n ≥ 0 ( − 1 ) n B 2 n 2 2 n x 2 n ( 2 n ) ! . 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 :
cot x = ( log ( sin x ) ) ′ . \cot x = (\log (\sin x))'. Applying this observation to the Euler-Weierstrass product formula for the sine function (see there for a proof):
sin ( π x ) = π x ∏ n = 1 ∞ ( 1 − x 2 n 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:
π cot ( π x ) = 1 x + ∑ n = 1 ∞ ( 1 x + n + 1 x − n ) \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
π x cot π x = 1 − 2 ∑ k ≥ 1 ζ ( 2 k ) x 2 k \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
π x cot ( π x ) = x ( 1 x + ∑ n = 1 ∞ ( 1 x + n + 1 x − n ) ) = 1 + ∑ n ≥ 0 2 x 2 x 2 − n 2 = 1 − 2 ∑ n ≥ 1 x 2 / n 2 1 − x 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
1 − 2 ∑ n ≥ 1 ∑ k ≥ 1 ( x 2 n 2 ) k = 1 − 2 ∑ k ≥ 1 x 2 k ∑ n ≥ 1 1 n 2 k 1 - 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 1 − 2 ∑ k ≥ 1 ζ ( 2 k ) x 2 k 1 - 2\sum_{k \geq 1} \zeta(2k)x^{2k}
References
Last revised on June 7, 2023 at 10:39:03.
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