Cotangent and the Herglotz trick Yang Han Christian Wude May 0, 0 The following script introduces the partial fraction expression of the cotangent function and provides an elegant proof, using the Herglotz trick This expression of the cotangent is used in the second part of this script to solve an actual IMC problem As the complex logarithm function which is needed in order to fully understand the IMC problem may not be familiar to everyone, its most important features will be summarized Partial Fraction Expansion of Cotangent Function The partial fraction expression of the cotangent function was proven by Euler in his Introductio in Analysin Infinitorum from 748 and is regarded as one of the most interesting formula involving elementary functions: π cotπx) = x + + ) x n = We will prove this equation through the Herglotz trick Let fx) := π cotπx) gx) := x R\Z) First, we derive four common properties of those two functions In a second step, we reason that fx) and gx) must be identical
A) The functions fx) and gx) are defined for all non-integral values and are continuous there It is obvious that f is defined for all non-integral values, as fx) = π cotπx) = π cosπx) sinπx) The function is continuous, as it is the quotient of two continuous functions, which denominator does not equal 0 The function gx) can be rearranged as gx) = x + + ) = x n x + x n x We have to prove that the term x converges uniformly in a neighborhood of n x x The first term for n = and the first n x terms do not provide any problems since they are finite, whereas for n and n > x, that is n x > n ) > 0, the summands are bounded by 0 < n x < n ) which is also true for values in the neighborhood of x As the sum to π 6, the function g is defined and continuous B) The functions fx) and gx) are periodic of period The function f is periodic of period, as the cotangent has period π Let Then g N x + ) = = x + + n g N x) = N+ + n ) converges = g N x) + x + N + x + N + Therefore gx + ) = g N x + ) = g N x) = gx) C) The functions fx) and gx) are odd functions For the function f we get fx) = π cotπx) = π cosπx) sinπx) = π cosπ x)) sinπ x)) = f x) The function g is also odd, because g N x) = g N x)
D) The functions fx) and gx) satisfy the functional equation f x ) + fx+ ) = fx) and g x x+ ) + g ) = gx) The function fx) satisfies the functional expression because of the addition theorems for sine and cosine: x ) [ x + cos πx f + f = π ) sin πx ) sin πx ) ] πx cos cos πx ) = π + πx ) sin πx + πx ) = fx) As a result of x + n + x+ + n = x + n + x + n + the functional equation is satisfied for g: x ) x + g + g = g N x) + x + N + Therefore, the two functions f and g satisfy the four properties Lets consider the function hx) := fx) gx) = π cotπx) x + ) x n x This function is continuous on R\Z and satisfies the properties B, C and D As a result of cot x ) x cos x sin x = = 0 π cotπx) ) = 0 x 0 x x 0 x sin x x 0 x ) and x 0 x n x = 0, the function h is continuously continuable with x n hx) = 0 for all n Z Since h is a periodic continuous function, it possesses a maximum m Let x 0 be a point in the interval [0, ] with hx 0 ) = m It follows from property D that x0 ) x0 + h + h = hx 0 ) = m, and hence that h x 0 ) = m Iteration gives h x 0 ) = m for all n Z and therefore h0) = m n by continuity But h0) = 0 and so m = 0, that is hx) 0 for all x R As h is an odd function, hx) < 0 is impossible Therefore hx) = 0 for all x R 3
Complex Logarithm Function Analog to the real definition, the complex logarithm of z is a complex number w = w e iϕ, such that e w = z Since the equation e kπi =, k Z holds, every nonzero complex number z has infinitely many logarithms Therefore, for any logarithm w of z, also w = w +kπi is a logarithm of z To assure uniqueness, we define the principal value logz) with argz) π, π] and logz) := ln z + i arg z logz) is discontinuous at each negative real number, but continuous everywhere else in C To obtain a continuous logarithm defined on complex numbers, we define the branches of the logarithm The logarithms w of z of the k-th branch hold the equation w = ln z + i arg z + kπ) k Z The identities that hold for the noncomplex logarithm don t have to hold for the complex logarithm, ie: log ) + log ) = πi 0 = log = log ) )) 3 Application of the Cotangent Function Problem For every complex number z {0, } define fz) := log z) 4 where the sum is over all branches of the complex logarithm Show that there are two polynomials P and Q such that fz) = P z)/qz) for all z C\{0, } and show that for all z C\{0, } fz) = z z + 4z + 6z ) 4 Solution From the well-known series for the cotangent function, and k= N k= N log z + πi k = i i log z cot w + πi k = i cot iw = i i log z ) + iexpi expi i log z ) = + z 4
Taking derivatives we obtain log z) = z d dx + ) = z log z) 3 = z d z dx z ) = log z) 4 = z 3 d zz + ) dx z ) 3 z z ), zz + ) z ) 3 = zz + 4z + ) 6z ) 4 5