The functions tan and cot are in some respected more complicated than sin and cos (and csc and sec). One thing is their Taylor series which involve the Bernoulli numbers. I thought this could be made into a little problem.
The Bernoulli numbers B(n) can be defined by x/(eˣ – 1) = Σ [n=0,∞] B(n)xⁿ/n!. Using this find the Taylor series around x = 0 (i e the Maclaurin series) for x·cot x. Then, using the fact that this function is real for real x, prove that B(1) = -1/2 but that B(n) = 0 for all odd n except 1.
x·cot x = ix·[eix + e-ix]/[eix – e-ix] =
= ix·[e2ix + 1]/[e2ix – 1] =
=ix·[e2ix – 1 + 2]/[e2ix – 1] =
= ix + 2ix/[e2ix – 1] = ix + Σ [n=0,∞] B(n)(2ix)ⁿ/n! =
= ix + B(0) + B(1)·2ix + Σ [n=2,∞] B(n)(2ix)ⁿ/n! =
= ix + B(0) + B(1)·2ix + Σ [n=2,∞] B(2n)(2ix)²ⁿ/(2n)! +
+ Σ [n=1,∞] B(2n+1)(2ix)²ⁿ⁺¹/(2n+1)! =
= ix + B(1)·2ix + Σ [n=0,∞] B(2n)(2x)²ⁿ·(-1)ⁿ/(2n)! +
+ i·Σ [n=1,∞] B(2n+1)(2x)²ⁿ⁺¹·(-1)ⁿ/(2n+1)!
In order that this be real for real x the last sum must vanish for all real x so B(2n+1) = 0 for n = 1, 2, 3, …, and ix + B(1)·2ix must be real so B(1) = -1/2. The series expansion then becomes
x·cot x = Σ [n=0,∞] B(2n)(-4)ⁿ/(2n)!·x²ⁿ