The exception case is normal

Posted on by John

Sine and cosine have power series with simple coefficients, but tangent does not. Students are shocked when they see the power series for tangent because there is no simple expression for the power series coefficients unless you introduce Bernoulli numbers, and there’s no simple expression for Bernoulli numbers.

The perception is that sine and cosine are the usual case, and that tangent is exceptional. That’s because in homework problems and examples, sine and cosine are the usual case, and tangent is exceptional. But as is often the case, the pedagogically convenient case is the exception rather than the rule.

Of the six common trig functions—sine, cosine, tangent, secant, cosecant, cotangent—sine and cosine are the only ones with power series not involving Bernoulli numbers. The power series for most of the common trig functions require Bernoulli numbers.

Details

The functions cosecant and cotangent have a singularity at 0, so the power series for these functions at 0 are Laurant series rather than Taylor series, i.e. they involve negative as well as positive exponents. We will look at z csc(z) and z cot(z) because multiplying by z removes the singularity.

\begin{align*} \sin(z) &= \sum_{n=0}^\infty (-1)^n \frac{ z^{2n+1} }{ (2n+1)! } \\ \cos(z) &= \sum_{n=0}^\infty (-1)^n \frac{ z^{2n} }{ (2n)! } \\ \tan(z) &= \sum_{n=1}^\infty (-1)^{n-1} 4^n (4^n-1) B_{2n} \frac{ z^{2n-1} }{ (2n)! } \\ \text{sec}(z) &= \sum_{n=0}^\infty (-1)^n E_{2n} \frac{ z^{2n} }{ (2n)! } \\ z\text{cot}(z) &= \sum_{n=0}^\infty (-4)^n B_{2n} \frac{ z^{2n} }{ (2n)! } \\ z\text{csc}(z) &= \sum_{n=0}^\infty (-1)^{n-1} (2^{2n} - 2) B_{2n} \frac{ z^{2n} }{ (2n)! } \end{align*}

The series for secant doesn’t use Bernoulli numbers directly but rather Euler numbers. However, Bernoulli and Euler numbers are closely related.

\begin{align*} B_n &= \sum_{k=0}^{n-1} \binom{n-1}{k} \frac{n}{4^n - 2^n} E_k \\ E_n &= \sum_{k=1}^{n} \binom{n}{k-1} \frac{2^n - 4^n}{k} B_k \end{align*}

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