0

By the asymptotes of $\tan x$, I came up with $$\tan x \stackrel{?}{=}(\sin x)\Bigg(\sum_{n=-\infty}^{ \infty}\cfrac{(-1)^n}{x-\frac{(2n-1)\pi}{2}}\Bigg)$$ So does $\sec x$ equal $\sum_{n=-\infty}^{ \infty}\cfrac{(-1)^n}{x-\frac{(2n-1)\pi}{2}}$? Or this is already a well known series representation?

xxxx036
  • 773

1 Answers1

0

It is well-known, yes. It can be obtained through the logarithmic derivative of Euler's product formula for sine (i.e. its Weierstrass factorization). And yes, it is a convenient approximation. See MSE Q1419280.

Jam
  • 10,325