Can someone show me how the $\displaystyle \tan x $ expansion is derived?
Thanks in advance.
$\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} = x-\frac{x^3}{6}+\frac{x^5}{120}-...$
$\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=1-\frac{x^2}{2}+\frac{x^4}{24}-...$
Now we use long division
The more terms we write out and use in the long division the more terms we get in the series of tangent.
$\displaystyle tan(x) = \sum_{n = 0}^{\infty}\frac{1}{n!}(tan(0))^{(n)}x^n$
where $\displaystyle (tan(0))^{(n)}$ is the nth derivative of tan(x) evaluated at x = 0.
$\displaystyle tan(0) = 0$
$\displaystyle (tan(0))^{(1)} = 1$
$\displaystyle (tan(0))^{(2)} = 0$
etc. The derivatives are something of a pain, but with a little effort you should be able to find that
$\displaystyle tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + ~...$
I'll leave it to you to find a general expression for the coefficient of the nth power of x.
-Dan