1. ## Maclaurin Expansion

Can someone show me how the $\displaystyle \tan x$ expansion is derived?

2. Originally Posted by Air
Can someone show me how the $\displaystyle \tan x$ expansion is derived?

$\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} = x-\frac{x^3}{6}+\frac{x^5}{120}-...$

$\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=1-\frac{x^2}{2}+\frac{x^4}{24}-...$

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.

3. Originally Posted by Air
Can someone show me how the $\displaystyle \tan x$ expansion is derived?

$\displaystyle tan(x) = \sum_{n = 0}^{\infty}\frac{1}{n!}(tan(0))^{(n)}x^n$
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