Maclaurin Expansion

**Simplicity** · June 18th 2008, 07:20 AM

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

Thanks in advance.

**TheEmptySet** · June 18th 2008, 07:42 AM

Originally Posted by Air

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

Thanks in advance.

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

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

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.

**topsquark** · June 18th 2008, 08:14 AM

Originally Posted by Air

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

Thanks in advance.

$tan(x) = \sum_{n = 0}^{\infty}\frac{1}{n!}(tan(0))^{(n)}x^n$
where $(tan(0))^{(n)}$ is the nth derivative of tan(x) evaluated at x = 0.

$tan(0) = 0$

$(tan(0))^{(1)} = 1$

$(tan(0))^{(2)} = 0$

etc. The derivatives are something of a pain, but with a little effort you should be able to find that
$tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + ~...$

I'll leave it to you to find a general expression for the coefficient of the nth power of x.

-Dan

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