Maclaurin Expansion

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June 18th 2008, 07:20 AM
Simplicity

Maclaurin Expansion

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

Thanks in advance.
June 18th 2008, 07:42 AM
TheEmptySet

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Quote:

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

Attachment 6857

The more terms we write out and use in the long division the more terms we get in the series of tangent.
June 18th 2008, 08:14 AM
topsquark

Quote:

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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