# Quick question on power series of secant

1. Jan 28, 2014

### AmagicalFishy

Hey, everyone.

I am trying to find the power series of secant from the known power series of cosin, but my answer does not match up with Wolfram and Wikipedia.

I know:
$cos(\theta) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 + ...$

So, using the first two terms (assuming a small angle), secant should equal...

$sec(\theta) = \frac{1}{cos(\theta)} = \frac{1}{1 - \frac{1}{2}x^2} = \frac{2}{2-x^2}$

But everywhere I go says the first two terms of the power series for secant are:

$sec(\theta) = 1 + \frac{1}{2}x^2$

I'm sure there's something (probably elementary) that I'm missing, but I have no idea what it is. Does it have something to do with the fact that I switched the sign of the exponent when taking the reciprocal?

2. Jan 28, 2014

### HallsofIvy

You are asking for a power series, right? But $\frac{2}{2- x^2}$ is NOT a power series. To get the corresponding power series, divide 2 by $2- x^2$. $2- x^2$ divides into $2+ 0x+ 0x^2+ 0x^3+ \cdot\cdot\cdot$ once and then $(2+ 0x+ 0x^2+ 0x^3+ \cdot\cdot\cdot)- (2- x^2)= x^2+ 0x^3+ \cdot\cdot\cdot$. $2- x^2$ divides into $x^2+ 0x^3+ \cdot\cdot\cdot[itex](1/2)x^2$ times and then $x^2+ 0x^3+ \cdot\cdot\cdot- (1/2)x^2(2- x^2)= (1/2)x^4+\cdot\cdot\cdot$ so to second power, the fraction is $1+ (1/2)x^2$