- #1
AmagicalFishy
- 50
- 1
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:
[itex]cos(\theta) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 + ...[/itex]
So, using the first two terms (assuming a small angle), secant should equal...
[itex]sec(\theta) = \frac{1}{cos(\theta)} = \frac{1}{1 - \frac{1}{2}x^2} = \frac{2}{2-x^2}[/itex]
But everywhere I go says the first two terms of the power series for secant are:
[itex]sec(\theta) = 1 + \frac{1}{2}x^2[/itex]
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?
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:
[itex]cos(\theta) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 + ...[/itex]
So, using the first two terms (assuming a small angle), secant should equal...
[itex]sec(\theta) = \frac{1}{cos(\theta)} = \frac{1}{1 - \frac{1}{2}x^2} = \frac{2}{2-x^2}[/itex]
But everywhere I go says the first two terms of the power series for secant are:
[itex]sec(\theta) = 1 + \frac{1}{2}x^2[/itex]
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?