# Expand tan(x+pi/4)

• Mar 29th 2011, 02:59 PM
StaryNight
Expand tan(x+pi/4)
What is the quickest method to do this Taylor expansion for the first three terms? Somebody previously suggested doing the expansion for tan(x) and then substituting in (x + pi/4) but if I do that I get the following working and run into problems:

http://latex.codecogs.com/gif.latex?...20+%20...%20=?

• Mar 29th 2011, 04:39 PM
ojones
First of all, the series for $tan$ is incorrect.

Substitution will give you the Taylor series about $-\pi/4$. Is it the Maclaurin series that you're looking for? If so, you need to differentiate and evaluation at 0.
• Mar 29th 2011, 06:54 PM
skeeter
$f(x) = f(c) + f'(c)(x - c) + \dfrac{f''(c)(x-c)^2}{2!} + ...$

$f(x) = \tan{x}$ centered at $c = \frac{\pi}{4} ...$

$f(x) = f\left(\frac{\pi}{4}\right) + f'\left(\frac{\pi}{4}\right)\left(x - \frac{\pi}{4}\right) + \dfrac{f''\left(\frac{\pi}{4}\right)\left(x - \frac{\pi}{4}\right)^2}{2!} + ...$

$\tan{x} = 1 + 2\left(x - \frac{\pi}{4}\right) + 2\left(x - \frac{\pi}{4}\right)^2 + ...$

$\tan\left(x + \frac{\pi}{4}\right) = 1 + 2x + 2x^2 + ...$
• Mar 29th 2011, 07:32 PM
ojones
Looks correct.