1. ## 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:

$\newline tan(x) = 1 + \frac{x^3}{3} + \frac{2x^5}{15} + ... \newline tan(x + \frac{\pi}{4}) = 1 + \frac{1}{3}(x + \frac{\pi}{4})^3 + \frac{2}{15}(x + \frac{\pi}{4})^5 +... \newline Considering \hspace{2} x^0 \hspace{2} terms: \newline 1 + \frac{1}{3} (\frac{\pi}{4})^3 + \frac{2}{15} (\frac{\pi}{4})^5 + ... = 1 + tan(\frac{\pi}{4}) = 2 \newline Considering \hspace{2} x^1 \hspace{2} terms: \newline \frac{1}{3} \binom{3}{1}(\frac{\pi}{4})^2 x + \frac{2}{15} \binom{5}{1}(\frac{\pi}{4})^4 x + ... =?$

2. First of all, the series for $\displaystyle tan$ is incorrect.

Substitution will give you the Taylor series about $\displaystyle -\pi/4$. Is it the Maclaurin series that you're looking for? If so, you need to differentiate and evaluation at 0.

3. $\displaystyle f(x) = f(c) + f'(c)(x - c) + \dfrac{f''(c)(x-c)^2}{2!} + ...$

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

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

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

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

4. Looks correct.

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# tan (x pi/4) by tayler theorem

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