# Mistake in Taylor expansion?

Given: The first derivative of $\tan x$ is $1/\cos^2 x$

So the derivative of $\tan x$ when $x=0$ should be $1$. This derivative times $x$ should be a term in the Taylor expansion (the term then being $x$).

However, in the answer it says that expansion is $1 - x^3/3\cdots$. Where did the $x$ go?

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It had a hangover and the $1$ covered for it, so the boss doesn't notice. It should be $x + \frac{x^3}{3} + \dotsc$. – Daniel Fischer Jan 7 '14 at 20:07
Taylor expansions never lie... – user88595 Jan 7 '14 at 20:09

$$f(x) = \tan(x)$$
The third order Taylor series at zero is: $$f(x)\approx f(0) + f'(0)x + f''(0)\frac{x^2}{2} + f'''(0)\frac{x^3}{6}$$
You can calculate that $$f(0) = \tan(0) = 0$$ $$f'(0) = \sec^2(0) = 1$$ $$f''(0) = \frac{8\sin(0)}{3\cos(0)+\cos(3\cdot0)} = 0$$ $$f'''(0) = (4\sin^2(0)+2)\sec^4(0) = 2$$ Plugging into the expansion, we get $$f(x)\approx x + \frac{x^3}{3}$$