Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
7  
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 at 20:07
    
Taylor expansions never lie... –  user88595 Jan 7 at 20:09
add comment

1 Answer 1

$$ f(x) = \tan(x) $$

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 $$ f(x)\approx x + \frac{x^3}{3} $$

That's the correct expansion, the one "in the answer" you stated is incorrect.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.