Question

How to calculate the Maclaurin series for tan x

Answer 1

Maclaurin series is the Taylor series expansion of an algebraic expression using derivatives, at point 0. 

$f(x) = f(0) + x [\frac{df}{dx}]^{x=0}+\frac{1}{2!}x^2[\frac{d^2f}{dx^2}]^{x=0}+[\frac{1}{3!}x^3 \frac{d^3f}{dx^3}]^{x=0}+....\\\\Let\ f(x) = tan\ x$

We find all the derivatives of f(x) and find their values at x = 0.  Substitute them in the above series expansion formula or polynomial.

$f'(x) = sec^2 x,\ \ \ f'(0) = sec^2 0 = 1$

$f''(x)=2sec\ x*sec\ x\ tan\ x=2sec^2x\ tan\ x,\ \ f''(0)=0,\ as\ tan0=0\\\\f'''(x)=2sec^2x*sec^2x+2*2sec\ x\ sec\ x\ tan\ x*tan\ x\\.\ \ \ \ =2sec^4x+4sec^2x\ tan^2x =2sec^4x+4sec^2x(sec^2x-1)\\.\ \ \ \ =6sec^4x-4sec^2x\\\\f'''(0)=6*1-4*1=2$

In the derviative f''''(x) we will have a term tan x,  hence f''''(0) = 0,  as tan0=0. All even numbered derivatives will be 0.

tan x = tan 0 + x * 1 + x²/2!  0 + x³/3! * 2 + x^4/4! * 0 + ....

tan  x = x + 2x³/3! + .....