Question

How do you find the Maclaurin series for `tan x` centered at 0?

Answer 1

Maclaurin series for `tanx` centered at `0` is

`x+x^3/3+(2x^5)/15+(17x^5)/315+.....`

Explanation:

Maclaurin series for `f(x)` centered at `a` is

`f(x)=f(a)+(x-a)f'(a)+(x-a)^2/(2!)f''(a)+(x-a)^3/(3!)f'''(a)+(x-a)^4/(2!)f''''(a)+...`

Hence for `f(x)=tanx`, and as `f'(x)=sec^2x`, we have `f'(0)=sec^2 0=1`;
as `f''(x)=2secx xx secxtanx=2sec^2xtanx`, `f''(0)=2sec^2 0tan0=0`,
as `f'''(xx)=2sec^4x+2sec^2xtan^2x` and `f'''(0)=2`

Similarly we can find subsequent terms.

Hence Maclaurin series for `tanx` centered at `0` is

`x+x^3/3+(2x^5)/15+(17x^5)/315+.....`