Question

How do you find the Maclaurin Series for `f(x)=sin(x^4)`?

Answer 1

`x^4-x^12/(3!)+x^20/(5!)-...+(-1)^(n-1)((x^4)^(2n+1))/((2n+1)!)+...`

Explanation:

Let `u=x^4` Then, fx)=g(u)=sin u#

Maclaurin series for `g(u)=u-u^3/(3!)+u^5/(5!)-..=f(x`)

Reverting to function of x,

`f(x)=x^4-x^12/(3!)+x^20/(5!)-...+(-1)^(n-1)(x^4)^(2n=1)/((2n+1)!)+...`

Maclaurin series is valid for any `x in (-oo, oo)`. Here, the mapping is `xtox^4`, for a subset of `(-oo. oo)` ...