Question

Question #13a86

Answer 1

`-1/6`

Explanation:

The typical Maclaurin series (a Taylor series about `x=0`) for `sin(x)` is:

`sin(x)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(2n+1)=x-x^3/(3!)+x^5/(5!)-x^7/(7!)+...`

So the same series for `sin(x^2)` can be found by replacing `x` with `x^2` in the series:

`sin(x^2)=sum_(n=0)^oo(-1)^n/((2n+1)!)(x^2)^(2n+1)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(4n+2)`

Writing out the first terms of this series gives:

`sin(x^2)=x^2-x^6/(3!)+x^10/(5!)-x^14/(7!)+...`

So the coefficient of the `x^6` term is `-1//3!` or `-1//6`.