# Question #13a86

##### 1 Answer

Mar 13, 2017

#### Explanation:

The typical Maclaurin series (a Taylor series about

#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)=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