How do you find the maclaurin series expansion of #sin(x^2)/x#?

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Answer 1 · 2015-11-09T10:55:39

#sum_{n=0}^infty (-1^n) x^(4n+1) / ((2n+1)!)#

Substitute #y=x^2#. Now we know the expansion of #sin(y)#, which is

#y-y^3/(3!)+y^5/(5!) - y^7/(7!)...=sum_{n=0}^infty (-1^n) y^{2n+1} / ((2n+1)!)#

Now substitute back #y=x^2#:

#x^2-x^6/(3!)+x^10/(5!) - x^14/(7!)...=sum_{n=0}^infty (-1^n) x^{2(2n+1)} / ((2n+1)!)#

#=sum_{n=0}^infty (-1^n) x^(4n+2) / ((2n+1)!)#

Now divide everything by #x#:

#x-x^5/(3!)+x^9/(5!) - x^13/(7!)...=sum_{n=0}^infty (-1^n) x^(4n+1) / ((2n+1)!)#