Find Maclaurin series for f(x)= 1-cos(x^9)/x^5?

Question

Need help finding the Maclaurin series for the function f(x)= #(1-cos(x^9))/x^5#

Very lost with how to begin. Still trying to understand the basics.

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Answer 1 · 2017-12-12T01:10:09

#(1-cos(x^9))/x^5 = sum_(n=1)^oo (-1)^(n+1) x^(18n-5)/((2n)!)#

Start from the MacLaurin series of #cost#:

#cost = sum_(n=0)^oo (-1)^n t^(2n)/((2n)!)#

substituting: #t= x^9# we get:

#cos(x^9) = sum_(n=0)^oo (-1)^n (x^9)^(2n)/((2n)!) = sum_(n=0)^oo (-1)^n x^(18n)/((2n)!) #

extract from the sum the term for #n=0#:

#cos(x^9)= 1+ sum_(n=1)^oo (-1)^n x^(18n)/((2n)!) #

so that:

#1-cos(x^9)= 1-1- sum_(n=1)^oo (-1)^n x^(18n)/((2n)!) = sum_(n=1)^oo (-1)^(n+1) x^(18n)/((2n)!)#

and finally divide by #x^5# term by term:

#(1-cos(x^9))/x^5 = sum_(n=1)^oo (-1)^(n+1) x^(18n)/(x^5(2n)!) = sum_(n=1)^oo (-1)^(n+1) x^(18n-5)/((2n)!)#