How do you find the Maclaurin series for #f(x)= x cos (1/2x^2)#?

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Answer 1 · 2017-01-04T13:07:38

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

We can start from the MacLaurin series for #cost#:

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

substitute #t=1/2x^2#

#cos(1/2x^2) = sum_(n=0)^oo (-1)^n (1/2x^2)^(2n)/((2n)!)=sum_(n=0)^oo (-1)^n 1/2^(2n)x^(4n)/((2n)!)#

and finally multiply by #x#:

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