What is the taylor series of #cos^2(x)#?

1 Answer
George C.
Mar 5, 2018

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

Explanation:

I will assume you mean the Taylor series about #0#, otherwise known as the Maclaurin series.

The Taylor series for #cos x# about #0# is:

#sum_(n=0)^oo (-1)^n x^(2n)/((2n)!) = 1/(0!) - x^2/(2!)+x^4/(4!)-x^6/(6!)+...#

Rather than trying to multiply this series by itself to get the series for #cos^2 x# note that:

#cos 2x = 2cos^2 x - 1#

and hence:

#cos^2 x = 1/2(1+cos 2x)#

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

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

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

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

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