How do you find the Maclaurin Series for #(cos(2x^(2))+1) / x^(2)#?

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Answer 1 · 2016-10-25T23:21:45

# (cos(2x^2)+1)/x^2 = 2/x^2-2x^2 +2/3x^6 -4/45x^10 + 2/315x^14 - ... #

There are several ways to do this; the "hardest" way is to use first principles and find the various derivatives, but there are other ways ...

Let s start by using the well known series for #cosx#:
# cosx=1-x^2/(2!) +x^4/(4!) -x^6/(6!) + x^8/(8!) - ... , x in RR#

So we can now just write down the series for # (cos(2x^2)+1)/x^2 #

# :. (cos(2x^2)+1)/x^2 = ( {1-(2x^2)^2/(2!) +(2x^2)^4/(4!) -(2x^2)^6/(6!) + (2x^2)^8/(8!) - ...} + 1 ) / x^2#

# :. (cos(2x^2)+1)/x^2 = {2-(4x^4)/(2) +(16x^8)/(24) -64x^12/(720) + 256x^16/(40320) - ... ) / x^2#

# :. (cos(2x^2)+1)/x^2 = {2-2x^4 +2/3x^8 -4/45x^12 + 2/315x^16 - ... ) / x^2#

# :. (cos(2x^2)+1)/x^2 = 2/x^2-2x^2 +2/3x^6 -4/45x^10 + 2/315x^14 - ... #