Question

Find the maclaurin series for the function. (use the table of power series for elementary functions.) f(x) = (cos(x 6))2

Answer 1

F'(x)=−2cos(x)sin(x) , f''(x)=2sin2(x)−2cos2(x)

f'''(x)=4sin(x)cos(x)+4cos(x)sin(x)=8cos(x)sin(x)

f''''(x)=−8sin2(x)+8cos2(x) , f'''''(x)=⋯=−32cos(x)sin(x) ,

f''''''(x)=32sin2(x)−32cos2(x) , etc...

Hence, f(0)=1 , f'(0)=0 , f''(0)=−2 , f'''(0)=0 , f''''(0)=8 , f'''''(0)=0 , f''''''(0)=−32 , etc...

Since 2!=2 , 84!=824=13 , and −326!=−32720=−245 , this much calculation leads to an answer of

1−x2+x43−245x6+⋯

This does happen to converge for all x and it does happen to equal cos2(x) for all x . You can also check on your own that the next non-zero term is +x8315

2) Use the well-known Maclaurin series cos(x)=1−x22!+x44!−x66!+⋯

=1−x22+x424−x6720+⋯ and multiply it by itself (square it).

To do this, first multiply the first term 1 by everything in the series to get

cos2(x)=(1−x22+x424−x6720+⋯)+⋯

Next, multiply −x22 by everything in the series to get

cos2(x)=(1−x22+x424−x6720+⋯)+(−x22+x44−x648+x81440+⋯)+⋯

Then multiply x424 by everything in the series to get

cos2(x)=(1−x22+x424−x6720+⋯)+(−x22+x44−x648+x81440+⋯)+(x424−x648+x8576−x1017280+⋯)+⋯

etc...

If you go out far enough and combine "like-terms", using the facts, for instance, that −12−12=−1 and 124+14+124=824=13 , etc..., you'll eventually come to the same answer as above:

1−x2+x43−245x6+⋯