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

Question

Emilia Aguilar · Accepted Answer

#1-x^2+x^4/3-2/45 x^6+x^8/315+\cdots#

There are two approaches.

1) Let #f(x)=cos^2(x)# and use #f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots#

According to the Product Rule and/or Chain Rule, we have

#f'(x)=-2cos(x)sin(x)#, #f''(x)=2sin^2(x)-2cos^2(x)#

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

#f''''(x)=-8sin^2(x)+8cos^2(x)#, #f'''''(x)=\cdots=-32cos(x)sin(x)#,

#f''''''(x)=32sin^2(x)-32cos^2(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#, #8/(4!)=8/24=1/3#, and #(-32)/(6!)=(-32)/720=-2/45#, this much calculation leads to an answer of

#1-x^2+x^4/3-2/45 x^6+\cdots#

This does happen to converge for all #x# and it does happen to equal #cos^2(x)# for all #x#. You can also check on your own that the next non-zero term is #+x^8/315#

2) Use the well-known Maclaurin series #cos(x)=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+\cdots#

#=1-x^2/2+x^4/24-x^6/720+\cdots# and multiply it by itself (square it).

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

#cos^2(x)=(1-x^2/2+x^4/24-x^6/720+\cdots)+\cdots#

Next, multiply #-x^2/2# by everything in the series to get

#cos^2(x)=(1-x^2/2+x^4/24-x^6/720+\cdots)+(-x^2/2+x^4/4-x^6/48+x^8/1440+\cdots)+\cdots#

Then multiply #x^4/24# by everything in the series to get

#cos^2(x)=(1-x^2/2+x^4/24-x^6/720+\cdots)+(-x^2/2+x^4/4-x^6/48+x^8/1440+\cdots)+(x^4/24-x^6/48+x^8/576-x^10/17280+\cdots)+\cdots#

etc...

If you go out far enough and combine "like-terms", using the facts, for instance, that #-1/2-1/2=-1# and #1/24+1/4+1/24=8/24=1/3#, etc..., you'll eventually come to the same answer as above:

#1-x^2+x^4/3-2/45 x^6+\cdots#

Ezra Adams · Answer

undefined To find the Maclaurin series for $ \cos^2(x) $, you can start with the Maclaurin series for $ \cos(x) $, which is $ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} $.

Now, to find $ \cos^2(x) $, you simply square each term of the series for $ \cos(x) $. Thus:

$$
\begin{split}
\cos^2(x) & = \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} \right) \cdot \left( \sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!}x^{2m} \right) \
& = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} \cdot \frac{(-1)^m}{(2m)!}x^{2m} \
& = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^{n+m}}{(2n)! (2m)!}x^{2(n+m)} \
& = \sum_{k=0}^{\infty} \left( \sum_{n=0}^{k} \frac{(-1)^{n+k-n}}{(2n)! (2(k-n))!} \right) x^{2k}
\end{split}
$$

Therefore, the Maclaurin series for $ \cos^2(x) $ is:

$$
\sum_{k=0}^{\infty} \left( \sum_{n=0}^{k} \frac{(-1)^{n+k-n}}{(2n)! (2(k-n))!} \right) x^{2k}
$$