why start the taylor series of $\cos^{2} x$ at $k=1$ and not just $k=0$ as I do not understand the problem with $2^{-1}$

Question

Im using $\cos^2 x=\frac{1}{2}(1+\cos(2x))$ and $\cos x = (-1)^k \frac{(2x)^{2k}}{(2k)!}$ to find the sum for the Taylor series of $\cos^2 x$. I thought I was getting it. When I find the answer $$\cos^{2} x =\frac{1}{2} + \sum_{k=0}^\infty (-1)^k \frac{2^{(2k-1)}x^{2k}}{(2k)!}$$ however, my textbook does one more step and starts at $k=1$ instead of $k=0$. Which they make $$\cos^{2} x = 1+\sum_{k=1}^\infty (-1)^k \frac{2^{2k-1}x^{2k}}{{2k}!}$$I understand that they do this to prevent $2^{-1}$ at $k=0$, but i dont understand what would be the problem of $2^{-1}$ as negative powers work fine. I dont understand why this last step is made. Thanks in advance.

Answer 1

Now some mathematical corrections and answers.

$$\cos (x) = \sum_{k=0}^\infty(-1)^k (x)^{2k}/(2k)!,$$

$$\cos (2x) = \sum_{k=0}^\infty(-1)^k (2x)^{2k}/(2k)!.$$

Hence

\begin{align}\cos^2 (x) &=\frac{1}{2}+\frac{1}{2}\cos (2x)=\frac{1}{2}+\frac{1}{2}\sum_{k=0}^\infty(-1)^k (2x)^{2k}/(2k)!=\\&=\frac{1}{2}+\sum_{k=0}^\infty(-1)^k 2^{2k-1}x^{2k}/(2k)!=\\&=1+\sum_{k=1}^\infty(-1)^k 2^{2k-1}x^{2k}/(2k)!. \end{align}

In the last step, in order to start the summation at $k=1$ I just evaluated the expression for $k=0$ (and got $1/2$).