<s>[INST] How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First term will contribute:
$$\left(\frac{z^2}{2!2!}-\frac{2z^4}{2!4!}+\frac{2z^6}{2!6!}-\cdots\right)$$
$$\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
Second term will contribute:
$$\left(-\frac{z^4}{4!}+\cdots\right)$$
This already seems wrong, so expanding via the first term seems good, and I hazard a guess that
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2=\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
 [/INST] Here is another way to square the series without multiplying out the terms.
Since $\cos z = 1 - \dfrac{z^2}{2!} + \dfrac{z^4}{4!} - \dfrac{z^6}{6!} + \cdots$, we have $\dfrac{1-\cos z}{z} = \dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots$. 
Thus, $\left(\dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots\right)^2 = \dfrac{(1-\cos z)^2}{z^2}$. 
Now, note that $(1-\cos z)^2 = 1-2\cos z + \cos^2 z = 1-2\cos z + \dfrac{1+\cos 2z}{2} = \dfrac{3}{2} - 2\cos z + \dfrac{1}{2}\cos 2z$. 
The Taylor series for $(1-\cos z)^2 = \dfrac{3}{2} - 2\cos z + \dfrac{1}{2}\cos 2z$ is given by: 
$\dfrac{3}{2} - 2\displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^nz^{2n}}{(2n)!} + \dfrac{1}{2}\sum_{n = 0}^{\infty}\dfrac{(-1)^n(2z)^{2n}}{(2n)!}$ $= \displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^n(-2+2^{2n-1})z^{2n}}{(2n)!}$. 
Now, divide by $z^2$ to get the series for $\dfrac{(1-\cos z)^2}{z^2}$.
EDIT: If you only need the first few terms of the series, then carefully multiplying out the first few terms of $\left(\dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots\right)\left(\dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots\right)$ will be easier. However, if you need all the terms of the series, the above method will be easier.
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