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
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2=\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
 A: Instead of working forward, work backward:
Calculate the $z^2$ term of the result, then the $z^3$ term, and so on.

Clearly the coefficients of the $z^0$ and $z^1$ terms are zero.  The $z^2$ term must be the result of squaring the $\frac z{2!}$ term, so the result is $\frac{z^2}{2!^2}$.
Now let's consider the $z^3$ term.  To obtain one  we'd need to multiply a $z^0$ term by a $z^3$ term, or a $z^1$ term by a $z^2$ term.  But there is no $z^0$ or $z^2$ term, so there is no $z^3$ term in the result.
The $z^4$ term is more interesting.  To obtain a $z^4$ term, we'd need to multiply a $z^0$ term by a $z^4$ term,  a $z^1$ term by a $z^3$ term, or a $z^2$ term by a $z^2$ term.  Only the second of these works, and we can take the $z^1$ term either from the left factor (getting $\frac z {2!}\cdot -\frac{z^3}{4!}$) or the right factor  (getting $ -\frac{z^3}{4!}\cdot \frac z {2!}$).  The $z^4$ term of the result is the sum of these two products, $$r_4 = -2\frac{z^4}{2!4!}.$$
Similarly, the $z^6$ term is obtained as $$\frac z{2!}\frac{z^5}{6!} + \left(-\frac{z^3}{4!}\right) \left(-\frac{z^3}{4!}\right) +\frac{z^5}{6!} \frac z{2!} \\
= \left(\frac{2}{2!6!}+\frac1{4!4!}\right)z^6 = \ldots
$$

In general, for any product of infinite (or finite!) power series, one has $$\left(\sum_{i=0}^\infty a_iz^i\right) \left(\sum_{i=0}^\infty b_iz^i\right) = \sum_{i=0}^\infty \left(\sum_{j=0}^i a_jb_{i-j}\right)z^i$$
where the important thing to notice is that the inner-sum on the right-hand side is finite and is a constant; it is the coefficient of the $z^i$ term in the product.
A: 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.
A: Your series in parentheses is
$$\frac{1-\cos z} {z}. $$
Expand the binomial. The series for $\cos^2z$ is a bit complicated, but can be written using binomial coefficients. 
A: Multiply each term inside the parenthesis by $z$. You'll find it is simply:
$$\frac{(1-\cos z)^2}{z^2}.$$
