Infinite number of projectors I am studying optics and I got to te part of the matrix treatment of polarization. Here I'll be talking only about linear polarization, hence it just a mere treatment of projectors in two dimensions. A linear polarizer is described by a 2x2 matrix like this:
$$
P_\theta=
  \begin{bmatrix}
    \cos^2 \theta & \sin\theta\cos\theta \\
    \sin\theta\cos\theta & \sin^2 \theta 
  \end{bmatrix}
$$
This is a matrix with null determinant, indicating its invertibility and hence the loss of information when polarized light goes through it. The curious thing about this is that if I have a vertical polarizer followed by a horizontal polarizer, then no light goes through. But if I place a third polarizer in the middle with its transmission axis at 45º then some light passes. It was staggering at first, but it actually makes sense. And then I got to thinking: What if I had an infinite number of projector one followed by the other, where its transmission axis differed only by an infinitesimal amount? Intuitively this amounts to a rotator matrix (rotates the polarization axis by $\beta$
$$
R_\beta=
  \begin{bmatrix}
    \cos \beta & -\sin\beta \\
    \sin\beta & \sin \beta 
  \end{bmatrix}
$$
So I tried to prove it:
Suppose I have an initial angle $\theta_0$ and a final angle $\theta_N$ such that $\theta_N-\theta_0=\beta$. Then I should be able to get the following relation:
$$
\lim_{N\to \infty}\prod_{i=0}^{N}P_{\theta_i}=R_\beta
$$
I tried starting it but I am very confused about the algebras because I think it will very quickly become intractable. Does anyone know how to prove this, if it possible to prove, or if there is an easier way to prove something of the sort?
 A: So, to make a long story short, we have the following problem:

Calculate $\lim_{\Delta \theta \to 0}\prod_{i=0}^{N}P_{\theta_i}$, where $\theta_0 = 0$, $\theta_N = \beta$, and $|\theta_{i+1} - \theta_i| \leq \Delta \theta$.

Interestingly, we want to take the limit of a "Riemann product" to produce a matrix-valued product-integral.
We first calculate the product $P_{\Delta \theta}P_0$:
$$
\begin{align}
P_{\Delta \theta}P_0 &= 
\pmatrix{\cos^2 \Delta \theta & \sin\Delta\theta\cos\Delta \theta \\
    \sin\Delta\theta\cos\Delta\theta & \sin^2 \Delta\theta}
\pmatrix{1&0\\0&0} =
\pmatrix{\cos^2 \Delta \theta & 0 \\
    \sin\Delta\theta\cos\Delta\theta & 0}
\\ & = 
\cos \Delta \theta \pmatrix{\cos \Delta \theta & 0\\ \sin \Delta \theta &  0} = \cos (\Delta \theta)\, R_{\Delta \theta}P_0. %= \cos (\Delta \theta)\, P_{\Delta \theta}R_{\Delta \theta}
\end{align}
$$
Now, note that $P_{\theta + \Delta \theta}R_\theta = R_{\Delta \theta} P_{\theta}.$ Using this, we can conclude (by induction, for instance) that
$$
\prod_{i=0}^N P_{\theta_i} = \left( \prod_{i=0}^{N-1} \cos(\theta_{i+1} - \theta_i)\right)R_{\beta}P_0.
$$
Now, it suffices to show that $\lim_{\Delta \theta \to 0}\prod_{i=0}^{N-1} \cos(\theta_{i+1} - \theta_i) = 1$.  Equivalently, we want to show that $\lim_{\Delta \theta \to 0}\sum_{i=0}^{N-1} \log(\cos(\theta_{i+1} - \theta_i)) = 0$.  We note that $\log(\cos(x))$ is analytic about $x=0$ with Maclaurin series $\log(\cos(x)) = - \frac 12 x^2 + O(x^4)$.  
If $\delta>0$ is sufficiently small, we can guarantee that $|x|< \delta$ implies that $0>\log(\cos(x)) > -x^2$. We find that for $\Delta \theta < \delta$,
$$
\begin{align}
0\geq \sum_{i=0}^{N-1} \log(\cos(\theta_{i+1} - \theta_i)) &\geq 
-\sum_{i=0}^{N-1} (\theta_{i+1} - \theta_i)^2
\\ & \geq 
- \Delta \theta \sum_{i=0}^{N-1} (\theta_{i+1} - \theta_i) 
= - \Delta \theta \beta > -\delta \beta.
\end{align}
$$
With that, we can indeed conclude that the sum approaches $0$ as $\Delta \theta \to 0$.
