Add a proof

_psets/3.md

@@ -217,13 +217,13 @@ So as long as the error rate isn't $$1/2$$ (i.e. zero information gets through),
 of majority voting the error rate can be made arbitrarily small.
 
 Let's prove this. Since $$f_0$$ is a polynomial, it's continuous. By inspection there are three
-solutions to $$x = 3 x^2 - 2 x^3$$: zero, one half, and one. Thus $$f_0$$ has three
-[fixed points](https://en.wikipedia.org/wiki/Fixed_point_(mathematics)). This suffices to show that
-for any $$x$$, if $$f_n(x)$$ converges it must converge to zero, one half, or one.
+[fixed points](https://en.wikipedia.org/wiki/Fixed_point_(mathematics)): zero, one half, and one.
+This suffices to show that for any $$x$$, if $$f_n(x)$$ converges it must converge to zero, one
+half, or one (I'll prove this below).
 
 Now fix any $$x$$ such that $$0 < x < 1/2$$. Because $$f_0$$ is a cubic polynomial, it can't cross
 the line $$y = x$$ more than three times. We've already noted that it does cross this line exactly
-three times (at zero, one half, and one). So the fact that
+three times, namely at its fixed points. So the fact that
 
 $$
 3 \cdot \left( \frac{1}{4} \right)^2 - 2 \cdot \left( \frac{1}{4} \right)^3
@@ -257,6 +257,23 @@ $$
 
 This symmetry establishes the claim.
 
+For completeness let's prove the claim that a sequence generated by recursive application of a
+continuous function $$f$$ can only converge to a fixed point of $$f$$. I'll do this in two steps.
+
+First, we must establish that if a sequence $$x_0, x_1, \ldots$$ in the domain of $$f$$ converges to
+a point $$x_\infty$$ (also in the domain of $$f$$), then the sequence $$f(x_0), f(x_1), \ldots$$
+converges to $$f(x_\infty)$$. Fix any $$\epsilon > 0$$. Since $$f$$ is continuous, there is some
+$$\delta$$ such that $$\lvert x - x_\infty \rvert < \delta$$ implies $$\lvert f(x) - f(x_\infty)
+\rvert < \epsilon$$. Since $$x_0, x_1, \ldots$$ converges to $$x_\infty$$, there is some $$N$$ such
+that $$n > N$$ implies $$\lvert x_n - x_\infty \rvert < \delta$$. Thus $$n > N$$ implies $$\lvert
+f(x_n) - f(x_\infty) \rvert < \epsilon$$ which establishes the claim.
+
+Second, we can show that any sequence $$x_0, x_1, \ldots$$ generated by successive application of
+$$f$$ that converges must converge to a fixed point of $$f$$. Say $$x_0, x_1, \ldots$$ converges to
+$$x_\infty$$. Since $$x_{n + 1} = f(x_n)$$ for all $$n$$, this means that $$f(x_0), f(x_1), \ldots$$
+converges to $$x_\infty$$. But by the result just proven, $$f(x_0), f(x_1), \ldots$$ converges to
+$$f(x_\infty)$$. Since limits are unique this means $$x_\infty = f(x_\infty)$$.
+
 ### Behavior of leading term
 
 For small $$\epsilon_0$$, the cubic term in $$f_0(\epsilon_0)$$ is approximately zero. So let's