How does one come up with this construction of approximating polynomials of $|x|$?

Question

I am reading Mathematical Analysis: An Introduction by Andrew Browder. In section 7.1, the author gave a proof of the real version of Weierstrass's Approximation Theorem, which I think is conceptually much simpler than the conventional proof based on Bernstein polynomials.

Essentially, the author first reduced the theorem to the special case where the function to be approximated is $|x|$ on $[-1,1]$. Then he constructed a sequence of polynomials $\{q_n\}_{n\in\mathbb N}$ that decreases to $|x|$ pointwise, so that the convergence is uniform, by Dini's Theorem.

The sequence is constructed recursively by $q_0=1$ and $q_{n+1}=\frac12(x^2+2q_n-q_n^2)$. It follows from mathematical induction that $1\geq q_n\geq q_{n+1}\geq|x|$ because

• $q_n-q_{n+1}=\frac12(q_n^2-x^2)\geq0$ and
• $q_{n+1}-|x|=\frac12[(1-|x|)^2-(1-q_n)^2]\geq0$.

Since $\{q_n\}$ is a decreasing sequence that is bounded below by $|x|$, it converges to some function $f$. Hence $f$ is a nonnegative solution to the equation $f=\frac12(x^2+2f-f^2)$. Therefore $f(x)=|x|$.

I like this proof. However, I don't understand how the author came up with the construction of $\{q_n\}$ above. That is, given a polynomial $p$ that satisfies $1\geq p\geq|x|$ on some interval, how can one envision that the inequality $1\geq p\geq q\geq|x|$ is satisfied by the polynomial $q=\frac12(x^2+2p-p^2)$?

(I have seen another construction in another question, but I don't see any apparent relation between the two constructions.)

Answer 1

The Core Idea Is To Approximate $\sqrt{t}$ on $[0, 1]$

The trick is realizing that $|x| = \sqrt{x^2}$. If we can find a sequence of polynomials $p_n(t)$ that converges nicely to $\sqrt{t}$ for $t \in [0, 1]$, then $q_n(x) = p_n(x^2)$ will be polynomials in $x$ converging to $\sqrt{x^2} = |x|$ on $[-1, 1]$.

How to Find $\sqrt{t}$ Iteratively? We Can Do So by Using Newton's Method (Kinda)

We want to solve $y = \sqrt{t}$, or $y^2 - t = 0$. Newton's method for finding a root of $f(y) = y^2 - t = 0$ gives the iteration: $y_{k+1} = y_k - \frac{f(y_k)}{f'(y_k)} = y_k - \frac{y_k^2 - t}{2y_k} = \frac{1}{2} \left( y_k + \frac{t}{y_k} \right)$

Problem: If $p_k(t)$ is our polynomial approximation at step $k$, the next step $p_{k+1}(t) = \frac{1}{2} (p_k(t) + t/p_k(t))$ involves division by $p_k(t)$, so it's generally a rational function, not a polynomial. We need polynomials.

So we will modify it to get a polynomial out of it:

The problematic term is the $2y_k$ in the denominator of Newton's update $\frac{y_k^2 - t}{2y_k}$. We need an iteration that only uses addition, subtraction, and multiplication.

Let's consider the update step $y_{k+1} = y_k - \text{correction}$. The correction term is $\frac{y_k^2 - t}{2y_k}$. What if we replace the $2y_k$ in the denominator with a constant? A simple, convenient constant, especially if our approximations stay near $1$, is $2$.

Let's define a new iteration using this modified correction term: $p_{k+1} = p_k - \frac{p_k^2 - t}{2}$

Now, simplify this expression: $p_{k+1} = p_k - \frac{1}{2}p_k^2 + \frac{t}{2}$ $p_{k+1} = \frac{1}{2} (t + 2p_k - p_k^2)$

If $p_k(t)$ is a polynomial in $t$, then $p_{k+1}(t)$ is clearly also a polynomial in $t$. This looks like the formula we want.

Let $p(t)$ be our current polynomial approximation, satisfying $1 \ge p(t) \ge \sqrt{t}$ on $[0, 1]$. Let the next approximation be $q(t) = \frac{1}{2}(t + 2p(t) - p(t)^2)$. We need to show $p(t) \ge q(t)$ and $q(t) \ge \sqrt{t}$.

1. Show $p(t) \ge q(t)$: $p(t) - q(t) = p(t) - \frac{1}{2}(t + 2p(t) - p(t)^2)$ $= \frac{1}{2} (2p(t) - t - 2p(t) + p(t)^2)$ $= \frac{1}{2} (p(t)^2 - t)$ Since we assumed $p(t) \ge \sqrt{t} \ge 0$, we have $p(t)^2 \ge (\sqrt{t})^2 = t$. So, $p(t)^2 - t \ge 0$, which means $p(t) - q(t) \ge 0$. Thus $p(t) \ge q(t)$.

2. Show $q(t) \ge \sqrt{t}$: $q(t) - \sqrt{t} = \frac{1}{2}(t + 2p(t) - p(t)^2) - \sqrt{t}$ $= \frac{1}{2} (t - 2\sqrt{t} + 2p(t) - p(t)^2)$ Now, let's rearrange cleverly by completing squares involving 1: $= \frac{1}{2} [ (t - 2\sqrt{t} + 1) - (p(t)^2 - 2p(t) + 1) ]$ $= \frac{1}{2} [ (\sqrt{t} - 1)^2 - (p(t) - 1)^2 ]$ $= \frac{1}{2} [ (1 - \sqrt{t})^2 - (1 - p(t))^2 ]$ Since we assumed $1 \ge p(t) \ge \sqrt{t}$, we have $0 \le 1 - p(t) \le 1 - \sqrt{t}$. Because these are non-negative quantities, squaring preserves the inequality: $(1 - p(t))^2 \le (1 - \sqrt{t})^2$. Therefore, $(1 - \sqrt{t})^2 - (1 - p(t))^2 \ge 0$. This means $q(t) - \sqrt{t} \ge 0$. Thus $q(t) \ge \sqrt{t}$.

We found an iteration $p_{k+1}(t) = \frac{1}{2}(t + 2p_k(t) - p_k(t)^2)$ that takes a polynomial $p_k(t)$ satisfying $1 \ge p_k(t) \ge \sqrt{t}$ and produces a new polynomial $p_{k+1}(t)$ satisfying $1 \ge p_k(t) \ge p_{k+1}(t) \ge \sqrt{t}$ (we also need $p_{k+1} \le 1$, which is easily checked: $1-p_{k+1} = \frac{1}{2}((p_k-1)^2 + (1-t)) \ge 0$).

Starting with $p_0(t) = 1$, this generates a sequence of polynomials $p_n(t)$ decreasing pointwise to $\sqrt{t}$ on $[0, 1]$.

Now, substitute $t = x^2$ and let $q_n(x) = p_n(x^2)$. The iteration becomes: $q_{n+1}(x) = p_{n+1}(x^2) = \frac{1}{2}(x^2 + 2p_n(x^2) - p_n(x^2)^2)$ $q_{n+1}(x) = \frac{1}{2}(x^2 + 2q_n(x) - q_n(x)^2)$

And the property $1 \ge p_n(t) \ge p_{n+1}(t) \ge \sqrt{t}$ becomes: $1 \ge q_n(x) \ge q_{n+1}(x) \ge \sqrt{x^2} = |x|$ for $x \in [-1, 1]$.

Answer 2

This is an idea which may help.

The author's $q_{n+1}$ is $$q_{n+1}=\frac 12(x^2+2q_n-q_n^2)\tag1$$ I noticed that $(1)$ can be written as $$q_{n+1}=(1-k)q_n+k|x|\tag2$$ where $0\le k=\dfrac{q_n+|x|}{2}\le 1$.

(see internal divisions)

From this, I think it may be possible that the author started with the form $q_{n+1}=(1-m)q_n+m|x|$ and chose $m=k$ to get $(1)$.

Answer 3

It is hard to guess how Andrew Browder devised this sequence of approximations. Yet, when examined pointwise, the sequence’s monotonicity is more apparent.

Fix $a\in[-1,1]$. Define $f_a\colon[0,1]\rightarrow\mathbb{R}$ by $$ f_a(x)=\frac12\bigl(a^{2}+2x-x^{2}\bigr). $$

This is a quadratic map. Its critical point is $1$, and $f_a$ is monotone increasing on $[0,1]$.
The minimum value on $[0,1]$ is $a^{2}/2$, and the maximum is $(a^{2}+1)/2$. Hence $$ f_a([0,1])=\left[\frac{a^{2}}{2},\frac{a^{2}+1}{2}\right]\subset[0,1]. $$

Note that $|a|$ is a fixed point of $f_a$, that is, $f_a(|a|)=|a|$. In fact, it is the unique fixed point of $f_a$ on $[0,1]$ (the other fixed point is $-|a|$).

Because $f_a\colon[0,1]\rightarrow[0,1]$ is continuous and increasing, for any $x_0\in[0,1]$, if we define recursively $$ x_{n+1}=f_a(x_n), $$ then the sequence $x_0,x_1,x_2,\dots$ is monotone and converges to the unique fixed point $|a|$.

In particular, if $x_0=1$, then $f_a(1)\le 1$, so the sequence is non-increasing: $$ 1\ge x_1\ge x_2\ge\cdots. $$ and $\lim_n x_n=|a|.$

So if you define $p_0(a)=1$ and $$p_{n+1}(a)=f_a(p_n(a))$$

It follows that $0\leq |a|\leq p_{n+1}(a)\leq p_n(a)\leq 1$ and $\lim_n p_n(a)=|a|$.