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\section{Approximation by polynomials}

Taylors Theorem is not the entire answer. 2
examples:

\begin{lemma*}
  If $E : \RR \to \RR$ is given by
  \[ E(x) = \begin{cases}
    \exp \left( -\frac{1}{x^2} \right) & x \neq 0
    \\
    0 & x = 0
  \end{cases} \]
  then $E$ is infinitely differentiable
  everywhere, with $E^{(r)}(0) = 0$ for all $r$.
  So the Taylor expansion about $0$ is
  \[ \sum_r 0 x^r = 0 \not\to E(r) \]
  for $x \neq 0$.
\end{lemma*}

\begin{proof}
  If $x \neq 0$ then standard differentiation
  theorems give that $E$ is infinitely
  differentiable with $E^{(r)}(x) = Q_r(1 / x)
  E(x)$ where $Q_r$ is a polynomial (proved by
  induction). At $0$, $E(0)$. Suppose $E^{(r)}$
  exists with $E^{(r)}(0) = 0$. Then
  \[ \frac{E^{(r)}(t) - E^{(r)}(0)}{t} =
  \frac{1}{t} Q_r \left( \frac{1}{t} \right) E(t)
  \to 0 .\]
  (Exponential beats polynomials). So $E^{(r +
  1)}(0)$ exists and equals $0$.
\end{proof}

The second problem is practical. It is true that
\[ \exp x = \sum_{r = 0}^\infty \frac{x^r}{r!}
\qquad \forall x \]
but computing $\exp(-20)$ by using
\[ \exp(-20) \approx \sum_{r = 0}^N
\frac{(-20)^r}{r!} \]
involves large $N$ and ridiculous cancelation.

One attempt to get polynomial approximation is to
use interpolation:

\begin{lemma*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\mathcal{P}_n$ the collection of
      polynomials of degree at most $n$ is a
      vector space of dimension $n + 1$.
    \item If $f \in C[a, b]$, $x_0 < x_1 < \cdots
      < x_n$, then there is at most one polynomial
      $P \in \mathcal{P}_n$ such that $f(x_j) =
      P(x_j)$.
    \item If $e_j(x) = \prod_{x \neq j} \frac{x -
      x_j}{x_i - x_j}$ then these form a basis of
      $\mathcal{P}_n$, and writing
      \[ P(x) = \sum_j f(x_j) e_j(x) \]
      we have $P \in \mathcal{P}_n$, $P(x_j) =
      f(x_j)$, so in fact by (ii) there is exactly
      one polynomial.
  \end{enumerate}
\end{lemma*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $1, t, t^2, \ldots, t^n$ is a basis for
      $\mathcal{P}_n$.
    \item Suppose $f(x_j) = P(x_j) = Q(x_j)$, $[0
      \le j \le n]$, $P, Q \in \mathcal{P}_n$.
      Then $P - Q \in \mathcal{P}_n$ and has $n +
      1$ zeroes (given by $x_0, \ldots, x_n$), so
      $P - Q = 0$.
    \item $e_j(x_i) = \delta_{ij}$, $e_j \in
      \mathcal{P}_n$. So if $P(x) = \sum_j f(x_j)
      e_j(x)$, then $P(x_i) = f(x_i)$ for $0 \le i
      \le n$.
  \end{enumerate}
\end{proof}

Practice shows interpolation to be an unreliable
friend.

Chebychev introduced an interesting polynomial
$T_n$ which shows how oddly polynomials can behave
(borhter noticed a companion $U_n$ Chebychev
polynomial of the second kind).

Chebychev says look at:
\begin{align*}
  (\cos n\theta + i \sin n\theta)
  &= (\cos \theta + i \sin\theta)^n \\
  &= \sum_r {n \choose r} (\cos\theta)^{n - r} i^r
  (\sin\theta)^r \\
  &= \sum_r (-1)^r {n \choose 2r} (\cos\theta)^{n
  - 2r} (\sin\theta)^{2r} \\
  &~~~~+ i \sum_r {n \choose 2r
  + 1} (\cos\theta)^{n - 2r - 1} (\sin\theta)^{2r
  + 1} \\
  &= \sum_r (-1)^e {n \choose 2r} (\cos\theta)^{n
  - 2r} (1 - \cos^2 \theta)^r \\
  &~~~~+ i \sum_r (-1)^r {n
  \choose 2r + 1} (\cos\theta)^{n - 2r - 1}
  (\sin\theta) (1 - \cos^2 \theta)^r
\end{align*}
so taking real and imaginary parts we get
\[ \cos n\theta = T_n(\cos\theta), \qquad \sin
n\theta = (\sin\theta) U_n(\cos\theta) \]
with
\begin{align*}
  T_n(t)
  &= \sum_r (-1)^r {n \choose 2r} t^{n - 2r} (1 -
  t^2)^r \\
  U_n(t)
  &= \sum_r (-1)^r {n \choose 2r + 1} t^{2n
  - 2r - 1} (1 - t^2)^r
\end{align*}
$T_n \in \mathcal{P}_n$, $|T_n(t)| \le 1$ for all
$t \in [-1, 1]$. Leading coefficient of $T_n$ is
\[ \sum_{0 \le 2r \le n} {n \choose 2r} (-1)^r =
\half ((1 + 1)^n - (1 - 1)^n) = 2^{n - 1} \]
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/e23a654cc0084ff3.png}
\end{center}
zeroes of $T_n$ are $\cos^{-1}(r\pi / n)$
($T_n(\cos\theta) = \cos n\theta$).

$\sin n\theta = U_n(\cos\theta) \sin\theta$, so
$U_n(\cos\theta) = \frac{\sin n\theta}{\sin
\theta}$. (if $\sin \theta = 0$, use continuity to
get $U_n = \pm n$ at this point).
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/fcfdd52425d8479a.png}
\end{center}

In spite of all this, Weierstrass showed:

\begin{flashcard}[weierstrass-thm]
\begin{theorem*}[Weierstrass approximation]
  \cloze{The polynomials are uniformly dense in $C[a,
  b]$, i.e. given $f : [a, b] \to \RR$ continuous
  and $\eps > 0$, there exists polynomial $P$ such
  that $|f(t) - P(t)| \le \eps$ for all $t \in [a,
  b]$.}
\end{theorem*}
\end{flashcard}

Bernstein:

\begin{flashcard}[bernstein-thm]
\begin{theorem*}[Bernstein]
  \cloze{If $f : [0, 1] \to \RR$ then
  \[ \sum_{r = 0}^n f \left( \frac{r}{n} \right)
  {n \choose r} t^r (t - t)^{n - r} \to f(t) \]
  uniformly on $[0, 1]$ as $n \to \infty$. (Can
  get from $[0, 1]$ to $[a, b]$ by a
  scaling transformation).}
\end{theorem*}
\end{flashcard}

One proof depends on reinterpreting the theorem
probabilistically. Let $X_1, X_2$ be independent
such that $\PP(X_j = 1) = p$, $\PP(X_j = 0) = 1 -
p$. Then
\[ \sum_{r = 0}^n f \left( \frac{r}{n} \right) {n
\choose r} p^r (1 - p)^{n - r} = \EE f(\ol{X}) \]
where $\ol{X} = \frac{X_1 + \cdots + X_n}{n}$.