Function Approximations

Taylor Series

\(f\in C^n(\mathbb R)\), then \(P_{n,a}(x):= \sum_{i=0}^n \frac{f^{(i)}(a)(x-a)^i}{i!}\) is the Taylor polynomial around \(a\)

Def'n. Taylor Form remainder

\(f\in C^n(\mathbb R)\) and \(f^{(n+1)}\) exists, then \(f(x) = P_{n,a}(x) + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}, c\in [a,x]\)

Thrm 1. Uniqueness

Claim. Taylor polynomial is unique

proof. Suppose \(f(x) = \sum a_k(x-a)^k + \epsilon_1(x-a)^n = \sum b_k(x-a)^k + \epsilon_2(x-a)\),
then \(\sum(a_k-b_k)(x-a)^k + (\epsilon_1-\epsilon_2)(x-a)^n = 0\),
by independence of the system of equations, \(a_n = b_n, \epsilon_1 = \epsilon_2\)

Claim. For \(x,a\in [-R,R], P_{a,n}\rightarrow^{u.c.}f\)

proof.

\[\begin{align*} \|f(x)-P_{a,n}(x)\| &\leq \frac{|f^{(n+1)(c)}|}{(n+1)!}|x-a|^{n+1}\\ &\leq \frac{|f^{(n+1)(c)}|}{(n+1)!}|2R|^{n+1} \\ &\leq \frac{M}{(n+1)!}(2R)^{n+1} &\text{by EVT}\\ &\leq \frac{M}{n^ne^{-n}\sqrt{2\pi n}}(2R)^{n+1} &\text{Stirling's formula}\\ &\leq M(\frac{2Re}{n+1})^{n+1}\rightarrow 0 \text{ for large }n \end{align*}\]

Thrm 2.Weierstrass Theorem

For any \(a<b, f\in C[a,b]\), exists \(p_n\rightarrow^{u.c.}f\)

proof. Define \(g(x):[0,1]\rightarrow \mathbb R := f(a+x(b-a))\).
Take \(q_n\rightarrow^{u.c.}g\), consider \(y=a+x(b-a)\), i.e. \(x = \frac{y-a}{b-a}\)
Define \(p_n(y):=q_n(\frac{y-a}{b-a}) = q_n(x)\rightarrow^{u.c.}f(y)\)

Thrm 2.

If \(f\in C^1[0,1]\), then \(\exists \{p_n\}\) over \([0,1]\), \(p_n\rightarrow^{u.c}f\land p_n'\rightarrow^{u.c}f'\)

proof. Let \(f\in C^1[0,1]\), then \(f'\in C[0,1]\), take \(q_n\rightarrow^{u.c}f'\),
then let \(p_n = \int_0^x q_n(x)dx + f(0)\),

\[\begin{align*} p_n &= \int_0^x q_n(x)dx + f(0) + f(0) \\ &\rightarrow^{u.c.}\int_0^x f'(x)dx +f(0)&\text{ICT} \\ &=f(x)-f(0) + f(0)&\text{FTC II} \\ &=f(x) \end{align*}\]

Thrm 3.

If \(f\in C[0,1]\) and \(f(0)=0\), then you can find \(p_n\rightarrow^{u.c.}f\) s.t. \(p_n(0)=0, p'_n(0)=0\)

lemma 1 If \(f\in C^1[-1,1]\) is even, then exists \(\{p_n\}\) is even and \(p_n\rightarrow^{u.c.}f\)

proof. Take \(q_n \rightarrow^{u.c.}f\) over \([-1,1]\) by Weierstrass Theorem.
Define \(p_n(x):= \frac{q_n(x)+q_n(-x)}{2}\), and by \(f\) even, known \(f(x)=\frac{f(x)+f(-x)}{2}\)

\[\begin{align*} \|p_n-f\|_\infty &= \sup_{x\in[-1,1]}|\frac{q_n(x)+q_n(-x)}{2} - \frac{q_n(x)+q_n(-x)}{2}|\\ &= \frac{1}{2}\sup|q_n(x)-f(x)+(q_n(-x)-f(-x))| \\ &\leq \frac{1}{2}(\sup|q_n(x)-f(x))| + \sup|q_n(x)-f(x))|) &\text{tri.ineq}\\ &\leq \frac{\epsilon + \epsilon}{2} = \epsilon \end{align*}\]

proof. extend \(f\) to \(F=\begin{cases}f(x) &[0,1]\\f(-x)&[-1,0]\end{cases}, F\) is even. by lemma 1, take even \(q_n\rightarrow F\).
Then, take \(p_n(x) := q_n(x)-q_n(0)\) and
(i) by \(f(0)=0, q_n\rightarrow F\Rightarrow q_n(0) < \epsilon/2\), \(\|p_n-f\| \leq \|q_n-F\|+ \epsilon/2 = \epsilon\)
(ii) \(p_n(0)=q_n(0)-q_n(0)=0\)
(iii) \(p_n\) is even, hence \(p_n(x)=\frac{p_n(x)+p_n(-x)}{2}\), \(p_n'(x)=\frac{p_n'(x)-p_n'(-x)}{2}\Rightarrow p'_n(0)=0\)

Thrm 4.

If \(f\in C(\mathbb R), p_n\rightarrow f\) over \(\mathbb R\), then \(f\) is a polynomial

proof. Since \(p_n\) uniform converge, \(p_n\) is also Cauchy. Take \(N\) large so that \(\forall n,m \geq N. \|p_m-p_n\|_\infty \leq 1\)
Then \(f(x)=(f(x)-p_m(x))+(p_m(x)-p_n(x))+p_n(x)\).
Consider \(q_{n,m}(x) = p_m(x)-p_n(x)\), it will also be a polynomial, and since \(\|q_{n,m}\|\leq 1\), it must converges to some \(a_{n,m}\in[-1,1]\).
By BW Theorem, take subsequence \(a_{n,m_k}\rightarrow a_n\) for each \(n\).
\(f(x)=\lim_{m_k\rightarrow\infty}f(x) = \lim_k(f(x)-p_{m_k}(x)) + \|p_m - p_n\| + p_n(x) = 0 + a_n + p_n(x)\)