1.6
Strong Taylor's Theorem
Let \(f : (a-\delta, a+\delta) \to \mathbb{R}\) be \(N+1\) times differentiable in \((a-\delta, a+\delta)\). Then
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\(f(x) - T_{N,a}(x) = \frac{f^{(N+1)}(c)}{N!} (x-a) (x-c)^N \quad \text{for some } c \text{ between } a \text{ and } x\)
(Error of using \(T_{N,a}\) instead of \(f\).)
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\(f(x) - T_{N,a}(x) = \frac{f^{(N+1)}(c)}{(N+1)!} (x-a)^{N+1} \quad \text{for some } c \text{ between } a \text{ and } x\)
Notation: \(E_{N,a}=R_{N,a}=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}\quad\)Remainder term
Proof
Let \(x\) fixed in \((a-\delta,a+\delta)\), then we have \(E_{N,t}\) where \(t\) is variable
Then \(\frac{\mathrm{d}E_{n,t}\left(x\right)}{\mathrm{d}t}=-\frac{f^{\left(N+1\right)}\left(t\right)}{N!}\left(x-t\right)^{N}\), let's use mean value theorem
\(\frac{E_{N,x}\left(x\right)-E_{N,a}\left(x\right)}{x-a}=\frac{\mathrm{d}E_{N,t}\left(x\right)}{\mathrm{d}t}|_{t=c}\Rightarrow(1)\)
GMVT: \(g(t)=(x-t)^{N+1}\)
Then \(\frac{E_{N,x}\left(x\right)-E_{N,a}\left(x\right)}{g\left(x\right)-g\left(a\right)}=\frac{\frac{\mathrm{d}E_{N,t}\left(x\right)}{\mathrm{d}t}|_{t=c}}{\frac{\mathrm{d}g\left(x\right)}{\mathrm{d}t}|_{t=c}}=\frac{-\frac{f^{\left(N+1\right)}\left(t\right)}{N!}\left(x-c\right)^{N}}{{\left(N+1\right)}\left(x-c\right)^{N}\left(-1\right)}=\frac{f^{\left(N+1\right)}\left(c\right)}{\left(N+1\right)!}\)
Then we also put
\(\frac{E_{N,x}\left(x\right)-E_{N,a}\left(x\right)}{g\left(x\right)-g\left(a\right)}=\frac{-E_{N,a}\left(x\right)}{-g\left(a\right)}=\frac{E_{N,a}\left(x\right)}{\left(x-a\right)^{N+1}}\)
Then we combine this two we get \((2)\)
Corollary
Let \(f: [\alpha, \beta] \to \mathbb{R}\), \(C^\infty\), infinitely many times differentiable.
If \(\exists M\) such that \(|f^{(n)}(x)| < M \quad \forall x \in [\alpha, \beta], \; \forall n \in \mathbb{N}\), then \(f(x) = \lim_{N \to \infty} T_{N, a}(x) \quad \forall x \in [\alpha, \beta], \; \forall a \in [\alpha, \beta]\)
Where \(f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x - a)^n\) (This is Taylor's Series.)
Proof
We must prove \(\lim_{N \to \infty} \left(f(x) - T_{N,a}(x)\right) = 0\) where \(f(x) - T_{N,a}(x) = E_{N,a}(x)\).
\(0 \leq \left|E_{N,a}(x)\right| = \left|\frac{f^{(N+1)}(c)}{(N+1)!} (x-a)^{N+1}\right| \leq \frac{M |x-a|^{N+1}}{(N+1)!}.\)(By hypothesis: \(|f^{(N+1)}(c)| \leq M\).)
Since \((N+1)!\) grows faster than \(|x-a|^{N+1}\)
Then \(\lim_{N \to \infty} E_{N,a}(x) = 0 \quad \forall x \in [\alpha, \beta], \; \forall a \in [\alpha, \beta]\)
Example
Let \(f(x) = \cos(x)\). \(\alpha,\beta\) are any, then \(\left|f^{(N)}(x)\right|=\begin{cases}|\sin(x)|\\ |\cos(x)|\end{cases}\leq1.\)
\(T_{N,a}(x) \to \cos(x) \quad \text{as} \quad N \to \infty \quad \forall x \in \mathbb{R}, \; \forall a \in \mathbb{R}.\)
Example 2
\(f(x) = \frac{1}{1+x^2}.\) \(f'(x) = \frac{-2x}{(1+x^2)^2}.\) \(f''(x)=\frac{-2+6x^2}{(1+x^2)^3}.\)
After hard work, you can see that: \(f(x) = \lim_{N \to \infty} T_{N,0}(x), \quad \forall x \in [-1, 1].\)
\(f(0) = 1, \quad f'(0) = 0, \quad f''(0) = -2, \quad f^{(3)}(0) = 0, \quad f^{(4)}(0) = 4!.\)
Example 3
\(f(x) = \arctan(x)\), \(f'(x) = \frac{1}{1+x^2}.\)
\(f(0) = 0,f'(0) = 1\), \(f^{(N)}(0)=\begin{cases}0, & \text{if }N\equiv0\pmod{2},\\ (N-1)!, & \text{if }N\equiv1\pmod{4},\\ -(N-1)!, & \text{if }N\equiv3\pmod{4}.\end{cases}\)
Then \(\arctan(x)=\lim_{N\to\infty}\left(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots+\frac{\left(N-1\right)!x^{N}}{N!}\right)\) where \(N\) is odd
For \(\frac{\pi}{4} = \arctan(1)\), we have \(\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \dots\)
This leads to \(\pi = 4 \left( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots \right).\)
Example 4:
\(f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & x \neq 0, \\ 0, & x = 0. \end{cases}\) 
\(f(x) = 0 \iff x = 0\). \(f(x) \approx 0\) as \(x \to 0\).
\(f\) is infinitely differentiable on \(\mathbb{R}\). \(f^{\prime}(0)=\lim_{x\to0}\frac{e^{-\frac{1}{x^2}}-0}{x-0}=0\)
\(f^{\prime}(x)=\begin{cases}2\frac{1}{x^3}e^{-\frac{1}{x^2}}, & x\neq0,\\ 0, & x=0.\end{cases}\)
We can prove that \(f^{(n)}(0) = 0 \quad \forall n \in \mathbb{N}.\)
Proof:
\(e^{-\frac{1}{x^2}}=\frac{1}{e^{\frac{1}{x^2}}}\leq\frac{1}{\frac{1}{N!}\left(\frac{1}{x^2}\right)^{N}}=N!x^{2N}\) for all \(N\in \N\)
\(\Rightarrow \lim_{x \to 0} \left| \frac{e^{-\frac{1}{x^2}}}{x} \right| \leq \lim_{x \to 0} \left| \frac{1! \cdot x^2}{x} \right| = 0\) \(\Rightarrow f'(0) = 0\)
For higher derivatives \(f^{(n-1)}(x) = e^{-\frac{1}{x^2}} \cdot \sum_{i=1}^{M(n)} \frac{C_i}{x^i}\)
Using the inequality: \(N > M(n) \Rightarrow f^{(n)}(0) = 0.\) (induction)
\(T_{N,0}(x) = \sum_{n=0}^N \frac{f^{(n)}(0)}{n!} (x-0)^n = 0.\)(Since \(f^{(n)}(0) = 0\) for all \(n\).)
\(\Rightarrow\lim_{N\to\infty}T_{N,0}(x)=0\neq f\left(x\right),\forall x\neq0\)
Uniform Convergence of Functions
Suppose \(\{f_n\}\text{ is sequence of function}, f: [a, b] \to \mathbb{R}\). What does it mean \(\lim_{n \to \infty} f_n = f\)?
If \(\{a_n\}\) is a sequence of numbers and \(a_n \to \ell\), then \(\forall \varepsilon > 0, \; \exists n_0 \; \text{s.t.} \; n > n_0 \implies |a_n - \ell| < \varepsilon.\)
For functions \(\text{dist}(f_n - f) = ?\)
Definition of version 1: Pointwise
We say \(\lim_{n \to \infty} f_n = f\) if \(\lim_{n \to \infty} f_n(x) = f(x), \quad \forall x \in [a, b].\) In this case, we say that \(f_n \to f\) pointwise as \(n \to \infty\).
That means: \(f_n \to f \; \text{pointwise if } \; \forall x \in [a, b] \; \text{and} \; \forall \varepsilon > 0, \; \exists n_0(\varepsilon, x) \; \text{such that} \; n > n_0 \implies |f_n(x) - f(x)| < \varepsilon.\)
None of the function will be inside the window
Example
Let \(f_n(x) = x^n\), where \(x \in [0, 1]\).
Then \(\lim_{n\to\infty}f_{n}(x)=\lim_{n\to\infty}x^{n}=\begin{cases}0, & \text{if }x\neq1,\\ 1, & \text{if }x=1.\end{cases}\)
Thus, if \(f(x) = \begin{cases} 0, & 0 \leq x < 1, \\ 1, & x = 1, \end{cases}\) then \(\lim_{n \to \infty} f_n = f\)
(We are interested in \(\lim_{N \to \infty} T_{N,a} \overset{?}{=} f\).)
Observation: The limit of continuous functions may not be continuous.
Definition of version 2: Uniform Convergence
Let \(f_n, f : [a, b] \to \mathbb{R}\). We say that \(f_n \xrightarrow{u} f\) (uniformly) as \(n \to \infty\), if:
\(\forall \varepsilon > 0, \; \exists n_0(\varepsilon) \; \text{such that} \; n > n_0 \implies |f_n(x) - f(x)| < \varepsilon \quad \forall x \in [a, b]\)
The entire function will be in the window
Main Theorem
If \(f_n, f : [a, b] \to \mathbb{R}\) and \(f_n\) are continuous and \(f_n \xrightarrow{u} f\) (uniformly) as \(n \to \infty\), then \(f\) is continuous.
Note that in example above \(f_n(x)=x^n\) are continuous on \([0,1]\) but \(f_n\) is pointwise to \(f\) (\(f_n \to f\) as \(n\to\infty\))
But not uniform converges to \(f\) (\(f_n \xrightarrow{u} f\))
Analysis
Since \(|f(x) - f(x_0)| = |f(x) - f_n(x) + f_n(x) - f_n(x_0) + f_n(x_0) - f(x_0)|\)
Using the triangle inequality: \(\leq |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)|\)
For the first part, use the definition of uniform convergence, choose \(n>n_0,\frac{\varepsilon}{3}\)
For the third part, use the definition of uniform convergence, choose \(n>n_0,\frac{\varepsilon}{3}\)
For the second part, use the definition of continuous, choose \(|x-x_0|<\delta,\frac{\varepsilon}{3}\)
Proof
Proof: We want to prove that \(f\) is continuous on \(x_0 \in [a, b]\).
Let \(\epsilon > 0\). We need \(\delta > 0\) such that \(|x - x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon\), \(\forall x \in [a, b]\).
Since \(f_n\) is uniform converges, let \(n_0\) such that \(n > n_0 \implies \left|f_n(x) - f(x)\right| < \frac{\epsilon}{3}, \forall x \in [a, b]\).
Since \(f_n\) is continuous, take \(n = n_0 + 1 > n_0\).
Let \(\delta > 0\) such that \(|x - x_0| < \delta \implies \left|f_{n_0+1}(x) - f_{n_0+1}(x_0)\right| < \frac{\epsilon}{3}\)
Let us see \(\delta\) works, \(|f(x) - f(x_0)| = |f(x) - f_{n_0+1}(x) + f_{n_0+1}(x) - f_{n_0+1}(x_0) + f_{n_0+1}(x_0) - f(x_0)|\)
\(\leq|f(x)-f_{n_0+1}(x)|+|f_{n_0+1}(x)-f_{n_0+1}(x_0)|+|f_{n_0+1}(x_0)-f(x_0)|\)
\(\leq\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon\)