104 lines
2.6 KiB
Markdown
104 lines
2.6 KiB
Markdown
# Math4121 Lecture 13
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## New book Chapter 2
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Riemann's motivation: Fourier series
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$$
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F(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty} a_k \cos(kx) + b_k \sin(kx)
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$$
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$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(kx) dx$
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$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx) dx$
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To study the convergence of the Fourier series, we need to study the convergence of the sequence of partial sums. (Riemann integration)
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Why Riemann integration?
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Let
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$$
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((x)) = \begin{cases}
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x-\lfloor x \rfloor & x \in [\lfloor x \rfloor, \lfloor x \rfloor + \frac{1}{2}) \\
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0 & x=\lfloor x \rfloor + \frac{1}{2}\\
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x-\lfloor x \rfloor - 1 & x \in (\lfloor x \rfloor + \frac{1}{2}, \lfloor x \rfloor + 1] \end{cases}
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$$
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We define
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$$
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f(x) = \sum_{n=1}^{\infty} \frac{((nx))}{n^2}=\lim_{N\to\infty}\sum_{n=1}^{N} \frac{((nx))}{n^2}
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$$
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(i) The series converges uniformly over $x\in[0,1]$.
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$$
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\left|f(x)-\sum_{n=1}^{N} \frac{((nx))}{n^2}\right|\leq \sum_{n=N+1}^{\infty}\frac{|((nx))|}{n^2}\leq \sum_{n=N+1}^{\infty} \frac{1}{n^2}<\epsilon
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$$
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As a consequence, $f(x)\in \mathscr{R}$.
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(ii) $f$ has a discontinuity at every rational number with even denominator.
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$$
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\begin{aligned}
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\lim_{h\to 0^+}f(\frac{a}{2b}+h)-f(\frac{a}{2b})&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))}{n^2}-\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}))}{n^2}\\
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&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
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&=\sum_{n=1}^{\infty}\lim_{h\to 0^+}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
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&>0
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\end{aligned}
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$$
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### Back to the fundamental theorem of calculus
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Suppose $f$ is integrable on $[a,b]$, then
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$$
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F(x)=\int_a^x f(t)dt
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$$
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$F$ is continuous on $[a,b]$.
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if $f$ is continuous at $x_0$, then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$.
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#### Theorem (Darboux's theorem)
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If $\lim_{x\to a^-}f(x)=L^-$, then $\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=L^-$.
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Proof:
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$$
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h\sup_{x\in [0,h]}f(x)\geq F(a+h)\geq \inf_{x\in [0,h]}f(x)h
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$$
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Consequently,
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$$
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f(x)=\sum_{n=1}^{\infty} \frac{((nx))}{n^2}
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$$
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then
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$$
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F(x)=\int_0^x f(t)dt
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$$
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is continuous on $[0,1]$.
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However, since $\lim_{x\to 0^+}f(x)\neq \lim_{x\to 0^-}f(x)$ holds for all the rational numbers with even denominator, $F$ is not differentiable at all the rational numbers with even denominator.
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Moral: There exists a continuous function on $[0,1]$ that is not differentiable at any rational number with even denominator. (Dense set)
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#### Weierstrass function
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$$
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g(x)=\sum_{n=0}^{\infty} a^n \cos(b^n \pi x)
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$$
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where $0<a<1$ and $ab>1+\frac{3}{2}\pi$.
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$g(x)$ is continuous on $\mathbb{R}$ but nowhere differentiable.
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_If we change our integral, will be differentiable at most points?_
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