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