pluto_fourier_series.jl

### A Pluto.jl notebook ###
# v0.19.36

using Markdown
using InteractiveUtils

# This Pluto notebook uses @bind for interactivity. When running this notebook outside of Pluto, the following 'mock version' of @bind gives bound variables a default value (instead of an error).
macro bind(def, element)
    quote
        local iv = try Base.loaded_modules[Base.PkgId(Base.UUID("6e696c72-6542-2067-7265-42206c756150"), "AbstractPlutoDingetjes")].Bonds.initial_value catch; b -> missing; end
        local el = $(esc(element))
        global $(esc(def)) = Core.applicable(Base.get, el) ? Base.get(el) : iv(el)
        el
    end
end

# ╔═╡ 581f8ff0-d1fc-11ec-3cab-5f335ec587ad
begin
	using PlutoUI 
	using Plots
	using Colors
	using LaTeXStrings

	# Small patch to make images look more crisp:
	# https://github.com/JuliaImages/ImageShow.jl/pull/50
	Base.showable(::MIME"text/html", ::AbstractMatrix{<:Colorant}) = false

	nbsp = html"&nbsp"
end

# ╔═╡ 79521e83-f32a-48aa-bdf1-27f3f95ecb9e
md"""
## Fourier Series
"""

# ╔═╡ c299148e-4b85-4bf3-a514-20d1dff0dc59
md"""
#### 1. Definition of Fourier Series
Throughout this note we consider piecewise continuous $1$-periodic functions $f(x)$ on $\mathbb{R}$, that is, 

-  $f(x+1)=f(x)$ for all $x\in\mathbb{R}$.

- There exist numbers $0=a_0<a_1<\dotsb<a_m=1$ such that $f(x)$ is continuous on each open interval $(a_{j-1},a_j$, $j=1,\dotsc,m$, and limits exist:
$f(a_0+0), \quad f(a_1-0), \quad f(a_1+0), \quad \dotsc \quad f(a_m-0).$

We remark that if we restrict $f(x)$ on each closed interval $[a_{j-1}+k,a_{j}+k]$ for $j=1,\dotsc,m$ and $k\in\mathbb{Z}$, $f(x)$ can be regarded as a continuous function. We denote by $\mathcal{L}$ the set of all piecewise continuous $1$-periodic functions on $\mathbb{R}$. 

All the $e^{2\pi i nx}$, $n\in\mathbb{Z}$ are $1$-periodic smooth functions. In the spirit of Joseph Fourier (1768-1830), we consider whether $f(x)\in\mathcal{L}$ can be expressed by a trigonometric series of the form

$\sum_{n=-\infty}^{\infty}c_ne^{2\pi i nx}.$

If $f(x)=\displaystyle\sum_{n=-\infty}^{\infty}c_ne^{2\pi i nx}$ holds, then we have by formal computation 

$\begin{aligned}
  \int_0^1
  f(x)
  e^{-2\pi i mx}
  dx
& =
  \int_0^1
  \sum_{n=-\infty}^{\infty}
  c_n
  e^{2\pi i (n-m)x}
  dx
\\
& =
  \sum_{n=-\infty}^{\infty}
  c_n
  \int_0^1
  e^{2\pi i (n-m)x} 
  dx
\\
& =
  \sum_{n=-\infty}^{\infty}
  c_n
  \delta_{mn}
  =
  c_m.
\end{aligned}$

Note that the condition $f(x)\in\mathcal{L}$ guarantees the existence of the left hand side of the above. The formal trigonometric series defined by 

$\sum_{n=-\infty}^{\infty}c_ne^{2\pi i nx},
\quad
c_n=\int_0^1f(x)e^{-2\pi i nx}dx$

is said to be the Fourier series of $f(x)$, and we denote this by 

$f(x)\sim\sum_{n=-\infty}^{\infty}c_ne^{2\pi i nx}.$

 $c_n$ is called the $n$-th Fourier coefficient of $f(x)$.

"""

# ╔═╡ 6f6479b6-265a-4474-8a90-3df0ff98cfac
md"""
#### 2. Real Form of the Fourier Series
If $f(x)$ is real valued, then the Fourier series can be written by the real-valued trigonometric series. In this case $\overline{c_n}=c_{-n}$ holds, and $c_0$ is a real number. If we set for $n=0,1,2,\dotsc$ 

$\overline{c_n}=c_{-n}=\frac{a_n+ib_n}{2},$

$a_n
=
2
\int_0^1
f(x)
\cos(2\pi nx)
dx,$

$b_n
=
2
\int_0^1
f(x)
\sin(2\pi nx)
dx.$

then $b_0=0$, and 

$f(x)
\sim
\frac{a_0}{2}
+
\sum_{n=1}^\infty
\bigl(
a_n\cos(2\pi nx)
+
b_n\sin(2\pi n x)
\bigr).$

It is convenient to compute 

$\frac{a_0}{2}
=
c_0
=
\int_0^1
f(x)
dx,$

$a_n+ib_n
=
2c_{-n}
=
2
\int_0^1
f(x)
e^{2\pi i nx}
dx,
\quad
n=1,2,3,\dotsc.$


For $\lambda=\mu+\nu\in\mathbb{C}$, $\mu,\nu\in\mathbb{R}$, we have 

$\begin{aligned}
  e^{\lambda x}
& =
  e^{\mu x + i\nu x}
  =
  e^{\mu x}
  \{\cos(\nu x) + i\sin(\nu x)\},
\\
  \frac{d}{dx}
  e^{\lambda x}
& =
  \mu
  e^{\mu x}
  \{\cos(\nu x) + i\sin(\nu x)\}
  +
  \nu
  e^{\mu x}
  \{-\sin(\nu x) + i\cos(\nu x)\}
\\
& =
  \mu
  e^{\mu x}
  \{\cos(\nu x) + i\sin(\nu x)\}
  +
  i\nu
  e^{\mu x}
  \{i\sin(\nu x) + \cos(\nu x)\}
\\
& =
  (\mu+i\nu)
  e^{\mu x}
  \{\cos(\nu x) + i\sin(\nu x)\}
  =
  \lambda
  e^{\lambda x},
\end{aligned}$

which is useful for computing Fourier coefficients. 

"""

# ╔═╡ fd9270b6-bcb1-444a-b2e8-bc07efee3a24
md"""
#### 3. Example of Fourier Series
Let $f(x)$ be a function defined by 

$f(x):=\min\{x-[x],1-x-[1-x]\}, \quad x\in\mathbb{R},$

$[s]:=\max\{m\in\mathbb{Z} : m \leqq s\}, \quad x\in\mathbb{R}.$

In other words, $f(x)$ is a $1$-periodic Lipshitz continuous function such that 

$f(x)
=
\begin{cases}
x, &\ x\in[0,1/2),
\\
1-x, &\ x\in[1/2,1). 
\end{cases}$

We shall confirm that the Fourier series of $f(x)$ is

$f(x)
\sim
\frac{1}{4}
-
\sum_{k=1}^\infty
\frac{2}{(2k-1)^2\pi^2}
\cos\bigl(2(2k-1)\pi x\bigr).$

Indeed

$\frac{a_0}{2}
=
\int_0^{1/2}xdx+\int_{1/2}^1(1-x)dx
=
\frac{2}{2^3}
=
\frac{1}{4},$

and for $n=1,2,3,\dotsc$, 

$\begin{aligned}
  a_n+ib_n
& =
  2
  \int_0^{1/2}xe^{2\pi i nx}dx
  +
  2
  \int_{1/2}^1(1-x)e^{2\pi i nx}dx
\\
& =
  \left[
  \frac{xe^{2\pi i nx}}{n\pi i}
  \right]_0^{1/2}
  +
  \left[
  \frac{(1-x)e^{2\pi i nx}}{n\pi i}
  \right]_{1/2}^1
  -
  \int_0^{1/2}
  \frac{e^{2\pi i nx}}{n\pi i}
  dx
  +
  \int_{1/2}^1
  \frac{e^{2\pi i nx}}{n\pi i}
  dx
\\
& =
  0
  -
  \left[
  \frac{e^{2\pi i nx}}{2(n\pi i)^2}
  \right]_0^{1/2}
  +
  \left[
  \frac{e^{2\pi i nx}}{2(n\pi i)^2}
  \right]_{1/2}^1
\\
& =
  \frac{e^{\pi i n}-1-1+e^{\pi i n}}{2n^2\pi^2}
  =
  \frac{\cos(\pi n)-1}{n^2\pi^2}
  =
  -
  \frac{1-(-1)^n}{n^2\pi^2}
\\
& =
  \begin{cases}
  -
  \dfrac{2}{(2k-1)^2\pi^2},
  &\ n=2k-1,
  \\
  0
  &\ n=2k,
  \end{cases}
  \quad
  k=1,2,3,\dotsc.
\end{aligned}$



"""

# ╔═╡ a94cb318-c3e9-4285-af58-154d6955f2d5
md"""
#### 4. Dirichelet Kernel
Suppose $f(x)\in\mathcal{L}$ and $f(x)\sim\displaystyle\sum_{n=-\infty}^\infty c_ne^{2\pi i nx}$. Our basic problems are the following.

- Q1. Does the formal series $\displaystyle\sum_{n=-\infty}^\infty c_ne^{2\pi i nx}$ converge?

- Q2. If the formal series $\displaystyle\sum_{n=-\infty}^\infty c_ne^{2\pi i nx}$ converges, $f(x)=\displaystyle\sum_{n=-\infty}^\infty c_ne^{2\pi i nx}$ holds?

We introduce the partial $N$-the sum of the series

$S_N(s):=\sum_{n=-N}^{n=N}c_ne^{2\pi i nx}.$

We have 

$S_N(x)
=
\int_0^1D_N(x-y)f(y)dy
=
\int_{-1/2}^{1/2}D_N(t)f(x+t)dt,$

$D_N(t)
=
\sum_{n=-N}^{n=N}
e^{2\pi i nt}
=
1+
2\sum_{n=1}^N
\cos(2\pi n x)$

by the definition of the Fourier coefficients. $D_N(t)$ is called the $N$-th Dirichelet kernel. $D_N(t)$ is $1$-periodic smooth function and 

$\int_0^1D_N(t)dt=1.$

Moreover we obtain 

$D_N(t)
=
\begin{cases}
\dfrac{\sin\bigl(\pi(2N+1)t\bigr)}{\sin(\pi t)},
&\ t\not\in\mathbb{Z},
\\
2N+1,
&\ t\in\mathbb{Z}.
\end{cases}$
 
This expression is useful to study the convergence and divergence of Fourier series. Here we observe the behavior of the Dirichelet kernel in a one-period interval $[-1/2,1/2]$. Roughly speaking 

- The mass is concentrating on an interval $I_N:=[-1/(2N+1),1/(2N+1)]$ as $N\rightarrow\infty$.

-  $D_N(t)$ oscillates more violently in $[-1/2,1/2]\setminus{I_N}=[-1/2,-1/(2N+1))\cup(1/(2N+1),1/2]$ as $N\rightarrow\infty$.

"""

# ╔═╡ 3edf319a-48e4-48f0-82ab-f81e5f2c581c
begin
	M0=100; 
	t = range(-0.6, 0.6, length = 301);
	S0 = ones(M0+1,301);
	for m=2:M0+1
		for k=1:301
		    S0[m,k]=S0[m-1,k]+2*cos(2*pi*(m-1)*t[k]);
		end
	end
end

# ╔═╡ a1aa2ce7-db42-44d2-a9d8-4ae67976f76f
md"""
 $N$ = $(@bind N0 Slider(0:M0, show_value=true))
"""

# ╔═╡ cceffb37-cb1f-4436-aab6-17f325504276
begin
    plot(t,S0[N0+1,:],
		 grid=false,
		 linewidth=2,
		 ylim=(-M0/2,2*M0+M0/10),
         title="Dirichelet Kernel \$D_N(t)\$",
         xticks = ([-0.5 0 0.5;], [-0.5,0,0.5]),
		 yticks = ([0 100 200;], [0 100 200]),
		 xlabel="t",
         label=false,
	     legend=:false)
end

# ╔═╡ 897a6954-9e77-41d8-a88d-d43b4c9902c6
md"""
#### 5. Fourier Series of Periodic Continuous Functions 
We have the following results on the convergence and divergence of the Fourier series of continuous function. 

1. Suppose that $f(x)$ is $1$-periodic Hoelder continuous of degree $\alpha\in(0,1]$, that is, there exists a constant $L>0$ such that $\lvert{f(x)-f(y)}\rvert \leqq L\lvert{x-y}\rvert^\alpha$ for any $x,y\in\mathbb{R}$. Then
$\max_{x\in[0,1]}\lvert{S_N(x)-f(x)}\rvert \rightarrow 0 \quad (N\rightarrow\infty).$

2. Moreover if $\alpha>1/2$, then
$\sum_{n=-\infty}^\infty\lvert{c_n}\rvert<\infty.$

3. There exists a is -periodic continuous function $f(x)$ such that 
$S_N(0) \rightarrow \infty \quad (N\rightarrow\infty).$

Here we have some remarks on Hoelder continuity. The condition $\lvert{f(x)-f(y)}\rvert \leqq L\lvert{x-y}\rvert^\alpha$ is a restriction only for small $\lvert{x-y}\rvert$ since the absolute value of a periodic continuous function has a maximum. Roughly speaking, the exponent $\alpha$ expresses that $f(x)$ is differentiable of order $\alpha$. When $\alpha=1$, the Hoelder continuity is said to be the Lipshitz continuity. The Rademach theorem shows that the Lipshitz continuous function is differentiable almost everywhere with respect to the Lebesgue measure on $\mathbb{R}$. Roughly speaking, Part 1 and Part 3 assert that if $f(x)$ is a little bit smoother than continuous functions, its Fourier series uniformly converges to , and otherwise, its Fourier series does not necessarily converge even pointwisely. 

We show the outline of the proof of Part 1. Note that $S_N(x)-f(x)$ can be written as 

$S_N(x)-f(x)
=
\int_{-1/2}^{1/2}
D_N(t)\{f(x+t)-f(x)\}
dt.$

The Hoelder condition implies that for any small $\delta>0$

$S_N(x)-f(x)
=
\int_{\lvert{t}\rvert<\delta}
\mathcal{O}(\lvert{t}\rvert^{-1+\alpha})
dt
+
\int_{\delta<\lvert{t}\rvert<1/2}
\sin\bigl(\pi(2N+1)t\bigr)
\times\dotsb
dt.$

The second term of the right hand side is vanishing as $N\rightarrow\infty$ due to the oscillation and $\displaystyle\int_0^1\sin(2\pi t)dt=0$. Indeed for any $\varphi(t) \in C^1[0,1]$, 

$\begin{aligned}
  \int_0^1\sin(2\pi Nt)\varphi(t)dt
& =
  \left[-\frac{\cos(2\pi Nt)\varphi(t)}{2\pi N}\right]_0^1
  +
  \int_0^1\frac{\cos(2\pi Nt)\varphi^\prime(t)}{2\pi N}dt
\\
& =
  \mathcal{O}(N^{-1})
  \quad
  (N\rightarrow\infty).
\end{aligned}$

The differentiability of  is not required for this. In fact one can also prove 

$\int_0^1\sin(2\pi Nt)\varphi(t)dt \rightarrow 0 \quad (N\rightarrow\infty)$

for $\varphi(t) \in C[0,1]$ by using the uniform continuity of $\varphi(t)$ and  $\displaystyle\int_0^1\sin(2\pi t)dt=0$. So we have for any small $\delta>0$

$S_N(x)-f(x)
=\mathcal{O}(\delta^\alpha)+o(1)
\quad
(N\rightarrow\infty).$

"""

# ╔═╡ e956146c-d9f1-4c0f-8422-32c29e6d5b69
md"""
#### 6. Fourier Series of Triangular Function
Consider a $1$-periodic function  defined by 

$f(x)
:=
\begin{cases}
x, &\ x\in[0,1/2),
\\
1-x, &\ x\in[1/2,1).
\end{cases}$

This is the same as the example in Section 3. Here we observe the uniform convergence of the Fourier series of $f(x)$, which is 

$\begin{aligned}
  f(x)
& =
  \frac{1}{4}
  -
  \sum_{k=1}^\infty
  \frac{2}{(2k-1)^2\pi^2}
  \cos\bigl(2(2k-1)\pi x\bigr),
\\
  S_{2K-1}[f](x)
& =
  \frac{1}{4}
  -
  \sum_{k=1}^K
  \frac{2}{(2k-1)^2\pi^2}
  \cos\bigl(2(2k-1)\pi x\bigr).
\end{aligned}$

"""

# ╔═╡ b13742b2-45ca-42ee-bdcd-649b036da15f
begin
    M1=20;
	x = range(-0.1, 1.1, length = 481);
	
	f = zeros(481);
	for l=1:481
	f[l] = min(x[l]-Float64(floor(x[l])),1-x[l]-Float64(floor(1-x[l])));
	end
	
	S1=ones(M1+1,481)/4;
    for k=2:M1+1
		for l=1:481
		    S1[k,l]=S1[k-1,l]-2*cos(2*(2*k-3)*pi*x[l])/(2*k-3)^2/pi^2;
		end
	end
end

# ╔═╡ c05b1f88-2750-4f76-8af0-10091651289a
md"""
 $K$ = $(@bind N1 Slider(0:M1, show_value=true))
"""

# ╔═╡ cb7d861a-8d69-4e08-a69d-7638de638e3f
begin
    plot(x,S1[N1+1,:],
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1/16,9/16),
		 title="Triangular Function and its Fourier Series",
		 xticks = ([0 1/2 1;], [0,1/2,1]),
		 yticks = ([0 1/2;], [0,1/2]),
		 xlabel="\$x\$",
		 label="\$S_{2K-1}[f](x)\$",
		 legend=:topright,
		 legendfont=font(12))
	plot!(x,f,
		  linewidth=2,
	      label="\$f(x)\$")
end

# ╔═╡ 729497d1-2a1e-4fd3-bef4-fd3280ffd673
md"""
#### 7. A continuous funtion whose fourier series does not converges even pointwisely.

Define a function

$f(x)
=
\sum_{k=1}^\infty
\frac{2}{3^k}
\sin\{2\pi(2k+1)n(k)x\}
\sum_{l=1}^{n(k)}
\frac{\sin(2\pi lx)}{l},$

$n(k)=3^{3^k},
\quad\text{i.e.,}\quad
\log{n(k)}=3^k\log{3},$

which converges uniformly on $\mathbb{R}$, and is a $1$-periodic continuous function. 
Consider its Fourier series and the patrial sum

$f(x) \sim \sum_{n=-\infty}^\infty C_ne^{2\pi nx}, 
\quad
S_N(x):=\sum_{n=-N}^N C_ne^{2\pi nx}.$

It is well-known that the sequence $\{S_N(0)\}$ does not converge, more precisely, we have 

$S_{(2k+1)n(k)}(0)-S_{2kn(k)}(0)
=
\frac{1}{3^k}\sum_{j=1}^{n(k)-1}\frac{1}{j}
>
\log{3}.$

We tried to see the graph of $f$ by using 

$f_K(x)
:=
\sum_{k=1}^K
\frac{2}{3^k}
\sin\{2\pi(2k+1)n(k)x\}
\sum_{l=1}^{n(k)}
\frac{\sin(2\pi lx)}{l}.$

Unfortunately, we could not do it well since 

$n(1)=9, \quad n(2)=19683, \quad n(3)=7625597484987.$

We could compute $f_2$, and not $f_3$, and show the graph of $f_2$ below.
"""

# ╔═╡ adf3292f-4b05-4779-a316-dfe9662c055a
begin
	x5 = range(-0.6, 0.6, length = 101);

	f5=zeros(length(x5))
	for p=1:length(x5)
		for k=1:2
		    g5=zeros(length(x5),3^(3^k))
			for l=1:3^(3^k)
			    g5[p,l] += sin(2*pi*l*x5[p])
		    end
		f5[p] += 2/3^k*sin(2*pi*(2*k+1)*3^(3^k)*x5[p])*g5[p,3^(3^k)]
		end
	end
end

# ╔═╡ 9956a3d2-9ae9-4ef4-9f1c-55064001a025
begin
    plot(x5,f5,
		 grid=false,
		 linewidth=2,
		 xlim=(-0.6,0.6),
		 title="The continuous function \$f_2\$",
		 xticks = ([-0.5 0 0.5;], [-0.5,0,0.5]),
		 yticks = ([-1/2 0 1/2;], [-1/2,0,1/2]),
		 xlabel="\$x\$",
		 legend=false,
		 legendfont=font(12))
end

# ╔═╡ 596f1b0f-7a83-4f48-9316-17ef15d88cbe
md"""
#### 8. Step Function and Sawtooth Function
Let $Y(x)$ be the heaviside function, that is, 

$Y(x)
=
\begin{cases}
1, &\ x\geqq0,
\\
0, &\ x<0.
\end{cases}$

The value $Y(0)=1$ does not matter essentially, and is not needed to be defined. Consider a step function $g(x)$ and a sawtooth function $h(x)$ defined by 

$g(x)
:=
\sum_{n=-\infty}^\infty
Y(x+n) \times Y(1/2-x-n),
\quad
h(x):=x-[x].$

They are $1$-periodic function such that for $x\in[0,1)$

$g(x)
=
\begin{cases}
1, &\ x\in[0,1/2],
\\
0, &\ x\in(1/2,1),
\end{cases}
\quad
h(x)=x.$

Their Fourier series are 

$\begin{aligned}
  g(x)
& \sim
  \frac{1}{2}
  +
  \sum_{k=1}^\infty
  \frac{2}{(2k-1)\pi}
  \sin\bigl(2(2k-1)\pi x\bigr),
\\
  S_{2K-1}[g](x)
& =
  \frac{1}{2}
  +
  \sum_{k=1}^K
  \frac{2}{(2k-1)\pi}
  \sin\bigl(2(2k-1)\pi x\bigr),
\\
  h(x)
& \sim
  \frac{1}{2}
  -
  \sum_{n=1}^\infty
  \frac{1}{n\pi}
  \sin(2\pi nx),
\\
  S_N[h](x)
& =
  \frac{1}{2}
  -
  \sum_{n=1}^N
  \frac{1}{n\pi}
  \sin(2\pi nx).
\end{aligned}$

On one hand, at the discontinuous points 

$\begin{aligned}
  S_{2K-1}[g](k)
  =
  \frac{1}{2}
& \rightarrow
  \frac{1}{2}
  =
  \frac{g(k-0)+g(k+0)}{2}
  \quad
  (K\rightarrow\infty),
\\
  S_{2K-1}[g](k+1/2)
  =
  \frac{1}{2}
& \rightarrow
  \frac{1}{2}
  =
  \frac{g(k+1/2-0)+g(k+1/2+0)}{2}
  \quad
  (K\rightarrow\infty),
\\
  S_N[h](k)
  =
  \frac{1}{2}
& \rightarrow
  \frac{1}{2}
  =
  \frac{h(k-0)+g(h+0)}{2}
  \quad
  (N\rightarrow\infty)
\end{aligned}$

for all $k\in\mathbb{Z}$. On the other hand, we can prove that for any small $\delta>0$

$\begin{aligned}
  \max_{x\in[\delta,1/2-\delta]\cup[1/2+\delta,1-\delta]}
  \lvert{S_{2K-1}[g](x)-g(x)}\rvert
& \rightarrow
  0
  \quad
  (K\rightarrow\infty),
\\
  \max_{x\in[\delta,1-\delta]}
  \lvert{S_N[h](x)-h(x)}\rvert
& \rightarrow
  0
  \quad
  (N\rightarrow\infty).
\end{aligned}$

The partial sums oscillate violately as $N$ increases near the discontinuous points. This is called the Gibbs Phenomenon.
"""

# ╔═╡ 1bdd1071-c4a0-474a-a394-cd5ff77d5481
begin
    M2=80;
	
	g = zeros(481)
	for l=41:241
	    g[l] = 1;
	end
	for l=441:481
		g[l]=1;
	end
	
	S2=ones(M2+1,481)/2;
    for k=2:M2+1
		for l=1:481
		    S2[k,l]=S2[k-1,l]+2*sin(2*(2*k-3)*pi*x[l])/(2*k-3)/pi;
		end
	end
end

# ╔═╡ e4f95f78-cd34-4ce6-96f4-65c0151c55cc
md"""
 $K$ = $(@bind N2 Slider(0:M2, show_value=true))
"""

# ╔═╡ 2c0a572e-1da4-40cb-a5d4-ff510175d7fb
begin
    plot(x,S2[N2+1,:],
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1/8,9/8),
		 title="Step Function and its Fourier Series",
		 xticks = ([0 1/2 1;], [0,1/2,1]),
		 yticks = ([0 1/2 1;], [0,1/2,1]),
		 xlabel="\$x\$",
		 label="\$S_{2K-1}[g](x)\$",
		 legend=:topright,
		 legendfont=font(12))
	plot!(x,g,
		  linewidth=2,
	      label="\$g(x)\$")
end

# ╔═╡ 83ee20aa-e518-442b-af61-d1473d2d3d1a
begin
    h = zeros(481);
	for l=1:481
		h[l]=x[l]-Float64(floor(x[l]));
	end
		
	S3=ones(M2+1,481)/2;
    for k=2:M2+1
		for l=1:481
		    S3[k,l]=S3[k-1,l]-sin(2*(k-1)*pi*x[l])/(k-1)/pi;
		end
	end
end

# ╔═╡ 7cabd643-b904-416f-9382-7a4703e19b22
md"""
 $N$= $(@bind N3 Slider(0:M2, show_value=true))
"""

# ╔═╡ 36c46073-1425-43fc-93e8-fea1b2e1f192
begin
    plot(x,S3[N3+1,:],
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1/8,9/8),
		 title="Sawtooth Function and its Fourier Series",
		 xticks = ([0 1/2 1;], [0,1/2,1]),
		 yticks = ([0 1/2 1;], [0,1/2,1]),
		 xlabel="\$x\$",
		 label="\$S_N[h](x)\$",
	     legend=:top,
		 legendfont=font(12))
	plot!(x,h,
		  linewidth=2,
	      label="\$h(x)\$")
end

# ╔═╡ 47454abd-5b71-465e-bb2a-a4e3c544d350
md"""
#### 9. Decomposition of Piecewise Smooth Functions
Finally we consider piecewise smooth $1$-periodic functions $f(x)$ on $\mathbb{R}$, that is, 

-  $f(x+1)=f(x)$ for all $x\in\mathbb{R}$.

- There exist numbers $0=a_0<a_1<\dotsb<a_m=1$ such that $f(x)$ is continuously differentiable on each open interval $(a_{j-1},a_j$, $j=1,\dotsc,m$, and limits exist:
$f(a_0+0), \quad f(a_1-0), \quad f(a_1+0), \quad \dotsc \quad f(a_m-0),$ 
$f^\prime(a_0+0), \quad f^\prime(a_1-0), \quad f^\prime(a_1+0), \quad \dotsc \quad f^\prime(a_m-0).$ 

We remark that if we restrict  on each closed interval $[a_{j-1}+k,a_j+k]$ for $j=1,\dotsc,m$ and $k\in\mathbb{Z}$, $f(x)$ can be regarded as a continuously differentiable function. We can split $f(x)$ into a Lipshitz continuous function, some step functions, and a sawtooth function. We shall illustrate this for a simplest example. 

Let $f(x)$ be a piecewise smooth $1$-periodic functions on $\mathbb{R}$ with discontinuities at most $0,c,1$ in the interval $[0,1]$ with some $c\in(0,1)$. Note that $f(0\pm0)=f(1\pm0)$. We denote by $\chi_{[c,1]}(x)$ the characteristic function of the interval $[c,1]$. We define $f_0(x)$, $f_1(x)$, and $f_2(x)$ by 

$\begin{aligned}
  f_0(x)
& :=
  f(x)-f_1(x)-f_2(x),
\\
  f_1(x)
& :=
  \bigl(f(c+0)-f(c-0)\bigr)\chi_{[c,1]}(x),
\\
  f_2(x)
& :=
  -
  \bigl\{
  \bigl(f(0+0)-f(0-0)\bigr)
  +
  \bigl(f(c+0)-f(c-0)\bigr)
  \bigr\}
  h(x).
\end{aligned}$

Then $f_0(x)$ is Lipshitz continuous. So the piecewise smooth function $f(x)$ is splitted into the Lipshitz function part $f_0(x)$, the step function part $f_1(x)$, and the sawtooth function part $f_2(x)$. From the view point of Fourier series, $f_0(x)$ is harmless, and $f_1(x)+f_2(x)$ contains discontinuities where the Gibbs phenomenon occurs. 

We illustrate this decomposition visually. 

"""

# ╔═╡ 252f9cc5-5c34-49f1-96a7-bf5108f30fbd
begin
	Z=100*ones(481);
    f0=sin.(2*pi*x);
	f1=(ones(481)-g)/2;
	f2=h/3;
	F=f0+f1+f2;

	plot(x,F,
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1.1,1.7),
		 title="An example of Piecewise Smooth Function",
		 xticks = ([0 1/2 1;], [0,"\$c\$",1]),
		 xlabel="\$x\$",
		 label="\$f(x)\$",
		 legend=:topright,
		 legendfont=font(12))
end

# ╔═╡ d8503bba-e3f3-4f9b-ac4e-bc4f1ffd7fbf
begin
	plot(x,F,
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1.1,1.7),
		 title="Decomposition of Piecewise Smooth Function",
		 xticks = ([0 1/2 1;], [0,"\$c\$",1]),
		 xlabel="\$x\$",
		 label="\$f(x)\$",
		 legend=:topright,
		 legendfont=font(12))
	plot!(x,Z,
		  linewidth=2,
	      label=:false)
	plot!(x,Z,
		  linewidth=2,
	      label=:false)
	plot!(x,Z,
		  linewidth=2,
	      label=:false)
	plot!(x,f0,
		  linewidth=2,
	      label="\$f_0(x)=f(x)-f_1(x)-f_2(x)\$")
	plot!(x,f1+f2,
		  linewidth=2,
	      label="\$f_1(x)+f_2(x)\$")
end

# ╔═╡ d51cfe76-e1a3-4894-a2f5-43949ece5892
begin
	plot(x,F,
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1.1,1.7),
		 title="How to resolve discontinuity at \$x=c\$",
		 xticks = ([0 1/2 1;], [0,"\$c\$",1]),
		 xlabel="\$x\$",
		 label="\$f(x)\$",
		 legend=:topright,
		 legendfont=font(12))
	plot!(x,f1,
		  linewidth=2,
	      label="\$f_1(x)\$")
end

# ╔═╡ 692f012c-ec6d-4a2e-b76e-e87cda3155fb
begin
	plot(x,F,
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1.1,1.7),
		 title="Discontinuity at \$x=c\$ is resolved",
		 xticks = ([0 1/2 1;], [0,"\$c\$",1]),
		 xlabel="\$x\$",
		 label="\$f(x)\$",
		 legend=:topright,
		 legendfont=font(12))
	plot!(x,Z,
		  linewidth=2,
	      label=:false)
	plot!(x,F-f1,
		  linewidth=2,
	      label="\$f(x)-f_1(x)\$")
end

# ╔═╡ d9e2f9f3-b162-46fd-abde-9a1797f20d3a
begin
	plot(x,Z,
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1.1,1.7),
		 title="How to resolve discontinuity at \$x=0\$ mod \$1\$",
		 xticks = ([0 1/2 1;], [0,"c",1]),
		 xlabel="\$x\$",
		 label=:false,
		 legend=:topright,
		 legendfont=font(12))
	plot!(x,Z,
		  linewidth=2,
	      label=:false)
	plot!(x,F-f1,
		  linewidth=2,
	      label="\$f(x)-f_1(x)\$")
	plot!(x,f2,
		  linewidth=2,
	      label="\$f_2(x)\$")
end

# ╔═╡ a24d7900-ecb0-4a42-bc1e-785b182fe677
begin
	plot(x,Z,
		 grid=false,
		 linewidth=2,
		 xlim=(-0.15,1.15),
		 ylim=(-1.1,1.7),
		 title="Discontinuity at \$x=0\$ mod \$1\$ is resolved",
		 xticks = ([0 1/2 1;], [0,"\$c\$",1]),
		 xlabel="\$x\$",
		 label=:false,
		 legend=:topright,
		 legendfont=font(12))
	plot!(x,Z,
		  linewidth=2,
	      label=:false)
	plot!(x,F-f1,
		  linewidth=2,
	      label="\$f(x)-f_1(x)\$")
	plot!(x,Z,
		  linewidth=2,
	      label=:false)
	plot!(x,f0,
		  linewidth=2,
	      label="\$f_0(x)=f(x)-f_1(x)-f_2(x)\$")
end

# ╔═╡ ed6e4c4d-a686-4271-b964-0fe31c0361be
md"""
#### 10. Weyl's equidistribution theorem

"""

# ╔═╡ e6c5f708-b224-42de-996c-e7ed7b3e9125
begin
	t10 = range(0, 1, length = 101);
	c101=cos.(2*pi*t);
	c102=sin.(2*pi*t);

    γ=sqrt(5);
	x10=zeros(300);
	y10=zeros(300);

	for n=1:300
		x10[n]=cos(2*pi*γ*n);
		y10[n]=sin(2*pi*γ*n);
	end

end

# ╔═╡ 95c91522-88db-4f8f-9c2f-4e2e9cab84b6
md"""
 $N$ = $(@bind N10 Slider(1:300, show_value=true))
"""

# ╔═╡ 85863bee-e916-4d1e-98ab-52c6bc080fac
begin
    scatter(x10[1:N10],y10[1:N10],
		    title="Weyl's Equidistribution Theorem",
	        grid=false,
	        xlim=(-1.1,1.1),
	        ylim=(-1.1,1.1),
	        aspect_ratio=1.0,
	        xticks=false,
	        yticks=false,
	        label=false,
	        xaxis=false,
	        yaxis=false,
	        annotations=[(0,0.1,"\$e^{2πiγn}, γ=√5\$",20),(0,-0.15,"\$n=1,2,3,...\$",20)],   
		    annotationguide=:auto)
	plot!(c101,c102,label=false,linecolor=:magenta)
end

# ╔═╡ 75e78914-4f93-4f98-8137-6512ba96e868
md"""
#### 11. Continuous but nowhere differentiable functions

For any $\alpha \in (0,1)$ we set 

$f_\alpha(x):=\sum_{n=0}^\infty\frac{\exp(2\pi i2^nx)}{2^{n\alpha}}.$

The right hand side of the above converges uniformly since $2^\alpha>1$, and $f_\alpha$ becomes a $1$-periodic continuous function on $\mathbb{R}$. 
To show the graph of $f_\alpha$ below, set

$S_{\alpha,N}(x):=\sum_{n=0}^N\frac{\exp(2\pi i2^nx)}{2^{n\alpha}}.$

"""

# ╔═╡ 7bcc6226-cfcf-4e27-b70a-5e88bbf271aa
M6=20;

# ╔═╡ 76b7bc7b-d29e-43b8-bd38-0772479c5fda
md"""
 $N$ = $(@bind N6 Slider(0:M6, show_value=true, default=0)) 

 $\alpha$ = $(@bind α Slider(0.4:0.1:0.9, show_value=true, default=0.5))
"""

# ╔═╡ 90937af9-9230-4676-8b6c-34559a96d635
begin
	x6 = range(-0.1, 1.1, length = 121);
	
	f6 = zeros(length(x6),M6+1);
	g6 = zeros(length(x6),M6+1);
	for l=1:length(x6)
		f6[l,1]=cos(2*pi*x6[l]);
		g6[l,1]=sin(2*pi*x6[l]);
		for n=2:M6+1
		    f6[l,n] = f6[l,n-1] + cos(2*pi*(2^n)*x6[l])/(2^(α*n));
		    g6[l,n] = g6[l,n-1] + sin(2*pi*(2^n)*x6[l])/(2^(α*n));
		end
	end
	
end

# ╔═╡ 7f3c0c03-a149-49b6-bbe8-a57556dce69d
begin
    plot(x6,f6[:,N6+1],
		 grid=false,
		 linewidth=2,
		 ylim=(-2,3.5),
		 title="Continuous but nowhere differentiable functions",
		 xticks = ([0 1/2 1;], [0,1/2,1]),
		 #yticks = ([0 1/2 1;], [0,1/2,1]),
		 xlabel="\$x\$",
		 label="real part",
		 legend=:top,
		 legendfont=font(12))
	plot!(x6,g6[:,N6+1],linewidth=2,label="imaginary part",linecolor=:magenta)
end

# ╔═╡ 99a69d04-aed2-436c-85f9-5f0ca3fb1ace


# ╔═╡ 00000000-0000-0000-0000-000000000001
PLUTO_PROJECT_TOML_CONTENTS = """
[deps]
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LaTeXStrings = "b964fa9f-0449-5b57-a5c2-d3ea65f4040f"
Plots = "91a5bcdd-55d7-5caf-9e0b-520d859cae80"
PlutoUI = "7f904dfe-b85e-4ff6-b463-dae2292396a8"

[compat]
Colors = "~0.12.10"
LaTeXStrings = "~1.3.0"
Plots = "~1.39.0"
PlutoUI = "~0.7.52"
"""

# ╔═╡ 00000000-0000-0000-0000-000000000002
PLUTO_MANIFEST_TOML_CONTENTS = """
# This file is machine-generated - editing it directly is not advised

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[[deps.libfdk_aac_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg"]
git-tree-sha1 = "daacc84a041563f965be61859a36e17c4e4fcd55"
uuid = "f638f0a6-7fb0-5443-88ba-1cc74229b280"
version = "2.0.2+0"

[[deps.libinput_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg", "eudev_jll", "libevdev_jll", "mtdev_jll"]
git-tree-sha1 = "ad50e5b90f222cfe78aa3d5183a20a12de1322ce"
uuid = "36db933b-70db-51c0-b978-0f229ee0e533"
version = "1.18.0+0"

[[deps.libpng_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg", "Zlib_jll"]
git-tree-sha1 = "94d180a6d2b5e55e447e2d27a29ed04fe79eb30c"
uuid = "b53b4c65-9356-5827-b1ea-8c7a1a84506f"
version = "1.6.38+0"

[[deps.libvorbis_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Ogg_jll", "Pkg"]
git-tree-sha1 = "b910cb81ef3fe6e78bf6acee440bda86fd6ae00c"
uuid = "f27f6e37-5d2b-51aa-960f-b287f2bc3b7a"
version = "1.3.7+1"

[[deps.mtdev_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg"]
git-tree-sha1 = "814e154bdb7be91d78b6802843f76b6ece642f11"
uuid = "009596ad-96f7-51b1-9f1b-5ce2d5e8a71e"
version = "1.1.6+0"

[[deps.nghttp2_jll]]
deps = ["Artifacts", "Libdl"]
uuid = "8e850ede-7688-5339-a07c-302acd2aaf8d"
version = "1.52.0+1"

[[deps.p7zip_jll]]
deps = ["Artifacts", "Libdl"]
uuid = "3f19e933-33d8-53b3-aaab-bd5110c3b7a0"
version = "17.4.0+2"

[[deps.x264_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg"]
git-tree-sha1 = "4fea590b89e6ec504593146bf8b988b2c00922b2"
uuid = "1270edf5-f2f9-52d2-97e9-ab00b5d0237a"
version = "2021.5.5+0"

[[deps.x265_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg"]
git-tree-sha1 = "ee567a171cce03570d77ad3a43e90218e38937a9"
uuid = "dfaa095f-4041-5dcd-9319-2fabd8486b76"
version = "3.5.0+0"

[[deps.xkbcommon_jll]]
deps = ["Artifacts", "JLLWrappers", "Libdl", "Pkg", "Wayland_jll", "Wayland_protocols_jll", "Xorg_libxcb_jll", "Xorg_xkeyboard_config_jll"]
git-tree-sha1 = "9c304562909ab2bab0262639bd4f444d7bc2be37"
uuid = "d8fb68d0-12a3-5cfd-a85a-d49703b185fd"
version = "1.4.1+1"
"""

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