# Understanding these basic definition and statement~

• Partitions, Upper and Lower Sums: Let $f$ be a bounded function on $[a, b] .$ A partition of $[a, b]$ is a finite ordered set
$$P=\{a=x_{0}<x_{1}<x_{2}<\ldots<x_{n}=b\} .$$
For each subinterval $\left[x_{k-1}, x_{k}\right]$ of $P$ set
$$m_{k}=\inf \{f(x): x \in$x_{k-1}, x_{k}$}$$
and
$$M_{k}=\sup \{f(x): x \in\left[x_{k-1}, x_{k}\right]\}$$
The lower sum of $f$ with respect to $P$ is given by
$$U(f ; P)=\sum_{k=1}^{n} m_{k}\left(x_{k}-x_{k-1}\right)$$
Similarly, we define the upper sum of $f$ with respect to $P$ by
$$L(f ; P)=\sum_{k=1}^{n} M_{k}\left(x_{k}-x_{k-1}\right)$$

Since $f$ is a bounded function on $[a, b]$, there exist numbers $M$ and $m$ such that $m \leq f(x) \leq M$ is true for all $x \in[a, b]$. Thus for any partition $P$ of $[a, b]$ we have
$$m(b-a) \leq L(f ; P) \leq U(f ; P) \leq M(b-a) .$$

• Upper and Lower Integrals: Let $\mathcal{P}$ be the collection of all possible partitions of the interval $[a, b] .$ The lower integral of $f$，

• $\int_{a}^{b} f(x) d x=L(f)=\sup {L(f ; P): P \in \mathcal{P}}$
Likewise the upper integral of $f$，
$$\overline{\int_{a}^{b}} f(x) d x=U(f)=\inf {U(f ; P): P \in \mathcal{P}} .$$
Clearly for a bounded function $f$ on $[a, b]$ we always have $U(f) \geq L(f)$
• Riemann Integral: A bounded function $f$ on $[a, b]$ is Riemann integrable if $U(f)=L(f) .$ In this case, we define $\int_{a}^{b} f(x) d x$ to be
$$\int_{a}^{b} f(x) d x=U(f)=L(f) .$$
• Riemann Sum: Let $P=\left{a=x_{0}<x_{1}<x_{2}<\ldots<x_{n}=b\right}$ be a partition of $[a, b] .$ A tagged partition is one where in addition to $P$ we have chosen points $x_{k}^{}$ in each of the subintervals $\left[x_{k-1}, x_{k}\right]$. Suppose $f:[a, b] \rightarrow \mathbb{R}$ and a tagged partition $\left(P, x_{k}^{}\right)$ is given. The Riemann sum generated by this partition is given as
$$R(f ; P)=\sum_{k=1}^{n} f\left(x_{k}^{*}\right)\left(x_{k}-x_{k-1}\right) .$$
It is clear that
$$L(f ; P) \leq R(f ; P) \leq U(f ; P)$$
true for any bounded function.
• Improper Integral: Let $f$ be defined on $[a, \infty)$ and integrable on $[a, c]$ for every $c>a$. If $\lim {c \rightarrow \infty} \int{a}^{c} f(x) d x$ exists, then the improper integral of $f$ on $[a, \infty)$, denoted by $\int_{a}^{\infty} f(x) d x$, is given by
$$\int_{a}^{\infty} f(x) d x=\lim {c \rightarrow \infty} \int{a}^{c} f(x) d x \text {. }$$
• Mean Value Theorem for Integrals: If $f$ is a continuous function on $[a, b]$, then there is a point $c \in(a, b)$ such that
$$\frac{1}{b-a} \int_{a}^{b} f(x) d x=f(c) .$$
• Fundamental Theorem of Calculus:

a) Let $f:[a, b] \rightarrow \mathbb{R}$ be integrable, and define $F(x)=\int_{a}^{x} f$ for all $x \in[a, b]$. Then
(a) $F$ is continuous on $[a, b]$.
(b) If $f$ is continuous at some point $x_{0} \in[a, b]$, then $F$ is differentiable at $x_{0}$ and
$$F^{\prime}\left(x_{0}\right)=f\left(x_{0}\right) \text {. }$$
The above theorem expresses the fact that every continuous function is the derivative of its indefinite integral.
b) If $g:[a, b] \rightarrow \mathbb{R}$ is integrable, and $G:[a, b] \rightarrow \mathbb{R}$ satisfies $G^{\prime}(x)=g(x)$ for all $x \in[a, b]$, then
$$\int_{a}^{b} g=G(b)-G(a) .$$

# Try the following problem to test yourself！

Problem 1.

Show that
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\frac{\pi}{2} \int_{0}^{\pi} e^{\sin (x)} d x .$$

Proof .

Let us use the substitution $u=\pi-x .$ Then we have
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\int_{\pi}^{0}-(\pi-u) e^{\sin (\pi-u)} d u=\int_{0}^{\pi}(\pi-u) e^{\sin (u)} d u$$
which implies
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\pi \int_{0}^{\pi} e^{\sin (u)} d u-\int_{0}^{\pi} u e^{\sin (u)} d u$$
Since
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\int_{0}^{\pi} u e^{\sin (u)} d u$$
we get
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\frac{\pi}{2} \int_{0}^{\pi} e^{\sin (x)} d x .$$

Problem 2.

Let $T:(C[0,1],|\cdot|) \rightarrow(C[0,1],|\cdot|)$ be a function defined as
$$T f(x)=\int_{0}^{x} f(t) d t$$
where by $C[0,1]$ we mean the vector space of all continuous real-valued functions defined on $[0,1]$ and $|f|=\sup _{0 \leq t \leq 1}|f(t)| .$ Show that:
a) $T$ is not a contraction.
b) $T$ has a unique fixed point.
c) $T^{2}$ is a contraction.

Proof .

a) If $T$ were a contraction, then there exist $0<\lambda<1$ such that
$$|T f-T g| \leq \lambda|f-g| \quad \forall f, g \in C[0,1]$$
Taking $f(t)=1, g(t)=0$ in $C[0,1]$, we have
$$\begin{gathered} |T f-T g|=\sup \left|\int_{0}^{x} d t-\int_{0}^{x} 0 d t\right|=\sup {0 \leq x \leq 1}|x|=1, \ |f-g|=\sup {0 \leq x \leq 1}|1-0|=1 \end{gathered}$$
and hence we will have $1 \leq \lambda \cdot 1$ which is a contraction since $\lambda<1$. Therefore $T$ is not a contraction.
b) Consider $f(x)=0$, then $T f=f$, therefore we have the existence of the fixed point. To show the uniqueness of the fixed point, assume not, suppose we have another fixed point say $h$ such that $T h=h$ or equivalently
$$\int_{0}^{x} h(t) d t=h(x) .$$
From the Fundamental Theorem of Calculus we have $\frac{d h}{d x}=h(x)$, and the solution to this differential equation is $h(x)=C e^{x}$. Since $h(0)=0$, we have $C=0$ and therefore $h(x)=0=$ $f(x)$.

c) First observe that
$$T^{2} f(x)=T(T f(x))=\int_{0}^{x}\left(\int_{0}^{t} f(s) d s\right) d t$$
and
$$\begin{gathered} \left|T^{2} f-T^{2} g\right|=\sup {0 \leq x \leq 1}\left|\int{0}^{x}\left(\int_{0}^{t}(f(s)-g(s)) d s\right) d t\right| \leq \sup {0 \leq x \leq 1} \int{0}^{x}\left(\int_{0}^{t}|f(s)-g(s)| d s\right) d t, \ \left|T^{2} f-T^{2} g\right| \leq|f-g| \sup {0 \leq x \leq 1} \int{0}^{x} t d t=|f-g| \sup _{0 \leq x \leq 1} \frac{x^{2}}{2}=\frac{1}{2}|f-g| \end{gathered}$$
We showed:
$$\left|T^{2} f-T^{2} g\right| \leq \frac{1}{2}|f-g|$$
and hence $T^{2}$ is a contraction.

Problem 3.

## Tchebycheff polynomials

1. For any $n \in \mathbb{N}$, find a polynomial $T_{n}$ such that $T_{n}(\cos (x))=\cos (n x)$. Find the degree of $T_{n}, n \in \mathbb{N}$
2. Show that for any $n \geq 1$, we have
$$T_{n+1}=2 x T_{n}-T_{n-1}$$
3. Find the integrals
$$\int_{-1}^{1} \frac{T_{n}(x) T_{m}(x)}{\sqrt{1-x^{2}}} d x$$
for any $n, m \in \mathbb{N}$.
4. Show that $\cos \left(\frac{(2 k-1) \pi}{2 n}\right), k \in[1, n]$, are n different roots of $T_{n}(x)$, for any $n \geq 1$.

Proof .

1. In order to prove the existence of such polynomials, we will need Euler’s formula, i.e., $e^{i \theta}=$ $\cos (\theta)+i \sin (\theta)$. Indeed we have
$$e^{i n \theta}=\cos (n \theta)+i \sin (n \theta)=\left(e^{i \theta}\right)^{n} .$$.

But
$$\left(e^{i \theta}\right)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l} n \ k \end{array}\right) i^{k} \sin ^{k}(\theta) \cos ^{n-k}(\theta) .$$
So the real part of the two complex numbers must be equal which gives
$$\cos (n \theta)=\sum_{0 \leq 2 k \leq n}\left(\begin{array}{c} n \ 2 k \end{array}\right)(-1)^{k} \sin ^{2 k}(\theta) \cos ^{n-2 k}(\theta) .$$
But $\sin ^{2 k}(\theta)=\left(\sin ^{2}(\theta)\right)^{k}=\left(1-\cos ^{2}(\theta)\right)^{k}$, which implies
$$\cos (n \theta)=\sum_{0 \leq 2 k \leq n}\left(\begin{array}{c} n \ 2 k \end{array}\right)(-1)^{k}\left(1-\cos ^{2}(\theta)\right)^{k} \cos ^{n-2 k}(\theta) .$$
Set
$$T_{n}(x)=\sum_{0 \leq 2 k \leq n}\left(\begin{array}{c} n \ 2 k \end{array}\right)(-1)^{k}\left(1-x^{2}\right)^{k} x^{n-2 k}$$
Then $T_{n}(\cos (\theta))=\cos (n \theta)$. Clearly $T_{n}$ is a polynomial function with degree $n .$

1. We have the trigonometric identity
$$\cos (n+1) x+\cos (n-1) x=2 \cos (x) \cos (n x) \text {. }$$
Hence
$$T_{n+1}(\cos (x))+T_{n-1}(\cos (x))=2 \cos (x) T_{n}(\cos (x))$$
which implies $T_{n+1}=2 x T_{n}-T_{n-1}$, for any $n \geq 1$.
2. In order to find the integrals
$$\int_{-1}^{1} \frac{T_{n}(x) T_{m}(x)}{\sqrt{1-x^{2}}} d x, n, m \in \mathbb{N},$$
we use the change of variable $x=\cos (t)$. Hence
$$\int_{-1}^{1} \frac{T_{n}(x) T_{m}(x)}{\sqrt{1-x^{2}}} d x=\int_{\pi}^{0} \frac{T_{n}(\cos (t)) T_{m}(\cos (t))}{\sqrt{1-\cos ^{2}(t)}}(-\sin (t)) d t=\int_{0}^{\pi} T_{n}(\cos (t)) T_{m}(\cos (t)) d t$$
Using the main property of the polynomial functions $T_{n}$, we get
$$\int_{-1}^{1} \frac{T_{n}(x) T_{m}(x)}{\sqrt{1-x^{2}}} d x=\int_{0}^{\pi} \cos (n t) \cos (m t) d t \text {. }$$
If $n \neq m$, we use the identity
$$\cos (n t) \cos (m t)=\frac{1}{2}(\cos (n-m) t+\cos (n+m) t),$$
to get
$$\int_{0}^{\pi} \cos (n t) \cos (m t) d t=\frac{1}{2}\left[\frac{\sin (n-m) t}{n-m}+\frac{\sin (n+m) t}{n+m}\right]_{0}^{\pi}=0 .$$

If $n=m$, then
$$\int_{-1}^{1} \frac{T_{n}^{2}(x)}{\sqrt{1-x^{2}}} d x=\int_{0}^{\pi} \cos ^{2}(n t) d t=\frac{1}{2} \int_{0}^{\pi}(1+\cos (2 n t)) d t=\frac{\pi}{2},$$
if $n \neq 0$. If $n=0$, then
$$\int_{-1}^{1} \frac{T_{0}^{2}(x)}{\sqrt{1-x^{2}}} d x=\pi$$

Fourier analysis代写

## 离散数学代写

Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析