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) .
$$
Try the following problem to test yourself！
Show that
$$
\int_{0}^{\pi} x e^{\sin (x)} d x=\frac{\pi}{2} \int_{0}^{\pi} e^{\sin (x)} d x .
$$
Let us use the substitution $u=\pi-x .$ Then we have
$$
\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 .
$$
Let $T:(C[0,1],|\cdot|) \rightarrow(C[0,1],|\cdot|)$ be a function defined as
$$
T f(x)=\int_{0}^{x} f(t) d t
$$
where by $C[0,1]$ we mean the vector space of all continuous real-valued functions defined on $[0,1]$ and $|f|=\sup _{0 \leq t \leq 1}|f(t)| .$ Show that:
a) $T$ is not a contraction.
b) $T$ has a unique fixed point.
c) $T^{2}$ is a contraction.
a) If $T$ were a contraction, then there exist $0<\lambda<1$ such that
$$
|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|
$$
and hence $T^{2}$ is a contraction.
Tchebycheff polynomials
- For any $n \in \mathbb{N}$, find a polynomial $T_{n}$ such that $T_{n}(\cos (x))=\cos (n x)$. Find the degree of $T_{n}, n \in \mathbb{N}$
- Show that for any $n \geq 1$, we have
$$
T_{n+1}=2 x T_{n}-T_{n-1}
$$ - Find the integrals
$$
\int_{-1}^{1} \frac{T_{n}(x) T_{m}(x)}{\sqrt{1-x^{2}}} d x
$$
for any $n, m \in \mathbb{N}$. - Show that $\cos \left(\frac{(2 k-1) \pi}{2 n}\right), k \in[1, n]$, are n different roots of $T_{n}(x)$, for any $n \geq 1$.
- In order to prove the existence of such polynomials, we will need Euler’s formula, i.e., $e^{i \theta}=$ $\cos (\theta)+i \sin (\theta)$. Indeed we have
$$
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}
$$
Then $T_{n}(\cos (\theta))=\cos (n \theta)$. Clearly $T_{n}$ is a polynomial function with degree $n .$
- We have the trigonometric identity
$$
\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))
$$
which implies $T_{n+1}=2 x T_{n}-T_{n-1}$, for any $n \geq 1$. - In order to find the integrals
$$
\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
$$
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Fourier analysis代写
离散数学代写
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