数学代写|MTH 710 Fourier analysis

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MTH710课程简介

Textbook: The main source we will follow are Bruce Driver’s excellent Probability notes:
Probability Tools with Examples, by Bruce Driver
Here are a few other textbooks we recommend as auxiliary sources; all are freely available to UCSD personnel.
Probability: Theory and Examples (5th Edition), by Rick Durrett
A Probability Path by Sidney Resnick
AModern Approach to Probability Theory by Bert Fristedt and Lawrence Gray
Probability Theory: A Comprehensive Course by Achim Klenke
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Prerequisites 

In mathematics, Fourier analysis (/ˈfʊrieɪ, -iər/)[1] is the study of the way general functions may be represented or approximated by sums of simpler trigonometric functions. Fourier analysis grew from the study of Fourier series, and is named after Joseph Fourier, who showed that representing a function as a sum of trigonometric functions greatly simplifies the study of heat transfer.

Weekly Contact: Lecture: 3 hrs.GPA Weight: 1.00Course Count: 1.00Billing Units: 1

Prerequisites
MTH 108 or MTH 141 and MTH 231 or MTH 310 or MTH 240
Co-Requisites
None
Antirequisites
None
Custom Requisites
None

MTH710 Fourier analysis HELP（EXAM HELP， ONLINE TUTOR）

Let $\phi: \mathbb{T} \rightarrow \mathbb{C}^$ be a continuous group homomorphism. Since $\mathbb{T}$ is a compact abelian group and $\mathbb{C}^$ is a multiplicative subgroup of $\mathbb{C}$, we know that $\phi(\mathbb{T})$ is a compact abelian subgroup of $\mathbb{C}^$. Since $\mathbb{C}^$ is isomorphic to $\mathbb{R}^+ \times \mathbb{T}$, we know that $\phi(\mathbb{T})$ is isomorphic to $\mathbb{Z} \times \mathbb{T}$ for some integer $n$.

Let $z = e^{2\pi i x} \in \mathbb{T}$. Then $\phi(z) = \phi(e^{2\pi i x}) = \phi(e^{2\pi i {x}})$, where ${x}$ denotes the fractional part of $x$. Since $\phi$ is a group homomorphism, we have $\phi(e^{2\pi i {x}}) = \phi(e^{2\pi i ({x} + k)})$ for any integer $k$. Since $\phi(\mathbb{T})$ is isomorphic to $\mathbb{Z} \times \mathbb{T}$, we can write $\phi(e^{2\pi i ({x} + k)}) = e^{2\pi i (nx + m_k)}$ for some integer $m_k$ that depends on $k$. Since $\phi$ is continuous, we have $m_k \rightarrow m$ as $k \rightarrow \infty$, where $m \in \mathbb{Z}$ is some fixed integer. Thus, we have $\phi(z) = e^{2\pi i nx + m}$ for all $z \in \mathbb{T}$.

Since $e^{2\pi i m} = 1$ for any integer $m$, we can write $\phi(z) = e^{2\pi i nx}$ for all $z \in \mathbb{T}$. Thus, we have shown that every continuous group homomorphism from $\mathbb{T}$ into $\mathbb{C}^*$ takes the form $x \mapsto e^{2\pi i nx}$ with $n$ an integer.

For the analogous statement for continuous homomorphisms $\mathbb{R} \rightarrow \mathbb{C}^$, we can use the fact that $\mathbb{R}$ is a divisible abelian group, which implies that any continuous group homomorphism from $\mathbb{R}$ into $\mathbb{C}^$ takes the form $x \mapsto e^{2\pi i nx}$ for some real number $n$. Here, we do not necessarily have $n \in \mathbb{Z}$, since $\mathbb{R}$ is not a compact group.

To show that ${\psi_n(x)}_{n\geq 0}$ form an orthogonal sequence of eigenfunctions for the Fourier transform on $L^2(\mathbb{R})$, we need to show that:

1. $\psi_n(x)$ are eigenfunctions of the Fourier transform, i.e., $\hat{\psi}_n(\xi) = C_n \psi_n(\xi)$ for some constant $C_n$ and all $\xi \in \mathbb{R}$, where $\hat{\psi}_n(\xi)$ denotes the Fourier transform of $\psi_n(x)$.
2. ${\psi_n(x)}{n\geq 0}$ are orthogonal with respect to the $L^2(\mathbb{R})$ inner product, i.e., $\int{-\infty}^{\infty} \psi_m(x) \overline{\psi_n(x)} dx = 0$ for $m\neq n$.
3. ${\psi_n(x)}{n\geq 0}$ form a complete set, i.e., every $f\in L^2(\mathbb{R})$ can be written as a linear combination of ${\psi_n(x)}{n\geq 0}$.

First, we note that $\$ \backslash p s i _0(x)=e^{\wedge}\left{-\backslash p i x^{\wedge} 2\right} \$$, which is known to be its own Fourier transform up to a constant:
$$
\hat{\psi}0(\xi)=\frac{1}{\sqrt{2 \pi}} \int{-\infty}^{\infty} e^{-\pi x^2} e^{-2 \pi i x \xi} d x=\frac{1}{\sqrt{2 \pi}} e^{-\pi \xi^2}
$$
Therefore, we have $\$ \backslash$ hat ${\backslash p s i} _0(\backslash x i)=C_{-} 0 \backslash p s i _0(\backslash x i) \$$ for $\$ C _0=\backslash$ frac ${1}$ {sqrt{2\pi}}\$, and $\$\left{\backslash p s i _0(x)\right} \$$ is an eigenfunction of the Fourier transform. of degree $\$ n \$$ in $\$ \backslash \operatorname{left}(\backslash \operatorname{frac}{d}{d x}$ – $2 \backslash$ pi $x \backslash$ right $) \$$, i.e.,
$$
\psi_n(x)=\sum_{k=0}^n a_{n, k}\left[\frac{d}{d x}-2 \pi x\right]^k e^{-\pi x^2}
$$

where $a_{n,k}$ are coefficients that depend only on $n$ and $k$. Then, applying the Fourier transform to both sides and using the fact that the Fourier transform commutes with differentiation, we have: \begin{align*} \hat{\psi}n(\xi) &= \sum{k=0}^{n} a_{n,k} \frac{d^k}{d x^k} \hat{e}^{-\pi x^2}(\xi) \ &= \sum_{k=0}^{n} a_{n,k} \frac{(-1)^k}{\sqrt{2\pi}} \hat{H}_k(\xi) \hat{e}^{-\pi \xi^2} \ &= C_n \psi_n(\xi), \end{align*} where $\hat{H}_k(\xi)$ denotes the $k$-th Hermite polynomial, and $C_n$ is a constant given by $C_n = \frac{(-1)^n}{\sqrt

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