MY-ASSIGNMENTEXPERT™可以为您提供sydney MATH4069 Fourier analysis傅里叶分析的代写代考和辅导服务！
这是悉尼大学 傅里叶分析课程的代写成功案例。
MATH4069课程简介
Measure theory is the study of fundamental ideas as length, area, volume, arc length and surface area. It is the basis for Lebesgue integration theory used in advanced mathematics ever since its development in about 1900. Measure theory is also a key foundation for modern probability theory. The course starts by establishing the basics of measure theory and the theory of Lebesgue integration, including important results such as Fubini’s Theorem and the Dominated Convergence Theorem which allow us to manipulate integrals. These ideas are applied to Fourier Analysis which leads to results such as the Inversion Formula and Plancherel’s Theorem. The Radon-Nikodyn Theorem provides a representation of measures in terms of a density. Key ideas of this theory are applied in detail to probability theory to provide a rigorous framework for probability which takes in and generalizes familiar ideas such as distributions and conditional expectation. When you complete this unit you will have acquired a new generalized way of thinking about key mathematical concepts such as length, area, integration and probability. This will give you a powerful set of intellectual tools and equip you for further study in mathematics and probability.
Prerequisites
At the completion of this unit, you should be able to:
- LO1. explain and apply the fundamentals of abstract measure and integration theory
- LO2. explain what an outer measure and use outer measures to construct other measures. Apply these concepts to Lesbesgue measures and other related measures.
- LO3. explain and apply the limit theorems including the dominated convergence theorem and theorems on continuity and differentiability of parameter integrals.
- LO4. explain the properties of Lp spaces
- LO5. use inequalities such as Holder’s, Minkowsi’s, Jensen’s and Young’s inequalities to solve problems
- LO6. explain and apply the properties of the Fourier series on normed spaces
- LO7. generate the measure theoretic foundations of probability theory
- LO8. recall and explain the definition and basic properties of conditional expectation
- LO9. create proofs and apply measure theory in diverse applications.
MATH4069 Fourier analysis HELP（EXAM HELP， ONLINE TUTOR）
Let $\varphi(x)=|x|$ for $x \in[0,1]$. Extend $\varphi$ to $\mathbb{R}$ as a 2-periodic function. Prove that the formula
$$
f(x)=\sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^n \varphi\left(4^n x\right)
$$
defines a continuous function on $\mathbb{R}$.
Solution. Since $|\varphi(x)| \leqslant 1$ for all $x \in \mathbb{R}$, the series defining $f$ may be compared to the absolutely converging geometric series $\sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^n$, thereby establishing that the series defining $f$ converges uniformly.
We continue to study properties of $\varphi$ defined in the first exercise. Fix $x_0 \in \mathbb{R}$. For every non-negative integer $m$ define a real number $\delta_m$ by the conditions $\left|\delta_m\right|=4^{-m} / 2$ and the sign is fixed by the condition that there are no integers between $4^m x_0$ and $4^m\left(x_0+\delta_m\right)$. Consider the quotient
$$
\gamma_n=\frac{\varphi\left(4^n\left(x_0+\delta_m\right)\right)-\varphi\left(4^n x_0\right)}{\delta_m} .
$$
Prove that
a) If $n>m$ then $\gamma_n=0$.
b) If $0 \leqslant n \leqslant m$, then $\left|\gamma_n\right| \leqslant 4^n$ and $\left|\gamma_m\right|=4^m$.
Solution. a) This follows from the fact that $\varphi$ is 2-periodic and the subsequent observation that when $n>m$, we have
$$
\varphi\left(4^n\left(x+\delta_m\right)\right)=\varphi\left(4^n x \pm 2^{2(n-m)-1}\right)=\varphi\left(4^n x\right)
$$
b) For $0 \leqslant n<m$ the inequalities follow directly from the fact that $\varphi$ is Lipschitz and in fact
$$
|\varphi(x)-\varphi(y)| \leqslant|x-y|
$$
for all $x, y \in \mathbb{R}$. Finally, the equality $\left|\gamma_m\right|=4^m$ follows from the observation that since the sign of $\delta_m$ was chosen in such a way that there are no integers between $4^m\left(x_0+\delta_m\right)$ and $4^m x_0$, the part of the graph of $\varphi$ between those values looks like a line segment of slope \pm 1 , and thus
$$
\left|\gamma_m\right|=\left|\frac{\varphi\left(4^n\left(x_0+\delta_m\right)\right)-\varphi\left(4^n x_0\right)}{\delta_m}\right|=\left|\frac{4^m\left(x_0+\delta_m\right)-4^m x_0}{\delta_m}\right|=4^m .
$$
(a) Find the Fourier series for the function:
$$
f(x)=x^2 \quad \text { if }-1 \leq x<1
$$
(extended periodically over the real line)
(b) Graph the finite trigonometric sums for $N=2,5,20$.
Solution. We use the symbols $\gamma_0, \gamma_1, \ldots$ defined in the previous exercise. Since $\left|\gamma_0\right| \leqslant 4^0=1$ and $\left|\gamma_m\right|=4^m$, we have
$$
\left|\left(\frac{3}{4}\right)^m \gamma_m+\gamma_0\right| \geqslant\left|\left(\frac{3}{4}\right)^m\right| \gamma_m|-| \gamma_0|| \geqslant 3^m-1
$$
Similarly, we may estimate
$$
\left|\sum_{n=1}^{m-1}\left(\frac{3}{4}\right)^n \gamma_n\right| \leqslant \sum_{n=1}^{m-1}\left(\frac{3}{4}\right)^n\left|\gamma_n\right| \leqslant \sum_{n=1}^{m-1} 3^n=3 \cdot \frac{3^{m-1}-1}{3-1}=\frac{3^m-3}{2} .
$$
Combining the above estimates with the part a) of the previous exercise gives the inequality
$$
\begin{aligned}
& \left|\frac{f\left(x_0+\delta_m\right)-f\left(x_0\right)}{\delta_m}\right|=\left|\sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^n \gamma_n\right|=\left|\sum_{n=0}^m\left(\frac{3}{4}\right)^n \gamma_n\right| \
\geqslant & ||\left(\frac{3}{4}\right)^m \gamma_m+\gamma_0|-| \sum_{n=1}^{m-1}\left(\frac{3}{4}\right)^n \gamma_n|| \geqslant|| 3^m-1|-| \frac{3^m-3}{2}||=\frac{3^m+1}{2} .
\end{aligned}
$$
Finally, we conclude that the function $f$ can not be differentiable at $x_0$ since the difference quotient $\frac{f\left(x_0+h\right)-f\left(x_0\right)}{h}$ can not tend to a finite limit when $h$ tends to zero.
Compute the Fourier transform of the characteristic function of the interval $[-a, a]$, where $a>0$.
Solution. For any $\xi \in \mathbb{R} \backslash{0}$, we have
$$
\begin{aligned}
\widehat{\chi_{[-a, a]}}(\xi) & =\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \chi_{[-a, a]}(x) e^{-i x \xi} \mathrm{d} x \
& =\frac{1}{\sqrt{2 \pi}} \int_{-a}^a e^{-i x \xi} \mathrm{d} x=\frac{e^{-i a \xi}-e^{i a \xi}}{-\sqrt{2 \pi} i \xi}=\sqrt{\frac{2}{\pi}} \cdot \frac{\sin a \xi}{\xi} .
\end{aligned}
$$
Since $\chi_{[-a, a]} \in \mathscr{L}^1(\mathbb{R})$, the Fourier transform $\widehat{\chi_{[-a, a]}}$ is continuous and we must have
$$
\widehat{\chi_{[-a, a]}}(0)=\lim {\xi \longrightarrow 0} \widehat{\chi{[-a, a]}}(\xi)=\lim _{\xi \longrightarrow 0} \sqrt{\frac{2}{\pi}} \cdot \frac{\sin a \xi}{\xi}=\sqrt{\frac{2}{\pi}} \cdot a .
$$
MY-ASSIGNMENTEXPERT™可以为您提供STANFORD EE261 FOURIER ANALYSIS傅里叶分析的代写代考和辅导服务！
