### 数学代写|勒贝格积分代写Lebesgue Integration代考|Math720

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|勒贝格积分代写Lebesgue Integration代考|Continuity and Differentiability

The fourth big question asks for the relationship between continuity and differentiability. We know that a function that is differentiable at a given value of $x$ must also be continuous at that value, and it is clear that the converse does not hold. The function $f(x)=|x|$ is continuous but not differentiable at $x=0$. But how nondifferentiable can a continuous function be?

Throughout the first half of the nineteenth century, it was generally believed that a continuous function would be differentiable at most points. ${ }^4$ Mathematicians recognized that a function might have finitely many values at which it failed to have a derivative. There might even be a sparse infinite set of points at which a continuous function was not differentiable, but the mathematical community was honestly surprised when, in 1875 , Gaston Darboux and Paul du-Bois Reymond ${ }^5$ published examples of continuous functions that are not differentiable at any value.
The question then shifted to what additional assumptions beyond continuity would ensure differentiability. Monotonicity was a natural candidate. Weierstrass constructed a strictly increasing continuous function that is not differentiable at any algebraic number, that is to say, at any number that is the root of a polynomial with rational coefficients. It is not differentiable at $1 / 2$ or $\sqrt{2}$ or $\sqrt[3]{5}-2 \sqrt[21]{35}$. Weierstrass’s function is differentiable at $\pi$. Can we find a continuous, increasing function that is not differentiable at any value? The surprising answer is No. In fact, in a sense that later will be made precise, a continuous, monotonic function is differentiable at “most” values of $x$. There are very important subtleties lurking behind this fourth question.

## 数学代写|勒贝格积分代写Lebesgue Integration代考|Term-by-term Integration

Returning to Fourier series, we saw that the heuristic justification relied on interchanging summation and integration, integrating an infinite series of functions by integrating each summand. This works for finite summations. It is not hard to find infinite series for which term-by-term integration leads to a divergent series or, even worse, a series that converges to the wrong value.

Weierstrass had shown that if the series converges uniformly, then term-by-term integration is valid. The problem with this result is that the most interesting series, especially Fourier series, often do not converge uniformly and yet term-by-term integration is valid. Uniform convergence is sufficient, but it is very far from necessary. As we shall see, finding useful conditions under which term-by-term integration is valid is very difficult so long as we cling to the Riemann integral. As Lebesgue would show in the opening years of the twentieth century, his definition of the integral yields a simple, elegant solution, the Lebesgue dominated convergence theorem.

1.1.1. Find the Fourier expansions for $f_1(x)=x$ and $f_2(x)=x^2$ over $[-\pi, \pi]$.
1.1.2. For the functions $f_1$ and $f_2$ defined in Exercise 1.1.1, differentiate each summand in the Fourier series for $f_2$. Do you get the summands in the Fourier series for $2 f_1$ ? Differentiate each summand in the Fourier series for $f_1$. Do you get the summand in the Fourier series for $f_1^{\prime}(x)$ ?
1.1.3. Using the Fourier series expansion for $x^2$ (Exercise 1.1.1) evaluated at $x=\pi$, show that
$$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6} .$$
1.1.4. Show that if $k$ is an integer $\geq 1$, then
$$\int_{-\pi}^\pi \cos (k x) d x=\int_{-\pi}^\pi \sin (k x) d x=0 .$$

# 勒贝格积分代考

## 数学代写|勒贝格积分代写Lebesgue Integration代考|Term-by-term Integration

Weierstrass 已经表明，如果级数一致收敛，则逐项积分是有效的。这个结果的问题在于，最有趣的级 数，尤其是傅立叶级数，通常不会一致收敛，但逐项积分是有效的。均匀收敛就足够了，但远非必要。 正如我们将要看到的，只要我们坚持黎曼积分，就很难找到使逐项积分有效的有用条件。正如勒贝格在 20 世纪初所表明的那样，他对积分的定义产生了一个简单、优雅的解决方案，即勒贝格支配的收敛定 理。
1.1.1. 求傅立叶展开式 $f_1(x)=x$ 和 $f_2(x)=x^2$ 超过 $[-\pi, \pi]$.
1.1.2. 对于函数 $f_1$ 和 $f_2$ 在练习 1.1.1 中定义，区分傅立叶级数中的每个被加数 $f_2$. 你得到傅立叶级数中的 被加数了吗 $2 f_1$ ? 区分傅里叶级数中的每个被加数 $f_1$. 你得到傅立叶级数中的被加数了吗 $f_1^{\prime}(x)$ ?
1.1.3. 使用傅里叶级数展开 $x^2$ (练习 1.1.1) 评估于 $x=\pi$ ， 显示
$$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$
1.1.4. 证明如果 $k$ 是一个整数 $\geq 1$ ， 然后
$$\int_{-\pi}^\pi \cos (k x) d x=\int_{-\pi}^\pi \sin (k x) d x=0$$

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## MATLAB代写

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