## 数学代写|实分析作业代写Real analysis代考|MATH2023

2023年3月27日

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## 数学代写|实分析作业代写Real analysis代考|Simpson’s Rule

The trapezoidal approximation $T_n(f)$ amounts to approximating the function $f$ with a piecewise linear function $g_n$ that passes through the points $\left{\left(x_i, f\left(x_i\right)\right)\right}, i=0, \ldots, n$. Our intuition should convince us that one way to obtain a better approximation to the integral of $f$ over $[a, b]$ is to use smoother curves. This is exactly what is done in Simpson’s rule which uses parabolas to approximate the integral. To use quadratic approximations we will need to use three successive points of the partition of $[a, b]$. This is due to the fact that three points are required to uniquely determine a parabola.

Prior to deriving Simpson’s rule, we first establish the following formula: If $p(x)=A x^2+B x+C$ is the quadratic function passing through the points $\left(0, y_0\right),\left(h, y_1\right),\left(2 h, y_2\right)$, then
$$\int_0^{2 h} p(x) d x=\frac{h}{3}\left[y_0+4 y_1+y_2\right]$$
One way to derive this formula would be to first determine the coefficients $A, B, C$ so that $p(0)=y_0, p(h)=y_1, p(2 h)=y_2$, and then integrate $p(x)$. This however is not necessary. By integrating first,
\begin{aligned} \int_0^{2 h} p(x) d x & =\frac{A}{3}(2 h)^3+\frac{B}{2}(2 h)^2+C(2 h) \ & =\frac{h}{3}\left[8 A h^2+6 B h+6 C\right] \ & =\frac{h}{3}[p(0)+4 p(h)+p(2 h)]=\frac{h}{3}\left[y_0+4 y_1+y_2\right] \end{aligned}
Let $f \in \mathcal{R}[a, b]$ and let $n \in \mathbb{N}$ be even. Set $h=(b-a) / n$. On each of the intervals $[a+2(i-1) h, a+2 i h], i=1, \ldots, n / 2$, we approximate the integral of $f$ by the integral of the quadratic function that agrees with $f$ at the points
$$y_0=f(a+2(i-1) h), \quad y_1=f(a+(2 i-1) h), \quad y_2=f(a+2 i h)$$

## 数学代写|实分析作业代写Real analysis代考|Proof of Lebesgue’s Theorem

In this final section, we present a self-contained proof of Lebesgue’s characterization of the Riemann integrable functions on $[a, b]$. Recall that a subset $E$ of $\mathbb{R}$ has measure zero if for any $\epsilon>0$, there exists a finite or countable collection $\left{I_n\right}$ of open intervals with $E \subset \bigcup_n I_n$ and $\sum_n \ell\left(I_n\right)<\epsilon$, where $\ell\left(I_n\right)$ denotes the length of the interval $I_n$. We begin with several preparatory lemmas.

LEMMA 6.7.1 A finite or countable union of sets of measure zero has measure zero.

Proof. We will prove the lemma for the case of a countable sets of measure zero. The result for a finite union is an immediate consequence.

Suppose $\left{E_n\right}_{n \in \mathbb{N}}$ is a countable collection of sets of measure zero. Set $E=\bigcup_n E_n$ and let $\epsilon>0$ be given. Since each set $E_n$ is a set of measure zero, for each $n \in \mathbb{N}$ there exists a finite or countable collection $\left{I_{n, k}\right}_k$ of open intervals such that $E_n \subset \bigcup_k I_{n, k}$ and $\sum_k \ell\left(I_{n, k}\right)<\epsilon / 2^k$. Since we can always take $I_{n, k}$ to be the empty set, there is no loss of generality in assuming that the collection $\left{I_{n, k}\right}_k$ is countable. Then $\left{I_{n, k}\right}_{n, k}$ is again a countable collection of open intervals with $E \subset \bigcup_{n, k} I_{n, k}$.

Since $\mathbb{N} \times \mathbb{N}$ is countable, there exists a one-to-one function $f$ from $\mathbb{N}$ onto $\mathbb{N} \times \mathbb{N}$. For each $m \in \mathbb{N}$, set $J_m=I_{f(m}$. Then $\left{J_m\right}_{m \in \mathbb{N}}$ is a countable collection of open intervals with $E \subset \bigcup_m J_m$. Since $f$ is one-to-one, for each $N \in \mathbb{N}$, the set $F_N=f({1, \ldots, N})$ is a finite subset of $\mathbb{N} \times \mathbb{N}$. Hence there exists positive integers $N_1$ and $K_1$ such that for all $(n, k) \in F_N$ we have $1 \leq n \leq N_1$ and $1 \leq k \leq K_1$. Hence
$$\sum_{m=1}^N \ell\left(J_m\right)=\sum_{(n, k) \in F_N} \ell\left(I_{n, k}\right) \leq \sum_{(n, k) \in N_1 \times K_1} \ell\left(I_{n, k}\right)$$

# 实分析代写

## 数学代写|实分析作业代写Real analysis代考|Simpson’s Rule

$$\int_0^{2 h} p(x) d x=\frac{h}{3}\left[y_0+4 y_1+y_2\right]$$

$$\int_0^{2 h} p(x) d x=\frac{A}{3}(2 h)^3+\frac{B}{2}(2 h)^2+C(2 h)$$

$$y_0=f(a+2(i-1) h), \quad y_1=f(a+(2 i-1) h)$$

## 数学代写|实分析作业代写Real analysis代考|Proof of Lebesgue’s Theorem

$\sum_k \ell\left(I_{n, k}\right)<\epsilon / 2^k$. 因为我们总能拿 $I_{n, k}$ 是空集，假设 集合不失一般性 \left{1_{n, k}\right } } k \text { 是可数的。然后 } $E \subset \bigcup_{n, k} I_{n, k}$

$$\sum_{m=1}^N \ell\left(J_m\right)=\sum_{(n, k) \in F_N} \ell\left(I_{n, k}\right) \leq \sum_{(n, k) \in N_1 \times K_1} \ell\left(I_{n, k}\right)$$

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## MATLAB代写

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