### 数学代写|实变函数作业代写Real analysis代考|MATH 350

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• Statistical Computing 统计计算
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|实变函数作业代写Real analysis代考|More Abstract Proofs and Even Trickier POMIs

The induction step can have a lot of algebraic manipulation so we need to look at some of those. Let’s start with this one.
Theorem 2.3.1
$$2^{n} \geq n^{2}, \quad \forall n \geq 4$$
Proof 2.3.1
BASIS $P(4)$ is the statement $2^{4}=16 \geq 4^{2}=16$ which is true. So the basis step is verified. INDUCTIVE. We assume $P(k)$ is true for an arbitrary $k>4$. Hence, we know $2^{k} \geq k^{2}$. Now look at $P(k+1)$. We note
$$2^{k+1}=2 \times 2^{k} \geq 2 \times k^{2}$$

We need to show $2^{k+1} \geq(k+1)^{2}$ so we must show $2 \times k^{2} \geq(k+1)^{2}$. We can simplify this by multiplying both sides out to get
$$2 k^{2} \geq k^{2}+2 k+1 \Rightarrow k^{2} \geq 2 k+1$$
We can answer this question by doing another POMI inside this one or we can figure it out graphically. Draw the graph of $x^{2}$ and $2 x+1$ together and you can clearly see $k^{2}>2 k+1$ when $k>3$.
Thus $P(k+1)$ is true and we have verified the inductive step. Hence, by the POMI, $P(n)$ holds for all $n \geq 4$.
Here is another one that is quite different.

Use the POMI to prove the following propositions.
Exercise 2.3.1 $\frac{1}{1 \cdot 2}+\frac{1}{2-3}+\cdots+\frac{1}{n-(n+1)}=\frac{n}{n+1}$.
Exercise 2.3.2 $\frac{d}{d x} x^{n}=n x^{n-1}, \quad \forall x, \forall n \in \mathbb{N}$. You can assume you know the powers $f(x)=x^{n}$ are differentiable and that you know the product rule: if $f$ and $g$ are differentiable, then $(f g)^{\prime}=$ $f^{\prime} g+f g^{\prime}$
Exercise 2.3.3 $1+x+\cdots+x^{n}=\frac{1-x^{n+1}}{1-x}, \quad \forall x \neq 1, \forall n \in \mathbb{N}$.
Exercise 2.3.4 $\int x^{n} d x=\frac{1}{n+1} x^{n+1}, \quad \forall n \in \mathbb{N}$. You can assume you know integration by parts. The basis step is $\int x d x=x^{2} / 2$ which you can assume you know. After that it is integration by parts.
Another type of proof is one that is done by contradiction.
Theorem 2.4.1 $\sqrt{2}$ is not a Rational Number
$\sqrt{2}$ is not a rational number.
Proof 2.4.1
We will prove this technique using a technique called contradiction. Let’s assume we can find positive integers $p$ and $q$ so that $2=(p / q)^{2}$ with $p$ and $q$ having no common factors. When this happens we say $p$ and $q$ are relatively prime. This tells us $p^{2}=2 q^{2}$ which also tells us $p^{2}$ is divisible by 2. Thus, $p^{2}$ is even. Does this mean $p$ itself is even? Well, if $p$ was odd, we could write $p=2 \ell+1$ for some integer $\ell$. Then, we would know
$$p^{2}=(2 \ell+1)^{2}=4 \ell^{2}+4 \ell+1 .$$
The first two terms, $4 \ell^{2}$ and $4 \ell$ are even, so this implies $p^{2}$ would be odd. So we see $p$ odd implies $p^{2}$ is odd. Thus, we see $p$ must be even when $p^{2}$ is even. So we now know $p=2 k$ for some integer $k$ as it is even. But since $p^{2}=2 q^{2}$, we must have $4 k^{2}=2 q^{2}$. But this says $q^{2}$ must be even.

The same reasoning we just used to show $p$ odd implies $p^{2}$ is odd, then tells us $q$ odd implies $q^{2}$ is odd. Thus $q$ is even too,

Now here is the contradiction. We assumed $p$ and $q$ were relatively prime; i.e. they had no common factors. But if they are both even, they share the factor 2 . This is the contradiction we seek.

## 数学代写|实变函数作业代写Real analysis代考|More Abstract Proofs and Even Trickier POMIs

$$2^{n} \geq n^{2}, \quad \forall n \geq 4$$

$$2^{k+1}=2 \times 2^{k} \geq 2 \times k^{2}$$

$$2 k^{2} \geq k^{2}+2 k+1 \Rightarrow k^{2} \geq 2 k+1$$

$\mathrm{~ 练 习 ~ 2 . 3 . 3 1 + x +}$

$2.4$ 一些矛盾证明

$\sqrt{2}$ 不是有理数。

$p^{2}$ 甚至。意思是不是 $p$ 本身是偶数? 好吧，如果 $p$ 很奇怪，我们可以写 $p=2 \ell+1$ 对于某个整数 $\ell$. 那么，我们就会 知道
$$p^{2}=(2 \ell+1)^{2}=4 \ell^{2}+4 \ell+1$$

。这就是我们寻求的矛盾。

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