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

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## 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 $a l l n \geq 4$.
Here is another one that is quite different.
Theorem 2.3.2
$$(1+x)^{n} \geq 1+n x, \quad \forall n \in \mathbb{N}, \quad \forall x \geq-1$$
Proof 2.3.2
BASIS: When $n=1$, we are asking if $1+x \geq 1+x$ when $x \geq-1$ which is actually true for all $x$. So the basis step is verified.
INDUCTIVE. We assume the proposition is true for any $x \geq-1$ and for any $k>1$. Thus, we assume $(1+x)^{k} \geq 1+k x$. Now look at the proposition for $k+1$. We have
$$(1+x)^{k+1}=(1+x)(1+x)^{k} \geq(1+x)(1+k x)$$
We must show $(1+x)(1+k x) \geq(1+(k+1) x)$ for $x \geq-1$. We have to show
$$1+(k+1) x+k x^{2} \quad \stackrel{?}{\geq} 1+(k+1) x .$$
We can cancel the $1+(k+1) x$ on both sides which tells us we must check if $k x^{2} \geq 0$ when $x \geq-1$. This is true. Thus $P(k+1)$ is true and we have verified the inductive step. Hence, by the POMI, $P(n)$ holds for all $n$.

A totally different kind of induction proof is the one below. We want to count how many subsets a set with a finite number of objects can have. We let the number of objects in a set $S$ be denoted by $|S|$. We call this the cardinality of the set $S$. For example, the cardinality of the set ${$ Jim, Pauli, Qait, Quinn $}$ is 4 . Given a set $S, S$ how many subsets does it have?

For example, ${J i m, P a u l i}$ is a subset of the original $S$ defined above. Since the original set has just 4 objects in it, there are 4 subsets with just one object, There are $\left(\begin{array}{l}4 \ 2\end{array}\right)$ ways to choose subsets of 2 objects. Recall $\left(\begin{array}{l}4 \ 2\end{array}\right)$ is $\frac{4 !}{2 ! 2 !}=\frac{24}{4}=6$. There are then $\left(\begin{array}{l}4 \ 3\end{array}\right)$ ways to choose 3 objects which gives $\frac{4 !}{3 ! 1 !}=\frac{24}{6}=4$. Finally there is just one way to choose 4 objects. So the total number of subsets is $1+4+6+4=15$. We always also add in the empty set $\emptyset={}$ to get the total number of subsets is 16 . Note this is the same as $2^{4}$. Hence, we might conjecture that if $S$ had only a finite number of objects in it, the number of subsets of $S$ is $2^{|S|}$. The collection of all subsets of a set $S$ is denoted by $2^{S}$ for this reason and the cardinality of $2^{S}$ is thus $2^{|S|}$. We can prove this using an induction argument.

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

Use the POMI to prove the following propositions.
Exercise 2.3.1 $\frac{1}{1 \cdot 2}+\frac{1}{2-3}+\cdots+\frac{1}{n \cdot(n+1)}=\frac{n}{n+1}$.
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

2n≥n2,∀n≥4

2ķ+1=2×2ķ≥2×ķ2

2ķ2≥ķ2+2ķ+1⇒ķ2≥2ķ+1

(1+X)n≥1+nX,∀n∈ñ,∀X≥−1

(1+X)ķ+1=(1+X)(1+X)ķ≥(1+X)(1+ķX)

1+(ķ+1)X+ķX2≥?1+(ķ+1)X.

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

2不是有理数：

p2=(2ℓ+1)2=4ℓ2+4ℓ+1.

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