Real Analysis – Physicspupil

September 27, 2020November 9, 2022

202009272336 Homework 1 (Q3)

Verify the following statements by definition of limit.

[Hint: Use the -language.]

i. $\displaystyle{\lim_{n\to \infty}\frac{2n^2-1}{4n^2+2}=\frac{1}{2}}$ ;

ii. $\displaystyle{\lim_{n\to \infty}\frac{\sqrt{n^2+n}}{n}=1}$ ;

iii. $\displaystyle{\lim_{n\to \infty}\sqrt[n]{n+2}=1}$ .

Verification.

i. Given

$\displaystyle{\lim_{n\to \infty}\frac{2n^2-1}{4n^2+2}=\frac{1}{2}}$ ,

or,

$\displaystyle{\lim_{n\to \infty}f(n)=\frac{1}{2}}$ ,

in which

$\displaystyle{f(n)=\frac{2n^2-1}{4n^2+2}}$ .

Informally speaking of it, the function $f(n)$ shall go to the limit $1/2$ if its variable $n$ goes to the ( $+\textrm{ve}$ ) infinity.

Remark.

In other words, if a function $f$ tends to its limit $L$ , by that it is to say, there exists some output value(s) $f(x)$ sufficiently close to the number $L$ .

Make-up arguments.

As needs

,

so let

Roughwork. (Instructive)

,

s.t.

(to be continued)

Definition. (Limit) Let $f$ be a real-valued function defined on a subset $D$ of the real numbers $\mathbb{R}$ . Let $c$ be a limit point of $D$ and let $L$ be a real number. Symbolically:

$\begin{aligned} & \qquad\enspace \lim_{x\to c} f(x) = L \\ & \Longleftrightarrow \bigg( \forall\, \epsilon >0,\, \exists\, \delta >0,\, \forall\, x\in D, 0<|x-c|<\delta \Rightarrow \big| f(x)-L \big| < \epsilon \bigg) \\ \end{aligned}$

_{Wikipedia on -definition of limit}

(continue)

The presentation below is based on the Suggested Solution:

First,

$\bigg| \displaystyle{\frac{2n^2-1}{4n^2+2}-\frac{1}{2}}\bigg| = \bigg| \frac{1}{2n^2+1}\bigg| <\bigg| \frac{1}{n^2} \bigg|$ .

Then, just do let

$N=N(\epsilon )=\bigg[ \displaystyle{\frac{1}{\sqrt{\epsilon}}} \bigg] +1$ .

Lastly, $\forall\, \epsilon >0$ , it stands that

$\bigg| \displaystyle{\frac{2n^2-1}{4n^2+2}-\frac{1}{2}}\bigg| <\epsilon$

whenever $n\ge N$ .

QED

Part ii. and part iii. are noteworthy exercises that have yet to be done.

September 27, 2020May 15, 2023

202009270104 Homework 1 (Q1)

i. Prove that $\sqrt{6}$ is irrational; and

ii. Determine whether $\sqrt{2}+\sqrt{3}$ is rational or not.

Solution.

i. Assume on the contrary that $\sqrt{6}$ be rational, may I rewrite it in terms of a quotient (or a fraction) where the numerator and the nonzero denominator are a pair of coprime integers.

I.e., $\sqrt{6}=\displaystyle{\frac{p}{q}}$ where $p,q \in\mathbb{Z}$ and $(p,q)=1$ .

Roughwork.

$\begin{aligned} \sqrt{6} & = \frac{p}{q} \\ 6 & = \frac{p^2}{q^2} \\ p^2 & = 6q^2 \\ & = 2(3q^2) \\ \therefore \qquad 2\big| p^2 & \Rightarrow 2\big| p \textrm{\qquad(why?)} \\ \end{aligned}$

From $p^2=6q^2$ we have $2\big| p$ .

So let $p=2r$ for some integer $r$ .

Roughwork.

We have also seen $2\big| q$ .

If and when both statements $2\big| p$ and $2\big| q$ meet, it implies $p$ and $q$ are no more coprime, and thus a contradiction.

Definition. (Coprime) Two integers $a$ and $b$ are said to be coprime if the only positive integer that divides them both is one. Equivalently speaking, the greatest common divisor (gcd) of $a$ and $b$ is 1,i.e., $\textrm{gcd}(a,b)=1$ , or written simply, $(a,b)=1$ . Synonymous with coprime' are relatively prime’, `mutually prime’, and the like.

ii.

To prove or disprove from scratch, assume that $\sqrt{2}+\sqrt{3}$ be rational, and see what happens. Let

$\sqrt{2}+\sqrt{3}=\displaystyle{\frac{p}{q}}$

for some coprimes $p,\, q\in\mathbb{Z}$ s.t. $(p,q)=1$ .

$\begin{aligned} \sqrt{3} & = \frac{p}{q} - \sqrt{2} \\ 3 & = \frac{p^2}{q^2} - 2\bigg(\frac{p}{q}\bigg)(\sqrt{2}) +2 \\ 2(\sqrt{2})\bigg(\frac{p}{q}\bigg) & = \frac{p^2}{q^2} - 1 = \frac{p^2-q^2}{q^2} \\ \sqrt{2} & = \frac{p^2-q^2}{2pq} \\ \sqrt{2} & = \frac{p}{2q} - \frac{q}{2p} \in \mathbb{Q} \\ \textrm{As is proven,\enspace} & \sqrt{2}\in \mathbb{R}\backslash\mathbb{Q} \\ \end{aligned}$

Contradiction arises $( \Rightarrow \Leftarrow )$ .

$\therefore$ $\sqrt{2}+\sqrt{3}$ is irrational.

(to be continued)

(doing another way around)

To make use of part i., one might find the Lemma below useful:

Lemma. The product of any two rational numbers is again one rational number.

Proof. Let $\displaystyle{\frac{a}{b}}$ and $\displaystyle{\frac{c}{d}}$
be rational numbers in their simplest forms reducible.
$\begin{aligned} \bigg( \frac{a}{b}\bigg) \bigg( \frac{c}{d}\bigg) & = \bigg( \frac{ac}{bd}\bigg) \\ \because \quad a,b,c,d\in\mathbb{Z} & \Rightarrow ac,\, bd\in\mathbb{Z}\\ \therefore \bigg( \frac{a}{b}\bigg) , \bigg( \frac{c}{d}\bigg) \in \mathbb{Q} & \Rightarrow \bigg( \frac{ac}{bd}\bigg) \in \mathbb{Q} \\ \end{aligned}$

If $\sqrt{2}+\sqrt{3}$ were rational, again were its square rational in view of the aforementioned Lemma. Just make an experiment in so doing:

$\begin{aligned} (\sqrt{2}+\sqrt{3})^2 & = (\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2 \\ & = 5 + 2\sqrt{6} \\ \end{aligned}$

On logic,

$\begin{aligned} & \quad (\sqrt{2}+\sqrt{3})^2 \textrm{\quad is rational} \\ & \Rightarrow 5+2\sqrt{6} \textrm{\quad is rational} \\ & \stackrel{\textrm{why?}}{\Longrightarrow} \sqrt{6} \textrm{\quad is rational} \\ \end{aligned}$

However, as shown in part i., $\sqrt{6}$ is irrational. The assumption that $\sqrt{2}+\sqrt{3}$ be rational has been contradicted.

The contrary is true that $\sqrt{2}+\sqrt{3}$ is $\textrm{\scriptsize{NOT}}$ rational.

August 19, 2020November 9, 2022

202008190019 Exercise 1.28

In each case, determine all real $x$ and $y$ which satisfy the given relation.

(a) $x+\mathrm{i}y=|x-\mathrm{i}y|$ ,

(b) $x+\mathrm{i}y=(x-\mathrm{i}y)^2$ ,

(c) $\displaystyle{\sum_{k=0}^{100}}\mathrm{i}^k=x+\mathrm{i}y$ .

Solution.

(a)

$\begin{aligned} x + \mathrm{i}y & = |x-\mathrm{i}y| \\ x + \mathrm{i}y & = \sqrt{(x)^2+(-y)^2} \\ x + \mathrm{i}y & = \big( \sqrt{(x)^2+(-y)^2} \big) + \mathrm{i}(0) \end{aligned}$

$\therefore y=0$ and $x\enspace (\geqslant 0)\, \in\mathbb{R}$ is any non-negative real number.

(b)

$\begin{aligned} x + \mathrm{i}y & = (x-\mathrm{i}y)^2 \\ x + \mathrm{i}y & = x^2 - 2xy\mathrm{i} + (\mathrm{i}y)^2 \\ x + \mathrm{i}y & = (x^2-y^2) + \mathrm{i}(-2xy) \end{aligned}$

Comparing the imaginary parts:

$\begin{aligned} y=-2xy & \Rightarrow (2x+1)y = 0 \\ & \Rightarrow x = -\frac{1}{2}\qquad \textrm{or}\qquad y=0 \end{aligned}$

Comparing the real parts:

$\begin{aligned} x & = x^2-y^2 \\ x^2 - x - y^2 & = 0 \\ x & = \frac{-(-1)\pm \sqrt{(-1)^2-4(1)(-y^2)}}{2(1)} \\ & = \frac{1\pm \sqrt{1+4y^2}}{2} \end{aligned}$

When $y=0$ , $x=\displaystyle{\frac{1\pm \sqrt{1+4(0)^2}}{2}}=0,\, 1$ .

When $x=-\frac{1}{2}$ ,

$\begin{aligned} -\frac{1}{2} & =\frac{1\pm\sqrt{1+4y^2}}{2} \\ \Rightarrow 2 & = \sqrt{1+4y^2} \\ \Rightarrow y & = \pm \displaystyle{\frac{\sqrt{3}}{2}} \end{aligned}$

$\therefore \begin{pmatrix} x \\ y \end{pmatrix}\in \Bigg\{ \begin{pmatrix} 0 \\ 0 \end{pmatrix},\,\begin{pmatrix} 1 \\ 0 \end{pmatrix},\, \begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix},\, \begin{pmatrix} -\frac{1}{2} \\ -\frac{\sqrt{3}}{2} \end{pmatrix} \Bigg\}$ .

(c)

$\begin{aligned} \sum_{k=0}^{k=100}\mathrm{i}^k & = x+\mathrm{i}y \\ \mathrm{i}^{0} + \mathrm{i}^{1} + \mathrm{i}^{2} + \dots + \mathrm{i}^{100} & = x+\mathrm{i}y \\ \sum_{n=0}^{50} \mathrm{i}^{(2n)} + \sum_{n=0}^{49} \mathrm{i}^{(2n+1)} & = x+\mathrm{i}y \\ \sum_{n=0}^{50} (-1)^n + \mathrm{i}\sum_{n=0}^{49} (-1)^{n} & = x+\mathrm{i}y \\ \end{aligned}$

$\begin{aligned} \therefore \qquad x & = \sum_{n=0}^{50}(-1)^n \\ & = (-1)^{0} + (-1)^{1} + (-1)^{2} +\dots + (-1)^{50} \\ & = 1-1+1+\dots +1 \\ & = 1 \end{aligned}$

$\begin{aligned} \therefore \qquad y & = \sum_{n=0}^{49}(-1)^n \\ & = (-1)^{0} + (-1)^{1} + (-1)^{2} +\dots + (-1)^{49} \\ & = 1-1+1+\dots -1 \\ & = 0 \end{aligned}$

Sol. $\bigg\{\begin{aligned} & x = 1 \\ & y = 0 \end{aligned}\bigg\}$ .

August 18, 2020November 9, 2022

202008182229 Exercise 1.27

Express the following complex numbers in the form $a+b\,\mathrm{i}$ .

(a) $(1+\mathrm{i})^3$ ,

(b) $(2+3\mathrm{i})(3-4\mathrm{i})$ ,

(c) $\mathrm{i}^5+\mathrm{i}^{16}$ ,

(d) $\frac{1}{2}(1+\mathrm{i})(1+\mathrm{i}^{-8})$ .

Solution.

(a)

$\begin{aligned} & \quad\enspace (1+\mathrm{i})^3 \\ & = 1^3 + 3(1)^2(\mathrm{i}) +3(1)(\mathrm{i})^2 + \mathrm{i}^3 \\ & = 1 + 3\mathrm{i} - 3 -\mathrm{i} \\ & = -2+2\mathrm{i} \end{aligned}$

(b)

$\begin{aligned} & \quad\enspace \frac{2+3\mathrm{i}}{3-4\mathrm{i}} \\ & = \bigg(\frac{2+3\mathrm{i}}{3-4\mathrm{i}}\bigg) \bigg( \frac{3+4\mathrm{i}}{3+4\mathrm{i}} \bigg) \\ & = \frac{(2+3\mathrm{i})(3+4\mathrm{i})}{(3)^2-(4\mathrm{i})^2} \\ & = \frac{6+8\mathrm{i}+9\mathrm{i}+12\mathrm{i}^2}{9+16} \\ & = -\frac{6}{25} + \frac{17}{25}\mathrm{i} \end{aligned}$

(c)

$\begin{aligned} & \quad\enspace \mathrm{i}^5+\mathrm{i}^{16} \\ & = (\mathrm{i}^2)^2\mathrm{i} + (\mathrm{i}^2)^8 \\ & = (-1)^2\mathrm{i} + (-1)^8 \\ & = 1+\mathrm{i} \end{aligned}$

(d)

$\begin{aligned} & \quad\enspace \frac{1}{2}(1+\mathrm{i})(1+\mathrm{i}^{-8}) \\ & = \frac{1}{2}(1+\mathrm{i}^{-8}+\mathrm{i} + \mathrm{i}^{-7}) \\ & = \frac{1}{2} \bigg( 1+ (\mathrm{i}^{-2})^4 + \mathrm{i} + (\mathrm{i}^{-2})^4\mathrm{i} \bigg) \\ & (\, \dots \enspace \mathrm{i}^{-2} = \frac{1}{\mathrm{i}^2} = -1 \enspace \dots\, ) \\ & = \frac{1}{2} \bigg( 1+(-1)^4 + \mathrm{i} + (-1)^4\mathrm{i} \bigg) \\ & = \frac{1}{2} (2+2\mathrm{i}) \\ & = 1+\mathrm{i} \end{aligned}$

April 24, 2020October 24, 2022

202004241649 Problem 2, Ch. 1 Sec. 1

Prove the following equalities:

(a) $|ab|=|a|\cdot |b|$ ;

(b) $|a|^2=a^2$ ;

(c) $\displaystyle{\bigg|\frac{a}{b}\bigg|=\frac{|a|}{|b|}}$ (where $b\neq 0$ );

(d) $\sqrt{a^2}=|a|$ .

Proof.

(a) (proof by cases)

i. When $a,b\geqslant 0$ :

$|ab|=ab=|a||b|$ .

ii. When $a,b< 0$ :

$|ab|=ab=(-a)(-b)=|a||b|$ .

iii. When $a\geqslant 0$ and $b<0$ :

$|ab|=-ab=a(-b)=|a||b|$ .

iv. When $a<0$ and $b\geqslant 0$ :

$|ab|=-ab=(-a)b=|a||b|$ .

QED

(b) (proof by induction)

By making a stronger claim

$P(n)$ : For any positive integer $n$ , $|a^n|=|a|^n$ .

Proof.

The trivial cases $n=0$ and $n=1$ are evident.

Consider the case $n=2$ ,

$\begin{aligned} |a^2| & = |a||a| \qquad \textrm{(by equality (a))} \\ & = |a|^2 \end{aligned}$

$P(2)$ is true.

Assume now that $P(n)$ is true,

$\begin{aligned} P(n+1): \qquad |a^{n+1}|& = |a^n\cdot a| \\ & = |a^n||a| \qquad \textrm{(by equality (a))}\\ & = |a|^n|a| \qquad \textrm{(by the assumption }P(n)\textrm{ is true)} \\ & = |a|^{n+1} \end{aligned}$