September 2020 – Physicspupil

September 29, 2020November 3, 2022

202009290227 Exercise 2.3.5

Prove that $10^{n+1}+10^n+1$ is divisible by $3$ for $n\in\mathbb{N}$ .

_{Extracted from T. W. Judson. (2021). Abstract Algebra Theory and Applications.}

Proof.

Let $P(n)$ be the statement:

$P(n):\qquad 3\,\Big| (10^{n+1}+10^n+1)$ for any $n\in\mathbb{N}$

Determine whether or not $P(n)$ is true when $n=1$ :

$P(1): \qquad 3\,\Big| (10^{(1)+1}+10^{(1)}+1)$

As $10^{(1)+1}+10^{(1)}+1 = 111 = 37\cdot 3$ is divisible by $3$ , $P(1)$ is true.

Suppose $P(n)$ is true for some $n\, (\geqslant 1) \in\mathbb{N}$ , try and prove the statement $P(n+1)$ :

$P(n+1): \qquad 3\,\Big| (10^{(n+1)+1}+10^{(n+1)}+1)$

$\begin{aligned} &\quad 10^{(n+1)+1}+10^{(n+1)}+1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +10 - 9 \\ = &\quad 10\cdot (10^{n+1}+10^n+1) - 9 \\ \end{aligned}$

As $P(n)$ is true and by the fact that three divides nine, $10\cdot (10^{n+1}+10^n+1) - 9$ is therefore divisible by $3$ . That is,

$P(n)\textrm{\enspace is true\enspace}\Rightarrow P(n+1)\textrm{\enspace is true\enspace}$

That $P(n)$ is true for $n=1$ , by the principle of mathematical induction, I have thus proven $10^{n+1}+10^n+1$ is divisible by $3$ for $n\in\mathbb{N}$ .

September 29, 2020May 8, 2023

202009290114 Exercise 2.3.1

Prove that

$1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$

for $n\in\mathbb{N}$ .

Proof.

Let $P(n)$ be the statement

$1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$ for $n\in\mathbb{N}$ .

When $n=1$ , check for the validity of $P(1)$ :

$\begin{aligned} \textrm{LHS} & = 1^2 = 1 \\ \textrm{RHS} & = \frac{(1)\big( (1)+1\big) \big( 2(1)+1\big) }{6} = 1 \\ \end{aligned}$

$\because \textrm{LHS}=\textrm{RHS}$

$\therefore P(n)$ is true for $n=1$ .

Suppose $P(n)$ holds true, let’s see if $P(n+1)$ holds too:

$P(n+1):\qquad 1^2+2^2+\cdots + n^2 + (n+1)^2 \stackrel{?}{=} \displaystyle{\frac{(n+1)\big( (n+1) +1 \big) \big( 2(n+1) +1 \big) }{6}}$

Beginning with the left hand side,

$\begin{aligned} \textrm{LHS} & = 1^2+2^2+\cdots + n^2 + (n+1)^2 \\ & = P(n) + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\ & = \frac{(n+1)\big[ (n)(2n+1) + 6(n+1) \big]}{6} \\ & = \frac{(n+1)(2n^2+n+6n+6)}{6} \\ & = \frac{(n+1)(2n^2+7n+6)}{6} \\ & = \frac{(n+1)(2n+3)(n+2)}{6} \\ \end{aligned}$

Then turn to the right hand side,

$\begin{aligned} \textrm{RHS} & = \frac{(n+1)\big( (n+1) +1 \big) \big( 2(n+1) +1 \big) }{6} \\ & = \frac{(n+1)(n+2)(2n+3)}{6} \\ \end{aligned}$

$\because \textrm{LHS} = \textrm{RHS}$

$\therefore P(n+1)$ holds when $P(n)$ holds.

As is proven $P(1)$ is true, by the principle of mathematical induction, $P(n)$ is also true for $n\geqslant 1$ , i.e.,

$\forall\, n\in\mathbb{N},\qquad 1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$

September 28, 2020October 26, 2022

202009281158 Logic the Basics

Notation and notion

$\begin{array}{ccc} \hline \textrm{\textbf{Statement connective}} & \qquad \qquad & \textrm{\textbf{Abbreviation}} \\ \hline \\ \textrm{\small{AND}} & & \wedge \\ \textrm{\small{OR}} & & \vee \\ \textrm{\small{IMPLIES}} & & \Rightarrow \\ \textrm{\small{IF AND ONLY IF}} & & \Leftrightarrow \\ \textrm{\small{IT IS NOT THE CASE}} & & \neg \\ \end{array}$

$\begin{array}{ccc} \hline & \qquad\qquad & \textrm{\textbf{Parenthesized expressions}} \\ \hline \\ (\textrm{\textbf{highest precedence}}) & & \textrm{\small{NOT}}\enspace (\neg ) \\ & & \textrm{\small{AND}}\enspace (\wedge )\textrm{,\,} \textrm{\small{OR}}\enspace (\vee ) \\ & & \textrm{\small{IMPLIES}}\enspace (\Rightarrow ) \\ (\textrm{\textbf{lowest precedence}}) & &\textrm{\small{IF AND ONLY IF}} \enspace (\Leftrightarrow ) \\ \end{array}$

Truth tables

$\begin{array}{c|c|c|c|c|c} \hline & & \textrm{\textbf{\scriptsize{AND}}} & & & \textrm{\textbf{\scriptsize{OR}}} \\ \hline P & Q & P\wedge Q & P & Q & P\vee Q \\ \hline \textrm{T} & \textrm{T} & \textrm{T} & \textrm{T} & \textrm{T} & \textrm{T} \\ \textrm{T} & \textrm{F} & \textrm{F} & \textrm{T} & \textrm{F} & \textrm{T} \\ \textrm{F} & \textrm{T} & \textrm{F} & \textrm{F} & \textrm{T} & \textrm{T} \\ \textrm{F} & \textrm{F} & \textrm{F} & \textrm{F} & \textrm{F} & \textrm{F} \\ \end{array}$

$\begin{array}{c|c|c|c|c} \hline & \textrm{\textbf{\scriptsize{NOT}}} & & & \textrm{\textbf{\scriptsize{IMPLIES}}} \\ \hline P & \neg P & P & Q & P\Rightarrow Q \\ \hline \textrm{T} & \textrm{F} & \textrm{T} & \textrm{T} & \textrm{T} \\ \textrm{F} & \textrm{T} & \textrm{T} & \textrm{F} & \textrm{F} \\ & & \textrm{F} & \textrm{T} & \textrm{T} \\ & & \textrm{F} & \textrm{F} & \textrm{T} \\ \end{array}$

$\begin{array}{c|c|c|c|c|c} \hline & & \textrm{\textbf{\scriptsize{IF AND ONLY IF}}} & & & \textrm{\textbf{\scriptsize{EXCLUSIVE-OR}}} \\ \hline P & Q & P\Leftrightarrow Q & P & Q & P\veebar Q \\ \hline \textrm{T} & \textrm{T} & \textrm{T} & \textrm{T} & \textrm{T} & \textrm{F} \\ \textrm{T} & \textrm{F} & \textrm{F} & \textrm{T} & \textrm{F} & \textrm{T} \\ \textrm{F} & \textrm{T} & \textrm{F} & \textrm{F} & \textrm{T} & \textrm{T} \\ \textrm{F} & \textrm{F} & \textrm{T} & \textrm{F} & \textrm{F} & \textrm{F} \\ \end{array}$

$\begin{array}{c|c|c|c|c|c} \hline & & \textrm{\textbf{\scriptsize{NOR}}} & & & \textrm{\textbf{\scriptsize{NAND}}} \\ \hline P & Q & P\downarrow Q & P & Q & P\uparrow Q \\ \hline \textrm{T} & \textrm{T} & \textrm{F} & \textrm{T} & \textrm{T} & \textrm{F} \\ \textrm{T} & \textrm{F} & \textrm{F} & \textrm{T} & \textrm{F} & \textrm{T} \\ \textrm{F} & \textrm{T} & \textrm{F} & \textrm{F} & \textrm{T} & \textrm{T} \\ \textrm{F} & \textrm{F} & \textrm{T} & \textrm{F} & \textrm{F} & \textrm{T} \\ \end{array}$

Logical equivalence

a. (rule of double negation)

$\neg\neg\, P\equiv P$

b. (or-form of an implication)

$P\Rightarrow Q\equiv \neg\, P\vee Q$

c. (contrapositive of an implication)

$P\Rightarrow Q\equiv \neg\, Q\Rightarrow \neg\, P$

d. (de Morgan’s laws)

$\neg\, (P\vee Q) \equiv \neg\, P\wedge \neg\, Q$ ;

$\neg\, (P\wedge Q)\equiv \neg\, P\vee \neg\, Q$

e. (rule for direct proof)

$(P\wedge R\Rightarrow Q)\equiv (R\Rightarrow (P\Rightarrow Q))$

f. (rule for proof by contradiction)

$(P\wedge \neg\, Q\Rightarrow O) \equiv (P\Rightarrow Q)$

g. (rule for proof by cases)

$(P\vee R\Rightarrow Q) \equiv \big[ (P\Rightarrow Q)\wedge (R\Rightarrow Q)\big]$

Boolean laws of logic

$\begin{array}{cc} \textrm{1a} \qquad & P\vee Q \equiv Q\vee P \\ \textrm{1b} \qquad & P\wedge Q\equiv Q\wedge P\\ \textrm{2a} \qquad & (P\vee Q)\vee R \equiv P\vee (Q\vee R)\\ \textrm{2b} \qquad & (P\wedge Q) \wedge R \equiv P\wedge (Q\wedge R) \\ \textrm{3a} \qquad & P\vee (Q\wedge R) \equiv (P\vee Q)\wedge (P\vee R) \\ \textrm{3b} \qquad & P\wedge (Q\vee R) \equiv (P\wedge Q)\vee(P\wedge R) \\ \textrm{4a} \qquad & P\vee O \equiv P \\ \textrm{4b} \qquad & P\wedge I\equiv P \\ \textrm{5a} \qquad & P\vee \neg\, P \equiv I\\ \textrm{5b} \qquad & P\wedge \neg\, P \equiv O \\ \end{array}$

(1a) and (1b) are known as communicative laws, (2a) and (2b) as associate laws, and (3a) and (3b) as distributive laws.

Predicate validity

$\begin{array}{cc} \textrm{Q1} \qquad & \exists\, y\,\forall\, x\enspace P(x,y) \Leftrightarrow \forall\, x\,\exists\, y\enspace P(x,y) \\ \textrm{Q2} \qquad & \neg\,\forall\, x\enspace P(x) \Leftrightarrow \exists\, x\enspace \neg\, P(x) \\ \textrm{Q3} \qquad &\neg\,\exists\, x\enspace P(x) \Leftrightarrow \forall\, x\enspace \neg\, P(x) \\ \end{array}$

Boolean laws for set theory

$\begin{array}{cccc} & \textrm{Union} & & \textrm{Intersection} \\ \textrm{S1a} & \quad A\cup B = B\cup A & \textrm{S1b} &\quad A\cap B=B\cap A \\ \textrm{S2a} & \quad (A\cup B)\cup C = A\cup (B\cup C) & \textrm{S2b} & \quad (A\cap B)\cap C=A\cap (B\cap C) \\ \textrm{S3a} & \quad A\cup (B\cap C) = (A\cup B)\cap (A\cup C) & \textrm{S3b} & \quad A\cap(B\cup C) = (A\cap B) \cup (A\cap C)\\ \textrm{S4a} & \quad A\cup \emptyset =A & \textrm{S4b} & \quad A\cap \mathrm{U}=A \\ \textrm{S5a} & \quad A\cup A' = \mathrm{U} & \textrm{S5b} & \quad A\cap A' = \emptyset \\ \textrm{S6a} & \quad A\cup \mathrm{U} = \mathrm{U} & \textrm{S6b} & \quad A\cap \emptyset = \emptyset \\ \textrm{S7a} & \quad A\cup A = A & \textrm{S7b} & \quad A\cap A =A \\ \textrm{S8a} & \quad \mathrm{U}' = \emptyset & \textrm{S8b} & \quad \emptyset' = \mathrm{U} \\ \textrm{S9a} & \quad (A')' = A & & \\ \textrm{S10a} & \quad (A\cup B)' = A' \cap B' & \textrm{S10b} &\quad (A\cap B)' = A' \cup B' \\ \textrm{S11a} & \quad A\cap (A\cup B) = A & \textrm{S11b} &\quad A\cup (A\cap B) = A \\ \textrm{S12a} & \quad A\subseteq A\cup B & \textrm{S12b} & \quad A\cap B\subseteq B \\ \textrm{S13} & \quad A\subseteq B\Leftrightarrow B'\subseteq A' & & \\ \textrm{S14} & \quad A\subseteq B \Leftrightarrow A'\cup B = \mathrm{U} & & \\ \textrm{S15} & \quad A\subseteq B \Leftrightarrow A\cap B' = \emptyset & & \\ \textrm{S16} & \quad A\subseteq B\Leftrightarrow A\cup B & & \\ \textrm{S17} & \quad A\subseteq B\Leftrightarrow A\cap B = A & & \\ \textrm{S18} & \quad A\subseteq B\,\wedge\, B\subseteq D \Rightarrow A\subseteq D & & \\ \end{array}$

Remarks. Statements (S1a) and (S1b) are called the commutative laws; (S2a) and (S2b) the associative laws; (S3a) and (S3b) the distributive laws; (S4a) and (S4b) the identity laws; (S5a) and (S5b) the complement laws; (S7a) and (S7b) the idempotent laws; (S8a) and (S8b) the universal/empty set complement law; (S10a) and (S10b) the De Morgan's laws; (S11a) and (S11b) the absorption laws; and (S18) the transitive law.

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.

September 26, 2020October 24, 2022

202009260344 Exercise 2.5.1

Show that $x=e^{2\theta}\sin\theta$ satisfies the equation $x''-4x'+5x=0$ .

Solution.

(bottom-up)

Let $x'$ denote $\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}\theta}}$ , $x''$ denote $\displaystyle{\frac{\mathrm{d}^2x}{\mathrm{d}\theta^2}}$ .

$\begin{aligned} x' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \sin\theta \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( \sin\theta \big) + \sin\theta \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = e^{2\theta} \cos\theta + 2e^{2\theta}\sin\theta \\ & = e^{2\theta}(\cos\theta + 2\sin\theta ) \\ \end{aligned}$

$\begin{aligned} x'' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( x' \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} (\cos\theta +2\sin\theta ) + (\cos\theta + 2\sin\theta )\cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = (e^{2\theta})(-\sin\theta + 2\cos\theta ) + (\cos\theta + 2\sin\theta )(2e^{2\theta }) \\ & = e^{2\theta}(3\sin\theta + 4 \cos\theta ) \\ \end{aligned}$

To know whether or not $x=e^{2\theta}\sin\theta$ is a solution, I simply do substitution in the equation $x''-4x'+5x=0$ .

$\begin{aligned} \textrm{LHS} & = x''-4x'+5x \\ & = e^{2\theta}(3\sin\theta +4\cos\theta )-4\big[ e^{2\theta}(\cos\theta +2\sin\theta )\big] + 5\big(e^{2\theta}\sin\theta \big) \\ & = \dots \\ & = 0 \\ & = \textrm{RHS} \\ \end{aligned}$

Revision.

(top-down)

We are given a second-order linear homogeneous ordinary differential equation (ODE):

$x''-4x'+5x = 0$

with some independent variable $\theta$ and some dependent variable $x(\theta )=e^{2\theta }\sin\theta$ , the coefficients of $x''$ , $x'$ , and $x$ being constants $1$ , $-4$ , and $5$ .

Let the primed $[ *]'$ be the function derived wrt. to $\theta$ . Following the routine procedures,

$\begin{aligned} x & = e^{r\theta} \\ x' & = re^{r\theta }\\ x'' & = r\cdot ( e^{r\theta })' + r' \cdot ( e^{r\theta }) \\ & = r^2e^{r\theta} \end{aligned}$

rewrite it,

$\begin{aligned} x''-4x'+5x & = 0 \\ r^2e^{r\theta} - 4re^{r\theta} + 5e^{r\theta} & = 0 \\ \dots \textrm{\quad excepting\quad} & x=e^{r\theta}=0\textrm{\quad \dots} \\ r^2 -4r+5 & = 0 \\ \end{aligned}$

and we shall obtain the auxiliary equation (aka. the characteristic equation) on the very last line.

$r=\displaystyle{\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(5)}}{2(1)}}=2\pm \textrm{i}$ .

Note.

If the roots of the auxiliary equation $ar^2+br+c=0$ are the complex numbers $r_1=\alpha +\textrm{i}\beta$ and $r_2=\alpha -\textrm{i}\beta$ , the general solution of $ay''+by'+cy=0$ is

$y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)$

_{(please refer to J. Stewart’s Calculus, Second-Order Linear Differential Equations)}

Now that $r_1=2+\textrm{i}$ and $r_2=2-\textrm{i}$ and $c_1$ , $c_2$ are arbitrary constants, to this $2^{\textrm{nd}}$ -order ODE, the general solution is

$x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta )$ .

Examination.

It is a good practice, no matter how much time one would allow oneself, to countercheck one’s solution…

Here it goes,

$\begin{aligned} x & =e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (c_2e^{2\theta })(\sin\theta ) + (c_1e^{2\theta })(\cos\theta ) \\ \end{aligned}$ .

$\begin{aligned} x' & = e^{2\theta}(-c_1\sin\theta +c_2\cos\theta ) + 2e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) + (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \\ \end{aligned}$

$\begin{aligned} x''& = \big[ (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) \big]' + \big[ (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \big]' \\ & = \big[ (2c_2e^{2\theta}-c_1e^{2\theta })(\cos\theta )+(4c_2e^{2\theta }-2c_1e^{2\theta })(\sin\theta ) \big] \\ & \qquad\quad + \big[ (2c_1e^{2\theta }+c_2e^{2\theta })(-\sin\theta )+(4c_1e^{2\theta }+2c_2e^{2\theta })(\cos\theta ) \big] \\ & = (-4c_1e^{2\theta }+3c_2e^{2\theta })(\sin\theta ) + (3c_1e^{2\theta }+4c_2e^{2\theta })(\cos\theta ) \\ \end{aligned}$

Then,

$\begin{aligned} \textrm{LHS} & = x'' - 4x' + 5x \\ & = \big( (-4c_1e^{2\theta }+3c_2e^{2\theta })-4(2c_2e^{2\theta}-c_1e^{2\theta})+5c_2e^{2\theta}\big)(\sin\theta ) \\ & \qquad\quad + \big( (3c_1e^{2\theta}+4c_2e^{2\theta}) - 4(2c_1e^{2\theta}+c_2e^{2\theta})+5c_1e^{2\theta} \big)(\cos\theta ) \\ & = 0 \\ & = \textrm{RHS} \end{aligned}$

In conclusion, $x=e^{2\theta}\sin\theta$ is a particular solution, the general solution being $x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta )$ .

September 25, 2020June 5, 2024

202009251201 Homework 1 (Q1)

Solve the system $\mathbf{A}\mathbf{x} = \mathbf{0}$ for each matrix $\mathbf{A}$ below.

(a)

$\mathbf{A}= \begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \\ \end{bmatrix}$

(b)

$\mathbf{A}= \begin{bmatrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & -2 & 0 \\ \end{bmatrix}$

(c)

$\mathbf{A}= \begin{bmatrix} 4 & 6 & 0 & 1 & -9 \\ 1 & 2 & -4 & 5 & 7 \\ 2 & 3 & 6 & 4 & 2 \\ 1 & 0 & 3 & 2 & -5 \\ \end{bmatrix}$

(d)

$\mathbf{A}= \begin{bmatrix} 1 & 2 & 3 & 1 & 1 \\ 1 & 4 & 0 & 1 & 2 \\ 0 & 2 & -3 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$

Recall

Observe that $\mathbf{A}$ ‘s in (a) and (b) are 3-by-4 matrices, and in (c) and (d) are 4-by-5 matrices. Multiplication of two matrices $\mathbf{A}$ and $\mathbf{B}$ results in a matrix product $\mathbf{A}\mathbf{B}$ . By convention, should an $m$ -by- $n$ matrix $\mathbf{A}$ be written on the left, $\mathbf{A}$ is meant the multiplicand, and should an $n$ -by- $p$ matrix $\mathbf{B}$ be written on the right, $\mathbf{B}$ is meant the multiplier. The matrix product $\mathbf{AB}$ will become an $m$ -by- $p$ matrix.

Solution.

(a)

Let a $4$ -by- $1$ column vector $\mathbf{x}$ be

$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ .

Then,

$\begin{aligned} \mathbf{A}\mathbf{x} & = \mathbf{0} \\ \begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} & = \mathbf{0} \\ \begin{bmatrix} (1)(x_1)+(3)(x_2)+(5)(x_3)+(7)(x_4) \\ (3)(x_1)+(5)(x_2)+(7)(x_3)+(9)(x_4) \\ (5)(x_1)+(7)(x_2)+(9)(x_3)+(1)(x_4) \\ \end{bmatrix} & = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \end{aligned}$

Here we have four unknowns but three equations.

$\begin{Bmatrix} 15x_1+ 45x_2+75x_3+105x_4 & = 0 \\ 15x_1+25x_2+35x_3+45x_4 & = 0 \\ 15x_1+21x_2+27x_3+3x_4 & = 0 \\ \end{Bmatrix}$

It is feasible to carry through all the calculations, but do let us not work in so awkward a manner.

Performing row operations of matrices and using shorthand notations (e.g., $R_1$ stands for Row 1; $5R_2$ stands for 5 times each entry in Row 2; $R_1-R_3$ means every entry in Row 3 is to be subtracted from the corresponding entry in Row 1.)

(1) $R_3-5R_1$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 0 & -8 & -16 & -34 \\ \end{bmatrix}$

(2) $R_2 - 3R_1$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & -8 & -16 & -34 \\ \end{bmatrix}$

(3) $R_3 - 2R_2$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & 0 & 0 & -58 \\ \end{bmatrix}$

(4) $R_3 \div (-58)$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 12 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(5) $R_2-12R_3$

$\begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(6) $R_1-7R_3$

$\begin{bmatrix} 1 & 3 & 5 & 0 \\ 0 & -4 & -8 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(7) $R_2 \div (-4)$

$\begin{bmatrix} 1 & 3 & 5 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

(8) $R_1-3R_2$

$\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

The matrix above is called the reduced row-echelon form of $\mathbf{A}$ .

Now that the simplification has come handy:

$\begin{aligned} x_1 -x_3 &=0 \\ x_2 +2x_3 & = 0 \\ x_4 & = 0 \\ \end{aligned}$

and the answer is $\mathbf{x}=\begin{bmatrix} x \\ -2x \\ x \\ 0 \end{bmatrix}$ .

Questions (b), (c), and (d) are left the readers.

September 25, 2020October 24, 2022

202009250041 Exercise 5.8.1

Find out if any the asymptote(s) of the curve given by:

$y=\displaystyle{\frac{x^2-6x+3}{x-3}}$ .

Attempts.

$\begin{aligned} & & x & -3 & & \\\cline{3-5} x & -3 \Big) & x^2 & -6x & +3 & \\ & & x^2 & -3x & & \\ \cline{3-5} & & & -3x & +3 \\ & & & -3x & +9 \\\cline{4-5} & & & & -6 \end{aligned}$

The curve is given by

$\begin{aligned} y & =\displaystyle{\frac{x^2-6x+3}{x-3}} \\ & = x-3 - \displaystyle{\frac{6}{x-3}} \end{aligned}$ .

Ans. Of the given curve, $x=3$ is a vertical asymptote and $y=x-3$ an oblique asymptote.

Retrospectively, for $f(x)=\displaystyle{\frac{x^2-6x+3}{x-3}}$ , maybe one could also have checked the conditions as follow:

i. $\displaystyle{\lim_{x\to 0}}f(x)$ ;

ii. $\displaystyle{\lim_{x\to 3^{+}}}f(x)$ ;

iii. $\displaystyle{\lim_{x\to 3^{-}}}f(x)$ ;

iv. $\displaystyle{\lim_{x\to +\infty}}f(x)$ ; and

iv. $\displaystyle{\lim_{x\to -\infty}}f(x)$ .

i.

$\begin{aligned} \lim_{x\to 0}f(x) & = \lim_{x\to 0} \frac{x^2-6x+3}{x-3} \\ & = \frac{(0)^2-6(0)+3}{(0)-3} \\ & = -1 \end{aligned}$

iii.

$\begin{aligned} \lim_{x\to 3^{-}}f(x) & = \lim_{x\to 3^{-}} \frac{x^2-6x+3}{x-3} \\ & = \lim_{x\to 3^{-}} \bigg( x-3-\frac{6}{x-3} \bigg) \\ & = 3^{-} -3-\frac{6}{3^{-}-3} \\ & = (3^{-}-3) + \bigg( \frac{6}{3-3^{-}} \bigg) \\ & \\ \hline & \\ & \because 3^{-} < 3 \\ & \therefore 3^{-}-3<0\qquad \textrm{and}\qquad 3-3^{-}>0 \\ & \dots \textrm{excuse me for writing with no more ado,}\dots \\ & \\ \hline & \\ & = 0^{-} + \frac{6}{0^{+}} \\ & = +\infty \\ \end{aligned}$

ii., iv., and iv. are not checked.

Just work with it like whom I was taught:

First, find the slope of tangent to the curve,

$\begin{aligned} f(x) & = \frac{x^2-6x+3}{x-3} \\ f'(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x) \big) \\ & = \frac{(x-3)(2x-6)-(x^2-6x+3)(1)}{(x-3)^2} \\ & = \frac{(2x^2-12x+18)-(x^2-6x+3)}{(x-3)^2} \\ & = \frac{x^2-6x+15}{x^2-6x+9} \\ & = 1 + \frac{6}{x^2-6x+9} \end{aligned}$

The purpose is to find out any crests or troughs, because at the tip or bottom of a plotted curve, the slope of tangent will be zero, i.e., $f'(x)=0$ .

Secondly, find the change of slope of tangent to the curve,

$\begin{aligned} f''(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big[ f'(x) \big] \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( 1+\frac{6}{x^2-6x+9} \bigg) \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \big( 1 \big) + \frac{\mathrm{d}}{\mathrm{d}x} \bigg( \frac{6}{x^2-6x+9} \bigg) \\ & = 0 + \frac{(x^2-6x+9)(0)+(6)(2x-6)}{(x^2-6x+9)^2} \\ & = \frac{12x-36}{(x^2-6x+9)^2} \\ \end{aligned}$

The purpose is to find out if there were any inflexion points (i.e., $f''(x)=0$ ) upon which curvature changes sign.

It begins, assuming $x\neq 3$ , I will set $f(x)=0$ :

$\begin{aligned} f(x) & = 0 \\ \Leftrightarrow \qquad\qquad x^2-6x+3 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(3)}}{2(1)} \\ & = 3 \pm \sqrt{6} \\ \end{aligned}$

The $x$ -intercepts are thus two, $3+\sqrt{6}$ and $3-\sqrt{6}$ .

Then I needs set $f'(x)=0$ with the same assumption $x\neq 3$ .

$\begin{aligned} f'(x) & = 0 \\ \Leftrightarrow \qquad \qquad 1+\frac{6}{x^2-6x+9} & = 0 \\ & \\ \textrm{\dots \quad by assumption }x\neq 3 & \enspace \textrm{jump legitimately to the next step\quad \dots} \\ & \\ x^2-6x+9+6 & = 0 \\ x^2 -6x +15 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(15)}}{2(1)} \\ & = 3 \pm \sqrt{6}\, \textrm{i} \end{aligned}$

As curvature here permits of no nonreal points, there are neither convex point(s) nor concave point(s).

Besides, there are none any one point of inflexion because if on one hand $f''(x)=0$ and on the other $x\neq 3$ be assumed,

$\begin{aligned} \frac{12x-36}{(x^2-6x+9)^2} & = 0 \\ \Rightarrow 12x-36 & = 0 \\ \Rightarrow x & =3\qquad\qquad\qquad \bot \\ \end{aligned}$

it would have resulted in contradiction.

In conclusion, this exercise was overdone.

September 24, 2020October 24, 2022

202009240421 Problem 2.1.1

For $x,\, y\in \mathbb{K}$ ( $\mathbb{R}$ or $\mathbb{C}$ ), define

$d(x,y)=\mathrm{min}\big\{ 1,\, |x-y| \big\}$ .

Prove that $d$ is a metric on $\mathbb{K}$ .

Motivation.

A ruler is marked by rules for the sake of measuring things. Not so much common to a ruler, but well worth the rule for the general, that a metric must measure its metric space, with the metric defined below:

Definition. (metric) Let $X$ be a non-empty set. A function $d: X\times X \rightarrow \mathbb{R}$ is said to be a metric on $X$ if it satisfies the following conditions:

i. $d(x,y)\ge 0\qquad \forall\, x,\, y\in X$ ;
ii. $d(x,y)=0 \Leftrightarrow x=y\qquad x,\, y\in X$ ;
iii. $d(x,y)=d(y,x) \qquad \forall\, x,\, y\in X$ ;
iv. $d(x,y)\le d(x,z)+d(z,y) \qquad \forall\, x,\, y,\, z\in X$ .

Remark.

i. As is known, distance should be either positive or zero (i.e., non-negative); ii. We are in one only by discrimination; iii. Fair and just from a symmetric point of view; and iv. The straighter the path, the shorter the distance (also known as the Triangle Inequality).

Proof.

Assume $x\, ,y\in\mathbb{C}$ for convenience. (Provided $\mathbb{R}\subset \mathbb{C}$ , the assumption is ready for reduction or extension.)

i.

If the minimum $\mathrm{min}\big\{ 1, |x-y|\big\} = 1$ , condition (i) $d(x,y)=1 \ge 0$ is seen. If the minimum $\mathrm{min}\big\{ 1,|x-y|\big\} = |x-y|$ , be it called the absolute value, the magnitude, the norm, the modulus, or whatsoever, a complex number is non-negative in norm.

Recall. (Norm of complex conjugate)

A complex number $z=x+\textrm{i}\, y$ contains two parts, the real part $x$ and the imaginary part $y$ . The norm $|z|$ (and the norm $|\bar{z}|$ of its complex conjugate $\bar{z}=x-\textrm{i}\, y$ )
is defined by the formula:

$|z|=|x+\textrm{i}\, y|=\sqrt{\big[\textrm{Re}(z)\big]^2+\big[\textrm{Im}(z)\big]^2}=\sqrt{x^2+y^2}$ .

$\dagger$ It turns out that $|z|=\sqrt{(x)^2+(y)^2}=\sqrt{(x)^2+(-y)^2}=|\bar{z}|$ (where $x,\, y\in\mathbb{R}$ ) is positive iff $x\enspace\textrm{\scriptsize OR}\enspace y\neq 0$ , and zero iff $x=y=0$ , but never negative.

$\ddagger$ The sum, the difference, the product, and the quotient of two complex numbers is one another complex number for the complex number field is algebraically closed.

ii.

(only-if) Giving a try straightforth:

$\begin{aligned} d(x,y) & = 0 \\ \textrm{min}\big\{ 1, |x-y| \big\} & = 0 \\ |x-y| & = 0\qquad \textrm{\scriptsize OR\qquad\quad } 1 = 0 \qquad \textrm{(rejected for\enspace }1\neq 0) \\ |x-y| & = 0 \\ x-y & = 0 \\ x & = y \end{aligned}$

(if) In reverse from backward:

$\begin{aligned} x & = y \\ x-y & = 0 \\ |x-y| & = 0 \\ \textrm{min}\big\{ 1, |x-y| \big\} & = 0 \\ d(x,y) & = 0 \end{aligned}$

If provided with appropriate explanation, the proof can be shortened by use of the two-way if-and-only-if.

iii.

Suffice it to check whether $d(x,y)=d(y,x)$ is true or not.

$\begin{aligned} \textrm{LHS} & = d(x,y) \\ & = \textrm{min}\big\{ 1,\, |x-y| \big\} \\ & = \textrm{min}\big\{ 1,\, |y-x| \big\} \\ & = d(y,x) \\ & = \textrm{RHS} \end{aligned}$

Roughwork.

$\forall\, x,\,y \in\mathbb{C}$ , let $x=a+\textrm{i}\, b$ and $y=c+\textrm{i}\, d$ where $a$ , $b$ , $c$ , and $d$ are real numbers. Then,
$\begin{aligned} x-y & = (a+\textrm{i}\, b)-(c+\textrm{i}\, d) \\ x-y & = (a-c) + \textrm{i}\, (b-d) \\ |x-y| & = \sqrt{(a-c)^2+(b-d)^2} \\ |x-y| & = \sqrt{(c-a)^2+(d-b)^2} \\ |x-y| & = |(c-a) + \textrm{i}\, (d-b)| \\ |x-y| & = |(c+\textrm{i}\, d) - (a+\textrm{i}\, b)| \\ |x-y| & = |y-x| \\ \end{aligned}$

iv.

Given here are some equations, I write out all them lest I might forget any:

$d(x,y)=\textrm{min}\big\{ 1, |x-y| \big\}$ ,
$d(x,z)=\textrm{min}\big\{ 1, |x-z| \big\}$ ,
$d(z,y)=\textrm{min}\big\{ 1, |z-y| \big\}$ .

RTP (i.e. required to prove):

$d(x,y)\le d(x,z) + d(z,y)$

$\begin{aligned} \textrm{RHS} & = d(x,z) + d(z,y) \\ & = \textrm{min}\big\{ 1,|x-z|\big\} + \textrm{min}\big\{ 1,|z-y|\big\} \\ & = \textrm{min}\big\{ 2,\, 1+|z-y|,\, 1+|x-z|,\, |x-z|+|z-y|\big\} \\ \end{aligned}$

If $d(x,y)=\textrm{min}\big\{ 1, |x-y|\big\} \stackrel{?}{=}1$ , so what have I done with?

WTS (i.e. wish to show):

$1\le \textrm{min}\big\{ 2, 1+|z-y|, 1+|x-z|, |x-z|+|z-y|\big\}$

The following inequalities hold evidently:

$\begin{aligned} 1 & \le 2 \\ 1 & \le 1+|z-y| \\ 1 & \le 1+|x-z| \\ \end{aligned}$

But $1\le |x-z|+|z-y|$ has not yet been ascertained.

The Argand diagram above replaces the usual $x$ – and $y$ -axes of the Cartesian plane with the real and the imaginary axes of Argand plane.

Owing to my giving too raw and rude maybe a proof, the problem should have otherwise been treated case-by-case. I.e., they are in either case:

$1\le |x-y| \qquad \qquad \textrm{\scriptsize OR}\qquad\qquad 1> |x-y|$

Just do it by rote:

$\begin{aligned} & \quad\, d(x,y) \\ & =\textrm{min}\big\{ 1, |x-y| \big\} \\ & =\textrm{min}\big\{ 1, |x-z+z-y| \big\} \\ \dots \, & \textrm{by the Triangle Inequality}\, \dots \\ & \leqslant \textrm{min} \big\{ 1, |x-z| + |z-y| \big\} \\ & \leqslant \textrm{min} \big\{ 1, |x-z| \big\} + \textrm{min} \big\{ 1, |z-y| \big\} \\ & \leqslant d(x,z) + d(z,y) \\ \end{aligned}$

$d(x,y) \le d(x,z) + d(z,y)$

QED