The Physics Language – Page 14

November 9, 2020October 24, 2022

202011091423 Exercise 1 (Q2)

If $f(x)=(x-1)(x+5)$ , find the values of $f(2)$ , $f(1)$ , $f(0)$ , $f(a+1)$ , $f\bigg(\displaystyle{\frac{1}{a}}\bigg)$ , $f(-5)$ .

Solution.

$\begin{aligned} f(2) & = (2-1)(2+5)=7 \\ f(1) & = (1-1)(1+5)=0 \\ f(0) & = (0-1)(0+5)=-5 \\ f(a+1) & = ((a+1)-1)((a+1)+5) = a(a+6) \\ f\bigg( \frac{1}{a}\bigg) & = \Bigg( \bigg( \frac{1}{a}\bigg) -1\Bigg) \Bigg( \bigg( \frac{1}{a}\bigg) +5 \Bigg) \\ f(-5) & = (-5-1)(-5+5) = 0 \end{aligned}$

On reflection.

Suppose you are given an unknown quadratic function $f(x;a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ such that

$\begin{aligned} 0 & \mapsto -5 \\ 1 & \mapsto 0 \\ 2 & \mapsto 7 \\ \end{aligned}$

and you are asked to make polynomial interpolation.

$\begin{cases} a_0 + a_1 (0) + a_2(0)^2 = f(0) = -5 \\ a_0 + a_1 (1) + a_2(1)^2 = f(1) = 0 \\ a_0 + a_1 (2) + a_2(2)^2 = f(2) = 7 \\ \end{cases}$
$\begin{cases} a_0=-5 \\ a_0 + a_1 + a_2 = 0\\ a_0 +2a_1 + 4a_2 = 7\\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 + a_2 = 5 \\ a_1 + 2a_2 = 6 \\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 = 4 \\ a_2 = 1 \\ \end{cases}$

And then is the interpolated polynomial $f(x)=x^2+4x-5\qquad \textrm{\scriptsize OR}\qquad (x+5)(x-1)$ .

November 5, 2020October 24, 2022

202011051527 Exercise 1 (Q1)

If $f(x)=2x^2-4x+1$ , find the values of $f(1)$ , $f(0)$ , $f(2)$ , $f(-2)$ , $f(a)$ , $f(x+\delta x)$ .

Solution.

Given $f(x)=2x^2-4x+1$ .

$\begin{aligned} f(1) & =2(1)^2-4(1)+1=-1 \\ f(0) & = 2(0)^2 - 4(0) +1 = 1 \\ f(2) & = 2(2)^2-4(2)+1 =1 \\ f(-2) & = 2(-2)^2-4(-2)+1=17\\ f(a) & = 2a^2 - 4a +1 \\ f(x+\delta x) & = 2(x+\delta x)^2 - 4 (x+\delta x) +1 \end{aligned}$

This exercise is done.

On reflection.

Suppose you are given the following conditions:

$\begin{aligned} x_0 = 0 & \qquad f(x_0) = 1 \\ x_1 = 1 &\qquad f(x_1) = -1 \\ x_2 = 2 &\qquad f(x_2) =1 \end{aligned}$

and you are asked to interpolate by Lagrange polynomials over the range $[0,2]$ .

$\begin{aligned} \mathcal{L}(x) & = (1)\bigg( \displaystyle{\frac{x-1}{0-1}} \bigg)\bigg( \displaystyle{\frac{x-2}{0-2}} \bigg) + (-1)\bigg( \displaystyle{\frac{x-0}{1-0}} \bigg) \bigg( \displaystyle{\frac{x-2}{1-2}} \bigg) + (1)\bigg( \displaystyle{\frac{x-0}{2-0}} \bigg) \bigg( \displaystyle{\frac{x-1}{2-1}} \bigg) \\ & = \displaystyle{\frac{(x-1)(x-2)}{2}} + x(x-2) + \displaystyle{\frac{x(x-1)}{2}} \\ & = \displaystyle{\frac{(x-1)(x-2)+2x(x-2)+x(x-1)}{2}} \\ & = \displaystyle{\frac{x^2-3x+2+2x^2-4x+x^2-x}{2}} \\ & = \displaystyle{\frac{4x^2-8x+2}{2}} \\ & = 2x^2-4x+1\\ \end{aligned}$

The interpolating polynomial $\mathcal{L}(x)$ checks with the original function $f(x)$ .

October 23, 2020October 20, 2022

202010230206 Sidenote of Clopen

Observe that both $\emptyset$ and $X$ are open and closed in $X$ , i.e., clopen.

Proof. Pastime.

Remark. ( $S$ ‘s in several symbols )

$S$ is a set of points.
$S^{0}$ , the interior of set $S$ , contains all interior points.
$S'$ , the derived set of set $S$ , contains all accumulation/cluster/limit points.
$\bar{S}$ , the closure of set $S$ , contains all adherent points.
$\partial S$ (also denoted by $b(S)$ or $S^{b}$ ), the boundary of set $S$ , contains all boundary points.

October 23, 2020October 20, 2022

202010230028 Problem 2.4.11

Let $(X,d)$ be a metric space, $a\in X$ and $0<r<r'$ .

Prove that the set

$\{ x\in X:\enspace r<d(x,a)<r' \}$

is open in $(X,d)$ .

Setup.

Understand the definition given to each of the following:

First, what is meant by whether a set is open or not in some metric space?

Definition. (open set) Let $(X,d)$ be a metric space. A set $G\subset X$ is said to be an open set if it is a neighborhood of each of its points. (Equivalently, a set $G\subset X$ is said to be an open set
if for each $x\in G$ , there exists an $r>0$ such that $S_r(x)\subset G$ .)

_{Please refer to pg. 20, Jain and Ahmad’s Metric Spaces.}

Second, what is referred to as a neighborhood of some point(s)?

Definition. (neighborhood) Let $(X,d)$ be a metric space and $x\in X$ . A set $N\subset X$ is said to be a neighborhood (nbd) of $x$
if there exists an open sphere centred at $x$ and contained in $N$ ,
i.e., if $S_r(x)\subset N$ for some $r>0$ .

_{Please refer to pg. 19, Jain and Ahmad’s Metric Spaces.}

Third, what is an open sphere?

Definition. (open sphere) Let $(X,d)$ be a metric space. Let $x\in X$ and $r>0$ be a real number. The open sphere with centre $x$ and radius $r$ , denoted by $S_r(x)$ , the subset of $X$ given by $S_r(x)=\{ y\in X:\enspace d(x,y)<r \}$ N.b. An open sphere is always non-empty since it contains its centre at least.

_{Please refer to pg. 16, Jain and Ahmad’s Metric Spaces.}

October 22, 2020October 24, 2022

202010220448 Exercises 2.1A (Q1)

Suppose $|S|=19$ , $|T|=11$ and $|S\cap T|=8$ . Find $|S\cup T|$ and $|S\backslash T|$ .

Hint. (Verbal translation)

You are given that the number of elements in set $S$ is $19$ ,the number of elements in set $T$ is $11$ , and the number of elements in the intersection of set $S$ and set $T$ is $8$ .

You are asked:

What is the number of elements in the union of set $S$ and set $T$ ?What is the number of elements in the relative complement $S\backslash T$ of set $T$ with respect to set $S$ ?

Definition. The relative complement of $T$ with respect to $S$ is the set

$S\backslash T=\{ x\, |\enspace x\in S\textrm{ and }x\notin T\}$ .

Can you try drawing a Venn diagram?

Attempts.

(constructive)

Let

$S=\{ a,\, b,\, c,\, d,\, e,\, f,\, g,\, h,\, i,\, j,\, k,\, l,\, m,\, n,\, o,\, p,\, q,\, r,\, s\}$ ,

and also

$T=\{ l,\, m,\, n,\, o,\, p,\, q,\, r,\, s,\, t,\, u,\, v\}$ ,

so that $S\cap T=\{ l,\, m,\, n,\, o,\, p,\, q,\, r,\, s \}$ .

The union $S\cup T$ of set $S$ and set $T$ must as follows be:

$S\cup T=\{ a,\, b,\, c,\, d,\, e,\, f,\, g,\, h,\, i,\, j,\, k,\, l,\, m,\, n,\, o,\, p,\, q,\, r,\, s,\, t,\, u,\, v\}$ ,

such that $|S\cup T|=22$ .

The relative complement of set $T$ w.r.t. set $S$ is

$S\backslash T=\{ a,\, b,\, c,\, d,\, e,\, f,\, g,\, h,\, i,\, j,\, k\}$

and the number of its elements is

$|S\backslash T| = 11$ .

(analytic)

By observation of the Venn diagram,

you are writing out

$|S\cup T|=|S|+|T|-|S\cap T|$ ,

keeping in mind that

$|S\cap T|=|S|+|T|-|S\cup T|$

shall answer another question of a different subject.

October 13, 2020October 24, 2022

202010132314 Homework 1 (Q1)

The radius $r$ of a right circular cylinder is decreasing at a rate of $12\,\mathrm{cm\, s^{-1}}$ , while its height $h$ is decreasing at a rate of $25\,\mathrm{cm\, min^{-1}}$ . How is the volume changing when $r=180\,\mathrm{cm}$ and $h=500\,\mathrm{cm}$ ? Is the volume increasing or decreasing?

Solution.

The volume of a right circular cylinder is calculated by the formula

$V=\pi r^2h$ .

The volume $V$ (a dependent variable) of a cylinder varies with its radius $r$ and height $h$ (both independent variables). The change of volume, simply put it, is a derivative of volume $V$ with respect to time $t$ :

$\displaystyle{\frac{\mathrm{d}V}{\mathrm{d}t}}$ .

Differentiate $V(r,h,t)$ wrt. time $t$ :

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(r,h,t) & = \frac{\mathrm{d}}{\mathrm{d}t} (\pi r^2 h) \\ & = \pi \frac{\mathrm{d}}{\mathrm{d}t} (r^2h) \\ & = \pi \bigg[ r^2 \frac{\mathrm{d}}{\mathrm{d}t}(h) + h \frac{\mathrm{d}}{\mathrm{d}t} (r^2) \bigg] \\ & = \pi \bigg[ \big( r(t) \big)^2 \frac{\mathrm{d}}{\mathrm{d}t}\big( h(t) \big) + \big( h(t)\big) \bigg( \frac{\mathrm{d}}{\mathrm{d}t}\big( r(t) \big)^2 \bigg) \bigg] \\ & = \pi (r^2h' +2hrr' ) \\ & = \pi r^2h' + 2\pi hrr' \\ \end{aligned}$

Given $r=180\,\mathrm{cm}$ , $h=500\,\mathrm{cm}$ , $r'=-12\,\mathrm{cm\, s^{-1}}=-720\,\mathrm{cm\, min^{-1}}$ , and $h'=-25\,\mathrm{cm\, min^{-1}}$ , you would have it.

In some formalism of partial derivatives,

$\begin{aligned} \frac{\mathrm{d}V(r,h)}{\mathrm{d}t} & = \frac{\partial V}{\partial r}\frac{\mathrm{d}r}{\mathrm{d}t} + \frac{\partial V}{\partial h}\frac{\mathrm{d}h}{\mathrm{d}t} \\ & = (2\pi hr)\frac{\mathrm{d}r}{\mathrm{d}t} + (\pi r^2)\frac{\mathrm{d}h}{\mathrm{d}t}\\ & = 2\pi hrr' +\pi r^2h' \\ \end{aligned}$

you could have it also.

Afterthought.

It just so happens that there are two lines of attack, by taking total/ordinary derivatives and by taking partial derivatives. Is here anyhow the difference? Is there anything the matter?

October 13, 2020October 24, 2022

202010130604 Example 1, Chapter 1.1, Methods in Physics II (2015-2016 Lectures)

Find the trajectory of a cannon ball fired by a cannoneer at $45^\circ$ above his eye level with an initial speed $v$ .

Ans.

Imagine you are in his place, like the picture below:

Then, suppose you are opening fire at such positive $\textrm{(+ve)\enspace }x$ -axis direction as that follows along the meridian. In addition, assume that the earth is to be lying flat beneath the cannon ball during its flight, that every pole ( $z$ -axis) upheld would be so much right-angled (/normal /perpendicular) to the ground as parallel (-transported) anywhere.

We neglect air friction.

1. Along the $x$ -direction, the speed is kept $v\cos 45^\circ$ .

2. Along the $z$ -direction, the speed is initially $v\sin 45^\circ$ . By the laws of gravity, the cannon ball will experience a net force $m\mathbf{g}$ due to gravitational pull by the Earth. If upward direction is taken the positive sign, an equation of motion due to the Galilean transformation (i.e., $v=u+at$ ) will depict that $v_z(t)=v\sin 45^\circ -gt$ .

3. One another equation $s(t)=ut+\displaystyle{\frac{1}{2}}at^2$ depicts how distance $s$ (i.e., the magnitude of displacement $\mathbf{s}$ ) varies with time $t$ during linear motion in constant acceleration $a$ .

One might have already noticed I am setting the original question aside, as it were. Let’s pinpoint the answer now.

The trajectory should be obtained in the form:

$\mathbf{s}(t)=s_x(t)\,\hat{\mathbf{i}} + s_z(t)\,\hat{\mathbf{k}}$ .

where

$\begin{aligned} s_x(t) & = u_x t+\frac{1}{2}a_xt^2 \\ & = (v\cos 45^\circ )\, t + \frac{1}{2}(0)(t^2) \\ & = (v\cos 45^\circ )\, t \\ \end{aligned}$

and

$\begin{aligned} s_z(t) & = u_zt + \frac{1}{2}a_zt^2 \\ & = (v\sin 45^\circ )\, t + \frac{1}{2}(-g)(t^2) \\ \end{aligned}$

October 2, 2020June 12, 2025

202010022249 Problem 2.1.10

In $C[0,1]$ , determine the values of $d_{\infty}(x,y)$ and $d_{1}(x,y)$ , when

(a) $x(t)=t^3+t+1$ and $y(t)=t^3+t^2+\frac{1}{2}t+1$ ;

(b) $x(t)=\sin t$ and $y(t)=t$ ;
(c) $x(t)=\sin t$ and $y(t)=t-\displaystyle{\frac{t^3}{6}}$ ;
(d) $x(t)=\mathrm{exp}(t)$ and $y(t)=\displaystyle{\sum_{m=0}^n}\frac{t^m}{m!}$ .

Recall.

Definition. (uniform metric) Let $C[a,b]$ be the set of all real-valued continuous functions defined on $[a,b]$ . For any $x,y\in C[a,b]$ , define the uniform metric $d_{\infty}$ :

$d_{\infty}(x,y)=\displaystyle{\max_{t\in [a,b]}}|x(t)-y(t)|$ .

N.b. If we let $B[a,b]$ be the set of all real-valued functions defined and bounded on $[a,b]$ , the uniform metric is then defined

$d_{\infty}(x,y)=\displaystyle{\sup_{t\in [a,b]}}|x(t)-y(t)|$ .

_{(cited from Examples 14 and 15, pg. 13, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Definition. For any $x, y\in C[a,b]$ , define

$d_1(x,y)=\displaystyle{\int_a^b}|x(t)-y(t)|\,\mathrm{d}t$

N.b. $d_1(x,y)$ represents the absolute area between the functions $x$ and $y$ as a measure of the distance between these two functions.

_{(cited from Example 16, pg. 14, Pawan K. Jain and Khalil Ahmad’s Metric Spaces (2e) on Introductory Concepts)}

Solution.

(a)

$\begin{aligned} & \quad\, d_{\infty} (x,y) \\ &= \max_{t\in [0,1]} |x(t)-y(t)| \\ \end{aligned}$

Roughwork.

Approach.

To know the maximum value of $|x(t)-y(t)|$ , apply differentiation to

$f(t)=-t^2+\displaystyle{\frac{1}{2}}t$

and attain

$\begin{aligned} f(t) & = -t^2 + \frac{1}{2}t \\ f'(t) & = -2t+\frac{1}{2} \\ f'(t) & = 0 \Leftrightarrow t=\frac{1}{4} \\ \end{aligned}$

If the quadratic function $f(t)$ is plotted in a graph, a parabola admits of no inflexion points, needless to check on $f''(x)=0$ . So,

$\begin{array}{c|c|c|c|c|c} & t=0 & 0<t<\frac{1}{4} & t=\frac{1}{4} & \frac{1}{4}<t<1 & t=1 \\ &&&&&\\ \hline &&&&&\\ f(t) & 0 & \dots & \displaystyle{\frac{1}{16}} & \dots & -\displaystyle{\frac{1}{2}} \\ &&&&&\\ \hline &&&&&\\ f'(t) & \displaystyle{\frac{1}{2}}\enspace (>0) & \dots & 0\enspace (=0) & \dots & -\displaystyle{\frac{3}{2}}\enspace (<0) \\ &&&&&\\ \hline &&&&&\\ \textrm{plot} & \diagup & \dots & --- & \dots & \diagdown \\ &&&&&\\ \end{array}$

The continuous function $f(t)$ in the closed interval $[0,1]$ attains its maximum value $f(\frac{1}{4})=\frac{1}{16}$ when $t=\frac{1}{4}$ .

$\begin{aligned} & d_{\infty} (x,y) \\ = & \max_{t\in [0,1]} |x(t)-y(t)| \\ = & \max_{t\in [0,1]} |f(t)| \\ = & \frac{1}{16} \qquad\qquad\qquad \checkmark\\ \end{aligned}$

and should you think of what follows as quite right

$\begin{aligned} d_1(x,y) & = \int_0^1 |x(t)-y(t)| \,\mathrm{d}t \\ & = \int_0^1 f(t)\,\mathrm{d}t \\ & = \int_0^1 \bigg| -t^2 + \frac{1}{2}t \bigg| \, \mathrm{d}t \\ & = \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_0^1 \\ & = -\frac{1}{12}\qquad\qquad\qquad \times \\ \end{aligned}$

you might have rather mistaken calculus.

Correction.

Get back to the basics,

$\begin{aligned} f(t) & = 0 \\ \bigg| -t^2+\frac{1}{2}t \bigg| & = 0 \\ t^2-\frac{1}{2}t &= 0 \\ (t-\frac{1}{2})t & = 0 \\ t & = 0\quad \textrm{\scriptsize{OR}}\quad \frac{1}{2} \\ \end{aligned}$

From the previous graph of $C[0,1]$ , $f(t)$ is found to be positive when $t\in (0,0.5)$ , zero when $t\in \{ 0\} \cup\{ 0.5\}$ , and negative when $t\in (0.5,1]$ .

Doing it step-by-step,

$\begin{aligned} d_1(x,y) & = \int_0^1 \bigg| -t^2+\frac{1}{2}t \bigg| \,\mathrm{d}t \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg| -t^2+\frac{1}{2}t \bigg| + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg| -t^2+\frac{1}{2}t \bigg| \\ & = \int_{0}^{0.5}\mathrm{d}t\enspace \bigg( - t^2 + \frac{1}{2}t \bigg) + \int_{0.5}^{1}\mathrm{d}t \enspace \bigg( t^2-\frac{1}{2}t\bigg) \\ \end{aligned}$

Evaluating term-by-term, the first term being

$\begin{aligned} & \int_{0}^{0.5} \bigg( -t^2 + \frac{1}{2}t \bigg) \,\mathrm{d}t \\ = & \bigg[ -\frac{t^3}{3} + \frac{t^2}{4} \bigg] \bigg|_{0}^{0.5} \\ = & \bigg[ -\frac{(0.5)^3}{3} + \frac{(0.5)^2}{4} \bigg] - \bigg[ -\frac{(0)^3}{3} + \frac{(0)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \frac{1}{48} \\ \end{aligned}$

and the second term being

$\begin{aligned} & \int_{0.5}^{1}\bigg( t^2-\frac{1}{2}t\bigg) \,\mathrm{d}t \\ = & \bigg[ \frac{t^3}{3} - \frac{t^2}{4} \bigg]\bigg|_{0.5}^{1} \\ = & \bigg[ \frac{(1)^3}{3} - \frac{(1)^2}{4} \bigg] - \bigg[ \frac{(0.5)^3}{3} - \frac{(0.5)^2}{4} \bigg] \\ \dots & \enspace \textrm{by arithmetic}\enspace \dots \\ = & \bigg( \frac{1}{12} \bigg) - \bigg( -\frac{1}{48} \bigg) \\ = & \frac{5}{48} \\ \end{aligned}$

In sum,

$d_{1}(x,y)=\displaystyle{\frac{1}{48}+\frac{5}{48}=\frac{6}{48}=\frac{1}{8}}=0.125$ .

Part (b), (c), and (d) are not chosen.

October 2, 2020October 24, 2022

202010020718 Problem 2.1.2

Let $(X,d)$ be a metric space and let $k$ be a fixed positive real number. For $x,\, y\in X$ , define

$d^{*}=kd(x,y)$ .

Prove that $d^{*}$ is a metric on $X$ .

Recall.

Definition. (metric) Let $X$ be a non-empty set. A metric on $X$ is a real-valued function $d:\enspace X\times X\rightarrow \mathbb{R}$ satisfying the following conditions i– iv:

i. $d(x,y)\ge 0$ ;
ii. $d(x,y)=0\Leftrightarrow x=y$ ;
iii. (Symmetry) $d(x,y)=d(y,x)$ ;
iv. (Triangle Inequality) $d(x,y)\le d(x,z)+d(z,y)$ for any $x,\, y,\, z\in X$ .

N.b. Given $x,\,y\in X$ , $d(x,y)$ is sometimes called the distance between $x$ and $y$ with respect to $d$ .

Proof.

i.

WTS (wish to show)

$d^{*}(x,y)\ge 0$

By definition $d^{*}(x,y)=kd(x,y)$ and in that the metric $d$ is let clear ( $\therefore d(x,y)\ge 0$ ) and $k$ a fixed positive real number ( $\therefore k>0$ ),

one can see

$\begin{aligned} d^{*}(x,y) & = kd(x,y) \\ \textrm{\dots because\enspace} & k>0 \enspace \textrm{and}\enspace d(x,y)\ge 0\textrm{\enspace \dots}\\ d^{*}(x,y) & \geqslant 0 \\ \end{aligned}$

$\therefore$ Condition i. is made.

ii.

$\begin{aligned} d^{*}(x,y) & = 0 \\ \Leftrightarrow kd(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace k>0 \enspace & \textrm{so}\enspace k\neq 0\enspace \dots \\ \Leftrightarrow d(x,y) & = 0 \\ \dots\enspace \textrm{as}\enspace d \enspace \textrm{was} &\enspace\textrm{foretold to be a metric}\enspace \dots \\ \Leftrightarrow x & = y \\ \end{aligned}$

$\therefore$ Condition ii. is made.

iii.

NTS (need to show)

$d^{*}(x,y)=d^{*}(y,x)$

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \enspace & \textrm{by the symmetric property of }d\enspace \dots \\ & =kd(y,x) \\ & = d^{*}(y,x) \\ & = \textrm{RHS} \\ \end{aligned}$

$\therefore$ Condition iii. is made.

iv.

RTP (required to prove)

$d^{*}(x,y)\leqslant d^{*}(x,z)+d^{*}(z,y)$

One starts with the left hand side,

$\begin{aligned} \textrm{LHS} & = d^{*}(x,y) \\ & = kd(x,y) \\ \dots \textrm{as does}\enspace & d(x,y)\le d(x,z)+d(z,y)\enspace\textrm{the metric}\enspace d\enspace \textrm{do} \dots \\ & \le k\Big( d(x,z) + d(z,y) \Big) \\ & = kd(x,z) + kd(z,y) \\ & = d^{*}(x,z) + d^{*}(z,y) \\ & = \textrm{RHS} \\ \end{aligned}$

Condition iv. is made.

In conclusion, $(X,d^{*})$ is a metric space metered by a well-defined metric $d^{*}$ . This metric space shall simply be called $X$ hence.

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 $n\in\mathbb{N}$ .