Physicspupil – Page 26

December 2, 2021October 24, 2022

202112021444 Solution to 1979-CE-AMATH-I-1

Let $\displaystyle{y=x+\frac{1}{x^2}}$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ from first principles.

Solution.

From first principles,

$\begin{aligned} y:=f(x) & =x+\frac{1}{x^2} \\ f(x+\Delta x) & =(x+\Delta x) + \frac{1}{(x+\Delta x)^2} \\ f(x+\Delta x)-f(x) & = \Delta x + \frac{1}{(x+\Delta x)^2} - \frac{1}{x^2} \\ & = \Delta x + \frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2} \\ & = \Delta x - \frac{2x\Delta x+(\Delta x)^2}{x^2(x+\Delta x)^2} \\ \frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + \Delta x}{x^2(x+\Delta x)^2} \\ \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + 0}{x^2(x+0)^2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = 1 - \frac{2}{x^3} \\ \end{aligned}$

Playground.

To make fun of calculus, do let

$\begin{aligned} y & =x+\frac{1}{x^2} \\ x^2y & =x^3+1 \\ 0 & = x^3-x^2y+1 := f(x,y) \\ \end{aligned}$

where $f(x,y)$ is a degree $3$ polynomial in two variables $x$ and $y$ .

The set of solutions to $f(x,y)=0$ is

$\begin{cases} \enspace x \neq 0 \\ \enspace y = \frac{x^3+1}{x^2} \end{cases}$

Computing $\partial_xf$ and $\partial_yf$ as follow:

$\begin{aligned} \partial_xf & = \frac{\partial}{\partial x}f(x,y) \\ & = 3x^2-2xy \\ \end{aligned}$

and

$\begin{aligned} \partial_yf & = \frac{\partial}{\partial y}f(x,y) \\ & = -x^2 \\ \end{aligned}$ ,

we oversee the relation

$\mathrm{d}f = (\partial_xf)(\Delta x)+(\partial_yf)(\Delta y)$ .

Definition. (Singularity and Smoothness)

A point $p=(a,b)$ on a curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:f(x,y)=0\}$ is said to be singular if

$\begin{aligned} \frac{\partial f}{\partial x}(a,b) & = 0 \\ \frac{\partial f}{\partial y}(a,b) & = 0 \\ \end{aligned}$ .

A point that is not singular is called smooth. If there is at least one singular point on $\mathcal{C}$ , then curve $\mathcal{C}$ is called a singular curve. If there are no singular points on $\mathcal{C}$ , the curve $\mathcal{C}$ is called a smooth curve.

_{C.f. Definition 1.9.1, T. Carrity, et al., Algebraic Geometry: A Problem Solving Approach}

That said, in our scenario

$f(x,y)=x^3-x^2y+1$ ,

the set of candidates for singularity

$\{ (a,b)\in\mathbb{C}^2\enspace \mathrm{s.t.}\enspace f(a,b)=0\textrm{ and }\partial_xf(a,b)=\partial_yf(a,b)=0\}$ .

is empty.

$\therefore$ The curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:x^3-x^2y+1=0\}$ is smooth, thus everywhere differentiable.

Wait, the function in question should be the curve

$F(x)=\displaystyle{x+\frac{1}{x^2}}$ ;

still, for better or worse, $F(x)$ is $\textrm{\scriptsize{NOT}}$ a differentiable function because the derivative does $\textrm{\scriptsize{NOT}}$ exist at $x=0$ , for that point is a discontinuity.

December 2, 2021October 24, 2022

202112021138 SQL Table 004

Background. (Equilibrium)

Equilibrium is established by the exchange of energy, volume, or particle number between different systems or subsystems.

_{Section 2.2. Arovas. (2019). Thermodynamics and Statistical Mechanics}

CREATE TABLE SQL_Table_004
    (ID int,
    Object_of_exchange varchar(255),
    Constancy varchar(255)
    Subject_of_equilibrium varchar(255)
    );
INSERT INTO SQL_Table_004
    (Object_of_exchange, Constancy, Subject_of_equilibrium)
    VALUES
    ('energy', 'temperature (T)', 'thermal')
    ('volume', 'momentum/temperature (p/T)', 'mechanical')
    ('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

CREATE TABLE SQL_Table_004

(ID int,

Object_of_exchange varchar(255),

Constancy varchar(255)

Subject_of_equilibrium varchar(255)

);

INSERT INTO SQL_Table_004

(Object_of_exchange, Constancy, Subject_of_equilibrium)

VALUES

('energy', 'temperature (T)', 'thermal')

('volume', 'momentum/temperature (p/T)', 'mechanical')

('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

Output.
+------+---------------------+-------------------------------+--------------------------+
|  ID  | Object_of_exchange  |  Constancy                    |  Subject_of_equilibrium  |
+------+---------------------+-------------------------------+--------------------------+
|  1   |  energy             |  temperature (T)              |  thermal                 |
|  2   |  volume             |  momentum/temperature (p/T)   |  mechanical              |
|  3   |  particle           |  chemical potential/          |  chemical                |
|      |                     |    temperature (\mu /T)       |                          |
+------+---------------------+-------------------------------+--------------------------+

Output.

+------+---------------------+-------------------------------+--------------------------+

+------+---------------------+-------------------------------+--------------------------+

+------+---------------------+-------------------------------+--------------------------+

December 2, 2021December 12, 2022

202112021037 Exercise 1.4 Lagrangian actions

For a free particle an appropriate Lagrangian is

Eq. (1.8):

$\mathcal{L}(t,x,v)=\displaystyle{\frac{1}{2}mv^2}$ .

Suppose that $x$ is the constant-velocity straight-line path of a free particle, such that $x_a=x(t_a)$ and $x_b=x(t_b)$ . Show that the action on the solution path is

Eq. (1.9):

$\displaystyle{\frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a}}$ .

_{Extracted from Structure and Interpretation of Classical Mechanics, SICP, 2e}

Background. (Lagrangians and Lagrangian actions)

The function $\mathcal{L}$ is called a Lagrangian for the system, and the resulting action,

Eq. (1.4):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}\circ\Gamma [q]}$ ,

is called the Lagrangian action. For Lagrangians that depend only on time, positions, and velocities the action can also be written

Eq. (1.5):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}(t,q(t),\mathrm{D}q(t))\,\mathrm{d}t}$ .

_{(Section 1.3 The Principle of Stationary Action)}

Working. (roughly)

$\begin{aligned} S & = \int_{t_a}^{t_b}\frac{1}{2}mv^2\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}(\mathbf{v}\cdot\mathbf{v})\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}\bigg( \frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\cdot\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\bigg) \,\mathrm{d}t \\ & = \frac{m}{2}\int_{x_a,t_a}^{x_b,t_b}\frac{(\mathrm{d}x)^2}{(\mathrm{d}t)} \\ & = \frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a} \\ \end{aligned}$

November 25, 2021October 24, 2022

202111251402 Exercise 2.59 in Lisp

Implement the union-set operation for the unordered-list representation of sets.

_{Extracted from Structure and Interpretation of Computer Programs, SICP}

Attempts.

Define a function element-of-set?by

(define (element-of-set? x set)
(cond ((null? set) false)
((equal? x (car set)) true)
(else (element-of-set? x (cdr set)))))

(define (element-of-set? x set)

(cond ((null? set) false)

((equal? x (car set)) true)

(else (element-of-set? x (cdr set)))))

to determine whether an object xis an element of a set set.

Define a function intersection-set by

(define (intersection-set set1 set2)
(cond ((or (null? set1) (null? set2)) ’())
((element-of-set? (car set1) set2)
(cons (car set1)
(intersection-set (cdr set1) set2)))
(else (intersection-set (cdr set1) set2))))

(define (intersection-set set1 set2)

(cond ((or (null? set1) (null? set2)) ’())

((element-of-set? (car set1) set2)

(cons (car set1)

(intersection-set (cdr set1) set2)))

(else (intersection-set (cdr set1) set2))))

to draw the intersection intersection-set of two sets set1 and set2.

Define a function union-set by

(define (union-set set1 set2)
(cond ((and (null? set1) (null? set2)) ’())
((null? set1) set2)
((null? set2) set1)
((element-of-set? (car set1) set2)
(cons (car set1)
(union-set (cdr set1) set2)))
(else (union-set (cdr set1) set2))))

(define (union-set set1 set2)

(cond ((and (null? set1) (null? set2)) ’())

((null? set1) set2)

((null? set2) set1)

((element-of-set? (car set1) set2)

(cons (car set1)

(union-set (cdr set1) set2)))

(else (union-set (cdr set1) set2))))

to draw the union union-set of two sets set1 and set2.

November 25, 2021October 24, 2022

202111251031 Solution to 1995-IB-MATH-HL-I-3

Find the value of $x$ such that

$\begin{array}{|cc|} x & 6 \\ x & x \end{array} = \begin{array}{|ccc|} 2 & 1 & 3 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}$ .

Working.

$\begin{aligned} \textrm{LHS} & = x^2 - 6x \\ \textrm{RHS} & = 2 \cdot \begin{array}{|cc|} 2 & 3 \\ 1 & 1 \end{array} - 1 \cdot \begin{array}{|cc|} 1 & 3 \\ 3 & 1 \end{array} + 3 \cdot \begin{array}{|cc|} 1 & 2 \\ 3 & 1 \end{array} \\ & = 2\big( (2)(1)-(1)(3) \big) - 1\big( (1)(1)-(3)(3) \big) + 3\big( (1)(1)-(3)(2) \big) \\ & = 2(-1) - 1(-8) + 3(-5) \\ & = -9 \\ \end{aligned}$

$\begin{aligned} \textrm{LHS} & = \textrm{RHS} \\ x^2-6x & = -9 \\ x^2-6x+9 & = 0 \\ (x-3)^2 & = 0 \\ x & = 3 \qquad\textrm{(double roots)} \\ \end{aligned}$

Answers.

$x=3$ .

November 25, 2021October 24, 2022

202111250959 Solution to 1995-IB-MATH-HL-I-2

Calculate $x$ and $y$ exactly, as rational numbers, if

$\begin{aligned} 5^{3x}\times 25^{y}& =\frac{1}{5} \\ \textrm{and} \qquad\qquad 7^{x}\times 49^{2y} & = 1 \\ \end{aligned}$ .

Working.

From the first equation,

$\begin{aligned} 5^{3x}\times 25^{y} & = \frac{1}{5} \\ 5^{3x} \times 5^{2y} & = 5^{-1} \\ 5^{3x+2y} & = 5^{-1} \\ 3x+2y & = -1 \\ \end{aligned}$ .

From the second equation,

$\begin{aligned} 7^x\times 49^{2y} & = 1 \\ 7^x\times 7^{4y} & = 7^{0} \\ 7^{x+4y} & = 7^0 \\ x+4y & = 0 \\ \end{aligned}$ .

We have two linear equations with two unknowns:

$\begin{cases} 3x+2y = -1 \\ x+4y = 0 \\ \end{cases}$

Substituting $-4y$ for $x$ ,

$\begin{aligned} 3(-4y)+2y & =-1 \\ -12y+2y & = -1 \\ -10 y & = -1 \\ y & = \frac{1}{10} \\ \end{aligned}$

Then,

$x=-4y=\displaystyle{-4\bigg(\frac{1}{10}\bigg)=-\frac{2}{5}}$ .

Answers.

$\begin{cases} \enspace x = - \displaystyle{\frac{2}{5}} \\ \enspace y = \displaystyle{\frac{1}{10}} \\ \end{cases}$

November 25, 2021October 24, 2022

202111250947 Solution to 1995-IB-MATH-HL-I-1

Given the complex numbers $z_1=3+\mathrm{i}$ and $z_2=2-\mathrm{i}$ , find

(a) $z_1z_2$ :

(b) $\displaystyle{\frac{z_1}{z_2}}$ .

Working.

(a)

$\begin{aligned} z_1z_2 & = (3+\mathrm{i})(2-\mathrm{i}) \\ & = 6-3\mathrm{i}+2\mathrm{i}-\mathrm{i}^2 \\ & = 6-\mathrm{i}-(-1) \\ & = 7-\mathrm{i} \\ \end{aligned}$

(b)

$\begin{aligned} \frac{z_1}{z_2} & = \frac{3+\mathrm{i}}{2-\mathrm{i}} \\ & = \frac{(3+\mathrm{i})(2+\mathrm{i})}{(2-\mathrm{i})(2+\mathrm{i})} \\ & = \frac{6+3\mathrm{i}+2\mathrm{i}+\mathrm{i}^2}{(2)^2-(\mathrm{i})^2} \\ & = \frac{5+5\mathrm{i}}{5} \\ & = 1 + \mathrm{i} \\ \end{aligned}$

Answers.

(a) $7-\mathrm{i}$ ; (b) $1 + \mathrm{i}$ .

November 25, 2021October 24, 2022

202111250910 Solution to 2020-DSE-PHY-1A-10

The diameter of Neptune is about $4$ times that of the Earth and its mass is about $17$ times that of the Earth. Estimate the acceleration due to gravity on Neptune’s surface.

Given: acceleration due to gravity on Earth’s surface $g=9.81\,\mathrm{m\, s^{-2}}$

Background.

From Newton’s law of gravitation, Eq. (B7):

$F=\displaystyle{\frac{Gm_{1}m_{2}}{r^2}}$

_{(pg. 15, List of data, formulae and relationship)}

Solution.

On the Earth, for some object of mass $m$ , we see that:

$\begin{aligned} F_{E} & = \frac{GmM_{E}}{r_{E}} \\ mg_{E} & = \frac{GmM_{E}}{r_{E}} \\ g_{E} & = \frac{GM_{E}}{r_{E}} \\ \end{aligned}$

Similarly, for some object of mass $m$ on Neptune, we have

$\begin{aligned} F_{N} & = \frac{GmM_{N}}{r_{N}} \\ mg_{N} & = \frac{GmM_{N}}{r_{N}} \\ g_{N} & = \frac{GM_{N}}{r_{N}} \\ \end{aligned}$

We compare $g_{N}$ to $g_{E}$ by the assumption:

$\begin{aligned} g_{N} & = \frac{GM_{N}}{r_{N}} \\ & = \frac{G(17M_E)}{(4r_{E})} \\ & = \frac{17}{4}\bigg(\frac{GM_{E}}{r_{E}}\bigg) \\ & = \frac{17}{4}g_{E} \\ \end{aligned}$

And the answer is D.

November 24, 2021October 24, 2022

202111241152 Solution to 1971-CE-AMATH-3

Prove by mathematical induction that $n(n^2+2)$ is divisible by $3$ for all positive integral values of $n$ .

Solution.

Let $P(n)$ be the statement that

$P(n):\enspace n(n^2+2)\textrm{ is divisible by }3\textrm{ for }n\in\mathbb{Z}^{+}$

First, we wish to show $P(1)$ holds true.

$\begin{aligned} &\quad (1)\big( (1)^2 + 2\big) \\ & = 1(1+2) \\ & = 3\qquad (\textrm{divisible by }3)\\ \end{aligned}$

$\therefore$ $P(1)$ is true.

Once we assume that $P(n)$ is true for some positive integral values, we wish to show $P(n+1)$ is true.

$\begin{aligned} &\quad (n+1)\big( (n+1)^2 + 2\big) \\ & = (n+1)(n^2+2n+3) \\ & = n^3 + 3n^2 + 5n +3 \\ \end{aligned}$

Let $f(x)=x^3+3x^2+5x+3$ .

Suppose we can divide $f(x)$ by $3$ , i.e.,

$f(x)=3\cdot g(x)$

for some $g(x)=Ax^3+Bx^2+Cx+D$ where $A,B,C,D\in\mathbb{R}$ .

Then, let’s see

$\begin{aligned} \frac{f(x)}{g(x)} & = 3 \\ \frac{\mathrm{d}}{\mathrm{d}x}\bigg( \frac{f(x)}{g(x)}\bigg) & = \frac{\mathrm{d}}{\mathrm{d}x}(3) \\ \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} & = 0 \\ g(x)f'(x)-f(x)g'(x) & = 0 \\ \end{aligned}$

Plugging in $f(x)$ , $f'(x)$ , $g(x)$ , and $g'(x)$ , we have

$\begin{aligned} &\quad (Ax^3+Bx^2+Cx+D)(3x^2+6x+5) \\ & = (x^3+3x^2+5x+3)(3Ax^2+2Bx+C) \\ \end{aligned}$

$\begin{aligned} &\quad \enspace \textrm{LHS} \\ & = (3A)x^5 + (6A+3B)x^4 + (5A+6B+3C)x^3 \\ &\qquad\quad + (5B+6C+3D)x^2 + (5C+6D)x + (5D) \\ \end{aligned}$

$\begin{aligned} &\quad \enspace \textrm{RHS} \\ & = (3A)x^5 + (9A+2B)x^4 + (15A+6B+C)x^3 \\ &\qquad\quad + (9A+10B+3C)x^2 + (6B+5C)x + (3C) \\ \end{aligned}$

We have a set of five equations below:

$\begin{aligned} 6A + 3B & = 9A + 2B \\ 5A+6B+3C & = 15A+6B+C \\ 5B+6C+3D & = 9A+10B+3C \\ 5C +6D & = 6B+5C \\ 5D & = 3C \\ \end{aligned}$

which can be summarized to

$\displaystyle{A=\frac{B}{3}=\frac{C}{5}=\frac{D}{3}}$ ,

that is equivalent to saying

$g(x)=\displaystyle{\frac{f(x)}{3}=\frac{x^3}{3}+x^2+\frac{5x}{3}+1}$ ,

viz. I have been wasting time over a circular reasoning of no use.

I should have tried to attest otherwise that $g(x)$ is of some positive integral values, even though this is $\textrm{\scriptsize{NOT}}$ proving the statement $P(n)$ by mathematical induction firsthand.

Reformulation.

To show that the following statement $Q(x)$ holds true:

$Q(x):\quad \displaystyle{\frac{x^3}{3}+x^2+\frac{5x}{3}+1}\textrm{ is an integer for any }x\in\mathbb{Z}^{+}$ .

$Q(1)$ is true because

$\displaystyle{\frac{(1)^3}{3}+(1)^2+\frac{5(1)}{3}+1}=4$

is an integer.

Assume $Q(x)$ is true for some $x\in\mathbb{Z}^{+}$ .

Then,

$\begin{aligned} &\qquad \frac{(x+1)^3}{3}+(x+1)^2+\frac{5(x+1)}{3}+1 \\ & = \frac{(x^3+3x^2+3x+1)}{3} + (x^2+2x+1)+\frac{(5x+5)}{3} + 1 \\ & = \bigg( \frac{x^3}{3}+x^2+\frac{5x}{3}+1\bigg) + \bigg( \frac{3x^2+3x+1}{3}+2x+1+\frac{5}{3}+1 \bigg) \\ \end{aligned}$

The first term is an integer for $Q(x)$ is assumed to be true. We then need to check whether the second term is of integral value(s).

The second term, after simplification, is

$x^2+3x+4$ ,

obviously an integer.

$\therefore$ $Q(x+1)$ thus holds.

Because $Q(1)$ is true, by mathematical induction, $Q(x)$ is true for all positive integers $x$ .

Escape to the original statement.

$3|n^3+3n^2+5n+3\Longleftrightarrow P(n+1)\textrm{ holds true.}$

Because $P(1)$ is true, by mathematical induction, $P(n)$ is true for any positive integers $n$ .

November 24, 2021October 24, 2022

202111241104 Solution to 1969-CE-AMATH-3

Prove by mathematical induction that if $n$ is a positive integer, $f(n)\equiv 3^{2n}-1$ is divisible by $8$ .

Solution.

We wish to show that the statement is true for $n=1$ , and also true for $n+1$ .

When $n=1$ , $f(1)=3^{2(1)}-1=8$ is divisible by $8$ .

Assume $f(n)$ is true for some positive integer $n$ , i.e.,

$\{f(n):3^{2n}-1=8m\, |\, \exists \, m, n\in\mathbb{Z}^{+}\}$ .

For $f(n+1)=3^{2(n+1)}-1$ , we have

$\begin{aligned} & \quad 3^{2(n+1)}-1 \\ & = 3^{2n+2} - 1 \\ & = 3^{2n}\cdot 3^2 - 1 \\ & = 9(3^{2n})-1 \\ & = 3^{2n} -1 + 8(3^{2n}) \\ \end{aligned}$

As is assumed $3^{2n}-1$ divisible by $8$ , the statement is thus also true for $n+1$ .

We have proven by mathematical induction that for any positive integer $n\enspace (\in\mathbb{Z}^+)$ , $f(n)$ is divisible by $8$ .

Afterword.

Try to prove $\textrm{\scriptsize{NOT}}$ by mathematical induction $\textrm{\scriptsize{BUT}}$ by direct proof.

$\begin{aligned} &\quad \frac{3^{2n}-1}{8} \\ & = \frac{3^{2n}-1}{9-1} \\ & = \frac{3^{2n}-1}{3^2-1} \\ & = \dots \end{aligned}$

Let $x=3^2$ , then

$\begin{aligned} &\quad \dots \\ & = \frac{x^n-1}{x-1} \\ & = 1+x+x^2+\cdots + x^{n-2} + x^{n-1} \enspace (\in\mathbb{Z}^{+}) \\ \end{aligned}$

$\therefore$ $f(n)\equiv 3^{2n}-1$ is divisible by $8$ .