Physicspupil – Page 30

September 10, 2021October 24, 2022

202109101556 Exercises 1.2.C (Q16)

For Exercises 16-21, assuming that $f'(x)$ exists, prove the given formula.

$f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(x+2h)-f(x-2h)}{4h}}$

Proof.

Renaming by dummy variables.

Let $y=x-2h$ , then $x+2h=(x-2h)+4h=y+4h$ .

Rewrite it as

$f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(y+4h)-f(y)}{4h}}$ .

Note that

$\displaystyle{\lim_{h\to 0}}[\,\cdots ]\Rightarrow \displaystyle{\lim_{4h\to 0}}[\,\cdots ]$ .

So,

$\begin{aligned} f'(x) & = \lim_{4h\to 0}\frac{f(y+4h)-f(y)}{4h} \\ & = \lim_{\Delta y\to 0}\frac{f(y+\Delta y)-f(y)}{\Delta y} \\ & = \lim_{\Delta y\to 0}\frac{\Delta f}{\Delta y}\\ & = \frac{\mathrm{d}f}{\mathrm{d}y}\\ & = \dots\enspace \textrm{(discontinued)}\enspace \dots \\ \end{aligned}$

Do you spot the flaw in the ~~Proof~~?

(revised)

As left-hand limit and right-hand limit are equivalent,

i.e., $f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}}$ ,

in our scenario, do write

$\begin{aligned} f'(x) & = \lim_{2h\to 0}\frac{f(x+2h)-f(x)}{2h}=\lim_{2h\to 0}\frac{f(x)-f(x-2h)}{2h} \\ \frac{1}{2}f'(x) & =\lim_{2h\to 0}\frac{f(x+2h)-f(x)}{4h}=\lim_{2h\to 0}\frac{f(x)-f(x-2h)}{4h}\\ \end{aligned}$

Then

$\begin{aligned} & \quad \lim_{h\to 0}\frac{f(x+2h)-f(x-2h)}{4h} \\ & = \lim_{h\to 0}\frac{\big( f(x+2h)-f(x)\big) + \big( f(x)-f(x-2h) \big) }{4h} \\ & = \lim_{h\to 0}\frac{f(x+2h)-f(x)}{4h} + \lim_{h\to 0}\frac{f(x)-f(x-2h)}{4h} \\ & = \lim_{2h\to 0}\frac{f(x+2h)-f(x)}{4h} + \lim_{2h\to 0}\frac{f(x)-f(x-2h)}{4h} \\ & = \frac{1}{2}\cdot f'(x)+\frac{1}{2}\cdot f'(x) \\ & = f'(x) \\ \end{aligned}$

QED

September 10, 2021October 24, 2022

202109101417 Exercises 1.1.A (Q5)

By equation (1.1), $\pi =4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots )$ , where the $n^{\textrm{th}}$ term in the sum inside the parenthesis is $\frac{(-1)^{n+1}}{2n-1}$ (starting at $n=1$ ). So the first approximation of $\pi$ using this formula is $\pi\approx 4(1)=4.0$ , and the second approximation is $\pi\approx 4(1-\frac{1}{3})=8/3\approx 2.66667$ . Continue like this until two consecutive approximations have $3$ as the first digit before the decimal point. How many terms in the sum did this require? Be careful with rounding off in the approximations.

Attempts.

$1^{\textrm{st}}$ approximation:

$\pi\approx 4(1)=4.0$

$2^{\textrm{nd}}$ approximation:

$\pi\approx 4(1-\frac{1}{3})=\frac{8}{3}\approx 2.66667$

$3^{\textrm{rd}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5})=\frac{52}{15}\approx 3.466667\quad (\textrm{5 d.p.})$

$4^{\textrm{th}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7})=\frac{304}{105}\approx 2.89523\quad (\textrm{5 d.p.})$

$5^{\textrm{th}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9})=\frac{1052}{315}\approx 3.33968\quad (\textrm{5 d.p.})$

$6^{\textrm{th}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11})=\frac{10312}{3465}\approx 2.97605\quad (\textrm{5 d.p.})$

$7^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}\bigg) \\ & =\frac{147916}{45045} \\ & \approx 3.28374\quad (\textrm{5 d.p.}) \end{aligned}$

$8^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}\bigg) \\ & =\frac{135904}{45045} \\ & \approx 3.01707\quad (\textrm{5 d.p.}) \end{aligned}$

$9^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}\bigg) \\ & =\frac{2490548}{765765} \\ & \approx 3.25237\quad (\textrm{5 d.p.}) \end{aligned}$

$10^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}\bigg) \\ & =\frac{44257352}{14549535} \\ & \approx 3.04184\quad (\textrm{5 d.p.}) \end{aligned}$

$11^{\textrm{th}}$ approximation:

$12^{\textrm{th}}$ approximation:

$13^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}\bigg) \\ & =\frac{5385020324}{1673196525} \\ & \approx 3.21840\quad (\textrm{5 d.p.}) \end{aligned}$

$14^{\textrm{th}}$ approximation:

$15^{\textrm{th}}$ approximation:

$16^{\textrm{th}}$ approximation:

$17^{\textrm{th}}$ approximation:

$18^{\textrm{th}}$ approximation:

$19^{\textrm{th}}$ approximation:

$20^{\textrm{th}}$ approximation:

$21^{\textrm{st}}$ approximation:

$22^{\textrm{nd}}$ approximation:

$23^{\textrm{rd}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\bigg) \\ & =\frac{937558224301357108}{294362129962575675} \\ & \approx 3.18505\quad (\textrm{5 d.p.}) \end{aligned}$

$24^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}\bigg) \\ & =\frac{42887788022313481376}{13835020108241056725} \\ & \approx 3.09994\quad (\textrm{5 d.p.}) \end{aligned}$

$25^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}\bigg) \\ & =\frac{308120241932332116332}{96845140757687397075} \\ & \approx 3.18158\quad (\textrm{5 d.p.}) \end{aligned}$

$26^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}\bigg) \\ & =\frac{300524544618003693032}{96845140757687397075} \\ & \approx 3.10315\quad (\textrm{5 d.p.}) \end{aligned}$

$27^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}\bigg) \\ & =\frac{16315181427784945318996}{5132792460157432044975} \\ & \approx 3.17862\quad (\textrm{5 d.p.}) \end{aligned}$

$28^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}\bigg) \\ & =\frac{15941887430682586624816}{5132792460157432044975} \\ & \approx 3.10589\quad (\textrm{5 d.p.}) \end{aligned}$

$29^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}\bigg) \\ & =\frac{16302083392798897645516}{5132792460157432044975} \\ & \approx 3.17607\quad (\textrm{5 d.p.}) \end{aligned}$

$30^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}\bigg) \\ & =\frac{941291750334505232905544}{302834755149288490653525} \\ & \approx 3.10827\quad (\textrm{5 d.p.}) \end{aligned}$

$31^{\textrm{st}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}\bigg) \\ & =\frac{58630135791001973169852284}{18472920064106597929865025} \\ & \approx 3.17384\quad (\textrm{5 d.p.}) \end{aligned}$

$32^{\textrm{nd}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}-\frac{1}{63}\bigg) \\ & =\frac{57457251977407903460019584}{18472920064106597929865025} \\ & \approx 3.11035\quad (\textrm{5 d.p.}) \end{aligned}$

$33^{\textrm{rd}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}-\frac{1}{63}+\frac{1}{65}\bigg) \\ & =\frac{4507234389098153984166548}{1420993851085122917681925} \\ & \approx 3.17189\quad (\textrm{5 d.p.}) \end{aligned}$

$34^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}-\frac{1}{63}+\frac{1}{65}-\frac{1}{67}\bigg) \\ & =\frac{296300728665235825268431016}{95206588022703235484688975} \\ & \approx 3.11219\quad (\textrm{5 d.p.}) \end{aligned}$

Summing without aim, I forgot my purpose. Where am I?

(discontinued)

(refreshed)

Please scroll up to the $7^{\textrm{th}}$ and $8^{\textrm{th}}$ approximation.

This required seven or eight terms in the sum for having $3$ as the first digit before the decimal point.

September 10, 2021October 24, 2022

202110091141 Exercises 1.1.A (Q1-Q4)

For Exercises 1-4, suppose that an object moves in a straight line such that its position $s$ after time $t$ is the given function $s=s(t)$ . Find the instantaneous velocity of the object at a general time $t\ge 0$ . You should mimic the earlier example for the instantaneous velocity when $s=-16t^2+100$ .

1. $s=t^2$

2. $s=9.8t^2$

3. $s=-16t^2+2t$

4. $s=t^3$

Ans.

1. $2t$

2. $19.6t$

3. $-32t+2$

4. $3t^2$

Solution.

1.

The average velocity of the object over the interval $[t,t+\Delta t]$ is $\frac{\Delta s}{\Delta t}$ , so since $s(t)=t^2$ :

$\begin{aligned} \frac{\Delta s}{\Delta t} & = \frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \frac{(t+\Delta t)^2 - t^2}{\Delta t} \\ & = \frac{(t^2+2t\Delta t+(\Delta t)^2)-(t^2)}{\Delta t} \\ & = \frac{2t\Delta t+(\Delta t)^2}{\Delta t} \\ & = \frac{\Delta t(2t+\Delta t)}{\Delta t} \\ & = 2t + \Delta t \end{aligned}$

Now let the interval $[t,t+\Delta t]$ get smaller and smaller indefinitely—that is let $\Delta t$ get closer and closer to $0$ . Then the average velocity $\frac{\Delta s}{\Delta t}=2t+\Delta t$ gets closer and closer to $2t+0=2t$ . Thus, the object has instantaneous velocity $2t$ at time $t$ . This calculation can be interpreted as taking the limit of $\frac{\Delta s}{\Delta t}$ as $\Delta t$ approaches $0$ , written as follows:

$\begin{aligned} & \qquad \textrm{instantaneous velocity at }t \\ & = \textrm{limit of average velocity over }[t,t+\Delta t]\textrm{ as }\Delta t\textrm{ approaches to }0 \\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t} \\ & = \lim_{\Delta t\to 0}(2t+\Delta t) \\ & = 2t+(0) \\ & = 2t \end{aligned}$

2.

$\begin{aligned} &\qquad \textrm{instantaneous velocity at }t\\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}\\ & = \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{9.8(t+\Delta t)^2 - 9.8t^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{9.8(t^2+2t(\Delta t)+(\Delta t)^2) - 9.8t^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{19.6t(\Delta t)+9.8(\Delta t)^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}19.6t+9.8(\Delta t) \\ & = 19.6t+9.8(0) \\ & = 19.6t \end{aligned}$

3.

$\begin{aligned} &\qquad \textrm{instantaneous velocity at }t\\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}\\ & = \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{\big( -16(t+\Delta t)^2+2(t+\Delta t)\big) - (-16t^2+2t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{-16(t^2+2t(\Delta t)+(\Delta t)^2)+2(t+\Delta t)+16t^2-2t}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{-32t(\Delta t)-16(\Delta t)^2+2(\Delta t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}(-32t-16(\Delta t)+2) \\ & = -32t-16(0)+2 \\ & = -32t+2 \end{aligned}$

4.

$\begin{aligned} s(t) & = t^3 \\ s(t+\Delta t) & = (t+\Delta t)^3 \\ & = t^3+3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \\ s(t+\Delta t)-s(t) & = \big( t^3+3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \big) - (t^3) \\ & = 3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \\ \frac{s(t+\Delta t)-s(t)}{\Delta t} & = \frac{3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3}{\Delta t} \\ & = 3t^2+3t(\Delta t)+(\Delta t)^2 \\ \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} & = 3t^2+3t(0)+(0)^2 \\ \frac{\mathrm{d}s}{\mathrm{d}t} & = 3t^2\\ \end{aligned}$

August 25, 2021July 25, 2022

202108251241 Photographs in ’00s

In my seventh year, the new millennium.

(colour)

(black-and-white)

August 5, 2021July 25, 2022

202108050932 Project Euler Problem 1 solved by C

Problem 1 of Project Euler

If we list all the natural numbers below $10$ that are multiples of $3$ or $5$ , we get $3$ , $5$ , $6$ and $9$ . The sum of these multiples is $23$ .

/* running a check on the first two statements by a direct computation */
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    printf("The sum of all the natural numbers below 10 that are multiples of 3 and 5 is %d.", 3 + 5 + 6 + 9);
    system("pause");
    return 0;
    }

/* running a check on the first two statements by a direct computation */

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

printf("The sum of all the natural numbers below 10 that are multiples of 3 and 5 is %d.", 3 + 5 + 6 + 9);

system("pause");

return 0;

}

Find the sum of all the multiples of $3$ or $5$ below $1000$ .

Ans. $233168$

(step-by-step approach, slow but sure logic)

I wish to find all the multiples of $3$ (excluding multiples of $5$ ) below $1000$ :

/* finding all the multiples of 3 (excluding multiples of 5) below 1000*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 3)
        {
        if (i % 5 != 0)
            {
            sum += i;
            }
        }
    printf("The sum of multiples of three (excluding multiples of five) below one thousand is %d.", sum);
    }

/* finding all the multiples of 3 (excluding multiples of 5) below 1000*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 3)

{

if (i % 5 != 0)

{

sum += i;

}

printf("The sum of multiples of three (excluding multiples of five) below one thousand is %d.", sum);

}

Output.
The sum of multiples of three (excluding multiples of five) below one thousand is 133668.

1 2	Output. The sum of multiples of three (excluding multiples of five) below one thousand is 133668.

Then I wish to find all the multiples of $5$ (excluding multiples of $3$ ) below $1000$ :

/* finding all the multiples of 5 (excluding multiples of 3) below 1000:*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 5)
        {
        if (i % 3 != 0)
            {
            sum += i;
            }
        }
    printf("The sum of multiples of five (excluding multiples of three) below one thousand is %d.", sum);
    }

/* finding all the multiples of 5 (excluding multiples of 3) below 1000:*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 5)

{

if (i % 3 != 0)

{

sum += i;

}

printf("The sum of multiples of five (excluding multiples of three) below one thousand is %d.", sum);

}

Output.
The sum of multiples of five (excluding multiples of three) below one thousand is 66335.

1 2	Output. The sum of multiples of five (excluding multiples of three) below one thousand is 66335.

Lastly I wish to find all the multiples of $15=3\times 5$ below $1000$ :

/* finding all the multiples of 15 below 1000:*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 15)
        {
        sum += i;
        }
    printf("The sum of multiples of fifteen below one thousand is %d.", sum);
    }

/* finding all the multiples of 15 below 1000:*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 15)

{

sum += i;

}

printf("The sum of multiples of fifteen below one thousand is %d.", sum);

}

Output.
The sum of multiples of fifteen below one thousand is 33165.

1 2	Output. The sum of multiples of fifteen below one thousand is 33165.

In sum,

$\begin{aligned} & \quad \textrm{The sum of all multiples of 3 or 5 below 1000} \\ & = 133\small,668 + 66\small,335 + 33\small,165 \\ & = 233\small,168 \end{aligned}$

Ans. $233168 \quad\checkmark$

(to be continued)

Going off at a tangent, the sum of all numbers from $1$ to $1000$ is $500500$ . See:

#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int i,sum;
    i = 1;
    sum = 0;
    do
    {
    sum += i++;
    }
    while (i <= 1000);
    printf("sum = %d", sum);
    system("pause");
    return 0;
    }

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int i,sum;

i = 1;

sum = 0;

{

sum += i++;

}

while (i <= 1000);

printf("sum = %d", sum);

system("pause");

return 0;

}

Output.
sum = 500500

1 2	Output. sum = 500500

(proof)

$\begin{aligned} & \quad 1+2+3+\cdots +1000 \\ & = \sum_{k=1}^{1000}k \\ & = \frac{(1000)(1000+1)}{2} \\ & = 500500 \\ \end{aligned}$

(continue)

Solution. (fast and furious attempt)

/* The sum of all the multiples of 3 or 5 below 1000 */
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    int i = 0;
    while(i < 1000)
    {
    if(i % 3 == 0 || i % 5 == 0)
    sum += i;
    i++;
    }
    printf("The sum of all the multiples of 3 or 5 below 1000 is %d.", sum);
    }

/* The sum of all the multiples of 3 or 5 below 1000 */

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

int i = 0;

while(i < 1000)

{

if(i % 3 == 0 || i % 5 == 0)

sum += i;

i++;

}

printf("The sum of all the multiples of 3 or 5 below 1000 is %d.", sum);

}

Output.
The sum of all the multiples of 3 or 5 below 1000 is 233168.

1 2	Output. The sum of all the multiples of 3 or 5 below 1000 is 233168.

August 4, 2021October 24, 2022

202108041142 Arithmetic Operators in C

Given two operands $a$ and $b$ , let them be integers:

int a;
int b;
printf("Please input two integers \n==>");
scanf("%d %d", &a, &b);

int a;

int b;

printf("Please input two integers \n==>");

scanf("%d %d", &a, &b);

Addition Operator $(+)$ adds two operands $a$ , $b$ :

sum = a + b;
printf("The sum is %d\n", sum);

1 2	sum = a + b; printf("The sum is %d\n", sum);

Subtraction Operator $(-)$ subtracts the second operand $b$ from the first operand $a$ :

difference = a - b;
printf("The difference is %d\n", difference);

1 2	difference = a - b; printf("The difference is %d\n", difference);

Multiplication Operator $(*)$ multiplies both operands $a$ , $b$ :

product = a * b;
printf("The product is %d\n", product);

1 2	product = a * b; printf("The product is %d\n", product);

Division Operator $(/)$ divides the first operand $a$ (numerator) by the second operand $b$ (denominator):

quotient = a / b;
printf("The quotient is %d\n", quotient);

1 2	quotient = a / b; printf("The quotient is %d\n", quotient);

Modulo Operator $(\% )$ returns the remainder after integer division of the first operand $a$ (dividend) by the second operand $b$ (divisor):

s = a % b;
printf("a mod b is %d\n", s);

1 2	s = a % b; printf("a mod b is %d\n", s);

Increment Operator $(++)$ increases the integer value by one.

int a;
int b;
//pre-increment
b = ++a;
//post-increment
b = a++;

int a;

int b;

//pre-increment

b = ++a;

//post-increment

b = a++;

Decrement Operator $(--)$ decreases the integer value by one.

int a;
int b;
//pre-decrement
b = --a;
//post-decrement
b = a--;

int a;

int b;

//pre-decrement

b = --a;

//post-decrement

b = a--;

Cf.

The pre-increment and pre-decrement operators increment (decrement resp.) their operand by 1, and the value of the expression is the resulting incremented (decremented resp.) value. The post-increment and post-decrement operators increase (decrease resp.) the value of their operand by 1, but the value of the expression is the operand’s value prior to the increment (decrement resp.) operation.

_{Wikipedia on Increment and decrement operators}

May 25, 2021October 24, 2022

202105251602 Homework 1 (Q1)

Recall that classical wave in one spatial dimension is described by the wave equation Eq. (1):

$\displaystyle{\frac{\partial^2u}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2u}{\partial t^2}}$ .

(a) Find conditions on the constants $a$ , $f$ , $k$ and $\phi$ so that $u=a\cos (2\pi ft+kx+\phi )$ is a solution of the Eq. (1).

(b) What are the physical meanings of $a$ , $f$ , $k$ and $\phi$ ?

(c) Show that if $u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1), then so is $c_1u_1(x,t)+c_2u_2(x,t)$ where $c_1$ , $c_2$ are constants. (Mathematically, we say that the solutions form a linear space or vector space.)

(d) Consequently, $u=\sum_{i=1}^2a_i\cos (2\pi f_it+k_ix+\phi_i)$ is a solution of Eq. (1) provided that each of the two terms is also a solution of Eq. (1). What could be said about the frequency of the wave described by this linear-superpositioned solution?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2265 Modern Physics Homework 1 Solution.)

(a)

$\begin{aligned} \frac{\partial^2u}{\partial x^2} & = \frac{\partial}{\partial x}\bigg(\frac{\partial u}{\partial x}\bigg) \\ & = \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}\Big( a\cos (2\pi ft+kx+\phi )\Big)\bigg) \\& = \frac{\partial}{\partial x}\Big( -ak\sin (2\pi ft+kx+\phi) \Big) \\ & = -ak^2\cos (2\pi ft+kx+\phi) \\ \frac{\partial^2u}{\partial t^2} & = \frac{\partial}{\partial t}\bigg(\frac{\partial u}{\partial t}\bigg) \\ & = \frac{\partial}{\partial t}\bigg(\frac{\partial}{\partial t}\Big( a\cos (2\pi ft+kx+\phi )\Big)\bigg) \\& = \frac{\partial}{\partial t}\Big( -a(2\pi f)\sin (2\pi ft+kx+\phi) \Big) \\ & = -4a\pi^2f^2\cos (2\pi ft+kx+\phi ) \end{aligned}$

$-ak^2\cos (2\pi ft+kx+\phi )=\displaystyle{-\frac{4a\pi^2f^2}{v^2}\cos (2\pi ft+kx+\phi )}$ ,

we have $k=\pm \displaystyle{\frac{2\pi f}{v}}$ .

There are no constraints on $a$ and $\phi$ .

(b)

$a$ : amplitude
$f$ : frequency
$k$ : wave number
$\phi$ : phase shift of the wave at $x=0$ and $t=0$

(c)

Set $u(x,t)=c_1u_1(x,t)+c_2u_2(x,t)$ .

Condition $\textrm{(i)}:\enspace u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1).

$\begin{aligned} \frac{\partial^2u}{\partial x^2} & = c_1\frac{\partial^2u_1}{\partial x^2} + c_2\frac{\partial^2u_2}{\partial x^2} \\ & \stackrel{\textrm{(i)}}{=} \frac{c_1}{v^2}\frac{\partial^2u_1}{\partial t^2} + \frac{c_2}{v^2}\frac{\partial^2u_2}{\partial t^2} \\ & = \frac{1}{v^2}\bigg( c_1\frac{\partial^2u_1}{\partial t^2}+c_2\frac{\partial^2u_2}{\partial t^2} \bigg) \\ \frac{\partial^2u}{\partial t^2} & = c_1\frac{\partial^2u_1}{\partial t^2}+c_2\frac{\partial^2u_2}{\partial t^2}\\ \therefore \enspace & \frac{\partial^2u}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2u}{\partial t^2} \end{aligned}$

That $u_1(x,t)$ and $u_2(x,t)$ are solutions of Eq. (1) implies $c_1u_1(x,t)+c_2u_2(x,t)$ is also a solution.

(d) Frequency is not well-defined for linear-superpositioned waves.

May 25, 2021June 24, 2023

202105251532 Homework 2 (Q1)

Prove that the electric field is always perpendicular to equipotential surface.

Solution.

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 2 Solution.)

$\mathbf{E}$ is the electric field vector; $\mathbf{dr}$ is a line element vector on the equipotential surface.

For any two arbitrary points $a$ and $b$ on the equipotential surface, we have the same potential there (i.e., $V(a)=V(b)$ ). From $-\int_a^b\mathbf{E}\cdot\mathbf{dr}=V(b)-V(a)=0$ . Thus $\mathbf{E}\cdot \mathbf{dr}=0$ , or, $\mathbf{E}\perp\mathbf{dr}$ .

May 24, 2021October 24, 2022

202105241204 Homework 1 (Q3)

Prove

(a) $\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A})-A\times (\nabla f)$

(b) $\nabla \times (\mathbf{A}\times \mathbf{B})=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}$

Attempts. (brute force)

(a)

$\begin{aligned} & \quad \nabla\times (f\mathbf{A}) \\ & = \nabla \times (fA_x,fA_y,fA_z) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ fA_x & fA_y & fA_z \end{vmatrix} \\ & = \bigg( \frac{\partial}{\partial y}(fA_z)-\frac{\partial}{\partial z}(fA_y),\, \frac{\partial}{\partial z}(fA_x) - \frac{\partial}{\partial x}(fA_z),\, \frac{\partial}{\partial x}(fA_y) - \frac{\partial}{\partial y}(fA_x) \bigg) \\ & = \Bigg( \bigg( f\frac{\partial A_z}{\partial y} + \frac{\partial f}{\partial y}A_z - f\frac{\partial A_y}{\partial z} - \frac{\partial f}{\partial z}A_y \bigg) , \\ & \quad \qquad \bigg( f\frac{\partial A_x}{\partial z} + \frac{\partial f}{\partial z}A_x - f\frac{\partial A_z}{\partial x} - \frac{\partial f}{\partial x}A_z \bigg) , \\ & \qquad \qquad \bigg( f\frac{\partial A_y}{\partial x}-\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}A_x - f\frac{\partial A_x}{\partial y} \bigg) \Bigg) \\ & = f\Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg) ,\, \bigg( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial z} \bigg),\, \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\bigg) \Bigg) \\ & \quad \qquad + \bigg( \frac{\partial f}{\partial y}A_z - \frac{\partial f}{\partial z}A_y,\, \frac{\partial f}{\partial z}A_x - \frac{\partial f}{\partial x}A_z,\, \frac{\partial f}{\partial x}A_y - \frac{\partial f}{\partial y}A_x \bigg) \\ & = f(\nabla \times \mathbf{A}) + \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = f(\nabla \times \mathbf{A}) + (\nabla f)\times\mathbf{A} \\ & = f(\nabla \times \mathbf{A}) - \mathbf{A}\times (\nabla f) \\ \end{aligned}$

(b)

$\begin{aligned} \textrm{LHS}\enspace & = \nabla \times (\mathbf{A}\times \mathbf{B}) \\ & = \nabla \times (A_yB_z-A_zB_y,\, -A_xB_z+A_zB_x,\, A_xB_y-A_yB_x) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_yB_z-A_zB_y & -A_xB_z + A_zB_x & A_xB_y - A_yB_x \end{vmatrix} \\ & = \Bigg( \bigg( \frac{\partial}{\partial y}(A_xB_y) - \frac{\partial}{\partial y}(A_yB_x) - \frac{\partial}{\partial z}(A_xB_z) + \frac{\partial}{\partial z}(A_zB_x) \bigg) ,\, \\ & \quad \qquad \bigg( -\frac{\partial}{\partial x}(A_xB_y) + \frac{\partial}{\partial x}(A_yB_x) + \frac{\partial}{\partial z}(A_yB_z) - \frac{\partial}{\partial z}(A_zB_y) \bigg) ,\, \\ & \qquad \qquad \bigg( \frac{\partial}{\partial x}(-A_xB_z) - \frac{\partial}{\partial x}(A_zB_x) - \frac{\partial}{\partial y}(A_yB_z) + \frac{\partial}{\partial y}(A_zB_y) \bigg) \Bigg) \\ & = \Bigg( \bigg( A_x\frac{\partial B_y}{\partial y} + \frac{\partial A_x}{\partial y}B_y - A_y\frac{\partial B_x}{\partial y} - \frac{\partial A_y}{\partial y}B_x - A_x\frac{\partial B_z}{\partial z} - \frac{\partial A_x}{\partial z}B_z + A_z\frac{\partial B_x}{\partial z} + \frac{\partial A_z}{\partial z}B_x \bigg) ,\, \\ & \quad \qquad \bigg( -A_x\frac{\partial B_y}{\partial x} - \frac{\partial A_x}{\partial x}B_y + A_y\frac{\partial B_x}{\partial x}+\frac{\partial A_y}{\partial x}B_x + A_y\frac{\partial B_z}{\partial z} + \frac{\partial A_y}{\partial z}B_z - A_z\frac{\partial B_y}{\partial z} - \frac{\partial A_z}{\partial z}B_y \bigg) ,\, \\ & \qquad \qquad \bigg( A_x\frac{\partial B_z}{\partial x} + \frac{\partial A_x}{\partial x}B_z - A_z\frac{\partial B_x}{\partial x} - \frac{\partial A_z}{\partial x}B_x - A_y\frac{\partial B_z}{\partial y} - \frac{\partial A_y}{\partial y}B_z + \frac{\partial A_z}{\partial y}B_y + A_z\frac{\partial B_y}{\partial y} \bigg) \Bigg) \\ \end{aligned}$

$\textrm{RHS}=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}$

Inspect these four terms on the right hand side by expanding one after the other.

The first term being

$\begin{aligned} (\mathbf{B}\cdot\nabla )\mathbf{A} & = \bigg( B_x\frac{\partial}{\partial x} + B_y\frac{\partial}{\partial y} + B_z\frac{\partial}{\partial z} \bigg) \mathbf{A} \\ & = \bigg( B_x\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_x}{\partial y} + B_z\frac{\partial A_x}{\partial z},\, \\ & \quad \qquad B_x\frac{\partial A_y}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_y}{\partial z},\, \\ & \qquad \qquad B_x\frac{\partial A_z}{\partial x} + B_y\frac{\partial A_z}{\partial y} + B_z\frac{\partial A_z}{\partial z} \bigg)\end{aligned}$

the second term being

$\begin{aligned} (\nabla \cdot \mathbf{B})\mathbf{A} & = \bigg( \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}\bigg)\mathbf{A} \\ & = \bigg( \frac{\partial B_x}{\partial x}A_x + \frac{\partial B_y}{\partial y}A_x + \frac{\partial B_z}{\partial z}A_x ,\, \\ & \quad \qquad \frac{\partial B_x}{\partial x}A_y + \frac{\partial B_y}{\partial y}A_y + \frac{\partial B_z}{\partial z}A_y , \, \\ & \qquad \qquad \frac{\partial B_x}{\partial x}A_z + \frac{\partial B_y}{\partial y}A_z + \frac{\partial B_z}{\partial z}A_z \bigg) \\ \end{aligned}$

the third term being

$\begin{aligned} -(\mathbf{A}\cdot\nabla )\mathbf{B} & = - \bigg( A_x\frac{\partial}{\partial x} + A_y\frac{\partial}{\partial y} + A_z\frac{\partial}{\partial z} \bigg) \mathbf{B} \\ & = -\bigg( A_x\frac{\partial B_x}{\partial x} + A_y\frac{\partial B_x}{\partial y} + A_z\frac{\partial B_x}{\partial z},\, \\ & \quad\qquad A_x\frac{\partial B_y}{\partial x} + A_y\frac{\partial B_y}{\partial y} + A_z\frac{\partial B_y}{\partial z},\, \\ & \qquad\qquad A_x\frac{\partial B_z}{\partial x} + A_y\frac{\partial B_z}{\partial y} + A_z\frac{\partial B_z}{\partial z} \bigg) \\\end{aligned}$

and the fourth and last term being

$\begin{aligned} -(\nabla \cdot \mathbf{A})\mathbf{B} & = -\bigg( \frac{\partial A_x}{\partial x} +\frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} \bigg)\mathbf{B} \\ & = - \bigg( B_x\frac{\partial A_x}{\partial x} + B_x\frac{\partial A_y}{\partial y} + B_x\frac{\partial A_z}{\partial z} ,\, \\ & \quad\qquad B_y\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_y\frac{\partial A_z}{\partial z} ,\, \\ & \qquad \qquad B_z\frac{\partial A_x}{\partial x} + B_z\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_z}{\partial z}\bigg) \\ \end{aligned}$

One can check that $\textrm{LHS}=\textrm{RHS}$ .

Solution. (proof)

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 1 Solution.)

Using Einstein summation (/notation) and the Levi-Civita symbol $\varepsilon_{ijk}$ ,

(a)

$\begin{aligned} & \quad \nabla \times (f\mathbf{A}) \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(f\mathbf{A}_k)\cdot\varepsilon_{ijk}\qquad\qquad\qquad i,j,k\in\{ x,y,z\} \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\bigg(\frac{\partial}{\partial j}f\bigg)\cdot A_k\cdot\varepsilon_{ijk}+\sum_{i,j,k}\hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_k \bigg)\cdot f\cdot \varepsilon_{ijk} \\ & = (\nabla f)\times \mathbf{A} + f\cdot (\nabla\times\mathbf{A}) \\ & = (\nabla f)\times \mathbf{A} - A\times (\nabla f) \end{aligned}$

(b)

$\begin{aligned} \mathbf{A}\times\mathbf{B} & = \sum_{k,l,m}\hat{\mathbf{e}}_kA_lB_m\varepsilon_{klm}\\ \nabla\times (\mathbf{A}\times\mathbf{B}) & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(\sum_{l,m}A_lB_m\varepsilon_{klm})\varepsilon_{ijk} \\ & = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] \varepsilon_{klm}\varepsilon_{ijk} \\ \textrm{by } & \varepsilon_{klm}\varepsilon_{ijk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \\ \textrm{Thus, }& = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \\ & = \sum_{i,j,k,l,m}\bigg[ \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m (-\delta_{im}\delta_{jl}) \\ & \quad\qquad + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}B_m\bigg) A_l\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i \bigg( \frac{\partial}{\partial j}B_m \bigg) A_l (-\delta_{im}\delta_{jl}) \bigg] \\ & = (\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B} \end{aligned}$

QED

and the proof is more concise.

May 24, 2021January 19, 2023

202105241144 Homework 1 (Q4)

$\displaystyle{\int \delta (f(x))g(x)\,\mathrm{d}x}=\, ?$

Attempts.

$\begin{aligned} &\quad \int \delta(f(x))g(x)\,\mathrm{d}x \\ & = \int\delta (f(x))g(f^{-1}f(x))\frac{\mathrm{d}f(x)}{f'(x)} \\ & = \frac{g(f^{-1}f(x))}{f'(x)}\bigg|_{f(x)=0} \\ & = \frac{g(f^{-1}(0))}{f'(f^{-1}(0))} \end{aligned}$

Solution.

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 1 Solution.)

Let the roots to $f(x)=0$ be $a_i$ where $i=1,2,3,\dots$ .
From $y=f(x)$ (or, $x=f^{-1}(y)$ ), we have $\mathrm{d}y=f'(x)\,\mathrm{d}x$ .

$\begin{aligned} &\quad \int \delta (f(x))g(x)\,\mathrm{d}x \\ & = \sum_i \int\delta (f(x))g(x)\,\mathrm{d}x\\ & = \sum_i \int\frac{\delta (y)g(f^{-1}(y))}{f'(f^{-1}(y))}\,\mathrm{d}y \\ & = \sum_i \frac{g(f^{-1}(0))}{f'(f^{-1}(0))} \\ & = \sum_i \frac{g(a_i)}{f'(a_i)} \\ \end{aligned}$

QED