Miscellaneous – Physicspupil

February 15, 2024February 15, 2024

202402151329 Solution to 1985-CE-AMATH-II-5

If the equation

$x^2+y^2+kx-(2+k)y=0$

represents a circle with radius $\sqrt{5}$ ,

(a) find the value(s) of $k$ ;
(b) find the equation(s) of the circle(s).

As if sitting an exam in additional mathematics, it certainly not being showing off, we should perhaps involve ourselves with some sort of calculus.

Roughwork.

As always, have in mind some pictures. Hence, we draw:

and also

after the stage got set, we write

$\begin{aligned} 0 & = x^2+y^2+kx-(2+k)y \\ 0 & = 2x\,\mathrm{d}x+2y\,\mathrm{d}y + k\,\mathrm{d}x - (2+k)\,\mathrm{d}y \\ y' & = \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x+k}{k-2y+2} \\ \end{aligned}$

and furthermore,

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ & \Rightarrow 2x+k=0 \\ & \Rightarrow x = -\frac{k}{2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \infty \\ & \Rightarrow k-2y+2 = 0 \\ & \Rightarrow y = \frac{k}{2}+1 \\ \end{aligned}$

In the case of $x=-\frac{k}{2}$ :

$\begin{aligned} 0 & = \bigg( -\frac{k}{2}\bigg)^2+y^2+k\bigg( -\frac{k}{2}\bigg) -(2+k)y \\ 0 & = y^2 - (2+k)y - \frac{k^2}{4} \\ \Delta & = \big( -(2+k)\big)^2 - 4(1)\bigg( -\frac{k^2}{4}\bigg) \\ & = 2(k^2+2k+2)\qquad (> 0) \\ y_1,y_3 & = \frac{-\big( -(2+k)\big) \pm \sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = y_3-y_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

and the case of $y=\frac{k}{2}+1$ :

$\begin{aligned} 0 & = x^2 + \bigg( \frac{k}{2}+1\bigg)^2 + kx - (2+k)\bigg( \frac{k}{2}+1\bigg) \\ 0 & = x^2+kx - \bigg(\frac{k}{2}+1\bigg)^2\\ \Delta & = (k)^2 - 4(1)\bigg( -\bigg(\frac{k}{2}+1\bigg)^2\bigg) \\ & = 2(k^2+2k+2) \qquad (> 0) \\ x_1,x_3 & = \frac{-(k)\pm\sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = x_3 - x_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

Targeting the centre $C(a,b)$ :

$\begin{aligned} a_{i=1,2} & = \frac{x_3 - x_1}{2} = -\frac{k}{2}\bigg|_{k=2,-4} = -1,2 \\ b_{i=1,2} & = \frac{y_3-y_1}{2} = \frac{-\big( -(2+k)\big)}{2}\bigg|_{k=2,-4} = 2,-1 \\ (a,b) & = (-1,2)\cup (2,-1) \\ \end{aligned}$

Of the equation of circle, the standard form is

$\left\{ \begin{aligned} (x+1)^2 + (y-2)^2 & = (\sqrt{5})^2 \\ (x-2)^2 + (y+1)^2 & = (\sqrt{5})^2 \\ \end{aligned}\right\}$

and the general form

$\left\{ \begin{aligned} x^2+y^2+2x-4y & = 0 \\ x^2+y^2-4x+2y & = 0 \\ \end{aligned}\right\}$

This problem is not to be attempted.

January 31, 2024January 31, 2024

202401311548 Pastime Exercise 008

Let there be a cone $R\,\mathrm{(unit)}$ in radius of its base and $h\,\mathrm{(unit)}$ in height.

Find the volume $V\,\mathrm{(cubic\, unit)}$ of the cone, using what are so-called i. the disc (/washer) method and ii. the (cylindrical) shell method, if the names my memory serves me right.

Answer. $\displaystyle{V=\frac{1}{3}\pi R^2h}$ .

Roughwork.

Visualize the cone.

i. By disc method,

$\begin{aligned} \frac{R}{h} & = \frac{r}{h-z} = \tan\theta \\ r & = R\bigg( 1-\frac{z}{h}\bigg) \\ V & = \pi R^2\int_{0}^{h}\bigg( 1-\frac{z}{h}\bigg)^2\,\mathrm{d}z \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ \textrm{let }& z'=1-\frac{z}{h} \\ \frac{\mathrm{d}z'}{\mathrm{d}z} & = -\frac{1}{h} \\ \mathrm{d}z & = -h\,\mathrm{d}z' \\ z=0 & \Leftrightarrow z'=1 \\ z=h & \Leftrightarrow z'= 0 \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ V& = \pi R^2\int_{1}^{0} z'^2(-h\,\mathrm{d}z') \\ & = -\pi R^2h \bigg[ \frac{z'^3}{3}\bigg]\bigg|_{1}^{0} \\ & = -\pi R^2h \bigg(\frac{(0)^3}{3} - \frac{(1)^3}{3}\bigg) \\ & = \frac{\pi R^2h}{3}\\ \end{aligned}$

ii. By shell method,

we are obtaining the volume of a solid of revolution about the $z$ -axis by integrating the slices (/cross sections) ranging between $\theta\in [0,2\pi ]$ .

It is left the reader as an exercise.

This problem is not to be attempted.

October 13, 2023October 16, 2023

202310130920 Exercise 14.1.3

If $\cos A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$ where $A$ and $B$ are both acute angles, find, without using tables or a calculator,

(a) $\sin (A+B)$ ,
(b) $\cos (A-B)$ ,
(c) $\tan (A+B)$ .

_{Extracted from J. F. Talbert & H. H. Heng. (1995). Additional Mathematics Pure and Applied (6e).}

Roughwork.

(a)

Not yet to approach a solution, I wish to get a feel of the problem from three perspectives— i. algebraic, ii. geometric, and iii. differential:

(i. algebraic)

$\begin{aligned} \cos A & =\frac{4}{5} \\ \cos B & = \frac{12}{13} \\ & \\ \cos^2A & = \frac{16}{25} \\ \cos^2B & = \frac{144}{169} \\ & \\ \sin^2A & = 1-\cos^2A\\ & = \frac{9}{25} \\ \sin^2B & = 1-\cos^2B\\ & = \frac{25}{169} \\ & \\ \sin A & = \sqrt{\frac{9}{25}} \\ & = \frac{3}{5} \\ \sin B & = \sqrt{\frac{25}{169}} \\ & = \frac{5}{13} \\ \end{aligned}$

(ii. geometric)

(iii. differential)

Let $f(x)=\sin x$ . Then, recall that

$\displaystyle{f(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }$

because of Taylor series/expansion of a function $f(x)$ at some point $a$ :

$\displaystyle{f(x)\big|_{x=a} = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\bigg|_{x=a}}$

such that in our situation, you know, we have $a=0$ (the Maclaurin series):

$\displaystyle{f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n}$

Now that we have enough, here begins the

Solution.

(i. algebraic)

$\begin{aligned} \sin (A+B) & = \sin A\cos B + \cos A\sin B \\ & = \bigg(\frac{3}{5}{\bigg)\bigg(\frac{12}{13}\bigg) + \bigg(\frac{4}{5}\bigg)\bigg(\frac{5}{13}\bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(ii. geometric)

$\begin{aligned} \frac{h}{14} & = \sin (90^\circ -B) \\ h & = 14\cos B \\ & \\ \sin (A+B) & = \frac{h}{15} \\ & = \frac{14\cos B}{15} \\ & = \frac{14}{15}\bigg( \frac{12}{13} \bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(iii. differential)

That angles $A$ and $B$ are acute angles, i.e.,

$0<B<A<90^\circ$

does not imply $A+B$ is also an acute angle, but that

$0<A+B<180^\circ$ .

such that

$\begin{aligned} 0 & <\sin (A+B)<1 \\ -1 & <\cos (A+B)<1 \\ 0 & <\sin B<\sin A<\cos A<\cos B < 1\\ \end{aligned}$

Letting $f(x)=\sin x$ ,

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A+B] \\ & = \int_{0}^{A+B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A+B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A+B} \\ & = 1-\cos (A+B) \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A] \\ & = \int_{0}^{A}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A} \\ & = 1-\cos A \\ & = 1-\frac{4}{5} \\ & = \frac{1}{5} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,B] \\ & = \int_{0}^{B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{B} \\ & = 1-\cos B \\ & = 1-\frac{12}{13} \\ & = \frac{1}{13} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [B,A] \\ & = \frac{1}{5}-\frac{1}{13} \\ & = \frac{8}{65} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [A,A+B] \\ & = 1-\cos (A+B) - \frac{1}{5} \\ & = \frac{4}{5}-\cos (A+B) \\ \end{aligned}$

then letting $g(x)=\cos x$ , and noting in a right-angled triangle the cosine of an angle admits of no negative values, so can it further be said

$\begin{aligned} & \quad\enspace 0<g(x)<1 \\ &\Rightarrow 0<g(A+B)<1 \\ &\Rightarrow 0<A+B<90^\circ \\ \end{aligned}$

I lost my way…

Let $w(u(t),v(t))=u(t)-v(t)$ be the difference in degrees between time-varying angles $u$ and $v$ where assumed is $u(t)>v(t)$ without loss of generality.

$\begin{aligned} f(u,v,w) & \stackrel{\textrm{def}}{=} \sin (u+v) \\ & = \sin (2v+w ) \\ & = \sin (2u-w ) \\ & \\ g(u,v,w) &\stackrel{\textrm{def}}{=} \cos (u+v) \\ & = \cos (2v+w ) \\ & = \cos (2u-w ) \\ \end{aligned}$

Taking partial derivatives wrt $u$ , $v$ , $w$ , and $t$ :

$\begin{aligned} \partial_uf(u,v,w) & = 2\cos (2u-w) \\ \partial_vf(u,v,w) & = 2\cos (2v+w) \\ \partial_wf(u,v,w) & = 0 \\ \partial_tf(u,v,w) & = (u'+v')\cos (u+v)\\ \end{aligned}$

$\begin{aligned} \mathrm{d}f(u,v,w) & = \partial_uf\,\mathrm{d}u + \partial_vf\,\mathrm{d}v + \partial_wf\,\mathrm{d}w \\ & = 2\cos (2u-w )\,\mathrm{d}u + 2\cos (2v+w )\,\mathrm{d}v + 0 \\ \dots\,\mathrm{d}u & = \mathrm{d}v+\mathrm{d}w\,\dots \\ & = 4\cos (u+v)\,\mathrm{d}v + 2\cos (u+v)\,\mathrm{d}w \\ & = 2\big(\cos (u+v)\big)(2\,\mathrm{d}v+\mathrm{d}w) \\ \end{aligned}$

I lost my faith…

$f(x,y)\stackrel{\textrm{(1)}}{=}\sin (x+y)$

then

$f(x'=x+y,0)\stackrel{\textrm{(2)}}{=} f(x,y)\stackrel{\textrm{(3)}}{=} f(0,x+y=y')$ .

By equality $\textrm{(1)}$ it states that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values of two arguments $(\texttt{var1},\texttt{var2})=(x,y)$ ;

by equality $\textrm{(2)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(x+y,0)$ , i.e., by one significant argument $\texttt{var1}=x+y$ , and another trivial argument $\texttt{var2}=0$ ; and

by equality $\textrm{(3)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(0,x+y)$ , i.e., by one trivial argument $\texttt{var1}=0$ , and another significant argument $\texttt{var2}=x+y$ .

Hence, assuming a partial variation

$\begin{aligned} &\quad\enspace \sin (\texttt{var1}+\texttt{var2}) \\ & = k_1(\texttt{var1},\texttt{var2})\sin (\texttt{var1}) + k_2(\texttt{var1},\texttt{var2})\sin (\texttt{var2}) \\ \end{aligned}$

we WTS the following

$\begin{aligned} k_1(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var2}) \\ k_2(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var1}) \\ \end{aligned}$

LHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}\sin (\texttt{var1}+\texttt{var2}) \\ & = \big[\cos (\texttt{var1}+\texttt{var2})\big] (\texttt{var1}'+\texttt{var2}') \\ \end{aligned}$

RHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}(k_1\sin (\texttt{var1})+k_2\sin (\texttt{var2})) \\ & = \big(k_1'\cos (\texttt{var1})\big) \texttt{var1}' + \big(k_2'\cos (\texttt{var2})\big) \texttt{var2}' \\ \end{aligned}$

and LHS=RHS implies:

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}k_1 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var1})} \\ \frac{\mathrm{d}}{\mathrm{d}t}k_2 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var2})} \\ \dots & \dots \dots \dots \\ k_1' & = \bigg( \frac{\cos (\texttt{var2})}{\cos (\texttt{var1})}\bigg) k_2' \\ \end{aligned}$

(to be continued)

May 11, 2023May 11, 2023

202305111603 Exercise 3.1

Solve the following equations:

(a) $13x-4=3x+16$ .

(b) $\displaystyle{\frac{3x-9}{18}+\frac{x}{27}-\frac{2x-5}{4}=\frac{4}{3}-x}$ .

(c) $\displaystyle{\frac{3}{x-1}-\frac{2}{x+4}=\frac{4}{2-2x}}$ .

(d) $\displaystyle{\frac{2x}{x-3}-\frac{4x+1}{2x-1}=\frac{21}{2x^2-7x+3}}$ .

(e) $\displaystyle{\frac{1}{4}(3y-2)-\bigg[ y-\frac{1}{y}(7-3y)\bigg] =-\frac{1}{4}y-7}$ .

_{Extracted from K. L. Nielsen. (1958). College Mathematics.}

Roughwork.

(a)

$\begin{aligned} 13x-4 & = 3x+16 \\ 13x-3x & = 16+4 \\ 10x & = 20 \\ x & = 2 \\ \end{aligned}$

(b)

$\begin{aligned} \frac{3x-9}{18} + \frac{x}{27} - \frac{2x-5}{4} & = \frac{4}{3}-x \\ \frac{3x-9}{2\cdot 3^2} + \frac{x}{3^3} - \frac{2x-5}{2^2} & = \frac{4}{3}-x \\ 6(3x-9) + 4x-27(2x-5) & =36(4)-108x \\ 18x-54+4x-54x+135 & =144-108x \\ 18x+4x-54x+108x & = 144+54-135 \\ 76x & = 63 \\ x & = \frac{63}{76} \\ \end{aligned}$

(c)

$\begin{aligned} \frac{3}{x-1}-\frac{2}{x+4} & = \frac{4}{2-2x} \\ \frac{3}{x-1}-\frac{2}{x+4} & = -\frac{2}{x-1} \\ 3(x+4) - 2(x-1) & = -2(x+4) \\ 3x+12-2x+2 & = -2x-8 \\ 3x & = -22 \\ x & = -\frac{22}{3} \\ \end{aligned}$

(d) Not to be attempted.

(e) Not to be attempted.

January 17, 2023May 4, 2023

202301170910 Exercise 3.2

Solve these simultaneous equations:

1. $\begin{cases} x+y=5 \\ xy=6 \\ \end{cases}$

Roughwork. Read More

Substituting $x=5-y$ for $x$ :

$\begin{aligned} (5-y)y & = 6 \\ y^2-5y+6 & = 0 \\ (y-2)(y-3) & = 0 \\ y & = 2,3 \\ \begin{pmatrix}x \\ y \end{pmatrix} = \bigg\{ \begin{pmatrix} 2 \\ 3 \end{pmatrix},\begin{pmatrix} 3 \\ 2 \end{pmatrix} \bigg\} & \\ \end{aligned}$

2. $\begin{cases} x-2y=1\\ x^2+y^2=29 \\ \end{cases}$

Roughwork. Read More

Substituting $x=1+2y$ for $x$ :

$\begin{aligned} (1+2y)^2 + y^2 & =29 \\ (1+4y+4y^2)+y^2 & = 29 \\ 5y^2 +4y -28 & = 0 \\ (5y+14)(y-2) & = 0 \\ y & = -\frac{14}{5},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}5\\2\end{pmatrix}, \begin{pmatrix}-23/5\\-14/5\end{pmatrix} \bigg\} & \\ \end{aligned}$

3. $\begin{cases} 2x+y=5 \\ x^2-xy=12 \\ \end{cases}$

Roughwork. Read More

Substituting $y=5-2x$ for $y$ :

$\begin{aligned} x^2-x(5-2x) & = 12 \\ 3x^2 -5x -12 & = 0 \\ (3x+4)(x-3) & = 0 \\ x & = -\frac{4}{3},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-4/3\\7/3\end{pmatrix},\begin{pmatrix}3\\-1\end{pmatrix}\bigg\} &\\ \end{aligned}$

4. $\begin{cases} 3x-y=7 \\ x^2-xy+y^2=7\\ \end{cases}$

Roughwork. Read More

Substituting $y=3x-7$ for $y$ :

$\begin{aligned} x^2-x(3x-7)+(3x-7)^2 & = 7 \\ 7x^2 -35x +42 & = 0 \\ (7x+7)(x-6) & = 0 \\ x & = -1,6 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-1\\-10\end{pmatrix} , \begin{pmatrix}6\\11\end{pmatrix}\bigg\} & \\ \end{aligned}$

5. $\begin{cases} 5x+y=9 \\ 3xy+y^2=-5 \\ \end{cases}$

Roughwork. Read More

Substituting $y=9-5x$ for $y$ :

$\begin{aligned} 3x(9-5x)+(9-5x)^2 & = -5 \\ 10x^2-63x +86 & = 0 \\ (10x-43)(x-2) & = 0 \\ x & = \frac{43}{10},2 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}43/10\\-25/2\end{pmatrix} & , \begin{pmatrix}2\\-1\end{pmatrix}\bigg\} \\ \end{aligned}$

6. $\begin{cases} 3x+2y=13 \\ 3x^2+y^2=31 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y = \frac{13-3x}{2}}$ for $y$ :

$\begin{aligned} 3x^2 + \bigg(\frac{13-3x}{2}\bigg)^2 = 31 \\ 7x^2 - 26x+15 & = 0 \\ (7x-5)(x-3) & = 0 \\ x & = \frac{7}{5},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}7/5\\22/5\end{pmatrix},\begin{pmatrix}3\\-13\end{pmatrix} \bigg\} & \\ \end{aligned}$

7. $\begin{cases} 2x-3y+11=0 \\ 2x^2-xy=36 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y=\frac{2x+11}{3}}$ for $y$ :

$\begin{aligned} 2x^2-x\bigg( \frac{2x+11}{3}\bigg) & =36 \\ 4x^2-11x-108 &= 0 \\ (4x-27)(x+4) & = 0 \\ x & = -4,\frac{27}{4} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}-4\\1\end{pmatrix},\begin{pmatrix}27/4\\49/6\end{pmatrix}\bigg\} \\ \end{aligned}$

8. $\begin{cases} 2x-5y=1\\ x^2-xy+3y^2=9 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{y=\frac{2x-1}{5}}$ for $y$ :

$\begin{aligned} x^2-x\bigg(\frac{2x-1}{5}\bigg)+3\bigg(\frac{2x-1}{5}\bigg)^2 & = 9 \\ 27x^2 -7x -222 & = 0 \\ (27x+74)(x-3) & = 0 \\ x & = -\frac{74}{27},3 \\ \begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}-74/27\\-35/27\end{pmatrix} , \begin{pmatrix}3\\1\end{pmatrix} \bigg\} & \\ \end{aligned}$

9. $\begin{cases} 2x+3y=7 \\ y^2=26-x^2 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{x=\frac{7-3y}{2}}$ for $x$ :

$\begin{aligned} y^2 & = 26-\bigg(\frac{7-3y}{2}\bigg)^2 \\ 0 & = 13y^2 -42y -55 \\ 0 & = (13y-55)(y+1) \\ y & = -1,\frac{55}{13} \\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{ \begin{pmatrix}5\\-1\end{pmatrix} , \begin{pmatrix}74/26\\55/13\end{pmatrix}\bigg\} \\ \end{aligned}$

10. $\begin{cases} 5x+3y+7=0 \\ 3y^2=x^2-4y+3 \\ \end{cases}$

Roughwork. Read More

Substituting $\displaystyle{x=\frac{-7-3y}{5}}$ for $x$ :

$\begin{aligned} 3y^2 & = \bigg(\frac{-7-3y}{5}\bigg)^2-4y+3 \\ 0 & = 33y^2+29y-62 \\ 0 & = (33y+62)(y-1) \\ y & = 1,-\frac{62}{33}\\ \begin{pmatrix}x\\y\end{pmatrix} & = \bigg\{\begin{pmatrix} -2 \\1\end{pmatrix},\begin{pmatrix} -3/11\\-62/33\end{pmatrix}\bigg\} \\ \end{aligned}$

_{Extracted from A. Godman & J. F. Talbert. (1975). Additional Mathematics Pure and Applied in SI Units.}

This problem is not to be attempted.

October 28, 2022November 2, 2022

202210280936 Problem 1.24

Expand $f(t)=\sin^2t\cos^3t$ in Fourier series.

_{Extracted from Hwei Piao Hsu. (1984). HBJ College Outline of Applied Fourier Analysis.}

Roughwork.

One can make use of the identities below:

$\begin{aligned} e^{\pm\mathrm{i}n\theta} & = \cos n\theta\pm\mathrm{i}\sin n\theta \\ \cos n\theta & = \frac{e^{\mathrm{i}n\theta}+e^{-\mathrm{i}n\theta}}{2} \\ \sin n\theta & = \frac{e^{\mathrm{i}n\theta}-e^{-\mathrm{i}n\theta}}{2\mathrm{i}} \\ \end{aligned}$

and thus

$\begin{aligned} & \begin{cases} \sin^2t = \displaystyle{\bigg(\frac{e^{\mathrm{i}t}-e^{-\mathrm{i}t}}{2\mathrm{i}}\bigg)^2 }\\ \cos^3t = \displaystyle{ \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^3 }\\ \end{cases} \\ \Longrightarrow & \begin{cases} \sin^2t = \displaystyle{\frac{e^{2\mathrm{i}t}-2e^{\mathrm{i}t}e^{-\mathrm{i}t}+e^{-2\mathrm{i}t}}{-4}} \\ \cos^3t = \displaystyle{\frac{e^{3\mathrm{i}t}+3e^{\mathrm{i}t}+3e^{-\mathrm{i}t}+e^{-3\mathrm{i}t}}{8}} \\ \end{cases} \\ \end{aligned}$

Or one can make use of the identity

$\sin^2\theta + \cos^2\theta = 1$ .

and hence

$\begin{aligned} f(t) & = (1-\cos^2t)\cos^3t \\ & = \cos^3t-\cos^5t \\ \end{aligned}$

where $\cos^5t$ is as expanded

$\begin{aligned} \cos^5t & = \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^5 \\ & = \frac{e^{5\mathrm{i}t}+5e^{3\mathrm{i}t} +10e^{\mathrm{i}t} +10e^{-\mathrm{i}t}+5e^{-3\mathrm{i}t}+ e^{-5\mathrm{i}t}}{32}\\ \end{aligned}$

so that

$\begin{aligned} f(t) & = -\frac{e^{5\mathrm{i}t}}{32}+\bigg(\frac{1}{8}-\frac{5}{32}\bigg)e^{3\mathrm{i}t} + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{\mathrm{i}t} \\ & \qquad + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{-\mathrm{i}t} + \bigg(\frac{1}{8}-\frac{5}{32}\bigg) e^{-3\mathrm{i}t} - \frac{e^{-5\mathrm{i}t}}{32}\\ & = -\frac{e^{5\mathrm{i}t}}{32}-\frac{e^{3\mathrm{i}t}}{32} + \frac{e^{\mathrm{i}t}}{16} + \frac{e^{-\mathrm{i}t}}{16} - \frac{e^{-3\mathrm{i}t}}{32} -\frac{e^{-5\mathrm{i}t}}{32} \\ & = \frac{1}{16}\Bigg( 2\bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg) - \frac{e^{3\mathrm{i}t}+e^{-3\mathrm{i}t}}{2} - \frac{e^{5\mathrm{i}t}+e^{-5\mathrm{i}t}}{2} \Bigg)\\ & = \cdots \\ \end{aligned}$

Answer. $\displaystyle{\frac{1}{16}(2\cos t-\cos 3t-\cos 5t)}$ .

No slightest disrespect to de Moivre’s and Euler’s formulae but in due respect of Fourier’s analysis, this problem had need be treated again.

Background.

A function $f(t)$ satisfying $f(t+T)=f(t)$ is called a periodic function, and the smallest $T$ its period. A periodic function $f(t)$ can be represented by the trigonometric Fourier series

$\displaystyle{f(t)=\frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\cos n\omega_0t+b_n\sin n\omega_0t)}$

$\displaystyle{f(t)=C_0+\sum_{n=1}^{\infty}C_n\cos (n\omega_0t-\theta_n)}$

where $\omega_0=2\pi /T$ . The coefficients of the Fourier series are found by using the orthogonality properties of sine and cosine functions over a period:

$\begin{aligned} \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{2}{T}\int_{-T/2}^{T/2}f(t)\begin{Bmatrix}\cos n\omega_0t \\ \sin n\omega_0t \end{Bmatrix}\,\mathrm{d}t \\ C_0 & = \frac{1}{2}a_0 \\ C_n & = \sqrt{a_n^2+b_n^2} \\ \theta_n & = \tan^{-1}\bigg(\frac{b_n}{a_n}\bigg) \\ \end{aligned}$

_{Text on pg. 11}

For $f(t)=\sin^2t\cos^3t$ ,

$\begin{aligned} T & = 2\pi \\ \omega_0 & = 1 \\ \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\begin{Bmatrix}\cos nt \\ \sin nt\end{Bmatrix}\,\mathrm{d}t\\ \end{aligned}$

When $n=0$ :

$\begin{aligned} a_0 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\,\mathrm{d}t \\ \cdots\enspace\textrm{Let }& u=\sin t\textrm{ s.t. }\mathrm{d}t=\frac{\mathrm{d}u}{\cos t}\enspace\cdots \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2\cos^2t\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2(1-u^2)\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}(u^2-u^4)\,\mathrm{d}u \\ & = \frac{1}{\pi}\bigg[ \frac{u^3}{3}-\frac{u^5}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = \frac{1}{\pi}\bigg[ \frac{\sin^3t}{3}-\frac{\sin^5t}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = 0 \\ \end{aligned}$

When $n=1$ :

$\begin{aligned} a_1 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^4t\,\mathrm{d}t \\ \dots\enspace &\textrm{By WolframAlpha}\enspace\dots \\ & = \frac{1}{\pi}\bigg[\frac{1}{192}(12t+3\sin (2t)-3\sin (4t)-\sin (6t))\bigg]\bigg|_{-\pi}^{\pi} \\ & = \frac{1}{\pi}\bigg(\frac{24\pi}{192}\bigg) \\ & = \frac{1}{8} \\ \end{aligned}$

a bit tiring an exercise.

(to be continued)