# A Mind Unleashed

In 1665, following an outbreak of the bubonic plague in England, Cambridge University closed its doors, forcing Issac Newton, then a college student in his 20s, to go home.

Away from university life, and unbounded by curriculum constraints and tests, Newton thrived. The year-plus he spent in isolation was later referred to as his annus mirabilis, the “year of wonders.”

First, he continued what he had begun at Cambridge: “forging the sword” in mathematical problem solving; Within a year, he gave birth to differential and integral calculus.

Next, he acquired a few glass prisms and made a hole in his window shutter so only a small beam could come through. What he saw after placing a prism in the sunbeam sprung his theories of optics.

And yes, there was an apple tree in the garden! One fateful day in 1666, while contemplating celestial body movements under that tree, Newton was bonked by a falling apple. It dawned on him that the force pulling the apple to the ground might be the same force that holds celestial bodies in orbit. The epiphany led him to discover the law of universal gravitation.

Newton returned to Cambridge in 1667 after the plague had ended. Within six months, he was made a fellow of Trinity College; two years later, the prestigious Lucasian Chair of Mathematics.

# Have we a new proof ?

For a right triangle:

On one hand, its area is

$\frac{1}{2}ab.$

On the other hand, according to Heron’s formula (see “An Algebraic Proof of Heron’s Formula“),

$\sqrt{s(s-a)(s-b)(s-c)}\quad\quad\quad(*)$

where $s=\frac{a+b+c}{2}.$

Hence,

$\frac{1}{2} ab=\sqrt{s(s-a)(s-b)(s-c)}.$

Squaring it gives

$\frac{1}{4}a^2b^2=s(s-a)(s-b)(s-c).$

Using a CAS, we obtain

$\frac{(c^2-b^2-a^2)^2}{16} = 0 \implies c^2-b^2-a^2=0$

from which the Pythagorean theorem emerges:

$a^2+b^2=c^2$

# Seek-Lock-Strike!

A guided missile is launched to destroy a fighter jet (Fig. 1).

Fig. 1

We introduce a coordinate axes such that at $t=0,$ the missile is at origin $(0, 0)$ and the jet at $(a,b)$. The jet flies parallel to the x-axis with constant speed $v_a$. The missile has locked onto the jet so it is always pointing at the jet as it moves. Its speed is $v_m.$

Find the time and position the missile strikes its target.

“The missile has locked onto the jet so it is always pointing at the jet as it moves” means that the tangent to the missile’s path at any point $(x, y)$ will pass through the position of the jet. The equation of the tangent is

$\frac{dy}{dx}\cdot(a + v_a t -x) = b-y.\quad\quad\quad(1)$

$(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2=v_m^2.\quad\quad\quad(2)$

Notice $x, y$ are functions of time $t : x=x(t), y=y(t).$

Differentiate (1) with respecte to $t:$

$\frac{d}{dt}(\frac{dy}{dx})\cdot(a+v_at-x) + \frac{dy}{dx}\cdot\frac{d}{dt}(a+v_at-x) = -\frac{dy}{dt},$

$\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dx}(\frac{dy}{dx})\cdot\frac{dx}{dt}=\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}$

$\left(\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\right)\cdot(a+v_at-x) + \frac{dy}{dx}\cdot(v_a -\frac{dx}{dt}) = -\frac{dy}{dt},$

$\left(\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\right)\cdot(a+v_at-x) + \frac{dy}{dx}\cdot v_a -\underbrace{\frac{dy}{dx}\cdot\frac{dx}{dt}}_{\frac{dy}{dt}}= -\frac{dy}{dt},$

$\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\cdot(a+v_at-x) + \frac{dy}{dx}\cdot v_a = 0,$

$\frac{d^2y}{dx^2}= \frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dy}{dx}$

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dy}{dx}\cdot\frac{dx}{dt}(a+v_at-x) + \frac{dy}{dx}v_a=0,$

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dx}{dt}\cdot\underline{\frac{dy}{dx}(a+v_at-x)} + \frac{dy}{dx}v_a=0.$

By (1), substituting $b-y$ for $\frac{dy}{dx}\cdot(a + v_a t -x)$,

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dx}{dt}\cdot(b-y) + \frac{dy}{dx}v_a=0.\quad\quad\quad(3)$

Let $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt},$ we express (2) as

$(\frac{dx}{dt})^2 + (\frac{dy}{dx}\cdot\frac{dx}{dt})^2=v_m^2,$

$(\frac{dx}{dt})^2\cdot(1+(\frac{dy}{dx})^2)= v_m^2.$

Suppose $a>0$, which implies that $\frac{dx}{dt} > 0$ (see Exercise-4). Solving for $\frac{dx}{dt}$ gives

$\frac{dx}{dt} = \frac{v_m}{\sqrt{1+(\frac{dy}{dx})^2}}.\quad\quad\quad(4)$

Submitting (4) into (3),

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{v_m}{\sqrt{1+(\frac{dy}{dx})^2}}\cdot(b-y) + \frac{dy}{dx}\cdot v_a=0,$

$\frac{d}{dy}(\frac{dy}{dx})\cdot(b-y) + \frac{v_a}{v_m}\cdot\frac{dy}{dx}\cdot\sqrt{1+(\frac{dy}{dx})^2}=0,\quad\quad\quad(5)$

Let $p = \frac{dy}{dx}, r=\frac{v_a}{v_m},$

(5) becomes

$\frac{dp}{dy}(b-y) + r\cdot p\cdot\sqrt{1+p^2}=0.$

We solve this non-linear differential equation as follows:

For $y < b$,

$\frac{1}{p\sqrt{1+p^2}}\frac{dp}{dy} = \frac{-r}{b-y},$

$\int (\frac{\sqrt{1+p^2}}{p}-\frac{p}{\sqrt{1+p^2}})\frac{dp}{dy} dy = \int \frac{-r}{b-y} dy,$

$\int\frac{\sqrt{1+p^2}}{p}dp-\int\frac{p}{\sqrt{1+p^2}}dp = r\log(b-y)+k_2.\quad\quad\quad(6)$

Since $\int\frac{\sqrt{1+p^2}}{p}dp=\sqrt{1+p^2} - \mathrm{arcsinh}\left(\frac{1}{|p|}\right)$ (See “I vs. CAS“), $\int\frac{p}{\sqrt{1+p^2}}dp=\sqrt{1+p^2},$

(6) gives

$-\mathrm{arcsinh}\left(\frac{1}{|p|}\right) = r\log(b-y)+k_1.\quad\quad\quad(7)$

At $t=0, y=0, p=\frac{dy}{dx}=\frac{b}{a},$

(7) yields

$k_1 = -\mathrm{arcsinh}\left(\frac{a}{b}\right) - r\log(b).$

Moreover, since $p = \frac{dy}{dx} > 0$, we have

$\frac{dy}{dx} = \frac{1}{-\sinh(r\log(b-y)+k_1)},$

$-\sinh(r\log(b-y)+k_1)\cdot \frac{dy}{dx} = 1,$

$\displaystyle\int -\sinh(r\log(b-y) + k_1) \cdot \frac{dy}{dx}\;dx = \displaystyle \int 1\; dx = x +k_2.$

Using Omega CAS Explorer:

we obtain

$\frac{1}{2}\left(\frac{e^{k_1}(b-y)^{r+1}}{r+1} - \frac{e^{-k_1}(b-y)^{1-r}}{1-r}\right) = x + k_2.\quad\quad\quad(8)$

Since $y=0, x=0$, (8) gives

$k_2 = \frac{1}{2}\left(\frac{e^{k_1}b^{r+1}}{r+1} - \frac{e^{-k_1}b^{1-r}}{1-r}\right).\quad\quad\quad(9)$

Suppose when $t = t_*,$ the missile hits the target. Then the striking coordinates $(x_*, y_*)$ are the same as that of the fighter jet, i.e.,

$x_* = a + v_a t_*, \;\;y_*=b.\quad\quad\quad(10)$

Hence,

$\lim\limits_{t \rightarrow t_*} x = x_* \overset{(10)}{=} a+v_a t_*, \;\;\lim\limits_{t \rightarrow t_*} y = y_* \overset{(10)}{=}b \implies \lim\limits_{t \rightarrow t_*} b-y = 0.\quad\quad\quad(11)$

For $r \le 1$ (i.e., $v_a \le v_m$), as $t \rightarrow t_*,$ (8) gives (see Exercise-1)

$0 = a +v_a t_* + k_2 \implies -k_2-a = v_a t_*.$

That is,

$-\underbrace{\frac{1}{2}\left(\frac{e^{k_1}b^{1+r}}{1+r}-\frac{e^{-k_1}b^{1-r}}{1-r}\right)}_{(9):\;\;k_2}-a=v_a t_*.$

By $(\star)$ (see below),

$-\frac{1}{2}\left(\frac{1}{b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot\frac{b^{1+r}}{1+r}-b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)\cdot\frac{b^{1-r}}{1-r}\right) - a = v_a t_*,$

$-\frac{1}{2}\left(\frac{1}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot\frac{b}{1+r}-\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)\cdot\frac{b}{1-r}\right) - a = v_a t_*,$

$\frac{\frac{-b}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$

$\frac{\frac{-b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{-1(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$

$\frac{\frac{b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{1+r} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$

$\frac{b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{2(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{2(1-r)} -a = v_a t_*,$

$\frac{a-\sqrt{b^2+a^2}}{2(1+r)} + \frac{a+\sqrt{b^2+a^2}}{2(1-r)} -a = v_a t_*,$

$\frac{(1-r)(a-\sqrt{b^2+a^2}) +(1+r)(a+\sqrt{b^2+a^2})-2a(1-r^2)}{2(1+r)(1-r)} = v_a t_*,$

$\frac{a-\sqrt{b^2+a^2}-ar+r\sqrt{b^2+a^2}+ar+r\sqrt{b^2+a^2}+a+\sqrt{b^2+a^2}-2a + 2ar^2}{2(1-r^2)}=v_a t_*,$

$\frac{2r\sqrt{b^2+a^2} + 2ar^2}{2(1-r^2)} = v_a t_*,$

$\frac{ar^2+\sqrt{b^2+a^2}r}{1-r^2} = v_a t_*,$

$\frac{r(ar + \sqrt{b^2+a^2})}{v_a(1-r^2)} = t_*.$

Since $r = \frac{v_a}{v_m}$, we have

$t_* = \frac{v_a}{v_m}\cdot\frac{ar + \sqrt{b^2+a^2}}{v_a(1-r^2)}=\frac{1}{v_m}\cdot\frac{ar + \sqrt{b^2+a^2}}{1-r^2}.$

Namely,

$t_* = \frac{\sqrt{b^2+a^2} + a\cdot r}{(1-r^2)\cdot v_m}, \quad\quad r=\frac{v_a}{v_m}.$

It follows that the guided missile strikes the fighter jet at $(a+r\cdot\frac{\sqrt{b^2+a^2}+a\cdot r}{1-r^2}, b).$

$e^{k_1} = e^{-\mathrm{arcsinh}(\frac{a}{b})-r\log(b)}$

$= \frac{1}{e^{\mathrm{arcsinh}(\frac{a}{b})+r\log(b)}}$

$= \frac{1}{e^{\mathrm{arcsinh}(\frac{a}{b})}\cdot e^{r\log(b)}}$

$\mathrm{arcsinh}(\blacksquare) = \log\left(\blacksquare + \sqrt{1+\blacksquare^2}\right)$ (see “Deriving Two Inverse Functions“)

$= \frac{1}{e^{\log\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot e^{r\log(b)}}$

$e^{r\log(b)}=e^{\log(b^r)}=b^r$ (see “Introducing Lady L” and “Two Peas in a Pod, Part 3“)

$= \frac{1}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b^r}.$

i.e.,

$e^{k_1} = \frac{1}{b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}.\quad\quad\quad(\star)$

Exercise-1 Show that for $r < 1$, (8) gives $0 = a +v_a t_* + k_2 \implies -k_2-a = v_a t_*.$ (Hint: (10))

Exercise-2 For $r < 1$, what is the total distance traveled by the missile when it strikes the fighter jet? (hint: Don’t make things harder than they are)

Exercise-3 Show that if $r \ge 1$ (i.e., $v_a \ge v_m$), the missile will not strike the fighter jet. Explain.

Excercise-4 Show that $a>0 \implies \frac{dx}{dt} >0$.

Excercise-5 For $a <0$, find the time and position the missile strikes its target .

# Faster Pi

There is a more astonishing algorithm than what is described in “Fast Pi” for rapidly calculating $\pi.$

Let us consider the three-term iteration with initial values

$a_0 = \sqrt{2}, \quad b_0 =0, \quad \pi_0 = 2+\sqrt{2}$

given by

$a_n = \frac{1}{2}\left(\sqrt{a_{n-1}} + \frac{1}{\sqrt{a_{n-1}}}\right), \quad b_{n} = \sqrt{a_{n-1}}\left(\frac{b_{n-1}+1}{b_{n-1} + a_{n-1}}\right), \quad \pi_{n} = \pi_{n-1}b_n\left(\frac{1+a_{n}}{1+b_{n}}\right).$

Then $\pi_n$ converges exponentially to $\pi$. In fact,

$|\pi-\pi_n|< \frac{1}{10^{2^n}}.$

Implemented in Omega CAS Explorer, $4$ iterations yield $40$ digits of $\pi$:

The $8$th iteration gives $\pi$ correctly to $694$ digits:

Exercise-1 Show that $20$ iterations will provide over $2$ million digits of $\pi.$

# Fast Pi

There is a recipe for rapidly calculating the digits of $\pi$.

Consider the two-term iteration with initial values

$a_0 = \sqrt{2}, \quad b_0 = 1\quad\quad\quad(1)$

given by

$a_n = \frac{a_{n-1}+b_{n-1}}{2}, \quad b_n = \sqrt{a_{n-1} b_{n-1}}, \quad n \ge 1.\quad\quad\quad(2)$

Then

$\pi_{n} = \frac{a_n^2}{1-\sum\limits_{i=0}^{n}2^{i-1}\left(a_n^2-b_n^2\right)}\quad\quad\quad(3)$

converges to $\pi.$

We have implemented (1), (2) and (3) in Omega CAS Explorer (see Fig. 1). With merely 4 iterations, it gives astounding 20 decimal places of $\pi$!

Fig. 1

# Sandwich Theorems and Their Proofs

Prove:

$\begin{cases} \exists N\in \mathbb{N} \ni n > N, a_n \le b_n \le c_n\quad\quad(1)\\ \lim\limits_{n \rightarrow \infty}a_n=\lim\limits_{n\rightarrow \infty}c_n=L \quad\quad\quad\quad\quad\quad(2) \end{cases}\implies \lim\limits_{n\rightarrow \infty}b_n = L.$

We express (1) as

$n > N \implies a_n \le b_n \le c_n$

and (2):

$\forall \epsilon >0, \exists n_1 \ni n >n_1 \implies |a_n-L|< \epsilon.$

$\forall \epsilon >0, \exists n_2 \ni n >n_2 \implies |c_n-L|< \epsilon.$

Let

$n^* = \mathrm{\bold{max}}(N, n_1, n_2).$

We have

$n>n^* \implies a_n \le b_n \le c_n.$

$\forall \epsilon >0, n>n^* \implies |a_n-L|<\epsilon.$

$\forall \epsilon >0, n>n^* \implies |a_n-L|<\epsilon.$

It means

$n>n^* \implies \underline{a_n \le b_n \le c_n}.$

$\forall \epsilon > 0, n >n^* \implies \underline{-\epsilon +L < a_n} < \epsilon + L.$

$\forall \epsilon > 0, n >n^* \implies -\epsilon +L < \underline{c_n < \epsilon + L}.$

Hence,

$\forall \epsilon >0, n>n^* \implies -\epsilon + L < \underline{a_n \le b_n \le c_n} < \epsilon+L.$

That is,

$\forall \epsilon >0, \exists n^* \ni n > n^* \implies -\epsilon + L < b_n < \epsilon + L.$

i.e.,

$\forall \epsilon >0, \exists n^* \ni n > n^* \implies |b_n-L| < \epsilon.$

It follows that

$\lim\limits_{n \rightarrow \infty} b_n = L.$

Exercise 1 Prove:

$\begin{cases} \exists \delta \ni |x-a| < \delta, f(x) \le g(x) \le h(x) \quad\quad\quad(1) \\ \lim\limits_{x \rightarrow a }f(x)=\lim\limits_{x\rightarrow a}h(x)=L \quad\quad\quad\quad\quad\quad\quad\quad(2) \end{cases}\implies \lim\limits_{x\rightarrow a}g(x) = L.$

“Keep hitting the square root button of a calculator after entering any positive number, “1” will be the eventual result displayed.”

This is observed by many when playing around with the calculator.

To explain this “phenomena”, we shall show that

$\forall a >0, \lim\limits_ {n \rightarrow \infty} a^{\frac{1}{n}} = 1.$

There are three cases to consider.

Case-1 For $a=1, \lim\limits_{n\rightarrow \infty} a^{\frac{1}{n}} = \lim\limits_{n\rightarrow \infty} 1^{\frac{1}{n}} = \lim\limits_{n\rightarrow \infty} 1 = 1$ since $1^{\frac{1}{n}} = 1$ (See “These Are No Jokes”).

Case-2 For $a>1,$ we have (see Exercise-1)

$a^{\frac{1}{n}} > 1 \implies a^\frac{1}{n}-1 >0.$

We want to show that

$\forall \epsilon>0, \exists n^{*} \ni (n>n^* \implies |a_n^\frac{1}{n}-1| < \epsilon)$.

To this end, let

$a_n = a^\frac{1}{n} - 1.\quad\quad\quad(1-1)$

It gives

$(a_n+1)^n = (a^\frac{1}{n})^n = a.$

i.e.,

$(a_n+1)^n = a.$

Expanding $(a_n + 1)^n$ (see “Double Feature on Christmas Day”), we have

$a_n^n+n\cdot a_n^{n-1} + ... + \underline{n\cdot a_n}+1=a\implies \underline{n\cdot a_n} =a-\underbrace{(a_n^n+n\cdot a_n^{n-1}+...+1)}_{>0}.$

Hence,

$n\cdot a_n < a.$

Consequently,

$n\cdot a_n

From (1-2), we see that

$\forall \epsilon >0,$ when $n > n^*=\frac{a}{\epsilon}, |a^\frac{1}{n}-1|< \frac{a}{n} \overset{n > n^*}{<} \frac{a}{n^*} = \frac{a}{\frac{a}{\epsilon}}=\epsilon$

i.e.,

$\forall \epsilon >0, \exists n^* = \frac{a}{\epsilon} \ni (n > n*\implies |a^\frac{}{}-1| < \epsilon).$

Therefore,

$\forall a >1, \lim\limits_ {n \rightarrow \infty} a^{\frac{1}{n}} = 1.\quad\quad\quad(1-3)$

Case-3 For $0 we write

$a^{\frac{1}{n}} = \left(\frac{1}{\frac{1}{a}}\right)^{\frac{1}{n}}=\frac{1}{\left(\frac{1}{a}\right)^\frac{1}{n}}.$

And,

Let $A=\frac{1}{a} \overset{01. \quad\quad\quad$

It follows that

$\lim\limits_{n\rightarrow \infty}a^{\frac{1}{n}} = \lim\limits_{n\rightarrow \infty}\frac{1}{A^\frac{1}{n}}=\frac{\lim\limits_{n\rightarrow \infty}1}{\lim\limits_{n\rightarrow \infty}A^{\frac{1}{n}}} \underset{(1-3)}{=} \frac{1}{1}=1.$

Recently, a childhood friend of mine shared with me a discovery:

“Pick any integer greater than 1: If the number is even, divide it by 2; if it’s odd, multiply it by 3 and add 1. Take that new number and repeat the process, again and again. You’ll eventually get 1.”

Fig. 1 $n=10$

Fig. 2 $n = 100$

Fig. 3 $n = 501$

What my friend came upon is the Collatz conjecture, named after Lothan Collatz, who introduced the idea in 1937. As of today, it is still an unsolved problem: we don’t have a proof showing that the claim is true for all numbers, even though it has been verified for every number less than $2^{68}.$

Exercise-1 Show that for $(a>1, n \in \mathbb{N}), a^\frac{1}{n}>1.$ This is not a joke.

# arcsin

Consider set

$S = \{(x,y) | \sin(y)=x, -1

Since

$\sin(\underbrace{\frac{\pi}{2}}_{y}) = \underbrace{1}_{x}, \quad \sin(\underbrace{\frac{5\pi}{2}}_{y}) = \underbrace{1}_{x},$

it does not define a function.

However, redefine $S$ as

$S = \{(x,y) | \sin(y)=x, -1 < x < 1, \boxed{-\frac{\pi}{2} \le y \le \frac{\pi}{2}}\},\quad\quad\quad(1)$

we have

$\forall (x, y_1), (x, y_2) \in S, \sin(y_1)-\sin(y_2) = \left( \frac{d}{dy}\sin(y)\bigg|_{y=\xi}\right)\cdot(y_1-y_2)$

where $-\frac{\pi}{2} < \xi < \frac{\pi}{2}.$

That is,

$\sin(y_1)-\sin(y_2) = \cos(\xi)\cdot(y_1-y_2), \quad-\frac{\pi}{2} < \xi < \frac{\pi}{2}.\quad\quad\quad(2)$

(see “A Sprint to FTC“)

From (1),

$\sin(y_1)-\sin(y_2) = x - x =0$

and it simplifies (2) to

$0 = \cos(\xi)\cdot(y_1-y_2), \quad-\frac{\pi}{2} < \xi < \frac{\pi}{2}.\quad\quad\quad(3)$

Since for $-\frac{\pi}{2} < \xi < \frac{\pi}{2}, \cos(\xi) \ne 0,$ (3) gives $y_1 = y_2.$ It means

$\forall (x, y_1), (x, y_2) \in S \implies y_1=y_2.$

i.e.,

It is true that $\forall (x_1, y_1), (x_2, y_2) \in S, x_1=x_2\implies y_1=y_2.$

And so,

The set $S$ defines a function.

In fact, $S$ defines $\arcsin$, the inverse function of $\sin.$

Let us now examine $\arcsin$ qualitatively.

Differentiate $\sin(y) = x$ gives

$\cos(y)\frac{dy}{dx} = 1.$

Since

$\cos(y) > 0$ for $-\frac{\pi}{2} < y <\frac{\pi}{2}$,

we have

$\frac{dy}{dx} = \frac{1}{\cos(y)} =\frac{1}{\sqrt{1-(\sin(y))^2}}\overset{\sin(y)=x}{=}\frac{1}{\sqrt{1-x^2}}$.

That is,

$\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$.

It follows that

$\arcsin(x)$ is an increase function on $-1 < x < 1$.

Moreover,

$\frac{d^2}{dx^2}\arcsin(x) = \frac{d}{dx}\left(\frac{d}{dx}\arcsin(x)\right) = \frac{d}{dx}\frac{1}{\sqrt{1-x^2}}=-\frac{1}{2}\frac{-2x}{\sqrt{1-x^2}}=\frac{x}{\sqrt{1-x^2}}.$

i.e.,

$\frac{d^2}{dx^2}\arcsin(x) = \frac{x}{\sqrt{1-x^2}}, -1

And so,

for $0 \le x < 1, \frac{x}{\sqrt{1-x^2}} >0 \implies \arcsin(x)$ is concave on $0 \le x < 1$. It is convex on $-1< x < 0.$

Fig. 1 $\arcsin$ illustrated qualitatively

To compute $\arcsin(x)$ for any given $x$, see “A Cautionary Tale of Compute Inverse Trigonometric Functions“, “A Mathematical Allegory“.

Exercise-1 Define $\arccos$, the inverse function of $\cos.$

# O.M.G !

Problem:

Find positive whole values for $a,b,c$ such that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}=4. \quad(*)$

Solution:

Given

$y^2=x^3 +(4n^2+12n-3)x^2+32(n+3)x\quad\quad\quad(1)$

where $n$ is an integer; $x,y$ are rationals, the map

$\begin{cases} a=\frac{8(n+3)-x+y}{2(4-x)(n+3)}, \\ b=\frac{8(n+3)-x-y}{2(4-x)(n+3)}, \\ c=\frac{-4(n+3)-(n+2)x}{(4-x)(n+3)} \end{cases}\;(2)\implies \;\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} - n=0.$

i.e.,

for a fixed $n$ and $(x,y)$ on $(1), (a,b,c)$ generated by $(2) \implies \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\quad(3)$

Notice also

$f(\square,\blacksquare,\circ)\overset{\Delta}{=}\frac{\square}{\blacksquare+\circ} + \frac{\blacksquare}{\square+\circ} + \frac{\circ}{\square+\blacksquare} \implies f(t\cdot a, \; t\cdot b, \; t\cdot c) = f(a, b, c).\quad\quad\quad(4)$

Moreover, starting with $(x_1, y_1)$ on

$y^2 = x^3+px^2+qx+r,\quad\quad\quad(5)$

the following map generates successive rational points $(x_i, y_i)$ on (5):

$\begin{cases} i=2:x_2 = (\frac{3x_1^2+2px_1+q}{2y_1})^2-p-2x_1, \; y_2 = -y_1 + \frac{(3{x_1}^2+2px_1+q)}{2y_1}(x_1-x_2) \\ i>2: x_i = \left(\frac{y_{i-1}-y_1}{x_{i-1}-x_1}\right)^2-p-x_1-x_{i-1}, \; y_i= -y_1+\frac{y_{i-1}-y_1}{x_{i-1}-x_1}(x_1-x_i)\; \end{cases}(6)$

Let $n=4$, we have

$y^2=x^3+109x+224x\quad\quad(7)$

$\begin{cases} a=\frac{56-x-y}{14(4-x)} \\ b=\frac{8(56-x-y)}{14(4-x)} \\ c=\frac{-28-6x}{7(4-x)} \end{cases}\quad\quad\quad(8)$

and

$\begin{cases} i=2:x_2 = (\frac{3x_1^2+218x_1+224}{2y_1})^2-109-2x_1, \; y_2 = -y_1 + \frac{(3{x_1}^2+218x_1+224)}{2y_1}(x_1-x_2) \\ i>2: x_i = \left(\frac{y_{i-1}-y_1}{x_{i-1}-x_1}\right)^2-109-x_1-x_{i-1}, \; y_i= -y_1+\frac{y_{i-1}-y_1}{x_{i-1}-x_1}(x_1-x_i)\; \end{cases}(9)$

We solve (*) using the Omega Computer System Explorer (see Fig. 3).

Staring with the point $(x_1, y_1) = (-100, 260)$ on (7) (see Fig. 1),

Fig. 1

(8) yields

$a=\frac{2}{7}, b=\frac{-1}{14}, c=\frac{11}{14}$.

By (4), $a,b,c$ are modified by multiplying their common denominator:

$a=4,b=-1,c=11.$

Even though

$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = \frac{4}{-1+11} + \frac{-1}{4+11} + \frac{11}{4+(-1)} = 4$

as expected, it is not a solution of (*) since $b = -1 <0.$

Using (9), the CAS gives:

$(x_2, y_2) = (\frac{8836}{25}, \frac{-950716}{125}).$

The corresponding values of $a, b, c$ are

$a=9499, b=-8784, c=5165.$

But alas – it is still not a solution!

Undeterred, the CAS generates $(x_i, y_i)$ and calculates $a,b,c$ again and again for $i = 3, 4, 5, ..., 9.$

At last, the $a, b, c$ are all positive integers:

Fig. 2

Fig. 3

Exercise-1 Show that (3) is true.

Exercise-2 Show that (4) is true.

Exercise-3 Deriving (6).

Exercise-3 Solving (*) for values other than 4.

# Why Complex Numbers Are Not Ordered

In mathematics, if a structure is a field and has an order $\le$, two additional axioms need to hold for it to be an ordered field. These axioms express the notion that the ordering is compatible with the field structure:

1) For any three $a,b,c$ elements, $a \le b$ implies $a+c \le b + c.$

2) For any two $a, b$ elements, $0 \le a$ and $0 \le b$ implies $0 \le a\cdot b.$

We can show that such an order does not exist in the set of Complex number.

Let us restate the two axioms:

$\forall a,b,c, (a \le b) \implies a+c \le b+c.\quad\quad\quad(1)$

$\forall a,b, (0 \le a, 0 \le b) \implies 0 \le a\cdot b.\quad\quad\quad(2)$

If $0 \le \bold{i}$, we write

$0\le \underbrace{\bold{i}}_{a}, 0\le \underbrace{\bold{i}}_{b}$.

By (2),

$0 \le \underbrace{\bold{i}}_{a}\cdot \underbrace{\bold{i}}_{b}$.

i.e.,

$0 \le \bold{i}^2 \implies 0 \le -1,$ a contradiction.

Otherwise ($\bold{i} \le 0$), we have

$\underbrace{\bold{i}}_{a} \le \underbrace{0}_{b}$.

By (1),

$\underbrace{\bold{i}}_{a} + \underbrace{(-\bold{i})}_{c} \le \underbrace{0}_{b} + \underbrace{(-\bold{i})}_{c} \implies 0 \le -\bold{i}.$

We write it as

$0 \le \underbrace{-\bold{i}}_{a}, 0 \le \underbrace{-\bold{i}}_{b}.$

By (2),

$0 \le \underbrace{(-\bold{i})}_{a}\cdot\underbrace{(-\bold{i})}_{b} = \bold{i}^2 = -1.$

i.e.,

$0 \le -1$, a contradiction again.

Therefore, the complex numbers do not possess an order.