“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 $\cos.$