A Mind Unleashed

December 25, 2022January 10, 2023 / Michael Xue / Leave a comment

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 ?

December 20, 2022January 10, 2023 / Michael Xue / Leave a comment

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:

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

Generating Integrals

December 1, 2022March 31, 2023 / Michael Xue / Leave a comment

It is well known that

$\displaystyle\int \frac{1}{a^2+x^2}\;dx =\frac{1}{a}\arctan\left(\frac{x}{a}\right).\quad\quad\quad(1)$

Let

$\displaystyle\int \frac{1}{(a^2+x^2)^k}\; dx=I_k, \;\;k=1, 2, 3, \cdots\quad\quad\quad(2)$

we have

$I_1 = \frac{1}{a}\arctan\left(\frac{x}{a}\right).\quad\quad\quad(3)$

Differentiate (2) with respect to $a:$

$\frac{\partial}{\partial a}\displaystyle\int\frac{1}{(a^2+x^2)^k}\;dx = \frac{\partial}{\partial a}I_k.$

By “Differentiation under the integral sign”(see Playing “Feynman’s Trick” on Indefinite Integrals – Tongue in Cheek),

$\displaystyle \int \frac{\partial}{\partial a}\left(\frac{1}{(a^2+x^2)^k}\right)\;dx = \frac{\partial}{\partial a}I_k.$

That is,

$\displaystyle \int \frac{-k\cdot 2a}{(a^2+x^2)^{k+1}}\;dx = \frac{\partial}{\partial a}I_k.$

It follows that

$\displaystyle \int \frac{1}{(a^2+x^2)^{k+1}}\;dx = \frac{-1}{k\cdot 2a}\cdot\frac{\partial}{\partial a}I_k.\quad\quad\quad(4)$

What emerges from (3) and (4) is an algorithm for evaluating $I_{k+1} = \displaystyle \int \frac{1}{(a^2+x^2)^{k+1}}\;dx:$

$\begin {cases} I_1 = \frac{1}{a}\arctan\left(\frac{x}{a}\right), \\ I_{k+1} = \frac{-1}{2ka} \cdot \frac{\partial}{\partial a}I_{k}, k\ge 1 \end {cases}\quad\quad\quad(*)$

We generate $I_i, i=2, 3, 4, 5$ from (*):

The results are verified using Omega CAS Explorer‘s ‘integrate’:

Examine (*)’s derivation, we see nothing prevents generating $I_{r+1}, r\in \mathbb{R}$ from a known $I_r.$ For example, knowing $I_{\frac{1}{2}} = \mathrm{arcsinh}\left(\frac{x}{|a|}\right)$ (see Exercise-2), we can obtain $I_{\frac{3}{2}}, \; I_{\frac{5}{2}}, I_{\frac{7}{2}}, ...$

Exercise-1 Show that $\displaystyle\int \frac{1}{a^2+x^2}\;dx =\frac{1}{a}\arctan\left(\frac{x}{a}\right).$

Exercise-2 Show that $\displaystyle\int\frac{1}{\sqrt{a^2+x^2}}\;dx=\mathrm{arcsinh\left(\frac{x}{|a|}\right)}.$

Exercise-3 Given $I_{\frac{1}{2}}= \mathrm{arcsinh\left(\frac{x}{|a|}\right)}$ , generate $I_{\frac{3}{2}}, \; I_{\frac{5}{2}}, I_{\frac{7}{2}}, I_{\frac{9}{2}}.$