A Worthy Indefinite Integral

April 5, 2022April 4, 2022 / Michael Xue / Leave a comment

Evaluate $\displaystyle\int \frac{1+\frac{\alpha}{2}x}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx, \; \alpha_1 \ne \alpha_2$ .

$\displaystyle\int \frac{1+\frac{\alpha}{2}x}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx + \frac{\alpha}{2}\displaystyle\int\frac{x}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx + \frac{\alpha}{2}\cdot\left(\frac{1}{-2}\right)\displaystyle\int\frac{-2x+(\alpha_1+\alpha_2)-(\alpha_1+\alpha2)}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1\alpha_2}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx - \frac{\alpha}{4}\left(\displaystyle\int\frac{-2x+(\alpha_1+\alpha_2)}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1\alpha_2}}\;dx - \displaystyle\int\frac{\alpha_1+\alpha2}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1\alpha_2}}\;dx\right)$

$=\boxed{ \displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx} - \frac{\alpha}{4}\displaystyle\int\frac{-2x+(\alpha_1+\alpha_2)}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1\alpha_2}}\;dx +\boxed{\frac{\alpha(\alpha_1+\alpha_2)}{4}\displaystyle\int\frac{1}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1\alpha_2}}\;dx}$

$= \boxed{(1+\frac{\alpha(\alpha_1+\alpha_2)}{4})\displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx}-\frac{\alpha}{4}\displaystyle\int\frac{-2x + (\alpha_1+\alpha2)}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1\alpha_2}}\;dx$

$= \left(1+\frac{\alpha(\alpha_1+\alpha_2)}{4}\right)\displaystyle\int\frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx -\frac{\alpha}{4}\cdot\left(\frac{1}{1-\frac{1}{2}}\right)\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1\alpha_2}$

$= \left(1+\frac{\alpha(\alpha_1+\alpha_2)}{4}\right)\displaystyle\int\frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx -\frac{\alpha}{2}\sqrt{-(x-\alpha_1)(x-\alpha_2)}$

By $\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx=\arcsin\left(\frac{2x-(\alpha_1+\alpha_2)}{|\alpha_2-\alpha_1|}\right)$ (see “Integral: The CAS & I“),

$= \left(1+\frac{\alpha(\alpha_1+\alpha_2)}{4}\right) \arcsin\left(\frac{2x-(\alpha_1+\alpha_2)}{|\alpha_2-\alpha_1|}\right)-\frac{\alpha}{2}\sqrt{-(x-\alpha_1)(x-\alpha_2)}$

Exercise-1 Evaluate $\displaystyle\int \frac{1+\frac{\alpha}{2}x}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx, \alpha_1 \ne \alpha_2$ using a CAS.

Integral: The CAS & I

March 31, 2022April 4, 2022 / Michael Xue / 1 Comment

Evaluate $\displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx, \alpha_2 \ne \alpha_1.$

$\displaystyle\int \frac{1}{\sqrt{-(x-\alpha_1)(x-\alpha_2)}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{-(x^2-(\alpha_1+\alpha_2)x+\alpha_1 \alpha_2)}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\alpha_1 \alpha_2}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{-x^2+(\alpha_1+\alpha_2)x-\left(\frac{\alpha_1+\alpha_2}{2}\right)^2 + \left(\frac{\alpha_1+\alpha_2}{2}\right)^2-\alpha_1 \alpha_2}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{\left(\frac{\alpha_1+\alpha_2}{2}\right)^2-\alpha_1 \alpha_2-x^2+(\alpha_1+\alpha_2)x-\left(\frac{\alpha_1+\alpha_2}{2}\right)^2}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{\left(\frac{\alpha_1+\alpha_2}{2}\right)^2-\alpha_1 \alpha_2-\left(x^2-(\alpha_1+\alpha_2)x+\left(\frac{\alpha_1+\alpha_2}{2}\right)^2\right)}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{\frac{\alpha_1^2+\alpha_2^2+2\alpha_1\alpha_2-4\alpha_1\alpha_2}{4}-\left(x^2-(\alpha_1+\alpha_2)x+\left(\frac{\alpha_1+\alpha_2}{2}\right)^2\right)}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{\frac{\alpha_1^2+\alpha_2^2-2\alpha_1\alpha_2}{4}-\left(x^2-(\alpha_1+\alpha_2)x+\left(\frac{\alpha_1+\alpha_2}{2}\right)^2\right)}}\;dx$

$= \displaystyle\int \frac{1}{\sqrt{\frac{\left(\alpha_1-\alpha_2\right)^2}{4} - \frac{4x^2-4x(\alpha_1+\alpha_2)+(\alpha_1+\alpha_2)^2}{4}}}\;dx$

$= \displaystyle\int \frac{2}{\sqrt{\left(\alpha_2-\alpha_1\right)^2-\left(2x-(\alpha_1+\alpha_2)\right)^2}}\;dx$

For $a \ne 0, \int\frac{1}{\sqrt{a^2-x^2}}\;dx = \arcsin(\frac{x}{|a|})$ (see Exercise-1)

$=\arcsin\left(\frac{2x-(\alpha_1+\alpha_2)}{|\alpha_2-\alpha_1|}\right)$

Exercise-1 Show that for $a \ne 0, \int\frac{1}{\sqrt{a^2-x^2}}\;dx = \arcsin(\frac{x}{|a|})$ .

A Complex Number Aided Proof in Plane Geometry

March 25, 2022March 26, 2022 / Michael Xue / Leave a comment

The proof in my post “A Computer Algebra Aided Proof in Plane Geometry” can be significantly simplified if complex numbers are used.

Consider complex numbers $v_1, v_2, w_1, w_2, p$ in FIg. 1.

Fig. 1

Since

$v_1=p-(-a)= p+a \overset{p=x+\bold{i}y}{=} x+\bold{i}y+a \implies v_1 = x+a+\bold{i}y,\quad\quad\quad(1)$

we have,

$\bold{i}v_1 = w_1-(-a) \implies w_1=-a+\bold{i} v_1 \overset{(1)}{=} -a +\bold{i}(x+a+\bold{i}y)=-a-y+\bold{i}(x+a).$

i.e.,

$\begin{cases} x_1 = -a-y \\ y_1=x+a\end{cases}\quad\quad\quad(*)$

Similarly,

$v_2 = p-a = (x+\bold{i}y)-a=x-a+\bold{i}y\quad\quad\quad(2)$

and

$-\bold{i}v_2=w_2-a\implies w_2 = a + (-\bold{i} v_2) \overset{(2)}{=} a + (-\bold{i}(x-a+\bold{i}y)) = a+y+\bold{i}(a-x)$

give

$\begin{cases} x_3 = a+y \\ y_3 = a-x \end{cases}\quad\quad\quad(**)$

Let $k_1, k_3$ denote the slope of line connecting $(0, a)$ and $(x_1, y_1), (x_3, y_3)$ respectively.

From

$k_1=\frac{y_1-a}{x_1-0} \overset{(*)}{=} \frac{x+a-a}{-a-y} = \frac{-x}{a+y},$

$k_3=\frac{y_3-a}{x_3-0} \overset{(**)}{=} \frac{a-x-a}{a+y-0} = \frac{-x}{a+y},$

we see that

$k_1 = k_3\implies (x_1, y_1), (0, a), (x_3, y_3)$ lie on the same line.

Notice that $a+y \ne 0$ since $y>0, a>0$ .

See also “Treasure Hunt with Complex Numbers“.

A Computer Algebra Aided Proof in Plane Geometry

March 22, 2022March 22, 2022 / Michael Xue / Leave a comment

Given $\Delta ABC$ and two squares $ABEF, ACGH$ in Fig. 1. The squares are sitting on the two sides of $\Delta ABC, AB$ and $AC$ , respectively. Both squares are oriented away from the interior of $\Delta ABC$ . $\Delta BCP$ is an isosceles right triangle. $P$ is on the same side of $A$ . Prove that points $E, P$ and $G$ lie on the same line.

Fig. 1

Introducing rectangular coordinates show in Fig. 2:

Fig. 2

We observe that

$y>0\quad\quad\quad(1)$

$x_1 < -a\quad\quad(2)$

$x_3 > a\quad\quad\quad(3)$

$CG=CA\implies (x-a)^2+y_3^2 = (x-a)^2 +y^2\quad\quad(4)$

$AB=AE\implies (x+a)^2+y^2=(x_1+a)^2+y_1^2\quad\quad(5)$

$CG\perp CA \implies y_3 y = -(x-a)(x_3-a)\quad\quad\quad(6)$

$BE\perp AB \implies y_1y = -(x_1+a)(x+a)\quad\quad\quad(7)$

Fig. 3

Solving system of equations (4), (5), (6), (7), we obtain four set of solutions:

$x_1=-y-a, y_1=x+a, x_3=y+a, y_3=a-x\quad\quad(8)$

$x_1=y-a, y_1=-x-a, x_3=y+a, y_3=a-x\quad\quad(9)$

$x_1=-y-a, y_1=x+a, x_3=a-y, y_3=x-a\quad\quad(10)$

$x_1=y-a, y_1=-x-a, x_3=a-y, y_3=x-a\quad\quad(11)$

Among them, only (10) truly represents the coordinates in Fig. 2.

Fig. 4

By Heron’s formula (see “Had Heron Known Analytic Geometry…“), the area of triangle with vortex $(-y-a, x+a), (0, a), (y+a, a-x)$ is

$\frac{1}{2}\begin{vmatrix} -y-a & x+a & 1 \\ 0 & a &1 \\ y+a & a-x & 1\end{vmatrix}$ .

From Fig. 4, we see that it is zero.

Therefore,

$E, P, G$ lie on the same line.

The reason we do not consider (9), (10), (11) is due to the fact that

(9) contradicts (2) since $y>0, a>0 \implies x_1=y-a=-a+y>-a$ .

And,

by (1), (10) and (11) indicate $x_3=a-y <a$ which contradicts (3).

Exercise-1 Prove “ $E, P, G$ lie on the same line” with complex numbers (hint: see “Treasure Hunt with Complex Numbers“).

A Proof without Trigonometric Function

March 15, 2022March 28, 2022 / Michael Xue / Leave a comment

Problem: $A$ and $B$ are squares. Without invoking trigonometric functions, show that the area of triangle $C$ equals that of $D.$

Solution: Introducing a rectangular coordinate system and complex number $V_1, V_2, P, W_1, W_2, Q:$

Let $A_1, A_2$ denote the area of triangle $C$ and $D$ respectively.

We have

$P = V_2-V_1.\quad\quad\quad(1)$

Let $s_1 = \frac{|V_1| + |V_2| + |P|}{2} \overset{(1)}{=} \frac{|V_1| + |V_2| + |V_2 - V_1|}{2}.$

By Heron’s formula derived without invoking trigonometric function (see “An Algebraic Proof of Heron’s Formula“),

$A_1 = \sqrt{s_1(s_1-|V_1|)(s_1-|V_2|)(s_1-|P|)}\overset{(1)}=\sqrt{s_1(s_1-|V_1|)(s_1-|V_2|)(s_1-|V_2-V_1|)}.\quad(*)$

We also have

$W_1 = \bold{i}V_1, \quad W_2 = -\bold{i}V_2\quad\quad\quad(2)$

(see “Treasure Hunt with Complex Numbers“) and,

$Q = W_2-W_1\overset{(2)}{=}-\bold{i}V_2-\bold{i}V_1.\quad\quad\quad(3)$

Similarly,

Let $s_2 = \frac{|W_1| + |W_2| + |Q|}{2} \overset{(2), (3)}{=} \frac{|iV_1| + |-\bold{i}V_2| +|-\bold{i}V_2-\bold{i}V_1|}{2}=\frac{|V_1| + |V_2| +|V_2+V_1|}{2}.$

$A_2 = \sqrt{s_2(s_2-|W_1|)(s_2-|W_2|)(s_2- |Q|)}$

$\overset{(2), (3)}{=}\sqrt{s_2(s_2-|\bold{i}V_1|)(s_2-|-\bold{i}V_2|)(s_2- |-\bold{i}V_2-\bold{i}V_1|)}.$

That is,

$A_2=\sqrt{s_2(s_2-|V_1|)(s_2-|V_2|)(s_2- |V_2+V_1|)}\quad\quad\quad(**)$

It is shown by Omega CAS Explorer that the expression under the square root of $A_1$ is the same as that of $A_2:$

Therefore,

$A_1 = A_2.$

Exercise-1 The two squares with area 25 and 36 ( see figure below) are positioned so that $AB=7.$ Find the area of triangle TSC.

My Pi

March 14, 2022April 12, 2022 / Michael Xue / Leave a comment

Besides “Wallis’ Pi“, there is another remarkable expression for the number $\pi$ as an infinite product. We derive it as follows:

From the trigonometric identity

$\sin(2x) = 2\sin(x)\cos(x)$ ,

we have

$\sin(x) = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$

$= 2\cdot 2\sin(\frac{x}{4})\cos(\frac{x}{4}) \cdot\cos(\frac{x}{2})$

$= 2\cdot 2\cdot 2\sin(\frac{x}{8})\cos(\frac{x}{8})\cdot \cos(\frac{x}{4})\cdot \cos(\frac{x}{2})$

$\ddots$

$= 2^n\sin(\frac{x}{2^n})\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i})$ .

That is,

$\sin(x) = 2^n\sin(\frac{x}{2^n})\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i}).$

Dividing both sides by $x$ yields

$\frac{\sin(x)}{x} = \frac{2^n\sin(\frac{x}{2^n})}{x}\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i})=\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i})$

or,

$\prod\limits_{i=1}^{n} \cos(\frac{x}{2^i})=\frac{\sin(x)}{x} /\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}$ .

It follows that since $\lim\limits_{n \rightarrow \infty}\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}=1$ ,

$\lim\limits_{n \rightarrow \infty}\prod\limits_{i=1}^{n} \cos(\frac{x}{2^i})=\lim\limits_{n \rightarrow \infty}\frac{\sin(x)}{x}/\lim\limits_{n \rightarrow \infty}\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}=\frac{\sin(x)}{x}$ ;

i.e.,

$\frac{\sin(x)}{x}=\lim\limits_{n \rightarrow \infty}\prod\limits_{i=1}^{n} \cos(\frac{x}{2^i}).\quad\quad\quad(1)$

Let

$x = \frac{\pi}{2},$

(1) becomes

$\frac{2}{\pi} =\lim\limits_{n\rightarrow \infty}\prod\limits_{i=1}^{n}\cos(\frac{\pi}{2\cdot2^i})=\lim\limits_{n\rightarrow \infty}\cos(\frac{\pi}{4})\cos(\frac{\pi}{8})...\cos(\frac{\pi}{2\cdot 2^n}).\quad\quad\quad(2)$

We know

$\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}.$

Applying the half-angle formula

$\cos(\frac{x}{2}) = \pm \sqrt{\frac{1+\cos(x)}{2}}$

gives

$\cos(\frac{\pi}{8}) = \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} = \frac{\sqrt{2+\sqrt{2}}}{2};$

$\cos(\frac{\pi}{16}) = \sqrt{\frac{1+\cos(\frac{\pi}{8})}{2}} = \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2};$

$\ddots$

Hence,

$\frac{2}{\pi} = \frac{\sqrt{2}}{2} \cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}...\quad\quad\quad(3)$

We compute the value of $\pi$ according to (3):

Fig. 1

Exercise-1 Compute $\pi$ from (1) by letting $x = \frac{\pi}{6}.$

Gotta Catch ‘Em All !

March 14, 2022March 14, 2022 / Michael Xue / Leave a comment

Solving $\frac{d^2y}{dx^2} = \frac{y}{x}\cdot\frac{dy}{dx}.$

Even though ‘contrib_ode’, Maxima’s ODE solver choked on this equation (see “An Alternate Solver of ODEs“), it still can be solved as demonstrated below:

multiplied by $x,$

$x\frac{d^2y}{dx^2} = y\frac{dy}{dx},$

i.e.,

$x\frac{d}{dx}\left(\frac{dy}{dx}\right) = y\frac{dy}{dx}.$

Integrate it, we have

$\int x\frac{d}{dx}\left(\frac{dy}{dx}\right)\;dx = \int y \frac{dy}{dx} \;dx \implies \int \left(\frac{dy}{dx}\right)'\cdot x\;dx = \frac{1}{2}y^2+c.\quad\quad\quad(1-1)$

By $\int u'\cdot v\;dx = u\cdot v - \int u\cdot v'\;dx$ (see “Integration by Parts Done Right“),

$\int \left(\frac{dy}{dx}\right)'\cdot x\;dx = \frac{dy}{dx}\cdot x-\int\frac{dy}{dx}\cdot x'\;dx \overset{x'=1}{=} x\frac{dy}{dx}-\int \frac{dy}{dx}\;dx = x\frac{dy}{dx}-y.$

As a result, (1-1) yields a new ODE

$x\frac{dy}{dx} -y = \frac{1}{2}y^2 + c\quad\quad\quad(1-2)$

or,

$2x\frac{dy}{dx}=2y+y^2+c_1,\quad c_1=2c.\quad\quad\quad(1-3)$

Upon submitting (1-3) to Omega CAS Explorer in non-expert mode, the CAS asks for the range of $c_1.$

Fig. 1

Depending on the range provided, ‘ode2’ gives three different solutions (see Fig. 2, 3 and 4).

Fig. 2 $c_1>1$

Fig. 3 $c_1<1$

Fig. 4 $c_1=1$

Let’s also solve (1-3) manually:

If $y^2+2y+c_1=0,$ (1-3) has a constant solution

$y=c_2.\quad\quad\quad(2-1)$

In fact, this solution can be observed from $\frac{d^2y}{dx^2} = \frac{y}{x}\cdot\frac{dy}{dx}$ right away.

Otherwise ( $y^2+2y+c_1 \ne 0$ ),

$\frac{1}{y^2+2y+c_1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

That is,

$\frac{1}{y^2+2y+1+c_1-1}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\frac{1}{(y+1)^2+c_1-1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

For $c_1-1>0,$ let $k=\sqrt{c_1-1},$ we have

$\frac{1}{(y+1)^2+k^2}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

Divide both numerator and denominator on the left side by $k^2$ ,

$\frac{1}{k^2}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

Write it as

$\frac{1}{k}\cdot\frac{1}{k}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}=\frac{1}{2x}.$

Multiply both sides by $k,$

$\frac{1}{k}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}=k\cdot\frac{1}{2x}$

Integrate it,

$\int\frac{1}{k}\cdot\frac{1}{(\frac{y+1}{k})^2+1}\cdot\frac{dy}{dx}\;dx=\int k\cdot\frac{1}{2x}\;dx,$

we obtain

$\arctan(\frac{y+1}{k}) = \frac{k}{2}\log(|x|)+c_2.$

i.e.,

$\arctan(\frac{y+1}{\sqrt{c_1-1}}) = \frac{\sqrt{c_1-1}}{2}\log(|x|)+c_2$

or,

$y = 2k_1\tan(k_1\log|x| +k_2) -1\quad\quad\quad(2-2)$

where $k_1=\frac{\sqrt{c_1-1}}{2}, k_2=c_2.$

For $c_1-1<0,$ let $k=\sqrt{1-c_1},$

$\frac{1}{(y+1)^2-k^2}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\frac{1}{(y+1-k)\cdot(y+1+k)}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\frac{1}{2k}\left(\frac{1}{y+1-k} - \frac{1}{y+1+k}\right)\cdot\frac{dy}{dx}= \frac{1}{2x}$

$\int \left(\frac{1}{y+1-k} - \frac{1}{y+1+k}\right)\cdot\frac{dy}{dx}\;dx= \int\frac{k}{x}\;dx$

$\log|y+1-k| - \log|y+1+k| = k\cdot\log|x| + c_2$

$\log\bigg|\frac{y+1-k}{y+1+k}\bigg| = k\cdot\log|x| + c_2$

$\log\bigg|\frac{y+1-\sqrt{1-c_1}}{y+1+\sqrt{1-c_1}}\bigg| = \sqrt{1-c_1}\cdot\log|x| + c_2$

$y= -\frac{k_2(1+k_1)|x|^{k_1}+k_1-1}{k_2|x|^{k_1}-1}, \quad k_1=\sqrt{1-c_1}, k_2=e^{c_2}.\quad\quad\quad(2-3)$

For $c_1-1=0 \quad(c_1=1),$

$\frac{1}{(y+1)^2}\cdot\frac{dy}{dx}=\frac{1}{2x}$

$\int \frac{1}{(y+1)^2}\cdot\frac{dy}{dx}\;dx=\int \frac{1}{2x}\;dx$

$-\frac{1}{y+1} = \frac{1}{2}\log|x| +c_2$

$y+1 = \frac{-1}{\frac{1}{2}\log|x|+c_2}$

$y = -1 - \frac{1}{\frac{1}{2}\log|x|+c_2}.\quad\quad\quad(2-4)$

Notice when $c_1 = 0, c=0.$ (1-2) becomes

$\frac{dy}{dx} - \frac{1}{x}y = \frac{1}{2x}y^2.$

This is a Bernoulli’s Equation $\frac{dy}{dx}+f(x)y=g(x)y^{\alpha}$ with $f(x)=\frac{1}{x}, g(x)=\frac{1}{2x}$ and $\alpha=2.$ Solving it (see “Meeting Mr. Bernoulli“),

$y^{1-2} = e^{(2-1)\int -\frac{1}{x}\;dx}\cdot\left((1-2)\int e^{-(2-1)\int-\frac{1}{x}\;dx} \cdot \frac{1}{2x}\;dx+c\right)$

$= e^{-\log|x|}\cdot\left(-\int e^{\log|x|} \frac{1}{2x}\;dx +c\right)$

$= \frac{1}{|x|}\cdot\left(-\int |x|\cdot\frac{1}{2x}\;dx + c\right)$

$= \frac{1}{|x|}\cdot\left(-\int (\pm x)\cdot\frac{1}{2x}\;dx+c\right)$

$= \frac{1}{|x|}\cdot\left(-(\pm)\int \frac{1}{2}\;dx + c\right)$

$= \frac{1}{|x|}\cdot\left(-\frac{1}{2}(\pm x) + c\right)$

$= \frac{1}{|x|}\cdot\left(-\frac{1}{2}|x|+c\right)$

$= -\frac{1}{2} + \frac{c}{|x|}$

$\frac{1}{y} = -\frac{1}{2} + \frac{c}{|x|}\implies y = \frac{-2|x|}{|x|-2c}.\quad\quad\quad(3-1)$

Since $c_1 =0 \implies c_1-1=0-1 = -1 <0,$ we can verify (3-1) as follows:

substitute $c_1=0$ into (2-3),

$y = -\frac{k2(1+\sqrt{1-0})|x|^{\sqrt{1-0}}+\sqrt{1-0}-1}{k_2|x|-1}=-\frac{2k_2|x|}{k_2|x|-1}\overset{k_2=\frac{1}{2c}}{=}\frac{-2|x|}{|x|-2c}.$

Unsurprisingly, this is the same as (3-1).

Exercise-1 Mathematica solves $\frac{d^2y}{dx^2} = \frac{y}{x}\cdot\frac{dy}{dx}:$

But it only return one solution. Show that it is equivalent to (2-2).

Exercise-2 Solving (1-2) using ‘contrib_ode’.

Exercise-3 Show that (2-2), (2-3) and (2-4) are equivalent to results shown in Fig. 2, 3 and 4 respectively.

Treasure Hunt with Complex Numbers

February 15, 2022March 27, 2022 / Michael Xue / 2 Comments

Fig. 1

In his popular book “One, Two, Three … Infinity“, physicist George Gamow told a story:

Once upon a time, there was a young man who found among his great grandfather’s papers a piece of parchment that revealed the location of a hidden treasure. It read:

“Sail to ____ North Latitude and ____ West longitude where you will find a deserted island. There is a large meadow on the north shore of the island where stand an oak and a pine. You will see also an old gallows on which we once used to hang traitors. Start from the gallows and walk to the oak counting your steps. At the oak, you must turn right by a right angle and take the same number of steps. Put a spike in the ground there. Now you must return to the gallows and walk to the pine counting your steps. At the pine, you must turn left by a right angle and see that you take the same number of steps, and put another spike into the ground. Dig half way between the spikes; the treasure is there.”

So the young man charted a ship and sailed to the South Seas. He found the island, the meadow, the oak and the pine, but to his great sorrow the gallows was gone. Unlike the living trees, the gallows has long since disintegrated in the weather, and not a trace of it or its location remains.

Unable to carry out the rest of the instructions (or so he believes), the young man fell into despair. In an angry frenzy he began to dig at random all over the field. But all his efforts were in vain; the island was too big!

Needless to say, the young man sailed back empty handed. And the treasure is still there.

This is a sad story, but what is sadder still is the fact that the young man might have gotten the treasure, if only he had known some mathematics, and specifically the use of complex numbers.

How come?

Consider the island as a plane of complex numbers; Place the origins of three rectangular coordinate systems at the location of oak( $o_1$ ), pine ( $o_2$ ) and half way between them ( $o$ ). $\Gamma, x_1, x_2, y_1, y_2$ and $d$ are complex numbers. Notably, $d$ is the half way point between the spikes.

Fig. 2

From Fig. 2, we see that

$-1 + x_1 =\Gamma\implies x_1=\Gamma+ 1,\quad\quad\quad(1)$

$1 + x_2 = \Gamma \implies x_2=\Gamma-1.\quad\quad\quad(2)$

By the fact (see Exercise-1) that

(1) The multiplication by $\bold{i}$ is geometrically equivalent to a counterclockwise rotation by a right angle

and

(2) The multiplication by $\bold{-i}$ is geometrically equivalent to a clockwise rotation by a right angle,

$y_1-(-1)=\bold{i}\cdot x_1\implies y_1=-1 + \bold{i}\cdot x_1\overset{(1)}{=}-1+ \bold{i}\cdot(\Gamma+1)=-1 + \bold{i}\cdot\Gamma + \bold{i},\quad\quad\quad(3)$

$y_2-1=-\bold{i}\cdot x_2\implies y_2=1+\bold{(-i)}\cdot x_2\overset{(2)}{=}1-\bold{i}\cdot(\Gamma-1)= 1-\bold{i}\cdot\Gamma + \bold{i}.\quad\quad\quad(4)$

Since the treasure is halfway between the spikes, we have

$y_1-d = \frac{1}{2}s,\quad\quad\quad(5)$

$d-y_2 = \frac{1}{2}s.\quad\quad\quad(6)$

Subtracting (6) from (5) gives

$y_1+y_2 -2d=0.$

It means

$2d = y_1+y_2 \overset{(3), (4)}{=} (-1 + \bold{i}\cdot\Gamma+\bold{i})+ (1-\bold{i}\cdot\Gamma + \bold{i}) = 2\bold{i}.\quad\quad\quad(7)$

Therefore,

$d = \bold{i}.\quad\quad(8)$

We see that the unknown position of gallows denoted by $\Gamma$ fell out in (7), and (8) tells regardless where the gallows stood, the treasure must be located at the point $\bold{i}$ of rectangular coordinate system with origin $o$ .

And so, had the young man done the simple math shown above, he could have looked for the treasure at the point indicated by the cross in Fig. 1 and found it there.

Exercise-1 Prove

(1) The multiplication by $\bold{i}$ is geometrically equivalent to a counterclockwise rotation by a right angle.

(2) The multiplication by $\bold{-i}$ is geometrically equivalent to a clockwise rotation by a right angle.

Exercise-2 Locate the treasure using a computer algebra system (hint: see “A Computer Algebra Aided Proof in Plane Geometry“).

Integral: I vs. CAS

February 11, 2022March 7, 2022 / Michael Xue / Leave a comment

Evaluate $\displaystyle\int\frac{\sqrt{1+p^2}}{p}\;dp$

Let

$u = \sqrt{1+p^2},\quad\quad\quad(1)$

we have

$u^2=1+p^2 \implies p^2=u^2-1 \implies p =\pm\sqrt{u^2-1} \quad\quad\quad(2)$

and

$\frac{dp}{du}= \pm\frac{1}{2}\cdot\frac{2u}{\sqrt{u^2-1}} =\pm \frac{u}{\sqrt{u^2-1}}.\quad\quad\quad(3)$

Consequently,

$\int\frac{\sqrt{1+p^2}}{p}\;dp$

$\overset{(1), (2)}{=} \int \frac{u}{\pm\sqrt{u^2-1}}\cdot\frac{dp}{du}\;du$

$\overset{(3)}{=}\int \frac{u}{\pm\sqrt{u^2-1}}\cdot(\pm\frac{u}{\sqrt{u^2-1}})\;du$

$= \int\frac{u^2}{u^2-1}\;du$

$= \int \frac{u^2-1+1}{u^2-1}\;du$

$= \int du + \int \frac{1}{u^2-1}\;du$

$= u +\int \frac{1}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)\;du$

$= u+\frac{1}{2}\log\frac{u-1}{u+1}$

$= u + \frac{1}{2}\log\frac{(u-1)^2}{u^2-1}$

$= \sqrt{1+p^2} + \frac{1}{2}\log\frac{\sqrt{p^2+1}-1)^2}{p^2+1-1}$

$=\sqrt{1+p^2}+\frac{1}{2}\log\frac{(\sqrt{p^2+1}-1)^2}{p^2}$

$=\sqrt{1+p^2}+\log\frac{\sqrt{|p|^2+1}-1}{|p|}$

$= \sqrt{1+p^2} +\log\frac{(\sqrt{p^2+1}-1)(\sqrt{p^2+1}+1)}{|p|(\sqrt{p^2+1}+1)}$

$= \sqrt{1+p^2} + \log\frac{p^2+1-1}{|p|(\sqrt{p^2+1}+1)}$

$= \sqrt{1+p^2} + \log\frac{p^2}{|p|(\sqrt{1+p^2}+1)}$

$\overset{p^2=|p|^2}{=} \sqrt{1+p^2} + \log\frac{|p|}{\sqrt{p^2+1}+1}$

$= \sqrt{1+p^2} + \log\left(\frac{\sqrt{1+p^2}+1}{|p|}\right)^{-1}$

$= \sqrt{1+p^2} - \log\frac{\sqrt{1+p^2}+1}{|p|}$

$=\sqrt{1+p^2}-\log(\sqrt{\frac{1}{|p|^2} + \frac{p^2}{|p|^2}} +\frac{1}{|p|})$

$\overset{|p|^2=p^2}{=}\sqrt{1+p^2} - \log(\sqrt{(\frac{1}{|p|})^2 + 1} + \frac{1}{|p|})$

From “Deriving Two Inverse Functions“:

$\mathrm{arcsinh}(x) = \log(\sqrt{x^2+1} + x), x\in (-\infty, \infty).$

Therefore,

$\int\frac{\sqrt{1+p^2}}{p}\;dp = \sqrt{1+p^2} - \mathrm{arcsinh}\left(\frac{1}{|p|}\right).\quad\quad\quad(*)$

Had we written $\int\frac{1}{u^2-1}\;du$ as $\int-\frac{1}{2}(\frac{1}{u+1}-\frac{1}{u-1})\;du,$ we would have

$\int du + \int\frac{1}{u^2-1}\;du$

$= u -\frac{1}{2}\int(\frac{1}{u+1}-\frac{1}{u-1})\;du$

$= u-\frac{1}{2}(\log(u+1)-\log(u-1))$

$= u-\frac{1}{2}\log\frac{u+1}{u-1}$

$= u-\frac{1}{2}\log\frac{(u+1)(u+1)}{(u-1)(u+1)}$

$= u-\frac{1}{2}\log\frac{(u+1)^2}{u^2-1}$

$= \sqrt{1+p^2}-\frac{1}{2}\log\frac{(\sqrt{1+p^2} + 1)^2}{1+p^2-1}$

$= \sqrt{1+p^2}-\frac{1}{2}\log\frac{(\sqrt{1+p^2}+1)^2}{p^2}$

$= \sqrt{1+p^2}-\log\frac{\sqrt{1+p^2}+1}{|p|}$

$= \sqrt{1+p^2}-\log(\sqrt{1+(\frac{1}{|p|})^2}+\frac{1}{|p|})$

$= \sqrt{1+p^2}-\mathrm{arcsinh}(\frac{1}{|p|}),$

the same as (*).

How i Was Born

February 3, 2022July 28, 2022 / Michael Xue / Leave a comment

In Memory of Johann Weilharter (1953-2021)

Girolamo Cardano (1501-76) was an Italian intellect whose interests and proficiencies ranged through those of mathematician, physician, biologist, physicist, chemist, astrologer, astronomer, philosopher, writer, and gambler.

While conducting research on solving algebraic equations, Cardano discovered that by means of a suitable substitution, the general cubic equation

$y^3+by^2+cy+d=0$

can be simplified. His substitution is $y = x-\frac{b}{3}$ , which yields

$(x-\frac{b}{3})^3 + b(x-\frac{b}{3})^2+c(x-\frac{b}{3})+d =0.$

Upon expanding and rearrange the terms, this becomes

$x^3+px=q,$

a depressed cubic (without the $x^2$ term) where $p = c-\frac{b^2}{3}, q = -d+\frac{bc}{3}-\frac{2b^3}{27}.$

Cadano applied this substitution in solving cubic equation $y^3-15y^2+81y-175=0:$

Substituting $y=x-(-15/3) = x+5$ into the cubic in $y$ , he obtained a depressed cubic in $x.$ namely,

$x^3+6x=20.$

Without a formula for this simplified equation, Cardano proceeded to solve it by way of ad hoc factoring:

$x^3+6x-20$

$= x^3-2x^2+2x^2+6x-20$

$= x^3-2x^2+2x^2-4x+10x-20$

$= x^2(x-2)+2x(x-2)+10(x-2)$

$= (x-2)(x^2+2x+10).$

Clearly, $x=2$ is a solution to $x^3+6x=20.$

Applying the quadratic formula to $x^2+2x+10=0$ gave $\frac{-2 \pm \sqrt{-36}}{2}.$ But this expression was immediately dismissed (for Cardano knew $x^2+2x+10=0$ has no real solution).

Therefore, $y=x+5 = 2+5=7$ is the only solution to the original cubic equation.

$x^2+2x+10=0$ has no solution

Cardano also solved $y^3+3y^2-12y-18=0$ in a similar fashion:

Obtaining first the depressed cubic (with $y=x-1$ )

$x^3-15x=4.$

Next is the ad hoc factoring again:

$x^3-15x-4$

$= x^3-4x^2+4x^2-15x-4$

$= x^3-4x^2+4x^2-16x+x-4$

$= x^2(x-4)+4x(x-4)+(x-4)$

$= (x-4)(x^2+4x+1).$

Surely,

$(x-4)(x^2+4x+1)=0 \implies x=4$ is a solution.

Furthermore, two additional solutions: $x = \frac{-4\pm\sqrt{4^2-4}}{2} = -2\pm\sqrt{3}$ were obtained by applying the quadratic formula to $x^2+4x+10=0.$

This image has an empty alt attribute; its file name is screen-shot-2022-01-06-at-1.26.26-pm.png

$x^3-15x=4$ has three solutions: $4, -2+\sqrt{3}, -2-\sqrt{3}$

But Cardano did not like the ad hoc factoring. He wanted a formula that readily solves the depressed cubic $x^3+px=q$ , one that resembles the formula for the quadratics (see “Deriving the quadratic formula without completing the square“).

His relentless search for such a formula took many years (see William Dunham’s “Journey through genuis“) but, lo and behold, he found one:

$x = \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}\quad\quad\quad(*)$

To be clear, (*) is not Cardano’s own making. The formula bears the name ‘Cardano’s formula’ today only because Cardano was the one who published it in his 1545 book “Ars Magna” but without its derivation. However, in a chapter titled “On the Cube and First Power Equal to the Number”, Cardano did give acknowledgement to Scipio del Ferro and Niccolo Fontana, who had independently derived (*) around 1515, but had kept the knowledge away from the public.

We derive (*) as follows:

Consider an algebraic identity that reminiscent of the depressed cubic $x^3+px=q.$ Namely,

$(u-v)^3+\underbrace{3uv}_{p}(u-v)=\underbrace{u^3-v^3}_{q}.$

It suggests that if we can determine the quantity $u$ and $v$ in terms of $p$ and $q$ from

$\begin{cases} 3uv = p \\u^3-v^3=q\end{cases}\quad\quad\quad(1-1, 1-2)$

then $u-v$ is a solution to $x^3+px=q.$

Asume $u \ne 0$ , (1-1) gives

$v = \frac{p}{3u}.$

Substituting this into (1-2) yields

$u^3-\frac{p^3}{27u^3}=q.$

Multply both sides by $u^3$ and rearrange terms, we have a sixth-degree equation:

$u^6-qu^3-\frac{p^3}{27} =0.$

But it is also quadratic in $u^3$ .

$(u^3)^2-qu^3-\frac{p^3}{27} = 0.$

Therefore, using the formula for quadratics,

$u^3 = \frac{q}{2}\pm\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}\implies u =\sqrt[3]{\frac{q}{2}\pm\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$

There are two cases to consider.

For

$u =\sqrt[3]{\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},\quad\quad\quad(1-3)$

we have $v^3 \overset{(1-2)}{=} u^3-q = \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}-q,$ i.e.,

$v = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.\quad\quad\quad(1-4)$

It follows that

$x \overset{(*)}{=} u-v \overset{(1-3), (1-4)}{=} \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}\quad\quad\quad(1-5)$

For $u =\sqrt[3]{\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}, \;v=\sqrt[3]{-\frac{q}{2} -\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$

$x = u-v = \sqrt[3]{\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}-\sqrt[3]{-\frac{q}{2} -\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}$

$= -\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}$

$= \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},$

the same as (1-5).

If $u=0,$ we see that on the one hand,

$\left(p \overset{(1-1)}{=} 0 \implies v \overset{(1-2)}{=} -\sqrt[3]{q}\right) \implies u-v = 0-(-\sqrt[3]{q}) =\sqrt[3]{q}.$

On the other hand, letting $p =0$ in (*) yields

$u-v = \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + 0}} -\sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + 0}}$

$= \sqrt[3]{\frac{q}{2}+\frac{|q|}{2}} -\sqrt[3]{-\frac{q}{2} + \frac{|q|}{2}}$

$= \begin{cases} \sqrt[3]{\frac{q}{2}+\frac{q}{2}} - \sqrt[3]{-\frac{q}{2}+\frac{q}{2}} = \sqrt[3]{q}, \;q \ge 0\\ \sqrt[3]{\frac{q}{2}-\frac{q}{2}} - \sqrt[3]{-\frac{q}{2}-\frac{q}{2}} = \sqrt[3]{q},\;q<0\end{cases}$

Cardano first tested (*) on cubic $x^3+6x=20$ by letting $p=6, q=20.$ . The formula yields

$x = \sqrt[3]{\frac{20}{2}+\sqrt{\frac{20^2}{4} + \frac{6^3}{27}}} - \sqrt[3]{-\frac{20}{2} + \sqrt{\frac{20^2}{4} + \frac{6^3}{27}}}= \sqrt[3]{10+\sqrt{108}} - \sqrt[3]{-10+\sqrt{108}}.$

It came as a surprise to Cardano initially. But he quickly realized that this sophisticated looking expression is nothing more than “2”, the unique solution of $x^3+6x=20$ , in disguise.

Today, this is easily checked by a CAS:

For a mathematical proof, see “A Delightful Piece of Mathematics“.

Cardano then tested the formula on cubic $x^3-15x=4.$ Substituting $p = -15, q=4$ into it gave

$x = \sqrt[3]{2+\sqrt{-121}} - \sqrt[3]{-2+\sqrt{-121}}\quad\quad\quad(**)$

He was startled by the result!

This image has an empty alt attribute; its file name is tmp-1.gif

The presence of $\sqrt{-121}$ alone did not surprise him for he had seen negative number under the square root before (while solving $x^2+2x+10=0,$ a quadratic clearly has no solution). What really perplexed Cardano this time was the fact that square root of negative number appearing in the result for a cubic that has three real solutions!

Cardano thus sought the value of $\sqrt[3]{2+\sqrt{-121}} - \sqrt[3]{-2+\sqrt{-121}}$ to see which solution, amongst $4, -2+\sqrt{3}$ and $-2+\sqrt{3}$ it represents.

He started with $\sqrt[3]{2+\sqrt{-121}}$ . At once, Cardano noticed that $2+\sqrt{-121}$ is a number in the form of

$a+\sqrt{-b}.$

And he speculated that the result of calculating $\sqrt[3]{2+\sqrt{-121}}$ has the same manner.

So Cardano wanted to find $a$ and $b$ such that

$\sqrt[3]{2+\sqrt{-121}} = a+\sqrt{-b}.$

He proceeded as follows:

Cubing both sides gives

$2+\sqrt{-121} = a^3 +3a^2\sqrt{-b}-3ab-b\sqrt{-b} = a^3-3ab+3a^2\sqrt{-b}-b\sqrt{-b}.$

Equating the similar parts on both sides yields a system of nonlinear algebraic equations

$\begin{cases} a^3-3ab=2 \\ 3a^2\sqrt{-b}-b\sqrt{-b} = \sqrt{-121} \end{cases}\quad\quad\quad(1-6, 1-7)$

Squaring both (1-6) and (1-7) gives:

$\begin{cases} a^6-6a^4b+9a^2b^2=4 \\ -9a^4b+6a^2b^2-b^3=-121\end{cases}\quad\quad\quad(1-8, 1-9)$

and subtracting (1-9) from (1-8) results in

$a^6+3a^4b+3a^2b^2+b^3 =125 \implies (a^2+b)^3 = 125\implies a^2+b=5$

or,

$b = 5-a^2.$

Substituting it back into (1-6) yields

$4a^3-15a=2 \implies a^3 = \frac{15}{4}a + \frac{1}{2}.$

And so,

$a^3-\frac{15}{4}a = \frac{1}{2}$

This is a depressed cubic with $p=-\frac{15}{4}, q=\frac{1}{2}.$ By Cardano’s formula,

$a=\sqrt[3]{\frac{1}{4} + \sqrt{\frac{1}{4}(\frac{1}{2})^2 + \frac{1}{27}(\frac{-15}{4})^3}}-\sqrt[3]{-\frac{1}{4} + \sqrt{\frac{1}{4}(\frac{1}{2})^2 + \frac{1}{27}(\frac{-15}{4})^3}}$

$=\sqrt[3]{\frac{1}{4} + \sqrt{\frac{1}{16} - \frac{3375}{27\cdot64}}}-\sqrt[3]{-\frac{1}{4} + \sqrt{\frac{1}{16} - \frac{3375}{27\cdot64}}}$

$=\sqrt[3]{\frac{1}{4} + \sqrt{\frac{1}{16}-\frac{3375}{1728}}}-\sqrt[3]{-\frac{1}{4} + \sqrt{\frac{1}{16}-\frac{3375}{1728}}}$

That is,

$a =\sqrt[3]{\frac{1}{4} + \sqrt{\underline{-\frac{121}{64}}}}-\sqrt[3]{-\frac{1}{4} + \sqrt{\underline{-\frac{121}{64}}}}$

So solving $4a^3-15a=2$ for $a$ by Cardano’s formula resulting in having to calculate another square root of a negative number. Cardano was put right back to where he had started. With the frustration he called the cubic “irreducible” and pursued the matter no further.

It would be another generation before Rafael Bombelli (1576-72) took upon the challenge of calculating $\sqrt[3]{2+\sqrt{-121}}$ again.

Bombelli’s was an engineer who knew how to drain the swampy marshes, and only between his engineering projects was he actively engaged in mathematics. Being practical and sound minded, he read the near-mystical $\sqrt{-121}$ not as the square root of a negative number but a symbolic representation for a new type of number that extends the real number. He imagined a set for a new type of number that

[1] Has every real number as its member.

[2] The arithmetic operations ( $+, \cdot$ ) are so defined that the commutative, associative and distributive law are obeyed.

[3] There is a member $\bold{i}$ such that $\bold{i}\cdot\bold{i} = -1.$ i.e.,

$\bold{i}^2=-1.\quad\quad\quad(2-1)$

Bombelli sanity checked his idea by consider any quadratic equation

$ax^2+bx+c=0.$

That is

$x^2+2hx+g=0$

where $g=\frac{c}{a}, h=\frac{b}{2a}$ which can be written as

$(x+h)^2+g-h^2=0\quad\quad\quad(2-2)$

or,

$(x+h)^2=h^2-g.$

if $h^2-g$ is positive, then it has a square root, and $-h + \sqrt{h^2-g}$ is a solution of the equation (so is the number $-h-\sqrt{h^2-g}).$ If $h^2-g$ is not positive, then $g-h^2$ is, and therefore has a square root $\sqrt{g-h^2}.$

Let

$x = -h + \sqrt{g-h^2}\cdot \bold{i}.\quad\quad\quad(2-3)$

The left side of (2-2) becomes

$(\underline{-h+\sqrt{g-h^2}\cdot \bold{i}}+h)^2+g-h^2$

$= (g-h^2)\cdot \bold{i}^2 +g-h^2$

$=(g-h^2)\cdot(\bold{i}^2+1)$

$\overset{[3]}{=}(g-h^2)\cdot (-1+1)$

$=(g-h^2)\cdot 0$

$= 0$

That is, (2-3) is a solution of (2-2).

Bombelli was elated as it suggested that by considering his set which contains the real numbers and $\bold{i},$ all quadratic equations have solutions!

It also gave him much needed confidence in showing what $\sqrt[3]{2+\sqrt{-121}} - \sqrt[3]{-2+\sqrt{-121}}$ really is.

Right away, Bombelli saw

$(\sqrt{121}\cdot\bold{i})^2\overset{[1],[2]}{=}(\sqrt{121})^2\cdot\bold{i}^2 \overset{[3]}{=} 121\cdot (-1) = -121$

so he replaced $\sqrt{-121}$ in $\sqrt[3]{2+\underline{\sqrt{-121}}}$ with $\sqrt{121}\cdot\bold{i}:$

$\sqrt[3]{2+\sqrt{-121}}=\sqrt[3]{2+\sqrt{121}\cdot\bold{i}} = \sqrt[3]{2+11\cdot\bold{i}}.$

He then anticipated that the value of $\sqrt[3]{2+11\cdot\bold{i}}$ is a new type of number $a+b\cdot\bold{i}$ where $a$ and $b$ are real numbers. i.e.,

$\sqrt[3]{2+11\cdot\bold{i}} = a + b\cdot \bold{i},\quad a,b \in R.\quad\quad\quad(3-1)$

And finally, he proceeded to find $a$ and $b$ from (3-1).

As an illustration, we solve (3-1) for $a, b$ as follows:

Cubing it gives

$2+11\cdot\bold{i} = (a+b\cdot\bold{i})^3.$

Since

$(a+b\cdot \bold{i})^3 = a^3+3a^2(b\bold{i})+3a(b\bold{i})^2+(b\cdot\bold{i})^3$

$=a^3+3a^2b\bold{i}+3ab^2\bold{i}^2+b^3\bold{i}^2\bold{i}$

$\overset{[3]}{=}a^3+3a^2b\bold{i}-3ab^2-b^3\bold{i}$

$= a^3-3ab^2 + (3a^2b-b^3)\cdot\bold{i},$

this is

$2+11\cdot\bold{i} = a^3-3ab^2 + (3a^2b-b^3)\cdot\bold{i}.$

Equating similar terms on both sides yields a system of nonlinear equations:

$\begin{cases} a^3-3ab^2=2 \\ 3a^2b-b^3=11 \end{cases}$

After factoring, it becomes

$\begin{cases} a(a^2-3b^2)=2 \\ b(3a^2-b^2)=11 \end{cases}\quad\quad\quad(3-2, 3-3)$

Assuming $a$ and $b$ are both integers, then $a$ and $a^2-3b^2$ on the left side of (3-2) are two integer factors of $2.$ Since $2$ has only two factors, namely, $1$ and $2.$ If $a=1$ then from (3-1), $1\cdot(1-3b^2) =2 \implies b^2 < 0,$ a contradiction. However, $(a=2, 2(4-b^2)=2)$ yields $b=-1$ or $+1.$ While $(a=2, b=-1) \implies b(3a^2-b^2)=(-1)\cdot(3\cdot 2^2-(-1)^2)=-11$ contradicts (3-3), $(a=2,b=1)$ gives $1\cdot(3\cdot 2^2-1^2)=11.$ Therefore, $a=2, b=1$ is the solution to (3-2, 3-3). i.e.,

$\sqrt[3]{2+\sqrt{-121}} = 2+\bold{i}.\quad\quad\quad(3-4)$

It is also easy to see that (3-4) is true as follows:

$(2+\bold{i})^3=2^3+3\cdot 2^2\cdot \bold{i}+ 3\cdot 2\cdot \bold{i}^2+\bold{i}^3$

$= 8 + 12\cdot \bold{i} +6\cdot \bold{i}^2 + \bold{i}^2\cdot \bold{i}$

$\overset{[3]}{=} 8 + 12\cdot \bold{i}-6-\bold{i}$

$= 2 + 11\cdot \bold{i}$

$= 2 + \sqrt{121}\cdot \bold{i}$

$= 2+\sqrt{-121}.$

And so

$2 + \sqrt{-121} = (2+\bold{i})^3 \implies \sqrt[3]{2+\sqrt{-121}} = 2 + \bold{i}.$

Similarly, Bombelli obtained (see Exercise-1)

$\sqrt[3]{-2+\sqrt{-121}} = -2+\bold{i}.\quad\quad\quad(3-5)$

By (3-4) and (3-5) Bombelli was able to reproduce the solution to cubic $x^3-15x=4$ :

$x = \sqrt[3]{2+\sqrt{-121}}-\sqrt[3]{-2+\sqrt{-121}}=2+\bold{i}-(-2+\bold{i})=4.$

Thus, with $\bold{i}$ and the ordinary rules of real numbers’ arithmetic, Bombelli broke the mental logjam concerning negative number under the square root.

Satisfied with his work that unlocked what seemed to be an impassable barrier, Bombelli moved on without constructing his set for the new type of number in a logically unobjectionable way. The world had to wait another two hundred years for that (see “Mr. Hamilton does complex numbers”). Still, Bombelli deserves the credit for not only recognizing numbers of a new type have a role to play in algebra, but also giving $\bold{i}$ its initial impetus and now undisputed legitimacy.

Exercise-1 Show that $\sqrt[3]{-2+\sqrt{-121}} = -2+\bold{i}.$