Integral: The CAS & I

March 31, 2022September 27, 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, 2022January 11, 2024 / 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 $k_1$ and $k_3$ denote the slopes of the lines connecting points $(0, a)$ to $(x_1, y_1)$ and $(x_3, y_3)$ 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, 2022October 6, 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, 2022January 11, 2024 / 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 areas 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, 2022March 2, 2023 / 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

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

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,

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

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, 2022October 6, 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“),