The Artistry of Mastery

February 29, 2024February 29, 2024 / Michael Xue / Leave a comment

Evaluate $\displaystyle \int\limits_{0}^{1} \frac{\log(1+x)}{1+x^2}\;dx$

The CAS says:

However, we can do better!

Let’s begin by defining

$I(\beta) = \displaystyle\int\limits_{0}^{1}\frac{\log(1+\beta x)}{1+x^2}\;dx.$

Next, we find the derivative with respect to $\beta:$

$\frac{d}{d\beta} I(\beta)= \displaystyle\int\limits_{0}^{1}\frac{\partial}{\partial \beta}\left(\frac{\log(1+\beta x)}{1+x^2}\right)\;dx$

$= \displaystyle\int\limits_{0}^{1} \frac{x}{(1+x^2)(1+\beta x)}\;dx$

$=\frac{1}{1+\beta^2}\displaystyle\int\limits_{0}^{1}\frac{x + \beta}{1+x^2}-\frac{\beta}{1+\beta x}\;dx$

$= \frac{1}{1+\beta^2}\displaystyle\int\limits_{0}^{1}\frac{x}{1+x^2} + \frac{\beta}{1+x^2}-\frac{\beta}{1+\beta x}\;dx$

$= \frac{1}{1+\beta^2}\left(\frac{1}{2}\log(1+x^2)\bigg|_{0}^{1} + \beta \arctan(x)\bigg|_{0}^{1} - \log(1+\beta x)\bigg|_{0}^{1}\right)$

$= \frac{1}{1+\beta^2}\left(\frac{1}{2}\log(2) + \beta\cdot\frac{\pi}{4}-\log(1+\beta)\right).$

That is,

$\frac{d}{d\beta}I(\beta) = \frac{1}{2}\log(2)\cdot\frac{1}{1+\beta^2}+\frac{\pi}{4}\cdot\frac{\beta}{1+\beta^2}-\frac{\log(1+\beta x)}{1+\beta^2}.$

Integrate it from $0$ to $\beta,$

$\displaystyle\int\limits_{0}^{\beta}\frac{d}{d \beta}I(\beta)\;d\beta = \int\limits_{0}^{\beta} \left(\frac{1}{2}\log(2)\cdot\frac{1}{1+\beta^2}+\frac{\pi}{4}\cdot\frac{\beta}{1+\beta^2}-\frac{\log(1+\beta x)}{1+\beta^2}\right)\;d\beta$

gives

$\displaystyle I(\beta) - I(0) = \frac{\log(2)}{2}\int\limits_{0}^{\beta}\frac{1}{1+\beta^2} \;d\beta+ \frac{\pi}{4}\int\limits_{0}^{\beta}\frac{\beta}{1+\beta^2}\;d\beta - \int\limits_{0}^{\beta}\frac{\log(1+\beta)}{1+\beta^2}\; d\beta$

$= \displaystyle\frac{\log(2)}{2}\arctan(\beta)\bigg|_{0}^{\beta} + \frac{\pi}{8}\log(1+\beta^2)\bigg|_{0}^{\beta}-\int\limits_{0}^{\beta}\frac{\log(1+\beta)}{1+\beta^2}\;d\beta$

$\overset{\int\limits_{0}^{\beta}\frac{\log(1+\beta)}{1+\beta^2}\;d\beta =\int\limits_{0}^{\beta}\frac{\log(1+x)}{1+x^2}\;dx}{=}\displaystyle\frac{\log(2)}{2}\arctan(\beta) + \frac{\pi}{8}\log(1+\beta^2) - \int\limits_{0}^{\beta}\frac{\log(1+x)}{1+x^2}\;dx.$

Since $I(0) = \int\limits_{0}^{1}\frac{\log(1 + 0\cdot x)}{1+x^2} \;dx =\int\limits_{0}^{1}\frac{\log(1)}{1+x^2} \;dx =0,$ we have

$\displaystyle\int\limits_{0}^{1}\frac{\log(1+\beta x)}{1+x^2}\;d\beta =\frac{\log(2)}{2}\arctan(\beta) + \frac{\pi}{8}\log(1+\beta^2) - \int\limits_{0}^{\beta}\frac{\log(1+x)}{1+x^2}\;dx.$

And so,

$\displaystyle \int\limits_{0}^{1}\frac{\log(1+\beta x)}{1+x^2} \;dx + \int\limits_{0}^{\beta}\frac{\log(1+x)}{1+x^2}\;dx = \frac{\log(2)}{2}\arctan(\beta) + \frac{\pi}{8}\log(1+\beta^2).$

When $\beta = 1,$ we obtain

$\displaystyle 2 \int\limits_{0}^{1}\frac{\log(1+ x)}{1+x^2} \;dx =\frac{\log(2)}{2}\arctan(1) + \frac{\pi}{8}\log(1+1)$

which simplifies to

$\displaystyle \int\limits_{0}^{1}\frac{\log(1+x)}{1+x^2} \;dx = \frac{1}{2}\left(\frac{\log(2)}{2}\cdot \frac{\pi}{4}+ \frac{\pi}{8}\log(2)\right) = \frac{\pi}{8}\log(2).$

Thus,

$\displaystyle \int\limits_{0}^{1}\frac{\log(1+x)}{1+x^2} \;dx = \frac{\pi}{8}\log(2).$

See also “The beat of a different drum“.

Exercise-1 Explain: $\displaystyle\int\limits_{0}^{\beta}\frac{\log(1+\beta)}{1+\beta^2}\;d\beta = \int\limits_{0}^{\beta}\frac{\log(1+x)}{1+x^2}\;dx.$

A Symmetric Equality

February 20, 2024February 21, 2024 / Michael Xue / Leave a comment

Given

$(f\star g)(t) \overset{\Delta}{=}\displaystyle\int\limits_{0}^{t}f(\tau)g(t-\tau)\; d\tau,\quad\quad\quad(1)$

show that $(f\star g)(t) = (g\star f)(t).$

By definition,

$(g \star f)(t) =\displaystyle\int\limits_{0}^{t}g(\tau)f(t-\tau)\;d\tau\quad\quad\quad(2)$

Let $\eta = t-\tau,$ we have $(\tau = t-\eta, \frac{d\tau}{d\eta}=-1).$

Since $(\tau=0 \implies \eta= t, \quad\tau=t \implies \eta =0),$

$(2) = \displaystyle\int\limits_{t}^{0}g(t-\eta)f(\eta)\cdot\frac{d\tau}{d\eta}\; d\eta$

$= \displaystyle\int\limits_{t}^{0}g(t-\eta)f(\eta)\cdot (-1)\; d\eta$

$= -\displaystyle\int\limits_{t}^{0}g(t-\eta)f(\eta)\;d\eta$

$= \displaystyle\int\limits_{0}^{t} g(t-\eta)f(\eta)\;d\eta$

$= \displaystyle\int\limits_{0}^{t} f(\eta)g(t-\eta)\;d\eta$

$\overset{\eta \text{ is a dummy variable}}{=} \displaystyle\int\limits_{0}^{t} f(\tau)g(t-\tau)\;d\tau\overset{(1)}{=} (f\star g)(t).$

Solving Minimization and Maximization Problems without Calculus [3]

February 16, 2024February 17, 2024 / Michael Xue / Leave a comment

A rectangular green space is to be built on an oval open space. The sides of the rectangle are parallel to the axis of symmetry of the ellipse. How should the design be done to maximize the green space area? What is the maximum area?

To maximize the green space area, we need to find the dimensions of the rectangle that can be inscribed within the oval such that its area is maximized.

Let’s denote the semi-major axis of the ellipse as $a$ and the semi-minor axis as $b$ . The equation of the ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The sides of the rectangle will be parallel to the axes of the ellipse. Let’s assume the width of the rectangle is $2x_0$ and the height is $2y_0$ .

Fig. 1

As illustrated in Fig. 1,

$\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1 \implies y_0 = \frac{b}{a}\sqrt{a^2-x_0^2}.\quad\quad\quad(1)$

The area of the rectangle is given by $A=4 x_0 y_0$ .

To maximize $A$ , we express $y_0$ in terms of $x_0$ so that

$A = 2x_0 \cdot 2y_0 \overset{(1)}{=} 2x_0\cdot 2\frac{b}{a}\sqrt{a^2-x_0^2}=\frac{4b}{a}x_0\sqrt{a^2-x_0^2}$

This expression reveals that maximizing $A$ is equivalent to maximizing $x_0\sqrt{a^2-x_0^2}$ :

$x_0\sqrt{a^2-x_0^2}$

$= \sqrt{a^2x_0^2-x_0^4}$

$= \sqrt{-x_0^4+a^2x_0^2-(\frac{a^2}{2})^2 + (\frac{a^2}{2})^2}$

$= \sqrt{(\frac{a^2}{2})^2-\left(x_0^4-a^2x_0^2+(\frac{a^2}{2})^2\right)}$

$= \sqrt{(\frac{a^2}{2})^2-\left(x_0^2-\frac{a^2}{2}\right)^2}.\quad\quad\quad(2)$

Since $\left(x_0^2-\frac{a^2}{2}\right)^2 \ge 0$ , (2) reaches its maximum when $\left(x_0^2-\frac{a^2}{2}\right)^2 = 0$ . Consequently,

$x_0^2-\frac{a^2}{2} = 0 \implies x_0 = \frac{a}{\sqrt{2}}\implies \left(x_0\sqrt{a^2-x_0^2}\right)_{max} = \frac{a^2}{2}.$

Thus,

$A_{max} = \frac{4b}{a}\cdot \left(x_0\sqrt{a^2-x_0^2}\right)_{max}=\frac{4b}{a}\cdot \frac{a^2}{2} = 2ab.$

Exercise-1 Maximizing $x_0\sqrt{a-x_0^2}$ using the theorem stated in “Solving Kepler’s Wine Barrel Problem without Calculus” (hint: $x_0\sqrt{a^2-x_0^2} = (x_0^2)^\frac{1}{2}(a^2-x_0^2)^\frac{1}{2}$ and $x_0^2+(a^2-x_0^2)=a^2$ , a constant)

A Stroll in the Woods

February 12, 2024February 13, 2024 / Michael Xue / Leave a comment

Solve $z^4=(z-1)^4$ for $z \in \mathbb{C}$

Solution-1

We define $\sqrt[n]{X}$ first:

$\sqrt[n]{X} : Y \in \mathbb{C} \ni Y^n = X.\quad\quad\quad(1)$

When $n=4,$ (1) becomes

$\sqrt[4]{X} : Y \in \mathbb{C} \ni Y^4=X\quad\quad\quad(2)$

Since $(z)^4 = z^4, (-z)^4 = (-1)^4z^4=z^4, (iz)^4 = i^4z^4 = (i^2)^2 z^4 = (-1)^2z^4=z^4, (-iz)^4 = (-i)^4z^4 = i^4z^4 = z^4,$

we have (by (2))

$\sqrt[4]{z^4} = z$ or $-z$ or $iz$ or $-iz.\quad\quad\quad(3)$

Similarly,

$\sqrt[4]{(z-1)^4} = z-1$ or $-(z-1)$ or $i(z-1)$ or $-i(z-1).\quad\quad\quad(4)$

From observing that

$z^4 = (z-1)^4 \implies \sqrt[4]{z^4} = \sqrt[4]{(z-1)^4},$

we now proceed to solve $z^4=(z-1)^4$ for $z \in \mathbb{C}$ by examining the following sixteen cases:

[1] $z = z-1 \implies 0=-1$ dismissed!

[2] $z =-(z-1) \implies 2z = 1 \implies z = \frac{1}{2}$

[3] $z=i(z-1) \implies z=\frac{-i}{1-i} = \frac{1-i}{2}$

[4] $z=-i(z-1) \implies z=\frac{i}{1+i}=\frac{1+i}{2}$

[5] $-z=z-1 \implies 2z=1 \implies z = \frac{1}{2}$

[6] $-z=-(z-1) \implies 0 = 1$ dismissed!

[7] $-z=i(z-1) \implies z = \frac{i}{1+i} = \frac{1+i}{2}$

[8] $-z = -i(z-1) \implies z = \frac{-i}{1-i}=\frac{1-i}{2}$

[9] $iz =z-1 \implies z=\frac{1}{1-i} = \frac{1+i}{2}$

[10] $iz = i(z-1) \implies 0 = -1$ dismissed!

[11] $iz=-i(z-1) \implies 2iz = i \implies z = \frac{1}{2}$

[12] $-iz = z-1 \implies z=\frac{1}{1+i} = \frac{1-i}{2}$

[13] $-iz = z-1 \implies z=\frac{1}{1+i}=\frac{1-i}{2}$

[14] $-iz = -(z-1) \implies z=\frac{1}{1-i} = \frac{1+i}{2}$

[15] $-iz = i(z-1) \implies 2z=1 \implies z=\frac{1}{2}$

[16] $-iz = -i(z-1) \implies -iz=-iz + i \implies 0 = i$ dismissed!

Hence,

$z=\frac{1}{2}$ or $\frac{1-i}{2}$ or $\frac{1+i}{2}.$

The number of cases can be reduced if we express $z^4=(z-1)^4$ as

$(z^2)^2 = ((z-1)^2)^2$

(see Exercise-1)

Solution-2

$z^4=(z-1)^4 \implies (\frac{z-1}{z})^4=1 \implies \left((\frac{z-1}{z})^2\right)^2= 1.$

This leads to

$(\frac{z-1}{z})^2 = 1 \implies (z-1)^2=z^2 \implies z^2-2z+1=z^2 \implies z=\frac{1}{2}.$

Or,

$(\frac{z-1}{z})^2 =-1 \implies (z-1)^2=-z^2 \implies z^2-2z+1=-z^2\implies 2z^2-2z+1=0$

$\implies z=\frac{1 - i}{2}$ or $\frac{1+i}{2}$

Solution-3

$z^4-(z-1)^4 =0$

$\implies ((z^2+(z-1)^2)(z^2-(z-1)^2) = 0$

$\implies (2z^2-2z+1)(z+z-1)(z-z+1)=0$

$\implies (2z^2-2z+1)(2z-1)=0$

$\implies z = \frac{1}{2}$ or $\frac{1-i}{2}$ or $\frac{1+i}{2}$

Solution-4

$z^4 = \underbrace{z^4-4z^3+6z^2-4z+1}_{(z-1)^4 \; \text{expanded}}$

$\implies 4z^3-6z^2+4z-1=0$

$\implies 4z^3\underbrace{-2z^2-4z^2}_{-6z^2}+4z-1=0$

$\implies 2z^2(2z-1)-(2z-1)^2=0$

$\implies (2z-1)(2z^2-2z+1)=0$

$\implies z=\frac{1}{2}$ or $\frac{1-i}{2}$ or $\frac{1+i}{2}$

Solution-5

Exercise-1 Solve $(z^2)^2=((z-1)^2)^2$ (hint: $\sqrt{X^2}=X$ or $-X$ )

Prove it analytically

February 8, 2024February 9, 2024 / Michael Xue / Leave a comment

Given a circle with diameter AB, let C be any point on the circle. If a perpendicular line CD is drawn from point C to diameter AB, and the tangent lines at C and A intersect at point Q, Prove that the intersection point E of lines CD and BQ is the midpoint of segment CD.

Fig. 1 $A(-r, 0), B(r, 0), C(x_1, y_1), D(x_1, 0), E(x_1, y_E), Q(-r, y_Q),$

As depicted in Fig. 1, the tangent line passing through $C(x_1, y_1)$ is

$x_1x + y_1y = r^2$

(see “Constructing the tangent line of circle without calculus” ) and,

$x_1^2+y_1^2=r^2 \implies r^2 - x_1^2 = y_1^2\quad\quad\quad(1)$

At $Q(-r, y_Q)$ where the tangent lines intersect,

$x_1(-r) +y_1 y_Q = r^2 \implies y_Q = \frac{r^2+r x_1}{y_1}.$

The slope of line $BQ,$ denoted by $k_{BQ}$ , is calculated as follows:

$k_{BQ} = \frac{y_Q-0}{-r-r} = \frac{y_Q}{-2r} =\frac{1}{-2r}\cdot\frac{r(r+x_1)}{y_1} = -\frac{r+x_1}{2y_1}.$

Subsequently, the line $BQ$ can be represented by

$y = k_{BQ}x+b = -\frac{r+x_1}{2y_1}x + b.$

Since $B(r, 0)$ is on line $BQ$ , we have

$0 = -\frac{(r+x_1)}{2y_1}\cdot r+b \implies b=\frac{(r+x_1)r}{2y_1}.$

Hence, the equation of line $BQ$ is

$y = -\frac{(r+x_1)}{2y_1}x + \frac{(r+x_1)r}{2y_1} = \frac{(r-x)(r+x_1)}{2y_1}.$

Given that $E(x_1, y_E)$ lies on line $BQ,$ we deduce that

$y_E = \frac{(r-x_1)(r+x_1)}{2y_1}=\frac{r^2-x_1^2}{2y_1} \overset{(1)}{=} \frac{y_1^2}{2y_1}= \frac{y_1}{2}.$

From Observation to Affirmation

February 7, 2024February 7, 2024 / Michael Xue / Leave a comment

A triangle possesses three altitudes, each of which connects a side to the opposite vertex perpendicularly. In an acute triangle, all three segments reside entirely within the triangle’s confines. However, in the case of an obtuse triangle, the altitudes from the acute angles intersect only the extensions of the opposite sides extended from the vertex of the obtuse angle, lying completely outside the triangle.

Case-1 $\Delta ABC$ is acute

Assume there exists an altitude AD outside the triangle:

Fig. 1 $AD \perp BD \;(\alpha_6=\frac{\pi}{2})$

From Fig. 1, it is evident that

$\alpha_5 = \pi - \alpha_3 \overset{\Delta ABC \text{is acute:\;\;} \alpha_3 < \frac{\pi}{2}}{\implies} \alpha_5 > \frac{\pi}{2}.\quad\quad\quad(1)$

In $\Delta ACD$ , it follows that

$\alpha_4 + \alpha_5 + \alpha_6 \overset{\alpha_6 = \frac{\pi}{2}, (1)}{>} \alpha_4 + \frac{\pi}{2} + \frac{\pi}{2} = \alpha_4 + \pi > \pi \implies \alpha_4 + \alpha_5 + \alpha_6 > \pi,$

contradicting $\alpha_4+\alpha_5 + \alpha_6 = \pi.$

Case-2 $\Delta ABC$ is obtuse

If an altitude $AD$ exists within the triangle:

Fig. 2 $AD \perp BC \; (\alpha_5 = \frac{\pi}{2})$

Then

$\alpha_5 = \alpha_2 + \alpha_4 \overset{\Delta ACD \text{ is obtuse:}\;\; \alpha_2>\frac{\pi}{2}}{\implies} \alpha_5 >\frac{\pi}{2},$

contradicting $AD \perp BC.$

Similarly, if there were an altitude outside the triangle connects to the extension of opposite side not extended from the vertex of the obtuse angle:

Fig. 3 $AD \perp BD \; (\alpha_6 = \frac{\pi}{2})$

We would have

$\alpha_4+ \alpha_5 + \alpha_6 \overset{\alpha_5 = \alpha_1 + \alpha_2, \alpha_6=\frac{\pi}{2}}{=} \alpha_4 + \alpha_1 + \alpha_2 + \frac{\pi}{2} \overset{\alpha_2 > \frac{\pi}{2}}{>}\alpha_4 + \alpha_1 + \frac{\pi}{2} + \frac{\pi}{2} = \alpha_4 + \alpha_2 +\pi >\pi,$