KindiakMath

Category: Uncategorized

Substitution Magic

Big Idea

The substitution $u = g(x)$ allows us to simplify integrals:

$\displaystyle \int f'(g(x)) \cdot g'(x)\, \mathrm dx = \int f(u)\, \mathrm du.$

In particular, we recover two useful integration forms:

$\displaystyle \int f'(ax+b)\, \mathrm dx = \frac 1a \cdot f'(ax+b) +C,\quad \int \frac{f'(x)}{f(x)}\, \mathrm dx = \ln|f(x)| +C.$

Questions

Question 1. Evaluate the integral $\displaystyle \int xe^{-x^2}\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $u = -x^2 \Rightarrow \mathrm du = -2x\, \mathrm dx$ ,

$\begin{aligned} \int xe^{-x^2}\, \mathrm dx &= -\frac 12 \int e^{-x^2} \cdot (-2x)\, \mathrm dx\\ &= -\frac 12 \int e^u\, \mathrm du \\ &= -\frac 12 \cdot e^u + C \\ &= -\frac 12 \cdot e^{-x^2} + C. \end{aligned}$

Question 2. Evaluate the integral

$\displaystyle \int \sin(\cos^2(x)) \cos(1 - \sin^2(x)) \sin(2x)\, \mathrm dx.$

(Click for Solution)

Solution. Make the substitution $u = \cos^2(x) = 1-\sin^2(x) = \frac 12(1 + \cos 2x)$ so that

$\displaystyle \frac{ \mathrm du }{\mathrm dx} = \frac 12 (0 + (-\sin(2x)) \cdot 2) = - \sin (2x) \quad \Rightarrow \quad \mathrm du = - \sin(2x)\, \mathrm dx.$

Hence,

$\begin{aligned} \int \sin(\cos^2(x)) \cos(1 - \sin^2(x)) \sin(2x)\, \mathrm dx &= \int \sin(u) \cos(u) \cdot (-1)\, \mathrm du \\ &= -\frac 12 \int 2 \sin(u) \cos(u) \, \mathrm du \\ &= -\frac 12 \int \sin(2u) \, \mathrm du \\ &= -\frac 12 \cdot \frac 12 \cdot (-\cos(2u)) + C \\ &= \frac 14 \cdot \cos(2u) + C \\ &= \frac 14 \cdot \cos(2 \cos^2(x)) + C.\end{aligned}$

Question 3. Evaluate the integral $\displaystyle \int x^x(\ln x + 1)\, \mathrm dx$ .

(Click for Solution)

Solution. Using the laws of exponents $a^b = e^{b \ln a}$ ,

$\displaystyle \int x^x(\ln x + 1)\, \mathrm dx = \int e^{x \ln x }(\ln x + 1)\, \mathrm dx.$

Then make the substitution $u = x \ln x$ so that

$\begin{aligned} \frac{\mathrm d u}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx} (x \ln x) \\ &= \ln x +1 \\ \mathrm du &= (\ln x + 1)\, \mathrm dx.\end{aligned}$

Since $e^u = e^{x \ln x} = x^x$ ,

$\displaystyle \int x^x(\ln x + 1)\, \mathrm dx = \int e^u\, \mathrm du = e^u + C = x^x + C.$

—Joel Kindiak, 4 Sept 25, 1644H

January 1, 2026
Tedious Integration

Big Idea

We regard integration, at the pre-university level, as the reverse process of differentiation:

$\displaystyle F(x) = \int f(x)\, \mathrm dx \quad \iff \quad \frac{\mathrm d}{\mathrm dx} (F(x)) = f(x).$

We call $f(x)$ the integrand. Integration is linear in the following sense:

$\displaystyle \int (af(x) + bg(x))\, \mathrm dx = a \int f(x)\, \mathrm dx + b \int g(x)\, \mathrm dx.$

The integrals of powers requires a bit more care: if $n \neq -1$ then nothing weird happens:

$\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} +C.$

However, if $n = -1$ , then we cannot divide by zero, and instead recover the logarithm:

$\displaystyle \int x^{-1}\, \mathrm dx = \int \frac 1x \, \mathrm dx = \ln|x| + C.$

Other commonly used integrals are free for use. Furthermore, in the context of pre-university mathematics, the definite integral is an application of the vanilla integral:

$\displaystyle \int_a^b f(x)\, \mathrm dx := F(b) - F(a) \equiv [F(x)]_a^b.$

Questions

You may not use integration by substitution in any of these problems.

Question 1. Evaluate $\displaystyle \int (3^{2x} + \sqrt{25x} - \ln(x^7))\, \mathrm dx$ .

(Click for Solution)

Solution. Simplifying the integrand then integrating term-wise,

$\begin{aligned} \int (3^{2x} + \sqrt{25x} - \ln(x^7))\, \mathrm dx &= \int (9^x + 5x^{1/2} - 7 \ln(x))\, \mathrm dx \\ &= \int 9^x\, \mathrm dx + 5 \int x^{1/2}\, \mathrm dx - 7 \int \ln(x)\, \mathrm dx \\ &= \frac{9^x}{\ln(9)} + 5 \cdot \frac {x^{3/2}}{3/2} - 7(x \ln(x) - x) + C \\ &= \frac {1}{\ln(9)} \cdot 9^x+ \frac{10}3 \cdot x^{3/2} - 7x \ln(x) + 7x + C. \end{aligned}$

Question 2. Evaluate $\displaystyle \int (2+3x)^4\, \mathrm dx$ .

(Click for Solution)

Solution. We first slowly expand the integrand:

$\begin{aligned} (2+3x)^4 &= ( (2+3x)^2 )^2 \\ &= ( 4 + 12x + 9x^2 )^2 \\ &= 16 + 96x + 216x^2 + 216x^3 + 81x^4. \end{aligned}$

Using linearity to integrate term-wise,

$\begin{aligned} \int (2+3x)^4 \, \mathrm dx &= \int (16 + 96x + 216x^2 + 216x^3 + 81x^4)\, \mathrm dx \\ &= \int 16\, \mathrm dx + 96 \int x^1\, \mathrm dx + 216 \int x^2\, \mathrm dx + 216 \int x^3\, \mathrm dx + 81 \int x^4\, \mathrm dx \\ &= 16x + 96 \cdot \frac{x^2}{2} + 216 \cdot \frac{x^3}{3} + 216 \cdot \frac{x^4}{4} + 81 \cdot \frac{x^5}{5} + C \\ &= 16x + 48x^2 + 72x^3 + 54 x^4 + \frac{81}{5} \cdot x^5 + C. \end{aligned}$

Question 3. Evaluate $\displaystyle \int \left( \frac 2x - 7 \sqrt x \right)^2\, \mathrm dx$ .

(Click for Solution)

Solution. We first slowly expand the integrand:

$\begin{aligned} \left( \frac 2x - 7 \sqrt x \right)^2 &= \left( \frac 2x \right)^2 - 2 \cdot \frac 2x \cdot 7\sqrt x + (7 \sqrt x)^2 \\ &= \frac 4{x^2} - 28x^{-1/2} + 49x \\ &= 4x^{-2}- 28x^{-1/2} + 49x^1. \end{aligned}$

Using linearity to integrate term-wise,

$\begin{aligned} \int \left( \frac 2x - 7 \sqrt x \right)^2\, \mathrm dx &= \int (4x^{-2}- 28x^{-1/2} + 49x^1)\, \mathrm dx \\ &= 4 \int x^{-2}\, \mathrm dx - 28 \int x^{-1/2}\, \mathrm dx + 49 \int x^1\, \mathrm dx \\ &= 4 \cdot \frac{x^{-1}}{-1} - 28 \cdot \frac{x^{1/2}}{1/2} + 49 \cdot \frac{x^2}{2} + C \\ &= - 4x^{-1}- 56 x^{1/2} + \frac{49}{2} \cdot x^2 + C. \end{aligned}$

Remark 1. In general, when feasible, final answers ought to take the form

$\displaystyle c_1 \cdot f_1(x) + c_2 \cdot f_2(x) + \cdots + c_n \cdot f_n(x) + C,$

where $c_1,\dots, c_n$ are numerical constants, $f_1(x),\dots, f_n(x)$ are expressions in terms of $x$ , and $C$ denotes the arbitrary constant of integration (if and only if needed), for easy recognition.

—Joel Kindiak, 4 Sept 25, 1334H

December 4, 2025
Optimised Geometry
Big Idea

To maximise (or minimise) a quantity $y\equiv y(x)$ , we apply the zero derivative condition

$\displaystyle \frac{\mathrm dy}{\mathrm dx} \equiv y'(x) = 0,$

and check that the corresponding value $x = c$ yields to a maximum (resp. minimum) via the second derivative test:
- a local maximum is obtained at $c$ when $y''(c) <0$ ,
- a local minimum is obtained at $c$ when $y''(c) > 0$ .
Recall that we denote

$\displaystyle y'' \equiv y''(x) := (y')'(x) \equiv \frac{\mathrm d}{\mathrm dx}(y'(x)) = \frac{\mathrm d}{\mathrm dx}\left(\frac{\mathrm dy}{\mathrm dx}\right) =: \frac{\mathrm d^2 y}{\mathrm dx^2}$

for brevity.

Questions

Question 1. Evaluate the area of the largest rectangle contained inside the ellipse graphed below.

(Click for Solution)

Solution. Given $x, y > 0$ , the base of the rectangle is $2x$ and its height is $2y$ , yielding a total area of $A = 4xy$ . Differentiating with respect to $x$ ,

$\displaystyle \frac{\mathrm dA}{\mathrm dx} = 4y + 4x \cdot \frac{\mathrm dy}{dx}.$

On the other hand, given the ellipse $\displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1$ , differentiate with respect to $x$ on both sides to obtain

$\displaystyle \frac{2x}{9} + \frac{2y}{4} \cdot \frac{\mathrm dy}{\mathrm dx} = 0 \quad \Rightarrow \quad \frac{\mathrm dy}{\mathrm dx} = -\frac{4x}{9y}.$

Therefore,

$\displaystyle \frac{\mathrm dA}{\mathrm dx} = 4y + 4x \cdot \frac{\mathrm dy}{dx} = 4y - \frac{16x^2}{9y}.$

By the zero-derivative condition, if $A$ is maximised, then $\displaystyle \frac{\mathrm dA}{\mathrm dx} = 0$ :

$\displaystyle 4y - \frac{16x^2}{9y} = 0 \quad \Rightarrow \quad y^2 = \frac{4}{9}x^2.$

Plugging back into the equation of the ellipse,

$\begin{aligned} \frac{x^2}{9} + \frac{1}{4} \cdot \frac{4}{9}x^2 &= 1 \\ x^2 &= 18\\ x &= \sqrt{18} \\ &= 3\sqrt{2}, \end{aligned}$

since $x > 0$ , yielding $y = \frac 23 x = \frac 23 \cdot 3\sqrt 2 = 2\sqrt 2$ . Thus, the area of the corresponding rectangle is $A = 4 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = 12\, \text{units}^2$ .

To apply second derivative test, we differentiate $A$ again:

$\begin{aligned} \frac{\mathrm d^2 A}{\mathrm dx^2} &= 4 \cdot \frac{\mathrm dy}{\mathrm dx} - \frac{16 \cdot 2x \cdot 9y - 16x^2 \cdot \frac{\mathrm dy}{\mathrm dx}}{81y^2} \\ &= 4 \cdot 0 - \frac{16 \cdot 2x \cdot 9y - 16x^2 \cdot 0}{81y^2} \\ &= - \frac{32x }{9y} < 0,\end{aligned}$

so $A$ is indeed maximised when $x = 3\sqrt 2$ and $y = 2\sqrt 2$ .

Question 2. The diagram below shows a movie theatre with a screen $9\, \text{m}$ tall, whose base is $4\, \text{m}$ above the ground.

The permissible viewing area is $50\, \text{m}$ long with an incline of up to $6\, \text{m}$ . Determine the position that a movie-goer $P$ should sit in order to maximise his viewing angle $\theta$ , justifying your answer.

(Click for Solution)

Solution. Let $x$ denote the horizontal distance of the movie-goer from the screen. His height $y$ is then determined using similar triangles:

$\displaystyle \frac{y}{x} = \frac6{50} = \frac 3{25} \quad \Rightarrow \quad y = \frac{3}{25}x.$

By observation,

$\begin{aligned} \theta &= \tan^{-1} \left( \frac{13-y}{x} \right) - \tan^{-1} \left( \frac{4-y}{x} \right) \\ &= \tan^{-1} \left( \frac{13}{x} - \frac 3{25} \right) - \tan^{-1} \left( \frac{4}{x} - \frac 3{25} \right). \end{aligned}$

We remark that if $y > 4$ , then $4-y < 0$ and the formula still works. Differentiating with respect to $x$ ,

$\begin{aligned} \frac{\mathrm d\theta}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx}\tan^{-1} \left( \frac{13}{x} - \frac 3{25} \right) - \frac{\mathrm d}{\mathrm dx}\tan^{-1} \left( \frac{4}{x} - \frac 3{25} \right) \\ &= \frac{1}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} \left( -\frac{13}{x^2} \right) - \frac{1}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} \left( -\frac{4}{x^2} \right) \\ &= -\frac 1{x^2} \left( \frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} - \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} \right), \\ x^2 \frac{\mathrm d\theta}{\mathrm dx} &= \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} -\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2}. \end{aligned}$

By the zero-derivative condition, if $\theta$ is maximised, then $\frac{\mathrm d\theta}{\mathrm dx} = 0$ :

$\begin{aligned}x^2 \cdot 0 &= \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} -\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2}. \end{aligned}$

We need to (painfully) solve this equation:

$\begin{aligned} \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} &=\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} \\ 4 \cdot \left( 1 + \left( \frac{13}{x} - \frac 3{25} \right)^2 \right) &=13 \cdot \left( 1 + \left( \frac{4}{x} - \frac 3{25} \right)^2 \right) \\ 4 \cdot \left( 1 + \frac{169}{x^2} -\frac{78}{25x} + \frac{9}{625} \right) &=13 \cdot \left( 1 + \frac{16}{x^2} -\frac{24}{25x} + \frac{9}{625} \right) \\ 4 + \frac{676}{x^2} -\frac{312}{25x} + \frac{36}{625} &= 13 + \frac{208}{x^2} -\frac{312}{25x}+ \frac{117}{625} \\ \frac{468}{x^2} &= \frac{5706}{625} \\ x^2 &= \frac{16\, 250}{317} \\ x &= \sqrt{\frac{16\, 250}{317}} \approx 7.160\, \text{m}, \end{aligned}$

since $x > 0$ .

To apply second derivative test, we differentiate again:

$\begin{aligned} 2x \frac{\mathrm d\theta}{\mathrm dx} + x^2 \frac{\mathrm d^2\theta}{\mathrm dx^2} &= \frac{\mathrm d}{\mathrm dx} \left( \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} \right) -\frac{\mathrm d}{\mathrm dx} \left( \frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} \right) \\ &= -\frac {4 \cdot 2\left( \frac{4}{x} - \frac 3{25} \right) \cdot -\frac{4}{x^2}}{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} + \frac {13 \cdot 2\left( \frac{13}{x} - \frac 3{25} \right) \cdot -\frac{13}{x^2}}{\left(1 + \left( \frac{13}{x} - \frac 3{25} \right)^2\right)^2} \\ &= \frac 2{x^2} \cdot \left( \frac {16 \cdot \left( \frac{4}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} - \frac {169 \cdot \left( \frac{13}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{13}{x} - \frac 3{25} \right)^2\right)^2} \right).\end{aligned}$

When $\displaystyle \frac{\mathrm d\theta}{\mathrm dx} = 0$ , $\displaystyle \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} =\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2}$ , so that

$\begin{aligned} 2x \cdot 0 + x^2 \frac{\mathrm d^2\theta}{\mathrm dx^2} &= \frac 2{x^2} \cdot \left( \frac {16 \cdot \left( \frac{4}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} - \frac {16 \cdot \left( \frac{13}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} \right) \\ &= \frac 2{x^2} \cdot \frac {16}{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} \cdot \left( \left( \frac{4}{x} - \frac 3{25} \right) - \left( \frac{13}{x} - \frac 3{25} \right) \right) \\ &= \frac 2{x^2} \cdot \frac {16}{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} \cdot -\frac 9x < 0, \end{aligned}$

so indeed $\theta$ will attain its maximum at $x = \sqrt{\frac{16\, 250}{317}}$ .

Question 3. Determine the smallest perimeter of a rectangle with area $1$ .

(Click for Solution)

Solution. Let $x, y$ denote the base and height of the rectangle respectively. Since $xy = 1$ implies $y = 1/x$ , the perimeter $P$ of the rectangle is given by

$\displaystyle P(x) = 2x + \frac 2x.$

Differentiating twice,

$\displaystyle P'(x) = 2 - \frac 2{x^2},\quad P''(x) = \frac 2{x^2} > 0.$

Solving $P'(x) = 0$ yields $x = 1$ so that $y = 1$ . Since $P''(1) > 0$ automatically, $x = 1$ yields a (local) minimum for $P$ . Hence, the rectangle has a minimum perimeter of $P(1) = 4$ , which corresponds to the rectangle being a square with base length $1$ .

Remark 1. The result remains true even if we consider rectangles of other areas; the rectangle with area $A$ whose perimeter is minimum must be a square with side length $\sqrt{A}$ .

Question 4. Given that $0 < x < 1$ , maximise $x^2(1-x)^3$ .

(Click for Solution)

Solution. Define $f(x) := x^2(1-x)^3$ . Differentiating twice,

$\begin{aligned} f'(x) &= 2x \cdot (1-x)^3 + x^2 \cdot 3(1-x)^2 \cdot (-1) \\ &= 2x \cdot (1-x)^3 - x^2 \cdot 3(1-x)^2 \\ &= x \cdot (1-x)^2 \cdot (2-5x), \\ f''(x) &= (1-x)^2\cdot (2-5x) + x \cdot 2(1-x) \cdot(-1) \cdot (2-5x) \\ &\phantom{==} + x \cdot (1-x)^2 \cdot (-5) \\ &= (1-x)^2\cdot (2-5x) - 2x(1-x) \cdot (2-5x) - 5x \cdot (1-x)^2. \end{aligned}$

Solving $f'(x) = 0$ , since $0 < x < 1$ , $2-5x = 0 \Rightarrow x = 2/5$ . Substituting into $f''$ ,

$f''(2/5) = 0 - 0 - 5 \cdot 2/5 \cdot (1-2/5)^2 < 0,$

so that $x = 2/5$ yields a local (global) maximum. Hence, $f$ has a maximum value of

$\displaystyle \left(\frac 25 \right)^2 \left(1 - \frac 25 \right)^3 = \frac{2^2 \cdot 3^3}{5^5} = \frac{108}{3125} = 0.03456.$

Remark 2. Using the same technique, we can prove that for any $0 < s < r$ and $s < x < r$ ,

$\displaystyle 0 < (x-s)^m (r-x)^n \leq m^m n^n \cdot \frac{(r-s)^{m+n}}{(m+n)^{m+n}}.$

with equality at $x = (rm+sn)/(m+n)$ . Setting $(s,r,m,n) = (0,1,2,3)$ yields Question 4.

Question 5. Given that $x > 0$ , find the value of $x$ that minimises $\displaystyle 6x^3 + 7/x^5$ .

(Click for Solution)

Solution. Define $f(x) := 6x^3 + 7/x^5 = 6x^3 + 7x^{-5}$ . Differentiating twice,

$\begin{aligned} f'(x) &= 18x^2 - 35x^{-6}, \\ f''(x) &= 36x + 210x^{-7} > 0. \end{aligned}$

Therefore, $f$ is minimised when $f'(x) = 0$ :

$\displaystyle 18x^2 - 35x^{-6} = 0 \quad \Rightarrow \quad x^8 = \frac{35}{18} \quad \Rightarrow \quad x = \sqrt[8]{ \frac{ 35 }{ 18 } },$

since $x > 0$ .

Remark 3. In general, the value of $x > 0$ that minimises $Ax^m + B/x^n$ is given by $x^{m+n} = (Bn)/(Am)$ .

—Joel Kindiak, 4 Sept 25, 1213H
November 27, 2025
Common Continuous Probabilities

Having ventured far into the measure-theoretic world and taking great pains to verify that taking the length $\lambda([a,b]) = b-a$ of a compact interval $[a,b]$ is mathematically legitimate, we turn our attention to modelling continuous data. If the uniform counting measure considered the “basic” distribution in discrete probability, then the uniform continuous distribution would naturally be considered the “basic” distribution in continuous probability.

Definition 1. Let $\Omega$ be a sample space. The uniform continuous distribution on $[a, b]$ is the random variable $X : \Omega \to \mathbb R$ with a distribution defined as follows: for any measurable $K \subseteq [a, b]$ ,

$\displaystyle \begin{aligned} \mathbb P_X(K) &= \frac{1}{b-a} \cdot \int_K \mathbb I_{[a, b]} \, \mathrm d\lambda \\ &\equiv \frac{1}{b-a} \cdot \int_a^b \mathbb I_K(x) \, \mathrm dx. \end{aligned}$

Here, the probability density function is given by $\displaystyle f_X = \frac{1}{b-a} \cdot \mathbb I_{[a,b]}$ . Henceforth also we will disregard the sample space unless we need to undertake some theoretic construction.

As usual, $\lambda$ here denotes the Lebesgue measure on $\mathbb R$ , and in this case, we write $X \sim \mathcal U(a, b)$ .

The expected value and variance tends to be useful statistical quantities for our purposes. Nevertheless, there is a special expected value and a special variance that causes the rest of our calculations to become trivial.

Lemma 1. Suppose $\mathbb E[X] = \mu$ and $\mathrm{Var}(X) = \sigma^2$ . Define the standardised random variable $\tilde X$ by

$\displaystyle \tilde X := \frac{X - \mu}{\sigma}.$

Then $\mathbb E[\tilde X] = 0$ and $\mathrm{Var}(\tilde X) = 1$ . Conversely, for any $X = a + b \cdot U$ with $b \neq 0$ , if the p.d.f. $f_U$ of $U$ is continuous, then the p.d.f. $f_X$ of $X$ can be written in terms of the p.d.f. of $U$ :

$\displaystyle f_X(x) = \frac 1b \cdot f_U\left( \frac{x-a}{b} \right).$

Furthermore, $\mathbb E[X] = a + b \cdot \mathbb E[U]$ and $\mathrm{Var}(X) = b^2 \cdot \mathrm{Var}(U)$ .

Proof. For the p.d.f. result, by the fundamental theorem of calculus,

$\begin{aligned} f_X(x) &= \frac{\mathrm d}{\mathrm dx} \int_{-\infty}^x f_X(t)\, \mathrm dt \\ &= \frac{\mathrm d}{\mathrm dx} (\mathbb P(X \leq x)) \\ &= \frac{\mathrm d}{\mathrm dx} (\mathbb P(a + b \cdot U \leq x)) \\ &= \frac{\mathrm d}{\mathrm dx} \left( \mathbb P\left( U \leq \frac{x - a}{b} \right) \right) \\ &= \frac{\mathrm d}{\mathrm dx} \int_{-\infty}^{(x-a)/b} f_U(t)\, \mathrm dt \\ &= \frac 1b \cdot f_U\left( \frac{x-a}{b} \right). \end{aligned}$

The other results follow from expectation and variance properties.

Theorem 1. If $X \sim \mathcal U(a, b)$ , then $\mathbb E[X] = (a+b)/2$ and $\mathrm{Var}(X) = (b-a)^2/12$ .

Proof. Let $U \sim \mathcal U(0, 1)$ for simplicity, so that

$\displaystyle f_X(x) = \frac{1}{b-a}\cdot \mathbb I_{[a, b]}(x) = \mathbb I_{[0, 1]} \left( \frac{x-a}{b-a} \right) \cdot \frac{1}{b-a}$

implies that

$X = a + (b-a) \cdot U.$

By definition,

$\begin{aligned} \mathbb E[U] &= \int_{-\infty}^{\infty} x \cdot \mathbb I_{[0,1]}(x)\, \mathrm dx = \int_{0}^{1} x\, \mathrm dx =\frac 12.\end{aligned}$

Similarly,

$\begin{aligned} \mathbb E[U^2] &= \int_{0}^{1} x^2\, \mathrm dx = \frac 13,\end{aligned}$

so that

$\displaystyle \mathrm{Var}(U) = \mathbb E[U^2] - \mathbb E[U]^2 = \frac 13 - \left(\frac 14\right)^2 = \frac 1{12}.$

Then the general case follows from

$\begin{aligned} \mathbb E[X] &= a + (b-a) \cdot \frac 12 = \frac{a+b}{2}, \\ \mathrm{Var}(X) &= (b-a)^2 \cdot \frac 1{12} = \frac{(b-a)^2}{12}. \end{aligned}$

We can generalise Lemma 1 to any differentiable, invertible transformation of $U$ .

Lemma 2. For any $X = g(U)$ where $g$ is a invertible differentiable transformation, if the p.d.f. $f_U$ of $U$ is continuous, then the p.d.f. $f_X$ of $X$ can be written in terms of the p.d.f. of $U$ :

$\displaystyle f_X(x) = f_U (g^{-1}(x)) \cdot (g^{-1})'(x).$

Proof. Apply the same arguments as in Lemma 1, but apply the chain rule.

The continuous analog of the geometric distribution would be the exponential distribution.

Theorem 2. The random variable $X$ follows an exponential distribution with rate parameter $\lambda > 0$ , denoted $X \sim \mathrm{Exp}(\lambda)$ , if is defined by the probability density function

$f_X(x) = \lambda e^{-\lambda x} \cdot \mathbb I_{[0,\infty)}.$

$\mathbb E[X] = 1/\lambda$ and $\mathrm{Var}(X) = 1/\lambda^2$ .

Proof. Let $U \sim \mathrm{Exp}(1)$ for simplicity, so that

$\displaystyle f_X(x) = f_U(\lambda x) \cdot \lambda \quad \Rightarrow \quad X = \frac 1\lambda \cdot U$

gives the final result. Integrating by parts,

$\begin{aligned}\mathbb E[U] &= \int_0^\infty xe^{-x}\, \mathrm dx \\ &= [-xe^{-x} - e^{-x}]_0^\infty \\ &= (0 - 0) - (0 - 1) = 1, \\ \mathbb E[U^2] &= \int_0^\infty x^2 e^{-x}\, \mathrm dx \\ &= \left[-x^2 e^{-x} \right]_0^\infty + 2\int_0^\infty x \cdot e^{-x}\, \mathrm dx \\ &= (0 - 0) + 2 \cdot \mathbb E[U] = 2,\end{aligned}$

so that $\mathrm{Var}(U) = \mathbb E[U^2] - \mathbb E[U]^2 = 2- 1^2 = 1$ .

Another crucial distribution in probability and statistics is the normal distribution. It is a useful model to represent all kings of continuous data—heights of humans, test scores for exams, and even month-on-month returns on investment. Actually more is true—as long as we have a continuous nonzero function $f$ that is integrable, we obtain a corresponding probability distribution.

Lemma 3. For any continuous and integrable nonzero function $f \geq 0$ , there exists a continuous random variable $X$ with probability density function

$\displaystyle f_X = C_f \cdot f,$

where $C_f := (\int_{-\infty}^{\infty} f(x)\, \mathrm dx)^{-1}$ is called the normalising constant of $f$ .

Theorem 3. The random variable $X$ follows an normal distribution with mean $\mu$ and variance $\sigma^2$ , denoted $X \sim \mathcal{N}(\mu, \sigma^2)$ , if is defined by the probability density function

$\displaystyle f_X(x) = \frac 1{\sigma\sqrt{2\pi}}\exp\left( - \frac{ (x - \mu)^2 }{2\sigma^2} \right).$

$\mathbb E[X] = \mu$ and $\mathrm{Var}(X) = \sigma^2$ .

Proof. Denoting $Z \sim \mathcal N(0, 1)$ ,

$\displaystyle f_X(x) = \frac 1{\sigma} \cdot f_Z\left( \frac{x-\mu}{\sigma} \right) \quad \Rightarrow \quad X = \mu + \sigma \cdot Z,$

so it suffices to prove that $\mathbb E[Z] = 0$ and $\mathrm{Var}(Z) = 1$ . We remark that the normalising constant of $e^{-z^2/2}$ is $1/\sqrt{2\pi}$ by the Gaussian integral

$\displaystyle \int_{-\infty}^{\infty} e^{-z^2}\, \mathrm dz = \sqrt{\pi}$

and a change of variables. The former is immediate since the function $ze^{-z^2/2}$ is odd and its integral over $\mathbb R$ converges absolutely, so that

$\displaystyle \sqrt{2\pi} \cdot \mathbb E[Z] = \int_{-\infty}^{\infty} ze^{-z^2/2}\, \mathrm dz = 0.$

The latter requires integration by parts (here, as with the exponential distribution calculation, all integrals converge absolutely so we do not need to worry about the limits):

$\begin{aligned}\sqrt{2\pi} \cdot \mathbb E[Z^2] &= \int_{-\infty}^{\infty} z^2 e^{-z^2/2}\, \mathrm dz \\ &= \int_{-\infty}^{\infty} -z \cdot (-z e^{-z^2/2})\, \mathrm dz \\ &= \left[ -z \cdot e^{-z^2/2} \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} (-1) \cdot e^{-z^2/2} \, \mathrm dz \\ &= (0 - 0) - (-1) \cdot \sqrt{2\pi} = \sqrt{2\pi}. \end{aligned}$

Therefore,

$\mathrm{Var}(Z) = \mathbb E[Z^2] - \mathbb E[Z]^2 = 1 - 0^2 = 1.$

What makes the normal distribution so ubiquitous is how it arises from samples of virtually any distribution.

Theorem 4 (Central Limit Theorem). Let $X_1,X_2,\dots$ be i.i.d. random variables with mean $0$ and variance $1$ . Define

$\displaystyle \bar X_n := \frac{1}{n} \sum_{i=1}^n X_i.$

Then $\sqrt{n} \cdot \bar X_n \to Z$ in the following sense: for any $\epsilon > 0$ , there exists $n > 0$ such that for any $z \in \mathbb R$ ,

$\displaystyle |\mathbb P(\sqrt{n} \cdot \bar X_n \leq z) - \mathbb P(Z \leq z)| < \epsilon.$

We say that $\sqrt{n} \cdot \bar X_n$ converges in distribution to the standard normal distribution $Z \sim \mathcal N(0, 1)$ .

Proof. Delayed.

Our main goal in these posts on probability is to prove this statement rigorously. The central limit theorem is responsible for our intuition about probability arising from repeated experiments.

A more immediate application comes from simulation. It turns out that the humble uniform random variable $U \sim \mathcal U(0, 1)$ (implemented using pseudorandom numbers) can be used to generate all other random variables.

Theorem 5. Suppose $U \sim \mathcal U(0, 1)$ . For any random variable $X \sim \mathbb Q$ , there exists a measurable function $F$ such that $F(U) \sim \mathbb Q$ .

Proof. Let $F_X := \mathbb Q((-\infty, \cdot ])$ denote the c.d.f. of $X$ . We observe that $F_X(x) \in [0, 1]$ by definition. Therefore, the idea is to generate $U \in [0, 1]$ , then define the variate $V := F_X^{-1}(U)$ , so that $F_X(V) = U$ .

To that end, define the measurable map $F$ by

$\displaystyle F(u) := \sup \{x \in \mathbb R : F_X(x) \leq u\}.$

Then define $V := F(U)$ . Observe that $\{V \leq v\} = \{U \leq F_X(v)\}$ . Therefore

$\begin{aligned} \mathbb P(V \leq v) &= \mathbb P(U \leq F_X(v)) = F_X(v) = \mathbb P(X \leq v). \end{aligned}$

Therefore, $F(U) = V \sim \mathbb Q$ .

But we have a slightly more urgent question to answer: what’s the distribution of $X +Y$ in general? We need to construct the relevant sample space, and prove relevant properties in the multivariable setting, before we can legitimately continue on our quest to prove the central limit theorem. In fact, once we have at least done so for normal distributions, we will be in a sufficiently good place to discuss the principle of statistical hypothesis testing—a quantitative implementation of the scientific method.

—Joel Kindiak. 19 Jul 25, 2258H

November 18, 2025
An Unnatural Logarithm

Problem 1. Prove that the sequence $\{ \gamma_n \}$ defined by

$\displaystyle \gamma_n := \sum_{k=0}^n \frac 1{k+1} - \ln(n+1).$

converges to some real number $\gamma$ , called the Euler-Mascheroni constant. Numerically, $\gamma \approx 0.577$ .

(Click for Solution)

Solution. Write the sum as an integral as follows:

$\begin{aligned} \gamma_n = \sum_{k=0}^n \frac 1{k+1} - \ln(n+1) &= \sum_{k=0}^n \frac 1{k+1} - \int_0^{n} \frac 1{x+1}\, \mathrm dx \\ &= \sum_{k=0}^n \int_{k}^{k+1} \frac 1{k+1}\, \mathrm dx - \sum_{k=0}^n \int_{k}^{k+1} \frac 1{x+1}\, \mathrm dx \\&= \sum_{k=0}^n \int_{k}^{k+1} \left( \frac 1{k+1} - \frac 1{x+1} \right) \, \mathrm dx. \end{aligned}$

We observe that for any $x \in [k, k+1]$ ,

$\displaystyle 0 \leq \frac 1{k+1} - \frac 1{x+1} \leq \frac 1{k+1} - \frac 1{k+2}.$

Since each summand is non-negative, $\{\gamma_n\}$ is non-decreasing. Using the upper-bound,

$\begin{aligned} 0 \leq \gamma_n &\leq \sum_{k=0}^n \left( \frac 1{k+1} - \frac 1{k+2} \right) = 1 -\frac 1{n+1} \leq 1. \end{aligned}$

Hence, $\{ \gamma_n \}$ is a non-decreasing sequence that is bounded above by $1$ . By the monotone convergence theorem, $\gamma_n \to \gamma$ for some $\gamma \in [0, 1]$ .

Problem 2. Given that $\{a_n\}$ is a non-negative $\mathbb R$ -sequence that converges to $0$ , prove that the series $\sum_{k=0}^\infty (-1)^k a_k$ converges. This result is called the alternating series test.

(Click for Solution)

Solution. For any $n \in \mathbb N$ , define the $n$ -th partial sum by

$\displaystyle s_n := \sum_{k=0}^n (-1)^k a_k.$

It is clear that $s_0 = a_0 \geq 0$ and $s_1 = a_0 - a_1 \geq 0$ since $\{ a_n \}$ is decreasing.

For general $m$ ,

$s_{2(m+1)} = s_{2m+2} = s_{2m} - \underbrace{ (a_{2m+1} - a_{2m+2}) }_{\geq 0} \leq s_{2m}.$

Similarly, $s_{2(m+1)+1} \geq s_{2m+1}$ . Combining these results, for any $m \in \mathbb N$ ,

$\displaystyle s_1 \leq s_{2m+1} \leq s_{2m} \leq s_0.$

Therefore, the sequence $\{s_{2m}\}$ is decreasing and bounded below by $s_1 \geq 0$ . By the monotone convergence theorem, $s_{2m} \to s$ for some $s \in \mathbb R$ . Similarly, $s_{2m+1} \to s'$ for some $s \in \mathbb R$ . Furthermore,

$\displaystyle s' = \lim_{m \to \infty} s_{2m+1} = \lim_{m \to \infty} (s_{2m} + (-1)^{2m+1} a_{2m+1}) = s + 0 = s.$

Finally, we observe that for any $n$ , $n = 2 i + j$ for some $i \in \mathbb N$ and $j \in \{0, 1\}$ . Hence,

$\displaystyle s_{2i+1} \leq s_{2i + j} \leq s_{2i}.$

Taking $n \to \infty$ yields $i = (n-j)/2 \to \infty$ , so that $s_{2i} \to s$ and $s_{2i+1} \to s$ . By the squeeze theorem, $s_n = s_{2i+j} \to s$ . Therefore, the desired series $s_\infty$ converges to $s$ .

Problem 3. Evaluate the series $1 - 1/2 + 1/3 + 1/4 + \cdots$ .

(Click for Solution)

Solution. Write the series as

$\displaystyle 1 - \frac 12 + \frac 13 - \frac 14 + \cdots = \sum_{k=0}^\infty (-1)^k \frac 1{k+1}.$

Since the sequence $\{1/(k+1)\}$ is non-negative and converges to $0$ , by the alternating series test, the series converges. For any natural number $n$ ,

$\begin{aligned} \sum_{k=0}^{2n-1} (-1)^k \frac 1{k+1} &= \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} \\ &= \sum_{k=0}^{2n-1} \frac{1}{k+1} - \sum_{k=0}^{n-1} \frac{1}{k+1} \\ &= (\gamma_{2n-1} + \ln(2n)) - (\gamma_{n-1} + \ln(n)) \\ &= \gamma_{2n-1} - \gamma_{n-1} + \ln(2).\end{aligned}$

Taking $n \to \infty$ ,

$\begin{aligned} \sum_{k=0}^{\infty} (-1)^k \frac 1{k+1} &= \lim_{n \to \infty} \sum_{k=0}^{2n-1} (-1)^k \frac 1{k+1} \\ &= \lim_{n \to \infty} (\gamma_{2n-1} - \gamma_{n-1} + \ln(2)) \\ &= \gamma - \gamma + \ln(2) \\ &= \ln(2). \end{aligned}$

—Joel Kindiak, 12 Sept 25, 1732H

September 12, 2025
Principle of Real Induction
Problem 1. Fix $K \subseteq [a, b]$ . Suppose the following three properties are required:
- $a \in K$ ,
- if $[a, x] \subseteq K$ , then there exists $\delta > 0$ such that $[x, x + \delta] \subseteq K$ ,
- if $[a, x) \subseteq K$ , then $x \in K$ .
Then $K = [a, b]$ . You could think of this result as the principle of “real induction”.

(Click for Solution)

Solution. Define $L := [a, b] \backslash K$ . We claim that $L = \emptyset$ . Suppose otherwise that $L \neq \emptyset$ . Since $L$ is bounded below, it has a greatest lower bound (i.e. infimum), which we will denote $m$ . By the first hypothesis, $a \in K$ , so that $m \neq a$ , and therefore, $m > a$ . By the second hypothesis, since $[a,a] = \{a\} \subseteq K$ , there exists $\delta > 0$ such that $[a, a+\delta] \subseteq K$ . In particular, if $m \leq a + \delta$ , then $m \in K$ , a contradiction. Therefore, $m > a + \delta$ . Finally, for any $n < m$ , if $n \notin K$ , then $m \leq n < m$ , a contradiction. Therefore, $n \in K$ , which implies $[a, m) \subseteq K$ . By the third hypothesis, $m \in K$ , a contradiction. Therefore, $L = \emptyset$ .

According to this thread, the principle of “real induction” can be used to prove many important results in calculus and real analysis.

—Joel Kindiak, 5 Sept 25, 2341H
September 5, 2025
An Inequality Puzzle

Problem 1. Suppose $f : \mathbb R \to \mathbb R$ satisfies the inequality

$\displaystyle |f(x) - f(y)| \leq (x-y)^2, \quad x ,y \in \mathbb R,$

and $f(0) = 1$ . Given $x \in \mathbb R$ , evaluate $f(x)$ .

(Click for Solution)

Solution. Fix $c \in \mathbb R$ . For any $h \in \mathbb R$ ,

$\displaystyle |f(c+h) - f(c)| \leq ((c+h)-c)^2 = h^2.$

Dividing by $|h|$ on both sides,

$\displaystyle \left| \frac{f(c+h) - f(c)}{h} \right| \leq |h|.$

Taking $h \to 0$ , by the squeeze theorem,

$\displaystyle f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} = 0.$

Since $c \in \mathbb R$ is arbitrary, $f' = 0$ . Thus, $f$ is differentiable on $\mathbb R$ with derivative $f' = 0$ . Fix $x \in \mathbb R$ . By the mean value theorem, there exists $\xi$ between $0$ and $x$ such that

$f(x) = f(0) + f'(\xi) \cdot (x-0) = 1 + 0 \cdot x = 1.$

—Joel Kindiak, 23 Aug 25, 2237H

August 23, 2025
Kindiak Gauntlet (Differential Equations)

Warning. All questions below are non-trivial.

Section A (48 marks)

Each question in this section is worth 8 marks.

Question 1. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{1 + 2y^2}{\sqrt{9 - 4x^2}}.$

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \frac{1}{1 +2y^2} \cdot \frac{\mathrm dy}{\mathrm dx} &= \frac 1{\sqrt{9-4x^2}} \\ \frac 12 \cdot \frac{1}{\left( \frac 1{\sqrt 2} \right)^2 +y^2} \cdot \frac{\mathrm dy}{\mathrm dx} &= \frac 1 2 \cdot \frac 1{\sqrt{ \left( \frac 32 \right)^2 - x^2}} \\ \int \frac{1}{\left( \frac 1{\sqrt 2} \right)^2 +y^2}\, \mathrm dy &= \int \frac 1{\sqrt{ \left( \frac 32 \right)^2 - x^2}}\, \mathrm dx \\ \frac 1{\frac 1{\sqrt 2}} \tan^{-1} \left( \frac{y}{\frac 1{\sqrt 2}} \right) &= \sin^{-1}\left( \frac{x}{\frac 32} \right) + C \\ \sqrt{2} \tan^{-1}(y \sqrt 2) &= \sin^{-1}\left( \frac {2x}3\right) + C. \end{aligned}$

Question 2. Solve the differential equation

$\displaystyle x \left( \frac{\mathrm dy}{\mathrm dx} - \cos(x) \right) + 2y = 0.$

(Click for Solution)

Solution. By algebraic manipulation,

$\displaystyle \frac{\mathrm dy}{\mathrm dx} + \frac{2y}{x} = \cos(x).$

Compute the integrating factor

$\displaystyle I(x) = e^{\int \frac 2x\, \mathrm dx} = e^{2 \ln|x|} = |x|^2 = x^2.$

Thus, the general solution is given by

$\displaystyle \begin{aligned} x^2 y &= \int x^2 \cos(x)\, \mathrm dx \\ &= x^2 \sin(x) - \int 2x \sin(x)\, \mathrm dx \\ &= x^2 \sin(x) + 2 \int x\cdot(-\sin(x))\, \mathrm dx \\ &= x^2 \sin(x) + 2\left(x \cos(x) - \int \cos(x)\, \mathrm dx \right) \\ &= x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C. \end{aligned}$

where we integrated by parts twice on the right-hand side.

Question 3. Use the substitution $u = xy$ to solve the differential equation

$\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2} - \left( 5 - \frac 2x \right) \frac{\mathrm dy}{\mathrm dx} - \left( \frac 5x - 6 \right) y = 0.$

(Click for Solution)

Solution. Writing $y = u/x$ , use the quotient rule to deduce

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{xu' - u}{x^2} = \frac{u'}{x} - \frac{u}{x^2},\\ \frac{\mathrm d^2y}{\mathrm dx^2} &= \frac{xu'' - u'}{x^2} - \frac{x^2u' - u \cdot 2x}{x^4} = \frac{u''}{x} - \frac{2u' }{x^2} + \frac{2u}{x^3}.\end{aligned}$

Making the substitutions,

$\begin{aligned} \left( \frac{u''}{x} - \frac{2u' }{x^2} + \frac{2u}{x^3} \right) - \left( 5 - \frac 2x \right) \left( \frac{u'}{x} - \frac{u}{x^2} \right) - \left( \frac 5x - 6 \right) \frac ux &= 0 \\ \left( \frac{u''}{x} - \frac{2u' }{x^2} + \frac{2u}{x^3} \right) - \left( \frac{5u'}{x} - \frac{5u}{x^2} -\frac{2u'}{x^2} + \frac{2u}{x^3} \right) - \left( \frac {5u}{x^2} - \frac{6u}{x} \right) &= 0 \\ \frac{u''}{x} - \frac{5u'}{x} + \frac{6u}{x} &= 0 \\ u'' - 5u' + 6u &= 0. \end{aligned}$

Since the auxiliary equation $m^2 - 5m + 6 = 0$ has real and distinct roots $m = 2,3$ , the general solution is given by

$\displaystyle xy = u = C_1 e^{2x} + C_2 e^{3x}.$

Question 4. Evaluate $\displaystyle \mathcal L\{(7+8t)^2 * (4 \sin 3t \cos 3t)\}$ .

(Click for Solution)

Solution. First carry out some simplifications:

$\begin{aligned} (7+8t)^2 &= 49 + 112t + 64t^2, \\ 4\sin 3t \cos 3t &= 2 \sin 6t.\end{aligned}$

Then taking Laplace transforms,

$\begin{aligned} \mathcal L\{(7+8t)^2 * (4 \sin 3t \cos 3t)\} &= \mathcal L\{(7+8t)^2 \} \cdot \mathcal L\{ 4 \sin 3t \cos 3t \} \\ &= \mathcal L\{ 49 + 112t + 64t^2 \} \cdot \mathcal L\{ 2 \sin 6t \} \\ &= \left( 49 \cdot \frac 1s + 112 \cdot \frac{1}{s^2} + 64 \cdot \frac{2}{s^3} \right) \cdot \left( 2 \cdot \frac{6}{s^2 + 6^2} \right) \\ &= \frac{49s^2 + 112s + 128}{s^3} \cdot \frac{12}{s^2 + 36} \\ &= \frac{12(49s^2 + 112s + 128)}{s^3 ( s^2 + 36 )}. \end{aligned}$

Question 5. Evaluate $\displaystyle \mathcal L^{-1} \left\{ \frac{1}{s^3+1} \right\}$ .

(Click for Solution)

Solution. We first note that $(-1)^3 + 1 = 0$ , so that by the factor theorem, $s+1$ is a factor of $s^3+1$ . Hence, we factorise $s^3+1 = (s+1)(s^2 - s + 1)$ . This suggests us to employ the partial fraction decomposition

$\displaystyle \frac 1{s^3 + 1} = A \cdot \frac{1}{s+1} + B \cdot \frac{s-1/2}{(s-1/2)^2+ (\sqrt{3}/2)^2} + C \cdot \frac{1}{(s-1/2)^2+ (\sqrt{3}/2)^2},$

where we will obtain $A,B,C$ later and completed the square in the last term. We recall the following well-known inverse Laplace transforms:

$\displaystyle \begin{aligned} \mathcal L^{-1} \left\{ \frac{1}{s+1} \right\} &= e^{-(-1)t} = e^t, \\ \mathcal L^{-1} \left\{ \frac{s-1/2}{(s-1/2)^2+ (\sqrt{3}/2)^2} \right\} &= e^{t/2} \cos\left( \frac{t\sqrt 3}{2} \right), \\ \mathcal L^{-1} \left\{ \frac{1}{(s-1/2)^2+ (\sqrt{3}/2)^2} \right\} &= \frac 2{\sqrt 3} e^{t/2} \sin\left( \frac{t\sqrt 3}{2} \right). \end{aligned}$

Since taking inverse Laplace transforms is linear,

$\displaystyle \mathcal L^{-1}\left\{ \frac{1}{s^3+1} \right\} = Ae^t + Be^{t/2} \cos\left( \frac{t\sqrt 3}{2} \right) + \frac {2C}{\sqrt 3} e^{t/2} \sin\left( \frac{t\sqrt 3}{2} \right).$

All that remains is for us to obtain $A, B, C$ . To that end, clear denominators by multiplying $x^3+1$ on all sides:

$\displaystyle \begin{aligned} 1 &= A(s^2 - s + 1) + B(s-1/2)(s+1) + C(s+1) \\ &= (A + B)s^2 + (-A +B/2 + C)s + (C-B/2). \end{aligned}$

Comparing coefficients,

$\left\{ \begin{aligned}A + B &= 0, \\ -A + \frac 12 B + C &= 0, \\ -\frac 12 B + C &= 1.\end{aligned} \right.$

Solving simultaneous linear equations, $A=1/2,B = -1/2,C = 3/4$ . Substituting and simplifying,

$\displaystyle \mathcal L^{-1}\left\{ \frac{1}{s^3+1} \right\} = \frac 12 e^t - \frac 12 e^{t/2} \cos\left( \frac{t\sqrt 3}{2} \right) + \frac{\sqrt 3}2 e^{t/2} \sin\left( \frac{t\sqrt 3}{2} \right).$

Question 6. Evaluate $\displaystyle \frac {20 + 23 \mathcal D^2}{(4 + 9 \mathcal D^2)(16 + 25 \mathcal D^2)}\left( \sin \frac 23 x \right).$

(Click for Solution)

Solution. Apply partial fractions again (left as an exercise) to obtain

$\displaystyle \frac {20 + 23 \mathcal D^2}{(4 + 9 \mathcal D^2)(16 + 25 \mathcal D^2)} = \frac{2}{4 + 9 \mathcal D^2} - \frac{3}{16 + 25 \mathcal D^2}.$

Therefore,

$\begin{aligned} \frac {20 + 23 \mathcal D^2}{(4 + 9 \mathcal D^2)(16 + 25 \mathcal D^2)} \left( \sin \frac 23 x \right) &= \left(\frac{2}{4 + 9 \mathcal D^2} - \frac{3}{16 + 25 \mathcal D^2}\right) \left( \sin \frac 23 x \right) \\ &= \frac 29 \cdot \frac{1}{\mathcal D^2 + \left(\frac 23\right) ^2}\left( \sin \frac 23 x \right) - 3 \cdot \frac{1}{16 + 25 \mathcal D^2}\left( \sin \frac 23 x \right) \\ &= -\frac 29 \cdot \frac{x \cos \left(\frac 23 x\right)}{2 \cdot \frac 23} - 3 \cdot \frac 1{16 + 25 \cdot \left(-\frac 49\right)} \cdot \sin \left(\frac 23 x \right) \\ &= -\frac 16 x \cos \left(\frac 23 x\right) - \frac{27}{44} \sin \left(\frac 23 x \right).\end{aligned}$

Section B (52 marks)

Each question in this section is worth 13 marks.

Question 1. Solve the differential equation

$\displaystyle 3y'' - 24y' + 48y = e^{5x} \cos(6x).$

(Click for Solution)

Solution. The auxiliary equation $3m^2 - 24m + 48 = 0$ has repeated real roots $m = 4$ . Therefore, the complementary function is given by

$y_{\mathrm C} = (C_1x + C_2)e^{4x}.$

For the particular integral,

$\begin{aligned} y_{\mathrm P} &= \frac{1}{3 \mathcal D^2 - 24 \mathcal D + 48}(e^{5x} \cos(6x)) \\ &= \frac 13 \cdot \frac{1}{\mathcal D^2 - 8 \mathcal D + 16}(e^{5x} \cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{(\mathcal D + 5)^2 - 8 (\mathcal D + 5) + 16}(e^{5x} \cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{(\mathcal D^2 - 10 \mathcal D + 25) - (8\mathcal D + 40) + 16}(\cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{\mathcal D^2 - 18 \mathcal D + 1}(\cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{-6^2 - 18 \mathcal D + 1}(\cos(6x)) \\ &= -\frac 13 \cdot e^{5x} \cdot \frac{1}{18 \mathcal D + 35}(\cos(6x)) \\ &= -\frac 13 \cdot e^{5x} \cdot \frac{18 \mathcal D - 35}{324 \mathcal D^2 - 1225}(\cos(6x)) \\ &= \frac 1{38\, 667} \cdot e^{5x} \cdot (18 \mathcal D(\cos(6x)) - 35\cos(6x)) \\ &= \frac 1{38\, 667} \cdot e^{5x} \cdot (-108 \sin(6x) - 35\cos(6x)) \\ &= -\frac 1{38\, 667} \cdot e^{5x} \cdot (108 \sin(6x) + 35\cos(6x)). \end{aligned}$

Therefore, the general solution is given by

$y = -\frac 1{38\, 667} e^{5x} (108 \sin(6x) + 35\cos(6x)) + (C_1x + C_2)e^{4x}.$

Question 2. Given the function $f$ defined by

$\displaystyle f(t) = \begin{cases}2, & t < 1, \\ 3, & 1 \leq t < 2, \\ 5, & 2 \leq t < 3, \\ 7 , & t \geq 3,\end{cases}$

evaluate $\displaystyle \mathcal L\left\{ \sqrt{f(\sqrt{t})} \cdot U(t - 5)\right\}$ .

(Click for Solution)

Solution. We first observe that

$\displaystyle \sqrt{ f(\sqrt{t}) } = \begin{cases}\sqrt 2, & \sqrt{t} < 1, \\ \sqrt 3, & 1 \leq \sqrt{t} < 2, \\ \sqrt 5, & 2 \leq \sqrt{t} < 3, \\ \sqrt 7 , & \sqrt{t} \geq 3\end{cases} = \begin{cases}\sqrt 2, & t < 1, \\ \sqrt 3, & 1 \leq t < 4, \\ \sqrt 5, & 4 \leq t < 9, \\ \sqrt 7 , & t \geq 9.\end{cases}$

Therefore,

$\displaystyle \sqrt{ f(\sqrt{t}) } \cdot U(t-5) = 0 + \sqrt{5} \cdot U(t-5) + (\sqrt{7} - \sqrt{5}) \cdot U(t-9).$

Taking Laplace transforms,

$\displaystyle \mathcal L\left\{ \sqrt{ f(\sqrt{t}) } \cdot U(t-5) \right\} = \frac 1s \left( \sqrt 5 e^{-5s} + (\sqrt 7 - \sqrt 5) e^{-9s} \right).$

Question 3. Use Laplace transforms to solve the initial value problem

$\displaystyle \left\{ \begin{aligned} y'' - 2y' + 3y &= t^2 \delta(t - 4), \\ y(0) &= 5, \\ y'(0) &= 6.\end{aligned} \right.$

(Click for Solution)

Solution. Taking Laplace transforms on both sides,

$\mathcal L\{y''\} - 2 \mathcal L\{y'\} + 3 \mathcal L\{y\} = \mathcal L\{t^2 \delta(t-4)\}.$

To evaluate the left-hand side,

$\begin{aligned} \mathcal L\{y''\} &= s^2 Y(s) - sy(0) - y'(0) = s^2 Y(s) - 5s - 6, \\ \mathcal L\{y'\} &= s Y(s) - y(0) = s Y(s) - 5, \\ \mathcal L\{y\} &= Y(s). \end{aligned}$

Therefore, the left-hand side simplifies to

$\begin{aligned} & \mathcal L\{y''\} - 2 \mathcal L\{y'\} + 3 \mathcal L\{y\} \\ &= (s^2 Y(s) - 5s - 6) - 2(s Y(s) - 5) + 3Y(s) \\ &= (s^2 - 2s + 3)Y(s) - 5s +4. \end{aligned}$

To evaluate the right-hand side,

$\begin{aligned} \mathcal L\{t^2 \delta(t-4) \} &= -\frac{\mathrm d}{\mathrm ds} \mathcal L\{t \delta(t-4) \} \\ &= -\frac{\mathrm d}{\mathrm ds} \left( -\frac{\mathrm d}{\mathrm ds} \mathcal L\{ \delta(t-4) \} \right) \\ &= -\frac{\mathrm d}{\mathrm ds} \left( -\frac{\mathrm d}{\mathrm ds} (e^{-4s}) \right) = 16e^{-4s}. \end{aligned}$

Putting it all together,

$(s^2 - 2s + 3)Y(s) - 5s +4 = 16e^{-4s}.$

By algebraic manipulation,

$\begin{aligned} Y(s) &= \frac{16 e^{-4s}}{s^2 - 2s + 3} + \frac{5s-4}{s^2 - 2s + 3} \\ y(t) &= 16 \cdot \mathcal L^{-1}\left\{ e^{-4s} \cdot \frac{1}{s^2 - 2s + 3} \right\} + \mathcal L^{-1} \left\{ \frac{5s-4}{s^2 - 2s + 3} \right\}. \end{aligned}$

It remains to (painfully) compute each of the following inverse Laplace transforms, by observing that $s^2 - 2s + 3 = (s-1)^2 + 2$ .

Define $G(s-1) = (5s-4)/((s-1)^2 + 2)$ so that

$\displaystyle G(s) = \frac{5(s+1)-4}{s^2 + 2} = \frac{5s+1}{s^2 + 2} = 5 \cdot \frac s{s^2+2} + \frac 1{s^2+2}.$

By the shift theorems,

$\begin{aligned}\mathcal L^{-1} \left\{ \frac{5s-4}{s^2 - 2s + 3} \right\} &= \mathcal L^{-1}\{ G(s-1)\} \\ &= e^t g(t) \\ &= e^t \cdot \mathcal L^{-1}\left\{ 5 \cdot \frac s{s^2+2} + \frac 1{s^2+2} \right\} \\ &= e^t \cdot \left(5 \cos (t\sqrt 2) + \frac 1{\sqrt 2} \sin(t \sqrt 2) \right).\end{aligned}$

Similarly, defining $H(s) = 1/(s^2 + 2)$ and $H_1 := H(\cdot - 1)$ so that

$\displaystyle h_1(t) = \mathcal L^{-1}\{ H_1(s)\} = \mathcal L^{-1} \{H(s-1)\} = e^t \cdot h(t) = e^t \cdot \frac{1}{\sqrt 2} \sin(t \sqrt 2),$

we have

$\begin{aligned} \mathcal L^{-1}\left\{ e^{-4s} \cdot \frac{1}{s^2 - 2s + 3} \right\} &= \mathcal L^{-1} \{e^{-4s} H_1(s)\} \\ &= h_1(t-4) \cdot U(t-4) \\ &= \frac 1{\sqrt 2} e^{t-4} \sin((t-4)\sqrt 2) \cdot U(t-4). \end{aligned}$

Piecing them all together,

$\displaystyle y(t) = 8\sqrt 2 \cdot e^{t-4} \sin((t-4)\sqrt 2) \cdot U(t-4) + e^t \cdot \left(5 \cos (t\sqrt 2) + \frac 1{\sqrt 2} \sin(t \sqrt 2) \right).$

Question 4. Evaluate the Fourier series of the function $f$ defined by

$f(t) = \begin{cases} 1, & -1 \leq t \leq 0, \\ 2t, & 0 \leq t \leq 1, \end{cases}$

and periodic extension $f(t) = f(t+2)$ .

(Click for Solution)

Solution. We need to evaluate each of the Fourier coefficients $a_0, a_n, b_n$ with $T = 2$ . For $a_0$ ,

$\begin{aligned} a_0 = \int_{-1}^1 f(t)\, \mathrm dt &= \int_{-1}^0 1\, \mathrm dt + \int_0^1 2t\, \mathrm dt \\ &= (0 - (-1)) + (1^2 - 0^2) = 2. \end{aligned}$

For $a_n$ , we need to integrate by parts:

$\begin{aligned} a_n &= \int_{-1}^1 f(t) \cos(n \pi t) \, \mathrm dt \\ &= \int_{-1}^0 \cos(n \pi t)\, \mathrm dt + \int_0^1 2t \cos(n \pi t)\, \mathrm dt \\ &= \left[ \frac{\sin(n \pi t)}{n \pi} \right]_{-1}^0 + \left[ \frac{2 t \sin(n \pi t)}{n \pi} \right]_0^1 + \left[ \frac{2\cos(n \pi t)}{n^2 \pi^2} \right]_0^1 \\ &= 0 + 0 + \frac 2{n^2 \pi^2} ((-1)^n - 1) \\ &= \frac {2 ((-1)^n - 1) }{n^2 \pi^2} . \end{aligned}$

Similarly for $b_n$ , we need to integrate by parts:

$\begin{aligned} b_n &= \int_{-1}^1 f(t) \sin(n \pi t) \, \mathrm dt \\ &= \int_{-1}^0 \sin(n \pi t)\, \mathrm dt + \int_0^1 2t \sin(n \pi t)\, \mathrm dt \\ &= \left[ \frac{-\cos(n \pi t)}{n \pi} \right]_{-1}^0 + \left[ \frac{-2 t \cos(n \pi t)}{n \pi} \right]_0^1 + \left[ \frac{2\sin(n \pi t)}{n^2 \pi^2} \right]_0^1 \\ &= -\frac 1{n\pi} (1 - (-1)^n) - \frac 2{n \pi} (-1)^n \\ &= -\frac 1{n\pi} ( (-1)^n + 1). \end{aligned}$

Therefore, the Fourier series of $f$ is given by

$\displaystyle f(t) = 1 + \sum_{n=1}^\infty \left( \frac {2 ((-1)^n - 1) }{n^2 \pi^2} \cos(n \pi t) -\frac 1{n\pi} ( (-1)^n + 1) \sin(n \pi t) \right).$

—Joel Kindiak, 19 Aug 25, 1613H

August 19, 2025
The Hunt for Metrizability

The crucial property that gives us massive control over a topological space is its compactness. For metrizable spaces, compactness is equivalent to sequential compactness. This therefore raises a crucial question: when is a topological space metrizable?

We first make some observations on the necessary conditions for a space to be metrizable.

Lemma 1. If $K$ is a metrizable space, then it is first-countable, i.e. for any $x \in K$ , there exists a countable basis $\mathcal B_x$ of neighbourhoods of $x$ .

Proof. Define $\mathcal B_x := \{ B_d(x, r) : r \in \mathbb Q^+ \}$ , where $d$ is the assumed metric underlying $K$ .

Rather than having a countable basis for just near any fixed $x$ , it is sometimes useful to have a countable basis for the whole topological space. This is known as second-countability.

Definition 1. A topological space $K$ is second-countable if it is generated by a countable basis.

Lemma 2. Second-countable spaces are first-countable.

Proof. If $K$ is second-countable with countable basis $\mathcal B$ , then for any $x \in K$ , $\mathcal B_x := \{B \in \mathcal B : x \in B\}$ yields a countable basis near $x$ .

Example 1. $\mathbb R^n$ is second-countable.

Proof. Assuming the standard metric, since $\overline{\mathbb Q} = \mathbb R$ , $\mathbb R^n$ has the countable basis

$\{B_d(\mathbf x, r) : \mathbf x \in \mathbb Q^n, r \in \mathbb Q_{\geq 0}\}.$

This fact also demonstrates that $\overline{\mathbb Q^n} = \mathbb R^n$ .

We might think, therefore, that there might be some connection between second-countability and metrizability, and even first-countability. None of these notions are identical to each other, and we prove them using several counterexamples.

Example 2. Equip the set $K := \{0, 1\}$ with the topology $\mathcal T := \{ \emptyset, \{ 0 \}, K \}$ . Then $K$ is first-countable but not metrizable.

Proof. First-countability is obvious, and non-metrizability follows from non-Hausdorffness.

Example 3. Consider $\mathbb R$ equipped with the discrete metric. Then any basis must contain the uncountable set $\{\{x\} : x \in \mathbb R\}$ .

Example 2 kills two birds with one stone, exhibiting a metrizable space, and thus a first-countable space, that are both not second-countable. Can we find a second-countable space that is not metrizable?

This question remains elusive, and will require us to consider the separability of the topological space. We have discussed one type of separability—Hausdorffness. It’s relatives we will discuss in the next post.

—Joel Kindiak, 16 Apr 25, 2255H

July 25, 2025
Cheryl’s Birthday Problem
This problem is an absolute classic in creative usage of modus ponens, and I’m definitely not its originator.

Problem 1. Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates:
- May 15, May 16, May 19
- June 17, June 18
- July 14, July 16
- August 14, August 15, August 17
Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.
- Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard doesn’t know too.
- Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.
- Albert: Then I also know when Cheryl’s birthday is.
When is Cheryl’s birthday?
(Click for Solution)

Solution. The key phrase to parse is the phrase “I don’t know”. We know that Albert is given the month and that Bernard is given the number.

Lemma 1. Albert will know Cheryl’s birthday if and only if given his month, there is one and only one possible date number.

Lemma 2. Bernard will know Cheryl’s birthday if and only if given his date number, there is one and only one possible month.

Both lemmas are of the form $p \leftrightarrow q$ .

Now, we parse Albert’s first sentence:
- I don’t know when Cheryl’s birthday is
- I know that Bernard doesn’t know too
Albert is given one of the four options: May, June, July, or August. He says “I don’t know when Cheryl’s birthday is”, so that by Lemma 1, his month has multiple date numbers. This phrase doesn’t reveal much, but the next one does.

Albert says “I know that Bernard doesn’t know”. By Lemma 2, Bernard’s date number can be matched by more than one month. If, however, his numbers are 18 or 19, then each of these numbers can only match one month, a contradiction:

$(18 \lor 19 \rightarrow c) \rightarrow \neg(18 \lor 19).$

Therefore, Bernard’s date numbers cannot be 18 or 19, and we are left with the following possibilities:
- May 15, May 16
- June 17
- July 14, July 16
- August 14, August 15, August 17
If Albert was given either May or June, then 18 or 19 are Bernard’s possibilities. This can be phrased as the conditional proposition

$\text{May} \lor \text{June} \rightarrow 18 \lor 19.$

Since $\neg(18 \lor 19)$ , modus tollens yields

$\neg (18 \lor 19) \rightarrow \neg(\text{May} \lor \text{June}) = \neg \text{May} \wedge \neg \text{June}.$

Thus, Albert was not given May nor June, leaving us with the following possibilities:
- July 14, July 16
- August 14, August 15, August 17
Now, we parse Bernard’s sentence:
- At first I don’t know when Cheryl’s birthday is
- But I know now.
Bernard says “At first I don’t know when Cheryl’s birthday is”. This means by Lemma 2, prior to Albert’s sentence, Bernard’s number can correspond to multiple months, but after Albert’s sentence, Bernard’s number can correspond only to one month. If Bernard was given the number 14, then there are still two possible months, a contradiction:

$(14 \to c) \to \neg 14.$

Thus, Bernard was given either 15, 16, or 17, and we are left with the following dates:
- July 16
- August 15, August 17
Finally, we parse Albert’s second sentence:
- Then I also know when Cheryl’s birthday is.
If Albert was given August, he will still be uncertain. Since he is certain, Albert must have been given July:

$(\neg \text{July} \to c) \to \neg(\neg \text{July}) = \text{July}.$

Therefore, Bernard was given 16.

Therefore, Cheryl’s birthday is July 16.

To summarise the argument, the premises yield:
- $(\text{May} \lor \text{June} \lor \text{July} \lor \text{August}) \wedge (14 \lor 15 \lor 16 \lor 17 \lor 18 \lor 19)$
- $18 \lor 19 \to c$
- $\therefore (\text{May} \lor \text{June} \lor \text{July} \lor \text{August}) \wedge (14 \lor 15 \lor 16 \lor 17)$
- $\text{May} \to 18\ \text{possible} \to c$
- $\text{June} \to 19\ \text{possible} \to c$
- $\therefore (\text{July} \lor \text{August}) \wedge (14 \lor 15 \lor 16 \lor 17)$
- $14 \to c$
- $\therefore (\text{July} \lor \text{August}) \wedge (15 \lor 16 \lor 17)$
- $\text{August} \to c$
- $\therefore \text{July} \wedge (15 \lor 16 \lor 17)$
- $\therefore \text{July} \wedge 16$
—Joel Kindiak, 21 Jun 25, 1402H
June 21, 2025