Category: Exercises

Tricky Limits

Big Idea

We write $\displaystyle \lim_{x \to c} f(x) = L$ to mean that $f(x) \to L$ as $x \to c$ . In particular, if $g$ is continuous at $c$ and $f(x) = g(x)$ for $x \neq c$ , then

$\displaystyle \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = g(c).$

Furthermore, the usual limit laws hold.

Questions

Question 1. Evaluate the limit $\displaystyle \lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8}$ .

(Click for Solution)

Solution. Since $2$ is a root of $x^3 - 8$ , $x-2$ is a factor of $x^3 - 8$ :

$\begin{aligned} x^3 - 8 &= (x-2)(x^2 + kx + 4) \\ &= x^3 + (k-2)x^2 +(4-2k)x - 8. \end{aligned}$

Comparing coefficients, $k = 2$ . Hence,

$\begin{aligned} \lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8} &= \lim_{x \to 2} \frac{(x-2)(x+2)}{(x-2)(x^2 + 2x + 4)} \\ &= \lim_{x \to 2} \frac{x+2}{x^2 + 2x + 4} \\ &= \frac{2+2}{2^2 + 2\cdot 2 + 4} = \frac 13.\end{aligned}$

Alternate Solution. Make the substitution $x = 2 +t$ , so that $x \to 2$ leads to $t \to 2$ . Then

$\begin{aligned} \lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8} &= \lim_{t \to 0} \frac{(2+t)^2 - 4}{(2+t)^3 - 8} \\ &= \lim_{t \to 0} \frac{ (4 + 4t + t^2) - 4}{ (8 + 12t + 6t + t^3) - 8} \\ &= \lim_{t \to 0} \frac{ 4t + t^2 }{ 12t + 6t^2 + t^3 } \\ &= \lim_{t \to 0} \frac{ 4 + t }{ 12 + 6t + t^2 } \\ &= \frac{ 4 + 0 }{ 12 + 6 \cdot 0 + 0^2 } = \frac 13.\end{aligned}$

Question 2. Evaluate the limit $\displaystyle \lim_{x \to 4} \frac{\sqrt{1+2x} - 3}{\sqrt x-2}$ .

(Click for Solution)

Solution. Rationalising both numerator and denominator,

$\begin{aligned} \lim_{x \to 4} \frac{\sqrt{1+2x} - 3}{\sqrt x-2} &= \lim_{x \to 4} \left( (\sqrt{1+2x} - 3) \cdot \frac{1}{\sqrt x-2} \right) \\ &= \lim_{x \to 4} \left( (\sqrt{1+2x} - 3) \cdot \frac{\sqrt{1+2x} + 3}{\sqrt{1+2x} + 3} \cdot \frac{1}{\sqrt x-2} \cdot \frac{\sqrt x + 2}{\sqrt x + 2} \right) \\ &= \lim_{x \to 4} \left( \frac{(1+2x) - 9}{\sqrt{1+2x} + 3} \cdot \frac{\sqrt x + 2}{x-4} \right) \\ &= \lim_{x \to 4} \left( \frac{ 2(x-4) }{\sqrt{1+2x} + 3} \cdot \frac{\sqrt x + 2}{x-4} \right) \\ &= 2\cdot \lim_{x \to 4} \frac{ \sqrt x + 2 }{\sqrt{1+2x} + 3} = 2 \cdot \frac{\sqrt 4+2}{\sqrt{1 + 2 \cdot 4} + 3} = \frac 43. \end{aligned}$

Question 3. Evaluate the limit $\displaystyle \lim_{x \to \infty} \frac{\sqrt{4x^2 + 5}}{\sqrt[3]{x^3 - 1}}$ .

(Click for Solution)

Solution. By factoring $x = \sqrt{x^2} = \sqrt[3]{x^3}$ on both the numerator and the denominator,

$\begin{aligned} \lim_{x \to \infty} \frac{\sqrt{4x^2 + 5}}{\sqrt[3]{x^3 - 1}} &= \lim_{x \to \infty} \frac{\sqrt{x^2 \cdot (4 + 5 / x^2)}}{\sqrt[3]{x^3 \cdot (1 - 1/ x^3)}} \\ &= \lim_{x \to \infty} \frac{\sqrt{x^2} \cdot \sqrt{ 4 + 5 / x^2 }}{\sqrt[3]{x^3} \cdot \sqrt[3]{ 1 - 1/x^3 } } \\ &= \lim_{x \to \infty} \frac{x \cdot \sqrt{ 4 + 5/ x^{2} }}{x \cdot \sqrt[3]{ 1 - 1/x^{3} } } \\&= \lim_{x \to \infty} \frac{\sqrt{ 4 + 5 /x^{2} }}{\sqrt[3]{ 1 - 1/x^{3} } } = \frac{\sqrt{ 4 + 0 }}{\sqrt[3]{ 1 - 0 } } = 2. \end{aligned}$

—Joel Kindiak, 3 Sept 25, 1804H

October 23, 2025
Some Normal Distribution Trivia
Problem 1. Let $f$ be a continuously differentiable function. Evaluate $\displaystyle \int x f'(x^2)\, \mathrm dx$ . Hence, evaluate

$\displaystyle \int x e^{-x^2}\, \mathrm dx,\quad \int \frac{x}{1+x^4}\, \mathrm dx.$

(Click for Solution)

Solution. Make the substitution $u = x^2 \Rightarrow \frac 12 \mathrm du = x\, \mathrm dx$ so that

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

Particularise the result to $f(x) = -e^{-x}$ and $f(x) = \tan^{-1}(x)$ to obtain

$\displaystyle \int x e^{-x^2}\, \mathrm dx = -\frac 12 e^{-x^2} + C,\quad \int \frac{x}{1+x^4}\, \mathrm dx = \frac 12 \tan^{-1}(x) + C.$

Assume the following integral identity

$\displaystyle \left( \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx \right)^2 = \int_0^{\infty} 2 \pi x e^{-x^2}\, \mathrm dx.$

Problem 2. Evaluate $\displaystyle \int_{-\infty}^{\infty} x^n e^{-x^2}\, \mathrm dx$ for $n = 0, 1, 2$ .

(Click for Solution)

Solution. For $n = 0$ , we use the given identity

$\begin{aligned} \left( \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx \right)^2 &= \int_0^{\infty} 2 \pi x e^{-x^2}\, \mathrm dx \\ &= 2\pi \int_0^{\infty} xe^{-x^2}\, \mathrm dx \\ &= 2\pi \left[-\frac 12 e^{-x^2}\right]_0^\infty \\ &= 2\pi \left(-\frac 12 \cdot 0 + \frac 12 \cdot 1\right) = \pi. \end{aligned}$

Hence, $\displaystyle \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx = \sqrt{\pi}$ . For $n = 1$ ,

$\begin{aligned}\int_0^{\infty} x e^{-x^2}\, \mathrm dx &= \left[-\frac 12 e^{-x^2}\right]_{-\infty}^\infty = -\frac 12 \cdot 0 + \frac 12 \cdot 0 = 0. \end{aligned}$

For $n = 2$ , we first integrate by parts to get

$\begin{aligned} \int x^2 e^{-x^2}\, \mathrm dx &= \int x \cdot x e^{-x^2}\, \mathrm dx \\ &= \underbrace{ -\frac 12 e^{-x^2} }_{\mathrm I} \underbrace{ x }_{\mathrm S} - \int \underbrace{ -\frac 12 e^{-x^2} }_{\mathrm I} \underbrace{ 1 }_{\mathrm D}\, \mathrm dx \\ &= -\frac 12 xe^{-x^2} + \frac 12 \int e^{-x^2}\, \mathrm dx. \end{aligned}$

Plugging in the limits,

$\displaystyle \begin{aligned} \int_{-\infty}^{\infty} x^2 e^{-x^2}\, \mathrm dx &= \left[ -\frac 12 xe^{-x^2} \right]_{-\infty}^{\infty} + \frac 12 \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx \\ &= (0 - 0) + \frac 12 \sqrt{\pi} = \frac 12 \sqrt{\pi}.\end{aligned}$

Problem 3. Prove that for any $C > 0$ and $n \in \mathbb N$ ,

$\displaystyle \int_{-\infty}^{\infty} Cxe^{-x^2/n}\, \mathrm dx = 0.$

Compute $C, n$ so that

$\begin{aligned} \int_{-\infty}^{\infty} Ce^{-x^2/n}\, \mathrm dx &= \int_{-\infty}^{\infty} Cx^2 e^{-x^2/n}\, \mathrm dx = 1. \end{aligned}$

(Click for Solution)

Solution. Make the substitution $x^2/n = t^2 \Rightarrow \mathrm dx = \sqrt n\, \mathrm dt$ so that

$\begin{aligned} \int_{-\infty}^{\infty} Cxe^{-x^2/n}\, \mathrm dx &= Cn \int_{-\infty}^{\infty} t e^{-t^2}\, \mathrm dt = Cn \cdot 0 = 0 \end{aligned}$

by Problem 2. Similarly,

$\begin{aligned} 1 = \int_{-\infty}^{\infty} Ce^{-x^2/n}\, \mathrm dx &= C \cdot \int_{-\infty}^{\infty} e^{-x^2/n}\, \mathrm dx \\ &= C\sqrt{n} \cdot \int_{-\infty}^{\infty} e^{-t^2}\, \mathrm dt \\ &= C\sqrt{n} \cdot \sqrt{\pi}\end{aligned}$

so that $C = 1/\sqrt{n \pi}$ . Furthermore,

$\begin{aligned} 1 = \int_{-\infty}^{\infty} Cx^2 e^{-x^2/n}\, \mathrm dx &= Cn\sqrt n \int_{-\infty}^{\infty} t^2 e^{-t^2}\, \mathrm dt \\ &= Cn \sqrt n \cdot \frac 12 \sqrt{\pi}. \end{aligned}$

Comparing results yields $n = 2$ , implying $C = 1/\sqrt{2\pi}$ .

Define $\phi(t) := Ce^{-t^2/n}$ , where $C,n$ are determined from Problem 3. It is obvious that $\phi(-t) = \phi(t)$ for any $t \in \mathbb R$ . Define

$\displaystyle \Phi(z) := \int_{-\infty}^z \phi(t)\, \mathrm dt.$

Here, $\phi, \Phi$ are the probability density function and cumulative distribution function respectively of the standard normal distribution, commonly denoted $\mathcal N(0, 1)$ .

Problem 4. Prove the following properties:
- For $z \geq 0$ , $\Phi(-z) = 1 - \Phi(z)$ .
- $\Phi(0) = 1/2$ .
- For any $a < b$ , $\displaystyle \int_a^b \phi(t)\, \mathrm dt = \Phi(b) - \Phi(a)$ .
- $\displaystyle \lim_{T \to \pm \infty} \Phi(T) = 1/2 \pm 1/2$ .
(Click for Solution)

Solution. We observe that

$\begin{aligned} \Phi(-z) = \int_{-\infty}^{-z} \phi(t)\, \mathrm dt &= \int_{-\infty}^{-z} \phi(-t)\, \mathrm dt \\ &= \int_{z}^{\infty} \phi(t)\, \mathrm dt \\ &= 1 - \int_{-\infty}^z \phi(t)\, \mathrm dt = 1 - \Phi(z). \end{aligned}$

Setting $z = 0$ , $\Phi(0) = 1 - \Phi(0)$ yields $\Phi(0) = 1/2$ . For the third result,

$\begin{aligned} 1 = \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt &= \int_{-\infty}^a \phi(t)\, \mathrm dt + \int_a^b \phi(t)\, \mathrm dt + \int_b^\infty \phi(t)\, \mathrm dt \\ &= \Phi(a) + \int_a^b \phi(t)\, \mathrm dt + (1- \Phi(b)) \\ &= 1 - (\Phi(b) - \Phi(a)) + \int_a^b \phi(t)\, \mathrm dt. \end{aligned}$

Algebra yields the desired result. For the final result, the case $T \to \infty$ is immediate since

$\displaystyle \lim_{T \to \infty} \Phi(T) = \lim_{T \to \infty} \int_{-\infty}^{T} \phi(t)\, \mathrm dt = \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt = 1.$

Furthermore,

$\displaystyle \lim_{T \to -\infty} \Phi(T) = \lim_{T \to \infty} \Phi(-T) = 1 - \lim_{T \to \infty} \Phi(T) = 1 - 1 = 0.$

Problem 5. For any $\mu, \sigma \in \mathbb R$ with $\sigma > 0$ , define $\phi_{\mu, \sigma}$ by

$\displaystyle \phi_{\mu, \sigma}(x) := K\phi\left( \frac{x-\mu}{\sigma} \right),$

where $K > 0$ is a normalising constant i.e. $\displaystyle \int_{-\infty}^{\infty} \phi_{\mu, \sigma}(x)\, \mathrm dx = 1$ .

Evaluate $K$ . Furthermore, evaluate

$\displaystyle \int_{-\infty}^{\infty} Kx\phi_{\mu, \sigma}(x)\, \mathrm dx , \quad \int_{-\infty}^{\infty} Kx^2\phi_{\mu, \sigma}(x)\, \mathrm dx .$

(Click for Solution)

Solution. Make the substitution $t = (x-\mu)/\sigma \Rightarrow \mathrm dx = \sigma\, \mathrm dt$ so that $\phi_{\mu,\sigma}(x) = K\phi(t)$ , and

$\begin{aligned} 1 = \int_{-\infty}^{\infty} \phi_{\mu, \sigma}(x)\, \mathrm dx &= K\sigma \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt = K \sigma \cdot 1 = K \sigma. \end{aligned}$

Hence, $K = 1/\sigma$ . Similarly,

$\begin{aligned}\int_{-\infty}^{\infty} x\phi_{\mu, \sigma}(x)\, \mathrm dx &= \int_{-\infty}^{\infty} K\cdot (\mu + \sigma t)\cdot \phi(t) \cdot \sigma \, \mathrm dt \\ &= \mu \cdot K\sigma \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt + K\sigma^2 \int_{-\infty}^{\infty} t \phi(t)\, \mathrm dt \\ &= \mu \cdot 1 \cdot 1 + K \sigma^2 \cdot 0 = \mu. \end{aligned}$

Likewise,

$\begin{aligned} \int_{-\infty}^{\infty} x^2 \phi_{\mu, \sigma}(x)\, \mathrm dx &= \int_{-\infty}^{\infty} K \cdot (\mu + \sigma t)^2 \cdot \phi(t) \cdot \sigma\, \mathrm dt \\ &= \int_{-\infty}^{\infty} (\mu^2 + 2 \mu \sigma t + \sigma^2 t^2) \cdot \phi(t) \, \mathrm dt \\ &= \mu^2 \int_{-\infty}^{\infty} \phi(t)\, \mathrm dt + 2 \mu \sigma \mu^2 \int_{-\infty}^{\infty} t\phi(t)\, \mathrm dt + \sigma^2 \mu^2 \int_{-\infty}^{\infty} t^2 \phi(t)\, \mathrm dt \\ &= \mu^2 \cdot 1 + 2 \mu \sigma \cdot 0 + \sigma^2 \cdot 1 \\ &= \mu^2 + \sigma^2. \end{aligned}$

Problem 6. Prove that $\displaystyle \phi_{\mu, \sigma}(x) = \frac 1{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ .

(Click for Solution)

Solution. By the previous problems,

$\displaystyle \begin{aligned} \phi_{\mu, \sigma}(x) &= K \phi\left( \frac{x - \mu}{\sigma} \right) \\ &= KC e^{-\frac{(x-\mu)^2/\sigma^2}{n}} \\ &= \frac 1{\sigma} \cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-\mu)^2/\sigma^2}{2}} \\ &= \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}. \end{aligned}$

Remark 1. Problem 6 gives us the standard formula for the probability distribution function of a normal distribution $\mathcal N(\mu, \sigma^2)$ with mean $\mu$ and variance $\sigma^2$ :

$\displaystyle \begin{aligned} \phi_{\mu, \sigma}(x) &= \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \end{aligned}$

—Joel Kindiak, 20 May 25, 1710H
October 20, 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
Classical Optimisation
Problem 1. Let $f$ be a non-negative function satisfying the following conditions:
- $f$ is continuous on $\mathbb R$ ,
- $f$ is differentiable on $\mathbb R \backslash \{0\}$ ,
- $f$ is strictly increasing on $(-\infty, 0)$ ,
- $f$ is strictly decreasing on $(0, \infty)$ , and
- $f(|x|) \to 0$ as $x \to \infty$ .
Let $A(x)$ denote the rectangle under the curve $y=f(x)$ whose base that intersects the positive $x$ -axis has length $x$ . If $A$ has a minimum prove that $A$ is minimised when

$\begin{aligned} \left( 1 - \frac { f'(x) }{(g' \circ g^{-1} \circ f)(x)} \right) \cdot f(x) + (x - (g^{-1} \circ f)(x)) \cdot f'(x) &= 0, \end{aligned}$

where $g = f|_{(-\infty, 0)}$ .

(Click for Solution)

Solution. Fix $c > 0$ . If $f(c) = 0$ , then $f$ being strictly decreasing implies $f(c+1) < 0$ , a contradiction. Therefore, $f(0) > f(c) > 0$ . Since $f(-x) \to 0$ as $x \to \infty$ , there exists $b \in (-\infty, 0)$ such that $f(b) < f(c) < f(0)$ . Use the intermediate value theorem to obtain $a \in (b, 0)$ such that $f(a) = f(c)$ . It is not hard to verify that $f(x) > f(c)$ for $x \in [a, c]$ . Therefore, the rectangle has an area $A(c)$ of $A(c) = (c - a) \cdot f(c)$ .

In fact, more is true. Since $g := f|_{(-\infty, 0)}$ is differentiable and strictly decreasing,

$g(a) = f(a) = f(c)\quad \Rightarrow \quad a = g^{-1}(f(c)).$

Hence, $A(x) = (x - (g^{-1} \circ f)(x) ) \cdot f(x)$ . Differentiating,

$\begin{aligned} A'(x) &= (1 - (g^{-1})'(f(x)) \cdot f'(x)) \cdot f(x) + (x - (g^{-1} \circ f)(x) ) \cdot f'(x) \\ &= \left( 1 - \frac { f'(x) }{g'((g^{-1} \circ f)(x))} \right) \cdot f(x) + (x - (g^{-1} \circ f)(x)) \cdot f'(x).\end{aligned}$

Setting $A' = 0$ yields the desired result.

Problem 2. Suppose further in Problem 1 that $f$ is even. Prove that the $x$ -value $x_0$ that maximises $A$ satisfies the equation

$f(x_0) + x_0 \cdot f'(x_0) = 0$

and the inequality

$2f'(x_0) + x_0 \cdot f''(x_0) < 0.$

Furthermore, the maximum area is given by $A(x_0) = 2x_0 \cdot f(x_0)$ .

(Click for Solution)

Solution. If $f$ is even, then for any $x > 0$ ,

$\displaystyle (g^{-1} \circ f)(x) = g^{-1}(f(x)) = f^{-1}(f(-x)) = -x,$

Furthermore, $f'$ is odd:

$\displaystyle f'(-x)\cdot(-1) = \frac{\mathrm d}{\mathrm dx} (f(-x)) = \frac{\mathrm d}{\mathrm dx}(f(x)) = f'(x),$

so that $f'(-x) = -f'(x)$ . Therefore, for any $x < 0$ , $g(x) = f(x) = f(-x)$ and

$g'(x) = f'(-x) \cdot (-1) = f'(x).$

Hence, for any $x > 0$ ,

$\begin{aligned} (g' \circ g^{-1} \circ f)(x) &= f' ((g^{-1} \circ f)(x)) \\ &= f'(-x) = -f'(x) \neq 0. \end{aligned}$

Substituting into the result in Problem 1,

$\begin{aligned} \left( 1 - \frac { f'(x) }{-f'(x)} \right) \cdot f(x) + (x - (-x)) \cdot f'(x) &= 0, \end{aligned}$

yielding $f(x) + x \cdot f'(x) = 0$ . For the inequality, apply the second derivative test.

Problem 3. Evaluate the maximum area of the rectangle defined in Problem 1 given that $f(x) = 1/( 1+x^2 )$ .

(Click for Solution)

Solution. For $x > 0$ , $f(x) = 1/( 1+x^2 )$ so that

$\displaystyle f'(x) = -\frac 1{( 1+x^2 )^2} \cdot 2x = -\frac{2x}{( 1+x^2 )^2}.$

Since $f$ is even, by Problem 2,

$\displaystyle \begin{aligned} \frac 1{1+x_0^2 } + x_0 \cdot \left( -\frac{2x_0}{ (1+x_0^2)^2 } \right) &= 0 \\ 1+x_0^2 + x_0 \cdot (-2x_0) &= 0\\ 1 - x_0^2 &= 0\\ x_0 &= 1, \end{aligned}$

since $x_0 = 1$ . To verify that this value gives the maximum, we evaluate the second derivative:

$\displaystyle f''(x) = 2 \cdot \frac{x^2 -1 }{(1+x^2)^4}\quad \Rightarrow \quad f''(x_0) = 0.$

Then $f'(x_0) = -1$ and $f''(x_0) = 0$ , so that

$2 \cdot f'(x_0) + f''(x_0) = -2 < 0,$

as required. Therefore, $f(x_0) =1/2$ and the maximum area is given by

$A(x_0) = 2 \cdot 1 \cdot f(1) = 1.$

—Joel Kindiak. 9 Aug 25, 1510H
August 9, 2025
Odd and Even Functions
Recall that $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$ for any $x \in \mathbb R$ .

Definition 1. For any function $f : \mathbb R \to \mathbb R$ , define $\tilde f : \mathbb R \to \mathbb R$ by $\tilde f(x) := f(-x)$ .

Problem 1. Given functions $f, g: \mathbb R \to \mathbb R$ , define $h_1 := f + g$ and $h_2 := f \cdot g$ . Prove that $\tilde h_1 = \tilde f + \tilde g$ and $\tilde h_2 = \tilde f \cdot \tilde g$ .

(Click for Solution)

Solution. By definition,

$\begin{aligned} \tilde h_1(x) = h_1(-x) &= (f+g)(-x) \\ &= f(-x) + g(-x) \\ &= \tilde f(x) + \tilde g(x) \\ &= (\tilde f + \tilde g)(x), \end{aligned}$

and

$\begin{aligned} \tilde h_2(x) = h_2(-x) &= (f \cdot g)(-x) \\ &= f(-x) \cdot g(-x) \\ &= \tilde f(x) \cdot \tilde g(x) \\ &= (\tilde f \cdot \tilde g)(x). \end{aligned}$

Definition 2. A function $f : \mathbb R \to \mathbb R$ is odd (resp. even) if there exists an odd (resp. even) integer $n$ such that $\tilde f = (-1)^n \cdot f$ .

Example 1. For any integer $n$ , the function $f_n$ defined by $f_n(x) = x^n$ is an odd (resp. even) function if and only if $n$ is an odd (resp. even) number. Furthermore, $\sin$ is an odd function while $\cos$ is an even function.

Problem 2. For any function $f: \mathbb R \to \mathbb R$ and real number $\alpha \in \mathbb R$ , define the function $f_\alpha$ by $f_\alpha(x) := f(\alpha x)$ . In particular, $f_{-1} = \tilde f$ . Prove that $f_\alpha$ is odd (resp. even) if $f$ is odd (resp. even).

(Click for Solution)

Solution. We observe that if $f$ is odd or even, then $\tilde f = (-1)^n f$ , so that

$\begin{aligned} \tilde f_\alpha(x) &= f_\alpha(-x) = f(\alpha(-x))= f(-(\alpha x)) \\ &= \tilde f(\alpha x) = (-1)^n \cdot f(\alpha x) = (-1)^n \cdot f_\alpha(x), \end{aligned}$

hence $\tilde f_\alpha= (-1)^n \cdot f_\alpha$ .

Problem 3. Prove the following properties:
- if $f, g$ are odd, then $f + g$ is odd and $f \cdot g$ is even,
- if $f, g$ are even, then $f + g$ is even and $f \cdot g$ is even,
- if $f$ is odd and $g$ is odd (resp. even), then $g \circ f$ is odd (resp. even) whenever it exists.
- if $f$ is even, then $g \circ f$ is even.
(Click for Solution)

Solution. Suppose $\tilde f = (-1)^m \cdot f$ and $\tilde g = (-1)^n \cdot g$ , and assume $m \leq n$ for simplicity. Denote $h_1 = f+g$ and $h_2 = f \cdot g$ as per Problem 1. Then

$\tilde h_1 = \tilde f + \tilde g = (-1)^m \cdot f + (-1)^n \cdot g = (-1)^m \cdot (f + (-1)^{n-m} \cdot g).$

If $m$ and $n$ are both odd or both even, then $n-m$ is even, so that

$\tilde h_1 = (-1)^m \cdot (f + g) = (-1)^m \cdot h_1.$

Hence, $h_1$ is odd (resp. even) if and only if $f$ is odd (resp. even). Similarly,

$\tilde h_2 = \tilde f \cdot \tilde g = (-1)^m \cdot f \cdot (-1)^n \cdot g = (-1)^{m+n} \cdot h_2.$

If $m$ and $n$ are both odd or both even, then $m+n$ is always even, so that $\tilde h_2 = h_2$ unconditionally, i.e. $h_2 = f \cdot g$ is even. For the composite function, define $h_3 := g \circ f$ . Then

$\tilde h_3(x) = h_3(-x) = (g \circ f)(-x) = g(f(-x)) = g(\tilde f(x)) = (g \circ \tilde f)(x).$

Applying the odd or even condition on $f$ ,

$\begin{aligned} g \circ \tilde f &= g \circ (-(-1)^{m+1} \cdot f) \\ &= \tilde g \circ ((-1)^{m+1} \cdot f) \\ &= (-1)^n \cdot g \circ ((-1)^{m+1} \cdot f). \end{aligned}$

If $f$ is odd, then $m + 1$ is even, so that

$\tilde h_3 = g \circ \tilde f = (-1)^n \cdot (g \circ f) = (-1)^n \cdot h_3.$

Thus, $h_3$ is odd (resp. even) if and only if $g$ is odd (resp. even). If $f$ is even, then

$\tilde h_3 = g \circ \tilde f = g \circ f = h_3,$

so that $h_3$ is even.

Problem 4. For any function $f: \mathbb R \to \mathbb R$ , prove that there exist a unique odd function $f_1$ and a unique even function $f_2$ such that $f = f_1 + f_2$ .

(Click for Solution)

Solution. For existence, define the functions $f_1, f_2$ by

$f_1 := \frac 12 \cdot (f - \tilde f),\quad f_2 := \frac 12 \cdot (f + \tilde f).$

It is obvious that $f = f_1 + f_2$ . Since $\tilde{ \tilde f } = f$ ,

$\begin{aligned} \tilde f_1 &= \textstyle \frac 12 \cdot (\tilde f - \tilde{ \tilde f }) = \frac 12 \cdot (\tilde f - f) = -f_1, \\ \tilde f_2 &= \textstyle \frac 12 \cdot (\tilde f + \tilde{ \tilde f }) = \frac 12 \cdot (\tilde f + f) = f_2, \end{aligned}$

so that $f_1$ is odd and $f_2$ is even. For uniqueness, suppose $f = f_1 + f_2 = \hat f_1 + \hat f_2$ , where $\hat f_1$ is odd and $\hat f_2$ is even. Then $g := f_1 - \hat f_1 = \hat f_2 - f_2$ , so that $g$ is both odd and even. Hence,

$g = \tilde g = -g \quad \Rightarrow \quad g = 0\quad \Rightarrow \quad \hat f_1 = f_1\quad \Rightarrow \quad \hat f_2 = f_2.$

Problem 5. Fix $a > 0$ , any odd continuous function $f_1$ , and any even continuous function $f_2$ . Prove that

$\displaystyle \int_{-a}^a f_1 (x)\,\mathrm dx = 0, \quad \int_{-a}^a f_2 (x)\,\mathrm dx = 2 \cdot \int_0^a f_2 (x)\,\mathrm dx.$

In particular, for any continuous function $f : \mathbb R \to \mathbb R$ ,

$\displaystyle \int_{-a}^a f (x)\,\mathrm dx = 2 \cdot \int_0^a f_2 (x)\,\mathrm dx.$

(Click for Solution)

Solution. Using a substitution,

$\displaystyle \int_{-a}^0 f_1(x)\, \mathrm dx = \int_{a}^0 f_1(-t)\cdot (-1)\, \mathrm dt = \int_0^a -f_1(t)\, \mathrm dt = - \int_0^a f_1(t)\, \mathrm dt.$

Therefore,

$\displaystyle \begin{aligned} \int_{-a}^a f_1 (x)\,\mathrm dx &= \int_{-a}^0 f_1(x)\, \mathrm dx + \int_0^a f_1(x)\, \mathrm dx = 0.\end{aligned}$

Similarly, we can prove that

$\displaystyle \int_{-a}^0 f_2(x)\, \mathrm dx = \int_0^a f_2(t)\, \mathrm dt$

so that

$\displaystyle \begin{aligned} \int_{-a}^a f (x)\,\mathrm dx &= \int_{-a}^0 f_2(x)\, \mathrm dx + \int_0^a f_2(x)\, \mathrm dx = 2 \cdot \int_0^a f_2(x)\, \mathrm dx.\end{aligned}$

Problem 6. For any continuous function $f : \mathbb R \to \mathbb R$ and $T > 0$ , define

$\displaystyle a_n := \frac 2T \int_{-T/2}^{T/2} f(t)\, \mathrm \cos \left( \frac{2n \pi t}{T} \right)\, \mathrm dt,\quad b_n := \frac 2T \int_{-T/2}^{T/2} f(x)\, \mathrm \sin \left( \frac{2n \pi t}{T} \right)\, \mathrm dt.$

Prove that $a_n = 0$ (resp. $b_n = 0$ ) if $f$ is odd (resp. even).

(Click for Solution)

Solution. By Example 1 and Problem 2, $\cos_{2n\pi/T}$ is even and $\sin_{2n\pi/T}$ is odd.

Thus, if $f$ is odd, then $f \cdot \cos_{2n\pi/T}$ is odd by Problem 3, so that $a_n = 0$ by Problem 5.

Similarly, if $f$ is even, then $f \cdot \sin_{2n\pi/T}$ is odd by Problem 3, so that $b_n = 0$ by Problem 5.

—Joel Kindiak, 5 Aug 25, 2044H
August 5, 2025
Some Calculus of Variations

Problem 1. Let $f : (a, b) \to \mathbb R$ be continuous. Prove that if $f \neq 0$ , then there exists $c, d \in (a, b)$ such that $\pm f(x) > 0$ whenever $x \in (c, d)$ .

(Click for Solution)

Solution. Suppose $f(x_0) \neq 0$ for some $x_0 \in (a, b)$ . Then $f(x_0) > 0$ or $f(x_0) < 0$ . In the case $f(x_0) > 0$ , use the continuity of $f$ to find $\delta \in (0, \min\{x_0-a, b-x_0\})$ such that

$\displaystyle |x - x_0| < \delta \quad \Rightarrow \quad |f(x) - f(x_0)| < \frac{f(x_0)}{2}.$

By expanding the inequality,

$\displaystyle f(x) - f(x_0) > -\frac{f(x_0)}{2} \quad \Rightarrow \quad f(x) > \frac{f(x_0)}{2} > 0.$

Defining $c := x_0 - \delta$ and $d:= x_0 + \delta$ ,

$x \in (c, d) \quad \Rightarrow \quad |x - x_0| < \delta \quad \Rightarrow \quad f(x) > 0.$

In the case $f(x_0) < 0$ , we have $(-f)(x_0) > 0$ . Applying the first case, there exists $c, d \in (a, b)$ such that

$x \in (c, d) \quad \Rightarrow \quad -f(x) = (-f)(x) > 0.$

Problem 2. Let $f : (a, b) \to \mathbb R$ be continuous and suppose $f(x) \geq 0$ for any $(a, b)$ . Prove that

$\displaystyle \int_a^b f(x)\, \mathrm dx = 0 \quad \Rightarrow \quad f = 0.$

(Click for Solution)

Solution. Suppose instead that $f \neq 0$ . Then there exists $(c, d) \subseteq (a, b)$ such that for any $x \in (c, d)$ , $f(x) > 0$ . Hence,

$\begin{aligned} \int_a^b f(x)\, \mathrm dx &= \int_a^c f(x)\, \mathrm dx + \int_c^d f(x)\, \mathrm dx + \int_d^b f(x)\, \mathrm dx \\ &\geq 0 + \int_c^d f(x)\, \mathrm dx + 0\\ &= \int_c^d f(x)\, \mathrm dx > 0 . \end{aligned}$

Definition 1. Call a function $h \in C^\infty$ compactly supported if there exists real numbers $c < d$ such that $h(x) = 0$ for $x \leq c$ and $x \geq d$ .

Problem 3. Let $f : (a, b) \to \mathbb R$ be continuous. Suppose that for any compactly supported $h$ ,

$\displaystyle \int_a^b f(x)h(x)\, \mathrm dx = 0.$

Then $f = 0$ .

(Click for Solution)

Solution. By way of contradiction, suppose $f \neq 0$ . By Problem 1, there exists $(c, d) \subseteq (a, b)$ such that without loss of generality, $f(x) > 0$ whenever $x \in (c, d)$ . Define $h \in C^\infty$ by

$h(x) = \begin{cases} e^{ -\frac{1}{ (x-c)(d-x) } }, & x \in (c, d), \\ 0, & \text{otherwise}.\end{cases}$

Then it is clear that $h$ is compactly supported, and that for any $x \in (a, b)$ , $f(x)h(x) \geq 0$ . By Problem 2, $f \cdot h = 0$ . However for $m := (c+d)/2$ ,

$0 = (f \cdot h)(m) = f(m) \cdot h(m) > 0 \cdot 0 = 0,$

a contradiction.

—Joel Kindiak, 25 Jun 25, 1420H

June 25, 2025
Curious Continuity

Problem 1. Define the function $f : [-1, 1] \to \mathbb R$ by

$f(x) = \begin{cases} x^2, & x \in [-1,1] \cap \mathbb Q, \\ -x^2, & x \in [-1,1] \backslash \mathbb Q. \end{cases}$

Prove that $f$ is continuous at $0$ .

(Click for Solution)

Solution. We observe that $-x^2 \leq f(x) \leq x^2$ . Since $\pm x^2 \to 0$ as $x \to 0$ , by the squeeze theorem, $f(x) \to 0 = 0^2 = f(0)$ as $x \to 0$ , as required.

Problem 2. For any function $f$ that is continuous at $c$ , prove that if $f(c) > 0$ , then there exists $\delta > 0$ such that for $x \in (c - \delta, c + \delta)$ , $f(x) > 0$ .

(Click for Solution)

Solution. Suppose $f(c) > 0$ . Fix $\epsilon > 0$ that will be chosen later. Since $f$ is continuous at $c$ , for any $k > 0$ , there exists $\delta > 0$ such that

$|x-c| < \delta \quad \Rightarrow \quad |f(x) - f(c)| < k \cdot \epsilon.$

Expanding the right-hand side, $f(x) > f(c) - k \cdot \epsilon$ . Setting $\epsilon = f(c) > 0$ and $k = 1/2$ ,

$\displaystyle f(x) > f(c) - \frac{1}{2} \cdot f(c) = \frac{f(c)}{2} > 0.$

Problem 3. Let $f : [a, b] \to \mathbb R$ be continuous. Suppose that for any $x \in [a, b]$ , $f(x) \neq 0$ . Prove that there exists $C > 0$ such that either $f < -C$ or $f > C$ . In particular, $|f| > C$ .

(Click for Solution)

Solution. The function $g := | \cdot | \circ f : [a, b] \to \mathbb R$ is a composition of continuous functions and thus continuous. By the extreme value theorem, there exists $c \in [a, b]$ such that for any $x \in [a, b]$ ,

$|f(x)| = g(x) \geq g(c) = |f(c)|.$

Since $f(c) \neq 0$ , $|f(c)| > 0$ , so that defining $C := |f(c)| / 2 > 0$ yields

$|f(x)| \geq |f(c)| > C.$

Now suppose $f(c_0) > 0$ without loss of generality. In the case $f(c_0) < 0$ , apply the first case to the continuous map $-f$ . We claim that $f > 0$ .

Suppose, for a contradiction, there exists $c_1 \in (a, b)$ such that $f(c_1) < 0$ . Assume $c_1 \in (c_0, b)$ without loss of generality. By the intermediate value theorem, there exists $c_2 \in (c_0, c_1)$ such that $f(c_2) = 0$ , a contradiction. Therefore, $f > 0$ . For any $x \in [a, b]$ , $f(x) = |f(x)| > C$ , yielding $f > C$ , as required.

Problem 4. Let $f : \mathbb R \to \mathbb R$ and suppose that for any $x, y \in \mathbb R$ ,

$f(x + y) = f(x) + f(y).$

Prove that if $f$ is continuous at some $c \in \mathbb R$ , then $f$ is continuous on $\mathbb R$ .

(Click for Solution)

Fix $x \in \mathbb R$ . Since $f$ is continuous at $c$ ,

$\begin{aligned} f(c) &= \lim_{t \to 0} f(c+t) \\ &= \lim_{t \to 0} (f(c)+f(t)) \\ &= \lim_{t \to 0} f(c)+\lim_{t \to 0} f(t) \\ &= f(c) \cdot 1 + \lim_{t \to 0} f(t), \end{aligned}$

so that $\displaystyle \lim_{t \to 0}f(t) = 0$ . Therefore,

$\begin{aligned} \lim_{t \to 0} f(x+t) &= \lim_{t \to 0} (f(x)+f(t)) \\ &= \lim_{t \to 0} f(x) + \lim_{t \to 0} f(t) \\ &= f(x) \cdot 1 + 0 = f(x). \end{aligned}$

Problem 5. Let $f : [0, 1] \to \mathbb R$ be continuous and $f(0) = f(1)$ . Prove that for any $n \in \mathbb N^+$ , there exists $\xi \in [0,1]$ such that $f(\xi+1/n) = f(\xi)$ .

(Click for Solution)

Solution. For each $n$ , define the function $g := f(\cdot + 1/n) - f$ . We aim to show that $g$ has at least one root. We claim that there exists $x_0 \in [0,1]$ such that $g(x_0) \geq 0$ . Otherwise, $f < f(\cdot + 1/n)$ . In particular,

$\displaystyle f(0) < f(1/n) < f(2/n) < \cdots < f(1) = f(0),$

a contradiction. Therefore, there exists $x_0 \in [0,1]$ such that $g(x_0) \geq 0$ . A similar argument shows that there exists $x_1 \in [0,1]$ such that $g(x_1) \leq 0$ .

If at least one equality holds, then set $\xi = x_0$ or $\xi = x_1$ . Otherwise, $g(x_0) > 0$ and $g(x_1) < 0$ . Use the intermediate value theorem to find $c$ between $x_0$ and $x_1$ such that $g(c) = 0$ , as required.

—Joel Kindiak, 16 Aug 25, 2039H

June 18, 2025
Exponential Fourier Series

The function $f$ is defined by $f(t) = e^t$ for $-1 < t \leq 1$ and extended periodically by $f(t) = f(t+2)$ .

Problem 1. For any nonnegative integer $n$ , evaluate the integrals

$\displaystyle \int f(t) \cos (n \pi t)\, \mathrm dt,\quad \int f(t) \sin(n \pi t)\, \mathrm dt.$

(Click for Solution)

Solution. Integrating by parts twice,

$\begin{aligned} \int e^{t} \cos (n \pi t)\, \mathrm dt &= \underbrace{e^t}_{\mathrm I} \cdot \underbrace{\cos (n \pi t)}_{\mathrm S} - \int \underbrace{e^t}_{\mathrm I} \cdot \underbrace{(-n\pi \cdot \sin(n \pi t))}_{\mathrm D}\, \mathrm dt \\ &= e^t \cos(n\pi t) + n \pi \int e^t \sin(n \pi t)\, \mathrm dt \\ &= e^t \cos(n \pi t) + n\pi \left( \underbrace{e^t}_{\mathrm I} \cdot \underbrace{\sin (n \pi t)}_{\mathrm S} - \int \underbrace{e^t}_{\mathrm I} \cdot \underbrace{(n\pi \cdot \cos(n \pi t))}_{\mathrm D}\, \mathrm dt \right) \\ &= e^t ( \cos(n \pi t) + n\pi \sin(n \pi t)) - n^2 \pi^2 \int e^t \cos(n\pi t)\, \mathrm dt. \end{aligned}$

By careful algebruh,

$\displaystyle \int e^{t} \cos (n \pi t)\, \mathrm dt = \frac{e^t ( \cos(n \pi t) + n\pi \sin(n \pi t))}{1 + n^2 \pi^2} + C_1.$

In a similar though tedious manner,

$\displaystyle \int e^{t} \sin (n \pi t)\, \mathrm dt = \frac{e^t ( \sin(n \pi t) - n\pi \cos(n \pi t))}{1 + n^2 \pi^2} + C_2.$

Remark 1. For an alternate solution, recall Euler’s formula which states that

$e^{i n \pi t} = \cos(n \pi t) + i \sin(n \pi t).$

Multiplying by $f(t)$ then integrating on the left-hand side yields

$\begin{aligned} \int f(t) e^{i n \pi t} \, \mathrm dt &= \int e^t \cdot e^{i n \pi t}\, \mathrm dt \\ &= \int e^{(1 + i n \pi )t}\, \mathrm dt \\ &= \frac {e^{(1 + i n \pi )t}}{1 + i n \pi} + C \\ &= e^t \cdot \frac{1 - i n \pi}{1 + n^2 \pi^2} (\cos (n \pi t) + i \sin(n \pi t)) + C \\ &= e^t \cdot \frac{(\cos (n \pi t) + n \pi \sin(n \pi t)) + i (\sin(n \pi t) - n \pi \cos(n \pi t))}{1 + n^2 \pi^2} + C \\ &= e^t \cdot \frac{\cos (n \pi t) + n \pi \sin(n \pi t)}{1 + n^2 \pi^2} + i \cdot e^t \cdot \frac{\sin(n \pi t) - n \pi \cos(n \pi t)}{1 + n^2 \pi^2} + C . \end{aligned}$

The real and imaginary parts then agree with our solutions in Problem 1.

Problem 2. Denoting $T = 2$ , the Fourier series of $f$ is defined by

$\displaystyle f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left( \frac{2n\pi t}{T} \right) + b_n \sin\left( \frac{2n\pi t}{T} \right) \right),$

where the Fourier coefficients $a_n, b_n$ are defined by

$\displaystyle a_n = \frac 2{T} \int_{-T/2}^{T/2} f(t) \cos\left( \frac{2n\pi t}{T} \right)\, \mathrm dt,\quad b_n = \frac 2{T} \int_{-T/2}^{T/2} f(t) \sin\left( \frac{2n\pi t}{T} \right)\, \mathrm dt.$

Use Problem 1 to evaluate the Fourier series of $f$ .

(Click for Solution)

Solution. By Problem 1,

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

Similarly,

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

By the definition of the Fourier series,

$\displaystyle \begin{aligned} f(t) &= \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left( \frac{2n\pi t}{T} \right) + b_n \sin\left( \frac{2n\pi t}{T} \right) \right) \\ &= \frac{a_0}{2} + \sum_{n=1}^\infty a_n (( \cos (n \pi t) - \pi n \sin (n \pi t) ) \\ &= \frac{e - e^{-1}}{2} + \sum_{n=1}^\infty \frac{(-1)^n ( e - e^{-1} )}{1 + n^2 \pi^2} (( \cos (n \pi t) - \pi n \sin (n \pi t) ). \end{aligned}$

Remark 2. If instead we used the complex-exponential form of the Fourier series

$\displaystyle f(t) = \sum_{n = -\infty}^\infty c_n e^{i \cdot 2\pi n t/T},\quad c_n := \frac 1{T} \int_{-T/2}^{T/2} f(t) \cdot e^{i \cdot 2\pi n t/T}\, \mathrm dt,$

then complex-valued integration yields

$\displaystyle c_n = \frac 12 \left[ \frac{e^{(1 + i \cdot \pi n) t}}{1 + i \cdot \pi n} \right]_{-1}^1 = \frac {(-1)^n (e - e^{-1})(1 - (\pi n) \cdot i)}{2(1 + n^2 \pi^2)},$

recovering the results in Problem 2 via the relationships

$a_n = c_n + \bar{c}_n,\quad b_n = -i \cdot (c_n - \bar{c}_n) =-n\pi a_n.$

—Joel Kindiak, 14 May 25, 1911H

May 14, 2025
The Wallis Product

Problem 1. By considering the integral $\displaystyle I(n) := \int_0^{\pi} \sin^n(x)\, \mathrm dx$ , prove the Wallis product

$\displaystyle \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45 \cdot \cdots := \lim_{n \to \infty} \prod_{k=1}^n \frac{2n \cdot 2n}{(2n-1)(2n+1)} = \frac{\pi}{2}.$

(Click for Solution)

Solution. We leave it as an exercise to check that $I(0) = \pi$ and $I(1) = 2$ . These observations motivate an expression of $I(n+2)$ in terms of $I(n)$ . To that end, we integrate by parts to obtain

$\begin{aligned} & \int \sin^{n+2}(x)\, \mathrm dx = \int \sin(x) \sin^{n+1}(x)\, \mathrm dx \\ &= \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{\sin^{n+1}(x)}_{\mathrm S} - \int \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{(n+1) \sin^n(x) \cos(x) }_{\mathrm D}\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \cos^2(x) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int (1-\sin^2(x)) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \left( \int \sin^n(x)\, \mathrm dx - \int \sin^{n+2}(x)\, \mathrm dx\right). \end{aligned}$

By algebruh,

$\displaystyle (n+2) \int \sin^{n+2}(x)\, \mathrm dx = -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \sin^n(x)\, \mathrm dx.$

Plugging in the limits of $[0, \pi]$ ,

$\displaystyle (n+2) \int_0^{\pi} \sin^{n+2}(x)\, \mathrm dx = (n+1) \int_0^{\pi} \sin^n(x)\, \mathrm dx.$

Hence,

$\displaystyle \frac{I(n+2)}{I(n)} = \frac{n+1}{n+2}.$

In particular,

$\displaystyle I(2k+2) = \frac{2k+1}{2k+2} \cdot I(2k) \quad\Rightarrow \quad I(2n+2) = \pi \cdot \prod_{k=1}^n \frac{2k-1}{2k}$

Similarly,

$\displaystyle I(2n+1) = 2 \cdot \prod_{k=1}^n \frac{2k}{2k+1}.$

Dividing the terms appropriately,

$\displaystyle \prod_{k=1}^n \frac{2k \cdot 2k}{(2k-1) (2k+1)} = \frac{\pi}{2} \cdot \frac{I(2n+1)}{I(2n+2)}.$

It suffices to prove that $I(2n+1)/I(2n+2) \to 1$ . Since $0 \leq \sin(x) \leq 1$ for $x \in [0, \pi]$ ,

$0 \leq \sin^{n+1}(x) \leq \sin^n(x) \leq \sin^{n-1}(x) \leq 1.$

By the monotonicity of integration, $I(n+1) \leq I(n)$ so that

$\displaystyle 1 \leq \frac{I(2n+1)}{I(2n+2)} \leq \frac{I(2n)}{I(2n+2)} = \frac{2n+2}{2n+1} \to 1.$

By the squeeze theorem, $I(2n+1)/I(2n+2) \to 1$ , as required.

—Joel Kindiak, 21 Apr 25, 2129H

April 21, 2025
The Intermediate Value Derivative

Problem 1. Let $f : [a, b] \to \mathbb R$ differentiable. Assume $f'(a) < f'(b)$ . Prove that for any $m \in (f'(a), f'(b))$ , there exists $c \in [a, b]$ such that $f'(c) = m$ .

(Click for Solution)

Solution. We observe that $f'(c) = m \iff (f(x)-mx)'(c) = 0$ . Define the differentiable function $g : [a, b] \to \mathbb R$ by $g(x) = f(x) - mx$ . Then it suffices to find $c \in [a, b]$ such that $g'(c) = 0$ .

Since $g$ is differentiable on $[a, b]$ , it is continuous on $[a, b]$ . By the extreme value theorem, there exists $c \in [a, b]$ such that $g(x) \geq g(c)$ for any $x \in [a, b]$ . We claim that $g'(c) = 0$ .

Firstly, we note that $g'(a) = f'(a) - m < 0$ , so that there exists $\delta_1 > 0$ such that $g(c) \leq g(a+\delta_1) < g(a)$ . Similarly, $g'(b) = f'(b) - m > 0$ , so that there exists $\delta_2 > 0$ such that $g(c) \leq g(b-\delta_2) < g(b)$ . Hence, $g(c) \neq g(a), g(b)$ . Therefore, $c \in (a, b)$ , which implies $g'(c) = 0$ , as required.

Remark 1. Any function that satisfies the conclusion of the intermediate value theorem is called a Darboux function. The intermediate value theorem demonstrates that continuous functions are Darboux functions. This problem demonstrates that derivatives are automatically Darboux functions, even if they are not continuous.

—Joel Kindiak, 22 Jan 25, 1234H

April 3, 2025