KindiakMath

Category: Calculus

Real-Analytic Inequalities

Fix $p > 1, q > 1$ such that $\displaystyle \frac 1p + \frac 1q = 1$ .

Problem 1. Verify that $p+q = pq$ .

(Click for Solution)

Solution. By algebruh,

$\displaystyle 1 = \frac 1p + \frac 1q = \frac{p+q}{pq}\quad \Rightarrow \quad p+q = pq.$

Problem 2. Show that for $x \geq 0$ , $\displaystyle x \leq \frac{x^p}{p} + \frac 1q$ with equality if and only if $x = 1$ .

(Click for Solution)

Solution. Define the function $f(x) := x^p/p + 1/q - x$ . By observation,

$\displaystyle f(1) = \frac 1p + \frac 1q - 1 = 0.$

Furthermore,

$f'(x) = x^{p-1} - 1 \quad \Rightarrow \quad f''(x) = (p-1)x^{p-2} > 0.$

By the second derivative test, solving for $f'(x) = 0$ yields a global minimum:

$\displaystyle x^{p-1} - 1 = 0 \quad \Rightarrow \quad x = 1.$

Therefore, $f(x) \geq f(1) = 0$ for any $x \geq 0$ , as required.

Problem 3. Use Problem 2 to prove Young’s inequality: for any $a \geq 0, b \geq 0$ ,

$\displaystyle ab \leq \frac{a^p}{p} + \frac{ b^q }{q}$

with equality if and only if $a^p = b^q$ .

(Click for Solution)

Solution. Setting $x = a/b^{q-1}$ ,

$\begin{aligned} \frac a{b^{q-1}} &\leq \frac{\left( \frac a{b^{q-1}} \right)^p}{p} + \frac 1q \\ \frac a{b^{q-1}} \cdot b^q &\leq \frac{1}{p} \cdot \frac{a^p}{b^{p(q-1)}} \cdot b^q + \frac {b^q}q \\ ab &\leq \frac{a^p}{p} \cdot b^{p+q-pq} + \frac {b^q}q \\ ab &\leq \frac{a^p}{p} + \frac {b^q}q . \end{aligned}$

Furthermore, since $p(q-1) = pq - p = q$ , equality holds if and only if

$\displaystyle x = 1 \quad \iff \quad a = b^{q-1} \quad \iff \quad a^p = b^{p(q-1)} \quad \iff \quad a^p = b^q.$

Problem 4. Use Young’s inequality to prove Hölder’s inequality for two-dimensional vectors: given non-negative numbers $a_1,a_2,b_1,b_2$ ,

$\displaystyle a_1 b_1 + a_2 b_2 \leq \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q}.$

(Click for Solution)

Solution. If $\left( a_1^p + a_2^p \right)^{1/p} = 0$ , then $0 \leq a_1^p + a_2^p = 0$ implies $a_1 = a_2 = 0$ , and the inequality holds trivially. Without loss of generality, suppose the right-hand side is non-zero. It suffices to prove that

$\displaystyle \frac{ a_1 b_1 + a_2 b_2 }{ \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q} } \leq 1.$

By one application of Young’s inequality,

$\begin{aligned} \frac{ a_1 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_1 }{\left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right)^{p/p} } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right)^{q/q} } \\ &\leq \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right) }. \end{aligned}$

By a second application of Young’s inequality,

$\begin{aligned} \frac{ a_2 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_2 }{\left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) }. \end{aligned}$

Summing the inequalities,

$\begin{aligned} \frac{ a_1 b_1 + a_2 b_2 }{ \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{ a_1 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_1 }{\left( b_1^q + b_2^q \right)^{1/q} } + \frac{ a_2 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_2 }{\left( b_1^q + b_2^q \right)^{1/q} } \\ &= \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right) } + \frac{a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) } \\ &= \frac{a_1^p + a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q + b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) } = \frac 1p + \frac 1q = 1.\end{aligned}$

Problem 5. Use Hölder’s inequality to prove Minkowski’s inequality for two-dimensional vectors: given real numbers $a_1,a_2,b_1,b_2$ ,

$\displaystyle (|a_1 + b_1|^p + |a_2 + b_2|^p)^{1/p} \leq (|a_1|^p + |a_2|^p )^{1/p} + (|b_1|^p + |b_2|^p )^{1/p}.$

(Click for Solution)

Solution. By the vanilla triangle inequality,

$\begin{aligned} &|a_1 + b_1|^p + |a_2 + b_2|^p \\ &= |a_1 + b_1| |a_1 + b_1|^{p-1} + |a_2 + b_2| |a_2 + b_2|^{p-1} \\ &\leq (|a_1| + |b_1|) |a_1 + b_1|^{p-1} +(|a_2| + |b_2|) (a_2 + b_2)^{p-1} \\ &= |a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} + |b_1| |a_1 + b_1|^{p-1} + |b_2| |a_2 + b_2|^{p-1}. \end{aligned}$

By Hölder’s inequality,

$\begin{aligned} & |a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} \\ &\leq (|a_1|^p + |a_2|^p)^{1/p} \cdot (|a_1 + b_1|^{q(p-1) }+ |a_2 + b_2|^{q(p-1)})^{1/q} \\ &\leq (|a_1|^p + |a_2|^p)^{1/p} \cdot (|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}, \end{aligned}$

so that

$\displaystyle \frac{|a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} }{(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|a_1|^p + |a_2|^p)^{1/p}.$

Similarly,

$\displaystyle \frac{|b_1| |a_1 + b_1|^{p-1} + |b_2| |a_2 + b_2|^{p-1} }{(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|b_1|^p + |b_2|^p)^{1/p}.$

Combining the displays,

$\displaystyle \frac{ |a_1 + b_1|^p + |a_2 + b_2|^p} {(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|a_1|^p + |a_2|^p)^{1/p} + (|b_1|^p + |b_2|^p)^{1/p}.$

The result follows by the observation

$\displaystyle \frac{ |a_1 + b_1|^p + |a_2 + b_2|^p} {(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} = (|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/p}.$

Remark 1. Denoting

$\left\| \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} \right\|_p := (|a_1|^p + |a_2|^p )^{1/p}$

and defining $\mathbf a := \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$ , Hölder’s inequality reduces to

$\| \mathbf a \cdot \mathbf b \|_1 \leq \| \mathbf a \|_p \| \mathbf b \|_q,$

and Minkowski’s inequality reduces to

$\| \mathbf a + \mathbf b \|_p \leq \| \mathbf a \|_p + \| \mathbf b \|_p.$

Setting $p = q = 2$ reduces to the usual Cauchy-Schwarz inequality and triangle inequality for two-dimensional vectors. Furthermore, these results hold for $n$ -dimensional vectors, and even “infinite-dimensional” (sufficiently nice) functions.

January 5, 2026
Nontrivial Rates of Change

Big Idea

The rate of change of a quantity $x(t)$ in terms of time $t$ is $\displaystyle \frac{\mathrm dx}{\mathrm dt}$ . Furthermore, given $y \equiv y(x)$ , the chain rule yields

$\displaystyle \frac{\mathrm dy}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt}.$

Questions

Question 1. The point $P(x, y)$ above the $y$ -axis at time $t$ is defined by the equations

$x = \cos (t), \quad y = \sin (t),\quad 0 \leq t < \pi.$

Define $A(-1, 0)$ . Let $Q$ denote the intersection of $AP$ with the $y$ -axis.

Evaluate the rate of change of the $y$ -coordinate of $Q$ at the point the gradient of the tangent to the curve at $P$ is $-1/\sqrt{3}$ .

(Click for Solution)

Solution. Differentiating with respect to $t$ ,

$\displaystyle \frac{\mathrm dx}{\mathrm dt} = -\sin(t),\quad \frac{\mathrm dy}{\mathrm dt} = \cos(t).$

By the chain rule,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dt} &= \frac{\mathrm dy}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt} \\ \cos(t) &= -\frac 1{\sqrt 3} \cdot -\sin(t)\\ \tan(t) &= \sqrt{3} \\ t &= \pi/3. \end{aligned}$

Let $y_Q$ denote the $y$ -coordinate of $Q$ . The equation of $AP$ is given by

$\displaystyle \frac{ y - 0 }{ x- (-1) } = \frac{\sin(t) - 0}{\cos(t) - (-1)}\quad\Rightarrow \quad y = \frac{\sin(t)}{1 + \cos(t)} \cdot (x+1).$

At the $y$ -axis, $x = 0$ , so that

$\displaystyle y_Q = \frac{\sin(t)}{1 + \cos(t)}.$

Applying the quotient rule,

$\begin{aligned} \frac{\mathrm dy_Q}{\mathrm dt} &= \frac{\mathrm d}{\mathrm dt} \left( \frac{\sin(t)}{1 + \cos(t)} \right) \\ &= \frac{\displaystyle (1 + \cos(t)) \cdot \frac{\mathrm d}{\mathrm dt}(\sin(t)) - \sin(t) \cdot \frac{\mathrm d}{\mathrm dt}(1 + \cos(t))}{( 1 + \cos(t) )^2} \\ &= \frac{(1 + \cos(t)) \cdot \cos(t) - \sin(t) \cdot (0 + (-\sin(t))}{( 1 + \cos(t) )^2} \\ &= \frac{(\cos(t) + \cos^2(t)) + \sin^2(t) }{( 1 + \cos(t) )^2} \\ &= \frac{\cos(t) + (\cos^2(t) + \sin^2(t)) }{( 1 + \cos(t) )^2} \\ &= \frac{ \cos(t) + 1 }{( 1 + \cos(t) )^2} = \frac{ 1 }{ 1 + \cos(t) }. \end{aligned}$

Therefore, at $t = \pi/3$ , since $\cos(t) = 1/2$ ,

$\begin{aligned} \frac{\mathrm dy_Q}{\mathrm dt} &= \frac{ 1 }{ 1 + 1/2 } = \frac 23\, \text{units}/\text{s}. \end{aligned}$

Alternate Solution. Using double-angle formulae,

$\displaystyle y_Q = \frac{\sin(t)}{1 + \cos(t)} = \frac{2 \sin(t/2) \cos (t/2)}{2 \cos^2(t/2)} = \tan(t/2).$

Differentiating,

$\displaystyle \frac{ \mathrm dy_Q }{ \mathrm dt } = \frac 12 \sec^2(t/2).$

The rest of the question follows as per usual.

Question 2. A man of height $1.6\, \text{m}$ metres is currently $300\, \text{m}$ away from a pole of height $3.6\, \text{m}$ . He runs in a straight line towards the pole at a speed of $3\, \text{m}/\text{s}$ . Let $\theta$ denote the angle of elevation from the man to the top of the pole.

Evaluate the rate of change of $\theta$ when the man is at the pole.

(Click for Solution)

Solution. The man’s distance from the pole at time $t$ is given by $(300 - 3t)$ metres. By considering the height difference between the pole and the man,

$\displaystyle \tan(\theta(t)) = \frac{3.6-1.6}{300 - 3t}\quad \Rightarrow \quad \theta(t) = \tan^{-1}\left( \frac{2}{3(100- t)} \right).$

Applying the chain rule,

$\begin{aligned} \frac{\mathrm d\theta}{\mathrm dt} &= \frac{ \mathrm d }{ \mathrm dt } \left( \tan^{-1}\left( \frac{2}{3(100- t)} \right) \right) \\ &= \frac 1{1 + \left( \frac{2}{3(100- t)} \right)^2} \cdot \frac{\mathrm d}{\mathrm dt}\left( \frac{2}{3(100- t)} \right) \\ &= \frac 1{1 + \frac{4}{9(100- t)^2}} \cdot \frac 23 \cdot \frac{\mathrm d}{\mathrm dt}\left( (100-t)^{-1} \right) \\ &= \frac {9(100- t)^2}{4 + 9(100- t)^2} \cdot \frac 23 \cdot (-1)(100-t)^{-2} \cdot \frac{\mathrm d}{\mathrm dt}(100 - t) \\ &= \frac {9(100- t)^2}{4 + 9(100- t)^2} \cdot \frac 23 \cdot (-1)(100-t)^{-2} \cdot (-1) \\ &= \frac {6}{4 + 9(100- t)^2}. \end{aligned}$

At the pole, $t = 100$ , so that

$\displaystyle \frac{\mathrm d\theta}{\mathrm dt} = \frac 6{4 + 9(100-100)^2} = 1.5\, \text{rad}/\text{s}.$

—Joel Kindiak, 4 Sept 25, 1149H

November 20, 2025
Exploiting the Trifecta

Big Idea

Three powerful results in differentiation handle more exotic kinds of differentiation, the product rule:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(uv) = v \cdot \frac{\mathrm du}{\mathrm dx} + u \cdot \frac{\mathrm dv}{\mathrm dx},$

the quotient rule:

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left(\frac uv \right) = \frac{\displaystyle v \cdot \frac{\mathrm du}{\mathrm dx} - u \cdot \frac{\mathrm dv}{\mathrm dx}}{v^2},$

and the chain rule: if $y = f(u)$ , then

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f(u)) = f'(u) \cdot \frac{\mathrm du}{\mathrm dx} \quad \iff \quad \frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm du} \cdot \frac{\mathrm du}{\mathrm dx}.$

Questions

Question 1. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} (2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8))$ .

(Click for Solution)

Solution. We first simplify the function

$\begin{aligned} 2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8) &= 4^x \cdot 4x^{1/2} \cdot 8 \ln(x) \\ &= 32 \cdot x^{1/2} \cdot 4^x \cdot \ln(x).\end{aligned}$

Apply the product rule once to obtain

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^{1/2} \cdot 4^x) &= 4^x \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \frac{\mathrm d}{\mathrm dx}(4^x).\end{aligned}$

Apply the product rule on the original function to obtain

$\begin{aligned} & \frac{\mathrm d}{\mathrm dx} (2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8)) \\ &= 32 \cdot \frac{\mathrm d}{\mathrm dx} ((x^{1/2} \cdot 4^x) \cdot \ln(x))\\ &= 32 \cdot \left( \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2} \cdot 4^x) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( \ln(x) \cdot \left( 4^x \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \frac{\mathrm d}{\mathrm dx}(4^x) \right) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( 4^x \cdot \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(4^x) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( 4^x \cdot \ln(x) \cdot \frac 12 x^{-1/2} + x^{1/2} \cdot \ln(x) \cdot 4^x \ln (4) + x^{1/2} \cdot 4^x \cdot \frac 1x \right) \\ &= 16 \cdot 4^x \cdot x^{-1/2} \cdot ( \ln(x) + 2x \cdot \ln(x) \cdot \ln (4) + 2 ).\end{aligned}$

Question 2. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left(\sin \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \right)$ .

(Click for Solution)

Solution. Noting that $\sin^{-1} \neq \csc = 1/{\sin}$ , applying the chain rule repeatedly and carefully,

$\begin{aligned} &\frac{\mathrm d}{\mathrm dx} \left(\sin \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \frac{\mathrm d}{\mathrm dx} \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( \frac{\mathrm d}{\mathrm dx} (\csc(x^2)) - \frac{\mathrm d}{\mathrm dx} (\sin^{-1}(x^2)) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( -\csc(x^2) \cot(x^2) \cdot \frac{\mathrm d}{\mathrm dx}(x^2) - \frac 1{\sqrt{1 - (x^3)^2}} \cdot \frac{\mathrm d}{\mathrm dx}(x^3) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} -\sin^{-1}(x^3) \right) \cdot \left( -\csc(x^2) \cot(x^2) \cdot 2x - \frac 1{\sqrt{1 - x^6}} \cdot 3x^2 \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( -2x\csc(x^2) \cot(x^2) - \frac {3x^2}{\sqrt{1 - x^6}} \right). \end{aligned}$

Question 3. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x)$ .

(Click for Solution)

Solution. Using the laws of exponents, the chain rule, the product rule, linearity, and common derivatives,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^x) &= \frac{\mathrm d}{\mathrm dx} (e^{x \ln(x)})\\ &= e^{x \ln(x)} \cdot \frac{\mathrm d}{\mathrm dx}( x \ln(x) ) \\ &= x^x \cdot \left( \frac{\mathrm d}{\mathrm dx}(x) \cdot \ln(x) + x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= x^x \cdot \left( 1 \cdot \ln(x) + x \cdot \frac{1}{x} \right) \\ &= x^x \cdot (\ln(x) + x). \end{aligned}$

—Joel Kindiak, 3 Sept 25, 1905H

November 13, 2025
Tedious Differentiation

Big Idea

The derivative of a function $f$ at a point $a$ , denoted $f'(c)$ , intuitively measures the gradient of the tangent to $y = f(x)$ at $(c, f(c))$ . For any general $x$ , we make the notation

$\displaystyle f'(x) \equiv \frac{\mathrm d}{\mathrm dx}(f(x)) \equiv \frac{\mathrm dy}{\mathrm dx}.$

For instance, given any real number $n$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = nx^{n-1}.$

Several other derivatives of commonly used functions exist. Furthermore, differentiation is linear in the following sense:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(af(x) + bg(x)) = a \cdot \frac{\mathrm d}{\mathrm dx}(f(x)) + b \cdot \frac{\mathrm d}{\mathrm dx}(g(x)).$

Questions

You may not use the chain rule in any of these problems.

Question 1. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(2 - 3 x + 4\sqrt x - \frac 5x\right)$ .

(Click for Solution)

Solution. Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\left(2 - 3 x + 4\sqrt x - \frac 5x\right) &= \frac{\mathrm d}{\mathrm dx}(2) - 3 \cdot \frac{\mathrm d}{\mathrm dx}(x) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(\sqrt x) - 5 \cdot \frac{\mathrm d}{\mathrm dx} \left( \frac 1x \right) \\ &= \frac{\mathrm d}{\mathrm dx}(2) - 3 \cdot \frac{\mathrm d}{\mathrm dx}(x^1) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) - 5 \cdot \frac{\mathrm d}{\mathrm dx} \left( x^{-1}\right) \\ &= 0 - 3 \cdot 1x^0 + 4 \cdot \frac 12 x^{-1/2} - 5 \cdot (-1)x^{-2} \\ &= -3 + 2 x^{-1/2} + 5x^{-2} \\ &= -3 + \frac 2{\sqrt x} + \frac 5{x^2}.\end{aligned}$

Question 2. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} (2^{2x} + \sqrt{16x} - \ln(x^8))$ .

(Click for Solution)

Solution. Simplifying then using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( 2^{2x} + \sqrt{16x} - \ln(x^8) ) &= \frac{\mathrm d}{\mathrm dx} ( (2^2)^x + \sqrt{16} \cdot \sqrt{x} - 8 \cdot \ln(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}( 4^x + 4x^{1/2} - 8 \ln(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}(4^x) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) - 8 \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \\ &= 4^x \ln(4) + 4 \cdot \frac 12 x^{-1/2} - 8 \cdot \frac 1x \\ &= 4^x \ln(4) + 2 x^{-1/2} - \frac 8x \\ &= 4^x \ln(4) + \frac 2{\sqrt x} - \frac 8x. \end{aligned}$

Question 3. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} ((5+2x)^4)$ .

(Click for Solution)

Solution. We first slowly expand the function

$\begin{aligned} (5+2x)^4 &= ((5+2x)^2)^2 \\ &= (25 + 20x + 4x^2)^2 \\ &= 625 + 1000x + 600x^2 + 160x^3 + 16x^4. \end{aligned}$

Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} ((5+2x)^4) &= ((5+2x)^2)^2 \\ &= \frac{\mathrm d}{\mathrm dx} (625 + 1000x + 600x^2 + 160x^3 + 16x^4) \\ &= \frac{\mathrm d}{\mathrm dx}(625) + 1000 \cdot \frac{\mathrm d}{\mathrm dx}(x) + 600 \cdot \frac{\mathrm d}{\mathrm dx}( x^2 ) + 160 \cdot \frac{\mathrm d}{\mathrm dx}(x^3) + 16 \cdot \frac{\mathrm d}{\mathrm dx}( x^4) \\ &= 0 + 1000 \cdot 1 + 600 \cdot 2x + 160 \cdot 3x^2 + 16 \cdot 4x^3 \\ &= 1000 + 1200x + 480x^2 + 64x^3.\end{aligned}$

Question 4. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \left( \frac 2x-5 \sqrt x \right)^2 \right)$ .

(Click for Solution)

Solution. We first expand and simplify the function

$\begin{aligned} \left( \frac 2x-5 \sqrt x \right)^2 &= \left( \frac 2x \right)^2 - 2 \cdot \frac 2x \cdot 5\sqrt x + (5 \sqrt x)^2 \\ &= \frac{4}{x^2} - 20 \cdot x^{-1} \cdot x^{1/2} + 25 x \\ &= 4x^{-2} - 20 x^{-1/2} + 25 x. \end{aligned}$

Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( \left( \frac 2x-5 \sqrt x \right)^2 \right) &= \frac{\mathrm d}{\mathrm dx} (4x^{-2} - 20 x^{-1/2} + 25 x) \\ &= 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{-2}) - 20 \cdot \frac{\mathrm d}{\mathrm dx} (x^{-1/2}) + 25 \cdot \frac{\mathrm d}{\mathrm dx}(x) \\ &= 4 \cdot (-2)x^{-3} - 20 \cdot \left(-\frac 12\right)x^{-3/2} + 25 \cdot 1x^0 \\ &= -8x^{-3} + 10x^{-3/2}+25 \\ &= -\frac 8{x^3} + \frac{10}{\sqrt{x^3}}+25. \end{aligned}$

—Joel Kindiak, 3 Sept 25, 1826H

October 30, 2025
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 $u = x^2 \Rightarrow \mathrm du = 2x\, \mathrm dx$ so that

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

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
Constructing Real Trigonometry

Let’s properly discuss classical trigonometry. For a novel approach using rational trigonometry, see this post. Assume that the notion of an angle is well-defined.

Definition 1. Consider the following right-angled triangle with acute angle $\theta \in (0, \pi/2)$ . Here and subsequently, we adopt the radian notation for angles that stipulates $2\pi = 360^{\circ}$ .

We abbreviate the words opposite, adjacent, and hypotenuse. We define the sine, cosine, and tangent of $\theta$ as follows:

$\begin{aligned} \sin(\theta) := \frac{\text{opp}}{\text{hyp}}, \quad \cos(\theta) := \frac{\text{adj}}{\text{hyp}}, \quad \tan(\theta) := \frac{\text{opp}}{\text{adj}}. \end{aligned}$

Example 1. By considering the $1$ – $\sqrt{3}$ – $2$ and $1$ – $1$ – $\sqrt 2$ right triangles, we have the following trigonometric ratios for special angles:

$\begin{aligned}\theta = \pi/6:&\quad \sin(\pi/6) = \frac{1}2,\quad \cos(\pi/6) = \frac {\sqrt{3}}2,\quad \tan(\pi/6) = \frac 1{\sqrt 3}. \\ \theta = \pi/4:&\quad \sin(\pi/4) = \frac 1{\sqrt 2},\quad \cos(\pi/4) = \frac 1{\sqrt 2} ,\quad \tan(\pi/4) = 1. \\ \theta = \pi/3:&\quad \sin(\pi/3) = \frac {\sqrt 3}2,\quad \cos(\pi/3) = \frac 12,\quad \tan(\pi/3) = \sqrt 3.\end{aligned}$

The case $\theta = \pi/4$ will play a crucial role for us later. Denote $\sin^2(\theta) := (\sin(\theta))^2$ for brevity. We will not care too much about the tangent function, since it is connected to sine and cosine in the following way:

Theorem 1. Let $\theta \in (0, \pi/2)$ . Then

$\displaystyle \sin^2(\theta) + \cos^2(\theta) = 1,\quad \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)},\quad \cos(\theta) = \sin(\pi/2 - \theta).$

Proof. For the first identity, use Pythagoras’ theorem to obtain

$\displaystyle \sin^2(\theta) + \cos^2(\theta) = \frac{\text{opp}^2}{\text{hyp}^2} + \frac{\text{adj}^2}{\text{hyp}^2} = \frac{\text{opp}^2 + \text{adj}^2}{\text{hyp}^2} = \frac{\text{hyp}^2}{\text{hyp}^2} = 1.$

For the second identity, we observe that

$\displaystyle \frac{\sin(\theta)}{\cos(\theta)} = \frac{\text{hyp} \cdot \sin(\theta)}{\text{hyp} \cdot \cos(\theta)} = \frac{\text{opp}}{\text{adj}} = \tan(\theta).$

For the last identity, since the complementary angle is $\pi/2-\theta$ ,

$\displaystyle \sin(\pi/2 - \theta) = \frac{\text{adj}}{\text{hyp}} = \cos(\theta).$

Strictly speaking, the cosine function is effectively a mutation of the sine function, and so we could technically do all of trigonometry in terms of sine. However, cosine does have its uses and will play a crucial role in our discussions moving forward.

There are many trigonometric identities built off the first two identities, and we will leave them as exercises in algebraic manipulation. For a serious study of trigonometry, we have to ask the all-important question: what is $\sin(\theta)$ if $\theta$ is not acute? In particular, what is a sensible definition for $\sin(\pi/2)$ ? The answer to the latter question turns out to answer the former question.

Theorem 2. Let $A,B$ be acute angles. If $A + B$ is acute, then

$\begin{aligned} \sin(A+B) &= \sin(A) \cos(B) + \cos(A) \sin(B), \\ \cos(A + B) &= \cos(A) \cos(B) - \sin(A) \sin(B). \end{aligned}$

Proof. Consider the following diagram for the proof of both identities.

The first identity corresponds to finding an alternate expression for $\sin(A+B) = m/h$ , and the second identity corresponds to finding an alternate expression for $\cos(A+B) = d/h$ . The ratios of interest are

$\displaystyle \sin(A) = \frac ah,\quad \cos(A) = \frac lh,\quad \sin(B) = \frac b{c+d},\quad \cos(B) = \frac l{c+d}.$

By considering the area of the whole triangle using the bases $(a+b)$ and $(c+d)$ respectively,

$\frac 12 \cdot (c + d)\cdot m = \frac 12 \cdot a\cdot l + \frac 12 \cdot b\cdot l.$

Dividing by $\frac 12 (c+d)h$ on both sides,

$\displaystyle \frac mh = \frac ah \cdot \frac l{c+d} + \frac lh \cdot \frac b{c+d}.$

By basic trigonometric ratios,

$\sin(A+B) = \sin(A) \cos(B) + \cos(A) \sin(B).$

For the second identity, by the Pythagorean theorem,

$(a+b)^2 - c^2 = m^2 = h^2 - d^2.$

By algebruh,

$c^2 = ((c+d) - d)^2 = (c+d)^2 - 2(c+d)d + d^2.$

Therefore,

$(a^2 + 2ab + b^2) - (c+d)^2 + 2(c+d)d - d^2 = h^2 - d^2.$

Simplifying the equation,

$2(c+d)d = (h^2 - a^2) + ((c+d)^2 - b^2) - 2ab.$

By Pythagoras’ theorem, $h^2 - a^2 = (c+d)^2 - b^2 = l^2$ , so that

$2(c+d)d =2l^2 - 2ab.$

Dividing by $2(c+d)h$ on both sides,

$\displaystyle \frac dh = \frac{l}{h} \cdot \frac{l}{c+d} - \frac{a}{h} \cdot \frac{b}{c+d}.$

By basic trigonometric ratios,

$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B).$

Corollary 1. Let $\theta$ be acute. If $2\theta$ is acute, then

$\begin{aligned} \sin(2\theta) = 2\sin(\theta) \cos(\theta), \quad \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta). \end{aligned}$

Proof. Set $A = B = \theta$ in Theorem 2.

The key insight is the following: as long as the right-hand side is well-defined, so is the left-hand side. This is our strategy to define $\sin$ and $\cos$ on all of $\mathbb R$ . Let’s systematise our plan.

Definition 1. For any subset $I \subseteq \mathbb R$ , let $\phi(I)$ be the proposition that for any $A, B \in I$ , $\sin(A+B)$ and $\cos(A + B)$ are well-defined, and

$\begin{aligned} \sin(A+B) &= \sin(A) \cos(B) + \cos(A) \sin(B), \\ \cos(A + B) &= \cos(A) \cos(B) - \sin(A) \sin(B). \end{aligned}$

Write $\phi(I)$ to abbreviate the proposition “ $\phi(I)$ is true”. Our overarching goal is to provide sensible definitions for $\sin$ and $\cos$ such that $\phi(\mathbb R)$ . By Theorem 1, we have established $\phi((0, \pi/8])$ and Corollary 1 will help us “double” our results to achieve the massive sub-goal of $\phi(\mathbb R^+)$ .

Theorem 3. For any $a > 0$ , suppose $\sin$ and $\cos$ are well-defined on $(0, 2a]$ . For any $\theta \in (0, 4a]$ , denote $\theta_2 := \theta/2 \in (0, 2a]$ and define

$\sin(\theta) := 2 \sin(\theta_2) \cos(\theta_2),\quad \cos(\theta) := \cos^2(\theta_2) - \sin^2(\theta_2).$

If $\phi((0, a])$ , then $\phi((0, 2a])$ .

Proof. Fix $A,B \in (0, 2a]$ . Then $A+B \in (0, 4a]$ . Observe that

$(A+B)_2 = A_2 + B_2 \in (0, 2a],$

so that $A_2,B_2 \in (0, a]$ and

$\begin{aligned} \sin(A+B) &= 2 \sin(A_2 + B_2) \cos(A_2 + B_2).\end{aligned}$

Since $\phi((0, a])$ , expanding the right-hand side yields

$\begin{aligned}\sin(A_2 + B_2) &= \sin(A_2) \cos(B_2) + \cos(A_2) \sin(B_2), \\ \cos(A_2 + B_2) &= \cos(A_2) \cos(B_2) - \sin(A_2) \sin(B_2).\end{aligned}$

On the other hand,

$\begin{aligned} \sin(A)\cos(B) &= 2 \sin(A_2)\cos(A_2) (\cos^2(B_2) - \sin^2(B_2)), \\ \sin(B)\cos(A) &= 2 \sin(B_2)\cos(B_2) (\cos^2(A_2) - \sin^2(A_2)). \end{aligned}$

With careful algebraic expansion, we will obtain

$\sin(A + B) = \sin(A)\cos(B) + \cos(A) \sin(B).$

Similarly, we will obtain

$\cos(A + B) = \cos(A) \cos(B) - \sin(A) \sin(B).$

Corollary 2. $\phi(\mathbb R^+)$ . In particular, we have the following special angles:

$\begin{aligned} \theta = \pi/2:&\quad \sin(\pi/2) = 1,\quad \cos(\pi/2) = 0 \\ \theta = \pi:&\quad \sin(\pi) = 0,\quad \cos(\pi) = -1. \\ \theta = 2\pi:&\quad \sin(2\pi) = 0,\quad \cos(2\pi) = 1.\end{aligned}$

Proof. Fix $A,B \in \mathbb R^+$ . Then there exists $N \in \mathbb N$ such that $A+B \in (0, 2^N \cdot \pi/8]$ . Using induction, we can prove that $\phi((0, 2^N \cdot \pi/8])$ . Therefore, $\sin, \cos$ are well-defined on $\mathbb R^+$ and the identities

$\begin{aligned} \sin(A+B) &= \sin(A) \cos(B) + \cos(A) \sin(B), \\ \cos(A + B) &= \cos(A) \cos(B) - \sin(A) \sin(B), \end{aligned}$

hold for any $A, B \in \mathbb R^+$ . Hence, $\phi(\mathbb R^+)$ , as required.

Corollary 3. For any $\theta > 0$ ,

$\sin(\theta + 2\pi) = \sin(\theta),\quad \cos(\theta + 2\pi) = \cos(\theta).$

Furthermore, for any positive integer $n$ ,

$\sin(\theta + n \cdot 2\pi) = \sin(\theta),\quad \cos(\theta + n \cdot 2\pi) = \cos(\theta).$

We have done a remarkable task: defining $\sin, \cos$ on $\mathbb R^+$ and proving that they satisfy the desired addition formulae. But we haven’t proven this case for all of $\mathbb R$ , since $\mathbb R = \mathbb R^- \sqcup \{0\} \sqcup \mathbb R^+$ . Surprisingly, though, Corollary 3 gives us a unique insight. Observe that for $\theta \in (-2\pi, 0]$ , $\sin(\theta +2\pi)$ and $\cos(\theta + 2\pi)$ are well-defined expressions. This means we can do a “reverse” definition for non-positive $\theta$ .

Theorem 4. For any $\theta \leq 0$ , let $n_\theta$ denote any integer such that $\theta + n_\theta \cdot 2\pi > 0$ . The notions

$\sin(\theta) := \sin(\theta + n_\theta \cdot 2\pi),\quad \cos(\theta) := \cos(\theta + n_\theta \cdot 2\pi)$

are well-defined by Corollary 3. Then $\phi(\mathbb R)$ . In particular,

$\sin(0) = \sin(2\pi) = 0,\quad \cos(0) = \cos(2\pi) = 1.$

Proof. Fix $A, B \in \mathbb R$ . Find $n_A, n_B \in \mathbb N$ such that $A +n_A \cdot 2\pi > 0$ and $B+n_B \cdot 2\pi > 0$ . Then $(A+B) + (n_A + n_B) \cdot 2\pi > 0$ so that $\phi(\mathbb R^+)$ allows

$\begin{aligned} \sin(A+B) &= \sin((A+B) + (n_A + n_B) \cdot 2\pi) \\ &= \sin((A + n_A \cdot 2\pi) + (B + n_B \cdot 2\pi)) \\ &= \sin(A + n_A \cdot 2\pi) \cos(B + n_B \cdot 2\pi) \\ &\phantom{==} + \cos(A + n_A \cdot 2\pi) \sin(B + n_B \cdot 2\pi) \\ &= \sin(A) \cos(B) + \cos(A) \sin(B). \end{aligned}$

A similar calculation yields

$\cos(A + B) = \cos(A) \cos(B) - \sin(A) \sin(B).$

Thus, we have properly defined $\sin$ and $\cos$ on all of $\mathbb R$ , and even proved that the identities

$\begin{aligned} \sin(A+B) &= \sin(A) \cos(B) + \cos(A) \sin(B), \\ \cos(A + B) &= \cos(A) \cos(B) - \sin(A) \sin(B), \end{aligned}$

are well-defined and hold for any $A, B \in \mathbb R$ . From these two key identities, we obtain all other trigonometric identities commonly obtained in tables of mathematical formulas.

—Joel Kindiak, 4 Jun 25, 1758H

August 31, 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