Tag: ai

Several Differentiation Drills

For any integer $n$ , you may freely use the results

$\begin{aligned} (x^n)' &= nx^{n-1},\\ \sin'(x) &= \cos(x),\\ \cos'(x) &= -{\sin(x)},\\ \tan'(x) &= \sec^2(x), \\ (e^x)' &= e^x, \\ \ln'(x) &= \frac 1x, \end{aligned}$

from either previous posts or from real analysis. You may also apply the chain rule, product rule, and quotient rule however you wish.

Problem 1. For any $a > 0$ , prove that

$\displaystyle \begin{aligned} (a^x)' = a^x \ln(a), \quad \log_a'(x) = \frac{1}{x \ln(a)}. \end{aligned}$

(Click for Solution)

Solution. By exponential properties,

$a^x = (e^{\ln(a)})^x = e^{x \ln(a)}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (a^x)' &= (e^{x \ln(a)})'\\ &= e^{x \ln(a)} \cdot \frac{\mathrm d}{\mathrm dx}(x \ln a) \\ &= e^{x \ln(a)} \cdot \ln(a) \\ &= a^x \ln(a). \end{aligned}$

By logarithm properties and linearity,

$\displaystyle \begin{aligned} \log_a'(x) &= \left( \frac{\ln(x)}{\ln(a)} \right)'\\ &= \frac{1}{\ln(a)} \ln'(x) \\ &= \frac{1}{\ln(a)} \cdot \frac 1x = \frac 1{x \ln (a)}. \end{aligned}$

Up till now, we have proven that $(x^n)' = nx^{n-1}$ only for integer values of $n$ . We extend this result to all real values of $n$ .

Problem 2. For any real $r$ , use Problem 1 to prove that $(x^r)' = rx^{r-1}$ .

(Click for Solution)

Solution. By exponential properties,

$x^r = (e^{\ln x})^r = e^{r \ln x}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (x^r)' &= (e^{r \ln x})'\\ &= e^{r \ln x} \cdot \frac{\mathrm d}{\mathrm dx}(r \ln x) \\ &= x^r \cdot r \cdot \frac 1x = rx^{r-1}. \end{aligned}$

Example 1. By Problem 2,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{\sqrt 2}) = \sqrt 2 \cdot x^{\sqrt 2 - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^\pi) = \pi \cdot x^{\pi - 1}.$

In particular,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}\neq e^x = \frac{\mathrm d}{\mathrm dx}(e^x).$

Problem 3. Prove that

$\displaystyle \sec'(x) = \sec(x)\tan(x),\quad \csc'(x) = -{\csc(x)\cot(x)}.$

(Click for Solution)

Solution. Writing $\displaystyle \sec(x) = \frac 1{\cos(x)}$ , apply the quotient rule to obtain

$\displaystyle \begin{aligned} \sec'(x) &= \frac{ (1)' \cdot \cos(x) - 1 \cdot \cos'(x) }{ \cos^2(x) } \\ &= \frac{ 0 \cdot \cos(x) - 1 \cdot (-\sin(x)) }{ \cos^2(x) } \\ &= \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x) \tan(x). \end{aligned}$

We recall that $\csc(x) = \sec(\pi/2-x)$ so that

$\displaystyle \begin{aligned}\csc'(x) &= (\sec(\pi/2-x))' \\ &= \sec'(\pi/2-x)) \cdot (-1) \\ &= \sec(\pi/2-x) \tan(\pi/2-x) \cdot (-1) \\ &= \csc(x) \cot(x) \cdot (-1) \\ &= - \csc(x) \cot(x). \end{aligned}$

Problem 4. Given that $t > 0$ , $x \equiv x(t) > 0$ , and $y \equiv y(t) > 0$ , evaluate $\displaystyle \frac{\mathrm d}{\mathrm dt}(x^y)$ .

(Click for Solution)

Solution. Writing $x^y = e^{y \ln(x)}$ and using the chain rule,

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

Remark 1. If $x = t$ and $y$ is constant, we recover the power rule. If $x$ is constant and $y = t$ , we recover Problem 1. Furthermore, if $y = x = t$ , we obtain

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x) = x^x (\ln(x) + 1).$

In the language of partial derivatives, denoting $z = x^y \equiv { x(t) }^{ y(t) }$ ,

$\displaystyle \frac{\mathrm d z}{\mathrm dt} = \frac{\partial z}{\partial y} \cdot \frac{\mathrm dy}{\mathrm dt} + \frac{\partial z}{\partial x} \cdot \frac{\mathrm dx}{\mathrm dt}.$

Problem 5. Suppose $f, f^{-1}$ are differentiable on $\mathbb R$ . Evaluate $(f^{-1})'(x)$ .

(Click for Solution)

Solution. By definition of the inverse function,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} (x) &= \frac{\mathrm d}{\mathrm dx} (f(f^{-1}(x))) \\ 1 &= f'(f^{-1}(x)) \cdot \frac{\mathrm d}{\mathrm dx}(f^{-1}(x)) \\ &= f'(f^{-1}(x)) \cdot f'(f^{-1}(x)) \\ (f^{-1})'(x) &= \frac 1{f'(f^{-1}(x))}. \end{aligned}$

—Joel Kindiak, 9 Feb 25, 2126H

February 9, 2025

ai, artificial-intelligence, math, mathematics, physics
Some Integration Drills

Problem 1. Evaluate the integral $\displaystyle \int(x^2 + \tan x - 3)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned}\int(x^2 + \tan x - 3)\, \mathrm dx &= \int x^2\, \mathrm dx + \int \tan x \, \mathrm dx - \int 3\, \mathrm dx \\ &= \frac 13 x^3 - \ln|{\cos x}| - 3x + C.\end{aligned}$

Problem 2. Evaluate the integral $\displaystyle \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned} \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx &= \int \frac{1}{9 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{4-x^2} } \, \mathrm dx \\ &= \int \frac{1}{3^2 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{2^2-x^2} } \, \mathrm dx \\ &= \frac 13 \tan^{-1} \left( \frac x3 \right) + \sin^{-1} \left( \frac x2 \right) + C. \end{aligned}$

Problem 3. Evaluate the integral $\displaystyle \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx$ .

(Click for Solution)

Solution. By the substitution $\displaystyle u = x^2 \Rightarrow \mathrm dx = \frac 1{2x}\, \mathrm du$ ,

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

Hence, substituting the limits,

$\displaystyle \begin{aligned} \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx &= \frac 12 [\sec(x^2)]_0^1\\ &= \frac 12 (\sec(1) - \sec(0)) \\ &= \frac 1{2 \cos(1)} - \frac 12 \\ &\approx 0.425, \end{aligned}$

when approximated to 3 decimal places.

Problem 4. Evaluate the integral $\displaystyle \int \tan^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. Using the Pythagorean identity $\sec^2 x = \tan^2 x +1$ ,

$\displaystyle \begin{aligned} \int \tan^3 x\, \mathrm dx &= \int \tan x \tan^2 x\, \mathrm dx \\ &= \int \tan x (\sec^2 x - 1)\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx - \int \tan x\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx -(- \ln|{\cos x}|) + C.\end{aligned}$

For the first integral, make the substitution $\displaystyle u = \tan x \Rightarrow \mathrm dx = \frac{1}{\sec^2 x}\, \mathrm du$ . Then

$\displaystyle \begin{aligned}\int \tan x \sec^2 x \, \mathrm dx &= \int u \sec^2 x \cdot \frac{1}{\sec^2 x}\, \mathrm du \\ &= \int u\, \mathrm du \\ &= \frac 12 u^2 + C\\ &= \frac 12 \tan^2 x + C.\end{aligned}$

Therefore, $\displaystyle \int \tan^3 x\, \mathrm dx = \frac 12 \tan^2 x + \ln|{\cos x}| + C.$

Problem 5. Evaluate the integral $\displaystyle \int \sin(\ln \theta)\, \mathrm d\theta$ .

(Click for Solution)

Solution. Make the substitution $u = \ln \theta \Rightarrow \mathrm d\theta = e^u\, \mathrm du$ . Then

$\displaystyle \int \sin(\ln \theta)\, \mathrm d\theta = \int e^u \sin(u)\, \mathrm du.$

Integrating by parts twice,

$\displaystyle \begin{aligned}\int e^u \sin(u)\, \mathrm du &= \underbrace{e^u}_{\mathrm I}\, \underbrace{\sin(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm D}\, \mathrm du \\ &= e^u \sin(u) - \int e^u \cos(u)\, \mathrm du\\ &= e^u \sin(u) - \left(\underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{-\sin(u)}_{\mathrm D}\, \mathrm du\right) \\ &= e^u \sin(u) - \left(e^u \cos(u) + \int e^u \sin(u) \, \mathrm du\right) \\ &= e^u \sin(u) - e^u \cos(u) - \int e^u \sin(u)\, \mathrm du \\&= e^u (\sin(u) - \cos(u)) - \int e^u \sin(u)\, \mathrm du.\end{aligned}$

By algebruh,

$\displaystyle \begin{aligned} \int e^u \sin(u)\, \mathrm du + \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ 2 \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ \int e^u \sin(u)\, \mathrm du &= \frac 12 e^u (\sin(u) - \cos(u)) + C\\ \int \sin(\ln \theta)\, \mathrm d\theta &= \frac 12 \theta (\sin(\ln \theta) - \cos(\ln \theta)) + C.\end{aligned}$

Problem 6. Evaluate the integral $\displaystyle \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36-x^2}} \right)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned} \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36 -x^2}} \right)\, \mathrm dx &= \int \frac{1}{36 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{36-x^2} } \, \mathrm dx \\ &= \int \frac{1}{6^2 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{6^2-x^2} } \, \mathrm dx \\ &= \frac 1{2 \cdot 6} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C\\ &= \frac 1{12} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C. \end{aligned}$

Problem 7. Evaluate the integral $\displaystyle \int_0^2 x \sin(x^2 + 5)\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $u = x^2 + 5$ yields

$\displaystyle \mathrm du = 2x\, \mathrm dx,\quad u(0) = 5, \quad u(2) = 9.$

Therefore,

$\displaystyle \begin{aligned} \int_0^2 x \sin(x^2 + 5)\, \mathrm dx &= \frac 12 \int_0^2 \sin(x^2 + 5) \cdot 2x\, \mathrm dx \\ &= \frac 1{2}\int_5^9 \sin(u)\, \mathrm du \\ &= \frac 12 [-\cos(u)]_5^9 \\ &= \frac 12 (-\cos(9) - (-\cos(5))) \\ &= \frac 12 (\cos(5) - \cos(9)) \approx 0.597, \end{aligned}$

when approximated to 3 decimal places.

Problem 8. Evaluate the integral $\displaystyle \int \cot^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. Recall that $\cot(x) = \tan(\pi/2 - x)$ . Making the substitution $u = \pi/2-x \Rightarrow \mathrm du = -\mathrm dx$ then using the result in Problem 4,

$\displaystyle \begin{aligned} \int \cot^3 x\, \mathrm dx &= \int \tan^3 (\pi/2-x)\, \mathrm dx \\ &= \int \tan^3(u) \cdot -1\, \mathrm du \\ &= -\int \tan^3(u)\, \mathrm du \\ &= -\left( \frac 12 \tan^2 u + \ln|{\cos u}| \right) + C \\ &= -\left( \frac 12 \tan^2 (\pi/2-x) + \ln|{\cos (\pi/2-x)}| \right) + C \\ &= -\left( \frac 12 \cot^2 x + \ln|{\sin x}| \right) + C \\ &= - \frac 12 \cot^2 x - \ln|{\sin x}| + C \end{aligned}$

Problem 9. Evaluate the integral $\displaystyle \int e^{2\theta} \sin (e^\theta) \, \mathrm d\theta$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = e^{\theta} \Rightarrow \mathrm d\theta = \frac 1{u}\, \mathrm du$ then integrating by parts,

$\displaystyle \begin{aligned} \int e^{2\theta} \sin(e^\theta) \, \mathrm d\theta &= \int u^2 \sin(u)\cdot \frac 1{u}\, \mathrm du \\ &= \int u \sin(u)\, \mathrm du\\ &= \underbrace{-\cos(u)}_{\mathrm I} \underbrace{u}_{\mathrm S} - \int \underbrace{-\cos(u)}_{\mathrm I}\underbrace{1}_{\mathrm D}\, \mathrm du \\ &= -u \cos u + \int \cos u\, \mathrm du\\ &= -u \cos u + \sin u + C\\ &= -e^\theta \cos(e^\theta) + \sin(e^\theta) + C. \end{aligned}$

Problem 10. Given that

$f(x) = \begin{cases} 1 - |x|, & |x| \leq 1, \\ 0, &\text{otherwise},\end{cases}$

evaluate $\displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx$ .

(Click for Solution)

Solution. By the definition of $f$ ,

$f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x^2 - 1| \leq 1, \\ 0, &\text{otherwise}.\end{cases}$

Parsing the inequality $|x^2 - 1| \leq 1 \iff |x| \leq \sqrt 2$ ,

$f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}$

By definition of the absolute value,

$|x^2 - 1| = \begin{cases} x^2 - 1, & |x| \geq 1, \\ 1 - x^2, &|x| \leq 1.\end{cases}$

Parsing the inequalities again,

$f(x^2 - 1) = \begin{cases} x^2, & 0 \leq |x| \leq 1, \\ 2-x^2 , & 1 \leq |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}$

Therefore, we can evaluate the integral as

$\begin{aligned} \displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx &= \int_0^1 f(x^2 - 1)\, \mathrm dx + \int_1^{\sqrt 2} f(x^2 - 1)\, \mathrm dx + \int_{\sqrt 2}^{2} 0\, \mathrm dx \\ &= \int_0^1 x^2\, \mathrm dx + \int_1^{\sqrt 2} (2 - x^2)\, \mathrm dx+ 0 \\ &= \left[ \frac{x^3}{3}\right]_0^1 + \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt 2} \\ &= \left( \frac 13 - 0 \right) + \left( 2\sqrt 2 - \frac{2\sqrt 2}{3} \right) \\ &= \frac 13 + \frac{4}{3}\sqrt 2. \end{aligned}$

—Joel Kindiak, 20 Jan 25, 0925H

January 20, 2025

ai, Calculus, math, mathematics, physics
Dividing Your Power

This problem is such an interesting one that, just like a previous problem, was given in an exam, but broken down into more palatable steps.

Problem 1. Prove that there are infinitely many positive even integers $n$ such that $n \mid (2^n + 2)$ and $(n - 1) \mid (2^n+1)$ .

(Click for Solution)

Solution. Consider $n_1 = 2$ , which satisfies the property since $2 \mid (2^2 + 2)$ and $(2-1) \mid (2^2+1)$ trivially. The core idea is to inductively define $n_{i + 1}$ in terms of $n_i$ so that both integers satisfy $n_i \mid (2^{n_i} + 2)$ and $(n_i - 1) \mid (2^{n_i} +1)$ .

To that end, for any even $n_i = n$ such that $n \mid (2^n + 2)$ and $(n - 1) \mid (2^n+1)$ , define $n_{i+1} = m := 2^n + 2$ . We claim that $m \mid (2^m + 2)$ and $(m - 1) \mid (2^m+1)$ . By definition of $m$ , we observe that

$\displaystyle 2^m + 2 = 2\times (2^{m-1} + 1) = 2 \times (2^{2^n + 1} + 1).$

Since $(n - 1) \mid (2^n +1)$ , find an integer $k$ such that $2^n + 1 = k(n-1)$ . Since the left-hand side is odd, the right-hand side must be odd, so that $k$ is odd. Thus,

$2^{2^n + 1} + 1 = 2^{(n-1)k} + 1 = (2^{n-1})^k + 1.$

Consider the function $x^k + 1$ . Since $(-1)^k + 1 = 0$ , by the factor theorem, $(x+1)$ is a factor of $(x^k+1)$ . In particular, $(2^{n-1} + 1) \mid ((2^{n-1})^k + 1)$ . Let $l$ denote an integer such that

$(2^{n-1})^k + 1 = (2^{n-1} + 1)l.$

Combining our results,

$2^m + 2 = 2 \times (2^{2^n + 1} + 1) = 2 \times (2^{n-1} + 1)l = (2^n + 2) l = ml,$

so that $m \mid (2^m + 2)$ . For the second result,

$2^m + 1 = 2^{2^n + 2} + 1.$

Since $n \mid (2^n + 2)$ , find an integer $r$ such that $2^n + 2 = nr$ so that

$2^{2^n + 2} + 1 = 2^{nr} + 1 = (2^n)^r + 1.$

Since $n$ is even, find an integer $s$ such that $n = 2s$ . Then

$\displaystyle 2^n + 2 = nr = 2sr \quad \Rightarrow \quad 2^{n-1} + 1 = sr.$

Since the left-hand side is odd, so is the right-hand side, and in particular, so is $r$ . Therefore,

$m-1 = (2^n + 1) \mid ((2^n)^r + 1) = 2^m + 1,$

as required. Finally, we have found some infinitely many integers $n = n_1,n_2,\dots$ such that $n \mid (2^n + 2)$ and $(n - 1) \mid (2^n+1)$ .

—Joel Kindiak, 13 Nov 24, 1813H

November 27, 2024

ai, cryptography, math, mathematics, number-theory
Profit Maximisation for Math People
When I had to learn economics in high school, I struggled tremendously due to its language-based emphasis (which meant tons of essay-writing with arguably subjective marking schemes).

Yet, I remember how one teaching intern scribbled some equations to prove them mathematically, and those equations were my only takeaways from the entire two years of economics. I’m convinced that economics without equations is gravely incomplete. Equivalently, what this means is that when we include the mathematics with the economic theory, the concepts therein start to illuminate meaning.

Here’s one of my favourite results in economics.

Theorem 1. A firm maximises profit when its marginal revenue (MC) equals its marginal cost (MC) and when the marginal cost is increasing.

Here’s a layperson’s justification for this theorem.

If the firm is producing at a quantity where MR > MC, then it can increase profit by increasing output. The reason is since the marginal revenue exceeds the marginal cost, additional output is adding more to profit than it is taking away. If the firm is producing at a quantity where MC > MR, then it can increase profit by reducing output.
Adapted from Lumen Learning

Technically, this explanation can make sense, but I was absolutely confused to pieces by it. Don’t even get me started on having to regurgitate this paragraph compounded with my atrocious handwriting when answering essay-based questions. Let’s define our terms then prove this mathematically.

Definition. Let $x$ denote the quantity of a product that a firm produces. Make the following notations:
- $R(x)$ : the revenue the firm earns by selling $x$ units of such a product,
- $C(x)$ : the cost the firm incurs by producing $x$ units of such a product,
- $\pi(x)$ : the profit the company earns by selling $x$ units of the product.
It is intuitive that $\pi = R - C$ . Note that we are adopting economists’ convention to denote profit by $\pi$ , not to be confused with the circle constant $3.14159\dots$ .

Equivalently, we define the marginal revenue function and marginal cost function by $\mathrm{MR} := R'$ and $\mathrm{MC} := C'$ respectively.

Now, onto the proof, using the derivative tests that we have previously developed. We will assume that $\mathrm{MR}$ is strictly decreasing, though that actually can be proven from more fundamental principles of economics.

Proof of Theorem 1. To maximise profit at the output level $c$ , we need $\pi'(c) = 0$ . Since $\pi = R - C$ , this implies that

$\displaystyle 0=\pi'(c) = (R-C)'(c) = R'(c) - C'(c) = \mathrm{MR}(c) - \mathrm {MC}(c).$

By algebra, $\mathrm{MR}(c) = \mathrm{MC}(c)$ . Assuming that $\mathrm{MC}$ is decreasing so that $\mathrm{MC}'(c) \leq 0$ ,

$\pi''(c) = \mathrm{MR}'(c) - \mathrm{MC}'(c) < 0 - 0 = 0.$

By the second derivative test, $c$ is a local maximum for $\pi$ .

We could have many more economics-contextualised applications of differential calculus, but perhaps another time.

For now, we shift gears and discuss the overrated l’Hôpital’s rule.

—Joel Kindiak, 31 Oct 24, 1402H
October 31, 2024

ai, machine-learning, math, mathematics, Probability
The Classic Secant Integral

No course in calculus is complete without the famous problem of integrating $\sec^3$ .

Problem 1. Evaluate $\displaystyle \int \sec^3(x)\, \mathrm dx$ .

(Click for Solution)

Solution. Write $\sec^3(x) = \sec^2(x) \sec(x)$ . Since $\sec^2(x)$ integrates to $\tan(x)$ , we integrate by parts to obtain

$\displaystyle \begin{aligned} \int \sec^3(x)\, \mathrm dx &= \int \sec^2(x) \sec(x)\, \mathrm dx \\ &= \underbrace{\tan(x)}_{\text I} \underbrace{\sec(x)}_{\text S} - \int \underbrace{\tan(x)}_{\text I} \cdot \underbrace{\sec(x)\tan(x)}_{\text D}\,\mathrm dx. \end{aligned}$

Writing $\tan^2(x) = \sec^2(x) - 1$ ,

$\displaystyle \begin{aligned} \int \sec^3(x)\, \mathrm dx &= \tan(x) \sec(x) - \int \tan(x) \cdot \sec(x)\tan(x)\,\mathrm dx \\ &= \tan(x) \sec(x) - \int (\sec^2(x) - 1) \cdot \sec(x)\,\mathrm dx \\ &= \tan(x) \sec(x) - \int \sec^3(x)\, \mathrm dx + \int \sec(x)\,\mathrm dx. \end{aligned}$

By algebra and recalling our result for the integral of $\sec$ ,

$\displaystyle \begin{aligned} \int \sec^3(x)\, \mathrm dx &= \frac 12 \tan(x) \sec(x) + \frac 12 \int \sec(x)\,\mathrm dx \\ &= \frac 12 \tan(x) \sec(x) + \frac 12 \ln|{\sec(x) + \tan(x)}| + C. \end{aligned}$

—Joel Kindiak, 28 Oct 24, 1042H

October 28, 2024

ai, geometry, machine-learning, math, mathematics
The Meanest Theorem in Calculus

Any self-respecting study of differential calculus must include the mean value theorem, which has ubiquitous uses in all of calculus. We will state it here and prove it in two steps: the easier step, followed by the harder step.

Theorem 1 (Mean Value Theorem). Let $f : [a,b] \to \mathbb R$ be continuous, and $f'$ be differentiable on $(a, b)$ . Then there exists $c \in (a, b)$ such that

$f(b) = f(a) + f'(c)(b-a).$

Proof. We will first prove the special case when $f(a) = f(b)$ , so that $f'(c) = 0$ . This is known as Rolle’s theorem. Suppose for simplicity that $f$ is non-constant.

Without loss of generality, there exists $c_0 \in (a, b)$ such that $f(c_0) > f(a)$ . Since $f$ is continuous, by the extreme value theorem, there exists $c \in (a, b)$ such that for any $x \in [a,b]$ ,

$f(x) \leq f(c).$

In particular, $f(c) \geq f(c_0) > f(a)$ so that $c \neq a,b$ . Since $f$ is differentiable at $c$ , we will compute its derivative $f'(c)$ in two steps.

For the first step, by considering $x$ -values to the left of $c$ , since $f(x) \leq f(c)$ and $x < c$ ,

$\displaystyle f'(c) = \lim_{x \to c^-} \frac{f(x) - f(c)}{x-c} \geq 0.$

For the second step, by considering $x$ -values to the right of $c$ , since $f(x) \leq f(c)$ and $x > c$ ,

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

Since $0 \leq f'(c) \leq 0$ , we have $f'(c) = 0$ , as desired.

Now that we have established Rolle’s theorem, we prove the mean value theorem. Define the continuous function $g : [a,b] \to \mathbb R$ that is differentiable on $(a,b)$ by

$\displaystyle g(x) := f(x) - f(a) - \frac{f(b)-f(a)}{b-a} \cdot (x-a).$

This helps us ensure that $g(a) = g(b)$ . By Rolle’s theorem, there exists $c \in (a, b)$ such that $g'(c) = 0$ . Alternatively, by computing $g'$ directly,

$\displaystyle g'(x) = f'(x) - \frac{f(b) - f(a)}{b-a}.$

Setting $x = c$ yields

$\displaystyle 0 = f'(c) - \frac{f(b) - f(a)}{b-a}\quad \iff \quad f(b) = f(a) + f'(c)(b-a),$

as required.

The mean value theorem is incredibly useful in calculus, and we will be using it repeatedly as we slowly but surely define rigorously the usual functions that we have been taking for granted.

For now, we evaluate $(f^{-1})'$ when $f$ is differentiable and bijective. We do that the next time.

—Joel Kindiak, 20 Oct 24, 1611H

October 20, 2024

ai, machine-learning, math, mathematics, physics