v1 · padrão canônico

Lesson 120 — Final program workshop

Capstone. 40 integrating problems spanning Years 1–3. Theme: real application in ML, finance, engineering, science.

Used in: Age 18 (final-year secondary) · German Leistungskurs Abitur equivalent · Singapore H2 Math equivalent

\mathcal{M} = \{\text{Calculus},\; \text{Linear Algebra},\; \text{Probability},\; \text{Modeling}\}

The final workshop integrates the four pillars of the program in 40 problems demanding the synthesis of techniques from multiple terms. Each problem combines at least two concepts: derivative with algebra, integral with statistics, ODE with linear algebra. Completing this workshop prepares you for quantitative university coursework.

Choose your door

Rigorous notation, full derivation, hypotheses

Formal synthesis — the four pillars

Structure of the completed program

"A mathematical theory is not to be considered complete until you have made it so clear that you can explain it to the first man whom you meet on the street." — David Hilbert, cited in Active Calculus §1.1

Definition· Central cross-domain connections

The connections below appear in at least two workshop problems:

Concept	Calculus	Linear Algebra	Probability
Derivative	instantaneous rate	gradient of $f: \mathbb{R}^n \to \mathbb{R}$	score of likelihood
Integral	area under curve	inner product in $L^2$	expected value $\mathbb{E}[X] = \int x\,f(x)\,dx$
Eigenvalue	ODE $y'' + \lambda y = 0$	diagonalization $A = PDP^{-1}$	principal variance (PCA)
Normal $N(0,1)$	$\int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi}$	space of functions $L^2$	iid error distribution

Flow of the four pillars of the program converging into the final workshop.

Worked examples

Example— 1· Optimization with derivative — profit maximization

Problem: A firm has revenue $R(q) = 120q - 2q^2$ and cost $C(q) = 200 + 40q + q^2$ (in dollars, $q$ units/day). Find the quantity $q^*$ that maximizes profit $L(q) = R(q) - C(q)$ .

Strategy: Set up $L(q)$ , compute $L'(q)$ , set to zero, verify with second derivative that it is a maximum.

Solution:

$L(q) = (120q - 2q^2) - (200 + 40q + q^2) = -3q^2 + 80q - 200$ .
$L'(q) = -6q + 80$ . Setting to zero: $q^* = 80/6 = 40/3 \approx 13.3$ units/day.
$L''(q) = -6 < 0$ , confirming maximum.
$L(40/3) = -3(40/3)^2 + 80(40/3) - 200 = -\frac{4800}{9} + \frac{3200}{3} - 200 = \frac{1400}{9} \approx 155.6$ dollars/day.

Verification: $L'(13) = -78 + 80 = 2 > 0$ and $L'(14) = -84 + 80 = -4 < 0$ , confirming the maximum lies between $q = 13$ and $q = 14$ .

Source. Active Calculus §3.3, Exercise 3.3.3 — CC-BY-NC-SA.

Example— 2· Definite integral by substitution — area under curve

Problem: Compute $\displaystyle\int_0^{\pi/2} \sin^2 x \cos^3 x \; dx$ .

Strategy: Rewrite $\cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x)\cos x$ and use substitution $u = \sin x$ .

Solution:

Write $\cos^3 x = (1 - \sin^2 x)\cos x$ .
Let $u = \sin x$ , so $du = \cos x\,dx$ . Limits: $x=0 \Rightarrow u=0$ ; $x=\pi/2 \Rightarrow u=1$ .
$\displaystyle\int_0^1 u^2(1 - u^2)\,du = \int_0^1 (u^2 - u^4)\,du = \left[\frac{u^3}{3} - \frac{u^5}{5}\right]_0^1 = \frac{1}{3} - \frac{1}{5} = \frac{2}{15}$ .

Verification: $0 < 2/15 \approx 0.133$ . The integrand $\sin^2 x \cos^3 x \geq 0$ on $[0, \pi/2]$ with maximum around $x \approx 0.63$ rad where $\sin x \approx 0.59$ , $\cos x \approx 0.81$ ; maximum value $\approx 0.21$ . Area $\approx 0.21 \times 1.57/2 \approx 0.16$ . Magnitude consistent with $2/15$ .

Source. Active Calculus §5.3, Exercise 5.3.5 — CC-BY-NC-SA.

Example— 3· Diagonalization of 2×2 matrix — eigenvalues and eigenvectors

Problem: Diagonalize $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ .

Strategy: Compute $\det(A - \lambda I) = 0$ , find eigenvalues, then eigenvectors. Assemble $P$ with eigenvectors as columns and verify $P^{-1}AP = D$ .

Solution:

$\det(A - \lambda I) = (3-\lambda)(2-\lambda) - 0 = (3-\lambda)(2-\lambda)$ . Eigenvalues: $\lambda_1 = 3$ , $\lambda_2 = 2$ .
For $\lambda_1 = 3$ : $(A - 3I)\mathbf{v} = 0 \Rightarrow \begin{pmatrix}0 & 1\\ 0 & -1\end{pmatrix}\mathbf{v} = 0 \Rightarrow v_2 = 0$ . Eigenvector: $\mathbf{v}_1 = (1, 0)^T$ .
For $\lambda_2 = 2$ : $(A - 2I)\mathbf{v} = 0 \Rightarrow \begin{pmatrix}1 & 1\\ 0 & 0\end{pmatrix}\mathbf{v} = 0 \Rightarrow v_1 = -v_2$ . Eigenvector: $\mathbf{v}_2 = (1, -1)^T$ .
$P = \begin{pmatrix}1 & 1 \\ 0 & -1\end{pmatrix}$ , $D = \begin{pmatrix}3 & 0 \\ 0 & 2\end{pmatrix}$ .

Verification: $AP = P \begin{pmatrix}3&0\\0&2\end{pmatrix}$ . Column 1: $A(1,0)^T = (3,0)^T = 3(1,0)^T$ . Column 2: $A(1,-1)^T = (2,-2)^T = 2(1,-1)^T$ . Correct.

Source. Linear Algebra Done Right, 4th ed., §5C Exercise 5 — CC-BY-NC.

Example— 4· Second-order ODE — simple harmonic oscillator

Problem: Solve $y'' + 4y = 0$ with $y(0) = 1$ , $y'(0) = 0$ .

Strategy: Linear homogeneous second-order equation with constant coefficients. Characteristic equation $r^2 + 4 = 0$ has complex conjugate roots.

Solution:

Characteristic equation: $r^2 + 4 = 0 \Rightarrow r = \pm 2i$ .
General solution: $y(x) = C_1 \cos(2x) + C_2 \sin(2x)$ .
Initial condition $y(0) = 1$ : $C_1 \cdot 1 + C_2 \cdot 0 = 1 \Rightarrow C_1 = 1$ .
$y'(x) = -2C_1 \sin(2x) + 2C_2 \cos(2x)$ . Condition $y'(0) = 0$ : $2C_2 = 0 \Rightarrow C_2 = 0$ .
Solution: $y(x) = \cos(2x)$ .

Verification: $y'' = -4\cos(2x)$ . So $y'' + 4y = -4\cos(2x) + 4\cos(2x) = 0$ . Conditions: $y(0) = 1$ , $y'(0) = -2\sin(0) = 0$ . Correct.

Source. Notes on Diffy Qs §2.2, Exercise 2.2.1 — CC-BY-SA.

Example— 5· Simple linear regression — coefficient estimation

Problem: Given points $(1,2)$ , $(2,3)$ , $(3,5)$ , $(4,4)$ , $(5,6)$ , find the regression line $\hat y = \hat\beta_0 + \hat\beta_1 x$ by least squares.

Strategy: Use formulas $\hat\beta_1 = \frac{\sum(x_i - \bar x)(y_i - \bar y)}{\sum(x_i - \bar x)^2}$ and $\hat\beta_0 = \bar y - \hat\beta_1 \bar x$ .

Solution:

$n = 5$ . $\bar x = (1+2+3+4+5)/5 = 3$ . $\bar y = (2+3+5+4+6)/5 = 4$ .
$\sum(x_i - \bar x)^2 = (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 4+1+0+1+4 = 10$ .
$\sum(x_i - \bar x)(y_i - \bar y) = (-2)(-2)+(-1)(-1)+(0)(1)+(1)(0)+(2)(2) = 4+1+0+0+4 = 9$ .
$\hat\beta_1 = 9/10 = 0.9$ . $\hat\beta_0 = 4 - 0.9 \times 3 = 4 - 2.7 = 1.3$ .
Line: $\hat y = 1.3 + 0.9x$ .

Verification: For $x = 3$ : $\hat y = 1.3 + 2.7 = 4.0 = \bar y$ . The regression line always passes through $(\bar x, \bar y)$ . Correct.

Source. OpenIntro Statistics 4th ed., §8.1 Exercise 8.1 — CC-BY-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 23Understanding 3Modeling 6Challenge 5Proof 3

Sources

This synthesis lesson integrates the four pillars of the program. Primary sources by axis:

Calculus (60% of problems):

Active Calculus 2.0 — Matt Boelkins · Grand Valley State University · 2024 · EN · CC-BY-NC-SA. §1–8 (limits, derivatives, integrals, series, intuitive ODEs): primary source for the calculus questions in this workshop.

Linear Algebra (15% of problems):

Linear Algebra Done Right (4th ed) — Sheldon Axler · 2024 · EN · CC-BY-NC. Chs. 1–10 (vector spaces, operators, eigenvalues, spectral decomposition, SVD): modern rigor without determinants-first; foundation of diagonalization and eigenvalue problems.

Probability and Statistics (15% of problems):

OpenIntro Statistics (4th ed) — Diez, Çetinkaya-Rundel, Barr · 2019 · EN · CC-BY-SA. Chs. 1–8 (distributions, CI, regression, tests): probability and inferential statistics with real data.

Applied Modeling (10% of problems):

Notes on Diffy Qs — Jiří Lebl · Oklahoma State University · 2024 · EN · CC-BY-SA. Chs. 1–4 (applied ODEs, reduced Black–Scholes to heat equation): foundation for quantitative finance and dynamic modeling exercises.