W6 · Toolkit— gates27–31 hrs

Week 6 of 28 · Toolkit · due 2026-07-04 · 27–31 hrs

Analysis: Integration + FTC + Algebra: Functional Equations + LA: Inner Products + Comb: Generating Functions + NT: Congruences + Hidden Tools: ODE Recognition + Generating Functions

Analysis: Integration + FTC + Algebra: Functional Equations + LA: Inner Products + Comb: Generating Functions + NT: Congruences + Hidden Tools: ODE Recognition + Generating Functions gate complete

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This week's lesson · 10 min read

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Integrals with leverage: FTC, symmetry tricks, and the generating function idea

Putnam integrals are almost never exercises in finding antiderivatives — they are exercises in refusing to. The three moves that actually appear (reflection, differentiation under the integral, exploiting FTC as a bridge between size and slope) all work AROUND the integrand instead of through it. The same philosophy powers this week's other big opening: generating functions, where you refuse to count directly and let coefficients do the counting.

Choose the representation in which the problem is easy: reflect the integral, encode the sequence, take the inner product. Representation is a decision, not a given.

Key ideas — the week on one card

  1. The reflection $x \mapsto a + b - x$ fixes $\int_a^b$; when the integrand has partner symmetry, ADDING the two copies removes the hard factor.
  2. Functional equations are interrogations: substitute $0$, $x = y$, swaps — every substitution extracts one fact.
  3. $\langle f, g \rangle = \int fg$ is a genuine inner product on $C[a,b]$: Cauchy–Schwarz turns integral constraints into integral bounds.
  4. Generating functions make parts into factors: ordered compositions from allowed parts $S$ are counted by $\frac{1}{1 - \sum_{s \in S} x^s}$.

1The reflection trick and its relatives

The single most profitable line in contest integration: $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$. Substituting $x \mapsto a+b-x$ costs nothing and often produces an integral that ADDS nicely to the original — the sum simplifies even when neither piece does. If $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$, reflection swaps sin and cos, so $2I = \int_0^{\pi/2} 1\,dx$ and $I = \pi/4$ with no antiderivative in sight.

The Fundamental Theorem of Calculus earns its name on this exam as a BRIDGE: it converts hypotheses about $f$ into conclusions about $\int f$ and back. If $f$ is continuous and $\int_0^1 f = 0$, then $F(x) = \int_0^x f$ satisfies $F(0) = F(1) = 0$ — and Rolle (last week) gives a point where $F' = f$ vanishes. Integration hypotheses plus MVT-family theorems is a standing combo.

Differentiation under the integral sign — $\frac{d}{dt}\int f(x,t)\,dx = \int \frac{\partial f}{\partial t}\,dx$ under mild conditions — turns a hard integral into a differential equation for it. Introduce a parameter where none exists ($\int_0^1 \frac{x^t - 1}{\ln x} dx$: differentiate in $t$, get $\frac{1}{t+1}$, integrate back). Feynman popularized it; Week 18 drills it.

mirror line x = π/4f(x)f(π/2−x)
The reflection x ↦ π/2 − x sends the solid curve to the dashed one; together they fill the box, so each integral is half of it.

Worked example

Evaluate $\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x}\,dx$.

  1. Nudge 1

    Apply $x \mapsto \pi - x$ and write the SECOND expression for $I$.

    Reveal step 1

    Reflect with $x \mapsto \pi - x$: $\sin$ and $\cos^2$ are invariant, but $x$ becomes $\pi - x$. So $I = \int_0^\pi \frac{(\pi - x)\sin x}{1 + \cos^2 x} dx$.

  2. Nudge 2

    Add the two copies of $I$ — the lonely $x$ disappears.

    Reveal step 2

    Add the two expressions for $I$: $2I = \pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x}\,dx$ — the $x$ is GONE, which was the whole obstruction.

  3. Nudge 3

    $u = \cos x$ turns what remains into an arctangent.

    Reveal step 3

    Substitute $u = \cos x$, $du = -\sin x\,dx$: $2I = \pi \int_{-1}^{1} \frac{du}{1 + u^2} = \pi \left[\arctan u\right]_{-1}^{1} = \pi \cdot \frac{\pi}{2}$. Hence $I = \frac{\pi^2}{4}$.

  4. Answer

    $\frac{\pi^2}{4}$ — reflection kills the linear factor, then $u = \cos x$ finishes.

Pitfall. Reflexively hunting an antiderivative. When the integrand mixes a polynomial factor with a symmetric trig body, reflect FIRST — the polynomial factor is usually the removable part.

2Functional equations: substitution is interrogation

A functional equation is a machine you interrogate with inputs. The standard opening sequence: plug $x = y = 0$; plug $y = 0$ with $x$ free; plug $y = x$ and $y = -x$; hunt for values forcing $f(\text{something}) = 0$ or a fixed constant. Each substitution is a QUESTION, and you should know what you are asking — 'is $f(0)$ forced?', 'is $f$ even?', 'can I isolate $f(x)$?'.

Cauchy's equation $f(x+y) = f(x) + f(y)$ is the ancestral form. Over $\mathbb{Q}$ its solutions are exactly $f(x) = cx$ — provable by building up from $f(1)$ through integers and fractions. Over $\mathbb{R}$, linearity needs ONE regularity crumb: continuity at a single point, monotonicity on an interval, or boundedness near zero all force $f(x) = cx$. Contest problems supply exactly one crumb; your job is to spot which and cite it.

Always finish with the verification loop: candidate solutions found by substitution must be CHECKED in the original equation, and the write-up must argue no others exist. Half credit lives in the difference between 'we found $f(x) = x$' and 'we showed every solution is $f(x) = x$'.

Worked example

Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + y) + f(x - y) = 2f(x)$ for all $x, y$.

  1. Nudge 1

    Feed the equation cheap inputs — $x = y = 0$, then $x = y$ — and normalize with $g = f - f(0)$.

    Reveal step 1

    Interrogate: $x = y = 0$ gives $2f(0) = 2f(0)$ — nothing. Try $x = y$: $f(2x) + f(0) = 2f(x)$. Now set $g(x) = f(x) - f(0)$, so $g(0) = 0$ and $g(2x) = 2g(x)$.

  2. Nudge 2

    Change variables so the equation says $g$ is additive (Jensen's shape).

    Reveal step 2

    The original equation in $g$: $g(x+y) + g(x-y) = 2g(x)$ (constants cancel). With $x = \frac{u+v}{2}, y = \frac{u-v}{2}$: $g(u) + g(v) = 2g(\frac{u+v}{2}) = g(u + v)$ using the doubling identity — Cauchy's equation for $g$.

  3. Nudge 3

    State the general solution honestly — regularity is what pins $g$ down to linear.

    Reveal step 3

    Without regularity, solutions over $\mathbb{R}$ are all additive functions: $f(x) = A(x) + c$ with $A$ additive. If the problem grants continuity (or any crumb), $A(x) = ax$ and $f(x) = ax + c$. Verify: $(a(x+y)+c) + (a(x-y)+c) = 2ax + 2c$ ✓. State which regularity you used — it is the difference between the two answers.

  4. Answer

    Exactly the additive-plus-constant functions; with any regularity crumb (continuity at a point suffices), $f(x) = ax + c$.

Pitfall. Announcing $f(x) = ax + c$ from Cauchy WITHOUT a regularity hypothesis. Pathological additive solutions exist (via a Hamel basis); the crumb is not optional, and grading rubrics know it.

3Inner products: geometry you can compute with

An inner product turns a vector space into geometry: lengths ($\|v\|^2 = \langle v, v\rangle$), angles, and orthogonality. The Cauchy–Schwarz inequality $|\langle u, v\rangle| \le \|u\| \|v\|$ you met as a sum inequality is really THIS statement — and choosing a clever inner product space is how it cracks integral inequalities: on $C[0,1]$ with $\langle f, g\rangle = \int_0^1 fg$, Cauchy–Schwarz reads $(\int fg)^2 \le \int f^2 \int g^2$.

Orthogonality is a computation device: if $\{e_i\}$ are orthonormal, coefficients are inner products ($v = \sum \langle v, e_i\rangle e_i$) and Pythagoras generalizes to $\|v\|^2 = \sum \langle v, e_i \rangle^2$ — Bessel/Parseval. The contest use: bound a sum of squares of 'components' by the square of the 'length', often with the trig system on $[0, 2\pi]$ or with rows of a matrix.

The projection formula — the closest point to $v$ in a subspace $W$ is the orthogonal projection, and the error $v - \operatorname{proj}_W v$ is orthogonal to $W$ — converts minimization problems into linear algebra. 'Minimize $\int_0^1 (x^2 - a - bx)^2 dx$ over $a, b$' is not calculus; it is projecting $x^2$ onto the span of $\{1, x\}$.

Worked example

For continuous $f$ on $[0,1]$ with $\int_0^1 f(x)\,dx = 1$, prove $\int_0^1 f(x)^2\,dx \ge 1$.

  1. Nudge 1

    The constraint is an inner product of $f$ with the constant function 1.

    Reveal step 1

    Recognize the inner product: on $C[0,1]$, take $\langle f, g \rangle = \int_0^1 fg$. The hypothesis says $\langle f, 1 \rangle = 1$, where $1$ is the constant function.

  2. Nudge 2

    Apply Cauchy–Schwarz to $\langle f, 1 \rangle$ and compute $\|1\|$.

    Reveal step 2

    Cauchy–Schwarz: $1 = |\langle f, 1\rangle|^2 \le \|f\|^2 \|1\|^2 = \left(\int_0^1 f^2\right)\left(\int_0^1 1\right) = \int_0^1 f^2$.

  3. Nudge 3

    Equality analysis: proportional to a constant means constant.

    Reveal step 3

    Equality iff $f$ is proportional to $1$, i.e. $f \equiv 1$ — consistent with the constraint. One line once the space is named; the entire solution is the CHOICE of inner product.

  4. Answer

    Cauchy–Schwarz against the constant function: $1 = \langle f, 1\rangle^2 \le \int f^2 \cdot \int 1^2$, equality iff $f \equiv 1$.

Pitfall. Verifying inner-product axioms in the write-up but forgetting the actual argument — or the reverse. On the exam: name the space, cite Cauchy–Schwarz, state equality. Three sentences; do not pad, do not skip.

4Generating functions: sequences as coefficients

A generating function packs a sequence into a power series $A(x) = \sum a_n x^n$, converting sequence operations into algebra: shifting multiplies by $x$, convolution of sequences is multiplication of series, and linear recurrences become rational functions. The Fibonacci archetype: $F(x) = \frac{x}{1 - x - x^2}$, read directly from the recurrence, and partial fractions then yields Binet's closed form with zero cleverness.

The identities you need daily: $\frac{1}{1-x} = \sum x^n$, its derivative-powered upgrade $\frac{1}{(1-x)^{k}} = \sum \binom{n+k-1}{k-1} x^n$ (stars and bars, encoded!), and the binomial theorem for arbitrary exponents. Counting problems phrased as 'number of ways to pay $n$ cents with coins of sizes $a, b, c$' are literally asking for a coefficient of $\frac{1}{(1-x^a)(1-x^b)(1-x^c)}$.

The evaluation trick runs the machine backward: identities among coefficients follow from evaluating the SERIES at points. $\sum_k \binom{n}{k} = 2^n$ is $(1+x)^n$ at $x = 1$; alternating sums are $x = -1$; and next week's roots-of-unity filter is 'evaluate at all $n$-th roots of unity and average' to extract every $k$-th coefficient.

Worked example

How many ways can $n \ge 0$ be written as an ordered sum of 1s and 2s (compositions with parts 1, 2)?

  1. Nudge 1

    Encode one part as $x + x^2$; a composition with $k$ parts is that factor to the $k$-th power.

    Reveal step 1

    Encode one part: a part is 1 or 2, so its generating function is $x + x^2$. A composition with $k$ parts contributes $(x + x^2)^k$; summing over all $k \ge 0$ gives $\frac{1}{1 - x - x^2}$ — the geometric series over $x + x^2$.

  2. Nudge 2

    Sum the geometric series over $k$ and read the recurrence off the denominator.

    Reveal step 2

    Recognize the series: $\frac{1}{1 - x - x^2} = \sum c_n x^n$ where the denominator relation forces $c_n = c_{n-1} + c_{n-2}$, $c_0 = c_1 = 1$ — Fibonacci shifted.

  3. Nudge 3

    Which famous sequence obeys $c_n = c_{n-1} + c_{n-2}$? Verify $n = 4$ by hand.

    Reveal step 3

    So the count is $F_{n+1}$ (with $F_1 = F_2 = 1$): compositions of 4 number $F_5 = 5$, and listing confirms: $1111, 112, 121, 211, 22$. The generating function DERIVED the recurrence; nothing was guessed.

  4. Answer

    $F_{n+1}$ — the generating function $\frac{1}{1-x-x^2}$ carries Fibonacci coefficients.

Pitfall. Worrying about convergence. Contest generating functions are FORMAL power series — coefficient bookkeeping, no analysis needed — unless you actually evaluate at a real point, at which point convergence becomes your problem again. Know which mode you are in.

Before you open the gates

  • Before integrating anything, test the reflection $x \mapsto a + b - x$; if the integrand simplifies when ADDED to itself, you are done in three lines.
  • Interrogate functional equations in a fixed order ($x=y=0$; one variable zero; $y = \pm x$), and know what question each substitution asks.
  • Every found solution of a functional equation gets verified in the original, and the write-up must exclude others.
  • See an integral inequality between products and squares → name the inner product space and let Cauchy–Schwarz speak.
  • Translate counting to coefficients when parts/choices multiply: encode one object, then take products and geometric series.

Check yourself

1. $\int_0^{\pi/2} \frac{\cos x}{\sin x + \cos x}\,dx$ equals:

2. $f(x+y) = f(x)f(y)$ for all reals, $f$ continuous, $f$ not identically zero. Then $f$ is:

3. The number of solutions of $a + b + c = 20$ in nonnegative integers is the coefficient of $x^{20}$ in:

Practice ladder — three rungs, rising

Each rung: attempt cold, one hint if stuck, worked resolution only after a real try.

Rung 1 (functional equation, Cauchy reduction). Find all continuous functions $f:\mathbb{R}\to\mathbb{R}$ with $f(x+y) = f(x)+f(y)+2xy$ for all real $x,y$.

One hint

Peel off the piece responsible for the $2xy$: set $g(x) = f(x)-x^2$ and show $g$ solves Cauchy's equation, then spend the continuity hypothesis.

Worked resolution

Put $x=y=0$: $f(0) = 2f(0)$, so $f(0)=0$. Let $g(x) = f(x)-x^2$. Then $g(x+y) = f(x+y)-(x+y)^2 = [f(x)+f(y)+2xy]-x^2-2xy-y^2 = g(x)+g(y)$, so $g$ is additive. An additive function that is also continuous is linear, $g(x)=cx$, hence $f(x) = x^2 + cx$ for a constant $c$. Check: $(x+y)^2 + c(x+y) = [x^2+cx]+[y^2+cy]+2xy$. ✓ Continuity is exactly the crumb that excludes the pathological (Hamel-basis) additive solutions. [Source: this week's Track B — Algebra: Functional Equations; the Cauchy-reduction substitution toolkit, PnB section 3.4.1 + Engel Ch. 11.]

Rung 2 (inner-product inequality). Let $f:[0,1]\to\mathbb{R}$ be continuous with $\int_0^1 f(x)\,dx = 0$. Prove that $\left(\int_0^1 x\,f(x)\,dx\right)^2 \le \frac{1}{12}\int_0^1 f(x)^2\,dx$.

One hint

The constraint lets you replace $x$ by $x-\tfrac12$ for free. Then apply the Cauchy–Schwarz inequality $\left(\int gh\right)^2 \le \int g^2 \int h^2$ on $[0,1]$.

Worked resolution

Because $\int_0^1 f = 0$, adding a constant multiple of $f$ changes nothing: $\int_0^1 x f\,dx = \int_0^1 (x-\tfrac12) f\,dx$. Apply Cauchy–Schwarz for the inner product $\langle g,h\rangle = \int_0^1 gh$: $\left(\int_0^1 (x-\tfrac12)f\right)^2 \le \left(\int_0^1 (x-\tfrac12)^2\,dx\right)\left(\int_0^1 f^2\,dx\right)$. The first factor is $\int_0^1 (x-\tfrac12)^2\,dx = \tfrac{1}{12}$, so $\left(\int_0^1 x f\right)^2 \le \tfrac{1}{12}\int_0^1 f^2$. Subtracting the mean is the orthogonal projection off the constants — the geometry the constraint hands you. [Source: this week's Track D — Linear Algebra: Axler Ch. 6 inner products, with the integral Cauchy–Schwarz recognition item in PnB section 3.2.7.]

Rung 3 (generating function, closed form). Let $a_0 = 2$, $a_1 = 1$, and $a_n = a_{n-1} + 2a_{n-2}$ for $n \ge 2$. Form $A(x) = \sum_{n\ge0} a_n x^n$, show $A(x) = \frac{2-x}{1-x-2x^2}$, and read off a closed form for $a_n$.

One hint

Multiply the recurrence by $x^n$ and sum over $n \ge 2$; write the shifted sums back in terms of $A(x)$. Then factor $1-x-2x^2$ and use partial fractions.

Worked resolution

Summing $a_n x^n$ over $n\ge2$ gives $A(x) - a_0 - a_1 x = x\big(A(x)-a_0\big) + 2x^2 A(x)$, i.e. $A - 2 - x = x(A-2) + 2x^2 A$. Solving, $A(x)(1 - x - 2x^2) = 2 - x$, so $A(x) = \frac{2-x}{1-x-2x^2}$. Factor $1-x-2x^2 = (1-2x)(1+x)$; partial fractions give $\frac{2-x}{(1-2x)(1+x)} = \frac{1}{1-2x} + \frac{1}{1+x} = \sum_{n\ge0} 2^n x^n + \sum_{n\ge0}(-1)^n x^n$. Hence $a_n = 2^n + (-1)^n$. Check: $a_0 = 2$, $a_1 = 1$, $a_2 = 5 = a_1 + 2a_0$. ✓ [Source: this week's Hidden Tool — Generating Functions (Wilf, generatingfunctionology); this is the linear-recurrence canon assigned in the GF drill gate.]

Prove it — constructed response

Prove that $\int_0^{\pi/2} \frac{dx}{1 + (\tan x)^{\sqrt{2}}} = \frac{\pi}{4}$. Reflection is your only real move: use it, then explain why the exponent $\sqrt{2}$ never enters the answer.

The gates

Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load

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- Read: Rudin *PMA* Ch. 6 (~38 pp.) — Riemann-Stieltjes integral, integrability, linearity, FTC (both forms), integration by parts, change of variables - Watch: Michael Penn Interesting Integrals — 3 videos - Watch: Silver Integration Bee Intermediate — 2 videos - Do: Rudin Ch. 6 ex 1, 3, 5, 7, 9, 11, 13, 16 - Do: PnB section 3.2.7 probs 1–3 - Do: MIT sum_integrals.pdf probs 1–5

sources & assignments (4)

Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Drill

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unlocks after: Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load

- Read: Rudin *PMA* Ch. 6 (~38 pp.) — Riemann-Stieltjes integral, integrability, linearity, FTC (both forms), integration by parts, change of variables - Watch: Michael Penn Interesting Integrals — 3 videos - Watch: Silver Integration Bee Intermediate — 2 videos - Do: Rudin Ch. 6 ex 1, 3, 5, 7, 9, 11, 13, 16 - Do: PnB section 3.2.7 probs 1–3 - Do: MIT sum_integrals.pdf probs 1–5 (write each as an explicit Riemann sum — partition + sample points — before converting)

sources & assignments (7)
  • Source Rudin — Principles of Mathematical Analysis (PMA) — Drill reference — this gate trains the material of “Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Interesting Integrals (playlist) — Drill reference — this gate trains the material of “Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Silver — Integration Bee Training (playlist) — Drill reference — this gate trains the material of “Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source MIT 18.100A — Intuition companion for “Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Rudin — Rudin Ch. 6 ex 1, 3, 5, 7, 9, 11, 13, 16Putnam
  • Problems PnB — PnB section 3.2.7 probs 1–3 Also, recognition-level: (a) l'Hôpital for 0/0 and ∞/∞ (and one case where it loops or fails); (b) ∫₁^∞ dx/xᵖ converges iff p > 1; (c) one integral inequality (Cauchy–Schwarz for integrals, ∫fg ≤ √(∫f²)√(∫g²)); (d) the Gamma function Γ(n) = (n−1)! and the Beta–Gamma relation B(x,y) = Γ(x)Γ(y)/Γ(x+y).Putnam
  • Problems MIT sum_integrals.pdf — MIT sum_integrals.pdf probs 1–5 — for each sum-to-integral conversion, write the limit explicitly as a Riemann sum first (name the partition and the sample points), then one line on why integrability justifies passing to the limitPutnam

Hidden Tool

Hidden Tool: ODE Recognition / Integral Equation → ODE: Load

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*Prerequisites: FTC (Track A above) + Leibniz rule (Week 5)* - Read: Paul's Online Math Notes — "Separable Equations" section + "Linear First Order" section only (~20 pp., free online) - Watch: Professor Leonard "Separable Differential Equations" (first 20 min) + "Integrating Factor Method" (first 15 min) - Do — 3 targeted problems: 1. Integral-to-ODE: given F(x) = ∫₀ˣ f(t)eˣ⁻ᵗ dt + x, differentiate both sides (Leibniz) → F'(x)−F(x)=1 → integrating factor μ=e^{−x} → F(x)=Ce^x−1. Work from scratch. 2. Paul's Online — Separable Equations exercises 1, 3, 5. Write dy/f(y) = g(x)dx explicitly before integrating. 3. Paul's Online — Linear First Order exercises 1, 3, 5. Write μ(x) = e^{∫P(x)dx} and verify d/dx[μf] = μQ before integrating. - ODE Recognition: - f' = kf → f = Ce^{kx} (recognize immediately — do not re-derive) - Integral equation F(x) = ∫₀ˣ f(t)h(x,t) dt → differentiate via Leibniz → first-order ODE in F - Separable: dy/f(y) = g(x) dx → integrate both sides independently - Integrating factor: f' + P(x)f = Q(x) → μ = e^{∫P dx} → (μf)' = μQ → integrate - Power series ODE: f = Σaₙxⁿ → substitute → coefficient recurrence (Week 8) - **Derive it cold (required, self-attested — no notes):** before you tick this gate, reconstruct the integral-equation→ODE reduction blank-page: take f(x) = 1 + ∫₀ˣ f(t)dt, differentiate both sides via FTC to get f'(x) = f(x), read f(0) = 1 off the equation at x = 0, and solve f(x) = eˣ; then state the general move — differentiate an ∫₀ˣ-equation (FTC/Leibniz) to trade the cumulative integral for pointwise data, and *when the integral term closes back on F* this collapses to an ODE, whose constant you recover from the base point x = 0. Self-check: (1) you *established* the reduction and ran it, didn't quote it; (2) you can say WHY it loses nothing — FTC turns the cumulative integral into a pointwise derivative condition and the original equation at the lower limit is the initial condition, so ODE + IC ⟺ the integral equation. Can't rebuild it blank-page → not learned yet.

sources & assignments (5)

Hidden Tool: ODE Recognition / Integral Equation → ODE: Drill

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unlocks after: Hidden Tool: ODE Recognition / Integral Equation → ODE: Load

*Prerequisites: FTC (Track A above) + Leibniz rule (Week 5)* - Read: Paul's Online Math Notes — "Separable Equations" section + "Linear First Order" section only (~20 pp., free online) - Watch: Professor Leonard "Separable Differential Equations" (first 20 min) + "Integrating Factor Method" (first 15 min) - Do — 3 targeted problems: 1. Integral-to-ODE: given F(x) = ∫₀ˣ f(t)eˣ⁻ᵗ dt + x, differentiate both sides (Leibniz) → F'(x)−F(x)=1 → integrating factor μ=e^{−x} → F(x)=Ce^x−1. Work from scratch. 2. Paul's Online — Separable Equations exercises 1, 3, 5. Write dy/f(y) = g(x)dx explicitly before integrating. 3. Paul's Online — Linear First Order exercises 1, 3, 5. Write μ(x) = e^{∫P(x)dx} and verify d/dx[μf] = μQ before integrating. - ODE Recognition: - f' = kf → f = Ce^{kx} (recognize immediately — do not re-derive) - Integral equation F(x) = ∫₀ˣ f(t)h(x,t) dt → differentiate via Leibniz → first-order ODE in F - Separable: dy/f(y) = g(x) dx → integrate both sides independently - Integrating factor: f' + P(x)f = Q(x) → μ = e^{∫P dx} → (μf)' = μQ → integrate - Power series ODE: f = Σaₙxⁿ → substitute → coefficient recurrence (Week 8)

sources & assignments (4)
  • Source Paul's Online Math Notes — Differential Equations — Re-watch the worked-solution segment only AFTER attempting this gate's drill problems; note where your route diverged.
  • Source MIT 18.100A — Problem-session companion for “Hidden Tool: ODE Recognition / Integral Equation → ODE: Drill” — watch one worked problem, stop, finish it yourself on paper, then compare.
  • Source MathDoctorBob — Intuition companion for “Hidden Tool: ODE Recognition / Integral Equation → ODE: Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Targeted drill (built on this gate's reading) — (1) Integral-to-ODE: given F(x) = ∫₀ˣ f(t)eˣ⁻ᵗ dt + x, differentiate both sides (Leibniz) → F'(x)−F(x)=1 → integrating factor μ=e^{−x} → F(x)=Ce^x−1. Work from scratch. (2) Paul's Online — Separable Equations exercises 1, 3, 5. Write dy/f(y) = g(x)dx explicitly before integrating. (3) Paul's Online — Linear First Order exercises 1, 3, 5. Write μ(x) = e^{∫P(x)dx} and verify d/dx[μf] = μQ before integrating.Drill

Track B — Algebra

Track B — Algebra: Functional Equations (1.5 hrs/day): Load

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- Read: PnB section 3.4.1 (pp. 185–200) — functional equations fully with examples - Read: Engel Ch. 11 — intro + probs 1–10 - Watch: little fermat FE Tutorial — lessons 1–4 - Do: 100 Functional Equations #1, 5, 10, 15, 20, 25, 30 (record every substitution tried) - Do: CMU 05-FE exercise sheet probs 1–4 - Archive: 3 Putnam A1/A2 · Algebra/FE · any slot · 18 min each - FE Substitution: - x = y = 0: find f(0) — often forces f(0) = 0 or 1 - y = x: get f(2x) in terms of f(x) - y = −x: relate f(x) and f(−x) — parity - x = 1/x: use when domain is ℝ>0 - y = f(x): iterate — f(f(x)) appears - Injectivity/surjectivity: prove before using if domain argument requires it - Domain trap: does the FE force f onto all of ℝ, or only ℝ>0? FE substitution toolkit (work each problem through this list): try x=0, y=0, x=y, y=−x, and swapping x↔y; test injectivity (f(a)=f(b)⇒a=b) and surjectivity; iterate f(f(x)); guess f(x)=x, cx, x+c, then verify; pin down f(0) and f(1) first; only use continuity/monotonicity if the problem grants it.

why this gate: the lesson's §2 Functional equations: substitution is interrogation is what it trains

sources & assignments (7)

Track B — Algebra: Functional Equations (1.5 hrs/day): Drill

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unlocks after: Track B — Algebra: Functional Equations (1.5 hrs/day): Load

- Read: PnB section 3.4.1 (pp. 185–200) — functional equations fully with examples - Read: Engel Ch. 11 — intro + probs 1–10 - Watch: little fermat FE Tutorial — lessons 1–4 - Do: 100 Functional Equations #1, 5, 10, 15, 20, 25 + the iteration/fixed-point rep (tag techniques on every rep: special values · injective · surjective · symmetry · Cauchy reduction · fixed point · iteration · monotonicity/continuity · domain) - Do: CMU 05-FE exercise sheet probs 1–4 - Archive: 3 Putnam A1/A2 · Algebra/FE · any slot · 18 min each - FE Substitution: - x = y = 0: find f(0) — often forces f(0) = 0 or 1 - y = x: get f(2x) in terms of f(x) - y = −x: relate f(x) and f(−x) — parity - x = 1/x: use when domain is ℝ>0 - y = f(x): iterate — f(f(x)) appears - Injectivity/surjectivity: prove before using if domain argument requires it - Domain trap: does the FE force f onto all of ℝ, or only ℝ>0?

why this gate: the lesson's §2 Functional equations: substitution is interrogation is what it trains

sources & assignments (7)
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track B — Algebra: Functional Equations (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source little fermat FE Tutorial — Drill reference — this gate trains the material of “Track B — Algebra: Functional Equations (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Putnam Exam Solutions (playlist) — Drill reference — this gate trains the material of “Track B — Algebra: Functional Equations (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source little fermat — Olympiad Inequalities Tutorial (playlist) — Intuition companion for “Track B — Algebra: Functional Equations (1.5 hrs/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems 100 Functional Equations — 100 Functional Equations #1, 5, 10, 15, 20, 25 (#30's slot displaced by the iteration rep below). For EVERY rep, before reading any solution, tag which techniques you used from this fixed list: special values · injectivity · surjectivity · symmetry/swap · Cauchy reduction · fixed point · iteration · monotonicity/continuity · domain restriction (Z/Q/R). Record every substitution tried.Putnam
  • Problems CMU 05-FE exercise sheet — CMU 05-FE exercise sheet probs 1–3 (prob 4's slot displaced to the polynomial-FE micro-rep) — same technique-tagging requirement on eachPutnam
  • Problems Iteration / fixed-point rep (self-contained) — (a) Describe all f: R → R with f(f(x)) = x (involutions): show injectivity is forced, give a non-identity example, and note the role of the fixed-point set. (b) Prove no CONTINUOUS f: R → R satisfies f(f(x)) = −x: continuous + injective ⟹ strictly monotone ⟹ f∘f increasing, but −x is decreasing. Tag both with the technique list. (c) Polynomial FE micro-rep: find all polynomials with P(x²) = P(x)² (answer: P(x) = xⁿ for n ≥ 0, plus P ≡ 0 — argue via leading coefficient + a too-many-roots/degree count, not by inspection). (d) Jensen-equation recognition, one sentence: f((x+y)/2) = (f(x)+f(y))/2 reduces to Cauchy's equation via g(x) = f(x) − f(0); additive + measurable/monotone ⟹ linear.Putnam

Track B — Algebra: Functional Equations (1.5 hrs/day): Archive bridge

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unlocks after: Track B — Algebra: Functional Equations (1.5 hrs/day): Drill

- Read: PnB section 3.4.1 (pp. 185–200) — functional equations fully with examples - Read: Engel Ch. 11 — intro + probs 1–10 - Watch: little fermat FE Tutorial — lessons 1–4 - Do: 100 Functional Equations #1, 5, 10, 15, 20, 25, 30 (record every substitution tried) - Do: CMU 05-FE exercise sheet probs 1–4 - Archive: 3 Putnam A1/A2 · Algebra/FE · any slot · 18 min each - FE Substitution: - x = y = 0: find f(0) — often forces f(0) = 0 or 1 - y = x: get f(2x) in terms of f(x) - y = −x: relate f(x) and f(−x) — parity - x = 1/x: use when domain is ℝ>0 - y = f(x): iterate — f(f(x)) appears - Injectivity/surjectivity: prove before using if domain argument requires it - Domain trap: does the FE force f onto all of ℝ, or only ℝ>0?

why this gate: the lesson's §2 Functional equations: substitution is interrogation is what it trains

sources & assignments (5)
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track B — Algebra: Functional Equations (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source little fermat FE Tutorial — Drill reference — this gate trains the material of “Track B — Algebra: Functional Equations (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Putnam Exam Solutions (playlist) — Drill reference — this gate trains the material of “Track B — Algebra: Functional Equations (1.5 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source little fermat — Olympiad Inequalities Tutorial (playlist) — Intuition companion for “Track B — Algebra: Functional Equations (1.5 hrs/day): Archive bridge” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Archive Browser — Putnam 2008 A1, 2001 A1, 2000 A1, 1999 A1 · Algebra/FE/Algebra · 18 min eachPutnam

archive pull: 4 problems · Functional Equations/Algebra · A1/A2 · Challenge · 18 min →

Track D — Linear Algebra

Track D — Linear Algebra: Axler Ch. 5C + Ch. 6 (1 hr/day): Load

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- Read: Axler Ch. 5C (~20 pp.) — real vector spaces, complexification - Read: Axler Ch. 6A–6B (~40 pp.) — inner products, norms, Gram-Schmidt, orthogonal complements - Watch: Sheldon Axler LADR lectures — lectures 11–13 - Watch: MIT 18.06 Strang — orthogonality lecture (OCW) - Do: Axler 5C: 1, 3, 5 · 6A: 1, 5, 9, 12, 15, 18, 21 · 6B: 1, 5, 9, 13 - Do: MIT linalg.pdf probs 3–4

sources & assignments (4)

Track D — Linear Algebra: Axler Ch. 5C + Ch. 6 (1 hr/day): Drill

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unlocks after: Track D — Linear Algebra: Axler Ch. 5C + Ch. 6 (1 hr/day): Load

- Read: Axler Ch. 5C (~20 pp.) — real vector spaces, complexification - Read: Axler Ch. 6A–6B (~40 pp.) — inner products, norms, Gram-Schmidt, orthogonal complements - Watch: Sheldon Axler LADR lectures — lectures 11–13 - Watch: MIT 18.06 Strang — orthogonality lecture (OCW) - Do: Axler 5C: 1, 3, 5 · 6A: 1, 5, 9, 12, 15, 18, 21 · 6B: 1, 5, 9, 13 - Do: MIT linalg.pdf probs 3–4

sources & assignments (6)
  • Source Axler — Linear Algebra Done Right — Drill reference — this gate trains the material of “Track D — Linear Algebra: Axler Ch. 5C + Ch. 6 (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Sheldon Axler — Linear Algebra Done Right (lecture videos) — Drill reference — this gate trains the material of “Track D — Linear Algebra: Axler Ch. 5C + Ch. 6 (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source MIT 18.06 Strang — Drill reference — this gate trains the material of “Track D — Linear Algebra: Axler Ch. 5C + Ch. 6 (1 hr/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source little fermat — Olympiad Inequalities Tutorial (playlist) — Intuition companion for “Track D — Linear Algebra: Axler Ch. 5C + Ch. 6 (1 hr/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems Axler 5C — Axler 5C: 1, 3, 5 · 6A: 1, 5, 9, 12, 15, 18, 21 · 6B: 1, 5, 9, 13Putnam
  • Problems MIT linalg.pdf — MIT linalg.pdf probs 3–4Putnam

Hidden Tool

Hidden Tool: Generating Functions: Load

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*Prerequisites: sequences (Track A Week 3) + bijection/recursion (Track C Week 5)* - Read: Wilf *generatingfunctionology* intro through section 1.4 (~30 pp.) — OGF notation, coefficient extraction - Watch: Mohamed Omar Generating Series — lessons 1–2 - Do: MIT genfunc.pdf probs 1–6 - Do: CMU 08-Recursions exercise sheet probs 1–3 - Generating Function: - aₙ → F(x) = Σaₙxⁿ; extract: aₙ = [xⁿ]F(x) - Shift: aₙ₊₁ ↔ (F(x)−a₀)/x - Convolution: aₙ = Σbₖcₙ₋ₖ ↔ F(x) = B(x)C(x) - Recurrence → algebraic equation for F(x) → solve → partial fractions → [xⁿ] - EGF: aₙ/n! → use for labeled counting - **Derive it cold (required, self-attested — no notes):** before you tick this gate, prove two things blank-page. (a) A linear recurrence has a *rational* generating function: set A(x) = Σaₙxⁿ, and for Fₙ = Fₙ₋₁ + Fₙ₋₂ (F₀=0, F₁=1) form (1 − x − x²)A(x) — its xⁿ coefficient is Fₙ − Fₙ₋₁ − Fₙ₋₂ = 0 for n ≥ 2, leaving only x, so A(x) = x/(1 − x − x²); then state the general shape (1 − c₁x − c₂x² − ⋯)A(x) = a polynomial of degree below the recurrence's order, fixed by the initial values. (b) Convolution ↔ product: A(x)B(x) = Σₙ(Σₖ₌₀ⁿ aₖbₙ₋ₖ)xⁿ, by collecting xᵏ·xⁿ⁻ᵏ. Self-check: (1) you *established* both, didn't quote it; (2) you can say WHY the GF is rational — multiplying the recurrence by xⁿ and summing kills every coefficient past the initial segment, so the denominator IS the reversed characteristic polynomial (its roots are the reciprocals of the characteristic roots), which partial fractions turns back into the closed form aₙ = Σ cᵢrᵢⁿ (the coefficients become polynomials in n when a root repeats). Can't rebuild it blank-page → not learned yet.

why this gate: the lesson's §4 Generating functions: sequences as coefficients is what it trains

sources & assignments (4)

Hidden Tool: Generating Functions: Drill

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unlocks after: Hidden Tool: Generating Functions: Load

*Prerequisites: sequences (Track A Week 3) + bijection/recursion (Track C Week 5)* - Read: Wilf *generatingfunctionology* intro through section 1.4 (~30 pp.) — OGF notation, coefficient extraction - Watch: Mohamed Omar Generating Series — lessons 1–2 - Do: MIT genfunc.pdf probs 1–6 - Do: CMU 08-Recursions exercise sheet probs 1–3 - Generating Function: - aₙ → F(x) = Σaₙxⁿ; extract: aₙ = [xⁿ]F(x) - Shift: aₙ₊₁ ↔ (F(x)−a₀)/x - Convolution: aₙ = Σbₖcₙ₋ₖ ↔ F(x) = B(x)C(x) - Recurrence → algebraic equation for F(x) → solve → partial fractions → [xⁿ] - EGF: aₙ/n! → use for labeled counting

why this gate: the lesson's §4 Generating functions: sequences as coefficients is what it trains

sources & assignments (6)
  • Source Wilf — generatingfunctionology (full free PDF) — Wilf §1–2 — the generating-function method this drill trains. (Replaced a wrong-resource carryover: Paul's ODE notes had been attached to this GF gate.)
  • Source Michael Penn — Interesting Integrals (playlist) — Drill reference — this gate trains the material of “Hidden Tool: ODE Recognition / Integral Equation → ODE: Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source MIT 18.100A — Intuition companion for “Hidden Tool: Generating Functions: Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems MIT genfunc.pdf — MIT genfunc.pdf probs 1–6Putnam
  • Problems CMU 08-Recursions exercise sheet — CMU 08-Recursions exercise sheet probs 1–2 (prob 3's slot displaced to the recurrence canon below)Putnam
  • Problems Linear-recurrence canon (self-contained) — Solve a_n = a_{n-1} + 2a_{n-2}, a_0 = 2, a_1 = 1, THREE ways and reconcile: (1) characteristic equation x² = x + 2 ⟹ roots 2, −1 ⟹ a_n = A·2^n + B·(−1)^n, fit constants; (2) generating function A(x) = Σ a_n x^n: derive A(x) = (2 − x)/(1 − x − 2x²), partial-fraction it over the roots, and read off the same closed form; (3) a nonhomogeneous variant a_n = a_{n-1} + 2a_{n-2} + 1 — find one particular solution (constant) and add the homogeneous part. Then one line: state the Fibonacci GF Σ F_n x^n = x/(1 − x − x²) and where the golden ratio enters as the dominant root (ties to the W8 asymptotics gate). Also note telescoping: Σ_{k=1}^n (b_k − b_{k-1}) = b_n − b_0 as the simplest 'recurrence' you should always spot. Also: (a) state the product/convolution rule on formal power series (convergence irrelevant) — if A(x)=Σaₙxⁿ and B(x)=Σbₙxⁿ then A(x)B(x) has coefficients cₙ = Σₖ aₖ bₙ₋ₖ (convolution); (b) the Catalan generating function C(x) = (1 − √(1−4x))/(2x) satisfies C = 1 + xC² — derive that functional equation from the Catalan recurrence; (c) matrix recurrence: write the Fibonacci recurrence as the power [[1,1],[1,0]]ⁿ and read F_n off the top-right entry.Putnam

Track C — Combinatorics

Track C — Combinatorics: Generating Functions + Recurrences (1 hr/day)

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- Do: 102 Combinatorial Problems #45, 50, 55 - Do: Stanford 06wk2 probs 1–6

why this gate: the lesson's §4 Generating functions: sequences as coefficients is what it trains

sources & assignments (6)
  • Source Rudin — Principles of Mathematical Analysis (PMA) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Michael Penn — Interesting Integrals (playlist) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Silver — Integration Bee Training (playlist) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Michael Penn — Real Analysis (playlist) — Intuition companion for “Track C — Combinatorics: Generating Functions + Recurrences (1 hr/day)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems 102 Combinatorial — 102 Combinatorial Problems #45, 50, 55Putnam
  • Problems Stanford 06wk2 — Stanford 06wk2 probs 1–6Putnam

Track E — Number Theory

Track E — Number Theory: Modular Arithmetic (0.75 hrs/day): Load

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- Read: PnB section 5.2.1–5.2.3 (pp. 257–265) — residue classes, CRT, Fermat, Euler - Read: Engel Ch. 13 probs 9–15 - Watch: Mu Prime Math modular arithmetic — 2 videos - Do: 104 NT Problems #12, 14, 16, 18 - Do: MIT cong.pdf probs 1–3 - Do: Stanford 07wk2 probs 1–2

sources & assignments (7)

Track E — Number Theory: Modular Arithmetic (0.75 hrs/day): Drill

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unlocks after: Track E — Number Theory: Modular Arithmetic (0.75 hrs/day): Load

- Read: PnB section 5.2.1–5.2.3 (pp. 257–265) — residue classes, CRT, Fermat, Euler - Read: Engel Ch. 13 probs 9–15 - Watch: Mu Prime Math modular arithmetic — 2 videos - Do: 104 NT Problems #12, 14, 16, 18 - Do: MIT cong.pdf probs 1–3 - Do: Stanford 07wk2 probs 1–2

sources & assignments (8)
  • Source Putnam and Beyond — Drill reference — this gate trains the material of “Track E — Number Theory: Modular Arithmetic (0.75 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Modular Arithmetic and Linear Congruences (playlist) — Drill reference — this gate trains the material of “Track E — Number Theory: Modular Arithmetic (0.75 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Michael Penn — Number Theory v2 (playlist) — Drill reference — this gate trains the material of “Track E — Number Theory: Modular Arithmetic (0.75 hrs/day): Load”. Stuck mid-drill? The tool lives here; go back, find the move, return and finish in writing.
  • Source Math Geeks Method of Infinite Descent — Intuition companion for “Track E — Number Theory: Modular Arithmetic (0.75 hrs/day): Drill” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.
  • Problems 104 NT — 104 NT Problems #12, 14, 16, 18Putnam
  • Problems MIT cong.pdf — MIT cong.pdf probs 1–3Putnam
  • Problems Stanford 07wk2 — Stanford 07wk2 probs 1–2Putnam
  • Problems AoPS Floor Function article — Prove Hermite's identity sum_{k=0}^{n-1} floor(x + k/n) = floor(nx) using the article's splitting method, then write the fractional-part version of the statement. Also run the extended Euclidean algorithm on gcd(252, 198) and write explicit Bézout coefficients x, y with 252x + 198y = gcd (Bézout's identity).AMC

AMC/AIME Warmup

AMC/AIME Warmup (keep speed-solving active during heavy reading weeks)

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- Archive: 2 AMC-band · any topic · any slot · 8 min each

sources & assignments (5)
  • Source Rudin — Principles of Mathematical Analysis (PMA) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Michael Penn — Interesting Integrals (playlist) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source Silver — Integration Bee Training (playlist) — Week reference — this gate applies this week's lead material (“Track A — Analysis: Rudin Ch. 6 (1.5 hrs/day): Load”). If an attempt stalls, the repair source is here.
  • Source MIT 18.100A — Intuition companion for “AMC/AIME Warmup (keep speed-solving active during heavy reading weeks)” — the picture behind the machinery; afterwards write one sentence connecting the visual to this gate's exercises.

Hidden Tool

Hidden Tool: Chinese Remainder Theorem

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CRT was unnamed in the curriculum (≈1% of Putnam NT, but a basic, fast tool). This names + drills it. - **Derive it cold (required, self-attested — no notes):** before you tick this gate, reconstruct ℤ/mn ≅ ℤ/m × ℤ/n (gcd(m,n)=1) blank-page. Bézout gives m·m' + n·n' = 1; then n·n' ≡ 1 (mod m) and ≡ 0 (mod n), while m·m' ≡ 0 (mod m) and ≡ 1 (mod n) — two indicator elements. So for any target (a mod m, b mod n), x = a·n·n' + b·m·m' satisfies x ≡ a (mod m) and x ≡ b (mod n): surjectivity, with the reconstruction formula explicit. Uniqueness mod mn: m∣(x−y) and n∣(x−y) with gcd(m,n)=1 ⟹ mn∣(x−y). Self-check: (1) you *established* the isomorphism and wrote the reconstruction formula, didn't quote it; (2) you can say WHY coprimality is essential — m·m' + n·n' = 1 is solvable iff gcd(m,n)=1, and that Bézout combination is exactly what builds the ≡1/≡0 indicators. Can't rebuild it blank-page → not learned yet.

sources & assignments (5)
  • Source CRT — statement & reconstruction — State CRT for coprime moduli; reconstruct x mod mn from x mod m and x mod n via the standard formula.
  • Source 104 Number Theory Problems — CRT / simultaneous-congruence problems — read the congruences section.
  • Source Michael Penn — Chinese Remainder Theorem — CRT statement, proof, and a worked reconstruction.
  • Problems 104 NT — 104 Number Theory Problems — CRT problems 1–2 (simultaneous congruences)Putnam
  • Problems CRT drill — Reconstruct x: x≡2 (mod 3), x≡3 (mod 5), x≡2 (mod 7); then state the general reconstruction recipe.AMC

archive pull: 1 problems · Number Theory · Putnam · 20 min →

Reach Contact — A3 official-solution autopsy

Reach Contact — A3 official-solution autopsy (lemma only)

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Introduces A3 contact immediately after the main FE launch so reach problems do not suddenly appear in Mastery.

Ritual: Goal is one certified first lemma, not a full solve. Stop at 35 minutes total.

sources & assignments (2)
  • Source Putnam archive — A3 autopsy model — Use Putnam 2016 A3 as a functional-equation flavored reach example: read after a 15-minute classification attempt, then extract the first forced substitution/lemma.
  • Problems Putnam A3 autopsy — Putnam 1991 B2 — 15-minute classification attempt, then solution autopsy. Output: first lemma/substitution and why it is forced.Nightmare

Problem-Solving Reps — weekly homework

Problem-Solving Reps — weekly homework

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Standing weekly homework — the problem-solving book stack, every week. Solve ALL listed; write ONE full clean solution (the rest may stay scratch). Up to 2 due re-solves from your review queue surface first.

sources & assignments (3)
  • Problems Engel — Engel Ch. 11 probs 1–8 (functional equations)Putnam
  • Problems Wilf — Generatingfunctionology Ch. 1–2 exercises 1–4Putnam
  • Problems 104 NT — 104 Number Theory Problems — Introductory probs 1–4 (congruences)Putnam

Reflect

Reflect: reconstruct and log the pattern

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- Reconstruct: 1 Rudin Ch. 6 exercise → close → rewrite → pattern note - Reflect: FTC both forms from memory · FE substitution card verified · GF coefficient extraction procedure - Verify: one FE — is injectivity/surjectivity claim proved, not assumed? ---

Ritual: 1 Rudin Ch. 6 exercise → close → rewrite → pattern note | FTC both forms from memory · FE substitution card verified · GF coefficient extraction procedure

sources & assignments (4)

Verify: audit one proof before closing

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unlocks after: Reflect: reconstruct and log the pattern

- Reconstruct: 1 Rudin Ch. 6 exercise → close → rewrite → pattern note - Reflect: FTC both forms from memory · FE substitution card verified · GF coefficient extraction procedure - Verify: one FE — is injectivity/surjectivity claim proved, not assumed? ---

Ritual: one FE — is injectivity/surjectivity claim proved, not assumed?

sources & assignments (4)

Lerma Training

Lerma Training: Combinatorics set (Northwestern)

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Source: Miguel A. Lerma, Northwestern Putnam team training problems (2023). Local PDF: /library/Lerma/putnam-training-2023.pdf. Hints + full solutions are in the PDF's later parts.

sources & assignments (5)
  • Problems Lerma Training 2023 §6 Generating Functions — Work problems 6.1–6.7 (4–6 per session). Encode a count as a coefficient; manipulate the closed form. Self-check against the PDF's hints/solutions parts.Putnam
  • Problems Lerma Training 2023 §9 Pigeonhole Principle — Work problems 9.1–9.11 (4–6 per session). Name the boxes and the objects before claiming a collision. Self-check against the PDF's hints/solutions parts.Putnam
  • Problems Lerma Training 2023 §12 Inclusion–Exclusion — Work problems 12.1–12.3 (4–6 per session). Alternate over- and under-counts; watch the sign pattern. Self-check against the PDF's hints/solutions parts.Putnam
  • Problems Lerma Training 2023 §13 Combinatorics and Probability — Work problems 13.1–13.8 (4–6 per session). Linearity of expectation, symmetry, complementary counting. Self-check against the PDF's hints/solutions parts.Putnam

Complex numbers keep-warm

Complex numbers keep-warm (20 min)

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Retention keep-warm (2026-07-02 audit): complex numbers went silent for 3-6 week stretches. ~20 minutes: two Lerma problems + one memory recall. Source: Lerma Training 2023 §5 (local PDF, hints/solutions in later parts).

Ritual: Recall (closed book): State De Moivre's formula and compute (1+i)^8 mentally.

sources & assignments (2)
  • Problems Lerma Training 2023 §5 Complex Numbers — Problems 5.1–5.2 (~15 min at conversion speed). Then the recall prompt below — closed book.Putnam

Exit contract — Week 6

Verification remaining

  • reading progress…

Carry-forward repairs

  • reading queue…

Next week opens with

W7 · Analysis: Uniform Convergence + Algebra: Advanced Inequalities + LA: Spectral Theorem + Comb: Extremal + NT: Diophantine + Geometry begins + Hidden Tools: Auxiliary Functions + Trace/Det Identities + Graph Modeling
the exam follows the final taper week

Train Week 6 in the trainer →