Deep-Dive Week 2: Weak Topics Continue + Cross-Topic Synthesis

Define $T_n$ by $T_n(\cos\theta) = \cos n\theta$: so $T_0 = 1$, $T_1 = x$, and the recurrence $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$ (which is just the sum-to-product identity $\cos(n{+}1)\theta + \cos(n{-}1)\theta = 2\cos\theta \cos n\theta$). Leading coefficient $2^{n-1}$, degree $n$, and on $[-1, 1]$ the polynomial oscillates between $\pm 1$ exactly $n+1$ times — the equioscillation that makes it extremal.

The extremal property is the Putnam-grade fact: among monic degree-$n$ polynomials, $\frac{T_n}{2^{n-1}}$ has the SMALLEST maximum absolute value on $[-1,1]$, namely $\frac{1}{2^{n-1}}$. Any claim of the form 'a polynomial with such-and-such coefficients must get this large somewhere on an interval' is a Chebyshev problem; the proof template is counting sign alternations of the difference against the equioscillation points.

Recognition cues: $\cos n\theta$ expansions, expressions like $x + \frac{1}{x} = 2\cos\theta$ (so $x^n + x^{-n} = 2\cos n\theta = 2T_n(\cos\theta)$), bounds on polynomials over intervals, and nested radicals like $\sqrt{2 + \sqrt{2 + \cdots}}$ (half-angle cascades). When a problem constrains a polynomial on an interval and asks about its size or coefficients, Chebyshev is the extremal witness — and usually the answer.

Pitfall. Forgetting that the alternation argument needs the challenger's bound to be STRICT at the equioscillation points, and needs the leading terms to cancel (both monic). Both hypotheses do real work — cite both.

Nonlinear recurrences on this exam almost always fall to a substitution from a short menu. $a_{n+1} = a_n^2 - 2$: try $a_n = x^n + x^{-n}$ (or $2\cos\theta_n$ / $2\cosh t_n$ depending on whether $|a_0| \le 2$), since $(x + x^{-1})^2 - 2 = x^2 + x^{-2}$ — squaring DOUBLES the exponent, and the recurrence becomes $\theta_{n+1} = 2\theta_n$. This is Chebyshev again: $a_{n+1} = T_2(a_n)$ in disguise (up to the factor-2 normalization $a_n = 2\cos\theta_n$).

Fraction forms $a_{n+1} = \frac{\alpha a_n + \beta}{\gamma a_n + \delta}$ (Möbius) linearize through the matrix $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$ — iterating the recurrence is taking matrix powers, so eigenvalues give the closed form and fixed points of the map are the eigen-directions. Tangent-addition forms $a_{n+1} = \frac{a_n + c}{1 - c\, a_n}$ are the special case $a_n = \tan\theta_n$, $\theta_{n+1} = \theta_n + \arctan c$ — periodicity questions become rationality-of-angle questions.

The meta-move: compute the first few terms NUMERICALLY and look at what is invariant or periodic. Period 6 whispers $\pi/3$ and tangent; doubling maps whisper $\cos$ or $\cosh$; a conserved quantity (check $a_{n+1}^2 - k a_{n+1} a_n + a_n^2$ for small $k$) converts the recurrence into a conic the orbit lives on — the Markov-equation trick.

Pitfall. Choosing $\cos$ when the orbit escapes $[-2, 2]$. The substitution $a_n = 2\cos\theta_n$ only represents $|a_n| \le 2$; outside that band you need $x + x^{-1}$ with real $x$ (equivalently $2\cosh$). Check the seed's size FIRST.

List the seams you have already crossed without naming them: complex numbers proving sum bounds (Week 9's $\sum \frac{\sin n}{n}$), linear algebra counting combinatorial structures, generating functions doing number theory (the filter), induction carrying analysis (Week 3's monotone recurrences). The synthesis gates this week make the crossing deliberate: each problem is chosen so the STATEMENT lives in one track and the SOLUTION in another.

The trained reflex: when a problem resists its home track for ninety seconds, ask 'what would this look like as algebra / as counting / as analysis / as geometry?' Translation is cheap — a recurrence becomes a matrix, a sum becomes an integral bound, a divisibility becomes a group-order statement — and the act of restating often exposes the structure the original phrasing hides.

Keep a seam ledger this week: for every synthesis problem, one line recording home track, solving track, and the translation that connected them. By Sunday you will have a personal map of which crossings feel native and which need reps — that map feeds directly into next week's toolkit audit.

Pitfall. Translating and then FORGETTING to translate back. The graded answer must live in the problem's original language — after the algebra closes, spend the last two minutes converting the conclusion and its conditions back to the original statement's terms.