--- title: "Polynomials" book: "University Mathematics — Year 1" subject: math language: en chapter: 8 exercises: 12 source: https://one-course.com/books/math/3/en/chapter/8-polynomials --- # Chapter 8 — Polynomials [Polynomials](#def-b1-poly-def) are the algebraist’s favorite functions — except that they are not treated as functions here, but as formal expressions in an indeterminate $X$, added and multiplied by the rules of a commutative [ring](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-ring). The theory runs strikingly parallel to [Chapter 6](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#ch-b1-arith): a Euclidean division, a gcd and Bézout relations, irreducible elements and a unique factorization. Throughout, $K$ denotes $\Q$, $\R$ or $\C$. ## 8.1 The ring $K[X]$ **Definition 8.1 (Polynomial, degree).** A *polynomial* with coefficients in $K$ is a formal sum $$ P = a_0 + a_1 X + a_2 X^2 + \dots + a_n X^n = \sum_{k} a_k X^k, $$ with $a_k \in K$ all zero from some index on. With the natural addition and the product $$ \Bigl(\sum_i a_i X^i\Bigr)\Bigl(\sum_j b_j X^j\Bigr) = \sum_k \Bigl(\sum_{i+j=k} a_i b_j\Bigr) X^k, $$ the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $K[X]$ is a commutative [ring](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-ring). The *degree* $\deg P$ of $P \neq 0$ is the largest $n$ with $a_n \neq 0$; $a_n$ is the *leading coefficient* ($P$ is *monic* when $a_n = 1$), and by convention $\deg 0 = -\infty$. Every polynomial defines a function $x \mapsto P(x)$ on $K$ by substitution. **Proposition 8.2 (Degree rules; integral domain).** For $P, Q \in K[X]$: $$ \deg(P + Q) \leq \max(\deg P, \deg Q), \qquad \deg(PQ) = \deg P + \deg Q . $$ Consequently $K[X]$ is an [integral domain](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field), and its units are the nonzero constants. **Proof.** The sum rule is clear (coefficients beyond the max vanish). For the product, let $a_m$ and $b_n$ be the leading coefficients: the coefficient of $X^{m+n}$ in $PQ$ is $a_m b_n \neq 0$ ($K$ is a [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field), hence an [integral domain](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field)), and all higher coefficients vanish. If $P, Q \neq 0$ then $\deg PQ = \deg P + \deg Q \geq 0$, so $PQ \neq 0$: [integral domain](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field). If $PQ = 1$ then $\deg P + \deg Q = 0$ forces $\deg P = \deg Q = 0$: invertible elements are the invertible constants, i.e. all of $K^*$. ∎ **Theorem 8.3 (Euclidean division).** Let $A, B \in K[X]$ with $B \neq 0$. There is exactly one pair $(Q, R)$ of [polynomials](#def-b1-poly-def) with $$ A = BQ + R, \qquad \deg R < \deg B . $$ **Proof.** *Existence*, by strong induction on $\deg A$. If $\deg A < \deg B$, take $(Q, R) = (0, A)$. Otherwise write $A = a X^m + \dots$, $B = b X^n + \dots$ with $m \geq n$; the [polynomial](#def-b1-poly-def) $A_1 = A - \frac ab X^{m-n} B$ has degree $< m$ (the leading terms cancel), so by induction $A_1 = BQ_1 + R$ with $\deg R < \deg B$, and $A = B(Q_1 + \frac ab X^{m-n}) + R$. *Uniqueness*: if $BQ + R = BQ' + R'$, then $B(Q - Q') = R' - R$ with $\deg(R' - R) < \deg B$; by the degree rule this forces $Q - Q' = 0$, then $R = R'$. ∎ **Example 8.4.** Divide $A = X^4 + X^3 - 2X + 1$ by $B = X^2 + 1$: $$ X^4 + X^3 - 2X + 1 = (X^2 + 1)(X^2 + X - 1) + (-3X + 2). $$ (Compute: subtract $X^2 B$, then $X B$, then $-B$; the remainder $-3X + 2$ has degree $1 < 2$.) **Method 8.5 (Horner’s scheme).** To evaluate $P = a_nX^n + \dots + a_0$ at $x$, or to divide $P$ by $X - x$, avoid computing powers: read the coefficients left to right, and iterate *multiply by $x$, add the next coefficient*: $$ b_n = a_n, \qquad b_{k} = a_{k} + x\,b_{k+1} \quad (k = n-1, \dots, 0) . $$ Then $b_0 = P(x)$, and the earlier $b_k$ are the quotient’s coefficients: $P = (X - x)(b_nX^{n-1} + \dots + b_1) + b_0$ (expand and compare). Example: $P = X^4 - 5X^3 + 6X^2 + 4X - 8$ at $x = 2$: the $b$’s are $1, -3, 0, 4, 0$, so $P(2) = 0$ and $P = (X-2)(X^3 - 3X^2 + 4)$ — one line instead of a long division, and $n$ multiplications instead of the $\approx n^2/2$ of the naive evaluation. Iterating the scheme at the same point extracts multiplicities (compare [Example 8.12](#ex-b1-poly-multexample)). **Remark 8.6 (Arithmetic of K[X]K[X]K[X]).** With Euclidean division in hand, the entire arithmetic of [Chapter 6](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#ch-b1-arith) transfers to $K[X]$, with the same proofs, degree playing the role of absolute value: gcd (normalized to be [monic](#def-b1-poly-def)), extended [Euclidean algorithm](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#met-b1-arith-euclid), Bézout’s identity, Gauss’s lemma, irreducible [polynomials](#def-b1-poly-def) and unique factorization. We freely use these transferred results, and [Exercise 8.6](#exo-b1-poly-6) rehearses one of them. ## 8.2 Roots **Theorem 8.7 (Factor theorem).** Let $P \in K[X]$ and $a \in K$. The remainder of $P$ upon division by $X - a$ is the constant $P(a)$. In particular $$ P(a) = 0 \iff (X - a) \mid P . $$ More generally, distinct roots $a_1, \dots, a_r$ of $P$ give the factorization $P = (X - a_1)\cdots(X - a_r)\, Q$. **Proof.** Divide: $P = (X - a) Q + R$ with $\deg R < 1$, so $R$ is a constant $c$; substituting $X = a$ (substitution respects sums and products) gives $P(a) = c$. The equivalence follows. For several roots, induct on $r$: the case $r = 1$ is the equivalence just proved. Assume the [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) for $r - 1$ roots and let $a_1, \dots, a_r$ be distinct roots of $P$. Write $P = (X - a_1)Q_1$; for each $i \geq 2$, substituting $a_i$: $$ 0 = P(a_i) = (a_i - a_1)\,Q_1(a_i), \qquad a_i - a_1 \neq 0 , $$ and since $K$ has no zero divisors, $Q_1(a_i) = 0$: the $r - 1$ distinct points $a_2, \dots, a_r$ are roots of $Q_1$. The induction hypothesis factors $Q_1 = (X - a_2)\cdots(X - a_r)\,Q$, and substituting back gives the claim. ∎ **Corollary 8.8 (A polynomial of degree nnn has at most nnn roots).** A nonzero $P \in K[X]$ of degree $n$ has at most $n$ distinct roots in $K$. Consequently, a [polynomial](#def-b1-poly-def) (of degree $\leq n$) vanishing at $n + 1$ distinct points is the zero [polynomial](#def-b1-poly-def), and two [polynomials](#def-b1-poly-def) of degree $\leq n$ agreeing at $n+1$ points are equal. **Proof.** If $a_1, \dots, a_r$ are distinct roots, [Theorem 8.7](#thm-b1-poly-factor) gives $P = (X-a_1)\cdots(X-a_r) Q$, so $n = \deg P \geq r$. The two consequences follow by contradiction and by difference. ∎ **Example 8.9 (The auxiliary-polynomial trick).** Let $P$ be the [polynomial](#def-b1-poly-def) of degree $\leq n$ with $$ P(k) = \frac{k}{k+1} \qquad (k = 0, 1, \dots, n) ; $$ it exists and is unique by [Lagrange interpolation](#thm-b1-poly-lagrange) below. What is $P(n+1)$? Clear denominators: the [polynomial](#def-b1-poly-def) $Q = (X+1)P - X$ has degree $\leq n + 1$ and vanishes at the $n + 1$ points $0, 1, \dots, n$, so by [Theorem 8.7](#thm-b1-poly-factor) $$ Q = c\,X(X-1)(X-2)\cdots(X-n) $$ for some constant $c$. Evaluate where $Q$ is known independently: at $X = -1$, $Q(-1) = 0 \cdot P(-1) + 1 = 1$, while the product equals $(-1)(-2)\cdots(-1-n) = (-1)^{n+1}(n+1)!$; hence $c = \frac{(-1)^{n+1}}{(n+1)!}$. Now evaluate at $X = n + 1$: $$ (n+2)\,P(n+1) - (n+1) = Q(n+1) = c\,(n+1)! = (-1)^{n+1} , $$ so $P(n+1) = \dfrac{(n+1) + (-1)^{n+1}}{n+2}$: equal to $1$ for odd $n$, and to $\frac{n}{n+2}$ for even $n$ — the interpolating [polynomial](#def-b1-poly-def) does *not* continue the pattern $\frac{n+1}{n+2}$. The trick to remember: encode the data as roots of an auxiliary [polynomial](#def-b1-poly-def), identify the unknown constant at a point outside the data, and harvest. **Definition 8.10 (Derivative, multiplicity).** The *formal derivative* of $P = \sum a_k X^k$ is $P' = \sum_{k \geq 1} k\,a_k X^{k-1}$; it satisfies the usual rules $(P+Q)' = P' + Q'$, $(PQ)' = P'Q + PQ'$ (checked on monomials and extended by linearity). A root $a$ of $P$ has *multiplicity* $m \geq 1$ when $(X-a)^m \mid P$ but $(X-a)^{m+1} \nmid P$; the root is *simple* if $m = 1$, *multiple* if $m \geq 2$. **Proposition 8.11 (Multiplicity via derivatives).** $a$ is a root of $P$ of [multiplicity](#def-b1-poly-derivative) $\geq m$ if and only if $$ P(a) = P'(a) = \dots = P^{(m-1)}(a) = 0 . $$ In particular, $a$ is a multiple root of $P$ if and only if $P(a) = P'(a) = 0$. **Proof.** Write $P = (X - a)^m Q + R$ where $R$ is the remainder of the division by $(X-a)^m$, $\deg R < m$. Differentiating $k \leq m - 1$ times and evaluating at $a$: the first term contributes $0$ (each derivative retains a factor $(X-a)$), so $P^{(k)}(a) = R^{(k)}(a)$. Now a [polynomial](#def-b1-poly-def) $R$ of degree $< m$ is determined by $R(a), R'(a), \dots, R^{(m-1)}(a)$: writing $R = \sum_{k < m} c_k (X - a)^k$ (possible: expand powers of $X = (X - a) + a$), one finds $R^{(k)}(a) = k!\, c_k$. Hence: all $P^{(k)}(a) = 0$ for $k < m$ $\iff$ all $c_k = 0$ $\iff$ $R = 0$ $\iff$ $(X-a)^m \mid P$. ∎ **Example 8.12 (Computing a multiplicity).** What is the [multiplicity](#def-b1-poly-derivative) of the root $2$ in $P = X^4 - 5X^3 + 6X^2 + 4X - 8$? Evaluate the successive derivatives at $2$: $$ P(2) = 16 - 40 + 24 + 8 - 8 = 0, \qquad P'(2) = 32 - 60 + 24 + 4 = 0, $$ $$ P''(2) = 48 - 60 + 12 = 0, \qquad P'''(2) = 48 - 30 = 18 \neq 0 $$ (with $P' = 4X^3 - 15X^2 + 12X + 4$, $P'' = 12X^2 - 30X + 12$, $P''' = 24X - 30$). Three vanishing values then a nonzero one: [multiplicity](#def-b1-poly-derivative) exactly $3$. Dividing, $P = (X - 2)^3(X + 1)$ — which one checks by expanding $(X-2)^3 = X^3 - 6X^2 + 12X - 8$ and multiplying by $X + 1$. The insight: multiplicities are read off *evaluations*, no factorization needed — which is exactly how one detects them when the factorization is out of reach. **Example 8.13 (Detecting multiple roots with a gcd).** When no root is known, [Proposition 8.11](#prop-b1-poly-multiplicity) still gives a *global* multiple-root detector: $a$ is a multiple root of $P$ iff it is a common root of $P$ and $P'$, so $P$ has a multiple root (in $\C$) iff $\gcd(P, P') \neq 1$ — computable by the [Euclidean algorithm](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#met-b1-arith-euclid) without solving anything. Sample: $P = X^3 - 3X + 2$, $P' = 3X^2 - 3 = 3(X - 1)(X + 1)$. Testing the roots $\pm1$ of $P'$ inside $P$: $P(1) = 0$ but $P(-1) = 4$, so $$ \gcd(P, P') = X - 1 : $$ the root $1$ is multiple; dividing twice, $P = (X - 1)^2(X + 2)$. The gcd even reports the full [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of multiple roots, each with [multiplicity](#def-b1-poly-derivative) lowered by one — the fact every computer-algebra system exploits to “square-free factorize” before any root-hunting, and the [polynomial](#def-b1-poly-def) twin of [Exercise 8.9](#exo-b1-poly-9)’s no-multiple-root arguments. **Theorem 8.14 (Fundamental theorem of algebra).** Every nonconstant [polynomial](#def-b1-poly-def) of $\C[X]$ has a root in $\C$. **Proof.** *Admitted at this level.* ∎ **Remark 8.15.** Despite its name, the theorem is a [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) of *analysis*: every known proof uses the completeness of $\R$ in some form, and none is purely algebraic — the honest proof is given in the Year 3 volume, once either complex integration or compactness arguments are available. What this chapter genuinely proves is the *reduction*: granted one root for every nonconstant [polynomial](#def-b1-poly-def), the full factorizations over $\C$ and $\R$ below follow by pure algebra. **Corollary 8.16 (Factorization over C\CC and over R\RR).** 1. Every nonzero $P \in \C[X]$ factors as $$P = c\, (X - a_1)^{m_1} \cdots (X - a_r)^{m_r},$$ with $c$ the leading coefficient, $a_i$ the distinct complex roots, $\sum m_i = \deg P$: *counted with [multiplicity](#def-b1-poly-derivative), a [polynomial](#def-b1-poly-def) of degree $n$ has exactly $n$ complex roots*. 2. Every nonzero $P \in \R[X]$ factors over $\R$ as $$P = c \prod_i (X - a_i)^{m_i} \prod_j (X^2 + p_j X + q_j)^{n_j},$$ the quadratic factors being distinct with $p_j^2 - 4q_j < 0$ (no real roots). **Proof.** (1) Induction on the degree, splitting off one root at a time by [Theorem 8.7](#thm-b1-poly-factor); the count of degrees matches at each step. (2) Let $P$ have real coefficients. If $z$ is a complex root of [multiplicity](#def-b1-poly-derivative) $m$, so is $\conj z$: conjugating $P(z) = 0$ gives $P(\conj z) = \conj{P(z)} = 0$ (the coefficients are their own conjugates), and the same applies to the derivatives ([Proposition 8.11](#prop-b1-poly-multiplicity)). Group the non-real roots in conjugate pairs: each pair contributes $$ (X - z)(X - \conj z) = X^2 - 2\Re(z)\, X + \abs z^2 , $$ a real quadratic with negative discriminant. The real roots contribute the linear factors. ∎ **Example 8.17.** $X^4 + 4$ was factored over $\R$ in [Exercise 3.5](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#exo-b1-complex-5) by pairing the four complex roots $\pm 1 \pm \iu$: $X^4 + 4 = (X^2 - 2X + 2)(X^2 + 2X + 2)$. Neither quadratic splits over $\R$ (discriminants $-4$). Note: an *irreducible* real [polynomial](#def-b1-poly-def) has degree $1$ or $2$ — this is exactly what the factorization theorem says. The same conjugate-pairing run on $X^4 + 1$, whose roots are $\eu^{\pm\iu\pi/4}$ and $\eu^{\pm3\iu\pi/4}$: each pair contributes $X^2 - 2\cos\theta\,X + 1$, so $$ X^4 + 1 = \bigl(X^2 - \sqrt2\,X + 1\bigr) \bigl(X^2 + \sqrt2\,X + 1\bigr) , $$ an identity invisible to naive factoring attempts over $\Q$ — the price of insisting on real (here even irrational) coefficients, and a standard input for integrating $\frac1{x^4 + 1}$ in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration). ![The Chebyshev polynomial T_5 = 16X5 - 20X3 + 5X on ∈tcc-11: it oscillates exactly between -1 and 1, touching the bounds at six points (marked). This equioscillation is what makes 2-4T_5 the monic quintic of smallest sup-norm on the interval ( and the weekend problem).](https://one-course.com/images/onecourse/chapters/math-3/b1-poly/fig-ce798e953ef6.svg) *The Chebyshev [polynomial](#def-b1-poly-def) $T_5 = 16X^5 - 20X^3 + 5X$ on $\intcc{-1}1$: it oscillates exactly between $-1$ and $1$, touching the bounds at six points (marked). This *equioscillation* is what makes $2^{-4}T_5$ the [monic](#def-b1-poly-def) quintic of smallest sup-norm on the interval ([Exercise 8.10](#exo-b1-poly-10) and the weekend problem).* **Remark 8.18 (Common pitfalls with polynomials).** 1. *[Polynomial](#def-b1-poly-def) versus function.* Over $K = \Q, \R, \C$ the two notions match (equal functions have equal coefficients, by [Corollary 8.8](#cor-b1-poly-nroots) and infiniteness of $K$ ), but conceptually a [polynomial](#def-b1-poly-def) is its coefficient list: over the two-element [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) $\Z/2\Z$ of [Chapter 7](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#ch-b1-structures) , $X^2 + X$ vanishes at both points, yet is not the zero [polynomial](#def-b1-poly-def) . 2. *Degrees under addition.* $\deg(P + Q)$ can drop below $\max(\deg P, \deg Q)$ when leading terms cancel; writing “ $\deg(P + Q) = \max(\dots)$ ” is only safe for distinct degrees. 3. *Roots counted correctly.* “ $n$ roots” in [Corollary 8.16](#cor-b1-poly-factorization) means *with [multiplicity](#def-b1-poly-derivative), in $\C$* : $X^2 + 1$ has no real roots, and $(X-1)^2$ has one distinct root but two with [multiplicity](#def-b1-poly-derivative) . [Statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) mixing the three counts are the commonest source of false proofs. 4. *Irreducibility depends on the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field).* $X^2 - 2$ is irreducible over $\Q$ , splits over $\R$ ; $X^2 + 1$ is irreducible over $\R$ , splits over $\C$ . The bare word “irreducible” means nothing until the coefficient [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) is named. ## 8.3 Coefficients and roots **Theorem 8.19 (Vieta’s formulas).** Let $P = X^n + c_{n-1} X^{n-1} + \dots + c_0$ be [monic](#def-b1-poly-def) with roots $a_1, \dots, a_n \in \C$ (with [multiplicity](#def-b1-poly-derivative)). Then $$ \sum_i a_i = -c_{n-1}, \qquad \sum_{i < j} a_i a_j = c_{n-2}, \qquad \dots, \qquad a_1 a_2 \cdots a_n = (-1)^n c_0 , $$ the $k$-th symmetric function of the roots being $(-1)^k c_{n-k}$. **Proof.** By [Corollary 8.16](#cor-b1-poly-factorization), $P = (X - a_1)\cdots(X - a_n)$ ([monic](#def-b1-poly-def), all roots listed). Expanding the product distributively produces one term per way of choosing, in each factor, either $X$ or the root term $-a_i$: choosing the roots in the factors indexed by $i_1 < \dots < i_k$ and $X$ in the $n - k$ others contributes $(-a_{i_1})\cdots(-a_{i_k})\,X^{n-k}$. Grouping by the power of $X$: $$ P = \sum_{k=0}^{n} (-1)^k \Bigl(\sum_{i_1 < \dots < i_k} a_{i_1}\cdots a_{i_k}\Bigr) X^{n-k} , $$ and identifying with $P = \sum_k c_{n-k}X^{n-k}$ (coefficients are unique, [Definition 8.1](#def-b1-poly-def)) gives $c_{n-k} = (-1)^k \sigma_k$, i.e. $\sigma_k = (-1)^kc_{n-k}$, where $\sigma_k$ denotes the $k$-th symmetric function displayed above. The three displayed cases are $k = 1$, $k = 2$ and $k = n$. ∎ **Example 8.20.** For the quadratic $X^2 - sX + p$: sum of roots $s$, product $p$ — already used repeatedly ([Exercise 3.8](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#exo-b1-complex-8)). For a [monic](#def-b1-poly-def) cubic $X^3 + aX^2 + bX + c$ with roots $\alpha, \beta, \gamma$: $$ \alpha + \beta + \gamma = -a, \quad \alpha\beta + \beta\gamma + \gamma\alpha = b, \quad \alpha\beta\gamma = -c , $$ which allows computing symmetric quantities like $\alpha^2 + \beta^2 + \gamma^2 = a^2 - 2b$ without solving. **Example 8.21 (Transforming the roots without finding them).** Let $\alpha, \beta$ be the roots of $X^2 - 3X + 1$. Which [monic](#def-b1-poly-def) quadratic has roots $\alpha^2, \beta^2$? By Vieta, $\alpha + \beta = 3$ and $\alpha\beta = 1$, so $$ \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 7, \qquad \alpha^2\beta^2 = (\alpha\beta)^2 = 1 : $$ the answer is $X^2 - 7X + 1$ — obtained without computing $\alpha = \frac{3 + \sqrt5}2$. (Check: $\alpha^2 = \frac{7 + 3\sqrt5}2$ and indeed $\alpha^2 + \beta^2 = 7$.) The same strategy handles reciprocals ($X^2 - \frac ba X + \frac ca$-type transformations), shifts, and any symmetric data: Vieta converts questions about *unknown* roots into algebra on *known* coefficients. It will serve constantly when the roots are eigenvalues ([Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det)). **Example 8.22 (Palindromic equations).** Solve $X^4 + X^3 - 4X^2 + X + 1 = 0$. The coefficients read the same in both directions, so $0$ is not a root and dividing by $X^2$ loses no solutions: $$ X^2 + X - 4 + \frac1X + \frac1{X^2} = 0 . $$ Set $y = X + \frac1X$: then $X^2 + \frac1{X^2} = y^2 - 2$, and the equation collapses to $$ y^2 + y - 6 = 0 \iff (y + 3)(y - 2) = 0 . $$ Unfold each value through $X^2 - yX + 1 = 0$: for $y = 2$, $X^2 - 2X + 1 = (X - 1)^2$ gives the double root $1$; for $y = -3$, $X^2 + 3X + 1 = 0$ gives $X = \frac{-3 \pm \sqrt5}2$. Four roots with [multiplicity](#def-b1-poly-derivative) for a quartic, as [Corollary 8.16](#cor-b1-poly-factorization) requires — obtained by solving two quadratics. The trick covers every *palindromic* [polynomial](#def-b1-poly-def): their roots come in reciprocal pairs $\{x, 1/x\}$ (replace $X$ by $1/X$ and clear denominators), and $y = X + \frac1X$ is precisely the quantity constant on such pairs, halving the degree. **Theorem 8.23 (Lagrange interpolation).** Let $x_0, \dots, x_n$ be distinct points of $K$ and $y_0, \dots, y_n \in K$. There is exactly one $P \in K[X]$ of degree $\leq n$ with $P(x_i) = y_i$ for all $i$, namely $$ P = \sum_{i=0}^{n} y_i\, L_i, \qquad L_i = \prod_{j \neq i} \frac{X - x_j}{x_i - x_j} . $$ **Proof.** Each $L_i$ has degree $n$ and satisfies $L_i(x_i) = 1$, $L_i(x_j) = 0$ for $j \neq i$ (each factor vanishes at the corresponding $x_j$). So the displayed $P$ has degree $\leq n$ and interpolates. Uniqueness: two interpolating [polynomials](#def-b1-poly-def) of degree $\leq n$ agree at the $n+1$ points $x_i$, hence are equal ([Corollary 8.8](#cor-b1-poly-nroots)). ∎ **Remark 8.24 (Interlude: polynomials are also vectors).** A change of viewpoint that [Chapter 18](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#ch-b1-vspaces) will make official: the [polynomials](#def-b1-poly-def) of degree $\leq n$ form a space in which addition and scaling behave exactly like coordinates — a [polynomial](#def-b1-poly-def) *is* its list of $n + 1$ coefficients. Three [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) of this chapter are secretly linear algebra. [Lagrange interpolation](#thm-b1-poly-lagrange) ([Theorem 8.23](#thm-b1-poly-lagrange)) says that the evaluation data $(P(x_0), \dots, P(x_n))$ determines $P$ uniquely: evaluation at $n + 1$ points is a linear bijection, and the $L_i$ are the basis adapted to it. The expansion $R = \sum c_k(X - a)^k$ in the proof of [Proposition 8.11](#prop-b1-poly-multiplicity) says that the powers of $(X - a)$ form another coordinate system, with $c_k = R^{(k)}(a)/k!$ as coordinates. And [Corollary 8.8](#cor-b1-poly-nroots) — more roots than the degree forces the zero [polynomial](#def-b1-poly-def) — is the engine of all uniqueness: it will become “a linear [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) on a space of dimension $n + 1$” in [Chapter 19](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#ch-b1-findim). When those chapters arrive, the space $K_n[X]$ will be their favorite example; it is worth arriving there already fluent in it. **Remark 8.25 (Where this chapter is used).** Factorization over $\R$ and $\C$ ([Corollary 8.16](#cor-b1-poly-factorization)) is the engine of partial fractions in [Chapter 9](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#ch-b1-fractions), hence of a large class of integrals in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration). The expansion of a [polynomial](#def-b1-poly-def) in powers of $(X - a)$, met in the proof of [Proposition 8.11](#prop-b1-poly-multiplicity), is the algebraic shadow of the Taylor formulas of [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor). Characteristic [polynomials](#def-b1-poly-def) already appeared for differential equations ([Chapter 5](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#ch-b1-diffeq)) and return for matrices in [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det); [Lagrange interpolation](#thm-b1-poly-lagrange) is the first existence-and-uniqueness theorem of numerical analysis, and the [Chebyshev polynomials](#pb-b1-poly-1) of [Exercise 8.10](#exo-b1-poly-10) — whose optimality the weekend problem below establishes — tell that discipline *where* to interpolate. Finally the whole arithmetic of $K[X]$, copied from [Chapter 6](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#ch-b1-arith), feeds the study of $K[X]$-ideals and quotient [rings](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-ring) in the Year 2 volume. ## 8.4 Exercises **Exercise 8.1 ★.** Carry out the Euclidean divisions: $X^5 - 1$ by $X^2 + X + 1$; then $2X^4 + X^3 - X + 3$ by $X^2 - 2$. **Solution of Exercise 8.1.** $X^5 - 1 = (X^2 + X + 1)(X^3 - X^2 + 1) + (-X - 2)$. Steps: subtract $X^3 B$, then $-X^2 B$, then $B$; the remainder $-X - 2$ has degree $1 < 2$. *Check at $X = 1$:* $\;0 = 3 \times 1 + (-3)$. $2X^4 + X^3 - X + 3 = (X^2 - 2)(2X^2 + X + 4) + (X + 11)$. *Check at $X = 0$:* $\;3 = (-2)(4) + 11$. **Exercise 8.2 ★.** For which $n \in \N$ does $X^2 + X + 1$ divide $X^{2n} + X^n + 1$? *Hint: the roots of $X^2 + X + 1$ are $j$ and $j^2$ with $j = \eu^{2\iu\pi/3}$; discuss $n$ mod $3$.* **Solution of Exercise 8.2.** $X^2 + X + 1 = (X - j)(X - j^2)$ with $j = \eu^{2\iu\pi/3}$, $j^3 = 1$. It [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $Q_n = X^{2n} + X^n + 1$ iff $j$ and $j^2$ are roots of $Q_n$; since $Q_n$ has real coefficients, $Q_n(j^2) = \conj{Q_n(j)}$, so the condition is just $Q_n(j) = 0$. Now $Q_n(j) = j^{2n} + j^n + 1$ depends on $n$ mod $3$: - $n \equiv 0$ : $Q_n(j) = 1 + 1 + 1 = 3 \neq 0$ ; - $n \equiv 1$ : $Q_n(j) = j^2 + j + 1 = 0$ ; - $n \equiv 2$ : $Q_n(j) = j^4 + j^2 + 1 = j + j^2 + 1 = 0$ . So $X^2 + X + 1 \mid X^{2n} + X^n + 1$ exactly when $3 \nmid n$. **Exercise 8.3 ★.** Determine the real $a, b$ so that $(X-1)^2$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $P = X^4 + aX^3 + bX^2 + 1$, then factor $P$ over $\R$ for these values. **Solution of Exercise 8.3.** By [Proposition 8.11](#prop-b1-poly-multiplicity), $(X-1)^2 \mid P$ iff $P(1) = P'(1) = 0$: $$ P(1) = 2 + a + b = 0, \qquad P'(1) = 4 + 3a + 2b = 0 . $$ Solving: $b = -a - 2$ and $4 + 3a - 2a - 4 = a = 0$, so $a = 0$, $b = -2$: $P = X^4 - 2X^2 + 1 = (X^2 - 1)^2 = (X-1)^2 (X+1)^2$, which is the real factorization. **Exercise 8.4 ★.** Factor over $\C$ and over $\R$: $X^3 - 1$; $\;X^4 + X^2 + 1$; $\;X^6 - 1$. **Solution of Exercise 8.4.** $X^3 - 1 = (X - 1)(X - j)(X - j^2)$ over $\C$ ($j = \eu^{2\iu\pi/3}$), and $(X - 1)(X^2 + X + 1)$ over $\R$. $X^4 + X^2 + 1 = (X^2 + X + 1)(X^2 - X + 1)$ over $\R$ (multiply out, or note $X^4 + X^2 + 1 = (X^2+1)^2 - X^2$); over $\C$, each quadratic splits: roots $j, j^2$ and $-j, -j^2$, i.e. $\eu^{\pm 2\iu\pi/3}, \eu^{\pm\iu\pi/3}$. $X^6 - 1 = \prod_{k=0}^{5} (X - \eu^{\iu k\pi/3})$ over $\C$, and over $\R$: $$ X^6 - 1 = (X-1)(X+1)(X^2 + X + 1)(X^2 - X + 1), $$ grouping the conjugate pairs $\eu^{\pm 2\iu\pi/3}$ and $\eu^{\pm \iu\pi/3}$. **Exercise 8.5 ★★.** Let $P = X^3 - 6X^2 + 11X - 6$. 1. Find the rational roots *(a rational root $p/q$ in lowest terms of a [monic](#def-b1-poly-def) integer [polynomial](#def-b1-poly-def) is an integer dividing the constant term — prove it)* , and factor $P$ . 2. Without solving, compute the sum of the squares and the sum of the inverses of the roots via Vieta, and check on the factorization. **Solution of Exercise 8.5.** 1. Let $p/q$ (lowest terms) be a root of the [monic](#def-b1-poly-def) integer [polynomial](#def-b1-poly-def) $X^3 + \dots + c_0$ : clearing denominators in $P(p/q) = 0$ gives $p^3 = -q\,(\text{integer})$ , so $q \mid p^3$ ; coprimality forces $q = \pm 1$ : the root is an integer $p$ , and $p \mid c_0$ (isolate $c_0$ ). Here the candidates divide $6$ : testing, $P(1) = 0$ , $P(2) = 0$ , $P(3) = 0$ . So $P = (X-1)(X-2)(X-3)$ . 2. Vieta: $s_1 = 6$ , $s_2 = 11$ , $s_3 = 6$ . Sum of squares: $s_1^2 - 2s_2 = 36 - 22 = 14 = 1 + 4 + 9$ , as expected. Sum of inverses: $\frac{s_2}{s_3} = \frac{11}{6} = 1 + \frac12 + \frac13$ , as expected. **Exercise 8.6 ★★.** Compute $\gcd(X^4 - 1,\; X^3 - X^2 + X - 1)$ by the [Euclidean algorithm](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#met-b1-arith-euclid), and write it as a combination $AU + BV$ of the two [polynomials](#def-b1-poly-def). **Solution of Exercise 8.6.** First division step of the [Euclidean algorithm](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#met-b1-arith-euclid): $$ (X + 1)(X^3 - X^2 + X - 1) = X^4 - X^3 + X^2 - X + X^3 - X^2 + X - 1 = X^4 - 1 , $$ so the division of $X^4 - 1$ by $X^3 - X^2 + X - 1$ is exact (quotient $X + 1$, remainder $0$), and the algorithm stops at once: $$ \gcd(X^4 - 1,\; X^3 - X^2 + X - 1) = X^3 - X^2 + X - 1 $$ (already [monic](#def-b1-poly-def)). The Bézout relation is the trivial one: $\gcd = 0 \cdot (X^4 - 1) + 1 \cdot (X^3 - X^2 + X - 1)$. Consistency check by factoring: $X^3 - X^2 + X - 1 = (X - 1)(X^2 + 1)$, which is indeed the product of the common irreducible factors of $X^4 - 1 = (X-1)(X+1)(X^2+1)$. **Exercise 8.7 ★★.** Let $P \in \R[X]$ with $P(x) \geq 0$ for all $x \in \R$. Prove that $P$ is a sum of two squares of real [polynomials](#def-b1-poly-def): $P = A^2 + B^2$. *Hint: in the real factorization, real roots have even [multiplicity](#def-b1-poly-derivative); write the quadratic factors as $(X - z)(X - \conj z)$ and use $\abs{\,\cdot\,}^2 = (\Re)^2 + (\Im)^2$ on the product of the $(X - z)$’s.* **Solution of Exercise 8.7.** Since $P \geq 0$ on $\R$, its real roots have even [multiplicity](#def-b1-poly-derivative) (at a root of odd [multiplicity](#def-b1-poly-derivative), $P$ changes sign). Using [Corollary 8.16](#cor-b1-poly-factorization) and pairing, write $$ P = c \prod_i (X - a_i)^{2k_i} \prod_j \bigl((X - z_j)(X - \conj z_j)\bigr)^{n_j}, $$ with $c > 0$ (behavior at $+\infty$). Let $$ S = \sqrt c\, \prod_i (X - a_i)^{k_i} \prod_j (X - z_j)^{n_j} \in \C[X], $$ so that $P = S\,\conj S$ where $\conj S$ has the conjugated coefficients. Split $S = A + \iu B$ with $A, B \in \R[X]$: then $$ P = (A + \iu B)(A - \iu B) = A^2 + B^2 . $$ **Exercise 8.8 ★★.** Find the [polynomial](#def-b1-poly-def) $P$ of degree $\leq 2$ with $P(0) = 1$, $P(1) = 3$, $P(2) = 2$, first by Lagrange’s formula, then by solving the linear system on the coefficients. Verify both answers agree. **Solution of Exercise 8.8.** Lagrange ([Theorem 8.23](#thm-b1-poly-lagrange)) with nodes $0, 1, 2$: $$ P = 1\cdot\frac{(X-1)(X-2)}{(0-1)(0-2)} + 3\cdot\frac{X(X-2)}{1\cdot(1-2)} + 2\cdot\frac{X(X-1)}{2\cdot 1} = \frac{(X-1)(X-2)}{2} - 3X(X-2) + X(X-1). $$ Expanding: $\frac{X^2 - 3X + 2}{2} - 3X^2 + 6X + X^2 - X = -\frac{3}{2}X^2 + \frac{7}{2}X + 1$. System: $P = aX^2 + bX + c$ with $c = 1$; $a + b + 1 = 3$; $4a + 2b + 1 = 2$. Subtracting twice the second from the third: $2a - 1 = -4$, so $a = -\frac32$, $b = \frac72$. Same [polynomial](#def-b1-poly-def): $P = -\frac32 X^2 + \frac72 X + 1$. (Check $P(2) = -6 + 7 + 1 = 2$.) **Exercise 8.9 ★★.** Prove that $P = X^{2n+1} - 1$ has exactly one real root, and that for every $n \geq 1$ the [polynomial](#def-b1-poly-def) $1 + X + \frac{X^2}{2!} + \dots + \frac{X^n}{n!}$ has no multiple root *(compare $P$ and $P'$)*. **Solution of Exercise 8.9.** $P = X^{2n+1} - 1$: $P' = (2n+1)X^{2n} \geq 0$, so the [polynomial](#def-b1-poly-def) function is increasing (strictly except at $0$), with limits $\mp\infty$: it vanishes exactly once on $\R$ (at $x = 1$). Let $E_n = \sum_{k=0}^{n} \frac{X^k}{k!}$. Then $E_n' = E_{n-1} = E_n - \frac{X^n}{n!}$. A multiple root $a$ would satisfy $E_n(a) = E_n'(a) = 0$ ([Proposition 8.11](#prop-b1-poly-multiplicity)), hence $\frac{a^n}{n!} = E_n(a) - E_n'(a) = 0$, so $a = 0$; but $E_n(0) = 1 \neq 0$. No multiple root. **Exercise 8.10 ★★★.** ([Chebyshev polynomials](#pb-b1-poly-1)) Define $T_0 = 1$, $T_1 = X$ and $T_{n+1} = 2X\,T_n - T_{n-1}$. 1. Prove by induction that $T_n(\cos\theta) = \cos n\theta$ for all $\theta$ . 2. Deduce the $n$ roots of $T_n$ and its leading coefficient. 3. Prove that $\sup_{x \in \intcc{-1}{1}} \abs{T_n(x)} = 1$ , attained at $n + 1$ points of $\intcc{-1}{1}$ . **Solution of Exercise 8.10.** 1. Induction (two base cases hold). Using $\cos(n+1)\theta + \cos(n-1)\theta = 2\cos\theta\cos n\theta$: $$T_{n+1}(\cos\theta) = 2\cos\theta \cos n\theta - \cos(n-1)\theta = \cos(n+1)\theta .$$ 2. $T_n(\cos\theta) = 0$ iff $n\theta \equiv \frac\pi2 \pmod \pi$: the numbers $$x_k = \cos\Bigl(\frac{(2k+1)\pi}{2n}\Bigr), \qquad k = 0, 1, \dots, n-1,$$ are $n$ distinct points of $\intoo{-1}{1}$ (the angles lie in $\intoo{0}{\pi}$ where $\cos$ is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj)), all roots of $T_n$; since $\deg T_n = n$ (from the recurrence, with leading coefficient $2^{n-1}$ for $n \geq 1$, by induction), these are *all* the roots, each simple. 3. For $x = \cos\theta \in \intcc{-1}{1}$ : $\abs{T_n(x)} = \abs{\cos n\theta} \leq 1$ , with equality iff $n\theta \equiv 0 \pmod\pi$ , i.e. at the $n+1$ points $y_k = \cos\frac{k\pi}{n}$ , $k = 0, \dots, n$ , where $T_n(y_k) = (-1)^k$ . (This equioscillation makes $2^{1-n}T_n$ the [monic](#def-b1-poly-def) degree- $n$ [polynomial](#def-b1-poly-def) of smallest sup-norm on $\intcc{-1}{1}$ — proved in this chapter’s weekend problem.) **Exercise 8.11 ★★★.** Let $P \in \C[X]$ be nonconstant with distinct roots $a_1, \dots, a_r$ (multiplicities $m_1, \dots, m_r$). Prove the identity of rational functions $$ \frac{P'(X)}{P(X)} = \sum_{i=1}^{r} \frac{m_i}{X - a_i}, $$ and deduce the Gauss–Lucas theorem: every root of $P'$ lies in the convex hull of the roots of $P$ *(evaluate the identity at a root $w$ of $P'$ which is not a root of $P$, take conjugates, and read the result as $w$ being a weighted average of the $a_i$)*. **Solution of Exercise 8.11.** Write $P = c\prod_i (X - a_i)^{m_i}$. The product rule (extended to several factors) gives $$ P' = c\sum_{i} m_i (X - a_i)^{m_i - 1} \prod_{k \neq i} (X - a_k)^{m_k}, $$ and dividing by $P$: $\frac{P'}{P} = \sum_i \frac{m_i}{X - a_i}$ (as rational functions, i.e. away from the roots). Let $w$ be a root of $P'$. If $w$ is one of the $a_i$, it lies in the convex hull trivially. Otherwise, evaluating at $w$: $$ 0 = \sum_i \frac{m_i}{w - a_i} = \sum_i m_i\, \frac{\conj w - \conj a_i}{\abs{w - a_i}^2} . $$ Conjugating: $\sum_i \lambda_i (w - a_i) = 0$ where $\lambda_i = \frac{m_i}{\abs{w - a_i}^2} > 0$. Hence $$ w = \frac{\sum_i \lambda_i a_i}{\sum_i \lambda_i} : $$ a convex combination (positive weights summing to $1$ after normalization) of the roots $a_i$. So every root of $P'$ lies in the convex hull of the roots of $P$. **Exercise 8.12 ★★.** (Roots-of-unity filter) Let $n \in \N^*$ and $j = \eu^{2\iu\pi/3}$. Evaluating $(1 + X)^n$ at $1$, $j$ and $j^2$, prove that $$ \sum_{k \geq 0} \binom{n}{3k} = \frac{2^n + 2\cos\frac{n\pi}{3}}{3} , $$ and check the formula for $n = 3$ and $n = 6$. *Hint: $1 + j^m + j^{2m}$ equals $3$ if $3 \mid m$ and $0$ otherwise; and $1 + j = \eu^{\iu\pi/3}$.* **Solution of Exercise 8.12.** Sum the evaluations of $(1 + X)^n$ at the three cube [roots of unity](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-unity): $$ 2^n + (1 + j)^n + (1 + j^2)^n = \sum_{k=0}^n \binom nk\,\bigl(1 + j^k + j^{2k}\bigr) = 3\sum_{k\,:\,3\mid k}\binom nk , $$ since $1 + j^k + j^{2k}$ is a geometric sum equal to $3$ when $3 \mid k$ and to $\frac{j^{3k} - 1}{j^k - 1} = 0$ otherwise. Now $1 + j = \frac12 + \iu\frac{\sqrt3}2 = \eu^{\iu\pi/3}$ and $1 + j^2 = \conj{1 + j} = \eu^{-\iu\pi/3}$, so $(1+j)^n + (1+j^2)^n = 2\cos\frac{n\pi}3$ and $$ \sum_{k\geq0}\binom n{3k} = \frac{2^n + 2\cos\frac{n\pi}3}{3} . $$ Checks: $n = 3$: $\frac{8 + 2\cos\pi}3 = 2 = \binom30 + \binom33$; $n = 6$: $\frac{64 + 2}3 = 22 = 1 + 20 + 1$. ## 8.5 Problem: Chebyshev polynomials and the flattest polynomial **Problem 8.1.** Among all *[monic](#def-b1-poly-def)* [polynomials](#def-b1-poly-def) of degree $n$, which one stays closest to zero on $\intcc{-1}1$? The answer — Chebyshev’s theorem, the birth certificate of approximation theory — is $2^{1-n}T_n$, where $T_n$ is the Chebyshev [polynomial](#def-b1-poly-def) of [Exercise 8.10](#exo-b1-poly-10), and no [monic](#def-b1-poly-def) competitor can beat its deviation $2^{1-n}$. This problem develops the algebra of the family $(T_n)$ (composition law, explicit coefficients, the second-kind family $U_n$, a differential equation), proves the extremality theorem with its equality case, and collects applications: optimal interpolation nodes, the exact value of $\cos 36^\circ$, and a [congruence](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-congruence) $T_p \equiv X^p \pmod p$. Throughout, $T_0 = 1$, $T_1 = X$, $T_{n+1} = 2X\,T_n - T_{n-1}$, and we freely use $T_n(\cos\theta) = \cos n\theta$ from [Exercise 8.10](#exo-b1-poly-10). **Part I — The family $(T_n)$.** 1. Compute $T_2, T_3, T_4, T_5$ from the recurrence. (Compare $T_3$ with the identity $\cos3\theta = 4\cos^3\theta - 3\cos\theta$ of [Example 3.9](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ex-b1-complex-cos3) .) 2. Prove by induction: $\deg T_n = n$ with leading coefficient $2^{n-1}$ for $n \geq 1$ , and $T_n$ has the parity of $n$ (only even or only odd powers appear). 3. Prove the uniqueness principle: $T_n$ is the *only* [polynomial](#def-b1-poly-def) satisfying $P(\cos\theta) = \cos n\theta$ for all $\theta$ . (Two [polynomials](#def-b1-poly-def) agreeing on $\intcc{-1}1$ agree everywhere: [Corollary 8.8](#cor-b1-poly-nroots) .) 4. Deduce the composition and product laws: $$T_m \circ T_n = T_{mn}, \qquad 2\,T_m T_n = T_{m+n} + T_{\abs{m-n}} .$$ 5. Recall from [Exercise 8.10](#exo-b1-poly-10) the roots $x_k = \cos\frac{(2k+1)\pi}{2n}$ and the equioscillation points $y_k = \cos\frac{k\pi}n$ with $T_n(y_k) = (-1)^k$ . Write down the complete factorization of $T_n$ over $\R$ , and justify that the $y_k$ interlace: $y_n < x_{n-1} < y_{n-1} < \dots < x_0 < y_0$ . 6. Prove that $T_n(\cosh t) = \cosh(nt)$ for all $t \in \R$ (same induction, using [Proposition 4.18](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-hyprules)), and deduce for $x \geq 1$ the closed form $$T_n(x) = \frac{\bigl(x + \sqrt{x^2 - 1}\bigr)^n + \bigl(x - \sqrt{x^2 - 1}\bigr)^n}{2} ,$$ so $T_n(x) > 1$ for $x > 1$: outside $\intcc{-1}1$ the [polynomial](#def-b1-poly-def) escapes at once. **Part II — Coefficients, the family $U_n$, a differential equation.** 7. From de Moivre’s formula ([Corollary 3.8](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#cor-b1-complex-demoivre)), prove the explicit expression $$T_n(x) = \sum_{0 \leq 2j \leq n} \binom{n}{2j}\, x^{\,n-2j}\,(x^2 - 1)^j ,$$ and verify it for $n = 3$. 8. Compute $T_n(1)$ , $T_n(-1)$ and $T_n(0)$ for all $n$ . 9. Define $U_n$ ( *second kind* ) by $U_0 = 1$ , $U_1 = 2X$ , $U_{n+1} = 2X\,U_n - U_{n-1}$ . Prove that $U_n(\cos\theta) = \frac{\sin(n+1)\theta}{\sin\theta}$ for $\theta \notin \pi\Z$ , and that $T_n' = n\,U_{n-1}$ for $n \geq 1$ . 10. Prove that $\abs{\sin n\theta} \leq n\,\abs{\sin\theta}$ for all $\theta$ (induction), and deduce the Markov-type bound $$\abs{T_n'(x)} \leq n^2 \quad\text{on } \intcc{-1}1, \qquad\text{with } T_n'(\pm1) = (\pm1)^{n-1}\,n^2 .$$ 11. Show that $y = T_n$ satisfies the differential equation $$(1 - x^2)\,y'' - x\,y' + n^2\,y = 0 ,$$ by differentiating the identity $\sin\theta\, T_n'(\cos\theta) = n\sin n\theta$ with respect to $\theta$; verify directly for $T_2$. **Part III — Chebyshev’s extremality theorem.** Let $\widetilde T_n = 2^{1-n}\,T_n$ ([monic](#def-b1-poly-def) by question 2) and write $\norm{P}_\infty = \sup_{x \in \intcc{-1}1}\abs{P(x)}$. 12. Justify $\norm{\widetilde T_n}_\infty = 2^{1-n}$ , attained with alternating signs at the $n + 1$ points $y_n < \dots < y_0$ . 13. Suppose some [monic](#def-b1-poly-def) $P$ of degree $n$ had $\norm P_\infty < 2^{1-n}$ , and set $D = \widetilde T_n - P$ . Show $\deg D \leq n - 1$ , and that $D(y_k)$ has the strict sign of $(-1)^k$ for each $k = 0, \dots, n$ . 14. Deduce that $D$ has at least $n$ distinct real roots (one in each gap, by the intermediate value property, used here at High School level and proved in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity)), and conclude *Chebyshev’s theorem*: every [monic](#def-b1-poly-def) $P$ of degree $n$ satisfies $$\norm{P}_\infty \geq 2^{1-n} .$$ 15. (Equality case, first step) Suppose now $\norm P_\infty = 2^{1-n}$ exactly, $P$ [monic](#def-b1-poly-def) of degree $n$ , and let $D = \widetilde T_n - P$ . Show that $(-1)^kD(y_k) \geq 0$ for all $k$ , and that if $D(y_k) = 0$ for an *interior* point $y_k$ ( $0 < k < n$ ), then $D'(y_k) = 0$ as well. *(At an interior $y_k$, both $\widetilde T_n$ and $P$ attain an extremum of absolute value $\norm{\cdot} _\infty$; a differentiable function has zero derivative at an interior extremum — used at High School level, proved in [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative).)* 16. (Equality case, conclusion) Count roots of $D$ with [multiplicity](#def-b1-poly-derivative) to show $D = 0$ : the minimizer is *unique* , $P = \widetilde T_n$ . 17. Transport to an arbitrary segment $\intcc ab$ : show that the minimal sup-norm of a [monic](#def-b1-poly-def) degree- $n$ [polynomial](#def-b1-poly-def) on $\intcc ab$ is $2\bigl(\frac{b-a}4\bigr)^n$ , attained by a rescaled Chebyshev [polynomial](#def-b1-poly-def) . *(Substitute $x = \frac{a+b}2 + \frac{b-a}2\,t$ and track the leading coefficient.)* **Part IV — Applications.** 18. Work out the case $n = 3$ by hand: locate the extrema of $\widetilde T_3 = X^3 - \frac34X$ on $\intcc{-1}1$ , verify the four-fold equioscillation with value $\frac14$ , and conclude that no [monic](#def-b1-poly-def) cubic does better. 19. (Optimal interpolation nodes) For $n + 1$ nodes $x_0, \dots, x_n \in \intcc{-1}1$ , the interpolation error is governed by $\omega(X) = \prod_i (X - x_i)$ (as [Chapter 16](https://one-course.com/books/math/3/en/chapter/16-taylor-formulas-and-asymptotic-expansions#ch-b1-taylor) will quantify). Prove that the choice minimizing $\norm\omega_\infty$ is the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of $n + 1$ roots of $T_{n+1}$ , with $\norm\omega_\infty = 2^{-n}$ : Chebyshev nodes are the right places to interpolate. 20. Using $T_5$, prove that $c = \cos 36^\circ$ satisfies $16c^5 - 20c^3 + 5c + 1 = 0$, factor this [polynomial](#def-b1-poly-def) as $(x + 1)(4x^2 - 2x - 1)^2$, and conclude $$\cos 36^\circ = \frac{1 + \sqrt5}4 .$$ Check the consistency with $\cos 72^\circ = \frac{\sqrt5 - 1}4$ from [Exercise 3.8](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#exo-b1-complex-8). 21. Estimate $T_{10}(1.1)$ with the closed form of question 6 (two significant digits suffice), and interpret: a [polynomial](#def-b1-poly-def) bounded by $1$ on $\intcc{-1}1$ can already exceed $40$ at $x = 1.1$ . (That $T_n$ grows *fastest* among such [polynomials](#def-b1-poly-def) is another extremal property of the family, beyond this problem.) 22. Prove the [congruence](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-congruence) : for every odd [prime](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-prime) $p$ , all coefficients of $T_p - X^p$ are divisible by $p$ . *(Use question 7 and $p \mid \binom p{2j}$ for $0 < 2j < p$, from the proof of [Theorem 6.23](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#thm-b1-arith-fermat).)* Verify on $T_3$ and $T_5$ . **Part V — Synthesis.** 23. Compute explicitly the [monic](#def-b1-poly-def) quadratic of minimal sup-norm on $\intcc01$ and its deviation. (Question 17 with $n = 2$ .) 24. Where exactly did the problem use: (i) the rigidity of [polynomials](#def-b1-poly-def) ( [Corollary 8.8](#cor-b1-poly-nroots) ); (ii) the trigonometry of [Chapter 3](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ch-b1-complex) and [Chapter 4](https://one-course.com/books/math/3/en/chapter/4-standard-functions#ch-b1-functions) ; (iii) the arithmetic of [binomial coefficients](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-objects) from [Chapter 6](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#ch-b1-arith) ? One sentence each. 25. Synthesis, in a short paragraph: the theorem says that the flattest [monic polynomial](#def-b1-poly-def) is the one that *equioscillates* , and the proof converts optimality into root-counting. Comment on this mechanism, on the role of the substitution $x = \cos\theta$ as a bridge between algebra and trigonometry, and name the two places where the problem needed analysis facts (IVT, interior extremum) that later chapters prove. **Solution of Problem 8.1.** **1.** $T_2 = 2X^2 - 1$; $T_3 = 2X(2X^2 - 1) - X = 4X^3 - 3X$; $T_4 = 2X\,T_3 - T_2 = 8X^4 - 8X^2 + 1$; $T_5 = 2X\,T_4 - T_3 = 16X^5 - 20X^3 + 5X$. The identity $T_3(\cos\theta) = \cos3\theta$ is exactly $\cos3\theta = 4\cos^3\theta - 3\cos\theta$ from [Example 3.9](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ex-b1-complex-cos3). **2.** True for $n = 1, 2$. If $T_{n-1}$, $T_n$ have degrees $n-1$, $n$ and leading coefficients $2^{n-2}$, $2^{n-1}$, then $2X\,T_n$ has degree $n+1$ and leading coefficient $2^n$, while $T_{n-1}$ has lower degree: $T_{n+1}$ has degree $n + 1$, leading coefficient $2^n$. Parity: if $T_{n-1}$ has the parity of $n - 1$ and $T_n$ that of $n$, then $2X\,T_n$ and $T_{n-1}$ both have the parity of $n + 1$, hence so does $T_{n+1}$. **3.** If $P(\cos\theta) = \cos n\theta$ for all $\theta$, then $P$ and $T_n$ agree at every point of $\intcc{-1}1$ — an infinite [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) — so $P - T_n$ has infinitely many roots and is the zero [polynomial](#def-b1-poly-def) ([Corollary 8.8](#cor-b1-poly-nroots)). **4.** For $x = \cos\theta$: $T_m(T_n(\cos\theta)) = T_m(\cos n\theta) = \cos(mn\theta) = T_{mn}(\cos\theta)$, and $2T_mT_n(\cos\theta) = 2\cos m\theta\cos n\theta = \cos(m+n)\theta + \cos\abs{m - n}\theta$. Both identities hold on $\intcc{-1}1$, hence as [polynomial](#def-b1-poly-def) identities by question 3’s argument. **5.** The $x_k$ are $n$ distinct simple roots and the leading coefficient is $2^{n-1}$: $$ T_n = 2^{n-1}\prod_{k=0}^{n-1} \Bigl(X - \cos\frac{(2k+1)\pi}{2n}\Bigr) . $$ Interlacing: the angles $0 < \frac{\pi}{2n} < \frac\pi n < \frac{3\pi}{2n} < \frac{2\pi}n < \dots < \pi$ alternate between the $y$-angles $\frac{k\pi}n$ and the $x$-angles $\frac{(2k+1)\pi}{2n}$; since $\cos$ is strictly decreasing on $\intcc0\pi$, the values interlace in the reverse order: $y_n < x_{n-1} < y_{n-1} < \dots < x_0 < y_0$. Between two consecutive extrema sits exactly one root, as a picture of $\cos n\theta$ suggests. **6.** Induction with $2\cosh a\cosh b = \cosh(a + b) + \cosh(a - b)$ ([Proposition 4.18](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-hyprules)): $T_{n+1}(\cosh t) = 2\cosh t\cosh nt - \cosh(n-1)t = \cosh(n+1)t$. For $x \geq 1$, write $x = \cosh t$ with $t \geq 0$; then $\eu^t = x + \sqrt{x^2 - 1}$ and $\eu^{-t} = x - \sqrt{x^2 - 1}$, so $$ T_n(x) = \cosh(nt) = \frac{(x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^n}2 . $$ For $x > 1$ the first term exceeds $\frac12(1)^n$ strictly and grows geometrically: $T_n(x) > 1$. **7.** De Moivre: $\cos n\theta = \Re\bigl((\cos\theta + \iu\sin\theta)^n\bigr) = \sum_{2j \leq n}\binom n{2j} \cos^{n-2j}\theta\,(\iu\sin\theta)^{2j}$, and $(\iu\sin\theta)^{2j} = (-\sin^2\theta)^j = (\cos^2\theta - 1)^j$. Substituting $x = \cos\theta$ and invoking question 3: $$ T_n(x) = \sum_{0\leq 2j\leq n}\binom n{2j}x^{n-2j}(x^2 - 1)^j . $$ For $n = 3$: $\binom30 x^3 + \binom32 x(x^2 - 1) = x^3 + 3x^3 - 3x = 4x^3 - 3x$, as in question 1. **8.** $T_n(1) = \cos(n\cdot0) = 1$; $T_n(-1) = \cos(n\pi) = (-1)^n$; $T_n(0) = \cos\frac{n\pi}2$, which is $0$ for odd $n$ and $(-1)^{n/2}$ for even $n$. **9.** Induction for $U_n(\cos\theta) = \frac{\sin(n+1)\theta}{\sin\theta}$: true for $U_0 = 1$ and $U_1 = 2X$ ($\sin2\theta = 2\sin\theta\cos\theta$); the step is the sum-to-product identity $\sin(n+2)\theta = 2\cos\theta\, \sin(n+1)\theta - \sin n\theta$. Now differentiate $T_n(\cos\theta) = \cos n\theta$ in $\theta$: $-\sin\theta\,T_n'(\cos\theta) = -n\sin n\theta$, so for $\theta \notin \pi\Z$: $$ T_n'(\cos\theta) = n\,\frac{\sin n\theta}{\sin\theta} = n\,U_{n-1}(\cos\theta) , $$ and the [polynomials](#def-b1-poly-def) $T_n'$ and $nU_{n-1}$, agreeing on $\intoo{-1}1$, are equal. **10.** $\abs{\sin(n+1)\theta} = \abs{\sin n\theta\cos\theta + \cos n\theta\sin\theta} \leq \abs{\sin n\theta} + \abs{\sin\theta}$, and induction gives $\abs{\sin n\theta} \leq n\abs{\sin\theta}$. Hence $\abs{U_{n-1}} \leq n$ on $\intoo{-1}1$ and $\abs{T_n'} = n\abs{U_{n-1}} \leq n^2$ there; at $\pm1$ the bound extends by taking limits (or directly: $U_{n-1}(1) = n$ from the recurrence, $U_n(1) = n + 1$ by induction, and parity gives $U_{n-1}(-1) = (-1)^{n-1}n$). Thus $T_n'(1) = n^2$ and $T_n'(-1) = (-1)^{n-1}n^2$: the bound $n^2$ is attained at the endpoints. **11.** Differentiate $\sin\theta\,T_n'(\cos\theta) = n\sin n\theta$ (question 9) with respect to $\theta$: $$ \cos\theta\,T_n'(\cos\theta) - \sin^2\theta\,T_n''(\cos\theta) = n^2\cos n\theta = n^2\,T_n(\cos\theta) . $$ With $x = \cos\theta$ and $\sin^2\theta = 1 - x^2$: $x\,T_n' - (1 - x^2)T_n'' = n^2T_n$ on $\intcc{-1}1$, hence everywhere: $(1 - x^2)y'' - xy' + n^2y = 0$ for $y = T_n$. Check for $T_2 = 2x^2 - 1$: $(1 - x^2)(4) - x(4x) + 4(2x^2 - 1) = 4 - 4x^2 - 4x^2 + 8x^2 - 4 = 0$. **12.** $\widetilde T_n$ is [monic](#def-b1-poly-def) (question 2) and $\abs{\widetilde T_n} = 2^{1-n}\abs{T_n} \leq 2^{1-n}$ on $\intcc{-1}1$, with $\widetilde T_n(y_k) = (-1)^k2^{1-n}$ at the $n + 1$ points $y_k$ ([Exercise 8.10](#exo-b1-poly-10)): the norm is exactly $2^{1-n}$, attained with alternating signs. **13.** $\widetilde T_n$ and $P$ are both [monic](#def-b1-poly-def) of degree $n$, so the leading terms cancel: $\deg D \leq n - 1$. At $y_k$: $D(y_k) = (-1)^k2^{1-n} - P(y_k)$, and $\abs{P(y_k)} \leq \norm P_\infty < 2^{1-n}$ forces the sign of $D(y_k)$ to be that of $(-1)^k2^{1-n}$, strictly. **14.** $D$ changes sign between $y_{k+1}$ and $y_k$ for each $k = 0, \dots, n-1$: by the intermediate value property, $D$ has a root in each of these $n$ pairwise disjoint open intervals — $n$ distinct roots for a nonzero [polynomial](#def-b1-poly-def) of degree $\leq n - 1$, impossible. And $D = 0$ is impossible too (the norms differ). Contradiction: no [monic](#def-b1-poly-def) $P$ of degree $n$ has $\norm P_\infty < 2^{1-n}$, which is Chebyshev’s theorem. **15.** Now $\abs{P(y_k)} \leq 2^{1-n}$ only, so $(-1)^k D(y_k) = 2^{1-n} - (-1)^kP(y_k) \geq 2^{1-n} - \abs{P(y_k)} \geq 0$. Suppose $D(y_k) = 0$ at an interior $y_k$ ($0 < k < n$): then $P(y_k) = (-1)^k2^{1-n}$, so $\abs P$ attains its supremum $2^{1-n}$ at the interior point $y_k$, whence $P'(y_k) = 0$ (interior extremum); and $T_n'(y_k) = nU_{n-1}(y_k) = 0$ since $\sin(n\cdot\frac{k\pi}n) = 0$ — so $\widetilde T_n'(y_k) = 0$ too, and $D'(y_k) = 0$: $y_k$ is a root of $D$ of [multiplicity](#def-b1-poly-derivative) at least $2$. **16.** Count roots of $D$ with [multiplicity](#def-b1-poly-derivative). Let $z$ be the number of interior points $y_k$ with $D(y_k) = 0$ (each a double root, by question 15) and $e \in \{0, 1, 2\}$ the number of endpoints ($y_0$ or $y_n$) with $D = 0$ (each a simple root at least). A gap $(y_{k+1}, y_k)$ whose two endpoints both have $D \neq 0$ carries strictly alternating signs, hence an interior root. Each vanishing interior point spoils at most its two adjacent gaps, each vanishing endpoint at most one gap: at least $n - 2z - e$ gaps still contribute one root each, all distinct from the $y$-roots. Total: at least $(n - 2z - e) + 2z + e = n$ roots with [multiplicity](#def-b1-poly-derivative), for a [polynomial](#def-b1-poly-def) of degree $\leq n - 1$: so $D = 0$ and $P = \widetilde T_n$. The minimizer is unique. **17.** The affine [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $t \mapsto x = \frac{a+b}2 + \frac{b-a}2\,t$ is a bijection $\intcc{-1}1 \to \intcc ab$. If $P$ is [monic](#def-b1-poly-def) of degree $n$, then $Q(t) = P(x(t))$ is a [polynomial](#def-b1-poly-def) in $t$ with leading coefficient $\bigl(\frac{b-a}2\bigr)^n$, and $\sup_{\intcc ab}\abs P = \sup_{\intcc{-1}1}\abs Q$. The [monic polynomial](#def-b1-poly-def) $Q/\bigl(\frac{b-a}2\bigr)^n$ has sup-norm $\geq 2^{1-n}$ (questions 13–14), so $$ \sup_{\intcc ab}\abs P \geq \Bigl(\frac{b-a}2\Bigr)^n 2^{1-n} = 2\Bigl(\frac{b-a}4\Bigr)^n , $$ with equality exactly for $P(x) = \bigl(\frac{b-a}2\bigr)^n \widetilde T_n\bigl(t(x)\bigr)$ (question 16). **18.** $\widetilde T_3 = \frac{T_3}4 = X^3 - \frac34X$; $\widetilde T_3{}' = 3X^2 - \frac34$ vanishes at $\pm\frac12$. Values: $\widetilde T_3(-1) = -\frac14$, $\widetilde T_3(-\tfrac12) = \frac14$, $\widetilde T_3(\tfrac12) = -\frac14$, $\widetilde T_3(1) = \frac14$: four alternating extrema of absolute value $\frac14$ — so $\norm{\widetilde T_3}_\infty = \frac14$, and by Chebyshev’s theorem no [monic](#def-b1-poly-def) cubic has smaller sup-norm on $\intcc{-1}1$. **19.** $\omega$ is [monic](#def-b1-poly-def) of degree $n + 1$, so $\norm\omega_\infty \geq 2^{-n}$ by Chebyshev’s theorem (degree $n+1$), with equality if and only if $\omega = \widetilde T_{n+1} = 2^{-n}T_{n+1}$ (question 16), i.e. if and only if the nodes are the $n + 1$ roots of $T_{n+1}$. With Chebyshev nodes the error factor $\norm\omega_\infty$ is $2^{-n}$ — the least possible. **20.** $5 \times 36^\circ = 180^\circ$, so $T_5(c) = \cos180^\circ = -1$: $16c^5 - 20c^3 + 5c + 1 = 0$. Testing $x = -1$: $-16 + 20 - 5 + 1 = 0$, and expanding confirms $$ 16x^5 - 20x^3 + 5x + 1 = (x + 1)\bigl(4x^2 - 2x - 1\bigr)^2 . $$ Since $c = \cos36^\circ \neq -1$, $c$ is a root of $4x^2 - 2x - 1$, whose roots are $\frac{1 \pm \sqrt5}4$; as $c > 0$, $$ \cos36^\circ = \frac{1 + \sqrt5}4 . $$ Consistency: $\cos72^\circ = T_2(c) = 2c^2 - 1 = 2\cdot\frac{3 + \sqrt5}8 - 1 = \frac{\sqrt5 - 1}4$, the value found in [Exercise 3.8](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#exo-b1-complex-8). **21.** $\sqrt{1.1^2 - 1} = \sqrt{0.21} \approx 0.458$, so $x + \sqrt{x^2-1} \approx 1.558$ and $(1.558)^{10} \approx 84.5$, while $(1.1 - 0.458)^{10} \approx 0.01$: $T_{10}(1.1) \approx \frac{84.5 + 0.01}2 \approx 42$. A [polynomial](#def-b1-poly-def) trapped in $\intcc{-1}1$ on the interval has already grown past $40$ one tenth beyond its edge: boundedness on a segment says nothing an inch outside it. **22.** In question 7’s formula for $T_p$, the term $j = 0$ is $X^p$; every other term carries $\binom p{2j}$ with $0 < 2j < p$ (note $2j \neq p$ since $p$ is odd), which is divisible by $p$ by the first step of the proof of [Theorem 6.23](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#thm-b1-arith-fermat). Hence every coefficient of $T_p - X^p$ is a multiple of $p$. Checks: $T_3 - X^3 = 3X^3 - 3X = 3(X^3 - X)$; $T_5 - X^5 = 15X^5 - 20X^3 + 5X = 5(3X^5 - 4X^3 + X)$. **23.** By question 17 with $\intcc ab = \intcc01$ and $n = 2$: minimal deviation $2\bigl(\frac14\bigr)^2 = \frac18$, attained by $\bigl(\frac12\bigr)^2\widetilde T_2(2x - 1) = \frac14\bigl((2x-1)^2 - \frac12\bigr) = x^2 - x + \frac18$. The [monic](#def-b1-poly-def) quadratic closest to zero on $\intcc01$ is $x^2 - x + \frac18$, with sup-norm $\frac18$. **24.** (i) Rigidity — a [polynomial](#def-b1-poly-def) with more roots than its degree is zero — powered the uniqueness principle (question 3), the transfer of trigonometric identities to [polynomial](#def-b1-poly-def) identities (questions 4, 7, 9, 11), and both root-counting arguments of the extremality proof (questions 14, 16). (ii) The trigonometry of [Chapter 3](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#ch-b1-complex) (de Moivre, sum-to-product) and the [hyperbolic functions](https://one-course.com/books/math/3/en/chapter/4-standard-functions#def-b1-functions-hyperbolic) of [Chapter 4](https://one-course.com/books/math/3/en/chapter/4-standard-functions#ch-b1-functions) supplied every identity behind the family; the substitution $x = \cos\theta$ is the bridge. (iii) The [divisibility](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $p \mid \binom p{2j}$ from [Chapter 6](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#ch-b1-arith) turned the coefficient formula into the [congruence](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-congruence) of question 22. **25.** Chebyshev’s theorem converts an optimization over an infinite-dimensional family (all [monic polynomials](#def-b1-poly-def)) into finite combinatorics: a competitor better than $\widetilde T_n$ would differ from it by a low-degree [polynomial](#def-b1-poly-def) forced to change sign $n$ times — one more root than its degree allows. The equioscillation pattern is thus not a curiosity but the very certificate of optimality, and the equality case sharpens root-counting with multiplicities. The substitution $x = \cos\theta$ deserves the last word: it transports the rigid, discrete world of [polynomials](#def-b1-poly-def) into the periodic world of trigonometry, where roots and extrema of $T_n$ are simply the regular grid of $\cos n\theta$. The two analysis facts borrowed — the intermediate value property (question 14; proved in [Chapter 13](https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity#ch-b1-continuity)) and the vanishing derivative at an interior extremum (question 15; proved in [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative)) — are exactly the tools those later chapters will hand back, closing the loop.