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Lagrange’s polynomial Nguyen Trung Tuan November 16, 2016 Abstract In this article, I will use Lagrange polynomial to solve some problems from Mathematical Olympiads Contents Lagrange’s interpolation polynomial 2 Examples 3 Problems 10 1 Lagrange’s interpolation polynomial Theorem Let n be a positive integer and x0 , x1 , · · · , xn ; y0 , y1 , · · · , yn are complex numbers such that xi = xj if i = j Then there exists precisely one polynomial P (x) of degree not greater than n such that P (xi ) = yi ∀i = 0, n Proof If P and Q are polynomials of degree not greater than n such that P (xi ) = Q(xi ) = yi ∀i = 0, n then the polynomial P − Q of degree not greater than n and has at least n + distinct roots, therefore P − Q is the zero polynomial and hence P = Q n x − xj then P (x) is a polynomial of degree not Now if P (x) = yi x − x i j i=0 j=i greater than n and P (xi ) = yi ∀i = 0, n ✷ n The polynomial yi i=0 j=i x − xj is called Lagrange’s interpolation polynoxi − xj mial or Lagrange’s polynomial at nodes x0 , x1 , · · · , xn Corollary If P (x) is a polynomial of degree not greater than n (n ∈ N∗ ) and x0 , x1 , · · · , xn are complex numbers such that xi = xj if i = j then n x − xj P (x) = P (xi ) x − x i j i=0 j=i 2 Examples Example Let A(x) = x81 + x49 + x25 + x9 + x and B(x) = x3 − x be polynomials Find the remainder in the division of A(x) by B(x) Solution Asume Q(x) and R(x) are the quotient and remainder in the division of A(x) by B(x), respectively We have A(x) = B(x)Q(x) + R(x), and deg R < (1) Because B(0) = B(1) = B(−1) = and (1) we have R(0) = 0, R(1) = and R(−1) = −5, therefore use Lagrange’s interpolation polynomial at nodes 0; and −1 we have (x − 1)(x + 1) (x − 0)(x + 1) (x − 0)(x − 1) + R(1) + R(−1) (0 − 1)(0 + 1) (1 − 0)(1 + 1) (−1 − 0)(−1 − 1) 5 = x(x + 1) − x(x − 1) 2 = 5x R(x) = R(0) Solution We have x3 ≡ x (mod B(x)) and therefore A(x) = (x3 )27 + (x3 )16 x + (x3 )8 x + (x3 )3 + x ≡ x27 + x17 + x9 + x3 + x = (x3 )9 + (x3 )5 x2 + (x3 )3 + x3 + x ≡ x9 + x7 + 2x3 + x = (x3 )3 + (x3 )2 x + 2x3 + x ≡ 4x3 + x ≡ 5x (mod B(x)) Because deg(5x) = < deg B(x), we have 5x is the remainder in the division of A(x) by B(x) Example (USAMO 1975) Let P be a polynomial of degree n (n ∈ k ∀k = 0, n Determine P (n + 1) Z, n > 0) satisfying P (k) = k+1 Solution Use Lagrange’s interpolation polynomial at nodes 0, 1, · · · , n we have n P (x) = P (k) k=0 n = k=0 n = k=0 x−j k−j j=k k x−j k + j=k k − j (−1)n−k k (x − j) (k + 1)!(n − k)! j=k Therefore n P (n + 1) = k=0 n = k=0 (−1)n−k k (n + − j) (k + 1)!(n − k)! j=k (−1)n−k k (n + 1)! (k + 1)!(n − k)! n + − k = n+2 = n+2 = n+2 = n k+1 k(−1)n−k Cn+2 k=0 n k+1 k+1 (k + 1)(−1)n−k Cn+2 + (−1)n−k+1 Cn+2 k=0 n n+1 (−1) n−k (n + i (−1)n+2−i Cn+2 + i=1 k=0 n + + (−1) n+2 k 2)Cn+1 n+1 Solution We have the polynomial Q(x) = (x + 1)P (x) − x of degree n + and 0, 1, 2, · · · , n are its roots, therefore Q(x) = Cx(x − 1)(x − 2) · · · (x − n), for some constant C Because Q(−1) = 1, we have C(−1)(−2)(−3) · · · (−n − 1) = (−1)n+1 (n + 1)!C = 1, therefore C = (−1)n+1 and (n + 1)! Q(x) = (−1)n+1 x(x − 1)(x − 2) · · · (x − n) (n + 1)! (−1)n+1 (n + 1)! = (−1)n+1 (n + 1)! (−1)n+1 + n + Q(n + 1) + n + = Note that P (n + 1) = n+2 n+2 n Therefore P (n + 1) is equal to if n is odd and equal to if n is even n+2 Hence Q(n + 1) = Example (IMO Longlist 1977) Let n be a positive integer Suppose x0 , x1 , , xn are integers and x0 > x1 > · · · > xn Prove that at least one of the numbers |F (x0 )|, |F (x1 )|, |F (x2 )|, , |F (xn )|, where F (x) = xn + a1 xn−1 + · · · + an ( ∈ R, i = 1, , n) is greater n! than or equal to n n! Solution Assume that |F (xi )| < n ∀i = 0, n Use Lagrange’s interpola2 tion polynomial at nodes x0 , x1 , · · · , xn we have n n x + a1 x n−1 + · · · + an = F (xk ) k=0 x − xj , x k − xj j=k see the coefficient of xn we have n 1= F (xk ) xk − xj k=0 j=k n ≤ |F (xk )| |xk − xj | k=0 j=k n n! 2n < k=0 j=k n ≤ k=0 j=k n = 2n |xk − xj | n! 2n |k − j| Cnk = 1, a contradiction, k=0 and therefore at least one of the numbers |F (x0 )|, |F (x1 )|, |F (x2 )|, , |F (xn )| n! is greater than or equal to n Example P (x) is a polynomial of degree 2n (n ∈ N) such that P (0) = P (2) = · · · = P (2n) = 0, P (1) = P (3) = · · · = P (2n − 1) = 2, and P (2n + 1) = −30 Determine n Solution Use Lagrange’s interpolation polynomial at nodes 0, 1, · · · , 2n we have 2n x−j P (x) − = = (P (k) − 1) k − j k=0 j=k 2n (−1)k+1 k=0 x−j , k − j j=k 2n k C2n+1 = − 22n+1 , but by hypothesis and therefore P (2n + 1) − = − k=0 P (2n + 1) = −30, hence we have −31 = − 22n+1 , so n = Example (Tepper’s identity) Prove that for any real number a we have the following identity n (−1)k k=0 n (a − k)n = n! ∀n ∈ N k Solution Assume that a = and n ≥ 3, then we need to prove n (−1)n+k k=0 n (k)n = n! (∗) k By Lagrange’s Interpolation polynomial at nodes 1, 2, · · · , n we have n n kn · x − (x − 1)(x − 2) · · · (x − n) = k=1 i=k x−i (∗∗) k−i Now, in (∗∗), setting x = we have n (−1)n+1 · n! = kn · k=1 (−1)n−1 · n! , · k (k − 1)! · (n − k)! · (−1)n−k and we are done Example Let x, y, z and t be real numbers satisfying  x2 2 + 22y−32 + 22z−52 + 22t−72 =  22 −12    x2 + y2 + z2 + t2 = 42 −12 42 −32 42 −52 42 −72 y2 z2 x2  + 62 −32 + 62 −52 + 62t−72 =  62 −12   x2 2 + 82y−32 + 82z−52 + 82t−72 = 82 −12 Determine x2 + y + z + t2 Solution Setting li = (2i − 1)2 , ci = (2i)2 ∀i = 1, and 4 (x − li ) − f (x) = i=1 (x − ci ) i=1 We have deg f ≤ 3, and therefore by Lagrange’s Interpolation polynomial at nodes l1 , l2 , l3 , l4 we obtain f (x) = f (li ) i=1 j=i x − lj (1) li − lj (cj − li ) ∀j = 1, we have From (1) and f (cj ) = i=1 4 (cj − li ) = f (li ) i=1 i=1 k=i cj − lk ∀j = 1, 4, li − lk and hence α1 α2 α3 α4 + + + = ∀j = 1, 4, cj − l1 cj − l2 cj − l3 cj − l4 where αi = f (li ) ∀i = 1, j=i (li − lj ) See coeficients of x in both sides of (1) we have (ci − li ) = 36, αi = i=1 i=1 2 2 and therefore x + y + z + t = αi = 36 i=1 Solution From hypothesis we have x2 y2 z2 t2 + + + = ∀w ∈ {4, 16, 36, 64}, w − 12 w − 32 w − 52 w − 72 and therefore 4, 16, 36 and 64 are roots of the polynomial (w − (2i − 1)2 ) − x2 (w − 32 )(w − 52 )(w − 72 ) − · · · P (w) = i=1 = w − (x2 + y + z + t2 + 12 + 32 + 52 + 72 )w3 + · · · hence P (w) = (w − 4)(w − 16)(w − 36)(w − 64) By see coeficients of w3 we have −(x2 + y + z + t2 + 12 + 32 + 52 + 72 ) = −(4 + 16 + 36 + 64), so x2 + y + z + t2 = 36 Example Let ABC be a triangle with BC = a, CA = b and AB = c Prove that for any points P, Q in the plane (ABC) we have a.P A.QA + b.P B.QB + c.P C.QC ≥ abc Solution In the complex plane (ABC) we assume that A = x1 , B = x2 , C = x3 , P = p, Q = q By Lagrange’s Interpolation polynomial at nodes x1 , x2 , x3 we have (x − p)(x − q) = (xi − p)(xi − q) i=1 j=i x − xj , xi − xj see coeficients of x2 in the both sides we obtain (x1 − p)(x1 − q) (x2 − p)(x2 − q) (x3 − p)(x3 − q) + + = 1, (x1 − x2 )(x1 − x3 ) (x2 − x3 )(x2 − x1 ) (x3 − x1 )(x3 − x2 ) and therefore 1≤ |x1 − p|.|x1 − q| |x2 − p|.|x2 − q| |x3 − p|.|x3 − q| + + , |x1 − x2 |.|x1 − x3 | |x2 − x3 |.|x2 − x1 | |x3 − x1 |.|x3 − x2 | but |x2 − x1 | = AB, · · · and |x1 − p| = AP, · · · , therefore 1≤ P A.QA P B.QB P C.QC + + , AB.AC BC.BA CA.CB and we are done Example Find all polynomials P (x) with real coefficients such that for every positive integer n there exists a rational r with P (r) = n Solution on AoPS Assume that P (x) ∈ R[x] is a polynomial satisfying for every positive integer n there exists a rational r with P (r) = n Clearly P (x) can’t be constant, so d := deg P ≥ For all n ∈ N take rn ∈ Q such that P (rn ) = n P (r1 ) = 1, P (r2 ) = 2, , P (rd+1 ) = d + gives P (x) ∈ Q[x] using the Lagrange’s interpolation polynomial at nodes r1 , r2 , · · · , rd+1 Then for some t ∈ N we have tP (x) ∈ Z[x] with leading coefficient m ∈ Z \ {0} But then tP (x) − tn ∈ Z[x] has rational root rn and also leading Z for all coefficient m So the denominator of rn divides m, i.e rn ∈ m n ∈ N P (x) Now assume d = deg P ≥ Then → +∞ for |x| → +∞, and we x P (x) find N ∈ N large enough with > 2|m| for all |x| ≥ N x But Z ∩ (−N, N ) contains exactly 2|m|N − elements So among the m 2|m|N different numbers r1 , r2 , , r2|m|N ∈ Z we must find |rk | ≥ N for m k 2|m|N P (rk ) = some k = 1, , 2|m|N This gives 2|m| < ≤ = 2|m|, rk |rk | N a contradiction So we must have P (x) ∈ Q[x] with deg P = 1, which is indeed a solution for the problem Problems Problem Let P be a polynomial of degree at most n satisfying P (k) = ∀k = 0, n Determine P (n + 1) k Cn+1 Problem A polynomial P (x) has degree at most 2k, where k = 0, 1, 2, · · · Given that for an integer i, the inequality −k ≤ i ≤ k implies |P (i)| ≤ 1, prove that for all real numbers x, with −k ≤ x ≤ k, the following inequality holds: |P (x)| ≤ 22k Problems Prove that at least one of the numbers |f (1)|, |f (2)|, · · · , n! Where |f (n + 1)| is greater than or equal to 2n f (x) = xn + a1 xn−1 + · · · + an ( ∈ R, i = 1, , n, n ∈ N.) Problem Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients is the average of two monic polynomials of degree n with n real roots Problem Let p be a prime number and f an integer polynomial of degree d such that f (0) = 0, f (1) = and f (n) is congruent to or modulo p for every integer n Prove that d ≥ p − Problem Let P be a polynomial of degree n ∈ N satisfying P (k) = 2k ∀k = 0, n Prove that P (n + 1) = 2n+1 − Problem P (x) is a polynomial of degree 3n (n ∈ N) such that P (0) = P (3) = · · · = P (3n) = 2, P (1) = P (4) = · · · = P (3n − 2) = 1, P (2) = P (5) = · · · = P (3n − 1) = 0, and P (3n + 1) = 730 Determine n Problem Let S = {s1 , s2 , s3 , , sn } be a set of n distinct complex numbers n ≥ 9, exactly n − of which are real Prove that there are at most two quadratic polynomials f (z) with complex coefficients such that f (S) = S (that is, f permutes the elements of S) Problem Let t and n be fixed integers each at least Find the largest positive integer m for which there exists a polynomial P , of degree n and 10 with rational coefficients, such that the following property holds: exactly one of P (k) P (k) and k+1 k t t is an integer for each k = 0, 1, , m Problem 10 Let f (x) = xn + an−2 xn−2 + an−3 xn−3 + + a1 x + a0 be a polynomial Prove that we have an i ∈ {1, 2, , n} | |f (i)| ≥ n! (ni) Problem 11 Let (Fn )n≥1 be the Fibonacci sequence F1 = F2 = 1, Fn+2 = Fn+1 + Fn (n ≥ 1), and P (x) the polynomial of degree 990 satisfying P (k) = Fk , for k = 992, , 1982 Prove that P (1983) = F1983 − Problem 12 Let P (x) be a polynomial of degree n with real coefficients and let a ≥ Prove that max 0≤j≤n+1 aj − P (j) ≥ Problem 13 Let P (x) be a polynomial of degree n ≤ 10 with integral coefficients such that for every k ∈ {1, 2, , 10} there is an integer m with P (m) = k Furthermore, it is given that |P (10) − P (0)| < 1000 Prove that for every integer k there is an integer m such that P (m) = k Problem 14 Given is a natural number n ≥ What is the smallest possible value of k if the following statements are true For every n points Ai = (xi , yi ) on a plane, where no three points are collinear, and for any real numbers ci (1 ≤ i ≤ n) there exists such polynomial P (x, y), the degree of which is no more than k, where P (xi , yi ) = ci for every i = 1, , n (The degree of a nonzero monomial ai,j xi y j is i + j, while the degree of polynomial P (x, y) is the greatest degree of the degrees of its monomials.) Problem 15 Find all P (x) ∈ R[x] such that ∀r ∈ Q ∃x ∈ Q : P (x) = r Problem 16 Is it true that for any two increasing sequences a1 < a2 < · · · < an and b1 < b2 < · · · < bn we can find a strictly increasing polynomial P [X] ∈ R[X] s.t P (ai ) = bi for i = 1, 2, · · · , n ? Problem 17 Prove that, for infinitely many positive integers n, there exists a polynomial P of degree n with real coefficients such that P (1), P (2), · · · , P (n+ 11 2) are different whole powers of Problem 18 Suppose q0 , q1 , q2 , is an infinite sequence of integers satisfying the following two conditions: (i) m − n divides qm − qn for m > n ≥ 0, (ii) there is a polynomial P such that |qn | < P (n) for all n Prove that there is a polynomial Q such that qn = Q(n) for all n Problem 19 Let P ∈ R[x] such that for infty of integer number c : Equation P (x) = c has more than one integer root Prove that P (x) = Q((x−a)2 ), where a ∈ R and Q is a polynomial Problem 20 Find all the polynomials P(x) with odd degree such that P (x2 − 2) = P (x) − Problem 21 http://artofproblemsolving.com/community/c6h83250p482076 Suppose p(x) is a polynomial with integer coefficients assumes at n distinct integral values of x that are different form and in absolute value less than (n − [ n2 ])! n 2n−[ ] Prove that p(x) is irreducible Prove that the bound may be replaced by n n d ( )n−[ ] (n − [ ])! 2 where d is minimum distance between any two of the n integral values of x where p(x) assumes the integral values considered Problem 22 Six members of the team of Fatalia for the IMO are selected from 13 candidates At the TST the candidates got a1 , a2 , , a13 points with = aj if i = j The team leader has already candidates and now wants to see them and nobody other in the team With that end in view he constructs a polynomial P (x) and finds the creative potential of each candidate by the formula ci = P (ai ) For what minimum n can he always find a polynomial P (x) of degree not exceeding n such that the creative potential of all candidates is strictly more than that of the others? Nguyen Trung Tuan Email: tuan.nguyentrung@gmail.com 12