Lagrange s polynomial

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 1 Lagrange s interpolation polynomial 2 2 Examples 3 3 Problems 10 1

1 Lagrange s interpolation polynomial Theorem. Let n be a positive integer and x 0, x 1,, x n ; y 0, y 1,, y n are complex numbers such that x i x j if i j. Then there exists precisely one polynomial P (x) of degree not greater than n such that P (x i ) = y i i = 0, n. Proof. If P and Q are polynomials of degree not greater than n such that P (x i ) = Q(x i ) = y i i = 0, n then the polynomial P Q of degree not greater than n and has at least n + 1 distinct roots, therefore P Q is the zero polynomial and hence P = Q. x x j Now if P (x) = y i then P (x) is a polynomial of degree not x i x j i=0 j i greater than n and P (x i ) = y i i = 0, n. x x j The polynomial y i is called Lagrange s interpolation polynomial or Lagrange s polynomial at nodes x 0, x 1,, x n x i=0 i x j j i. Corollary. If P (x) is a polynomial of degree not greater than n (n N ) and x 0, x 1,, x n are complex numbers such that x i x j if i j then P (x) = P (x i ) x x j. x i=0 i x j j i 2

2 Examples Example 1. Let A(x) = x 81 + x 49 + x 25 + x 9 + x and B(x) = x 3 x be polynomials. Find the remainder in the division of A(x) by B(x). Solution 1. 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 < 3. (1) Because B(0) = B(1) = B( 1) = 0 and (1) we have R(0) = 0, R(1) = 5 and R( 1) = 5, therefore use Lagrange s interpolation polynomial at nodes 0; 1 and 1 we have (x 1)(x + 1) 0)(x + 1) R(x) = R(0). + R(1).(x (0 1)(0 + 1) (1 0)(1 + 1) + R( 1). (x 0)(x 1) ( 1 0)( 1 1) = 5 2 x(x + 1) 5 x(x 1) 2 = 5x. Solution 2. We have x 3 x (mod B(x)) and therefore A(x) = (x 3 ) 27 + (x 3 ) 16 x + (x 3 ) 8 x + (x 3 ) 3 + x x 27 + x 17 + x 9 + x 3 + x = (x 3 ) 9 + (x 3 ) 5 x 2 + (x 3 ) 3 + x 3 + x x 9 + x 7 + 2x 3 + x = (x 3 ) 3 + (x 3 ) 2 x + 2x 3 + x 4x 3 + x 5x (mod B(x)). Because deg(5x) = 1 < deg B(x), we have 5x is the remainder in the division of A(x) by B(x). Example 2. (USAMO 1975) Let P be a polynomial of degree n (n Z, n > 0) satisfying P (k) = k k = 0, n. Determine P (n + 1). k + 1 Solution 1. Use Lagrange s interpolation polynomial at nodes 0, 1,, n 3

we have P (x) = = = P (k) j k k k + 1 j k x j k j x j k j ( 1) n k k (k + 1)!(n k)! (x j). j k Therefore P (n + 1) = = ( 1) n k k (k + 1)!(n k)! ( 1) n k k (k + 1)!(n k)! = 1 = 1 = 1 (n + 1 j) j k k( 1) n k C k+1 n+2. (n + 1)! n + 1 k [ ] (k + 1)( 1) n k Cn+2 k+1 + ( 1) n k+1 Cn+2 k+1 ( ) n+1 ( 1) n k ()Cn+1 k + ( 1) n+2 i Cn+2 i = n + 1 + ( 1)n+1. Solution 2. We have the polynomial Q(x) = (x + 1)P (x) x of degree n + 1 and 0, 1, 2,, n are its roots, therefore for some constant C. Because Q( 1) = 1, we have Q(x) = Cx(x 1)(x 2) (x n), C( 1)( 2)( 3) ( n 1) = ( 1) n+1 (n + 1)!C = 1, therefore C = ( 1)n+1 (n + 1)! and Q(x) = ( 1)n+1 x(x 1)(x 2) (x n). (n + 1)! 4

Hence Q(n + 1) = ( 1)n+1 (n + 1)! (n + 1)! = ( 1)n+1. Q(n + 1) + n + 1 Note that P (n + 1) = = ( 1)n+1 + n + 1. n Therefore P (n + 1) is equal to 1 if n is odd and equal to if n is even. Example 3. (IMO Longlist 1977) Let n be a positive integer. Suppose x 0, x 1,..., x n are integers and x 0 > x 1 > > x n. Prove that at least one of the numbers F (x 0 ), F (x 1 ), F (x 2 ),..., F (x n ), where F (x) = x n + a 1 x n 1 + + a n ( a i R, i = 1,..., n) is greater than or equal to n! 2. n Solution. Assume that F (x i ) < n! i = 0, n. Use Lagrange s interpolation polynomial at nodes x 0, x 1,, x n we 2n have x n + a 1 x n 1 + + a n = F (x k ) j k x x j x k x j, see the coefficient of x n we have F (x k ) 1 = x k x j j k F (x k ) x k x j < j k j k j k n! 2 n x k x j n! 2 n k j = 1 C k 2 n n = 1, a contradiction, and therefore at least one of the numbers F (x 0 ), F (x 1 ), F (x 2 ),..., F (x n ) is greater than or equal to n! 2 n. 5

Example 4. 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. we have Use Lagrange s interpolation polynomial at nodes 0, 1, 2, 2n P (x) 1 = 2 (P (k) 1) j k x j 2 k j = ( 1) k+1 x j k j, j k and therefore P (2n + 1) 1 = 2 C k 2n+1 = 1 2 2n+1, but by hypothesis P (2n + 1) = 30, hence we have 31 = 1 2 2n+1, so n = 2. Example 5. (Tepper s identity) Prove that for any real number a we have the following identity ( ) n ( 1) k (a k) n = n! n N. k Solution. Assume that a = 0 and n 3, then we need to prove ( ) n ( 1) n+k (k) n = n! ( ). k By Lagrange s Interpolation polynomial at nodes 1, 2,, n we have x n (x 1)(x 2) (x n) = k n x i k i ( ). k=1 i k Now, in ( ), setting x = 0 we have ( 1) n+1 n! = k=1 k n 1 k ( 1) n 1 n! (k 1)! (n k)! ( 1), n k and we are done. 6

Example 6. Let x, y, z and t be real numbers satisfying x 2 + y2 + z2 + t2 = 1 2 2 1 2 2 2 3 2 2 2 5 2 2 2 7 2 x 2 + y2 + z2 + t2 = 1 4 2 1 2 4 2 3 2 4 2 5 2 4 2 7 2 x 2 + 6 y2 + z2 + t2 = 1 2 1 2 6 2 3 2 6 2 5 2 6 2 7 2 x 2 + y2 + z2 + t2 = 1. 8 2 1 2 8 2 3 2 8 2 5 2 8 2 7 2 Determine x 2 + y 2 + z 2 + t 2. Solution 1. Setting l i = (2i 1) 2, c i = (2i) 2 i = 1, 4 and f(x) = 4 (x l i ) 4 (x c i ). We have deg f 3, and therefore by Lagrange s Interpolation polynomial at nodes l 1, l 2, l 3, l 4 we obtain f(x) = 4 f(l i ) j i x l j l i l j. (1) From (1) and f(c j ) = 4 (c j l i ) j = 1, 4 we have 4 (c j l i ) = 4 f(l i ) k i c j l k l i l k j = 1, 4, and hence where α i = α 1 c j l 1 + α 2 c j l 2 + α 3 c j l 3 + α 4 c j l 4 = 1 j = 1, 4, f(l i ) j i (l i l j ) i = 1, 4. See coeficients of x 3 in both sides of (1) we have and therefore x 2 + y 2 + z 2 + t 2 = 4 α i = 36. Solution 2. From hypothesis we have x 2 w 1 2 + y2 w 3 2 + z2 w 5 2 + 4 α i = 4 (c i l i ) = 36, t2 = 1 w {4, 16, 36, 64}, w 72 7

and therefore 4, 16, 36 and 64 are roots of the polynomial P (w) = 4 (w (2i 1) 2 ) x 2 (w 3 2 )(w 5 2 )(w 7 2 ) = w 4 (x 2 + y 2 + z 2 + t 2 + 1 2 + 3 2 + 5 2 + 7 2 )w 3 + hence P (w) = (w 4)(w 16)(w 36)(w 64). By see coeficients of w 3 we have (x 2 + y 2 + z 2 + t 2 + 1 2 + 3 2 + 5 2 + 7 2 ) = (4 + 16 + 36 + 64), so x 2 + y 2 + z 2 + t 2 = 36. Example 7. 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 = x 1, B = x 2, C = x 3, P = p, Q = q. By Lagrange s Interpolation polynomial at nodes x 1, x 2, x 3 we have (x p)(x q) = 3 (x i p)(x i q) j i x x j x i x j, see coeficients of x 2 in the both sides we obtain (x 1 p)(x 1 q) (x 1 x 2 )(x 1 x 3 ) + (x 2 p)(x 2 q) (x 2 x 3 )(x 2 x 1 ) + (x 3 p)(x 3 q) (x 3 x 1 )(x 3 x 2 ) = 1, and therefore 1 x 1 p. x 1 q x 1 x 2. x 1 x 3 + x 2 p. x 2 q x 2 x 3. x 2 x 1 + x 3 p. x 3 q x 3 x 1. x 3 x 2, but x 2 x 1 = AB, and x 1 p = AP,, therefore and we are done. 1 P A.QA AB.AC + P B.QB BC.BA + P C.QC CA.CB, 8

Example 8. 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 1. For all n N take r n Q such that P (r n ) = n. P (r 1 ) = 1, P (r 2 ) = 2,..., P (r d+1 ) = d + 1 gives P (x) Q[x] using the Lagrange s interpolation polynomial at nodes r 1, r 2,, r d+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 r n and also leading coefficient m. So the denominator of r n divides m, i.e. r n 1 Z for all m n N. Now assume d = deg P 2. Then P (x) x + for x +, and we find N N large enough with P (x) x > 2 m for all x N. But 1 Z ( N, N) contains exactly 2 m N 1 elements. So among the m 2 m N different numbers r 1, r 2,..., r 2 m N 1 m Z we must find r k N for some k = 1,..., 2 m N. This gives 2 m < P (r k ) r k = k r k 2 m N N = 2 m, a contradiction. So we must have P (x) Q[x] with deg P = 1, which is indeed a solution for the problem. 9

3 Problems Problem 1. Let P be a polynomial of degree at most n satisfying P (k) = 1 k = 0, n. Determine P (n + 1). C k n+1 Problem 2. 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) 2 2k. Problems 3. Prove that at least one of the numbers f(1), f(2),, f(n + 1) is greater than or equal to n! 2n. Where f(x) = x n + a 1 x n 1 + + a n ( a i R, i = 1,..., n, n N.) Problem 4. 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 5. Let p be a prime number and f an integer polynomial of degree d such that f(0) = 0, f(1) = 1 and f(n) is congruent to 0 or 1 modulo p for every integer n. Prove that d p 1. Problem 6. Let P be a polynomial of degree n N satisfying P (k) = 2 k k = 0, n. Prove that P (n + 1) = 2 n+1 1. Problem 7. 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, Determine n. P (2) = P (5) = = P (3n 1) = 0, and P (3n + 1) = 730. Problem 8. Let S = {s 1, s 2, s 3,..., s n } be a set of n distinct complex numbers n 9, exactly n 3 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 9. Let t and n be fixed integers each at least 2. 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) and P (k) t k t k+1 is an integer for each k = 0, 1,..., m. Problem 10. Let f (x) = x n + a n 2 x n 2 + a n 3 x n 3 +... + a 1 x + a 0 be a polynomial. Prove that we have an i {1, 2,..., n} f (i) n! ( n i). Problem 11. Let (F n ) n 1 be the Fibonacci sequence F 1 = F 2 = 1, F n+2 = F n+1 + F n (n 1), and P (x) the polynomial of degree 990 satisfying Prove that P (1983) = F 1983 1. P (k) = F k, for k = 992,..., 1982. Problem 12. Let P (x) be a polynomial of degree n with real coefficients and let a 3. Prove that max a j P (j) 1. 0 j n+1 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 3. What is the smallest possible value of k if the following statements are true. For every n points A i = (x i, y i ) on a plane, where no three points are collinear, and for any real numbers c i (1 i n) there exists such polynomial P (x, y), the degree of which is no more than k, where P (x i, y i ) = c i for every i = 1,..., n. (The degree of a nonzero monomial a i,j x i 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 a 1 < a 2 < < a n and b 1 < b 2 < < b n we can find a strictly increasing polynomial P [X] R[X] s.t. P (a i ) = b i 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 2. Problem 18. Suppose q 0, q 1, q 2,... is an infinite sequence of integers satisfying the following two conditions: (i) m n divides q m q n for m > n 0, (ii) there is a polynomial P such that q n < P (n) for all n Prove that there is a polynomial Q such that q n = 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 (x 2 2) = P 2 (x) 2. 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 0 and in absolute value less than (n [ n 2 ])! 2 n [ n 2 ] Prove that p(x) is irreducible. Prove that the bound may be replaced by ( d 2 )n [ n 2 ] (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 a 1, a 2,..., a 13 points with a i a j if i j. The team leader has already 6 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 c i = P (a i ). For what minimum n can he always find a polynomial P (x) of degree not exceeding n such that the creative potential of all 6 candidates is strictly more than that of the 7 others? Nguyen Trung Tuan Email: tuan.nguyentrung@gmail.com 12