Lagrange s polynomial Nguyen Trung Tuan November 13, 2016 Abstract...In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points x j and numbers y j, the Lagrange polynomial is the polynomial of the least degree that at each point x j assumes the corresponding value y j (i.e. the functions coincide at each point). The interpolating polynomial of the least degree is unique, however, and it is therefore more appropriate to speak of the Lagrange form of that unique polynomial rather than the Lagrange interpolation polynomial, since the same polynomial can be arrived at through multiple methods. Although named after Joseph Louis Lagrange, who published it in 1795, it was first discovered in 1779 by Edward Waring and it is also an easy consequence of a formula published in 1783 by Leonhard Euler... Wikipedia. In this article, I will use Lagrange polynomial to solve some polynomial problems from Mathematical Olympiads. Contents 1 Lagrange s interpolation polynomial 2 2 Examples 3 3 Problems 9 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 Z, n > 0) 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 + 5 + 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) = 5x. 2 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 7 + x 17 + x 9 + x 3 + x = (x 3 ) 9 + (x 3 ) 5 + (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. we have P (x) = Use Lagrange s interpolation polynomial at nodes 0, 1,, n P (k) j k x j k j = k x j k + 1 k j = j k ( 1) n k k (k + 1)!(n k)! (x j). j k Therefore P (n+1) = ( 1) n k k (k + 1)!(n k)! (n+1 j) = j k 3 ( 1) n k k (n + 1)!. (k + 1)!(n k)! n + 1 k
= 1 = 1 k( 1) n k C k+1 n+2 = 1 ( ) n+1 ( 1) n k ()Cn+1 k + ( 1) n+2 i Cn+2 i [ ] (k + 1)( 1) n k Cn+2 k+1 + ( 1) n k+1 Cn+2 k+1 = 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)! 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 ( ),..., 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 = 4 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 j k n! 2 n k j = 1 2 n j k F (x k ) x k x j < Cn k = 1, j k n! 2 n x k x j a contradiction, and therefore at least one of the numbers F (x 0 ), F (x 1 ), F ( ),..., F (x n ) is greater than or equal to n! 2 n. 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 2 P (x) 1 = (k) 1) (P x j 2 k j = j k ( 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 ( ). 5 k=1 i k
Now, in ( ), setting x = 0 we have and we are done. ( 1) n+1 n! = k=1 k n 1 k ( 1) n 1 n! (k 1)! (n k)! ( 1), n k Example 6. Let x, y, z and t be real numbers satisfying + y2 + z2 + t2 = 1 2 2 1 2 2 2 3 2 2 2 5 2 2 2 7 2 + y2 + z2 + t2 = 1 4 2 1 2 4 2 3 2 4 2 5 2 4 2 7 2 + 6 y2 + z2 + t2 = 1 2 1 2 6 2 3 2 6 2 5 2 6 2 7 2 + y2 + z2 + t2 = 1. 8 2 1 2 8 2 3 2 8 2 5 2 8 2 7 2 Determine + y 2 + z 2 + t 2. Solution 1. Setting l i = (2i 1) 2, c i = (2i) 2 i = 1, 4 and f(x) = (x l i ) (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 From (1) and f(c j ) = and hence where α i = f(x) = f(l i ) j i x l j l i l j. (1) (c j l i ) j = 1, 4 we have (c j l i ) = f(l i ) k i c j l k l i l k j = 1, 4, 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 6 α i = (c i l i ) = 36,
and therefore + y 2 + z 2 + t 2 = α i = 36. Solution 2. From hypothesis we have w 1 2 + y2 w 3 2 + z2 w 5 2 + t2 = 1 w {4, 16, 36, 64}, w 72 and therefore 4, 16, 36 and 64 are roots of the polynomial P (w) = (w (2i 1) 2 ) (w 3 2 )(w 5 2 )(w 7 2 ) = w 4 ( + 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 ( + y 2 + z 2 + t 2 + 1 2 + 3 2 + 5 2 + 7 2 ) = (4 + 16 + 36 + 64), so + 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 =, C = x 3, P = p, Q = q. By Lagrange s Interpolation polynomial at nodes x 1,, 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 in the both sides we obtain (x 1 p)(x 1 q) (x 1 )(x 1 x 3 ) + ( p)( q) ( x 3 )( x 1 ) + (x 3 p)(x 3 q) (x 3 x 1 )(x 3 ) = 1, and therefore 1 x 1 p. x 1 q x 1. x 1 x 3 + p. q x 3. x 1 + x 3 p. x 3 q x 3 x 1. x 3, 7
but 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, 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. 8
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). 9