Tìm tòi sáng tạo - SOME POLYNOMIAL PROBLEMS ARE BUILT ON IDENTITIES
1 like436 views
The document discusses three outstanding results from identities used to create polynomial problems. The first result finds all polynomials satisfying an equality for triples satisfying ab + bc + ca = 0. The solution is polynomials of the form ax^2. The second result finds polynomials satisfying an equality involving binomial coefficients, with the solution kx^n. The third result finds polynomials satisfying an equality for triples satisfying (a+b)(b+c)(c+a) = 8c^3, with solution kx^3.
Tìm tòi sáng tạo - SOME POLYNOMIAL PROBLEMS ARE BUILT ON IDENTITIES
- 2. Supervisor: Assoc. Prof. Dr Nguyen Vu Luong Authors: Nguyen Thi Ngoc Anh Nguyen Thi Nhat Linh Le Manh Cuong Do Thuy Phuong Lien Nguyen Thanh Tung Classes: 11A1 Math and 11A2 Math Hanoi, October 2012
- 3. Outline Introduction Basic knowledge • Problem • Comments Results • 3 outstanding results
- 4. Outline Introduction Basic knowledge • Problem • Comments Results • 3 outstanding results
- 5. INTRODUCTION In this seminar, using an IMO problem as an example, we will discuss our findings from some identities and the conditions we set for them. We then used them to create some polynomial problems. The main difficulty of this method is to find an integer root of an equation.
- 6. Outline Introduction Basic knowledge • Problem • Comments Results • 3 outstanding results
- 7. Basic knowledge Problem. (IMO 2004) Find all polynomials P(x) with real coefficients that satisfy the equality: P(a – b) + P(b – c) + P(c – a) = 2P(a + b + c) for all triples a, b, c of real numbers such that ab + bc + ca = 0. Comment. For all triples a, b, c belonging to set of real numbers such that ab + bc + ca = 0, we have: (a – b)2 + (b – c)2 + (c – a)2 = 2(a + b + c)2 (a – b)4 + (b – c)4 + (c – a)4 = 2(a + b + c)4
- 8. Problem. (IMO 2004) Find all polynomials P(x) with real coefficients that satisfy the equality: P(a – b) + P(b – c) + P(c – a) = 2P(a + b + c) (1) for all triples a, b, c of real numbers such that ab + bc + ca = 0. An integer root of the equation ab + bc + ca = 0 is (a,b,c)=(6,3,-2). ⇒ For all x ⋲ R, the triple (a, b, c) = (6x, 3x,−2x) satisfies the condition ab + bc + ca = 0. Basic knowledge Substituting this triple into (1), we have: P(3x) + P(5x) + P(-8x) = 2P(7x)
- 9. Problem. (IMO 2004) Find all polynomials P(x) with real coefficients that satisfy the equality: P(a – b) + P(b – c) + P(c – a) = 2P(a + b + c) (1) for all triples a, b, c of real numbers such that ab + bc + ca = 0. Comparing lead coefficients, we have (3i + 5i + (−8)i − 2 · 7i )ai = 0 for i running from 0 to n. Basic knowledge Denote: K(i) = 3i + 5i +(−8)i −2.7i By using the inductive method, we prove that: K(i) = 0 i = 2 or i = 4.⇔ Hence ai = 0 for i odd, i = 0 and for i even ≥ 6; ai ≠ 0 for i= 2, i= 4. It follows that P(x) = mx4 + nx2 for all real numbers m, n. Verifying and we have all such P(x) satisfy the required condition.
- 10. Outline Introduction Basic knowledge • Problem • Comments Results • 3 outstanding results
- 11. Results In this section, we will present some identities and some expanded problems from these. Our findings include three outstanding results and each result has its own form.
- 12. Result 1. Find all polynomials P(x) with real coefficients that satisfy the equality: P(2a+2b-c) + P(2b+2c-a) + P(2c+2a–b) = 9P(a+b+c) for all triples a, b, c of real numbers such that ab + bc + ca = 0. Results Results Result 2. Find all polynomials P(x) with real coefficients that satisfy the equality: P(a+b) = P(a)+(2n -1)P(b) for all a, b ⋲ R, n ⋲ N* and Result 3. Find all polynomials P(x) with real coefficients that satisfy the equality: P(a + b+c) = P(a) + P(b) + 25P(c) for all triples a, b, c belonging to R such that (a+b)(b+c)(c+a) = 8c3 .
- 13. Result 1. Find all polynomials P(x) with real coefficients that satisfy the equality: P(2a+2b-c) + P(2b+2c-a) + P(2c+2a–b) = 9P(a+b+c) for all triples a, b, c of real numbers such that ab + bc + ca = 0. Comment. This problem is built on the following identity for all x, y, z belonging to R such that xy+yz+zx=0: (2x+2y-z)2 + (2y+2z-x)2 + (2z+2x-y)2 = 9(x2 +y2 +z2 ) = 9(x+y+z)2 (*) with xy+yz+zx=0 • For all x belonging to R, the triple (a, b, c) = (6, 3,−2) satisfies the condition ab + bc + ca = 0. Substituting the triple (6x, 3x, -2x) into (*), we have: P(20x) + P(-4x) + P(5x) = 9P(7x) for all x ⋲ R Answer: P(x) = ax2 (a R).⋲ Results
- 14. Result 2. Find all polynomials P(x) with real coefficients that satisfy the equality: P(a+b) = P(a)+(2n -1)P(b) for all a, b⋲ R; n ⋲ N* and Comment. This problem is built on the binomial theorem for all a, b ⋲ R: Results If we put: , we will obtain the identity that create the above problem: (a + b)n = an + (2n – 1)bn .
- 15. Result 2. Find all polynomials P(x) with real coefficients that satisfy the equality: P(a+b) = P(a)+(2n -1)P(b) (*) for all a, b⋲ R; n N*⋲ Solution: Substituting this couple into (*), we have: P(2x) = P(x) + (2n -1) P(x) for all x ⋲ R. Comparing lead coefficients, we have: ai(2i – 2n ) = 0. => It follows that P(x) = kxn (k belongs to R). Results
- 16. Result 3. Find all polynomials P(x) with real coefficients that satisfy the equality: P(a + b+c) = P(a) + P(b) + 25P(c) for all triples a, b, c belonging to R such that (a+b)(b+c)(c+a) = 8c3 . Remark. We have built the above problem on the following identity for three variables x,y,z belonging to R and having the condition 3(x+y)(y+z) (z+x)=24z3 : (x+y+z)3 = x3 + y3 + 25z3 *Finding an integer root of the equation (a+b)(b+c)(c+a) = 8c3 , we obtain a=1, b=1, c=1. → The triple (x; x; x) satisfies the condition (a+b)(b+c)(c+a) = 8c3 Answer: P(x) = kx3 ( k belongs to R) Results
- 17. We'd like to thank you all for your attention Supervisor: Assoc. Prof. Dr Nguyen Vu Luong Authors: Nguyen Thi Ngoc Anh Nguyen Thi Nhat Linh Le Manh Cuong Do Thuy Phuong Lien Nguyen Thanh Tung Classes: 11A1 Math and 11A2 Math
Editor's Notes
- #16: Basic Knownedge Basic Knownedge