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0 AMATYC Conference 2008 Washington Hilton Washington DC Dr. Joseph E. Cicero November 21, 2008 A New Solution to a Well Known Problem.

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Presentation on theme: "0 AMATYC Conference 2008 Washington Hilton Washington DC Dr. Joseph E. Cicero November 21, 2008 A New Solution to a Well Known Problem."— Presentation transcript:

1 0 AMATYC Conference 2008 Washington Hilton Washington DC Dr. Joseph E. Cicero November 21, 2008 A New Solution to a Well Known Problem

2 1 The Problem: Given: a + b + c = 1 a^2 + b^2 + c^2 = 2 a^3 + b^3 + c^3 = 3 Find a^4 + b^4 + c^4. Note: This presentation will concentrate on three variables "a", "b", "c" although all results generalize to n variables, "x1", "x2", "x3",..., "xn"

3 2 The elementary symmetric polynomials S1, S2 and S3 are defined as follows: S1 = a + b + c S2 = ab + ac + bc S3 = abc The power symmetric polynomials P1, P2, P3,...Pn,... are defined as follows: P1 = a + b + c P2 = a^2 + b^2 + c^2 P3 = a^3 + b^3 + c^3... Pn = a^n + b^n + c^n... The Solution: A) Some useful notations:

4 3 If f3(x) = (x - a) (x - b) (x - c) Then f3(x) = x^3 - S1 x^2 + S2 x - S3. Let f4(x) = x f3(x) f5(x) = x f4(x) = x^2 f3(x)... fn(x) = x^(n-3) f3(x) (Proof by Mathematical Induction) B) Well known fact:

5 4 f3(a) = f3(b) = f3(c) = 0 f4(a) = a f3(a) = a 0 = 0 f4(b) = b f3(b) = b 0 = 0 f4(c) = c f3(c) = c 0 = 0 f4(x) = x^4 - S1 x^3 + S2 x^2 - S3 x f4(a) = 0 = a^4 - S1 a^3 + S2 a^2 - S3 a f4(b) = 0 = b^4 - S1 b^3 + S2 b^2 - S3 b f4(c) = 0 = c^4 - S1 c^3 + S2 c^2 - S3 c C) Preliminary conclusions:

6 5 Therefore, by adding the above and transposing (a^4 + b^4 + c^4) = S1 (a^3 +b^3 +c^3) - S2 (a^2 +b^2+c^2) + S3 (a + b + c) ie. P4 = P3 S1 - P2 S2 + P1 S3 Similarly, P5 = P4 S1 - P3 S2 + P2 S3 and Pn = Pn-1 S1 - Pn-2 S2 + Pn-3 S3

7 6 S1 = P1 ( P1 1 ) S2 = 1/(2!) Det ( ) ( P 2 P1 ) ( P1 1 0 ) S3 = 1/(3!) Det ( P2 P1 2 ) ( P3 P2 P1) D) Some results from [D E Littlewood, "A University Algebra" London, Heinemann © 1958, pp 81 - 89]

8 7 ( P1 1 0 0 ) P4 = 1/(3!) Det ( P2 P1 2 0 ) ( P3 P2 P1 3 ) ( 0 P3 P2 P1) ( P1 1 0 0 ) P5 = 1/(3!) Det ( P2 P1 2 0 ) ( P3 P2 P1 3 ) ( 0 P4 P3 P2) E) Theorem: (New Solution to the Problem)

9 8 In general, ( P1 1 0 0 ) Pn = 1/(3!) Det ( P2 P1 2 0 ) ( P3 P2 P1 3 ) ( 0 Pn-1 Pn-2 Pn-3) Also, ( P1 1 0 0 0 ) ( P2 P1 2 0 0 ) P5 = 1/(3!) Det ( P3 P2 P1 3 0 ) ( 0 P3 P2 P1 - 1 ) ( 0 0 P3 P2 P1 )

10 9 Proof: Expand by last column ( P1 1 0 0 ) 1/(3!) Det ( P2 P1 2 0 ) = ( P3 P2 P1 3 ) ( 0 P3 P2 P1) [ (P1 1 0 ) (P1 1 0 ) ] 1/(3!) [ P1 Det (P2 P1 2 ) - 3 Det (P2 P1 2 ) ] = [ (P3 P2 P1) ( 0 P3 P2) ] (P1 1 0 ) (P1 1 ) (P1 1 ) P1 / (3!) Det (P2 P1 2 ) - P2 / (2!) Det ( ) + 2/(2!) Det ( ) (P3 P2 P1) (P2 P1 ) ( 0 P3 ) = P1 S3 - P2 S2 + P3 P1 = P3 S1 - P2 S2 + P1 S3 = P4

11 10 Given P1 = 1, P2 = 2 and P3 =3, Use Pivotal Condensation, [ CHIO, F "Memoire sur les fonctions... " Turin E. Pons 1835], to evaluate ( 1 1 0 0 ) (-1 2 0 ) P4 = 1/3! Det ( 2 1 2 0 ) = 1/6 Det (-1 1 3 ) ( 3 2 1 3 ) ( 3 2 1 ) ( 0 3 2 1 ) = 1/6 [ -1 + 18 + 6 + 2] = 25/6 = 4 1/6. ( 1 1 ) ( 1 1 0 ) S1 = 1, S2 = 1/2! DET ( ) = -1/2, S3 = 1/3! DET ( 2 1 2 ) = 1/6 ( 2 1 ) ( 3 2 1 ) Note: It is not necessary to calculate S1, S2 and S3 to find P4. Set up the determinant and quickly find the answer. S1, S2 and S3 will be used to find a, b, c. F) The Answer:

12 11 Note that 3, 5, 8 are taken from the Fibonacci Sequence { 1, 1. 2, 3, 5, 8. 13, 21, 34, 55, … }. Use Pivotal Condensation to evaluate ( 3 1 0 0 ) ( 1 3 0 0 ) P4 = 1/3! Det ( 5 3 2 0 ) = -1/6 Det ( 3 5 2 0 ) ( 8 5 3 3 ) ( 5 8 3 3 ) ( 0 8 5 3 ) ( 8 0 5 3 ) ( -4 2 0 ) -1/6 Det ( -7 3 3 ) = -1/6 [ -36 - 144 + 60 + 42] ( -24 5 3 ) = -1/6 [ -180 + 102] = -1/6 [ -78 ] = 13. That is the next member of the Fibonacci sequence. Will P5 = 21? G) Let P1 = 3, P2 = 5 and P3 = 8. Find P4.

13 12 Since f3(x) = (x - a) (x - b) (x - c) = x^3 - S1 x^2 + S2 x - S3 find the roots, a, b and c. Note: S1 = 1, S2 = -1/2, S3 = 1/6 Using synthetic division, 1 -1 -.5 -.1666666666 | 1 1 0 -.5 _____________________________________ 1 0 -.5 < 0 ie. f3(1) < 0 1 -1 -.5 -.1666666666 | 2 2 2 3 _____________________________________ 1 1 1.5 > 0 ie. f3(2) > 0, it follows that a root of f3(x) lies between 1 and 2. H) Find the value of a, b and c when P1 = 1, P2 = 2 and P3 = 3.

14 13 f3(x) = x^3 - S1 x^2 + S2 x - S3 f3'(x) = 3 x^2 - 2 S1 x + S2 x+ = x - f3(x) / f3'(x), ( Note: x+ represents the next iterate. ) x+ = ( 2 x^3 - S1 x^2 + S3 ) / ( 3 x^2 - 2 S1 x + S2) x0 = 2 x1 = 1.622222222222 x2 = 1.463280914325 x3 = 1.432027839949 x4 = 1.430851206961 x5 = 1.430849566246 x6 = 1.430849566243 x7 = 1.430849566243 x8 = 1.430849566243 x9 = 1.430849566243 Use Newton-Raphson Iteration to locate the root between 1 and 2.

15 14 1 -1 -.5 -.1666666666 | 1.430849566243 1.430849566243.6164809149744.1666666666 ________________________________________________ 1.430849566243.1164809149744 0 Then f3(x) = (x - 1.430849566243) (x^2 +.430849566243 x +.1164809149744) It follows that a = 1.430849566243 b = -.215424783121 +.264713199125 i c = -.215424783121 -.264713199125 i Note that the problem always has a solution for any real values of P1, P2 and P3 and the approximate values of "a", "b" and "c" can be found by the above method. /// If “k” is the number of variables, a factor of (-1)^(k+1) is needed to find P(k+1), etc. In this presentation k = 3, so (-1)^(k + 1) = (-1)^4 = 1, Thus the factor was omitted.


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