Symmetric Polynomials and Power Sum Derivations Notes

Initial System of Equations

The mathematical transcript provides a set of three equations involving three variables (aa, bb, and cc) and a final expression to be evaluated. The given values are as follows:

  • First Order Sum: a+b+c=4a + b + c = 4
  • Second Order Sum (Power of 2): a2+b2+c2=10a^2 + b^2 + c^2 = 10
  • Third Order Sum (Power of 3): a3+b3+c3=22a^3 + b^3 + c^3 = 22
  • Target Expression: The transcript lists the final query as "a² + b + c 4 =?". In the context of the mathematical progression shown (n=1,2,3n=1, 2, 3), this is systematically interpreted as the calculation of the sum of the fourth powers: a4+b4+c4a^4 + b^4 + c^4.

Theoretical Background: Symmetric Polynomials and Newton-Girard Sums

To solve for higher-order power sums when the lower-order sums are known, we utilize the theory of Symmetric Polynomials. Specifically, the Newton-Girard identities provide a direct recursive relationship between power sums and elementary symmetric polynomials.

Definitions of Elementary Symmetric Polynomials (for 3 Variables)
  • First Elementary Symmetric Polynomial (e1e_1):     e1=a+b+ce_1 = a + b + c
  • Second Elementary Symmetric Polynomial (e2e_2):     e2=ab+bc+cae_2 = ab + bc + ca
  • Third Elementary Symmetric Polynomial (e3e_3):     e3=abce_3 = abc
Definition of Power Sums (pnp_n)
  • p1=a1+b1+c1p_1 = a^1 + b^1 + c^1
  • p2=a2+b2+c2p_2 = a^2 + b^2 + c^2
  • p3=a3+b3+c3p_3 = a^3 + b^3 + c^3
  • p4=a4+b4+c4p_4 = a^4 + b^4 + c^4

Calculation of Elementary Symmetric Polynomials

Before determining the fourth power sum, we must find the values of e1e_1, e2e_2, and e3e_3 based on the provided data.

Calculating e1e_1

From the first given equation: e1=a+b+c=4e_1 = a + b + c = 4

Calculating e2e_2

We use the relationship between p2p_2, e1e_1, and e2e_2: p2=e1p12e2p_2 = e_1 p_1 - 2e_2 Substituting the known values (p2=10p_2 = 10, p1=4p_1 = 4, e1=4e_1 = 4): 10=(4)(4)2e210 = (4)(4) - 2e_210=162e210 = 16 - 2e_26=2e2-6 = -2e_2e2=3e_2 = 3

Calculating e3e_3

We use the Newton-Girard identity for the third power sum: p3=e1p2e2p1+3e3p_3 = e_1 p_2 - e_2 p_1 + 3e_3 Substituting the known values (p3=22p_3 = 22, p2=10p_2 = 10, p1=4p_1 = 4, e1=4e_1 = 4, e2=3e_2 = 3): 22=(4)(10)(3)(4)+3e322 = (4)(10) - (3)(4) + 3e_322=4012+3e322 = 40 - 12 + 3e_322=28+3e322 = 28 + 3e_36=3e3-6 = 3e_3e3=2e_3 = -2

Derivation of the Fourth Power Sum

The target expression corresponds to p4p_4. For a system of three variables (a,b,ca, b, c), the elementary symmetric polynomial en=0e_n = 0 for all n>3n > 3. The recursive formula for p4p_4 is:

p4=e1p3e2p2+e3p1p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1

Step-by-Step Numerical Substitution
  1. Identify Constants:
    • e1=4e_1 = 4
    • e2=3e_2 = 3
    • e3=2e_3 = -2
  2. Identify Power Sums:
    • p1=4p_1 = 4
    • p2=10p_2 = 10
    • p3=22p_3 = 22
  3. Execute Calculation:
    • p4=(4)(22)(3)(10)+(2)(4)p_4 = (4)(22) - (3)(10) + (-2)(4)
    • p4=88308p_4 = 88 - 30 - 8
    • p4=588p_4 = 58 - 8
    • p4=50p_4 = 50

Summary of Final Results

By applying algebraic identities and Newton's Sums to the provided transcript data, the following values are derived:

  • Elementary Sum (e1e_1): 44
  • Pairwise Sum (e2e_2): 33
  • Product of Variables (e3e_3): 2-2
  • Sum of Fourth Powers (p4p_4): 5050

The answer to the query "a² + b + c 4 =?" (interpreted as a4+b4+c4a^4 + b^4 + c^4) is 50.