All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
A PowerSum Puzzle (Posted on 2014-08-27) Difficulty: 3 of 5
Suppose a, b, c, d are complex numbers such that

a + b + c + d = 1
a^2 + b^2 + c^2 + d^2 = 0
a^3 + b^3 + c^3 + d^3 = 1
a^4 + b^4 + c^4 + d^4 = 0

What is a^10 + b^10 + c^10 + d^10?

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 3 of 3 |
Having the values for the symmetric polynomials would immediately lead to a fourth power polynomial for which a, b, c, d are the roots of.  Given the first four power sums, the Newton-Girard Formulas can be used to find the values for the symmetric polynomials.

Let the given power sums be S(1), S(2), S(3), and S(4).  Define the symmetric polynomials as P(1), P(2), P(3), and P(4).  The Newton-Girard Formulas state these equations are identities:
S(1) - P(1) = 0
S(2) - S(1)*P(1) + 2*P(2) = 0
S(3) - S(2)*P(1) + S(1)*P(2) - 3*P(3) = 0
S(4) - S(3)*P(1) + S(2)*P(2) - S(1)*P(3) + 4*P(4) = 0

Given S(1) = 1, S(2) = 0, S(3) = 1, and S(4) = 0, then P(1) = 1, P(2) = 1/2, P(3) = 1/2, and P(4) = 3/8.

By use of the symmetric polynomials, a, b, c, d are the roots of x^4 - P(1)*x^3 + P(2)*x^2 - P(3)*x + P(4) = 0.  Then x^4 - x^3 + (1/2)*x^2 - (1/2)x + 3/8 = 0.  The roots of this polynomial are x = -0.24504+/-0.70494i and x = 0.74504+/-0.34378i.  These are consistent with Jer's two digit approximations earlier.

Multipying each side by powers of x and rearranging yields: x*n = x^(n-1) - (1/2)*x^(n-2) + (1/2)*x^(n-3) - (3/8)*x^(n-4).

By adding all four x^n values (one for each of a, b, c, d) then a recursion for S(n) is formed: S(n) = S(n-1) - (1/2)*S(n-2) + (1/2)*S(n-3) - (3/8)*S(n-4)

Then the sequence for S(n) continues as -7/8, -3/8, -5/16, -9/16, -17/64, 0

Therefore a^10 + b^10 + c^10 + d^10 = 0

  Posted by Brian Smith on 2016-06-21 12:12:11
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information