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 A PowerSum Puzzle (Posted on 2014-08-27)
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

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 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

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