wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Sum of fourth powers
        
(Message started by: pcbouhid on Nov 26^th, 2005, 5:17am)

Title: Sum of fourth powers
Post by pcbouhid on Nov 26^th, 2005, 5:17am

Given:

a + b + c = 6
a^2 + b^2 + c^2 = 8
a^3 + b^3 + c^3 = 5

What is the value of a^4 + b^4 + c^4?

Title: Re: Sum of fourth powers
Post by JocK on Nov 26^th, 2005, 8:27am

on 11/26/05 at 05:17:15, pcbouhid wrote:

Given:

a + b + c = 6
a^2 + b^2 + c^2 = 8
a^3 + b^3 + c^3 = 5

What is the value of a^4 + b^4 + c^4?

I don't think the plane

a + b + c = 6

has an intersection with the sphere

a² + b² + c² = r²

for r² < 12 ... ???

Title: Re: Sum of fourth powers
Post by Icarus on Nov 26^th, 2005, 9:03am

I get [hide]
a⁴ + b⁴ + c⁴ = 0
a⁵ + b⁵ + c⁵ = 118/3
a⁶ + b⁶ + c⁶ = 913/3
[/hide]

or more generally
[hide] If A_n = aⁿ + bⁿ + cⁿ,
then A_n = 6A_n-1 - 14A_n-2 + (41/3)A_n-3.

This is the first time I can recall starting with the general term, and figuring out the recursion.[/hide]

Title: Re: Sum of fourth powers
Post by Icarus on Nov 26^th, 2005, 9:04am

on 11/26/05 at 08:27:41, JocK wrote:

I don't think the plane

a + b + c = 6

has an intersection with the sphere

a² + b² + c² = r²

for r² < 12 ... ???

Who says a, b, c have to be real?

Title: Re: Sum of fourth powers
Post by Grimbal on Nov 26^th, 2005, 11:39am

I can just confirm Icarus's answer.

But it would be too long to generalize.

[hide]
S₁ = (a+b+c) = 6
S₂ = (a²+b²+c²) = 8
S₃ = (a³+b³+c³) = 5
S₄ = (a⁴+b⁴+c⁴) = ?

S₁·S₁ = (a+b+c)(a+b+c) = (a²+b²+c²) + 2(ab+ac+bc) = S₂ + S₁₁
S₁₁ = 2(ab+ac+bc) = S₁·S₁ - S₂

S₂·S₁ = (a²+b²+c²)(a+b+c) = (a³+b³+c³) + (a²b+a²c +b²a+b²c +c²a+c²b) = S₃ + S₂₁
S₂₁ = (a²b+a²c +b²a+b²c+c²a +c²b) = S₂·S₁ - S₃

S₁₁·S₁ = 2(ab+ac+bc)(a+b+c) = 6abc + 2(a²b+a²c+b²a +b²c+c²a +c²b) = S₁₁₁ + 2·S₂₁
S₁₁₁ = 6abc = S₁₁·S₁ - 2·S₂₁
S₁₁₁ = S₁·S₁·S₁ - 3·S₂·S₁ + 2·S₃

S₁₁₁·S₁ = 6(abc)(a+b+c) = 6(a²bc+b²ac+c²ab) = 3·S₂₁₁
S₂₁₁ = 2(a²bc+b²ac+c²ab) = S₁₁₁·S₁/3
S₂₁₁ = S₁·S₁·S₁·S₁/3 - S₂·S₁·S₁ + 2/3·S₃·S₁

S₂·S₁₁ = 2(a²+b²+c²)(ab+ac+bc) = 2(a³b+a³c +ab³+b³c +ac³+bc³) + 2(a²bc+ab²c+abc²) = 2·S₃₁ + S₂₁₁
S₃₁ = (a³b+a³c +b³a+b³c+c³a +c³b) = (S₂·S₁₁-S₂₁₁)/2
S₃₁ = S₂·S₁·S₁ - S₂·S₂/2 - S₁·S₁·S₁·S₁/6 - S₃·S₁/3

S₃·S₁ = (a³+b³+c³)(a+b+c) = (a⁴+b⁴+c⁴) + (a³b+a³c+b³a +b³c +c³a+c³b) = S₄ + S₃₁
S₄ = (a⁴+b⁴+c⁴) = S₃·S₁ - S₃₁
S₄ = S₁·S₁·S₁·S₁/6 + 4/3·S₃·S₁ + S₂·S₂/2 - S₂·S₁·S₁

S₄ = 6⁴/6 + 4/3·5·6 + 8²/2 - 8·6² = 0
[/hide]

Title: Re: Sum of fourth powers
Post by Icarus on Nov 26^th, 2005, 11:57am

Not that long. The recursion formula I offered is just

A_n = S₁A_n-1 - (S₁₁/2)A_n-2 + (S₁₁₁/6)A_n-3,

using your notation. Once you have those three values (one of which is given), you can find all the powers by induction.

Title: Re: Sum of fourth powers
Post by pcbouhid on Nov 26^th, 2005, 1:04pm

Great work. The solution I have:

a + b + c = 6
a^2 + b^2 + c^2 = 8
a^3 + b^3 + c^3 = 5

[hide]If cubic equation x^3 - Ax^2 + Bx - C = 0 has roots a, b, c, then, expanding
(x - a)(x - b)(x - c), we find:

A = a + b + c

B = ab + bc + ca

C = abc

Then:

B = ab + bc + ca = 1/2[(a + b + c)^2 - (a^2 + b^2 + c^2)] = 14.

Hence a, b, c are roots of x^3 - 6x^2 + 14x - C = 0, and we have:

a^3 - 6a^2 + 14a - C = 0

b^3 - 6b^2 + 14b - C = 0

c^3 - 6c^2 + 14c - C = 0

Adding, we have:

(a^3 + b^3 + c^3) - 6(a^2 + b^2 + c^2) + 14(a + b + c) - 3C = 5 - 6×8 + 14×6 - 3C = 0.

Hence C = 41/3, and

x^3 - 6x^2 + 14x - 41/3 = 0.

Multiplying the polynomial by x, we have:

x^4 - 6x^3 + 14x^2 - 41x/3 = 0.

Then:

a^4 - 6a^3 + 14a^2 - 41a/3 = 0

b^4 - 6b^3 + 14b^2 - 41b/3 = 0

c^4 - 6c^3 + 14c^2 - 41c/3 = 0

Adding, we have:

(a^4 + b^4 + c^4) - 6(a^3 + b^3 + c^3) + 14(a^2 + b^2 + c^2) - 41(a + b + c)/3 = 0.

Hence:

a^4 + b^4 + c^4 = 6×5 - 14×8 + (41/3)×6 = 0.[/hide]

Title: Re: Sum of fourth powers
Post by JocK on Nov 27^th, 2005, 3:51am

It is a nice problem. Just wonder how many people came up with the requested answer without realising they are dealing not just with ordinary numbers, but with elements of a more generic division ring... :o

Anyone dare to represent a, b and c as complex numbers, quaternions or octonions?

Title: Re: Sum of fourth powers
Post by Barukh on Nov 27^th, 2005, 4:17am

You may want also to visit the following thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1078518934) with exactly the same name.

Title: Re: Sum of fourth powers
Post by Icarus on Nov 27^th, 2005, 9:08am

My approach was slightly different. I recalled the result that a linear recursion with constant coefficients

A_n+k+1 = a_kA_n+k + a_k-1A_n+k-1 + ... + a₀A_n

has general term of the form

A_n = p₀(n)r₀ⁿ + ... + p_m(n)r_mⁿ,

where the r_i are the distinct roots of the characteristic polynomial x^k+1 - a_kx^k - ... - a₀, and for all i, p_i is a polynomial whose degree is 1 less than the multiplicity of r_i as a root.

In particular, if the characteristic polynomial does not have any multiple roots, the general term is given by

A_n = b₀r₀ⁿ + ... + b_kr_kⁿ.

---------

Looking at the given information, I saw that it was, assuming that a, b, c were distinct, the first 3 terms of a 3rd-order recursion, so all I needed to do was to find the recursion relation. The characteristic polynomial is (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+ac+bc)x - abc.

A calculation similar to Grimbal's showed this to be x³ - 6x^{2[/sub] + 14x[sup]3} - 41/3. So the recursion formula is A_n = 6A_n-1 - 14A_n-2 + (41/3)A_n-3.

Applying the recursion for n=4 gives A₄ = 0. And once you have the recursion, calculating sums of higher powers is almost gratis.

Title: Re: Sum of fourth powers
Post by Icarus on Nov 27^th, 2005, 9:14am

on 11/27/05 at 03:51:55, JocK wrote:

Nothing more than complex numbers is needed. Any consistent system of polynomial equations with real or complex coefficients will be completely solvable within the complex numbers.

As for representing a, b, c, just find the roots of x³ - 6x² + 14x - 41/3. Cartan will oblige.
______________________________________

Barukh's link to NickH's previous posting of this riddle makes me curious if NickH is not the originator, at least of the form given here. Since in the previous version, Nick originally had different values, but then changed them so the answer would be 0.