Roots

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this question is loads of fun:

let the roots of the following equation be positive, find all coefficients:

x^6-6x^5+ax^4+bx^3+cx^2+dx+1=0

Since it's 6^th order polynomial then it has 6 roots. There are only 4 coefficients to be determined so the number of possible solutions is infinite. If we "force" the positive roots to be, say, 1, 2, 3 and 4, then we are left with the following system for the coefficients:

a + b + c + d = 4
16a + 8b + 4c + 2d = 127
81a + 27b + 9c + 3d = 728
256a + 64b + 16c + 4d = 2047

Solving them by any suitable method yields

a = -119 / 24
b = 1075 / 12
c = -4189/24
d = 1127 / 12

That's wrong though, it has 6 roots. 4 roots are positive, what about the other 2?

Quote:

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Originally posted by Yonatan
That's wrong though, it has 6 roots. 4 roots are positive, what about the other 2?

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I don't quite get what you mean, just WHAT is wrong?

And who cares about the other 2 roots? All you are supposed to calculate is the coefficients, not the roots.

The assumption is that the 6 roots are positive. You found the coefficients assuming 4 of the 6 roots are positive. If one or both of the other two is/are negative, the coefficients you found are an incorrect answer to the question.

I checked it out, and both of the other roots are negative.

The roots, given your coefficients, are:

x=1
x=2
x=3
x=4
x=(-24-sqrt(570)/12
x=-1/2(24+sqrt(570))

Yes, I have to admit you're quite right... I had wrongly taken the statement about 6 positive roots rather as a hint than as a requirement to meet...
Well, so much the better, that should mean more fun when I have some time to re-think it all over.

Wait, bugz, you can't have all positive roots, because the sum of the roots is -6, according to your original equation.

actually, the sum is 6, not -6

Well,

I guess I found the solution while sitting in front of a boring TV program...

The first 2 and the last term suggest the polynome could be the expansion of (x-1)⁶. Thus there would be 6 degenerate roots all equal to 1 and the unknown coefficients would be equal to the corresponding numbers in the so called triangle of Tartaglia, that is to say:

a = 15
b = -20
c = 15
d = -6

and (x-1)⁶ = x⁶ - 6x⁵ + 15x⁴ - 20x³ + 15x² - 6x + 1

Nice :)

prove all other cases doesn't work

What do you exactly mean?

can you prove that for for any positive roots r1,r2,r3,r4,r5,r6, only 1,1,1,1,1,1 would give you the equation x^6-6x^5+ax^4+bx^3+cx^2+dx+1=0? ie prove if all the roots to this equation is positive, they MUST all be ones -- if they are not ones they don't work! (hint hint)

This is a suggestion, though I haven't actually found the proof.

If r₁, r₂, r₃, r₄, r₅, r₆ are the roots then

x⁶ - 6x⁵ + ax⁴ + bx³ + cx² + dx + 1 = (x - r₁)(x - r₂)(x - r₃)(x - r₄)(x - r₅)(x - r₆)

Now, if these products are calculated and we look at the coefficients for the x⁵ and x⁰ terms we have:

r₁ + r₂ + r₃ + r₄ + r₅ + r₆ = 6

r₁r₂r₃r₄r₅r₆ = 1

At this stage it seems kind of obvious that these 2 equations can only be fulfilled by all ones if the r_i's are to be all positive. However I could go no farther than this and don't see the way to prove this statement...

you just need one more step to complete the answer. have you ever heard of AM-GM Inequality? (arithmetic mean-geometric mean inequality)?

Gorblimey!!!! How come I never thought of that!!!!????? :) :( :p :D :rolleyes: :eek: :mad: :confused:

I suppose the proof can be now completed as follows:

( r₁ + r₂ + r₃ + r₄ + r₅ + r₆ ) / 6 = 1 = AM

r₁r₂r₃r₄r₅r₆ = 1 = 1^(1/6) = (r₁r₂r₃r₄r₅r₆)^(1/6) = GM

... and since GM <= AM and the equality only holds when all the numbers are equal, then all must be 1's.

is AM>=GM, and equality occurs if and only if all the numbers are equal. of course, the restriction to the positive roots is because AM-GM works only for positive numbers