Roots

[bugzpodder]

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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

[krtxmrtz]

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

[Yonatan]

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

[krtxmrtz]

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

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.

[Yonatan]

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))

[krtxmrtz]

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.

[Kalkewl8ter]

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

[bugzpodder]

actually, the sum is 6, not -6

[krtxmrtz]

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

[Yonatan]

Nice

[bugzpodder]

prove all other cases doesn't work

[krtxmrtz]

What do you exactly mean?

[bugzpodder]

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)

[krtxmrtz]

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...

[bugzpodder]

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

[krtxmrtz]

Gorblimey!!!! How come I never thought of that!!!!?????

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.

[bugzpodder]

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