Finding the minimum number of positive roots where a₂ + a₃ is negative

[zorro]

Homework Statement

If equation https://www.physicsforums.com/attachment.php?attachmentid=32867&stc=1&d=1299509982 has n positive roots, then least value of n for which a₂ + a₃ is negative, is

The Attempt at a Solution

Time allotted is 1 minute.
I have no idea of solving such questions.

 

[zorro]

Sorry I missed out the equation. PF was slow at that time and I could not check my post.

It is xⁿ - nx^n-1 + a₂x^n-2 + a₃x^n-3 + ...a_n-1x + (-1)ⁿ = 0

 

[Mentallic]

I'm unsure myself, but I'll see if I can shed some light on the problem and hopefully someone else can carry on from there. By the way, I'm not even sure if this will be helping at all with the answer - I'm pretty much just throwing some ideas and relationships out there to see if it will lead to anything.

$$x^n-nx^{n-1}+a_2x^{n-2}+a_3x^{n-3}+...+a_{n-1}x+(-1)^n=0$$

Since this polynomial of n^th degree has n positive roots, then it is equivalent to

$$(x-r_1)(x-r_2)...(x-r_n)=0$$

where r_i are the roots and $r_i>0, i=1,2,...,n$

If we were to expand out the constant from the factored form, we would get

$$(-1)^nr_1r_2r_3...r_n$$

and this is equivalent to the constant in the polynomial, so we have

$$(-1)^nr_1r_2r_3...r_n\equiv (-1)^n$$

thus

$$r_1r_2r_3...r_n=1$$

Similarly if you expanded the factored form to get the coefficient of x^n-1 you'll get

$$r_1+r_2+...+r_n=n$$

Now I'll sit here and think some more about it...

 

[epenguin]

Awful, not very well defined question. I mean, am I allowed to say it's a quadratic? - i.e. a₃ doesn't exist - which is different from saying it's zero. Am I allowed to say a₂ doesn't exist either?

If so it's fairly easy. But if this a₃ is meant to imply it's at least a cubic it's a bit more difficult. Try and answer whether it can be quadratic and whether can be cubic too, or if not try the next number, too and submit more than one with "if is meant..." Anyway I think you're meant to work up from a small number like that.

Descartes' rule tells you something about whether it might or could not be or whether it might be though not whether it is. It would take me more than a minute to tell whether it is for n=3. The first and second terms of the poly are suggesting you compare with (x - 1)ⁿ .

Maybe with these hints you can construct a polynomial with the properties asked.

 

[zorro]

I figured out a 30 second method to work this out with someone else's help. Let me post it so that you both can learn from it

Let us assume that the n positive roots are r1, r2, r3...rn

We have r1 + r2 + r3...+rn = n
Using A.M.>=G.M.

(r1 + r2 + r3...+rn)/n >=(r1.r2.r3...rn)^1/n
which implies that (r1.r2.r3...rn)<=1

From the equation we can see that the product of the roots is 1, which is also proved by Mentallic.
So equality will hold when all of them are equal to 1.

This leaves us with ⁿC₃ > ⁿC₂
which gives 6 as the answer.

 

[Dick]

I figured out a 30 second method to work this out with someone else's help. Let me post it so that you both can learn from it

Let us assume that the n positive roots are r1, r2, r3...rn

We have r1 + r2 + r3...+rn = n
Using A.M.>=G.M.

(r1 + r2 + r3...+rn)/n >=(r1.r2.r3...rn)^1/n
which implies that (r1.r2.r3...rn)<=1

From the equation we can see that the product of the roots is 1, which is also proved by Mentallic.
So equality will hold when all of them are equal to 1.

This leaves us with ⁿC₃ > ⁿC₂
which gives 6 as the answer.

That's clever. Not only to solve the problem like that but to craftily engineer a problem which is so exactly solved by the boundary case of AM-GM. I hate problems like that. Thanks for reporting back!