Polynomial Min Value: Find a & b Real #s

[VertexOperator]

But why do the roots have to be 1 and -1?
How can I write this in a proper way?

 

[epenguin]

Shouldn't you try to work it out for yourself? There's hints galore here.

Yes I think so. I've even recovered from last night and think I've got it.

To save you time - you don't need to go into the detail of the solution of these equations.

 

[Dick]

But why do the roots have to be 1 and -1?
How can I write this in a proper way?

You have to start at the beginning. The first thing to work out is the reason why if a and b minimize a^2+b^2 then no root can be a simple root, i.e. multiplicity one.

 

[Dick]

This question is from the IMO btw.

There is really no reason to post Olympiad problems if you just want to watch other people working on them and not participate yourself, now is there? You could just look up past solutions. Do you have any real interest in this problem?

 

[VertexOperator]

I don't know which paper the question is from but my teacher said it is relevant to the maths course we are doing in school... it is doable by the skills I learned in this course but meant to be very challenging. Off course I am interested in knowing how to do it but there are many different solutions here no of which is final so I am just confused.

 

[Dick]

I don't know which paper the question is from but my teacher said it is relevant to the maths course we are doing in school... it is doable by the skills I learned in this course but meant to be very challenging. Off course I am interested in knowing how to do it but there are many different solutions here no of which is final so I am just confused.

I outlined one solution here. Just try following through on that or, if you have other solutions pick one of them. You can ask questions about it here as long as the question isn't "What's the answer?". You have to put some work into it.

 

[VertexOperator]

The solution I did (thanks to your hint) is "almost" really easy. If there is a single real root of multiplicity 2 then it must be 1 or -1. Then you just have to extremize a^2+b^2 subject to the linear constraint you get from putting 1 or -1 in as a root. That's easy. The glitch I ran into was excluding the case that at the minimal value of a^2+b^2 there might be two real roots of multiplicity 2. That made things a little ugly. But doable.

Thank you a lot everyone!
I went through all the posts now and this is what I got so far:
We divide the equation by x^2 to get Z^2+aZ+b=0 but why did epenguin say the discriminant is a^2-4b+8? Where did the 8 come from. My second question is why does x have to equal 1/x? If you can explain to me why x=1/x I will then understand why the solutions must be either 1 or -1.
But if the solutions are 1 or -1 we can simply sub into the original equation and solve them simultaneously and get a=0 and b=2 which is too large as you said so could you please explain the bold part :)
I am not used to the rules here, I thought I can ask for the answer but I get it now!

 

[Dick]

Thank you a lot everyone!
I went through all the posts now and this is what I got so far:
We divide the equation by x^2 to get Z^2+aZ+b=0 but why did epenguin say the discriminant is a^2-4b+8? Where did the 8 come from. My second question is why does x have to equal 1/x? If you can explain to me why x=1/x I will then understand why the solutions must be either 1 or -1.
But if the solutions are 1 or -1 we can simply sub into the original equation and solve them simultaneously and get a=0 and b=2 which is too large as you said so could you please explain the bold part :)
I am not used to the rules here, I thought I can ask for the answer but I get it now!

I can tell you about the second part. If f(x)=x^4+ax^3+bx^2+ax+1, so if x is a root, then f(x)=0. Now show that f(x)/x^4=f(1/x). Work it out, it's just simple algebra. So if f(x)=0 then f(x)/x^4=f(1/x)=0. So if x is a root then 1/x is also a root.

 

[VertexOperator]

wow, that is easy, I get 1+a/x+b/x^2+a/x^3+x^4 which is f(1/x)
Very nice so far, I am just still confused about the bold part!

 

[Dick]

wow, that is easy, I get 1+a/x+b/x^2+a/x^3+x^4 which is f(1/x)
Very nice so far, I am just still confused about the bold part!

First go back to my question in post 33. Suppose r is a simple multiplicity 1 root and a^2+b^2 is at it's minimal value. The graph of the polynomial near the root the just looks like a line crossing the x-axis, right? That means if you move the graph just a little by varying a and b, there will still be a real root because the graph will still cross the x-axis, just in a slightly different spot. That in turn means you can make a or b just a little smaller so you decrease a^2+b^2 and still have a root. So that would contradict a^2+b^2 being the smallest value where you still have a real root.

That means at the minimal value, all of the real roots must have multiplicity 2 or greater.

 

[epenguin]

We divide the equation by x^2 to get Z^2+aZ+b=0 but why did epenguin say the discriminant is a^2-4b+8? Where did the 8 come from. My second question is why does x have to equal 1/x? If you can explain to me why x=1/x I will then understand why the solutions must be either 1 or -1.

First qestion, you do not get that - it looks like you are taking (x² + 1/x²) to be the same thing as
(x + 1/x)²

 

[VertexOperator]

oh yes it should be x^2+1/x^2.

 

[epenguin]

oh yes it should be x^2+1/x^2.

My typo, I meant looks like you are taking (x² + 1/x²) to be the same thing as
(x + 1/x)²

 

[VertexOperator]

So can you guys tell me what to do next :)

 

[epenguin]

So can you guys tell me what to do next :)

Tell us what you did about what we have already suggested. E.g. When you told a very little in #37 I was able to put you right. Do something using these suggestions. Take a step.

I now think the approach of Dick explained in #40 and elsewhere is the easiest. Only personally I would start at a condition that you now know where there is a double root and vary from that as was explained. Personally I would not worry at the start about whether it has two double roots.

But it should also be doable and would be beneficial to also do via the solution of the equation about which you have been told in #6, 20, 22.

 

[Dick]

Tell us what you did about what we have already suggested. E.g. When you told a very little in #37 I was able to put you right. Do something using these suggestions. Take a step.

I now think the approach of Dick explained in #40 and elsewhere is the easiest. Only personally I would start at a condition that you now know where there is a double root and vary from that as was explained. Personally I would not worry at the start about whether it has two double roots.

But it should also be doable and would be beneficial to also do via the solution of the equation about which you have been told in #6, 20, 22.

That's a good idea. Just assuming you have one real root of multiplicity 2 not only eliminates some of the harder complications but it also leads pretty directly to the answer for the minimum of a^2+b^2. That should be satisfying. You can worry about why you can make that assumption later.

 

[VertexOperator]

When I assume we have one real root of multiplicity 2 I get a^2=-4(b+2), is that right?

 

[Dick]

When I assume we have one real root of multiplicity 2 I get a^2=-4(b+2), is that right?

Why do you think that?

 

[VertexOperator]

Because when it has multiplicity 2 the discriminant is 0 :(

 

[Dick]

Because when it has multiplicity 2 the discriminant is 0 :(

You should really say what quadratic that is supposed be the discriminant of. I'm guessing it's the one involving x+1/x. Then no, that doesn't work. x is a double root of the original equation. That doesn't mean x+1/x is a double root of that equation.

 

[VertexOperator]

Then how do I find the discriminant of the quartic equation? Do I even need the discriminant?

 

[Dick]

Then how do I find the discriminant of the quartic equation? Do I even need the discriminant?

No, you don't need any kind of discriminant. If you have one real root, say r, it turns out there aren't many possiblities for the value of that root. Look around at past posts and see if you can find a fact that can help you prove that.

 

[epenguin]

When I assume we have one real root of multiplicity 2 I get a^2=-4(b+2), is that right?

Then how do I find the discriminant of the quartic equation? Do I even need the discriminant?

That was explained and discussed in #22. Your equation is quite related to my expression there, but they are not in agreement so one of us has made a mistake.

Necessary? It is one way, worth pursuing in a complete job, but you have also been pointed to another way starting from conditions that give a double root.