If the roots of a polynomial p are real, then the roots of p' are real.

[tylerc1991]

Homework Statement

Let p be a polynomial. Show that the roots of p' are real if the roots of p are real.

Homework Equations

The Attempt at a Solution

So we start with a root of p', call it r. We want to show that r is real. Judging by the condition given, I am assuming that we have to relate this root of p' to a root of p so that we can be guaranteed r is real. I am pretty stuck on this one. I've tried to see how Taylor's theorem could help here (since this is an exercise from the section where Taylor's theorem is introduced), but no such luck. Maybe some application of l'Hopital's, since this relates a function to it's derivative? I've tried this but have come up blank. Any words of wisdom would be seriously appreciated!

 

[voko]

If all the roots are real, how many minima and maxima will the polynomial have?

 

[tylerc1991]

If all the roots are real, how many minima and maxima will the polynomial have?

If all the roots are real, then the number of minima and maxima depend on the multiplicity of the roots. After thinking about a couple of examples, I'll try to generalize here. I'm not 100% sure this is right.

If all the roots of a polynomial of degree n are real, then there are either n-1 maxima / minima, or just 1 maximum / minimum.

With that, say we have a polynomial p of degree n. Then p' is a polynomial of degree n-1. From the fundamental theorem of algebra, there are n roots of p counted with multiplicity. By assumption, all of these n roots are real. From above (excluding the case of just 1 maximum / minimum), there are n-1 maxima / minima of p. So there are n-1 points at which p' vanishes, which means none of the roots of p' can be complex.

I think that is sort of the idea you were shooting for? There are 2 things now. First, the case of where there is only one maximum / minimum is worrysome. Second, it seems like the problem has now transformed into: Show that there are n-1 maxima / minima of a polynomial of degree n (excluding those special cases).

 

[voko]

If any root has multiplicity > 1, then will the derivative have the same root? What multiplicity? And between two adjacent roots, there is one extremum.

 

[tylerc1991]

If any root has multiplicity > 1, then will the derivative have the same root? What multiplicity? And between two adjacent roots, there is one extremum.

If some root A has multiplicity m > 1, then A will be a root of p' with multiplicity m - 1. Also, I understand that between two simple roots is a root of p'. Now it becomes about counting how many roots p' has. We know that p has n roots by the FTA, and by assumption these are all real. Say there are s simple roots and m roots with multiplicity > 1, so that s + m = n. From above, this means that there are s-1 roots of p'. Also, there are m roots of p'. So s-1+m = n-1, and we are done.

Of course when I write it I will be more formal, but does this sound right? I con't see anything horribly wrong. Thank you so much for your help!

 

[voko]

If some root A has multiplicity m > 1, then A will be a root of p' with multiplicity m - 1. Also, I understand that between two simple roots is a root of p'.

Between ANY two distinct roots is at least one extremum. So you need to order by value. If the smallest root r₁ has multiplicity m₁, the next smallest root r₂ multiplicity m₂, then the derivative will have at least multiplicity m₁ - 1 at r₁ and at least multiplicity m₂ - 1 at r₂, and at least one root in between, so all together m₁ - 1 + 1 + m₂ - 1 = m₁ + m₂ - 1. And so on.