Polynomial Roots

**sehrishqau** · November 23rd 2007, 05:37 PM

If A Polynomial Of Degree Greater Than Or Equal To Two With Coefficient From The Set Of Real Numbers Has All Roots In Set Of Real Numbers.then Prove That All The Roots Of Its Derivatives Are Also Real Numbers.

**Opalg** · November 24th 2007, 12:34 AM

Originally Posted by sehrishqau

If A Polynomial Of Degree Greater Than Or Equal To Two With Coefficient From The Set Of Real Numbers Has All Roots In Set Of Real Numbers.then Prove That All The Roots Of Its Derivatives Are Also Real Numbers.

Hint: Rolle's theorem. Between each pair of roots of a real polynomial there must be a root of the derivative.

**CaptainBlack** · November 24th 2007, 12:34 AM

Originally Posted by sehrishqau

If A Polynomial Of Degree Greater Than Or Equal To Two With Coefficient From The Set Of Real Numbers Has All Roots In Set Of Real Numbers.then Prove That All The Roots Of Its Derivatives Are Also Real Numbers.

A ploynomial of degree n has n roots, and here they are all real.

Rolle's theorem tells us that here there is a zero of the derivative between
each pair or real roots. Hence for our polynomial we have n-1 zeros of the
derivative.

But the derivative of our polynomial is a polynomial of degree n-1, and we
have shown that it has n-1 zeros.

RonL

**ThePerfectHacker** · November 24th 2007, 07:02 PM

There is one more issue that needs to be addressed. CaptainBlank is assuming that f(x) has exactly n roots. Then by using Opalg's argument with CaptainBlank's commentary we see there are n-1 real roots for f'(x) and so these are its only roots for deg f'(x) = n-1. But the situation changes if there is not exactly n real roots. But that (this problem) is not really so bad. Say that f(x) = (x-a)^n*(x-b)^m where 'a' is a root of multiplicity n and 'b' of multiplicity m. Then $f'(x) = n(x-a)^{n-1}(x-b)^m + m(x-a)^{n-1}(x-b)^{m-1}$ . Thus, 'a' is a zero (of f'(x)) of multiplicity n-1 and 'b' is a zero of multiplicity 'b' is a zero of multiplicity m-1. Now using Opalg's argument there is another zero between 'a' and 'b'. So far we have (n-1)+(m-1)+1 = n+m-2 zeros (counting multiplicity) but since the number of complex zeros must remain even it means the last zero cannot be complex. Thus, there are n+m-1 zeros altogether (counting multiplicity). Thus, f'(x) has only real zeros. Now in general suppose we have $f(x)=(x-a_1)^{n_1}...(x-a_k)^{n_k}$ then $a_1,...,a_k$ are zeros of multiplicity $(n_1-1),...,(n_k-1)$ . Using Opalg's hint we also have zeros between $n_1,...,n_k$ and so there are $k-1$ zeros. Thus in total with multiplicity we have $(n_1+...+n_k)-k-2$ since the number of complex solutions must be even it means the last remaining zero must be real. Thus $f'(x)$ has only real zeros.

**CaptainBlack** · November 24th 2007, 11:45 PM

Originally Posted by ThePerfectHacker

There is one more issue that needs to be addressed. CaptainBlank is assuming that f(x) has exactly n roots. Then by using Opalg's argument with CaptainBlank's commentary we see there are n-1 real roots for f'(x) and so these are its only roots for deg f'(x) = n-1. But the situation changes if there is not exactly n real roots. But that (this problem) is not really so bad. Say that f(x) = (x-a)^n*(x-b)^m where 'a' is a root of multiplicity n and 'b' of multiplicity m. Then $f'(x) = n(x-a)^{n-1}(x-b)^m + m(x-a)^{n-1}(x-b)^{m-1}$ . Thus, 'a' is a zero (of f'(x)) of multiplicity n-1 and 'b' is a zero of multiplicity 'b' is a zero of multiplicity m-1. Now using Opalg's argument there is another zero between 'a' and 'b'. So far we have (n-1)+(m-1)+1 = n+m-2 zeros (counting multiplicity) but since the number of complex zeros must remain even it means the last zero cannot be complex. Thus, there are n+m-1 zeros altogether (counting multiplicity). Thus, f'(x) has only real zeros. Now in general suppose we have $f(x)=(x-a_1)^{n_1}...(x-a_k)^{n_k}$ then $a_1,...,a_k$ are zeros of multiplicity $(n_1-1),...,(n_k-1)$ . Using Opalg's hint we also have zeros between $n_1,...,n_k$ and so there are $k-1$ zeros. Thus in total with multiplicity we have $(n_1+...+n_k)-k-2$ since the number of complex solutions must be even it means the last remaining zero must be real. Thus $f'(x)$ has only real zeros.

Fundamental theorem of algebra and the stated condition that all the roots
are real implies n real roots for the nth degree polynomials in question.

RonL

**ThePerfectHacker** · November 25th 2007, 07:34 AM

Originally Posted by CaptainBlank

Fundamental theorem of algebra and the stated condition that all the roots
are real implies n real roots for the nth degree polynomials in question.

What about $x^n$ it only has 1 root not n. We need to show that $f'(x)$ has real roots. The situation can happen that $1+i,1-i$ are roots in that case $f'(x)$ has real coeffcients, but not real roots.

Consider for example $f(x) = (x-1)^3(x+1)^4$ this polynomial has real roots only $1,-1$ . We want to show $f'(x)$ has only real roots, we know that there is a root between -1 and 1 but since degree is 6 it can still be that the other roots are complex conjugates in that case f'(x) will have real coefficients but not real roots. So we need to use what I mentioned above.

Now consider $f(x)=(x-1)(x-2)(x-3)$ it has 3 real roots as much as its degree so its derivative has zeros between 1 and 2 , 2 and 3 hence it is two real solutions equal to its degree and we can stop right there. But this is just a special case because this polynomial has a many roots as its degree. But in the above case this does not work. Thus we need to consider this possibility in the proof.

**CaptainBlack** · November 25th 2007, 09:03 AM

Originally Posted by ThePerfectHacker

What about $x^n$ it only has 1 root not n. We need to show that $f'(x)$ has real roots. The situation can happen that $1+i,1-i$ are roots in that case $f'(x)$ has real coeffcients, but not real roots.

Consider for example $f(x) = (x-1)^3(x+1)^4$ this polynomial has real roots only $1,-1$ . We want to show $f'(x)$ has only real roots, we know that there is a root between -1 and 1 but since degree is 6 it can still be that the other roots are complex conjugates in that case f'(x) will have real coefficients but not real roots. So we need to use what I mentioned above.

Now consider $f(x)=(x-1)(x-2)(x-3)$ it has 3 real roots as much as its degree so its derivative has zeros between 1 and 2 , 2 and 3 hence it is two real solutions equal to its degree and we can stop right there. But this is just a special case because this polynomial has a many roots as its degree. But in the above case this does not work. Thus we need to consider this possibility in the proof.

The problem as I see it is that roots are being counted with multiplicities. Which can of course be sorted out by perturbing the polynomials, the using a suitable limiting argument. But not a good idea

RonL

**Opalg** · November 25th 2007, 09:32 AM

I agree that multiplicity must be taken into account, for the reasons that ThePerfectHacker gives. However, if a polynomial has a root of multiplicity n, its derivative will have that same root with multiplicity n–1: $\frac d{dx}\bigl((x-a)^nq(x)\bigr) = (x-a)^{n-1}\bigl(nq(x)+(x-a)q'(x)\bigr)$ . So multiplicity ought not to create any problems.

**ThePerfectHacker** · November 25th 2007, 09:37 AM

Originally Posted by Opalg

So multiplicity ought not to create any problems.

It does not create any serious problem (post #4 shows it is easily correctable).

What about the following problem? Let $\bar F$ be the algebraic closure of $F$ with $[\bar F : F] < \infty$ and $f(x)\in F[x]$ has roots in $F$ then does $f'(x)$ (the algebraic derivatice) have roots in $F$ also?

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