3
$\begingroup$

Show that if the polynomial that has all real zeros but without multiple roots, has the properties that all its derivatives till the derivative of order $n-1$ have also real zeros.

Can somebody help me with this problem? I thought that if we take $f(x)$ that satisfies the given conditions we can write it as $f(x)=c(x-x_1)(x-x_2) \cdots (x-x_n)$

Where $x_1,x_2, \cdots ,x_n$ are the real zeros of $f(x)$

Then we have $$f'(x)=\frac{cf(x)}{x-x_1}+\frac{cf(x)}{x-x_2}+ \cdots +\frac{cf(x)}{x-x_n}$$

Where $f'(x_i) \ne0$ for $i \in {1,2, \cdots , n }$

But I don't know how to prove that $f'(x)$ has real zeros. Or should I approach it differently?

I'm really sorry for any mistakes in my English. It's not my native language.

$\endgroup$
  • 7
    $\begingroup$ Hint: Rolle's theorem. $\endgroup$ – Robert Israel Jun 21 '19 at 14:36
  • $\begingroup$ It is often useful to draw a diagram. $\endgroup$ – Mark Bennet Jun 21 '19 at 14:54
3
$\begingroup$

The derivative of a smooth (all its derivatives exist) function between two zeros of degree $1$ must change sign. Why?

Apply this argument inductively (the second derivative is the derivative of the first derivative, and so on).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ A small comment to address the repeated root(s) caveat would complete the argument. $\endgroup$ – Deepak Jun 21 '19 at 14:42
  • $\begingroup$ Can you explain what you mean by the repeated roots caveat, @Deepak? $\endgroup$ – Alexander Geldhof Jun 21 '19 at 15:08
  • $\begingroup$ I was just wondering why the question bothered to specifically exclude multiple roots. But having thought about your argument, it can be applied even in this case (the root(s) of the derivative will coincide with the multiple root(s) of the original polynomial rather than just lying in between them). So please disregard my comment. $\endgroup$ – Deepak Jun 21 '19 at 15:37
1
$\begingroup$

A polynomial with no multiple roots alternates in sign and has extrema between two successive roots. Hence its derivative has at least as many roots, minus one.

On the other hand, it cannot have more, as the number of roots cannot exceed the degree.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Since $f$ is continuous, its first derivative must reach zero at least once inside the interval $[x_{I}, x_{I+1}]$. This accounts for all its zeros since there's exactly $n-1$ of them. An induction argument completes the proof for higher order derivatives.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.