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.
Best Answer
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).