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For a polynomial $P(x) = (x-x_1)(x-x_2)\cdots (x-x_n)$ with distinct real zeroes, $x_1 < x_2<\cdots < x_n$, prove or disprove that all zeroes of $f(x) := P'(x) - kP(x)$ are real and that for any two distinct zeroes $y_0, y_1$ of $f(x)$, there is a zero of $P(x)$ between them.

I think it would be useful to apply Rolle's theorem to the function $e^{-kx} P(x)$, which has the same zeroes as $P(x)$. The derivative is $e^{-kx}f(x)$. The issue is that Rolle's theorem only ascertain $n-1$ real and distinct zeroes of $e^{-kx}f(x)$, while if $k\neq 0$, there may be another zero. How can I prove that this extra zero is real and that it is smaller than the smallest zero of $P(x)$ or larger than the largest zero of $P(x)$ (this would satisfy the requirement since the $n-1$ zeroes of $e^{-kx}P(x)$ guaranteed by Rolle's theorem are in-between two consecutive zeroes of $P(x)$).

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  • $\begingroup$ If you have an extra zero which is complex you would have to have two extra zeroes, as the coefficients and all the other zeroes are real. So if it exists it must be real. $\endgroup$
    – Peter
    May 14 at 2:22
  • $\begingroup$ Alt. hint: since there are no multiple roots, $\,P'\,$ changes sign at each $\,x_k\,$, so $\,f\,$ changes sign at each $\,x_k\,$. $\endgroup$
    – dxiv
    May 14 at 2:30

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Assuming $n>1.$

$P$ and $P'$ have no common zeroes so $0=P'(x)-kP(x)\iff k=\frac {P'(x)}{P(x)}=\sum_{j=1}^n\frac {1}{x-x_j}.$

Let $1\le j<n.$ For each $i$ such that $j\ne i\ne j+1,$ the function $\frac {1}{x-x_i}$ is continuous, and hence bounded, on the interval $ [x_j,x_{j+1}],$ while the function $g_j(x)=\frac {1}{x-x_j}+\frac {1}{x-x_{j+1}}$ is continuous on $(x_j,x_{j+1})$ with $\lim_{x\to x_j^+}g_j(x)=+\infty$ and $\lim_{x\to x_{j+1}^-}g_j(x)=-\infty.$ Therefore $\{P'(x)/P(x):x\in (x_j,x_{j+1})\}=\Bbb R.$ So there exists $y_j\in (x_j,x_{j+1})$ with $P'(y_j)/P(y_j)=k.$

The polynomial $f(x)=P'(x)-kP(x)$ has degree $n$ or less and has real co-efficients and has $n-1$ zeroes $y_1,...,y_{n-1}$ so the $n$th zero of $f(x),$ if it exists, must be in $\Bbb R.$

Because if $z\in \Bbb C\setminus \Bbb R$ and $f(z)=0$ then $\bar z\ne z$ and $0=\overline {f(z)}=f(\bar z)$ (because $f$ has real co-efficients), but then $f$ is a polynomial of degree $n$ or less with at least $n+1$ zeroes $z,\bar z,y_1,...,y_{n-1},$ implying $f$ is identically $0.$ But $f(x_1)=\prod_{j=2}^n(x_1-x_j)\ne 0,$ a contradiction.

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  • $\begingroup$ Could you take a look at the following partial differential equation and integration problem if you have time: math.stackexchange.com/questions/4449990/…? Seeing as you're good at calculus, perhaps you might know how to solve it? $\endgroup$
    – user3472
    May 15 at 19:26

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