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Let $$f(x) = a+b(x+3)^r+c(x+3)^r(x+2)^r+d(x+3)^r(x+2)^r(x+1)^r,$$ where $r>1$ and $a,b,c,d$ are real numbers. I need to prove that $f(x)$ can't have more than $3$ positive zeros.

Attempts -

  1. By Mathematica, we see that it can have at most 2 positive zeros.
  2. This problem seems to be related to the Descartes' sign problem, where the number of positive zeros terns to be not more than the number of change of signs of coefficient. I tried the concepts of proof of Descartes' sign theorem, but it didn't work for me.
  3. The induction and Roll'e theorem approach also didn't work as it needs the proof of this: $$Z(r g(x)+(x+3)g'(x))\le Z(g(x)),$$ for any $r>1$ and a function $g(x)=c(x+2)^r+d(x+2)^r(x+1)^r$, where $Z$ denotes the number of positive roots.
  4. This can also be done by proving that the wronskian of the functions $$1, (x+3)^r, (x+3)^r(x+2)^r, (x+3)^r(x+2)^r(x+1)^r$$ is non-zero for all $x>0$.

Any help is appreciated.

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  • $\begingroup$ Zeros or roots? $\endgroup$ Commented Oct 25, 2025 at 13:36
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    $\begingroup$ Both are same. Whats the difference $\endgroup$ Commented Oct 25, 2025 at 13:44
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    $\begingroup$ Functions have zeros. Polynomials have roots. Functions can be polynomial but they are not the same as polynomials $\endgroup$ Commented Oct 25, 2025 at 14:12

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Let $$f_1(x):=\frac{f'(x)}{r(3+x)^{r-1}} =b+c (2 x+5) (x+2)^{r-1}+d \left(3 x^2+12 x+11\right) (x+1)^{r-1} (x+2)^{r-1},$$ $$f_2(x):=\frac{f_1'(x)}{(2+x)^{r-2}(2 r x+5 r-1)} \\ =c+\frac{d \left(6 r x^3+33 r x^2+58 r x+33 r-3 x^2-10 x-9\right) (x+1)^{r-2}}{2 r x+5 r-1},$$ $$f_3(x):=f_2'(x)\frac{(2 r x+5 r-1)^2}{d(1+x)^{r-3} } \\ =s^3 \left(12 x^4+96 x^3+281 x^2+356 x+165\right)+s^2 \left(36 x^4+264 x^3+699 x^2+784 x+311\right)+s \left(36 x^4+240 x^3+574 x^2+576 x+202\right)+12 \left(x^2+3 x+2\right)^2 $$ if $d\ne0$, where $s:=r-1>0$.

If $f$ had $>3$ positive zeroes, then $f_1$ would have $>2$ positive zeroes, $f_2$ would have $>1$ positive zeroes, and $f_3$ would have $>0$ positive zeroes -- but $f_3(x)$ is manifestly $>0$ for $x>0$ if $d\ne0$. So, $f$ has $\le3$ positive zeroes if $d\ne0$.

If now $d=0$, then $f_2=c$. So, if $c\ne0$, then $f_2$ has no positive zeroes and hence $f$ has $\le2\le3$ positive zeroes -- if $d=0$ but $c\ne0$.

If now $c=d=0$, then $f_1=b$. So, if $b\ne0$, then $f_1$ has no positive zeroes and hence $f$ has $\le1\le3$ positive zeroes -- if $c=d=0$ but $b\ne0$.

If now $b=c=d=0$, then $f=a$. So, if $a\ne0$, then $f$ has $0\le3$ positive zeroes -- if $b=c=d=0$ but $a\ne0$.

If finally $a=b=c=d=0$, then $f=0$.

Thus, $f$ has $\le3$ positive zeroes -- unless $a=b=c=d=0$, in which case all real numbers are zeroes of $f$. $\quad\Box$

It is not true "that it can have at most 2 positive zeros". Indeed, any polynomial of degree $3$ can be written as $f$ for $r=1$. For instance, $f(x)=(x-1)(x-2)(x-3)$ if $r=1$ and $(a,b,c,d)=(-120, 60, -12, 1)$. So, $f$ can have $3$ positive zeroes if $r=1$. By continuity, $f$ can have $3$ positive zeroes at least in the case when $r$ is slightly greater $1$ (if you want to insist that $r>1$). For instance, here is the graph $\{(x,f(x))\colon0<x<4\}$ for $r=21/20$ and $(a,b,c,d)=(-120, 60, -12, 1)$:

enter image description here

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    $\begingroup$ @losif, how are you constructing $f_2(x)$ and $f_3(x)$? I want to prove it in general. This idea won't work for the case of five term and six term cases. I want to prove it by induction (by approach 3). We have $f(x) = a+(x+3)^rg(x)$. Can we show the number of positive zeros of $f(x)$ is not more than that of $g(x)$? $\endgroup$ Commented Oct 26, 2025 at 2:06
  • $\begingroup$ @VSP : The same idea should work, in principle, for any number of terms. Taking $f'$, we have $f'(x)=bg_{11}(x)+cg_{12}(x)+dg_{13}(x)$ for certain function $g_{11},g_{12},g_{13}$. Dividing both sides of this equation by $g_{11}(x)$, we get $f_1(x)=b+ch_{12}(x)+dh_{13}(x)$, thus killing $a$ and getting three remaining coefficients $b,c,d$ instead of the initial four coefficients $a,b,c,d$. Continuing thus, we kill all the coefficients $a,b,c,d$ and get $f_3$. Anyhow, your question has been fully answered. If you have additional questions, consider posting them separately. $\endgroup$ Commented Oct 26, 2025 at 2:25
  • $\begingroup$ To use this proof we must know $h_{12}(x)$, $h_{13}(x)$ and so on. For more terms case, its difficult to find these functions. That’s why this proof is not that useful. $\endgroup$ Commented Oct 26, 2025 at 2:34
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    $\begingroup$ @VSP : Your post did not say anything about allowed methods. It also did not mention more general settings. If this method does not work in a more general case, unstated in your post, it is all the more of a reason to post the more general questions separately. That would be better in a number of respects. One of them would be giving better exposure to what is really of interest to you. $\endgroup$ Commented Oct 26, 2025 at 2:52

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