2

I wonder if the answer to this question is true:

Having two functions $f(x)$, $g(x)$ where $f(x)$ has $N$ real roots, and $g(x)$ is positive for all $x$ (no real roots), does the product of $f(x)g(x)$ also have exactly $N$ roots?

For example. Let $f(x)$ is a polynomial and $g(x)=e^x$, then clearly roots of $$f(x)e^x=c(x-a_1)(x-a_2)\dots(x-a_N)e^x=0$$ are again only $a_1,a_2,\dots,a_N$.

Does this hold for general functions $f(x)$, $g(x)$ ($f(x)$ not necessarily a polynomial, $g(x)$ any positive function)? Thanks.

pisoir
  • 1,521
  • 2
    Of course. Because $g$ is positive you can divide the equation $fg = 0$ by $g$ and deduce that the roots of $fg$ are precisely the roots of $f$. – Frank Mar 19 '14 at 12:02
  • More generally, $x$ is a root of $f(x)g(x)$ exactly if $x$ is either a root of $f$ or a root of $g$. This works because over any field, because for a fields, $ab = 0$ exactly if either $a=0$ or $b=0$. – fgp Mar 19 '14 at 12:04
  • 1
    Note that in the example described, $g(x)=e^x$ has neither real roots nor complex roots. So the conclusion that roots of $f(x)g(x)=0$ are precisely the roots of $f(x)=0$ holds in both the field of real numbers and the field of complex numbers. – hardmath Mar 19 '14 at 12:08
  • @hardmath. Are you saying that the statement holds also for complex numbers? I.e., $f(x)$ has $N$ roots (complex or real), $g(x)$ has no roots (complex nor real), then $f(x)g(x)$ has $N$ roots. What about combination, e.g., $g(x)$ has only complex roots, no real roots? – pisoir Mar 19 '14 at 12:12
  • @pisoir: I was addressing the example, where $g(x)=e^x$, but the first part of your Comment correctly generalizes this. Since the roots of $f(x)g(x)=0$ are precisely the union of roots of $f(x)=0$ and of $g(x)=0$, when $g(x)=0$ has no real roots, the only real roots of $f(x)g(x)=0$ are the real roots of $f(x)=0$. – hardmath Mar 19 '14 at 12:23

2 Answers2

1

Let $h(x)=f(x)\cdot g(x)\forall x\in\Bbb R$

Now if $h(x_0)$ is $0$, i.e, $x_0$ is a root of $h(x)$,

$$f(x_0)\cdot g(x_0)=0$$

$$f(x_0)=0 \text{ or }g(x_0)=0$$

But we are given that $g(x_0)\ne0$

$\therefore f(x_0)=0$

Therefore $h(x')=0 \iff f(x')=0\implies$ $h$ and $f$ have the same roots.

Guy
  • 8,857
  • 1
  • 28
  • 57
0

Consider $fg=0$. Since $g\neq 0$ hence dividing by $g$ makes sense and therefore you'll get the roots of $fg$ to be same as roots of $f$.

wanderer
  • 2,928