general theorem on roots of a polynomial needed to show it's identically zero.

Question

Polynomial degree k, one variable, if it's zero at k+1 values, then it's identically zero. Can someone point me to a proof of this?
I know derivatives at points can count as these roots (if k-degree f(x) vanishes at x=a and k derivatives vanish at x=a, it vanishes). So, can someone point me to a more general theorem incorporating the derivatives?
What if, for example, there's a k degree polynomial (in one variable) that vanishes at x=a, also at x=b, and k-1 derivatives vanish at x=a. This is a total of k+1 "roots"; does this count? What if some of those x=a vanishings are other x values?

I understand the first one, it could be linked to the fundamental theorem of algebra (any $n$ degree real polynomial has at most $n$ roots). I don't get the other two, though: what does "derivatives at points can count as these roots" mean? Maybe, that if $f(x_0)=0$ then $f^{(k)}(x_0)=0$ too for $k\leq n$? — yellon, Apr 16 '15 at 13:18
Please give more context, What type of numbers are the coefficients? Are they in a field like $,\Bbb Q,$ or $,\Bbb C,$ or integral domain like $,\Bbb Z,,$ or do you allow more general coefficient rings? — Bill Dubuque, Apr 16 '15 at 14:13
If $f$ is a polynomial with degree $k > 1$ satisfying $f(a) = f'(a) = 0$, then $f(x) = (x-a)^2 g(x)$ for some polynomial $g$ (of degree $k-2$); you can iterate this down to as many derivatives as you need. Proving this is actually easy: if $f(x) = (x-a)h(x)$, then $f'(x) = h(x) + (x-a)h'(x)$, so that $h(a) = f'(a) = 0$, and hence $h(x) = (x-a)g(x)$. — Pedro M., Apr 16 '15 at 14:15
All real numbers here, only. And what I meant by #2 is: if f(x) is a degree k polynomial, and vanishes at a, and k derivatives of f vanish at a, f is identically zero (proof please?). I believe also, that if f(x) is say, a 3rd degree polynomial, and: f(a)=f'(a)=f(b)=f'(b)=0 for some a, b, f is identically zero. Also pretty sure that for a 2nd degree polynomial, f(a)=f(b)=f'(a)=0 implies f(x) identically 0. Continuing other possibilities, is there some theorem that collects these all? — Nights, Apr 16 '15 at 15:58
Thank you IIham for the link and Pedro for that nice statement/proof. — Nights, Apr 16 '15 at 22:13
@PedroM. I am trying to show by induction that your statement mentioned holds for n derivatives, that is, that if $f(a)=f'(a)= \cdots =f^{n}(a)=0$, then $f(x)=(x-a)^{n+1}g(x)$ where $g(x)$ is degree $k-(n+1)$ but I can't seem to get the inductive step. — Nights, Apr 17 '15 at 01:38

score 0 · Answer 1 · answered Apr 17 '15 at 13:46

Proposition: Let $f$ be a polynomial, and let $a \in \mathbb{R}, n \in \mathbb{N}$. If for each $0 \leq k \leq n-1$ we have $f^{(k)}(a) = 0$, then $$f(x) = (x-a)^n g(x)$$ for some polynomial $g$.

Obs: $g$ may be identically zero (which implies that $f = 0$ as well). Since we either have $f = 0$ or $\deg(g) = \deg(f) - n$, this result should be enough for most of your needs.

Proof (by induction on $n$): The case $n = 0$ is easy. Suppose the result holds for $n-1$, and suppose $f(a) = f'(a) = \ldots = f^{(n-2)}(a) = f^{(n-1)}(a) = 0$. By induction, we know in particular that $f(x) = (x-a)^{n-1}h(x)$, so that $f'(x) = (x-a)^{n-2}((n-1)h(x) + (x-a)h'(x))$.

On the other hand, the polynomial $f'(x)$ also satisfies the hypothesis for $n-1$, so by induction we have $f'(x) = (x-a)^{n-1}q(x)$ for some polynomial $q$. Hence, $$(x-a)^{n-2}((n-1)h(x) + (x-a)h'(x)) = (x-a)^{n-1}q(x), $$ or $(n-1)h(x) = (x-a)(q(x) - h'(x)),$ and thus $h(x) = (x-a)g(x)$ for $g(x) = \frac{q(x) - h'(x)}{n-1}.$ Therefore,

$$f(x) = (x-a)^{n-1}h(x) = (x-a)^ng(x).\tag*{$\blacksquare$}$$

general theorem on roots of a polynomial needed to show it's identically zero.

1 Answers1