Multiplicity of a root of a polynomial

Question

:)

It's true that, if a polynomial has a root (let's say, k, for example) with multiplicity n (n>1, for n integer), then it's true that the derivate polynomial have k as a root with multiplicity n-1.

The reciprocal is always true? I.e., it's true that if a derivative polynomial have k as a root with multiplicity n-1, then the original polynomial have k as a root with multiplicity n? If this is not always true, in which conditions this is true?

Thanks every one! :)

I need a translation request but here you have good things https://fr.wikipedia.org/wiki/Racine_d'un_polyn%C3%B4me#Crit.C3.A8re_diff.C3.A9rentiel_pour_la_multiplicit.C3.A9_d.27une_racine — ParaH2, Jun 29 '15 at 01:17
The converse is true if you assume that the original polynomial actually vanishes at the point in question. — David C. Ullrich, Jun 29 '15 at 01:24

score 2 · Answer 1 · answered Jun 29 '15 at 01:58

2

The complete story is this:

The following conditions are equivalent for a polynomial $f(x)$:

(i) $\;f(x)$ is divisible by $(x-a)^n$;

(ii) $\;f(a)=f'(a)=\dots=f^{(n-1)}(a)=0$.

It is a generalisation of the well-known result:

$f(x)$ has $a$ as a root if and only if $f(x)$ is divisible by $x-a$.

answered Jun 29 '15 at 01:58

Bernard

175,478

It will have $n$ roots – Aakash Kumar Jul 13 '16 at 16:19
@Aakash Kumar: What do you mean by $n$ roots here? – Bernard Jul 13 '16 at 16:38
There will be n roots of f(x) – Aakash Kumar Jul 13 '16 at 16:48
@Aakash Kumar: Yes, all equal to $a$. – Bernard Jul 13 '16 at 19:44
@Bernard I need reference of this result. I will be very thankful to you. – Tusif Ahmed Nov 13 '16 at 18:43

score 1 · Answer 2 · answered Jun 29 '15 at 02:02

Consider the simple quadratic $y=x^2$. This has a root at $x=0$ of multiplicity 2.

Its derivative, $y'=2x$, has a root at $x=0$ of multiplicity 1. So far so good.

Now consider $y=x^2 + 1$. This has no real roots at all; it does not have a root of any multiplicity at $x=0$. However its derivative is again $y'=2x$, as above.

The point is that for a given derivative function $y'=f'(x)$, there are infinitely many functions of which that is the derivative, differing only by a constant term.

For any root (with multiplicity n) of the derivative function $f'(x)$, we can always choose a constant term so that the base function $f(x)$ is also zero at that root, and that root will indeed have multiplicity $n+1$.

However we may have a derivative function with two or more roots, where it is not possible to choose a single constant term for the antiderivative such it has roots at all the same places.

For example if $f'(x)=3x^2 - 3$, this has roots at +/-1 of multiplicity 1. Its antiderivatives are of form $f(x)=x^3 - 3x + C$. In order for -1 to be a root, C must be -2 (and the root does have multiplicity 2); whereas for +1 to be a root, C must be +2. There is no choice of C for which each root of f'(x) corresponds to a root of f(x).

TL;DR: if f'(x) has a root of multiplicity n at x=a, and x=a is also a root of f(x), then that root will have a multiplicity of n+1. But it needn't be a root of f at all.

score 0 · Answer 3 · answered Jun 29 '15 at 02:25

suppose $f$ is a polynomial of degree $m$ and $f'$ has a root $k$ of multiplicity $k.$ then we can write $$f'(t) = (t-a)^{n-1}q(t), q(a) \neq 0. \tag 1 $$ integrating $(1)$ from $a$ to $x,$ we get $$\begin{align}f(x)-f(a) &= \int_a^x(t-a)^{n-1}q(t)\, dt \\ &=\int_0^{x-a}t^{n-1}q(t+a)\, dt\\ &=\int_0^{x-a} t^{n-1}\left(q(a)+a_1t+a_2t^2 +\cdots+a_{m-n+1}t^{m-n+1}\right)\, dt\\&=(x-a)^n\left(\frac{q(a)}n+\frac{a_1}{n+1}(x-a)+\cdots+a_{m-n+1}(x-a)^m\right)\end{align}$$ therefore if $f(a) = 0,$ then the zero $x=a$ of $f$ has multiplicity $n.$

Multiplicity of a root of a polynomial

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