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The fundamental theorem of algebra states that:

Every non-zero, single-variable, degree $n\,$ polynomial with complex coefficients has, counted with multiplicity, exactly $n\,$ complex roots.

The term I want to understand is multiplicity. Now, I already know the following:

The multiplicity of a root $x_0$ of a polynomial equation $p(x) = 0\,$ tells us how many times the factor $(x - x_0)\,$ divides the polynomial $p(x)$. That's fine.

But what I want to have, is an intuition. The definition of a root is: $x_0\,$ is a root of $p(x) = 0\,$ if and only if $p(x_0) = 0$. So, I don't understand how a root could have a property called multiplicity. For example, the multiplicity of a root $x_0$ cannot be the number of times $x_0\,$ solves the equation $p(x) = 0\,$ because a root cannot really solve an equation more than once.

Nor does it make sense as the number of times the graph of the function $y = p(x)\,$ meets the $x$-axis at the $x$-coordinate that is $x_0$, because they meet at a given point only once [because of the definition of a function]. But I do know of the geometric intuition: If multiplicity is odd, then the axis is crossed, otherwise it touches and comes back, and the higher the multiplicity, the closer $p(x)\,$ stays near the $x$-axis in the neighborhood of $x = x _0$, and so on.

What I want to have is an intuition, an answer to the question:

What is the multiplicity of a root of a polynomial equation the multiplicity of?

Can somebody plz help me on this one?

NOTE: Here, I am using multiplicity the same way I would use the term frequency. Plz correct me if this usage is wrong.

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  • $\begingroup$ We do have $p(x) = k\prod_{i}(x-r_i)$ $\endgroup$ – King Tut Mar 31 '18 at 16:25
  • $\begingroup$ The multiplicity can be determined also with the derivates, but this is not intuitive either. $\endgroup$ – Peter Mar 31 '18 at 16:27
  • $\begingroup$ A root can solve an equation more than once if you ask for "how often" rather than for a yes/no answer. $\endgroup$ – Hurkyl Apr 1 '18 at 1:21
  • $\begingroup$ @Hurkyl How can one ask "how often" does a root solve an equation? Can you explain, plz? $\endgroup$ – Truth-seek Apr 1 '18 at 2:21
  • $\begingroup$ Do not think of the multiplicity of a root as a way of counting "how often" that root "solves the equation". Rather, think of the following, related question: how many roots does the polynomial have? The correct answer is found by counting the roots with multiplicity. The multiplicity of a particular root is a weight we give to that root when counting roots, so that the answers come out nice and everything works algebraically. It's a form of book-keeping which happens to also be quite geometrically intuitive. $\endgroup$ – Will R Apr 1 '18 at 2:46
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For polynomials over real (and complex) numbers, I have always found the following intuitive pictures helpful. They show different polynomials intersecting the line y=0. enter image description here

(own work) For simplicity, assume all roots are real. Consider polynomials with roots of multiplicity 1,2,3 and 4 as shown in the upper row of graphs in the picture. By perturbing the coefficients of these polynomials a tad bit (adding small numbers to them) you can see how the roots split into groups of 1 (just a shift),2,3, and 4 distinct roots. The perturbed polynomials are shown in the lower row. When solving for polynomial roots using numerical methods, the roots you will find will in practice correspond to such perturbed polynomials since you cannot represent polynomial coefficients with arbitrary precision.

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The fundamental theorem of algebra is stated in the complex domain ($\mathbb C$). This reminded us that maybe we could have an answer with complex ($\mathbb C$) viewpoint.


Let $f(z)$ be a polynomial over $\mathbb C$, and $u\in\mathbb C$ be a root of $f(z)$ of multiplicity $m$. This is expressed by an equation: $$f(z)=(z-u)^m g(z)$$ where $g(z)$ is a polynomial that does not have $u$ as its root.

Clearly $f$ brings $u$ to $0$, from one complex plane to another. This characterizes the notion of root geometrically:

illustrate root

Since $f$ is continuous, points near $u$ are mapped to some points near $0$. By complex analysis, a small loop around $u$ should be mapped to another loop winding $m$ times around the origin. The following picture shows the case when $m=2$:

illustrate winding

Example:

For every complex number $a$ we can assign a color for it according to its angle $\operatorname{arg} a$. The brighter color means that the angle is closer to $0^\circ$:

reference of color <-> arg

Let $f(z)=(z+1)^3(z-2)$. For every point $z$ on the complex plane we paint the corresponding color for $\operatorname{arg}f(z)$:

illustrate quartic f

The horizontal axis is real and the vertical one is imaginary. The small green squares indicate the unit length.

When we travel along a path $\gamma$ around $z=-1$, which is the root of multiplicity $3$, the color changes (white to black) three times. This shows that the corresponding path $f(\gamma)$ loops three times around the origin of the codomain. The discussion of the other root $z=2$ is similar.

illustrate color change

This characterizes the notion of multiplicity geometrically, so I will try to prove this property (or, at least, give an idea of how it happens, because I'm not really familiar with complex analysis). But so far you can just think that the multiplicity is (or coincides with) how many times the color changes. That's the viewpoint that I wanted to provide.

(In fact, you can count the change of the color around a circle that contains all roots. The result coincides with the degree of $f$. This is related to a proof of the fundamental theorem of algebra.)


Proof (or explanation):

I will work on this statement:

Let $f$ has a root $u$ of multiplicity $m$, and $f$ is described as $f(z)=(z-u)^mg(z)$ where $g(z)$ is nonzero at $u$.

Let $\gamma:[0,2\pi]\to\mathbb C$ defined by $\theta\mapsto u+e^{i\theta}$ be a small circle that loops once around $u$ and does not contain other roots inside it. Then $\gamma$ is mapped to a curve $\gamma_2$ by $f$: $$\begin{matrix}\gamma_2:=f\circ\gamma:&[0,2\pi]&\to&\mathbb C\\ & \theta &\mapsto & f(u+re^{i\theta}).\end{matrix}$$ which loops $m$ times around the origin, just as the second picture of this answer.

Since $\gamma$ does not contain other root of $f$, $\gamma$ contains no root of $g$. This means if we let the radius $r$ tend to zero, then $\gamma$ shrinks to $u$ gradually without passing through any root of $g$. Thus the corresponding curve $g\circ \gamma$ can shrink to a point without passing through zero.

(I implicitly used the continuity of $g$.)

enter image description here

Now, let's simplify $\gamma_2$: $$\begin{aligned} \gamma_2(\theta)=f(u+re^{i\theta}) &= ((u+re^{i\theta})-u)^mg(u+re^{i\theta})\\[0.7em] &= re^{im\theta}g(u+re^{i\theta}). \end{aligned}$$

Since $g(u+re^{i\theta})$ can shrink continuously to a point without passing through the origin, the net change of angle of the loop $g(u+re^{i\theta})$ w.r.t the origin is zero.

(I implicitly use some property of the homotopy)

change of angle

So the net change of angle of $\gamma_2$ is only caused by $re^{im\theta}$, which winds $m$ times around the origin.

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If $a,b\in\mathbb C$, let $p(z)=(z-a)(z-b)$. It has two and only two roots, which are $a$ and $b$, right? Well, not quite. It has two and only two roots whe $a\neq b$, but only one when $a=b$. However, if we count the roots with multiplicity, them yes, it has always two roots. So, if, in the case in which $a=b$, we count twice the root $a$ in $(x-a)(x-a)=(x-a)^2$, there will always be two roots.

Besides, counting the roots with multiplicity, it's not only true that a $n$th polynomial has $n$ roots, as it it is true that the number of roots of $p(x)q(x)$ is always equal to the number of roots of $p(x)$ plus the number of roots of $q(x)$.

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The multiplicity of $x_0$ as a root of polynomial $P$ is the number of times $(x-x_0)$ divides into $P$.


I've always been satisfied with:

Think of dividing a polynomial that has a root at $x_0$ with $(x-x_0)$ as removing that root (you also change the polynomial completely at a most other points, but that's not the point here), but if it is a root with multiplicity higher than $1$ it will also be a root of the result of the division.


I guess you could also see the multiplicity as some kind of indication of how flat the graph is at the root. If the multiplicity is $1$,it just crosses the $x$-axis, for higher multiplicities it stays near (to use your own words) the $x$-axis in a larger interval.

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  • $\begingroup$ That's more or less how I used to think of it too, but now somehow it doesn't seem like intuition $\endgroup$ – Truth-seek Mar 31 '18 at 16:48
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Quite simply, the multiplicity of $r$ as a root of $p(x)$ is the multiplicity of $(x-r)$ in a factorisation of $p(x)$ into linear factors (assuming we are working over the complex numbers, when such a factorisation will exist) or as a linear factor in a factorisation into general factors in other contexts.

Using the division algorithm we can always write $$p(x)=(x-r)p_1(x)+r_1$$ where $r_1$ is a constant. Setting $x=r$ we obtain $r_1=0$. This can be used to confirm for us that $(x-r)$ is a factor of $p(x)$ if and only if $r$ is a root of $p(x)$ ie $p(r)=0$.

We continue to divide by $x-r$ until we reach $$p(x)=(x-r)^mp_m(x)$$ and $r$ is not a root of $p_m(x)$. Then $m$ is the multiplicity.

To get some intuition for what is going on here, look at the graphs of $p(x)=x, p(x)=x^2, p(x)=x^3 \dots$ and examine the behaviour at the sole root $x=0$. Then try $q(x)=(x-1)p(x)$ for these examples and see what happens. If find that intuition is built more out of such examples, than out of wordy explanations.

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I think the best intuition we can have about multiplicity of a root $x_0$ of a polynomial $p(x)$ is the number of time $(x - x_0)$ divides $p(x)$.

Not only I see this as an enough "concrete" intuition, but it also has the advantage of being defined in an only algebraic way, thus it still works in any field, not only in the complex field.

Anyway there is another, more analytic, way to see multiplicity. $x_0$ has multiplicity $k$ as a root of $p(x)$ if $p^{(n)}(x_0) = 0$ for every $n < k$ and $p^{(k)}(x_0) \neq 0$. As you see this definition involves derivatives of $p(x)$ and so is more about $p(x)$ as a function then $p(x)$ as a polynomial (since the complex field has infinite elements we don't distinguish these two things).

From that we can also see that $x_0$ has multiplicity $k$ as a root of $p(x)$ if near $x_0$ we have that $p(x)$ goes to $0$ at "the same speed" $x^k$ goes to $0$ near $0$. So multiplicity of a root involves the speed a polynomial reach the $0$. But again this is an analytic intuition and has the disadvantage we can't use it in other fields (like finite fields).

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