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Is multiplicity of root of polynomial meaningful in any way?

I encounter this problem, when I find roots of some polynomial and there are fewer roots found than the order of the polynomial. Which means that some are repeated.

But I wonder if it's necessary to know that "these roots are repeated"? Does it serve any practical purpose?

mavavilj
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    What is a "practical purpose"? – Angina Seng Jun 04 '19 at 19:05
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    Is there a specific context you are thinking in? For example, the multiplicities of the eigenvalues (which are roots of a matrix's characteristic polynomial) are important in determining whether a given matrix has an eigenbasis. – paulinho Jun 04 '19 at 19:06
  • @paulinho Well just in general. I was finding roots of a 5th order complex polynomial today and I noticed that my factorization showed only 4. Thus I was thinking, do I need to factor it more to discover multiplicities. – mavavilj Jun 04 '19 at 19:07
  • Yes, in a big way. Roots of multiplicity greater than one often cause numerical issues. – copper.hat Jun 04 '19 at 19:12

3 Answers3

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Yes, there are lots of situations where it's important:

  1. The degree of the polynomial is the number of roots (over the complex field). So if you don't include multiplicities, you can end up with a polynomial of lower degree. Polynomials are defined up to a constant scaling by their roots (that is, if you take two polynomials that are the same except that one is the other times a constant, then the two will have the same roots), so if you don't include multiplicities, you don't get the same polynomial.

  2. The multiplicity of the root is one more than the number of derivatives with a root. For instance, if the multiplicity is 3, then the first 2 derivatives will have a root there.

  3. If the root has odd multiplicity, the graph will intersect the x-axis. If it is even, it will be tangent.

  4. When you're working with the characteristic polynomial of a matrix, multiple roots can indicate that the eigenspace for that eigenvalue is multidimensional. If that's too advanced for you, the simplified version is that you can have things associated with each root, and when you have different things associated with the same root, that root tends to be multiple. Conversely, if a root is multiple, that may mean that you have several of the things associated with the root.

  5. There are wide variety of more esoteric applications; as just one example, elliptic curve math requires distinguishing between single and double roots.

Acccumulation
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In addition to the given solution, when using a numerical method such as Newton-Raphson, it is important to take the number of repeated roots in consideration to get a better speed. If you know that there are repeated roots you'd use the modified version of the algorithm. See for example newton-raphson-method-for-double-roots.

Also, if you are constructing the polynomial from a graph, it is important to recognize repeated roots so that you could create the correct polynomial function, for example $(x-1)$ is a different function from $(x-1)^2$.

NoChance
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(I don't know if this counts as an answer to the question, but it's too long for a comment.)

If a polynomial $$ p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0 $$ has a root $w$ of multiplicity $m$, and $C$ is a closed disc with centre $w$ containing no other roots of $p$, then there is a number $\delta > 0$ such that any polynomial $$ p^*(z) = z^n + a^*_{n-1}z^{n-1} + \cdots + a^*_1z + a^*_0 $$ has exactly $m$ roots, counted with multiplicity, in the disc $C$, so long as $$ |a^*_i - a_i| < \delta \quad (i = 0, 1, \ldots, n-1). $$ Conversely, if $p(z) = (z - w)^mq(z)$, where $q(w) \ne 0$, then, for any $\delta > 0$, there is a disc $C$ with centre $w$ such that the coefficients of any polynomial of the form $$ p^*(z) = (z - w_1)\cdots(z - w_m)q(z) $$ differ from the corresponding coefficients of $p(z)$ by less than $\delta$, so long as $$ w_1, \ldots, w_m \in C. $$ So multiple roots behave like distinct roots with small changes in the polynomial.