I was wondering if the following properties of the Legendre polynomials are true in general. They hold for the first ten or fifteen polynomials.

  1. Are the roots always simple (i.e., multiplicity $1$)?

  2. Except for low-degree cases, the roots can't be calculated exactly, only approximated (unlike Chebyshev polynomials).

  3. Are roots of the entire family of Legendre Polynomials dense in the interval $[0,1]$ (i.e., it's not possible to find a subinterval, no matter how small, that doesn't contain at least one root of one polynomial)?

If anyone knows of an article/text that proves any of the above, please let me know. The definition of these polynomials can be found on Wikipedia.

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    $\begingroup$ According to a Corollary (page 114) in the book linked below, the Legendre polynomial of degree $k$ has $k$ distinct roots in the interval $(-1,1)$. tinyurl.com/29c89tu $\endgroup$ Nov 28, 2010 at 3:32
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    $\begingroup$ 1. Yes. It is a deep theorem of the theory of orthogonal polynomials that all their roots within their support interval are simple. 2. There are no explicit closed forms for the general roots of a Legendre polynomial, but there are asymptotic expansions for the roots. $\endgroup$ Nov 28, 2010 at 3:39
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    $\begingroup$ 3. Note that the roots of successive Legendre polynomials are interlacing (they form a Sturm sequence). $\endgroup$ Nov 28, 2010 at 3:40
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    $\begingroup$ For 2) I mean are these polynomials solvable by radicals (can the roots be written in a finite amount of space using ration numbers and radicals). $\endgroup$
    – user3180
    Nov 28, 2010 at 3:41
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    $\begingroup$ @user3971: what a specific definition of "calculated exactly"... what makes a radical so easy to calculate compared to other functions? Even to solve a cubic equation by the cubic formula "by radicals" requires one to compute cosines in the general case and I see no reason this is essentially easier than computing any other function whose Taylor series decays rapidly. $\endgroup$ Nov 28, 2010 at 3:44

5 Answers 5


To resolve the second question, note first that the Legendre polynomials are odd functions for odd order (0 then is one root of the polynomial), and even functions for even order. Thus, with regards to solubility in terms of radicals, you should be able to derive (possibly complicated!) radical expressions at least up until $P_9(x)$. To use that as an example, note that


is a quartic; thus, one can use the quartic formula to derive explicit expressions for its roots, and then you can easily derive the roots of $P_9(x)$ .

$P_{10}(x)$ is where your trouble starts. If we take a look at the polynomial


we have a quintic to contend with. I'll skip the relatively tedious details, but you can verify that its Galois group is not a solvable group, and thus the solution cannot be expressed in terms of radicals (you can use theta or hypergeometric functions, though).

So, not much hope in the symbolic front. In the numeric front, things are much easier. The slickest way of getting accurate values of the roots of the Legendre polynomial is to use the Jacobi matrix in my previous answer. Since there exist stable and efficient algorithms (e.g. QR algorithm or divide-and-conquer) for the symmetric eigenproblem (in LAPACK, for instance), and things can be set such that only eigenvalues are returned, you have a good way of generating good approximate values of Legendre polynomial roots. (In the context of Gaussian quadrature, where the roots of orthogonal polynomials play a pivotal role, the scheme is referred to as the Golub-Welsch algorithm.)

Alternatively, as I mentioned in the comments, there exist asymptotic approximations for the roots, which can then be subsequently polished with a few applications of Newton-Raphson. One such asymptotic approximation is due to Francesco Tricomi. Letting $\xi_{n,k}$ be the $k$-th root of $P_n(x)$, ordered in decreasing order, we have


and $O(n^{-4})$ and further terms are omitted. Other asymptotic approximations due to Luigi Gatteschi use roots of Bessel functions, but I won't say more about those.

  • $\begingroup$ Do you have the reference for Tricomi's approximation? I couldn't find it. $\endgroup$
    – user14717
    Apr 8, 2017 at 17:26
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    $\begingroup$ @user, see this. $\endgroup$ Apr 8, 2017 at 18:06
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    $\begingroup$ The asymptotic proof is in the paper "Una formula asintotica per l'approssimazione degli zeri dei polinomi di Legendre" which is available at bdim.eu/item?id=BUMI_1949_3_4_3 $\endgroup$
    – Coolwater
    Sep 11, 2018 at 19:12

I'll answer question 1 only for now, but I might edit this to address the others later.

One should note that corresponding to any set of orthogonal polynomials, there exists a symmetric tridiagonal matrix, called a Jacobi matrix, whose characteristic polynomial is the monic (leading coefficient is 1) version of the set of orthogonal polynomials considered. To use the Legendre polynomials as an explicit example, we first note that the monic Legendre polynomials satisfy the following two-term recurrence relation:

$$\hat{P}_{n+1}(x)=x \hat{P}_n(x)-\frac{n^2}{4 n^2-1}\hat{P}_{n-1}(x)$$

where $\hat{P}_n(x)=\frac{(n!)^2 2^n}{(2n)!}P_n(x)$ is the monic Legendre polynomial.

From this, we can derive an explicit expression for the corresponding Jacobi matrix (here I give the 5-by-5 case):


(the general pattern is that you have $\frac{n}{\sqrt{4 n^2-1}}$ in the $(n,n+1)$ and $(n+1,n)$ positions, and 0 elsewhere.)

We now note that $\frac{n}{\sqrt{4 n^2-1}}$ can never be 0, and then use the fact that if a symmetric tridiagonal matrix has no zeroes in its sub- or superdiagonal, then all its eigenvalues have multiplicity 1. (A proof of this fact can be found in Beresford Parlett's The Symmetric Eigenvalue Problem.) Thus, all the roots of the Legendre polynomial are simple roots.

A more conventional proof of this fact is in page 27 of Theodore Chihara's An Introduction to Orthogonal Polynomials. Briefly, the argument is that $P_n(x)$ changes sign at least once within $[-1,1]$ (and thus has at least one zero of odd multiplicity within the support interval) since

$$\int_{-1}^1 P_n(u)\mathrm du=0$$

Now, the polynomial


where the $\xi_j$ are the distinct zeroes of odd multiplicity within $[-1,1]$, should be greater than or equal to zero within $[-1,1]$, and thus its integral over $[-1,1]$ should be greater than zero. However, since

$$\int_{-1}^1 P_n(u) u^k\mathrm du=0\qquad\text{if}\qquad k < n$$

we have a contradiction, and thus all the roots of the Legendre polynomial are simple (and within the support interval $[-1,1]$).

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    $\begingroup$ If you can obtain a copy of Chihara's book, do so; it's a very good way to get a feel for the subject of orthogonal polynomials. $\endgroup$ Nov 28, 2010 at 14:27
  • $\begingroup$ Golub and Welsch used QR to get the eigenvalues of the Jacobi matrix, but this generates a dense matrix from a sparse one. Do you know of a better algorithm to get the eigenvalues which preserves sparsity? $\endgroup$
    – user14717
    Apr 13, 2017 at 1:08
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    $\begingroup$ @user, Tridiagonal QR only needs the diagonal and subdiagonal elements to generate the quadrature nodes. You may be thinking about how it is implemented in most languages these days, where the entire matrix of eigenvectors has to be computed even tho only the first components are needed. The FORTRAN code in the original Golub-Welsch paper actually modifies the eigenvector computations so that only the first components (and thus the weights) are generated; you might want to look into it. $\endgroup$ Apr 13, 2017 at 1:12
  • $\begingroup$ All your answers are on point; thanks so much for your assistance. $\endgroup$
    – user14717
    Apr 13, 2017 at 1:15

The density of the roots of any family of orthogonal polynomials follows from this result:

If $\{p_n\}$ is a family of orthogonal polynomials with roots in $[-1,1]$ and $N(a,b,n)$ represents the number of roots of $p_n$ in $[\cos(b),\cos(a)]$ then

$$\lim_{n\to \infty} \frac{N(a,b,n)}{n} = \frac{b-a}{\pi}$$


I think there is a simpler proof that the roots are simple.

The Legendre polynomial $\displaystyle P_n(x)$ satisfies the differential equation

$$\displaystyle (1-x^2) y'' -2x y' + n(n+1) y = 0$$

Note that, we scale the polynomials so that $\displaystyle P_n(1) = 1$, so if $\displaystyle \alpha$ is a root, then $\displaystyle \alpha \neq 1$.

Suppose $\displaystyle \alpha$ is a root of multiplicity $\displaystyle \gt 1$.

Then we must have that $P_n(\alpha) = P_n'(\alpha) = 0$.

The above equation implies that $P_n''(\alpha) = 0$.

By induction, by differentiating the above equation we can show that

$$\displaystyle (1-x^2) y^{(k+2)'} - f_k(x) y^{(k+1)'} + g_k(x) y^{k'} = 0$$

where $\displaystyle y^{l'}$ is the $\displaystyle l^{th}$ derivate of $\displaystyle y$

Since $\displaystyle \alpha \neq 1$ we see that $\displaystyle P_n^{k'}(\alpha) = 0 \ \ \forall k$ and thus the only root of $\displaystyle P_n(x)$ is $\displaystyle \alpha$.

I believe it is easy to show that $\displaystyle P_n(x) \neq C(x-\alpha)^n$.


The Abramowitz–Stegun Handbook of Mathematical Functions claims on page 787 that all the roots are simple: http://convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150&Page=787


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