I need a proof to show that the inequality $m < n$ leads to a contradiction and $P_n$ has $n$ distinct real roots, all of which lie in the open interval $(-1, 1)$.

  • $\begingroup$ What is the meaning of $m$? $\endgroup$ Oct 15, 2013 at 14:02

3 Answers 3


From the wording of your question I guess you are looking for something like this:

The Legendre polynomial $P_n$ with $n>0$ has $n$ simple roots in $(-1, 1).$

Proof by contradiction: Assume $P_n$ has $m$ with $0 \le m \lt n\;$ pairwise different zeroes $x_1, x_2, \dots x_m$ of odd multiplicity in $(-1,1),\;$ i.e. $P_n$ changes sign in $x_i$, and consider the polynomial $Z_n(x) = (x-x_1)\cdot (x-x_2) \dots (x-x_m).\;$ Because $P_n$ is orthogonal to $P_0$ the intgeral $\int_{-1}^{1} P_n(x) dx$ vanishes and $P_n$ has at least one zero of odd multiplicity in $(-1,1),\,$ i.e. $m>0.\;$

The polynomial $Z_n(x) P_n(x)$ does not change sign in $(-1,1)$ and therefore $I_n = \int_{-1}^{1} Z_n(x) P_n(x) dx \ne 0.\;$ On the other hand $Z_n$ has degree $m$ and can be written as a linear combination $Z_n = \sum_{k=0}^m c_k P_k(x)$. Thus we get the contradiction by means of the orthogonality of the $P_k$: \begin{align} 0 \ne I_n &= \int_{-1}^{1} Z_n(x) P_n(x) dx\\ &= \int_{-1}^{1} P_n(x) \sum_{k=0}^m c_k P_k(x) dx\\ &= \sum_{k=0}^m c_k \int_{-1}^{1} P_n(x) P_k(x) dx\\ &= 0 \end{align}

This shows that $P_n$ has $n$ pairwise different zeroes of odd multiplicity, and because $P_n$ has at most $n$ zeroes, the zeroes $x_1,\dots, x_n$ are simple.

Note that this kind of proof can be applied to general orthogonal polynomials over intervals $(a,b)$ with weighting function $w(x) \ge 0$.

  • $\begingroup$ I don't see why "The polynomial $Z_n(x) P_n(x)$ does not change sign in $(-1,1)$". Suppose for example $P_n(x)$ has a root of multiplicity 3 in $(-1,1)$. Then $Z_n(x)P_n(x)$ will change sign at that point. $\endgroup$ Dec 18, 2013 at 14:32
  • $\begingroup$ @Goos: A root of multiplicity 3 is not simple. $\endgroup$ Dec 18, 2013 at 15:19
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    $\begingroup$ I am aware of that. Let $a$ be the root of multiplicity $3$. $Z_n(x)$ will not change sign at $a$ (since $a$ is not simple), and $P_n(x)$ will change sign (since $a$ has multiplicity 3). Hence $Z_n(x)P_n(x)$ will change sign. $\endgroup$ Dec 18, 2013 at 15:29
  • $\begingroup$ @Goos: I see what you mean and I refined the proof by showing that $P_n$ has $n$ roots of odd multiplicity. $\endgroup$ Dec 19, 2013 at 7:55
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    $\begingroup$ Nice approach. But the proof only works if you have shown that there is at least one odd zero? Otherwise $Z_n = 0$ and there is no contradiction. $\endgroup$
    – andreas
    Mar 24, 2014 at 12:07

By Rodrigues formula for Legendre polynomials, $$\displaystyle P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n} (x^2 - 1)^{n}\tag{*1}$$ $P_n(x)$ is the $n^{th}$ derivative of a polynomial with roots at $-1$ and $1$. Repeat applying Gauss-Lucas theorem $n$ times, we know all roots of $P_n(x)$ lie on the closed line segment $[-1,1]$ in the complex plane. By direct expansion of $(*1)$, one can check that $P_n(\pm 1) \ne 0$, so the roots of $P_n(x)$ are all real and belongs to $(-1,1)$.

Since Legendre polynomials are solutions of the Legendre's differential equation: $$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P_n(x)\right] + n(n+1)P_n(x) = 0$$ which is a $2^{nd}$ order ODE, $P_n(x)$ cannot have any double roots. To see this, let's say $P_n(x)$ has a double root at $\alpha \in (-1,1)$. $P_n(x)$ will then be a solution of following initial value problem:

$$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}y(x)\right] + n(n+1)y(x) = 0,\quad \begin{cases}y(\alpha) = 0,\\y'(\alpha) = 0\end{cases}$$ Apply Picard-Lindelöf theorem to this $2^{nd}$ order ODE, the "uniqueness" part of the theorem tell us $P_n(x)$ vanishes over some neighbor of $\alpha$. Since $P_n(x)$ is not identically zero, this is impossible. As a consequence, all roots of $P_n(x)$ are simple.

  • $\begingroup$ @Sam: If you think the author of the Accepted Answer made a substantive error, you should point out your concern in a Comment rather than making a unilateral change by Editing. $\endgroup$
    – hardmath
    Apr 22, 2018 at 22:31

Let Legendre polynomial be $$ f_n(x)=\frac{1}{2^n n!}\frac{d^{n}}{d x^n}(x^2-1)^n .$$

Note that $ (x^2-1)^n $ has roots $ -1, 1 $ each with multiplicity $ n $, and $$ (x^2-1)\Big|\frac{d^m}{d x^{m}}(x^2-1)^n, \, \text{when}\ m<n .$$

Thus, $ -1, 1 $ are always roots of $ \frac{d^{m}}{d x^m}(x^2-1)^n $. And by Rolle's theorem, we know that there are $ n-1 $ distinct zeros $ x_1, x_2, \cdots, x_{n-1} $ in the interval $ (-1, 1) $ for the polynomial $ \frac{d^{n-1}}{d x^{n-1}}(x^2-1)^n $. Since $ -1, 1 $ are also zeros of $ \frac{d ^{n-1}}{d x^{n-1}}(x^2-1)^n $. Use Rolle's theorem once again, we get $ n $ distinct zeros in $ (-1, 1) $ and since $ \deg (\frac{d^{n}}{d x^{n}}(x^2-1)^n)=n $, we have found all the distinct roots of $ f_n(x) $ in $ (-1, 1) $.

  • $\begingroup$ why there are $n-1$ distinct roots by Rolle's Theorem? $\endgroup$ May 25, 2023 at 12:59

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