# Proof the Legendre polynomial $P_n$ has $n$ distinct real zeros

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 $(-1, 1)$.

• What is the meaning of $m$? – gammatester Oct 15 '13 at 14:02

## 3 Answers

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.

• @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. – hardmath Apr 22 '18 at 22:31

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 $w(x) \ge 0$.

• 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. – 6005 Dec 18 '13 at 14:32
• @Goos: A root of multiplicity 3 is not simple. – gammatester Dec 18 '13 at 15:19
• 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. – 6005 Dec 18 '13 at 15:29
• @Goos: I see what you mean and I refined the proof by showing that $P_n$ has $n$ roots of odd multiplicity. – gammatester Dec 19 '13 at 7:55
• 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. – andreas Mar 24 '14 at 12:07

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

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 $$(-1, 1)$$.