I am trying to prove that $\frac{1}{n!\space2^n}\frac{d^n}{dx^n}\{(x^2-1)^n\}=P_n(x)$, where $P_n(x)$ is the Legendre Polynomial of order n.

I've been told that the proof uses complex analysis, of which I know nothing, isn't there a proof with elementary methods? (If there isn't, I'm still interested in the other one).

  1. Check that the left side indeed defines an $n$th order polynomial.

  2. Check that $\displaystyle \int_{-1}^{1}P_n(x)P_m(x)dx$ vanishes for $m\neq n$ (integration by parts).

  3. Check the normalization condition $P_n(1)=1$ (Leibniz rule).

Added: As you almost correctly write in the comment below, the result of integration by parts (assuming that $m<n$ and transferring the derivatives from $P_n$ to $P_m$) can be written as $$\int_{-1}^1P_m(x)P_n(x)dx=\sum_{k=1}^{n}c_{mnk}\left[\frac{d^{m+k-1} (x^2-1)^m}{dx^{m+k-1}}\frac{d^{n-k}(x^2-1)^n}{dx^{n-k}}\right]_{-1}^{1},$$ where $c_{mnk}$ is some irrelevant constant. Consider the second factor in the square brackets. There you have a polynomial having $n$th order zeros at $x=\pm 1$ which we differentiate $n-k$ times. The result will therefore have $k$th order zeros at these points, which implies vanishing of the integral.

  • $\begingroup$ Wow, I hadn't even thought of this, thanks! However, is that enough? Aren't there other polynomials that satisfy condition number 2? And shouldn't I also check that $\int_{-1}^1 P_n(x)P_n(x)dx=\frac{2}{2n+1}$? I will start trying this right now! $\endgroup$ – Alubeixu Jun 18 '14 at 20:32
  • $\begingroup$ @Alubeixu it's enough, there are no other polynomials satisfying 2 (since the weight is positive). As for the 3rd question, this property is a consequence but not a part of the standard definition (weight + value at 1 fixing the normalization). A hint for the point 2: e.g if $m<n$, then use integration by parts to transfer the derivatives in $P_n$ to $P_m$. $\endgroup$ – Start wearing purple Jun 18 '14 at 20:38
  • $\begingroup$ I'm sorry, could you do point 2 explicitly? I'm stuck at $\int_{-1}^1 \frac{d^n}{dx^n}(x^2-1)^n\frac{d^m}{dx^m}(x^2-1)^mdx=\sum_{i=o}^n(\frac{d^{n+i}}{dx^{n+i}}(x^2-1)^n\frac{d^{m-1-i}}{dx^{m-1-i}}(x^2-1)^m(-1)^{i} \space \Big]_{-1}^1$ $\endgroup$ – Alubeixu Jun 19 '14 at 11:03
  • $\begingroup$ @Alubeixu I've added some explanations. $\endgroup$ – Start wearing purple Jun 19 '14 at 17:34

If $n$ is integer \begin{eqnarray*} \frac{d^n}{d x^n} (x^2-1)^n &=& \frac{d^n}{d x^n} \left [ \sum_{k=0}^n (-1)^k \frac{n!}{k!(n-k)!} x^{2n-2k} \right ] \\ &=& \sum_{k=0}^n (-1)^k \frac{n!}{k! (n-k)!} \frac{(2n-2k)!}{(n-2k)!} x^{n-2k}. \end{eqnarray*} The sum above does not go up to $k=n$, since after $k=[n/2]$, the derivatives are 0 then we write

\begin{eqnarray*} \frac{d^n}{d x^n} (x^2-1)^n &=& \sum_{k=0}^{[n/2]} (-1)^k \frac{n!}{k! (n-k)!} \frac{(2n-2k)!}{(n-2k)!} x^{n-2k}. \end{eqnarray*} It follows from the infinite series truncated to the Legendre polynmial that

\begin{eqnarray*} P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n. \end{eqnarray*}

The approach followed here is in reverse order. We started with Rodriguez's formula and showed that it corresponds to a Legendre polynomial. A more intuitive approach is to start at the polynomials

\begin{eqnarray*} y(x)= (1-x^2)^n. \end{eqnarray*} and take derivates, and verifty that the derivatives taken $n$ times will get you to the Legendre differential equation. That is, we have that

\begin{eqnarray*} y' = -2 n x (1-x^2)^{n-1} \end{eqnarray*} which we can write as

\begin{eqnarray} (1-x^2) y' + 2n x y = 0. \label{tole} \end{eqnarray} and starts looking a bit like a Legendre differential equation.

We want to differentiate this equation $k$ times and use the Leibniz rule. That is, if we call $u=1-x^2$,

\begin{eqnarray*} \frac{ d^k}{dx^k} [u y'] = \sum_{j=0}^{k} \binom{k}{j} u^{(j)} y^{(k-j+1)} \end{eqnarray*} Given that $u$ is a second order polynomial only three terms of this sum will survive. That is

\begin{eqnarray*} \frac{ d^k}{dx^k} [u y'] &=& u y^{(k+1)} + k u' y^{(k)} + k(k-1) u^{(2)} y^{(k-1)} \\ &=& (1-x^2)y^{(k+1)} - 2 k x y^{(k)} -2 \frac{k(k-1)}{2} y^{(k-1)} = 0 \end{eqnarray*} Likewise we use the Leibniz rule for the product $2nxy$ where only two terms will survive. That is

\begin{eqnarray*} \frac{ d^k}{dx^k} [2 n x y] &=& 2 n x y^{(k)} + 2 n k y^{(k-1)}, \end{eqnarray*} we combine the two results above to find

\begin{eqnarray*} (1-x^2)y^{(k+1)} - 2 k x y^{(k)} - k(k-1) y^{(k-1)} + 2 n x y^{(k)} + 2 n k y^{(k-1)} = 0 \end{eqnarray*} At this point we observe that if $k=n+1$, we find

\begin{eqnarray*} (1-x^2)y^{(n+2)} - 2(n+1) x y^{(n+1)} - n(n+1) y^{(n)} + 2 n x y^{(n+1)} + 2 n (n+1) y^{(n)} = 0 \end{eqnarray*} which simplifies to

\begin{eqnarray*} (1-x^2) y^{(n+2)} - 2 x y^{(n+1)} + n(n+1) y^{(n)}=0. \end{eqnarray*}

and this is the Legendre differential equation with $y^n=P_n$. We then showed that

\begin{eqnarray*} \frac{d^n}{dx^n}(1-x^2)^n \end{eqnarray*} satisfies the Lagrange differential equation. The factor $1/(2^n n!)$ is included to make $P(1)=1$.

  • 1
    $\begingroup$ You meant Legendre not Lagrange ? +1 $\endgroup$ – Aryadeva May 10 '20 at 13:16
  • $\begingroup$ @Aryadeva: Thanks. $\endgroup$ – Herman Jaramillo May 10 '20 at 18:22

(The general formula of Legendre Polynomials is given by following equation: $$ P_k(x)=\sum_{m=0}^{\frac{k}{2}|\frac{k-1}{2}}{\frac{(-1)^m(2k-2m)!}{2^km!(k-m)!}}\frac{1}{(k-2m)!} x^{k-2m} $$

The Rodrigues' formula is:

$$ \frac{1}{2^kk!}\frac{d^k}{dx^k}[(x^2-1)^k] $$

The Binomial theorem is as follow: $$(x+y)^k=\sum_{i=0}^{k}\frac{k!}{i!(k-i)!}x^{k-i}y^{i}$$

Then $$(x^2-1)^k=\sum_{i=0}^{k}\frac{k!}{i!(n-i)!}{(x^{2})}^{k-i}(-1)^{i}$$


So $$ =\frac{1}{2^kk!}\sum_{i=0}^{k}\frac{k!}{i!(n-i)!}\frac{d^k}{dx^k}(x)^{2k-2i}(-1)^{i} .....(1)$$


By compensate (2) into (1), we get:

$$ P_k(x)=\sum_{m=0}^{\frac{k}{2}|\frac{k-i}{2}}{\frac{(-1)^i(2k-2i)!}{2^ki!(k-i)!}}\frac{1}{(k-2i)!} x^{k-2i} $$

Change dummy variable (i) to (m), we get the same general formula of Legendre Polynomials :

$$ P_k(x)=\sum_{m=0}^{\frac{k}{2}|\frac{k-1}{2}}{\frac{(-1)^m(2k-2m)!}{2^km!(k-m)!}}\frac{1}{(k-2m)!} x^{k-2m} $$

  • 2
    $\begingroup$ Welcome to math.se! I have to ask you: what does notation $\frac{k}{2}\Big| \frac{k-1}{2}$ mean? $\endgroup$ – Nik Pronko Aug 8 '18 at 13:57
  • $\begingroup$ Imagine you are looking for $$P_5(x)$$ (odd polynomial) then the the sum start from m=0 to 2: $$\sum_{m=0}^\frac{k-i}{2}$$, while $$\frac{k}{2}$$ isvalid for even polynomial such as $$P_4$$. $\endgroup$ – Mohamed Abugammar Aug 9 '18 at 14:46

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