I learned of the Legendre polynomials for the first time, in the context of finding an orthogonal basis for $\text{span} \{1, x, x^2, ... \}$.

According to Wolfram, the Legrndre Polynomials are $$P_0 = 1$$ $$P_1 = x$$ $$P_2 = \dfrac 12 (3x^2-1)$$ $$P_3 = \dfrac 12(5x^3-3x)$$ and so on.

Now, I did Gram-Schmidt (without normalzation) on the monomials, and I got $1, x, x^2 - \dfrac 13, x^3 - \dfrac 35 x, ...$.

So I thought that perhaps the Legendre Polynomials are in that for for normalization. But I worked it out and they do not have a norm of one. So my question is, why are the Legendre Polynomials different from what you get by pure Gram-Schmidt? I understand perhaps wanting integer coefficients, but I am not sure about the $\dfrac 12$.

I suspect that probably the answer will be "because in another derivation of them, the come out with coefficient $\dfrac 12$. But I'm not sure.


As Hans Lundmark mentioned, the 'normalizing' requirement is that $P_n(1) = 1$. You may wonder what silly kind of normalizing this is; normally some integral is required to be normalized to $1$.

The reason is that, apart from the ensuing very practical consequences, originally Legendre came to defining these polynomials by looking at the series expansion of the reverse of the distance between two points in space, $\vec{a} = \left(\begin{matrix}0\\ 0 \\ 1\end{matrix}\right)$ and $\vec{r} = \left(\begin{matrix}x\\ y \\ z\end{matrix}\right)=\left(\begin{matrix}r\sin\theta\cos\phi\\ r\sin\theta\sin\phi \\ r\cos\theta\end{matrix}\right)$

This function amounts to $$\frac{1}{| {\vec{r} -\vec{a}}|} = \frac{1}{\sqrt{r^2-2r\cos\theta+1}} = \sum_{n=0}^{\infty} P_n(\cos\theta)r^n$$ And, as you can see, when $\theta = 0$, this becomes $$\frac{1}{|r-1|} = \frac{1}{\sqrt{r^2-2r+1}} = \sum_{n=0}^{\infty} P_n(1)r^n$$ Note that the radius of convergence for this series is $1$ (independent of the value of $\theta$: in the complex plane the singular points are $z_{1,2}=\cos \theta \pm i\sqrt{1-\cos^2\theta}$, the norm of which is $1$).

Hence $r<1$, for which $$\frac{1}{1-r} = \sum_{n=0}^{\infty} P_n(1)r^n$$ Hence $P_n(1)=1$ for all $n$.


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