Considering the Legendre polynomials: $$P_n(x) = \sum_{m=0}^{n}a_{n,m}{x^m}$$ I know that: $P_0=1$ and $P_1=x$.

However given $P_0$ if I want to find $P_1$ by hand: $$\langle P_1 | P_0 \rangle = 0 = \int_{-1}^{1}(a_0+a_1x)dx = 2a_0 = 0 \longrightarrow a_0=0$$ Imposing normalization condition: $$\langle P_1 | P_1 \rangle = 1 = \int_{-1}^{1}a_1^2x^2 dx = \frac{2}{3}a_1^2 = 1 \longrightarrow a_1= \sqrt{ \frac{3}{2}} $$ So from my computations $P_1= \sqrt{ \frac{3}{2}} x$.

Where am I wrong?

  • $\begingroup$ The normalization condition for Legendre polynomials $P_n(z)$ is $\int_{-1}^1 P_n(x) P_m(x) dx = \frac{2}{2n+1} \delta_{n,m}$. i.e $P_n(z)$ are not orthonormalized, they are only orthogonal. $\endgroup$ – achille hui Nov 18 '20 at 9:08

$P_1' = x$ is not normalized since

$$\langle P_1' | P_1' \rangle = \int_{-1}^{1} x^2 dx = \dfrac{2}{3} $$

$P_1 = \sqrt{\dfrac{3}{2}}x$ is normalized.

$$\langle P_1 | P_1 \rangle = 1 $$

Both are correct upto scalar multiple. It depends which one you need.


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