How to proof that $(n+1)P_{n}(x)-(2n+1)x P_{n}(x)+n P_{n-1}(x)=0$?

Question

How to proof that $(n+1)P_{n+1}(x)-(2n+1)x P_{n}(x)+n P_{n-1}(x)=0$

where

$P_{n}(x)=\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}$.

Do you want to work directly with that definition of $P_n$, or through some other procedure (e.g. Gram-Schmidt)? — Ian, May 10 '16 at 16:09
@Ian, I am not sure but this question came in an exam yesterday and I couldn't proof it. In my lecture notes $P_{n}(x)$ is define as above. — user3879021, May 10 '16 at 16:17
See https://en.wikipedia.org/wiki/Legendre_polynomials#Recursive_definition. — lhf, May 10 '16 at 16:20
The convenient way to proceed really depends on what you already know. To me the convenient way to proceed is to define $P_0=1$, then $P_n$ is (up to your choice of normalization) $x^n$ minus the projection of $x^n$ onto $P_0,P_1,\dots,P_{n-1}$, where the inner product is that given by $L^2[-1,1]$. If you work this way, then you can start by following http://www.math.hkbu.edu.hk/ICM/LecturesAndSeminars/08OctMaterials/1/Slide2.pdf slides 12-14. This tells you that there is some three term recurrence relation but does not explicitly tell you what the coefficients are. — Ian, May 10 '16 at 16:29
To find the coefficients you need some kind of inductive argument specific to the case of Legendre polynomials. (You also need to be careful about normalization; the Legendre polynomials are, by convention, not normalized to be monic.) — Ian, May 10 '16 at 16:29
@Ian, So the above definition for $P_n (x)$ can't be used to prove. Is that what you saying? — user3879021, May 10 '16 at 16:37
It probably can, but I haven't done it that way. The "obvious" place to start is to observe $\frac{d^{n+1}}{dx^{n+1}}((x^2-1)^{n+1})=\frac{d}{dx}(\frac{d^n}{dx^n}((x^2-1)^n(x^2-1)))$. Then do the product rule on the outer derivative. So if $Q_n(x)=\frac{d^n}{dx^n}((x^2-1)^n)$ then $Q_{n+1}(x)=Q_n'(x)(x^2-1)+Q_n(x)(2x)$. Now you need to translate that into the desired statement, which takes some work. — Ian, May 10 '16 at 16:38

score 0 · Accepted Answer · edited Jun 12 '20 at 10:38

You may use Rodrigues' formula to prove that Legendre polynomials have to following generating function: $$ \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n\geq 0}P_n(x)\,t^n = g(t)\tag{1}$$ The relation $(n+1) P_{n+1}(x)-(2n+1) x P_n(x) + n P_{n-1}(x)=0 $ translates into:

$$ g'(t) - \left( x g(t) + 2tx g'(t)\right)+t\cdot\frac{d}{dt}\left( t g(t)\right) =0 \tag{2}$$

or:

$$(1-2tx+t^2)\, g'(t) = (x-t)\, g(t) \tag{3} $$

or:

$$ \frac{d}{dt}\log g(t) = \frac{x-t}{1-2tx+t^2} \tag{4}$$

that is straightforward to prove, given the LHS of $(1)$.

How to proof that $(n+1)P_{n}(x)-(2n+1)x P_{n}(x)+n P_{n-1}(x)=0$?

1 Answers1