# Legendre Polynomials notation

Im looking at the Legendre Polynomials and trying to recreate the first few terms

$${d \over dx} \left[ (1-x^2) {d \over dx} P_n(x) \right] + n(n+1)P_n(x) = 0.$$

What exactly is P_n(x)? and could you give an example of how you would produce the Legendre polynomial if we set n to 3 or 4?

Also, maybe a separate question but does $nP_n$ make any sense as an idividual statement in this context?

## 1 Answer

This is the differential equation the Legendre polynomials satistky; you may use the recursive relation, known as Bonnet's recursion formula, in order to recreate them:

$(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x)$

You can use this recursion in order to get them, as long as you get the first 2 terms: you can find them using explicitly the differential equation:

For $n=0$, the equation becomes $\frac{d}{dx}\Big[ (1-x^2) \frac{d}{dx}P_0(x) \Big] = 0$ This clearly has $P_0(x)=$ constant as a solution, and if you use the normalization

$P_n(1)=1$ for every $n$

you get $P_0(x)=1$.

For $n=1$, the equation becomes $\frac{d}{dx}\Big[ (1-x^2) \frac{d}{dx}P_1(x) \Big] +2P_1(x)= 0$. This can seem a little messy: let's just try polynomials of the simplest form (non-constant), that is, of the form $Ax+B$ (linear); we get

$-2Ax + 2Ax + 2B=0$

which clearly tells us that the polynomial $P_1(x) = Ax$ is a solution; using again the normalization condition, we get $P_1(x) = x$.

From now on, is trivial to get the other ones, using the above recursion formula: for example for the 2nd polynomial we have

$2P_2(x) = 3xP_1(x)-P_0(x) = 3x^2 - 1$

• Do i have to use the recurrence relation? Can you get n=5 say using differentiation? – MathsPro Nov 5 '14 at 19:51
• You can also use the equation $P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \Big[(1-x^2)^n \Big]$, I have given the recursive equation because I read you wanted to recreat all the polynomials one by one, from 0 to $n=4 or 5...$ – Klaramun Nov 5 '14 at 20:01
• Thank you, the only thing i dont understand is how you get from $${d \over dx} \left[ (1-x^2) {d \over dx} P_n(x) \right] + n(n+1)P_n(x) = 0$$ to $$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \Big[(1-x^2)^n \Big]$$ – MathsPro Nov 5 '14 at 20:18
• You can try using the standard method of power series: say $P_n(x) = \sum_{k=0}^{+\infty} a_k(n) x^k$ and incert this into the differential equation: you will get a recurrence relation for the coefficients. You do not need to worry about the convergence, this is only a trick: you can start by assuming $|x| < 1$, using the method and then try to find a "cute" expression for the $P_n(x)'s$; by obtaining these polynomials you only need to check that these are indeed solutions, and you are done. (The normalization condition will be still $P_n(1)=1$, it's the standard one) – Klaramun Nov 5 '14 at 20:30