Renormalizing Legendre polynomials to $P_n(0)=1$ One way to define the Legendre polynomials is with the recurrence relation
$$(n+1)P_{n+1} (x) = (2n+1)xP_{n} (x)-nP_{n-1} (x),$$
with $P_0(x)=1$ and $P_1(x)=x$. This standardization is normalized so that $P_n(1)=1$ for all $n$, and so that the polynomials are orthogonal on $[-1,1]$. I'm trying to form an iterative method to solve $Ax=b$, and one thing that came up is that I need the Legendre polynomials renormalized such that $P_n(0)=1$ instead of $P_n(1)=1$.
However  I don't know how to do this. Simply dividing the polynomial by $P_n(0)$ for example is not very helpful, because half the degrees of the Legendre polynomial already have $P_n(0)=0$ and so I'd be dividing by zero.
 A: Legendre polynomials are solutions of the Legendre differential equation $$[(1-x^2)y']' + n(n+1)y = 0$$
Applying the series method by substituting $y=\sum_{j=0}^n a_jx^j$, one finds a recurrence relation between the coefficients $a_j$ that lead to the elementary solutions $P_n(x)$ that are polynomials of degree $n$ which contain only odd-numbered or only even-numbered powers of $x$. Obviously then, $P_n(0)=0$ if $n$ is odd, and it is easy to derive from the recurrence relation that $P_n(0)\not=0$ if $n$ is even. Solutions equal to non-terminating series also exist and are usually written as $Q_n(x)$, which are technically equal to a polynomial of degree $n-1$ of again only odd or only even-powers of $x$, plus a term containing a factor $ln(\frac{1+x}{1-x})$. As this factor evaluates to $0$ when $x=0$, it is again obvious that $Q_n(0)=0$ if $n$ is even, and again it is simple to see that $Q_n(0)\not=0$ if $n$ is odd.
A general solution of the Legendre differential equation then is a linear combination of $P_n(x)$ and $Q_n(x)$, as the equation itself is a linear one:
$$y=f(x)=\alpha P_n(x) +\beta Q_n(x)$$
for which it is clear that by choosing appropriate values for $\alpha$ or $\beta$, it is possible to achieve $f(0)=1$.
Thus, you will not be able to use $P_n(x)$ (nor $Q_n(x)$), but if you can use more general solutions to the Legendre differential equation instead, then functions $R_n(x)$ are available which are multiples of either $P_n(x)$ or $Q_n(x)$ (alternatively P or Q for each next value of n), and have $R_n(0)=1$.
But, obviously, the recurrence relation above will not hold for the $R_n(x)$, as from it follows for instance that $2R_2(0) = -R_0(0)$.
If you need this recurrence relation to hold and can also live with the fact that just $R_n(0) \not=0$, then you can use
$$R_0(x) = P_0(x) + Q_0(x) = 1 + \frac{1}{2}ln(\frac{1+x}{1-x})$$
$$R_1(x) = P_1(x) + Q_1(x) = x + \frac{x}{2}ln(\frac{1+x}{1-x}) -1$$
$R_n(x)$ for $n>=2$ to be derived by applying the recurrence relation
$$(n+1)R_{n+1}(x) = (2n+1)xR_n(x) - nR_{n-1}(x)$$
which you will quickly understand leads to the same as saying that
$$R_n(x) = P_n(x) + Q_n(x)$$
I hope this helps!
