Choice of the First Term in Legendre Polynomials

Question

The two solutions of the Legendre's Differential Equation obtained by series solution method are :

and

Now according to my textbook, for the useful polynomial for n equal to a positive integer, the constant $a_{0}$ in the first case is chosen as $a_{0}={\frac{1.3.5...(2n-1)}{n!}}$ and the solution is then called Legendre's polyinomial or coffecient or Zonal Harmonics of the first kind.

For n being the negative integer the constant $a_{0}$ in the second case is chosen as $a_{0}={\frac{n!}{1.3.5...(2n+1)}}$ and the corresponding polynomial is defined as Legendre's function of the second kind.

Now the choice of $a_{0}$ seems rather arbitrary and exactly how is such a choice useful?

P.S.: While solving the Hermite Equation we choose $a_{0}$ in a similar (and apparently) arbitrary manner. Is the reason behind the choice somehow related?

The complete solution of Legendre's equation using series solution method can be found here

Answer 1

It is in fact not $a_0$ for which a value is chosen. For Legendre polynomials $P_n$, the "normalization" is to set $P_n(1) = 1$. That choice leads to Rodrigues' formula $$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$ from which we derive \begin{equation} \begin{split} P_n(x) &= \frac{1}{2^nn!}[(x+1)^n\cdot(x-1)^n]^{(n)} \\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}[(x+1)^n]^{(k)} \cdot [(x-1)^n]^{(n-k)} \\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}\frac{n!}{(n-k)!}(x+1)^{n-k} \cdot \frac{n!}{k!}(x-1)^k \\ &= \frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2(x+1)^{n-k}(x-1)^k \end{split} \end{equation}

so \begin{equation} \begin{split} a_0 &=\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2 = \frac{1}{2^n}\binom{2n}{n} = \frac{(2n)!}{2^nn!^2} \\ &= \frac{(2n)(2n-1)(2n-2)\cdot\cdot\cdot2\cdot 1}{2^n\cdot n(n-1)(n-2)\cdot\cdot\cdot2\cdot 1 \cdot n!} \\ &= \frac{(2n-1)(2n-3)\cdot\cdot\cdot3\cdot 1}{n!} \end{split} \end{equation} which is exactly the value you found in your textbook.