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I'm doing some self-study on Legendre's Equation. I have seen and understand the proof of Bonnet's Recursion Formula for the Legendre Polynomials, $P_n(x)$.

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

Additionally, I have read that this formula also applies to Legendre functions of the second kind $Q_n(x)$, however I have not found any source that does more than simply state this fact. My question is: how does one prove that Bonnet's Recursion Formula extends to Legendre functions of the second kind?

For context, the proof I am familiar with for Bonnet's Recursion Formula relies on the generating function for the Legendre Polynomials $$ \frac{1}{\sqrt{1-2xt+t^2}} = \sum_{k=0}^{\infty} P_n(x){t^n} $$ where you differentiate with respect to $t$ and then collect coefficients of ${t^n}$.

Chris Duerschner
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  • What is your definition of $Q_n(x)$? – metamorphy Dec 23 '22 at 05:17
  • @metamorphy $Q_n(x)$ is: $$Q_n(x)=\frac{1}{2} ln\left(\frac{1+x}{1-x}\right)P_n(x) - \sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\frac{2n-4k-1}{(2k+1)(n-k)}P_{n-2k-1}(x).$$ You could also use the equivalent integral $$Q_n(x) = P_n(x) \int\frac{1}{(1-x^2)P_n(x)}dx.$$ – Chris Duerschner Dec 25 '22 at 17:45
  • Adding a comment to point out that in my previous comment $P_n(x)$ should be squared in the integrand. – Chris Duerschner Dec 28 '22 at 17:31

1 Answers1

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The simplest way I see to get one from another is to use the representation $$Q_n(x)=\frac12\int_{-1}^1\frac{P_n(t)}{x-t}\,dt\tag{*}\label{intrep}$$ (with the integral understood in the principal-value sense). Indeed, for $n>0$ $$(n+1)P_{n+1}(t)=(2n+1)tP_n(t)-nP_{n-1}(t)$$ gives, after division by $2(x-t)$ and integration, $$(n+1)Q_{n+1}(t)=\frac{2n+1}2\int_{-1}^1\frac{tP_n(t)}{x-t}\,dt-nQ_{n-1}(t),$$ and it remains to use $\displaystyle\frac{t}{x-t}=\frac{x}{x-t}-1$ and $\displaystyle\int_{-1}^1 P_n(t)\,dt=0$ (for $n>0$).

A proof of \eqref{intrep} may be found elsewhere.

metamorphy
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