Analytical expressions and recurrence relations for the $P_{n-1}(t) - P_{n + 1}(t)$ function, derivative and integral

Abstract

In this paper, we discuss some methods for the calculation of the Legendre polynomial difference $P_{n - 1}(t) - P_{n + 1}(t)$. Such differences frequently appear when calculating the spherical harmonic coefficients of truncated spatial filters. By examining the relation between the Legendre polynomials and other classical orthogonal polynomials (i.e., Gegenbauer and Jacobi polynomials), we provide analytical expressions for $P_{n - 1}(t) - P_{n + 1}(t)$ in terms of the latter. We also derive a recurrence relation that avoids the direct evaluation of the Legendre polynomials. Analogous analytical expressions and recurrence relations are provided for the calculation of the $P_{n - 1}(t) - P_{n + 1}(t)$ derivative and integral. We show that these recurrence relations can be derived from a more general recurrence relation; thus, the former are special cases of the latter. The expressions given in this study can be useful for the more efficient calculation of truncated kernels in the spherical harmonic domain. We demonstrate the applicability of these new expressions for the Pellinen (i.e., rectangular) filter kernel that is regularly used in regional gravity field modeling.

Appendices

Appendix A Differentiation of Eq. (32)

The differentiation of Eq. (32) with respect to t yields the following recurrence relation:

$$\begin{aligned} \begin{aligned} k_{n}'&= \frac{2 n + 1}{n + 1} k_{n - 1} + \frac{2 n + 1}{n + 1} t \, k_{n - 1}' \\&\quad - \frac{(2 n + 1) (n - 2)}{(2 n - 3)(n + 1)} k_{n - 2}'. \end{aligned} \end{aligned}$$

(A.1)

Setting $n = n - 1$ in Eq. (39) and substituting the resulting expression for $k_{n - 1}$ in Eq. (A.1) leads to the recurrence relation of Eq. (40).

Appendix B Integration of Eq. (32)

Integrating Eq. (32) with respect to t yields the following recurrence relation:

$$\begin{aligned} Ik_{n}&=\frac{(2 n + 1) (2 n - 1) (t - t^{3})}{(n + 1) [n(n - 1) - 6]} P_{n - 1}\nonumber \\&\quad + \frac{(2 n + 1) (1 - 3 t^{2})}{(n + 1) [n(n - 1) - 6]} k_{n - 1} \nonumber \\&\quad - \frac{(2 n + 1) (n - 2)}{(2 n - 3)(n + 1)} Ik_{n - 2}. \end{aligned}$$

(B.1)

Appendix C A proof of the equivalence of Eqs. (71) and (72)

Applying the transformation formula (Gradshteyn and Ryzhik 2014, §9.131, eq. 1)

$$\begin{aligned} F (a,b;c;t) = (1 - t)^{c - a - b} F (c - a,c - b;c;t) \end{aligned}$$

(C.1)

on Eq. (71), we derive

$$\begin{aligned} F (-n,n + 1;2;\frac{1 - t}{2})&= \frac{t + 1}{2} \nonumber \\&\quad \times F (n + 2,1 - n;2;\frac{1 - t}{2}). \nonumber \\ \end{aligned}$$

(C.2)

From the analytical expression of the hypergeometric function (Gradshteyn and Ryzhik 2014, §9.14, eq. 1)

$$\begin{aligned} F (a,b;c;t) = \sum _{j = 0}^{\infty } \frac{(a)^{\bar{j}} (b)^{\bar{j}}}{(c)^{\bar{j}}} \frac{t^{j}}{j!}, \end{aligned}$$

(C.3)

it is easy to infer that $F (a,b;c;t) = F (b,a;c;t)$, and therefore, the right-hand side of Eq. (C.2) equals Eq. (72).