Some Particular Cases

We may now use the general formula in Eq. (50) for $\sigma(p,\nu)$ in order to write explicitly a few cases which are not found in the current literature. With a little help from the free-software algebraic manipulation program maxima, we get the following two results, thus completing the sequence of known results up to $p=8$,

$$\sigma(6,\nu) = \frac {21\nu^{3}+181\nu^{2}+513\nu+473} {2^{11}(\nu+1)^{6}(\nu+2)^{3}(\nu+3)^{2}(\nu+4)(\nu+5)(\nu+6)},$$ (66)

$$\sigma(7,\nu) = \frac {33\nu^{3}+329\nu^{2}+1081\nu+1145} {2^{12}(\nu+1)^{7}(\nu+2)^{3}(\nu+3)^{2}(\nu+4)(\nu+5)(\nu+6)(\nu+7)}.$$ (67)

We now point out that our results for $\sigma(p,\nu)$ are valid for all real values of $\nu$, not just for the integers. Therefore, exchanging $\nu$ for $\nu+1/2$ we may obtain formulas that are valid for the zeros of the regular spherical Bessel functions $j_{\nu}(\xi)$, since we have the well-known relation

$$j_{\nu}(\xi) = \sqrt{\frac{\pi}{2\xi}}\, J_{\nu+1/2}(\xi)$$ (68)

between these two families of functions. In particular, using the value $\nu=1/2$ we obtain the results for $j_{0}(\xi)$, whose zeros are given by $k\pi$, since this particular function is proportional to $\sin(\xi)$ [7]. In this way we obtain a direct relation between our results and the Riemann zeta function, for certain real integer arguments of $\zeta(z)$. In fact, we have

$$\sigma(p,1/2) = \sum_{k=1}^{\infty}\frac{1}{\pi^{2p}k^ {2p}} \Rightarrow$$

$$\zeta(2p) = \pi^{2p}\sigma(p,1/2).$$ (69)

Using the formulas we obtained here for $\sigma(p,\nu)$ in the case $\nu=1/2$, we obtain for example the values

$$\zeta(12) = \frac{691\pi^{12}}{3^{6}\times 5^{3}\times 7^{2}\times 11\times 13},$$ (70)

$$\zeta(14) = \frac{2\pi^{14}}{3^{6}\times 5^{2}\times 7\times 11\times 13}.$$ (71)

Finally, using our general formula for the case $p=9$ and solving for $\sigma(9,\nu)$, once more with some help from the free-software program maxima, we obtain

$$\sigma(9,\nu) = \frac{P_{9}(\nu)}{Q_{9}(\nu)},$$ (72)

where the two polynomials are given by

$$P_{9}(\nu) = 715\,\nu^{7} + 22287\,\nu^{6} + 291104\,\nu^{5} + 2066618\,\nu^{4} +$$

$$+ 8671671\,\nu^{3} + 21789479\,\nu^{2} + 29485822\,\nu + 15144368,$$ (73)

$$Q_{9}(\nu) = 2^{17} (\nu+1)^{9} (\nu+2)^{4} (\nu+3)^{3} (\nu+4)^{2} \times$$

$$\times (\nu+5) (\nu+6) (\nu+7) (\nu+8)^{2} (\nu+9).$$ (74)

Up to the case $p=8$ it is possible and in fact fairly easy to verify the resulting formulas numerically with good precision, with the use of standard computational facilities. However, in this $p=9$ case it is just too difficult to verify this formula by numerical means, except for $\nu=0$, using the usual double-precision floating-point arithmetic. In order to do this one would have to use quadruple precision or better numerical arithmetic. The difficulty seems to lie in the direct numerical calculation of the sum $\sigma(9,\nu)$, not in the evaluation of the ratio of polynomials. Hence, the results discussed here acquire an algorithmic, numerical significance, enabling one to easily calculate the values of the sums.