Proof of a General Formula

It is possible to systematize the resolution process described above to the point where a general formula for the linear combination of the sums $\sigma(p,\nu)$ can be written. This is based on a systematization of the general formula for the ratio of Bessel functions, which is found to be

$$\frac{J_{\nu+p}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \... ...ma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-1)-2q},$$ (49)

where $q_{M}=(p-1)/2$ for odd $p$ and $q_{M}=(p-2)/2$ for even $p$, and for which we will provide proof in what follows. The use of the general formula in Eq. (35) then produces a corresponding general formula for the linear combination of the sums $\sigma(p,\nu)$,

$$\frac{\Gamma(\nu+1)}{2^{p}\Gamma(\nu+p+1)} = \sum_{q=0}^{... ...nu+p-q)}{[(p-1)-2q]!\;q!\;\Gamma(\nu+q+1)}\, \sigma(p-q,\nu).$$ (50)

Note that the left-hand side of this equation can be written as the inverse of a polynomial on $\nu$, with integer coefficients, with the simple use of the properties of the gamma function. On the other hand, the coefficients on the right-hand side can all be written as polynomials on $\nu$, with integer coefficients, since we have for the arguments of the two gamma functions, in the numerator and in the denominator,

$$(\nu+p-q) = (\nu+q+1) + p-2q-1,$$ (51)

where $p-2q-1$ is an integer whose minimum value is $0$ for odd $p$, and $1$ for even $p$. It follows that, once the equation is solved for $\sigma(p,\nu)$, resulting in

$$\frac{\Gamma(\nu+p)}{\Gamma(\nu+1)}\, \sigma(p,\nu) = \frac{\Gamma(\nu+1)}{2^{2p}\Gamma(\nu+p+1)} +$$

$$- \sum_{q=1}^{q_{M}} \frac{(-1)^{q}}{2^{2q}}\, \frac{[(p-1)-q]!\;\Gamma(\nu+p-q)}{[(p-1)-2q]!\;q!\;\Gamma(\nu+q+1)}\, \sigma(p-q,\nu),$$ (52)

and assuming that the previous sums all have this same property, the expression for this sum will have the form of the ratio of two polynomials on $\nu$, with integer coefficients. Hence, since we saw that this is valid for the first three sums, by finite induction it is valid for all the sums.

This set of equations, taken for all strictly positive integer values of $p$, forms an infinite linear system of equations in triangular form, that can be solved iteratively in order to obtain closed forms for $\sigma(p,\nu)$ in a purely algebraic way, in principle for arbitrary integer values of $p$, although for large values of $p$ the algebraic work involved can be very large. However, it is straight, direct algebraic work, well suited for a computer-algebra approach.

We will now prove these two general formulas. Since the general formula in Eq. (50) follows from the general formula in Eq. (49), it suffices to prove the latter. We can do this by finite induction. Since the upper limits of the summations involved depend on the parity of $p$, it is necessary to consider the two cases separately. The first step is to verify that our general formula reproduces the correct results for the first three cases, which we have already derived individually. Applying the general formula in Eq. (49) for $p=1$, in which case we have $q_{M}=0$, we obtain at once

$$\frac{J_{\nu+1}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = 1,$$ (53)

which is obviously the correct result. Applying now the same general formula for $p=2$, for which we also have $q_{M}=0$, we get

$$\frac{J_{\nu+2}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \frac{2(\nu+1)}{\xi_{\nu k}},$$ (54)

which is also the correct result. Finally, applying the general formula for $p=3$, in which case we have $q_{M}=1$, we obtain

$$\frac{J_{\nu+3}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \frac{2^{2}(\nu+2)(\nu+1)}{\xi_{\nu k}^{2}} - 1,$$ (55)

which once more is the correct result. It suffices now to use the recurrence formula of the Bessel functions to show that the formula for $p$ follows from the previous formulas, for $p-1$ e $p-2$. We start with the case in which $p$ is even, and writing explicitly the upper limits of the sums, we have

$$\frac{J_{\nu+p-1}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \sum_{q=0}^{(p-2)/2} (-1)^{q}\, \frac{[(p-2)-q]!\;\Gamma(\nu+p-1-... ...-2)-2q]!\;q!\;\Gamma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-2)-2q},$$ (56)

$$\frac{J_{\nu+p-2}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \sum_{q=0}^{(p-4)/2} (-1)^{q}\, \frac{[(p-3)-q]!\;\Gamma(\nu+p-2-... ...-3)-2q]!\;q!\;\Gamma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-3)-2q}.$$ (57)

Writing now the recurrence formula which gives the function $J_{\nu+p}(\xi_{\nu k})$ in terms of $J_{\nu+p-1}(\xi_{\nu k})$ and $J_{\nu+p-2}(\xi_{\nu k})$, and substituting Eq. (56) and (57), we get, after some manipulation of the indices of the sums,

$$J_{\nu+p}(\xi_{\nu k}) = \frac{2(\nu+p-1)}{\xi_{\nu k}}\,J_{\nu+p-1}(\xi_{\nu k}) - J_{\nu+p-2}(\xi_{\nu k}) \Rightarrow$$

$$\frac{J_{\nu+p}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \frac{2(\nu+p-1)}{\xi_{\nu k}}\, \frac{J_{\nu+p-1}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} - \frac{J_{\nu+p-2}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})}$$

$$= \sum_{q=0}^{(p-2)/2} (-1)^{q}\, \frac {[(p-1)-q]!\;\Gamma(\nu+p-q)} {[(p-1)-2q]!\;q!\;\Gamma(\nu+q+1)} \times$$

$$\hspace{3em} \times \frac {(\nu+p-1)[(p-1)-2q]+q(\nu+q)} {[(p-1)-q][\nu+(p-1)-q]} \left(\frac{2}{\xi_{\nu k}}\right)^{(p-1)-2q}.$$ (58)

It is easy to verify that we have for the second fraction in this sum,

$$\frac {(\nu+p-1)[(p-1)-2q]+q(\nu+q)} {[(p-1)-q][\nu+(p-1)-q]} = 1.$$ (59)

We therefore conclude that

$$\frac{J_{\nu+p}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \... ...ma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-1)-2q},$$ (60)

thus proving the general formula for even $p$. For odd $p$, once more writing explicitly the upper limit of the sums, we start from

$$\frac{J_{\nu+p-1}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \sum_{q=0}^{(p-3)/2} (-1)^{q}\, \frac{[(p-2)-q]!\;\Gamma(\nu+p-1-... ...-2)-2q]!\;q!\;\Gamma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-2)-2q},$$ (61)

$$\frac{J_{\nu+p-2}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \sum_{q=0}^{(p-3)/2} (-1)^{q}\, \frac{[(p-3)-q]!\;\Gamma(\nu+p-2-... ...-3)-2q]!\;q!\;\Gamma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-3)-2q}.$$ (62)

Writing once again the recurrence formula which gives $J_{\nu+p}(\xi_{\nu k})$ in terms of $J_{\nu+p-1}(\xi_{\nu k})$ and $J_{\nu+p-2}(\xi_{\nu k})$, and substituting Eq. (61) and (62), we get, after some similar manipulation of the indices of the sums,

$$J_{\nu+p}(\xi_{\nu k}) = \frac{2(\nu+p-1)}{\xi_{\nu k}}\,J_{\nu+p-1}(\xi_{\nu k}) - J_{\nu+p-2}(\xi_{\nu k}) \Rightarrow$$

$$\frac{J_{\nu+p}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \frac{2(\nu+p-1)}{\xi_{\nu k}}\, \frac{J_{\nu+p-1}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} - \frac{J_{\nu+p-2}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})}$$

$$= \sum_{q=0}^{(p-3)/2} (-1)^{q}\, \frac {[(p-1)-q]!\;\Gamma(\nu+p-q)} {[(p-1)-2q]!\;q!\;\Gamma(\nu+q+1)} \times$$

$$\hspace{3em} \times \frac {(\nu+p-1)[(p-1)-2q]+q(\nu+q)} {[(p-1)-q][\nu+(p-1)-q]} \left(\frac{2}{\xi_{\nu k}}\right)^{(p-1)-2q}$$

$$+ (-1)^{(p-1)/2}.$$ (63)

The second fraction within the last summation, which does not involve factorials, is the same as before, and therefore is equal to $1$. It follows that we have

$$\frac{J_{\nu+p}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \sum_{q=0}^{(p-3)/2} (-1)^{q}\, \frac {[(p-1)-q]!\;\Gamma(\nu+p-q... ...p-1)-2q]!\;q!\;\Gamma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-1)-2q}$$

$$+ (-1)^{(p-1)/2}.$$ (64)

It is not difficult to verify that the additional term that we have here is in fact equal to the argument of the summation in the case $q=(p-1)/2$, so that we may merge it with the summation and this obtain

$$\frac{J_{\nu+p}(\xi_{\nu k})}{J_{\nu+1}(\xi_{\nu k})} = \... ...ma(\nu+q+1)}\, \left(\frac{2}{\xi_{\nu k}}\right)^{(p-1)-2q},$$ (65)

thus proving the general formula in this case. This completes the proof of the general formula in Eq. (49), from which follows the general formula in Eq. (50) for the linear combination of the sums $\sigma(p,\nu)$, which is therefore proven as well.