Let $J_\nu(x):=\displaystyle\sum^\infty_{k=0}\frac{(-1)^k(x/2)^{\nu+2k}}{k!~\Gamma(\nu+k+1)}$ denote a Bessel function. When $\nu\geq0$, let $0<j_{\nu,1}<j_{\nu,2}<\cdots$ denote the positive zeroes of $J_\nu(x)$. My questions are:
$(a)$ Keeping $\nu$ fixed, is it known how the $j_{\nu,k}$'s are distributed on the real line (that is, how fast they increase, whether they accumulate somewhere, etc)?
$(b)$ Also, is there a lower or upper bound on the expression $J_{\nu+1}(j_{\nu,k})$ when $\nu$ is held fixed?
Any help is highly appreciated, thanks!
To be honest, I am absolutely stuck on this. I just know the definition of the Bessel function. I have no clue how to make any conclusions about the zeros of a function from an infinite series. The reason I am interested in this is, the terms $j_{\nu,k}$ and $J_{\nu+1}(j_{\nu,k})$ appear in a formula for an $n$-dimensional Bessel process.
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$\begingroup$I did some googling, and found that a good $($slightly old$)$ reference for this kind of questions is G.N. Watson's “A treatise on the theory of Bessel functions”. In the $1922$ edition that I currently have access to, the relevant theory appears in Chapter XV. See particularly $15.22$, $15.4$ and $15.81$. This does not answer all the questions, of course, but it's a decent amount of information.
In particular, $15.81$ gives an answer to how fast the zeros of $J_\nu(x)$ grow. $15.22$ tells us that the zeros of $J_\nu$ and $J_{\nu + 1}$ are interlaced. But this does not particularly answer question $(b)$ above.
$\endgroup$ 1 $\begingroup$I can help with part (a) of your questions. The distribution of zeroes for the Bessel functions (at least for First Kind, unsure of other ones) on the real line is known. Rather than calculate each zero, I used the scipy module in python to plot the zeroes, from which the pattern can be recognized. Also note that the rate of dampening for fixed nu decreases as x approaches infinity, at which point the spacing between consecutive zeroes approaches pi. For comparison purposes, pi = 3.1415926535897931 (16 decimal digits). From the same python code:
zeros for nu = 0 :
--------------------
[ 2.40482556 5.52007811 8.65372791 11.79153444 14.93091771 18.07106397 21.21163663 24.35247153 27.49347913 30.63460647 33.77582021 36.91709835 40.05842576 43.19979171 46.34118837 49.4826099 52.62405184 55.76551076 58.90698393 62.04846919 65.1899648 68.33146933 71.4729816 74.61450064 77.75602563 80.89755587 84.03909078 87.18062984 90.32217264 93.46371878]
consecutive differences in zeros:
- - - - - - - - - - - - - - - - -
[ 3.11525255 3.1336498 3.13780653 3.13938327 3.14014626 3.14057266 3.1408349 3.1410076 3.14112734 3.14121375 3.14127814 3.14132741 3.14136595 3.14139666 3.14142153 3.14144194 3.14145891 3.14147317 3.14148526 3.14149561 3.14150453 3.14151227 3.14151904 3.14152499 3.14153024 3.14153491 3.14153907 3.14154279 3.14154614]
zeros for nu = 1 :
--------------------
[ 3.83170597 7.01558667 10.17346814 13.32369194 16.47063005 19.61585851 22.76008438 25.90367209 29.04682853 32.18967991 35.33230755 38.47476623 41.61709421 44.759319 47.90146089 51.04353518 54.18555364 57.32752544 60.46945785 63.6113567 66.75322673 69.89507184 73.03689523 76.17869958 79.32048718 82.46225991 85.60401944 88.74576714 91.88750425 95.02923181]
consecutive differences in zeros:
- - - - - - - - - - - - - - - - -
[ 3.1838807 3.15788147 3.1502238 3.14693811 3.14522846 3.14422587 3.14358771 3.14315645 3.14285138 3.14262764 3.14245868 3.14232798 3.14222478 3.14214189 3.1420743 3.14201846 3.1419718 3.14193241 3.14189885 3.14187004 3.1418451 3.14182339 3.14180436 3.14178759 3.14177274 3.14175952 3.14174771 3.14173711 3.14172756]
zeros for nu = 2 :
--------------------
[ 5.1356223 8.41724414 11.61984117 14.79595178 17.95981949 21.11699705 24.27011231 27.42057355 30.5692045 33.71651951 36.86285651 40.00844673 43.15345378 46.29799668 49.44216411 52.58602351 55.72962705 58.87301577 62.01622236 65.15927319 68.30218978 71.44498987 74.58768817 77.73029706 80.87282695 84.01528671 87.15768394 90.30002515 93.44231602 96.58456145]
consecutive differences in zeros:
- - - - - - - - - - - - - - - - -
[ 3.28162184 3.20259703 3.17611061 3.16386771 3.15717756 3.15311526 3.15046124 3.14863095 3.14731501 3.146337 3.14559022 3.14500704 3.1445429 3.14416743 3.1438594 3.14360355 3.14338872 3.14320659 3.14305083 3.14291659 3.14280008 3.14269831 3.14260888 3.14252989 3.14245976 3.14239723 3.14234122 3.14229087 3.14224543]
zeros for nu = 3 :
--------------------
[ 6.3801619 9.76102313 13.01520072 16.22346616 19.40941523 22.58272959 25.7481667 28.90835078 32.06485241 35.21867074 38.37047243 41.52071967 44.66974312 47.81778569 50.96502991 54.11161557 57.2576516 60.40322414 63.54840218 66.69324167 69.83778844 72.9820804 76.12614918 79.27002139 82.41371955 85.55726287 88.70066784 91.84394868 94.98711773 98.13018573]
consecutive differences in zeros:
- - - - - - - - - - - - - - - - -
[ 3.38086123 3.25417759 3.20826544 3.18594907 3.17331437 3.16543711 3.16018408 3.15650163 3.15381833 3.1518017 3.15024724 3.14902345 3.14804257 3.14724421 3.14658566 3.14603603 3.14557253 3.14517804 3.14483949 3.14454677 3.14429196 3.14406878 3.14387221 3.14369816 3.14354332 3.14340497 3.14328084 3.14316905 3.14306801]At least in the limit of large $j_{\nu,m}$, one can answer both questions.
From the Digital Library of Mathematical Function, 10.21(iv), we see that for large $m$, $$ j_{\nu,m}\simeq \left(m+\frac\nu2-\frac14\right)\pi+\mathcal O(m^{-1}), $$which works already quite well for $\nu=0$ already for $m=1$!
Concerning $J_{\mu}(j_{\nu,m})$, one can use the expansion of the Bessel functions for large argument,$$ J_{\mu}(z)\simeq \sqrt{\frac{2}{\pi z}}\cos\left(z-\frac{\mu\pi}2-\frac\pi4\right)+\mathcal O(z^{-1}), $$to obtain$$ J_{\mu}(j_{\nu,m})\simeq\frac{2 \sqrt{2}}{\pi } \frac{(-1)^m}{\sqrt{4 m+2 \nu -1}} \sin \left(\frac{\pi}{2} ( \nu -\mu)\right) +\mathcal O(m^{-1}). $$
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