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I must precise that I am a very limited user of Mathematica (I can only run it from time when going at university).

Working this problem, I found that $$\sigma_n=(1)^n\frac{\pi}{2} \big( j_{0,n+1} \,\pmb{H}_0\left(j_{0,n+1}\right)\, J_1\left(j_{0,n+1}\right)- j_{0,n}\, \pmb{H}_0\left(j_{0,n}\right)\, J_1\left(j_{0,n}\right)\big)$$ The OP already found $$\sigma_n=\frac{2\sqrt2}{\pi\sqrt n}\left(1-\frac1{8n}+O\left(\frac1{n^2}\right)\right)$$ but wondered about the next terms.

Very tedious regression works made me thinking that the expansion could be something like $$\sigma_n=\frac{2\sqrt2}{\pi\sqrt n}\left(1+\sum_{k=1}^p\frac{a_k}{(8n)^k}+O\left(\frac1{n^{p+1}}\right)\right)$$ and, in my answer, I proposed the values of the first coefficients. But I must confess that I am not totally sure about the third and fourth I suggested.

My question

Is it possible with Mathematica to generate the $a_k$ coefficients (without using stupid regressions as I did with my own tools) the coefficients ?

If this is feasible, I would enjoy learning about the process and, hopefully, get as many of the $a_k$ as possible.

In advance, thanks for your help.

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  • $\begingroup$ You mean the general expression depending on $k$? $\endgroup$
    – Turgon
    Commented May 6, 2019 at 7:47
  • $\begingroup$ @Turgon. No, just a few $a_k$'s (if many, better !). But almost, I am curious to see how it can be done. I still enjoy to learn ! Cheers :-) $\endgroup$
    – Claude Leibovici
    Commented May 6, 2019 at 7:53
  • $\begingroup$ Simply using Series won't work? $\endgroup$
    – Turgon
    Commented May 6, 2019 at 8:03
  • $\begingroup$ @Turgon. I tried in vaine. There is the problem of the expansion of the two functions where sines and cosines appear plus the problem of the expansion of the roots of $J_0$. I would even be happy if this could work with the asymptotics of Bessel and Struve function and with the known asymptotics of the roots of $J_0$. I tried and .... monstreous nightmares. This is why I worked using stupid and tedious regressions. $\endgroup$
    – Claude Leibovici
    Commented May 6, 2019 at 8:29
  • $\begingroup$ The questions arise: What for? Where is that asymptotics applied? $\endgroup$
    – user64494
    Commented May 6, 2019 at 9:28

1 Answer 1

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10
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Exact expression for $\sigma_n$:

b[n_] = BesselJ[1, BesselJZero[0, n]]*BesselJZero[0, n]*StruveH[0, BesselJZero[0, n]];
σ[n_] = π/2*(-1)^n*(b[n + 1] - b[n]);

Check:

Array[σ, 10] // N

{0.801454, 0.599323, 0.49905, 0.436535, 0.392823, 0.360057, 0.334321, 0.313415, 0.295996, 0.281191}

Series expansion of $b_n$ for large $n$:

With[{m = 10},
  bs[n_] = Assuming[n >= 1 && Element[n, Integers], 
    Normal[Series[
      Normal[Series[b[n], {n, ∞, m}]] /.
        {Cos[n*π+z_] -> (-1)^n*Cos[z], Sin[n*π+z_] -> (-1)^n*Sin[z]} /.
        (-1)^(2n) -> 1, {n, ∞, m}]] // FullSimplify]]

2/π + (1/(450971566080 Sqrt[2] n^(19/2) π^10))(-1)^(1 + n) (2523662421852160 (17 + 8 n) - 403696254976 (1105 + 8 n (195 + 16 n (13 + 8 n))) π^2 + 63150080 (21879 + 8 n (6435 + 32 n (429 + 8 n (99 + 16 n (9 + 8 n))))) π^4 - 53760 (36465 + 8 n (15015 + 16 n (3003 + 8 n (1155 + 32 n (105 + 8 n (35 + 16 n (5 + 8 n))))))) π^6 + 105 (12155 + 8 n (6435 + 64 n (429 + 8 n (231 + 16 n (63 + 8 n (35 + 32 n (5 + 8 n (3 + 16 n (1 + 8 n))))))))) π^8)

Assemble approximation for $\sigma_n$:

σs[n_] = Assuming[n >= 1 && Element[n, Integers],
  π/2 (-1)^n (bs[n + 1] - bs[n]) // FullSimplify]

(1/(901943132160 Sqrt[2] π^9))((1/(n^(19/2)))(2523662421852160 (17 + 8 n) - 403696254976 (1105 + 8 n (195 + 16 n (13 + 8 n))) π^2 + 63150080 (21879 + 8 n (6435 + 32 n (429 + 8 n (99 + 16 n (9 + 8 n))))) π^4 - 53760 (36465 + 8 n (15015 + 16 n (3003 + 8 n (1155 + 32 n (105 + 8 n (35 + 16 n (5 + 8 n))))))) π^6 + 105 (12155 + 8 n (6435 + 64 n (429 + 8 n (231 + 16 n (63 + 8 n (35 + 32 n (5 + 8 n (3 + 16 n (1 + 8 n))))))))) π^8) + (1/((1 + n)^(19/2)))(2523662421852160 (25 + 8 n) - 403696254976 (5353 + 8 n (995 + 16 n (37 + 8 n))) π^2 + 63150080 (21879 + 8 (1 + n) (6435 + 32 (1 + n) (3397 + 8 n (771 + 16 n (33 + 8 n))))) π^4 - 53760 (36465 + 8 (1 + n) (15015 + 16 (1 + n) (3003 + 8 (1 + n) (1155 + 32 (1 + n) (2049 + 8 n (579 + 16 n (29 + 8 n))))))) π^6 + 105 (12155 + 8 (1 + n) (6435 + 64 (1 + n) (429 + 8 (1 + n) (231 + 16 (1 + n) (63 + 8 (1 + n) (35 + 32 (1 + n) (5 + 8 (1 + n) (3 + 16 (1 + n) (9 + 8 n))))))))) π^8))

Extract the series coefficients of $\sigma_n/\sqrt{\frac{8}{n\pi^2}}$:

Table[SeriesCoefficient[σs[n]/Sqrt[8/(n π^2)], {n, ∞, i}], {i, 0, 9}] // FullSimplify

{1, -(1/8), 15/128 - 1/π^2, -(65/1024) + 5/(8 π^2), 1435/32768 + (3524 - 525 π^2)/(384 π^4), -((3 (902144 - 116480 π^2 + 2541 π^4))/(262144 π^4)), -(219997/(960 π^6)) + (165 (902144 - 36736 π^2 + 511 π^4))/(4194304 π^4), 2859961/(7680 π^6) - (143 (11727872 - 325248 π^2 + 3279 π^4))/(33554432 π^4), 481350407/(43008 π^8) + (65 (-57670893568 + 33 π^2 (73975808 - 1308160 π^2 + 9843 π^4)))/(2147483648 π^6), -((17 (504732484370432 + 91 π^2 (-749721616384 + 33 π^2 (654956544 - 8394240 π^2 + 49205 π^4))))/(360777252864 π^8))}

numerically:

% // N

{1., -0.125, 0.0158663, -0.000150823, -0.000520557, -5.75437*10^-6, -0.00035554, 0.000711042, -0.000148991, -0.000680426}

It looks like your coefficients were slightly off. The third coefficient, for example, is $15/128 - 1/\pi^2$, not $1/64$.

update: a few more terms

{1,
 -1/8,
 15/128 - 1/π^2,
 -65/1024 + 5/(8 π^2),
 1435/32768 + 881/(96 π^4) - 175/(128 π^2),
 -7623/262144 - 2643/(256 π^4) + 1365/(1024 π^2),
 84315/4194304 - 219997/(960 π^6) + 145365/(4096 π^4) - 47355/(32768 π^2),
 -468897/33554432 + 2859961/(7680 π^6) - 1637779/(32768 π^4) + 363363/(262144 π^2),
 21113235/2147483648 + 481350407/(43008 π^8) - 14299805/(8192 π^6) + 77479545/(1048576 π^4) - 5480475/(4194304 π^2),
 -119617355/17179869184 - 8182956919/(344064 π^8) + 632051381/(196608 π^6) - 777441093/(8388608 π^4) + 39856245/(33554432 π^2),
 1363730225/274877906944 - 18227078923/(20160 π^10) + 777380907305/(5505024 π^8) - 37874463523/(6291456 π^6) + 14852765785/(134217728 π^4) - 2273191635/(2147483648 π^2),
 -7810278567/2199023255552 + 18227078923/(7680 π^10) - 2021190358993/(6291456 π^8) + 782430990341/(83886080 π^6) - 133430737011/(1073741824 π^4) + 15909108215/(17179869184 π^2),
 179637759119/70368744177664 + 371900453534749361/(3406233600 π^12) - 419222815229/(24576 π^10) + 146614039117723/(201326592 π^8) - 10857038287459/(805306368 π^6) + 9212336420001/(68719476736 π^4) - 219560566225/(274877906944 π^2),
 -1036369087075/562949953421312 - 371900453534749361/(1089994752 π^12) + 27249482989885/(589824 π^10) - 2163451065029815/(1610612736 π^8) + 116112943720085/(6442450944 π^6) - 230261614114675/(1649267441664 π^4) + 1496970058675/(2199023255552 π^2),
 11992280893875/9007199254740992 - 36261032568064143407/(1968046080 π^14) + 1859502267673746805/(645922816 π^12) - 257822031365835/(2097152 π^10) + 19578338150476425/(8589934592 π^8) - 3136961678966505/(137438953472 π^6) + 1243496949712875/(8796093022208 π^4) - 40418495801775/(70368744177664 π^2),
 -69555209794785/72057594037927936 + 1051569944473860158803/(15744368640 π^14) - 140206470982600509097/(15502147584 π^12) + 4413158575915683/(16777216 π^10) - 1700204340700414365/(481036337152 π^8) + 30317801368098205/(1099511627776 π^6) - 9834708350336805/(70368744177664 π^4) + 270492331726575/(562949953421312 π^2),
 6468635111995395/9223372036854775808 + 164298065227103543578172039/(39675808972800 π^16) - 32598668278689664922893/(50381979648 π^14) + 13707878816837326697099/(496068722688 π^12) - 137561678475167915/(268435456 π^10) + 158215447115260608855/(30786325577728 π^8) - 563949306923810275/(17592186044416 π^6) + 304878253585009845/(2251799813685248 π^4) - 3593686841197875/(9007199254740992 π^2),
 -37670285661875955/73786976294838206464 - 164298065227103543578172039/(9618377932800 π^16) + 32598668278689664922893/(14092861440 π^14) - 40454959435056500740219/(601295421440 π^12) + 13593732002522826099/(15032385536 π^10) - 1740016292357419096735/(246290604621824 π^8) + 5075429157882741489/(140737488355328 π^6) - 2321759336846553945/(18014398509481984 π^4) + 23718326540021685/(72057594037927936 π^2)}

These can be calculated from a "magic list", whose first nine elements are

Z = {1, 64/3, 112768/315, 901107712/155925, 3943222534144/42567525,
     305799640140218368/206239658625, 389965889965653345959936/16436269594119375,
     1216718351554893671588429824/3205072570853278125,
     5512928257394410409952810366926848/907628475977085565828125}

with the formula

F[n_] := Sum[Z[[i+1]]*(-1)^i*(1+(-3)^(n-2i-1))/(2^(2n-1)*π^(2i+1/2))*
         Gamma[n-1/2]/Gamma[n-2i], {i, 0, (n-1)/2}]

Check:

Array[F, 18]
(* same result as above *)

If you can find a recipe for the "magic list" Z and how to extend it to infinity, you're all done.

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5
  • $\begingroup$ This is fantastic ! Thank you very much. $\endgroup$
    – Claude Leibovici
    Commented May 6, 2019 at 10:18
  • $\begingroup$ You made my day ! $\endgroup$
    – Claude Leibovici
    Commented May 6, 2019 at 10:49
  • 1
    $\begingroup$ Art for art's sake. Cheers! $\endgroup$
    – Roman
    Commented May 6, 2019 at 10:56
  • 2
    $\begingroup$ It even works very well for $n=2$ ! I learnt a lot today thanks to you. Thanks again and again and again (infinite loop). Cheers :-) $\endgroup$
    – Claude Leibovici
    Commented May 6, 2019 at 11:11
  • $\begingroup$ Beautiful would be a real understatement !! $\endgroup$
    – Claude Leibovici
    Commented May 7, 2019 at 7:49

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