5
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The following expression when summed from $k=0$ to $\infty$ converges but returns an absolutely hideous expression that I cannot fit in here due to the character limit. I want the numerical value up to some decimal places of the resulting sum. Using N does nothing even after I give Mathematica almost a day. Is there a way to make this thing faster? I need these values for plotting and I have other similar expressions. Any help would be appreciated. Thank you.

expr = -((110206758610268910878544017 \
(186834090059215607761467807732427161471120595214066963979369398128453\
01786169675263126632095592601850865558428520447057406873 + 
       94478400000 k \
(173774794663767077786739776733074395211307157623599746681764426443901\
3981048833102263101600365761661227370784450127 + 
          3110400000 k \
(851351732885176844209899280952048656884104234239429449012154874772745\
0387301925793146265381324235959716193 + 
             6998400000 k \
(654297280278431113651409491369696299387098803813553304690613022199100\
0592262049002232576940068297 + 
                437400000 k \
(392584160230322034373325703017429405584715822946246171162535785968686\
45339637150995662893 + 
                   31492800000 k \
(188596813460722406899470924590034429997036858863722270358111422985127\
5584278109 + 
                    1555200000 k \
(110764150346899809621911474265812494882337420511725914770631585787734\
7 + 6998400000 k \
(82934628836586353165682512404734790410406956507937216317763 + 
                    36960300000 k \
(567887925097805270104116079234293917525705130331 + 
                    82752203809620374118353467515436138958433600000 \
k))))))))) Gamma[5190562591/1166400000]^2 Gamma[49/10] Gamma[99/
      20]^2 Gamma[2886922591/583200000] Pochhammer[388787897/
      388800000, 
      k]^2)/(162188175416956504401847825156904932955566925808588735777\
600000000000000000 2^(
     58237409/
      291600000) (-388812103 + 388800000 k) (-12103 + 
       388800000 k) (758169973 + 3499200000 k) (1108089973 + 
       3499200000 k)^3 (2566208873 + 3499200000 k) (6065408873 + 
       3499200000 k) \[Pi] k! Gamma[12103/388800000]^2 Gamma[19/
      20]^2 Gamma[2857762591/583200000] Gamma[5773762591/
      1166400000]^2 Gamma[89/10] Gamma[2886922591/583200000 + k]))
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6
  • 1
    $\begingroup$ Trying with NSum[] with v12.3 crashed my PC, so I'd advise against it ;-) $\endgroup$
    – Hans Olo
    Commented Mar 22 at 8:47
  • $\begingroup$ @HansOlo it turns out that Sum[] actually gives a closed form (with HypergeometricPFQ[], though). Could one try and do something with N[] there? $\endgroup$
    – QFTheorist
    Commented Mar 22 at 8:55
  • $\begingroup$ N[Expand[expr], 20][[1]] results in -((2.5815278486077617480*10^65 Pochhammer[0.99996887088477366255, k]^2)/((-3.8881210300000000000*10^8 + 3.8880000000000000000*10^8 k) (-12103.000000000000000 + 3.8880000000000000000*10^8 k) (7.5816997300000000000*10^8 + 3.4992000000000000000*10^9 k) (1.1080899730000000000*10^9 + 3.4992000000000000000*10^9 k)^3 (2.5662088730000000000*10^9 + 3.4992000000000000000*10^9 k) (6.0654088730000000000*10^9 + 3.4992000000000000000*10^9 k) k! Gamma[ 4.9501416169410150892 + k])). $\endgroup$
    – user64494
    Commented Mar 22 at 13:10
  • 1
    $\begingroup$ 10^65 is not from the real world. Is not your question art for art's sake? $\endgroup$
    – user64494
    Commented Mar 22 at 13:11
  • $\begingroup$ @HansOlo Nsum does not crash here on 14.0 on Windows $\endgroup$ Commented Mar 25 at 7:55

1 Answer 1

5
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A big caveat emptor for the stuff below, as I'm playing fast and loose with the precision, but it seems to work.

Defining the function numerically as follows

expr[k_] := N[ your expression ,20]

and then crunching some numbers for 5000 terms, gives in a few secs

LaunchKernels[7]
data = ParallelTable[expr[k], {k, 0, 5000, 1}];
data // Total
(* 0.00005241198640366926742 *)

which seems to have converged to several decimals (but do check this). Indeed, successive terms seem to drop exponentially fast as powerlaws!!

enter image description here

I hope this helps.

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4
  • 1
    $\begingroup$ Just a small comment: You are plotting on a log-log scale, so the terms decrease with a power-law, not exponentially :) $\endgroup$
    – Domen
    Commented Mar 22 at 10:57
  • 1
    $\begingroup$ Simplify[Series[N[Expand[expr], 20], {k, Infinity, 1}] // Normal] outputs 2.60492368959135922*10^-7/k^3.950203875171467764 - 4.695454453782571963*10^-8/k^2.950203875171467764. $\endgroup$
    – user64494
    Commented Mar 22 at 13:08
  • $\begingroup$ @Domen You are of course right, apologies!! I'll fix this now :-) $\endgroup$
    – Hans Olo
    Commented Mar 22 at 15:38
  • $\begingroup$ @user64494 Elegant solution! $\endgroup$
    – Hans Olo
    Commented Mar 22 at 15:41

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