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Mathematics Wolfram Language Numerical Computation

An operation that sums irrational numbers to get a rational one. Maybe!

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

I know of two separate sequences which give approximations to the MRB constant:as partial sums Let 5^3+50 be just a sample large number f[x_, n_] := (Log[x]/x)^n/n!; g[x_] := (x^(1/x) - 1) then (NSum[(-1)^x (f[x, n]), {x, 1, Infinity}, {n, 1, 5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) and (NSum[(-1)^x g[x], {x, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) both give 100 digits of the constant. According to what I have found out it is unknown whether the constant is irrational. However, here is an operation with a twist that gives a seemingly rational result from the two series: In the following, my first suspicion is that a great number of the terms of the two series cancel each other out Let a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, 5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) (* 18.9738238886687791451003113394815962356462125849140174033724730930372\ 0038803068158297872) and b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, 5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) ( -18.598104603744644904603275471373049775534406395113739831028463724858\ 24592490136115559301) Then a/b + 101/99 gives 0. to a great deal of accuracy. Using another large number we have the dialog that follows: a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, 10^3}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) ( 18.9738238886687791451003113394815962356462125849140174033724730930372\ 0038803068158297872) b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, 10^3}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) ( -18.598104603744644904603275471373049775534406395113739831028463724858\ 24592490136115559301) a/b (\ -1.0202020202020202020202020202020202020202020202020202020202020202020\ 20202020202020202020) a/b + 101/99 ( 0.10^-88) And changing of the upper limit of x to match the upper limit of n gives the following dialog: a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 5^3 + 50}, {n, 1, 5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) (* 17.4647396051636702353546828200037984460733942552094333462726566948090\ 9435574173425497138) b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 5^3 + 50}, {n, 1, 5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) ( -17.118903177338647062377362368122535110507584467977563379019732799862\ 37961602407615091254) a/b + 101/99 (0.10^-88) However, the result is probably only close to being rational!!!!!! Because with a sample smaller number for the upper bound of the summation, u. a/b Is directly related to u. a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) (* 9.58084176556542313267441463676793473285105783990707809477223888856333\ 8809797670898335791) b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) ( -9.2051224806412888921773787686593882727392516501068005224282295203843\ 84346668350470950074) a/b + 51/49 ( 0.10^-89) And again changing of the upper limit of x to match the upper limit of n gives the following dialog: a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 50}, {n, 1, 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) (* 11.6400420049405567712416347589817738386513218016649258684668924395814\ 104491180570624594618) b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 50}, {n, 1, 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]) ( -11.183569769452691799820394180198174864586564083952575834409367245872\ 3355295447999227551691) a/b + 51/49 ( -0.10^-90) So we had an upper limit of 50 and a/b + (50+1)/(50-1)==0. That pattern holds for many small upper limits: Table[ a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 400, NSumTerms -> 100]); b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 400, NSumTerms -> 100]); Print[N[a/b + (u + 1)/(u - 1), 30]], {u, 10, 100, 10}] (* -4.2643703787989088820940334952810^-14 -3.4529697592309312828258770061010^-30 -9.1368337148513601899787158414510^-49 -8.7108841896464038696492075989510^-69 -7.1293101216171095134330107869310^-90 -8.2560200317856508828134965595210^-112 -1.8785111736101254140110312141710^-134 -1.0598026154446875304051045497610^-157 -1.7636731096421343970582774768510^-181 -9.9052940450762519739748614671610^-206) But the 0 is not found by the same pattern for larger upper limits: Table[a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]); b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 100]); Print[a/b + (u + 1)/(u - 1)], {u, 10^3, 10^4, 10^3}] ( -0.018200018200018200018200018200018200018200018200018200018200018200018200018200018200018 -0.019201519951895139488936387385611997918150994689263823831107472928383383610997417900870 -0.019535131239032539466017291959272619492692850478779121660081980188682422760448769118326 -0.019701895170762387566588616851182492592845180992217751407548856911197496343782915425826 $Aborted*) I will try to find out why this all happens. You can help me if you wish! (Even if you do I will no doubt make my own contribution too.) When I make these post I always hope that someone will build upon them and perhaps do better research! Like RICHARD J. MATHAR did with my integral analog of the MRB constant at http://arxiv.org/pdf/0912.3844v3.pdf . My first suspicion was that a great number of the terms of the two series cancel each other out., bu why then does the operation behave so differently with smaller numbers?

I know of two separate sequences which give approximations to the MRB constant:as partial sums

Let 5^3+50 be just a sample large number

f[x_, n_] := (Log[x]/x)^n/n!; g[x_] := (x^(1/x) - 1)

then

 (NSum[(-1)^x (f[x, n]), {x, 1, Infinity}, {n, 1, 5^3 + 50}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

and

 (NSum[(-1)^x g[x], {x, 1, Infinity}, Method -> "AlternatingSigns",
    WorkingPrecision -> 90, NSumTerms -> 100])

both give 100 digits of the constant.

According to what I have found out it is unknown whether the constant is irrational.

However, here is an operation with a twist that gives a seemingly rational result from the two series: In the following, my first suspicion is that a great number of the terms of the two series cancel each other out

Let

 a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, 
    5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

(*
18.9738238886687791451003113394815962356462125849140174033724730930372\
0038803068158297872*)

and

 b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, 
    5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

(*
-18.598104603744644904603275471373049775534406395113739831028463724858\
24592490136115559301*)

Then

   a/b + 101/99

gives 0. to a great deal of accuracy.

Using another large number we have the dialog that follows:

  a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, 
    10^3}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

 (*
18.9738238886687791451003113394815962356462125849140174033724730930372\
0038803068158297872*)

   b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, 
    10^3}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

 (*
-18.598104603744644904603275471373049775534406395113739831028463724858\
24592490136115559301*)

  a/b

 (*\
-1.0202020202020202020202020202020202020202020202020202020202020202020\
20202020202020202020*)

  a/b + 101/99

  (* 0.*10^-88*)

And changing of the upper limit of x to match the upper limit of n gives the following dialog:

   a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 5^3 + 50}, {n, 1, 
    5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

 (*
17.4647396051636702353546828200037984460733942552094333462726566948090\
9435574173425497138*)

  b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 5^3 + 50}, {n, 1, 
    5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

(*
-17.118903177338647062377362368122535110507584467977563379019732799862\
37961602407615091254*)

      a/b + 101/99

      (*0.*10^-88*)

However, the result is probably only close to being rational!!!!!!

Because with a sample smaller number for the upper bound of the summation, u. a/b Is directly related to u.

   a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, 
    50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

(*
9.58084176556542313267441463676793473285105783990707809477223888856333\
8809797670898335791*)

   b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, 
    50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

 (*
-9.2051224806412888921773787686593882727392516501068005224282295203843\
84346668350470950074*)

   a/b + 51/49

 (* 0.*10^-89*)

And again changing of the upper limit of x to match the upper limit of n gives the following dialog:

  a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 50}, {n, 1, 50}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

 (*
11.6400420049405567712416347589817738386513218016649258684668924395814\
104491180570624594618*)

   b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 50}, {n, 1, 50}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 100])

 (*
-11.183569769452691799820394180198174864586564083952575834409367245872\
3355295447999227551691*)

   a/b + 51/49

 (* -0.*10^-90*)

So we had an upper limit of 50 and a/b + (50+1)/(50-1)==0.

That pattern holds for many small upper limits:

 Table[
 a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 400, 
    NSumTerms -> 100]);
 b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 400, 
    NSumTerms -> 100]);
 Print[N[a/b + (u + 1)/(u - 1), 30]], {u, 10, 100, 10}]

(* -4.26437037879890888209403349528*10^-14

 -3.45296975923093128282587700610*10^-30

 -9.13683371485136018997871584145*10^-49

 -8.71088418964640386964920759895*10^-69

 -7.12931012161710951343301078693*10^-90

 -8.25602003178565088281349655952*10^-112

 -1.87851117361012541401103121417*10^-134

 -1.05980261544468753040510454976*10^-157

 -1.76367310964213439705827747685*10^-181

 -9.90529404507625197397486146716*10^-206*)

But the 0 is not found by the same pattern for larger upper limits:

Table[a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 90, 
    NSumTerms -> 100]); 
 b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 90, 
    NSumTerms -> 100]);
 Print[a/b + (u + 1)/(u - 1)], {u, 10^3, 10^4, 10^3}]

(*   -0.018200018200018200018200018200018200018200018200018200018200018200018200018200018200018

-0.019201519951895139488936387385611997918150994689263823831107472928383383610997417900870

-0.019535131239032539466017291959272619492692850478779121660081980188682422760448769118326

-0.019701895170762387566588616851182492592845180992217751407548856911197496343782915425826
  $Aborted*)

I will try to find out why this all happens. You can help me if you wish! (Even if you do I will no doubt make my own contribution too.) When I make these post I always hope that someone will build upon them and perhaps do better research! Like RICHARD J. MATHAR did with my integral analog of the MRB constant at http://arxiv.org/pdf/0912.3844v3.pdf .

My first suspicion was that a great number of the terms of the two series cancel each other out., bu why then does the operation behave so differently with smaller numbers?

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

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Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

Let m be the MRB constant and f and g as shown above. g[x_] := (x^(1/x) - 1); f[x_, n_] := (Log[x]/x)^n/ n!; m = (NSum[(-1)^x g[x], {x, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 300, NSumTerms -> 1000]); Besides a/b+(u+1)/(u-1) going quickly to 0,as u gets large. It makes more sense that (a-b)/(2u) =( (f_u+g_u)-(f_u-g_u))/(2u)==(2g_u)/(2u)==g which when summed =m, Here we see how they start off: Table[(a = NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 400, NSumTerms -> 100]); b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 400, NSumTerms -> 100]); Print[m - (a - b)/(2 u), N[a/b + (u + 1)/(u - 1), 30]], {u, 2, 20}] (2.985010^-295 0.0588044076747937067994021066547 2.985010^-295 0.00165475082899660297675193835286 2.985010^-295 0.0000562290785871773061494180345698 2.985010^-295 1.7864044594268082852996558564310^-6 2.985010^-295 4.7993313248871354999076679355610^-8 2.985010^-295 9.4180489153997025028912802438210^-10 2.985010^-295 4.8637877128230602667984505024010^-12 2.985010^-295 -6.7029449499303079405687659631310^-13 2.985010^-295 -4.2643703787989088820940334952810^-14 2.985010^-295 -1.7603146575111691980868262322510^-15 2.985010^-295 -5.9340223197281623147289360315010^-17 2.985010^-295 -1.7472174231052415627170474731110^-18 2.985010^-295 -4.6325735291187422030874767471910^-20 2.985010^-295 -1.1252584619383136454205049959010^-21 2.985010^-295 -2.5320807022584293168668415684110^-23 2.985010^-295 -5.3202741931032500697008785847110^-25 2.985010^-295 -1.0500829215513752797140420910710^-26 2.985010^-295 -1.9561901414904101296236978255610^-28 2.985010^-295 -3.4529697592309312828258770061010^-30) It also make sense that (a+b)/2 ==( (f+g)+(f-g))/2 == (2f)/2 ==f which when summed ==m Table[(a = NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 200, NSumTerms -> 100]); b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 200, NSumTerms -> 100]); N[m - (a + b)/(2 ), 30], {u, 2, 20}] ({0.00280295020690991487554302477903, \ 0.000207354973003898926467754641480, \ 0.0000118838220054139752268887360275, 5.3694966854861294334828769555610^-7, 1.8783347599579122375166283364110^-8, 4.5495547782464039071157042255210^-10, 2.7982351010293562193098311860710^-12, \ -4.4772012150629365847030050584810^-13, \ -3.2444675334822583901506685155610^-14, \ -1.5031458280944732567295317496510^-15, \ -5.6202650280363791995648108994610^-17, \ -1.8178983161060950034204219525910^-18, \ -5.2527228414018819239438060623010^-20, \ -1.3810855952693614750953656809710^-21, \ -3.3445952958611837902380550875610^-23, \ -7.5253820816304128185992133360710^-25, \ -1.5836252898685962298004966840010^-26, \ -3.1333288027394656880447353132510^-28, \ -5.8542798212231109701187058361710^-30}) BTW (a + b)/(b - a) + 1/u ->0 as u gets large. Table[(a = NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 200, NSumTerms -> 100]); b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, Method -> "AlternatingSigns", WorkingPrecision -> 200, NSumTerms -> 100]); N[(a + b)/(b - a) + 1/u, 30], {u, 2, 20}] ( {0.00746022447976257186735255754231, \ 0.000367925348035246279539761887242, \ 0.0000158147618212005980219200928810, 5.7164983549570479534725300836610^-7, 1.6664345211320806519505284616610^-8, 3.4596914397065323631810021094710^-10, 1.8619187338190397485224548835610^-12, \ -2.6480770172556290667420874857510^-13, \ -1.7270700034135249553503425561410^-14, \ -7.2740275103767264032368975625410^-16, \ -2.4931135440524570158120437175710^-17, \ -7.4437665363063545807208026683310^-19, \ -1.9972064449517026334309810099110^-20, \ -4.9011257453313216556067369666410^-22, \ -1.1127307773596613208887355289710^-23, \ -2.3563844177066297886564439043110^-25, \ -4.6832401902522755530456506607010^-27, \ -8.7784709950539180332143780529310^-29, \ -1.5581526038529577413751769990010^-30}*) I still haven't answered why (a + b)/(b - a) + 1/u ->0 and a/b + (u+1)/(u-1) -> 0, where u is the upper limit of the summations.

Let m be the MRB constant and f and g as shown above.

 g[x_] := (x^(1/x) - 1); 
 f[x_, n_] := (Log[x]/x)^n/
   n!; m = (NSum[(-1)^x g[x], {x, 1, Infinity}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 300, 
    NSumTerms -> 1000]);

Besides a/b+(u+1)/(u-1) going quickly to 0,as u gets large. It makes more sense that (a-b)/(2u) =( (f_u+g_u)-(f_u-g_u))/(2u)==(2g_u)/(2u)==g which when summed =m, Here we see how they start off:

Table[(a = 
   NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 400, 
    NSumTerms -> 100]); 
 b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 400, 
    NSumTerms -> 100]); 
 Print[m - (a - b)/(2 u), N[a/b + (u + 1)/(u - 1), 30]], {u, 2, 20}]

(*2.9850*10^-295     0.0588044076747937067994021066547

2.9850*10^-295      0.00165475082899660297675193835286

2.9850*10^-295     0.0000562290785871773061494180345698

2.9850*10^-295     1.78640445942680828529965585643*10^-6

2.9850*10^-295     4.79933132488713549990766793556*10^-8

2.9850*10^-295     9.41804891539970250289128024382*10^-10

2.9850*10^-295     4.86378771282306026679845050240*10^-12

2.9850*10^-295      -6.70294494993030794056876596313*10^-13

2.9850*10^-295     -4.26437037879890888209403349528*10^-14

2.9850*10^-295     -1.76031465751116919808682623225*10^-15

2.9850*10^-295     -5.93402231972816231472893603150*10^-17

2.9850*10^-295     -1.74721742310524156271704747311*10^-18

2.9850*10^-295     -4.63257352911874220308747674719*10^-20

2.9850*10^-295      -1.12525846193831364542050499590*10^-21

2.9850*10^-295     -2.53208070225842931686684156841*10^-23

2.9850*10^-295     -5.32027419310325006970087858471*10^-25

2.9850*10^-295     -1.05008292155137527971404209107*10^-26

2.9850*10^-295      -1.95619014149041012962369782556*10^-28

2.9850*10^-295     -3.45296975923093128282587700610*10^-30*)

It also make sense that (a+b)/2 ==( (f+g)+(f-g))/2 == (2f)/2 ==f which when summed ==m

Table[(a = 
    NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, 
     Method -> "AlternatingSigns", WorkingPrecision -> 200, 
     NSumTerms -> 100]);
  b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, 
     Method -> "AlternatingSigns", WorkingPrecision -> 200, 
     NSumTerms -> 100]);
  N[m - (a + b)/(2 ), 30], {u, 2, 20}] 


(*{0.00280295020690991487554302477903, \
0.000207354973003898926467754641480, \
0.0000118838220054139752268887360275, 
 5.36949668548612943348287695556*10^-7, 
 1.87833475995791223751662833641*10^-8, 
 4.54955477824640390711570422552*10^-10, 
 2.79823510102935621930983118607*10^-12, \
-4.47720121506293658470300505848*10^-13, \
-3.24446753348225839015066851556*10^-14, \
-1.50314582809447325672953174965*10^-15, \
-5.62026502803637919956481089946*10^-17, \
-1.81789831610609500342042195259*10^-18, \
-5.25272284140188192394380606230*10^-20, \
-1.38108559526936147509536568097*10^-21, \
-3.34459529586118379023805508756*10^-23, \
-7.52538208163041281859921333607*10^-25, \
-1.58362528986859622980049668400*10^-26, \
-3.13332880273946568804473531325*10^-28, \
-5.85427982122311097011870583617*10^-30}*)

BTW (a + b)/(b - a) + 1/u ->0 as u gets large.

   Table[(a = 
   NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 200, 
    NSumTerms -> 100]);
 b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u}, 
    Method -> "AlternatingSigns", WorkingPrecision -> 200, 
    NSumTerms -> 100]);
 N[(a + b)/(b - a) + 1/u, 30], {u, 2, 20}]

 (* {0.00746022447976257186735255754231, \
0.000367925348035246279539761887242, \
0.0000158147618212005980219200928810, 
 5.71649835495704795347253008366*10^-7, 
 1.66643452113208065195052846166*10^-8, 
 3.45969143970653236318100210947*10^-10, 
 1.86191873381903974852245488356*10^-12, \
-2.64807701725562906674208748575*10^-13, \
-1.72707000341352495535034255614*10^-14, \
-7.27402751037672640323689756254*10^-16, \
-2.49311354405245701581204371757*10^-17, \
-7.44376653630635458072080266833*10^-19, \
-1.99720644495170263343098100991*10^-20, \
-4.90112574533132165560673696664*10^-22, \
-1.11273077735966132088873552897*10^-23, \
-2.35638441770662978865644390431*10^-25, \
-4.68324019025227555304565066070*10^-27, \
-8.77847099505391803321437805293*10^-29, \
-1.55815260385295774137517699900*10^-30}*)

I still haven't answered why (a + b)/(b - a) + 1/u ->0 and a/b + (u+1)/(u-1) -> 0, where u is the upper limit of the summations.

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

In the original post the upper limits of 5^3 and 5^3 + 50 gave the same results; Mathematica didn't see the difference in them. The reason for the change in the rule (a/b + (u + 1)/(u - 1)==0) for large upper limits is "u" (the NSumTerms setting).. NSumTerms -> 500 helps produce a/b + (u + 1)/(u - 1)==0 for larger u. In[276]:= a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 5^3 + 50}, {n, 1, 5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 1000]) b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 5^3 + 50}, {n, 1, 5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, NSumTerms -> 1000]) a/b + (5^3 + 50 + 1)/(5^3 + 50 - 1) Out[276]= \ 30.4336056486020392220041997655511735297912612764045571182573027553108\ 970951539131608684145 Out[277]= \ -30.087769220777016049026879313669910194225451489172687151004378860364\ 1823554362550567676370 Out[278]= 0.*10^-90 OK now, all we have to do is find out why a/b + (u+1)/(u-1) == 0, where u is the upper limit of the summations.

In the original post the upper limits of 5^3 and 5^3 + 50 gave the same results; Mathematica didn't see the difference in them. The reason for the change in the rule (a/b + (u + 1)/(u - 1)==0) for large upper limits is "u" (the NSumTerms setting).. NSumTerms -> 500 helps produce a/b + (u + 1)/(u - 1)==0 for larger u.

In[276]:= a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 5^3 + 50}, {n, 1, 
    5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 1000])



b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 5^3 + 50}, {n, 1, 5^3 + 50},
    Method -> "AlternatingSigns", WorkingPrecision -> 90, 
   NSumTerms -> 1000])



a/b + (5^3 + 50 + 1)/(5^3 + 50 - 1)



Out[276]= \
30.4336056486020392220041997655511735297912612764045571182573027553108\
970951539131608684145

Out[277]= \
-30.087769220777016049026879313669910194225451489172687151004378860364\
1823554362550567676370

Out[278]= 0.*10^-90

OK now, all we have to do is find out why a/b + (u+1)/(u-1) == 0, where u is the upper limit of the summations.

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

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