Group Abstract

Message Boards

WOLFRAM COMMUNITY

7.4K Views

4 Replies

0 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Mathematics Mathematica Numerical Computation

Exploring the Upper End of the Partial Sums of the MRB constant in Mma

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

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

4 Replies

Sort By:

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

Attachments:

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

f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 400] As x grows without bound f[10^x] == 10^-x x Log[10]/2 leads to the following acceleration method for the partial sums of the MRB constant Here is an example: In[225]:= m = NSum[(-1)^n(n^(1/n) - 1), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 400]; standard[x_] := (NSum[(-1)^n (n^(1/n) - 1), {n, 1, 10^x}, Method -> "AlternatingSigns", WorkingPrecision -> 500]) plus[x_] := (standard[x] - 10^-x x Log[10]/2) Table[N[{(m - standard[x]), (m - plus[x])}, 30], {x, 0, 100, 10}] Out[228]= {{0.187859642462067120248517934054, 0.187859642462067120248517934054}, \ {-1.1512925477674327431942923944410^-9, \ -1.2704099011852966670971490324410^-18}, \ {-2.3025850929940456845369183402710^-19, \ -5.1892688558286957271227705642210^-38}, \ {-3.4538776394910685260269871821410^-29, \ -1.1759076866601842097462400015510^-57}, \ {-4.6051701859880913680359829093710^-39, \ -2.0979833932614187473844866408910^-77}, \ {-5.7564627324851142100449786367110^-49, \ -3.2851540053865731855508165996910^-97}, \ {-6.9077552789821370520539743640510^-59, \ -4.7374195230356475242452298779110^-117}, \ {-8.0590478254791598940629700914010^-69, \ -6.4547799462086417634677264755710^-137}, \ {-9.2103403719761827360719658187410^-79, \ -8.4372352749055559032183063926610^-157}, \ {-1.0361632918473205578080961546110^-88, \ -1.0684785509126389943496969629210^-176}, \ {-1.1512925464970228420089957273410^-98, \ -1.31974306488711438843037161851*10^-196}} That table looks like We see,,for x>0, the simple 10^-x x Log[10]/2 increases the accuracy exponentialy!!

f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 400]

As x grows without bound f[10^x] == 10^-x x Log[10]/2 leads to the following acceleration method for the partial sums of the MRB constant

enter image description here

Here is an example:

In[225]:= 
m = NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 400];

standard[x_] := (NSum[(-1)^n (n^(1/n) - 1), {n, 1, 10^x}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 500])

plus[x_] := (standard[x] - 10^-x x Log[10]/2)

Table[N[{(m - standard[x]), (m - plus[x])}, 30], {x, 0, 100, 10}]

Out[228]= {{0.187859642462067120248517934054, 
  0.187859642462067120248517934054}, \
{-1.15129254776743274319429239444*10^-9, \
-1.27040990118529666709714903244*10^-18}, \
{-2.30258509299404568453691834027*10^-19, \
-5.18926885582869572712277056422*10^-38}, \
{-3.45387763949106852602698718214*10^-29, \
-1.17590768666018420974624000155*10^-57}, \
{-4.60517018598809136803598290937*10^-39, \
-2.09798339326141874738448664089*10^-77}, \
{-5.75646273248511421004497863671*10^-49, \
-3.28515400538657318555081659969*10^-97}, \
{-6.90775527898213705205397436405*10^-59, \
-4.73741952303564752424522987791*10^-117}, \
{-8.05904782547915989406297009140*10^-69, \
-6.45477994620864176346772647557*10^-137}, \
{-9.21034037197618273607196581874*10^-79, \
-8.43723527490555590321830639266*10^-157}, \
{-1.03616329184732055780809615461*10^-88, \
-1.06847855091263899434969696292*10^-176}, \
{-1.15129254649702284200899572734*10^-98, \
-1.31974306488711438843037161851*10^-196}}

That table looks like

enter image description here

We see,,for x>0, the simple 10^-x x Log[10]/2 increases the accuracy exponentialy!!

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 previous reply we saw something interesting about f[10^10^x], f[10^10^x] being isomorphic to 10^x Log[10]/2: as x grows without bound f[10^10^x]10^10^x == 10^x Log[10]/2, but those arguments are so big and the results are so small, it is hard to conceptualize! I just now stumbled on to something interesting about the easier to imagine f[10^x] : As x grows without bound f[10^x] == 10^-x x Log[10]/2 . The former is just a special case of the later.. As always, your comments are welcome! f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 400] You can see as x grows without bound f[10^x] == 10^-x x Log[10]/2 in the following two pieces of code and results: First we will compute some values of f: In[126]:= Table[N[f[10^(x)], 30], {x, 0, 20}] Out[126]= {0.187859642462067120248517934054, \ 0.133553295001578355542577811018, -0.976341352833974415772015902634, \ 0.00346732160172991953948973985134, \ 0.000460749704412491717791977788008, \ 0.0000575682039917382492456392183551, 6.9078062002067804052655249682410^-6, 8.0590546982608351670189711165410^-7, 9.2103412638316423827916176272410^-8, 1.0361633030767459618328096116610^-8, 1.1512925478775619980975766589910^-9, 1.2664218013131896530726817546510^-10, 1.3815510558161800191366095116410^-11, 1.4966803104484420805788963268110^-12, 1.6118095650960995808678329693710^-13, 1.7269388197455649246597749051410^-14, 1.8420680743952400300325875896910^-15, 1.9571973290449392240134178637710^-16, 2.0723265836946411595727302959410^-17, 2.1874558383443434047089277185710^-18, 2.3025850929940456845594441912010^-19} Then look the the difference between f[10^x] and x/2 Log[10] 10^-x. Notice the negative magnitude is doubled already, as x gets to 20. In[133]:= Table[N[f[10^x] - x/2 Log[10] 10^-x, 30], {x, 0, 20}] Out[133]= {0.187859642462067120248517934054, \ 0.0184240403518760713416782382842, -0.999367203763914872612195817181, \ 0.0000134439622388510134627526693130, 2.3268581368258098837949707138610^-7, 3.5766668871071451894319879486310^-9, 5.0921224643353211550604183875810^-11, 6.8727816752729560010251456549610^-13, 8.9185545964671965180849886542610^-15, 1.1229425404024713457055735536610^-16, 1.3805391560885809316484155081410^-18, 1.6646452686278645457166756430010^-20, 1.9752608725814638831272844983710^-22, 2.3123859672018812604201406330510^-24, 2.6760205523895109554146677190410^-26, 3.0661646281314128252636031327110^-28, 3.4828181944259410512530818625710^-30, 3.9259812512728900413479799831610^-32, 4.3956537986722345078579082818010^-34, 4.8918358366239713819682477673510^-36, 5.41452736512810029562830508680*10^-38} Any comments as to why this is, would be welcome!

in the previous reply we saw something interesting about f[10^10^x], f[10^10^x] being isomorphic to 10^x Log[10]/2: as x grows without bound f[10^10^x]*10^10^x == 10^x Log[10]/2, but those arguments are so big and the results are so small, it is hard to conceptualize! I just now stumbled on to something interesting about the easier to imagine f[10^x] : As x grows without bound f[10^x] == 10^-x x Log[10]/2 . The former is just a special case of the later.. As always, your comments are welcome!

f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, 
      Method -> "AlternatingSigns", WorkingPrecision -> 400]

You can see as x grows without bound f[10^x] == 10^-x x Log[10]/2 in the following two pieces of code and results:

First we will compute some values of f:

In[126]:= Table[N[f[10^(x)], 30], {x, 0, 20}]

Out[126]= {0.187859642462067120248517934054, \
0.133553295001578355542577811018, -0.976341352833974415772015902634, \
0.00346732160172991953948973985134, \
0.000460749704412491717791977788008, \
0.0000575682039917382492456392183551, 
 6.90780620020678040526552496824*10^-6, 
 8.05905469826083516701897111654*10^-7, 
 9.21034126383164238279161762724*10^-8, 
 1.03616330307674596183280961166*10^-8, 
 1.15129254787756199809757665899*10^-9, 
 1.26642180131318965307268175465*10^-10, 
 1.38155105581618001913660951164*10^-11, 
 1.49668031044844208057889632681*10^-12, 
 1.61180956509609958086783296937*10^-13, 
 1.72693881974556492465977490514*10^-14, 
 1.84206807439524003003258758969*10^-15, 
 1.95719732904493922401341786377*10^-16, 
 2.07232658369464115957273029594*10^-17, 
 2.18745583834434340470892771857*10^-18, 
 2.30258509299404568455944419120*10^-19}

Then look the the difference between f[10^x] and x/2 Log[10] 10^-x. Notice the negative magnitude is doubled already, as x gets to 20.

In[133]:= Table[N[f[10^x] - x/2 Log[10] 10^-x, 30], {x, 0, 20}]

Out[133]= {0.187859642462067120248517934054, \
0.0184240403518760713416782382842, -0.999367203763914872612195817181, \
0.0000134439622388510134627526693130, 
 2.32685813682580988379497071386*10^-7, 
 3.57666688710714518943198794863*10^-9, 
 5.09212246433532115506041838758*10^-11, 
 6.87278167527295600102514565496*10^-13, 
 8.91855459646719651808498865426*10^-15, 
 1.12294254040247134570557355366*10^-16, 
 1.38053915608858093164841550814*10^-18, 
 1.66464526862786454571667564300*10^-20, 
 1.97526087258146388312728449837*10^-22, 
 2.31238596720188126042014063305*10^-24, 
 2.67602055238951095541466771904*10^-26, 
 3.06616462813141282526360313271*10^-28, 
 3.48281819442594105125308186257*10^-30, 
 3.92598125127289004134797998316*10^-32, 
 4.39565379867223450785790828180*10^-34, 
 4.89183583662397138196824776735*10^-36, 
 5.41452736512810029562830508680*10^-38}

Any comments as to why this is, would be welcome!

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

I think I found something interesting! First lets define f to give several digits from the upper end of the partial sums of the MRB constant : f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 2000] Define g such that g[u] is approximately f[10^x]: g[u_] := (f[10^1000]/10^(-997 + u))u: Discover that f[10^10^x]10^10^x is isomorphic to 10^x Log[10]/2, as x gets large. I mean the mantissa of f[10^10^x] becomes Log[10]/2, as x gets large: In[69]:= N[Log[10]/2, 20] Out[69]= 1.1512925464970228420 In[57]:= Table[N[g[10^x], 20], {x, 0, 5}] Out[57]= {0.11512925464970228420, 1.151292546497022842010^-9, 1.151292546497022842010^-98, 1.151292546497022842010^-997, 1.151292546497022842010^-9996, 1.151292546497022842010^-99995} Discover when f[10^10^x]10^10^x == 10^x Log[10]/2: In[62]:= Table[N[f[10^10^x]10^10^x - 10^x Log[10]/2, 20], {x, 1, 3}] Out[62]= {1.380539156088580931610^-8, 1.331205990352084616910^-96, 1.326049923892848014110^-994} Try it for f of great value: In[60]:= Table[N[g[10^x]10^10^x - 10^x Log[10]/2, 20], {x, 0, 8}] Out[60]= {1.326049923892848014110^-997, 1.326049923892848014110^-996, 1.326049923892848014110^-995, 1.326049923892848014110^-994, 1.326049923892848014110^-993, 1.326049923892848014110^-992, 1.326049923892848014110^-991, 1.326049923892848014110^-990, 1.326049923892848014110^-989} I think the last set of answers is governed to "limit out" in the neighborhood of 10^-996 by the formula for g. Any help here would be great!

I think I found something interesting!

First lets define f to give several digits from the upper end of the partial sums of the MRB constant :

f[x_] := NSum[(-1)^n (n^(1/n) - 1), {n, x, Infinity}, 
      Method -> "AlternatingSigns", WorkingPrecision -> 2000]

Define g such that g[u] is approximately f[10^x]:

g[u_] := (f[10^1000]/10^(-997 + u))*u:

Discover that f[10^10^x]*10^10^x is isomorphic to 10^x Log[10]/2, as x gets large. I mean the mantissa of f[10^10^x] becomes Log[10]/2, as x gets large:

In[69]:= N[Log[10]/2, 20]

Out[69]= 1.1512925464970228420

In[57]:= Table[N[g[10^x], 20], {x, 0, 5}]

Out[57]= {0.11512925464970228420, 1.1512925464970228420*10^-9, 
 1.1512925464970228420*10^-98, 1.1512925464970228420*10^-997, 
 1.1512925464970228420*10^-9996, 1.1512925464970228420*10^-99995}

Discover when f[10^10^x]*10^10^x == 10^x Log[10]/2:

In[62]:= Table[N[f[10^10^x]*10^10^x - 10^x Log[10]/2, 20], {x, 1, 3}]

Out[62]= {1.3805391560885809316*10^-8, 1.3312059903520846169*10^-96, 
 1.3260499238928480141*10^-994}

Try it for f of great value:

In[60]:= Table[N[g[10^x]*10^10^x - 10^x Log[10]/2, 20], {x, 0, 8}]

Out[60]= {1.3260499238928480141*10^-997, 
 1.3260499238928480141*10^-996, 1.3260499238928480141*10^-995, 
 1.3260499238928480141*10^-994, 1.3260499238928480141*10^-993, 
 1.3260499238928480141*10^-992, 1.3260499238928480141*10^-991, 
 1.3260499238928480141*10^-990, 1.3260499238928480141*10^-989}

I think the last set of answers is governed to "limit out" in the neighborhood of 10^-996 by the formula for g. Any help here would be great!

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

Reply to this discussion

Reply Preview

Attachments

Remove Add a file to this post

Follow this discussion

or Discard

Feedback