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Marvin Ray Burns

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19 years, 254 days
Contact Marvin Ray Burns

 I've been using Maple since 1997 or so.

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These are replies submitted by Marvin Ray Burns

I added a little theory behind the integ...

Marvin Ray Burns 550
July 14 2021

I added a little theory behind the integrated analog of the MRB constant.

 

Tell me Maple wrote it and I would be ev...

Marvin Ray Burns 550
July 09 2021

Tell me Maple wrote it and I would be even more impressed!

The connection from the MRB constant sum...

Marvin Ray Burns 550
July 03 2021

This connection from the MRB constant sum in the Reals, to the Complex M2 integral seems remarkable but it also follows from

 

@Marvin Ray Burns 

@acer Okay. I asked how simila...

Marvin Ray Burns 550
June 24 2021

@acer Okay.

 

I asked how similar can the Abel-Plana integrals be to the MRB constant summation formula. 

Also. if there are any special qualities of the base function in each, i.e. x^(1/x)   - 1? ( I think I am finding qualities of expressions that's value go to 0, as the independent variable goes to infinity. Not sure if there is anything else.)

Here is what I came up with. I call it The Abel-Plana formula on steroids.

It is provable from the first few messages and the fact that .

many related integrals...

Marvin Ray Burns 550
June 21 2021

 

Here are how many related integrals correspond to the MRB constant (CMRB).

 

 

From the MRB constant (CMRB)   we...

Marvin Ray Burns 550
June 17 2021

From the MRB constant (CMRB)eta equals

 

we have

 

 

       ...

Marvin Ray Burns 550
June 15 2021

 

 

 

 

   You can see it worked ...

Marvin Ray Burns 550
May 27 2021

I asked about the following.

So I found

QED.

 

I got the following feedback from

Diger (https://math.stackexchange.com/users/427553/diger), Analytically prove these Abel-Plana type integral equations, URL (version: 2021-05-29): https://math.stackexchange.com/q/4155426

I believe that analysis can be simplified to the following.

 

@Fereydoon_Shekofte, thanks for such enc...

Marvin Ray Burns 550
April 16 2021

@Fereydoon_Shekofte, thanks for such encouraging words! 

The MRB constant and its integrated analog were the results of a climax of frustration in my life. It truly fulfills me to see people absorbing some of what I worked so long on! You may build upon my foundation to your heart's content. If you haven't looked the MRB and MKB constants up already, search engines and Google Scholar can help you see all that I have produced. It may be important to note that some of what I have said in the past is woefully incomplete because I started my math journey with only a high school education.

Enjoy all my math!

Let's look at the MRB constant's...

Marvin Ray Burns 550
April 08 2021

Let's look at the MRB constant's integrated analog.

I discovered that for

g(x)=x^(1/x), 

hypothesis

which is the same as saying

because changing the upper limit to 2N  decreases MI by 2i/π.

 

It was proven kindly by Ariel Gershon.

I used this formula to compute and verify over 50,000 digits of M1.(A notebook explaining how is available upon request.)

Iimofg->1

Cauchy's Integral Theorem

Lim surface h gamma r=0

Lim surface h beta r=0

limit to 2n-1

limit to 2n-

Plugging in equations [5] and [6] into equation [2] gives us:

leftright

Now take the limit as N→∞ and apply equations [3] and [4] : QED He went on to note that

enter image description here

 

Here is a more concise catalog of  ...

Marvin Ray Burns 550
March 10 2021

Here is a more concise catalog of  Abel-Plana Formulas for MKB.

Someone might want to change my code, proving the above to a good degree of certainty, into maple!

Here are the results from another CAS used for checking those formulas:

  f[x_] = (-1)^x (x^(1/x) - 1);

I  CMRB = N[NSum[f[n], {n, 1, Infinity}, Method -> \
"AlternatingSigns", WorkingPrecision -> 100], 50]

0.18785964246206712024851793405427323005590309490014

I ReMKB = 
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] + 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

0.070776039311528803539528021830282001365754696203363

  ReMKB - 
 Im[N[NIntegrate[(f[(1 - I t)] + f[( 1 + I t)])/(Exp[2 Pi t] - 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

0.*10^-51

  NoMKB = 
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] - 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

0.117083603150538316708989912223991228690148398696776

 ReMKB + NoMKB - CMRB

0.*10^-51

  g[x_] = x^(1/x);

  MKB = 
 N[NIntegrate[Exp[I Pi t] g[t], {t, 1, 1 Infinity}, 
    WorkingPrecision -> 100], 50] - I/Pi

0.07077603931152880353952802183028200136575469620336 - 0.68400038943793212918274445999266112671099148265500 I

  ImMKB = -(N[
     I NIntegrate[(g[(1 - t I)] + g[(1 + t I)])/(2 Exp[Pi t]), {t, 0, 
        Infinity}, WorkingPrecision -> 100], 50] + I/Pi)

0.*10^-67 - 0.68400038943793212918274445999266112671099148265500 I

  ReMKB = 
 N[I NIntegrate[(g[(1 - t I)] - g[(1 + t I)])/(2 Exp[Pi t]), {t, 0, 
     Infinity}, WorkingPrecision -> 100], 50]

0.070776039311528803539528021830282001365754696203363

ImMKB + ReMKB - MKB

0.10^-51 + 0.10^-51 I

Forgive me; I've been using another ...

Marvin Ray Burns 550
March 01 2021

Forgive me; I've been using another CAS lately, but I did come up with some new analysis of the MKB constant.

CMKB=M1=

ReMKB=Re(CMKB).

CMRB

Then

g[x_] = x^(1/x); ReMKB = 
N[I NIntegrate[(g[(1-t I)] - g[(1+t I)])/(2 Exp[Pi t] ), {t, 0, Infinity}, WorkingPrecision -> 100], 50]

 
 (* 0.070776039311528803539528021830282001365754696203363.*)

Which is similar to the integral for the MRB constant (CMRB).

 I NIntegrate[(g[(-t I + 1)] - g[(t I + 1)])/(Exp[Pi t] - 
      Exp[-Pi t]), {t, 0, Infinity}, WorkingPrecision -> 100]
    (*0.18785964246206712024851793405427323005590309490013878617200468408947\
    72315646602137032966544331074969.*)

P.S.

enter image description here

f[x_] = (-1)^x (x^(1/x) - 1); ReMKB = 
 Im[N[NIntegrate[(f[(1-I t)] - f[( 1+I t)])/(Exp[2 Pi t] + 1), {t,
      0, Infinity}, WorkingPrecision -> 100], 50]]

(*0.070776039311528803539528021830282001365754696203363*)

 f[x_] = (-1)^x (x^(1/x) - 1);

  CMRB = N[NSum[f[n], {n, 1, Infinity}, Method -> \
"AlternatingSigns", WorkingPrecision -> 100], 50]

 (* 0.18785964246206712024851793405427323005590309490014*)

 ReMKB = 
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] + 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

 (*0.070776039311528803539528021830282001365754696203363*)

 NoMKB = 
 Im[N[NIntegrate[(f[(1 - I t)] - f[( 1 + I t)])/(Exp[2 Pi t] - 1), {t,
           0, Infinity}, WorkingPrecision -> 100], 50]]

 (* 0.117083603150538316708989912223991228690148398696776*)

 ReMKB + NoMKB - CMRB

 (* 0.*10^-51*)

Original code...

Marvin Ray Burns 550
September 24 2014

For translating to Maple it is probably best to use Crandall's original code: 

 

(*Fastest (at RC's end) as of 30 Nov 2012.*)prec = 500000;(*Number of \

required decimals.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, n, end, iprec, xvals, x, pc, cores = 4, 
    tsize = 2^7, chunksize, start = 1, ll, ctab, 
    pr = Floor[1.02 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["end ", end];
   Print[end*chunksize];
   d = N[(3 + Sqrt[8])^n, pr + 10];
   d = Round[1/2 (d + 1/d)];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr, pc = Min[3 pc, pr];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(*N[Exp[Log[ll]/ll],
        pr]*)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> "EvaluationsPerKernel" -> 1]];
    ctab = Table[c = b - c;
      ll = start + l - 2;
      b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
      c, {l, chunksize}];
    s += ctab.(xvals - 1);
    start += chunksize;
    Print["done iter ", k*chunksize, " ", SessionTime[] - T0];, {k, 0,
      end - 1}];
   N[-s/d, pr]];
 
t2 = Timing[MRBtest2 = expM[prec];];
MRBtest2 - MRBtest3
 
(*===============*).

Transcendental numbers that are not ...

Marvin Ray Burns 550
April 14 2014

Transcendental numbers that are not solitaire with respect to base 10:

Liouville's constant.

 

 

As you know the sin(x)~=x for small x, s...

Marvin Ray Burns 550
October 31 2013

As you know the sin(x)~=x for small x, so instead of (10 Mantissa[sin(10^(-100 - 1/x))])^x we are really confronted with (10 Mantissa[10^(-100 - 1/x)])^x.= (10 Mantissa[10^(- 1/x)])^x = 10^(x-1). Thus we have the roots of the powers of 10 that I was concerned about. If I missed anything that you notice let me know.

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