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A look at Liouville's number

LATEST EDIT: Below I thought I found one way the MRB constant and Liouville's number are closely related. I have my doubts now because the constants used in the formulas( MRB ,Pi and E) cancel themselves out. BTW, I found you don't need the FractionalPart if you remove the -9/10. If anyone can still use the use the connection between the near integer found by 78389363/MRB constant and the appearance of s(defined below) /MRB constant to calculate more digits of the MRB constant, I would be happy to see your work!

EDIT: This situation seems to describe Liouville's number more than the MRB constant, but there is still the connection with the MRB constant, with the near integer, that I bring up first.

A bit of background is that I previously noticed to following near integer:

mRB = NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 120];

 78389363/mRB

(*
4.17276227999999994569754457408249715437109801508228868441973891743091\
2647962437195343830064089476272991351989858628605*10^8*)

Consider the MRB constant (mRB) and Liouville's number (lusvl).

 mRB = 
 NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}, 
  Method -> "AlternatingSigns", WorkingPrecision -> 120]

(*
0.18785964246206712024851793405427323005590309490013878617200468408947
72315646602137032966544331074969038423458562593959*)


 lusvl = NSum[10^-n!, {n, 1, 10}, WorkingPrecision -> 120]



(*  0.11000100000000000000000100000000000000000000000000000000000000000000
    0000000000000000000000000000000000000000000000000001*)

Let s be the result from the following linear equation of Liouville's number:

 s = (3293 - 3569 lusvl)/37*10^7
(*7.83893629999999999999999035405405405405405405405405405405405405405405\
405405405405405405405405405405405405405405405405404*10^8*)

Look at 

78389363/mRB
 (*4.17276227999999994569754457408249715437109801508228868441973891743091\
 2647962437195343830064089476272991351989858628605*10^8*)

and

 s/mRB

(*
4.17276227999999994569753943942674058671514724997498814125929522496248\
4305587025225348719985350569843522622756397106627*10^9*)

They look very similar!!!!

To show how they are similar enter the following expression:

FractionalPart[78389363/mRB] - 9/10 - FractionalPart[s/mRB]/10 - 
 3569*10^-18/(37 mRB)

(* 0.*10^-110*)

However, if you round off Liouville's number just a little bit (by 10^-120), you only get 0 to 10^-16.

 lusvl2 = 0.110001`120

(* 0.1100010000000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000000*)

 s2 = (3293 - 3569 lusvl2)/37*10^7

(* 7.8389363000000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000*10^8*)

 FractionalPart[78389363/mRB] - 9/10 - FractionalPart[s2/mRB]/10 - 
 3569*10^-18/(37 mRB)

(*-5.134655756567655950765107300543160443692468428342375411969995110078738906429\
468729233461521978*10^-16*)
POSTED BY: Marvin Ray Burns A.G.S. (cum laude)
7 Replies
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It looks like you learned something new. Nothing to be sad about there.

(I have not tried to verify your claim about such sums but I'll assume it is most likely correct.)
POSTED BY: Daniel Lichtblau

Sadly for me, after all my work, it becomes obvious that there are an unlimited number of ways to sum up terms involving Liouville's number with infinite sums rational numbers to arrive at each and every integer! EDIT: The expressions don't have to be as precise in removing fractional parts as mine.

They can simply involve 

10 (Sum[10^-n!, {n, 1, Infinity}] - Sum[10^-n!, {n, 2, Infinity}])

which equals 1
POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Concerning the numerator of the 20th convergent of the MRB constant (109920468), I found an excellent answer to that part of my question, "what can be said about the other numerators of the convergents of [it] in terms of Liouville's number:"

Let l = Liouville's number

Then

 ((3293 + (387/95 + 280/57 l))*10^7/(300) - 4/3*10^-19) - 
  52/171*10^-19 - Sum[28/171*10^(-n! + 6), {n, 5, Infinity}=109920468

The partial sums of

((3293 + (387/95 + 280/57 l))10^7/(300) - 4/310^-19) - 52/17110^-19 - Sum[28/17110^(-n! + 6), {n, 5, b}

are accurate to, very very close to, (b+1)! .

Try

N[((3293 + (387/95 + 280/57 l))*10^7/(300) - 4/3*10^-19) - 
  52/171*10^-19 - Sum[28/171*10^(-n! + 6), {n, 5, 7}], 8! + 5]
POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Code for this message follows,

l = Sum[10^-n!, {n, 1, 9}];sU = (3293 - 3569 l);(sU*10^7/370 + 
  NSum[3569*10^(-n! + 6)/37, {n, 4, Infinity}, 
   WorkingPrecision -> 1000])

.

The previous message should read,

I think the main point to this message is, if you let

sU = (3293 - 3569 Liouville's number),

then

(sU*10^7/370 + Sum[3569*10^(-n! + 6)/37, {n, 4, Infinity}])==78389363; any given partial sum of which gives [delta'] less than the factorial of the upper limit digits of accuracy, where [delta'] is a small integer: such that *(unknown < [delta'] >6).; It's noted that 78389363 is the numerator of the 19th convergent of the MRB constant.

I now ask, "what can be said about the other numerators of the convergents of the MRB constant in terms of Liouville's number?"
POSTED BY: Marvin Ray Burns A.G.S. (cum laude)
Updating Name
Updating Name
Posted 10 years ago

I think the main point to this message is, if you let

sU = (3293 - 3569 Liouville's number);

Then

(sU10^7/370 + Sum[356910^(-n! + 6)/37, {n, 4, Infinity}])==78389363 ; and 78389363 is the numerator of the 19 convergent of the MRB constant.

I now ask, "what can be said about the other numerators of the convergents of the MRB constant in terms of Liouville's number?"
POSTED BY: Updating Name

To get 0 with infinite precision, you have to add additional terms:

In[222]:= 
mRBbig = NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 720];

In[244]:= lusvlBig = NSum[10^-n!, {n, 1, 10}, WorkingPrecision -> 720];

In[245]:= sBig = (3293 - 3569 lusvlBig)/37*10^7;

In[246]:= 
FractionalPart[78389363/mRBbig] - 9/10 - 
 FractionalPart[sBig/mRBbig]/10 - 3569*10^(-18)/(37 mRBbig) - 
 3569*10^(-114)/(37 mRBbig)

Out[246]= 0.*10^-709
POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

There's a lot of meat in my question because you can substitute about any other number in places of mRB and still get "exact looking" answers Try

  In[204]:= N[78389363/E, 120]

Out[204]= \
2.88378350542253373811142519943153245485045715798898133109587453930213\
253684699060839920363331628825224850494268455994347*10^7

In[205]:= N[s/E, 120]

Out[205]= \
2.88378350542253373811142165088632829031627938927456050338280450119422\
266555131126118421644888364416157986120847206767161*10^8

In[202]:= 
FractionalPart[78389363/E] - 9/10 - FractionalPart[s/E]/10 - 
 3569*10^-18/(37 E)

Out[202]= \
-0.8999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999996

and

In[208]:= 
FractionalPart[78389363/Pi] - 9/10 - FractionalPart[s/Pi]/10 - 
 3569*10^-18/(37 Pi)

Out[208]= \
-0.6999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999997

In[209]:= 
FractionalPart[78389363/(1/5)] - 9/10 - FractionalPart[s/(1/5)]/10 - 
 3569*10^-18/(37 (1/5))

Out[209]= \
-0.9999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999952

In[213]:= Table[
 FractionalPart[78389363/(mRB - a/10)] - 9/10 - 
  FractionalPart[s/(mRB - a/10)]/10 - 
  3569*10^-18/(37 (mRB - a/10)), {a, 1, 10}]

Out[213]= {0.*10^-109, \
-1.4000000000000000000000000000000000000000000000000000000000000000000\
000000000000000000000000000000000000000, \
-0.9000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000, \
-1.6000000000000000000000000000000000000000000000000000000000000000000\
000000000000000000000000000000000000000000, \
-1.6000000000000000000000000000000000000000000000000000000000000000000\
000000000000000000000000000000000000000000, \
-1.7000000000000000000000000000000000000000000000000000000000000000000\
000000000000000000000000000000000000000000, \
-1.6000000000000000000000000000000000000000000000000000000000000000000\
000000000000000000000000000000000000000000, \
-1.2000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000, \
-1.4000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000, \
-1.0000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000}

In[214]:= Table[
 FractionalPart[78389363/(E - a/10)] - 9/10 - 
  FractionalPart[s/(E - a/10)]/10 - 3569*10^-18/(37 (E - a/10)), {a, 
  1, 10}]

Out[214]= \
{-0.899999999999999999999999999999999999999999999999999999999999999999\
99999999999999999999999999999999999999999999996, \
-0.2999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999996, \
-0.6999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999996, \
-0.3999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999996, \
-0.2999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999996, \
-0.3999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999995, \
-0.3999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999995, 
 5.*10^-113, \
-0.6999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999995, \
-0.5999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999994}
POSTED BY: Marvin Ray Burns A.G.S. (cum laude)
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