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Did you know you can do this with Liouville's number?

Liouville's number is transcendental. Nonetheless one can still make an infinite series using it to produce an integer! I came across the procedure in a previous message. Can anyone figure out how I could have done it? Here is an example.

The infinite series is

(313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - 
  Sum[28/171*10^(-n! + 6), {n, 5, Infinity}] == 109920468.

A proviso is that this is not a type of series you want Mathematica to evaluate in total. That is because

 N[(313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - Sum[28/171*10^(-n! + 6), {n, 5, 5}]
   -
  ((313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - Sum[28/171*10^(-n! + 6), {n, 5, Infinity}])]

gives 0. instead of a small residue.

First make a good enough approximation to Liouville's number such that it produces no round-off error in the computation:

l = NSum[10^-n!, {n, 1, 10}, WorkingPrecision -> 8!];

Truncate your infinite series to a finite number of partial sums so that Matheamtica can do arithmetic with it:

f[x_] := (313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - 
   Sum[28/171*10^(-n! + 6), {n, 5, x}];

Finally, approximate your answers:

  N[109920468 - Table[f[x], {x, 5, 7}]]

(* {-1.637426900584795*10^-715, -1.637426900584795*10^-5035,  -2.*10^-40315}*)

I conjectured that one can find such infinite series equaling every integer.

There might be another

POSTED BY: Marvin Ray Burns
3 Replies

A direct proof proves the formulas are good for every x, element of C. And an induction proof gives us a pattern for an infinitude of such formulas. The proofs are simple exercises.

POSTED BY: Marvin Ray Burns

Hold the presses!

I have some doubts in what I did. I'll check it thoroughly tomorrow. If you have any comments before then let me know.

POSTED BY: Marvin Ray Burns

I found a pattern to produce an infinite series using Liouville's number in a nontrivial way to get every integer:

It appears that for any integer,x, x= (2 l + (39 + 50 (-1 + x))/50 - Sum[2*10^-n!, {n, 3, Infinity}])

l = NSum[10^-n!, {n, 1, 10}, WorkingPrecision -> 8!];
  Table[N[x - (2 l + (39 + 50 (-1 + x))/50 - 
     Sum[2*10^-n!, {n, 3, 5}])], {x, -100, 100}]

gives -2.000000000000000*10^-720 for every x.

Also,it appears that for any integer,x, x= (3 l + (67 + 100 (-1 + x))/100 - Sum[3*10^-n!, {n, 3, Infinity}])

Table[N[x - (3 l + (67 + 100 (-1 + x))/100 - 
     Sum[3*10^-n!, {n, 3, 5}])], {x, -100, 100}]

gives -3.000000000000000*10^-720 for every x.

It looks like there are infinite "independent?" such formulas.

Some induction should verify that the infinite series give zero for every integer.

POSTED BY: Marvin Ray Burns
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