Computing the MRB constant , using only an Addition and Multiplication loop of Integers
S=(-1)
∞
∑
n=1
n
(-1)
1/n
n
An two one-hour task using Mathematica and AI assistants.
Here is an example of what can be done using an AI assistant and Mathematica . Written as n1/n, nth roots of natural numbers have always intrigued me . I first used the to describe the MRB constant in 1999.
I noticed, the researcher at Wolfram pointed out that my discovery was related to "Power Towers" as I read
MathWorld,
I noticed, the researcher at Wolfram pointed out that my discovery was related to "Power Towers" as I read
MathWorld,
Galidakis,IoannisandWeisstein,EricW."Power Tower."FromMathWorld--AWolframWebResource.https://mathworld.wolfram.com/PowerTower.html
I wanted to make a Mathematica code to effectively use that generalized continued fraction to compute several digits of the MRB constant . I had my Copilot laptop with me so we had the following conversation .
I wanted to make a Mathematica code to effectively use that generalized continued fraction to compute several digits of the MRB constant . I had my Copilot laptop with me so we had the following conversation .
S=Sm[(-1),{n,1,Infinity}]
n
(-1)
1/n
n
Later that night, I had some time with my cell - phone and its Gemini App :
Later that night, I had some time with my cell - phone and its Gemini App :
I pasted Copilot’s code:
Out[]=
2(-1+x)
1+-(-1+)
2
x
2
(-1+x)
2
x
3x(1+x)-(-1+4)
2
(-1+x)
2
x
5x(1+x)--1+9
2
(-1+x)
2
x
7x(1+x)--1+16
2
(-1+x)
2
x
9x(1+x)--1+25
2
(-1+x)
2
x
11x(1+x)--1+36
2
(-1+x)
2
x
13x(1+x)--1+49
2
(-1+x)
2
x
15x(1+x)--1+64
2
(-1+x)
2
x
17x(1+x)--1+81
2
(-1+x)
2
x
19x(1+x)--1+100
2
(-1+x)
2
x
It gave
a[x_] := 2 (x - 1)/b[x, 1]
b[x_, k_] := x^2 + 1 - c[x, k]
c[x_, k_] := ((k^2 x^2 - 1) (x - 1)^2)/((k + 2) x (x + 1) - If[k < 100, c[x, k + 2], 0])
a[x_] := 2 (x - 1)/b[x, 1]
b[x_, k_] := x^2 + 1 - c[x, k]
c[x_, k_] := ((k^2 x^2 - 1) (x - 1)^2)/((k + 2) x (x + 1) - If[k < 100, c[x, k + 2], 0])
Through trail and error, I perfected that Gemini code. I color codded
to get
Now I used Wolfram Alpha to find the first several digits of the MRB constant.
Finally, using a rapidly convergent integral instead of the slowly converging MRB constant sum shown above, I documented that this generalized continued fraction can be used to effectively compute many digits of the MRB constant.