10,824 Pages

## REPLACED

This blog has been replaced. A new Alpha Function blog should be referred to instead. The blog can be found here.

## (previous) Alpha Function

The Alpha Function has one parameter: $$\alpha(r)$$ where r is any real number. It is derived from the Strong D Function with a variable number of input parameters. This blog replaces the old description of the Alpha Function.

## What is the Alpha Function

My motivation to create this function was to develop a finely grained number notation system for really big numbers. $$\alpha(1)$$ for example can be used to reference the number 2. Therefore 1 is the Alpha Index for the number 2. Alpha needs to reference big numbers very quickly to be useful, therefore it uses the Strong D Function for this purpose. Alpha should also be strictly hierarchical and every number a > b, must reference larger numbers, so that $$\alpha(a) >> \alpha(b)$$ in all cases. The function is finely grained. It accepts a real number input and offers some finesse to locate large big numbers.

The Alpha Function is defined recursively based on a series of binary decisions.  The logic can be explained by referring to the following notation.

## Some Notation

Following notation helps to explain the behaviour of Strong D Functions and the logic of the Alpha Function.

$$D(m_{[x]}) = D(m_1,m_2,...,m_x)$$

$$D(m_{[x]},n_{[y]}) = D(m_1,m_2,...,m_x,n_1,n_2,...,n_y)$$

$$D(1,0_{[y]}) = D(D(1_{[y]})_{[y]})$$

$$D(m,n_{[y]}) = D(m-1,D(m,n_{[y-1]},n_y-1)_{[y]})$$

$$D(m_{[x]},n,0_{[y]}) = D(m_1-1,D(m_{[x]},n-1,n_{[y]})_{[x+y]})$$ when x>0 and which is equal to

$$= D(m_{[x]},n-1,n_{[y-1]},n_y+1)$$

## Alpha Function Logic

The Alpha Function is defined using the following logic.

$$\alpha(r) = D(D(m_{[x]})_{[x]})$$ where $$2^{x-1} <= r < 2^x$$   and

$$\alpha(2^x) = D(1, 0_{[x]})$$

The values of $$m_{[x]}$$ are calculated based on the value of r but only legal values can be used which follow these restrictions:

• duplicate values should be avoided
• out of sequence results must be avoided

The second constraint is important to force $$\alpha(a) >> \alpha(b)$$ whenever a > b.  Additional logic is derived from the following rules:

Maximum Value Rule: M1

$$D(1,0_{[x]}) = D(D(1_{[x]})_{[x]})$$

therefore

$$D(m_{[x]}) < D(D(1_{[x]})_{[x]})$$

$$<= D(D(1_{[x]})_{[x]})-1$$ or alternatively

$$<= D(D(1_{[x]})_{[x-1]},D(1_{[x-1]},0))$$ Rule M1a

and

$$m_1 >= 1$$   Rule M1b

Maximum Value Rule: M2

$$D(m_{[x]},n+1,0_{[y]}) = D(m_{[x]},n,n_{[y-1]}+1,n_y+2)$$

therefore

$$D(m_{[x]},n,p_{[y]}) < D(m_{[x]},n,n_{[y-1]}+1,n_y+2)$$

$$<= D(m_{[x]},n,n_{[y-1]}+1,n_y+2)-1$$ or alternatively

$$<= D(m_{[x]},n,n_{[y]}+1)$$ Rule M2a

or

$$p_{[y]} <= n_{[y]}+1$$ Rule M2b

Maximum Value Rule: M3

It can also be shown that the only legal values for D functions in the form:

$$D(m_{[x]})$$

are when

$$m_i <= m_{i-1}+1$$   for all   $$3 <= i <= x$$

Maximum Value Rule: M4

The final rule is used for D functions of the form:

$$D(n+1,0_{[y]}) = D(n,D(n,n_{[y]}+1)_{[y]}$$

therefore

$$D(n,p_{[y]}) < D(n,D(n,n_{[y]}+1)_{[y]})$$   $$<= D(n,D(n,n_{[y]}+1)_{[y]}-1)$$   or alternatively

$$<= D(n,D(n,n_{[y]}+1)_{[y-1]},D(n,n_{[y-1]}+1,n))$$   Rule M4a

and

$$p_i <= D(n,n_{[y]}+1)$$   for all   $$1 <= i < y$$   Rule M4b

and

$$p_y <= D(n,n_{[y-1]}+1,n)$$   Rule M4c

## Some Calculations

Program Code was used to calculate these examples.

$$\alpha(0.00) = D() = 0$$

$$\alpha(0.50) = D(0) = 1$$

$$\alpha(1.00) = D(1) = 2$$

$$\alpha(2.00) = D(1,0) = 3$$

$$\alpha(2.08) = D(1,1) = 4$$

$$\alpha(2.16) = D(1,2) = 5$$

$$\alpha(2.56) = D(2,1) = 11$$

$$\alpha(e) = D(2,6) = 26$$

$$\alpha(3.00) = D(3,0) = 59$$

$$\alpha(3.04) = D(3,2) = 563$$

$$\alpha(\pi) = D(3,14) = 301327043$$

$$\alpha(3.52) = D(4,0) >>$$ Googol

$$\alpha(3.6) = D(4,1) >> f_{\omega}(3)$$

$$\alpha(3.64) = D(4,2) >>$$ Googolplex

$$\alpha(4.00) = D(1,0,0)$$

$$\alpha(4.0005) = D(1,0,1) >> g_2$$ where $$g_{64}$$ is Graham's number

$$\alpha(4.008304) = D(1,11,0) >> D(1,9,9) >>$$ Graham's number

$$\alpha(4.1250) = D(2,0,0) >> f_{\omega+1}^2(3)$$

$$\alpha(4.1255) = D(2,0,1) >> f_{\omega+2}(3)$$

$$\alpha(4.2501) = D(3,0,1) >> f_{\omega.2}(3)$$

$$\alpha(6) = D(D(1,0,0),0,0) = D(\alpha(4),0,0)$$

$$\alpha(6.5) = D(D(1,0,1),0,0) = D(D(D(4,4),D(4,4)),0,0)$$

$$\alpha(8) = D(1,0,0,0) = D(D(1,1,1),D(1,1,1),D(1,1,1))$$

$$\alpha(8.tba) = D(D(3,0,1),0,0) >> f_{\omega.2+1}(3)$$ This result is being checked. Work in progress

$$\alpha(9) = D(D(59,0,0),0,0,0) = D(\alpha(5),0,0,0)$$

$$\alpha(16) = D(1,0,0,0,0)$$

$$\alpha(32) = D(1,0,0,0,0,0)$$

$$\alpha(48) = D(D(1,0,0,0,0,0),0,0,0,0,0) = D(\alpha(32),0,0,0,0,0)$$

$$\alpha(125) = D(D(1,1,1,1,0,1,0),0,0,0,0,0,0)$$

$$\alpha(128) = D(1,0,0,0,0,0,0,0)$$

## Program Code and Description

Version 1 of the program code for the Alpha Function is available here. The code is not complete and various errors will be corrected in Version 2 (work in progress).

Look forward to comments and questions. I am learning heaps by writing these blogs and correcting all the mistakes the community finds in them !

Cheers B1mb0w.

## References

The Alpha Function