User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-1605058-20170915175750/@comment-25912386-20170916034247

Answers to Denis Maksudov and Alemagno12

1. Consider the function f(a,n)

2. In my work n is limited to 10

3. This means that we are dealing with f(a,10)

4. As a grows, so does the function f(a,10)

5. Now, let us consider the growth of a (the ordinal)  from 0 to Infinity,

6. a grows as such:

0,1,2,3,4,5,6,7,8,9,10=w, 11 = w+1, ..., 19 = w+9, 20 = w+w = w2, ..., w3,...,w^3,...,w^w,

...,w^w2, ..., w^^w,...,...............

7. As you can see from 6, w+9 is the 19-th ordinal as a increases by 1 from an initial value

of 0.

8. If you have time to count, you would see that w^w is the 10^10 or the 10000000000-th

term of the series or 10000000000-th ordinal from the initial value of a being 0

9. f(a,10) is not a function of a given growth rate as Alemagno12 believes wrongly

10. f(a,10) is a function where the ordinal a increases to infinity in step of 1. ( Just think of the         Slow growing hierarchy and yuo guys will understand my point)

11. Unfortunately you all here did not take time to understand the H function

12. H(10000000000) refers to f(a,10) for a = the 10000000000-th term in the series mentioned        in 6.

13. H(10000000000) = h(99999999990,10).

14. In the above point, 9999999990 refers to the 9999999990 -th term after w

15. The 9999999990-th after w in the series given in 6. is the 10000000000-th term

if  you start from the initial value of a being 0

16. So, H(10000000000) = h(9999999990,10) = f ( the 10000000000-th value of a ,10)

17. And this gives you H(10000000000) = f ( w^w,10)

18. Guys ! It is impossible to make things simpler.