(08/09/2014, 10:52 PM)tommy1729 Wrote: I have the feeling Jay's post/idea is not complete yet or still under development , probably there will be an edit or a followup.Yes, still brainstorming, and I'm not attached to any particular notation yet. But I needed something so I could start to formalize...

A little bit strange notation I think , but that might be due to the unfinished brainstorming too.

Quote:My first critical remark is that the analogue for exp does not seem convincing right now ; I do not see (1+1/n)^n.

But I might need to read again and Jay will probably do an edit or followup. Its just a first impression anyway.

(...)

I was referring to Jay's post 30 mainly. He posted nr 31 while I made this reply.

Take exp_1:

Code:

`k E_1(k) E_{1/2}(k) E_{1/3}(k)`

0 1 1 1

1 2 3/2 4/3

2 4 9/4 16/9

3 8 27/4 64/27

exp_1(1) = 2 = (1+1)^1

exp_{1/2}(1) = 2.25 = (1+1/2)^2

exp_{1/3}(1) = 2.370370... = (1+1/3)^3

exp_{1/n}(1) = (1+1/n)^n

I could have defined exp_h that way to begin with, but I defined it as a sequence with a recurrence relation similar to A(k), to help draw the analogy between a_1 = 1.08306 and e_1= e/2 = 1.35914.

(08/09/2014, 11:10 PM)tommy1729 Wrote: Considering your lastest posts Jay , are you still convinced that J(x)/1.08... is a very good approximation to A(x) ?

The reason I wanted to set 1.08306 as a_1 was to make clear what it was. By itself, J(x) is not a perfect approximation of A(k). To Tommy's question, do I consider J(x) a "very good approximation" to A(x)? A "good approximation", yes, but not a "very good approximation".

But by re-imagining A(k) as A_1(k), it makes clear why it's not such a good approximation. In reality, it's not so much an issue of J(x) being a poor approximation of A(k). It's that A_1(k) uses such a large step size, that it's a poor approximation of J(x). In this sense, J(x) is the right function to approximate A_1(k), but not the right function to approximate A(k). The distinction is not mathematical, but metaphysical.

~ Jay Daniel Fox