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[2014] tommy's lost notebook (Chapter II)
#1
As mentioned before my 2sinh method comes from my lost notebook.
Most of the notebook was number theory but there were also things about tetration.

I recall a big part of the notebook.
So its time to post yet another C^oo solution to tetration.

Due to lack of time and the huge amount of pages I cannot post all right now in a single post.

There are 2 ways to compute the tetration (method) and they are equivalent.

Method 1 assumes analyticity and analytic continuation.
(it uses Taylor series)

Method 2 depends strongly on convergeance assumptions.

The method is basicly a way to solve the equation

f ' (x) = Continuum product f(x)

Or f ' (x) ~ f(x) f(x-1) f(x-2) ... if you like.

It then follows f(x) should be tetration (base e) or at least close to it.

I will focus on method 2 here.

It is based on a real sequence a_n.

a_0 a_1 ...

Forgive any typos.

***
Method 2

k is a postive integer.
The larger k , the better the approximation.
So lim k -> +oo is the solution.

Let sexp(1) = sexp ' (1) = 1

Let a_(n+m) > a_n (a_n is strictly increasing).

g(k) ~ k

a_0 = 1
a_1 = 1 +1/g(k)
a_2
a_3
...
a_k = exp(a_0) = e
a_(k+1) = exp(a_1) = exp(1+1/g(k))

k(a_(k+1)-a_k) = a_1 a_k

k(a_(km + q) - a_(km + q -1)) = a_(km + q - 1)[k(a_(k(m-1) + q) - a_(k(m-1) + q -1))]

where n = km + q

and q = n mod k

then

sexp(1+x) = lim a_(x/k)

to be continued ...

regards

tommy1729
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#2
As for method 1 :

d sexp(x) dx = Continuum Product sexp(x).

(CP is short for Continuum product, CS is short for Continuum sum)

Remember from post 1 : sexp(1) = sexp ' (1) = 1.

So we have a parabolic fixpoint at 1.

The reason for this choice will become Obvious ...

for x > 1
sexp ' (x) dx / f ' (1) = CP^x_1 (sexp(t)) dt

This must be the case because sexp '(1) / 1 = the empty product = 1.

Otherwise we get a paradox near x = 1 : The LHS must = the RHS.

Now take ln on both sides :

ln(sexp ' (x)) = CS^x_1 (ln(sexp(t))) dt

This reduces to

ln(sexp ' (x)) = CS^x_0 (ln(sexp(t))) dt - CS^1_0 (ln(sexp(t))) dt

Let the CONSTANT C = CS^1_0 (ln(sexp(t))) dt

Then we get :

ln(sexp ' (x)) = CS^x_0 (ln(sexp(t))) dt - C

Now take the derivative on both sides :

sexp '' (x) / sexp ' (x) = [ CS^x_0 (ln(sexp(t))) dt ] '

SO we need to solve for f(x) :

f '' (x) / f ' (x) = [ CS^x_0 ( ln(f(t)) ) dt ] '

Write g(x) = f(ln(x))

then we get :

[D^2 g(exp(x))] / [D g(exp(x))] = D [ CS^x_0 ( ln(g(exp(t))) ) dt ]

The reason i work with this exp and ln here is to have a simple way of taking both the derivative and the continuum sum of a series expansion.

More specifically we use Mike3 's method for the continuum sum.

The coefficients can then be solved for but we need analytic continuation.

Once the equation is solved we can regain our sexp by remembering the C constant.

This method assumes a nonzero radius.

Also it is assumed that any analytic V(x) that satisfies V ' (x) = V(x) * V(x-1) * V(x-2) * ...

Or any analytic W(x) that satisfies W ' (x) = W(x) * ln(W(x)) * ln(ln(W(x))) * ...

must be tetration base e.

Another way to solve the equation

[D^2 g(exp(x))] / [D g(exp(x))] = D [ CS^x_0 ( ln(g(exp(t))) ) dt ]

Is by using the discrete analogue ...

... and that is method 2 from post 1.

Looks promising.

regards

tommy1729
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#3
About method 2 from post 1.

There is a difference between f(x) f(x-id(1)) f(x - id(2)) ... f(x - id(q))

Where q satisfies q is a positive integer such that x - i(q) >= 0

AND the continuum product.

example :

2^17.5 - 1 equals the continuum product of 2^n from 0 to 16.5.

2^17.5 - 1 =/= 2^16.5 + 2^15.5 + 2^14.5 + ... 2^0.5.

If f(z) is real-analytic for Re(z) > -1 then the ratio between the continuum product (CP) and the product f(x) f(x-id(1)) f(x - id(2)) ... f(x - id(q)) remains bounded on the real line ... " usually".

Therefore method 2 is conjectured to give a result that is

O ( sexp(x)^(1+o(1)) ) * O(C^x) if the parameter k goes to +oo.

Method 1 however is exact , but it might have a natural boundary , in which case the functional equation sexp(z+1) = exp(sexp(z)) starts too loose meaning.

Hope that was insightfull.

regards

tommy1729
Reply


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