[2014] tommy's lost notebook (Chapter II)

[tommy1729]

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