07/02/2014, 08:30 AM

07/02/2014, 12:27 PM

The bottom side represents the real valued part of sexp*(x) where the domain is [0,3]. The range is thus [sexp*(0),sexp*(3)]. sexp*(x) is the ordinary sexp but with sexp(0) = 0.

We assume strictly rising here.

This part is estimated by a truncated real Taylor series with alternating signs.

Now the upper left side is then sexp*(ix).

the upper right side is exp^[3](sexp*(ix)) = sexp*(ix + 3).

The diagonal has lenght 3sqrt(2) and is computed by continuation of sexp*(ix) and taking sexp*((-1)^{1/4} x).

Now the red triangle has contour 0 ( because of cauchy's integral theorem ).

Therefore all sides are defined thus solvable.

Ofcourse we also require

On the bottom side :

sexp*(1) = 1 , sexp*(2) = e ,

On the diagonal side

sexp*(i+1) = exp(sexp*(i)) , sexp*(2i+2) = exp(exp(sexp*(2i))).

Now we can solve the equations and then pick one that satisfies sexp*(z+1) = exp(sexp*(z)).

regards

tommy1729

We assume strictly rising here.

This part is estimated by a truncated real Taylor series with alternating signs.

Now the upper left side is then sexp*(ix).

the upper right side is exp^[3](sexp*(ix)) = sexp*(ix + 3).

The diagonal has lenght 3sqrt(2) and is computed by continuation of sexp*(ix) and taking sexp*((-1)^{1/4} x).

Now the red triangle has contour 0 ( because of cauchy's integral theorem ).

Therefore all sides are defined thus solvable.

Ofcourse we also require

On the bottom side :

sexp*(1) = 1 , sexp*(2) = e ,

On the diagonal side

sexp*(i+1) = exp(sexp*(i)) , sexp*(2i+2) = exp(exp(sexp*(2i))).

Now we can solve the equations and then pick one that satisfies sexp*(z+1) = exp(sexp*(z)).

regards

tommy1729

07/02/2014, 08:14 PM

Is that clear ?

Does everyone know what I mean ?

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For those possibly confused ; the analyticity does not come from the contour integral being 0 , rather the opposite way. We have analyticity because of the truncated Taylor series estimate.

We need to be careful not to approximate a Taylor series that has a more limited region of convergeance then sexp*.

Our domain of analyticity should be larger than the figure and it should not have a natural boundary anywhere.

---

I was thinking about adding a uniqueness condition.

It is not clear what our initial guess should be , that is if we want to use an initial guess.

My idea about methods related to integrals is that if you based your estimate on method A , then finally the result is equivalent to method A.

Perhaps a generalization is favorable.

The idea is imho intresting but not yet mature.

regards

tommy1729

Does everyone know what I mean ?

---

For those possibly confused ; the analyticity does not come from the contour integral being 0 , rather the opposite way. We have analyticity because of the truncated Taylor series estimate.

We need to be careful not to approximate a Taylor series that has a more limited region of convergeance then sexp*.

Our domain of analyticity should be larger than the figure and it should not have a natural boundary anywhere.

---

I was thinking about adding a uniqueness condition.

It is not clear what our initial guess should be , that is if we want to use an initial guess.

My idea about methods related to integrals is that if you based your estimate on method A , then finally the result is equivalent to method A.

Perhaps a generalization is favorable.

The idea is imho intresting but not yet mature.

regards

tommy1729

07/02/2014, 09:00 PM

An important concept is that the equations MUST also hold for :

sexp* ' (z)

This augments the number of equations.

If I am not mistaken by considering sexp* alone you get about N/sqrt(2) equations where N is the number of estimates on the bottom line.

by adding the requirement for the derivative you double the amount of equations so you get about sqrt(2) N equations.

The degree of the estimated truncated Taylor then has a degree somewhere between N / sqrt(2) - C and N sqrt(2) + C2 where the C's are constants.

This makes the idea more serious.

How to make the set of equations converging is another matter but it seems some kind of solvability , existance and uniqueness should be possible.

---

Btw convergance issues for systems of equations reminds me of an idea I had that my friend mick posted on MSE :

http://math.stackexchange.com/questions/...-growing-n

For those who are intrested.

Not sure if it helps here unless someone has a very general theory about convergance for systems of equations.

---

regards

tommy1729

sexp* ' (z)

This augments the number of equations.

If I am not mistaken by considering sexp* alone you get about N/sqrt(2) equations where N is the number of estimates on the bottom line.

by adding the requirement for the derivative you double the amount of equations so you get about sqrt(2) N equations.

The degree of the estimated truncated Taylor then has a degree somewhere between N / sqrt(2) - C and N sqrt(2) + C2 where the C's are constants.

This makes the idea more serious.

How to make the set of equations converging is another matter but it seems some kind of solvability , existance and uniqueness should be possible.

---

Btw convergance issues for systems of equations reminds me of an idea I had that my friend mick posted on MSE :

http://math.stackexchange.com/questions/...-growing-n

For those who are intrested.

Not sure if it helps here unless someone has a very general theory about convergance for systems of equations.

---

regards

tommy1729

07/03/2014, 08:23 AM

Ok to solve the underdetermined issue , I think it is rather better to use

sexp * ^ p (x).

Prime powers are better than derivatives.

The number of primes we solve for depends on the degree and the amount of estimates.

We might also use positive integers perhaps.

We can also solve the white triangle and/or the square to fight the underdetermined issue.

Combining these ideas makes me believe we can solve this.

regards

tommy1729

sexp * ^ p (x).

Prime powers are better than derivatives.

The number of primes we solve for depends on the degree and the amount of estimates.

We might also use positive integers perhaps.

We can also solve the white triangle and/or the square to fight the underdetermined issue.

Combining these ideas makes me believe we can solve this.

regards

tommy1729