05/09/2014, 11:19 PM

Perhaps this is trivial to Sheldon but some more comments.

Let p(x) be a polynomial.

The growth of f(x) and the growth of f(x) + p(x) are equal.

This could be intresting.

So the half-iterate of exp(x) and exp(x)+1 have the same growth rate.

SO growth works a bit like the concept of convergeance of a limit.

ONLY the " tail " of the expression matters.

Just like conv(lim a0 + a1 + a2 + a3 + ...) = conv(lim a2 + a3 + ...).

Hence when trying to find an asymptotic to exp^[0.5] we could as well use exp(x)-1+x = 0 + 2x + x^2/2 + ... .

This implies that we can play with the fixpoints and dummy variables.

So we can investigate C0 + C1 x + C2 x^2 + x^3/3! + x^4/4! + ...

I only see advantages for this.

However one disadvantage I see is that this might show that trying to find the half-iterate of exp(x) by taking the half-iterate of its truncated Taylor might not be such a good idea for studying the half-iterate for large x ( works fine for small x under some conditions ).

However thats a bit of topic here, but worth mentioning imho.

A further implication is that we can arrive at usefull results if we set f(0)=0.

I mean C0 + C1 x + C2 x^2 + x^3/3! + x^4/4! + ... with C0 = 0 and C1 > 1.

This allows us to use stuff like carleman matrices for example.

although C1,C2 appear as variables , 2 variables ( or finite ) are quite easy too handle.

This bring me to a few other remarks.

f(x) and f(a x) also have the same growth.

This implies that the Taylor coefficients t_n and a^n t_n imply the same growth rate !!

( Many conjectures about what gives an equal growth rate are possible , but its not immediate which ones are the most usefull )

These were algebraic ideas , but some calculus ideas occur too :

For instance use the Laplace transform instead of a Taylor series.

( with x = exp(-s) )

Then the theorems involving that transform can be used too !

For instance the Laplace transform of f ' (x) is very intresting.

It suggests that G(x) and G(x) * (ln(x)^k) have the same growth even when iterated.

HOWEVER one thing seems to put us on our feet again.

We want all derivatives to be nonnegative.

Its not clear those dummy variables can provide us with a solution that is both entire and has all Taylor coefficients nonnegative.

Why positive ?

Because then the Taylor is dominated by the largest coefficients.

Compare with exp vs 2sinh. The zero's coefficients of 2sinh do not effect its growth much. Because neither has negative coefficients !

This positivity removes unnessary up and down jumps in the sizes of the coefficients. So they are easier to approximate.

Also the positivity gives us insight about the functions values at complex imput... because of absolute convergeance !

I assume Sheldon was aware of all that.

But that information needed to be shared for all.

regards

tommy1729

Let p(x) be a polynomial.

The growth of f(x) and the growth of f(x) + p(x) are equal.

This could be intresting.

So the half-iterate of exp(x) and exp(x)+1 have the same growth rate.

SO growth works a bit like the concept of convergeance of a limit.

ONLY the " tail " of the expression matters.

Just like conv(lim a0 + a1 + a2 + a3 + ...) = conv(lim a2 + a3 + ...).

Hence when trying to find an asymptotic to exp^[0.5] we could as well use exp(x)-1+x = 0 + 2x + x^2/2 + ... .

This implies that we can play with the fixpoints and dummy variables.

So we can investigate C0 + C1 x + C2 x^2 + x^3/3! + x^4/4! + ...

I only see advantages for this.

However one disadvantage I see is that this might show that trying to find the half-iterate of exp(x) by taking the half-iterate of its truncated Taylor might not be such a good idea for studying the half-iterate for large x ( works fine for small x under some conditions ).

However thats a bit of topic here, but worth mentioning imho.

A further implication is that we can arrive at usefull results if we set f(0)=0.

I mean C0 + C1 x + C2 x^2 + x^3/3! + x^4/4! + ... with C0 = 0 and C1 > 1.

This allows us to use stuff like carleman matrices for example.

although C1,C2 appear as variables , 2 variables ( or finite ) are quite easy too handle.

This bring me to a few other remarks.

f(x) and f(a x) also have the same growth.

This implies that the Taylor coefficients t_n and a^n t_n imply the same growth rate !!

( Many conjectures about what gives an equal growth rate are possible , but its not immediate which ones are the most usefull )

These were algebraic ideas , but some calculus ideas occur too :

For instance use the Laplace transform instead of a Taylor series.

( with x = exp(-s) )

Then the theorems involving that transform can be used too !

For instance the Laplace transform of f ' (x) is very intresting.

It suggests that G(x) and G(x) * (ln(x)^k) have the same growth even when iterated.

HOWEVER one thing seems to put us on our feet again.

We want all derivatives to be nonnegative.

Its not clear those dummy variables can provide us with a solution that is both entire and has all Taylor coefficients nonnegative.

Why positive ?

Because then the Taylor is dominated by the largest coefficients.

Compare with exp vs 2sinh. The zero's coefficients of 2sinh do not effect its growth much. Because neither has negative coefficients !

This positivity removes unnessary up and down jumps in the sizes of the coefficients. So they are easier to approximate.

Also the positivity gives us insight about the functions values at complex imput... because of absolute convergeance !

I assume Sheldon was aware of all that.

But that information needed to be shared for all.

regards

tommy1729