Let the t ' th iteration of a real entire function f(x) with f(0) = 0 and f ' (0) = 1 be given by

f^[t](x) = 1/n ( f( t^a_1 * x)/(t^a_1) + f( t_a^2 * x)/(t^a_2) + ... + f( t^a_n * x)/(t^a_n) )

for some integer n > 0 and where the a_n are positive reals.

for some interval t element of [a,b].

Clearly it automaticly holds for (lim) t = 0 or t = 1.

Many related ideas can be made.

For instance solve for

***

I had the idea

lim

f^[-n]( ( f( t f^[n](x) )/t + f( t^2 f^[n](x) )/t^2 )/2 )

as a kind of koenings type function giving

f^[t](x) = lim f^[-n]( ( f( t f^[n](x) )/t + f( t^2 f^[n](x) )/t^2 )/2 )

for 0 < t =< 1.

regards

tommy1729

hmm

there is a problem.

if f(x) is entire then so is f( 1/2 x).

but half iterates are usually not entire.

So I guess we are left with special cases and approximations.

Still worth investigating though.

regards

tommy1729

Or perhaps better

with t in interval

the equation or asymptotic :

You know Im starting to get slightly serious when I start using tex.

regards

tommy1729

(06/12/2022, 05:38 PM)tommy1729 Wrote: [ -> ]For instance solve for

ok lets try.

f(0) = 0

f(1) = a

f(f(1)) = f(a) = f(2)/4 + f(4)/8.

f(f(2x)) = f(4x)/4 + f(8x)/8.

f(2) = b, f(4) = c,

f(f(2)) = f(b) = f(4)/4 + f(

/8.

f(a) = b/4 + c/8.

f(b) = c/4 + f(

/8.

f(a)/f(b) = (2b + c)/(2c + f(

)

hmmm

seems like a pattern but idk.

lets use differentiation

f(f(2x)) = f(4x)/4 + f(8x)/8.

f ' (f(2x) ) f ' (2x) 2 = f ' (4x) + f ' (8x).

2 f ' (f(x)) f ' (x) = f ' (2x) + f ' (4x)

hmm

taylor at 0

This gives us a system of equations.

Not sure were it leads us but ...

the first term :

f(x) = a x + ...

f(f(x)) = a^2 x + ...

f(f(x)) = a^2 x + ... = 2 x /4 + ... + 4 x / 8 + ... = 1/2 x + 1/2 x + ... = x + ...

so a = 1.

But we already knew this : f ' (0) = 1 as condition.

But anyways this is solvable.

I think I solved it before actually hmm

well im just thinking out loud.

regards

tommy1729

A little proof

f(0) = 0 , f ' (0) = 1 , f ' ' (0) < 0 , f ' ' (0) = k.

then

$$ f^{[t]}(x) = \lim f^{[-n]} ( f^{[n]}(x) + t k f^{[n]}(x)^2 ) = lim f^{[-n]} ( f^{[n]}(tx) / t ) $$

Proof :

Let the taylor of a general f(x) := a x + b x^2 + ...

(a x + b x^2 + ...)^[t] = a^t x + k a^(t-1) ( a^t - 1 )/(a-1) x^2 + ...

and as can be shown by the lim a to 1 or directly :

(x + b x^2 + ...)^[t] = x + k t x^2 + ... [1]

The taylor series for f( tx )/t = a (tx)/t + k (t x)^2 / t + ...

that reduces to a x + k t x^2 + ... [2]

Now if a = 1 then [1] = [2].

Iterations of f(x) go towards the fixpoints.

Therefore

$$ f^{[t]}(x) = \lim f^{[-n]} ( f^{[n]}(x) + t k f^{[n]}(x)^2 ) = lim f^{[-n]} ( f^{[n]}(tx) / t ) $$

QED

regards

tommy1729

hey tommy

just got home from uni

i took some time to solve for this:

By assumption, $$f^t(x)=\frac{f (t x)}{t},f(0)=0,f(0)=\sum_{n=1}^{\infty}\frac{f_n x^n}{n!},f_n=f^{(n)}(0)$$

Take derivative of x at 0, $$f'(0)^t=f'(0)\Longrightarrow f'(0)=1$$

We can write $$t f^t\left(x\right)=\sum _{n=1}^{\infty } \frac{f_n t^n x^n}{n!}=f (t x)$$

And recall a formula about the Julia function:$$\frac{d f^t(x)}{d t}=\frac{d f^t(x)}{d x}[t^1]f^t(x)\text{ where }[t^1]f^t(x)=\lambda_f(x)$$

Thus by the series and this equation we have, $$[t^1]f^t(x)=f_2\frac{x^2}{2}=\lambda_f(x)=\frac{\frac{\mathrm{d}f^t(x)}{\mathrm{d}t}}{\frac{\mathrm{d}f^t(x)}{\mathrm{d}t}}=\frac{\frac{\mathrm{d}\frac{f(tx)}{t}}{\mathrm{d}t}}{\frac{\mathrm{d}\frac{f(tx)}{t}}{\mathrm{d}t}}=\frac{tx-\frac{f(tx)}{f'(tx)}}{t^2}$$

use tx=y to reduce this, we see that $$f_2\frac{y^2}{2}=y-\frac{f(y)}{f'(y)}$$

solve it as ODE, we can show the only possible solutions are (C1 and f2 are constants):

$$f(x)=\frac{C_1x}{2-f_2x}$$

As f(x) is a rational function, we can solve it very easily now:

if C_1 is not 2, we put t=2 then we suddenly get C1=0 is the only way to equalize the 2 sides, which gives f(x)=0

if C_1 = 2, we suddenly get f_2 can be any constant, so that we have a series of solutions

So we conclude that $$f(x)=0\text{ or }f(x)=\frac{2x}{2-f_2x}$$ are the solutions of this equation.

regards

leo

(06/17/2022, 11:57 AM)Leo.W Wrote: [ -> ]hey tommy

just got home from uni

i took some time to solve for this:

By assumption, $$f^t(x)=\frac{f (t x)}{t},f(0)=0,f(0)=\sum_{n=1}^{\infty}\frac{f_n x^n}{n!},f_n=f^{(n)}(0)$$

Take derivative of x at 0, $$f'(0)^t=f'(0)\Longrightarrow f'(0)=1$$

We can write $$t f^t\left(x\right)=\sum _{n=1}^{\infty } \frac{f_n t^n x^n}{n!}=f (t x)$$

And recall a formula about the Julia function:$$\frac{d f^t(x)}{d t}=\frac{d f^t(x)}{d x}[t^1]f^t(x)\text{ where }[t^1]f^t(x)=\lambda_f(x)$$

Thus by the series and this equation we have, $$[t^1]f^t(x)=f_2\frac{x^2}{2}=\lambda_f(x)=\frac{\frac{\mathrm{d}f^t(x)}{\mathrm{d}t}}{\frac{\mathrm{d}f^t(x)}{\mathrm{d}t}}=\frac{\frac{\mathrm{d}\frac{f(tx)}{t}}{\mathrm{d}t}}{\frac{\mathrm{d}\frac{f(tx)}{t}}{\mathrm{d}t}}=\frac{tx-\frac{f(tx)}{f'(tx)}}{t^2}$$

use tx=y to reduce this, we see that $$f_2\frac{y^2}{2}=y-\frac{f(y)}{f'(y)}$$

solve it as ODE, we can show the only possible solutions are (C1 and f2 are constants):

$$f(x)=\frac{C_1x}{2-f_2x}$$

As f(x) is a rational function, we can solve it very easily now:

if C_1 is not 2, we put t=2 then we suddenly get C1=0 is the only way to equalize the 2 sides, which gives f(x)=0

if C_1 = 2, we suddenly get f_2 can be any constant, so that we have a series of solutions

So we conclude that $$f(x)=0\text{ or }f(x)=\frac{2x}{2-f_2x}$$ are the solutions of this equation.

regards

leo

Thank you !

I will look at the details later ...

regards

tommy1729