1. Nov 25, 2003

### phoenixthoth

what are all the functions $$f$$ that satisfy the following
relations that you can think of?

$$\lim_{x\rightarrow \infty}\frac{f\left( x\right) }{2^{x}}=0$$

$$\lim_{x\rightarrow \infty}\frac{x^{n}}{f\left( x\right) }=0$$ for all $$n\in N$$.

Last edited: Nov 25, 2003
2. Nov 25, 2003

### NateTG

Any function that satisfies
$$\lim_{x \rightarrow \infinity} x^{n+\epsilon} < f(x) < 2^{x*(1-\epsilon)}$$
works. There are $$\mathbb{R}^\mathbb{R}$$ such functions.

Perhaps you had something more specific in mind?

3. Nov 25, 2003

### phoenixthoth

how about infinitely differentiable functions defined on some set containing (0,oo)?

4. Nov 26, 2003

### uart

How about f(x) = a^x : For all a in (1..2).

I'm sure there are other functions, but hey that's a start. :)

5. Nov 26, 2003

### phoenixthoth

thank you. actually, i was thinking xq2bx where b is in (0,1) and q is anything. i think xq may possibly be replaced by any laurant series convergent on [0,oo) but i haven't thought much about that one yet. i can also add a few other 2cx in there as long as all the c's are in (0,1). the condition i'd really like to know if it's satisfiable is some function g and constant r such that g(2^x)=r g(x) that g is invertible with inverse G and f(x):=G(r1/2g(x)) satisfies those limit conditions.

6. Nov 26, 2003

### NateTG

g(2x)=r means that g(x)=r for positive x, so if G is g inverse, then G(r1/2g(x))=G(r3/2) for positive x.
So your f(x) will be poorly defined, undefined or less than or equal to zero for all x > 0, so it cannot meet your limit conditions.

7. Nov 26, 2003

### phoenixthoth

ok, how about if g(2^x)=r g(x), as originally stated? if g were constant, it wouldn't have an inverse, btw. in other words, if
h(x)=2^x, i want to find an invertible g with inverse G such that
g o h = rg on [0,oo) and such that f, defined to be
G o (r.5g), satisfies all the conditions above. (note that f o f = h and i'm really secretly thinking about the half-iterate of 2^x like i am in that other thread.) i think the second limit condition (since it's true for all n) implies that all derivatives of f go to infinity which would seem to suggest that for all n in N we have $$\lim_{x\rightarrow \infty }\frac{f^{\left( n\right) }\left( x\right) }{2^{x}\left( \log 2\right) ^{n}}=0$$ and $$\lim_{x\rightarrow \infty }\frac{f^{\left( n\right) }\left( x\right) }{2^{x}}=0$$.

Last edited: Nov 26, 2003
8. Nov 26, 2003

### NateTG

Then
$${g(x)}^{-1} * g(log_2 x) = k$$
if you differentiate both sides, rearange, and use your identity, you get:
$$ln2*2^x*g'(2^x)=c$$or$$g(x)=0$$
clearly the latter is not acceptable.
$$ln2*2^x*g'(2^x)=c$$
integrate both sides
$$g(2^x)=cx+C=r g(x)$$
so
$$g(x)=c'x+C'$$
Therefore there is no such non-zero function that is differentiable.

9. Nov 26, 2003

### phoenixthoth

if you differentiate both sides, rearange, and use your identity, you get:...

instead of a constant on the right, i get rg'(x) which ends in no conlcusions cuz it just leads us back to g(2^x)=rg(x). one explanation is that you forgot to use the chain rule once and another is that i'm not understanding.

Last edited: Nov 26, 2003
10. Nov 26, 2003

### NateTG

It's me being sloppy.

Let's say we've got some n with 2^a=a. a exists by the mean value theorem since 2^0>0 and 2^1>1. Now, g(2^a)=g(a) so r=1 or g is not well defined. That means that g is not invertible.

11. Nov 26, 2003

### phoenixthoth

a wouldn't be a real number if 2^a=a (2^x > x for all real x). there doesn't seem to be a problem even if 2^x had real fixed points. if x is a fixed point of h (in the equation g o h = rg), it just means that r=1 (in which case, if g is invertible then h is the identity function and so if h is not the identity function then g is not invertible) or g(x)=0. but g wouldn't have to be identically 0 and hence not necessarily not invertible. when i define f to be
G(r0.5g(x)), i guess i should add that r is not in {0,1} so that f is not the identity function f(x)=x or constant f(x)=G(0).

if h does have fixed points, the advantage of expanding a series about a fixed point is that the constant term in the series for a g is 0. that doesn't seem to be a big deal though.

12. Dec 13, 2003

### phoenixthoth

looking for an $$f$$ such that $$f\circ g=rf$$ for some constant $$r$$ that isn't 0 or 1. especially for $$g\left( x\right) =2^{x}$$ with domain R and $$f$$ an invertible function with differentiable inverse (diffeomorphism?).

if $$f$$ satisfies the equation, then so does $$kf$$ for any constant $$k$$ so without loss of generality, $$f\left( 0\right) =1.$$ i'm assuming here that for no $$f$$ is $$f\left( 0\right) =0$$ and i'm replacing $$f$$ with $$f/f\left( 0\right)$$.

since $$f\left( 1\right) =f\left( 2^{0}\right) =rf\left( 0\right) =r$$, $$r=f\left( 1\right)$$. also, under the right conditions, we have $$1=f\left( 0\right) =f\left( 2^{-\infty }\right) =f\left( 1\right) f\left( -\infty \right)$$, so $$f\left( -\infty \right) =1/f\left( 1\right)$$. we also have that $$f\left( g^{n}\left( 0\right) \right) =f\left( 1\right) ^{n}$$ where the first $$n$$ repersents iteration and the second one is exponentiation. from $$g^{n}\left( x\right) =f^{-1}\left( f\left( 1\right) ^{n}f\left( x\right) \right)$$, we have a result about $$g^{n}$$'s derivative:

$$\left( g^{n}\right) ^{\prime }\left( x\right) =\frac{f\left( 1\right) ^{n}f^{\prime }\left( x\right) }{f^{\prime }\left( f^{-1}\left( f\left( 1\right) ^{n}f\left( x\right) \right) \right) }$$.

if i could find either $$f\left( 1\right)$$ or the $$f$$ such that $$f\left( 0\right) =1$$ or determine that it can't exist, that'd be swell.

Last edited: Dec 13, 2003
13. Dec 14, 2003

### phoenixthoth

progress?

similar to what koenings found in 1884...

something that appears to solve some of the conditions above is $$f=\lim_{n\rightarrow \infty }\frac{g^{n}}{f\left( 1\right) ^{n}}$$ . i take it that $$f\left( x\right)$$ would be $$\lim_{n\rightarrow \infty } \frac{g^{n}\left( x\right) }{f\left( 1\right) ^{n}}$$. is that right? here, $$g^{n}$$ is the $$n$$-th iterate of g.

note that $$f\circ g=rf$$ where $$r=f\left( 1\right)$$ becomes $$\left( \lim_{n\rightarrow \infty }\frac{g^{n}}{f\left( 1\right) ^{n}}\right) \circ g=f\left( 1\right) \lim_{n\rightarrow \infty}\frac{g^{n}}{f\left( 1\right) ^{n}}$$ and since $$f\left( 1\right) >1$$ , we can divide by it to get $$\left( \lim_{n\rightarrow \infty }\frac{g^{n}}{f\left( 1\right) ^{n+1}}\right) \circ g=\lim_{n\rightarrow \infty }\frac{g^{n}}{f\left( 1\right) ^{n}}$$, which appears valid if the left hand
side reduces to $$\lim_{n\rightarrow \infty }\frac{g^{n+1}}{f\left( 1\right) ^{n+1}}$$. does it?

i would prefer it if this weren't the solution...

i'm working on a value for $$f\left( 1\right)$$ when $$g\left( x\right) =2^{x}$$. one thing i'm suspecting strongly is that $$\lim_{n\rightarrow \infty }\frac{g^{n}\left( x\right) }{f\left( 1\right)^{n}}$$ does not exist for any $$x$$.

Last edited: Dec 14, 2003
14. Dec 16, 2003

### Sting

Pardon my ignorance but what do "naked ladies" have to do with mathematics?

Is it math slang that I haven't come across yet?

15. Dec 16, 2003

### phoenixthoth

when i write a paper on this, i think i'll call solutions to schroder's equation naked ladies due to their highly desirbale nature. either that or it was just a dumb ploy to make people read it. see how many views this thread got for yourself... i did that so i could get as much help as possible.

16. Dec 16, 2003

### Sting

lol Phoenix,

I'll confess to the "highly desirable nature" but I was just confused by it.

17. Dec 16, 2003

### phoenixthoth

actually, the solution in that it depends on the iterates of g is highly undesirable. so this naked lady, while it appears to solve schroeder, it also appears to force f to be constant (oo, actually), which makes this naked lady not just a *****, or a biatch, or a biznitch, or even a bizniatch, but an ARCHbizniatch-atch , mathematically speaking, of course.

18. Dec 16, 2003

### Hurkyl

Staff Emeritus
For the record, I specifically avoided offering help on this thread because of the ploy.

19. Dec 16, 2003

### phoenixthoth

darn!

20. Dec 22, 2003

### phoenixthoth

finding r and neccessary conditions

i'm lead to believe that the following is a necessary condition on schroder's equation being solvable by arguments in nonstandard analysis (i also made a couple of assumptions):

necessary condion for the existence of a solution:
let $$h\left( x\right) :=\lim_{n\rightarrow \infty }\frac{g\left( x^{n}\right) }{x^{n}}$$. then the necessary condition is that h has a fixed point.

furthermore, this is what r can be.

so the full statement would be that if fog = rf, then $$\lim_{n\rightarrow \infty }\frac{g\left( r^{n}\right) }{r^{n}}=r$$.

what do you think?