Functions Satisfying the Given Relations

[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 $$n\in N$$.

 

[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?

 

[phoenixthoth]

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

 

[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. :)

 

[phoenixthoth]

thank you. actually, i was thinking x^q2^bx where b is in (0,1) and q is anything. i think x^q 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 2^cx 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(r^1/2g(x)) satisfies those limit conditions.

 

[NateTG]

g(2^x)=r means that g(x)=r for positive x, so if G is g inverse, then G(r^1/2g(x))=G(r^3/2) 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.

 

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

 

[NateTG]

Originally posted by phoenixthoth
ok, how about if g(2^x)=r g(x), as originally stated?

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
$$g(x)=c'x+C'$$
which contradicts your original assumption.
Therefore there is no such non-zero function that is differentiable.

 

[phoenixthoth]

question about something you wrote:
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 because 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.

 

[NateTG]

Originally posted by phoenixthoth
question about something you wrote:
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 because 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.

It's me being sloppy.

How about this:

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.

 

[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(r^0.5g(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.

 

[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.

 

[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 $$\lim_{n\rightarrow \infty }<br /> \frac{g^{n}\left( x\right) }{f\left( 1\right) ^{n}}$$. 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 $$\left( \lim_{n\rightarrow \infty }\frac{g^{n}}{f\left( 1\right)<br /> ^{n+1}}\right) \circ g=\lim_{n\rightarrow \infty }\frac{g^{n}}{f\left( 1\right) ^{n}}$$, 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 $$x$$.

 

[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?

 

[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.

 

[Sting]

lol Phoenix,

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

 

[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.

 

[Hurkyl]

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.

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

 

[phoenixthoth]

darn!

 

[phoenixthoth]

finding r and necessary 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?

 

[one_raven]

Originally posted by Hurkyl
For the record, I specifically avoided offering help on this thread because of the ploy.

I would say that too (if I actually could help with this thread).
It is like when I get spam that is supposed to sound like it came from some girl that must have inadvertently written down the wrong email address when she was chatting with Bob online and accidently emailed me a link to the naked pictures she promised him. Lucky me!

Even if I was going to go to that site to begin with, that email would have disuaded me.
Anyway, sorry for taklking it all way off topic.

On second thought, I guess it really isn't off topic at all.
The topic IS "naked ladies", and was a ploy.

 

[phoenixthoth]

this does work for g(x)=ax+b but doesn't seem to for g(x)=x^a.