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Naked ladies

  1. Nov 25, 2003 #1
    what are all the functions [tex]f[/tex] that satisfy the following
    relations that you can think of?

    [tex]\lim_{x\rightarrow \infty}\frac{f\left( x\right) }{2^{x}}=0[/tex]

    [tex]\lim_{x\rightarrow \infty}\frac{x^{n}}{f\left( x\right) }=0[/tex] for all [tex]n\in N[/tex].
     
    Last edited: Nov 25, 2003
  2. jcsd
  3. Nov 25, 2003 #2

    NateTG

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

    Perhaps you had something more specific in mind?
     
  4. Nov 25, 2003 #3
    how about infinitely differentiable functions defined on some set containing (0,oo)?
     
  5. Nov 26, 2003 #4

    uart

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    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. :)
     
  6. Nov 26, 2003 #5
    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.
     
  7. Nov 26, 2003 #6

    NateTG

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    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.
     
  8. Nov 26, 2003 #7
    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 [tex]\lim_{x\rightarrow \infty }\frac{f^{\left( n\right) }\left( x\right) }{2^{x}\left( \log 2\right) ^{n}}=0[/tex] and [tex]\lim_{x\rightarrow \infty }\frac{f^{\left( n\right) }\left( x\right) }{2^{x}}=0[/tex].
     
    Last edited: Nov 26, 2003
  9. Nov 26, 2003 #8

    NateTG

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    Then
    [tex]{g(x)}^{-1} * g(log_2 x) = k[/tex]
    if you differentiate both sides, rearange, and use your identity, you get:
    [tex]ln2*2^x*g'(2^x)=c[/tex]or[tex]g(x)=0[/tex]
    clearly the latter is not acceptable.
    [tex]ln2*2^x*g'(2^x)=c[/tex]
    integrate both sides
    [tex]g(2^x)=cx+C=r g(x)[/tex]
    so
    [tex]g(x)=c'x+C'[/tex]
    which contradicts your original assumption.
    Therefore there is no such non-zero function that is differentiable.
     
  10. Nov 26, 2003 #9
    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 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
  11. Nov 26, 2003 #10

    NateTG

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    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.
     
  12. Nov 26, 2003 #11
    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.
     
  13. Dec 13, 2003 #12
    looking for an [tex]f[/tex] such that [tex]f\circ g=rf[/tex] for some constant [tex]r[/tex] that isn't 0 or 1. especially for [tex]g\left( x\right) =2^{x}[/tex] with domain R and [tex]f[/tex] an invertible function with differentiable inverse (diffeomorphism?).

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



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

    [tex]\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) }[/tex].


    if i could find either [tex]f\left( 1\right) [/tex] or the [tex]f[/tex] such that [tex]f\left( 0\right) =1[/tex] or determine that it can't exist, that'd be swell.
     
    Last edited: Dec 13, 2003
  14. Dec 14, 2003 #13
    progress?

    similar to what koenings found in 1884...

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

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

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

    i'm working on a value for [tex] f\left( 1\right)[/tex] when [tex] g\left( x\right) =2^{x}[/tex]. one thing i'm suspecting strongly is that [tex]\lim_{n\rightarrow \infty }\frac{g^{n}\left( x\right) }{f\left( 1\right)^{n}}[/tex] does not exist for any [tex]x[/tex].
     
    Last edited: Dec 14, 2003
  15. Dec 16, 2003 #14
    Pardon my ignorance but what do "naked ladies" have to do with mathematics?

    Is it math slang that I haven't come across yet?
     
  16. Dec 16, 2003 #15
    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.
     
  17. Dec 16, 2003 #16
    lol Phoenix,

    I'll confess to the "highly desirable nature" but I was just confused by it.
     
  18. Dec 16, 2003 #17
    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.
     
  19. Dec 16, 2003 #18

    Hurkyl

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    For the record, I specifically avoided offering help on this thread because of the ploy.
     
  20. Dec 16, 2003 #19
  21. Dec 22, 2003 #20
    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 [tex]h\left( x\right) :=\lim_{n\rightarrow \infty }\frac{g\left( x^{n}\right) }{x^{n}}[/tex]. 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 [tex]\lim_{n\rightarrow \infty }\frac{g\left( r^{n}\right) }{r^{n}}=r[/tex].

    what do you think?
     
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