Conference rusure::math

    i'd love to see such a function !
    since a function cant have two values for same argument, this function,
    if it exist, must it not have then an infinite number of points of 
    discontinuities?

    i was going to say something like
    let D == d/dt == f
    then f(f(sin(t))= -sin(t)

    but sin(*) has values from -1..1 ,not over all real

    so, gave up..
    /nasser

    I came up with
    
    	f(x) = -1 * ABS(-1/x)
    
    spoilt only by the discontinuity where f(x) is undefined when x=0.
    
    James.

        Let f(0) = 0.  Then use the axiom of choice to produce a
        partition of the positive reals into disjoint pairs.  For
        each such pair {a,b} with 0 < a < b define
        
        	f(a) = b	f(b) = -a	f(-a) = -b	f(-b) = a
        
        Dan

        re .-1,
        
        It is fairly straightforward to see that any f satisfying
        the constraints of .0 must be of a form similar to .-1,
        specifically
        
        	f(0) = 0
        	
        	the positive reals are partitioned into disjoint
        	pairs {a,b}, a < b, such that f maps either
        
        		a -> b -> -a -> -b -> a
        	or
        		a <- b <- -a <- -b <- a
        
        But such partitions can be defined without AC, for
        example, let the a's be in (2n, 2n+1] for nonnegative
        integral n and let b = a + 1; or let the a's be in
        [2^2n, 2^(2n+1)) for signed integral n, and let b = 2a.
        
        Dan 
        

	This problem is already in the Notesfile somewhere, I'm sure, but I
haven't managed to track it down.

Cheers,
Andrew.

Here is the construction of an actual function that works.

Define f(0) = 0.

For non-zero x, use the following construction.

Define S(x) =	{ -1	if x < 0,
		{ 1	if x > 0.

Define [x] to be the truncated value of x. Formally, this means the unique
integer such that 0 <= S(x)(x - [x]) < 1.

So, for all y, 0 <= S(y)(y - [y]) < 1.

Substituting y = x + S(x), we get 0 <= S(x + S(x))(x + S(x) - [x + S(x)]) < 1.

Clearly S(x + S(x)) = S(x), and S(x)S(x) = 1.

Therefore 0 <= S(x)x + 1 - S(x)[x + S(x)] < 1.

Subtracting all terms from 1, and reversing the inequality, we get
	0 < S(x)([x + S(x)] - x) <= 1.

Define R(x) = S(x)([x + S(x)] - x).

Then 0 < R(x) <= 1.

Define R'(x) =	{ R(x) + 1/2	if R(x) <= 1/2,
		{ R(x) - 1/2	otherwise.

So R'(R'(x)) = x, and 0 < R'(x) <= 1.

We can now define the required function f(x) for all real x:

Define f(x) =	{ 0				if x = 0,
		{ [x + S(x)] - S(x)R'(x)	x <> 0 and R(x) <= 1/2,
		{ S(x)R'(x) - [x + S(x)]	otherwise.

Basically what we've done is split the real line into half-open intervals. Our
function maps the bottom half of each interval to the top half, and the top
half to the bottom half reversing the sign.

Nick.

What's wrong with the trivial answer ?  We want f(f(x) = -x.  If we start
with x, and subtract x, we get 0.  Subtract x again, we get -x.  So subtracting
x sounds like a satisfactory function to me.

/Eric

    re .10
    
    It doesn't work.
    
    So f(x) = x-x = 0
    thus f(f(x)) = f(0) = 0
    
    The problem is that the definition of your function is such that the
    second time it is invoked it needs to have remembered the value with
    which it was invoked the first time. This might be OK for a Pascal FUNCTION
    (bad programming practice though) but it's not OK for a mathematical
    function.

    >                 <<< Note 1637.5 by DESIR::BUCHANAN >>>
    >	This problem is already in the Notesfile somewhere, I'm sure, but I
    >   haven't managed to track it down.
    
    (To return a favour) possibly 731, 791 and shades of 971.
    
    I am looking for functions (multi-idempotent? :-) such that
    
    	f^n(x) = x   (but not for any lower n)
    
    Simple examples for n=2 are  -x, 1/x and -1/x
    
    Also if f(x) = (1+x)/(1-x) then f^2(x) = -1/x  so f^4(x) = x
    
    The discontinuous functions mentioned here that give f^2(x) = -x
    obviously also give f^4(x) = x
    
    How about n=3 or >4?
    

    Re .12:
    
    There's a theorem described in _Mathematics Magazine_ 65 Number 2,
    April 1992, pages 91 to 103, that might be of some use.  First, order
    the natural numbers:
    
    	3, 5, 7, . . . 2*3, 2*5, 2*7, . . . 2^2*3, 2^2*5, 2^2*7, . . .
    	16, 8, 4, 2, 1.
    
    That is, all the odd numbers except one, followed by two times the
    odds, followed by 2^2 times the odds, et cetera.  At the end of all
    that, list the powers of two in decreasing order, ending with 1.
    
    Call a point, x, n-periodic if f^n(x)=x, where f^n indicates f composed
    with itself n times.  So f^1(x)=f(x), f^2(x)=f(f(x)), and
    f^n+1(x)=f(f^n(x)).
    
    Sarkovskii's theorem (1964) says that if f is a continuous function
    from an interval I to the interval I and f has an l-periodic point and
    l is before m in the ordering above, then f has an m-periodic point
    too.  I can be any interval, finite or infinite, open or closed,
    semi-open or semi-closed.
    
    Some consequences are:  If f has a periodic point whose period is not a
    power of two, then f must have infinitely many periodic points. 
    Conversely, if f has only finitely many periodic points, then each
    period must be a power of two.
    
    Period 3 is the least period in the Sarkovskii ordering and therefore
    implies the existence of all other periods.
    
    
    				-- edp

>    Some consequences are:  If f has a periodic point whose period is not a
>    power of two, then f must have infinitely many periodic points. 

    Are any of these known to exist?
    If so, are there examples?

    Is there an existence proof that does not depend on the Axiom of Choice?

      John

    We seem to have diverged slightly here. The functions that I am looking
    for are (equally) periodic everywhere or almost everywhere.
    
    (That wasn't mean't to sound ungrateful, Sarkovskii's looks a
    fascinating theorem)
    
    For example (to answer my own previous question):
    
    f(x) = 1-1/x has period 3 everywhere
    
    f(x) = (1+2x)/(1-x) has period 6 everywhere
    
    (as long as I define them suitably at the finite number of points of
    discontinuity of f, f^2 etc).
    
    This does not contradict Sarkovskii (who would predict points of
    higher order) since I can't map continuously *from I to I* with those
    functions.
    
    The second f(x) can be termed a 6 Mars bar function. My son's maths
    teacher rashly promised n Mars bars to anyone finding a function with
    period n > 2. Out of deference to the teacher I am letting my son find
    them for himself!
    
    I've just been playing with simple functions (ratios of polynomials),
    so it becomes an exercise in recurrence relations.

    Re .15:
    
    I was thinking that you could use Sarkovskii's theorem to partially
    characterize the solutions.  For example, if a function is l-periodic
    everywhere and there is some interval over which the function is
    continuous and for which the image of the function over that interval
    is a subset of the interval, then you know the function must also have
    m-periodic points in the interval, for any m that follows l in the
    ordering.  Thus, the function f(x)=-x, which is continous and
    2-periodic everywhere, must have a point that is 1-periodic.  And it
    does, at the origin.
    
    Note that although you consider the function m-periodic everywhere if
    f^m(x)=x, for the purposes of the theorem, it is m-periodic at a point
    only if f^n(x) is not equal to x for any positive n less than m, so
    f(x)=-x is 2-periodic at 0 for you but not for the theorem.
    
    If a function were three-periodic, the Sarkovskii's theorem would say
    it must have a point that is five-periodic.  That meaning of
    five-periodic precludes being three-periodic, so we cannot let it
    happen -- so the function must not have any interval over which it is
    continuous and maps into a subset of the interval.  You get a similar
    result for any periodicity that is not a power of 2 -- it implies a
    higher periodicity at some point.
    
    That is, there is no function with a period of m everywhere such that m
    is not a power of 2 and there exists an interval I such that f(I) is a
    subset of I.
    
    
    				-- edp

    Re .14:
    
    The function                 x+1/2, 0<= x <= 1/2
    			f(x) = {
    				 2-2x,	1/2 < x <= 1
    
    is 3-periodic (at 0) and hence has n-periodic points for every positive
    integer n.
    
    
    				-- edp

    Thanks for the example.
    
    I mistook .13 to say f had to be n-periodic over an open set, so
    thought an example might be more difficult than that.
    
    Let's start at 1/4:  1/4 --> 3/4 --> 1/2 --> 1 --> 0  urgh: 3-p
    Let's start at 1/3:  1/3 --> 5/6 --> 1/3 urgh: 2-p
    Let's start at 2/3:  2/3 --> 2/3 urgh: 1-p
    Let's start at 5/9:  5/9 --> 8/9 --> 2/9 --> 13/18 --> 5/9 urgh: 4-p
    
    Looks like those odd-p points are hard to come by, given this sample.
    
      John

    Re .18:
    
    1/9 is five-periodic.
    
    
    				-- edp

Ages ago I published the following problem in the American Mathematical
Monthly (1977, 140, problem 6132):


	Find all functions f: R -> R with the Darboux property such
	that for some n >= 1,

		(1)  f^n(x) = -x for all x


(Here "Darboux" means "intermediate value property" and ^ stands for
composition of functions: f^2(x) = f(f(x)) ).

It is easy to see that under the assumption (1), Darboux means continuity.
The solution? n is odd and f = -id (i.e. f(x) = -x for all x).

As a matter of fact, under the assumption (1), either f = -id (and n is odd)
or f is very nasty.