Conference rusure::math

    What you've been playing with is called a linear recurrance relation.
    This is a recurrance of the general form

	z[n+1] = c[n]*z[n] + c[n-1]*z[n-1] + ... + c[0]*z[0]

    where the c[k]'s are constants, and the z[k]'s are the sequence
    you're generating.  This is related to the field of digital
    signal processing, where such a sequence is known as the impulse
    response of a recursive (or infinite impulse response) filter.

    You were lucky to have picked such a simple example, since most
    recurrances are chaotic in behaviour.  In addition, your first example
    was fortunate to have resulted in an undamped sine wave - most such
    sequences will result in a combination of exponentially decaying or
    increasing sequences, or exponentially decaying or increasing sinusoids.

    Let's analyze your example.  The linear recurrance relation is a linear
    operator, thus the eigenfunction will be an exponential.

    You have

	z[n+1] = c[1]*z[n] + c[0]*z[n-1]

    with coefficients c[0] = -1.0 and c[1] = 1.8,
    and initial conditions of z[0] = 5.0 and z[1] = 10.0.

    We assume the sequence is of the form z[k] = r^k, where r is some
    as yet undetermined number.

    Thus

	r^(k+1) = c1*r^k + c0*r^(k-1)

    Divide thru by r^(k-1) and rewrite this as a quadratic equation

	r^2 - c1*r - c0 = 0

    Solving for r by the grade school quadratic formula:

	r =  c1/2 +/- sqrt((c1/2)^2 - c0)

    Plugging in the c's,

	r = 0.9 +/- sqrt(-0.19)

    And we find that r is a complex number with a magintude of one!

	r = 0.9 +/- i*sqrt(0.19)

	|r| = 0.9*0.9 + 0.19 = 0.81 + 0.19 = 1.0

    This is why you got an undamped sinusoid in your example.  If you
    take the real part of successive powers of r, you'll get your sequence
    (within a scale factor.)

    To generate a sine wave of a given frequency set c1 to 2.0*cos(2*pi/T),
    where T is the period of the sinusoid in sample points, and leave c0 = -1.0.

    For more info, see any book on digital signal processing, such as
    Oppenheim and Shafer, "Digital Signal Procssing", or probably better
    Hamming's little book "Introduction to Digital Filters".

    - Jim

Similar patterns can easily be shown on a pocket calculator.

After turning it on, enter 0.5 and then alternately hit SIN and COS
buttons.  (I think those two buttons are right, but there are other
pairs that work too).  You converge on a single or pair of numbers.

What other pairs of buttons work ?

/Eric

    "On" and "Off"
    
    (Re: .2)

    Just hitting COS will do fine, if your calculator is doing radians.
    There's a theorem by Banach, quite simple to prove, that 
    
    "Every contraction of a complete metric space has a unique fixed point".
    
    A contraction (in the matehmatical sense, but then not many women
    read this file) can be loosely defined as a function f that satisfies
    | f(a) - f(b) | < | a - b | for every "a" and "b" in some domain D.

    Even more loosely, but more clearly, think of the function as
    taking an interval (a,b) and mapping it into a smaller interval
    (f(a),f(b)) INSIDE the interval (a,b).
    
    In the interval (0,1) there are lots of contractions, such as
    f(x) = x/47 and f(x) = x�.  EXP(1/x) is a 2-button function that
    seems to contract for proper values of x (which ones?).
    
    I think you can see that COS(X) is a contraction by looking at the
    power series expansion for COS(X), though being rigorous might take
    a bit of sweat.
    
    Evidently Eric has discovered that SIN(COS(X)) is also a contraction.
    
      John

    In even more of an abstract vein than .-1 in topology there are
    function defined as automorphisms that is maps that go from a set
    to itself and is bijective (one-to-one and onto or equivalently
    has an inverse).  Topology (or more precisely algebraic topology)
    proves that any such functions have a fixed point.  The consequences
    are numerous, the collic in ones hair, the area of calm in a hurricane,
    and why a polynomial always has at least one root (the fundamental
    theorem of algebra), etc. .
    

    I think you must have missed at least one condition - for example
    continuity, since obvious f(x)  ( (x - 0.5) {+1 if x < 0.5} doesn't
    have a fixed point yet is bijective on [0,1].
           
    /Bevin

    Your counterexample does not satisfy |f(a) - f(b)| < |a - b|.
    Prove (or disprove!) that that condition is strong enough to
    imply continuity for functions from one metric space to another.
    
    Dan

    Re .5:
    
    It is not true that "any such" functions have a fixed point, although
    it is true for a large class of functions in a large class of spaces.
    For example, the set {x | 0 <= x <= 1 or 2 <= x <= 3} can be mapped
    [even contracted] into itself bijectively without a fixed point;
    just switch halves.
    
    Dan

    I hadn't realised you intended to inherit the contraction requirement
    from earlier replies as one of the conditions.
    
    Sorry.
    
    /Bevin

    Re .-1
    
    Oops.  I responded as I did to .6 because I thought that you were
    replying to .4's claim about contractions.  If you were replying
    to .5, then yes, there are assumptions there that are either
    not stated or are assumed as included in his use of certain terms.
    I'm not sure which.
    
    Dan

     First, a retraction of a claim that I made in .8:

>>    For example, the set {x | 0 <= x <= 1 or 2 <= x <= 3} can be mapped
>>    [even contracted] into itself bijectively without a fixed point;
>>    just switch halves.

     The function that swicthes halves that I had in mind is not
     a contraction as defined in .4.  If f(1) and f(2) are in
     different halves then |f(1) - f(2)| >= 1 = |1 - 2|.
     It is a continuous bijective function with a continuous
     inverse in a complete metric space; but it has no fixed
     point.  A more obvious example of that is f(x) = x + 1 on
     the real line.

     But the definition of a contraction given in .4 is not
     correct:

>>    There's a theorem by Banach, quite simple to prove, that 
>>    
>>    "Every contraction of a complete metric space has a unique fixed point".
>>    
>>    A contraction (in the matehmatical sense, but then not many women
>>    read this file) can be loosely defined as a function f that satisfies
>>    | f(a) - f(b) | < | a - b | for every "a" and "b" in some domain D.

     The simple proof mentioned is this:  Let x be any point of
     the complete metric space.  Define a sequence x0, x1, ... by
     taking x0 = x and xn+1 = f(xn).  Use the fact that f is a
     contraction to show that this is a Cauchy sequence.  Since
     the space is complete, the sequence converges to some value,
     call it y.  Then, because f is a contraction, the only
     possible value that f(y) can have is y.  So f(y) = y and y
     is a fixed point of f.  The fixed point must be unique,
     otherwise if both a and b are fixed points then
     | f(a) - f(b) | = | a - b | which can't happen for a
     contraction.

     This proof does not work for the definition of contraction
     given.  The ratio | f(a) - f(b) | / | a - b | is less than
     one, but can get closer and closer to one as a and b move
     down the sequence.  Consider the following metric space:
     The points are the positive integers, and the metric is
     given by d(n,m) = 1 + | 1/n - 1/m |.  The only Cauchy
     sequences are eventually constant, so the space is complete.
     The function f(n) = n + 1 obviously has no fixed point, but
     it satisfies the definition d( f(a) , f(b) ) < d(a , b) in .4.
     [Actually, after I had worked all of this out I realized
     that no function satifies this definition:  if a=b it
     reduces to 0 < 0 which is false.  I mean here that my f
     satisfies it for "a" not equal to "b"]

     The correct definition of a contraction [for an arbitrary
     distance function d] is:

          There is a real number r such that 0 < r < 1 and that
          for every a and b in the metric space,
                     d(f(a),f(b)) <= r * d(a,b)

     Now the ratio of the distance between successive terms is
     bound away from one.

     With this definition, if x0 is arbitrary and xn+1 = f(xn)
     for some contraction f, and c = d(xo,x1), it is easy to
     prove by induction that d(xn,xn+1) <= c * r^n and that
     d(xn,xm) <= (c * r^min(n,m)) / (1-r).  Thus x0, x1, ... is a
     Cauchy sequence which by definition [of complete] must
     converge in a complete metric space.  The limit point will
     then be a fixed point of f.  Uniqueness follows as before.

     Dan

     P.S.

     For some particular spaces, such as the closed unit interval
     of the real numbers, *every* continuous function from the
     space into itself has a fixed point.  The same is true for a
     closed disk as a subset of the Euclidean plane.  The
     fundamental theorem of algebra may be proven from that, as
     stated in .5.  [In these examples, there may be more than
     one fixed point.]

>     This proof does not work for the definition of contraction
>     given.  The ratio | f(a) - f(b) | / | a - b | is less than
>     one, but can get closer and closer to one as a and b move
>     down the sequence.  Consider the following metric space:
>     The points are the positive integers, and the metric is
>     given by d(n,m) = 1 + | 1/n - 1/m |...
    
    -Noting- that I commented the definition of contraction was
    "loosely defined", it's nice to see someone nail it down more
    rigorously.  Unfortunately, the nail is a bit rusty.  My memory
    is also a bit rusty, but I recall the definition of a METRIC
    on a set S is a function "d" from S cross S into R (the reals)
    satisfying:
    
    (1)	d(s,s) = 0		   for all s in S
    (2)	d(s,t) = d(t,s)	           for all s,t in S
    (3)	d(s,u) <= d(s,t) + d(t,u)  for all s,t,u in S.  <"Triangle inequality">

    The non-counterexample supplied violates (1), since d(m,m) = 1.

      John

    Re .-1
    
>>    (1) d(s,s) = 0                 for all s in S
>>     The non-counterexample supplied violates (1), since d(m,m) = 1.
    
    I forgot about that case.  Let d(n,m) = 0 if n=m, otherwise
    d(n,m) = 1 + | 1/n - 1/m |.
    
    Dan

>    I forgot about that case.  Let d(n,m) = 0 if n=m, otherwise
>    d(n,m) = 1 + | 1/n - 1/m |.

    That seems to work, and shows why the "loose" definition of contraction
    is too loose.  BTW, metric spaces also have to satisfy two other
    properties not mentioned in .12:
    
        d(m,n) = 0 ==> m=n (not just d(m,m) = 0).
    	d(m,n) >=0 always.
    
    The above metric satisfies these criteria as well.  It's a useful
    counterexample to have in your bag of gooies.
    
    I wonder, he thought idly, if the loose definition of contraction
    would work in a normed space?  Norms yield metrics that are more
    nicely-behaved than D'Eramo's "d" above.  This would mean relaxing
    one set of criteria for Banach's theorem, while tightening another.
    As such, is it legitimate fodder for status as a separate theorem?

      John

     My favorite fixed point theorem is the recursion theorem.

     Let M0, M1, M2, ... be an "effective enumeration" of all
     Turing machines, say with two symbols -- "1" and blank --
     and with a single two-way infinite tape.  Let f0, f1, f2, ...
     be the functions from the nonnegative integers to the
     nonnegative integers computed by the Turing machines.

     [When I say "effective enumeration" I mean that it is
     possible to determine the state transition table for Mn
     given n, and vice versa.  There must also be a convention
     for when a machine has been given input k and computed
     output j.  The functions f0, f1, f2, ... corresponding to
     these conventions are not necessarily defined for all
     nonnegative integers.  They are called "partial recursive
     functions" and if by chance one is defined for all
     nonnegative integers then that is also a "total recursive
     function."]

     Now let g be a total recursive function from the nonegative
     integers to the nonegative integers.  Think of g as changing
     the Turing machine Mn into machine Mk where k = g(n).  Now
     this is quite a transformation.  If you encoded your last
     lisp program into an integer, cubed the integer, and then
     decoded back to a lisp program, you would not expect the
     result to run as well as the original.

     However, the recursion theorem states that there will be an
     integer j such that the function fj corresponding to Mj will
     be the same function -- defined and undefined for the same
     sets of numbers -- as the function fk corresponding to Mk
     where k = g(j).  Note that it is not machines j and k = g(j)
     that are identical, it is the functions that are identical.
     [You can effectively compute j given an index i for a Turing
     machine that computes g.]

     Does this mean that no matter how great a debugger you are,
     that there will be at least one program whose behavior you
     cannot change?

     Dan