Conference rusure::math

	Hmmm.  So f looks like
	             ___
	            /   \
	           /     \
	__________/       \__________
	          0  1  2  3

	except that the "corners" are very smooth.

	That should be easy enough to do "piecewise" with
	things involving

		     2
		 -1/x
		e

	Dan

    And the answer is found in any book on the theory of distributions.
    Such real functions which are C-infinity but not analytic are well
    known and loved as points in the space of test functions.
    
    CWS

It looks like you haven't specified what to do when x is between 0 and 1.

    re -1:
    
    It is for you to decide.  Had I told you what is between 0 and 1,
    I would have given you the answer.
    
    Eugene

     re .3
     
     The constraints there were 0 <= f(x) <= 1, and f is
     infinitely differentiable.  You have to specify f between 0
     and 1, and between 2 and 3, so that those conditions are
     satisfied.
     
     Dan

My real analysis is real creaky, but let me give it a shot:

Lets say you had such a function. Since the function is infinitely
differentiable, you can express f(1) in terms of the function and all its
derivatives at 0 (Taylor's Theorem?), as follows:

f(1) = f(0) + f'(0) + f''(0)/2! + f'''(0)/3! + ... = 1

If the sum is one, then at least one of the terms in the series must be
non-zero; pick the "leftmost" such term, f[n](0) - the nth derivative of
f at 0.

f[n](x) = 0 for all x < 0, so lim(f[n](x)) = 0 as x approaches zero from
below; however f[n](0) is not zero. I believe (its been a while) that
this means that f[n+1](0) doesn't exist, which is a contradiction.

So there is no such function.

     re .6
     
     No, if the function were analytic in an open set containing
     the origin [is that the precise condition?] then it would
     have a valid expansion as f(x) = f(0) + x f'(0) + ... for
     all x less than some positive constant.  Such a function
     can't satisfy all of the conditions in .0.  "Analytic" is
     kind of a strong differentiability for complex functions of
     a complex variable.
     
     But .0 only required that the function be infinitely
     differentiable on the real line, and that is a weaker
     condition.  For example, f(x) = e^(-1/x^2) for x /= 0,
     f(0) = 0 is infinitely differentiable.  And all of its
     derivatives at 0 have the value 0 (it is very flat there). 
     But if you approached "complex 0" along the imaginary axis
     this f diverges and so is not analytic there.  So no "Taylor
     series".
     
     Dan
     

    The function exp(-1/x^2) has an essential singularity at zero,
    an infinite number of terms in the Laurent series expansion with
    negative exponents.  It does converge in any neighborhood of zero
    in the complex plane, but not at zero itself.  In fact, you can find
    any complex number (except for two, I think - Picard's theorem) near zero.

    It just happens to be well behaved on the real line, but that's an
    exception.  Consider approaching along the imaginary axis - you get
    trig functions like sin(-1/x^2) - and that's pretty wild near zero!

    Every reference I've seen to the bump function uses this exp(-1/x^2)
    trick...

    - Jim

    In any arbitrarily small neighborhood of an essential singularity of an
    analytic function (of one variable) you can find every complex number
    except *one*.

	Which one?

	Dan

>	Which one?
    
    Why, the value to which it would converge, if it did converge,
    but since it doesn't, it won't.

    Looks to me like there is confusion between infinitly differentiablity
    and analyticity.  They are the same for functions of a complex
    variable.  They are not the same for functions of a real variable.
    An analytic function of a real variable is represented by its Taylor
    Series everywhere.  An infinitly differentiable function of a real
    variable, need only be represented locally by its Taylor Series.
    The function asked for here is one of the latter class.  By the
    way, Shilov's (I believe) Generalized Functions, or some such, gives
    several examples if I remember correctly.
    
    Chuck

    re -1
    
    Chuck,
    
    I don't think it is true that "an infinitely differentiable function
    of a real variable, need only be represented locally by its Taylor
    Series".  As a matter of fact, the solution to this problem cannot
    be locally represented by its Taylor Series at 0 1 2 and 3.
    
    Eugene

    Well, I might have put in a disclaimer - I can't remember details.
    However, normally we only talk about Taylor Series on open
    neighborhoods.  Clearly, the function is represented by its Taylor
    Series on every neighborhood not containing 0, 1, 2, or 3.  This
    is local rather than global.  Of course any C-infinity function
    with compact support has at least two points where the Taylor Series
    fails.  The function wanted here is analytic at every point except
    0, 1, 2, and 3 as well (and analytic also only makes sense on open
    neighborhoods).
    
    Chuck

>>	The function wanted here is analytic at every point except
>>	0, 1, 2, and 3 as well

	It could be constructed that way, or you could be perverse
	and construct such a function that is also not analytic at,
	say, 1/2.

	Dan

    There doesn't seem to be any confusion, though the base problem highlights
    the difference between infinite differentiability and analyticity.

    Note that analyticity is a very strong requirement - given even one little
    piece of an analytic function (its expansion about a point) you know
    it everywhere it exists by analytic continuation.  In general you'll need
    to leave the real line to piece together all the branches of a real
    analytic funcion though. No such relationship need exist between different
    parts of an infinitely differentiable function.  In fact, the analytic
    functions are of measure zero in the infinitely differentiable functions...

    Right on the at most one omitted value.  What I should have been thinking
    of is that if two values of an entire function are omitted, then it is
    a constant.  This is not to say that one value *must* be omitted - just
    that one can be, but two cannot.

    - Jim

    Re .16
    
    Don't understand measure zero in that context, i.e. what is the
    measure.  Do you mean not dense with respect to some topology?
    I even have trouble with that one because neither is a subset of
    the other so it would have to be in a relative topology induced
    on the intersection by a topology on one of them wouldn't it?
    
    Chuck

>>	I even have trouble with that one because neither is a subset of
>>	the other

	There is a subset relationship involved.  The larger set is
	the set of functions f:R -> R that are infinitely differentiable,
	and the smaller set is the set of functions f:R -> R which are
	restrictions of functions analytic in a region of the complex
	plane including the real line.  The second set is a subset of
	the first.

	I don't know what measure he is using, though.

	Dan

Select at random a function from the set of infintely differentiable functions.
The probability that "the selected function is analytic" is zero.

I'm not sure if that implies a measure that can't be described, but the
main point is that the analytics form a "very small" subset if the i-d's.

  John

>      There is a subset relationship involved.  The larger set is
>      the set of functions f:R -> R that are infinitely differentiable,

    I was still talking about the space of test functions, i.e. functions
    like the "bump function."  The real analytic functions are not a
    subset of these (a non-zero constant function is trivially analytic
    but it is not a test function).
    
    Chuck

    Re: .19
    
    I didn't notice .19 when I wrote .20.  That defines a measure in
    a way.  I missed it because I don't think of these things in terms
    of probabilities.  My comment about these sets not being subset
    related has to do with what I point out in .20.
    
    Chuck

    I don't see why the constant zero is not an analytic function?

    f(x) = 0 does not satisfy the constraints of .0

I haven't read alll the replies in here, so this must be commented in previous
replies.

Om ( m = 1,2,... ) ; Open sets in R^n, each Om are disjoint, so to say 
intersection of Oi and Oj is empty if i is not j.

There is a C-infinity function fm on Om.
	(f1 is C-infinity in O1, f2 is C-infinity in O2,...)

You can find C-infinity function F, such that
F = fm in Om (m=1,2,...)

I forget the name of this thorem, but I think its famouse.
May be Tiez or Teze ???

    Solution:

    Let
		{
		{ 0			if x <= 0
		{
		{    -1/(1-x)�
		{  -e         /x�
		{ e			if 0 < x < 1
		{
	f(x) =	{ 1			if 1 <= x <= 2
		{
		{    -1/(x-2)�
		{  -e         /(3-x)�
		{ e			if 2 < x < 3
		{
		{ 0			if 3 <= x
		{