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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1084.0. "The "bump" function." by HPSTEK::XIA () Sat May 20 1989 21:53

    Find a function f :R-->R (i.e. f takes real number to real number)
    such that f satisfies the following properties:
    
    1. f(x) = 0 for any x <= 0.
    2. f(x) = 1 for any 1 <= x <= 2.
    3. f(x) = 0 for any x >= 3.
    4. 0 <= f(x) <= 1 for any real number x.
    5. f is infinitely differentiable for all x in R.
       (That is if x is a real number, then f'(x), f''(x), f'''(x)...
        all exist.)
    
    Eugene
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1084.1The details are left as an exercise. :-)AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoSun May 21 1989 00:4317
	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
1084.2They are all around us!HIBOB::SIMMONSTristram Shandy as an equestrianSun May 21 1989 02:415
    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
1084.3what about between 0 and 1 ?HANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Tue May 23 1989 20:214
It looks like you haven't specified what to do when x is between 0 and 1.


1084.4HPSTEK::XIATue May 23 1989 21:396
    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
1084.5AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoTue May 23 1989 22:048
     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
1084.6A shot in the darkSSDEVO::LARYOld programmers never die, they just Tue May 23 1989 23:2217
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.
1084.7AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoTue May 23 1989 23:4721
     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
     
1084.8CTCADM::ROTHIf you plant ice you'll harvest windWed May 24 1989 10:5214
    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
1084.9A small nit on .8COOKIE::PBERGHPeter Bergh, DTN 435-2658Wed May 24 1989 11:553
    In any arbitrarily small neighborhood of an essential singularity of an
    analytic function (of one variable) you can find every complex number
    except *one*.
1084.10:-)AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoWed May 24 1989 12:243
	Which one?

	Dan
1084.11;-}POOL::HALLYBThe Smart Money was on GoliathWed May 24 1989 14:174
>	Which one?
    
    Why, the value to which it would converge, if it did converge,
    but since it doesn't, it won't.
1084.12Analytic is not infinite differentiabilityHIBOB::SIMMONSTristram Shandy as an equestrianWed May 24 1989 19:0911
    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
1084.13HPSTEK::XIAWed May 24 1989 19:4510
    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
1084.14Definition of local?HIBOB::SIMMONSTristram Shandy as an equestrianWed May 24 1989 21:4111
    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
1084.15AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoThu May 25 1989 02:008
>>	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
1084.16CTCADM::ROTHIf you plant ice you'll harvest windThu May 25 1989 10:4817
    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
1084.17Question, measure, re 16HIBOB::SIMMONSTristram Shandy as an equestrianThu May 25 1989 22:279
    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
1084.18AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoFri May 26 1989 00:1413
>>	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
1084.19Making liberal use of ZFC,POOL::HALLYBThe Smart Money was on GoliathFri May 26 1989 13:137
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
1084.20We are in different spacesHIBOB::SIMMONSTristram Shandy as an equestrianFri May 26 1989 13:159
>      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
1084.21Understand measureHIBOB::SIMMONSTristram Shandy as an equestrianFri May 26 1989 13:248
    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
1084.22Zero is my favorite analytical functionCOOKIE::PBERGHPeter Bergh, DTN 435-2658Fri May 26 1989 15:411
    I don't see why the constant zero is not an analytic function?
1084.23Theory/problem divergencePOOL::HALLYBThe Smart Money was on GoliathFri May 26 1989 17:011
    f(x) = 0 does not satisfy the constraints of .0
1084.24JRDV04::KOMATSUExistentialistFri Jul 14 1989 04:3914
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 ???
1084.25Solution4GL::GILBERTDon't Worry, Be Bobby McFerrinFri Jul 14 1989 21:2518