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Taylor series of real function with zero radius of convergence

  1. Mar 6, 2009 #1
    Can anyone please give me an example of a real function that is indefinitely derivable at some point x=a, and whose Taylor series centered around that point only converges at that point? I've searched and searched but I can't come up with an example:P

    Thank you:)
     
  2. jcsd
  3. Mar 12, 2009 #2
    I think I'm starting to believe that such a function doesn't exist:P Ima think about proving this . . .
     
  4. Mar 12, 2009 #3

    mathman

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    Try e(-1/x2)

    At x=0, the function and all its derivatives =0.
     
  5. Mar 12, 2009 #4
    Ahh mathman, thank you but no, that wasn't what I was looking for . . . that is an example of a function that is indefinately derivable at x=0 but not analytic at x=0 because its Taylor series at x=0 has all its terms equal to nil and so does not equal the function in any neighbourhood of the point . . . but the Taylor series itself does converge (in fact, its radius of convergence is infinity, if you can talk about a series with all its terms equal to zero converging, and it converges to the function f(x)=0), if not to the function e(-1/x2).

    What I am trying to find is a function whose Taylor series about a certain point has a radius of convergence equal to zero :P That is, it does not equal any function in any neighbourhood of the point it is constructed around. I know a power series in general can have zero radius of convergence, but I'm thinking that such is not the case for power series that are Taylor series... (o.o, what is the plural of series? o: I was about to say serieses, or serii :P:P)

    Thank you anyhow :)
     
  6. Mar 12, 2009 #5

    lurflurf

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    Two popular examples
    see _Counterexamples in Analysis_ Gelbaum and Olmsted
    Σn!x^n
    Σexp(-n)cos((n^2)* x)
    in each sum n=0,1,...
     
  7. Mar 12, 2009 #6
    *flails* omg lurflurf, this is so exciting! I've got to look for that book!!
    Thank you, thank you!! (although I was hoping that my suspicions were correct :(
     
  8. Mar 13, 2009 #7

    HallsofIvy

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    The Taylor series of that has infinite radius of convergence. It just isn't equal to the function except at 0.
     
  9. Mar 14, 2009 #8
    *nods*
    I agree with Hallsof
     
  10. Mar 14, 2009 #9
    aww but lurflurf,

    Σexp(-n)cos((n^2)* x)

    is not a power series is it? And though

    Σn!x^n

    is a power series, it's not the Taylor series of a function is it?, coz that would mean f(n)(x0=0)=(n!)^2 and I'm doubting such a function f exists.. :(

    Ima try and get hold of that book anyway and have a look:P

    Thank you,
    Bobby
     
  11. Mar 14, 2009 #10

    lurflurf

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    This is one of those many counterexamples in analysis are confusing to construct because they are messy looking and use a type of function we are not ussed to.


    Σexp(-n)cos((n^2)* x)
    is not in the form of a power series, but it has a power series form that diverges for all x except 0. Giving the function this way avoids the problem with

    Σn!x^n
    which is questionable as a definition of a function since it only converges when x=0. Defining any function by power series require additional conditions since by our previous example we cannot tell the difference between f and f+exp(-1/x^2) by looking at their Maclaurin expansion. We may be more formal and say Σn!x^n is an example of a function whose Maclurin expansion has radius of convergence 0 if there exist functions having that expansion. Indeed there are, the example given in _Counterexampple in Analysis_ is
    f=Σfn(x)
    where
    fn(x)=[n antiderivatives of]{n!^2) if 0<=|x|<=2^-(n+1)/(n+1)!^2, 0 otherwise}
     
  12. Mar 14, 2009 #11

    lurflurf

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  13. Mar 27, 2009 #12
    Thank you lurflurf, and thanks for the links too, wow there seems to be an answer to everything . . . or almost:P:P I've skimmed through these bedazzling examples but I shall read over more carefully as soon as I can. Thank you!
     
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