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Math Help - finite number of discontinuety points

  1. #1
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    finite number of discontinuety points

    i have a function
    f(x)= 1 for x=1/n
    f(x)=0 for otherwise

    n is a natural number
    prove that f(x) is defferentiable [0,1]
    ?

    first i tried to prove that there are finite number of discontenuety points

    n>1/a so 0<1/n<a
    then the book says that there at most n-1 discontinuty ponts
    dont know how they got it from the last innequality

    ?
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  2. #2
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    Re: finite number of discontinuety points

    Quote Originally Posted by transgalactic View Post
    i have a function
    f(x)= 1 for x=1/n
    f(x)=0 for otherwise
    n is a natural number
    prove that f(x) is differentiable [0,1]
    ?
    first i tried to prove that there are finite number of discontenuety points
    n>1/a so 0<1/n<a
    then the book says that there at most n-1 discontinuty ponts
    dont know how they got it from the last innequality
    I cannot figure out if you are translating these questions or if you are just unfortunately in a very poor course.
    I don't think I have ever seen a poorer written question.
    There is absolutely no part of that question that is correct.
    Even the definition of the function is nonsense.

    Moreover, if a function is differentiable at a point, then it must be continuous at that point. Do you see why it is all nonsense?
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  3. #3
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    Re: finite number of discontinuety points

    Quote Originally Posted by transgalactic View Post
    i have a function
    f(x)= 1 for x=1/n
    f(x)=0 for otherwise

    n is a natural number
    prove that f(x) is defferentiable [0,1]
    ?

    first i tried to prove that there are finite number of discontenuety points

    n>1/a so 0<1/n<a
    then the book says that there at most n-1 discontinuty ponts
    dont know how they got it from the last innequality

    ?
    The function you have defined is 1 at x= 1, 1/2, 1/3, 1/4, 1/5, 1/6, etc. 0 everywhere else. Since for any number that is NOT of the form 1, 1/2, 1/3, etc., there exist an interval around it such none of those are in the interval. As a result, the function is continuous and differentiable (with derivative 0) at all x except 1, 1/2, 1/3, 1/4, etc. It is not continuous at those points so there is an infinite number of points at which it is not continuous.

    But, since you say "there are at most n-1 discontinuity point", which only makes sense if n is a fixed number, did you mean to define the function f_n as f_n(x)= 1 if x= 1/n, f(x)= 0 otherwise, for a specific n? That is, f_2 is 0 everywhere except x= 1/2 where it is 1, f_3(x) is 0 everywhere except at x= 1/3 where it is 1, etc? That also doesn't quite make sense- any such function differentiable everywhere except at the single point x= 1/n.
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    Re: finite number of discontinuety points

    Quote Originally Posted by HallsofIvy View Post
    The function you have defined is 1 at x= 1, 1/2, 1/3, 1/4, 1/5, 1/6, etc. 0 everywhere else. Since for any number that is NOT of the form 1, 1/2, 1/3, etc., there exist an interval around it such none of those are in the interval. As a result, the function is continuous and differentiable (with derivative 0) at all x except 1, 1/2, 1/3, 1/4, etc. It is not continuous at those points so there is an infinite number of points at which it is not continuous.
    Even under your reading of this question, how can f be differentiable on [0,1]~?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: finite number of discontinuety points

    Quote Originally Posted by Plato View Post
    Even under your reading of this question, how can f be differentiable on [0,1]~?
    May be that the 'true' question is: is f differentiable in x=0?...

    Kind regards

    \chi \sigma
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  6. #6
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    Re: finite number of discontinuety points

    sorry i ment
    n-1 discontinuety point in [a,1]

    we need to prove defferentiability in [a,1]

    0<a<1

    cant imagine why there are at most n-1 discontinuety points
    Last edited by transgalactic; September 16th 2011 at 10:26 PM.
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  7. #7
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    Re: finite number of discontinuety points

    You appear to be writing the function incorrectly. You do NOT mean "f(1/n)= 1, for all n, f(x)= 0 otherwise" nor do you mean f_n(1/n)= 1, f(x)= 0 otherwise" for a specific n. You seem to be saying " f_n(1/i)= 1 for all [tex]i\le n[/itex], f(x)= 0 otherwise..

    If that is correct, look at some examples: f_2(x)= 0 for all x except 1/2- one point of discontinuity.

    f_3(x)= 0 for all x except 1/2 and 1/3- two points of discontinuity.

    f_4(x)= 0 for all x except 1/2, 1/3, and 1/4- three points of discontinuity.
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