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Math Help - Two cauchy sequence questions

  1. #1
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    Two cauchy sequence questions

    1. prove that the infinite decimal .123456789101112....
    is not eventually periodic, i.e. it represents an irrational number.

    2. Let x be a real number having the infinite decimal expansion
    x = 1.01001000100001... without the use of a calculator find the first four digits of x^(1/3)

    Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting x=1+\xi is \xi= .01001000100001\dots and now we remember that is...

    (1+\xi)^{\frac{1}{3}} = 1 + \frac{\xi}{3} - \frac{\xi^{2}}{9} + \dots (1)

    ... so that with four decimal digit accuracy is...

    x^{\frac{1}{3}} = 1.0033\dots

    Kind regards

    \chi \sigma
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  3. #3
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    thanks for your reply....

    i want to answer the question in a way that references Cauchy sequences however. I know that the decimal 1.010010001... is a Cauchy sequence because you can make
    a_1 = 1.01
    a_2 = 1.01001
    a_3 = 1.010010001
    ...

    and you can make the difference between two successive terms of this sequence arbitrarily small by going out far enough

    so is there some way to evaluate a Cauchy sequence to a certain degree of accuracy? I had trouble finding a sequence that represents this decimal.


    my shot at #1 (would appreciate it if someone can tell me yes it's valid/no it's not and why):

    1. prove that the infinite decimal .123456789101112....
    is not eventually periodic, i.e. it represents an irrational number.

    define a_n = n

    then if this decimal is periodic, for some m and k,

    a_m = a_(m+k)
    a_(m+1) = a_(m+1+k)
    ...
    a_(m+k-1)=a_(m+2k-1)

    but a_n is strictly increasing, so a_m <> a_(m+k) for k > 0

    so the decimal cannot be periodic.
    Last edited by minivan15; February 25th 2010 at 07:52 PM.
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