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Math Help - The continued fraction expansion for e

  1. #1
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    Question The continued fraction expansion for e

    The continued fraction expansion for the number e = 2.718281828459045.....
    is [2;1,2,1,1,4,1,1,6,1,1,8,1,1,...].

    How many terms are required to get 4 decimal places (2.718) correct?

    Please teach me how to solve this question. Thank you very much.
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  2. #2
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    Quote Originally Posted by beta12 View Post
    The continued fraction expansion for the number e = 2.718281828459045.....
    is [2;1,2,1,1,4,1,1,6,1,1,8,1,1,...].

    How many terms are required to get 4 decimal places (2.718) correct?
    That is a classical expansion by Euler (sadly not even I know the proof).

    But I hope you are asking not to prove it but to get 4 decimals?

    Using a well-known inequlality from continued fractions:
    |x-p_n/q_n|<=(1/q_n)^2
    In order to get 4 decimal points we require that,
    (1/q_n)^2<=.0001
    Equivalently,
    1/q_n<=.0316
    Thus,
    q_n>=31.62
    Thus,
    q_n>=32
    That is the smallest "n" that makes the denominator of the convergent at least 32.
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  3. #3
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    Hi Perfecthacker,

    Thank you very much.
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