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Math Help - prove irrational

  1. #1
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    prove irrational

    prove that sin(1) (one radian) is irrational.
    Hint start with taylor series expansion
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  2. #2
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    Without Taylor-series see this on the 3rd page (7***).
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  3. #3
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    Quote Originally Posted by scubasteve123 View Post
    prove that sin(1) (one radian) is irrational.
    Hint start with taylor series expansion
    you first need this basic and simple fact about alternating series that if 0 < a_{i+1} < a_i, \ \forall i \in \mathbb{N}, then \sum_{i=1}^{\infty}(-1)^{i-1} a_i > 0. \ \ \ \ \ \ \ \ \ (*)

    now suppose \sin 1 = \frac{m}{n}, for some m,n \in \mathbb{N}. let A=(2n-1)! \left (\sin 1 - \sum_{i=1}^n \frac{(-1)^{i-1}}{(2i-1)!} \right)=\frac{m(2n-1)!}{n} - \sum_{i=1}^n \frac{(-1)^{i-1} (2n-1)!}{(2i-1)!}.

    thus A is an integer. i'll show that this is false, i.e. A cannot be an integer: using the MacLaurine expansion of \sin 1 we have:

    A= \sum_{i=n+1}^{\infty} \frac{(-1)^{i-1}(2n-1)!}{(2i-1)!}=(-1)^n \sum_{i=n+1}^{\infty} \frac{(-1)^{n+i-1}}{(2n)(2n+1) \cdots (2i-1)}. it's clear by (*) that |A| > 0. we also have:

    |A| \leq \sum_{i=n+1}^{\infty} \frac{1}{(2n)(2n+1) \cdots (2i-1)} \leq \sum_{i=n+1}^{\infty} \frac{1}{(2n)^{2i-2n}}=\frac{1}{4n^2 -1} < 1. therefore 0 < |A| < 1 and so A is not an integer. Q.E.D.
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