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    I know that \sqrt 2 cannot be expressed as a fraction due to proof by infinite descent but is there another way to calculate \sqrt 2?

    Or is the only method an iterative process of getting closer with each estimate?
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    (Original post by Retsek)
    I know that \sqrt 2 cannot be expressed as a fraction due to proof by infinite descent but is there another way to calculate \sqrt 2?

    Or is the only method an iterative process of getting closer with each estimate?
    You can do it by using binomial theorem and using lots of terms to make the estimate better and better.
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    You could use the Taylor series of a suitable function. For example,

    \displaystyle \sin \frac \pi 4 = \sum_{n \ge 0} \frac{(-1)^n}{(2n+1)!} \left(\frac \pi 4\right)^{2n+1} = \frac \pi 4 - \frac{(\pi/4)^3}{3!} + \frac{(\pi/4)^5}{5!} - \frac{(\pi/4)^7}{7!} + \frac{(\pi/4)^9}{9!} \ldots = \frac{\sqrt 2} 2

    Or equally \cos\frac \pi 4. You can cut off this infinite expansion at a point, and get an increasingly accurate approximation for \displaystyle \frac{\sqrt 2} 2. The more terms you add, the more accurate your approximation. I entered the first 6 or so terms into my calculator and got a pretty accurate estimate within like 7 \cdot 10^{-12} iirc.

    Similarly, you can use the series expansion of \arctan to approximate \pi (4\arctan(1) = \pi)
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    (Original post by Retsek)
    I know that \sqrt 2 cannot be expressed as a fraction due to proof by infinite descent but is there another way to calculate \sqrt 2?

    Or is the only method an iterative process of getting closer with each estimate?
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    (Original post by Retsek)
    I know that \sqrt 2 cannot be expressed as a fraction due to proof by infinite descent but is there another way to calculate \sqrt 2?

    Or is the only method an iterative process of getting closer with each estimate?
    This article discusses various ways of computing roots:

    https://en.wikipedia.org/wiki/Method...g_square_roots

    In addition, you can do this for \sqrt{2}:

    Note that 1 < 2 < 4 \Rightarrow 1 < \sqrt{2} < 2 \Rightarrow \sqrt{2} = 1 + \epsilon, 0 < \epsilon < 1.

    Then we can write:

    x^2 = (1+\epsilon)^2 = 2 \Rightarrow 1+2\epsilon +\epsilon^2 = 2 \Rightarrow \epsilon = \dfrac{1}{2+\epsilon}

    so that \sqrt{2} = x = 1+\epsilon=1+\frac{1}{2+\frac{1}  {2+\cdots}}

    and you can compute the continued fraction as deep as you need for accuracy.
 
 
 
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