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Math Help - approximation

  1. #1
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    approximation

    What is your favorite to approximate  \sqrt[n]{x+1} when you are unable to use a calculator.
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    Quote Originally Posted by putnam120 View Post
    What is your favorite to approximate  \sqrt[n]{x+1} when you are unable to use a calculator.
    That depends on what size x is...

    -Dan
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    Quote Originally Posted by putnam120 View Post
    What is your favorite to approximate  \sqrt[n]{x+1} when you are unable to use a calculator.
    I derived a formula for a general n-th degree evaluation of \sqrt[n]{x}.
    It is based on Newton's method.
    You make a guess and keep improving it.
    You be supprised how simple it is.

    Maybe, I can post it later on.
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  4. #4
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    I use the Taylor expansion

    (1+x)^n=\sum_{k=0}^{\infty}{n\choose {k}}x^k

    and if n is not a positive integer we just use

    {n\choose{k}}=\frac{(n)(n-1)(n-2)\dots(n-k+1)}{k!}

    so

    {-3\choose{4}}=\frac{(-3)(-4)(-5)(-6)}{4!}=\frac{360}{24}=15
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  5. #5
    Senior Member TriKri's Avatar
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    Well, if you are going to calculate c = \sqrt{a}, then you can use the iterative formula c_{i\ +\ 1} = \frac{c_i + a/c_i}{2}, cause a/c_i is approx as much lower than \sqrt{a} as c_i is higher than it, or vice versa.



    When dealing with \sqrt[n]{a} there's not easy to find anything that is approximately as much lower than \sqrt[n]{a} as c_i is higher than it, or vice versa.

    If c_1 = \sqrt[n]{a} + h, where h \approx 0, we can get an approximate value of a/c_i^{n-1}, namely \sqrt[n]{a} - (n-1)\cdot h. I can show why:

    Set y = a/c^{n-1}, where c = \sqrt[n]{a}.

    y(c) = a\cdot c^{-n+1}

    y(c)' = a\cdot (-n+1)\cdot c^{-n+1-1} = -\frac{a\cdot(n-1)}{c^n}

    But since we know that c^n = a, we can put that into the equation:

    y(c)' = -\frac{a\cdot(n-1)}{a} = -(n-1)

    y(c+h)\ \approx\ y(c) - (n-1)\cdot h\ =\ \sqrt[n]{a} - (n-1)\cdot h, where h \approx 0

    Since we now know that c_i = \sqrt[n]{a} + h and that y(c_i) \approx \sqrt[n]{a} - (n-1)\cdot h, we can dissolve \sqrt[n]{a} as a linear combination of c_i and y(c_i).

    And we get \sqrt[n]{a} \approx \frac{(n-1)\cdot c_i\ +\ y(c_i)}{n} = \frac{(n-1)\cdot c_i + a/c_i^{n-1}}{n} Hence we can use that for the next better value of c.


    c_{i+1} = \frac{(n-1) c_i + \displaystyle\frac{a}{c_i^{n-1}}}{n}
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