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Thread: approximation

  1. #1
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    approximation

    What is your favorite to approximate $\displaystyle \sqrt[n]{x+1}$ 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 $\displaystyle \sqrt[n]{x+1}$ 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 $\displaystyle \sqrt[n]{x+1}$ when you are unable to use a calculator.
    I derived a formula for a general n-th degree evaluation of $\displaystyle \sqrt[n]{x}$.
    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

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

    and if n is not a positive integer we just use

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

    so

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



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

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

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

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

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

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

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

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

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

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


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