A method to compute roots other than sqrt.

  1. is there a method to compute roots other than sqrt?, like 10th root or 13th root of a number?

    and, what are they?
     
  2. jcsd
  3. VietDao29

    VietDao29 1,422
    Homework Helper

    This one is a good candidate for Newton's method. We choose an arbitrary value x0, then use:
    [tex]x_{n + 1} = x_n - \frac{f(x_{n})}{f'(x_{n})}[/tex].
    And then let n increases without bound to obtain the answer.
    [tex]x = \lim_{n \rightarrow \infty} x_n[/tex].
    For example:
    Find [tex]\sqrt[3]{4}[/tex]
    Let [tex]x = \sqrt[3]{4} \Rightarrow x ^ 3 = 4 \Rightarrow x ^ 3 - 4 = 0[/tex]
    We then define f(x) := x3 - 4.
    Say, we choose x0 = 1, plug everything into a calculator, use the formula:
    [tex]x_{n + 1} = x_n - \frac{x_{n} ^ 3 - 4}{3 x_{n} ^ 2}[/tex].
    and we'll have:
    x1 = 2
    x2 = 1.6666667
    x3 = 1.5911111
    x4 = 1.5874097
    x5 = 1.5874010
    x6 = 1.5874010
    ...
    So the value of xn will converge quite fast to [tex]\sqrt[3]{4}[/tex], as n tends to infinity.
     
  4. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    Of course, the most obvious question is 'what is this method you have for computing arbitrary square roots'
     
  5. jim mcnamara

    jim mcnamara 1,488
    Science Advisor
    Gold Member

    FWIW - Logarithms work well for this. Especially if you're a programmer, and are happy with the inherent imprecision of floating point numbers.

    If you take
    result = (log(x) / n)

    and then convert the result back ie.,

    nth_root = exp(result)

    you can generate all roots of x.
     
  6. If you're doing fractions then you can use Newton's binomial expansion

    For instance, work out the fifth root of 31 by expanding [tex](1+x)^{\frac{1}{5}}[/tex] with x = -1/32

    [tex](1+x)^{\frac{1}{5}} = 1 + \frac{1}{5}x + \frac{1}{5}\left(-\frac{4}{5}\right)\frac{1}{2!}x^{2} + ....[/tex]

    Put in x = -1/32 (which gives excellent convergence) to get

    [tex]\left( \frac{31}{32} \right)^{\frac{1}{5}} = 1 + \frac{1}{5}\left(-\frac{1}{32}\right) + \frac{1}{5}\left(-\frac{4}{5}\right)\frac{1}{2!}\left(-\frac{1}{32}\right)^{2} + .... = 1 - \frac{1}{160} - \frac{1}{12800} = \frac{12719}{12800}[/tex]

    [tex]\frac{(31)^{\frac{1}{5}}}{2} = \frac{12719}{12800}[/tex]

    [tex](31)^{\frac{1}{5}} = \frac{12719}{6400}[/tex]

    In decimal form this is 1.9873475. Raise it to the 5th power and get 31.00023361. A nice approximation for 2 minutes work.
     
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