A method to compute roots other than sqrt.

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Discussion Overview

The discussion revolves around methods for computing arbitrary roots of numbers, such as the 10th or 13th root, beyond the square root. Participants explore various mathematical techniques and approaches for this computation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire about methods for computing roots other than square roots, specifically mentioning the 10th and 13th roots.
  • One participant suggests using Newton's method for finding roots, providing a detailed example of calculating the cube root of 4.
  • Another participant mentions the use of logarithms to compute arbitrary roots, explaining the process of taking the logarithm, dividing by the root, and then exponentiating the result.
  • A different approach is proposed involving Newton's binomial expansion to compute the fifth root of 31, demonstrating convergence through a series expansion.

Areas of Agreement / Disagreement

Participants present multiple competing methods for computing roots, with no consensus on a single preferred approach. Each method has its own merits and applications, and the discussion remains open-ended.

Contextual Notes

Some methods rely on specific assumptions about the numbers involved, such as the choice of initial values in iterative methods or the conditions under which logarithmic approximations are valid. The convergence and accuracy of these methods may vary based on the context of their application.

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

and, what are they?
 
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loop quantum gravity said:
is there a method to compute roots other than sqrt?, like 10th root or 13th root of a number?

and, what are they?
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.
 
Of course, the most obvious question is 'what is this method you have for computing arbitrary square roots'
 
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.
 
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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