A method to compute roots other than sqrt.

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Methods for computing roots beyond square roots, such as the 10th or 13th root, include Newton's method and logarithmic approaches. Newton's method involves iterating a formula to converge on the root, exemplified by finding the cube root of 4 through a defined function and initial guess. Logarithms can also be used, where the nth root is computed by taking the logarithm of the number, dividing by n, and then exponentiating the result. Additionally, Newton's binomial expansion can approximate roots for fractions effectively. These techniques provide efficient ways to calculate various roots with good precision.
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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:
x_{n + 1} = x_n - \frac{f(x_{n})}{f'(x_{n})}.
And then let n increases without bound to obtain the answer.
x = \lim_{n \rightarrow \infty} x_n.
For example:
Find \sqrt[3]{4}
Let x = \sqrt[3]{4} \Rightarrow x ^ 3 = 4 \Rightarrow x ^ 3 - 4 = 0
We then define f(x) := x3 - 4.
Say, we choose x0 = 1, plug everything into a calculator, use the formula:
x_{n + 1} = x_n - \frac{x_{n} ^ 3 - 4}{3 x_{n} ^ 2}.
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 \sqrt[3]{4}, 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 (1+x)^{\frac{1}{5}} with x = -1/32

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

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

\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}

\frac{(31)^{\frac{1}{5}}}{2} = \frac{12719}{12800}

(31)^{\frac{1}{5}} = \frac{12719}{6400}

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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