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 x_{0}, 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) := x^{3} - 4. Say, we choose x_{0} = 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: x_{1} = 2 x_{2} = 1.6666667 x_{3} = 1.5911111 x_{4} = 1.5874097 x_{5} = 1.5874010 x_{6} = 1.5874010 ... So the value of x_{n} 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.