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A method to compute roots other than sqrt.

by MathematicalPhysicist
Tags: compute, method, roots, sqrt
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MathematicalPhysicist
#1
Mar11-06, 01:31 AM
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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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VietDao29
#2
Mar11-06, 02:30 AM
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Quote Quote by loop quantum gravity
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.
matt grime
#3
Mar11-06, 02:45 AM
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Of course, the most obvious question is 'what is this method you have for computing arbitrary square roots'

jim mcnamara
#4
Mar13-06, 02:46 PM
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P: 1,381
A method to compute roots other than sqrt.

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.
AlphaNumeric
#5
Mar13-06, 03:04 PM
P: 290
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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