How do you find a one-decimal place approximation for $root3 30$ ?

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Answer 1 · 2015-10-18T12:33:05

Use one step of Newton's method to find $root(3)(30) ~~ 3.1$

To find the cube root of a number $n$ , start with an approximation $a_0$ and apply the following iteration step:

$a_(i+1) = a_i + (n - a_i^3) / (3a_i^2)$

This is based on Newton's method for finding the zero of a function $f(x)$ using:

$a_(i+1) = a_i - f(x)/(f'(x))$

In our case $f(x) = x^3 - n$ and $f'(x) = 3x^2$

Since $3^3 = 27$ is not far off, let us use $a_0 = 3$ .

Then:

$a_1 = a_0 + (n - a_0^3)/(3a_0^2) = 3+(30-27)/(3*3^2) = 3+3/27 = 3+1/9 = 3.dot(1)$

So to one decimal place $root(3)(30) ~~ 3.1$

How do you find a one-decimal place approximation for root3 30root3 30?