How do you find the cubic root of n without using log keys?

[logics]

what is the quickest way to find $\sqrt[3]{n}$ [on a pocket calculator] whitout using any $\sqrt{}$ or log key?

 

[HallsofIvy]

If x is the cube root of a, then $x^3= a$ or $x^3- a= 0$.

Now use 'Newton's algorithm" to solve that equation- it is of the form f(x)= 0 so Newton's algorithm requires starting with some $x_0$ and the iterating $x_{n+1}= x_n- f(x_n)/f'(x_n)$.

In particular, with $f(x)= x^3- a$ $f'(x)= 3x^2$ so the algorithm becomes
$$x_{n+1}= x_n- \frac{x_n^3- a}{3x_n^2}= \frac{2x_n^3- a}{3x_n^2}$$

The starting value, $x_0$ doesn't matter a great deal but probably something like a/3 would be good.

 

[logics]

Now use 'Newton's algorithm" to solve that equation-...

how long does that take to find the cube root of a 30-digit number, [on a 10-digit-display calculator] ?
isn't there any better and simpler way?

 

[scurty]

how long does that take to find the cube root of a 30-digit number, [on a 10-digit-display calculator] ?
isn't there any better and simpler way?

How long on a calculator? Using the "ANS" button in your algorithm to continuously loop the algorithm and depending on how fast you can press the Enter button, 3-5 seconds seems like a good range.

Edit: Here is the formula you should use on your calculator. Type in your starting number and press the Enter button. Now type in (using Hall's formula):

$\displaystyle\frac{2ANS^3-a}{3ANS^2}$ and continuously press the Enter key and you will have your answer shortly.

 

[logics]

Here is the formula you should use on your calculator. Type in your starting number and press the Enter button. Now type in (using Hall's formula):...

Is it possible to get a result without using the n³ key?

In junior school we used to find the cube root of a 6- or even 9-digit number such as [635³] without using a pencil. Is that trick generally known?
For example, can you find x = $\sqrt[3]{377933067}$ using only logics, knowing that n = x³ ?

 

[TheDestroyer]

http://en.wikipedia.org/wiki/Taylor_series

 

[logics]

....without using a pencil. Is that trick generally known?
can you find x = $\sqrt[3]{377933067}$ using only logics, knowing that n = x³ ?

http://en.wikipedia.org/wiki/Taylor_series

Can you find 723 using no tool whatsoever? Only logics and your mind [knowing 2³...9³, of course]?

 

[logics]

what is the quickest way to find $\sqrt[3]{n}$

$$x_{n+1}= x_n- \frac{x_n^3- a}{3x_n^2}= \frac{2x_n^3- a}{3x_n^2}$$

Newton's method requires 7 operations per round : [(2 * x * x * x) - a] : (3 * x * x)
Babylon [adapted] only 4 [x + (a : x * x)] : 2.
Can you find a method that requires only 3?

The starting value, $x_0$ doesn't matter a great deal but probably something like a/3 would be good.

Why start with such a huge number?, if a is 377933067 a/3 is 125,977,689. Starting with this x₀, it takes 38 rounds to get x, which means 266 operations!

 

[logics]

http://www.infoocean.info/avatar2.jpg how long does that take to find the cube root of a 30-digit number, [on a 10-digit-display calculator] ?

It depends on the algorithm you choose and on the staring value (x₀)

As sar as I know, the best known algorithm would still be 'Babylon', if you adapt it to the cube: $\frac{1}{3}\, \left(2x+\frac{a}{x²}\right)$ as it requires 5 operations per round, but I started this thread to learn, one may find quicker methods that require less (3 or even 2) operations...
..as to (x₀) If a = 7123456789³ [3.6147...^29] and if you are able, as you certainly are, to guess the first digit [7], you have to press 3 times Enter, which makes 15 operations, in about one second. If you make an error (x₀ in the order 10^9 it will take 5 rounds, 2 seconds, and so on.

Being able to find directly the cube root of a 9-digit number is useful to solve [reduced] cubic funtions such as x³+x = 378,005.367: x = 72.3, or x³+8x = 377,354.667...
With pencil and paper one can solve in a few seconds cubic functions with 4/5-digit solution

 

[logics]

$\frac{2ANS^3 -a }{3ANS^2}$ and ... press the Enter key.

(Just for future readers)
There is a typo there, the right formula is 2* ans³ + a.

I am remarking this only because it is interesting to note that the formula works all the same, in spite of the mistake, and with same convergence, and gives a negative result.
The point is that if you start, as you should, with a positive x₀ you need more iterations, and you get the impression that the method is less powerful.
In reality it is as powerful as Babylon-adapted but requires more ops.

I have a question:
if we modify Newton's formula in a different way :
( ans³ +a) : 2 ans²
would you say this has the same quadratic convergence as Newton's?