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

  1. Apr 8, 2012 #1
    what is the quickest way to find [itex]\sqrt[3]{n}[/itex] [on a pocket calculator] whitout using any [itex]\sqrt{}[/itex] or log key?
     
  2. jcsd
  3. Apr 8, 2012 #2

    HallsofIvy

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    If x is the cube root of a, then [itex]x^3= a[/itex] or [itex]x^3- a= 0[/itex].

    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 [itex]x_0[/itex] and the iterating [itex]x_{n+1}= x_n- f(x_n)/f'(x_n)[/itex].

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

    The starting value, [itex]x_0[/itex] doesn't matter a great deal but probably something like a/3 would be good.
     
  4. Apr 8, 2012 #3
    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?
     
  5. Apr 8, 2012 #4
    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):

    [itex]\displaystyle\frac{2ANS^3-a}{3ANS^2}[/itex] and continuously press the Enter key and you will have your answer shortly.
     
    Last edited: Apr 8, 2012
  6. Apr 8, 2012 #5
    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 = [itex]\sqrt[3]{377933067}[/itex] using only logics, knowing that n = x³ ?
     
    Last edited: Apr 8, 2012
  7. Apr 8, 2012 #6
  8. Apr 9, 2012 #7
    Can you find 723 using no tool whatsoever? Only logics and your mind [knowing 2³...9³, of course]?
     
    Last edited: Apr 9, 2012
  9. Apr 9, 2012 #8
    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?
    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!
     
    Last edited: Apr 9, 2012
  10. Apr 10, 2012 #9
    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: [itex]\frac{1}{3}\, \left(2x+\frac{a}{x²}\right)[/itex] 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
     
    Last edited by a moderator: May 5, 2017
  11. Apr 25, 2012 #10
    (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?
     
    Last edited: Apr 25, 2012
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