20070908, 16:26  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Cube root
We all know how to get the square root of a number by a plain arithmetical method. I was wondering how would one work out the cube root of a number arithmetically? Take a number say 12,812,904 ? Mally 
20070908, 19:04  #2 
"William"
May 2003
New Haven
23·103 Posts 
Googling Cube root by hand turns up this as first choice
http://www.nist.gov/dads/HTML/cubeRoot.html 
20070909, 07:17  #3 
Jun 2003
3·23^{2} Posts 
Cube root or for that matter the nth root, all work by the same principle.
What is your difficulty? 
20070909, 07:56  #4  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Cube root.
Quote:
Thank you William. As usual you always come out with unusual pin point accuracy! I have never been content by just using algorithms without knowing the theory behind it. This reference even gives the explanation of why it works! They even take it for granted that even if one does not know the Binomial theorem one can still understand the method employed. This URL is fit for a printout for later reference. Mally :coffee; 

20070909, 08:09  #5  
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
Same principle
Quote:
Well Citrix that does not give much to build on, for any newbie reading your comment If you even mentioned what 'principle' it would be a clue, like the Binomial Theorem or De Moivre's for nth roots perhaps I would be satisfied with your answer. I would like to see you tackling even the fourth root arithmetically and getting it right! An extract from the URL. "Coming up with the next "divisor" is more involved than for square roots. First bring down 3 times the square of the number on top (3 × 3²=27) leaving room for two more digits (27__)." Thanks all the same Mally Last fiddled with by mfgoode on 20070909 at 08:11 

20070909, 11:37  #6 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
I expect Mally will run tearfully to the moderators again but...
How come a Bronze medallist who spends much time in front of a computer and is interested in algorithms can dismiss "programming" as a newfangled invention apparently beneath him??? David (whose Gaelic motto has disappeared for some reason) Last fiddled with by davieddy on 20070909 at 11:38 
20070909, 11:47  #7  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
I suspect the same may apply to cube roots, but when it comes to trying these things out, I am nearly as lazy as you are. 

20070909, 12:31  #8 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 

20070910, 18:05  #9 
Jul 2003
Behind BB
2^{2}×11×41 Posts 
I calculated this cube root on one side of a standard size sheet of paper using Newton's method in 4 steps using crude approximations along the way to keep the numbers easy to calculate:
Here's the sequence of approximate solutions: {100,500,300,250,234} This seems (to me, anyways) to be a lot easier than the method described in that link. 
20070911, 00:30  #10 
∂^{2}ω=0
Sep 2002
República de California
2^{3}·3·487 Posts 
The fullblown Newton's method for finding [inverse] cube root of a real input S is generated [analogously to that for inverse square root] by applying Newton's method to the function 1/x^{3}S, yielding the iteration
x_{n+1} = x_{n}*(4  S*x_{n}^{3})/3 . Then square result [if it converges], and multiply by S. 
20071005, 04:12  #11 
Nov 2005
2·7·13 Posts 
And here I was, thinking we'de just use the function with f(x) and f'(x) from any calculus book mentioned earlier. Create 1line algorythm that uses the previous answer as x. But a penandpaper method is 'cooler'.

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