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 2007-09-08, 16:26 #1 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22×33×19 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
 2007-09-08, 19:04 #2 wblipp     "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
 2007-09-09, 07:17 #3 Citrix     Jun 2003 3·232 Posts Cube root or for that matter the nth root, all work by the same principle. What is your difficulty?
2007-09-09, 07:56   #4
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
Cube root.

Quote:
 Originally Posted by wblipp Googling Cube root by hand turns up this as first choice http://www.nist.gov/dads/HTML/cubeRoot.html

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;

2007-09-09, 08:09   #5
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

40048 Posts
Same principle

Quote:
 Originally Posted by Citrix Cube root or for that matter the nth root, all work by the same principle. What is your difficulty?

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 2007-09-09 at 08:11

 2007-09-09, 11:37 #6 davieddy     "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 2007-09-09 at 11:38
2007-09-09, 11:47   #7
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by mfgoode "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__)." Mally
Long division and square roots are simpler in binary.
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.

2007-09-09, 12:31   #8
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by mfgoode This URL is fit for a printout for later reference. Mally :coffee;
How about reading it?

 2007-09-10, 18:05 #9 masser     Jul 2003 Behind BB 22×11×41 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.
 2007-09-11, 00:30 #10 ewmayer ∂2ω=0     Sep 2002 República de California 23·3·487 Posts The full-blown 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/x3-S, yielding the iteration xn+1 = xn*(4 - S*xn3)/3 . Then square result [if it converges], and multiply by S.
 2007-10-05, 04:12 #11 nibble4bits     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 1-line algorythm that uses the previous answer as x. But a pen-and-paper method is 'cooler'.

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