Cubic Root from Square Root

I dont know if this passed by this ng already, but I need to ask...

Can anyone calculate the cubic root from a number using a eletronic calculator which have only the square root key??

Not precisely (use Galois theory

and David Eppstein replied:

Another way to get an approximation to the cube root of x (as long as x is non-negative) would be to raise x to the 85th power, then take the 256th root of it. The former can be performed by multiplication; the latter can be performed by using the square root key 8 times. The result will be very close to the cube root.

Perhaps not the most efficient method, but ...

Yeah, that was my thought too.

To minimize multiplications:

REM CALCULATE Y=CUBEROOT(X) K=2 FOR I=1 TO 4 Y=X FOR J=1 TO K Y=SQRT(Y) NEXT J X=X*Y K=K*2 NEXT I Y=SQRT(X) Y=SQRT(Y) PRINT Y STOP

<http://www-rohan.sdsu.edu/~jmahaffy/courses/f00/ math122/lectures/newt...

Another page that uses Newton's method to do a cube root is:
http://mathforum.org/dr.math/problems/ walters.4.4.97.html

It does not give any of the theory behind the method like the other page.

Well you can take logs using a square root calculator so there is the lead. The log is:

Enter number eg 1000 press sqrt 10 times That takes the 1024th root => 1.006768659 record this number (pen or mem) compute inverse of number => 0.993276847 subtract this result from stored number => 0.013491812 multiply by 512 => 6.90780767

and that is the natural log of the original number

Note real answer => 6.907755279

Right now your cube root requires taking one third of the log

=> 2.302602557 and then antilogging that. This is a bit tricky.

first compute 1/512 of that => 0.004497271

and compute the value a such that

a = (No +sqrt(No^2+4))/2 => 1.002251163

and take this to the 1024th power ( by multimultiplication of course) ie if one squares it and squares the result etc 10 times 2^10=1024

All this gives => 10.00015523

Now if you think I made that up then check out the algorithm in Abramowitz and Stegun. I forget where it is but it is there.

The log algorithm is

Ln(A) = 512*(A^1/1024 - 1/(A^1024)).

All the rest is easy.

Note also that you can use different powers eg

Ln(A) = 256*(A^1/512 - 1/(A^512)) etc.

Aint maths such a sweet number!!!

Well, if it has only the square root key, you can only get the n-th root, where n is a power of 2.

Also it will be hard to get the number into the calculator.

Yes , assuming the calculator also has a division key. Here is the algorithm :

Let N be the argument , so N = a*a*a Let x be an approximation to a

Now compute

x1 = SQRT( N/x)

Then x1 will be a better approximation to a than x was.

Example :

Let a = 4 thus N=64 Let x = 5

Then x1 = SQRT(64/5) = SQRT(12.8) = 3.58

and the sequence of approximations is 5 , 3.58 , 4.23 , 3.89 , 4.06 , 3.97 , 4.015 , 3.9925 , 4.0038 , ..... which is clearly converging on the cube root 4.

Yes! That was great! Thanks for all answers!