May we generalize a formula to determine the cube roots of any arithmetic number?
Real mathematicians may laugh at this idea, but, this would be my starting point:
I assume that if you are asking about cube roots, you already have a general formula for calculating the square root of any real number, so, how about square rooting three times and cubing the answer (all algebraically, of course)?
Arithmetically how we calculate cube root? For example, how to Cube roots of 17, as square root?
Sorry Bijan. If I understand your suggestion correctly it appears to be a way of obtaining the 3/8 power of a number. While 3/8 is close to 1/3 it is nowhere near exact, and this will be a bad approximation in most situations.
If you want to numerically find a cube root by hand, why not use the first several terms of the generalized binomial formula (which, unlike the ordinary binomial formula is an infinite series, but the first few terms give a pretty good approximation, which can be improved by using more terms)? You can look it up in many undergraduate texts.
Or you can modify an old trick of Newton. There are infinitely many variants on this, but a particularly simple (though maybe not particularly efficient) way of doing this is to just pick a number you think might be close, let us say 3 is close to the cube root of 17 (It doesn't have to be very close. We could even start with 100 as a "cube root of 17" and the method will still work!). To get a better approximation we reason as follows: 3^3 is not exactly the cube root of 17, it's either too big or too small. But (17/9)*3^2 is exactly equal to 17. The "other" factor we've introduced, 17/9, is either too large or too small, corresponding to 3 being too small or too large (we don't need to know which is which!)
This means that the cube root of 17 is somewhere between 3 and 17/9. this means that both of these numbers approximate the cube root of 17 with an error of at most 3-17/9 = 10/9. Not great, but (in some sense) not too bad either.
No problem. We can divide the error (at least!) in half by taking the average of the two:
(17/9+3)/2 = 22/9. The error thus no more than 5/9. However, it's "complement" by the same reasoning as above is 17*81/22^2 = 1377/484 or, if you prefer decimals (since our fractions are going to involve large numbers in a couple of steps this is close to 2.845. Subtracting from 22/9 we see that we have trapped the cube root of 17 on an interval of width about 0.4, which is even better than we had predicted. So at this point you probably only want to hold onto 3 sig figs if you're using decimal arithmetic. Do it again. Etc.
There are, as I said, many such algorithms (infinitely many!), and the one I give here is extremely crude. You might reason that (17/9)*3*3 uses 17/9 once but 3 twice, so why not take the weighted average (17/9 + 2*3)/3 = 71/27? Indeed, if you compare the cube of this number to that of 22/9 you'll understand that this is a considerably better approximation. In fact, our method now shows that we have trapped the cube root on an interval of width about 0.16.
This "weighted" version is considerably faster, but you can speed it up immensely with a few simple tricks from numerical analysis, if necessary.
I can think of four ways of which 3 might be practical.
1. Impractical: binomial theorem
2. Taylor series
3. Pade expansion
4. Newton' Raphson method
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