# Calculating cube roots

there is a way of calculating the square root of any number (without using a calculator of course).
is there a similar way, or any way, in fact to calculate cube roots, fourth roots, etc. again without using a calculator??

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
"Newton's method" is good for that general kind of problem.

If $x_0$ is a good guess for a solution to f(x)= 0, then
$$x_1= x_0- \frac{f(x_0)}{f '(x_0)}[/itex] is a better one. (f ' is the derivative of f.) Now, do [tex]x_2= x_1- \frac{f(x_1)}{f '(x_1)}[/itex] and "repeat as desired". In the case of $x^3= a$, we take $f(x)= x^3- a$ so $f '(x)= 3x^2$. Thus [tex]x_{n+1}= x_n- \frac{x_n^3- a}{3x_n^2}= x_n- \frac{1}{3}x_n+ \frac{a}{3x_n^2}= \frac{2}{3}x_n+ \frac{a}{3x_n^2}$$

For $x^4= a$, take $f(x)= x^4- a$ so $f '(x)= 4x^3$ and do the same:
$$x_{n+1}= x_n- \frac{x_n^4- a}{4x^3}= x_n- \frac{1}{4}x_n+ \frac{a}{4x^3}= \frac{3}{4}x_n+ \frac{a}{4x^3}$$

I wouldn't be crazy about doing those calculations with pencil and paper, but you're the one who doesn't want to use a calculator!

NateTG
Science Advisor
Homework Helper
Another option would be to use a Taylor/McLaurin series starting from a nearby perfect cube, fourth power, or whatever. It might be a bit more work to get started than Newton's method, but it converges faster.

Gokul43201
Staff Emeritus
Science Advisor
Gold Member
I often use a Taylor approximation (+ some intuition from experience), when I play the famous drinking game "What's the cube root of...?". Sometimes I resort to logarithms if convenient. But that probably doesn't help unless you're a veteran of the also famous drinking game "What's the logarithm of...?"

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I have memorised a good majority of the lower roots $$x^{1/2}$$ for $$|x| < 10$$ to as much as 12 decimal places, i could go further.

Is there any need for an accuracy greater than 12d.p for any physical calculation?

What degree of uncertainty, in terms of physical dimensions does 12.dp approximation have? The scale of molecules? of Marbles?

Would it be sufficient for atomic simulations? or would that require more precision?

Why do you want to memorize those. If you're doing calculations to that precision you'll be using a computer anyways.

Because i can, and youll be supprised what insight knowing and recognising these values has when looking for patterns and problem solving.
I used to think the same, why? then i memorised them and wondered why i hadnt done it earlier.

I also memorise a bunch of log values, factorials, trigonometric values, cube roots, decimal fractions, binary tree alphabets etc. I find the lookup table faster than calculating. And its kinda fun.

So, Why not? Why bother thinking at all, if a computer can do it faster and cheaper and more accuratley and....

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I learned how to do square roots...like pull out decimals etc in 8th grade. That one is simple but I never did anything with cube roots. It's probably not what you want but you can get the EXACT FORM (worthless for any application) by simply factoring and pulling out 1/3 of every repeating factor...

Edit: I just tried you can't just do same thing and triple instead of double or add one more number...:(

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3trQN said:
[just] Because i can
doesn't mean that I should! :tongue2:

Why bother thinking at all, if a computer can do it faster and cheaper and more accuratley and....
There are a lot of things that computers can't do, and they're unreliable. But they are pretty good at calculating things to 12 decimal places!

I have a copy of the Handbook of Mathematical Functions sitting on my bookshelf in case I need to do something without a computer. Of course, a friend of mine has $\pi$ memorized to some 400 digits. To my knowledge he hasn't actually used it for anything yet, but if I ever need to know what the circumference of a perfect sphere with a diameter of 10^400m is to 1m precision, then I know who to ask.

Why not? Well, if I were an economist, I'd just say "opportunity cost."

Data said:
Why do you want to memorize those. If you're doing calculations to that precision you'll be using a computer anyways.
You asked why, i answered, it wasnt a personal attack.

As far as the "complicated method" for square roots, just remember that once upon a time, there were no computers. Roots were taken in college Physics then largely using slide rules. I had a math book in grade school that gave the hand method for square roots and also for cubes.

Square roots seem to be used a lot more than cubic ones, but, I guess, the feeling was that the student should not only learn square roots, but also top off their study of Arithmetic with cube roots for good measure.

I want to add that in the old, old days of my grandfather on the farm, math studies consisted of knowing arithmetic up to 12x12. Power of 2 went much higher, at least for the quick student, and that it was useful to know 12^3, a great gross, since they are 12 eggs in a dozen and 12 dozen in a gross. A farm boy might need to know that.

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3trQN said:
You asked why, i answered, it wasnt a personal attack.
I know! I was just giving my answer to "why not!"

There is in fact a way to calculate cube roots by hand. In fact, if you look at the steps closely, you will see that the method is very similar to some methods for calculating square roots by hand. Searching Google will yeild some quick results. Does anyone know how to calculate quartic roots by hand?

uart
Science Advisor
I have memorised a good majority of the lower roots $$x^{1/2}$$ for $$|x| < 10$$ to as much as 12 decimal places, i could go further.

Is there any need for an accuracy greater than 12d.p for any physical calculation?

What degree of uncertainty, in terms of physical dimensions does 12.dp approximation have? The scale of molecules? of Marbles?

Would it be sufficient for atomic simulations? or would that require more precision?
The majority of roots $$x^{1/2}$$ for (all real) $$|x| < 10$$ to as much as 12 decimal places. Ok I've got to admit that would be impressive. ;)

decimal precision

I don't know how precise on the small scale, but I know that knowing pi to the 15th decimal place is enough to accuratly get the circumfrance of the universe to within a thousandth of an inch. So I would say 12 is pretty solid.

I always use a calculator. I might of missed something in school because I don't remember doing it by hand. I have taken beginners algebra to calculus II and I don't recall doing it by hand. I guess I missed out. Sorry I couldn't help you but I'm asking myself the same question.