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I was reading Richard Feynman biography when I read that one time he was able to calculate the cube root of large number in his brain by just using simple facts of everyday life. So my question is there a way to calculate square root or cube root by using just a pen a paper, I mean a systematic way not by trial and error.

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  • $\begingroup$ Trial and error is the most effective way to understand and to improve one's mechanisms... $\endgroup$ – abiessu Nov 20 '13 at 22:46
  • $\begingroup$ This "pen and paper" question (for square roots) was asked at math.stackexchange.com/questions/90435/… $\endgroup$ – Barry Cipra Nov 21 '13 at 1:10
  • $\begingroup$ @abiessu I think amaach means a trial-and-error way of calculating the root simply by guessing at the root, cubing or squaring it, and adjusting the guess larger or smaller. If you keep doing that, instead of switching to a better algorithm, you will not improve your mechanisms. $\endgroup$ – Kaz Nov 21 '13 at 2:07
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One approach is to use Newton's Method, which provides a convergent numerical approximation (with the caveat that you choose an initial guess that is 'close enough' to the real solution; in this sense, doing so for large numbers gets more difficult).

Newton's Method is based upon finding roots of a function $f(x)$. To see how this applies to square or cube roots, suppose that $y=\sqrt{n}$ for some fixed $n$. Well, then this $y$ would be a root of the equation $f(x)=x^2-n$. Similarly, $f(x)=x^3-n$ would provide us with a way to calculate the cube root of $n$.

Newton's Method works as follows: start with an initial guess $x_0$, and then, for each $n$, define $x_{n+1}:= x_n-\frac{f(x_n)}{f'(x_n)}$.

Take the simple example of calculating $\sqrt{10}$. We use $f(x)=x^2-10$ with Newton's Method to find this. In this case, $f'(x)=2x$. Letting out initial guess be $x_0=3$ ($3^2=9$ which is close to $10$), we have $x_1=3-\frac{3^2-10}{2(3)}=3+1/6=19/6$. Then $x_2=19/6-\frac{(19/6)^2-10}{19/3}\approx 3.16228$. For comparison, $\sqrt{10}\approx 3.162277$, which is already very close to 3.16228.

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The way I use to get an approximate $n$-th roots is to use the first terms of the binomial theorem in the form $$\sqrt[n]{a^n+b} =a \sqrt[n]{1+\frac{b}{a^n}} \sim a(1+\frac{b}{n a^n}) = a +\frac{b}{n a^{n-1}} $$

For example, $$\sqrt[3]{130} =\sqrt[3]{125+5} =\sqrt[3]{5^3+5} \sim 5+\frac{5}{3\cdot 5^2} = 5+\frac{1}{15} = 5.0666... $$

The correct result is $5.06579701910089...$.

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There are certainly ways to calculate these using pen and paper - I mentioned one here in a previous answer - it is a method I learned at school. Whether it is innovative is doubtful, since it is rather old-fashioned!

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Have you tried using Google? https://en.wikipedia.org/wiki/Methods_of_computing_square_roots http://mathforum.org/library/drmath/view/52605.html

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  • $\begingroup$ I would like an innovative way. $\endgroup$ – amaach Nov 20 '13 at 22:50
  • $\begingroup$ A way which has not been proposed before is very unlikely. Better you look for known ways and learn one or more. $\endgroup$ – marty cohen Nov 20 '13 at 23:53
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In addition to the other answers, one way is to use the Maclaurin series of $\sqrt \frac{1+x}{1-x}$ or $\sqrt[3]{\frac{1+x}{1-x}}$ around $x=1$. This conveniently puts $x$ in terms of a polynomial which makes the value easier to approximate.

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If you're talking about Feynman's "duel" with the abacus guy, he got lucky on that one. The number the guy chose was $1729$, which is $12^3 + 1$. Feynman knew enough to rattle that off quickly to three or four decimals by applying a series expansion.

But there are algorithms for both square (cube) roots. These are straightforward to visualize in terms of area (volume) of a square (cube).

If you have just a pen and paper, you may "miss" on the multiplication once or twice before you find the correct next digit, but the algorithms themselves are well-defined. Google square root by hand and pick your favorite explanation.

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