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Suppose of the number sqrt(156934).

Any help is appreciated !

Thanks

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  • $\begingroup$ As noted in the title, $156934$ is not a square root. So $\sqrt{156934}$ is irrational. Are you looking to approximate it, and if so, to what level of accuracy? $\endgroup$ – paw88789 Sep 13 '14 at 13:24
  • $\begingroup$ newton's method $\endgroup$ – Rustyn Sep 13 '14 at 13:36
  • $\begingroup$ 2 decimal places $\endgroup$ – Byteology Sep 13 '14 at 13:36
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Start finding the number $n$ the square of which is the closest to your number. In this case, it should be $396$ but admit that I am lazy and that I found an estimate of $400$.

So now, we need to solve the equation $f(x)=x^2-a=0$ knowing an estimate $x_0$ of the solution. The simplest would be use Newton which will update the guess according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ which in this case would write $$x_{n+1}=\frac{x_n^2+a}{2 x_n}$$ Let us try with your number starting at $x_0=400$; the successive iterates will then be $396.1675000$, $396.1489623$, $396.1489619$ which is the solution for ten significant figures.

Let us consider the calculation of $\sqrt{123456789}$ and again be lazy and use $x_0=10000$. The same method will provide the following iterates : $11172.83945$, $11111.28158$, $11111.11106$ which is the solution for ten significant figures.

You could even go faster to the solution using Halley iterative scheme. For this problem, $$x_{n+1}=\frac{3x_n^4+6 a x_n^2-a^2}{8 x_n^3}$$ For the first example, the iterates are $396.1491399$ and $396.1489619$. For the second one, the iterates are $11104.06183$ and $11111.11106$

For sure, you can use the same way to compute $\sqrt[k] a$. The equation becomes $f(x)=x^k-a$ and Newton formula is $$x_{n+1}=\frac{(k-1)x_n^k+a}{k x_n^{k-1}}$$. Let us suppose that you want the fifth root of $12345678987654321$; in scientific notation, this number is $1.23457\times 10^{16}$ that is to say $12,3457\times 10^{15}$ so the fifth root is somewhere between $10^3$ and to $2\times 10^3$ since $1^5=1<12.3457<2^5=32$. So, let us start iterating at $x_0=1500$; the iterates are $1687.730528$, $1664.137407$, $1654.505863$, $1653.117069$, $ 1653.114731$ which is the solution for ten significant figures.

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Well, for a rough approximation you can do several things to rapidly estimate it.

First, roughest estimation, number of digits. If it has 1-2 digits, then the square root is 1 digit long, if it has 3-4 digits, then the square root is 2 digits long. If it has 2n-1 or 2n digits, then the square root is n digits long. (This is true due to the fact that the square root of 100 is 10)

Next, once you know how many digits long it is, you can use the leading digits to find the approximate number. The digits at the end are useless for now, so just divide by 100, and round down, and repeat this until you get a number 1 or 2 digits long. If the number is larger than 4, than the first digit of the root is at least 2, if the number is larger than 9, then the first digit of the root is at least 3, and if the number is larger than n^2 (from 1 to 9), than you know the first digit is at least n.

These arent exact at all. These will relatively easily give you scale of magnitude and leading digits. If you want more exact guesses, just retest for roots closer and closer to that said root.

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When I was in grade school, my smart teacher had her students memorize the square roots from 1 to 10. Of course, some of these are easy to memorize but these are all you know, plus a few rules about exponents, to get a reasonable calculation in your head of any number. I also memorized the square root of 98 since it has an interesting value: 9.8994949+ -- I like the grouping of the couplets: 98, 99, 49, 49. Of course, the next two digits, 37, that follow are easy to remember as it is the first number less than 49 that is prime and where both digits are prime.

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