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I wasn't at school when we were learning this, and I've forgot how to calculate a square root on paper using a formula?

Can anyone please help me? What is the formula?

I need this to write an algorithm for my college assignment home work. Thanks!

  • 1
    Do you mean Taylor series?2012-10-19
  • 6
    Try Newton's method.2012-10-19
  • 3
    Perhaps you mean this [Method of calculating square roots digit by digit](https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit-by-digit_calculation)?2012-10-19
  • 2
    I find the one MJD cites easier than the Newton series approach, but that is a matter of taste.2012-10-19
  • 1
    Also, take a look at http://math.stackexchange.com/questions/127310/root-calculation-by-hand/127316#127316, in particular the comments by robjohn & James Fennell.2012-10-19
  • 0
    Use a slide rule, raising to the $1/2$ power.2012-11-30

3 Answers 3

7

Let $ a \geq 0 $. If you want to find $ \sqrt{a} $, use the following iteration: \begin{align} x_{0} &:= a. \\ x_{n+1} &:= \frac{1}{2} \left( x_{n} + \frac{a}{x_{n}} \right). \end{align} Then $ \displaystyle \lim_{n \rightarrow \infty} x_{n} = \sqrt{a} $. This is an example of the Newton-Raphson method, applied to the function $ f(x) = x^{2} - a $. This particular iteration exhibits quadratic convergence, which means that the number of correct digits about doubles in size with each successive step. Hence, on an ordinary scientific calculator, one can obtain a good approximation after just a few steps.

  • 3
    I remember asking my teacher in about third grade, since we were doing the division step by hand as well, and a certain amount of careful work was involved, to how many digits should we calculate $a / x_n?$ I never got an answer.2012-10-19
  • 1
    I do not think it accepted your edit. Try to put in $\frac{a}{x_n}$ again and be careful to hit whatever button means "save edit." Never mind, I did it.2012-10-19
  • 3
    That is known as the [Babylonian method](https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method).2012-10-19
  • 0
    this I used! helped alot, thanks!2012-10-19
  • 1
    @WillJagy: Thanks! My keyboard is having some problems, so typing is a little difficult.2012-10-19
  • 2
    Haskell, I have always had good luck using my home version of Latex, composing there until it is what I actually want, then copying and pasting to window here. Not usually necessary for short posts unless your computer has serious problems. Very helpful with long posts if the computer insists on rendering the Latex/MathJax as you go along, thus slowing things down.2012-10-19
  • 1
    Thanks for the advice! It's even harder to post comments sometimes, especially when I have to use braces: '{' and '}'. Anyway, I'm glad that you made the edit.2012-10-20
  • 1
    @MJD: Yes, I just realized that the Babylonians discovered this method centuries before Newton discovered his iterative method.2012-10-20
  • 0
    Wait, if the number of digits doubles each step, that means the number of digits it gives grows exponentially. Is how fast the digits grow a different order of growth then how fast in converges.2013-06-26
0

Use $$(1+x)^{1/2}=\sum_{n=0} \binom{1/2}{n}x^n$$ for $|x|<1$, where $\binom{a}{n}=\frac{a(a-1)\dots (a-n+1)}{n!}$.

-1

Observe that $$\sin \left( {\frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}$$ so you can use Taylor series for either $\sin (\pi /4)$ or $\cos (\pi /4)$ to approximate ${\sqrt 2 }$ for example.

Or you could just use linear approximation from calculus.

I am not familiar with algorithms, but it seems like linear approximation should be easier.

  • 4
    How does this generalize to arbitrary $x$?2012-10-19
  • 1
    Well, the easier way is $\sqrt r = \sum\nolimits_{n = 0}^\infty {{{\left( {\ln \sqrt r } \right)}^n}/n!}$ for $r \in {\mathbb R}$.2012-10-19
  • 2
    But you need to calculate $ \ln(\sqrt{r}) $, and the $ \ln $ function, I believe, is more complicated than the square-root function.2012-10-19
  • 1
    @HaskellCurry I agree. So I think linear approximation is the easiest solution.2012-10-19
  • 2
    But you have to calculate powers of $\pi/4$. That does not sound very convenient to me.2012-10-19