If you were given a rad and it wasn't a perfect square like 129, how would you do it if you wanted to get the decimal?
Example of \(\sqrt{y} = \sqrt{130}\) Start with a guess. It helps if it a good guess. 130 is between \(10^2\) and \(14^2\) so we will use \(12 = \frac{10 + 14}{2}\) as our initial guess, \(x_0\). Iterate \(x_{n+1} = \frac{x_n + \frac{y}{x_n}}{2}\). So your guesses would be \( 12, \frac{137}{12}, \frac{37489}{3288}, \frac{2810847841}{246527664}, ...\) and already the calculation is very accurate. 2810847841/246527664 = 11.401754250995539389.... ( a rational number with period 18744) \(\sqrt{130}\) = 11.40175425099137979... ( an irrational number ) Why would this work? Lets say \(x_n = \sqrt{y}(1 + \Delta)\) Then \(x_{n+1} = \sqrt{y} \frac{1 + \Delta + \frac{1}{1 + \Delta}}{2} \approx \sqrt{y} \frac{(1 + \Delta) + (1 - \Delta + \Delta^2 + ... )}{2} = \sqrt{y} ( 1 + \frac{1}{2}\Delta^2 + O(\Delta^3) )\). So if your guess is 10% off, your next guess should only by 1% off, and your next guess 0.01% off. http://mathworld.wolfram.com/NewtonsIteration.html
thanks it's hard for me to understand the way your aroaching it (with the Newton algorithem) because I haven't learned those symbols yet but I will look more into that site you posted. thanks again.
Guess zero: 12 Guess one: (12 + 130/12)/2 = (72 + 65)/12 = 137/12 = 11.41666.... Guess two: ( (137/12) + 130/(137/12) )/2 = 37489/3288 = 11.4017639902676399026... Guess three: 2810847841/246527664 = 11.401754250995539389.... And so on...
Hey rpenner, I just wanted to point out that the algorithm you're using originally comes from the ancient Babylonians, I remember seeing it in one of my math history books. I suppose you can derive it using the Newton-Raphson method, but the original derivation was mostly just based on simple intuition. Suppose \(x_n<\sqrt{y}\). Then \(\frac{y}{x_n}>\sqrt{y}\). If we define a number \(x_{n+1}\) as \(x_{n+1}=\frac{x_n+\frac{y}{x_n}}{2}\), then we know \(x_{n+1}\) lies between \(x_n\) and \(\frac{y}{x_n}\), and hence so does \(\frac{y}{x_{n+1}}\). Now, repeating the previous steps, we call the average of these last two numbers \(x_{n+2}\), i.e. \(x_{n+2}=\frac{x_{n+1}+\frac{y}{x_{n+1}}}{2}\). Then we know, using the same reasoning as before, that the numbers \(x_{n+2}\) and \(\frac{y}{x_{n+2}}\) lie inbetween the numbers \(x_{n+1}\) and \(\frac{y}{x_{n+1}}\). So in this way we can produce a sequence of numbers trapping \(\sqrt{y}\) between increasingly tighter bounds. At this point the Babylonians were satisfied with this method and didn't even have a concept of rigorous proof at the time. But if you wanted to be rigorous about in the modern mathematical sense, all you have to do is invoke the nested intervals theorem to show that the sequence converges, and from there it's easy to show that the sequence must converge to \(\sqrt{y}\), without even needing to invoke any calculus.
He uses a proper method, my preferred method is this. for \(sqrt(130)\) \(11^2 < (130)<12^2\) \(11.5^2 > (130)\) \(11.25^2<(130)\) \(11.375^2 <(130)\) \(11.4^2 < (130)\) \( 11.42^2 > (130)\) Rpenner provided a very methodical and undeniable method. If your calculation needs to be quick more than accurate I prefer mine because after time the direction of the needed ratio becomes intuitive.
Since your search by subdivision gets about the same number of digits of accuracy per K steps, and once you get close Newton's method doubles the number of accurate digits on each step, then it is clear to see that Newton's method is preferred if you want many digits of accuracy.
Yeah, I'll have to Newtons method. Mine's just the way I made up when I had no calculator available. That's surprisingly easier than I thought it would be. Thanks for the tip. second edit: Looking up all of Newtons iteration processes are amazingly useful..and fascinating to chart.
