QUESTION 12:
Say that the number you want the root of is N, and that the current value of the computation is oldGuess. How can you tell when the results are close to perfect?
Yes. Each time through the outer loop the last statement changes the rate from the value that was just used. If the loop exits (because goal was reached), this new rate will be printed, not the rate used in the just completed calculation.
Enough about boring financial stuff.
The library java.lang.Math contains the square root function as well as many other functions like sine, cosine, and absolute value. The functions in this library have been carefully written and tested by programmers who are experts in numerical analysis. Ordinarily, you use a method from this library to compute a square root. Here, let us investigate how the square root function could be written.
Newton's method of computing a square root of a number N is to make a first guess, then to improve that guess to get a better guess, then to impove the better guess to get an even better guess, and so on. The guess improvement is given by a formula:
newGuess = N/(2*oldGuess) + oldGuess/2
(The reason this works is one of the subjects of Calculus I, but you don't need to know why it works to understand the program.) For example, say that you want the square root of N = 3.00. The first guess does not have to be very accurate; 1.0 is good enough.
| oldGuess | N/(2*oldGuess) | oldGuess/2 | newGuess |
|---|---|---|---|
| 1.0 | 1.5 | 0.5 | 2.0 |
| 2.0 | 0.75 | 1.0 | 1.75 |
| 1.75 | 0.85714 | 0.875 | 1.73214 |
| 1.73214 | 0.86598 | 0.86607 | 1.73205 |
The last new guess is pretty close; 1.73205*1.73205 = 2.99999 You can't always count on getting that close in four steps, however. A program that uses this method will have to keep improving the guess and stop when the results are close to perfect, however long that takes.
Say that the number you want the root of is N, and that the current value of the computation is oldGuess. How can you tell when the results are close to perfect?