Let's say we want to solve for the roots of f(x) = x^2 - 2. We know the roots are the plus and minus square roots of 2, but we need to start with an easy example. Newton's method, in short, asks you to guess a solution to f(x)=0, and calculate an improved guess using f(x) and its first derivative f'(x). The algorithm:

`x_{n+1} = x_{n} - \frac{f(x_{n})}{f^{\prime}(x_{n+1})}`

For our function, this means

`x_{n+1} = x_n - \frac{x_n^2-2}{2x_n}`

So, you first guess at the root, and that is your starting x. The next guess for x is calculated from the formula above. Let's say we guess 1 for the root. It is a bad guess, but it doesn't matter too much.

`x_0=1 \qquad \Longrightarrow \qquad x_1 = 1 - \frac{1^2-2}{2\cdot 1}=1.5`

Our next-best guess is then 1.5. We plug 1.5 into the equation above, and the next-next-best guess is 1.417. Not bad! After 3-4 iterations, you'll see the numbers start to converge.

How to do this on the TI? First, type your guess and hit [enter]

`(ans) - ((ans)-2)/(2*(ans))`

Don't forget the parentheses. Hit [enter], and you've found your second guess, which is now the new value of "ans." Keep hitting [enter] until the result starts to converge, and that's that! You've implemented an iterative root-finding algorithm.Doing this for a more complicated function, like the one on the homework, is not much harder. You just need to know the function and its derivative, otherwise you follow the same procedure.

## No comments:

## Post a Comment