Question

How do you use Newton's Method to approximate the positive root of the equation `sin(x)=x^2` ?

Answer 1

The answer is 0.8767262154.

Recall that Newton's Method uses the formula:

`x_(n+1)=x_n-(f(x_n))/(f'(x_n))`

So we need to change the equation into a function. This is done by moving all terms to one side:

`f(x)=sin x-x^2`

And we need the derivative:

`f'(x)=cos x - 2x`

The easiest way to iterate is to program your calculator. Enter `f(x)` into `Y_1` and `f'(x)` into `Y_2`. Then enter a very short program that does this:

`A-(Y_1(A))/(Y_2(A))->A`

You can go to my website for specific instructions for the TI-83 or the Casio fx-9750 .

Finally, you need a starting value, `x_1`. Since the question is asking for a positive root, we know that `sin x` has a maximum value of `1` at `x=pi/2` and `x^2` is `1` at `x=1`. So it is safe to say that the root is less than `1` and therefore will be the starting value. Note, we don't want a starting value close to `0` because `0` is also a root but not positive. So enter the following into your calculator,

`1->A`

Then execute the program until you get the desired accuracy:

`0.8913959953`
`0.8769848448`
`0.8767262985`
`0.8767262154`
`0.8767262154`

We get 3 digits of accuracy after 2 iterations, 7 after 3 iterations, and 10 after 4 iterations. So the answer converges very quickly for this root.

If you aren't getting these values or you are finding that the convergence is very slow for another function, it is most likely that your calculator is in DEGREE mode rather than RADIAN mode.