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

##### 1 Answer

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

#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,

#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.