Question 383758
<font face="Garamond" size="+2">


First write your equation as a function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \tan(x)\ -\ x]


Newton-Raphson says


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{n\,+\,1}\ =\ x_n\ -\ \frac{f(x_n)}{f'(x_n)}]


So take the first derivative of the function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \sec^2(x)\ -\ 1]


Then start with *[tex \Large x_0\ =\ 4.5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{1}\ =\ x_0\ -\ \frac{\tan(x_0)\ -\ x_0}{\sec^2(x_0)\ -\ 1}\ =\ 4.5\ -\ \frac{\tan(4.5)\ -\ 4.5}{\sec^2(4.5)\ -\ 1]


Once you have fat-fingered your calculator (in Degrees mode) to get a nice approximation of *[tex \Large x_1], plug that number into:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{2}\ =\ x_1\ -\ \frac{\tan(x_1)\ -\ x_1}{\sec^2(x_1)\ -\ 1}]


to calculate *[tex \Large x_2], that you then plug into:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{3}\ =\ x_2\ -\ \frac{\tan(x_2)\ -\ x_2}{\sec^2(x_2)\ -\ 1}]


Which is your third iteration and the approximation you need.  If you don't come up with something very close to *[tex \LARGE 4.49340946], then re-do your calculations.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>