Newton's method to find intersection of tanx and 2x

[kimchi]

Estimate the point of intersection of the graphs of y = tanx and y = 2x. Use Newton's method and continue the iterations until two successive approximations differ by less than 0.0001. Explain steps taken. Do not use the trivial case, where x=0.

What I've done so far...
f(x)=tanx-2x
f'(X)=sec^2(x)-2
x(n+1) = xn - (tanxn-2xn)/(sec^2(xn)-2)
I kind of dosed off when we learned this in class, can anyone help me with the next couple of steps?

 

[soroban]

Re: newton's method

Hello, kimchi!

Estimate the point of interseciton of the graphs of \(\displaystyle y\:=\:\tan x\) and \(\displaystyle y\:=\:2x.\)
Use Newton's method and continue the iterations
until two successive approximations differ by less than 0.0001.


What I've done so far ...

\(\displaystyle f(x)\:=\:\tan x\,-\,2x\)

\(\displaystyle f'(x)\:=\:\sec^2x\,-\,2\)

\(\displaystyle x_{n+1} \;= \;x_n\,-\,\frac{\tan(x_n)\,-\,2(x_n)}{\sec^2(x_n)\,-\,2}\;\) . . . Good!

Sketch the graphs and estimate the point of intersection.


My guess was: \(\displaystyle \:x_1\:=\:1.2\)
Then: \(\displaystyle \L\:x_2\;=\;1.2\,-\,\frac{\tan(1.2)\,-\,2(1.2)}{\sec^2(1.2)\,-\,2} \;=\;1.169236024\)

. . The difference is: \(\displaystyle \:1.2\,-\,1.169 \:=\:0.031\) ... too big



I used: \(\displaystyle \:x_2\:=\:1.169\)
Then: \(\displaystyle \L\:x_3\;=\;1.169\,-\,\frac{\tan(1.169)\,-\,2(1.169)}{\sec^2(1.169)\,-\,2} \;=\;1.165600938\)

. . The difference is: \(\displaystyle \:1.169\,-\,1.1656 \:=\:0.0034\) ... too big



I used: \(\displaystyle \:x_3 \:=\:1.1656\)
Then: \(\displaystyle \L\:x_4\;=\;1.1656\,-\,\frac{\tan(1.1656) - 2(1.1656)}{\sec^2(1.1656) - 2} \;=\;1.16556119\)

. . The difference is: \(\displaystyle \:1.1656\,-\,1.16556 \:=\:0.00004\) . . . Yes!