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

kimchi

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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?
 
Re: newton's method

Hello, kimchi!

Estimate the point of interseciton of the graphs of y=tanx\displaystyle y\:=\:\tan x and y=2x.\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 ...

f(x)=tanx2x\displaystyle f(x)\:=\:\tan x\,-\,2x

f(x)=sec2x2\displaystyle f'(x)\:=\:\sec^2x\,-\,2

xn+1  =  xntan(xn)2(xn)sec2(xn)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: x1=1.2\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: 1.21.169=0.031\displaystyle \:1.2\,-\,1.169 \:=\:0.031 ... too big



I used: x2=1.169\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: 1.1691.1656=0.0034\displaystyle \:1.169\,-\,1.1656 \:=\:0.0034 ... too big



I used: x3=1.1656\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: 1.16561.16556=0.00004\displaystyle \:1.1656\,-\,1.16556 \:=\:0.00004 . . . Yes!

 
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