How do you use Newton's method to find the approximate solution to the equation $tan x = e^{x}, 0 < x < \frac{π}{2}$ $tanx=e^x, 0<x<pi/2$ ?

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Answer 1 · 2017-01-02T11:50:29

$x = 1.30633$ $x=1.30633$ to 6dp

Let $f (x) = tan x - e^{x}$ $f(x) = tanx-e^x$ Then our aim is to solve $f (x) = 0$ $f(x)=0$ in the interval $0 < x < \frac{1}{2} π$ $0 lt x lt 1/2pi$

First let us look at the graphs:
graph{tanx-e^x [-1, 5, -15, 15]}

We can see there is one solution in the interval $0 < x < 1.57079 (= \frac{π}{2})$ $0 < x < 1.57079 (=pi/2)$ .

We can find the solution numerically, using Newton-Rhapson method

$f(x) = tanx-e^x => f'(x) = sec^2x-e^x$ , and using the Newton-Rhapson method we use the following iterative sequence

${ (x_0,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}$

Then using excel working to 6dp we can tabulate the iterations as follows:

enter image source here

And we conclude that the remaining solution is $x=1.30633$ to 6dp

How do you use Newton's method to find the approximate solution to the equation tanx=e^x, 0<x<pi/2tanx=ex,0<x<π2tanx=e^x, 0<x<pi/2?