Use Newtons method to find an approximation to the first pos

Use Newton’s method to find an approximation to the first positive zero of f(x)=xtan(x)-1.

(a) Solve the equation numerically using MATLAB:

        format long

        f=@(x) x*tan(x)-1

        sol = fzero(f,[0,1])

       Why are we guaranteed at least one solution in the interval [0,1]?

(b)   Find the derivative of f(x).

(c)    Using ½+10/n as your starting value, provide your iterative formula

(d)   Use MATLAB to write and execute a routine. How many iterations does it take for your answer to match the answer given in a? Do you have fast (quadratic) or slow (linear) convergence? Why? Display the intermediate approximate solutions as a table.

Solution

on script page

clear all

f=@(x)x*tan(x)-1

df=@(x)x*sec(x)^2+tan(x)

x0=input(\'enter initial guess\')

while abs(f(x0))>0.0001

x1=x0-(f(x0)/df(x0))

x0=x1

end

On work space

f =

    @(x)x*tan(x)-1


df =

    @(x)x*sec(x)^2+tan(x)

enter initial guess0.5

x0 =

    0.5000


x1 =

    1.1080


x0 =

    1.1080


x1 =

    0.9466


x0 =

    0.9466


x1 =

    0.8710


x0 =

    0.8710


x1 =

    0.8605


x0 =

    0.8605


x1 =

    0.8603


x0 =

    0.8603

>>

Use Newton’s method to find an approximation to the first positive zero of f(x)=xtan(x)-1. (a) Solve the equation numerically using MATLAB: format long f=@(x) x
Use Newton’s method to find an approximation to the first positive zero of f(x)=xtan(x)-1. (a) Solve the equation numerically using MATLAB: format long f=@(x) x

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