Question

Solve for a positive root of r- cos x = 0 by Regula Falsi Method.​

Answer 1

Answer:

Solve for a positive root of x-cos x =0 by regular falsi method.

Answer 2

Answer:

0.73908

Step-by-step explanation:

Concept

• Regula Falsi method or the method of false position is a numerical method for solving an equation in one unknown

Given

r-cos x=0

Find

positive roots of r-cos x=0

Solution

It is given the equation

$\cos x=x$

Let consider the function

$f(x)=r-\cos x$

then, each root  $\mathrm{f}(\mathrm{x})$ is a solution to the given equation. To approximate a root  $\mathrm{f}(\mathrm{x})$ we can use the Newton-Raphson Method. Since$|\cos x| \leq 1$ we have that$|x| \leq 1$, then $x_{0}=1$ is a convenient choice. Then

$x_{n+1} =x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$

$=x_{n}-\frac{\left(x_{n}-\cos x_{n}\right)}{\left(x_{n}-\cos x_{n}\right)^{\prime}}$      ....1

$=x_{n}-\frac{x_{n}-\cos \left(x_{n}\right)}{1+\sin \left(x_{n}\right)}$       .....2

Now, for $n=0,1,2 \ldots$

$x_{1}=1-\frac{1-\cos (1)}{1+\sin (1)}=0.7503$   ...3

$x_{2}=x_{1}-\frac{x_{1}-\cos \left(x_{1}\right)}{1+\sin \left(x_{1}\right)}=0.7391$ ...4

$x_{3}=x_{2}-\frac{x_{2}-\cos \left(x_{2}\right)}{1+\sin \left(x_{2}\right)}=0.73908$

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