solve sin(x)+e^(-x)=x to the 10th decimal places

Introduction : In this article, we will solve sin(x)+e^(-x)=x numerically using Newtons method. Solution of the given equation are points where function intersects x-axis also called roots.

Question: Write a Matlab program to find the root of \(\sin\left(x\right)+e^{-x}=x\) to the 10th decimal places using Newton’s Method.\[\]

Solution:

Given equation

\[\sin\left(x\right)+e^{-x}=x\]

can be written as

\[\sin\left(x\right)+e^{-x}-x = 0\]

Let \(f\left(x\right) =\sin\left(x\right)+e^{-x}-x\)

Newton’s Method of finding root of an equation \(f\left(x\right)=0\) is

\[x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^\prime\left(x_n\right)} , n\geq 0\hspace{5cm}(1)\]

for given initial guess \(x_0\) to the root.

% Matlab Program to find root of an equation using Newton's Method 
f=@(x) sin(x)+exp(-x)-x; % function
df=@(x) cos(x)-exp(-x)-1; % derivative of f
x(1) = 1; %initialGuess

Error_tolerance = 0.0000000001; % tolerance 1e^-10

N = 100; % steps
for i=1:N;
    functionValue(i)=feval(f,x(i));% function value evaluation at x
    derivativeValue(i)=feval(df,x(i)); % derivative evaluation at x
    x(i+1)=x(i)-functionValue(i)/derivativeValue(i); % Newton Mthod 
     
    if abs(x(i+1)-x(i))<Error_tolerance
    break;
    end
    iteration = i+1
    Root = x(i+1)
end

Output Results

Therefore \(x = 1.2350\) is the required root of the given equation.