y=(x.*exp(x)+0.3*x-0.5) ./(x-5);
y_prime=((x-5) .*(x.*exp(x)+exp(x)+0.3)-x.*exp(x)+0.3*x-0.5)) ./(x-5) .^2;
x_intial=-1;
epsilon=10^(-6);
newton(@(x)(x.*exp(x)+0.3*x-0.5) ./(x-5),@(x)((x-5).*exp(x)+exp(x)+0.3)-(x.*exp(x)+0.3*x-0.5)) ./(x-5).^2,-1,10(^6))
function [x]= newton(y,y_prime,x_initial,epsilon)
x(-1) =x_initial;
i = 1;
while abs(y(x(i))>epsilon)
x(i+1) = x(i) - y(x(i))/y_prime(x(i));
i = i+1;
end
fprintf('this answer is x=%.4f',x(i))
end

5 Comments

Rik
Rik on 21 Apr 2020
Why are you trying x(-1)=x_intitial? Your code looks like x could be a scalar instead.
correct me then please
In addition to the above comment...
This line:
y_prime=((x-5) .*(x.*exp(x)+exp(x)+0.3)-x.*exp(x)+0.3*x-0.5)) ./(x-5) .^2;
has a syntax error, one too many parentheses. Maybe you mean this instead?
y_prime=((x-5) .*(x.*exp(x)+exp(x)+0.3)-x.*exp(x)+0.3*x-0.5) ./(x-5) .^2;
This line:
newton(@(x)(x.*exp(x)+0.3*x-0.5) ./(x-5),@(x)((x-5).*exp(x)+exp(x)+0.3)-(x.*exp(x)+0.3*x-0.5)) ./(x-5).^2,-1,10(^6))
contains a few syntax errors. You are calling your function, newton, which takes four arguments, but it is a bit hard to tell what each of the four arguments should be. Maybe this is correct?
newton(@(x) (x.*exp(x)+0.3*x-0.5)./(x-5),... first arg
@(x) ((x-5).*exp(x)+exp(x)+0.3)-(x.*exp(x)+0.3*x-0.5)./(x-5).^2,... second arg
-1,... third arg
10^6) % fourth arg
This line:
abs(y(x(i))>epsilon)
MATLAB should be warning you that it is unexpected, because
y(x(i))>epsilon
will return a logical (1 or 0). Do you mean
abs(y(x(i)))>epsilon
instead?
the code is not runing
You need to click the green run triangle. That will run it. What else can I say given the incredibly detailed explanation you gave? ?‍♂️
Attach the entire script -- everything including the part where you define x - with the paper clip icon.

Sign in to comment.

 Accepted Answer

Deepak Gupta
Deepak Gupta on 21 Apr 2020
Edited: Rik on 21 Apr 2020
Hello Edward,
Looking at your code, seems like you are trying to solve an euqation using Newton's Mathod. I see many errors in your code. So i have written a new code to solve the equation, you are trying to solve. Have added comments along the lines for you to understand.
syms f(x) x; %Symbolic variables needs to be defined before using them
f(x) = (x*exp(x)+0.3*x-0.5)/(x-5); % Equation you want to solve
g = diff(f); %To find the derivative of function
epsilon=1e-6;%Max allowed deviation
x = zeros();
x(1)= -1 ;%In Matlab index starts from 1, so this is the initial value of x.
%No need to create another variable for it.
i =1;
while(abs(f(x(i)))>epsilon) %Loop should be stopped when tolerance is reached
y(i) = double(f(x(i))); %If you need function values too, otherwise, not needed
x(i+1) = x(i) - f(x(i))/g(x(i));% Calculating values for x iteratively
i = i+1;
end
Thanks,
Deepak

More Answers (0)

Categories

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!