The problem occured with the function f. MATLAB took f to be an array inside the while loop rather than as a function, hence this happened. The resolved code is given below:

Also note that as 0.5 is the answer, and you are checking from i = 3, hence in the first iteration p(i-1) remains the same and zero division error occurs in the next iteration. Hence try p(2) = 0.45.
can some one edit this code p(1)=0.4 p(2)=0.5 e=0.00001 *secant metod >> syms x; f=(1/x) -2;...
Matlab Regula Falsi Method
A zero of f(x) = x^2 -4x-12.2 is known to exist on the interval
[1.2 , 2.2 , 3.2,...9.2] and respective right endpoints x1 =[2.2
,3.2, 4.2....10.2], find the sub-interval in which the zero exists.
Set the left endpoint of this sub-interval =a and the right
endpoint=b
With tolerance of 10^-9, use the Regular Falsi method to compute
the root of (f) in [a,b]
User input is required at #####
CONTENTS Close Courses LMS Integration Documentation...
in
matlab
-Consider the equation f(x) = x-2-sin x = 0 on the interval x E [0.1,4 π] Use a plot to approximately locate the roots of f. To which roots do the fol- owing initial guesses converge when using Function 4.3.1? Is the root obtained the one that is closest to that guess? )xo = 1.5, (b) x0 = 2, (c) x.-3.2, (d) xo = 4, (e) xo = 5, (f) xo = 27. Function 4.3.1 (newton) Newton's method...
Translate the following code to java with comments x0 = 1 # The initial guess f(x) = x^2 - 2 # The function whose root we are trying to find fprime(x) = 2 * x # The derivative of the function tolerance = 10^(-7) # 7 digit accuracy is desired epsilon = 10^(-14) # Do not divide by a number smaller than this maxIterations = 20 # Do not allow the iterations to continue indefinitely solutionFound = false # Have...
Use the Newton-Raphson method to find the root of f(x) = e-*(6 - 2x) - 1 Use an initial guess of xo = 1.2 and perform 3 iterations. For the N-R method: Xi+1 = x; - f(x;) f'(x;)
Please code in MatLab or Octave
Output should match Sample Output in Second Picture
Thank you
5. In this problem we will investigate using the Secant Method to approximate a root of a function f(r). The Secant Method is an iterative approach that begins with initial guesses , and r2. Then, for n > 3, the Secant Method generates approximations of a root of f(z) as In-1-In-2 n=En-1-f (x,-1) f(Fn-1)-f(-2) any iteration, the absolute error in the approximation can be...
The first two parts should be solved by Matlab. This is from an
intro to Numerical Analysis Class and I have provided the Alog 3.2
in below. Please write the whole codes for me.
Alog3.2
% NEWTONS INTERPOLATORY DIVIDED-DIFFERENCE FORMULA ALGORITHM
3.2
% To obtain the divided-difference coefficients of the
% interpolatory polynomial P on the (n+1) distinct numbers
x(0),
% x(1), ..., x(n) for the function f:
% INPUT: numbers x(0), x(1), ..., x(n); values f(x(0)),
f(x(1)),
% ...,...
Please do this in Matlab.
Not sure if you need this code:
e cofunction [x, er, n] = FixedPoint(g, x1, maxtol, maxitr)
if nargin < 4, maxitr = 25; end
if nargin < 3, maxtol = 1e-3; end
k = 0 ;
er = 1;
x = x1;
while er >= maxtol && k < maxitr
k=k+1;
xold = x;
x=g(x);
er=abs((x-xold)/x);
fprintf('iter = %i, x = %e, er = %e ', k,x,er);
end
n=k;
if n ==...
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5.1.2 Open Methods - Newton-Raphson Method Xi+1= xi – FOTO Matlab Code Example:4 function mynewtraph (f, f1,x0,n) Xx0; for ilin x = x - f(x)/f1(x); disp (li if f(x) <0.01 f(x))) break end end end Matlab Code from Chapra function [root, ea, iter)=newtraph (func,dfunc, xr, es,maxit,varargin) newtraph: Newton-Raphson root location zeroes 8 [root, ea, iter)-newtraph (func, dfunc, xr, es,maxit,pl,p2, ...): $uses Newton-Raphson method to find the root of fune input: func- name of function 8dfunc = name of derivative of...
I need someone to fix my C Program Code
Firstly I will post the Q and the input and the outputs
Then the code
Secondly the code
===================================
#include <stdio.h>
#include <math.h>
int main()
{
double
a,b,c,d,l,x,x_0,x_1,x_mid,p_x,p_0,p_1,p_mid,E=1e-6;
int i1,i2;
printf("Enter a,b,c,d of ax^3+bx^2+cx+d=0: ");
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
x_0=-50;x_1=-50;x=-50;p_0=a*(pow(x_0,3))+b*(pow(x_0,2))+c*x_0+d;i1=0;i2=0;
do{
p_x=a*x*x*x+b*x*x+c*x+d;
l=p_x*p_0;
if(l<0){
x_1=x;}
i1=i1+1;x++;}
while(fabs(p_x)>=E &&
x<50);
if(fabs(p_x)<E){
printf("the polynominal has root...