# Newton Raphson method

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Durgga Rajendren on 19 Apr 2011
Answered: kalyan m on 15 Feb 2022
Given f(x) = x^3-8*x^2-200*x+1000
Locate the maximum of f(x) for x[-10,10]. The maximum must be located by finding the root of derivative of f(x).Use Newton Raphson method to perform root finding.
The question asks us to select the initial guess buy ourself after looking at the f(x) graphically. The solution must have a precision of 0.01%. Then, plot of f'(x) and the root must b done on the same figure.
I have the function for Newton Raphson method,but I think I'm missing an importance sentence because I can't get the iterations printed on the command window. Plus,I do not know how to plot the f'(x) and the root. I googled for hours but nothing helped.
Here is my function:
function root = newraph(f,df,xi,precision)
% comments on input and output
%Calculate f(xi) and f'(xi)
fxi=f(xi);
dfxi=df(xi);
%Iteration for Newton Raphson method begins
while abs(fxi)>precision
%Calculate the new estimated root,x_i_+_1
xi=xi-fxi/dfxi;
fxi=f(xi); %recalculate f(x_i_+_1)
dfxi =df(xi); %recalculate f'(x_i_+_1)
end
%The final xr value is the root
root = xi;
I hope someone can help me fix the mistakes. Any help is much appreciated. Thanks :)

Matt Tearle on 19 Apr 2011
2. Why do you want to display the iterations? But either way, just add a disp or fprintf statement in your loop, after updating xi. (Or even just leave off the semicolon.)
3. To plot the functions, make a vector of x values in [-10,10] (try linspace), then evaluate the functions at these x values. If f is a function handle, you can simply evaluate f(x).

Meysam Mahooti on 5 Dec 2019

kalyan m on 15 Feb 2022
function[]=Newton_Raphson(f,df,x0)
% INPUT f and df as inline functions
% where df is the dirivative of f
Maxit = 100;
x = zeros(1,Maxit);
x(1) = x0;
for i = 1:Maxit
if abs(df(x(i)))<10^-15
fprintf('The derivative is zero at x = %f\n',x(i))
break
end
x(i+1) = x(i)-f(x(i))/df(x(i));
err = abs(x(i+1)-x(i));
if err<10^-6
fprintf('\n The Number of iteration is %d, the root of function is x = %f\n',i,x(i+1))
break
end
end
if i == Maxit
fprintf('The Newton - Raphson method does not converge after %d iteration\n',i)
end
if err<10^-7
tt = -1:0.001:5;
y = f(tt);
plot(tt,y,'k',x(i+1),0,'*')
grid on
legend('function','Approximate root')
end
end
harmonics
x=[45 90 135 180 225 270 315 360];
y=[4 3.8 2.4 2 -1.5 0 2.8 3.4 ]
x=(pi/180)*x
syms t
T=2*pi;
w=2*pi/T;
h=2; % number of hormonics
a0=2*mean(y);
HS=a0/2;
for i=1:h
a(i)=2*mean(y.*cos(i*w*x));
b(i)=2*mean(y.*sin(i*w*x));
HS=HS+a(i)*cos(i*w*t)+b(i)*sin(i*w*t);
end
HS=vpa(HS,4);
disp(HS)
plot(x,y,'*')
hold on
t=linspace(x(1),x(end),1000);
y1=eval(HS);
plot(t,y1,'r')
function[] = Gauss_Seidal22(A,B)
[m,n]=size(A);
f=0;
%%% Diagonally Dominant
for j = 1:m
if(sum(abs(A(j,:)))-abs(A(j,j)))>abs(A(j,j))
f = 1;
fprintf('The system is not diagonally dominant\n')
end
end
if f~=1
fprintf('The given system is diagonally dominant\n')
%Gauss-Seidal Iteration Method
tol = 1e-6;
maxit = 100;
x = zeros(n,1); % Initial value As a column vector
for k = 1:maxit
xold = x;
for i = 1:n
Num = B(i)-A(i,1:i-1)*x(1:i-1)-A(i,i+1:n)*x(i+1:n);
x(i) = Num/A(i,i);
end
err = abs(x-xold);
iferr<tol
fprintf('Gauss-Seidal method is convergent\n')
break
end
end
fprintf('The %d iterations are required to get \n x=',k)
fprintf('\t\t%f\n',x)
end
end