The smallest positive guess for which the Newton method diverges for the function f(x)=atan(x) is 1.4. The Newton method was able to find the root for guesses up to 1.39 but produced a divide by zero error at 1.4, indicating that 1.4 is the smallest positive value where the method diverges.

1. EC 313: Numerical Methods Mayank Awasthi
MATLAB Assignment – 1 M. Tech, IIT Roorkee
(Communication Systems)
Ques1:
MATLAB routine which finds the coefficients of Lagrange polynomial is below:
%Implementing Lagrange Interpolating Formula
function l = lagranp(x,y)
N = length(x)-1; %the degree of polynomial
l = 0;
for m = 1:N + 1
P = 1;
for k = 1:N + 1
%Multiplication of two polynomials corresponds to convolution of their
coefficients.
if k ~= m
%”linconv” function created by own.
P = linconv(P,[1 -x(k)])/(x(m)-x(k));
end
end
l = l + y(m)*P; %Lagrange polynomial coefficients
end
MATLAB routine for Linear Convolution:
%Linear convolution of sequence using formula
function s=linconv(x,h)
N=length(x);
M=length(h);
for n=1:N+M-1
s(n)=0;
for k=1:M
a=n-k;
if a<0
y(k)=0;
elseif k>N
y(k)=0;
elseif a<M
y(k)=x(k)*h(a+1);
else
y(k)=0;
end
s(n)=s(n)+y(k);
end
end

2. MATLAB routine which finds the coefficients of Newton polynomial is below:
%Implementing the Newton divided difference formula
function n = newtonp(x,y)
N = length(x)-1;
f = zeros(N+1,1);
f(1:N+1,1) = y;
for k = 2:N + 1
for m = 1: N + 2 - k %Divided Difference
f(m,k) = (f(m + 1,k - 1) - f(m,k - 1))/(x(m + k - 1)- x(m));
end
end
a = f(1,:);
n = a(N+1);
for k =N:-1:1
%Nested multiplication form for better computation efficiency
%a_k=f(x0,x1,……….xk)
n = [n a(k)] - [0 n*x(k)]; %f(x)*(a_k*(x - x(k - 1))+a_k – 1)
end
>> x=[1 3 4]; Calculating the polynomial using both routines for
>> y=[1 27 64]; given data.
>> coefflag = lagranp(x,y)
coefflag =
8.0000 -19.0000 12.0000 So required polynomial we get 8x2 – 19x + 12
>> coeffnew = newtonp(x,y)
coeffnew =
8 -19 12

3. Ques2:
a)- For Straight Line: deg(Polynomial)=1
So using Newton or Lagrange routine we can get the coeff. Of polynomial as
below:
>> x= [1970 1990] ;
>> y= [3707475887 5281653820] ;
>> nc=newtonp(x,y)
nc =
1.0e+011 *
0.0008 -1.5135
>> v0=polyval(nc,1980)
v0 = % 1980 Population using straight line
4.4946e+009
b)- For parabola: deg(Polynomial)=2
>> x= [1960 1970 1990];
>> y= [3039585530 3707475887 5281653820];
>> nc= newtonp(x,y)
nc =
1.0e+012 *
0.0000 -0.0015 1.4063
>> v1=polyval(nc,1980) % 1980 Population parabola
v1 =
4.4548e+009
c)- For cubic curve deg(Polynomial)
>> x= [1960 1970 1990 2000];
>> y= [3039585530 3707475887 5281653820 6079603571];
>> nc=newtonp(x,y)
nc =
1.0e+013 *
-0.0000 0.0000 -0.0107 7.0777
>> v3=polyval(nc,1980) % 1980 Population using cubic curve
v3 =

4. 4.4729e+009
Ques4:
MATLAB routine which computes the divide difference coefficient is below:
%matalb script for computing dividedifference coefff.
function ddc = dividediff(x,y)
N = length(x)-1;
f = zeros(N+1,1);
f(1:N+1,1) = y;
for k = 2:N + 1
for m = 1: N + 2 - k %Divided Difference
f(m,k) = (f(m + 1,k - 1) - f(m,k - 1))/(x(m + k - 1)- x(m));
end
ddc=f(1,:);
end
end
Calculating the divide difference coefficients.
>> x=[1 2 4 5 7 9];
>> y=[2 3 8 6 14 6];
>> divide_diff_coeff = dividediff(x,y)
divide_diff_coeff =
2.0000 1.0000 0.5000 -0.5000 0.2000 -0.0518
Now Calculating the polynomial using Newton method routine used in Ques1.
>> x=[1 2 4 5 7 9];
>> y=[2 3 8 6 14 6];
>> nc= newtonp(x,y)
nc =
-0.0518 1.1839 -9.7875 35.6018 -53.4464 28.5000
Required Polynomial is
-0.0518x5 + 1.1839x4 -9.7875x3 + 35.6018x2-53.4464x +28.5000
v0=p(1.5), v1=p(3.7), v2=p(6.6), v3=p(9.5)
>> v0=polyval(nc,1.5) >> v2=polyval(nc,6.6)
v0 = v2 =
1.0020 10.6441
>> v1=polyval(nc,3.7) >> v3=polyval(nc,9.5)
v1 = v3 =

5. 8.3474 -21.6454
Plotting Graph
>> x=[1 2 4 5 7 9];
>> y=[2 3 8 6 14 6];
>> nc= newtonp(x,y)
nc =
-0.0518 1.1839 -9.7875 35.6018 -53.4464 28.5000
>> m=[0:10];
>> n=polyval(nc,m)
n=
28.5000 2.0000 3.0000 7.6286 8.0000 6.0000 7.0714 14.0000 20.7000
6.0000 -72.5714
>> plot(m,n)

6. Ques5 :
Bisection Method :
%Bisection method to solve Non-linear equations.
function [x] = bisct(f,a,b,z,MaxIter)
fa = feval(f,a);
fb = feval(f,b);
if fa*fb > 0
error('We must have f(a)f(b)<0!');
end
for k = 1:MaxIter
xx(k) = (a + b)/2;
fx =feval(f,xx(k));
err = (b-a)/2;
if abs(err)<z
break;
elseif fx*fa > 0
a = xx(k); fa = fx;
else b = xx(k);
end
x = xx(k);
end
fprintf('The Interval in which root lie is %d %dn',a,b)
fprintf('Iteration Required is %dn',k)
fprintf('The approx. error is %dn',err)
end
Regula Falsi Method :
%Implementing Regula Falsi algorithm.
function [] = falsp(f,a,b,z,MaxIter)
fa =feval(f,a);
fb=feval(f,b);
if fa*fb > 0
error('We must have f(a)f(b)<0!');
end
for k = 1: MaxIter
xx(k) = (a*fb-b*fa)/(fb-fa);
fx = feval(f,xx(k));
err = max(abs(xx(k) - a),abs(b - xx(k)));
if err<z
break;
elseif fx*fa > 0
a = xx(k);
fa = fx;
else b = xx(k);
fb = fx;
end
end
x = xx(k);
fprintf('The Interval in which root lie is %d %dn',a,b)
fprintf('Iteration Required is %dn',k)
fprintf('The approx. error is %dn',err)
end

7. Modified Regula Falsi Method :
%Implementing Modified Regula Falsi algorithm
function [] = mrfalsp(f,a,b,z,MaxIter)
fa =feval(f,a);
fb=feval(f,b);
F=fa;
G=fb;
if fa*fb > 0
error('We must have f(a)f(b)<0!');
end
xx(1)=a/2;
for k = 1: MaxIter
xx(k+1) = (a*G-b*F)/(G-F);
fx = feval(f,xx(k+1));
err = max(abs(xx(k) - a),abs(b - xx(k)));
if err<z
break;
end
if fx*fa > 0
a = xx(k+1);
fa = fx;
elseif feval(f,xx(k))*feval(f,xx(k+1))>0
F=F/2;
end
if fx*fa < 0
b = xx(k+1);
fb = fx;
elseif feval(f,xx(k))*feval(f,xx(k+1))>0
G=G/2;
end
end
fprintf('The Interval in which root lie is %d %dn',a,b)
fprintf('Iteration Required is %dn',k)
fprintf('The approx. error is %dn',err)
end

8. Newton Raphson Meyhod :
%Implementing Newton Raphson Method
function [] = newton(f,df,x0,z,MaxIter)
xx(1) = x0;
fx = feval(f,x0);
for k = 1: MaxIter
if ~isnumeric(df)
dfdx = feval(df,xx(k));end
dx = -fx/dfdx;
xx(k+1) = xx(k)+dx;
fx = feval(f,xx(k + 1));
a= xx;
if abs(dx) < z
break; end
end
l=length(a);
fprintf('The approx. Root is %dnn',x);
fprintf('Iteration Required is %dn',k)
end
Now calculating the roots of given polynomials
a)- exp(-x)-sin(x) assuming interval (0,1) & Max_Iteration=50
>> f=inline('exp(-x)-sin(x)'); %matlab command to define function
>> df=inline('-exp(-x)-cos(x)'); %derivative of f
1)- Bisection method:
>> bisct(f,0,1,1e-6,50)
The Interval in which root lie is 5.885315e-001 5.885334e-001
Iteration Required is 20
The approx. error is 9.536743e-007
2)- Regula Falsi:
>> rfalsi(f,0,1,1e-6,50)
The Interval in which root lie is 0 5.885327e-001
Iteration Required is 50
The approx. error is 5.885327e-001
3)- Modified Regula Falsi:
>> mrfalsi(f,0,1.0,1e-6,50)
The Interval in which root lie is 5.885315e-001 5.885342e-001
Iteration Required is 23
The approx. error is 3.976230e-006
4)- Newton Raphson Method:
>> newtonrap(f,df,.2,1e-6,50)
The approx. Root is 5.885327e-001
Iteration Required is 4

9. b)- x- exp(-(x^2)) assuming interval (0,1) & Max_Iteration=50
>> f=inline('x-exp(-(x^2))'); %matlab command to define function
>> df=inline('1+2*x*exp(-(x^2))'); %derivative of f
1)- Bisection method:
>> bisct(f,0,1,1e-6,50)
The Interval in which root lie is 6.529179e-001 6.529198e-001
Iteration Required is 20
The approx. error is 9.536743e-007
2)- Regula Falsi:
>> rfalsi(f,0,1.0,1e-6,50)
The Interval in which root lie is 6.529186e-001 6.534421e-001
Iteration Required is 50
The approx. error is 5.234929e-004
3)- Modified Regula Falsi:
>> mrfalsi(f,0,1.0,1e-6,50)
The Interval in which root lie is 6.529059e-001 6.529198e-001
Iteration Required is 25
The approx. error is 2.490377e-005
4)- Newton Raphson Method:
>> newtonrap(f,df,.5,1e-6,50)
The approx. Root is 6.529186e-001
Iteration Required is 4
c)- (x^3)-x-2 assuming interval (-1,2) & Max_Iteration=50
>> f=inline('(x^3)-x-2') %matlab command to define function
>> df=inline('3*(x^2)-1'); %derivative of f
1)- Bisection method:
>> bisct(f,-1,2,1e-6,50)
The Interval in which root lie is 1.521379e+000 1.521382e+000
Iteration Required is 22
The approx. error is 2.861023e-006
2)- Regula Falsi:
>> rfalsi(f,-1,2,1e-6,50)
The Interval in which root lie is 1.521380e+000 2
Iteration Required is 50
The approx. error is 4.786203e-001

10. 3)- Modified Regula Falsi:
>> mrfalsi(f,-1,2,1e-6,50)
The Interval in which root lie is 1.521356e+000 1.521386e+000
Iteration Required is 28
The approx. error is 6.021365e-005
4)- Newton Raphson Method:
>> newtonrap(f,df,1,1e-6,50)
The approx. Root is 1.521380e+000
Iteration Required is 6
Ques3:
For finding the the smallest positive guess for which Newton method diverges,
we can use Newton Raphson method check for various guess as the root start
diverging we can say that that is our required positive guess.
Assuming z-1e-006 % Tolerance parameter
>> f=inline('atan(x)')
f=
Inline function:
f(x) = atan(x)
>> df=inline('1/(1+x^2)')
df =
Inline function:
df(x) = 1/(1+x^2)
>> newtonrap(f,df,1,1e-6,1000)
The approx. Root is 0
Iteration Required is 5
>> newtonrap(f,df,1.1,1e-6,1000)
The approx. Root is -1.803121e-019
Iteration Required is 5

11. >> newtonrap(f,df,1.2,1e-6,1000)
The approx. Root is 0
Iteration Required is 6
>> newtonrap(f,df,1.3,1e-6,1000)
The approx. Root is 0
Iteration Required is 7
>> newtonrap(f,df,1.38,1e-6,1000)
The approx. Root is 0
Iteration Required is 9
>> newtonrap(f,df,1.39,1e-6,1000)
The approx. Root is 0
Iteration Required is 11
>> newtonrap(f,df,1.4,1e-6,1000)
Warning: Divide by zero.
> In newtonrap at 8
Warning: Divide by zero.
> In newtonrap at 8
The approx. Root is NaN
Iteration Required is 1000
From the above data we get
Since at 1.4 there is divide by zero error so beyond 1.4 function (‘atanx’) may
diverge.
Smallest positive guess will be 1.4.