Question

Problem 1. Consider the harmonic function 2 cos 3x 1xs5 Investigate the validity of the numerical differentiation process by considering two different values for the number of points in the domain: (a) 11, and (b) 101 Plot the exact derivative of function y vs approximate (ie numerically determined) derivative of function y for both cases Qi. What is exact value of " for x=1.6? dx dy for x Q2. What is approximate value of 1.6 when number of points is N=11, i.e. for point dx N1-5? dy for x 2? dx Q3. What is exact value of dy for x 2 when number of points is N=101, ie. for point dx Q4. What is approximate value of N1-51? USE MATLAB

Answer 1

%%%bMatlab code %%

clc;
close all;
syms x;
y=2*cos(3*x)/x;
dy=diff(y);
N1=11;
N2=101;
a=1;
b=5;
h1=(b-a)/(N1-1);
h2=(b-a)/(N2-1);
x1=a:h1:b;
x2=a:h2:b;
for n=1:length(x1)-1
D1y(n)=double((subs(y,x1(n+1))-subs(y,x1(n)))/h1);
end
for n=1:length(x2)-1
D2y(n)=double((subs(y,x2(n+1))-subs(y,x2(n)))/h2);
end

Dyy=double(subs(dy,x2));

figure;
plot(x2,Dyy);
hold on
plot(x1(1:end-1),D1y,'-*');
hold on
plot(x2(1:end-1),D2y,'-o');
legend('Exact derivative','Using N=11','Using N=101');
grid on
fprintf(' Exact Derivative at x=1.6 is = %f \n' ,double(subs(dy,1.6)));
fprintf(' Using N=11 Derivative = %f \n ',D1y(3));

fprintf(' Exact Derivative at x=2 is = %f \n',double(subs(dy,2)) );
fprintf(' Using N=101 Derivative is = %f \n ',D2y(26));

OUTPUT:

5 Exact derivative Using N-11 Using N 101 4 2 1 0 1 -2 -3 5 1 2,5 3 3.5 4 1.5 2 4.5 LC LC

Exact Derivative at x=1.6 is = 3.667259
Using N=11 Derivative = 0.396584   
Exact Derivative at x=2 is = 0.358161
Using N=101 Derivative is = 0.179929