Numerical analysis Experiment with the Midpoint Rule for Int

Numerical analysis

Experiment with the Midpoint Rule for Integration, Euler\'s Method for First Order Initial Value Problems, Newton\'s Method for finding zeros of functions, and the Sparse First Difference Matrix for computing first derivative approximations

please help

Solution

f=@(x) x;
g=@(x) x.^3;

disp(\'***** question 1, part 1*****\')
n=[10 20 40 80 160];
error=zeros(1,5);
for i=1:length(n)
h=(1-0)/n(i);
x=0:h:1;
y=f(x);
result=sum(y)*h;
error(i)=abs(result-0.5);
fprintf(\'\ integration with n=%d steps is %f\ \',n(i),result);
end

fprintf(\'\ exact result is 0.5\ \');

disp(\' n error \')
table(:,1)=n\';
table(:,2)=error\';
disp(table)

disp(\'***** question 1, part 2*****\')
n=[10 20 40 80 160];
error=zeros(1,5);
for i=1:length(n)
h=(1-0)/n(i);
x=0:h:1;
y=g(x);
result=sum(y)*h;
error(i)=abs(result-0.5);
fprintf(\'\ integration with n=%d steps is %f\ \',n(i),result);
end

fprintf(\'\ exact result is 0.5\ \');

disp(\' n error \')
table(:,1)=n\';
table(:,2)=error\';
disp(table)

disp(\'****question 2 part 1*****\')
n=[10 20 40 80 160];
for i=1:length(n)
h=1/n(i);
x=0:h:1;
y0=1;
y1=y0+f(x(1))*h;
for j=2:n(i)
y0=y1;
y1=y0+f(x(j))*h;
end
fprintf(\'\ soultion for n=%d is %f.\ \',n(i),y1);
end

disp(\'****question 2 part 2*****\')

for i=1:length(n)
h=1/n(i);
x=0:h:1;
y0=1;
y1=y0+g(x(1))*h;
for j=2:n(i)
y0=y1;
y1=y0+g(x(j))*h;
end
fprintf(\'\ soultion for n=%d is %f.\ \',n(i),y1);
end