Week 3 - Taylor table method and Matlab code

function y=f(x);

y=exp(x)*cos(x);

end

% Taylors Table Derivative

% Analytical Way

clc
clear all
close all
x=pi/3;
analytical=-2*exp(x)*sin(x);
n=10;
dx=linspace(0.01,1,10);
% Using Forward Difference Scheme
for i=1:length(dx )
forward(i)=((35*f(x))-(104*f(x+dx(i)))+(114*f(x+(2*dx(i))))-(56*f(x+(3*dx((i)))))+(11*f(x+(4*dx(i)))))/(12*(dx(i))^2);
backward(i)=((35*f(x))-(104*f(x-dx(i)))+(114*f(x-(2*dx(i))))-(56*f(x-(3*dx((i)))))+(11*f(x-(4*dx(i)))))/(12*(dx(i))^2);
cds(i)=((-30*f(x))+(16*f(x-dx(i)))+(16*f(x+(dx(i))))-(f(x+(2*dx((i)))))-(f(x-(2*dx(i)))))/(12*(dx(i))^2);
error_forward(i)=abs(forward(i)-analytical);
error_backward(i)=abs(backward(i)-analytical);
error_cds(i)=abs(cds(i)-analytical);
end
loglog(dx,error_forward,\'-kd\',\'markerfacecolor\',\'r\',\'markersize\',7)
grid on
hold on
loglog(dx,error_backward,\'-ko\',\'markerfacecolor\',\'b\',\'markersize\',7)
hold on
loglog(dx,error_cds,\'-ks\',\'markerfacecolor\',\'g\',\'markersize\',7)
fprintf(\' \\n Forward scheme= %g\',forward(n));
fprintf(\' \\n backward scheme = %g\',backward(n));
fprintf(\' \\n cds scheme = %g\',cds(n));
fprintf(\' \\n Actual derivative = %g\',analytical);
fprintf(\' \\n Forward Error = %g\',error_forward(n));
fprintf(\' \\n Backward Error = %g\',error_backward(n));
fprintf(\' \\n CDS Error= %g\',error_cds(n));
xlabel(\'step size\')
ylabel(\'error\')
legend(\'SKEWED RIGHT\',\'SKEWED LEFT\',\'CDS\',\'location\',\'northwest\')
title(\'step size vs relative error\');

fname={\'forward\',\'backward\',\'cds\'};
figure(2)
x=[1:3];
y=[error_forward(n) error_backward(n) error_cds(n)];
bar(x,y)
set(gca,\'xTick\',1:length(fname),\'xTickLabel\',fname);
xlabel(\'order\')
ylabel(\'error\')
legend(\'error prediction using 1 2 4 order approximation\')

Results:

Forward scheme= 47.5215
backward scheme = -2.59705
cds scheme = -5.17395
Actual derivative = -4.93575
Forward Error = 52.4573
Backward Error = 2.3387
CDS Error= 0.238207>>

a. Provide a plot that compares the absolute error between the above mentioned schemes

Answer:Explanation:1 We cab see that forward scheme results least accurate as error is higher. We can see from bar graph.

b. How do these errors compare with FDS and BDS approximations

Output: Error is least with cds scheme.

c. In your own words, describe why a skewed scheme is useful? What can a skewed scheme do that a CD scheme cannot do?

When we have values for end points then skewed schemes are useful. For such case we cant use CDS scheme.