4-point first derivative

 I am given data t=[0 1 2 3 4 5 6 7 8 9 10] and 
y(t)=[1 2.7 5.8 6.6 7.5 9.9 10.2 11.2 12.6 13.6 15.8]
 and have to evaluate the derivative of y at each given t value
 using the following finite difference schemes 
at 4-point first derivative

My code at finite difference is

t= 0: 1: 10;
y= [1 2.7 5.8 6.6 7.5 9.9 10.2 11.2 12.6 13.6 15.8];
n=length(y);
dfdx=zeros(n,1);
dfdx(t)=(y(2)-y(1))/(t(2)-t(1));
for i=2:n-1
    dfdx(1)=(y(i+1)-y(i-1))/(t(i+1)-t(i-1));
end
dfdx(n)=(y(n)-y(n-1))/(t(n)-t(n-1));

but I have interesting of method of 4-point first derivative for more accuracy thanks in avance!

[Aquatris]

Please format the question properly and include the "following scheme" for us to be able to help.

[alburary daniel]

I did

Answer 1

[Aquatris]

I found a source where the equations for the differentiation are shown, with some typos. Here is an example code, where d4 is 4 point d5 is 5 point differentiation.

dt = 1e-3;
t = 0:0.001:20;
x = sin(0.4*t)+exp(-0.9*t);         % sample signal
xd = 0.4*cos(0.4*t)-0.9*exp(-0.9*t);% sample signals exact derivative
n=length(x);
d4=zeros(size(x));
d5=zeros(size(x));

for j = 3:n-2;
  d4(j) = -1/6/dt*(-2*x(j+1)-3*x(j)+6*x(j-1)-x(j-2));
  d5(j)=   1/12/dt*(x(j-2) - 8.*x(j-1) + 8.*x(j+1) - x(j+2));
end
d4(1:2)=d4(3);
d4(n-1:n)=d4(n-2);
d5(1:2)=d5(3);
d5(n-1:n)=d5(n-2);
plot(t,xd,t,d4,'r--',t,d5,'m-.') % comparison plot

[alburary daniel]

I feeling confused because I do not how include the time data in the above representation for 4 and 5 points, any suggestion?

[Aquatris]

dt is your timestep information. Which part are you confused with?

Edit: My bad. There is a typo in the code. It should be;

dt = 1e-3;
t = 0:dt:20;

to make it clear dt is the time step. Also I thought it was clear t is the time.

[alburary daniel]

if true
  % code
end
t= 0:dt:10 
y= [1 2.7 5.8 6.6 7.5 9.9 10.2 11.2 12.6 13.6 15.8]; 
n=length(y);
dfdx=zeros(n,1);
dfdx=(y(2)-y(1))/(t(2)-t(1));
for i=2:n-1
  dfdx(1)=(y(i+1)-y(i-1))/(t(i+1)-t(i-1));
end
dfdx(n)=(y(n)-y(n-1))/(t(n)-t(n-1));
m=length(y);
d4=zeros(size(y));
d5=zeros(size(y));

for j = 3:m-2 d4(j) = -1/6/dt*(-2*y(j+1)-3*y(j)+6*y(j-1)-y(j-2)); d5(j)= 1/12/dt*(y(j-2) - 8.*y(j-1) + 8.*y(j+1) - y(j+2)); end

d4(1:2)=d4(3); d4(m-1:m)=d4(m-2); d5(1:2)=d5(3); d5(m-1:m)=d5(m-2); plot(t,dfdx,'blue',t,d4,'r--',t,d5,'m-.')

solutions are not equal

[Aquatris]

I recommend you look into limitations of numeric differentiation to understand the reason. Feels like you are not that familiar.

[Mohammad Ezzad Hamdan]

What do you mean by the terms below;

d5(1:2)=d5(3);

d5(n-1:n)=d5(n-2);

[Aquatris]

It is not possible to calculate the first 2 or last 2 elements for the d5, I just equate them to some value. For the last 2 elements I hold the last calculated value of d5 and for the first 2 elements I hold the first calculated value. Depending on the application this might be acceptable.