Matlab problem: How to calculate a rough, approximate, derivative vector by using the following formula?
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Jabir Al Fatah
on 27 May 2014
Edited: George Papazafeiropoulos
on 31 May 2014
Here is the complete task which I need to solve:
We know that the derivative is the coefficient k of a tangent line with mathematical expression, y=kx + m, that goes through a certain point, P, on a function curve. Lets start with a simple function, namely, y=x^2.
(a) Calculate and Plot this function for -10 < x < 10 with small enough step length to get a smooth curve.
(b) Now, In (a) you calculated two equally long vectors, one for the chosen x values and one with the calculated y values. From these two vectors, we can calculate a rough, approximate, derivative vector that we can call Yprimenum: Use a for-loop to calculate each element of Yprimenum according to the formula:
Yprimenum(i) = (y(i+1) – y(i)) / ∆x Where ∆x is the x step length, or equivallently x(i) – x(i-1)
NOTE: (a) is solved and for the (b) I wrote the following code but not working.
if true
x=-10:10;
for i=1: length(x);
for y=x.^2
Yprimenum=(y(i+1)-y(i))./(x(i)-x(i-1));
end;
end;
display Yprimenum;
end
3 Comments
John D'Errico
on 27 May 2014
Why do you think it necessary to add the "if true" conditional around your code? Silly.
Accepted Answer
George Papazafeiropoulos
on 27 May 2014
Edited: George Papazafeiropoulos
on 27 May 2014
x=-10:10;
y=x.^2;
Yprimenum=diff(y)./diff(x)
5 Comments
George Papazafeiropoulos
on 31 May 2014
Edited: George Papazafeiropoulos
on 31 May 2014
Try this:
% 1st way: for loop
Yprimenum1=zeros(1,2*10);
x=-10:10;
y=x.^2;
for I=2:length(x)
Yprimenum1(I-1)=(y(I)-y(I-1))/(x(I)-x(I-1));
end
% 2nd way: diff
Yprimenum2=diff(y)./diff(x);
% test the results
all(Yprimenum1==Yprimenum2)
Be careful that the length of the approximate derivative is equal to the length of the initial vectors minus 1
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