Transform Equation X(j+1)=X(j)+c1*(Y(j)-Y(j-1))+c2(Z(j)-Z(j-1)) into Diff Eq

[Aline Rocha]

I have to transform this equation X(j+1)=X(j)+c1*(Y(j)-Y(j-1))+c2(Z(j)-Z(j-1))
in a differential one.

I would like to know if is it possible multiply the equation by (1/z), like this:

(X(j+1)-X(j)) / Z = c1*((Y(j)-Y(j-1))/ Z)+c2((Z(j)-Z(j-1)) / Z)

Then approximate the difference by derivates:

dX(j)=c1*dY(j)+c2*dZ(j) with z->0 ??
dz dz dz

 

[HallsofIvy]

I have to transform this equation X(j+1)=X(j)+c1*(Y(j)-Y(j-1))+c2(Z(j)-Z(j-1))
in a differential one.

I would like to know if is it possible multiply the equation by (1/z), like this:

(X(j+1)-X(j)) / Z = c1*((Y(j)-Y(j-1))/ Z)+c2((Z(j)-Z(j-1)) / Z)

Then approximate the difference by derivates:

dX(j)=c1*dY(j)+c2*dZ(j) with z->0 ??
dz dz dz

What, exactly, is this "z" that you are dividing by? Is it the same as the function Z(j)?
And why would j+1 go to j as z goes to 0?

 

[Aline Rocha]

What, exactly, is this "z" that you are dividing by? Is it the same as the function Z(j)?
And why would j+1 go to j as z goes to 0?

Oh sorry! The "z" that I'm dividing isn't the same Z(j). And j+1=z+delta(z).

 

[Aline Rocha]

Ok, I'm going to try to write the problem again:
I have the following equation:

f(x+h)=f(x)+c1*(g(x)-g(x-h))+c2(h(x)-h(x-h)) (1)

I would like to now if is it possible multiply the equation by (1/h):

((f(x+h)-f(x))/ h)= c1*((g(x)-g(x-h))/z)+c2*((h(x)-h(x-h))/z)

Then approximate the differences by the derivates:

f '(x)=c1*g'(x)+c2*g'(2).

Or should I derivate the equation (1) in both sides, like this:

d[f(x+h)] = d [f(x)+c1*(g(x)-g(x-h))+c2(h(x)-h(x-h))]
dx dx

 

[tanujkush]

Ok, I'm going to try to write the problem again:
I have the following equation:

f(x+h)=f(x)+c1*(g(x)-g(x-h))+c2(h(x)-h(x-h)) (1)

I would like to now if is it possible multiply the equation by (1/h):

((f(x+h)-f(x))/ h)= c1*((g(x)-g(x-h))/z)+c2*((h(x)-h(x-h))/z)

Then approximate the differences by the derivates:

f '(x)=c1*g'(x)+c2*g'(2).

Or should I derivate the equation (1) in both sides, like this:

d[f(x+h)] = d [f(x)+c1*(g(x)-g(x-h))+c2(h(x)-h(x-h))]
dx dx

You could take the first approach provided the derivative of both g(x) and h(x) exists. This is reminiscent of the Taylor expansion series. Only, your equation does not have the higher order terms in it.