A Have I solved this iterative equation correctly?

[Tygra]

TL;DR Summary

Equation for computing rotations for a rigid frame

Hi all,

I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of:

$$ x_i = \frac {Qh_i + Qh_{i+1}}{4K} + \frac {C_{i+1}}{K}x_{i+1} + \frac {C_{i-1}}{K}x_{i-1} $$

The book says to use initial values x by using this formula:

$$x_i = \frac {Qh_i + Qh_{i+1}}{24G_i}$$

So, you can compute the rotations for each level and then using the iterative formula to improve the values of the rotations.

Now, I have used MATALB to do this.

x1(1) = (Q*h/4)/(24*G1)
x2(1) = (2*Q*h/4 + Q*h/4)/(24*G1)
x3(1) = (3*Q*h/4 + 2*Q*h/4)/(24*G1)
x4(1) = (4*Q*h/4 + 3*Q*h/4)/(24*G1)
x5(1) = (5*Q*h/4 + 4*Q*h/4)/(24*G1)
x6(1) = (6*Q*h/4 + 5*Q*h/4)/(24*G1)
x7(1) = (7*Q*h/4 + 6*Q*h/4)/(24*G1 + 2*C)
x8(1) = 0

for i = 1:10
     x1(i+1) = Q*h/(4*K) + C/K*x2(i);
     x2(i+1) = (Q*h + 2*Q*h)/(4*K) + C/K*x1(i) + C/K*x3(i);
     x3(i+1) = (2*Q*h + 3*Q*h)/(4*K) + C/K*x2(i) + C/K*x4(i);
     x4(i+1) = (3*Q*h + 4*Q*h)/(4*K) + C/K*x3(i) + C/K*x5(i);
     x5(i+1) = (4*Q*h + 5*Q*h)/(4*K) + C/K*x4(i) + C/K*x6(i);
     x6(i+1) = (5*Q*h + 6*Q*h)/(4*K) + C/K*x5(i) + C/K*x7(i);
     x7(i+1) = (6*Q*h + 7*Q*h)/(4*K) + C/K*x6(i);
     x8(i+1) = 0;
end

My questions is: have I solved this iterative equation correctly?

Note: C and G are the flexural stiffnesses of the columns and girders respectively; that is

$$ C = \frac {I}{h} $$
$$ G = \frac {I}{L} $$

$$K = (6G_i + C_i + C_{i+1})$$

EDIT: Q is the horizontal wind shear at each storey.

To give you a visual insight, here is a rigid frame illustrating the nodal loads and rotations at the joints:

 

[Tygra]

Sorry, I'm having real trouble with the latex.

EDIT: Latex Sorted

 

[Tygra]

Is there not anyone who can help? Should I have posted this in a different section?