That is sufficient information, thanks. So let me see if I can help. (Best of course, is to not write your own spline code, but to use code written by someone who does that for a living. In fact, you might even find some spline code on the File Exchange with my name on it. :) At the same time, it is ALWAYS good to learn how to write something like a spline tool, at least if you are interested.) Ok,...
Define two vectors, (x,y) pairs, so 4 points.
x = [x1 x2 x3 x4], y = [y1 y2 y3 y4]
One thing I recommend STRONGLY. Learn about the use of semi-colons at the end of a line. It suppresses junk output to the command line.
Look at each of those equations in the matrix equation given to you by your instructor. I'll explain what they mean, in context of the spline as you are building it.
So, first that matrix equation relates to a 2 segment spline. So a pair of cubic polynomials, passing through 3 points, connected at the intermediate point.
Given two cubic segments, we would have 8 coefficients in the representation as it is defined, thus 4 coefficients for each cubic segment.
When we multiply that first row of A times the coefficient vector, we will get
x(1)^3*a_13 + x(1)^2*a_12 + x(1)*a_11 + 1*a_10 = y(1)
So just expand the dot product for that first row. What does that equation mean? It states that the first cubic segment, evaluates at the point x(1), yields y(1). Essentially, it requires that the spline interpolate the first data point. So it says that f(x(1))==y(1), for the FIRST cubic segment.
Second row?
x(2)^3*a_13 + x(2)^2*a_12 + x(2)*a_11 + 1*a_10 = y(1)
These are the coefficients of the first cubic segment, used to evaluate the function at the second data point. So this row enforces f(x(2))==y(2).
Rows 3 and 4 are similar, but you can see they apply to the SECOND cubic segment.
x(2)^3*a_23 + x(2)^2*a_22 + x(2)*a_21 + 1*a_20 = y(2)
x(3)^3*a_23 + x(3)^2*a_22 + x(3)*a_21 + 1*a_20 = y(3)
Again, these are interpolation equations, in the sense that they force the second segment to pass through the points (x(2),y(2)) and (x(3),y(3)).
Note that that constraint for point number 2 is applied to both segments. You can think of that as continuity. The two segments are tied together, so the function is continuous at that point.
Had you a 4th point in the curve, then you would have 3 cubic segments, and therefore two more rows similar to the first set of rows in A. Effectively, all of those rows were talking about the function values of the spline at the data points.
What do lines 5,6,7, and 8 do? For this, we need to look at the derivative of those cubic segments. What is the first derivative of the first segment (I'll call it f_1) evaluated at some arbitrary point u? It is just a cubic polynomial
f_1(u) = u^3*a_23 + u^2*a_22 + u*a_21 + a_20
so the derivative is just
f_1'(u) = 3*u^2*a_13 + 2*u*a_12 + a_11
Similarly, the first derivative of f_2 at u is just:
f_2'(u) = 3*u^2*a_23 + 2*u*a_22 + a_21
How about the second derivatives of f_1 and f_2?
f_1''(u) = 6*u*a_13 + 2*a_12
f_2''(u) = 6*u*a_23 + 2*a_22
Ok, so in light of all that knowledge, what does row 5 tell us? Look at it in light of what I've just said. Row 5 is just:
Just as rows 2 and 3 implied continuity of the spline at x(2) as well as forcing the curve to pass through the point (x(2),y(2)), row 5 implies that the spline is differentiable across that point, with a continuous first derivative there. Effectively row 5 says, "While I don't know what that derivative is at x(2), I do know the derivative is continuous across that knot."
Row 6 is similar, but it applies to the second derivative. (Ok, your instructor divided by 2 in that line. Irrelevant.) Row 6 tells us that
f_1''(x2) - f_2''(x2) = 0
So it enforces second derivative continuity at x(2).
Finally, look at rows 7 and 8. EXPAND THOSE LINES AS DOT PRODUCTS, as we did above. Row 7 enforces the value of the first derivative of f_1, thus f_1', evaluated at the point x(1). Likewise, row 8 enforces the value of f_2', evaluated at x(3).
So rows 7 and 8 are the boundary conditions, those zero-clamp conditions that were given to you.
I honestly don't want to write the code for you, because I think you will gain more by writing it yourself. I'd rather offer suggestions as to what you need to have though.
In your code, for a 4 knot spline written out in the above form...
You will have 12 unknowns, 4 unknown coefficients for each of the 3 cubic segments.
You need 6 rows that correspond to the values of the function at each point. Thus...
f_1(x(1)) == y(1)
f_1(x(2)) == y(2)
f_2(x(2)) == y(2)
f_2(x(3)) == y(3)
f_3(x(3)) == y(3)
f_3(x(4)) == y(4)
Next, you need to enforce first AND second derivative continuity at the interior knot points, x(2) and x(3). So we would have 4 more rows in A, corresponding to:
f_1'(x2) - f_2'(x2) = 0
f_1''(x2) - f_2''(x2) = 0
f_2'(x3) - f_3'(x3) = 0
f_2''(x3) - f_3''(x3) = 0
Finally, the zero-clamp first derivative end-conditions.
In total, your matrix should have 12 rows, and 12 columns, since you have 12 coefficients to solve for.
The resulting matrix A will be theoretically non-singular as long as x(1), x(2), x(3), and x(4) are distinct.
So, having said all of this, this formulation of a cubic spline, is NOT what I would like to see done. It may suffer from numerical problems because those polynomials may get nasty. But that may well be the subject of your next assignment, and since I've now written a LOT, time to stop writing. :) But feel free to ask a question if I've not been clear in what I said above. Or, post what you write, and I can look it over to see if you got it right. I'll always try to answer as long as I see a spark of interest in a poster, as long as I see you making an effort.
