Cubic Spline what is the logic in constructing it?

Cubic Spline....what is the logic in constructing it?

Hey guys,
I am not trying to directly code a cubic spline computation, but I am writing a sub routine in VBA that takes input data, and outputs it to a text file in Maple syntax.

Anywho, my cubic spline actually produces LINEAR splines!!! I can't figure out what is wrong with my logic?!

Here is my logic:
Suppose I have n data points

Then I will have n-1 splines, defined for interval (x_i, x_(i+1)) where i = 1 to (n-1)
I am modeling the spline according to the equation ax^3 + bx^2 + cx + d,
and therefore IN TOTAL I will have 4*(n-1) different coefficients. The reasoning behind that is simple. The first spline has 4 coefficients, the second has 4, etc, and there are (n-1) splines in total.

I am also setting the first and second derivatives equal to zero (and hence equal to each other) at (x_i) and (x_(i+1))

So, here is what I am constraining each spline to:
a(x_i)^3 + b(x_i)^2 + c(x_i) + d = y(x_i)
a(x_(i+1))^3 + b(x_(i+1))^2 + c(x_(i+1)) + d = y(x_(i+1))
3a(x_i)^2 + 2b(x_i) + c = 3a(x_(i+1))^2 + 2b(x_(i+1)) + c
6a(x_i) + 2b = 6a(x_(i+1)) + 2b

So for example, suppose My points are: (-1,.5), (0,0) , (3,3)
I get the following function:

S(x) = (-.5x, -1<x<0
= ( x, 0<x<3

Where is my logic flawed? I would really appreciate any help.
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Re: Cubic Spline....what is the logic in constructing it?

EDIT: In my code I was forcing the derivatives to be equal to zero at the end points of S(x), i.e x_0 and x_n. I loosened that constraint up by just making them equal each other, and now I produce pretty nice solutions.

EDIT 2: Consider this solved. I set the 2nd derivatives of the end points to 0, and removed the constraint that the first derivatives be 0 or equal to each other at the end points. Not sure why I thought that was right.

I fixed my logic up...and now I'm getting empty solutions.
Here are my new constraints, which I figured would be correct

At the end points of S(x), my piecewise cubic spline function, S'(x) and S''(x) = 0.
I.e Once again, for n data points:
First Derivative at x_0: 3a(x_0)^2+2b(x_0)+c(x_0)+d = 0
First Derivative at x_n: 3a(x_n)^2+2b(x_n)+c(x_n)+d = 0
Second Derivative at x_0: 6a(x_0) + 2b = 0
Second Derivative at x_n: 6e(x_n) + 2f = 0

At each node of S(x), i.e, at the endpoint of each intersecting spline, the derivative and second derivative of each spline is equal.
The following shows what I mean:
If we have any node, (x_i,y_i),where 0<i<n(strictly) and two splines, az^3+bz^2+cz+d, ez^3+fz^2+gz+h, then we have:
First Derivative at x-value x_i: 3a(x_i)^2+2b(x_i)+c(x_i)+d = 3e(x_i)^2+2f(x_i)+g(x_i)+h
Second Derivative at x_i: 6a(x_i) + 2b = 6e(x_i) + 2f

The final constraint is that the Spline be continuous, and therefore for all x_i, 0<=i<=n,
S(x_i) = a(x_i)^3 + b(x_i)^2 + c(x_i) + d and S(x_(i+1)) = a(x_(i+1))^3 + b(x_(i+1))^2 + c(x_(i+1)) + d must hold to ensure both points in each sub interval are contained in the spline,
additionally S(x_(i+1)) = e(x_(i+1))^3 + f(x_(i+1))^2 + g(x_(i+1)) + h must hold to ensure continuity from spline to spline.
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