Constructing cubic spline interpolation polynomials

[bremenfallturm]

Homework Statement

Given the following points:
(-5,1.561), (-4,2) and (-3, 2.56).
Construct cubic spline interpolation polynomials between them.

Relevant Equations

General equations for a third-degree / cubic polynomial:
$y=ax^3+bx^2+cx+d$
$\frac{dy}{dx}=3ax^2+2bx+c$
$\frac{d^2y}{dx^2}=6ax+2b$

My attempt at a solution: We have three points, hence we will have two polynomials $p_1(x)$ for $x\in [-5,-4]$ and $p_2(x)$ for $x\in [-4,-3]$. Define: $p_1(x)=c_1x^3+c_2x^2+c_3x+c_4$ and $p_2(x)=c_5x^3+c_6x^2+c_7x+c_8$ where the $c_n$ stuff are coefficients for the polynomial.

I came up with the following equations:

1. For $p_1$, it must pass through the first two points:
   1. $1,5610=c_1(-5)^3+c_2(-5)^2+c_3(-5)+c_4$
   2. $2=c_1(-4)^3+c_2(-4)^2+c_3(-4)+c_4$
2. For $p_2$, it must pass through the middle point and the last point:
   1. $2=c_5(-4)^3+c_6(-4)^2+c_7(-4)+c_8$
   2. $2,56=c_1(-3)^3+c_2(-3)^2+c_3(-3)+c_8$
3. 1st and 2nd order derivatives of the polynomials should be equal at the middle point, i.e. $(-4,2)$
   1. $3\cdot (-4^2)c_1+2\dot(-4)c_2+c_3-3\cdot (-4^2)c_5-2\cdot (-4)c_6-c_7=0$
   2. $6c_1\cdot (-4)+2c_2-6c_5\cdot (-4)-2c_6=0$
4. Finally I make the decision that the 2nd derivatives should be equal to 0 at the endpoints.
   1. $6c_1\cdot (-5)+2c_2=0$
   2. $6c_5\cdot (-3)+2_6 =0$

I turned all equations into matrix form and plugged it into an online matrix calculator. The matrix form is this:

The matrix calculator gave me the following solutions:

If I try to plot it, it looks *almost* right but it's certainly not what I expected.
I assume it's a calculation problem, maybe it's even trival but I can not find it.
The plots look like below by the way, and you can find them using this link.

Many thanks for help!

 

[renormalize]

Many thanks for help!

Check your arithmetic.
Here's the 8x8 matrix I get in Mathematica using your method of solution:

Using this to solve your system results in the cubic-spline plot:

This looks correct to me.

 

[pasmith]

Why not start with the acutal expression for a cubic bezier curve with control points $\mathbf{x}_i$, $$\mathbf{x}(t) = (1-t)^3\mathbf{x}_0 + 3(1-t)^2 \mathbf{x}_1 + 3(1-t)t\mathbf{x_2} + t^3\mathbf{x}_3$$ which has the properties that $\mathbf{x}(0) = \mathbf{x}_0$, $\mathbf{x}(1) = \mathbf{x}_3$, $\mathbf{x}'(0) = 3(\mathbf{x}_1 - \mathbf{x}_0)$ and $\mathbf{x}'(1) = 3(\mathbf{x}_3 - \mathbf{x}_2).$ Fitting two of these to the curve in question gives you eight unknowns; fixing the end points immediately determines four of them. Continuity of the derivative gives you a fifth equation. For the other three, we have only three points so we should fit them with a quadratic. The conditions therefore are that the coefficients of $t^3$ each vanish and the coefficients of $t^2$ are equal.

 

[bremenfallturm]

Check your arithmetic.
Here's the 8x8 matrix I get in Mathematica using your method of solution:
View attachment 348354
Using this to solve your system results in the cubic-spline plot:
View attachment 348355
This looks correct to me.

Haha what I silly mistake I made, I forgot a parenthesis when feeding the equations into the matrix calculator. All worked out now, thank you!

 

[bremenfallturm]

Why not start with the acutal expression for a cubic bezier curve with control points $\mathbf{x}_i$, $$\mathbf{x}(t) = (1-t)^3\mathbf{x}_0 + 3(1-t)^2 \mathbf{x}_1 + 3(1-t)t\mathbf{x_2} + t^3\mathbf{x}_3$$ which has the properties that $\mathbf{x}(0) = \mathbf{x}_0$, $\mathbf{x}(1) = \mathbf{x}_3$, $\mathbf{x}'(0) = 3(\mathbf{x}_1 - \mathbf{x}_0)$ and $\mathbf{x}'(1) = 3(\mathbf{x}_3 - \mathbf{x}_2).$ Fitting two of these to the curve in question gives you eight unknowns; fixing the end points immediately determines four of them. Continuity of the derivative gives you a fifth equation. For the other three, we have only three points so we should fit them with a quadratic. The conditions therefore are that the coefficients of $t^3$ each vanish and the coefficients of $t^2$ are equal.

The reason I didn't start with that is because I didn't learn about it yet I am currently looking into more math regarding curve fitting so hopefully I will get there soon:)