Recipe

Construct a natural cubic spline

Use when: you have a table of (x, y) data points and want one smooth, piecewise-cubic curve through all of them, with no extra information available about the slope at either end

Goal: every coefficient a_j,b_j,c_j,d_j of every piece, assembled into the full piecewise definition of S(x)

Taught in Natural boundary conditions

1. 1

   Arrange the nodes x_0<x_1<\cdots<x_n with their values y_i. Compute each interval width h_j=x_{j+1}-x_j.

   Example

   For (1,3), (2,1), (4,3), (5,10): h_0=2-1=1, h_1=4-2=2, h_2=5-4=1.

   Why: every later formula uses these widths directly. Getting the nodes ordered correctly first is what makes the rest of the recipe mechanical, not error-prone.

2. 2

   For j=1,\ldots,n-1, compute \alpha_j=\dfrac3{h_j}(y_{j+1}-y_j)-\dfrac3{h_{j-1}}(y_j-y_{j-1}) -- how much the two secant slopes meeting at node j disagree.

   Example

   \alpha_1=\dfrac32(3-1)-\dfrac31(1-3)=3-(-6)=9; \alpha_2=\dfrac31(10-3)-\dfrac32(3-1)=21-3=18.

   Why: this packages exactly the mismatch the spline's curvature has to absorb at each interior node. The bigger the disagreement between nearby slopes, the more bend the curve needs there.

3. 3

   Write h_{j-1}c_{j-1}+2(h_{j-1}+h_j)c_j+h_jc_{j+1}=\alpha_j for each interior node. Then substitute the natural condition's c_0=0 and c_n=0 directly into the first and last of these equations.

   Example

   j=1: c_0+6c_1+2c_2=9 \;\to\; 6c_1+2c_2=9 (since c_0=0). j=2: 2c_1+6c_2+c_3=18 \;\to\; 2c_1+6c_2=18 (since c_3=0).

   Why: this is Section 3's equation (5), closed by the specific information (or lack of it) this section supplies. Every other boundary choice would close the same interior equations differently.

4. 4

   Solve the resulting system for c_1,\ldots,c_{n-1}. For a small system, use direct elimination; for a larger one, use the general Thomas algorithm from Section 7. c_0 and c_n are already 0.

   Example

   From 6c_1+2c_2=9 and 2c_1+6c_2=18: solving gives c_1=\dfrac9{16}, c_2=\dfrac{45}{16} (and c_0=c_3=0).

   ⚡ Shortcut

   For only two or three interior unknowns, plain substitution is faster: solve one equation for one variable, then substitute into the other. That's quicker than setting up the full Thomas-algorithm bookkeeping Section 7 introduces — save that machinery for a genuinely long chain of nodes.

   Why: c_j is exactly S''(x_j)/2 — the curvature the spline must carry at each node. It keeps both neighboring pieces' slope and curvature matching there.

5. 5

   Using every now-known c_j, compute b_j=\dfrac{y_{j+1}-y_j}{h_j}-\dfrac{h_j}3(2c_j+c_{j+1}) and d_j=\dfrac{c_{j+1}-c_j}{3h_j} for each j.

   Example

   b_0=\dfrac{1-3}1-\dfrac13\left(0+\tfrac9{16}\right)=-2-\tfrac3{16}=-\tfrac{35}{16}; similarly b_1=-\tfrac{13}8, b_2=\tfrac{41}8. d_0=\dfrac{9/16-0}{3}=\tfrac3{16}; similarly d_1=\tfrac38, d_2=-\tfrac{15}{16}.

   Why: these formulas came directly from Section 3's equations (3) and (4). Once every c_j is known, no further system-solving is needed to get b_j and d_j.

6. 6

   Write S_j(x)=a_j+b_j(x-x_j)+c_j(x-x_j)^2+d_j(x-x_j)^3 on [x_j,x_{j+1}] for each j, using a_j=y_j.

   Example

   S(x)=\begin{cases} 3-\tfrac{35}{16}(x-1)+\tfrac3{16}(x-1)^3, & 1\le x\le2\\[2pt] 1-\tfrac{13}8(x-2)+\tfrac9{16}(x-2)^2+\tfrac38(x-2)^3, & 2\le x\le4\\[2pt] 3+\tfrac{41}8(x-4)+\tfrac{45}{16}(x-4)^2-\tfrac{15}{16}(x-4)^3, & 4\le x\le5 \end{cases}

   Why: the spline is genuinely three separate formulas, one per subinterval. Evaluating S anywhere means finding which piece x falls in first, then using only that piece's own coefficients.