6.5.2b Cubic splines labs

Source: Lab ideas from Computational Physics (Landau, Páez, Bordeianu), §6.5.2–6.5.3. Rewritten as original drills. Continues 6.5.2 Cubic splines.

Build a natural cubic spline from scratch — no calling a library spline routine. Stubs fail until filled in.

Formula refresh

Warm up: pick the matching equation among lookalikes (same walk as the theory lesson).

Labs

Activity: Estimate g'' by forward difference

Implement the endpoint second-derivative estimate

$$g''(x_2)\simeq\frac{[g(x_3)-g(x_2)]/[x_3-x_2]-[g(x_2)-g(x_1)]/[x_2-x_1]}{[x_3-x_1]/2}.$$

Input: x1 g1 x2 g2 x3 g3. Print with 6 decimals.

Write-ups

0/4 solved

6.5.2b.W1

Using “value”, “first”, “second”, and “derivative”, explain:

Which quantities does a cubic spline force to be continuous across interval boundaries?

Minimum 3 words. Your answer must include: “value”, “first”, “second”, “derivative”.

0 words/ 3 min

Meet the writing requirements, then you can open the model answer.

Explorers

# spline_ref.py — natural cubic spline: solve, evaluate, integrate (original sketch)
def natural_spline_M(xs, ys):
    n = len(xs)
    h = [xs[i + 1] - xs[i] for i in range(n - 1)]
    A = [[0.0] * n for _ in range(n)]
    b = [0.0] * n
    A[0][0] = 1.0
    A[n - 1][n - 1] = 1.0
    for i in range(1, n - 1):
        A[i][i - 1] = h[i - 1]
        A[i][i] = 2.0 * (h[i - 1] + h[i])
        A[i][i + 1] = h[i]
        b[i] = 6.0 * ((ys[i + 1] - ys[i]) / h[i] - (ys[i] - ys[i - 1]) / h[i - 1])
    for i in range(n):
        piv = A[i][i]
        for j in range(i + 1, n):
            f = A[j][i] / piv
            for k in range(n):
                A[j][k] -= f * A[i][k]
            b[j] -= f * b[i]
    M = [0.0] * n
    for i in range(n - 1, -1, -1):
        s = b[i]
        for k in range(i + 1, n):
            s -= A[i][k] * M[k]
        M[i] = s / A[i][i]
    return M

Next: 6.6 Exponential decay and least-squares fitting.