Norms¶

This notebook accompanies Chapter 2, section on norms.

A single number has to stand for the size of an error function, and the plain integral fails in that role, because positive and negative gaps cancel. The mode $0.3\sin(16\pi x)$ integrates to zero exactly while reaching $0.3$ at every crest. The sup norm ignores cancellation by taking the largest magnitude, and the $L^2$ norm defeats it by squaring the gaps before integrating. The first lab measures the distance between the displacement and a copy of itself slid sideways, plus a small sixth mode. Yellow bars mark the pointwise gaps; squaring, averaging, and taking the root turns the bars into one number, and a finite set of measurement points approaches the integral as the set grows. A small slide produces a large distance, the same fact the chapter measures with the shifted spike.

The measure that sees slopes¶

The second lab plots a displacement $A\sin(2\pi n x)$ beside its derivative. Raising the frequency $n$ barely changes the sup and $L^2$ norms and multiplies the $H^1$ seminorm by $n$, because differentiation multiplies each oscillation by its frequency.

The book's numbers¶

The cell below reproduces the measured values of the chapter, the three norms of the $m = 5$ reconstruction error and the spike and wiggle on which the norms disagree.

In [1]:

import numpy as np

base = (0.8, 0.4, 0.2)
x = np.linspace(0.0, 1.0, 8001)
u = base[0]*np.sin(np.pi*x) + base[1]*np.sin(2*np.pi*x) + base[2]*np.sin(3*np.pi*x)
du = np.pi*(base[0]*np.cos(np.pi*x) + 2*base[1]*np.cos(2*np.pi*x) + 3*base[2]*np.cos(3*np.pi*x))

def l2(f):
    return np.sqrt(np.trapezoid(f**2, x))

m = 5
xi = np.linspace(0, 1, m)
ui = base[0]*np.sin(np.pi*xi) + base[1]*np.sin(2*np.pi*xi) + base[2]*np.sin(3*np.pi*xi)
v = np.interp(x, xi, ui)
slopes = np.diff(ui) / np.diff(xi)
dv = slopes[np.clip((x*(m-1)).astype(int), 0, m-2)]
e, de = u - v, du - dv
print(f"m = 5 error:   sup = {np.max(np.abs(e)):.3f},  L2 = {l2(e):.4f},  H1 = {l2(de):.3f}")
print(f"relative:      sup = {np.max(np.abs(e))/np.max(np.abs(u)):.1%}, "
      f"L2 = {l2(e)/l2(u):.1%}, H1 = {l2(de)/l2(du):.1%}")

for w in (0.04, 0.01, 0.0025):
    s_ = np.maximum(0.0, 1.0 - np.abs(x - 0.5)/(w/2))
    print(f"spike base {w}: sup = {np.max(s_):.0f},  L2 = {l2(s_):.4f},  "
          f"H1 = {2/np.sqrt(w):.0f},  energy = {2/w:.0f}")

spike = np.maximum(0.0, 1.0 - np.abs(x - 0.5)/0.02)
shifted = np.maximum(0.0, 1.0 - np.abs(x - 0.56)/0.02)
print(f"spike shifted by 0.06: L2 distance = {l2(spike - shifted):.4f}  (sqrt(2) x 0.1155)")

for n in (10, 20, 40):
    wig = 0.05*np.sin(n*np.pi*x)
    dwig = 0.05*n*np.pi*np.cos(n*np.pi*x)
    print(f"wiggle n = {n:2d}:  sup = {np.max(np.abs(wig)):.3f},  L2 = {l2(wig):.4f},  H1 = {l2(dwig):.3f}")

import numpy as np base = (0.8, 0.4, 0.2) x = np.linspace(0.0, 1.0, 8001) u = base[0]*np.sin(np.pi*x) + base[1]*np.sin(2*np.pi*x) + base[2]*np.sin(3*np.pi*x) du = np.pi*(base[0]*np.cos(np.pi*x) + 2*base[1]*np.cos(2*np.pi*x) + 3*base[2]*np.cos(3*np.pi*x)) def l2(f): return np.sqrt(np.trapezoid(f**2, x)) m = 5 xi = np.linspace(0, 1, m) ui = base[0]*np.sin(np.pi*xi) + base[1]*np.sin(2*np.pi*xi) + base[2]*np.sin(3*np.pi*xi) v = np.interp(x, xi, ui) slopes = np.diff(ui) / np.diff(xi) dv = slopes[np.clip((x*(m-1)).astype(int), 0, m-2)] e, de = u - v, du - dv print(f"m = 5 error: sup = {np.max(np.abs(e)):.3f}, L2 = {l2(e):.4f}, H1 = {l2(de):.3f}") print(f"relative: sup = {np.max(np.abs(e))/np.max(np.abs(u)):.1%}, " f"L2 = {l2(e)/l2(u):.1%}, H1 = {l2(de)/l2(du):.1%}") for w in (0.04, 0.01, 0.0025): s_ = np.maximum(0.0, 1.0 - np.abs(x - 0.5)/(w/2)) print(f"spike base {w}: sup = {np.max(s_):.0f}, L2 = {l2(s_):.4f}, " f"H1 = {2/np.sqrt(w):.0f}, energy = {2/w:.0f}") spike = np.maximum(0.0, 1.0 - np.abs(x - 0.5)/0.02) shifted = np.maximum(0.0, 1.0 - np.abs(x - 0.56)/0.02) print(f"spike shifted by 0.06: L2 distance = {l2(spike - shifted):.4f} (sqrt(2) x 0.1155)") for n in (10, 20, 40): wig = 0.05*np.sin(n*np.pi*x) dwig = 0.05*n*np.pi*np.cos(n*np.pi*x) print(f"wiggle n = {n:2d}: sup = {np.max(np.abs(wig)):.3f}, L2 = {l2(wig):.4f}, H1 = {l2(dwig):.3f}")

m = 5 error:   sup = 0.223,  L2 = 0.0930,  H1 = 1.200
relative:      sup = 20.1%, L2 = 14.4%, H1 = 42.2%
spike base 0.04: sup = 1,  L2 = 0.1155,  H1 = 10,  energy = 50
spike base 0.01: sup = 1,  L2 = 0.0577,  H1 = 20,  energy = 200
spike base 0.0025: sup = 1,  L2 = 0.0289,  H1 = 40,  energy = 800
spike shifted by 0.06: L2 distance = 0.1633  (sqrt(2) x 0.1155)
wiggle n = 10:  sup = 0.050,  L2 = 0.0354,  H1 = 1.111
wiggle n = 20:  sup = 0.050,  L2 = 0.0354,  H1 = 2.221
wiggle n = 40:  sup = 0.050,  L2 = 0.0354,  H1 = 4.443

What comes next¶

Each norm evaluates a description after it is chosen. The next notebook asks for the best description within a family, and the answer is a projection.