[NumCS] Finish quadrature, a few fixes

semester3/numcs/format.py

@@ -1,33 +1,4 @@
-import numpy as np
-import scipy as sp
-
-
-def fastbroyd(x0, F, J, tol=1e-12, maxit=20):
-    x = x0.copy()  # make sure we do not change the iput
-    lup = sp.linalg.lu_factor(J)  # LU decomposition of J
-    s = sp.linalg.lu_solve(lup, F(x))  # start with a Newton corection
-    sn = np.dot(s, s)  # squared norm of the correction
-    x -= s
-    f = F(x)  # start with a full Newton step
-    dx = np.zeros((maxit, len(x)))  # containers for storing corrections s and their sn:
-    dxn = np.zeros(maxit)
-    k = 0
-    dx[k] = s
-    dxn[k] = sn
-    k += 1  # the number of the Broyden iteration
-
-    # Broyden iteration
-    while sn > tol and k < maxit:
-        w = sp.linalg.lu_solve(lup, f)  # f = F (actual Broyden iteration x)
-        # Using the Sherman-Morrison-Woodbury formel
-        for r in range(1, k):
-            w += dx[r] * (np.dot(dx[r - 1], w)) / dxn[r - 1]
-            z = np.dot(s, w)
-            s = (1 + z / (sn - z)) * w
-            sn = np.dot(s, s)
-            dx[k] = s
-            dxn[k] = sn
-            x -= s
-            f = F(x)
-            k += 1  # update x and iteration number k
-    return x, k  # return the final value and the numbers of iterations needed
+def simpson(f, a, b, N):
+    x, h = np.linspace(a, b, 2 * int(N) + 1, retstep=True)
+    I = h / 3.0 * (np.sum(f(x[::2])) + 4.0 * np.sum(f(x[1::2])) + f(x[0]) - f(x[-1]))
+    return I