Solved Task 2b

Schenk_Brandenberger_S2_Aufg2.py

@@ -32,22 +32,31 @@ print("min f2: ", min(yf2), "max f2: ", max(yf2))
 xmin = -10 ** -14
 xmax = 10 ** -14
 xsteps = 10 ** -17
-def g(x):
+def g1(x):
     return x / (np.sin(1 + x) - np.sin(1))
 x2 = np.arange(xmin, xmax + xsteps, xsteps)
-yg = np.array([])
+yg1 = np.array([])
 for x_value in x2:
-    yg = np.append(yg, g(x_value))
+    yg1 = np.append(yg1, g1(x_value))
-plt.plot(x2, yg)
+plt.plot(x2, yg1)
-plt.legend(["g(x)"])
+print("min g1: ", min(yg1), "max g1: ", max(yg1))
-plt.figure()
-print("min g: ", min(yg), "max g: ", max(yg))
 
 # Die Berechnung des Grenzwertes für x --> 0 g(x) ist nicht stabil.
 # Der Grenzwert scheint unendlich gross / klein zu sein.
 
 # Aufgabe 2c
+# a = 1+x, b = 1
+def g2(x):
+    return x / (2 * np.cos((1 + x + 1) / 2) * np.sin((1 + x - 1) / 2))
+yg2 = np.array([])
+for x_value in x2:
+    yg2 = np.append(yg2, g2(x_value))
+plt.plot(x2, yg2)
+print("min g2: ", min(yg2), "max g2: ", max(yg2))
+plt.legend(["g1(x)", "g2(x)"])
 
+# Der Grenzwert für x = 0 beträgt 1.85. Die Funktion ist nun stabil?
+#
 
 
 plt.show()

Schenk_Brandenberger_S2_Aufg3.py

@@ -0,0 +1,39 @@
+import numpy as np
+import matplotlib.pyplot as plt
+
+def s2n(s1n):
+    return np.sqrt(2 - 2 * np.sqrt(1 - ((s1n ** 2) / 4)))
+
+def s2n_new(s1n):
+    return np.sqrt((s1n ** 2) / (2 * (1 + np.sqrt(1 - ((s1n ** 2) / 4)))))
+
+r = int(1) #Radius
+n = int(6) #Anzahl Ecken
+sn = r
+sn_new = r
+x = np.array([])
+y = np.array([])
+y_new = np.array([])
+for i in range(50):
+    sum_s = sn * n
+    sum_s_new = sn_new * n
+    pi = sum_s / 2
+    pi_new = sum_s_new / 2
+    print("n: ", n, " sn: ", sn_new, " pi: ", pi_new)
+    x = np.append(x, n)
+    y = np.append(y, pi)
+    y_new = np.append(y_new, pi_new)
+
+    n = n * 2
+    sn = s2n(sn)
+    sn_new = s2n_new(sn_new)
+
+
+
+plt.plot(x, y)
+plt.plot(x, y_new)
+plt.xscale('log', base=2)
+plt.xlim((2**3, 2**31))
+plt.ylim((3.125, 3.15))
+plt.legend(["pi", "pi_new"])
+plt.show()