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There are three methods that find approximately the roots of a non-linear function

Author: mavroudo

Bisection.py

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+import math
+
+
+def bisection(function, a, b):  # finds where the function becomes 0 in [a,b] using bolzano
+
+    start = a
+    end = b
+    if function(a) == 0:  # one of the a or b is a root for the function
+        return a
+    elif function(b) == 0:
+        return b
+    elif function(a) * function(b) > 0:  # if noone of these are root and they are both possitive or negative,
+        # then his algorith can't find the root
+        print("couldn't find root in [a,b]")
+        return
+    else:
+        mid = (start + end) / 2
+        while abs(start - mid) > 0.0000001:  # untill we achive percise equals to 10^-7
+            if function(mid) == 0:
+                return mid
+            elif function(mid) * function(start) < 0:
+                end = mid
+            else:
+                start = mid
+            mid = (start + end) / 2
+        return mid
+
+
+def f(x):
+    return math.pow(x, 3) - 2*x -5
+
+
+print(bisection(f, 1, 1000))

Intersection.py

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+import math
+
+def intersection(function,x0,x1): #function is the f we want to find its root and x0 and x1 are two random starting points
+    x_n = x0
+    x_n1 = x1
+    while True:
+        x_n2 = x_n1-(function(x_n1)/((function(x_n1)-function(x_n))/(x_n1-x_n)))
+        if abs(x_n2 - x_n1)<0.00001 :
+            return x_n2
+        x_n=x_n1
+        x_n1=x_n2
+
+def f(x):
+    return math.pow(x,3)-2*x-5
+
+print(intersection(f,3,3.5))

NeutonMethod.py

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+import math
+
+def newton(function,function1,startingInt): #function is the f(x) and function1 is the f'(x)
+  x_n=startingInt
+  while True:
+      x_n1=x_n-function(x_n)/function1(x_n)
+      if abs(x_n-x_n1)<0.00001:
+          return x_n1
+      x_n=x_n1
+      
+def f(x):
+    return math.pow(x,3)-2*x-5
+
+def f1(x):
+    return 3*math.pow(x,2)-2
+
+print(newton(f,f1,3))