Split into 4 scripts; increased usability of the scripts allowing for easier control over the variables involved (tolerance, Max iterations, function,etc.)

Author: fritzwill

bisection.py

@@ -1,15 +1,28 @@
 #!/usr/bin/env python2.7
 
 # 9/28/2017
-# Finds the root, or solution, of the form f(x) = 0. It uses:
+# Finds the root, or solution, of the form f(x) = 0. f(x) is a continuous
+# function defined on an interval [a,b] and where f(a) and f(b) have 
+# opposite signs [f(a)*f(b) < 0]
+# Also, uses:
 # abs((w(n)-w(n-q))/w(n)) <= 10^-5 as the tolerance
 
 import math
 
+# globals
+TOL = 10**-5
+A = 1
+B = 2
+N = 100
+
 # give python eqn of the formula using python math syntax 
 def f(x):
-    return (-32.17/(2*math.pow(x,2)))*(((math.exp(x)-math.exp(-x))/2) - math.sin(x)) - 1.7
+    return (math.pow(x,3) - x - 2)
 
+# INPUT: function (func), low guess (a), high guess (b), tolerance (tol), 
+#        MAX iterations (N)
+# CONDITIONS: a < b, f(a)*f(b) < 0
+# OUTPUT: value which differs from a root of f(x)=0 by less than 'tol' 
 def bisect(func, low, high, tol, N):
     # switch low and high if low is larger than high
     if low > high:
@@ -23,7 +36,7 @@ def bisect(func, low, high, tol, N):
     lastFuncVal = func(low)
     for i in range(0, N):
         mid = (high+low)/2.0
-        if abs(func(mid) - lastFuncVal)/abs(func(mid)) <= tol:
+        if func(mid) == 0 or (high-low)/2.0 < tol:
             return mid
         # same sigh
         if func(mid)*func(low) > 0:
@@ -34,4 +47,4 @@ def bisect(func, low, high, tol, N):
         lastFuncVal = func(mid)
     return "Method failed after {} iterations".format(N)
 
-print "Bisection method soln:, x =", bisect(f, -1, -.001, math.pow(10,-5), 100)
+print "Bisection method soln: x = ", bisect(f, A, B, TOL, N)

fixedPoint.py

@@ -1,20 +1,24 @@
 #!/usr/bin/env python2.7
 
-# Will Fritz
 # 9/28/2017
 # Finds the root of equation in the form of x = f(x) or g(x) = x - f(x) by using the fixed point method
 import math
 
+# globals
+APPROX = 4.6
+TOL = 10**-4
+N = 100
+
 # function to test using python syntax in form g(x) = x - f(x)
 def f(x):
     return (1/math.tan(x)) - (1/x) + x
 
-def fixedPoint(func,  approx, tol, N):
-    for i in range(0, N):
+def fixedPoint(func,  approx, tol, n):
+    for i in range(0, n):
         p = func(approx)
         if abs(p-approx) < tol:
             return p
         approx = p
-    return "Method failed after {} iterations".format(N)
+    return "Method failed after {} iterations".format(n)
 
-print "Fixed point solution: x =", fixedPoint(f, 4.6, 10**-4, 100) 
+print "Fixed point soln: x = ", fixedPoint(f, APPROX, TOL, N) 

newtons.py

@@ -1,45 +1,34 @@
 #!/usr/bin/env python2.7
 
-# Will Fritz
 # 9/29/2017
-# program to find solution to equation, namely f(x), using newton's method
-# and secant method
+# program to find solution to equation, namely f(x) = 0, using newton's 
+# method
+
 import math
 import time
 
+# globals
+APPROX = 0.1
+TOL = 10**-5
+N = 100
+
+# functions we are using, namely f and f'
 def f(x):
     return math.pow((1+x),204)-440*x-1
 
 def fprime(x):
-    return 204*pow((x+1),203) - 500
+    return 204*math.pow((x+1),203) - 440
 
-def newtons(approx, tol, N):
+# Actual Newton's Method
+def newtons(approx, tol, n):
     p0 = approx
-    for i in range(0, N):
-        p = p0 - (f(p0)/fprime(p0))
+    for i in range(0, n):
+        p = p0 - (f(p0)/fprime(p0)) # this is the calculation of the guess
         if abs(p - p0) < tol:
             return p
         p0 = p
-    return "The method failed after {} iterations".format(N)
-
-def secant(p0, p1, tol, N):
-    for i in range(0, N):
-        p = p1 - ((f(p1)*(p1-p0))/(f(p1) - f(p0)))
-        if abs(p-p1) < tol:
-            return p
-        p0 = p1
-        p1 = p
-    return "Them method failed after {} iterations">format(N)        
-
-startTime = time.time()
-output = newtons(.1, pow(10, -5), 100)
-elapsedN = time.time() - startTime
-startTime2 = time.time()
-output1 = secant(.1,.09,pow(10,-5),100)
-elapsedS = time.time() - startTime2
-
-print "Newtons method soln: x =", output
-print "Newtons time in seconds: {}".format(elapsedN)
+    return "The method failed after {} iterations".format(n)
 
-print "Secant method soln: x =", output1
-print "Secant time in seconds: {}".format(elapsedS)
+# Call the method and print its answer
+output = newtons(APPROX, TOL, N)
+print "Newton's method soln: x = ", output

secant.py

@@ -0,0 +1,28 @@
+#!/usr/bin/env python2.7
+
+# 9/12/2018
+# finds the solution to f(x) = 0 using secant method
+
+import math
+
+# globals
+APPROX = 0.1
+TOL = 10**-5
+N = 100
+
+# function we are using
+def f(x):
+    return math.pow((1+x),204)-440*x-1
+
+# Actual Secant Method
+def secant(p0, p1, tol, N):
+    for i in range(0, N):
+        p = p1 - ((f(p1)*(p1-p0))/(f(p1) - f(p0)))
+        if abs(p-p1) < tol:
+            return p
+        p0 = p1
+        p1 = p
+    return "Them method failed after {} iterations">format(N)        
+
+
+print "Secant method soln: x = ", secant(.1,.09,pow(10,-5),100)