Update README.md

Author: fritzwill

README.md

@@ -24,7 +24,7 @@ Bisection Method is a root finding method that solves the form: f(x) = 0.
 
 The current low guess, high guess, tol, and N for demonstration is (1, 2, 10^-5, 100) respectively
 * Note these can be changed under:
-```
+```python
 # globals
 TOL = 10**-5
 A = 1
@@ -34,13 +34,13 @@ N = 100
 
 The current fucntion for demonstration is: f(x) = x^3 - x - 2
 * Note this can be changed under:
-```
+```python
 def f(x):
  return (math.pow(x,3) - x - 2)
 ```
 
 Run from command line:
-```
+```python
 $ ./bisection.py
 Bisection method soln: x = 1.52138519287
 ```
@@ -52,7 +52,7 @@ The Fixed-Point Iteration Method finds the root of an equation in the form: x =
 
 The current approx, tol, and N for demonstration is (4.6, 10^-4, 100) respectively
 * Note these can be changed under:
-```
+```python
 # globals
 APPROX = 4.6
 TOL = 10**-4
@@ -61,13 +61,13 @@ N = 100
 
 The current fucntion for demonstration is: f(x) = (1/tan(x))- (1/x) + x
 * Note this can be changed under:
-```
+```python
 def f(x):
  return (1/math.tan(x)) - (1/x) + x
 ```
 
 Run from command line:
-```
+```python
 $ ./fixedPoint.py
 Fixed point solution: x = 4.49340945791
 ```
@@ -78,7 +78,7 @@ The Newton's Method finds the root of an equation: x : f(x) = 0
 
 The current approx, tol, and N for demonstration is (4.6, 10^-4, 100) respectively
 * Note these can be changed under:
-```
+```python
 # globals
 APPROX = 0.1
 TOL = 10**-5
@@ -87,20 +87,20 @@ N = 100
 
 The current fucntion for demonstration is: f(x) = (1 + x)^204 - 440x - 1
 * Note this can be changed under:
-```
+```python
 def f(x):
  return math.pow((1+x),204)-440*x-1
 ```
 
 For Newton's Method we also need to know f'(x). Currently, f'(x) = 204*(1 + x)^203 - 440
 * Note that this MUST be changed when f(x) is changed. Do this under:
-```
+```python
 def fprime(x):
  return 204*math.pow((x+1),203) - 440
 ```
 
 Run from command line:
-```
+```python
 $ ./newtons.py
 Newton's Method soln: x = 0.00681932148758
 ```
@@ -112,7 +112,7 @@ The Newton's Method finds the root of an equation: x : f(x) = 0. Considered an
 
 The current approx, tol, and N for demonstration is (4.6, 10^-4, 100) respectively
 * Note these can be changed under:
-```
+```python
 # globals
 APPROX = 0.1
 TOL = 10**-5
@@ -121,13 +121,13 @@ N = 100
 
 The current fucntion for demonstration is: f(x) = (1 + x)^204 - 440x - 1
 * Note this can be changed under:
-```
+```python
 def f(x):
  return math.pow((1+x),204)-440*x-1
 ```
 
 Run from command line:
-```
+```python
 $ ./secant.py
 Secant method soln: x = 0.00681932406799
 ```