Added newton-raphson material and inclused recursive vs iterative approaches

1 files changed, 55 insertions, 13 deletions

diff --git a/tutorials/module_3/2_roots_optimization.md b/tutorials/module_3/2_roots_optimization.md
index 2c94f92..3d126e0 100644
--- a/tutorials/module_3/2_roots_optimization.md
+++ b/tutorials/module_3/2_roots_optimization.md
@@ -54,12 +54,21 @@ This exact same concept can be applied to finding roots of functions. Instead of
 > The **Intermediate Value Theorem** says that if f(x) is a continuous function between a and b, and sign(f(a))≠sign(f(b)), then there must be a c, such that a<c<b and f(c)=0.
 
 Let's consider a continuous function $f(x)$ with an unknown root $x_r$ . Using the intermediate value theorem we can bracket a root with the points $x_{lower}$ and $x_{upper}$ such that $f(x_{lower})$ and $f(x_{upper})$ have opposite signs. This idea is visualized in the graph below.
-
-![Intermediate Value Theorem|350](https://pythonnumericalmethods.studentorg.berkeley.edu/_images/19.03.01-Intermediate-value-theorem.png)
-
+<img 
+    style="display: block; 
+           margin-left: auto;
+           margin-right: auto;
+           width: 50%;"
+    src="https://pythonnumericalmethods.studentorg.berkeley.edu/_images/19.03.01-Intermediate-value-theorem.png" 
+    alt="Intermediate Value Theorem">
 Once we bisect the interval and found we set the new predicted root to be in the middle. We can then compare the two sections and see if there is a sign change between the bounds. Once the section with the sign change has been identified, we can repeat this process until we near the root.
-
-![[bisection.png|500]]
+<img 
+    style="display: block; 
+           margin-left: auto;
+           margin-right: auto;
+           width: 50%;"
+    src="bisection.png" 
+    alt="Bisection">
 As you the figure shows, the predicted root $x_r$ get's closer to the actual root each iteration. In theory this is an infinite process that can keep on going. In practice, computer precision may cause error in the result. A work-around to these problems is setting a tolerance for the accuracy. As engineers it is our duty to determine what the allowable deviation is.
 
 So let's take a look at how we can write this in python.
@@ -128,23 +137,53 @@ plt.legend()
 plt.show()
 ```
 # Open Methods
-So far we looked at methods that require us to bracket a root before finding it. However, let's think outside of the box and ask ourselves if we can find a root with only 1 guess. Can we improve the our guess by if we have some knowledge of the function itself?
+So far we looked at methods that require us to bracket a root before finding it. However, let's think outside of the box and ask ourselves if we can find a root with only 1 guess. Can we improve the our guess by if we have some knowledge of the function itself? The answer - derivatives.
 
 ## Newton-Raphson
-Possibly the most used root finding method implemented in algorithms is the Newton-Raphson method. Unlike the bracketing methods, we use one guess and the slope of the function to find our next guess. We can represent this graphically below.
-
-![Newton-Raphson illustration|350](https://pythonnumericalmethods.studentorg.berkeley.edu/_images/19.04.01-Newton-step.png)
-
-
+Possibly the most used root finding method implemented in algorithms is the Newton-Raphson method. By using the initial guess we can draw a tangent line at this point on the curve. Where the tangent intercepts the x-axis is usually an improved estimate of the root. It can be more intuitive to see this represented graphically.
 <img 
     style="display: block; 
            margin-left: auto;
            margin-right: auto;
-           width: 60%;"
+           width: 50%;"
     src="https://pythonnumericalmethods.studentorg.berkeley.edu/_images/19.04.01-Newton-step.png" 
     alt="Newton-Raphson illustration">
+Given the function $f(x)$ we make our first guess at $x_0$. Using our knowledge of the function at that point we can calculate the derivative to obtain the tangent. Using the equation of the tangent we can re-arrange it to solve for $x$ when $f(x)=0$ giving us our first improved guess $x_1$.
 
-If we start with a single 
+As seen in the figure above. The derivative of the function is equal to the slope.
+$$
+f'(x_0) = \frac{f(x_0)-0}{x_0-x_1}
+$$
+Re-arranging for our new guess we get:
+$$
+x_{1} = x_0 - \frac{f(x_0)}{f'(x_0)}
+$$
+Since $x_0$ is our current guess and $x_0$ is our next guess, we can write these symbolically as $x_i$ and $x_{i+1}$ respectively. This gives us the *Newton-Raphson formula*.
+$$
+\boxed{x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)}}
+$$
+### Assignment 2
+Write a python function called *newtonraphson* which as the following input parameters
+ - `f` - function
+ - `df` - first derivative of the function
+ - `x0` - initial guess
+ - `tol` - the tolerance of error
+ That finds the root of the function $f(x)=e^{\left(x-6.471\right)}\cdot1.257-0.2$ and using an initial guess of 8.
+
+```python
+def newtonraphson(f, df, x0, tol):
+	# Output of this function is the estimated root of f
+	# using the Newton-Raphson method
+	
+	if abs(f(x0)) < tol:
+		return x0
+	else:
+		return newtonraphson(f,df,x0)
+
+f = lambda x:x**2-1.5
+f_prime = labmda x:2*x
+
+```
 
 ## Modified Secant
 This method uses the finite derivative to guess where the root lies. We can then iterate this guessing process until we're within tolerance.
@@ -163,6 +202,9 @@ x_{x+1} = x_i - \frac{f(x_i)\delta_x}{f(x_i+\delta x)-f(x_i)}
 $$
 
 
+## Issues with open methods
+Diverging guesses
+
 # Pipe Friction Example
 
 Numerical methods for Engineers 7th Edition Case study 8.4.
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