README.md

# Homework #2
## due 10/6/17 by 11:59pm


**1\.** Create a new github repository called '02_roots_and_optimization'. 

  a. Add rcc02007 and zhs15101 as collaborators.

  b. submit the clone repository URL to:
  [https://goo.gl/forms/svFKpfiCfLO9Zvfz1](https://goo.gl/forms/svFKpfiCfLO9Zvfz1)


**2\.** You're installing a powerline in a residential neighborhood. The lowest point on the
cable is 30 m above the ground, but 30 m away is a tree that is 35 m tall. Another
engineer informs you that this is a catenary cable problem with the following solution

  ![eq. 1](./equations/eq1.png)
  
  $y(x)=\frac{T}{w}\cosh\left(\frac{w}{T}x\right)+y_{0}-\frac{T}{w}$.

  where y(x) is the height of the cable at a distance, x, from the lowest point, $y_{0}$,
  T is the tension in the cable, and w is the weight per unit length of the cable. Your
  supervisor wants to know which numerical solver to use when they have to install these
  powerlines in similar places. 
    
  a. Use the three solvers `falsepos.m`, `bisect.m`, and `mod_secant.m`
  to solve for the tension neededi, T, to reach y(30 m)=35 m, with w=10 N/m, and $y_{0}$=30 m. 

  b. Compare the number of iterations that each function needed to reach an
  accuracy of 0.00001%. Include a table in your README.md with:

    ```
    | solver | initial guess(es) | ea | number of iterations|
    | --- | --- | --- | --- |
    |falsepos   |  |  |  |
    |mod_secant |  |  |  |
    |bisect     |  |  |  |
    ```


  c. Add a figure to your README that plots the final shape of the powerline
  (![eq2](./equations/eq2.png)) from x=-10 to 50 m. 

**3\.** The Newton-Raphson method and the modified secant method do not always converge to a
solution. One simple example is the function f(x) = (x-1)*exp(-(x-1)^2). The root is at 1, but
using the numerical solvers, `newtraph.m` and `mod_secant.m`, there are certain initial
guesses that do not converge. 

  a. Calculate the first 5 iterations for the Newton-Raphson method with an initial
  guess of x_i=3 for f(x)=(x-1)*exp(-(x-1)^2).

  b. Add the results to a table in the `README.md` with:

    ```
    ### divergence of Newton-Raphson method

    | iteration | x_i | approx error |
    | --- | --- | --- |
    | 0 | 3 | n/a |
    | 1 |   |     |
    | 2 |   |     |
    | 3 |   |     |
    | 4 |   |     |
    | 5 |   |     |
    ```

  c. Repeat steps a-b for an initial guess of 1.2. (But change the heading from
  'divergence' to 'convergence')

![Model of Gold chain, from molecular dynamics simulation](../08_optimization/Auchain_model.png)

**4\.** Determine the nonlinear spring constants of a single-atom gold chain. You can assume
the gold atoms are aligned in a one dimensional network and the potential energy is
described by the Lennard-Jones potential as such
  
  ![eq3](./equations/eq3.png)

  $E_{LJ}(x)=4\epsilon
  \left(\left(\frac{\sigma}{x}\right)^{12}-\left(\frac{\sigma}{x}\right)^{6}\right)$.

  Where x is the distance between atoms in nm, $\epsilon$=2.71E-4 aJ, and $\sigma$=0.2934
  nm. The energy term that must be minimized is 

  ![eq4](./equations/eq4.png)

  $E_{total}(\Delta x)=E_{LJ}(x_{0}+\Delta x)-F\Delta x$.

  Where ![x0](./equations/x0.png) is the distance between atoms with no force applied and
  ![dx](./equations/deltax.png) is the
  amount each gold atom has moved under a given force, F.

  a. Determine ![x0](./equations/x0.png) when F=0 nN using the golden ratio and parabolic methods. *Show
  your script and output in your README and include your functions*

  b. Solve for ![dx](./equations/deltax.png) is the
  amount each gold atom has mov for F=0 to 0.0022 nN with 30 steps. *Use the golden ratio
  solver or the matlab/octave `fminsearch`

  c. create a sum of squares error function `sse_of_parabola.m` that calculates the sum of
  squares error between a function ![F(x)](./equations/fx.png) $F(x)=K_{1}\Delta x+1/2K_{2}\Delta x^{2}$ and the
  Forces used in part B for each ![dx](./equations/deltax.png). 

  d. Use the `fminsearch` matlab/octave function to determine
  ![k1k2](./equations/k1k2.png). 

  e. Plot the force vs calculated ![dx](./equations/deltax.png) and the best-fit parabola using ![k1k2](./equations/k1k2.png) in part d.