This lab should only take you a single lab period. The assignment is due at the beginning of lecture on Monday, 6 Apr. In this lab you'll construct a bisection method (root finder) and a golden section search (1D optimization), and apply them to various problems. Your solutions should include 2 function code listings and a table of numerical results (including respective tolerances) for each problems.

1. Construct a MATLAB bisection method function (to find the a root of a function) that has the following inputs/outputs:
   • inputs: an interval [a,b], exit tolerance, function handle
   • outputs: either a final interval [a_f , b_f], the final midpoint, or an interpolated value across the final interval
   
   See the code rootBiSec.m for an example of how to accomplish this. Note that I also use a version of an interpolation function, borrowed from Lab #3, to better estimate the roots. That said, I report back the final interval, midpoint & interpolated root. If your exit tolerance is sufficiently small, you should not see much of a difference in any of these.


2. In celestial mechanics, Kepler's Equation relates the mean anomaly M to the eccentric anomaly E of an elliptical orbit of eccentricity e as: M = E - e sin(E).
   1. For M = 1 radian and e = 0.206 (Mercury), 0.0167 (Earth). Calculate the eccentric anomalies E for each planet.
   2. Are there values for the eccentric anomaly for which E = M? If so, what are they.
   
   See plot below and Lab #7 solution code for details. You will find it calls the function powerpointFig.m, a handy piece of code that formats figure windows for use with PowerPoint, or for display on the web. Note that the graphical roots & the ones found by the code are consistent: E_E ≈ 1.0142 and E_M ≈ 1.1913. For E = M, either e = 0 (the orbit is circular) or E = nπ.
   
3. In neutron transport theory the critical length of a fuel rod in a reactor is determined by the roots of this equation: cot(x) = (x² - 1) / (2x). Determine the smallest positive root of this equation.

See plot below and Lab #7 solution code for details. Note that the graphical root (smallest positive) & the one found by the code are consistent: x ≈ 1.3065. I had to zoom in on a plot of the function to observe the rough locale of the first positive root and use the appropriate interval.

4. Construct a MATLAB golden section search function (to find a local minimum of a function) that has the following inputs/outputs (you'll need this for HW #6):
   • inputs: an interval [a,b], exit tolerance, function handle
   • outputs: either a final interval [a_f , b_f] or the final midpoint

See the code minGoldSec.m for an example of how to accomplish this. In this case, only the final interval and a midpoint are reported.


5. Verify that your function works by finding the relative minimum of y = x³ - 2x² - 4x + 1. (If you can, compare the numerical result with the exact result.)

See plot below and Lab #7 solution code for details. Note that the minimum is found at x = 2; both by eye and the plot. You can also directly verify this by noting that y' = 3x² - 4x - 4.

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6. EXTRA CREDIT The Ideal Gas Law, PV = nRT, works well for low densities since it ignores the size of molecules and assumes perfectly elastic collisions. A second-order approximation is the van der Waals (non-ideal gas) Equation of State, which corrects for finite molecular size and works fairly well even for densities that are not low. It is given by (P + a(n/V)²)(V - nb) = nRT, where a is related to the attractive force between molecules and b is related to the size of each molecule. For P = 3 MPa, T = 300 K and n = 1 mol, find the volume occupied by the two gasses shown below in both the ideal gas and van der Waals approximations.
   
   

   Substance	a (J. m3/mole2)	b (m3/mole)
   Nitrogen (N2)	0.1361	3.85x10-5
   Freon (CCl2F2)	1.078	9.98x10-5