Water vapor splits into gaseous Hydrogen and Oxygen according to 2𝐻2𝑂 → 2𝐻2 + 𝑂2.

INSTRUCTIONS TO CANDIDATES

ANSWER ALL QUESTIONS

Exam 1 Take Home

 

Root Solving Methods

1) Bisection

Water vapor splits into gaseous Hydrogen and Oxygen according to 2𝐻2𝑂 → 2𝐻2 + 𝑂2. The mole fraction 𝑓 of 𝐻2𝑂 can be found from the equation

where 𝑘 is a reaction equilibrium constant and 𝑝𝑡 is the total pressure of the mixture (in atmospheres). Solving this equation for 𝑓 analytically is challenging, but because the mole fraction fundamentally varies from 0 to 1, a bracketed root finding method is well-suited to solve the problem.

Use a bisection method to solve for mole fraction 𝑓 given 𝑘 = 0.04 and 𝑝𝑡 values varying from 0.25 atm to 10 atm. Solve the problem at many 𝑝𝑡 values in that range and graph 𝑓 vs.

𝑝𝑡 (𝑓 should be on the vertical axis). Each data point in your graph will be built using the bisection method. You should use a loop to solve the bisection method many times for many different values of 𝑝𝑡

2) Newton-Raphson

Write a Newton-Raphson code for the function to find a solution of cos(𝑥) sin(𝑥) = 0.4. Write the code twice, running with 2 different starting guesses.

a) Choose a starting guess where the code returns an error or the method diverges (Please comment this out before submitting if it generates an error)

b) Choose a starting guess where the method converges and a root is found

 

Eigenvalue problem – The popularity problem

 

The google search algorithm, in its infancy, solved a problem in linear algebra called the popularity problem. It attempts to answer the following question in a straight forward yet sophisticated way:

 

If I ask a group of 𝑛 websites which among the other 𝑛 − 1 sites are the most popular, how can I rank each website’s popularity?

 

The simplest method is just count links. Every time a website is linked to, its popularity increases by 1. But this is problematic, and doesn’t always give results that people agree with. Maybe one site is linked to many times, but it is exclusively from sites that are unpopular themselves. Also, maybe certain sites contain a ton of links, so their individual links shouldn’t carry as much weight.

 

We would like to find the popularity of each website, where instead of simply counting links, we sum the popularity of each website that linked to each other website.

 

Read the following paragraph slowly and several times.

 

Let 𝑅𝑖𝑗 be a 1 or a 0 depending on whether or not the 𝑗𝑡ℎ website linked to the 𝑖𝑡ℎ website, and let 𝑝𝑖 be the popularity of the 𝑖𝑡ℎ website. In that case, a sensible expression for the popularity of the 1st website would be

 

𝑝1 = 𝑅12𝑝2 + ⋯ 𝑅1𝑛𝑝𝑛

So, for example, if website 1 was linked to by websites 3, 4, and 5, then website 1’s popularity

𝑝1 would be given by

 

𝑝1 = 𝑝3 + 𝑝4 + 𝑝5

We could write a similar equation for every single website 𝑖 ∈ 1 … 𝑛. This creates a system of equations of the form

 

𝑅̅𝑝⃗ = 𝑝⃗

 

where R is a matrix containing all of the 𝑅𝑖𝑗 entries, and 𝑝⃗ is an unknown column vector containing the popularities of each website.

 

This should be immediately recognizable as an eigenvalue problem. Remember, the eigenvalue equation is 𝐴̅𝑥⃗ = 𝜆𝑥⃗

It is not at all clear that the above equation (𝑅̅𝑝⃗ = 𝑝⃗) would, in general, have a solution. That’s because it isn’t a general eigenvalue equation. It specifically has an eigenvalue of 1. An arbitrary matrix 𝑅̅ will not necessarily have that eigenvalue. However, we can fix this problem if we enforce the following rule: every column of 𝑅 must sum to 1. If every column adds up to one,

the matrix is guaranteed to have 1 as an eigenvalue, and the above equation will have a solution. Essentially, each website is only allowed to give one total “link”, evenly distributed amongst all the different websites it links to. The above fact is not obvious or simple to prove.

The 𝑗𝑡ℎ column of the R matrix, again, is the number of times the 𝑗𝑡ℎ website linked to the others. If we take each column and divide it by its sum, it will then meet our requirement. That is to say, we normalize each column, individually, so they all sum to 1. If the 5th website linked to you, but it also linked to 9 other websites, you only get an 𝑅 value of 0.1. And that means that

website 5’s contribution to your popularity is not 𝑝5, but is only 0.1𝑝5.

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