Science:Math Exam Resources/Courses/MATH307/April 2012/Question 07 (c)
Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q4 (e) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q6 (c) • Q6 (d) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) • Q8 (c) •
Question 07 (c) 

Consider the MATLAB/Octave computation 1> A=[0 2 1; 1 0 0; 0 1 0] A = 0 2 1 1 0 0 0 1 0 2> [S D]=eig(A) S = 0.80902 0.57735 0.30902 0.50000 0.57735 0.50000 0.30902 0.57735 0.80902 D = Diagonal Matrix 1.61803 0 0 0 1.00000 0 0 0 0.61803 (c) Write down initial values which result in a nonzero periodic sequence where the x_{n} repeatedly cycle through the same values. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

Science:Math Exam Resources/Courses/MATH307/April 2012/Question 07 (c)/Hint 1 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. Let us begin by assuming that the cycle only repeats over a single value. It should be immediately obvious that the only possible value that this could be is 0 since if the initial values were less than 0, the values will become increasingly negative and if the initial values were greater than 0, the values will become increasingly positive. Since we want a nonzero periodic sequence, a cycle of 1 value is not possible. Therefore, we will look at a cycle which repeats over 2 values. This implies:
If we just take the first value that can use the recursion relation (), we see that we can arrive at the following through basic algebra (where in step two we use that :
We just showed that a recursive cycle of 2 values is possible! If we assign the first two values to be the negative of each other, we will have a repeating cycle of the two selected values. Therefore, the following are all possible choices for the initial three values (remember that we need three initial values!):
