Science:Math Exam Resources/Courses/MATH307/December 2010/Question 05 (b)

MATH307 December 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q6 (e)  •  Q6 (f)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  

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Question 05 (b)

Define a sequence x0, x1, ... by the initial conditions x0 = a, x1 = b and x2 = c together with the recursion relation $\displaystyle x_{n+3}=x_{n+2}+x_{n+1}+x_{n}$ for n = 0, 1, 2, ... (b) If the matrix A from part (a) is defined in MATLAB/Octave, we can do the following calculations >eig(A) ans = 1.83929 + 0.00000i -0.41964 + 0.60629i -0.41964 - 0.60629i >abs(eig(A)) ans = 1.83929 0.73735 0.73735 Describe how you could make further use of the eig command and other MATLAB/Octave commands to determine all (possibly complex) initial values a, b and c for which $x_{n}\to 0$ as $n\to \infty$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First, we notice that the recursion relation Xn+1 = AXn implies that $X_{n+1}=AX_{n}=A(AX_{n-1})=A(A(AX_{-2}))=\ldots =A^{n+1}X_{0}$

Hint 2
Write the vector of initial conditions X0 as a linear combination of the eigenvectors, then apply An+1 to each of those.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. First, we notice that the recursion relation Xn+1 = AXn implies that $X_{n+1}=AX_{n}=A(AX_{n-1})=A(A(AX_{-2}))=\ldots =A^{n+1}X_{0}$ Since A is a 3×3 matrix with three distinct eigenvalues λ1 = 1.83929, λ2 = -0.41964 + 0.60629i, λ3 = -0.41964 + 0.60629i there is a basis of corresponding eigenvectors v1, v2, v3. Let us rewrite X0 with respect to this basis: $\displaystyle X_{0}=\alpha v_{1}+\beta v_{2}+\gamma v_{3}$ Then ${\begin{aligned}X_{n+1}&=A^{n+1}X_{0}=A^{n+1}(\alpha v_{1}+\beta v_{2}+\gamma v_{3})\\&=\alpha A^{n+1}v_{1}+\beta A^{n+1}v_{2}+\gamma A^{n+1}v_{3}\\&=\alpha \lambda _{1}^{n+1}v_{1}+\beta \lambda _{2}^{n+1}v_{2}+\gamma \lambda _{3}^{n+1}v_{3}\end{aligned}}$ Now we see that the last two terms will vanish as n goes to infinity, because $|\lambda _{2}|<1$ and $|\lambda _{3}|<1$. However, since $|\lambda _{1}|>1$ the norm of the first vector will blow up as n approaches infinity, unless α = 0. This is the key to this question: As long as the vector with the initial conditions, $X_{0}={\begin{bmatrix}c\\b\\a\end{bmatrix}}$ lies in the plane that is spanned by v2 and v3 but has no component in direction of v1, then the sequence will converge to zero: xn → 0 as n → ∞. Eigenvectors v2 and v3 can be obtained in MATLAB/Octave with [V,D] = eig(A);  % V is a matrix with eigenvectors in its columns v_2 = V[:,2]; v_3 = V[:,3]; Then the initial values that imply xn → 0 as n → ∞ are $\left\{{\begin{bmatrix}c\\b\\a\end{bmatrix}}{\text{ such that }}{\begin{bmatrix}c\\b\\a\end{bmatrix}}=sv_{2}+tv_{3}{\text{ for some }}s,t\in \mathbb {C} \right\}$

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