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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[hide] 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 $n\to \infty$.

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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.

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

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

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[show] 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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