pset7-s09-soln

# pset7-s09-soln - 18.06 Problem Set 7 Solution Due Wednesday...

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Unformatted text preview: 18.06 Problem Set 7 Solution Due Wednesday, 15 April 2009 at 4 pm in 2-106. Total: 150 points. Problem 1: Diagonalize A = 2 1 1 2 and compute S Λ k S- 1 to prove this formula for A k : A k = 1 2 3 k + 1 3 k- 1 3 k- 1 3 k + 1 . Solution (15 points) Step 1: eigenvalues. det( A- λI ) = λ 2- trace( A ) + det( A ) = λ 2- 4 λ + 3 = 0. The solutions are λ 1 = 1 ,λ 2 = 3. Step 2: solve for eigenvectors. λ 1 = 1 , A- λI = 1 1 1 1 , v 1 = 1- 1 λ 2 = 3 , A- λI =- 1 1 1- 1 , v 2 = 1 1 . Step 3: let S = 1 1- 1 1 be the matrix whose columns are eigenvectors. Let Λ = 1 0 0 3 be the matrix for eigenvalues. Then we have A k = S Λ k S- 1 = 1 1- 1 1 1 0 0 3 k 1 1- 1 1- 1 = 1 1- 1 1 1 0 3 k 1 2 1- 1 1 1 = 1 2 3 k + 1 3 k- 1 3 k- 1 3 k + 1 . Problem 2: Consider the sequence of numbers f , f 1 , f 2 , ..., defined by the recur- rence relation f n +2 = 2 f n +1 + 2 f n , starting with f = f 1 = 1 (giving 1, 1, 4, 10, 28, 76, 208, ...). (a) As we did for the Fibonacci numbers in class (and in the book), express this process as repeated multiplication of a vector ~u k = ( f k +1 ,f k ) T by a matrix A : ~u k +1 = A~u k , and thus ~u k = A k ~u . What is A ? 1 (b) Find the eigenvalues of A , and thus explain that the ratio f k +1 /f k tends to- wards as k → ∞ . Check this by computing f k +1 /f k for the first few terms in the sequence. (c) Give an explicit formula for f k (it can involve powers of numbers, but not powers of matrices) by expanding f in the basis of the eigenvectors of A . (d) If we apply the recurrence relation in reverse , we use the formula: f n = f n +2 / 2- f n +1 (just solving the previous recurrence formula for f n ). Show that you get the same reverse formula if you just compute A- 1 . (e) What does | f k /f k +1 | tend towards as k → -∞ (i.e. after we apply the formula in reverse many times)? (Very little calculation required!) Solution (30 points = 5+5+10+5+5) (a) The recurrence relation gives f k +2 = 2 f k +1 + 2 f k f k +1 = f k +1 . Another way to write this is ~u k +1 = f k +2 f k +1 = 2 2 1 0 f k +1 f k = A~u k , ⇒ A = 2 2 1 0 . (b) Solving det( A- λI ) = λ 2- 2 λ- 2 = 0 gives λ 1 = 1 + √ 3 ≈ 2 . 732 and λ 2 = 1- √ 3 ≈ - . 732. Since | λ 1 | > | λ 2 | , f k +1 /f k tends towards λ 1 ≈ 2 . 732, as k → ∞ . Check first few terms: 1 , 1 , 4 , 10 , 28 , 76 , 208 , 568 ,... 1 / 1 = 1 4 / 1 = 4 10 / 4 = 2 . 5 28 / 10 = 2 . 8 76 / 28 ≈ 2 . 7143 208 / 76 ≈ 2 . 7368 568 / 208 ≈ 2 . 7308 . 2 REMARK: Another phenomenon one should notice is that, in the sequence, f k +1 /f k > λ 1 when k is odd and f k +1 /f k < λ 1 when k is even. This is because the second eigenvalue λ 2 is negative. We will see in the explicit formula of f k below....
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## This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Spring '08 term at MIT.

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pset7-s09-soln - 18.06 Problem Set 7 Solution Due Wednesday...

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