This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 18.06 Problem Set 7 Solution Due Wednesday, 15 April 2009 at 4 pm in 2106. 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....
View
Full
Document
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.
 Spring '08
 Strang

Click to edit the document details