and then using the solution to H N 1 v N 1 h N 1 to solve the N N system 4 H N

# And then using the solution to h n 1 v n 1 h n 1 to

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and then using the solution to H N - 1 v N - 1 = h N - 1 to solve the N × N system 4 H N v N = h N . Adding together the computational costs at each stage: Total cost = (some constant)(1 + 2 + · · · + N - 1 + N ) = O ( N 2 ) . General right-hand sides Solving for a general right-hand side is not much harder — it just takes twice the work. (And 2 N 2 still beats N 3 every day of the week.) To solve Hx = y we again subdivide it into sections: " H N - 1 J h N - 1 ( J h N - 1 ) T h 0 # w α = y N - 1 y N , and so H N - 1 w + α J h N - 1 = y N - 1 h T N - 1 J w + αh 0 = y N . Solving the first equation: w = H - 1 N - 1 y N - 1 - α J H - 1 N - 1 h N - 1 . 4 By definition, H N = H . 29 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019

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Now suppose we have the following solutions in hand: x N - 1 = H - 1 N - 1 y N - 1 , and v N - 1 = H - 1 N - 1 h N - 1 . Then we can again quickly solve for w and α : w = x N - 1 - α J v N - 1 , with α = y N - h T N - 1 J x N - 1 h 0 - h T N - 1 v N - 1 . So given the solutions to H N - 1 x N - 1 = y N - 1 H N - 1 v N - 1 = h N - 1 , the solution to H N x N = y N , (and also the solution to H N v N = h N ) can be computed in O ( N ) time. Moral: Solving the N × N symmetric Toeplitz system Hx = y can be done in O ( N 2 ) time. What we have done above is easily extended to non-symmetric Toeplitz systems as well. 30 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 13:22, November 13, 2019
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