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We know that ∥g(1)∥ ≤ α ∥g(0)∥ because it’s just a step of the power method, and the above equation says the same thing. The next iterate satisfies (j) P g ≤ ∥αβj−1Pjg(0)∥ + ∑∥ )∑β P g l=1 (5.15) (5.16) (5.17) (5.18) (5.19) (5.20) (5.21) (5.22) ∥g(2)∥ = ∥αβP2g(0) + (α − β)Pg(0)∥ ≤ αβ ∥g(0)∥ + (α − β) ∥g(0)∥ = ((α − 1)β + α) ∥g(0)∥ ≤ α ∥g(0)∥ , (5.11) (5.12) (5.13) (5.14) and so we monotonically decrease a bound on the error for the first two iterations. For any iterate, we have ∥g j−1 j (0) α−β j−1 l l (0) ∥=αβ +( β j−1 α−β ≤ αβj−1 ∥g(0)∥ + ∑ l=1 = αβj−1 +( )∑βl ⎝ βl=1⎠ ≡κj β ⎛ α−βj−1⎞ ∥g(0)∥. We can rearrange κ j to a more meaningful form using j−1 1 − β j−1 ∑βl =β l=1 , β j−1 α−β l=1 βlPlg(0)∥ βl ∥g(0)∥ 5.4 ⋅ convergence 111 1−β which, after substitution into (5.18), yields κj =αβ j−1 (α − β)(1 − βj−1) + = 1−β (α − β) + (1 − α)β j 1−β α−β . This bound is monotonically converging to 1−β and we further have κj ≤ α for j ≥ 1.PDF Image | Instagram Cheat Sheet
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