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Graph = 0.001 aa-stan 1.72 × 10−10 ee-stan 5.62 × 10−11 cs-stan 5.31 × 10−11 cnr-2000 1.79 × 10−10 = 0.01 1.72 × 10−9 5.62 × 10−10 5.31 × 10−10 1.79 × 10−9 = 0.1 4.30 × 10−8 5.62 × 10−9 2.90 × 10−10 5.35 × 10−9 Table 3.1 – Experimental validation of theorem 7. The table entries show the value of ∥y(γ) − x(w(γ), α)∥2 using the notation from section 3.3.1 with α = 0.85. x (α)i < Otherwise y(1 − α − ε)i > 1 for some small but positive ε.15 The same idea 15 In a less succinct statement, the idea is to assume that x(α)i = 0anduseγ = 1−αtobound the maximum of x′(α)i so that y(γ)i < 1. some small ε. Thus i 1−α ′ ∣x (α)∣ ≤ Using rather different techniques, Langville and Meyer [2006a, p.66] establish this same fact. As α → 1, these bounds become meaningless. What actually happens with the derivative at α = 1 is discussed next. 3.3.3 Limiting derivatives Section 2.7 establishes the PageRank vector when α = 1. Now, we differenti- ate the explicit PageRank function to establish the PageRank derivative when α=1.GivenP=XJX−1 withJ=[I then we have I 16Inthelastchapter,J=[ D1 J2] for a diagonal D1 with all simple eigenvalues of ∣λ∣ = 1 and J2 with all other eigenvalues. Thus, J1=[D1J2]. 1 1−α . ],16 X=[X0 X1 ],andX−1 =[Y0 Y1 ]T, x(α) = X0Y0v + (1 − α)X1(I − αJ1)−1Y1v J1 and x′(α) = (α − 1)X1(I − αJ1)−1J1(I − αJ1)−1Y1 − X1(I − αJ1)−1Y1v. This result matches Langville and Meyer [2006a, theorem 6.1.3], but with an explicit form for the group inverse of (I − P) using the Jordan form of P. 3.3 ⋅ analysis 57 denoted x(w(γ), α). Note that x(w(γ), α) is not a strongly preferential Page- Rank vector. The norm of the differences are listed in table 3.1 for a few graphs and values of γ. These norms are quite small, demonstrating experimental evidence for the theorem. 3.3.2 Bounds An immediate implication of the previous theorem is that ′ . works to show that x′(α) > − 1 , because otherwise y(1 − α − ε) < 0 for 1 1−αPDF Image | Instagram Cheat Sheet
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