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Because the limit of each component exists, we can move the summation inside the limit operation: x(α + ω) − x(α) eT x′(α) = lim eT . ω→0 ω But x(α + ω) and x(α) are both distribution vectors, which implies eT x(α + ω) = eT x(α) = 1. The difference of these scalars eT (x(α + ω) − x(α)) = 0. Consequently, the derivative sums to 0. Whether or not this result merits the full formal lemma-and-proof treatment is debatable.5 In the ensuing discussion of the derivative, it is used repeatedly and thus deserves more than passing attention. Finally, let’s see an equation for the derivative. 3.1.1 The linear system The PageRank derivative is easy to determine from the linear system. Re- call the system (I − αP)x = (1 − α)v from (2.4). Let’s make the dependence on α explicit: (I − αP)x(α) = (1 − α)v. Separating the left-hand side shows that, implicitly, x(α) = αPx(α) + (1 − α)v. Standard rules of matrix calculus give x′(α) = αPx′(α) + Px(α) − v, or more conveniently, (I − αP)x′(α) = Px(α) − v. (3.1) The PageRank derivative is extraordinarily close to a PageRank system! It is not a PageRank system because Px(α) − v has some components less than 0 and eT Px(α) − eT v = 0. That is good though. If it were a PageRank problem then eTx′(α) would be 1, but we know it’s zero (lemma 6). Indeed, (3.1) satisfies the property that eT x′(α) = 0 because it implies eT x′(α) = αeT x′(α). Only one solution is possible for α < 1 : eT x′(α) = 0. Langville and Meyer [2006a] write the derivative ex nihilo as x′(α) = (I − αP)−2(P − I)v. Our preference is to emphasize the PageRank-like structure of the derivative in (3.1). As we shall see after getting around to algorithms, looking at the problem in this manner is highly suggestive of algorithms. 5 Langville and Meyer [2006a] mention it incidentally, for example. 3.1 ⋅ formulations 49PDF Image | Instagram Cheat Sheet
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