By H. Ives

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**Extra info for Natural Trigonometric Functions [math tables]**

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Proof. Without loss of generality let us assume that S = {1, . . , i ∗ } for some i ∗ ∈ [m]. 5). Then, we have B(k) = B1 (k) 0 , 0 B2 (k) for any k ≥ 0. 6, it follows that {B(k)} is an 1 -approximation of {A(k)}. 5 it follows that i↔ B j for any i, j ∈ S. But by the block diagonal form of {B(k)}, this implies that i↔ B1 j for any i, j ∈ S. 1 it follows that any i ∈ S is an ergodic index for {B1 (k)} which implies that any i ∈ S is an ergodic index for 34 3 Ergodicity of Random Chains {B(k)}. 5, it follows that any i ∈ S is an ergodic index for {A(k)}.

Also, since g is a convex function defined on R, it has a sub-gradient ∇g(x) at each point x ∈ R. Furthermore, we have m m πi (k)∇g(π T (k)x)(xi − π T (k)x) = ∇g(π T (k)x) i=1 πi (k)(xi − π T (k)x) = 0. 4) i=1 where Dg (α, β) = g(α) − g(β) − ∇g(β)(α − β) . If g is a strongly convex function, Dg (α, β) is the Bregman divergence (distance) of α and β with respect to the convex function g(·) as defined in [9]. 4) shows that our comparison function is in fact a weighted average of the Bregman divergence of the points x1 , .

2 Let {B(k)} be an 1-approximation of a chain {A(k)}. Then, {B(k)} is ergodic if and only if {A(k)} is ergodic. Proof Note that a chain {A(k)} is ergodic if and only if i↔ A j for any i, j ∈ [m]. 5 implies that {A(k)} is ergodic if and only if {B(k)} is ergodic. 1. 1, the ergodicity of a chain {A(k)} implies the connectivity of the infinite flow graph of {A(k)}. 6 Let {A(k)} be a deterministic chain and let G ∞ be its infinite flow graph. Then, i↔ A j implies that i and j belong to the same connected component of G ∞.