Rank-3 scale theorems: Difference between revisions

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M_b <= F: Prove that F(x) describes a mos.

Say F has #s s's and #L L's across interval [m, m'] ~~in R/nZ = circle of circumference n~~. Say there is #s+t small steps and #L-t large steps on some k step [r, r'], t >= 2. This implies that the slope of the line b/a* x itself satisfies

Say F has #s s's and #L L's across interval [m, m']. Say there is #s+t small steps and #L-t large steps on some k step [r, r'], t >= 2. This implies that the slope of the line b/a* x itself satisfies

(F(m')-F(m)-1)/(m'-m) <= b/a <= (F(m')-F(m)+1)/(m'-m).

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Thus b/a#L <= b/a(#L-t), a contradiction.

M_b >= F: (bc it's a mos) Suppose there is an x-value ~~x_0~~ where M_b(~~x_0~~) <= F(~~x_0~~) - 1. Let m = min(n_0, n-n_0), n = scale size. Then find three different m-mossteps/average slopes by taking ~~one~~ interval before n_0, one interval containing n_0 and one interval after n_0. (We already know that mosses are slope-LQ.)

M_b >= F: (bc it's a mos) Suppose there is an x-value n_0 where M_b(n_0) <= F(n_0) - 1. Let k = min(n_0, n-n_0), n = scale size. Then find three different k-mossteps/average slopes by taking the interval [n_0-k, n_0] before n_0, one interval containing n_0 and one interval after n_0. (We already know that mosses are slope-LQ.)

(M_b(n_0) - M_b(n_0-k))/k <= (F(n_0) - 1 - M_b(n_0-k))/k

(M_b(n_0+k) - M_b(n_0))/k >= (M_b(n_0+k) - M_b(n_0-k) + 1)/k

==== MV3 Theorem 1 (WIP) ====

@@ Line 38: / Line 38: @@
 M_b <= F: Prove that F(x) describes a mos.
-Say F has #s s's and #L L's across interval [m, m'] in R/nZ = circle of circumference n. Say there is #s+t small steps and #L-t large steps on some k step [r, r'], t >= 2. This implies that the slope of the line b/a* x itself satisfies
+Say F has #s s's and #L L's across interval [m, m']. Say there is #s+t small steps and #L-t large steps on some k step [r, r'], t >= 2. This implies that the slope of the line b/a* x itself satisfies
 (F(m')-F(m)-1)/(m'-m) <= b/a <= (F(m')-F(m)+1)/(m'-m).
@@ Line 51: / Line 51: @@
 Thus b/a#L <= b/a(#L-t), a contradiction.
-M_b >= F: (bc it's a mos) Suppose there is an x-value x_0 where M_b(x_0) <= F(x_0) - 1. Let m = min(n_0, n-n_0), n = scale size. Then find three different m-mossteps/average slopes by taking one interval before n_0, one interval containing n_0 and one interval after n_0. (We already know that mosses are slope-LQ.)
+M_b >= F: (bc it's a mos) Suppose there is an x-value n_0 where M_b(n_0) <= F(n_0) - 1. Let k = min(n_0, n-n_0), n = scale size. Then find three different k-mossteps/average slopes by taking the interval [n_0-k, n_0] before n_0, one interval containing n_0 and one interval after n_0. (We already know that mosses are slope-LQ.)
+(M_b(n_0) - M_b(n_0-k))/k <= (F(n_0) - 1 - M_b(n_0-k))/k
+(M_b(n_0+k) - M_b(n_0))/k >= (M_b(n_0+k) - M_b(n_0-k) + 1)/k
 ==== MV3 Theorem 1 (WIP) ====