Rank-3 scale theorems: Difference between revisions

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Take the graph of the brightest mode of the mos, M_b(x) (right = L, up = s). We claim that this is the required graph of F(x) = floor(b/a*x).

M_b <= F: Prove that F(x) describes a mos. Two possibilities, for each k from 0 to a-1: (a) F(k + 1) = F(k) or (b) F(k + 1) = F(k) + 1. (b) happens b times, and (a) happens a-b times. For a given k, in case (a) (b/a)x hasn't crossed a y-gridline (i.e. one indicating a y-value such as y = m), during time k < x <= k+1, and in case (b), (b/a)x ''has'' crossed a y-gridline during time k < x <= k+1.

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

~~Suppose that~~ some r~~-step comes in 3 sizes~~, in x-~~coordinates~~

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

(~~m11, m12~~), (~~m21, m22~~), (~~m31, m32~~)~~. This means that there are three numbers of small steps and there are three numbers of y~~-~~gridline crossings in the r~~-~~step~~. ~~But~~ this is ~~impossible since it requires either~~ floor((b/a)~~mi1~~) < (b/a~~)mi1~~ - 1 ~~or floor(~~(b/a~~)mi2) > (~~b/a~~)mi2 for some i~~ (~~why?~~).

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

(bounded by "floor minus ceiling" and "ceiling minus floor" slopes; this is because x-x' <= x-floor(x') <= floor(x)+1-floor(x').)

Rearranging,

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

But F(m') -F(m) = #s and m'-m = #L. So #s -1 <= b/a*#L <= #s+1. Do the same thing for the "bad" interval [r', r] and you get

#s+t-1 <= b/a(#L-t) <= #s+t+1.

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). Let m = min(n_0, n-n_0), n = scale size. Then find three different m-mossteps by taking one interval before n_0, one interval containing n_0 and one interval after n_0.

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 Take the graph of the brightest mode of the mos, M_b(x) (right = L, up = s). We claim that this is the required graph of F(x) = floor(b/a*x).
-M_b <= F: Prove that F(x) describes a mos. Two possibilities, for each k from 0 to a-1: (a) F(k + 1) = F(k) or (b) F(k + 1) = F(k) + 1. (b) happens b times, and (a) happens a-b times. For a given k, in case (a) (b/a)x hasn't crossed a y-gridline (i.e. one indicating a y-value such as y = m), during time k < x <= k+1, and in case (b), (b/a)x ''has'' crossed a y-gridline during time k < x <= k+1.
+M_b <= F: Prove that F(x) describes a mos.
-Suppose that some r-step comes in 3 sizes, in x-coordinates
+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
-(m11, m12), (m21, m22), (m31, m32). This means that there are three numbers of small steps and there are three numbers of y-gridline crossings in the r-step. But this is impossible since it requires either floor((b/a)mi1) < (b/a)mi1 - 1 or floor((b/a)mi2) > (b/a)mi2 for some i (why?).
+(F(m')-F(m)-1)/(m'-m) <= b/a <= (F(m')-F(m)+1)/(m'-m).
+(bounded by "floor minus ceiling" and "ceiling minus floor" slopes; this is because x-x' <= x-floor(x') <= floor(x)+1-floor(x').)
+Rearranging,
+F(m') - F(m) - 1 <= b/a(m'-m) <= F(m') - F(m) + 1
+But F(m') -F(m) = #s and m'-m = #L. So #s -1 <= b/a*#L <= #s+1. Do the same thing for the "bad" interval [r', r] and you get
+#s+t-1 <= b/a(#L-t) <= #s+t+1.
+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). Let m = min(n_0, n-n_0), n = scale size. Then find three different m-mossteps by taking one interval before n_0, one interval containing n_0 and one interval after n_0.