Combinatory Analysis - Volume 2 by Percy A. MacMahon

By Percy A. MacMahon

Initially released in 1915-16. This quantity from the Cornell college Library's print collections was once scanned on an APT BookScan and switched over to JPG 2000 layout via Kirtas applied sciences. All titles scanned hide to hide and pages may well contain marks notations and different marginalia found in the unique quantity.

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The proof we may always assume no rowsum or columnsum case). in both cases. 7) ~(r i) H _ i=l~ ~(r i - I) < i= 1 ~(r i We first remark = 2. Now (all that it is easily checked = 1. Then, taking ~ {~(2)} e all > 2) . 7) may possibly be proved by induction We a s s u m e ~ ( 1 ) r. I) r . 7) for ! on ~ or on the r i s or both. ,~). ,r %) ~ + 1. If r I = r 2 = ... 7) holds. 8) holds = 3 I/3 ~(r I - l)~(r 2 - I) ... 8) for ~ + i instead side is increased by I) < ~(rl)~(r2) of % and replace ... ~(r~) .

1). 6) is d e r i v e d t) -I. 2 u -I + n l o g ( l + u ) < "6 + log ~ u < -6172 u-I + ( n - l)u + log u . 3. 2) takes no more work than the proof of itself[ Partitions, s e r i e s cmd products. 43 Problem 2 that p(n) ment using series partitions and products of n into parts n 2 2. It follows (cf. is convex. We give a proof of this state- [3]). Let c(0) := I and c(n) := the number of > 3. Then c(1) = c(2) = 0, c(3) = I and c(n) that for n k 2 tbe coefficient of x n in the expansion k c(n-I) for of co (I + x) -I is positive.

The r e s u l t log p(n) t I-t < ~1 value 72 u-1 + nu for u = ~(6n) -! 2. S u b s t i t u t i n g result. is much w e a k e r of the proof n. T h e r e f o r e that f(t) its m i n i m a l the r e q u i r e d inequality slight m o d i f i c a t i o n I ~2 < -6- Hence (0 < u < ~) and find expression of u we find this > p(n)t n for 0 < t < I, n ( ~. 1). 1). 6) is d e r i v e d t) -I. 2 u -I + n l o g ( l + u ) < "6 + log ~ u < -6172 u-I + ( n - l)u + log u . 3. 2) takes no more work than the proof of itself[ Partitions, s e r i e s cmd products.

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