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> (bonus) We say that A B if A(X) = ✓ anX" and B(X) = bnX" and an ≤ bn for all n ≥ 0. n≥0 n≥0 Consider subsets of {1,2,,q}, where some subsets are "special". We are not told what "special" means exactly, except that the subsets of a special set are also special (other subsets might be special too). For each n, the number of special subsets is Sn, and S(X) = Σn20 SnXn. Show that if there exists a special subset of size p then (1+X) P ≤ S(X) ≤ (1+X)ª.

> (bonus) We say that A B if A(X) = ✓ anX" and B(X) = bnX" and an ≤ bn for all n ≥ 0. n≥0 n≥0 Consider subsets of {1,2,,q}, where some subsets are "special". We are not told what "special" means exactly, except that the subsets of a special set are also special (other subsets might be special too). For each n, the number of special subsets is Sn, and S(X) = Σn20 SnXn. Show that if there exists a special subset of size p then (1+X) P ≤ S(X) ≤ (1+X)ª.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.6: Inequalities
Problem 79E
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> (bonus) We say that A B if A(X) = ✓ anX" and B(X) = bnX" and an ≤ bn for all n ≥ 0. n≥0 n≥0 Consider subsets of {1,2,,q}, where some subsets are "special". We are not told what "special" means exactly, except that the subsets of a special set are also special (other subsets might be special too). For each n, the number of special subsets is Sn, and S(X) = Σn20 SnXn. Show that if there exists a special subset of size p then (1+X) P ≤ S(X) ≤ (1+X)ª.
Transcribed Image Text:> (bonus) We say that A B if A(X) = ✓ anX" and B(X) = bnX" and an ≤ bn for all n ≥ 0. n≥0 n≥0 Consider subsets of {1,2,,q}, where some subsets are "special". We are not told what "special" means exactly, except that the subsets of a special set are also special (other subsets might be special too). For each n, the number of special subsets is Sn, and S(X) = Σn20 SnXn. Show that if there exists a special subset of size p then (1+X) P ≤ S(X) ≤ (1+X)ª.
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