all n N we have We apply this successively for n = bn+1 bn N, N+1, n 9 bN+1 < (m + €)bN bN+2 < (m+ €)bN+1 : : bn < (m+€)bn-1.
all n N we have We apply this successively for n = bn+1 bn N, N+1, n 9 bN+1 < (m + €)bN bN+2 < (m+ €)bN+1 : : bn < (m+€)bn-1.
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter4: Polynomials
Section4.6: Multiplying Polynomials
Problem 42WE
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I don't understand the part that says n=N,N+1,...n-1. I don't understand what the n-1 means, or why it changes from N to n.

Transcribed Image Text:all n N we have
We apply this successively for n =
bn+1
b₁
N, N+1,
n 1 to get
-
9
bN+1 < (m + €)bN
bN+2 < (m+ €)bN+1
:
:
bn
< (m + €)bn-1.
<m + €.
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