Explanation Given: The unit disk D = { ( x , y ) ∈ R 2 | x 2 + y 2 ≤ 1 } and the unit circle S 1 = { ( x , y ) ∈ R 2 | x 2 + y 2 = 1 } . The function q : R 2 → R 2 defined as q ( x , y ) = ( x , 0 ) , and the function p : D ∗ → S 1 , where D ∗ = D \ { ( 0 , 0 ) } . Draw: Let ( x , y ) ∈ D ∗ such that 0 < x < 1 and 0 < y < 1 . Then, p ∘ q | D ∗ ( x , y ) = p ( q | D ∗ ( x , y ) ) = p ( x , 0 ) = ( 1 , 0 )

3rd Edition

ISBN: 9781284056242

Author: David J. Hunter

Publisher: Jones & Bartlett Learning

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Chapter 2.3, Problem 32E

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To draw a well-labelled diagram to show that p∘q|D≠q|D∘p.

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Chapter 2.3, Problem 31E

Chapter 2.3, Problem 33E

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To draw a well-labelled diagram to show that p ∘ q | D * ≠ q | D * ∘ p .

Solution Summary: The author explains how to draw a well-labelled diagram to show that pcirc q|_D*ne

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To draw a well-labelled diagram to show that p∘q|D≠q|D∘p.

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To draw a well-labelled diagram to show that p ∘ q | D * ≠ q | D * ∘ p .

Solution Summary: The author explains how to draw a well-labelled diagram to show that pcirc q|_D*ne

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To draw a well-labelled diagram to show that p∘q|D*≠q|D*∘p.

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To draw a well-labelled diagram to show that p∘q|D≠q|D∘p.