[a, b] = N=1 (a -,b+-) To prove this, show the following: То Show that [a, b](a – ÷,b +- ) for every n >1. а п n • Show that [a, b] C N=1 (a -, b + ). • Show that N-1 (a -,b +) [a, b]. To prove this, use the n n n n method of contradiction. Let x E N=1 (a -,b +-). It implies that -! 1. suppose on contrary that x € n п

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Show that
a. [a, b] = N=1 (a -, b+) To prove this, show the following:
• Show that [a, b] (a - ,b+-) for every n > 1.
• Show that [a, b] C N=1 (a - -, b +-).
n
n
1
а
n
Show that N=1 (a -,b +-) s [a, b]. To prove this, use the
а
n
method of contradiction. Let x E N=1 (a -,b +-). It implies that
=D1
п
n
1
a - -< x < b +÷ for every n > 1. suppose on contrary that x €
п
n
[a, b]. It means either x < a or x > b. Case 1: Let x < a. We can
find an no E N such that x<÷<a = x < a
По
<а3х<а — —. Which is a
По
-
contradiction. Case 2: Let x > b. This case can be dealt similarly.
b. (а, со) %3D U#-1(а, а + п).
n=1
Transcribed Image Text:Show that a. [a, b] = N=1 (a -, b+) To prove this, show the following: • Show that [a, b] (a - ,b+-) for every n > 1. • Show that [a, b] C N=1 (a - -, b +-). n n 1 а n Show that N=1 (a -,b +-) s [a, b]. To prove this, use the а n method of contradiction. Let x E N=1 (a -,b +-). It implies that =D1 п n 1 a - -< x < b +÷ for every n > 1. suppose on contrary that x € п n [a, b]. It means either x < a or x > b. Case 1: Let x < a. We can find an no E N such that x<÷<a = x < a По <а3х<а — —. Which is a По - contradiction. Case 2: Let x > b. This case can be dealt similarly. b. (а, со) %3D U#-1(а, а + п). n=1
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