Rudin: theorem 1.21

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Summary:
Rudin theorem 1.21
Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?
 

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Answers and Replies

  • #2
mathman
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How is X (or x) defined?
 
  • #3
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How is X (or x) defined?
X is a real number and X>0
All the information is in the pitcure
 
Last edited by a moderator:
  • #4
pasmith
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Summary: Rudin theorem 1.21

Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?

t is the result of dividing x by a number which is strictly greater than 1. Hencee [itex]t < x[/itex].
 
  • #5
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Summary: Rudin theorem 1.21

Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?
To add to what @pasmith wrote, if x > 0, the graph of ##t = f(x) = \frac x {x + 1}## is always below the graph of ##t = g(x) = x##. From this we can conclude that ##t < x## for x > 0.

In addition, ##x - \frac x {x + 1} = \frac{x^2 + x - x}{x + 1} = \frac{x^2}{x + 1} > 0## for x > 0. This means that ##t = \frac x {x + 1} < x##, for x > 0.
 
  • #6
RPinPA
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Summary: Rudin theorem 1.21

Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?

Because ##x## is positive, so (x + 1) > 1, so ##x/(x+1) < x##.

Style note: ##X## and ##x## are different symbols. You should stick to one or the other and not treat them as interchangeable.
 
  • #7
WWGD
Science Advisor
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## t/x=1/(x+1) <1 \rightarrow t < x ##
 

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