Rudin 8.10

  • Thread starter ehrenfest
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  • #1
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[SOLVED] rudin 8.10

Homework Statement


Prove that [itex](1-x)^{-1} \leq \exp 2x[/tex] when [itex]0 \leq x \leq 1/2[/itex].

Homework Equations


[tex]e^x = \sum_{i=0}^{\infty}\frac{x^n}{n!}[/tex]

[tex]1/(1-x) = 1+x+x^2+\cdots[/tex]


The Attempt at a Solution


I tried working with the series and that failed miserable. Maybe I need to use calculus and find out whether the function is increasing or decreasing but I started that but I tried that a little and did not see how it would help.
 

Answers and Replies

  • #2
HallsofIvy
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1. What are their values when x= 0?


2. How do their derivatives compare?
 
  • #3
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1. What are their values when x= 0?
They are both 1.

2. How do their derivatives compare?
Let f(x) = 1/(1-x) and g(x) = e^{2x}.

Then [itex]f'(x) = 1/(1-x)^2[/itex] and [itex]g'(x) = 2 e^{2x}[/itex].

Is there an obvious inequality between f'(x) and g'(x) when [itex]0 \leq x \leq 1/2[/itex]? I do not see it.
 
  • #4
MathematicalPhysicist
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well I think it's best to evaluate:
(e^(2x)(1-x))=f(x)

and to find where is its minimum point in the interval: [0,1/2]
 
  • #5
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Yay, I got it thanks!
 

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