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Rudin 8.10

  1. May 7, 2008 #1
    [SOLVED] rudin 8.10

    1. The problem statement, all variables and given/known data
    Prove that [itex](1-x)^{-1} \leq \exp 2x[/tex] when [itex]0 \leq x \leq 1/2[/itex].

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

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

    3. 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.
  2. jcsd
  3. May 8, 2008 #2


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    1. What are their values when x= 0?

    2. How do their derivatives compare?
  4. May 8, 2008 #3
    They are both 1.

    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.
  5. May 8, 2008 #4
    well I think it's best to evaluate:

    and to find where is its minimum point in the interval: [0,1/2]
  6. May 8, 2008 #5
    Yay, I got it thanks!
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