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Wallis's Formula

  1. Aug 18, 2013 #1
    The portion highlighted in the document seems wrong.

    (n-1)/n < (n-2)/(n-1)
    n2 - 2n + 1 < n2 - 2n

    1<0 (WRONG)

    So how can I believe what the book is saying?

    Now if I let n be n - 2 then I still have 1<0. Therefore the element of (1) is greater than the corresponding element of (2), and not the other way around as the book says
     

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  3. Aug 18, 2013 #2

    Ray Vickson

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    It might not be saying what you claim (although it is badly worded); maybe it asserts that ##\sin^{n-1}(x) < \sin^{n}(x) ## for every ##x \in (0, \pi/2),## and that is certainly true.
     
  4. Aug 18, 2013 #3
    But it never said that. It says sin^{n-1}(x) > \sin^{n}(x)
     
  5. Aug 18, 2013 #4

    vanhees71

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    Yes and that's correct as long as [itex]\sin x[/itex] is not negative. Proof: Let [itex]0<q \leq 1[/itex]. Then you can multiply the inequality [itex]q \leq 1[/itex] by [itex]q^{n-1}>0[/itex], leading to [itex]q^n \leq q^{n-1}[/itex].
     
  6. Aug 18, 2013 #5
    Well why doesn't it work in the math in post #1?

    and is q equal to sin(theta)?
     
    Last edited: Aug 18, 2013
  7. Aug 18, 2013 #6
    Why would this argument not work..
    If n>0 then we have n>(n-1) and sin(x)n ≥ sin(x)n-1 for sin(x)≥0

    is it because for 0≤x≤pi/2, 0≤sin(x)≤1?
     
    Last edited: Aug 18, 2013
  8. Aug 18, 2013 #7

    Dick

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    Well, yes. If 0<=q<=1 then q^n=q*q^(n-1)<=q^(n-1). As vanhees71 already said. E.g. (1/2)^4<=(1/2)^3.
     
    Last edited: Aug 18, 2013
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