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Test for Convergence

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Test for convergence or divergence.


    2. Relevant equations
    [itex]\displaystyle\int_0^∏ {\frac{sin∅}{\sqrt{pi-∅}} d∅}[/itex]


    3. The attempt at a solution
    the solution manual does the following:

    pi -∅ = x

    [itex]\displaystyle\int_0^pi {\frac{sinx}{\sqrt{x}} d∅}[/itex]


    [itex]0 <= \frac{sinx}{\sqrt{x}} <= \frac{1}{\sqrt{x}} [/itex]

    but: sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏
     
  2. jcsd
  3. Mar 14, 2013 #2

    SammyS

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    (Use "\theta" in LaTeX for θ and "\pi" for pi.)

    Yes, but dx = -dθ , and [itex]\displaystyle \ \ -\int_{a}^{b}f(x)\,dx=\int_{b}^{a}f(x)\,dx[/itex]
     
  4. Mar 14, 2013 #3
    i'm sorry what do you mean?
     
  5. Mar 15, 2013 #4

    SammyS

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    I thought you were having difficulty understanding why the limits of integration for the integral in x don't go from π to 0 .

    OK, then what did you mean by the following?
    the solution manual does the following:

    pi -∅ = x

    [itex]\displaystyle\int_0^pi {\frac{sinx}{\sqrt{x}} d∅}[/itex]


    [itex]0 <= \frac{sinx}{\sqrt{x}} <= \frac{1}{\sqrt{x}} [/itex]

    but: sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏​
    Especially the statement following the word "but" .
     
  6. Mar 15, 2013 #5
    they are replacing (∏ - ∅) by x (which is greater than which is essentially a replacement for ∅.

    in other words:

    sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏

    but when (∅> ∏): sqrt( ∏ - ∅) > sqrt(∅)

    how does this apply on the integral of the problem. *checked* oopps.

    the interval is from 0 to pi, so sqrt( ∏ - ∅) > sqrt(∅) when ∅< ∏ applies.
     
  7. Mar 15, 2013 #6

    SammyS

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    Oops is right, but you continue writing incorrect statements. (Also, please use θ, not the symbol for the empty set, ∅ .)

    If θ = π/2 , then [itex]\displaystyle \ \sqrt{\pi-\theta\,}=\sqrt{\theta\,}\ .[/itex]

    If 0 < θ < π/2, then [itex]\displaystyle \ \sqrt{\pi-\theta\,}>\sqrt{\theta\,}\ .[/itex]

    If π/2 < θ < π, then [itex]\displaystyle \ \sqrt{\pi-\theta\,}<\sqrt{\theta\,}\ .[/itex]

    However, I don't see what any of this has to do with your problem.
     
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