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ML inequality question

  1. Sep 20, 2011 #1
    Having trouble with this question:


    The question is: establish the inequality

    |[itex]\int[/itex]eizzdz| [itex]\leq[/itex] [itex]\pi[/itex](1-e-R2)/4R

    on C {z(t) = Reit, t [itex]\in[/itex] [0,[itex]\pi[/itex]/4, R>0


    When i saw the modulus of an integral i thought ML inequality.

    I think the length will be R[itex]\pi[/itex]/4 but im struggling with finding the maximum of eizz. I tried changing to ei(r2(cos(2t)+isin(2t). but i dont feel any closer to the result.

    Am i on the right track, and can anyone help me with finding the max of the function.

    thanks
     
  2. jcsd
  3. Sep 20, 2011 #2
    Yes, you're on the right track. Convert it to sines and cosines:

    [tex]R\int_{0}^{\pi/4} e^{-R^2 \sin(2t)+iR^2 \cos(2t)} ie^{it}dt[/tex]

    and the absolute value of that is less than:

    [tex]R\int_0^{\pi/4} e^{-R^2 \sin(2t)}dt[/tex]

    now, draw the plot of sin(2x) over the range of 0 and pi/4 and can you see that:

    [tex]\left|e^{-R^2\sin(2t)}\right|<e^{-R^2(4t/\pi)}[/tex]

    Draw the line [itex]y=\frac{4}{\pi}t[/itex] over sin(2t) to see that.
     
  4. Sep 20, 2011 #3
    im now more confused...

    are you saying i dont need the ML inequality?

    I dont understand where the R in the first line comes from? and where do you go from the last line to get closer to the solution?

    im sorry i havnt understood
     
  5. Sep 20, 2011 #4
    Been out. I'm just placing an upper bound on the integrand. The R comes from letting [itex]z=Re^{it}[/itex] so when you substitute that into the integral, you get:

    [tex]\int_{0}^{\pi/4} e^{i R^2 e^{2it}} Ri e^{it} dt[/tex]

    right?

    Now, convert that all to sines and cosines to get:

    [tex]Ri\int_{0}^{\pi/4} e^{iR^2(cos 2t+i\sin 2t)}e^{it}dt[/tex]

    now the absolute value of that:

    [tex]\biggr|Ri\int_{0}^{\pi/4} e^{iR^2(cos 2t+i\sin 2t)}e^{it}dt\biggr|\leq R\int_0^{\pi/4} e^{-R^2 \sin(2t)}dt[/tex]

    right? Now, that's when we place an upper bound on the integrand by using the line [itex]y=4/\pi t[/itex]

    Draw that line and [itex]\sin(2t)[/itex] to see that it's underneath the sine function so that we can write:

    [tex]\biggr|Ri\int_{0}^{\pi/4} e^{iR^2(cos 2t+i\sin 2t)}e^{it}dt\biggr|\leq R\int_0^{\pi/4} e^{-R^2 \sin(2t)}dt\leq\biggr|\int_0^{\pi/4} e^{-R^2(4t/\pi)}dt\biggr|[/tex]

    and that you can integrate directly.
     
  6. Sep 21, 2011 #5
    Hey Jackmell,

    could you please explain how you would show that the magnitudes of some of the things in this integral are equal to one? i understand for eit but not for eiR2cos(2t)
     
  7. Sep 22, 2011 #6
    You know:

    [tex]|e^{iv}|=1[/tex]

    for any v right? Ok, then there you go:

    [tex]\left|e^{iR^2(\cos(2t)+i\sin(2t)}\right|=\left|e^{-R^2\sin(2t)+iR^2\cos(2t)}\right|[/tex]

    and the absolute value of that i-part is going to be one right?

    [tex]=\left|e^{-R^2 \sin(2t)\left(\cos(R^2\cos(2t))+i\sin(R^2 \cos(2t))\right)}\right|[/tex]

    [tex]=\left|e^{-R^2 \sin(2t)}\right|[/tex]
     
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