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Inequality quick question context cauchy fresnel integral

  1. Jan 20, 2017 #1
    1. The problem statement, all variables and given/known data

    please see attached, I am stuck on the second inequality.

    2. Relevant equations

    attached

    3. The attempt at a solution

    I have no idea where the ##2/\pi## has come from, I'm guessing it is a bound on ##sin \theta ## for ##\theta## between ##\pi/4## and ##0## ?

    I know ##sin \theta \approx \theta ## for small ##\theta## so we could bound it to ##\pi/4## since ##sin ## increases in this range ## 0, \pi /4 ##, however that would be a stronger approximation , loosing the actual integration over ##\theta ## which is clearly not what has been done.

    Many thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Jan 20, 2017 #2

    Charles Link

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    I think I managed to prove it. Let ## x=\frac{\pi}{4}-x' ##. Starting with ## sin(\frac{\pi}{4}-x') \geq \frac{\frac{\pi}{4}-x'}{\frac{\pi}{2}} ## ==>> (This must be proven to be the case), This leads to ## cos(x')-sin(x')-\sqrt{2}(\frac{1}{2}-(\frac{2}{\pi})x') \geq 0 ## for ## 0<x'< \frac{\pi}{4} ##. The function (on the left side of the second inequality) can be shown to be monotonically decreasing from ## 1 -\frac{\sqrt{2}}{2} ## to ## 0 ## as ## x' ## goes from ## 0 ## to ## \frac{\pi}{4} ## by looking at the first derivative. You can try to work through what I did and see if you agree. ## \\ ## Editing: I found a sign error when I took the derivative=scratch the proof=it is incomplete at present... ## \\ ## Additional editing: A little algebra on the derivative shows it indeed is monotonically decreasing=the proof appears to work. The proof of the first derivative being less than zero on the interval ## 0<x'< \frac{\pi}{4} ## has one step to it that may not be so obvious, but I believe it works. I would be happy to show you a couple of the steps if you get stuck.
     
    Last edited: Jan 20, 2017
  4. Jan 21, 2017 #3

    mmm okay thanks.

    I am stuck on making the final conclusion from what has been done,
     
  5. Jan 21, 2017 #4

    Charles Link

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    The first derivative is constantly negative over the interval and at the very right (## x'=\frac{\pi}{4} ##) the function has the value zero. The function is thereby positive for the interval ## 0<x'< \frac{\pi}{4} ##. The first derivative is ## f'(x')=-sin(x')-cos(x')+2 \sqrt{2}/\pi ##. To show this is less than zero over the interval ## 0<x'< \frac{\pi}{4} ##, it helps to see that ## sin(x')+cos(x')=\sqrt{2} cos(x'-\frac{\pi}{4}) ##. Then the point ## x'=0 ## is the point of most concern, because that is where ## cos(x'-\frac{\pi}{4}) ## has its minimum value on the interval ## 0<x'<\frac{\pi}{4} ##..
     
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