residues over simple poles

Discussion in 'Physics & Math' started by camilus, Jun 1, 2011.

  1. Tach Banned Banned

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    Sure, see here. I would still go with the less expensive one I mentioned earlier.


    You are right, your result in the OP is correct. This is not a very complicated field.
     
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  3. camilus the villain with x-ray glasses Registered Senior Member

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    i dont mean this problem in particular, I suspected I was right the whole time, i mean discussing the theory of RZF in general.

    I posted this because it seemed to be a rather persistent typo. Since Im doing independent study on this theory, I can assure you this will not be my last question. I also have some questions regarding 2*xi/(s(s-1)) as a transform.
     
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  5. Tach Banned Banned

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    Sure, no problem.
     
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  7. CptBork Valued Senior Member

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    A few years ago I took a number theory class, and RZF came up towards the end. It's interesting to see what you guys are doing here- we never got far enough to actually apply any residue calculus methods to the subject, possibly because not everyone in the class was actually familiar with complex analysis.
     
  8. Tach Banned Banned

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    It is a very powerful tool , with many applications (easy evaluation of definite integrals, sum of numerical series, etc).
     
  9. CptBork Valued Senior Member

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    When I took the course, I had this really hardass prof who always added an extra couple of layers of complication to the topic. You understand conformal mapping? Great, but first you gotta manipulate this question into a conformal mapping problem, otherwise you get zero points and ultimately a big fat fail. In retrospect, he probably should have chosen a more advanced textbook or something matching his specific approach. It's neat though to see what you can do with it when you really push it, making fancy variable changes or whatever just to squeeze a seemingly unrelated problem into that box we call complex analysis.
     
  10. camilus the villain with x-ray glasses Registered Senior Member

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    this might be a bit embarrassing but Im having trouble with a simple change of variable in this same integral..

    \(\psi (z) = {1 \over 2\pi i}\int_{1/2-i\infty}^{1/2+i\infty} {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}ds + {1 \over \sqrt{z}}\)

    Letting s=1/2 + it, ds=i*dt, and s(s-1)=(1/2+it)(it-1/2) = -(t^2+1/4)

    My problem is arising from that negative infront of t-squared + 1/4. Upon doing these substitutions, I got

    \(\psi (z) -{1 \over \sqrt{z}}= {z^{-1/4} \over 2\pi }\int_{-\infty}^{+\infty} {\xi (\frac{1}{2}+it) \over -(t^2+\frac{1}{4})} z^{{-it \over 2}}dt\)

    My book has this same integral without the negative sign infront of the t*t+1/4. I showed it to a few friends, as well as my professor and they are stumped as well as to what happened 2 that negative sign. Apparently they did not cancel it out by flipping the limits of integration, so idk wtf happaned 2 that -sign... try it yourself and tell me what you think..?
     
  11. CptBork Valued Senior Member

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    Wish I could help you, but I checked your integral manipulations and they look fine to me. Is it possible the book just has some typos? Is the integral used to derive further results, or repeated with the same signage?
     
  12. camilus the villain with x-ray glasses Registered Senior Member

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    well the residue computation wouldnt be affected by the 1/2 because they derive a Big O bound, so the 1/2 is irrelevant.

    But this negative sign could cause problems.. Im gonna post a picture of my book shortly.
     
  13. camilus the villain with x-ray glasses Registered Senior Member

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    Please Register or Log in to view the hidden image!



    as you can see, the negative sign is no where 2 be found. Temur where you at when I need you man!
     
  14. Tach Banned Banned

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    \( s=1/2 + it\), means that \(t=\frac{s-1/2}{i}\)

    so, for \(s=1/2-i\infty\) you have \(t=- \infty\) and
    for \(s=1/2+i\infty\) you have \(t=+ \infty\)

    The book is incorrect, there should be a minus sign.
     
    Last edited: Jun 8, 2011
  15. camilus the villain with x-ray glasses Registered Senior Member

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    \(t=\frac{s-1/2}{i}\)

    plug in s=1/2+i(inf)

    \(t=\frac{1/2+i\infty-1/2}{i} = \frac{i \infty}{i}=\infty\)....

    i really dont see how you got -infinity.., can you explain?
     
  16. Tach Banned Banned

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    You are right, the book is wrong.
     
  17. temur man of no words Registered Senior Member

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    I think your minus sign and the factor 2 are both correct. But none of this is going to matter (I borrowed the book from library). As z approaches i from the first quadrant, you want to bound the size of the integral that has the minus sign issue, so the sign goes away. This integral is equal to something like \(\omega(z)-C/\sqrt{z}\) and as z goes to i, the omega term will blow up like \(Re(z)^{-1/2}\), so whatever value the constant C has, does not affect the bigO bound.
     
  18. camilus the villain with x-ray glasses Registered Senior Member

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    actually the factor of 2 is wrong, well not wrong but it was canceled out earlier. And how I said, they took a big O bound so it is irrelevant. But the negative sign is another story. apparently the negative sign isnt there before they bound the integral..
     
  19. temur man of no words Registered Senior Member

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    How it was canceled? I would be interested to know.

    The negative sign is irrelevant because for instance in (5.15) something=O(d) means |something| <= constant*d.
     
  20. camilus the villain with x-ray glasses Registered Senior Member

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    do you see the (z^-s/2)/2 in front of the ds on the bottom of page 174? that /2 canceled the 2 in \(\frac{2\xi(s)}{s(s-1)}\).


    But the negative sign is missing even before they bound the integral...
     
  21. temur man of no words Registered Senior Member

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    Yes but what does that change? The first formula on page 175 already starts with the result of that calculation, and the error is on page 175 (not on 174), as it was the first question of this thread. Or are you talking about something different?

    It is missing before, but my point is that even if you put the minus sign back, you will still get (5.15) because you take absolute value before bounding it.
     
  22. camilus the villain with x-ray glasses Registered Senior Member

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    okay. At first I thought we were finding the residue of \(\frac{\xi(s)}{s(s-1)}z^{-s/2}\) which would be the same as finding the residue of \(\frac{1}{2}\pi^{-s/2}\Gamma (s/2) \zeta (s) z^{-s/2}\)
    by the def of xi, and which would indeed be 1/2. But we were finding the residue of \(\frac{2\xi(s)}{s(s-1)}\frac{z^{-s/2}}{2}\) which would be the same as \(\pi^{-s/2}\Gamma (s/2) \zeta (s) z^{-s/2}\) which equals 1.

    http://www.wolframalpha.com/input/?i=residue of pi^(-z/2)*gamma(z/2)*zeta(z) at z=1


    That makes sense. Its pretty odd that they missed that negative sign because it's so trivial, plug in s=1/2 + it into s(s-1) and you get -(t*t+1/4), that simple.
     
  23. CptBork Valued Senior Member

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    I absolutely hate it when a textbook makes mistakes like that, makes you puzzle over it for hours and hours. Sometimes the book is wrong, sometimes you're wrong. I have the same problem when I spot a caveat to something a textbook says, puzzle over it for ages trying to understand, and then find that a few pages later the book also discusses the same caveat- too much delay there.
     

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