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Post by mondo on Jun 15, 2023 20:47:42 GMT
There is a lot of small gaps in this section that I can't grasp:
1. At the bottom of page 436 is an example usage of residues theorem on a real integral of $ \int_{-\infty}^{\infty}(x^2+1)^{-2}$. So in the second to last integral (which at this time is already adjusted to a complex variable $z$) I don't understand why, after the second equal sign we do $\frac{1}{(z+1)^2}$ only? What about the whole denominator $(z+1)^2(z-i)^2$, why can we neglect it?
2. In the same chapter I don't understand the next example on page 437 - in the middle of the page "Consider the elementary result.. Differentiating this with respect to $a$ yields" first of all what is "this"? From the derivative calculation it looks like only the integrand of above integral got differentiated. Second, why do we do this and why in respect to $a$?
3. For the next example at the same page I wonder why on the figure [4b] and later in the text, at the bottom author says "$dz = izd\theta$" to me the length of $dz$ is just the arc length - $z*d\theta$ so we extra multiply by $i$ here?
Thank you.
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Post by Admin on Jun 16, 2023 15:09:46 GMT
Mondo
1. You must be reading these pages very fast as you seem not to see the explanations. Try reading subsection 3 on page 436 again. You have also misquoted the function as $(z+1)^2(z-i)^2$ when it should be $(z+i)^2(z-i)^2$. Nothing has been neglected here. Everything is still there. Why do you think something has been neglected?
2. Again I ask you to read page 437 again as everything is explained in the book.
3. I'm sure we have discussed this before. $dz$ is not a length it is a complex number or vector. Also $z*d\theta$ is not the arc length it is again a complex number or vector. You do not seem to understand figure 4 and you do not seem to understand what complex numbers are because you are not distinguishing between real numbers and complex numbers. Since $C$ in figure 4b is the unit circle then the arc length is $rd\theta=d\theta$ since $r=1$. Treating a complex number as a real number is not a trivial mistake, but a serious mistake as it will lead you to the wrong conclusions.
A general point
Why do you never do any of the exercises in the book at the end of each chapter? In my opinion this is the only way to be sure that you have understood the chapter and are ready to progress. You cannot expect to understand in a very short time what it sometimes took mathematicians hundreds if not thousands of years to understand.
Maybe you have a good reason for not doing the exercises, but in my opinion you are just making your life harder than it needs to be.
Please keep the questions coming.
Vasco
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Post by mondo on Jun 17, 2023 6:27:33 GMT
Thanks for the response Vasco.
1. When I said "neglected" I meant the fact that
$f(z) = \frac{1}{(z+i)^2(z-i)^2}$ however in the calculation of residues
Only part of this denomonator is taken, namely $(z+i)^2$ what happened to $(z-i)^2$?
There are two more aspects to that. First author says, and I agree that there are two singularities $z = \pm i$ but in residue calculation only one $z=i$ is taken.
Second, a residue only makes sense for a power of $-1$ as only this one gives non zero value integral. However in this example the integral is calculated for a power of $-2$ and yet the integral is non zero.
3. Yes, I did say several things wrong. I meant to say $dz$ is calculated based on the arc length generated by the vector $z$ and an angle $d\theta$. But your commet was very helpfull, I think I got my mistake.
As to your general point, I agree, doing exercises is what I should do more.
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Post by mondo on Jun 17, 2023 6:53:12 GMT
For the point #1, looks like this is using a shortcut method for a residue calculation derived in (10). I will study it again.
For the point #2, you said it is all explained, but can you give me some hints as to why this method works? Also, at the end author got $\pi/a$ and said "substituting $a =1$ confirms residue calculation". But this is not true. In the residue calculation the result is $\pi/2$ not $\pi$.
So I would need to know why he claims a derivative of the integrant in respect to the parameter is "easier" and why this method works, wby the derivative is needed?
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Post by Admin on Jun 17, 2023 9:15:04 GMT
Mondo
1. If we accept (10) on page 435 then in the example at the bottom of page 436 where
$\displaystyle f(z)=\frac{1}{(z^2+1)^2}=\frac{1}{(z+i)^2(z-i)^2}$ which has second order poles at $z=\pm i$ we have
$\displaystyle\oint _{L+J}f(z)dz=2\pi i\text{Res}[f(z),i]$
Since here $m=2$ (second order poles) and only one of them ($z=i$) is inside the loop then we have, using (10)
$\displaystyle \text{Res}[f(z),i]=\frac{1}{(1)!}\bigg[\frac{d}{dz}\bigg]^1[(z-i)^2f(z)]\bigg|_{z=i}$
Since $\displaystyle (z-i)^2f(z)=\frac{1}{(z+i)^2}$ we can write this as
$\displaystyle \text{Res}[f(z),i]=\bigg[\frac{d}{dz}\bigg]\bigg[\frac{1}{(z+i)^2}\bigg]\bigg|_{z=i}=\bigg[\frac{-2}{(z+i)^3}\bigg]\bigg|_{z=i}=\frac{-2}{(2i)^3}$
Substituting for $\text{Res}[f(z),i]$ we can see that
$\displaystyle\oint _{L+J}f(z)dz=2\pi i\text{Res}[f,i]=2\pi i\frac{-2}{(2i)^3}=\frac{\pi}{2}$
You can see that this is just a simple calculation using (10).
Vasco
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Post by Admin on Jun 17, 2023 20:34:40 GMT
Mondo
Just made a correction to my post #4 above. On line 3 and on the second to last line I had missed out $f(z)dz$ on the left hand side.
Vasco
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Post by mondo on Jun 18, 2023 6:41:12 GMT
Thanks Vasco, the point #1 is all clear now, I have missed some cancelation that author did on the fly. Point #3 seems to be also clear, but I am lost with the second point. As I said in my reply #3 I have no idea how this differentiation in respect to parameter $a$ helps in this integral evaluation? Plus, after the substitution $a=1$ the result is $\frac{\pi}{1}$ not $\frac{\pi}{2}$ as it should be to say this method gives exact same result.
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Post by Admin on Jun 18, 2023 7:00:04 GMT
Mondo
You have missed the fact that there is a 2 on the LHS which you need to divide by!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Vasco
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Post by Admin on Jun 18, 2023 7:13:13 GMT
Mondo
You make it sound like nothing but if you had done what any reasonable person would have done, which is get some paper (do you know what that is?) and a pencil (do you know what that is?) and worked through the calculation you would have understood this weeks ago. Same goes for $\pi$ and $\pi/2$
Stop being lazy and do some proper maths with paper and pencil!
Vasco
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Post by Admin on Jun 18, 2023 7:26:01 GMT
Mondo
It doesn't just help, it's an alternative method which works in this case because we can use the standard result in the middle of page 437 involving $\tan^{-1}$. It's a method which avoids using contour integration.
Vasco
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Post by Admin on Jun 18, 2023 11:12:51 GMT
Mondo
Integration of real functions is a simple idea but it is not always easy to achieve for specific functions and over many years a lot of different methods have been devised to solve specific integrals.
Contour integration is one of the ways in which we can solve real integrals even though it is based on complex number theory.
Some integrals can be solved using more than one of these techniques and the one shown on page 437 of differentiating with respect to a real parameter sometimes works as an alternative to contour integration.
Integration is an art as well as a science but nowadays most integrals can be solved using a robot (computer).
There is a link to one of these in the Useful Links section of this forum. Try it on this integral.
Vasco
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Post by mondo on Jun 18, 2023 18:03:42 GMT
Mondo You have missed the fact that there is a 2 on the LHS which you need to divide by!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Vasco I think the idea is to find a very similar integrand that we know how to integrate or is elementary and adjust it to be like the original integral we need to solve. You said it as well in your reply Reply #9: If I understand correctly we take $\frac{dx}{x^2 + a^2}$ and calculate it's derivative to get the denominator to the second power, because a derivative of this happens to give that! Is it the case? However, $\frac{d}{da} \frac{dx}{x^2+a^2} = - \frac{2a}{(x^2+a^2)^2}$ - this minus sign is missing in the book, a typo? So if my assumption is correct - the derivative was needed to adjust the integrand to our target one. What should be done if the target integrant has $(x^2 + 1)^3$ in it's denominator instead? Theoretically I can do a second derivative $\frac{d^2}{d^2a} \frac{1}{x^2 + a^2} = \frac{6a^2-2x^2}{(a^2+x^2)^3}$ but now I have much more $2x^2$ term in the nominator and this method won't help anymore right? Mondo You make it sound like nothing but if you had done what any reasonable person would have done, which is get some paper (do you know what that is?) and a pencil (do you know what that is?) and worked through the calculation you would have understood this weeks ago. Same goes for $\pi$ and $\pi/2$ Stop being lazy and do some proper maths with paper and pencil! Vasco Not sure I would call it laziness but I agree, I should have tried to write this equation down myself instead of only staring at it and trying to understand. I am sorry that I irritated so much here! |
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Post by Admin on Jun 18, 2023 18:25:13 GMT
Mondo
Yes
The minus sign is not missing.
When we differentiate with respect to $a$ we have to differentiate both sides of the equation. So when we differentiate the RHS ($\pi/a$) we also get a minus sign which cancels with the one on the LHS.
Yes, this just works in this particular case - that's the point.
Vasco
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Post by Admin on Jun 18, 2023 18:31:27 GMT
Mondo Don't worry about being irritating - it's part of the fun.  Vasco |
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Post by mondo on Jun 19, 2023 3:20:28 GMT
I appreciate the feedback/criticism, it really helps to improve the way I study. And as you said this material is not easy, it's not a novel that we can just read - one needs to get his hands dirty in order to really learn the matter.
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