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#11726 2013-07-11 17:50:49

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

That's great, how did you do it? What contour did you use?

#11727 2013-07-11 18:00:36

bobbym

Online

Re: Linear Interpolation FP1 Formula

Hi;

http://140.177.205.23/ContourIntegration.html

For these type they always use a half circle.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11728 2013-07-11 18:08:36

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

I tried that first but got the wrong answer (2π/3). The function is defined only for x ≥ 0 and f(x) ≥ 0 (i.e. the first quadrant)...

#11729 2013-07-11 18:10:06

bobbym

Online

Re: Linear Interpolation FP1 Formula

That is because you picked the right pole but forgot that you are using only half a circle. You would get half that answer.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11730 2013-07-11 18:17:36

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

I'm not sure I follow that, though... the reason you can normally halve your answer is because even functions have this property (if f(x) is even):

that's the reason I halved my answer, not because it is half a circle. In this example, wouldn't you use a quarter of a circle too? f(x) in this instance is neither odd nor even.

#11731 2013-07-11 18:23:56

bobbym

Online

Re: Linear Interpolation FP1 Formula

It is not where the function is defined but the limits of integration. When you go from -∞ to ∞ you multiply the sum of the residues that apply by 2 π i. Here we are going from 0 to ∞, so it is π i.

I hope you understand that when I say something like that I am just using you for a sounding board. I am learning this topic too. I might be wrong with my explanations of things but not with the answers.

I am watching a nice pair of videos that cover this in a basic way. I do not know if they will answer all our questions but would you like to watch them too?

I watched the videos, they are at

See you in a bit, need some rest.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11732 2013-07-11 22:48:48

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

Thanks for the videos -- I have watched them before although I do think it is missing something pretty important. The ML inequality to bound the integral over a path is essential I think, because they do not always tend to zero.

#11733 2013-07-11 22:56:33

bobbym

Online

Re: Linear Interpolation FP1 Formula

Using the drawing I chose the path (-1,0) to (1,0) back to (-1,0). That passes through the only pole, that is why I chose 1 / 2 the circle.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11734 2013-07-13 09:41:24

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

If you mean the right half of the circle, I do not think that is correct. Why are you only integrating from 0 to 1 -- how does it relate to the original integral? z = 0 is also a branch point, yet your contour passes through it (if you used half a circle). This is not allowed.

I think I have a solution that uses a keyhole contour. I will post it tomorrow -- on my iPod at the moment.

#11735 2013-07-13 10:21:52

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

Okay, I am really tired but I just couldn't sleep without writing the solution in my notebook. I can't believe it worked (sort of). I ended up getting -pi/3 (even though it should be positive)... I'll sort that out tomorrow though. You need a keyhole contour in four parts, I'll explain tomorrow morning.

#11736 2013-07-13 10:28:50

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

Just re-did the residue calculations and now there is no negative in the final answer, which is great! I think I am getting the hang of it.

#11737 2013-07-13 11:50:15

bobbym

Online

Re: Linear Interpolation FP1 Formula

Hi;

Okay, I will see you then.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11738 2013-07-13 18:26:06

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

First, observe this diagram (remove spaces):

o i 4 2 . t i n y p i c . c o m / 1 4 j n m o g . j p g

Call this curve C, traversed along ABDEFGHJ. The smaller circle (the keyhole) is of radius r, the larger circle is of radius R. In other words:

We have a branch point at z = 0 (because of the square root) and the poles are at z = eiπ, z = eiπ/3 and z = e-iπ/3. All of these lie in our contour. Summing the residues at these points and multiplying by 2iπ gives 2π/3.

So now we have

Note that:

Using the substitution z = Re in the second integral yields:

For the third integral, we replace z with ze2iπ, as we have gone through a rotation of 2π to get to this point:

For the fourth integral, we use the same substitution as we did for the second integral, except with z = re:

Putting it all together:

Now, letting r -> 0 and R -> ∞, the second and fourth integrals vanish, and we have:

which gives

#11739 2013-07-13 18:31:54

bobbym

Online

Re: Linear Interpolation FP1 Formula

Hi;

Okay, thanks. Any quick way on how you got that branch cut?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11740 2013-07-13 18:51:19

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

When we plot y2 = x in the real plane, we plot two functions: y = √x and y = -√x, each single-valued.

If we consider w = z1/2, and let z = Rei(θ + 2kπ), then w = Rei(θ + 2kπ)/2. Letting k = 0 and k = 1 gives w1 = R1/2eiθ/2 and w2 = -R1/2eiθ/2, respectively. This corresponds to our branch cut.

#11741 2013-07-13 19:06:50

bobbym

Online

Re: Linear Interpolation FP1 Formula

Okay I need to work on this keyhole one.

I have a book problem that uses your method and another method. Would you like to see the other method?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11742 2013-07-13 19:10:07

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

Sure -- I am guessing the other method uses the Cauchy integral formula?

#11743 2013-07-13 19:30:52

bobbym

Online

Re: Linear Interpolation FP1 Formula

Not exactly, I found this one.

Make the substitution x = x^2 and you get:

Taking advantage that the integrand is an even function:

The integral can be evaluated along the contour C along the real axis from  -R to R , then circle back, counterclockwise, along a circle in the upper-half-plane back to -R.

The poles are simple:

You only need the ones that are positive on the imaginary axis:

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11744 2013-07-13 19:41:37

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

Thanks, that is a better way of doing it -- although it is nice to see both ways, as if there is an instance where you can't use a substitution it is helpful to practice on simpler problems. In fact, if you let u = x3 the problem reduces to a simple arctan integral (which can then be calculated via trig substitution or a simple semicircular contour).

#11745 2013-07-13 19:49:53

bobbym

Online

Re: Linear Interpolation FP1 Formula

Hi;

Yes, I thank you for making some of those concepts a little clearer to me.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11746 2013-07-13 20:26:50

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

No problem -- have you worked through the problems in Spiegel's book? I am doing those now, the trig ones look interesting.

#11747 2013-07-13 20:32:34

bobbym

Online

Re: Linear Interpolation FP1 Formula

No, I have not because I am working on two other problems that are not related to complex analysis but are a complete mystery to me. Also, I must look at your keyhole method and try to understand it. Lastly, I am reading a book on complex analysis. It is tough for me but I am hoping to pick up a bit here and there.

If you have some problem just post it and I will work on that with you.

But first I am going to get some sleep as soon as I answer this last guy's questions.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11748 2013-07-13 21:37:59

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

Okay, see you later.

#11749 2013-07-13 23:35:51

bobbym

Online

Re: Linear Interpolation FP1 Formula

Hope you are having no problems. See you later.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#11750 2013-07-14 03:10:38

zetafunc.
Guest

Re: Linear Interpolation FP1 Formula

I looked at my solution again and the sum of the residues is actually i/3, so my contour actually ends up yielding -π/3, not π/3... why is this?