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You are not logged in. #11726 20130711 17:50:49
Re: Linear Interpolation FP1 FormulaThat's great, how did you do it? What contour did you use? #11727 20130711 18:00:36
Re: Linear Interpolation FP1 FormulaHi; 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 20130711 18:08:36
Re: Linear Interpolation FP1 FormulaI 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 20130711 18:10:06
Re: Linear Interpolation FP1 FormulaThat 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 20130711 18:17:36
Re: Linear Interpolation FP1 FormulaI'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 20130711 18:23:56
Re: Linear Interpolation FP1 FormulaIt 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. 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 20130711 22:48:48
Re: Linear Interpolation FP1 FormulaThanks 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 20130711 22:56:33
Re: Linear Interpolation FP1 FormulaUsing 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 20130713 09:41:24
Re: Linear Interpolation FP1 FormulaIf 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. #11735 20130713 10:21:52
Re: Linear Interpolation FP1 FormulaOkay, 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 20130713 10:28:50
Re: Linear Interpolation FP1 FormulaJust redid 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 20130713 11:50:15
Re: Linear Interpolation FP1 FormulaHi; 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 20130713 18:26:06
Re: Linear Interpolation FP1 FormulaFirst, observe this diagram (remove spaces): We have a branch point at z = 0 (because of the square root) and the poles are at z = e^{iπ}, z = e^{iπ/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^{iθ} in the second integral yields: For the third integral, we replace z with ze^{2iπ}, 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^{iθ}: Putting it all together: Now, letting r > 0 and R > ∞, the second and fourth integrals vanish, and we have: which gives #11739 20130713 18:31:54
Re: Linear Interpolation FP1 FormulaHi; 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 20130713 18:51:19
Re: Linear Interpolation FP1 FormulaWhen we plot y^{2} = x in the real plane, we plot two functions: y = √x and y = √x, each singlevalued. #11741 20130713 19:06:50
Re: Linear Interpolation FP1 FormulaOkay I need to work on this keyhole one. 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 20130713 19:10:07
Re: Linear Interpolation FP1 FormulaSure  I am guessing the other method uses the Cauchy integral formula? #11743 20130713 19:30:52
Re: Linear Interpolation FP1 FormulaNot 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 upperhalfplane 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 20130713 19:41:37
Re: Linear Interpolation FP1 FormulaThanks, 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 = x^{3} the problem reduces to a simple arctan integral (which can then be calculated via trig substitution or a simple semicircular contour). #11745 20130713 19:49:53
Re: Linear Interpolation FP1 FormulaHi; 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 20130713 20:26:50
Re: Linear Interpolation FP1 FormulaNo problem  have you worked through the problems in Spiegel's book? I am doing those now, the trig ones look interesting. #11747 20130713 20:32:34
Re: Linear Interpolation FP1 FormulaNo, 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. 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 20130713 21:37:59#11749 20130713 23:35:51
Re: Linear Interpolation FP1 FormulaHope 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 20130714 03:10:38
Re: Linear Interpolation FP1 FormulaI 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? 