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You are not logged in. #1 20121030 09:23:34
Fourier Series for e^xI've run into a problem with deriving the Fourier series for e^x. putting it all together: But, clearly this can't be valid, since plugging n = 1 into the RHS gives an undefined result for the first term. What can I do about this? #2 20121030 09:50:54
Re: Fourier Series for e^xHi; 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. #4 20121030 10:07:06
Re: Fourier Series for e^xI did not say that. I will have to look at more. Just do not expect it to be a very good approximation. 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. #6 20121030 10:16:34
Re: Fourier Series for e^xIn some interval you can possibly fit a non periodic function with a Fourier series. Fourier series are notorious for having problems at the endpoints due to the Runge effect just like polynomial fits. 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. #7 20121030 10:17:14
Re: Fourier Series for e^xIt has a period of 2*i*pi not 2*pi. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #8 20121030 10:21:32
Re: Fourier Series for e^xYes, he may have used the wrong equations for the coefficients. This yields: Plotting that and e^x we get the typical Fourier fit of a function. Notice how the purple line ( fourier ) hugs e^x. 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. #9 20121030 21:36:46
Re: Fourier Series for e^xSo the period is 2iπ since you're thinking about it in the complex plane? (2iπ being one revolution) #11 20121030 21:38:44
Re: Fourier Series for e^xHi; 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. #12 20121030 21:40:29
Re: Fourier Series for e^xOh, okay. I thought it was 2π since a lot of Fourier series tend to be defined that way (f(x) = f(x + 2π)). #13 20121030 21:43:31
Re: Fourier Series for e^xI would say that is a fairly useless fit. The good point about Fourier fits is that they are orthogonal and least squares at the same time. 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. #14 20121030 21:47:58
Re: Fourier Series for e^xWhere is the orthogonality in the diagram? Also, do you think there are any uses for this Fourier series? For example, finding the Fourier series for x^4 can get you the sum of 1/(n^4) (integer n). #15 20121030 21:49:48
Re: Fourier Series for e^xHi; 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. #16 20121030 21:54:55
Re: Fourier Series for e^xWhat do you mean by the basis? Are you referring to the peaks/troughs of the Fourier series, and the curve (e^x) it is approximating? And, why is this orthogonality useful? #17 20121030 21:57:37
Re: Fourier Series for e^xTo understand it as mathematicians do you have to understand basis vectors. 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. #18 20121030 22:00:42
Re: Fourier Series for e^xCould you tell me the geometric way? Maybe that could then help me understand basis vectors. #19 20121030 22:09:36
Re: Fourier Series for e^xYou already know about something about basis vectors but I will draw some pictures that will make it clear to you. 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. #21 20121030 22:29:01
Re: Fourier Series for e^xFirst some background. When I frst met you in here you wanted to end up at Cern. You might find when you get there that they are doing a completely different type of mathematics than you were taught. 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. #23 20121030 22:39:43
Re: Fourier Series for e^xIt is okay for now, you do not need to for us to get back on track. 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. #24 20121030 22:48:47
Re: Fourier Series for e^xHmm. So, AB, and CD, are orthogonal. But... I don't understand what's going on on the righthand side. So you are saying that EF and GH are orthogonal (as you showed with AB and CD) but the bit about the line joining EG and FH has confused me... #25 20121030 22:55:00
Re: Fourier Series for e^xForget orthogonality for a second. It is easy for you to eyeball the intersection of AB and CD. It is well defined.It is a tiny point. GH is on top of EF and therefore it is impossible to pinpoint an intersection. There are zillions of them. The third one is the interesting case, it is what we call nearly singular. It is close to being on top but it is not. The circle shows how wide the possible point of intersection is. It is hard to pick it out by eye. Do you follow? 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. 