TheFourierTransformAndItsApplications-Lecture11 Instructor (Brad Osgood):I wonder how long I can keep this up. All right. So the first thing – any questions? Any comments? All right. The first thing I want to do is fix a little mistake I made at the end of last time. I was upset about it for days afterwards, so I just want to correct it. This was [inaudible] central limit theorem, which I hope you will agree was a real treat. I mean, it’s just an amazing fact, and the way that it brings in convolution, the [inaudible] transform of the galaxy and so on is really, I think, a real highlight of the course to see those things emerge and to – the whole spooky nature of repeated convolutions approaching the galaxy, I think, is just an amazing thing. But I did screw it up at the end, so I just wanted to correct that. Everything was fine except for one final formula that I wrote, so I’m not going to redo it. I just want to point out where I wrote the wrong thing. It’s correct in the notes. I just wrote the wrong thing on the board. This was the setup – so X1 through XN are, as they say in the biz, independent and identically distributed random variables. You don’t have to write this down. I’m just recalling what the notation was. And P of little X is the distribution that goes along with all of them because they have the same distribution. There’s a single function, P of X, which describes how each one of them is distributed. So it’s a distribution for each. And then I formed P of N of X was the distribution for the sum scaled by square root of N. So it’s the average – excuse me. There was some assumption we made on the Xs on normalation, that is. We assume they had mean zero and we assume they had standard deviation or variance one, and then if you form the sum, the mean of the sum is zero but the standard deviation or the variant center deviation of the sum is the square root of N, so it’s scaled by the square root of N. You divide by square root of N to have this sum, SN, I called it, have mean zero and standard deviation one, and P of N of X was the distribution for this. What we found was that the Fourier transform – here we go. The Fourier Transform of P of N of S was the Fourier transform of P and S over the square root of N to the Nth power. And then the trick of the analysis – the way the proof worked was to compute the Fourier transform of P and S over the square root of N just using the definition of Fourier transform that uses a very sneaky thing of looking at the Taylor series of expansion with the complex exponential integrating the terms and so on, and it was really quite clever. What I found was that the Fourier transform of P at S over the square root of N was approximately if N is large was one minus two Pi squared S squared over N. That was fine. That was fine. And then for some reason, I raised this to the Nth power, so the Fourier transform P at S over the square root of N to the N was then approximately this thing raised to the Nth one minus two Pi squared S squared over N to the N and then inexplicably I had the wrong approximation for this in terms of an exponential in terms of the power of E. That is this one minus two Pi squared S squared over N is approximately E to the minus two Pi squared S squared. I’m going to look that up again to make sure I have it right thistime. Sure enough. All right. Then from there, and for some reason I wrote two Pi squared over S squared last time, but it’s two Pi squared times S squared. That’s an elementary to the N. That’s an elementary fact from calculus that you learned probably a long time ago when you were first learning about properties of the exponential function. And then from there you take the inverse Fourier transform and you get the result. FT of this inverse Fourier transform of this galaxy and E to the minus two Pi squared S squared to get the result, which I am afraid to write down because I’m afraid I’m just going to make one more mistake, so I’m not gonna do it to get the final result of the central limit theorem. There. I feel better now that I’ve corrected it. I’m not sure everybody else here feels better, but I feel better. Okay. Now to get on to the day’s events. Brothers and sisters, today we have to come to a reckoning with our failings. Today we have to confront things we have not confronted but must. Now first, the good news, and there is always good news. The good news is the formulas we have derived, the applications we have done are all correct. So relax. Nothing that we have done, strictly speaking, is wrong. However, much has been left unsaid and today, we have to say it. We have to confront the convergence of intervals. It is a sin to consider integrals that do not convert, and I’m afraid there are times when we have sinned. But in the end, nothing that we have done is wrong, so relax. But much has been left unsaid. So it is time now. Are you ready? Are you ready? Okay. Now we need to have a better, more robust definition of the Fourier transform in order to consider – I think I gotta stop this. There’s only so much of this I can do. We need a more robust understanding of the Fourier transform. We need a more robust definition of the Fourier transform that will allow us, verily, to work with the signals that society needs to function – sines, cosines, even the function one – constant functions. And the Fourier transform as we have defined it will not do the trick. More robust definition, and that’s really what it amounts to – definition of the Fourier transform – to deal with common signals, ones for which the classical definition does not work or will not – is not adequate. The issue is exactly the convergence of the integral or if not the convergence of the integral for the function, applying Fourier inversion. The classical definition, let me just say, will not do. There are two issues, really. One is the definition of the Fourier transform itself – the convergence of the integral defining the Fourier transform – when I say convergence of the integral, I mean just some way of evaluating, some way of making the integral make sense – convergence of the integral defining the Fourier transform. And the second and just as important, certainly for applications – it doesn’t do you any good to take the Fourier …