 So, we continue from where we left off. So, we discussed the least square approximation. What is the least square approximation? If you take a function f of x which is in L 2, you take the Fourier coefficients and it form the trigonometric polynomial which is Sn fx that is the nth partial sum of the Fourier series. It is a trigonometric polynomial of degree at most n. You take another trigonometric polynomial qnx of degree at most n and you compare the differences f minus Sn f and f minus qn and take the norms. Now you will realize that the difference f minus Sn that is less than or equal to the norm f minus qn which means among all possible trigonometric polynomials of degree at most n, the nth partial sum or the Fourier series gives you the best possible approximation, best possible approximation in L 2 norm remember. If you look at other norms, the story is very different. Now we shall use this principle of least square approximations to deduce the Bessel's inequality. So, theorem 19 which you see on the slide, the a naught a and bn have the usual meanings. They are the Fourier coefficients of an L 2 function. Then we have the inequality 2.4 mod a naught squared plus 1 half summation n from 1 to infinity mod a n squared plus mod b n squared less than or equal to the average value 1 upon 2 pi integral minus pi to pi mod fx squared dx. We have already seen that one cos x sin x divided by cos nx sin nx are orthogonal to fx minus the nth partial sum. This is a part of the proof of the least square approximation. So, we conclude that the linear combination of these things in particular Sn fx itself is orthogonal to this difference fx minus Sn fx. So, because fx minus Sn fx is orthogonal to Sn, I can apply the Pythagoras's theorem. So, norm of the squared plus norm of the squared will be equal to the norm squared of the sum. What is the sum of these two things? f and norm squared. So, this is the relation 2.5 we get and we simply knock off this first piece and we conclude that the L 2 norm of Sn f is less than or equal to the L 2 norm of f and that basically is the vessels inequality. Let us see why that is so. Let us compute this norm squared on the left hand side of 2.6 and let us see. We calculate it. It will be integral minus pi to pi mod a naught plus summation j from 1 to n a j cos jx plus b j sin jx squared dx. Just expand this. Expand this you are going to get the terms cos jx sin lx. The integral of that is going to be 0 cos squared jx that integral is going to contribute. Sin squared jx that integral is going to contribute cos jx cos kx k naught equal to j integral is 0 sin jx sin kx j naught equal to k integral is going to be 0. So, only the square terms are going to contribute and the value is going to be exactly 1 2 pi mod a naught squared plus pi times summation j from 1 to n mod a j squared plus mod b j squared. So, we are computed the L 2 norm of the partial sum Sn fx. So, the inequality norm Sn fx squared less than or equal to norm f squared simply translates to this inequality that we see displayed in the slide. Now, simply allow n to go to infinity and we get the result that we have stated. So, the Bessel's inequality has been established. And so, the idea is exactly the same as the idea of the least square approximation namely Pythagoras's identity. Now, the remarkable fact is that equality holds in Bessel's inequality. It turns out that the Bessel's inequality is actually an equality that is theorem 20. Suppose f is an L 2 of minus pi pi, then the average value 1 upon 2 pi integral minus pi 2 pi mod fx squared is exactly equal to mod a naught squared plus 1 half summation n from 1 to infinity mod a n squared plus mod b n squared 2.7 display. Formula 2.7 has a simple physical interpretation. You must think of this 2 pi periodic function f of x as some kind of a signal and the left hand side of 2.7 can be interpreted as the energy of the signal. And the right hand side also gives you the expression for the energy in terms of the Fourier coefficients. How much does the sin nx term contribute and how much does the cos nx term contribute to the total energy and that is displayed in the right hand side of 2.7. Now, proof of 2.7 namely how to make the Bessel's inequality into an equality that hinges upon a technical result. The technical result that we need is an approximation theorem. It is very similar to the Weierstrass approximation theorem. And the analog of the Weierstrass approximation theorem for trigonometric polynomials. The theorem says that trigonometric polynomials are dense in L2. That is the theorem that we need. How to prove that theorem? You can deduce that theorem from Weierstrass approximation theorem. There is a way to prove that result using Weierstrass approximation theorem. Instead, we shall prove it using Feier's theorem. Feier's theorem will be discussed in the next chapter when we discuss Cesareau's summability of Fourier series. As a corollary of Feier's theorem, we will deduce that the trigonometric polynomials are dense in L2. That is very essential to prove this theorem of Parseval. So, now we shall turn to a couple of examples illustrating Parseval formula. After we finish these examples, we will move on to two nice applications of the Parseval formula. The first example is a simple calculation. We start with f of x equal to x squared on minus pi pi, which is an even function. Do a 2 pi periodic extension. The 2 pi periodic extension is Holder continues. It is Lipschitz continues. So, the point wise convergence theorem can be applied. And the Fourier series converges to the given function on the interval minus pi pi. So, what remains is to compute the Fourier coefficients. a0 is easy to compute. It is pi squared by 3. bn's are 0 because the function that I have taken is an even function. And an's, you simply have to do a couple of integration by parts to calculate this integral from 0 to pi. x squared cos nx and the value is exactly 4 times minus 1 to the power n upon n squared. So, I have calculated the an's and the bn's. Now you should apply this to the Parseval's formula 2.7. a0, an's and bn's are known, bn is 0. And now f of x is x squared. So, mod f x squared is x to the power 4. So, I have to integrate x to the power 4 from minus pi to pi. And I have to see what this equality turns into in this special case. It turns into the following equation. The first display on the right hand side. 1 upon 2 pi integral minus pi to pi x to the power 4 dx equal to pi to the power 4 by 9 plus 1 half summation n from 1 to infinity 16 by n to the power 4. After a little simplification, after a little simplification, after computing this integral, you will get zeta 4 will 1 plus 1 upon 2 to the power 4 plus 1 upon 3 to the power 4 plus dot dot dot equal to pi to the power 4 by 9t. This is what you will get, right? Alright, now the next topic is the Bernoulli polynomials and Bernoulli numbers. We are just going to mention these things, not go deep into it. We will go deep into it later if time permits. So, we have computed zeta 2 and zeta 4. Now you might suspect that we can calculate zeta 6, etcetera. Maybe you may suspect that if I take f of x equal to mod x cube and do a two-periodic extension or something like that, you might get zeta 6. You should experiment with these things. You should experiment and see whether you can compute zeta 6. It turns out that the zeta values are known for even arguments. Zeta 2, zeta 4, zeta 6, etcetera are known numbers. What is not known is zeta 3, zeta 5, etcetera. Exact value of these things is not known. It is known that zeta 3 is irrational, for example, after I proved that in 1976. But we are not going to be looking into these number theoretic aspects of the zeta values. Zeta values are even arguments. So, what about zeta 2, zeta 4, zeta 6? How can you compute them in general? They have a closed form expression involving the so-called Bernoulli numbers. So, we must define the Bernoulli numbers first. So, we define them inductively. So, you take B naught of x to be 1, the 0th Bernoulli function and you define the nth Bernoulli function in terms of the n minus first Bernoulli function. Bn prime is n times bn minus 1, for n greater than or equal to 1. Of course, this will only determine bn up to a constant. To fix the value of the constant, we put a normalization condition 0 to 1, bn x dx equal to 0. The exercise is to calculate the first few Bernoulli polynomials. Obviously, these are going to be polynomials. So, calculate the Bernoulli polynomials and the values b and 0 are called the Bernoulli numbers. So, why are these important or interesting? Well, the zeta values for even arguments can be expressed in terms of Bernoulli numbers and that is equation 2.8 that you see in the slide. This was discovered by Euler. James Bernoulli in his hours conject on D in 1713 was able to prove that the sum of the p-th powers or the first n natural numbers can be expressed in terms of the Bernoulli polynomial. Remember that the Bernoulli bkx is a polynomial and so, bp plus 1 n plus 1 is a polynomial in n and so, that is a polynomial expression and James Bernoulli is able to obtain a closed form expression for the sum of the p-th powers of the first n natural numbers p equal to 1, 2, 3, etcetera. You may want to check the result for p equal to 1, 2, 3 using the calculation of the Bernoulli polynomials that you have already done so far. The brothers James and John Bernoulli tried hard to find the value of 1 plus 1 upon 2 square plus 1 upon 3 square plus dot dot dot. They did not succeed. It was discovered by Euler a few decades later. However, James Bernoulli did not live to see the last displayed formula in which the numbers that bear his name feature so prominently. So, here I have given you an exercise on calculating zeta 6. For the first time you are using an odd function pi square x minus x cube is an odd function but this time the a n's will be 0 and the b n's will be non-zero. And calculate this and verify the correctness with equation 2.8. So, these are some exercises that I am leaving it to you. Then now we come to a beautiful geometrical application of the Parseval formula. This is an important variational problem which goes back at least to Descartes. So, what is the problem? In fact it is a theorem of all piecewise smooth closed curves whose perimeter is prescribed the circle encloses the maximum area. Now you see take a thread, take a thread of fixed length and try to create a closed curve with this thread. I could make the closed curve like an oval shaped like an ellipse and I can make the ellipse thinner and thinner. The area will become smaller and smaller. Then experiment with an equilateral triangle and a square and a regular polygons. Try to find the areas of these objects when the perimeter is fixed. And you will probably convince yourself that when the shape is a circle, the area is maximized. You should perform these experiments and then you should convince yourself the correctness of this. This problem is known as Dido's puzzle. There is a very beautiful story behind the Dido's puzzle. I will leave it to you to look at the literature. Maybe you can Google it D-I-D-O Dido's puzzle. Just type Dido's puzzle and you will be able to read the story. It is a very amusing or a very interesting story and I will leave it to you. The dual problem is to fix the area and try to minimize the perimeter. In the earlier case, you fix the perimeter and you maximize the area. The dual problem is fix the area and minimize the perimeter. So what is the dual problem? Of all piecewise closed curves enclosing a fixed area, the circle has the least parameter. The theorem obviously generalizes to higher dimensions. There are no prices for guessing what will be the higher dimensional analog of this problem. But rather than tell you what the higher dimensional analog is, let us see what George Polia had to say about this isoparametric problem. This is a quotation from George Polia's magnificent book, Mathematics and Plausible Reasoning, Princeton University Press 1954 on page 170. With a little knowledge of physics of surface tension, we could learn the isoparametric theorem from a soap bubble. When a child blows soap bubbles, the soap bubbles assume a spherical shape. Why do the soap bubbles assume a spherical shape? Why not cubes? Physics will teach you that the surface tension should be minimized and only a sphere will achieve that. Yet even if you are ignorant of serious physics, we can be led to isoparametric theorem by quite primitive considerations. We can learn it from a cat, such a domestic creature, a beautiful domestic creature like a cat. I think you have seen what a cat does when he prepares himself for sleeping through a cold night. He pulls in his legs, curls up and in short makes his body as spherical as possible. He does so obviously to keep warm. To minimize the heat escaping through the surface of his body, the cat has no intention of decreasing his volume. He tries to decrease his surface area. He solves the problem of a body with a given volume and minimum surface in making himself as spherical as possible. He seems to have some knowledge of the isoparametric theorem. It's a marvelous quotation from Holia. So now let us look at the proof of this theorem using Fourier analysis. So you got a piecewise smooth closed curve with a fixed perimeter and I can choose the arc length parameterizations. I am parameterizing the curve by being to the arc length x of s y of s and s runs from 0 to l and the curve is traced counterclockwise. Then what are the area of the curve? The area enclosed by the curve is integral x dy the line integral or you parameterize and you write it as a Riemann integral 0 to l x dy by ds ds. dy can be written as dy by ds ds and I am going to make a change of variables. So that instead of working from 0 to l I will work from minus pi to pi for obvious reasons. So make a change of variables t equal to 2 pi s upon l minus pi. So t runs through the interval minus pi pi when s runs through the interval 0 to l. Then what are the area formula? The area formula is simply integral minus pi to pi x dy by dt dt right 2.9 that you see. For the perimeter for the perimeter you have dx by dt squared plus dy by dt squared equal to ds by dt the whole squared and ds by dt the whole squared is going to be l squared upon 4 pi. Now which may conveniently be written as 1 upon 2 pi integral minus pi to pi dx by dt the whole squared plus dy by dt the whole squared equal to l squared by 4 pi squared right. You know that ds squared is dx squared plus dy squared from calculus. So in equation 2.10 let us apply the Parseval formula to 2.9 and 2.10 let us apply the Parseval formula and you take the nth Fourier coefficient of x t call it a n and b n and take the Fourier series for y call it c n and d n. For the area integral we will get a equal to pi times summation n from 1 to infinity n into a n d n minus b n c n and for the second we get l squared equal to pi squared summation n from 1 to infinity n squared into a n squared plus b n squared plus c n squared plus d n squared. Now you might ask how to apply the Parseval formula to this. After I finish the proof of the theorem I will explain to you how from a square you can do it for a product that is called polarization and meanwhile you can try to figure it out yourself. Now for the perimeter part it is a straight away sum of 2 squares you can apply the Fourier Parseval formula to the first piece the Parseval formula to the second piece and add them up ok. So, you get this. So, we see l squared minus 4 pi a what is that going to be equal to 2 pi squared times summation n from 1 to infinity n squared a n squared plus b n squared plus c n squared plus d n squared minus 2 n times a n d n minus b n c n just complete the squares and we get this ok. So, what happens is that we get l squared is greater than or equal to 4 pi a l squared is greater than or equal to 4 pi a because everything on the right hand side is non negative. When will equality hold equality will hold when the summons are all 0 all these things must be 0 individually. That means of course when n equal to 1 this part anyway disappears when n equal to 2, 3, 4 onwards this must also be 0 and this must be 0 this must be 0 and for n equal to 1. This goes away this must be 0 and this must be 0 n equal to 1 we get a 1 equal to d 1 b 1 equal to minus c 1 all right. So, for a for n equal to 1 we get a 1 equal to d 1 b 1 equal to minus c 1 from n equal to 2 3 onwards n a n minus d n must minus d n must be 0, n b n plus c n must be 0, c n must be 0, d n must be 0. So, only thing that survives are the n equal to 1 terms and so we can write on x x t equal to a naught plus a 1 cos t plus b 1 sin t, y t equal to c naught minus b 1 cos t plus a 1 sin t which is a circle and the proof is complete. So, I think I will close this session here. I will just briefly recall next time the trick of passing from the Parseval formula for squares to a Parseval formula for a product 1 upon 2 pi integral minus pi 2 pi f of x into g of f x. When you have a product of 2 things what happens to the right hand side? You can probably guess the answer and I like you to think about it ok. I would also like you to think about this particular equation which I am not discussed and equation 2.10. The perimeter equation I would like you to think about these 2 things. We will come back to these things in the next lecture. We will briefly touch upon this and we will go to the next application the proof of the maximum modulus theorem. I think that is a good place to stop this capsule. Thank you very much.