Lecture 06

[Music] welcome back uh so let's recall we have defined the 4E coefficient for a reman integrable function which is 2 pi periodic and we find F hat of n which is equal to 1 by 2 piun 0 to 2 piun f of x e^ minus of I INX DX and we have computed for a coefficient of various functions and the four year series is defined to be summation Over N varies over z f hat of n e to the power I n x and uh we have no AI knowledge that given any reman integrable function the series is going to converge but what we have seen is that if F belongs to CK recall that CK is the consist of 2 pi periodic functions which K derivative exist and K derivative is continuous in that case what we have seen is that F hat of n is less or equal to some constants which may depend on F but not on N mod n to the power K so therefore for K bigger than 1 we have the foror series of this functions they converge but till now we do not know whether the conver to F or not okay so in order to talk about the forer uh series uh converges or not so first question what one would like to ask from our previous knowledge of convergence of series we know that if if CN is a sequence of complex number if the ser is converges then mod CN this goes to zero as mod than goes to Infinity this this is a necessary condition for a series to converge it's not sufficient all of us we know so now that means it makes sense to ask if CN is a f of n e^ i NX then mod CN is mod fat of n does mod fat of n converges to zero as mod n goes to Infinity what we have seen that only easily by triangle inequality that this is less or equal to some constant which is depend on it namely this constant is uh 1 by 2 pi integral 0 to 2 pi mod of f ofx DX now this quantity is going to come again and again so it is but let give a symbol to this let us give it the symbol this which I will be calling it that Norm of f okay to answer this question we have the reman Lebec LMA which states that let f b a 2 pi periodic reman integrable function then converges to zero as mod n goes to Infinity so at least what we can see from this result which is a celebrated result that the necessary condition for the series to converge is there okay so to write the proof for this so let's write down F hat of n this by definition is 1 by 2 pi then 0 to 2 pi f of x e the powerus of i n of X [Music] DX now this I can write 1 by 2 piun and I multiply a negative sign 0 to 2 piun f of x e^ minus of i n x into e to the power minus of I Pi DX because e the^ minus of I Pi is cos Pi which is -1 and -1 is here so this is - 1 by 2 piun now I this is 0 2 pi this is f of x e to the power minus i n uh x + 5 by n DX I will take n not equal to0 because we are interested for large value of n so we can assume this now if I make a change of variable then this is going to be minus of 1x 2 piun this is becomes F of xus of Y by n and then this is minus of I INX DX and the limit which is going to be whatever it is if we make a change of variable that X going to x - of 1 by n this is uh minus of 1 by n to 2 pi - 1 by by n but nevertheless both the function f and e the^ i INX is 2 pi periodic so it doesn't matter which 2 pi length interval I am integrating so this again I will keep it as 2 pi now this is also our fat of n now if I add this two then I am going to get 2 fat of n which is equal to 1 by 2 piun 0 to 2 pi F of xus f of x - of n e power minus of I NX DX now for n large xus of 1 by n it converges to X now if f is continuous then f of x - 1 by n will Converse to f of x so now if it is a continuous function it is uniformly continuous because of the compact interval 0 to 2 pi therefore for f continuous what we have got is that 2 F hat of n this converges to zero as mod n converges to Infinity so for f continuous function we have got our remen LEC LMA now how to get for arbitrary remen Integra function recall that we have seen this for f if we are using r or 0 to 2 pi this 2 pi periodic REM integrable function then there exist a sequence FK of continuous function such that integral 0 to 2 pi mod of f k of x minus of FX DX this goes to zero as K goes to Infinity now for FKS this sequence they are continuous each FK is continuous therefore REM and LEC LMA holds now look at FK hat of n minus of f hat hat of n this what we have seen is by triangle inequality is less equal to 1 by 2 pi 0 to 2 pi mod of f k of xus of f of x DX now this goes to zero this will imply now F hat of n because if K hat of n all of them they go to zero therefore F hat of n also goes to zero as mod n goes to Infinity therefore what we have got is the remen LEC LMA for remen integrable function with this approximation result okay so good so now we have at least the necessary condition so now we need to see that whether uh it it is true that for every remen integrable function uh the for year series converges or not in order to ask this question so let's understand what is the meaning of convergence the the convergence of this series means we Define the partial sum SN f of x this is equal to summation / minus of n to n f hat of n e power I INX now let us try to look closely how this partial sum looks like now this is equal to summation over minus of n to n 1 by 2 piun 0 to 2 pi F of Y e^ minus of i n y Dy e the power I in X this is equal to 1x 2 pi Y is scalar we can take it out now now this is a finite sum so we can interchange the sum and the integral by the linearity of the integral we can write this 0 to 2 pi F of y d uh this is summation minus N2 n e to the power minus of i n or e the^ i n x - y then this is dy now this is a function so let me denote d n of x this is equal to summation - n to n e power i n x therefore this is equal to 1 by 2 piun symbolically I can write this as F of y d n of x - of y d y this DN is called derle kernel now because to make significance of this let me put n d legel so this suggest the simple observation suggests us that in order to understand the conver convergence of the for year series we need to understand the behavior of DNX now let us see what exactly this function DNX or the dich kernel looks like so simple derivation will say DN of X this is equal to now if I take e^ minus of I NX outside then this becomes the first term is 1 + e^ IX and this goes up to e to the power I 2 N X now this is a geometric progression so all of us we know how to add if we are in a geometric progression the sum is going to be 1 - E power I this is 2 n + 1 of X ided by 1 - of I uh now this is = to e^ minus of i n x - e^ i n + 1 x ided by 1 - e^ IX now in order to have the Symmetry if I am taking out e to the power i x by 2 then this one is e the power- of i n + 12 of x - E power i n + 12 of X now similarly from the bottom if I take E power IX by 2 I get e^ minus of IX by 2 - e^ iix by 2 so this two gets cancelled now this is the symmetry so this is going to be - 2 I sin n + 12 x ided - 2 I sin x by 2 so therefore this is sin n + half X by sin x by 2 so this is what one form of the dich kernel will look like the N derel will look like now to see certain properties so what happens so now limit X goes to 0o s n +/ x by sin x by 2 now this is we know that sin x by X limit X goes to Z is 1 by doing that what we will get is so this is equal to if I write it as X goes to zero sin n + half of x / n + half X into n + half X and then X by 2 / sin x by 2 and then 1 1 / X by 2 so this term goes to one this term goes to 1 x x get cancel so this is n 2 n + 1 by 2 into 2 which is = 2 n + 1 so this means DN of 0 this is 2 n + 1 now let us try to draw let at least d one of X how this is going to look like now this is at zero this is going to take the value three now this is my Pi this is my - of Pi so for this so sin 1 + this has to be zero in order to have D1 X is zero this will imply that 3 by 2x is equal to plus or minus Pi this will will imply X is = + or - 2 piun by 3 so now this is 2 piun by 3 and then this is - 2 piun by 3 and uh D1 of Pi is a negative number so this is going to be then if I draw this graph this is something will look like this this is our D1 now look at for D2 so D2 at xal to 0 this is going to take the value of uh 2 n + 1 that is means five so let me take this as as five here D2 now let me find out what are the zeros of D2 can have so now basically here in this case 5 by [Music] 2 x this is equal to sin 5x 2x is equal to0 so therefore this is going to be equal to n Pi now now what are the kind of n so now this is X has to be 2 N by 5 piun now X we are considering on minus piun to Pi so 2 N by 5 piun has to be less than pi and which has to be greater than minus of Pi therefore this will tell us that uh n is equal to 1 2 and -1 - 2 so therefore for n = 1 this is 2x 5 this is 2x 5 Nal to 2 is 4 Pi by so similarly this is - 2x 5 piun and then this is - 4 < by 5 now here this is zero so D2 is going to look like this so similarly if we go for any arbitrary end then we can find between minus pi to Pi we are going to have 2 and + 1 1 to 2 N zeros so therefore for arbitary n this is something which is going to look like so now on this side so this is going to be something like 2 n + 1 so this is going to oscillate like this something like this so now as we can see the when we are taking in larger and larger then the dich kernel is going to pick up at zero and in the rest away from the zero is going to vanish and oscillates too much because there will number of zeros will be 2 N in that interval and this is going to happen that there will be too many times it is going to cross the x- axis and and it is going to pick take more value large value if we are considering n to be large okay so now what we need to see is that we need to get the SN of f of x minus of f of x that is our Target we want to show that this modulus is less than Epsilon giv an Epsilon positive for large n this has to be small now this is equal to let let's give an attempt and to understand how we are going to attack this problem this is equal to integral f of x minus of Y DN of y d y and minus of F ofx this 2 pi should be inside the modulus this is what is as for our definition so now this is integral so we would like to have one integral over here so now observe integral 1 by 2 piun - < 2 piun d n of y d y I this is equal 1 by 2 piun - > 2 piun summation of - n to n e the power i n y Dy again this is a finite sum we take this integral uh interchange the integral and the sum so what we are going to get this is - PK to piun e to the power i n y Dy now what we know is that this is the fundamental observation for this course is that - PK to Pi e^ I NX DX this as you can see is that if n this is equal to 1 if n is equal to 0 now if n not equal to Zer then e the power i n by i n minus pi to Pi and they are going to have the same value this is a 2 pi periodic function at the end point so this is going to be zero so this is very very important relationship easy but the fundamental concept of the forer analysis so in this case if we are coming back to the computation of the integral of D and Y then what we get here this is going to survive only when n is equal to 0 therefore this is going to be equal to 1 because because for rest of the n n not equal to 0 this is going to be0 except for n equal to 0 and we get that to be one therefore when we write this SN f of x minus of f of x modulus I can write this as 1X 2 piun and then mod of- pi to piun F of x- of y d n y and in the term F I can write 1 by 2 pi - < 2 pi f of x d n y because 1X 2 pi integral minus pi to Pi D and Y is 1 so this is equal to 1 by 2 pi f of x - of Yus of f of x then this is DN y Dy Now by pushing the modulus inside what we will be getting that this is less or equal to mod of this and mod of D and Y now let's say that f is a continuous function then for small y f of x - of y - FX is going to be small so now for and then we are left with absolute value of integral of DN of Y for small small change in if Y is small that is what we'll see in the next lecture that by taking breaking this integral into small y near zero and breaking and away from zero what we are going to get thank you [Music] [Music] [Music]