Lec 27 MIT 18.085 Computational Science and Engineering I

lectures 27 and 28 uh which will complete the discussion of wavelets this this idea and and especially in 28 their applications in Signal processing image processing um so this uh outlines the lecture as usual uh I'm in continuous time today um so wavelets are functions up to now uh we've been talking about vectors uh input signals was a sequence of samples which it is in reality and the key was the filters now like maybe a fresh start with the idea of multi-resolution for functions and that will lead us to the key functions which are fee the scaling function and W which is the wavelet so that's our goal but but uh I'll be uh like as I say a fresh start on in continuous time before I start could I just mention what I maybe haven't mentioned the uh course web page for the course 18 which is math number 085 actually it's a full year course 1808 5 and 086 out of the applied math book and then I should also mention uh the the course number for uh the wavelets course which uh probably doesn't start that way so I won't try to get its course uh web page right uh but what I wanted to say was this uh page uh which is maintained by who ever whoever is teaching the course at that semester and uh next fall it'll be me again uh includes quite a few old exams now I don't know who wants old exams but uh uh uh everyone to his own taste so there might be a high demand for old exams and in case there is I found um an old exam that somebody else friend of mine had given when he taught the course and uh he particularly he's a great teacher and he particularly called my attention to the last two problems on his exam so what I did this week was make a copy make copies of that last uh couple of problems on that final exam just thinking you might like to take them home and and uh have fun uh I I don't know the answers myself so I'd be happy to to uh um uh with any discussion actually those particular questions go back to calculus of variations um so that would be like review of a month ago here we are today with low pass and high pass filters but now making a fresh start okay so the here's the idea then of multi-resolution you know what the I mean I don't know that's sort of a created word I suppose but it's a very inw that's seeing seeing a function at different scales and in a way that reflects what our ears do and what our eyes do and what our minds probably do that is if I look at a scene like this I first of all take in the whole picture but then I focus on a person and then on uh face and so on we we just narrow we just change scales okay so the idea in multi-resolution is that I have a space of functions v0 so so these are will be spaces of functions and I'll say what that means again and and let's call the first space v0 and then there's going to be a space a bigger space that contains VZ Z so V1 and then of course after that will come a V2 and so on and that symbol means that this is like the coarser space functions that are that are at the co scale so this is course course scale finer scale even finer scale and so on and the idea is that as we have finer and finer scales we capture more and more of the function so that sort of in the limit we've got all functions so the UN so the union of all these VJs would be the whole Space all functions that we're trying to capture uh and in many applications that's the that's we have to I haven't said what the whole Space is so I better do it uh usually we're thinking about functions with finite energy that's the space that's often called L2 finite energy the energy of course is the integral of the function squared okay should be finite that's the famous Hilbert space that it's the sort of appropriate infinite dimensional version of ordinary two-dimensional three-dimensional four dimensional as the dimension goes up um you have more and more components and this is the space that has infinit many components but otherwise behaves like three-dimensional space it's got angles it's got dot products it's got lengths that's the length squared so on anyway it's the functions that we usually consider and we're building up to those from this space v0 Okay now what's in v0 okay that's actually that's the key Point what's our our space going to be our space is going v0 the the basis is going to be the basis meaning let me not even use it word basis course space is going to be all combination of and that's what of a particular function so this is like this function is a key to everything and F I use T as the variable I could use x and it's shifts so that's that's what's in the space v0 what's in the space V1 well you can guess it'll be all combinations of what will be you maybe just guessing it's like good to anticipate because it connects then what do you think if this is the finer space we're looking at functions on a finer scale and we're this is octave uh multi-resolution so the poers of two come in so the functions that give this space are the same fee but with the finer SC scale shifted by two uh dilated by two refined by two whatever is the good verb and uh the key is that refinement okay so um many many papers about wavelets we'll begin with these assumptions that Mala I think was the first to write down so let me put his name up here Stefan Mala they often what I'm hoping to do is like save you the trouble of going through the assumptions this is essentially what's going on we have a sequence of more and more refined spaces and we have a function with its translates that spans V 0 and then scale two here the scale would be four then the scale would be 8 it would be F of8 T minus and it shifts and and everybody recognizes if I look at F of 2T that squeezes the function by a factor of two so it so uh of course the the the model problem the the easiest example is har where F of T is the Box function and then we said what's v0 we we described v0 last time if if this is the Box function on the interval 01 I heard it described yesterday as the Box Car function so maybe that's also a useful word what's in v0 take that function take all its shifts integer shifts take all their combinations and what have you got step step functions right peie wise constant so peie wise constant on unit intervals on unit intervals and now let me ask about V1 so we draw we drew those pictures just just a constant on every interval but a jump between okay what about V1 V of two what's the graph of f of 2T it's of course the same thing but only going to a half and what's in the space V1 then if I take all combinations of those guys and their shifts that's what I meant all all combinations of this and all its its uh half integer shifts I'll again get peie wise constants but the pieces will be half a half a unit long so this is pie wise constant on on can I say half intervals and you could imagine that then the next one would be peie wise constant on quarter intervals and then any smooth function any whole any function in our whole space like whatever like that could get approximated first of all by something on unit intervals and then I could do better on the half intervals probably bring it down a little stop that one there and then I could do better on quarter intervals and I would approach in the limit as the scale gets finer and finer I'd get the function okay so are you restricting the values of k um to be integers yeah K should be integers these are all integers minus infinity or k equal 0 plus or minus one plus or minus two Etc yeah and and I'm working on the whole line here I'm not uh for for my functions second one k can be half also yes uh no I think I always have to take a look it it's I if if I could write it as two two uh two times that so you could say yeah then this could be 1/2 right let let me just draw a fee of 2T minus one just to be so 2T minus K is what we're allowing and let me draw draw F of 2T and below it which I did and below it let me draw V of 2T minus1 where does that sit just so we see when the that that one at first glance looks like we're shifting at a unit interval but then because it's 2 T minus one where does Fe of 2T minus one sit I'm I'm claiming it sits right here right next to it that it's a shift of a half because you see when T is a half this thing will be zero and the function will be starting and when T is one this will be one and the function will be ending so so yeah so that this this dotted guy here is is this guy yeah so it comes out right yeah yes just combinations uh well well okay that's a fine point or a finite Point uh yes um maybe I should allow infinite combinations but but but keep them keep them with finite energy so the so they tail off at Infinity it's uh yeah it's uh of course our in our focus is sort of on a function on a finite interval but if we want to be in these spaces and get all these functions on the whole line then probably I better let Let There Be infinitely many combinations infinitely many uh terms in the combination was another yes ah you have hit on the central question that's it that's the key but so the the this looked a little innocent here this saying v0 should be contained in V1 but it it was it's the key to everything because it means that this first guy in particular fft which is our like Champion member of V Zer because all the others are just shifts of it this first guy is also in V1 but the whole Space is so in particular are are are in person is f of T is is in V1 and now if it's in V1 it means it's supposed to be a combination of these guys so that's the key so that v0 and V1 leads to it leads immediately to the fact that F of T is supposed to be in V1 and what's in V1 combinations with some coefficients and I'm going to call them h of K of the basis functions for V1 and so that that I'm I'm uh jumping the gun you could say or leading the witness or whatever by uh using the uh notation H for the coefficients in fact I'll you going go even further and call them H Zer of K so you have no doubt that what those so the There's the link between this setup of a bunch of spaces with a basis and the previous discussion of filters and uh particularly H zeros so the so the answer was definitely yes that in this this is you could say the classical wavelet framework now in 2000 2001 in the new century there's like you might try to you might find this somewhat uh restrictive and you might be trying to create wavelets on spheres because you wanted to represent the GE potential as a combination of wavelets or wavelets on surfaces because you work for Pixar and they're modeling uh say in Toy Story or other uh I mean that's become a big application now of computerated geometry and and uh uh on and surfaces are involved but that's like second generation this is the place to to start and this is the equation to start with this uh it says that the function our function f of T is a combination of f of 2T minus K and of course you saw it happen here and the coefficients were one and one the coefficients were since the full box is OB which is somewhere here is obviously one of this half box box plus one of this half box so the coefficients are one and one in that particular example of course that's the easiest example so th this Fe of 2T then Fe of t i don't I can erase it and and and put the equation because we remember it's the Box F of T is then F of 2T plus v of 2T minus one in that example our example so the coefficients would be one and one oh and if we wanted to stick to our earlier convention that the coefficients add up to one well one and one add up to two so we re we I'll I left space here to insert a two so that I could stay with my convention that the coefficients uh low pass coefficients add to one that's convention and not everybody does that okay so the beauty is the beauty is that if if we have a function with which solves this equation with some numbers h0 that's that would be called This is often called the refinement equation this is the ref refinement equation or the book calls it the dilation equation so both words in use and that function so most functions won't satisfy such an equation most functions have fight well oh I I think I left last time with the question of the Hat function does it solve such an equation and I think I probably implied with uh my uh the way I said it that it did so let's get that example up there up here two the second case would be the Hat function which is on an interval of length two which suggests that I'm going to need three coefficients you'll see that uh the number of coefficients is matches the number of uh Z of numbers zero one and two so three coefficients and I think I asked what they were how could I express this hat let me blacken it let me yellow it in how could I express this hat as a combination of how much of of of of a half of a hat half of a hat shifted and half of a hat shifted once more do do you see what that I could get out of those three small hats by combining them with the right coefficients I could get the big hat what would be the coefficients then so for this one this is then the Hat example the Hat F of T is what combination of small hats one two one one two one is exactly right one of this first guy two of the second guy which will reach up that far and then one of the 3 right and so if I wanted to take the two out so so so I have Fe of 2T the Hat two Fe of 2 T minus one the middle hat and Fe of 2T minus 2 the right hand half those coefficients add to four so um what do I do I've I've got that factor two out there so my coefficient so my H KN what's my H KN of K then for this one well it's supposed to add to one I guess I can figure it out it has to be one quarter one half one one quarter right to be to be uh so what am I doing am I doing something dumb here no okay May oh this this it was one to one in proportion but this this was really I only want the what I drew there only had height one2 so I should have had a 1 half there and a one2 there that's the that's the thing that's the thing and then when I take the factor two away I'm I'm down to one quarter one2 one quart good okay but the main point is this that I could consider this equation for other coefficients and for those I'm not going to have any Simple Solution if I take these coefficients H KN of K some other set of numbers I'm not going to have a a simple picture and in fact I may not even have a solution at all so really the this has become the most important equation in wavelet theory is there a solution so let me just put put is there a solution at all maybe I'll put at all that that would be a first question not not an easy question then would come a question uh well formula for it how smooth is it how many now I have to mention a topic that's of of importance uh how many which pols can it reproduce it's approximation approximation uh Power that's and I have to explain what that is questions that I want to know about my function and my and my space that comes from that function so these are the questions these are the questions is there a solution at all what's the what is it how smooth is it how what good is it that's really and all answers have to come from these numbers because that's they all answers in terms of H not of K that's that's where we are that's the that's the posing the problem of uh the analysis problem of for wavelets well I haven't actually mentioned the word wavelet so I'll mention that in in now in a moment um but just to repeat that this equation has constant coefficients it's linear but it's got two time scales and that makes it so different from other equations that it didn't get studied for years for centuries amazing yes could there be more than could there be more than one solution right uniqueness right another question it turns out that question well certainly if Fe is a solution so is 25 times V because just we have that we have that so we would have to normalize it by they say the integral of Fe should be one that's the usual normalization so we'll normalize by the integral of Fe DT equal one for convention and then it is unique effectively unique and we'll have a formula for it so in a way that's or like our one way to prove uniqueness is to get a formula for it it won't be the most explicit simple formula in the world and in fact it won't be a formula that we would actually use to find the values of f of T probably we would and it won't be a formula that we'll find very convenient to answer these other questions but it will be a formula for the for the solution so that's like One Direction is to find a formula so here are some of the answers here are some of the answers solution at all and the answer is is not always not by any means and the formula well we'll actually the formula will be for the 48 transform of the of the of of of fee we're we're if we go into 48 transform space we got a better shot at this equation if we go into frequency into the frequency domain and it turns out to be an infinite product so you can see why that's not the world's most convenient form to get a to get a function Sol I mean I mean so remember I that's right I remember okay now so the input is now the hes you give me some hes that add to one I say okay those hes add to one that's a low pass filter it's got a chance is there a solution with your hes now why don't we just do a example so sorry so um and I'll finish later this uh that that solution but to come back to your so this is the fun question in the subject you give me some hes and how do we decide if there is a solution at all I mean a solution that's anywhere decent like a solution that has finite energy how oh let me tell you one case so I'll I'll Now erase the outline and because we're well into the lecture except I I'm uh well most fees there won't be an H because most function if you just give me a function fee if you happen to give me the hat function I smile because boy I know the H but if you give me some other function or even the Hat function on some you know lopsided hat function or practically anything uh there wouldn't be an H so really it's more this it usually goes in this order well I mean it's it's like and but we want to know the answers to these questions we want to know here's what we ultimately want to know what what is a good H what's a good filter to choose and we would like it to be symmetric that would be a good property to have we would like it to produce a nice solution in fact we would like it to produce a smooth reasonably smooth solution we would like that solution to be useful be to have this approximating power because what do we plan to do with the with the function we we hope that so here's what we hope we hope that this space v0 captures quite successfully our signal we're in continuous time now so captures a typical signal of course if it's a really difficult signal to capture then it's we're going to need more of the details but what we hope is that we hope we we hope to choose 80 so that so that our we can say it in many ways in continuous time I could say I hope to choose h0 so that these give a really good approximation to my signal they might be try to match my signal well actually you're you're specifying the H the function fee raises the tricky question that's really never got resolved suppose we wanted to suppose our the signal we're looking for for is like a tank or something shape of a tank okay so you might say well match that choose fee to look like a tank and then uh it'll be then by you by using these functions as as our basis we'll we'll we'll capture a tank if it's there but the the shape of a tank I I have no idea what hes would give such a thing so somehow that direction hasn't been very fruitful we we more likely ask what hes give good approximation to a reasonable shapes good approximation to reasonable shapes and then fine we say great so these are all great good questions I'm I'm happy to have more yeah Bob some reason why has to beite these H's or the function fe uh let's see well first of all if the hes are finite the Fe will automatically be finite in fact the length is just matches if I've got three hes as I did in this example the the uh fee will go from zero to two and that's kind of first surprise about the solutions to this equation that they do have finite length and then become zero mean the solutions of differential equations you know they go up exponentially or they Decay exponentially but they don't cut off that's uh that's only possible because we have these two scales in here so that uh somehow so depending on the number of terms so K will go from zero to n then our function will also go from zero to n that's pretty neat so um I I guess then the answer do we don't have to stay with finite length I mean but that cost the finite length functions corresponds to finite length filters and and is sort of the natural starting place oh but then I have you know papers keep coming in of constructions that break out of this mold and say well let's use infinite length filters and get get have more freedom in the function yeah it's uh uh let me just mention one other example oh I was going to mention two two or three other examples okay here here's a neat here's a challenge suppose I look at this particular dilation equation that's just the that one okay what's the filter that goes with it what's the H knot the what are the coefficients that you're seeing on the right hand side there well there's only one coefficient and it's one right and then that factor two is the two that I okay so there is a a lazy that's called The Lazy filter it doesn't do anything at all it just uh it's the identity right if I if I The Matrix is the Matrix of ones on the diagonal if I do the convolution nothing happens now tell me what's the scaling function that's the right name for f of T is scaling function I don't think I've written that down here let me these are scaling not not especially brilliant name but anyway right name uh anybody know what function solves that a Delta equation does yeah a Delta equation does which I was like news to me so F of T is Delta of T that's that's kind of fun I I never really thought about Delta of 2T or something and and maybe we should think about this a minute um oh why why whoever said Delta of T why how did well I won't ask how you came up with it but let's just see that it's true how do you how do you so we're claiming that we're claiming that Delta of 2 Delta of T is two Delta of 2T and I would like to know what the heck does that mean it's certainly clear that both sides are zero away from the origin clear totally clear and it's clear that they both blow up at the origin somehow whatever the Delta thing means but what does how could what does it mean why how do I know there's a factor two in there and what's what's going on I could take limits yes it has the same area that's true yeah that's that's maybe that's the the the key Insight that if I integrate if I everybody knows that the integral of the Delta function is one and then now suppose I integrate to Delta 2T do I really get one out of that integral change variables right let new variable let U be 2T so U is 2T and du is 2dt you see how that two paid off okay and that's one so the areas are the same so that's sort of why that Factor two is in there and more generally since I'm into this qu somewhat um direct question here um so the areas match of the two sides but the the full match would be to check that that the only way you really know what a Delta function is because it isn't really a function is that you know what it does to a smooth smooth function and what does it do to a smooth function if you if you take a smooth function and you multiply by this direct and you integrate this is really the definition of the direct and you get what F of zero right okay so that's really the definition I I shouldn't but I will put that okay now so that if this is all is same then we should check that it gives the same definition so I I should take F of T and multiply by this strange thing and integrate and now do I really get the same F of zero because then I would say sure they are the same um why do I get well what am I going to do here same CH change of variables right let 2 t v u 2dt is Du and of course T is U over 2 so looks a little different but it's now I'm taking this function I'm multiplying by Delta I'm integrating and what do I get out of that I again get this I get the value of this thing at U equals z which is f of Z good deal so the two sides gave me the right thing okay well so there is I have successfully solved one more um refinement equation one more very very special case uh so we tackle another one let me let me uh tackle one more that I know the answer to so I'll I'll erase this for the Delta I'm going to write up a new special case here um yeah my my coefficients are going to be 14 641 all over 16 to add up to one so my I don't have Delta anymore it's V of T so I I'm okay if I erase this this example and make space for the new one all right this is the next example so the coefficient are 14641 which is really it's the convolution of one 2 one with itself well one 121 over four I guess it was the convolution of 121 with 121 would be 14641 H how do I know that well I'm multiplying pols with coefficient one two one squaring it because this one has the same one two one and then you know if I remember those binomial numbers or Pascal's triangle or something I'm seeing those numbers so F of T then it turns out that if I look at the equations it would be the convolution of the Hat with the hat if I'm if I know the solution for that and the solution for that then if I convolve the coefficients you could just check that that I convolve the solution so what is the Hat convolve with a hat first of all tell me how far it stretches how far does the hat convolved with a hat stretch remember the Hat we've already got up here it stretches from zero to two now where will hat if I involve that with itself uh it'll go to four right it'll go to four 0 1 2 three and four and any idea what what it will look like now let me just say that's doing a convolution in your head is not a lot of fun especially when the function has different parts and you would have to in the in in the convolution integral you'd have to you'd have to keep track of which part you were in and actually it's going to produce four different parts here well let me say what it is it's the cubic spline it's the cubic spline but why we you would have think for a bit but it's the spline go then it changes to some other cubic then it changes to some other cubic and then changes to some other cubic and uh and it's otherwise zero and if this fact that it has finite length is pretty neat uh again yeah so that's a very important function and of course was discussed long before the word wavelet ever got thought of um and nobody paid any special attention or even probably noticed that it that that function solved one of these refinement equations with nice numbers but uh now of course we do notice that this we think hey yes the spline is a is a solution to this this equation with this with this very very special important lowth filter so there's another example so all the so there's a family of examples in other words box hat quadratic spline cubic spline on up that uh we can do exactly and that is about the limit of Exact Solutions I know yes Les well I just waved my hand and said that would will work uh it I'd have to I'd have to uh I'd have to I would have to erase and make space just to to do that but let me I leave that as an exercise this is a professor's uh privilege right so suppose I have one Lopez filter and and the function Fe of T that corresponds and then just to make the exercise simple I'll keep the same coefficients giving the same function and then I'll ask what about the coefficients that come from convolution of this and my claim is that the result would come from convolution of the fee with itself so I I'm not answering your question I'm I'm asking it back to you which is the no symmetry isn't built into this at all uh oh it's built into these examples but uh I'm not making any assumption about symmetry here uh that's right that's right and maybe I I'll give a a way you might tackle the proof first you could I'm pretty sure you could do the proof directly in the time domain but probably for convolution you're better off to get in the other domain right do it in frequency domain where it's a multiplication so let let's just see how I would get into frequency domain yeah okay good thinking all right so I could finish this formula which would be the formula in the frequency domain okay so let me finish it uh this is an infinite product well who needs infinite products but that's what it is of I'm in the frequency domain so you will not be surprised to see the function that we called Capital H of Omega appear right you remember what Capital H of Omega was it was the function in of Omega which had these as its coefficients so Capital H of Omega you remember was the sum of the coefficients times e to the minus J you see I'm I'm using J today J Omega K that that the the the the finite series it's just a polinomial because we assumed we're thinking here of firir filters you remember that h of Omega now h of Omega the the thing is when I take the when I go into the frequency domain I've got two frequency scales if if the if the trans for a transform of this gives me fat of Omega the 4A transform of this will involve fat of what anybody know it won't be two Omega because somehow it flips the other way around and it will be fat of Omega over two yeah yeah and we could we could patiently do that and of course the book does it and so Omega over two and then and then okay let me say it in words without writing it down but I'm willing to write it down when I take the 4A transform that oh I'll I'll write down yeah sure I'll write down what the 4A transform is of that equation the for a transform of this equation is fantastic I'll put it in yellow to highlight it so I'm taking the 4A transform so I get Fe had of Omega on the left and what I get on the right which is is I'll just write down the neat answer it's h this very H at Omega over two fat of Omega over two and the factor two disappears also isn't that neat that in in the transform domain well this was a convolution right this was a this was really a convolution followed by a change of scale and of course in the frequency domain it becomes a multiplication with a change of scale change in the opposite direction if it was 2T there it'll be Omega over two here I I I I really recommend that that's now that here's a serious suggested exercise is to take the for it's just terrific practice to take the for transform of that and see that you get that so do you see that fat of Omega of Omega every time we then then then let's take one more step this would be h of Omega over two now what could I substitute for fat of Omega over two using the same just recursively it would be h of Omega over 4 time V hat of Omega over 4 right let me pause a second because it's is so neat that that works so nice in the frequency domain but there that it's nice but it's not like self-evident because there's still this change of scale and then of course I carry on and then I get an H of Omega over 2 * h of Omega 4 * h of Omega over 8 and you can see that in the limit I'm getting this h of Omega over 2 to the J that's my formula oh what happened to this guy at the tail end okay it it was yeah good question so it became it it goes away some it goes to one why does it go to one yeah so there should be a fee of Omega over so I'm claiming that this this guy it no it goes to Fe hat at zero right because I keep dividing by two every this this is approaching that factor that's that's on the tail end so once I've got about 77 terms in that product I've got fee over 2 to the 77th and it's fee of Omega over 2 to the 77th so it's very near fee of Fe had of zero and that's one yes I don't think so but I don't don't know quite why not yeah maybe it's because we're in the frequency domain yeah it's it it Bears checking of course so let's see first of all why is fat of zero one because well that's the same as our little convention here that was that that that's we had of that's the integral right um so somewhere in here let's see I I know it does approach one but uh um I don't know the right explanation to give here we're at fat so this is if I take J terms here then I have so I I have going along and stopping at two to the J eventually then this would be this would be fat of Omega for each Omega yeah for each Omega will this will approach fat of zero so maybe this is the point that this gives us an answer a good answer for each Omega well a good answer in this we got an infinite product well that's not world's easiest answer answer and then if we wanted to ask the question what's fee you then I'd have an inverse for a transform still to do and uh be lucky to be able to do that so you you can see that it's extremely lucky to get a few to get to have an explicit solution not going to happen often and maybe what led us to the question is this form would give you a a a way to tackle my uh little convolution question you remember what what was the question that we started with that that if I had if I had the fee which solves this equation then now if I convolve H notot with itself and I look at that equation I'm claiming that the answer would be Fe involved with itself well we and why because in the trans in this domain in the in the frequency domain when I convolve this with itself what would it do to Capital H it would multiply it by itself right so then that would be then then I'd have H squared and I'd have fat times fat so I'd have fat squared in the frequency domain and when I go back I'd have Fe convolved with itself it's uh it's it's the convolution rule that that's coming up there but this is a little bit tangential to the central well no it isn't really because the what we discover is we got an answer here at each Omega so why do I say the solution is in doubt why do I say not always when I've produced the answer and it does this product does converge at each for each Omega because the H is if this is a low pass filter so I really need to build in this fact that the sum of the H KN of K's is one that's that's also part of my convention then this these things approach one and the and the product is converges but so can I ask again I got an answer why should I say there isn't always an answer yeah that's right it I got an answer for each Omega but now if I want to go back and find the function well well by the way what was the answer when for the lazy filter what was fat of Omega for the what was Capital H for the lazy filter remember the lazy filter just had one Co efficient which was one it's a constant in fact it's one so I have a product of ones so I get the answer one which is exactly right right fat of Omega should be one when Fe is the Delta function so that that very lazy case that's a lazy lazy um lecturer too uh to check that case okay um but no the main point is you give me a function of Omega here or we've discovered one but that's by no means sure that we can invert that and get anything decent as a function of T and what is the requirement or what would what would what would be a when would I know that yeah I yes I could invert and get a decent fee of T I mean what how would what kind of property of Fe hat tells me yes there is an inverse and um a decent function um so if fat is identically one there isn't a decent function it's the Delta function if fat is a s a sync function then there is um a fee and what is it it's the Box what's what's up there yeah I mean why is a sync function fine I can invert that and get a box where the constant function if I invert that I get a Delta which is not not an okay function and then all sorts of in between I mean of course everything's in between so so let me ask that question again what sort of for transforms the nice fun the good functions have so that I say yes I can that comes from a good function and and and the and there is a solution well but those would be really good if F had of Omega is Ban limited as for this as as for the uh s the like the one that gives me back the Box function great otherwise sorry no what what is it I'm not saying the right thing am I or or am I uh yes let me give my answer this function if it's if I'm going to be if it comes from a a good function it should decay it should tail off I mean it can wobble a while but I I would like it to tail off and and and of course the bad one is the constant function that doesn't tail off that that leads to Delta which is which is off the chart the uh the the band limited one not only decays but actually dies of course so that leads to um that leads to very good functions but uh gaussian what would the gaussian so what would the gaussian lead to gaussian right so exteme still still deser maybe if that's sensationally good and gaussian should qualify as very good and lots of others so others anyway you you see what I'm saying okay but my point is again we've got an answer for each Omega but we haven't got a very good way to tell do do those answers Decay as Omega gets large so that I can do the inverse transform and find a nice fee Let me Give an example where we won't have a nice fee yeah well no I it is the solution but it's I wouldn't say yeah yeah it wouldn't be in my it wouldn't be a legal it's not a function a solution I'm proud of right yes that's right that's right so the function the solutions that I can work with are and use I want their combinations you you you give me combinations of the Delta function well boy I want things like hat fun functions and splines and and and I'm willing to accept a few Wiggles but I don't want a Infinities yeah so that's so the Delta function would be a case that would be an example I would say there is no solution in my in my uh sort of standard finite energy space well it's pretty hard to decide so what I guess what I'm trying to say is some hes give a solution in my finite energy space others don't let me take one that surely doesn't because it's just sort of fun to to play with these things a little uh suppose I take the here here you go suppose I take this equation F of T now I'm going to get the coefficients to add to two so suppose I took 3 Fe of 2T minus 1 V of 2T minus one I claim that that choice so I guess my filter H then is what H KN here is three halves and minus a half and they add to one and looks you know sort of half okay what would the what would the graph of Capital H look like for this case let's see what is capital H H is three Hales of one minus a half of e the minus J Omega right I'm just like I'm going to claim that this is a bad guy but uh just to see first of all it does go through the point one so it's like it's okay to consider it anyway but then what does it do well I guess at at Pi it's uh no good so that I really know that that that this example is in serious trouble because what is it at Pi at Omega equal Pi this will be minus one so it'll be uh is it two at pi is that is that what's happening here it doesn't go to zero at Pi at at Omega equal Pi this is minus1 so I think it's two is it is it is a graph like that well gosh that's that's in serious trouble and now how do but how do I come back coming back to the equation how do I see that I'm somehow it should be supported on the interval 0 to one because I've got two coefficients and it is and it's not a Delta function but it's it's almost as scary I mean somehow it's I don't know exactly yeah oh okay go with the infinite product yeah right right that's I at least you're right and I could try to go with that infinite product and it would converge it would for each Omega for each Omega this will converge because this it's there's an Omega over two to the J here and that's that's that that that gives us good something good at each Omega but if I graph that I have no idea none if I graph this thing because of course I can't take an infinite product in my head to put it mildly uh but if I so here's my h of Omega here's my formula actually if somebody's got a computer handy that's sitting idle uh that would be an idea to go with it uh just graph this thing for a different omegas it it'll be complex admittedly but let not you know that's okay we could we got a real and imaginary part what do you think would happen or I took the log that seems like a smart idea then I've got to take the log of this guy so all right whatever the computer is happy with uh I think I don't think it'll be as bad as a Delta I guess so I think that the H of that that this F had of Omega won't I don't expect it to approach a constant but then nor do I expect it to Decay I I don't know what it does maybe it bounces around anyway there there's no decent function Fe of T how you you could still ask how am I how why do I think that well partly I think it from that yeah maybe I should look at particular values of Omega like Omega equal multiples of Pi I might learn something actually that might be a smart thing to do let Omega be multiples of Pi and then then oh no it would still still be pi over fractions there not too good well oh but somebody is observing that all the all the factors would be above one oh yes you're right sorry should should I just say that again which you you realized before me that H has magnitude bigger than one apparently so apparently fat has magnitude bigger than one so which makes it look as if be because it's a product of things bigger than one so but converging to one that's interesting isn't it ah how let's see that better oh how I mean sorry the factors converge to one yeah the factors converge to one uh let's see would we be able I mean here here how you want a little practice on infinite products okay what's the infinite product of just the numbers 1 plus Omega over 2 to the J I don't know I figur we ought to be able to do that but I'm not sure there's another exercise figure out that infinite product so that's 1 + Omega over 2 * 1 plus Omega over 4 * 1 + Omega over 8 and so forth my my point was that this every factor is bigger than one but the product is still finite because um you're going to make me comp okay this times some e to the okay I'll let that be the amplitude I don't know if it's okay I I just I just now now we're out of time here but let's see if I just took this thing as it stands without doctoring it would this is like practice on infinite products which we're all of course highly familiar with infinite products I'm I mean I realized that I hadn't seen an infinite product since graduate school and when I saw it there it was only one day and I thought ah luckily I'll never see these again and uh here they are uh but I believe that that infant product we and let's make omega B one hey let's why should we struggle here what's the factor what's the product of one of If I multiply three Hales by 54s by 9/8 and so forth do I get a finite number in the end all those factors are bigger than one but they're approaching one that's what I meant to say I I think I do if the if I take the log then right that's right that's a good point yeah yeah the good way to yeah the good way to to analyze infinite product is take the log good point and then we have an infinite sum and we know about those right right yeah so this converges to some limit that uh let's see if I actually multiply these things out well yeah I mean how are we why are we doing this but if I took this it would be 1 +4 + 12 plus 1/8 look at the beauty of that right right when I just take this product and now now if I multiply by 1 + 1/8 what do I have if I take all the terms well I've got a a lot of terms uh I I'm feeling might be two what it I don't know not two yeah I'm I'm not anyway back to this great C example somehow it looks to me like the U all the factors are larger than one but they're approaching one so that the infinite product will converge and will give us some function and that function will be bigger than one maybe not bigger than two or not bigger than something and it'll wander around and it'll be a a bad function it won't be a function it'll be a this would be a case of the not a not a solution wouldn't have finite energy how how do I know it wouldn't have finite energy I'll I'll I promise to stop at without question suppose I know which I do that fat of Omega is somewhere between one and let's say k how do I know that the energy is not that that when I take the inverse transform to get my fee itself how do I know that it hasn't got finite energy so that's let me write that down again so I I think we would learn from this that that F hat or its magnitude was somewhere between one and some number that that would be that would come from some some number K huh what and I claim that we know we're in trouble that if I have a 48 transform that's in this range I know the energy isn't fine finite in the function so I know I I have not got a finite Energy Solution and why is that yeah what's the relation between energy and a function and energy in the transform they're the same they're the same right so so the the one thing I do remember is that the integral of the function squared is always the integral of its transform squared with a maybe a two Pi or something okay and what's What's Happening Here with this if my function never even goes below one it hasn't got a chance I hasn't got a chance in heck of being finite so it's infinite and and so then I know the energy in the function is infinite and it's it would be a casee with no solution and I could uh I could try to figure out that function and I would see it it blew off to Infinity at certain points maybe not at all points that that's the beauty of these functions of these fees the fee that solves the equation it it'll blow up probably at all the at a whole lot of points whole dense set of points it might not blow up at other points it's it's just quite remarkable what um uh what this equation leads us to so my point here ending this lecture is that a very simple choice three halves minus a half can lead us to a totally strange function but of course that's not what the subject is about so can I in the in lecture 28 come back to sensible functions sensible filters where there is a solution and ask about how good it is how smooth it is how good it is in approximation and in applications and then move ahead to the applications so luckily we're electure ahead of schedule but I'm I do want to uh today to not just not just work with this equation and I still have to tell you about wavelets this I've given you the fee but not the W so that's coming in the next lecture so should we could we start the next lecture at 3 it's 3:20 now 10 minutes or 15 minutes maybe maybe 15 minutes 3:35