So, good afternoon. You had some exam today? What? Awful? Well, I hope for the best. What? It was not so good. Well, anyway, we'll see. Okay, maybe... So, you're already unhappy? Yeah. Okay, I will have a look. Okay, so maybe now I start all the same. So, we wanted to talk about... I said you wanted to talk now about finite fields. This was just some kind of interlude, so that we see that finite fields, at least we can understand completely. And so, the idea is we want to classify the finite fields up to isomorphism. So, we know... So, if p is a prime number, then if I take zp, which is a field I call fp, with a usual multiplication, so the number's modulo p, with an additional multiplication, is a field, and it has p elements, which I can identify with the numbers from 0 to p minus 1. And so, now we want to show, up to isomorphism, so for every prime pump power, p to the n, there is a unique field with p to the n elements. So, when there are q, there is a unique field with p to the n elements. And there are no other finite fields. So, if a field is finite, it has to have p to the n elements for some prime p. Okay. So, first, we know that the characteristic of a finite field must be a prime number p, because the characteristic, if the characteristic is 0, then the field must be infinite. And so, let's say f, so maybe this is called a lemma. So, anyway, for the rest of this section, so let p be a prime number, and n a positive integer, and q equal to p to the n prime power. And so, let we take a field of characteristic p. Then, if we look at this, the following polynomial, the polynomial x to the q minus 1, so q is p to the n as here, has precisely q simple roots. So, simple zeros in its splitting field. So, over fq, over f. So, if we look at the zeros in any extension of f of this thing, there will be q simple roots. So, as many roots as a degree and all are different. This is very simple. I just have to see that there are no multiple roots. So, if a is a multiple root, then we have seen last time that both the polynomial and the derivative of the polynomial have to be vanish at a. So, then x to the p minus 1. So, if I take x to the p minus 1 applied to a is equal to 0 and if I take the derivative of x to the p minus 1 applied to a, this is equal to 0. But this polynomial is the constant polynomial minus 1. This is equal, identically equal to minus 1. So, therefore, this will never happen. So thus, we have no multiple roots. So, we can see that, we see that this polynomial has always q simple roots. And we will see that actually we find all the finite fields as the splitting fields of this polynomial over f p for different values of n. So, let's start with that. First, we want to show that there exists a field with q elements. So, let's say q which was p to the n elements for every prime power p to the n. Well, so we just take the following. We look again at this thing.  And we put f to be inside the splitting field just a set of all the roots of all the zeros of this polynomial. Say a in, yeah, except I made a mistake and you should have noticed. I mean, there's a misprint in not in the notes but in what I've written here. Obviously, it's x to the p. I have x to the p minus x because you can see if I take the derivative of this one, this will be p which is 0. But we wanted to be, I wanted it to be minus 1 and this is if I have x here. If I take the derivative of x to the p minus x, this will be minus 1. So, this is the polynomial I want. But so, this maybe shows I wasn't too fast because you didn't notice. Anyway, so we take this polynomial here too. We take this splitting field of this. So, this is the set of all in k, a in k such that a to the q minus x to minus a is equal to 0. So, now we want to see that this is a field. So, claim f is a field. We know that it has q elements because that is just what we said here. There are q different roots. Well, you just have to see the field axioms are satisfied. So, obviously we have 1 to the q minus 1 is equal to 0. And we have, if we have a to the, so if a is an element of f and b is an element of f, then we have a to the q minus is equal to a and b to the q is equal to b. And so, if we take a b to the q, this is a times b by just applying this. So, it follows that a times b is an f. And if we take minus a to the q, we need to actually know, if we take a to the minus 1 to the q, this obviously is a to the q to the minus 1 is equal to a to the minus 1. And so, also a to the minus 1 is in f. And what is more interesting or what is, comes from the fact that we are over a finite field is that also the sum is there. So, if I take a plus b to the q. Now, you can compute this by the binomial coefficient obviously. So, this is a power r i sum i from 0 to q, q through i a to the i b to the q minus i. But it's a very easy fact to see that if q is a prime number that all these binomial coefficients are divisible by q or at least by p, they are I think also divisible by q if i is not equal to 0 or q. So, this is equal to 0. So, here we are computing in this is an element in this field. So, we have a field of characteristic p. So, p is equal to 0. Any multiple of p is equal to 0. So, this binomial coefficient will be equal to 0 unless i is equal to 0 or q in which case it's 1. So, this thing is just a equal to the q plus b to the q. And this we know is equal to a plus b. So, also a plus b is in f. And let me see what I have. And it's the same if we have instead of plus if you put the minus we get the same thing. So, here of i is equal to q. So, if q is equal to a power of 2 then the characteristic 2. So, a plus b is equal to a minus b. And if q is not that, is a power of an odd prime then q is odd. And then if I put the minus here I get the minus here. So, in any case I can put the minus here. The minus here. Okay. And so putting all this together we get thus f is a field with q elements. So, now what we want to see that f actually is equal to k that f is the splitting field of this polynomial. And as such it will be uniquely determined. So, at least we have now a field with q elements. And now we want to see. So, anyway it is clear just to remark this that if I already having said it f is a finite field. So, then it must have finite characteristic so assume of characteristic p. So, then I can look at the subfield generated by 1. So, then f contains the set of all n times 1 where n is in z. So, just this means you add 1 n times to itself as before as a power of the subset but it is a subfield because if you add or subtract such things or multiply them you stay in this realm as a subfield.  It is a field with p elements. We have seen that the field has characteristic p. So, p times 1 will be equal to 0. And that is the smallest number for which is equal to 0. So, this thing here is just equal isomorphic to f p.  And as it's a finite field it's a finite extension of f p. So, thus f is a finite extension of f p. So, in particular if it's a finite extension of f p it's a finite dimensional p vector space, f p vector space. So, in particular f is a finite dimensional f p vector space because we have always seen that the field extension and the bigger field is a vector space over the smaller one. And so, thus in particular the number of elements so say I write like this the number of elements of f is equal to the number of elements of f p which is p to the power the dimension of this to the n power n where n is the dimension of f as f p vector space which is just equal as we know to the degree of the field extension. Okay, so in particular we know that a finite field must have as an order the number p to the n for some n and the prime p and that it is a finite extension of it. And now we want to show that it actually must be as said the splitting field of this polynomial I've written their proposition.  So, which then in particular means that funnily enough this will show that our finite field with f to the p with q elements consist precisely of all the zeros of this polynomial. So, we have to see basically we want to see that every element in f must satisfy this equation a in f satisfies a to the q minus a is equal to zero. So, obviously so we start very slowly. So, if a is equal to zero then this is true. So, we have already checked it for one element. So, now we have only q minus 1 to go. So, but now we look at f without zero so with the multiplication so the multiplicative group of f this is a group. So, is a finite group with a you know one element less obviously q minus 1 elements. Now, you know generally if you know we already have seen this near the beginning of the course if g is a finite group and we take any element g in g. So, a finite group say of order k and if g is an element in g well then I claim if I take g to the k this is equal to the identity element in g to the unit element. So, the neutral element in g and why is this now I can look at the subgroup generated by g is a subgroup and so as it is a subgroup its order is a divisor of the number of elements of g. So, divides this number k on the other hand we also know that the subgroup is just given by taking all the powers of g. So, we have that the order of g which by definition is the smallest in z bigger than 0 such that g to the n is equal to 1 is equal to the order of this group because we had seen that the elements are just 1 gg squared until g to this order minus 1. So, we have that so thus we have that the order of g divides our number k and now if we take g to the k so I can write k equal to L times the order of g so this will be equal to g to the order of g to the power L this 1 to the L is equal to 1 so therefore we see that g to the k is equal to 1 so anyway this I don't remember whether we covered this before but it's always true if you find a group and any element then if you take the power of the group you get 1. So thus now we go back to our situation we take an element a in f without 0 if you have an element a in f without 0 then if you take a to the q minus 1 then by what we have just seen this is equal to 1 so if you multiply this by a you have a to the q is equal to a so thus we have shown that for all elements a and f identity holds so now obviously as this polynomial x to the q minus x has degree q it can have at most q roots in its splitting field so it means the polynomial splits into linear factors over f x to the q minus x splits over f into linear factors now because we have all the 0's and it's the product of the factors of all the linear factors for all the 0's and so and on the other hand we have but then f is the splitting field of this polynomial because it consists only of 0's of it if you remember we had the statement that if we have a field over which the polynomial splits so we start with the small field k and we have a big field l so in l we have a polynomial in kx which splits over l then if we join all the 0's of this polynomial to the smaller field we get precisely the splitting field and this we have done here we have joined all the 0's of this polynomial to fq it happens to be that these are also all the elements of the field but that doesn't make a difference this thing does not contain contains only roots of this thing it splits and it doesn't contain anything which you cannot express in terms of roots because in fact it only consists of roots okay so we have found so where are we? so thus it is the splitting field so we can put it together so we know that the splitting field of a polynomial for a field is uniquely determined up to isomorphism so therefore this the finite field of p to the n elements is uniquely determined up to isomorphism so we can just put what we have learned in a theorem together so finite fields with say p to the n elements exist only finite fields with q elements exist only if q is a prime power p to the n and for each prime power p to the n there is up to isomorphism a unique field with p to the n elements namely the splitting field of this stupid polynomial so this is now x to the p to the n minus x over fq over fp okay so this is what we proved and so one can say that we have understood everything that there is to say about the classification of finite fields so so there is not so much we have this kind of list for every prime power there is one up to isomorphism and anyway so it's not such a complicated story so in some sense from now on we can mostly restrict ourselves to fields which are not finite okay so that's what I wanted to say about this is more like an aside show that one can at least completely understand the finite fields now we want to go on with the general theory and we want to introduce the Galois groups so later we will want to understand field extensions in terms of their Galois groups and the Galois group of a field extension L over K is the group of K-automorphisms of L so let me do this now so Galois groups we want to understand a little bit about them and we want to introduce them so as I said so if L over K is a field extension then the Galois group so the Galois group after Galois the French mathematician is of L over K is I write it Gal L over K is the group of K-automorphisms of L so it's clear that these are automorphisms of L so they form a group by a composition as automorphisms always do and you recall that a K-automorphism is an automorphism of L which is the identity on K which sends every element of K to itself so these form a group and we want to use this group to understand this field extension now in the moment that doesn't look particularly good because this seems a rather unconcrete thing you have this field extension and you have this group of automorphisms smaller that seems to be something very complicated which is impossible to understand so we want to see that you can say something about it so in particular if L is a splitting field of some polynomial with coefficients in K then this Galois group will turn out to be a subgroup of the group of permutations of the roots of this polynomial so it's certainly in particular will be a finite field but you know it's concretely described as a permutation group but this we have to prove and there's also another description so we will find two different descriptions of this Galois group as a permutation group which depending on the context are useful so let's first say I recall something we said in the group theory part so if you have a finite so for a finite set we had introduced what I call S of M the group of permutations of M which was just a set of all bijections of M to itself and obviously if M has N elements then this is isomorphic to the symmetric group on N letters we also write it so if number of elements of M is equal to N then SM is isomorphic so SM so and recall also that we had talked about actions of groups on a set and so just one word which we had used an action of G on a set M was called simply transitive we only need that now if for any two elements here there is an element in G which sends one to the other so if for all M1 and M2 in M there is there exists a unique element G in G with if I apply the action of G to M1 I get M2 so the action is you recall that the action was called transitive if there always exists a G which sends any M1 to any M2 and if it's unique it's called simply transitive and it's easy to see and we had seen it that it then follows that the number of elements of G is equal to the number of elements of M just fixing one element of M we can identify G we find the bijection of G and M this was all things that we had before and so now we have the first statement about the Galois group for a simple algebra extension theorem so we take a simple algebra extension and we want to identify say something about the Galois group so we say of degree N so that you remember that and so we have our element A so we take the minimal polynomial so then we find that the Galois group will be subgroup of the group of permutations of the roots of the minimal polynomial and actually it will be subgroup which acts simply transitively so it will have precisely as many elements as this thing has zeros so let's see it so so let P the set of zeros of this polynomial so the set of all B in KA such that F of B is equal to zero so this is the set of roots of F in KA so then we find that the Galois group of KA over K acts simply transitively on R in particular it is isomorphic to a subgroup of the symmetric group of R and the number of elements of the Galois group is equal to the number of elements of R and F was a minimal polynomial of A the field extension has degree N so this polynomial has at most N zeros so it means that this number R is smaller equal to N so we know that this Galois group of KA over K has at most N elements and it can be identified with the subgroup of the permutations of the roots of this polynomial okay so this we want to see so we first have to see that the Galois group will kind of permute these roots so that an element of the Galois group will always send a root to another root I mean to another or the same root but it doesn't send a root to something which is not a root so proof so we take an element in the Galois group and so if we write F equal to sum AI so F was a polynomial polynomial X to the I then if we if B is an element is an element in R so is a root of this polynomial then if we take Phi of F of B F of B is equal to 0 so this is Phi of 0 and thus 0 we can write this 0 in a more complicated way namely we apply Phi to the polynomial so this is Phi of sum I equal to 0 to N AI A B to the I and now Phi is a ring homophism so it pulls into the sum the AI are in K so it is the identity on them so this is equal to Phi of sum I equal to 0 no, like this AI Phi to the B I so this is nothing else as F of Phi of B so we see that if B is in R then also Phi of B is in R for any element in the Galois group so thus Phi of B is in R so in particular we see thus so if I take Phi and restrict it to R maps R to itself and obviously Phi is an automorphism of fields so it is a bijective map particularly it is an injective so also the restriction will be injective because Phi is and as R is a finite set and they have a finite injective map of a set to itself it is bijective so in other words so Phi restricts it to R is bijective so it means Phi restricts it to R is an element of the symmetric group on R now I want to see that the map which sends Phi to this restriction is actually and that this element is simply transitive so so we can say thus we have a map we have a group homomorphism because in both cases the group structure is by composition so it is the same which I couldn't call the restriction to R from the Galois group of L of K A over K to S of R so as I said we want to show that this is you know so as I said this is a group homomorphism because here the you know the composition of automorphism is just the automorphisms of maps and here is also the composition so it is just the same thing the group structure so Phi composed with P C restricted to R is obviously equal to Phi restricted to R composed with P C so we have a result this was the unique extension of automorphisms to simple algebraic equations which we proved at the beginning of our field theory adventure that so that there is a that for every root if whenever I have a root the two roots then there is always a unique extension of the identity so a unique K isomorphism of K A which sends one root to the other so we have seen for unique extension field isomorphisms to simple extensions that for for any two roots that for P 1 P 2 in R there exists the unique so this means unique Phi in the Galois group of K A over K with Phi of P 1 is equal to P 2 this was the theorem that we had so this precisely says that the this group the Galois group acts simply transitively on R because that says it for any so the action is by Phi applied to any D is Phi of P and the action is simply transitive so thus Phi there is the Galois group of K A over K acts simply transitively on R but if this is the case then it means also that this map here so that this restriction map must be injective because the the kernel of the restriction map consists of all elements in the Galois group which act as the identity here and according to what we have said here in particular this sent one P 1 to itself and the thing that the only thing that does this is the identity according to the fact that this acts simply transitive so the kernel consists only of so the identity on K 8 so the map that sends every element in K 8 to itself so well and so we have found that the Galois group of K A over K isomorphic to subgroup of S of R which acts simply transitively on R as it acts simply transitively on R it follows that the number of elements in this group is equal to the number of elements in R and as R is a set of roots of this polynomial F the polynomial F has degree N so the number of roots that it can have is at most N now you have to remember that the degree of the field so we had K A over K was field extension of degree N and F was a minimal polynomial so the degree of the minimal polynomial is N so the minimal polynomial has at most N roots in K A and this number R is smaller equal to N so we have precisely proven what we claimed so I want to so I want to give some examples very simple so moment we only know that the number of elements in the Galois group is at most N the field extension so let's look at two cases to see that actually different things can happen so if we take the Galois group of C the second root of 5 of Q or whatever it can also take second root of 2 over Q then you can see the polynomial of this is X squared plus X squared minus 2 and this splits into X plus second root of 2 and X minus second root of 2 and it is clear we have inside the we I claim that the Galois group this will be equal to no first we can have the identity and then we have the map which exchanges these two roots so that means if I have an element A plus B squared of 2 this can be sent to A minus B squared of 2 so it's easy to see that this element this map is field isomorphism and it's the identity on Q obviously and so this certainly these elements are in the Galois group and by what we just saw we see that the Galois group has at most two elements so this is the Galois group okay it's a very simple okay so if it was always as simple as that wouldn't have to worry now we can look at a more different example which instead take the third root of 2 then this Galois group is very small namely it contains just one element the identity so we see that the minimal polynomial of this thing third root of 2 over Q is you know obviously x to the third minus 2 but you find that in Q third root of 2 you know this polynomial has no other you know has no other roots in fact you know we had already an example where we had seen that you can if you want to find the splitting field if you find another root you have to actually add a complex number to it this e to the 2 pi i divided by 3 I think so here there's no other 0 of this polynomial so now the theorem says that the Galois group is a subgroup of the of the symmetric group on the roots there's only one root the only sub you know the symmetric group on one element is one element so it follows that the Galois group can only be the identity so but in some sense we will soon we now want to restrict our attention to so called Galois extension which are separated and separable and normal extensions and so in that case we find that the Galois group will always have order equal to the extension to the degree of the field extension and it will be easier to study so let me see so okay so now we want to introduce Galois extensions so a field extension L over K so this is a definition after all is called normal is a Galois extension if it is what we need we need finite or something well maybe I don't know whether that's actually part of the definition but we only deal with finite extension so finite field extension is called the Galois extension if it is separable and normal and you have to remember that in characteristic p every field extension is separable so then the condition would only be normal so we get as a corollary so let L over K be a Galois extension of degree N so then the Galois group is isomorphic to a subgroup of Sn so which acts simply transitively on the you know numbers 1 to N so the difference of the previous cases that now we really have and so in particular the number of elements in the Galois group is N so I could also say it like this the number of elements in the Galois group of L over K is equal to the degree of L over K which after all is N so before we had seen this thing where we have it simply that it is asomorphic to a subgroup of a symmetric group of on the roots and now basically we only have to see that there always will be precisely N roots of this minimal polynomial well let's see minimal polynomial of what so let's have a look so proof so we have we can our Galois extension you know by the theorem of the primitive element we can always take it as a simple algebraic extension for every separable extension there is a primitive element for every so by the theorem of the primitive element there is an element A in L such that you can write L equal to K of A so we find so our extension is a simple algebraic extension so we are in the situation of the previous theorem so we know that as so let's say let F be the minimal polynomial polynomial of A over K so this field extension L over K is normal and we have and F is an irreducible polynomial which has a 0 in L it follows that it splits splits over L into linear factors and as the field extension is separable and F is again an irreducible polynomial which has a 0 here you know all its roots are distinct of F in L are distinct so the set R the set of roots has precisely N elements all that we need to know so thus R which was set of all B in L such that F of B is equal to 0 has R elements R equal to N and so therefore the Galois group of L over K is isomorphic to subgroup of the symmetric group of R which is the isomorphic to a symmetric group of on N letters which acts simply transitive that was what the previous theorem said okay so this is this so now we come now we have another proposition we want to see so this would be mostly lemma we want to use that so until now so we know that this Galois group acts on this root of this polynomial but we want to also have some positive results about existence of elements in the Galois group and so we have the following statement so again we have a Galois extension and we take two elements A and B and L so then the statement is that there is an element in the Galois group which sends A to B if and only if A and B have the same minimal polynomial so we know precisely when the Galois group sends an element to another one this is if and only if A and B have the same minimal polynomial K I mean obviously it's normally very easy to find out what the minimal polynomial of an element is so it actually will be I think mostly use the other round so if we somehow can prove that two things so yeah we well anyway maybe I will say that later so now let's see how to see that it's not particularly difficult it's a variation of things we had before you know again about extension of field isomorphisms and so on so if A and so we have obviously two directions so first we assume that A and B have the same minimal polynomial we have to find an isomorphism we have to find an element in the Galois group now certainly we have that we can we have this statement of the extension of field isomorphisms to the for simple algebraic extensions so by the extension actually the unique extension of field isomorphisms for simple algebraic extension statement was that if two elements have the same minimal polynomial then there will be an isomorphism from K of A to K of B which sends a unique isomorphism from K of A to K of B when it sends A to B but we just need that there exists one we don't need the uniqueness so there is a is a K isomorphism C from K of A to K of B with is equal C of A is equal to B so now we have that L over K is a Galois extension so L is the splitting field of some polynomial with coefficients in Kx because it's a normal extension we had seen that if you have a normal extension an extension is normal if and only if it is the splitting field of some polynomial with coefficients in this model so as L over K is normal we have that L is the splitting field of some polynomial and we call it F in Kx over K now we know that if if something is a splitting field of a polynomial over some field then it will also be the splitting field of the polynomial over any intermediate field of F over K and the splitting field of F over KB because these are intermediate fields and so it is also splitting field over these but then we had this other extension theorem if you have an isomorphism between two fields then it can be extend so and you take a polynomial here and you take the image under this morphism of the polynomial here then it extends to an isomorphism of the splitting of the polynomial here the splitting field of that so in this particular case the polynomial is defined over K so if I apply Psi to this polynomial to the coefficients of the polynomial I get the same polynomial so therefore the extension the statement about the extension of isomorphisms to the splitting field says there is an extension of Psi to L which is the given map here by the extension of isomorphisms to the splitting fields there exists an isomorphism Phi from L to itself with Phi restricted to KA is equal to Psi so I hope you remember the statement here we were it was a somewhat complicated statement so we have we have a field we have two fields so I mean just just to remind you we have so F is in KX we have an isomorphism Phi from K to K prime and we apply so we take Phi star of F is the polynomial where you apply to the coefficients of the polynomial this isomorphism and then the statements in then we take L the splitting field of F over K and L prime the splitting field of F prime over K prime so I call this F prime I call it Phi star F over K prime then we had seen there exists an extension so there exists Phi from L to L prime with Phi restricted to K is equal to small Phi this was a general statement and we have applied it here what is called K there is K of A and what is called K prime there is K of B so what we have found so anyway we have this that Phi restricted to K is equal to Phi but this Psi was a K isomorphism so it means if I restrict Psi to K if I restrict it further to K it is the identity on K so it means that Psi is an element in the Galois group so as Psi restricted to K A is equal to Psi so Phi Psi restricted to K is equal to identity on K it follows that Phi is an element in the Galois group over K and by definition sending A to B so one direction we showed have the same minimal polynomial then there is an element in the Galois group which sends one to the other so now we want to show the other direction which is a bit simpler maybe I hope you maybe I will wipe it out here so that you can remember the statement that's okay so we want to show the converse so we take an element in the Galois group which sends A to B so now we have to find that A and B have the same minimal polynomial well that's not so difficult we take F and G minimal polynomial of A over K and G minimal polynomial of B over K then I can use the same simple observation as before which I maybe write down once more so we have 0 is obviously equal to F of A therefore it's also equal to phi of F of A and by the same computation as before this is the same as F of phi of A which is F of B so we see that B is also 0 of F but G is the minimal polynomial of B so it follows that G divides F G divides F but obviously I could have done the other round I could exchange the role of A and B and say the same for G of B is a 0 so phi of G of B is a 0 of G and by taking phi to the minus 1 so in the same way exchanging the role of A and B we have that F divides G so as F and B F and G are both monochromes it means they are equal so this proves this proposition and so much time up, basically no time so let me see okay so I will maybe at least I can even prove it so the next result so maybe I first make a definition so with the the next result which is quite easy to prove will be one of the will be half of the main theorem of Galois theory and basically we have worked until now to prove it quite simple but anyway maybe I first make a definition so let I really want to make a definition the Galois group the Galois extension and let G be a subgroup of the Galois group then we can consider a fix set of this thing so then that fix of G be the set of all elements A in L such that V of A is equal to A for all so we have we'll see later that this is actually always a field but it doesn't matter now so we have the we can look at the set of all elements in L which are fixed under all elements in G where G is a subgroup so the moment we are only interested in this in the case G is a subgroup of the Galois group okay so maybe I call it maybe I'll also call it Phi because the Galois group is always called Phi okay so the theorem that we want to show we'll later see that it's an important step is so if you have a Galois extension we can ask ourselves what is this fixed field if we take the whole Galois group so then the fixed of the Galois group of L over K this is just K so an element in the Galois group will be an element in the field a big field will be fixed under the whole Galois group if and only if it is in the smaller field of the field extension let me see I think that's fairly we can still prove it it's fairly simple so we have two inclusions so this inclusion is clear because obviously every element in K is fixed under the Galois group by definition the elements in the Galois group are the identity on K and now conversely so we assume that Phi of A so A is an element in L that Phi of A is equal to A for all Phi in the Galois group K so we take the minimal polynomial of A over K that even makes sense if A lies in K because then it's just a linear polynomial so as L over K is a normal extension we know that the Galois group splits that F splits over in L so any other root of F in L then we have seen there exists an element in the Galois group so it sends A to B in the Galois group with Phi of A is equal to B it's actually here so we have made the assumption that our element that Phi of A is equal to A for all elements in the Galois group thus A is equal to B by our assumption that A is fixed under all elements in the Galois group so it means that all the roots of F are equal on the other hand L over K it was a Galois extension, it's also a separable extension and F was a minimal polynomial so F is irreducible in K and has obviously 0 so in L so it follows that it has no multiple roots in L so for one thing all the roots are equal on the other hand it has no multiple roots that means there's only one root this polynomial is a polynomial of degree 1 a linear polynomial and therefore A is in K and this proves this result okay so this maybe is enough as you see this was not so particularly difficult but one always you know one always kind of uses the same thing but always in a each time in a slightly different way so anyway so this is this and next time we will at least start to finish the proof of the principle theorem of Galois theory okay so any questions okay thanks


Everybody forced here ripping 21 22 NT basketball hobby two box number four It's a random team style of rake. You can always find their results at lane sports cards comm Thanks everybody for joining and good luck There's a Lincoln chat too if I didn't say that sometimes I feel like my spiel is like robotic at this point Or say the same thing It's like when you drive home the same way for like months and you like black out your tone You're like well then. Yeah, I don't know what happened, but I'm home We're like every single every single day blends together kind of Except today I Got here like double the time I normally get here. I had to go through like To like, you know, like one lane roads where they have like the road blocked off then I was behind the car that just Was I don't know what's going on. Anyway, Harrison Barnes for the Kings to 99 printing plates spurs No, sorry Celtics, I don't know why I said spurs. Hey, no that That's the wrong That's not Aaron Ness with that's Devon Vaselle, right? Yeah, it says Aaron Ness with on the back Rich are you still here? I'm just where where does this go? It's a Celtics on the back. It's the wrong plate So that's a Celtics card, right? That's clearly Aaron Ness or not. That's clearly a Devon Vaselle Yeah, so that'll be So that it this is So by rule guys at LSE we always go with what's on the car. This is a Celtics Aaron Nessmith Crossover RPA It's the same style. I think I'm not sure but this is gonna go Celtics We always go by the back of the card and whatever like low goes on it So we'll send an empty box and wrapper to the Celtics spot Rich is in chat right now. Thanks for confirming rich. So yeah Celtics getting this one And we're gonna send an empty box and wrapper because that is the incorrect plate Spurs Jersey card here. Kawhi Leonard to 99. Glad the Spurs got hit right after that actually Didn't feel great about that. Yeah, that's tough error There you go Spurs nice Kawhi game use Jersey Here we go treasured threads LeBron James game use Jersey Lakers 32 of 99 This is nasty one of one Hawks Nike swoosh patch auto D'Andre Hunter boom Love the neon swoosh Sick Hunter Congrats Atlanta That is freaking awesome. Oh card behind. It's nice to we got true RPA Kings Davion Mitchell 82 of 99 What a box what a case actually there's the same two left over from the last break as well like doing two Nice Davion. We'll get down Maggie for you Kings Jersey card Hornets PJ Washington, Jr. To 99 Cookin Jersey card for the Clippers Koran Butler with a loose I think that'll I'm not gonna mess with that That'll probably come out. It's part of the cut like with the the frame was cut This is part of that the clips dual patch rookie auto Kai Jones nice one Hornets a 25 pretty card Got thunder auto they will mall it on to 99 Could be let's see that next box now they will mall it on to 99 Five daytime spots left guys NT and prism two box mixer just five to go links in chat currently And then 5 p.m. We resume with breaking Brad everything else daytime is sold out other than that break Brad resumes at 5 p.m. His spots are filling up quick to guys make sure you check out the break schedule late in sports cards Com also make sure you check out the break schedule for the rest of the week Don Russ optic football releases Friday big Football product looking for some downtowns Beautiful rated rookies Friday is gonna be awesome. Good week, man. Good week. Good week to lead into the National I'd say There's a Tyler here at a 20 or to 99 for the heat 25 of 99 printing play Jason Tatum Celtics Here you go C's I'm sick Jason Tatum Game gear Buckster Z Chris Middleton to 99 nice thunder rookie patch here Jeremiah Robinson Earl 13 of 25 Thunder always have nice patches. I feel like That's gorgeous rockets Rookie dual patch auto Alperin shengun to 25 nice hit Houston there. That was going to Simon Beautiful shengun. We'll get that mags Other good box here true RP a Nix Quentin Grimes Look at that That must be the oh in New York, right? I Would think yeah I don't know what else it would be Yeah, it's got to be There you go Nix. I'm not saying that we're magging that yeah, that's beautiful man. What a box Mavericks patch card Sean Marion to 99 Best bet the jump shot form of all time Sean Marion Rookie jumbo jersey for the Pacers Isaiah Jackson 5 of 99 treasures of the hall bullets auto elvin Hayes To 99 It's Washington on that one and on the back is a T wolves auto Ricky Rubio to 99. He was a cab all year. It's kind of strange I was from last year. That's why sorry. Yes. It's from last year Ricky Rubio T wolves All right, let's start with the relics on our recap here Is that Jackson Jeremiah Robinson Earl Chris Middleton PJ Washington, Jr. Koran Butler as a clipper Nice LeBron for the Lakers there game use and a Kauai for the Spurs game use Jersey We're relic than the autos in the mag stuff. We had a Sean Marion Jersey card for Dallas auto's Ricky Rubio T wolves elvin Hayes Washington They all mall it on Thunder Kai Jones Hornets But that shengun a 25 Ricky dual patch auto Rockets RPA true RPA Nick's Quentin Grimes to 99 True RPA Kings Davion Mitchell to 99 and one of one Nike swoosh patch auto Deandre Hunter for the Hawks and it is game worn Great break Some seriously nice stuff there. Thanks again, everybody. We've got NT and prism coming up now


the SID display week and hello so who are you? Hi my name is Cheri Wada I'm located in Santa Clara California at our satellite office and who are you? Yes I'm Tony Yonezaki working in Silicon Valley with Cheri. So what is DNP? So we provide some components for display materials like flexible like what is this like a flexible what is that? Yeah this is a hard-coating film for passive displays. Hard-coating films? Yes. Yeah you can stand right there so that means it goes on the front to protect it? Yes exactly. Is it plastic? Yes so we coating we make a coating on the film. So it doesn't scratch it doesn't break? Yes we have a pencil hardness to this one is five-way hardness. So in the future every display is going to be flexible right the plastic display plastic OLED? Yes we believe it. Plastic LCD you will be in the top to make sure it doesn't scratch? Yes we believe it. Do you do other parts of the display what is this? This is a car filter for flexible buttons. So this is solar solar? I don't know so this car filter is used for display products and they make the many colors like LCD yes usually they use the grass but we change it for the immidial films to make the carpet the flexible. These products also contribute to the make display flexible. And what are you showing here? So we have a gas barrier here so you know by what the vapor oxygen components is deteriorated so to keep such a phenomenon we provide a barrier film to read. Okay let's walk over there let's check it out. What is this stuff here? This one so conductive film is used for the touch sensor Is it somewhere here? Yes sometimes the conventional ITO films is deteriorated by folding to five. Is this silver nanowire? Yes we keep low conductivity after folding. So are you one of the leaders in silver nanowire or? Yeah now we are mass producing the silver nanowire films. Right and what are you showing here? You say don't touch? Yes so this and this products are the same one so just a moment. Alexa, turn on the window. Okay. So we can change the transparency here. Nice. We can Alexa turn off the window. So it's transparency on demand. Yes exactly. How does it work? Do you mean? How does it work? This technology is uh it's Okay. So smart window. Okay and electronic shade it's the same? Yes almost same we can change the transparency continuously. Nice. Is it already in mass production? Now we are developing and searching for the output product. All right low reflection standard and what goes on in the car? So this is the point is appetite so we reduce this refraction all these frames so while you're driving sometimes that's kind of this refraction is very annoying. So it doesn't oh yeah it's less annoying up there and the company is based in where? Japan. Yes. How long does this company exist? Over 140 years. Yes we are very old company. Doing all kinds of materials? So we started with just printing, printing magazines or paper and we developed our technology with printing so now we're making some film products for right yeah. It's great.


A warm welcome to the thirtieth lecture on the subject of wavelets and multirate digital signal processing. We had begun building the piecewise linear multi resolution analysis in the previous lecture. That was a difficult task. Let us quickly recapitulate what we talked about last time. We had seen that the piecewise linear function obtained by convolving the Haar scaling function with itself which we called phi 1 t was not orthogonal to all its translates. I mean integer translates. The troublemaker translates were translation by 1 and minus 1 and this manifested in the sum of translated spectra not being a constant. But as expected since it was only 1 and minus 1 which were troublemakers, the sum of translated spectra was kind of constant within two positive bounds and as a consequence of that one could bring out or extract from that phi 1 t another piecewise linear function which was orthogonal to its own translates. And we were trying to use that function to build a multi resolution analysis based on piecewise linear scaling functions and wavelets. And therefore, today we take further in this lecture the construction of piecewise linear scaling functions and wavelets. Well, we need to put back a few ideas that we did the last time to bring continuity and those are as follows. The first thing is we noted that we could construct the sum of translated spectra of phi 1 with translations of 2 pi evaluated at all omega and this was given by summation k running from minus to plus infinity phi 1 omega plus 2 pi k mod squared. We saw that this sum of translated spectra was not a constant as required but it lies between two positive bounds. So, for phi 1 t looking like this the sum of translated spectra lies between two positive bounds. Let us call them a, a is strictly greater than 0 and b, b is strictly less than infinity and b by a was equal to 3. This is another way of looking at it. You know if you scale a function by a constant the sum of translated spectra is scaled by the square magnitude of that constant. So, that does not affect this property and that is why I have just emphasized the ratio b by a and the fact that both a and b are strictly positive quantities that is all that we really require. Anyway, what we said was that we could construct a phi 1 till day in terms of its Fourier transform and phi 1 till day cap as a function of omega was phi 1 cap omega divided by the sum of translated spectra of phi 1 with a translation of 2 pi evaluated at omega and a positive square would take. We noted that the sum of translated spectra of phi 1 till day. In other words, this quantity is a constant. In fact, it is of the form sum of translated spectra of phi 1 with translations of 2 pi at omega divided by s t s phi 1 2 pi at omega which is in fact equal to 1 and this cancellation of numerator and denominator is acceptable because the numerator and denominator each neither go to 0 nor infinity. Based on this, we noted that phi 1 till day is orthogonal to all its integer translates. These integer translates could be written as phi 1 till day t minus m for all m integer. Now, we were trying to get the nature or a feel for the function phi 1 till day t and we had noted that phi 1 till day cap omega is of the form phi 1 cap omega divided by the positive square root of 2 by 3 times 1 plus half cos omega. Now, we had also used the so called binomial expansion of the denominator. So, we said that one could expand the denominator in a binomial series a generalized binomial expansion or one could also think of it as an expansion in a Taylor series. So, there are two ways of looking at it you know what we had noted was that if you had something like 1 plus gamma to the power r r any real number and mod gamma strictly less than 1. One could use a generalized binomial expansion or one could also think of the same thing as a Taylor series expansion and essentially this expansion was of the form 1 plus gamma to the power r is 1 plus r times gamma plus r into r minus 1 by 2 factorial times gamma squared plus and so on. So, for the next term for example would be r into r minus 1 into r minus 2 by 3 factorial gamma cubed and so on so forth. Now, one can view this as you see to be a Taylor series expansion in the variable gamma around the point gamma equal to 0 or one could also view it as I said as a generalized binomial expansion. In whatever way one views it what one is saying is that ultimately phi 1 tilde t rather the Fourier transform of phi 1 tilde t can be written as the Fourier transform of phi 1 multiplied by essentially an expansion of this form. So, you know this would be divided by so you would have 2 by 3 inverse here and 1 plus half cos omega there to the power minus half. Now, one could of course write down the binomial expansion here I am not so much interested in writing down the expansion in all its gory detail as taking note of the fact that when we so expand you would get essentially powers of cos omega here.  In other words what we said was cos to the power n omega is essentially e raise the power j omega plus e raise the power minus j omega by 2 to the power n and therefore, what we have here is that ultimately phi 1 tilde cap omega is equal to summation let us say k running from let us use a different variable l running from minus to plus infinity let us call the constant c l tilde c l tilde times e raise the power j omega l as we said times phi 1 cap omega. Now, this was the critical step I took the trouble to repeat these steps because this is a very important idea here the actual calculations in this are cumbersome. In fact, probably it is not even a good idea to do these calculations manually it might be a good idea to write a small computer program to calculate these coefficients c l or c l tilde. You know incidentally if you look carefully these coefficients c l tilde each of them will involve writing out a series. So, to get c 0 for example c 0 tilde you would have to write a series to get c 1 tilde or c minus 1 tilde you would have to write a series. In fact, what is also interesting is that here c l tilde and c minus l tilde are going to be equal from the symmetry in expanding cos omega to the power l and therefore, what we have is essentially if we take the inverse Fourier transform here. So, if we take the inverse Fourier transform of both sides phi 1 tilde t is summation l running from minus to plus infinity c l tilde phi 1 t plus l and therefore, that phi 1 tilde t is a linear combination of phi 1 t and its integer translaves. Now, as I said calculating these c l tilde is a cumbersome job it is not difficult it is cumbersome. I leave it to you to write a small computer program to calculate them that is not so terribly important. What is important is the inference that we draw from here. The inference is that phi 1 tilde is piecewise linear. In fact, if you go a step further and if you just calculate a few of the coefficients one can by calculating a few of these coefficients may be for l equal to 0, l equal to plus minus 1, l equal to plus minus 2 and so on may be up to 3 or 4 and one would see that c l tilde decays. So, luckily as l tends to infinity c l tilde and c minus l tilde d k towards 0. Therefore, what we are saying is that phi 1 tilde t is going to have an appearance that looks something like this. It is essentially going to be well let us draw a small compressed diagram here. So, between 0 and 2 is going to lie phi 1 between 1 and 3 would be phi 1 t minus 1 and so on going backwards. So, this is phi 1 t here. So, you will have something like c 0 phi 1 t, c 1 phi 1 t minus 1 or c 1 phi 1 t plus 1 c 1 and c minus 1 are equal would be something like this. C 2 of course would be even lower and so on. These would decay fast and when you add these you would get the dot dash line that I am going to draw. So, it is very easy to add piecewise linear functions. For example, if I wish to add this piecewise linear segment and this see you know in the region between say 1 and 2 if I wish to add these two segments just look at the end points the sums at the end points. So, this linear segment goes to 0 here and this goes to c 0 tilde here. This linear segment goes to c 1 tilde here and this one has gone to 0. So, at this point it is going to be c 0 tilde at this point it is going to be c 1 tilde and therefore, this is the sum here. So, may be what I will do is instead of using a dot dash I will use a thick line like this.   So, it is piecewise linear a sum of piecewise linear functions. Secondly, phi 1 t as we know well it is orthogonal to all its integer translates. Third property and that is the most important property when we wish to build a multi resolution analysis is that phi 1 obeys a dilation equation leading to a dilation equation on phi 1 tilde. You see it is the dilation equation that ultimately builds the MRA. So, does phi 1 tilde obey a dilation equation? We need to show this because this is the most critical this is critical for MRA. It is critical because it is this dietic dilation equation that ensures that phi 1 tilde t when dilated by a factor of 2 and then translated by all the integers constructs a basis for the next subspace. So, you can talk about a subspace v 0 which is spanned by phi 1 t and its integer translates and then you have a subspace which is spanned by phi 1 t by 1 tilde t contracted by a factor of 2 and integer translated. So, we need to prove this dietic dilation equation obeyed by phi 1 tilde and we will do it again using the frequency domain. Now, you know let us go back to the basic dietic dilation equation that we have on the scaling function phi t. So, you see the time domain equation is as follows for any scaling function for a general scaling function phi t. Phi t is summation k overall the integers in general h k phi of 2 t minus k. So, if you look at the Fourier domain, we would have phi cap omega is summation of phi 1 tilde some let us call it capital H of omega by 2 times phi cap omega by 2. In fact, very strictly speaking you also have a factor of half here where H capital H is essentially the discrete time Fourier transform of the sequence h k. So, you know the critical idea here in establishing that there is a dietic dilation equation is to show that phi cap omega divided by phi cap omega by 2 is essentially a discrete time Fourier transform with the frequency variable essentially scaled omega replaced by omega by 2. So, what we are saying is to establish a dietic dilation equation on phi 1 tilde what we essentially need to do we essentially need to consider phi 1 tilde cap omega divided by phi 1 tilde cap omega by 2 and show that this is of the form half h omega by 2 where h omega is summation k running from minus to plus infinity h k e raise the power minus j k omega essentially a discrete time Fourier transform. A DTFT now when would something be a DTFT when would some function in capital omega be a DTFT well there are two things that characterize a DTFT. One is periodicity in omega. So, the function must be periodic with a period of 2 pi in capital omega. Secondly, it must be inverse discrete time Fourier transformable. So, in fact if the function is bounded all over the interval between 0 and 2 pi or on any contiguous interval of 2 pi then it is of course inverse Fourier inverse discrete time Fourier transformable because the inverse DTFT integral would converge. So, all that we need to establish is that the following is true. So, we need to establish essentially that phi 1 tilde cap omega divided by phi 1 tilde cap omega by 2 is a DTFT and that amounts to saying that is it is periodic with period 2 pi it is bounded by 2 pi on any interval of 2 pi. You know the reason why we wanted to be bounded on any interval of 2 pi is as I said inverse DTFT calculation. So, the boundedness is needed because we wish this integral to converge whatever that function is that function in terms of omega e raise the power j omega n d omega by 1 by 2 pi must converge where f omega is phi 1 tilde cap omega by phi 1 tilde cap omega by 2 quite a mouthful but not so difficult really. What we essentially need to do here is to write down the expression for this and immediately note the qualities that we are trying to assess. In fact, it is very easy to write down phi 1 tilde cap omega divided by phi 1 tilde cap omega by 2. Let us write that down by noting that phi 1 tilde cap omega is actually phi 1 cap omega divided by something like 2 by 3 into 1 plus cos omega multiplied by half and the whole raise the power of 1 by 2 here and the numerator we would have the same thing but with omega replaced by omega by 2 and the denominator here we have phi 1 cap omega by 2 we are trying to write phi 1 tilde cap omega by 2 that is phi 1 cap omega by 2 divided by this put in the numerator therefore, an omega replaced by omega by 2. Now, we are in a very comfortable situation look at this expression carefully for the moment forget about these terms and focus your attention only on these new terms that have come up. This term is obviously periodic with period 2 pi. In fact, if you look at this this is the same term but stretched by a factor of 2. So, if this is periodic with the period 2 pi this is going to be periodic with the period 4 pi. In fact, well let us let us look at it a little more carefully. So, you know focus on now if you look at this is a function of omega then it is obviously periodic. So, this is definitely of the form this is definitely of the form half h omega by 2 with the properties of h desired. Let me justify this what the properties desired forget this factor half that is just a constant. The properties desired are that if you replace omega by 2 omega here. So, that means in place of omega by 2 if you had just omega here and place of omega therefore, you had 2 omega here. So, you know this whole thing has to be of the form h omega by 2. So, another way of looking at it is replace omega by 2 omega omega by 2 by omega or omega by 2 omega to get essentially 2 by 3. 1 plus half cos omega here in the numerator and 2 by 3 1 plus half cos 2 omega in the denominator the all of that to the power half. Let me just put it back before you to show you. So, what I have done is I have replaced omega by 2 by omega here and omega by 2 omega there. Now look at this expression carefully this is periodic with period 2 pi all this is periodic with period 2 pi cos omega is periodic with period 2 pi and this is now periodic with period pi because you have double the frequency here. Now, if this is periodic with period pi it is also periodic with period 2 pi. So, both the numerator and the denominator are periodic with a period of 2 pi of course, the denominator is periodic also with a period of pi, but being periodic with a period of pi also means being periodic with a period 2 pi. So, this entire expression is periodic with a period 2 pi. So, the periodicity is established now for the boundedness to establish boundedness we need to establish when you have a ratio of two quantities and you wish to establish the boundedness of that quantity you essentially need to show that the numerator and the denominator are only between two positive bounds. If you can show that then you can also show the boundedness of the fraction. Let me go back to this fraction to illustrate. So, you know that the numerator is between two positive bounds say A and B. The denominator is also between two positive bounds again say A and B. So, let us write that down explicitly the numerator here is between two positive finite bounds and so is the denominator. In fact, the bounds are the same that is very easy to show. We have already proved this we know these bounds in fact B by A is 3 we know that. So, therefore, the numerator by denominator can at most be when the numerator is a maximum and the denominator a minimum and B by A is of course, less than infinity and it can at least be when the numerator is a minimum and the denominator a maximum and this is also greater than 0. The fraction is also between two positive finite bounds and therefore, the boundedness has also been established. So, if we go back to that ratio you know if we go back phi 1 tilde cap omega divided by phi 1 tilde cap omega by 2 is of the form phi 1 cap omega by 2 is of the form of omega by phi 1 cap omega by 2 multiplied by some q where q already obeys the periodicity and boundedness requirements. So, I put this expression back here this already has been shown to obey the periodicity and boundedness requirements. As far as this is concerned we already know that it does phi 1 cap omega divided by phi 1 cap omega by 2 is known to us we know that phi 1 obeys a dietic dilation equation. In fact, we also know the coefficients of the sequence h k there they are half 1 and half. So, if you look at this composite term this one obeys phi 1 cap omega by phi 1 cap omega by 2 obeys the requirements. This also obeys the requirements this is periodic with period 2 pi this is periodic with period 2 pi. So, of course, the product is periodic with period 2 pi this is bounded this is bounded. So, the product is bounded. So, all in all the ratio phi 1 tilde cap omega divided by phi 1 tilde cap omega by 2 obeys the requirements. And therefore, it is very clear that phi 1 tilde must obey a dilation equation. The dietic dilation equation is of the form phi 1 tilde t is summation k going from minus to plus infinity h k tilde phi 1 tilde t. And h k tilde is essentially obtained from the following considerations. If h k has the discrete time for well if h tilde k has the discrete time Fourier transform capital H tilde omega. In other words capital H tilde omega is summation k going from minus to plus infinity h k e raise the power minus j k omega. If this is true then half h tilde omega by 2 is essentially phi 1 tilde cap omega divided by phi 1 tilde cap omega by 2. And this is essentially phi 1 cap omega by phi 1 cap omega by 2 times that quantity q which we had calculated and let us write the full the complete thing in all its gory detail. So, this would essentially be half well you know you know what this ratio is it is essentially the discrete time Fourier transform of the sequence 1 to 1 or half 1 half. So, it is of the form some constant let us say some c 0 times 1 plus e raise the power minus j omega. Plus e raise the well 1 plus 2 e raise the power minus j omega plus e raise the power minus j 2 omega square well this does not have to be square this is, but omega has to be replaced by omega by 2 here because you know you do replace the variable omega by omega by 2 as you know. This one multiplied by q, q is of the form well if you forget about the two-thirds. So, if you want we can keep the two-thirds where they are two-thirds 1 plus half cos omega by 2 this whole thing is of the form half h omega by 2. So, this whole thing gives you the sequence h tilde. Now, with this working of course this is definitely a difficult discussion as we see it is non-trivial what we have done is to establish the existence of this h tilde k as you could see actually calculating h tilde is a difficult job not at all convenient. All that we have shown is the existence of this h tilde we have also shown how to construct h tilde. If you do manifest the courage to carry out this inverse discrete time Fourier transformation you could arrive at h tilde k it is a difficult job, but not impossible. Moreover, once you have h tilde k you have the impulse response of the low pass filter in the orthogonal MRA. Once you have a dyadic dilation equation the coefficients of the dyadic dilation equation give you the low pass filter response. Once you have the analysis low pass filter you could construct all the other filters the synthesis low pass filter the analysis high pass filter and the synthesis high pass filter ugly, but possible and many things need to be noted here. The frequency response of this low pass filter is not only infinite in length these h tilde k's are going to be infinite in length. Let me put back this expression before you you know if you look at the inverse discrete time Fourier transform of this you could calculate it by making a binomial expansion of the numerator and denominator here you could do that, but as you can see this expansion will last forever. So, it is going to have an infinite number of coefficients it is going to have all the vices which make it unrealizable. It is going to be infinitely non-causal it is going to be irrational and these are going to make this unrealizable as a filter. I mean one can write down the impulse response, but that filter may or may not be realizable. So, to get an orthogonal piecewise linear multi-resolution analysis one requires unrealizable filters. Now, we could get it so much more easily by using the bio orthogonal multi-resolution analysis which we did in the five three case. Having spent these two lectures on constructing the orthogonal M R A with piecewise linear phi we can now see the advantage of bio orthogonal M R A. There you get at least the piecewise linear phi on one side and the side where you have a length 3 low pass filter. Now, just to complete the lecture one little detail once you know the impulse response of the low pass filter on the analysis side you also know how to construct the wavelet because the analysis high pass filter coefficients impulse response coefficients would construct the wavelet from the scaling function you know that psi t can be expanded in terms of the dilates of phi t phi 2 t minus n with the coefficients in the linear combination given by the impulse response coefficients of the high pass filter. So, once you know the low pass filter you know the high pass filter impulse response those impulse response coefficients or samples give you the linear expansion of psi t. So, constructing psi t is again existentially possible though extremely cumbers. So, this is a clear illustration of how we could bring in the variant of introducing infinite length impulse response filters get piecewise linear multi resolution analysis which are orthogonal, but a better option is then to go for bio orthogonal M R A which we did earlier. Now, we understand why with that then we come to the end of this lecture and in the next lecture we shall take up one more variant of the multi resolution analysis the wave packet transform. Thank you.


Well, some of Cunning's best horses and riders cussled it out for the Open Classic Challenge Championship at the 2018 Summer Spectacular. I'm with the man of the moment, James Cain, who scored a massive 2-29. Congratulations to you and how does it feel to score a run like that? Really good, but had before I start, I gotta thank Nadine because last time I did an interview, I didn't thank her and they've been like giving me shit about it. So, my wife Nadine, I appreciate it. Okay, I'm sure you'll get rewarded when you get home. Now, tell us about your run tonight. Give us a play-by-play because it was such a great run. Tell us about, you know, what you talk about with your help, how each cow felt, that kind of thing. Well, in the first, I was in the first launch too and I cut too much cow and so it kind of, I wasn't real sure where to go with it and stuff like that. So, I wanted to kind of cut something in between too soft and not being able to mark something and being too risky to where it took me out. And so, I feel like that, you know, like John and Josh were great in helping me find cows and we cut what we wanted to cut, even the third cow that we chipped. It was something we talked about and we kind of needed to be funny and stuff. Well, and it really felt like it built each cow, the crowds, momentum, they got really into it. And by that third cow, they were going, wow, it was almost tangible. Does it lift you up? How does it feel? What do you experience? Just a big adrenaline rush. It's really cool. Does it make a difference to your showing? Does it help or does it, your sewing zone?  Just hope you don't mess it up. So, tell me, now this is the third time you've won in here in this pan. That's extraordinary. What's your trick? What's the skill here? What's the secret with this particular event? I guess it just takes me this long to get them where I can win on them. I got Serena Doola, she was six whenever I won Derby on her. And then once my boon was six and this one was six, I would like to, there's been several times I thought I could have won another event on him and just little stuff happened, little bubbles here and it was just meant to be here. I felt like I took a big risk cutting, chipping my third cow with 30 seconds and it was a risky cow and it paid off. So it bit me in the rear and when I get it, went home power and so it worked. Now, there was no semi-final for this event this year. In fact, you had to kind of get really a couple of 219s, which is pretty tough to progress. What's your thoughts on that? Does it add some pressure or is that just competition? I think it's just competition. I mean, I think it's the way it should be. Okay. Now, your horse, PG, heavily armed. How long have you had this horse? I've had him probably since the fall of his two-year-old year, I believe. So, 160,000, the horse is one. Five or six-year-old? He's six. Tell us what this horse has been like to train, how this horse has developed. Like to me, he's a typical clay gun and I've rode a lot of clay guns and I like clay guns. They're very strong, physical. They usually get better the longer they walk, the older they get, the more maturity.  If anything, as a young horse, he might be just too big of a move. But as he's gotten older, he's gotten smarter and just turned into a true blue cowl horse that you can kind of go over the tough cowling. Now, you did have another horse tonight in the first scent. It didn't work out. What's your strategy in that sense when you've got a second horse to just lay it all out there? Well, like me and Josh Towns and we set up a pop after I went and kind of watched what cows are being good at trying to get a feel for the cows. Because I think the cows are a little different feel, had a little different feel to them tonight and they're just trying to make sure we kept good cows. So on that, why do you think they feel differently? Do you think the temperatures come into play when it's 110 outside? Does that make a difference? Yes, I think so.  And also, it would have to be a challenge for getting your horse ready. I mean, they're athletes, but they're standing out in that heat for most of the day. How do you make sure that they've got everything to give when it comes to showing? Well, I mean, just try to make sure, kind of keep them cool, rinse them off, take care of them. Just a normal horsemanship. And what have you got for the derby? I've got the two horses. I've got a, I did guard this cat smart gilding for Kathleen Moore that I made the Super Six finals on. And I've got a cat team master's in America, John Johnson. Well, good luck with that and congratulations for tonight. Oh, thank you. Oh, thank you. Thank you. Plenty more cutting tomorrow and of course another final of the non-pro classic challenge. Make sure you're watching the webcast tomorrow. It's all mine. Good night.


Alright guys, I actually took one head apart fully, so I can give it to the machine shop. And so I go through this one more smoothly for the sake of the video. Now some heads you don't need to take the valves out, but in this case what I've noticed is, I don't know if you're going to see it, but the intake valves and this head are very, very close to the edge or the surface of the head. Obviously, we don't want to deck the valves. Make sure you've got a clean area, everything needs to go back to the same spot, I'm talking about each valve, okay, alright, I'm going to talk through it, just watch me do it, trying to trap these off today, anyways, let's go, wait, same kind of order, taking off these bolts to get the rocker assembly off, you don't want to be just taking bolts off, just like this, cylinder heads, there's a process, let's say number 1, 2, 3, 4, 5, 6, 7, 8, and when you put them on it's like, well, when you put them on it's 1, 2, 3, 4, 5, 6, and then reverse, so 8, 7, 6, and so on, 5, just another cross pattern. Now it's time to compress the springs, so we can take the valve keepers out and then remove the valves. So I'm running out of room, but this is super, super important, I've got to make a spot for just the valves, they're going to sit there until they get installed, a clean spot right there. So you can see that these are the exhaust valves, and this is the intake side, this is the front of the head, or the engine, and that's the back. So I'm going to start with the exhaust in this case, and I'm just going to go put them in order, 1, 2, 3, 4, 1, 2, 3, 4, alright, using this kind of tool, this is a valve spring compression, compressing thing tool, this is pretty much for, I guess almost all the heads out there, alright, so you do have a few different fitments for the top of the valve, choose wisely, this end will go here, and this pointy head will go on the surface of the valve. I guess I should actually, well, okay, I'm just going to do it in reverse, it'd be nice to have this turning thing, and this, actually I'm going to switch these two, actually what am I doing, I can just do this, the switch. So top of the valve, keep a magnet handy, I'm going to have to catch those keepers, 1, and that's 2, then you can slowly undo this, remove the tool, remove the spring, and then push on the valve, and that's your exhaust valve, now check this out, you want to be careful with this area here, this actually moves up and down on the valve stand seal, so if you scratch this, that's going to leak, and, or when these are old, at startup after your car sits for a while, that's when your car smokes for the first very, you know, few seconds or so, when these are bad, or this is bad, but check this out, check this out, okay, look at this, hopefully you're going to see these little dots kind of eaten away from the heat, I know this camera is not good at close-ups, but yeah, so these, these are, I mean they're not bad, you could just put them back and get, you know, thousands and thousands of miles from it, but got a dyke it, so might as well, well, I am, you know, removing these anyways, and I will let these, alright, so we took one out, now another exhaust valve, alright, and just keep going, alright guys, that's it for this one, now what's left is the valve stems, I'm not going to worry about them, I'm done for today, so these, I'm going to mark, I can't see left or right on these heads, so I'm going to just mark it, mark L, so this is the passenger side, which is the right one, I know it's on the left, but you know, as you look, as you sit inside, that's your left, that's your right, so this will get an R, that one will get an L, so these are going to the, to the machine shop as they are for decking and checking for leaks, once they get back, I'm going to, first thing we will do is, we're going to lap the valves, I do have a video on it, I'm going to try and put it up there, when I was doing my 06 STI, same process, everything on, everything else looks good, all these guys here, the camshafts, these guys, everything, the surfaces look very good, so we should gain a little bit of power and maybe smoothness by redoing the valves, and also I've been thinking, I actually spoke with the body of mine that has done a ton of Subarus, to go with the same cylinder head gasket, the single layer one, these are now not graphite any longer, I think Subaru got done with those in 99 or 2000, something like that, not that these don't have an issue, but if I were to put the MLS, the multi-layer gaskets on this, we're going to lose compression and power, so it's a big no-no for this race car, so we're going to go with the single layer, dealer head gasket, valve cover gasket, intake manifold gasket, exhaust gasket and all that stuff, next one I'll show you how to check the level of the block with a straight edge, we're also going to clean up these guys too, see you soon!


Today I have the privilege of speaking with Chris Paul from Bald Eagle Gold. How are you today? I'm doing well, Tracy. Thanks. How are you? The reason I paused everyone and invested Intel Land is that Bald Eagle Gold is really Bald Eagle Silver. Tell us, Silver, Idaho, great time to get introduced to this story.  Tell us why Chris Thompson loves this story so much. Well, thank you for that introduction, Tracy. So yes, my name is Chris Paul. I'm the CEO of Bald Eagle Gold Corp. And we trade on the TSX Venture under the symbol BIG. So Bald Eagle is focused on a high grade silver asset in western Idaho, which we just recently acquired. So this project has over 300 historical drill holes and a historical silver resource, but hasn't been explored in close to 40 years. So this is an asset we purchased privately last summer. And we are only just now in the process of reopening what was once a very important silver exploration account in the state of Idaho, which the Fraser Institute has ranked the number one jurisdiction in the world for mining. And even though it's early days, it seems to me that you're attracting some star power. For instance, your new chairman, Louis De Silva. Tell us a little bit more about him and also believe you have a new advisory board member that you just announced in the last month. Yeah, no, exactly. So first came the project and then came the people and the firepower. So Louis De Silva has now joined the company as our new chairman. And Louis comes from a very strong operational background. So Louis was formerly the president and CEO of a company called Andean Precious Metals. So Andean was actually a mid tier silver producer in Bolivia. So Louis is bringing that actual mining and operational and development experience with him. So he was a very strong addition to the company and it's going to help us as we build this resource and going to help us with some of the more sort of mining related background that he's got. So yes, and we also added a new advisor to the company. So that's Tom Henriksen. Tom has a very interesting background. So Tom has won several awards. He recently won an award in 2018 from AMEBC for discovery excellence. So his most recent discovery success was the Hot Maiden deposit in Turkey, which is a very large goal deposit. But if you look at his resume and the most recent news release, he's made close to a dozen discoveries that have become mines in his lifetime. So Tom, you know, like I mentioned, he's got a very interesting background. He actually started his career on the Hercules silver asset in 1970, which is where he did his PhD thesis. So this is a bit of a full circle for him. We actually got put in touch with him recently because he was working in Idaho. So him and his partner, Doug Elliott at Haywood, actually built Midas Gold, which is now perpetual resources. So he's an Idaho guy as well. We got put in touch with them. We started talking about the PhD work that he'd done in his early days on the Hercules. He was very excited to get involved and he's going to be helping us out on the technical side going forward as a geologist. So a company relatively new, but we have a lot of interesting parties or well-known players attracted to this company already. What would you say the compelling reason is for this?  Is that correct? Yeah, that's exactly it. So right now Bald Eagle is trading at what I consider to be a very undervalued price because this is not an early stage exploration story. This project brings a historical resource. It's had over 28,000 meters of historical drilling and it is primary silver. So the minimization is over 70% silver by value. So this really makes Bald Eagle one of the most pure play silver stories on the market, which is very rare to find especially in a jurisdiction like the United States, which is very safe. We also have a very large target size. So the historical drilling, which was over 300 drill holes, actually covered over three and a half kilometers of high grade silver. This is all within one continuous rock unit. So we think that there's very good potential for expanding on that historical resource and potentially reaching upwards of 100 to 200 million ounces of silver on this project. We are right now, we're in the final stages of permitting for a 3000 meter phase one drill program, which is going to be targeted that both drilling within the historical resource area, which is going to allow us to basically upgrade that to 43 one on one compliance and about half the program is going to be purely exploration. So we're going to be stepping out on that resource looking to build on that resource, as well as drilling some pure exploration new targets that we've generated from some work we did last fall. And so, of course, if someone wants to get in touch with you, a lot of people will be attending PDAC. Will will you be there and how can they reach you? Yes, absolutely. I will be at the PDAC conference this year. And the best way to reach me is actually by email, which is cpaul at bald eaglegold.com And of course, for those of you looking for more information, here's the website. Thank you, Chris, so much for joining us today. It is always a pleasure. Thanks very much, Tracy. I appreciate it.


find it interesting that you agree with me and say, yes, he was a real historical person. He just didn't arise from the dead. Well, why would you do that, Matt?  Because there's absolutely nothing extraordinary about the notion that there was an itinerary Jewish rabbi or a collection of them or stories about a person. There's something incredibly extraordinary about someone rising from the dead. Those two claims are not remotely in the same category and are not remotely establishable by the same sort of evidence. And certainly not second or third hand evidence, here say decades later, just because they may have known someone who knew someone.


This is Witch Based News for Friday the 15th of December 2023, I'm CMDR Buur. In Elite Dangerous News this week, we go 80's nostalgia heavy with some original Elite Deep Dives, there's free arcs for the live game in the run up to Christmas, the interim patch arrives in the game and more. You know how this bit goes, please like, subscribe and ding that little bell so that YouTube shows you all our content and if you'd like to help directly support our work here at the Burpit you can also support us through Patreon, links to that and everything else are in the description below. We start this week with a brief reminder that as it's the run up to Christmas, Frontier are doing their usual daily free arcs drops into the live version of the game. All you need to do is log in to your account once a day and get as far as the cockpit or on foot, don't stop at the menu, log all the way in and then an increasing amount of free arcs will be added to your account every day right up until the 25th of December. Also, just as we were going to press today it seems this year's Christmas in-game event started in the live version of the game. Check your mailbox and you'll see a message entitled Help Needed at Angel Station. We won't post any spoilers but suffice to say, following up these messages at Christmas invariably results in gifts in one form or another as well as being fun so they're absolutely worth doing. You have until the 4th of January to complete the quest. Two fantastic pieces of technical elite nostalgia passed across our desks this week that could help pad out a long super cruise journey or two at the very least with both pieces going a long way toward explaining why elite had such a cultural and technical impact on everything that came after it, an effect that continues to this day. The first is by UK based web developer, producer and journalist Mark Moxon. It's a series of very technical deep dives into how Ian Bell and David Braben managed to cram a game as huge as the original elite and its various versions onto some of the machines of the time, specifically the BBC Micro and the Nez console. The deep dives cover a multitude of subjects in fine detail from memory usage in the BBC Micro, the program flow of the main game loop and generating the galaxy and even all the way through to drawing satin on the game's loading screen. There are more Nez specific deep dives being released in the run up to Christmas all the way up to the 22nd of December, so be sure to check back for the latest ones. You can also follow Mark on the artist formerly known as Twitter as he's announcing the releases there when they go live. The second piece this week is a just as fascinating video essay from YouTuber Alexander the OK, entitled The Game That Couldn't Be Written. That similarly breaks down the creation of elite starting with the birth of its first host machine the BBC Micro and following through with some slightly more easily digestible chapter breakdowns of how procedural generation works, elite's many gameplay innovations and the closing chapters which I found particularly interesting regarding elite's lasting cultural impact. Alexander himself says that the story surrounding elite is much better known here in the UK but gaming culture on the whole tends to be viewed through the lens of the US and Japan so viewers from outside the UK may not be fully aware of the games initial impact and lasting legacy.  The promised interim and stability patch updates 1701 dropped into the live game this week bringing with it a number of bug fixes, gameplay tweaks and optimizations of particular note for me personally transferring cargo to fleet carriers and SRV's no longer needs a re-log to see the desired effects show up in the UI. After accepting a mission from the mission boards you will now be returned to the same previously selected mission category screen rather than the all category screen significantly reducing the amount of required UI navigation if you're picking up multiple missions of the same type and authorised class Thargoids engaged in scanning at spire sites will now attempt to jump out if attacked making that scenario somewhat less of a turkey shoot. As you'd perhaps expect there were also a few bugs and glitches around the still fairly newly implemented Thargoid spire sites that were stomped with 1701. You'll find the full patch notes posted by Paul Crowther to the FDF official forums linked in the description below this video if you want to check them out for yourself. As we reported last week Frontier did confirm that update 18 is still in our future and when we hear any more about that we will of course let you know right here. In a perhaps not entirely unexpected move this week community manager Paul Crowther took to the forums to announce that the annual Christmas live stream from Frontier for Elite Dangerous had been cancelled this year. As we've covered in the last few editions of this show Frontier is in some tricky financial waters right now and this situation has resulted in the company instigating a wave of cost cutting staff redundancies right across their titles including Elite Dangerous. In such circumstances it would make hosting a public celebration of the holiday season on a live stream somewhat inappropriate and insensitive at best. Clearly in very uncertain times for Frontier for Elite Dangerous and for the wider community right now and the cancellation of what has been a live stream calendar highlight every year is not going to ease that uncertainty. All I would say is to reiterate again that quite correctly FDF's first priority must be with the people at the sharp end who are being directly affected by what's happening in the company right now and in January FDF will be reporting back on how the cost cutting measures have progressed. Frontier did state just last week that they intend to continue to support and develop their existing library of titles including Elite Dangerous and we think it likely that following that report will start to get a better idea of what shape that future support and development will take. Indeed Paul's post does go on to say that they plan to return in January with more news on Elite Dangerous. Were you playing Elite on the BBC Micro on Nez in the 80s? Was your favourite bug squashed in the interim patch and what would you like to see for Elite Dangerous in 2024? Let us know in the comments below. That's it for now, thanks very much for watching, we'll be back later this week with more videos, until then 07 Commanders follow the Greens on the way out and do keep clear of the toast rack, we very much look forward to seeing you next time.


Hi, my name is Abigail Solis. I'm originally from Northborough, Massachusetts. I went to undergrad at George Mason University and completed my master's at University of Pittsburgh. I'm most excited to explore the mountains and learn from the world renowned medical team at the Stemit Clinic.


Oh, he's saddened a little bit with us. Mr. Meadloaf, 30 minutes after 12 o'clock on a Saturday afternoon, welcome back. It is Saturday Shuffle, Paul McDevitt and the final hour, as always, and we are live here just protesting out something here if you're watching us on Facebook. You probably think to yourself, God. Desperate, bad enough in the radio, we're on TV. And, hi Paul, I've just got another corny joking. My wife bet me a thousand euros that I couldn't turn spaghetti into a car. I should have seen her face when I drive past that. And Matt, thank you very much. Paul, my very first job, where I made my first money, was in a photocopying shop where I made my first money. I love it. My Alex, hello to you. Thanks for taking care of your corny joke, Paul. My uncle wore his heart on his sleeve. Lovely fella, but a terrible surgeon. Well, I went to McDonald's and asked for a burger the other day in fries. The man says, get off my farm, oh man. Oh, it's $660, $25,000. And as always, I'd love to hear from you. You're going to request a dedication. Oh, cah, hello to you and Derry, listening in. I've got your song on the way in the next couple of hours. Don't forget, coming up around a bit of half past 12, we will have a tune of Roe from Dean Martin, so why not? And I request Derry Roe. Paul, any chance for some Barry White in the program? Of course.


And in River State, where the situation seems to be worse, the River State Command of the Nigeria Security and Civil Defense Corps says it will continue to levy war against all theft and economic sabotage by elements who are bent on undermining the economic growth of the country. Details in this report. The commandant of the MSCDC River State Command, Michael Ogre, said these when he paraded three suspects arrested with one large wooden boat and two vehicles leading with illegal refined petroleum products. He informed that the MSCDC River State Command's Marine and Land Patrol team arrested the suspects and impounded the said items following the dissolutionary constitution of the anti-vandal unit of the call across the federation. He relayed that the suspects loaded the illegally refined petroleum products in cellophane bags and concealed them in sack bags. Commandant Ogre, while issuing a stand warning to those milking the economy of the nation through oil theft and illegal bank rein activities, asked them to relocate from River State, declaring that as the new sheriff in town, he will not compromise nor leave any student turned in dealing with them. He also charged the newly constituted state anti-vandal team to scale up their game in the war against crude oil theft and pipeline vandalism. The River State MSCDC both affirmed that the suspects will have their time in court appealing to the public to provide a command with credible and actionable intelligence that will enable the command clamped down on oil thieves in the state.