In this video we're going to apply Burnside's counting theorem to solve some combinatorial problems using group actions. Now before we do that it's important to never underestimate a theorem which counts something. Combinatorics is a very powerful discipline in mathematics. Counting things can be very very useful so we should always always pay attention to things that count things like Burnside's theorem, the fundamental counting theorem, when it comes to these group actions. Burnside's theorem is a very important combinatorial tool especially in the scenarios that we're going to do in this video because it can often be difficult to count the orbits of a generic group action. Now before we get to our combinatorial example let me present one more example of a group action that is I want to present a way of constructing a new group action from an old group action because these are the type of things we're going to count in these combinatorial problems. So imagine we have a group G that acts upon a set X, it's just some generic group action, I don't care what it is it's just we have a group action. Then if you take any other set Y, it might have nothing to do with X, I don't care, but if you take any set Y you can form the set of functions which is commonly denoted as Y to the X. So Y to the X will be the set of all functions of the form F takes as its domain X and as its codomain Y. So Y to the X is the set of functions. The set of functions Y to the X is itself a G set in a natural way. The set of functions inherits the G set structure from X and you do it in the following way. Well, since Y to the X is a collection of functions and G is a group, we have to make sense on what does it mean for a group element to act upon a function. It should produce a new function. So what is the function G dot F? So the image of the function F when it's acted upon by G. It should be a function that takes as its domain X and then lands in Y. But remember the domain of the function F is a G set so it makes sense to act on X by the element G and that's exactly what we do. So to define the function, I'm going to tell you what the rule is because a function is rule that assigns everything to its domain, something in the codomain. So the function given by the rule G dot F will be if you take an arbitrary element of the domain X, G dot F of X will then be F of G dot X. So in some regard, you're sticking the G inside of that. So the function that's acted upon by F will be that function which then always acts upon its input by that element G. This is a lot easier to do if you have a left action. If you have a right action, you can do the same thing, but you have to do it on the right and it's got to be like an inverse that's common when you use right actions. We always do left actions, of course, in this video. So we can construct an action on functions by just pushing the action on to the input. That's exactly the type of action we're going to see in just a second. So now let's get to applications of Burnside's theorem. How can we use Burnside's theorem to help us count combinatorial objects? Well, imagine for the moment a square such as square is visualized right here. Imagine we have the task where we want to color the vertices of a square. So like we want to color this one, we want to color this one, we want to color this vertex, and maybe we want to color this vertex where repetition could be allowed. We could allow color this a different vertex, the same color. So on the screen a moment ago with my circles, I did yellow, yellow, green and blue. That would be a coloring of this square. So imagine we're considering how many different ways we could color a square. Now for simplicity sake, let's just focus on two colors because if you all have one color, there's only one way to do it. You color all the corners the same color. Alright, we don't worry about that. But let's say we want to label or we want to place two colors on our square and we'll just make it simple to say black and white. How many ways can we do it?  So in essence, you have two options, two options, two options, two options. So the counting principle in play here is that if these decisions are made independent of each other, which they are, the coloring one vertex doesn't affect the other, you're going to get two times two times two times two, which is equal to 16. There are 16 ways to color that square. But that's not exactly true, is it? I mean, if the if the square is sort of stuck to the ground, you can't move it, then sure, there's gonna be 16 ways to color it. But what if the square is allowed to move? What if it can move freely in the plane or move freely in the space? You could rotate and flip it kind of like the Dihegel group does to a square. What about that situation? Well, I want you to look at these four squares for a moment. Here's a square which has a black corner, and then three white corners. Here's one where this of course is going to be in the northwest corner is black. But what if the northeast corner is black, all the others are white? Or what about the southeast is black, all the others are white? Or what about the southwest corner is black and everything else is white? These would be considered four different colorings by the strategy we were employing a moment ago. But do we really want to consider these different colorings? Like imagine this is some toy that a kid plays with. You start spinning around your square. If you started off by coloring the northwestern corner black, they want you spin it around. It's really the same square because once you pull that square out of the box, each and every one of those squares is a possible orientation. So the problem we had earlier is when we started coloring and we got two times two times two times two is we believed we could tell the corners apart. Like there was some labeling we thought they were distinguishable. So like oh, the first vertex could be black or white. The second vertex could be black or white. The third and the fourth could be black or white. But what happens when we can't tell the vertices apart? Well, when you start rotating the square, when you start flipping the square, you can't tell apart who was who was the original vertex one because it got moved around. So because of these equivalent orientations, it becomes a lot harder to count the number of square colorings we can have. So this is what Burnside's theorem comes into play because as I'm talking about moving the square, this is a D4 action on the square. The D4, the dihedral group there, it rotates the square, it reflects the square.  Any reflection of that coloring would be an equivalent coloring. So as D4 acts on the square with its colors, that creates equivalent colorings and we don't want to count those colorings as different ones. So our 16 number has actually over counted the numbers because two different colorings are really the same, but we counted them as if they're different. Now D4 does act on the square, but the G set that we're looking for is not going to be the square. It's going to be colorings of the square. So imagine we have our set Y right here for which Y is the square. Okay, so we have 1234 as the square right there. So the vertices of the square. We're not trying to count, we're not trying to count the orientations of the square. We're trying to count the colorings. A coloring of the square is a function from the set 1234 to the set BW where we can think of B is black and W is white. So as we think of this function, we have some function here F where you have Y over here and then black and white over here. We have to make a choice on these vertices.  The second one is white. The second one is white. The fourth one is black. Every function is a coloring. It's an assignment of the vertices to a color. So the colorings are what we're trying to look at the equivalent colorings. So the set that we actually care about is the set of functions from the set 1234 to the colors black and white. So we're looking at this function set right here. What are the possible colorings? But when you think of your function, you have two options for this first, two options for the second, two options for the third, two options for the fourth. So this number 16 that we got earlier is significant, but we were counting the wrong thing. 16 is counting the number of colorings of the square, but we're trying to count the number of equivalent colorings. So two different equivalent colorings are considered the same. And so we're counting the orbits of the coloring under the orient, the rotations and reflections of the square. So Burnside's theorem is relevant here with D4 acting on this set X because it's a function set. It's a function set on a G set on a D4 set. So we can do that. So let's look at the stabilizers for a moment. I should say the stable sets. Who is stabilized by the identity? That's an easy one. Every function, every coloring with its orientation is fixed and therefore you're going to get 16. The stable set of the identity is always the entire thing. So we get 16 right there. So let's think about rotations for a moment. Who gets fixed by a rotation? So if you have a square over here and you rotate it, well, if this color was white, when you rotate it, then this color has to be white, okay? And so this one will move over and become this one over here. But this has to be the same thing we started with. So if this one was white, when we rotate it, it will move down here and that would then be white. So this one has to be white. Similarly, if this one was white, because it starts and stops at the same spot, that the square, this will move over here and be white. And this one also be over here coming white. So basically, when you rotate the square, the only way you can have the same square when you're done is if all of the colors are the same. So they're all white or they're all black. I did the example with white, but we can do it all with all black. So there's two options.  By similar reasoning, if you rotate three times, that's the same thing as rotating counterclockwise once. And the only thing that will be left stable by that action would be the all black or the all white square. So you only get two options right there. Okay, what about R2? If we rotate twice, that one gives a little bit more flexibility here, because if this one was white and we rotate it twice, then that means this one's going to have to be white, but then when you rotate it back, that's going to be white again. So these two colors, these two corners have to be the same color. Okay, but this one right here, it could potentially be a different color because when you rotate and come over here, this one's going to have to agree with it, but then this will rotate back and it'll have to agree with it as well. So these ones on a different diagonal actually could disagree with each other. They don't have to be the same color. So basically what I'm saying here is when you rotate this thing, the corners that are opposite of each other, they will have to be the same color. There's two options there, but then these ones, they will have to be the same color too, and it could be the same color as this one, but it doesn't have to be. So you get two options for this color and two options for this color. So that gives you two squared options, which then turns out to be four. So that'll take care of the rotations. What about reflections now? Well, if we think of reflections here, let's do s. s is the horizontal reflection, so we reflect like that. Well, with the horizontal reflection, whatever this color on this corner has to be, it has to be this color too, because those two are just going to swap each other. But then this one, it could be a different color. It'll come down here and agree with this color. So those colors have to agree with each other for this permutation. So in other words, we're saying that this side of the square has to have the same color, which there's two options there. And this side of the square is going to have to have the same color too, which it could be a different color, but you'll have two options there. That gives you two squared, which is equal to four. Alright? Jumping down to r squared s, we get a very similar picture there, because in this situation, you have now a horizontal line. And by similar reasoning, these two vertices will have to be the same color, which gives you two options. And these two vertices will have to be the same color, which gives us two options. Because we're looking for what is a square that's not changed when you do these rotations. There's reflections in this situation. So like if these are black and these are white, when you reflect it, you won't be able to tell the difference on the coloring. It doesn't give you a different coloring. So those are stable. Those are stabilized by r squared s. So again, you're going to get two squared, which is equal to four. You'll notice I skipped rs, because that is a reflection, but it's a diagonal reflection that actually has a different effect on things. Because notice that when you look at this element right here, this corner, whatever its coloring is, let's say it's black, when you reflect it over, if this is the same coloring, this would have to also be black. Okay? Basically, everyone who's in the same orbit of the D4 action on the square has to have the same color. But what about this one right here? This one could be white because these two colors doesn't affect it. But this one could also be white or black. It doesn't matter what the other colorings are. These, if we look at the orbits of the vertices by the D4 action, these two color, these two vertices have to have the same color. There's two options there. But this one, because it's left fixed by the reflection, it could be its own color. You have two options. But also this corner down here could also be a different color. It doesn't have to coincide with the other vertices. So you actually get two times, two times two options, which is equal to eight. And so then if we do this whole exercise one more time with our cube desk, that's the diagonal, that's the reflection across this diagonal, I should say. Same thing. This could be any color. This could be any color. And let me use a different color to emphasize. It doesn't have to be the same color. But then these ones, they do have to be the same. So you get two, two, two, two times two times two is eight. And that then gives you the size of that right there. All right. So let me bring up my notes here. So then what we then see is we counted up all these stable sets. We then get that by Burnside's theorem, the number of orbits we equal one over the group, the group's order, this is the diagonal group, it's order eight. And then you add up all of these stable set sizes.  For which, while some of these numbers don't look like they're divisible by eight, notice if you take two plus four plus two, that's an eight. Four plus four is eight.  All of these are actually divisible by eight, which of course has to be the case. K is an integer, but yet you're dividing it by an integer. That means this has to add up to be something that's divisible by eight. It'll happen eventually. So sixteen plus eight plus eight plus eight plus eight is divisible by eight. You get two plus one plus one plus one plus one, which adds up to be six. This tells us, by Burnside's theorem, there are six colorings of the square. And I can give them to you real quick. There's the all-white coloring. There's the all-black coloring. We talked about those. We also saw the example where one of the corners is black and the other corners are white. While there are ways you could rotate that one, those aren't considered different squares. There's also its dual where you have a white corner and three black corners. But then what about the other two? There should be two more. Well, these other ones are going to be when you have two blacks and two whites. So what if like both the tops are black and both of the bottoms are white? That's one. And then the very last one is what if the colors are kitty corner of each other? What if the blacks are diagonal from each other and the whites are diagonal from each other? So up to equivalents, these six squares are the only squares that you can get by coloring, right? Every other square will be equivalent to them up to rotation and reflections, which is what we're trying to consider here. Before we do another example, I want to point out a very important observation here. Looking at the sizes of all these stable sets, what's in common with all of them? We end up with 16, 2, 4, 2, 4, 8, 4, 8. These are all powers of two. And notice where they came from, right? 16 was 2 to the fourth. 4 was 2 squared, 8 was 2 cubed. We also had a 2, which of course is just 2 to the first. They're all powers of two. Is that a coincidence? The answer, of course, is no. Each of them has a form. Each of the stable sets had the form 2 to the A. And where did where do these numbers come from? Where does the base come from? Where does the power come from? The base the base is easy to explain here. The base is the number of colors. Because whenever there was a decision that had to be made, that decision was either black or white. Every decision had two options you could appeal to. And then the letter A, it represented how many choices had to be made. That depended on the orbit structure that was going on with our square. And that comes directly from the dihedral action on the square. If we wrote our elements of the dihedral group as permutations on the set 1, 2, 3, 4, it would look something like this. The identity fixes everything. 1, 2, 3, 4 is what R is. R squared is 1, 3 and 2, 4. And then we can keep on going with each of every one of these. We could think of the dihedral group as a permutation group acting on the set 1, 2, 3, 4. Now I want you to look at the orbit structures here. The identity has four different cycles in its decomposition. We typically skip one cycles because they just leave the elements fixed. But for the sake of counting, I want to include all of the one cycles. So the identity has four one cycles. Remember what the size of X1 was? It was 2 to the fourth, 4 being the number of orbits, the number of cycles in this decomposition here. Then what about R? If you remember from the previous slide, R, when we looked at XR, it had just 2, which is 2 to the first. Your base 2, because you had two colors, two options. But then the exponent was 1 because you had one cycle in the permutation action there. What about the other ones? Well, for XR squared, we had a 2 squared. 2 came from the two colors. That's the base. And then the exponent, there's two cycles in the decomposition. So that's how all of these are going to work. So the how many for this one? You had 2 to the first, one cycle. For this one, you had 2 squared. 2 cycles. This one, R, S, you actually had 2 cubed. Where do you get that? You get 1, 2, 3 cycles in the permutation representation. And therefore, you get 2 to the third. The number of cycles tells you the number of options you had. And you had 2 options for each. That's where the 2 to the 2 cubed came from. So with this perspective about the permutation cycles and the number of colors, this idea of coloring up to equivalence becomes a whole lot easier. Let's look at another example of this. I want to redo this exact same example that we did a moment ago, but this time let's color the squares vertices with four different colors. If I have four different colors, how many possible ways could we color the square up to equivalence? Well, we're trying to compute k. What are the number of orbits of this group action up to equivalence? That's what orbits are measuring here. Well, the group is still the dihedral group. It still has order a. So you have this one eighth in front. So then look at a stable set. Let's see if I can get this to fit on the same screen. Voila! We have all of the permutation decompositions of the dihedral group right here. So the first one, you have now four colors, but you still have four cycles. So you get four to the four. That's going to be 256. For the next one, you have four colors, one cycle, in which case you're going to get four to the first. R squared has two cycles, so you get four squared. R cubed has one cycle, so you get four to the first. S has two cycles, so you get four squared. R s has three cycles, so you get four cubed, which is 64. You get four squared for R squared s has two cycles, and then r cubed s also has three cycles, so you get four cubed right there. So with that perspective, it's like, voila! I can pull this thing out really, really quickly, because I know the cycle structures for this permutation action. We have to take 1 eighth of 256 plus 8 plus 16 plus 16 plus 64 plus 16 plus 64. That simplifies just to be 55. So while the two colored example I probably could have done it by hand, this one's going to be a lot harder because there are 55 distinct colorings, and that's not even considering that some colorings are equivalent. These are 55 distinct colorings up to equivalents. Burnside's theorem is a very effective counting technique when we have to count functions of a permutation action. So to finish this lecture, let me give you one more example. We're going to switch things up a little bit. This time we're going to ask ourselves how many ways can we label a four-sided die? So like if you're really into Dungeon and Dragons and other role-playing games, six-sided die are not enough for you. You want your dice to have four sides and eight sides and 20 sides. You know, you take all the platonic solids there. So if we think of like a tetrahedron, right? Something like this. So let's color the vertices of the tetrahedron. So you have like there's one, two, three, four vertices. So what if we want to, we're not coloring this one, I'm sorry, we're just going to label this one. So we could label it like that. Could I label it something different? Like could I label like we'll do this one one, this one two, this one three, this one four.  It's going to move around. It's still the same die even though if we change our orientation, what labels actually give us a different labeling of this thing? That's an interesting question, but group theory provides a very slick way of counting the ways we can label the four vertices of a tetrahedron of a four-sided die here. And the important observation is to remember the symmetry group of the tetrahedron. If we look at just rotational symmetries, because as we roll our four-sided die, we can't reflect it through space. We're just rolling it. That's rotational symmetry. The rotational symmetry group of the tetrahedron was a four. This is something we talked about in the first semester of abstract algebra, which of course with a four remember you have the identity. You're going to have three two two cycles. So you have the as we listed these things, you have the identity, you're going to have the two cycles, two two cycles, one two three four, you have one three two four, and we have one four two three. We're also going to have eight three cycles. So we have like one two three and we have one three two. Okay, we're going to have one three four and we're going to have one four three. Let's see is there another one. We have one two four and we have one four two. And then lastly, we're going to have two three four and we're going to have two four three. So these are, these are eight three cycles. Okay, now the the three cycles actually show up in two different consciously classes and I don't claim, I don't claim that I have any orientation with those consciously classes all right now, but it turns out if you know the conjugates, you can actually simplify these things a lot more because conjugates in a permutation group will have to have the same cycle structure, but nonetheless it's really just the cycle structure we know. So we have one element which really that one element is going to look like one two three four okay. We're going to have these two two cycles which have two cycles each as the name suggests and then for the three cycles, well there are three cycles but there's a fixed element so we should really should think of it as like one two three and then four as a separate thing. So they have two orbits each, each of the three cycles. But what's left fixed right? How many things are left fixed going on with this with this right here? So with the tetrahedron, the idea leaves everything fixed. So you get the whole thing, you get all possible labels here which are going to get four factorial of this. This is a little bit different than the colorings we did before because we can't repeat colors anymore. A label this time is a permutation. We have to label this one one, this one two, this one three, this one four. So there's four factorial ways you can color or you could label the dye. Okay. Now let's ask ourselves what's left fixed by a two two cycle. So if you have a two two cycle like if you take one two three four you're going to swap those things around. You're going to swap those things around. The issue here is that with a with a two two cycle everything that's moved around but we can distinguish the vertices because they're labeled. This isn't like the coloring did before. When you do a two two cycle nothing is left fixed. Everything got moved around. Okay. So nothing's left fixed. So we get that gives us a different labeling. So it turns out if you take your three two two cycles each of their stable sets are empty. It has size zero. So that's actually just going to be ignored in the sum. The same thing is also going to happen with the three cycles. Okay. There's eight of them but each of their stable sets is empty. If we look at the stable set of one two three this is actually the empty set. There is no tetrahedron labeled where if we swap one two and three it would be the same labeling because one two and three moved. You would have to swap.  No. You're swapping a one with a two. That's noticeable. And as such these labels are you see them move the stable set is going to be empty here as well. So by Burnside's theorem the number of colorings I should say the number of labeling here is going to be one over the order of the group which is 12. That's a four. You have four factorial plus zero plus zero. So you get four factorial which is 24 over 12. That just gives you two. So it turns out there are only two labelings of a four-sided die. One of them is on the screen right now. Can you come up with the other one that is different? That's actually not equivalent to the one on the board right now. It's an interesting question.  How many ways can you label a six-sided die so that when you roll them they're actually different labels. You have you I'll give you of course the hint there are six sides to the six-sided die obviously you have to label one through six. But what's the symmetry group there? We proved previously that the symmetry group was it's isomorphic to S4. Use that to try to identify what are the number of labels you can put on a six-sided die. Thanks for watching this video and this lecture everyone. I hope you learned something. Give us a like if you did. If you want to see more videos like this please subscribe to the channel and as always if you have any questions or comments or answers to questions I gave to you in the video put those comments below and I'll answer them at my soonest convenience.


Every human being, it is said, is part of the social fabric, and indeed so obvious does this seem that if ever we should happen to hear of a man who was actually isolated, actually alone, we might well assume that he was a stranger among us, an intruder on the earth. Yet the case of at least one such man has been recorded. He was lying in a proud field when they found him. Gillum, the hired man, was the first to notice the body. It lay white and naked and terrible under the noonday sun. A dead man lying out in the field. He's naked as the diary was born. I never seen him in these parts before. You'd better come out and have a look, sir. The owner of the farm, having examined the corpse carefully, announced that he had never seen him before either. And so the Petersville police were called in. And the local inspectors decided that identification should not be too difficult. It's simply a matter of finding the tapped clothes, that's all. He must have been wearing them when he came out here. There are no fresh card tracks in the road, and he could scarcely have walked up and down naked without causing some comment. Though the clothes must be around here somewhere. But the clothes were not to be found. During the search, however, the hired man made one important discovery. Footprints, Inspector. They've been mined by bare feet, if you'll notice. Though they must be eggs. So they followed the stranger's footprint. They followed them through the road a half mile away. And across the road, it entered the field on the other side. There the footprint stopped. Why, it's almost as if the blow came from another planet. Just dropped down out of the ovens he did and died. But the inspector, who was a practical man, chose to ignore Mr. Gilliam's fantastic suggestion. It might prove a little more difficult than he thought at first, but in the end, the body would be identified. The inquest, the coroner reported that there were no wounds on the deceased. There were neither drugs nor poison in the contents of his stomach. He was obviously not the victim of foul play. His death was due simply to exposure. Exposure? How could he have joined him exposure? How could he have joined him exposure? He could see the hospital where he was. All he'd have had to do was knock on the door and walk him. What the gentleman says may be quite true. Nevertheless, death was due to exposure. To continue, the man was between 35 and 40 years of age. He was not a manual worker. His hands and nails indicate that. He was well-nourished, and I should say well-educated. A man of considerable substance. Would it not be difficult to learn the identity of such a person? The efforts that were exerted during the weeks that followed were more than adequate. A photograph of the man, together with the description of him, was published in practically every paper throughout England. Two months later, on April 16th, the London Daily News reported that the police were still without a clue. That there was no record of any missing person bearing the slightest resemblance to this man. If most Englishmen were puzzled by the affair, Mr. Gilliam, the hired man, was not. They're wasting their time with their photographs and whatnot, I tell you. The bloke was a stranger, and not just the renampure either. He was a stranger on the earth. They'll never find anyone who ever knew him. Today, the identity of the dead man still remains a mystery. The incident takes its rightful place in the annals of amazing stories. Stories incredible, but true.


course on dealing with materials data. In the present sessions, this is the fifth one in the series of this sessions on special random variables, we are talking about certain special distribution functions which we encounter very commonly in our day to day experimental work. Let us quickly review, we first introduce some continuous random variable, before that we introduce some discrete random variables, uniform discrete uniform random variables, then Bernoulli trials and three distributions derived out of Bernoulli trials, we gave a 3D APFIM example there. Then we introduce the uniform continuous random variables and we introduce uniform random variable or uniform distribution, then we talked about in details on normal distribution. We also discussed the some distributions which are derived out of normal distributions, they do not largely naturally occur, however they are useful to carry out certain inference on the random variable which originate from the normal distribution. And the last session, we discussed about the central limits theorem and that basically showed that how normal distribution is kind of all pervading distribution, primarily because it said that with the condition that it has a finite variance, a large number of random variables coming from independent random variables coming from identical distribution. And if you take its mean value, then as n tends to infinity roughly we can say that the mean value tends to normal distribution with the, sorry, the average value of the random variable follows the normal distribution with the common mean and the common sigma swell divided by n or the random, the average minus the mean divided by its variance follows a standard normal variable that is the correct way to put it. Now we are going to come to some of the distributions which are of importance to metallurgical and materials engineering and that is very specific but in general it is good for the engineering distributions or engineering data. The two distribution that we wish to introduce right now is log normal distribution and Bible distribution and then we will follow up with the exponential distribution which is a special case. Why this special treatment to some of this distribution particularly coming from a metallurgical engineering data? Well, normal distribution is a widely applied distribution, central theorem makes it almost all pervading distribution however it has one limitation with respect to engineering data. In normal distribution the random variable varies between minus infinity and plus infinity. It means that it can take any value from the real line while data from metallurgy and material science generally take on only positive values. Now you may assume the distribution to be normal and carry out your inference. Sometimes you encounter a very absurd result and it may amount to the fact that you have assumed normality which takes on all kinds of values from negative infinity to positive infinity and therefore it creates this absurdity. The only thing is that this metallurgical and material science data in particular it has been seen that they have, they are generally a skewed data. They are not beautifully symmetric bell shaped histogram they do not represent those such beautiful histogram, they generally represent the skewed distribution. Also to apply the central limit theorem as we say if the distributions are skewed you need at least 30 data points or at times you may need even more than that if the skewness is very high. In that case such a data availability in material science and in particularly in metallurgical engineering is very rare. It is very expensive to generate such data and therefore it is important that we learn about certain distributions which are defined over the positive part of the straight line that is for the positive random variables and we will also see that how they have been used and applied for the, in the area of metallurgy and material science. So first we introduce what is known as log normal distribution. As the name says it is a positive random variable which has the natural log of that random variable follows a normal distribution then you say that X has a log normal distribution. So there is a random variable whose natural log follows normal distribution then we say that X follows log normal distribution. Such a distributions pdf is given in this manner. You see that this X is replaced in the, this exponential part is same as a normal distribution where X is replaced by ln X and this part is also the same as normal distribution. However, we have a denominator multiplied by X is because the derivative of ln X is 1 over X dx and therefore it comes X here. Here also the two parameters are there one is a mu and the other is sigma. Mu is a mean value of ln X and sigma square is a variance of ln X therefore mu varies between minus infinity and plus infinity and sigma square is always positive. The notation we use is X is distributed as ln mu sigma square there are two parameters to it. Here I have shown you some of the plots I have picked it up from the Wikipedia. The reference is given here. Here what I want to show is that the mean value is kept as 0 and the variance of the normal part of the distribution is varied from 1 to 0.5 to 0.25 and you can say that with the change in sigma value remember that this is not sigma square this is sigma. So if you try to regenerate it please make sure that it is with sigma and not sigma square given here and then you can see that as the sigma changes the shape of the curve changes as the sigma becomes smaller the curve becomes more and more skewed. And also please notice that this is always skewed towards the right log normal distributions are always skewed towards right. As sigma you can see that when sigma is 0.25 it comes very I think I made a mistake in my first statement as the value of sigma increases the skewness decreases when sigma is large the skewness is large and when the sigma is small the skewness decreases and you can see that here at sigma equal to 0.25 this is almost like a normal distribution curve but there is you can see the tail part of it and you can make out that it is still a skewed distribution. The expected value is given thus I leave it for you to derive it and the variance is expressed in this manner exponential of mu is actually the median value of the distribution. So exponential of mu is the here it is 1. So this is the median value of the distribution for every distribution this is the median value. So this is all centered around the median value and median value a number of times is also denoted by L sub 50 and sometimes mu in the all the above equations these equations as well as these equation is replaced by L ln L 50 because remember that L 50 is exponential of mu therefore mu is log normal of it is a natural log of L 50. Now please remember here let us come back that sigma square decides the shape of the distribution and therefore here sigma is called the shape parameter unlike normal distribution it is not a scale parameter it is a shape parameter. Now if you want to introduce a location parameter then it is a third parameter and it makes it a three parameter log normal distribution where the x is replaced by x minus psi where your assumption now is not x greater than 0 but x is greater than psi mu is between minus infinity and infinity sigma square is positive and then the form of the distribution is this. It is denoted as x is distributed as log normal psi mu and sigma square. Let us look at some of this pdf plot here I have kept the to show the effect of the location parameter which is psi the first graph shows that psi is 0 you see this is when the location is sorry location is 0 and you see that this is where the x values are larger than 0. So, it starts from 0 0 and it goes up and down when you make psi is equal to 2 the whole graph is sort of shifted by 2 units and therefore it because starts from 2 and exactly the same trend it follows. So, that is why it is called a location parameter because it shifts the location of the plot or the of the pdf. This log normal distribution is most commonly used to estimate the particle size distribution. So, another way is also that it is heavily used sometimes to represent the data such as the height weight of a human being because generally height and weight is modeled as a normal distribution. But as I said height and weight cannot take a negative value at all and therefore it is actually a positive distribution it is a positive random variable and therefore it is number of times estimated the distribution of height and weight is estimated very well with the log normal distribution. Also I find that the density is also very well modeled as a log normal distribution. It also gives a good model to express the distribution of fracture toughness, yield strength etc. etc. So, it has a very typical very very specific and many applications in the field of metallurgy and material science. Now we move on to the next distribution which is a Weibull distribution. Weibull distribution is named after a Swedish physicist Valori Weibull in 1939 he first time demonstrated that a distribution that can represent the distribution of breaking strength of material and he must have got so excited with his discovery or his this particular finding of the distribution that he wrote a paper in 1951 which titles a statistical distribution of wide applicability. I have given the reference and I strongly recommend that everyone should go through this paper because it is a very interesting and account of how the physical phenomena can be looked into and can be used to derive the statistical distribution. It is a very beautiful distribution about two paragraphs of it but very very inspiring and very interesting. He in this paper he said that not just the breaking strength of the material in terms of yield strength of buffer steel or fatigue life of another steel STs 37. He even found that the fibre strength of the Indian cotton, height of the male born in British Isles, he also found certain length of a certain beans which were produced in a certain farm all of them tend to follow Weibull distribution. It appears that Rosen and Remler had in 1933 had also come up with a very similar distribution in connection with particle size distribution. However, the distribution became very more known or more popular when Weibull put it forward in his 1951 paper and it has been heavily referred. I again once again I said that I strongly suggest that you please go through this I have given the full reference of this paper, please go through this paper. It is a very interesting account of how the distribution is derived. So if X is a positive random variable and it has a PDF that looks like this then it is said to follow Weibull distribution. Here we have two parameters C and alpha. C is called a shape parameter and alpha is called a scale parameter and the cumulative distribution function has much easier expression which is given here which is 1 minus exponential of minus alpha X over alpha to the power C and the notation is generally used as X is distributed as Weibull, alpha and C. First comes the scale parameter and then comes the shape parameter. Please look at this particular expression and note that if you take double log of this CDF you will have a linear relationship between C and rather log C and alpha and you will be C and log alpha and this will be one of the ways of estimating the parameters when we come to that. So we will discuss it in more details at that time. Again I have taken some plots of Weibull distribution from Wikipedia. Please remember here the notation is slightly different. The scale parameter is given as lambda and the shape parameter is given as K and here again for different shape parameters the distributions are shown and please note that this distribution is not just skewed but it can be skewed in either direction. That is the beauty of Weibull distribution. It can be skewed in either direction. So here you see a kind of an exponential graph here. You see another distribution which is very typically right skewed distribution. This one, this is for shape parameter is equal to 1.5. This is another skewed distribution, right skewed distribution. However, when you take K is equal to 5 as such it looks like a symmetric curve but notice the tails and you realize that now it is becoming slowly the left skewed distribution. The longer tail is coming on the left side. That is the beauty of Weibull distribution. In log normal distribution only if the data is positively skewed you can fit the data but Weibull gives you an open field that any data which may be skewed or may have an exponential kind of expression can also be covered as a Weibull distribution, can also be modeled as a Weibull distribution. Weibull distribution also has its variant called 3 parameter Weibull distribution in the exactly the same manner as in the log normal distribution. So there is a location parameter psi and of course psi is greater than 0 and the variable for random variable X greater than psi, this is the expression of PDF. You please recall that the expression only changes that X has been replaced by X minus psi in all the places and the CDF is also generated in the same way and in this situation the notation is X is distributed as Weibull, psi, alpha, C it means that location, scale and shape. How does the 3 parameter Weibull differ? This is the 2 parameter Weibull because I have a psi is equal to 0 and here I have taken psi is equal to 1. You can see that since I have taken the shape parameter to be 8, it is left skewed distribution and it simply shifts is location from instead of starting from 0, I can say that if they need the X greater than 0, so X starts from 0 onwards, here it starts 1 onwards, otherwise it has simply shifted. I also want to tell you from the graph that as the value of shape parameter increases, the shape changes from the right skewness to the left skewness. So here also we have very specifically showed a left skewed distribution arising out of Weibull and how it shifts, it is actually a location, it shifts the location of the distribution. Now there is one distribution, exponential distribution which is very special case of Weibull distribution where the shape parameter is 1. You please recall shape parameter 1, the red color distribution is represented is a special distribution, it is called exponential distribution. The random variable is greater, X is greater than 0, then the distribution PDF has this form and the notation is given by X, X is distributed as exponential alpha. You can derive the CDF, I have purposefully left it out, please derive the CDF of exponential distribution, it will be a good mathematical calculating algebraic exercise for you. Also please note that before going any further to the memory less property, you can have a 3 or rather 2 parameter exponential distribution. Here you have only one parameter which is alpha, now you can introduce a psi which is greater than 0 and say that X is greater than psi follows a 3 parameter exponential distribution where you will replace X by X minus psi, okay. So you can generate a 3 parameter exponential, 2 parameter exponential distribution, please I stand corrected and then you will have X distributed as exponential, first you will have psi and then you will have alpha. This distribution has a wonderful property called a memory less property. What does it mean? It says that probability if X is distributed as exponential alpha, then probability that X greater than t1 plus t2 given that X is already greater than t1 is same as probability that X is greater than t2. This is a very important and very interesting property, let us understand this. Because this is a time line, this is a time t1 and this is a time t1 plus t2, okay. This is the value of t2. What it really says is that you already know that the random variable is larger than this. Then if you try to find out that the random variable is larger than this given that it is already larger than that, only means that it is beyond the time t2. What it has done in this range, it has already gone beyond the t1, it has forgotten. It only remembers that it has gone further t2 time, that is all it says. So this is called that it has forgotten the memory that it was already larger than t1 but now it only considers itself as larger than t2. This proof also I leave it to all of you to consider it. It is a very interesting proof, you should try it, it is not very difficult. So with this let us summarize what we did today. We considered all the distribution with the random variable which takes on only positive values. Why did we do that? Because in material science and metallurgical engineering, most of the data we encounter is positive in nature. You can assume normality and carry out the inference but at times you encounter certain absurdity in your results and that is likely to arise because you have considered a distribution which has which is supposed to have its random variable coming from both positive and negative side while in reality you have got the random variable which is only positive. So we have introduced specifically distributions which have a great applications in the field of metallurgy and material science and it has it takes on a positive value. So we introduced two distributions of this nature log normal distribution, two parameter and three parameter distribution, Weibull distribution also two parameter and three parameter and then exponential distribution as a special case of Weibull distribution. I would like to let every one of you know that these are not the only two distributions useful to with having a positive random variable and useful to metallurgy and material science. There are many such distributions, one of them is Bernbaum Sonder's distribution, there is a gamma distribution, there is beta distribution, there are many such distributions but right now we have introduced most commonly useful distribution in the field of material science and metallurgical engineering. With this we complete the sessions on introducing the special random variables. We can have a quick look at it, we started off by saying that there is a random variable which is a function from the probability space to a real line and then for such a random variable we talked about it expected value, its variance etc. Then we introduced the special random variables which take on very special distribution. So we also we talked about the discrete random variables of special kind where we had uniform distribution, Bernoulli trials, binomial distribution, geometric distribution, we also had a negative binomial distribution, hypergeometric distribution and all of this we showed through 3D, APFIM and other examples that they are all useful to us in the field of metallurgy and material science. Then we came to the continuous distribution in which we talked about the continuous uniform distribution and we showed that in random number generation it plays a very central role. Then we introduced normal distribution, we had a long discussion on normal distribution because through central limit theorem we saw that it is largely all pervading distribution if you have a large amount of data. We also got introduced to derivatives of normal distribution such as chi square distribution, t distribution and f distribution. We have not seen the applications of it directly but it is a preparation for the future work that we are going to do in estimation of parameters and testing of hypothesis. Then of course we talked about the central limit theorem and then we introduced ourselves to the distributions which are useful to the metallurgical engineering and material science because most of the time random variable takes on of only positive values. We introduced the commonly used such distribution as log normal distribution, Weibull distribution and we introduced exponential distribution which is a memory less property as a special case of Weibull distribution. Thank you.


What are the requirements for cryptocurrency to be considered a cryptocurrency? Is there a way of creating some sort of checklist to determine what a cryptocurrency is? There isn't really a canonical definition of a cryptocurrency simply because cryptocurrency is a marketing term and it has become a marketing term. Meaning that any currency that is based in some form on cryptography, some form of cryptography, or uses cryptography in some way, might be called by some people a cryptocurrency. What does that tell you about how it works? It doesn't tell you much. So the fact that somebody calls something a cryptocurrency doesn't mean much unless you ask some follow-up questions. So the interesting things are the checklist that you said is a series of follow-up questions you can ask to see if this is an interesting kind of cryptocurrency. And of course, that's highly subjective. What is interesting to you? What is interesting to me? I can tell you what my list is and it's a list that I've tried to popularize. So what's interesting to me is a cryptocurrency that is truly open and truly decentralized. Meaning that it differs from previous forms of currency that we have and various other closed and controlled blockchains and cryptocurrencies that have been built since by being open and decentralized. So nobody has control of the system, but it works on a basis of rules and everyone can participate. So I have a series of questions as a checklist. The first question is, is it open? What does open mean? It means, can anyone access it? Can anyone participate in the network without authorization, without vetting? Pretty much. Can any computer on the open internet simply connect to the network and be part of it and participate in the network? There are nuances that, of course, is it open to miners? If it has a mining algorithm, can anyone participate as a miner without authorization and vetting? Can anyone do a transaction without authorization and vetting? So open. Is it borderless? Does it operate across the entire world without concern about national borders, geographic borders, and various artificial limits on its propagation? Can you access this system from anywhere in the world, anytime, anywhere by anyone? So is it borderless? Is it neutral? Neutral meaning that the system does not discriminate transactions based on source, destination, value, or purpose. A transaction is propagated on the system, regardless of who sent it, who they sent it to, which source and destination addresses are involved in the transaction, how much was sent, and what the purpose, which is outside of the system, what the purpose of that transaction is. So a neutral system will transmit transactions regardless of source, destination, value, or purpose. It simply propagates transactions just like a neutral internet for its packets, regardless of source IP, destination IP, or application that is being served by that particular IP packet. Is it immutable? When you record something on this blockchain or this cryptocurrency, can that transaction be reversed? And of course, this is not a yes or no answer. After a certain time, does it become immutable? And how strong is that immutability? So for example, in the case of Bitcoin, we can say with a fairly high degree of certainty, after six confirmations, the data in that blockchain is immutable enough that you can put quite a significant amount of value into it. And you can think of immutability increasing over time, the more confirmations you have, the harder it is to reverse a transaction. So after 100 transactions, sorry, after 100 blocks or 100 confirmations, the Bitcoin blockchain is considered immutable enough that you can spend the coin-based transaction, which is the mining reward. Maybe after a day, 144 blocks, it has a very high degree of immutability. After a month, the amount of energy and money required to reverse the Bitcoin blockchain is just staggering. So immutable, that's another important criterion. So is it open, borderless, neutral, censorship resistant? Is it immutable? And you can add to that also, is it publicly verifiable? Meaning, can anyone verify the transactions that are happening? Can you audit it, basically, is it transparent? And you can keep adding other criteria, depending on the importance you give to these criteria. But I think those are the most important for me in order to determine whether a cryptocurrency is interesting in that it is encompassing some of the characteristics that we see in Bitcoin that really made it different from everything that came before. Now, that doesn't mean that something that doesn't fulfill these criteria isn't a cryptocurrency. It just means that it's not interesting to me because it doesn't represent the principles and characteristics that made Bitcoin interesting by being different. So, for example, if it's a controlled closed system that only operates within one country, sure, maybe that's called a cryptocurrency by some people, and maybe it's interesting to them. It just, to me, it's incremental innovation. To me, it's more like a database and not particularly interesting in its technology and certainly not very interesting in its political and social implications.


I'm M.M.F.A. Maddox from Spring Garden, Alabama. My reason for it is machine smith. I joined the Navy to really see places. I'm from a small town in Alabama. Not a whole lot there. This is what everyone dreads. This is what they know they have to start off with when they first get here cranking. They know that they have to go to the deep sink and they dread it. We wash the dishes after they're done using them. The big things that they use to put food out on the line is to get here at 05. I start washing dishes because the night crew starts breakfast, so there's already dishes back here. So I start washing the dishes. I'm normally in towards the end of breakfast, I'm done. Then they bring the rest of the stuff back here that was used for the food to breakfast. After a meal, it normally takes about an hour to an hour and a half working alone. The dirtiest thing I've had to do was clean the deck at the end of the day for inspection.  There's always constantly dishes coming back here. That's pretty rough too. Being back here has given me a lot of respect for C.S. because not only is it hot, the hours are horrible.  Everybody always talks about they want to be a C.S. They have to cook food, but you have to cook for all those people. You can't get behind. It's hot. It's long hours. I respect C.S. a lot. I definitely don't want this job.


During a bushfire, embers will reach your home long before the flames do. Ember attacks are the most common way that houses catch fire. Embers are burning leaves, twigs and pieces of bark. They help the bushfire spread by starting spot fires ahead of the main fire front. Short distance ember attacks happen when leaves and small pieces of bark are blown from burning trees. The intense shower of sparks that forms fills the air with hot burning embers which will land on nearby vegetation and properties.  Small fires will start all around you and quickly become uncontrollable. The resulting chaos creates confusion as the fire seems to come from many directions meaning it will be difficult to make good decisions about your safety. The experience will be physically exhausting and emotionally traumatic. Long distance ember attack is caused by large bushfires that generate intense heat. As the hot air rises it forms a column of smoke that sucks in air like a vacuum, increasing the intensity of the fire. The updraft in the column lifts embers like large pieces of burning ribbon bark hundreds of metres into the air where strong winds can carry them many kilometres beyond the fire front. On Black Saturday embers travelled more than 30 kilometres ahead of the main fire. When these embers land they often start fires where leaves naturally accumulate like in gutters, doorways and garden beds. So while you might think you're safe when a bushfire is far away embers can fall from the sky and land around your home long before you even know there's a fire. But remember, your home isn't the only thing under threat during an ember attack. By the time you realise the danger it might be too late as escape routes become jammed. Embers can also start fires on roads and block them completely making late evacuation dangerous or impossible. That's why leaving early before a fire starts is always your safest option.


I am just really excited to show Stevenson what we've been working on. Matt Grimm and Brett Adams, the AD and assistant AD really have given me the confidence to be my best self-coaching and to really portray that onto my athletes and prepare them for, you know, being successful to prepare them to represent Stevenson in the best way they can. I'm really excited to show them what this sport is about, what this sport can be here and lay out a really good foundation for us to build on in the future here. So acrobatic and tumbling is comprised of acrobatic lifts, tosses and then tumbling. I'm coming from a background of 15-plus years of competitive cheerleading, so I'm used to the tumbling aspect and the tosses aspect. However, the acrobatic lifts is completely new to me. I've met people I've never, I never would have met before. A lot of them are in different majors than I am and I've gotten really close with freshmen and seniors and juniors and people that I never would have talked to. So I've definitely, especially with COVID, it's helped me meet people. It's also physically helped me a lot. We're practicing all the time, we're lifting, so for my health it's amazing. I feel strong, like I wake up early in the morning to lift and to do all these things and I know if I had these struggles or anything I can ask any of the seniors that are nursing majors or anything, I can ask them, I feel comfortable enough to ask them all of my questions. And the trust that everyone, it's really helped me like to trust people as well because in acrobatic society you have to trust people an insane amount. Slipping off of people, holding people in the air, you have to trust that they're going to catch you, keep you up there and it's just like really helped me to just trust people blindly because I know as my teammates they're just, they're going to be there for me no matter what. Yes, I love it. I really like the trust that we put into each other with the lifts. Of course like you're trusting one person, there's normally like one person underneath you, whereas cheerleading there's normally three or four people directly under you the entire time. With acrobatic so it's normally one person under you and sometimes that person is being held up by one or two other people. So the trust definitely is by far my favorite part. I'm really excited for them to put out there what they've been working on. We've been watching a lot of meets and seen some other first year teams and I think we're on par with a lot of these other first year teams and I know I keep going back to the not having recruited students and not having people brought in knowing what they were getting into but they had to really get in that mindset of learning and picking up really fast and I think they're on par with a lot of the other first year teams and I think we'll do, I think we'll do pretty well.


Earlier in the year, Eleanor Swanson arrived at Sanford Health in Bemidji with some stomach pain that quickly turned into a much worse diagnosis. R. Malmire was there when Eleanor got to thank one of the staff members that helped save her life. Since Air Med took its first flight out of Bemidji in 2012, we have the potential to transport two patients. The Air Ambulance Service has responded to nearly 1,800 calls across Minnesota. One of those was a common transfer request. We landed at the hospital, we packaged the patient up and we moved them to another hospital. On March 12th, Eleanor Swanson went to the ER with some stomach pains. The next day, her colon suddenly ruptured. I can remember the initial pain but that might have been my last moment. She was flown to Ascension St. Mary's in Duluth for treatment with doctors. Not thinking that I would make it. She didn't wake up until her return to Bemidji confused about what had happened. So it's been quite a ride. It's a ride she's thankful she had taken. Happy that things turned out well. Thank you. I'm just so glad the service is available. Which is why Eleanor and her daughter were sure to drop by the Air Med Ice Cream Social. What's it like to be able to meet them? Fantastic. She wanted to meet the team that helped save her life, an unusual occurrence for Joe Rain who was on her flight. I guess I'm speechless. It's weird. You don't get to meet them. You don't get to meet your patients. You drop them off and unless you know people at the hospital, you might get a little bit of an update. It's a heartwarming surprise. Best served over ice cream. It confirms that what we're doing is a good thing and that you're helping people. Have many more good life saving flights. Eleanor says she's looking forward to thanking the rest of the Air Med team that helped save her life. In Bemidji, I'm Mel Meyer, Lakeland News. For Ambulance Service, Sampert Health is able to take patients from Bemidji to Minneapolis in just about an hour. If you've enjoyed this segment of Lakeland News, please consider making a tax-deductible contribution to Lakeland Public Television.


The construction sector is of huge value to the Scottish economy and should not be underestimated. The committee wanted to try and understand the challenges facing the sector and those which stop it reaching its full potential. We found innovative work is being done. For example, off-site manufacturing, new skills frameworks, but we feel that positive leadership is needed to cement positive change. Procurement, access to finance and cultural image appear to be limiting progress in the sector. The committee believes that industry and the Scottish government should work together to facilitate and create and ensure positive and progressive change going forward.


Welcome to WISPA Palooza 2016 at the Rio Hotel and Casino in Las Vegas. We're now on day two of the WISPA Palooza show. I have with me John Brown from CityLink Wireless. John's actually quite interesting. He found us on the internet. He kind of tested one of our antennas kind of by accident. It was quite blown away by it. So it's really powerful to hear his story. So I'll let you John kind of take the viewers through how you found the antennas and how things went along for you. Yeah, absolutely. It's great. I can't tell you how much I enjoy the product. We had a customer that was having some issues about two miles away from our access point. Actually, a friend of mine, Steve Grabio, who runs the ISP Wireless radio show, had a 30 degree horn and said, hey, you should try this thing out. It's like, okay, cool. I'll try it out. So Steve came into town. We bolted it up on the roof, tried it out. Long story short, we went from a customer that was at an egg 78, two miles away. After we put the horn up, they were an egg 63. Wow. Yeah. And we did zero change to the subscriber location. We never went out there or anything. Subscriber was running a ubiquity nanobridge, and we were running a rocket M5 into the 30 degree horn. And so we were blown away by that. We went out and bought 30 additional horns. Out of that 30, we've deployed six of them. We've seen similar night and day differences in the signals and the quality experience that the customer has. We're now in a process of getting ready to deploy pretty much down the entire Rio Grande Valley. And we're looking at deploying hundreds of these 30 degree sectors. We also bought a number of the different twist locks. Twist port adapters. Yep. So that we could try out other radios and other technologies. And that's one of the other cool parts that I really found quite compelling with the antenna is I can bolt this antenna to my mast, and if I don't like the rocket M5, and I need to change it, let's say, to the rocket M5 AC prism, I take the twist port, change it, boom, I'm done. I don't have to realign, move my antenna, I don't have any of that issue, or I want to put a cambion on it, I can, or a microtake. I had that flexibility. From a price perspective, compared to the traditional sector, I think the price value is a heck of a lot better. Excellent, excellent. Well, thanks for sharing it with the users out there. We appreciate you stopping by the booth. And again, sharing your experience and your feedback with us. Thank you very much. That's an excellent product, thank you.