Welcome back to our lecture series linear algebra done openly. As usual, I'll be your professor today, Dr. Andrew Missildine. Now we've mentioned in the previous section, section 3.1 about matrix operations, that the set M to the N by N, so this was the set of all N by N matrices with coefficients coming from F right here. We mentioned that this formed a vector space where the matrix addition and skill emultification satisfy the eight properties associated to definition 1.3.1. That is the definition of the vector space, those eight axioms. So this will tell us things like matrix addition commutes. Matrix addition is associative. There is an additive identity with respect to matrix addition, the so-called zero matrix. Every matrix has an additive inverse, which is really just negative one times that matrix. Scale emultification distributes over matrix addition. We have that matrix addition distributes over skill emultification. We have that matrix, our skill emultication is associative and that multiple by one gives us just the original matrix. So all of those nice properties we had before are true for these matrix operations of addition and skill emultification. But in the previous section, we introduced some new operations that didn't have counterparts for vector spaces that we studied in chapter two. Some examples would be the trace map, the transposition matrix multiplication. What happens to these? What type of properties can we stay about those operations? And so those are things I wanna talk about right here and right now. So let's begin with the transposition map. Suppose that A and B are two matrices with the same size. So let's say they're both in by in matrices just to be clear here. So then it makes sense that we could add together the matrices, at least for this first situation. So we could take A plus B in this situation. What if you take A plus B and then you take the transpose of that? How does transposition interact with matrix addition? Well, it turns out that transposition distributes over matrix addition. The A plus B transpose is equal to A transpose but B transpose. Remember the transposition operation, it switches rows to columns and columns to rows. When you add together two matrices and take the transpose, it'll just distribute the transpose, right? It's A transpose with B transpose. Likewise, if you interact with some scalar, let's say that C is some scalar from the field F. If you take C A transpose, this will be the same thing as C A transpose. That is the scalar essentially comes out of the transposition process and doesn't really interact with the transposition too much. So I just want to pause right here and just look at properties one and two. We'll get to three, four in a second. So what this tells us is that transposition preserves matrix addition and it preserves scalar multiplication. This right here is what we refer to as the linearity property. More specifically, this means that the transposition can be thought of as a linear transformation. What I mean is the following. We could think of the transposition as a map that takes matrices from the vector space F to the M by N and it transitions it to a matrix of the form N by M. So it switches rows and columns so those numbers are gonna switch around. And so the transposition map, it switches a matrix from M by N to N by M and this map, well, since this is a vector space and this is a vector space, we can think of these spaces and matrices as vector spaces. We can then ask is this map linear and properties of the transposition in fact are. The transposition is a linear map on these spaces of matrices. Well, what do we have here? Property three tells us that if you take the transpose of the transpose, so A transpose transpose, the double transpose gives you back the original matrix A. So taking the transpose twice sends you back. That's quite useful here. But what this also tells you is that the inverse, the inverse of a transpose is itself. The transpose undoes itself. It's like taking the reciprocal of the reciprocal. That's really nice here. And so then the fourth property here, we're gonna consider how does the transposition interact with matrix multiplication? Now in this situation, A we'd have to say is like an M by N matrix and then B is an N by P matrix in order for this product to be compatible, right? Cause the ends would cancel out and the product AB would be an M by P matrix, all right? So think about that. Think about that. I'll just erase this right here. We don't need it right now. So we have M by N times N by P. That'll give you an M by P matrix. This is what we think of for AB right here. So AB transpose, what it creates is the following. AB transpose is equal to the matrix B transpose A transpose. So you'll notice that when you take the transpose of a product, it actually switches things around. So B comes first now and A comes second. This is what we often refer to in this literature and this is not a joke. This is often referred to as the shoe sock principle. What are shoes and socks have to do with anything right here? Well, the idea is the following. In the morning you put your socks on then your shoes, but in the afternoon or evening when you return home you take your shoes off then your socks that the process reverses itself. So we first had shoe then, sorry, we first had sock then shoe, then we get shoe then sock. So transposition flips these things around and let's actually see why that actually has to be. So if B was a matrix of the form N by P, then B transpose is gonna be a matrix of the form P by N. And if A was a matrix that was M by N, then it's transposed will be matrix of the form N by M for which when you simplify that gives you a P by M matrix. Which as A B is an M by P matrix, it's transpose should be P by M. Where this right here is B transpose, A transpose. If you didn't flip these things around, I'm gonna draw this one in red so you think of death and blood right here because this is the wrong way to do it. If you did A transpose, B transpose, this would look like well N by M, that would be A transpose and B transpose would be P by N. We don't even know if such a matrix product would be defined or not. So you have to flip these things around in order for the matrix product to even make sense. So you hit the shoe sock principle when it comes to transposition. This is the thing that remember the most is that when you take the transposition of a product it switches things around by the shoe sock principle. This is actually alluding to or something we'll talk about in the next video is that matrix multiplication as it was defined is not a commutative operation that A times B does not equal B times A in general. And this oftentimes is the fact because like we see here, even if A times B is well-defined, B times A might not exist. The product might be incompatible but it turns out there's other issues going on there but we'll talk about that later. Let's talk about the trace for a moment. Remember the trace of a matrix was the sum of the main diagonal. So you add together all the terms on the diagonal here. That was the trace. So imagine we have two matrices that are in by in, right? We only define the trace for square matrices. So what can we say about the trace map? Well, if you take the trace of the sum of two matrices this is equal to the trace of A plus the trace of B. So the trace preserves addition and that kind of makes sense because how do you find the diagonal entries of A plus B? Well, you're just gonna add together the diagonal entries of A plus B. And since those scalars come from a field which we can associate and commute we can break apart those terms. And so the trace, the sum of the diagonals of A plus B will be the sum of the diagonals of A plus the diagonals of B. Well, what if we put a scalar multiple in there? So if you take the trace of C times A, well, if you times A by C you're gonna times every entry in the matrix by C particularly the diagonal entries will be multiplied by C as well. And so as you add up each of those terms of the diagonal of C A all of them are divisible by C, you can factor it out. And if you factor out what's left behind well, you'd be left behind with the sum of the diagonals of A. So you times that by C. So scalar multiplication is preserved by the trace. But then if I look at properties one and two again what is this telling us if we think of the trace as a map from F N by N which is an N squared dimensional vector space this would map down to just a scalar which a scalar we can think of as the one dimensional vector space, the square matrices and the field themselves are we can view those as vector spaces. And as such, if we have a map between vector spaces we should ask is it a linear map and properties one and two right here tell us that this is in fact a linear transformation. What a surprise that all these operations that we keep on considering our linear transformations hopefully it's not a surprise. We are in linear algebra. The only things we care about in this class are linear things. If the trace was not linear we probably wouldn't care about it. So yes, the transposition map is linear transformation so is the trace map. And I should mention that when it comes to the transpose map we could also say similar statements about the conjugate transpose for complex matrices. So property three here how does the trace interact with the transposition, right? So if you take the trace of A transpose that's actually the same thing as A transpose because as you switch rows to columns and columns to rows I want you to notice that for a transpose of a matrix the main diagonal is identical to what you started with the transposition doesn't change anything with respect to the diagonal. So as you add up to the other diagonal the A transpose will have the same trace as A. So transposition doesn't do anything to the trace. What about matrix multiplication? The transpose or the trace of AB is actually gonna equal the trace of BA. Now why is this property right here important? But like I was mentioning earlier when we talk about transposition here matrix multiplication is not necessarily commutative. A times B if it exists doesn't guarantee that B times A exists. Now in this setting right here I don't know the red pins out again for blood.  And in this situation A times B will be defined and B times A is also defined. So because these are square matrices both AB and BA will be defined. But what we are not saying here is that AB is equal to BA. In general these matrices will be distinct from each other and check out the next video to actually see a counter example to show you that matrix multiplication does not commute. So we're not saying that AB equals BA but what we are saying is that the trace of these two matrices which are likely distinct will still be the same number. And also notice here that we're not saying that the trace of A times B we're not saying that this is equal to the trace of A times the trace of B. So the trace map does not preserve matrix multiplication but it does tell us that as things commute the trace is still the same number which is actually quite impressive thing because when we look at some examples and I'll point this out in the next video I'll show you a counter example of A times B versus B times A that are different matrices but I'll point out that the trace still turns out to be the same. But even though the diagonal entries are not the same. Again that's what's kind of curious here this is sort of a non-elementary fact right here. Speaking of all of these properties of matrix multiplication so I've sort of given some bad news, right? Matrix multiplication will not commute we'll see some examples of that soon counter examples of that soon but there are some good algebraic properties for matrix multiplication that we can count on similar to multiplication we've seen in the past. So for example matrix multiplication is associative A times BC is the same thing as AB times C that you can either multiply the first two then the last or the last two then the first and those two products would be the same thing so matrix multiplication is associative. Matrix multiplication is also distributive if you have A times B plus C you can distribute A and get AB plus AC you get the left distributive law you also get the right distributive law A plus B times C you can distribute the C and you get AC plus BC. It's important that we mention both of these properties because as I keep on mentioning matrix multiplication is non-commutative so because it distributes on the left does not mean it automatically distributes on the right although matrix multiplication does work in that regard. What about scalar multiplication? How does matrix multiplication interact with that? If you take the product AP and you scale that by some scalar C right here this is the same thing as scaling A and then multiplying them together or you could scale B and then multiply them together. So even though matrix multiplication does not commute in general it turns out scalar multiplication does commute. You can either scale the product, the first factor or the second factor it doesn't matter you get all the same things. And then the last property I did wanna mention in this video right here is that matrix multiplication does have an identity matrix and that's it has an identity, a multiplicative identity that's what we call the identity matrix. If A is an M by N matrix then we see that IM which is the M by M identity matrix once along the diagonals zeroes everywhere else IM times A will equal A and likewise A times IN will also equal A. So if you times the identity on the left or the right you'll always get back A. Now the identity you choose does depend on the size of A. If A has M many rows you need IM on the left. If A has N many columns then you're gonna have to have an IN on the right. Now of course if A was an N by N matrix it's a square then this would be the same on both sides. But because there could be a mismatch between rows and columns you have to multiply on the left or right by the appropriately sized identity matrix. So this gives us some important properties of matrix multiplication, matrix transposes and matrix traces operations we learned about in the previous section. And so for the most part these things behave very, very nicely. But like I said there are some properties of matrix multiplication that don't turn out to work so nicely and we're gonna explore those in our next video.


Hey there friends. Thanks for checking in at the range. We've got the sticcato P DPO Dawson precision optics. Let's go ahead and take a few shots and we'll talk more about this very popular handgun All right people love Sticcato, that's what they know the double-stack nine millimeter 1911 Pattern handgun has become so popular Many people will tell you sticcato led the way and I'm not here to tell you that They're wrong But what we have here is this sticcato P. It's got a 4.4 inch bull barrel that is bushing list fiber optic front sight and Serrated black rear sights tall So they somewhat cold witness if you if you do that, you know a lot of people say well do they co-witness I find that that works better and Rifles then on handguns because you focus on the diet. You don't necessarily put it within the the sight themselves. Nonetheless Has a rail there forward serrations. It's a sticcato formerly known as STI. It's now sticcato P Midsize gun they also have a five inch barrel But this has a flared mag well very nice polymer grip Over steel and a polymer trigger guard and trigger that is fine You don't hear anybody complaining about that overall nice trigger. You're looking at a Two and a half pound trigger pull with that sweet 1911 reset you're looking at about twenty five hundred dollars and if you go with this you have to Order a specific plate. That's another hundred and thirty bucks to put your preferred optic now what we have here is a loophole delta point pro and It certainly works just fine one of the reasons why people love sticcato so much is The the performance it delivers and is it being modeled by many other manufacturers Yeah, a little bit, but many people say hey sticcato is where it is At it is so nice, and I'm not here to argue that I think it really is it comes with two seventeen in a 21 round mag And overall very nice shooting handgun. Let me load it up, and we'll take a few more shots Correction on the sticcato five inch barrel model that I mentioned. It's called the XL It's actually five and a half inches. This is the P model 4.4 inch barrel Nice nice feeling hang I like the heavy steel guns. It just Shoot so nice. Let me see how I do one-handed here. The dot shakes a bit Yeah, I like the heavy steel frame except when I'm shooting one-handed because that is Something you certainly feel Plus it's freezing cold out here. My my nose is running like a river But anyways like spring weight overall nice I like that the 21 round mag 17 round mags are great. These are interchangeable with other Handguns of this same type of pattern. I think the sticcato does a great job. It looks super cool And it shoots so nice very light trigger pull like spring easy disassemble people love sticcato This is what many consider the standard with the 1911 Platform of nine millimeter handguns. Oh, by the way, there's that flared-out mag. Well Very nice very nice one day I'm going to own this this belongs to rider from riders range Let's get him over here and we'll see what he likes about the sticcato P here. We have rider from riders range That's where we're at today. He's so kind to let me shoot his gun and use his range on a cold winter day But you have owned this for a little while. I know you've always been a huge sticcato fan. Why is that? You know, I started out with a sticcato C way back when and then I went to the the C2 and then the P You know if I were still working in uniform down I would be carrying this if I could if my department allowed this platform This is probably the for me the greatest duty gun. It also makes a good competition gun But it's the right size the right weight. It it works just extremely well. I've never had a problem with this gun It's never hitched on any kind of ammo the ambidextrous safety is certainly good for a for-duty use Lupo Delta Point Pro is maybe not the most sturdy Optic for duty use but the gun itself just performs extremely well. It really does and that's the one thing I noticed if you do your part as the shooter you should be all set and It's it's great. Now. How many sticcato's do you own? Right now I'm down to three He's down to three. I'm trying to get a hold of one. He's down to three. Okay, and is this your favorite? Well, yes, and no, I mean, it's my favorite for range of work. It would be my favorite for duty work Right now since I'm no longer sworn officer I'm happily retired, but I'm carrying a sticcato C and they don't make the C anymore Oh really and I've got the C2, but it's a little bulky for for me to carry They've got the CS now that'll be coming out which should be really interesting and I may have to get a hold of that All right. All right. Well, thanks for joining me. Thanks for the range again. I love this range You know and allow me to shoot this It was a lot of fun and maybe we'll do it and on one of those nice summer days again. Yes All right, if you like videos like this place subscribe and share I always appreciate thumbs up button Thanks for watching and you guys be safe


We will study some applications of recurrence relations in particular what we will see today is that many counting problems can be solved by using recurrence relations relatively easily the basic strategy of handling these problems is that given a problem we will try to build up a recurrence relation on certain numbers and then we will solve that recurrence relation. So our topic today is applications of recurrence relations let us look at the first example using only three letters a b c how many words can be formed so that consecutive is do not appear in those words here we have a counting problem now if somebody asked me that given three letters how many words can be formed of which are of length n then I would have considered n positions like this total number n and then I know that each position can be occupied by three ways since we have three symbols a b c and therefore here we have three here we have three here we have three and the last also three so the total number is 3 x 3 x and so on up to 3 again n times which is equal to 3 to the power n but here we have a slightly different condition because here the problem says that I cannot consider the words which have consecutive is now let us consider if n equal to 0 if n equal to 0 then that means in the words there is no letter so the total number of such words is let us call it a 1 which is equal to 1 now when n equal to 1 and I have got three letters then that sorry here I will write a 0 this is a 0 of course because this is this corresponds to n equal to 0 now we have n equal to 1 for which we have a 1 and a 1 is equal to 3 the reason is that there are three letters and I can write a word a another word b and the third word c so these are three distinct words so a 1 equal to 3 now if we consider a 2 for n equal to 2 then let us count how many ways we can write a 2 the number of ways we can build up two letter words out of the three symbols is let us start counting we can write a a a b a c and then we can write b a b b b c and lastly c a c b c c so there are all together 3 x 3 9 words now what we see here is that there is one word which contains two consecutive a's so we have to cancel this therefore now we have 8 words and a 2 equal to 8 now like that of like this we can go on but we would like to have a compact relation in the form of a recurrence relation therefore what we do is that we say that suppose a n is the total number of such words so a n equal to the total number of such words so I have some n positions where letters have been put in now we ask a question what about the last position the last position can be occupied by a b or c now then we ask that suppose the last position is occupied by a then what will happen if the last position is occupied by a then of course in the position previous to that a cannot appear again so either it will be b or it will be occupied by c so now let us count the positions this is the position number 1 this is position number 2 and so on this is position number n – 2 this is position number n – 1 and this is position number n similarly here this is position number 1 this is position number 2 and we move up in this way and then ultimately we arrive at position number a n – 2 position number a n – 1 and lastly position number a n therefore we see that if we have a word in the set of all words which are ending with a and not having 2 consecutive a then ending with a means the nth position has a then the n – 1th position of those words will not have a will have either b or c so we have these two configurations and if we look at the segment from n to n – 2 in both these words then we will see that this segment can be occupied by any word of length n – 2 consisting of the three symbols a b c and not having two consecutive s now according to our notation a n – 2 is equal to the total number of words of length n such that no consecutive is appear so we will at least have a n – 2 plus a n – 2 many words in the set of words of length n having no consecutive is now let us move to the next page here so what we have seen is that when we are considering length n segment ending with a when we are considering length n segment ending with a then the previous position can be either b or c and there is there are n – 2 many positions and those positions for each case b and b or c can be occupied by a n – 2 many words of length n – 2 therefore I have got a n – 2 plus a n – 2 now we look at the other situation when the word of length n is not ending with a then it may end with either b or c in either of these cases the initial segment consisting of n – 1 letters can be occupied by a n – 1 and a n – 1 words so total number of words that we have is a n – 1 plus a n – 1 therefore we come to the conclusion that if we have words of length n ending with a then there will be a n – 2 plus a n – 2 many words of length n ending with a and not having two consecutive is anywhere and if the n length word is ending with b or c then we will have n plus 1 plus a n plus 1 many words and these are the only possible words of length n consisting of three symbols a bc so that no two consecutive as appear therefore we can write a recurrence relation in this form a n equal to a n – 2 plus a n – 2 plus a n – 1 plus a n – 1 in other words we have the recurrence relation a n – 2 times a n – 1 – 2 times a n – 2 which is equal to 0 once we have done this we need to solve this recurrence relation we note that it is a second order recurrence relation with constant coefficients and of course it is a linear recurrence relation we consider the solution a n equal to cr raise to the power n and substitute it in the recurrence relation which we denote by 1 to obtain cr to the power n – 2 c r to the power n – 1 – 2 c r to the power n – 2 which is equal to 0 that is taking the common factor c and r to the power n – c into r to the power n – 2 out from all the terms we will get r square – 2 times r – 2 equal to 0 and solving this we get r equal to 2 plus or minus 4 plus 8 and this gives me 2 plus or minus square root of 12 by 2 which is 1 plus or minus root of 3 therefore we get the general solution a n equal to a constant capital A times 1 plus square root of 3 raise to the power n plus b into 1 – square root of 3 raise to the power n now we have to substitute the values of a 0 and a 1 as we have already noted that a 0 is 1 so let us let us go to the next page and write the general solution of the recurrence relation a n equal to a 1 plus square root of 3 raise to the power n plus b 1 – square root of 3 whole raise to the power n and for n equal to 0 1 equal to a 0 equal to a plus b this implies that b is equal to 1 – a if we put n equal to 1 we know that a 1 is equal to 3 which is equal to a 1 plus square root of 3 plus b 1 – square root of 3 if we substitute the value of b in this equation then we will get a 1 plus square root of 3 plus 1 – a 1 – square root of 3 which gives us a 1 plus square root of 3 plus 1 – square root of 3 – a 1 – square root of 3 well and proceeding further we have a plus a square root of 3 plus 1 – square root of 3 – a plus a square root of 3 now there is some there are some cancellations first of all a will get cancelled and let me write the equation here which is 3 equal to 1 – square root of 3 plus 2 times a square root of 3 that is 2 times a square root of 3 equal to 2 plus root 3 which gives me a equal to 2 plus root 3 divided by 2 root 3 from this we see that b is equal to 1 – a which is equal to 1 – 2 plus square root of 3 2 root 3 which is equal to 2 into root 3 in the denominator and in the numerator 2 root 3 – 2 – root 3 which is equal to 2 root 3 root 3 – 2 therefore we have the solution to our recurrence relation which is a n equal to 2 plus square root of 3 divided by 2 square root of 3 this is the value of the first constant capital A and this is multiplied by 1 plus square root of 3 raise to the power n plus square root of 3 – 2 divided by 2 root 3 into 1 – square root of 3 raise to the power n this is our final answer let us look at another problem now this problem states that if a first case of measles if a first case of measles in a certain school system is recorded and Pn denotes the probability that at least one case is reported during the nth week after the first recorded case the school record provide evidence that Pn equal to Pn – 1 – 1 by 4 Pn – 2 then which week probability will decrease below 0.01 for the first time. So what we are looking at here is that suppose a school system checks the reports of a certain disease let us say measles among children and it starts counting from a week when one recorded event has occurred that is some case of measles have come up now what the school record says that starting from a week where at least one measles attack has been recorded if we start counting the number of weeks then the probability that in the nth week there will be at least one reported case is given by this recurrence relation now we have got no control over this recurrence relation it is from some data source possibly from which the school record is from the records of different years this has been observed let us say and we are asked a question that after how many weeks for the first time this Pn is going to go below 0.01 for that we have to solve this recurrence relation in order to do that we first write down the recurrence relation in the form 4 Pn – 4 Pn – 1 plus Pn – 2 which is equal to 0 and now considering that Pn is of the form C raise to the power r to the power n we have the characteristic equation as 4r square – 4r plus 1 equal to 0 and this leads to 2r – 1 whole square equal to 0 and therefore r is half half so half is a repeated root of this equation now from the discussions of the previous days we have seen that this means that Pn equal to a half raise to the power n plus Bn half raise to the power n and now we try to look at the initial conditions if we put n equal to 0 that is in the 0th week then by definition P0 is 0 because we have not started our observations now if we start from P1 what we have told in the statement of the problem that we start counting n from the point of observing at least one disease disease case therefore P1 is 1 now in the equation 1 if we substitute these values we get 0 equal to a plus B into 0 which is equal to a so the constant a reduces to 0 and we have 1 equal to a times half plus b times half now a is 0 therefore we have only b times half therefore we have b is equal to 2 thus we have the probability Pn is equal to 2 times n into 1 by 2 to the power n which is essentially n divided by 2 to the power n – 1 now we want this probability to drop below 0.01 for the first time 0.01 is 1 by 100 therefore we would like this equation to hold that is 100 n less than 2 to the power n – 1 now if we check this inequality and keep on varying n from 1 onward like 1 2 and 3 4 like that then we will find that the cutoff is n equal to 12 so when n equal to 12 then for the first time this inequality will be satisfied therefore Pn will become less than 0.01 now we will look at a case of application of non-homogeneous recurrence relations now we will again discuss one example let me write down the example first for n greater than or equal to 2 suppose that there are n people at a party and that each of these people shakes hands exactly once all the other people of course of course nobody shakes hands himself or herself now our question is that what is the number of handshakes what is the number of handshakes when we think of the solution we again start denoting the total number of handshakes when n people are present under the given constraints as an so an is equal to the total number of handshakes under the given constraints under the given set of constraints now I would like to build a recurrence relation now let us try to see what happens if we start with let us say two people because we have seen that we do not want n to be to go below 2 because if n is equal to 1 then of course there is no handshakes so we start with n equal to 2 so suppose I have got person number 1 and person number 2 let us call them v1 and v2 and I will join them to signify that shaken hand exactly once so we see that a2 is going to be 1 now if we consider that another person is coming now what will happen so let us draw the situation over here we have got three people so we have now a party of three and they are shaking hands with with the given constraints now what we can always do is that we can remove for the time being one person suppose we have left out v3 then there is only one handshake so this is the number of handshakes for two people so I have got a too many handshakes now when with v3 comes in then he does not shake hands with himself therefore he has to shake hand with all the other people exactly once so therefore he will shake hand with this person and shake hand with this person so the total number of handshakes will be a2 plus 2 which is denoted by a3 now suppose we know a3 so there are three people who shakes hands with each other v1 v2 v3 they are shaking hand with each other and suppose a fourth person is coming in and let us call him v4 and so that means now we have a party of four so when v4 is out the remaining people shakes hands in a3 many ways and when v3 v4 comes in he has to shake hand with all the people except for himself he shakes hand with v3 he shakes hand with v2 and lastly or whatever he shakes hand with v1 so he has to shake another three many hands so total number of handshakes is a3 plus 3 which is equal to a4 now we can go to the general case that suppose we have got n people so we have got v1 v2 and so on and lastly we have got let us say vn minus 2 vn minus 3 and over here vn minus 1 and here we have vn so we have a party of n people and we know that they shake hand in an many ways now suppose that we have got one more person vn plus 1 and vn plus 1 so we have a party of n plus 1 people now suppose they shake hands in the designated manner now what we do is that we can choose any of the vertices now let us let us take that we are taking out vn plus 1 and removing all the edges from vn plus 1 then we have a party of n and of course they are shaking hands exactly once with the others and not shaking hand with himself or herself so we have got total of a n minus 1 handshakes and then when vn minus 1 vn plus 1 is considered then he is bound to shake hands with each of the people in the other group so the total number of handshakes will be n so this is increased by n and which is the total number of handshakes with n plus 1 people so we have built up a recurrence relation corresponding to the handshakes let us write down this recurrence relation and then let us try to solve it the homogeneous part of this equation is a n plus 1 minus a n equal to 0 and we take a trial solution let us call it a n equal to CRS to the power n then we get CRS to the power n plus 1 minus CRS to the power n which is equal to 0 and this leads us to R minus 1 equal to 0 because we can of course assume very reasonably that C not equal to 0 and R not equal to 0 therefore we have R equal to 1 therefore the homogeneous part written as a n superscript H in bracket is C times 1 to the power n or C which is a constant next we try to find out a particular solution here we choose that trial solution to be a n particular equal to a 1 n square plus a 0 n which is indeed a rather complicated expression and if we substitute this in the recurrence relation then we will get a 1 n plus 1 whole square plus a 0 n plus 1 equal to a n which is a 1 n square plus a 0 n plus n and if we do the usual manipulations then we will get n square a 1 minus a 1 plus 2 times a 1 plus a 0 minus a 0 minus 1 plus a 0 plus a 1 equal to 0 and this should hold for all n therefore we have got a 1 equal to a 1 by equating the coefficients to 0 and when I equate the coefficients of n to 0 I get a 1 minus 1 equal to 0 that is a 1 equal to half and when I equate the constant term to 0 that is a 0 plus a 1 equal to 0 then I get a 1 equal to or minus a 1 equal to a 0 which implies a 0 equal to minus half therefore we arrive at the general solution a a n equal to constant plus half n square minus half n and now if I put the value of n equal to 2 we know that 2 is equal to sorry 1 here we have got 1 a 2 is 1 so therefore 1 equal to a 2 which is equal to c plus half of 4 minus half into 2 which gives me c plus 2 minus 1 which is c plus 1 this implies c equal to 0 therefore finally we have the solution as a n equal to half of n square minus half of n for n greater than or equal to 2 thus we know now that if we have n people in a party and each person shakes hand exactly once with each of the other people present in the party then the total number of handshakes is half of n square minus half of n with this handshaking problem I will end today's lecture thank you


I'm Brian Sakawa, you're watching He Spoke Style and today I'm going to show you my desk setup. Want to give a shout out to Brock at the Modest Man for the inspiration for this. Appreciate that very much. So first things first, my desk is never as clean as you're going to see it today. I strive to keep it this clean because I find the neater things are at the desk and around the studio the less anxious I feel and the more productive I am. But you know this is a workspace and it's impossible to keep completely pristine all the time. I tend to be someone who's prone to making piles of things. I know it's only going to take like 10 to 15 minutes to neaten up but sometimes it's the last thing I want to do though I always feel so much better when I do put that time in. All right so moving on now to the desk setup and first the actual desk. This desk or table is by Blue Dot. Blue Dot is a sort of minimalist furniture company. It was founded by two architects and a sculptor so the form of everything is always very very well considered. Everything's got sort of a minimal modern feel with clean lines and I really like the aesthetic. This table is the strut. It's the large size at 75 inches long and 35 inches wide. It's a huge surface obviously bigger than I need for just working at the computer but it's great to have the extra space for when we're shooting outfit videos. I can line up all the accessories like pocket squares, watches, glasses, sunglasses and just have them all readily available. We've also used this surface for top-down video in the past and of course this area is where I sit for this main talking shot. It is also the area that tends to become the repository of piles of things but not now and I'm going to try to keep it that way. Next is my chair. It's very old. I honestly don't even know what brand this is. It doesn't say anywhere on the chair. It's comfortable enough. The height is adjustable. I personally like the armrests. It has wheels so it can easily roll across the floor or to the other side of the desk. Actually sometimes I set up my laptop on the other side of the desk sometimes for other work and the chair makes it easy to go back and forth between computers. Speaking of computers, my desktop computer is an iMac. It's very old. It's from late 2012 21 and a half inches. I completely maxed out the RAM so it runs as fast as it possibly can. I have the Magic Keyboard and the Magic Mouse as well. My mousepad is by McMurray and Blonde. It's a small company in Chicago. They make really nice high quality leather goods. This computer obviously not the fastest or newest model but because it's a job done for me since I have Ian who edits all the videos I'm not doing any of those things that require a real high speed processor. Though in terms of photos I do work from an external hard drive just to keep as much off the machine itself so it'll run as smoothly as possible. Probably upgrade, get something newer at some point but until this doesn't turn on it's totally fine. Mentioned that I work from an external hard drive and the hard drive I use and travel with all the time is the Lacy Rugged Mini. It's four terabytes not even half filled up with all the video footage and photos from the beginning of time. I personally don't worry too much about losing data. Rob has all the photos backed up in multiple places. We have copies of all the video footage in the cloud and Ian also keeps that as well. I'm really particular about media management though as are the other two guys so we have very comprehensive folder structures for everything. It makes things very easy to find and a lot less frustrating when you're looking for that one particular thing from that one particular time. Even though I personally don't do the photo or video editing I do have a card reader here. It's by Lexar. It's a dual card reader so it takes both SD and CF cards. A lot of times when we're shooting we'll just do a quick test shot for like lighting or framing and then just quickly pop the card into the card reader and check out the footage so we know we like what we're working with.  It's by Hein. We had them made when we launched the shop and sadly they are all sold out now. It's handmade. It has the HSS monogram underneath. It's a nice little desk accessory and right now I have my Mon Blanc Meisterstück ballpoint pen resting proudly in it. I am not a fan of making notes on my phone or making lists that live on my computer. I feel like if I do that they just disappear. I completely forget about them which totally defeats the purpose of writing things down in the first place. I need to see things in front of me.  So two ways that I do that. I have a desk pad here. It's by Exocompta also available in the shop. It's a very simple desk pad. It's metal. It has rubber feet on the back so it doesn't slide around or scratch the surface of the desk. One of the best things about this is that it's refillable. You just unscrew these two nuts right here and then you can load in another pad when you're finished with the one you're using. I'm also kind of obsessed and very particular about post-it notes. They have to be this canary yellow color. These I often stick on the bottom of the computer monitor. Generally I use these for keeping track of content ideas. If I have an idea for a video or a series of videos I will put them on a post-it note and then on the bottom of the computer screen right in front of me so it's impossible for me to forget the idea that I just had. It makes it easy to refer back to when I'm looking for new video topics. It's not a very modern or cutting edge system but it works really well for me and really helps keep me on track. Over here on the side of the desk I have the Anker Powerport USB charger. It has six different slots. I always try to keep things topped off including these two power banks. One is the Anker Powercore 13000 and the other one is also by Anker but it's much older. It's not available anymore actually and I really recommend the Powercore 13000 over that one. Fully charged you can get four full iPhone charges out of it. Super handy. All of these things here the Powerport and the two power banks are things that I always travel with. Definitely the Powercore 13000 is with me every single day. Behind the desk is a bookshelf slash cabinet by Ikea. It's the Calax model. On the top shelves I've got some rake magazines and some other random stuff that is not really organized in any special way. On the bottom I've got drawers on one side and doors on the other. The drawers have some miscellaneous gear extra office supplies stuff like that and then the doors hide a lot of the piles that accumulate so we're not going to show you those. On top of the bookcase is my printer. It's a Canon PIXMA IP110. Nothing fancy. It's all I really need. Next to that is a brass magazine stand by Firm Living. This is available in the shop. In front of that a five slot version of the Hein-Penn slash pencil tray. Got a letter opener made for us by Abby Horne going to be available in the shop but currently not listed. A couple random notebooks a few select books that I keep displayed up there. Then I have this brass business card holder which is something we used to carry in the shop. Super popular item but the brand stopped producing them for some reason so they can no longer be had. Then a small desk lamp. Got this at Target. It's very inexpensive. And then finally this brass cup which I use to keep all my extra pens and pencils in one place. So that's it. That is my desk setup here in the studio. It's where I spend a lot of my time doing a lot of the work for He Spoke Style that you don't see on camera. And speaking of we have a video in the works where I'll show you how we share our videos what our setup is so definitely stay tuned for that. I'm going to put links to everything I mentioned in this video down below in the description. As always thumbs up. Don't forget to subscribe and hit the bell so you get notified every time we publish a new video. Till next time thanks for watching everyone and stay tailored.


The study conducted in Zambia showed that poor healthcare waste management poses a serious risk to the health of healthcare workers, patients and communities. Despite the fact that the country has made progress towards improving healthcare waste management, there is still much work to be done. The study revealed that only 43% of facilities had a functioning incinerator on site, and that needle sticks were alarmingly high, with 31.3% of all respondents reporting a prior needle stick. Additionally, only 56.9% of facilities used an infectious waste bag, yellow, red or orange bin liner, and a black bag for general waste. Furthermore, only 37.3% of respondents reported receiving any sort of healthcare waste management training. These findings highlight the urgent need for improved healthcare waste management systems in Zambia. This article was authored by Colleen M. Leonard, Chipwela Chulwichuna, Justine M. Nkama, and others.


to be personally and for us as a musical organization. There are many things in our lives, also part in it, divide us and make it difficult to connect as a community. And this is one of the things I cherish the most. It's almost, to me, not quite Christmas. I'm totally pleased to have you all here with us. It's also very important for us that we repeat this work each year. And I can tell you that since we started this every single year in this year is no exception. The chorus, the orchestra, the solos, bring a greater understanding, a greater love, a greater knowledge of this music. And I can tell you that the rehearsals we've had the past few weeks putting this together have been a real joy for me. And the work we have to put into it seems to get smaller, but the awards we get out of it become even greater, and that's a real joy for me as a musician. I also would like to thank our hosts here at Holy Name, the people at this congregation who have been wonderful to help us in every way you can possibly imagine. There are quite a number of them, but I did want to recognize by the Dan Sanders, who not only loves his church and the community, but he also loves music very much. So Father Dan, please stand and think about it. I was mayor. That was not happening. Don and I were talking. That type of meeting was not happening. When I knew I had to reduce by two important things to five or six years, the relationship between Lakeland College and Sheboygan Symphony has grown into me, become a really wonderful relationship musically and personally for me. Most of the or many of the educational things that we do with the Sheboygan Symphony involve Lakeland College. All of the guest artists that come and sing or play with the Sheboygan Symphony give master classes at Lakeland College. We do our pop's concert there. Young artists can share their competition to name just a few. And I really enjoy the college. I enjoy the students very much. And also the faculty, there's one special faculty member. Janet Herrick is the conductor of the Lakeland College Chorus, as you know, as part of the Sheboygan Symphony Chorus, and she trains the Sheboygan Symphony Chorus. I could not be more blessed to have such a wonderful colleague both musically and personally. I'd like her to come up and conduct the next chorus.