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Advance Maths Problem 45Please don’t answer compulsory I will givedislikeProblem 45: Lagrange’s identity (✓✓✓) 1987 Paper II If y = ...

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Advance Maths Problem 45Please don’t answer compulsory I will givedislikeProblem 45: Lagrange’s identity (✓✓✓) 1987 Paper II If y = f(x), the inverse of f is given by Lagrange’s identity: f −1 (y) = y+ ∑∞ 1 1 n! d n−1 dy n−1 [ y − f (y) ]n , when this seriesconverges. (i) Verify Lagrange’s identity when f (x) = ax. (ii)Show that one root of the equation x − 1 4 x 3 = 3 4 is x = ∑∞ 0 32n+1 (3n)! n!(2n + 1)! 43n+1 . (†) (iii) Find a solution for x, asa series in Î

Advance Maths Problem 45 Please don’t answer compulsory I will give dislike Problem 45: Lagrange’s identity (✓✓✓) 1987 Paper II If y = f (x), the inverse of f is given by Lagrange’s identity: f −1 (y) = y + ∑∞ 1 1 n! d n−1 dy n−1 [ y − f (y) ]n , when this series converges. (i) Verify Lagrange’s identity when f (x) = ax. (ii) Show that one root of the equation x − 1 4 x 3 = 3 4 is x = ∑∞ 0 3 2n+1 (3n)! n!(2n + 1)! 43n+1 . (†) (iii) Find a solution for x, as a series in λ, of the equation x = eλx . [You may assume that the series in part (ii) converges and that the series in parts (i) and (iii) converge for suitable a and λ.] Comments This looks pretty frightening at first, because of the complicated and unfamiliar formula. However, its bark is worse than its bite. Once you have decided what you need to find the inverse of, you just substitute it into the formula and see what happens. Do not worry about the use of the word ‘convergence’; this can be ignored. It is just included to satisfy the legal eagles who will point out that the series might not have a finite sum. In part (ii) you can, as it happens, solve the cubic by normal means (find one root by inspection, factorise and use the usual formula to solve the resulting quadratic equation). The root found by Lagrange’s equation is the one closest to zero. Equation (†) turns out to be a very obscure way of writing a familiar quantity.29 Lagrange was one of the leading mathematicians of the 18th century; Napoleon referred to him as the ‘lofty pyramid of the mathematical sciences’. He attacked a wide range of problems, from celestial mechanics to number theory. In the course of his investigation of the roots of polynomial equations, he discovered group theory (in particular, his eponymous theorem about the order of a subgroup dividing the order of the group), though the term ‘group’ and the systematic theory had to wait until Galois and Abel in the first part of the 19th century. Lagrange’s formula, produced before the advent of the theory of integration in the complex plane, which allows a relatively straightforward derivation, testifies to his remarkable mathematical ability. It is practically forgotten now, but in its day it had a great impact. The applications given above give an idea how important it was, in the age before computers



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Solve each problem. Several years ago, mathematical ecologists created a model to analyze population dynamics of the endangered northern spotted owl in the Pacific Northwest. The ecologists divided the female owl population into three categories: juvenile (up to $1 \text { yr old }),$ subadult $(1$ to 2 yr old ) and adult (over 2 yr old). They concluded that the change in the makeup of the northern spotted owl population in successive years could be described by the following matrix equation. $$ \left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right]=\left[\begin{array}{rrr} 0 & 0 & 0.33 \\ 0.18 & 0 & 0 \\ 0 & 0.71 & 0.94 \end{array}\right]\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right] $$ The numbers in the column matrices give the numbers of females in the three age groups after $n$ years and $n+1$ years. Multiplying the matrices yields the following. $j_{n+1}=0.33 a_{n}$ Each year 33 juvenile females are born for each 100 adult females. $s_{n+1}=0.18 j_{n}$ Each year 18\% of the juvenile females survive to become subadults. $a_{n+1}=0.71 s_{n}+0.94 a_{n} \quad$ Each year $71 \%$ of the subadults survive to become adults, and $94 \%$ of the adults survive. (a) Suppose there are currently 3000 female northern spotted owls made up of 690 juveniles, 210 subadults, and 2100 adults. Use the matrix equation on the preceding page to determine the total number of female owls for each of the next 5 yr. (b) Using advanced techniques from linear algebra, we can show that in the long run, $$ \left[\begin{array}{c} j_{n+1} \\ s_{n+1} \\ a_{n+1} \end{array}\right] \approx 0.98359\left[\begin{array}{c} j_{n} \\ s_{n} \\ a_{n} \end{array}\right] $$ What can we conclude about the long-term fate of the northern spotted owl? (c) In the model, the main impediment to the survival of the northern spotted owl is the number 0.18 in the second row of the 3 $\times 3$ matrix. This number is low for two reasons. The first year of life is precarious for most animals living in the wild. In addition, juvenile owls must eventually leave the nest and establish their own territory. If much of the forest near their original home has been cleared, then they are vulnerable to predators while searching for a new home.
Suppose that, thanks to better forest management, the number 0.18 can be increased to $0.3 .$ Rework part (a) under this new assumption.

Hello. Shouldn't No way would conduct in this question. This is crushed worse and a support. There are currently 300 females, so 300 are three Teligent female northern sported calls 600 out of 3690 are juvenile. 200 then are so adult. So it'll and don't be 100 are dead abs. So after five years on the business division, given this is given no plus and 4 to 1 so day and plus one invented toe equal to zero point 33 a one into question. Oh, it is given our jato 0.23 even which is job. 133 and a one, Which is good Question 2100. So in 2100 which gives which gives our six 93 And after five years, so multiply with five. So this is equal to 3465 on this is Joe 0.18 in tow. First year, uh, this is don't be 100. So this is equal to it's a 36 straight and for five years were deployed with 55 major After five years. This is equal 18 40 summer jobs it gives 2100 23 Eder's on our provide. The Heiress No more adult. After five years, 2123 into five, just upper five years. This is a number of others and be we can conclude that the number off all's linearly decorating and see if, uh, forest management you can pick them, then their population will encourage.

Hello everyone welcome back to another differential equations problem. This is a long one. So let's get right into it. Here we have do I. D. X. Is equal to one minus four. X over X minus Y. Now the first part of this problem party asks to rewrite this equation. So all the variables are Y over X. Instead of Y. And X. Separately. So that what I noticed is that we can factor out an X. From both the top and the bottom. I'm gonna do that here. So we'll take X times Y. Rex from. This is just the numerator of this up here. Uh And subtract for if we multiply this X out and distribute it, we'll get the numerator of the last problem. We did the same thing. The denominator will get one minus Pyrex. And if you multiple if we distribute this X out, we'll get the same as the initial problem. And now that we have an excellent numerator and denominator we can just cancel that. Great. That's part of Herbie is to replace why wrecked with a variable V. This is just a temporary variable. But we're also going to try to find DVD X. So the derivative derivative of V with respect to X. All right. So let's just start off with the easy part replacing Y. Rex with v. So we'll have a V -4 Over 1 -50. Great. Now we have to find DVD ex. Uh This is a little bit more challenging. But if we realize that X R V is in terms of X such that Y equals X times via fax. This has given in the problem statement. Then we can just do the product rule conservatives and get with the DVD excess. We'll have do Y dx equal to X in terms of distributive V with respect to X and the same thing but flipped. Uh Plus the derivative effects with respect to X. Not just one great Part C says to conflate these two values. Um So we see that this is actually do I D X itself. Um since that's the original problem. And this is also do I D X right there. So they're really equal. All right. In fact, if we do some simplification, we can find that D V D X times X is equal to v squared minus four Over 1- ft. Just through all direct manipulation. All right. Let's move on. The part. D part D says to solve our differential equation explicitly getting V in terms of X without this DVD extra. All right. So to do this, we have to realize that this is a separable differential equation and we can get the dx is on the right and the d v s on the left. Someone to do that. We'll have T V Times 1 -3. All over V squared minus four. Then on the right side will just be D X or X. Great. And let's uh integrate the right side. The left side will have to do some partial fractions on but it won't be too bad. I'm just gonna do these partial fractions quickly. Uh skip all the labor intensive steps because that's not what this problem is really about. But it will be T. V times this quantity of negative, 3/4 times. Oh My bed. I could have 3/4 times B minus two. The videos too minus 1/4 and the V -2 is the denominator, recognize that this is just um uh separating these two right here. Great. And this is going to be equal to dx over X. Let's integrate both sides. Should be simple because these are all algorithms. So we will have um negative Well the fact that this negative 1 4th and we will have the natural log three times the natural log of V plus two. And add this to the natural log of b minus two. Great. Oh that will be equal to the natural log of X. Perfect. The next step is to actually let's multiply this negative four over to the right side and do the power rule. Um at the same time let's combine these two terms. So we will have the natural log of the quantity V plus two cubed. Remember if there's a constant on the outside that goes to the power inside the log rhythm. So we ve plus two cubed Times V -2. Great. And then this will be equal to natural log of X To the negative 4th power. And recognize that the same constant rule applies if you move it over to the right Alright. Since we both have logarithms on both sides. Oh I forgot there's A plus C. Quickly we'll have the plus C. Here. Um All the continents will just be added onto that. We'll have a plus C. Here. All right, let's do E. To the power of everything. So this will be a. v. plus two cubed Times of U -2. Um Actually these should be absolute value signs because we're taking absolute log. So let's do that. And this is going to be equal to the natural the natural log but just X. To the uh negative fourth at times of constant C. Because E. To the A. Plus B. Is equal to eat to the A. B. So these are just going to uh be most played together. All right. Um Sorry I meant E to the A to the B. Anyway. Um This city will just be multiplied it against this X to the -4. Um and that concludes part D. Let's move on to party here. We have to replace V. With Y. To the X. With Y over X. And simplify. All right. So let's do that. Um ignore this stuff. Said that incorrectly but we will have whatever X plus two times whatever X minus two. And the side will be cubed. That will be equal to X the minus fourth times C. We can simplify this a bit. Uh For example let's multiply both sides by X. And specifically this term. If we do that we're going to get Y -2 x. And the site will be X to the -3. Which what this really means is that every time we multiply this left side by uh by an ex uh this bottom part will be removed. So this is really to the first power. So we multiplied X. This term by X. One time to get rid of this numerator X. If we multiply this term by X three times we'll get rid of this uh denominator X. So let's just do it four times to get rid of all the exes it's all got Why Plus two X. Cubed Times. Why -2 x. Is equal to see? And that concludes that problem. Party part F. And H. Is silly graphing this is very difficult to do without um graphing calculator. So I'll just show some uh rough sketches of what it looks like to get some intuition. Um This will look like um strong curves up like this, never touching the origin. They're also be strong curves down like this and strong curves out like this. Lastly, that records like this. Um The there is a line of symmetry on this that is right here. Um If we folded over, there will be symmetry on the other side. Um There's also a lot of symmetry right there. Um That concludes the problem. Thank you for watching.

For this problem. We have been given a matrix that shows how the population of northern spotted owls is changing year by year. And we're watching three different subgroups of female owls juveniles, sub adults, which are 1 to 2 years old and then full adults two years and older. And we're looking at this. The Matrix A that I have here shows the percentages, um, that we need to use in order to find our population. So if I have Matrix B, that shows how many female owls in each of those three subgroups 900 juveniles, 500 sub adults, 2600 adults if I multiply a Times B, that shows me the number off owls in each of those subsections the following year. So after one year I go from my population has given to this new population, and I want to take a look at for each of the next five years. So let's go five years out. This is your one now for year two. I'll take this new matrix that we have and multiply it by a so I'll have a times what we just found a times B. So this is after year one after year, too. What about your three? Well, I'm going to take my Matrix A and then multiplied by this result that I just got so for years, three. I will get these numbers. Your one year, two year three. How about your four A. B and year five. Great five. Okay, now, let's just take a look at what we have here now. We started with 900 juveniles. Now we had a little bit of a dip, and then it increased. So this is not perfectly steady moving in one direction. But as we go on, those numbers do seem to be declining. What about our sub adults? What looks like we must have had a very large population of sub adults because it went down rapidly that first year and then kind of stabilized a little, but definitely stabilized lower. And how about our adults and went up a little bit again. We had a large group of sub adults, but now it is going back down again, and we are. So we went up really high, and it's starting to decline. In fact, we've been told that our, um the overall um Matrix so we can use to find long term growth of this population. If I want to find the end plus one year of juveniles, sub adults and adults, I'm going to take the current numbers that we have. And I'm gonna multiply that by 0.9835 n. Well, that number is less than one. So overall, every year that I check these populations I'm expecting from one year to the next to go down, not by a lot. You know, one would make a stable population the same number from one year to the next. It was bigger than one. Our population would be growing. But this means that my spotted our population is in decline. This is less than one. Well, what if we take a look at our original graph? I'm gonna come back up here and look at a so that's 0.18 number there in that second row first column that says that Onley 18% of juveniles become sub adults. The rest of them don't make it past that first year. Yeah, What could cause that number? Well, maybe it's the habitat. There isn't. Ah, good habitat for the babies. to grow in. Maybe the population of predator versus prey is a little bit out of whack. Maybe there are too many animals that eat small birds that that's causing a problem here. Maybe dogs or cats that are on the loose or foxes or something that might damage these babies. Maybe there's simply not enough food for them to, uh, go from egg to mature adults. Maybe there's something in the environment that's making them sick. We don't know, but there's a lot of things that could cause that problem. What would happen, though, if we targeted that number that we put the management habitat management in place may be reduced pollution. Um, took a look at the environment and try to make some changes, and we could increase that from 18% of 40%. What would that change? Will now take a look What happens to my numbers? Well, I start with 900 juveniles, and again, it's a little bit of a dip, But look at those numbers. 9 23 9 52 till your five. I'm over 1000 juveniles sub adults again, there's a small gap. I think there were more sub adults than usual in this first year. But now, once it kind of stabilizes, those numbers are increasing. And what about my adults? 27 99 28 87? I'm over 3100 by the end of this five year period. In fact, if I add up all of the birds at the end of my five year period, I have 1004 juveniles, 390 sub adults and 3130 adults. So I have gone from a population of 4000 at the beginning of my study, up to 4524. And that's with the 40% if I had added up those same numbers back when it was only 18%. The some of those numbers after five years was 3639 a decrease rather than increase. So that number is very important to the population of this spotted owl. Getting that number increased from 18 to 40 will make a huge impact on the overall health and well being of the spotted owl population


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