Problem 2Another measure of impurity of a node is the Gini index The formula of the Gini index for given node that contains individuals of k = 1,_ K classes is:Gini...

Solve each problem. Suppose a sample of a community shows two species with 50 individuals each. Use the formula to find the measure of diversity $H$, where $P_{1}, P_{2}, \ldots P_{n}$ are the proportions of a sample belonging to each of $n$ species. $$ H=-\left[P_{1} \log _{2} P_{1}+P_{2} \log _{2} P_{2}+\cdots+P_{n} \log _{2} P_{n}\right] $$

Were asked to show that forgiven an undirected graph with Enver theses, then no more than the four of an over to to the K trees remain after the first step of Solans algorithm has been carried out, and the second step of the album has been carried out K minus one times. So let G b. A connected, waited undirected graph with in verse, Eze and let PKB the statement Solans algorithm has no more in the floor event ever to K trees remaining after the first step and the next K minus. One steps have been carried out and we'll prove this by induction. So the base case we have the K is equal to one. So as part of the algorithm, we'll construct a forest of entries where you shriek contains one vertex. Then we know that every new tree that will be formed in the first step will contain at least two of the entries. This is because each of the entries us to be connected to a new tree, and so there are most and over to new trees. And since the number of trees is an energy er, this means that there are no more, then the floor event over two trees. When we see that in this case statement, P one is true for the inductive step, let P m be true for some m greater than or equal to one. So it follows that the algorithm has no more than n over to to the entries remaining after the first step and the next M minus one steps of the algorithm have been carried out. He wants prove that P M plus one is true. So start from a forest with utmost and over to the M trees at most, the floor of the never toothy entries dynasty. You know that every new tree that will be formed in the next step will contain at least two of yet most Florida and over to the M. Trees contain at least two of the at most floor and over to to the M trees. This is because each of the at most and over to the M floor trees has to be connected to a new tree. So there are at most the floor of the floor of an over to the em over to Well, I guess, first of just do there are at most the floor and over to the M, divided by two which is equal to the floor of n over to to the M plus one trees. So it follows that statement. P m plus one is true. PM is true. So by principle of mathematical induction it follows that's PN is going to be true for all then greater than.

For the given problem, we want to consider the logistic growth function. So it's gonna be Q. Of T. Is equal to a over one plus B, B to the negative GDP. Um So with that in mind we want to suppose that the population is Q one when T equals T. One um and Q two when T equals T. Two. So we would want to show that the value of K is going to be given to us. So um Q one is going to be our initial value. So based on that the value of K. That we get if this is T. One And we compared to T. two, what we end up showing is that K is equal to one over T two minus T one. And the natural log of hue to and a minus 21 over 21 times a minus youtube. So do we end up showing? And that's going to be our final results.

For the given problem, we want to consider the logistic growth function. So it's gonna be Q. Of T. Is equal to a over one plus B, B to the negative GDP. Um So with that in mind we want to suppose that the population is Q one when T equals T. One um and Q two when T equals T. Two. So we would want to show that the value of K is going to be given to us. So um Q one is going to be our initial value. So based on that the value of K. That we get if this is T. One And we compared to T. two, what we end up showing is that K is equal to one over T two minus T one. And the natural log of hue to and a minus 21 over 21 times a minus youtube. So do we end up showing? And that's going to be our final results.

Okay, so we want to find our index of diversity. Given that we have two species, that is RN value is two. And then we have 50 individuals in each species, so we have to. So that's going to be 50 plus 50 which gives us a total of 100. So then p one that's going to be our first species that's 50 over a total, and then p two is also going to be 50 which is our understory shoes over our total. So it's your at this. We have h is equal to negative. He won. So it's 50 over 100 actually have simplified out to OK, no, let's see this. And then we have log base to of 50 over 100. Okay. Actually, it might be useful to simplify this to 1/2 so actually we can get this. You're not. So we got 1/2 then we have log based two of 1/2 but we can hear it 1/2 a step too far out. Negative one. And then we have the same thing for p two. So here it is simplifies. This year is a negative one to have negative 1/2 plus a Are you sure that's a minus 1/2 on? But it's 1/2. So this is it quits negative, negative one, which is equal to positive one.