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2.1 Deviation of middle element value from average. Suppose 1 21. We dlefine the middlle element Value of' a5 Tmn * Definen-vectol, with2m - 1 andf() = CmTi; i...

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2.1 Deviation of middle element value from average. Suppose 1 21. We dlefine the middlle element Value of' a5 Tmn * Definen-vectol, with2m - 1 andf() = CmTi; i=lwhich is the difference hetween the midldle element value and the average of the coefficients in Express f in the form f() =(lI. where a is a n-vector 2.2 Nonlinear functions. Show that the following two functions f R3 4 R are not linearf(z) = (x1 T2 + 13)2 f(z) (81 + 272 13 + where for any real number &, (a) - is the positive p

2.1 Deviation of middle element value from average. Suppose 1 21. We dlefine the middlle element Value of' a5 Tmn * Define n-vectol, with 2m - 1 and f() = Cm Ti; i=l which is the difference hetween the midldle element value and the average of the coefficients in Express f in the form f() =(lI. where a is a n-vector 2.2 Nonlinear functions. Show that the following two functions f R3 4 R are not linear f(z) = (x1 T2 + 13)2 f(z) (81 + 272 13 + where for any real number &, (a) - is the positive part of &, defined as (a)+ max{a, 0}. Remark:. This function is widely used in neural networks, where is it called the relu function. shortening of rectified linear unit.



Answers

(a) Verify that the average of the left and right end-point approximations as given in Table 7.7 .1 gives Formula ( 2 ) for the trapezoidal approximation. (b) Suppose that $f$ is a continuous nonnegative function on the interval $(a, b]$ and partition $[a, b]$ with equally spaced points, $a=x_{0}<x_{1}<\cdots<x_{n}=b$. Find the area of the rapezoid under the $<x_{n}=b$ ent joining points $\left(x_{e}, f\left(x_{k}\right)\right)$ and $\left(x_{k+1}, f\left(x_{k+1}\right)\right)$ and $t$ above of Formula ( 2) is the sum of these trapezoidal areas (Figure $7.7 .1) .$

Using the data and exercise 24 three State I skis is given here again. We have some years, and for each year we have the amount given him billions of dollars. And for the year 1999 we have for 39,268 billions of dollars and this is a gross domestic product. And then we have for 911,817 in 2000 and one we have 10,128. 2002. We have 10,470 in 2000 and three There waas 10,961. For two 1004 we got 11. 11,686. In 2000 and five. We had amount of 12,434. 2000 and 6, 13,195. And in 2000 and seven there waas 30,841 billions of dollars. So we did the data. We're going to go struck a table this way. We're going to compute the average rates of change. Delta while we're Delta X on each of interval off length one year and then we're going to calculate the mid points of each preventable and put them in the first column of the table. And the average rate rates of change are going to be the second column off the table. So we will have a table off mid points off this issue interval. And it's corresponding rate of change in part B. We're going to fight it, find a linear regression following the lead squares met, and this will be a model for thes data and disable Here. In fact, we're gonna find a regression model from data on the table. We're going to construct in part A. Yeah, And then we're gonna use model the regression model linear regression model to approximate the rate of change in 1997. And in 2000 and one, we're going to compare the results in exercise 24. See where we used the regression for the original data and an approximation of the derivative. So we're doing a similar thing, but with another approach. Get so here we are using the average rate of change, and we know that the average rate of rate of change is an approximation to derivative on the midpoint of the sarin tipple. That is why we're going toe calculate the midpoint of sarin table and the average rate of change in that serene tools. And with that, we're gonna pass a model or the linear regression model off the data. And that is going to be the approximation to the derivative off the function given in the original table we have at the left. Well, so we're going to start with Bar A. So we're going to calculate for the first interval. The first interval look is the interval from 1999 to 2000. It's one year let's intervals, and there we're going to calculate first the midpoint, and we know the midpoint is the average off the end points of the interval that is 90 1999 plus 2000 over to So is 3009 Sorry, 3999 over to. And that is 1999.5, as we expected because he's, uh, the end point at the left, plus the half of the length of the interval that is 1/2. Well, this is the midpoint. Another rate the rate of change right, or the average rate of change is okay, the difference of the amount on the end points, starting with the lasting point writing point finest e amount of the left in point. So we get 9000 817 minus 9200 and 68. That is the amount at the 1000 at the year 2000, which is the writing point, minus the amount at the left hand 0.9 1999. And that's divided by themes span of time. In that interval, that is 2000 minus 1999. She's one, and the difference in the numerator is, um, 500 49/1 and studies 549. We're not. The units are billions of dollars over a year or over a year. Mhm. So this is the two values we get in the first column. So the first column or the first role rather the first column. We'll have the midpoint. He's one and the second column good, and that's the first roll off the table because it's the first interval, and we do that for all the intervals we have here in the table given. So we're gonna have the following table where we have mid points, move an average rates of change. So we get for the first midpoint, which is Theis one here. Yeah, Thea bridge rate of change is 549. Then for the second one, we will have the midpoint 2000.5 and the rate of change the average rate of change is 311. And this one is 2000 and 1.5, with an average rate of change of three, 142. Then we had 2000 and 2.5 with 400 91 in 2000 and 3.5 with seven 1025. So the stable continues here and we have 2000 and four 0.5 with an average rate of change off 700 48. And we have 2000 and 5.5 with 761 and 2000 and 6.5 correspond to the midpoint off he last interval Mhm, with an average rate of rate of change of 646. So this is our table first part here, second part here, and that's the result we obtain using a calculator. So now, uh, we're going to do in party. We're going to find a linear regression in your regression, which is calculated using these squares, which will we want thio at hand. But we're going to use a graphing calculator. In our case, we have used the scientific computer software Madeline for that, and we have found uh huh and, uh, end of X equal, 52 point 155 or 52.155 time six minus 100 three thousands 894 point 363 And now here we emphasize that we have used years. We're using years as they are. That is, with its magnitude 9 1999. We're using 2000, 2000 and one and so on. So we're not using the convention of using the zero for these or one forties and so on, as we did in the exercise. Uh, 24 waken do that. But we here we have used instead three original data. And for that reason, we have obtained conventions like this, which is a large number. But that's not any problems are all and Now, with this, we can plot the data, the points in the table here that is the mid points and the average rates of change at the same time with thes linear regression from with the calculator. And now we are going Thio. So what? This graph we have found he's in mud lab Mhm. And here we can see the graph off the regression model in blue here line L of Mexico or Y equals 52.155 x minus 103,800 94.363 It is blue line here. And the dots, the red dots are the points we got or we obtained in the bar in part a With that is table off mid points, an average rate of change. We can see the regression linear regression model. It's not as good as we could say, because we have it, not a linear pattern of the points, but we're going to use that as theorist Rate of change. Yeah, as we have said said to do so and in part c, we're going to use that to approximate your average rate of changing parts. Sorry. In years 1997 in 2000 and one. So we're going to do that first? Uh, 1997. And for that, we're going to evaluate Yeah, the regression model at that year. And remember, we have said in Barbie that the years that taking us they are so we're going to evaluate at 1997 directly. So we get 52 point 155 times 1997 minus minus. Let's see here. Yes, we have a minus 103,000 894 0.363 and that give us 467 point Yeah, 792 With the result, we can say that the rate of change in 1997 he is more or less or is about 467.792 billions of dollars per year. Okay, and we're going to do the same for year 2000 and one that is L at 2000 and one, which is 52.155 times 2000 and one minus 103,000 894.363 And that's equal to sorry. I made a mistake here. Sorry, it's not a result here. The rial result here using a calculator I wrote. The other one is to a 259 point 172 That's the rial result here and here we have the one we wrote above. So he's 460 seven 0.792 so we can say the same thing. The average or sorry. The rate of change in 2000 and one is more or less. 400 64. 67.792 billions, dollar mhm per year. And in exercise, um, 24 part C. What we obtained there was, um, 433 point 376 billions of dollars per year. Yeah, so we can say that this relatively close the result to this one. But it could be closer if we use another type of regression here. A better function that betters followed it the butter of the points. But taking into account that the units off the amount given in the original table are billions of dollars which are huge amounts. Big numbers. We can say that the results not asses bad as we can imagine. so it's not but result. But we can be better if we use a better regression for the data given or obtain in the bar in part a. So this is the result. We have the rate of change in the years 1997 90 and 2000 and one given a proximity approximately as C quantities, 259.1 point 72 and 400 67 0.79 to billions of dollars per year.

Who? So Chapter five, Section one Problem number 42. We have a few parts to this one. So for part A, we want to find the average of the critical numbers for exercise. 15. So if we go back to exercise 15 we find that the critical numbers, um, of exercise 15 are negative. Three and four. Okay. So we can easily find the average of those of the average is equal to your negative three plus four divided by two, which is equal to 0.5 or 1/2 and then for part B. So using the same exercise exercise 15. Where to find the average of the zeroes of the function. So the function and 15 was 2/3 x cubed, minus X squared minus 24 X minus four. So we're going to punch those and our front step function into twice of one of our graphing calculator. We're gonna grab it and we're gonna find the zeros and the zeros turn out to be negative. 5.2005 Negative zero 0.1680 and six point 8685 That's an eight. There we go so we can find the average. So the average, um, is going to be if we add them together. So negative 5.2005 plus five uh, plus than negative 0.1680 plus 6.8 six feet five oh, over three. And we add that all up. Divide by three and we get surprise price. We get 1/2 or 0.5 around close to five silver Part C were asked, What do we notice about thes two answers? And so we noticed the answers to Party A and party are the same. Yeah, which tells us the average of the critical numbers for the function is equal to the average of the zeroes of the function, which is an interesting result. So then they want us to do the same thing for another function and see if that comes up again. So let's do that. And we can just keep the same page here. You know, I do Part D, which was we're gonna find the average of the critical numbers for exercise 17. And so in 17 the critical numbers turn out to be three halves and four. So for exercise 17 we can go straight to finding the average so we can save the average, Uh, the critical numbers. Okay, See, um is gonna be at sea, so waas three halves and force if I add three halves plus four, I'm divide by two. I'm gonna get by force or 1.25 and then So we want to do what we did in part B for exercise 17. So we want Thio using the function. You're 17. Which, which f of X equaling were ex cued minus 15 X squared, minus 72 X plus five. And then we're gonna punch that in the calculator. Why? One equals, we're gonna zoom six to grab it. Then we're going to find our zeros. And so when we do that, we're gonna find our average three zeros. It turns up, right. So the three zeros are We get naked of two point 8112 Plus the 2nd 0 is point zero 685 Hoping to squeeze this in her 3rd 0 is six point for 9 to 7. Kind of squeezed it in. I hope you can read that. Okay. And then we're dividing by three. So if we punch that all in it's gonna turn out to be Guess what? 1.25 again? And so we can see that the answer to D and the answer to be just like the answers for A and B are equal to each other. So for part C, we would say that they are the same, which again shows that the critical the average of the critical numbers for the function is equal to the average of the actual zeros of the function. So interesting problem. And there we are.

For this given problem, we know that an instructor gives 20 point quizzes and 100 point exams in a math course and the average scores for the six students are given an ordered pairs. Acts will be the average quiz score, and what will be the average exam score. So we could use regression capabilities to find the least squares regression line, um and plot the points in the graphing window with the regression line. So our Russian line, it's going to be Y equals 18.8, 1 0.91 Plus 3.97. It's, yeah, that's the resulting regression line we have. So we see that if x equals 17, um if that's the average chris score, then we see that the average exam score, It's going to be an 86.4. So we can use this model to predict future other possibilities.

For this problem, we're going to be referencing to exercises from earlier in this chapter, we're going to be looking at exercise number 15 and 17. So if you're not sure where some of these numbers come form, you might want to go back and reference those two problems. So, first, we're going to start with the equation from Exercise 15 which is the one that I marking in red f of X equals two thirds X cubed minus X squared minus 24 X minus four. Now we found an exercise 15 that there are two critical numbers. Negative three and four. Now, what we're supposed to start with here is to find the average of these two critical numbers. So if I average them, add them up and divide by two. I get an average of one half. Now, let's take a look at where the roots of this function are, and I'm gonna use you can use a graphing calculator or a graphing application on the computer like desk most, which is what I have here. And you can see we have a cubic function. And if I want to know where my roots are, I have three routes, one at negative, 5.2011 it negative 0.168 and one it's 6.869 So let's take those three numbers, and we're going to average those. So again, my roots occur at the points. Negative. 5.201 I have negative 0.168 plus 6.869 I am going to add all of those up. Divide by three, and when I find the average of these three routes, it is also 0.5 thes two numbers match. The average of my roots in the average of my critical numbers are are a perfect match. Let's try a second equation, this equation that I'm marking in green on the side. This is from problems 17 and this function is four X cubed, minus 15 X squared minus 72 X plus five. So I'm gonna look at the same two numbers. First, let's average are critical numbers. If you remember from exercise, 17 are two critical numbers are negative. Three halfs and four. So if I add thumbs up and divide by two, that gives me an average of 1.25 is the average of my critical numbers. What about my roots? Well, again, I'm gonna go to my graphing application. And this blue equation right here is our quad are, ah cubic function that we're looking at right now. And again, you can see that there are three routes. Negative. 2.8110 point 068 and 6.493 Select average Those three numbers I have negative. 2.811 positive 0.68 and 6.493 So I'll add those up, divide by three to find the average. And this is 1.25 as well. So these two numbers are a match, so it doesn't matter which one of these we looked at. The average of our critical numbers equals the average of our roots.


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