Assn 7

course Mth 174

Sorry for the lateness of this assignment and #8 which should be in within the next few hours. To get back on track I will do 9&10 next week and then 11&12 togther in one week. I have an uncanny knack for spreading myself to thin... It's like when your out of peanut butter and still try and get the stuff off the lid to make a sandwich, you might as well have just eaten the bread. Also a question that has come up that you may be able to answer:After watching Cosmos and a Steven Hawking special about the information paradox. (black holes send information to parallel universes, it isn't destoyed because it is imprinted on matter which can't be destoyed, only scambled.)

Since everything can be described mathematically, it seems that most of all formulas or equations are derived from another, a combination of basic functions, or are represented by a physical process. When we create new formulas they are derived from something we already know, so in a sense we already know everything that we ever will, we just further break it down to try and understand it. So how can new things come about?

02-27-2008

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11:27:20

query problem 7.6.6 approx using n=10 is 2.346; exact is 4.0. What is n = 30 approximation if original approx used LEFT, TRAP, SIMP?

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11:40:15

If the approximation used LEFT then what is your estimate of the n = 30 approximation and how did you get it?

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Since we are increasing n by a factor of 3 and we need to increase by a factor of 10 in order to achieve another digit of accuracy by using only 3 we will acieve only 1/3 of a digit of accuracy so .3 closer than the prevous ""guess"".

2.646

This seems to go along with the theory to me... Although I'm somewhat confused due to that lack of what we're trying to integrate.

Good reasoning, but you are thinking in terms of digits of accuracy, and can be more precise that that.

LEFT and RIGHT approach the exact value in proportion to the number of steps used.

MID and TRAP approach the exact value in proportion to the square of the number of steps used.

SIMP approachs the exact value in proportion to the fourth power of the number of steps used.

Using these principles we can work out this problem as follows:

** The original 10-step estimate is 2.346, which differs from the actual value 4.000 by -1.654.

If the original estimate was done by LEFT then the error is inversely proportional to the number of steps and the n = 30 error is (10/30) * -1.654 = -.551, approximately. So the estimate for n = 30 would be -.551 + 4.000 = 3.449.

If the original estimate was done by TRAP then the error is inversely proportional to the square of the number of steps and the n = 30 error is (10/30)^2 * -1.654 = -.184, approximately. So the estimate for n = 30 would be -.184 + 4.000 = 3.816.

If the original estimate was done by SIMP then the error is inversely proportional to the fourth power of the number of steps and the n = 30 error is (10/30)^4 * -1.654 = -.020, approximately. So the estimate for n = 30 would be -.02 + 4.000 = 3.98. **

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11:44:28

If the approximation used TRAP then what is your estimate of the n = 30 approximation and how did you get it?

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with the trapazoid, by increasing n by a factor of three we get a higher accuracy of 3^2 or 9 which is equilvilant to getting .9 closer and almost adding another unit of accuracy.

3.246

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11:53:07

If the approximation used SIMP then what is your estimate of the n = 30 approximation and how did you get it?

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since we are using Simpsons rule and increasing n by a factor of 3 then we increase our accuracy by a factor of 3^4. 81 which is almost 2 digits of accuracy.

3.98

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11:53:23

This problem has been omitted from the present edition and may be skipped: query problem 7.6.10 If TRAP(10) = 12.676 and TRAP(30) = 10.420, estimate the actual value of the integral.

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Here's a solution for your reference:

** You need to use the inverse square proportionality of the error with the number of steps.

Trap(30) is approximately (10/30)^2 = 1/9 of TRAP(10). So the difference 10.420 - 12.676 = -2.256 between TRAP(10) and TRAP(30) is approximately 8/9 of the error of TRAP(10).

It follows that the error of TRAP(10) is 9/8 * -2.256 = -2.538. Our best estimate of the integral is therefore -2.538 + 12.676 = 10.138. **

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11:53:32

What is your estimate of the actual value and how did you get it?

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11:53:46

a < b, m = (a+b)/2. If f quadratic then int(f(x),x,a,b) = h/3 ( f(a) / 2 + 2 f(m) + f(b) / 2).

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11:53:52

How did you show that if f(x) = 1, the equation holds?

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11:53:57

How did you show that if f(x) = x, the equation holds?

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11:54:02

How did you show that if f(x) = x^2, the equation holds?

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11:54:08

How did you use your preceding results to show that if f(x) = A x^2 + B x + c, the equation must therefore hold?

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12:10:52

query problem 7.7.19 integrate 1 / (u^2-16) from 0 to 4 if convergent

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12:19:31

does your integral converge, and why or why not?

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no because as u -> 4 we get extremely close to dividing by zero. If we divide 1 by an infintismall number then we get close to an infinate number, and therefor the integral doesn't converge it diverges.

The integral from 0 to 1 of 1 / (x-1) diverges, but if the integrand is 1 / (x-1)^.999999 it converges.

Just because the function approaches infinity at one of the bounds doesn't mean the integral diverges.

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12:19:40

If convergent what is your result?

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it diverges

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12:22:03

Why is there a question as to whether the integral does in fact converge?

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because with a realitivly small n the numbers are ""normal"" sized and once we pass the u=4 then the integral is continuous again, it's just that on the limit defined here the continuity of the function is questionable near 4.

There's no question about whether the function is continuous at any given point; the function is undefined at x = 4 and at x = -4, it is defined everywhere else.

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12:27:02

Give the steps in your solution.

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By plugging in the numbers on the intreval with a n=1 we see that as we get closer to u=4 then we are given numbers closer to infinity. and thus the integral is divergent.

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12:28:43

If you didn't give it, give the expression whose limit showed whether the integral was convergent or divergent.

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1 / u^2 - 16 on the intreval if 0->4

You are right that the integral diverges, but your reason is not sufficient to establish this fact. Presumably you would have reached the same conclusion about 1 / (u^2 - 16)*.9999, which however converges.

1 / (u^2-16) = 1 / [(u+4)(u-4)] . Since for 0 < x < 4 we have 1/8 < 1 / (u+4) < 1/4, the integrand is at most 1/4 times 1/(u-4) and at least 1/8 of this quantity, so the original integral is at most 1/4 as great as the integral of 1 / (u-4) and at least 1/8 as great. That is,

1/8 int(1 / (u-4), u, 0, 4) < int(1 / (u^2-4), u, 0, 4) < 1/4 int(1 / (u-4), u, 0, 4).

Thus if the integral of 1 / (u-4) converges or diverges, the original integral does the same. An antiderivative of 1 / (u-4) is ln | u-4 |, which is just ln(4) at the limit u=0 of the integral but which is undefined at the limit u = 4.

We must therefore take the limit of the integral of 1/(u-4) from u=0 to u=x, as x -> 4.

The integral of 1 / (u-4) from 0 to x is equal to ln (4) - ln(x-4) = ln( 4 / (x-4) ).

As x approaches 4 the denominator approaches 0 so the fraction approaches infinity and the natural log approaches infinity. Thus the integral diverges. **

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12:29:27

query problem 7.7.44 (was #39) rate of infection r = 1000 t e^(-.5t)

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12:36:20

describe your graph, including asymptotes, concavity, increasing and decreasing behavior, zeros and intercepts

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The graph begins very high on the Y axis and exponentially comes downward towards the x axis and becomes asymptoteic to the x axis. it is concave up.

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12:39:36

when our people getting sick fastest and how did you obtain this result?

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people get sick the fastest at the begining of the epidemic this is found by graphing the function and also evauluating the function for t=1 and t=100 and comparing the difference.

If you mean graphing with a graphing calculator, that is not a valid justification.

If you obtain the graph by finding critical points, etc., then the graph could be used as justification.

dividing through by e^-.5 t we get t = 2, which with a first-or second-derivative test confirms that the t = 2 graph point is a relative maximum.

Concavity is determined by the second derivative r'' = e^(-.5 t) [ -1000 + 250 t ], which is 0 when t = 4. This is a point of inflection because the second derivative changes from negative to positive at this point. So the function is concave downward on the interval (-infinity, 4) and concave upward on (4, infinity).

The first derivative has a critical point where the second derivative is zero. This occurs at x = 4, which was identified in the preceding paragraph as the point of inflection for the original function. Since the second derivative goes from negative to positive, this point is a minimum of the first derivative. The first derivative is a decreasing function from t = 0 to t = 4 (2d derivative is negative) and is then an increasing function with asymptote y = 0, the x axis, which it approaches through negative values. Its maximum value for t >= 0 is therefore at t = 0. **

** The calculator is fine for checking yourself, but you need to use the techniques of calculus to determine inflection points, maxima, minima etc.. The careful use the calculator to enhance rather than replace mathematical understanding. I get a lot of students in these courses who are now at 4-year institutions and who have taken courses based on the graphing calculator or hand-held computers, and many of them tend to have a very difficult time in courses that don't permit them, and in courses where rigorous mathematical reasoning is required. **

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12:43:37

How many people get sick and how did you obtain this result?

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The number of people that get sick would be the area under the graph, for a inteval of 1 day (t= 1->2) we get an area of 671, or 671 people are infected.

You need to show the details of your integration.

** To find the number getting sick between day 1 and day 2 you would integrate the rate function over the interval. You can check your result with the antiderivative given below. I expect your result is right. For t = 1 we get r = 1000 e^-.5 = 1000 / sqrt(e) = 1000 / 1.7, approximately, and the result is clearly less than 1000 / 1.6. Since 1000 / 1.6 = 667, the t = 1 rate is less than 667 per day. However the maximum of the rate function does not occur until t = 2, so an average rate of 671 per day is not unreasonable.

To find the total number who get sick you would integrate the rate function r = 1000 t e^(-.5t) from t = 0 to t = infinity.

An antiderivative of the function is F(t) = 1000 int ( t e^(-.5 t)) = 1000 [ -2 t e^(-.5t) - int ( e^(-.5 t) ) ] = 1000 [ -2 t e^(-.5 t) - 4 e^(-.5 t) ].

Integrating from 0 to x gives F(x) - F(0) = 1000 [ -2 t e^(-.5 x) - 4 e^(-.5 x) ] - 1000 [ -2 * 0 e^(-.5 *0 ) - 4 e^(-.5 * 0 ) ] = 1000 e^-(.5 x) [ -2 t - 4 ] - (-4000).

As x -> infinity, e^-(.5 x) [ -2 t - 4 ] -> 0 since the exponential will go to 0 very much faster than (-2 x - 4) will approach -infinity. This leaves only the -(-4000) = 4000. **

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12:47:33

What improper integral arose in your solution and, if you have not already explained it, explain in detail how you evaluated the integral.

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I used the given equation which gave the number of people that got sick per day of epidemic.

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12:49:07

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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I understand how Simpsons rule works and finding the convergent / divergent integrals is very similar to finding limits. It's not what it is exactly at a point but what happens as we apporoach that point.

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You have the right ideas, but be sure you understand the details. See my notes and let me know if you have questions.