assignment 7

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course mth 174

The relative accuracy of the trapezoidal and midpoint rules depends on the nature of the function. As shown in my previous notes for a positive function which is concave down the midpoint rule is more accurate. A brief numerical example is y = x^2 on the interval (-1, 0), where trap gives 1/2, actual is 1/3 and mid is 1/4. 1/3 is closer to 1/4 than to 1/2. The same is true if the function is positive and concave up, for the same reasons.In fact I think that the argument can be extended to show that if concavity doesn't change on an interval, mid has to beat trap. I'd have to draw a picture or two to be sure, but it seems to be fairly obvious so I'll leave that to you. Use my previous notes as a guide.

Then you might want to draw a picture where trap is more accurate than mid. I'd recommend starting with something like y = x^3 on (-1,1), for which trap and mid agree with the accurate integral to give you 0. If you leave the points (-1, -1) and (1, 1) alone trap won't change but mid can be changed so that mid becomes less accurate (just pull (0,0) up or down a little and let the curve follow). How could the curve be manipulated to make trap less accurate than mid?

174

assignment # 7

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Question: Section 7.6 Problem 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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21:44:07

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Your solution:

I first founfd the difference this is done by taking the initial value of 2.346-4.0 is -1.654 I then used this to calculate the left value of (10/30)*-1.654 to be -.551. I then did the same with the trap formula and the Simp formula these resulted in values of -.184 and -.02. I then applied these values to the initial equation and found my final values to be 3.449, 3.816, 3.98.

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3

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Given Solution:

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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Self-critique (if necessary):

The first time I worked this problem I initially used the value of 4.0-2.346=positive 1.654 which caused my work to be wrong after seeing this mistake I was able to rework this correctly.

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Self-critique Rating:

3

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Question: Section 7.6 Problem 5

problem 7.6.5 (previously problem 7.6.10) If TRAP(10) = 12.676 and TRAP(30) = 10.420, estimate the actual value of the integral. **** What is your estimate of the actual value and how did you get it?

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Your solution:

I found the difference between the two values to be 12.676 and 10.42 to be -2.256. I then can ass this to the lower portion which will result in -2.256+10.42=10.138

confidence rating #$&*:

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3

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Given Solution:

** 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. **

Ok I notice that I multiplied by 8/9 and you multiplied by 9/8 where did this come from?????? And why???


If you know 8/9 of a number (as in this case, where you know 8/9 of the difference between the first approximation and the accurate value), then to get that number you take 9/8 of the number you know (so you take 9/8 of the difference between the n = 10 approximation and the n = 30 approximation).
In other words, (trap(30) - trap(10)) is 8/9 * (accurate value - trap(10)), so (accurate value - trap(10)) = 9/8 * (trap(30) - trap(10)).

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Self-critique (if necessary):

ok

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Self-critique Rating:

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Question: Section 7.7 Problem 2

problem 7.7.2 (previously 7.7.12) integrate 1 / (u^2-16) from 0 to 4 if convergent

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Your solution:

I am still having a lot of trouble with integration. I have watched the videos multiple times and worked and reworked the book problems to try to learn this but I am still strugiling. What should I do to try to better understand this concept????????????????

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Given Solution:

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(x-4) - ln(4). As x -> 4, ln(x - 4) approaches -infinity,

Thus the integral diverges.

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Self-critique (if necessary):

I am lost from the start of this problem I see where the integrand is (1/4) at its most but how can it be 1/8 at its least. Can you show me a step by step as to how we should have found this.

@& Since

1 / ( (u + 4) ( u - 4) ) = (1 / (u + 4) ) * ( 1 / (u - 4) ),

and since on this interval

1/8 < 1 / (u + 4) < 1/4

it follows that

1/8 | 1 / (u - 4) | < |1 / (u + 4) ) * ( 1 / (u - 4) | < 1/4 | 1 / (u - 4) |

and therefore that on this interval

1/8 integral | 1 / (u - 4) du | < integral | 1 / ( (u + 4) ( u - 4) ) du | < 1/4 integral | 1 / (u - 4) du |.

Since | 1 / (u - 4) | -> infinity as u -> 4, we proceed as indicated to show that this integral approaches -infinity.

Our previous inequality

1/8 integral | 1 / (u - 4) du | < integral | 1 / ( (u + 4) ( u - 4) ) du | < 1/4 integral | 1 / (u - 4) du |

thereby 'sandwiches' the magnitude of the desired integral between two values, both of which approach infinity.

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&#Your work looks good. See my notes. Let me know if you have any questions. &#