Assignment 6 Query

course Mth 174

蜐{{Цsyo ȵ쁾assignment #006

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Physics II

10-03-2007

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08:48:27

Query problem 7.5.13 (3d edition #10) graph concave DOWN and decreasing (note changes indicated by CAPS)

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RESPONSE -->

The exact values, from smallest to largest, are as follows:

Right(n) < Trap(n) < Exact < Mid(n) < Left(n).

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08:48:33

list the approximations and their rules in order, from least to greatest

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See above.

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08:48:53

between which approximations does the actual integral lie?

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The actual integral lies between Trap(n) and Mid(n).

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08:56:12

Explain your reasoning

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RESPONSE -->

In this example, the graph is concave down and decreasing.

The fact that it is decreasing tells me that the Left(n) function is an overestimate, while the Right(n) is an underestimate. This can be visualized since the Right(n) function uses the divisions on the right, making the estimation stay under the curve. The Left(n) function uses the divisions on the left, making the estimation stay over the curve.

The fact that it is concave down indicates that the Trap(n) function will be less since it takes the divisions and draws a straight line between them, creating a series of lines that is underneath the main curve. As for the Mid(n) function, this takes the midpoint and draws a tangent line there. Since the tangent line is over the main curve in this example, the Mid(n) function is an overestimate.

Good. Compare your reasoning to the following.

If the graph is decreasing then we know that the left end of an interval gives us a higher estimate than the right.

The trapezoidal estimate is the mean of the left and right estimates, so it will lie between the two. The trapezoidal estimate also corresponds to a straight-line estimate between the graph points at the left and right endpoing of each interval.

Since the graph is concave down, any straight line between two graph points will lie below the graph. In particular the value of the function at the midpoint will lie above the straight line. The trapezoidal estimate corresponds to the straight-line, estimate so in this case the midpoint estimate exceeds the trapezoidal estimate.

The straight-line trapezoidal area is also clearly less than the area beneath the curve, so the trapezoidal estimate is lower than the actual value of the integral.

The midpoint rule uses the function value at the midpoint for the altitude of the rectangle. Thus we imagine a rectangle whose top is a horizontal line segment through the midpoint value of the graph. You should sketch this.

If the function is concave downward, the part of the actual graph that lies between the endpoints and above the midpoint-value rectangle (this segment lies to the left of the midpoint) represents graph area under the actual graph which is left out of the midpoint approximation. The area under the horizontal segment at the midpoint which lies above the graph (this are lies to the right of the midpoint) represents area included in the midpoint approximation but which is not part of the area under the curve.

Since the function is concave down, the average altitude of the left-out area to the right is greater than that of the wrongly-excluded area to the left. Since the widths of the two regions are equal, the wrongly excluded area must be less than the wrongly-included area and the midpoint estimate must therefore be high. We conclude that the actual area is less than the midpoint area.

Since the trapezoidal approximation is less than the actual area, we have our final ordering:

RIGHT < TRAP < exact < MID < LEFT.

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08:57:42

if you have not done so explain why when a function is concave down the trapezoidal rule UNDERestimates the integral

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I believe I did.

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08:57:50

if you have not done so explain why when a function is concave down the midpoint rule OVERrestimates the integral

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Done.

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08:58:35

Query NOTE: this problem has been left out of the new edition of the text, which is a real shame; you can skip on to the next problem (was problem 7.5.18) graph positive, decreasing, concave upward over interval 0 < x < h

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RESPONSE -->

?

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09:03:25

why is the area of the trapezoid h (L1 + L2) / 2?

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RESPONSE -->

Taking a guess at it, the reason I see is that the Trap(n) function takes the average of two points with a linear function between them.

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09:04:41

Describe how you sketched the area E = h * f(0)

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Hmm. I am not quite sure.

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09:04:52

Describe how you sketched the area F = h * f(h)

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Lost again.

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09:05:05

Describe how you sketched the area R = h*f(h/2)

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RESPONSE -->

Another one bites the dust.

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09:05:27

Describe how you sketched the area C = h * [ f(0) + f(h) ] / 2

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This would be the Trap(n) area, right?

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09:05:46

Describe how you sketched the area N = h/2 * [ f(0) + f(h/2) ] / 2 + h/2 * [ f(h/2) } f(h) ] / 2

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Hmm.

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09:06:40

why is C = ( E + F ) / 2?

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Because E and F are equal to the Left and Right sums, and C is the Trapezoidal sum?

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09:07:01

Why is N = ( R + C ) / 2?

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I am not sure. It looks like another variation on the Trapezoidal function.

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09:07:12

Is E or F the better approximation to the area?

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I don't know.

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09:07:29

Is R or C the better approximation to the area?

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Again, not sure. Two different graphs, or two different sections on one graph?

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10:54:02

query problem 7.5.24 show trap(n) = left(n) + 1/2 ( f(b) - f(a) ) `dx

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This is 7.4.25 in 4th edition.

According to the book:

trap(n) = ( left(n) + right(n) ) / 2

I have used this to work on my transformations.

trap(n) = left(n) + (1/2) f(b) 'dx - (1/2) f(a) 'dx

(1/2) f(b) 'dx = (1/2) right(n)

(1/2) f(a) 'dx = (1/2) left(n)

trap(n) = left(n) + (1/2) right(n) - (1/2) left(n)

trap(n) = (1/2) left(n) + (1/2) right(n)

trap(n) = ( left(n) + right(n) ) / 2

However, I am not sure that these two equations:

(1/2) f(b) 'dx = (1/2) right(n)

(1/2) f(a) 'dx = (1/2) left(n)

are warranted, as I can't think of a way to prove it, since left(n) and right(n) are both based on n. Since n is the number of subintervals in the integral and the number of calculations in left(n) or right(n) depends on n, I can't think of a way to simplify it. Did I miss something?

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10:57:03

Explain why the equation must hold.

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I am not sure.

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11:17:14

In terms of a graph describe how trap(n) differs from left(n) and what this difference has to do with f(b) - f(a).

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Trap(n) creates the integral based on a trapezoid between each subinterval, whereas left(n) creates the integral based on rectangles based on the left side of the subinterval.

As for f(b) - f(a), this equation returns the overall difference in the rate function between b and a. When this is multiplied by (1/2) 'dx, this produces the integral of the difference between b and a.

Perhaps this is the relationship:

Left(n) is either an overestimate or an underestimate depending on whether f(x) is increasing or decreasing over the interval.

If f(x) is increasing, then Left(n) is an underestimate and (1/2) ( f(b) - f(a) ) 'dx adds to it. If f(x) is decreasing, then Left(n) is an overestimate and (1/2) ( f(b) - f(a) ) 'dx removes from Left(n).

In either case the small addition or removal by (1/2) ( f(b) - f(a) ) 'dx brings Left(n) closer to the true value.

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Good reasoning. See my notes and let me know if you have questions.

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Let me know if you have questions. &#