Assignment 8

10-10-2007

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14:01:04

query explain the convergence or divergence of series (no summary needed)

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With a series convergence is described when two lines on a graph are going towards one another but they never touch. The other divergence is when two lines on a graph go away from each other and never ever come even close to one another.

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14:01:27

explain how we know that the integral from 1 to infinity of 1 / x^p converges if p > 1, diverges for p < 1, diverges for p = 1

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do not understand the question here.

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14:03:14

explain how we know that the integral from 0 to 1 of 1 / x^p diverges if p > 1, converges for p < 1, diverges for p = 1

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RESPONSE -->

we know this by looking at the graph of the function of 1 / x^p because we see that the graph diverges when the function is greater than 1 and we see it converge when the function is less than 1. But when it equals 1 it neither diverges or converges.

You can't tell anything by looking at the graph. The graph of 1 / x^.99 looks almost the same as the graph of 1 / x^1.01 but one diverges and the other converges.

** There is very little difference in the rapidity with which 1/x^1.01 approaches zero and the rapidity of the approach of 1 / x^.99. You would never be able to tell by just looking at the partial sums which converges and which diverges. In fact just looking at partial sums you would expect both to converge, because they diverge so very slowly.

If you graph 1 / x^.99, 1 / x and 1 / x^1.01 you won't notice any significant difference between them. The first is a little higher and the last a little lower past x = 1. You will never tell by looking at the graphs that any of these functions either converge or diverge.

However if you find the antiderivatives you can tell pretty easily. The antiderivatives are 100 x^.01, ln(x) and -100 x^-.01.

On the interval from, say, x = 1 to infinity, the first antiderivative 100 x^.01 will be nice and finite at x = 1, but as x -> infinity is .01 power will also go to infinity. This is because for any N, no matter how great, we can find a solution to 100 x^.01 = N: the solution is x = (N / 100) ^ 100. For this value of x and for every greater value of x, 100 x^.01 will exceed N. The antiderivative function grows without bound as x -> infinity. Thus the integral of 1 / x^.99 diverges.

On this same interval, the second antiderivative ln(x) is 0 when x = 0, but as x -> infinity its value also approaches infinity. This is because for any N, no matter how large, ln(x) > N when x > e^N. This shows that there is no upper bound on the natural log, and since the natural log is the antiderivative of 1 / x this integral diverges.

On the same interval the third antiderivative -100 x^-.01 behaves very much differently. At x = 1 this function is finite but negative, taking the value -100. As x -> infinity the negative power makes x-.01 = 1 / x^.01 approach 0, since as we solve before the denominator x^.01 approaches infinity. Those if we integrate this function from x = 1 to larger and larger values of x, say from x = 1 to x = b with b -> infinity, we will get F(b) - F(1) = -100 / b^.01 - (-100 / 1) = -100 / b^.01 + 100. The first term approaches 0 and b approaches infinity and in the limit we have only the value 100. This integral therefore converges to 100.

We could look at the same functions evaluated on the finite interval from 0 to 1. None of the functions is defined at x = 0 so the lower limit of our integral has to be regarded as a number which approaches 0. The first function gives us no problem, since its antiderivative 100 x^.01 remains finite over the entire interval [0, 1], and we find that the integral will be 100. The other two functions, however, have antiderivatives ln(x) and -100 x^-.01 which are not only undefined at x = 0 but which both approach -infinity as x -> 0. This causes their integrals to diverge.

These three functions are the p functions y = 1 / x^p, for p = .99, 1 and 1.01. This example demonstrates how p = 1 is the 'watershed' for convergence of these series either on an infinite interval [ a, infinity) or a finite interval [0, a]. **

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The convergence of the integral of e^-(ax) from 0 to infinity is plausible because this is a decreasing function. However 1 / x^.99 is also decreasing on the same interval but it doesn't converge; as before you can’t tell anything by looking at the graph.

The reason e^(- a x) converges is that it antiderivative is -1/a e^(-a x), and as x -> infinity this expression approaches zero. Thus the integral of this function from 0 to b is -1/a e^(-a b) - (-1/a e^0) = -1/a e^(-ab) + 1. If we allow the upper limit b to approach infinity, e^-(a b) will approach 0 and we'll get the finite result 1.

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** If it wasn't for the 1 in the denominator this would be 1 / `sqrt(`theta^2) or just 1 / `theta, which is just 1 / `theta^p for p = 1. As we know this series would diverge (antiderivative of 1 / `theta in ln(`theta), which appropaches infinity as `theta -> infinity).

As `theta gets large the + 1 shouldn't matter much, and we expect that 1 /`sqrt (`theta^2 + 1) diverges. If this expression is shown to be greater than something that diverges, then it must diverge.

However, 1 / `sqrt ( `theta^2 + 1) < 1 / `theta, and being less than a function that diverges does not prove divergence.

We can adjust our comparison slightly. Since 1 / `theta gives a divergent integral, half of this quantity will yield half the integral and will still diverge. i.e., 1 / (2 `theta) diverges. So if 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we will prove the divergence of 1 / `sqrt(`theta^2 + 1).

We prove this. Starting with 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we square both sides to get

1 / (4 `theta^2) < 1 / (`theta^2+1). Multiplying by common denominator we get

`theta^2 + 1 < 4 `theta^2. Solving for `theta^2 we get

1 < 3 `theta^2

`sqrt(3) / 3 < `theta.

This shows that for all values of `theta > `sqrt(3) / 3, or about `theta > .7, the function 1 / `sqrt(`theta^2+1) exceeds the value of the divergent function 1 / (2 `theta). Our function 1 / `sqrt(`theta^2 + 1) is therefore divergent. **

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** 1 / `sqrt(theta^3 + `theta) < 1 / `sqrt(`theta^3) = 1 / `theta^(3/2).

= 1 / `theta^(3/2) is an instance of 1 / x^p for x =`theta and p = 3/2.

So the integral converges by the p test. **

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A Riemann sum for a function f(x) on an interval from a to b is formed by partitioning this interval into n equal subintervals, each of length `dx = (b - a) / n.

Subinterval number i has endpoints x_(i-1) and x_i, and c_i is a sample point within that interval. The partition is usually listed as x0 = a, x1, x2, ..., x_i, ..., x_n = b.

The area estimate for subinterval number i is

area = sample 'altitude' * width,

where the sample 'altitude' is f(c_i). So the area for this interval is

area = f(c_i) * `dx.

The Riemann sum is a sum of the form

sum{i = 1 to n} ( f(c_i) * `dx)

where c_i (read c-sub-i) is a sample point in subinterval number i and `dx_i is the length of that interval.

For the given function we have x^2 + y^2 = 10, so that y = sqrt(10 - x^2), where x can be any number between 0 and sqrt(10). We are thus approximating the area under the curve

y = f(x) = sqrt(10 - x^2)

on the interval from x = 0 to x = sqrt(10).

We focus attention on interval number i, which has width `dx and sample point c_i.

We approximate the area under the curve as a rectangle with width `dx and 'altititude' f(c_i). So we have

area of ith interval = f(c_i) * `dx.

Now there are n subintervals, so i can stand for be any number from 1 to n. We want to write an expression for the total area of all the subintervals, which is the sum of all the areas. The expression we write is

sum{i = 1 to n} ( f(c_i) * `dx).

The actual area under the graph is the limiting value of this sum, as n approaches infinity. We know that every sum is between the left- and right-hand approximations, and that as n approaches infinity the difference between these approximations approaches zero. So the limit

limit{n -> infinity} [ sum{i = 1 to n} ( f(c_i) `dx) ]

exists.

In the limit the sample points c_i cover the x axis densely, and since c_i represents the x coordinate, the expression f(c_i) will be replaced by f(x).

Since the finite interval `dx becomes 'infinitesimal' in the limit, we replace `dx by the differential dx.

Since x values on this interval vary from 0 to sqrt(10) our integral goes from 0 to 10, and we write

integral( f(x) dx, x from 0 to 10 ) =

integral ( sqrt(10 - x^2) dx, x from 0 to 10)

When you write this on paper you write an integral sign with limits 0 and 10, then write sqrt(10 - x^2) dx.

The dx in your integral tells you that x is the variable of integration. This is important because your expression might contain a variety of symbols, and you have to know with respect to which variable you are integrating.

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The exact value is the area of a quarter-circle of radius sqrt(10). The area of the is pi r^2 = pi ( sqrt(10) ) ^ 2 = 10 pi, and the area of a quarter circle is therefore 10 pi / 4 = 5 pi / 2.

You cannot express the exact value of this integral with a single decimal number.

You are to actually evaluate the integral, which requires a substitution like x = sqrt(10) sin(theta); this technique was covered in Chapter 7. You need to show the details.

You get dx = sqrt(10) cos(theta). Since

sqrt(10 - x^2) = sqrt(10 - (sqrt(10) sin(theta))^2 ) = sqrt(10) * sqrt(1 - sin^2(theta) = cos^2(theta)

and since x = 0 and x = sqrt(10) correspond to sin(theta) = 0 and sin(theta) = 1, the limits x = 0 and x = sqrt(10) become theta limits 0 and pi/2.

Thus we have

integral(sqrt(10 - x^2), x from 0 to sqrt(10))

= integral (sqrt(10) cos(theta) * sqrt(10) cos(theta), theta from 0 to pi/2)

= 10 integral(cos^2*(theta), theta from 0 to pi/2).

As seen in Chapter 7 this integral is pi / 4 and we find that our area is therefore

area = 10 integral(cos^2*(theta), theta from 0 to pi/2) = 10 * pi / 4 = 5 pi / 2.

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You have to do the Riemann sum, get the integral then perform the integration.

A slice of the region parallel to the y axis is a rectangle. If the y coordinate is y = c_i then the slice extends to an x coordinate such that x^2 + (c_i)^2 = 7^2, or x = sqrt(49 - (c_i)^2).

So if the y axis, from y = 0 to y = 7, is partitioned into subintervals y_0 = 0, y1, y2, ..., y_i, ..., y_n = 7, with sample point c_i in subinterval number i, the corresponding 'slice' of the solid has thickness `dy, width sqrt(49 - (c_i)^2) and length 10 so its volume is 10 * sqrt(49 - (c_i)^2) * `dy and the Riemann sum is

sum(10 * sqrt(49 - (c_i)^2) * `dy, i from 1 to n).

The limit of this sum, as x approaches infinity, is then

integral ( 10 * sqrt(49 - y^2) dy, y from 0 to 10) =

10 integral (sqrt(49 - y^2) dy, y from 0 to 10).

Using the trigonometric substitution y = 7 sin(theta), analogous to the substitution used in the preceding solution, we find that the integral is 49 pi / 4 and the result is 10 times this integral so the volume is

10 * 49 pi / 4 = 490 pi / 4 = 245 pi / 2.

A decimal approximation to this result is between 350 and 400. Compare this with the volume of a 'box' containing the figure:

The 'box' would be 7 units wide and 7 units high, and 10 units long, so would have volume 7 * 7 * 10 = 490.

The solid clearly fills well over half the box, so a volume between 350 and 400 appears very reasonable.

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Right idea, but you need to use f ' (x), not f(x).

The arc length is

arc length = integral(sqrt(1 + (f ' (x)) ^2) dx, x from a to b).

f(x) = x^(3/2) so f ' (x) = 3/2 x^(1/2), and (f ' (x))^2 = (3/2 x^(1/2))^2 = 9/4 x.

So our integral is

integral(sqrt(1 + 9/4 x) dx, x from 0 to 2).

We easily integrate this letting u = 1 + 9/4 x so that du = 9/4 dx; the limits x = 0 and x = 2 become u = 1 + 9/4 * 0 = 1 and u = 1 + 9/4 * 2 = 11/2 so we find

4/9 integral(sqrt(u), u from 1 to 11/2).

Our antiderivative of sqrt(u) is 2/3 u^(3/2) and the change in the antiderivative is

2/3 ( 11/2 ) ^ (3/2) - 3/2 ( 1 ) ^ (3/2) = 2/3 ( 11/2)^(3/2) - 1)

and the arc length is

4/9 * 2/3 ( 11/2)^(3/2) - 1) = 8/27 ( (11/2)^(3/2) - 1).

This expression can be simplified or approximated.

The result should also be compared to the estimated arc length of a sketch of the graph of the original function. The original function starts at (0, 0) and ends at (2, 2 sqrt(2)). A straight line segment between these points has length about 3.5, and the arc will be a little longer than this straight-line approximation. This is consistent with the decimal approximation of the above expression.

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x runs from 0 to 1.

At a given value of x the 'altitude' of the y vs. x graph is e^x, and a slice of the solid figure is a square with sides e^x so its area is (e^x)^2 = e^(2x).

If we partition the interval from x = 0 to x = 1 into subintervals defined by x_0 = 0, x_1, x_2, ..., x_i, ..., x_n, with interval number i containing sample point c_i, then the slice corresponding to interval number i has the following characteristics:

the thickness of the 'slice' is `dx

the cross-sectional area of the 'slice' at the sample point x = c_i is e^(2x) = e^(2 * c_i)

so the volume of the 'slice' is e^(2 * c_i) * `dx.

The Riemann sum is therefore

sum(e^(2 * c_i * `dx) and its limit is

integral(e^(2 x) dx, x from 0 to 1).

Our antiderivative is easily found to be 1/2 e^(2 x) and the change in our antiderivative is

1/2 e^(2 * 1) - 1/2 e^(2 * 0) = 1/2 e^2 - 1/2 = 1/2 (e^2 - 1).

The approximate value of this result is between 3 and 3.5.

A 'box' containing this region would have dimensions 1 by e by e, with volume e^2 = 7.3 or so. The figure fills perhaps half this box, so our results are reasonably consistent.

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end of document

You need to provide much more detail in your responses about what you are thinking, how you are approaching the problems, and the details of your solution.

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