1. Determine if the sequence a converge?
= converges as n —+ cxD. Does the series Z_ a
Solution: urn n-* + 1
= liiri n-c
3+
1
1/n
Because the limit is finite, it converges by definition.
=
1/3
The series does not converge because it is necessary for 1in a
=
0 for series convergence to occur.
2. Sustainability versus growth: Suppose a first-grader rulining a lemonade stand makes a 1-dollar profit on day one (i 1) and then takes half of the profit to re invest to prepare for the next day. On day two she makes half a dollar and re-invests
1/3-dollars, and so on for subsequent days on to infinity. Compose the sum of the net daily balance over all days and determine its total valile.
Solution: For the net balance for the first day is 1/1
— 1/2.
The net balance for the second day is 1/2 — 1/3. For the ith day the balance is 1/i — 1/(i + 1). The infinite sum is leaving
S 1/i — 1/(i ± 1) is telescoping, with subsequent terms canceling,
= only the first term, so
S
=
1.

3. Bouncing ball: Suppose a ball is dropped from 2 meters height and bounces to 1.5
meters on the second bounce, and 9/8ths of a meter on the second bounce, and so on indefinitely. What is the total vertical distance traveled by the ball?
Solution: The initial height is 2 meters, then bounces to 75% of the initial height, then 75% of the subsequent, and so on. The vertical elevation gained on each bounce is then
0.75 x 2 ± 0.75
x 0.75 x 2 + 0.75 x 0.75 x 2...
DC
2(O.75
2
DC x O.752(0.75)?
—
2
— x 0.75
1
—
0.75
The total distance traveled includes all down paths as well:
2 +0.75 x 2+0.75 x 0.75 x 2...
=
=
1
The sum of up and down paths is
1
2 x 0.75
—
2
—
2(7/4)
—
0.75
1
—
0.75
—
3/4
—
2
2
4.
Bouncing Sunbeams: A sunray with intensity I is directed at Earth (see below

Figure). A fraction Ra of the ray’s intensity is reflected back to space off the atmospheric layer (depicted as a cloud). Another fraction Aa is absorbed in the atmosphere, and the remaining fraction Ta is transmitted through the atmosphere to the Earth’s surface.
These three fractions sum to unity: Ra + Ta + Aa
=
1, meaning all the light must be accounted for. The fraction transmitted through the atmosphere undergoes a similar fractionation on the Earth’s surface, with a fraction R being reflected back, and the remaining fraction A
=
1
—
R is absorbed (no radiation transmits through the earth too thick). The fraction of radiation Re reflected back upward to the atmosphere layer undergoes a further reflection-transmission-absorption fractionation, with some radiation
(Ta) escaping to outer space, the remainder either being absorbed as heat or re-reflected back to earth iii an infinite cycle on the increasingly small fraction of remaining radiation intensity. As you will see, this back-and-forth process in our atmosphere can cause a greater amount of light to be absorbed compared to if we had no atmosphere.
(a) Assume tensity
Ta
=
0.4, Ra
=
0.5, and R,
=
0.6. Compute the fraction of the original in
1
=
1 that escapes back to space.
Hint: add up all the transmitted-tospace fractions and use your knowledge of series to calculate the infinite sum. Don’t plug in numerical values till the end.
(b) Easy question: With parameters as in (a), what is the fraction absorbed by the earth?
(c) Another easy question: If the atmosphere did not exist, then Ta
=
1, what intensity will be absorbed by the planet?
Compare to (b).
This is a toy model of the the atmosphere, causing greater temperature on earth.
Solution: (a) Just as in the bouncing ball example, we must find the sequence of all reflection-absorption-transmission interactions then compose the sum. Ultimately, after a number of reflection events, the radiation either escapes to space, or it is absorbed.
We will sum all the fractions of light escaping to space.
The first of such interactions is from light entering the atmosphere for the first time then begin reflected right back out: I
=
1 x 0.5
=
1/2. The get the next interaction resulting in light escaping to space, we compute the fraction of light transmitting to earth third escape event is from the fraction transmitted to earth, atomsphere, then back to earth, then out to space: TaReRaRe(TaIo). we sum all these fractions resulting from reflections to be bRa + TaRe(TaIü) + TaReRaRe(Ta1 +...
2 + TRe(RaRe) . .
.1
= bo[Ra + TRe (RaRe)]

This last line accounts for all escape fractions by summing all terms. The first term is the fraction representing tile reflection off the atmosphere. the remaining terms are always transmitted twice through the atmosphere T (once on the way down, then once back up), between these reflections the quantity is reflected at least one time off the earth R, resulting in the factor TRC.
The infinite sum merely adds up all the scenarios counting the ri times each fraction of radiation was reflected back and forth between the earth and atmosphere (RaRe)1.
Calculated numerically: 1/2 ± (4/25)(6/1O)/(1
—
6/20) 0.6371.
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