LESSON 17: GEOMETRIC SERIES AND CONVERGENCE MATH FALL 2018

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1 LESSON 17: GEOMETRIC SERIES AND CONVERGENCE MATH FALL 2018 ELLEN WELD 1. Solutions to In-Class Examples Example 1. A ball has the property that each time it falls from a height h onto the ground, it will rebound to a height of rh for some 0 < r < 1. Find the total distance traveled by the ball if r 1 and it is dropped from a height of 9 feet. Solution: We draw a picture to get a feel for what is going on. Notice that other than when we originally drop the ball, at each step the distance traveled by the ball is doubled because we must include the height the ball rebounds to and the distance the ball travels as it falls to the ground. Observe ) 1 (9) 1 ( 1 (9) 1 () 1 ( (9) 1 ) ( ) 1 9 ( 1 ) ( ) 1 (9) ( ) 2 1. From this we can determine a pattern: the distance the ball travels is described by ( ) n (9) ( ) n 1.

2 2 ELLEN WELD This is clearly a geometric series so we use the geometric series formula to compute this sum. But our series starts at n 1 (not n 0), so we can t apply our formula just yet. Instead, write Hence, ( 1 ) n n0 ( ) n+1 1 n0 ( 1 ( ) ( n ) ( ) n 1 1 n0 ) n ( 1 n0 ( ) ( ) ( ) n0 ( ) ) n 1 ( ) n 1. Example 2. Suppose that in a country, 75% of all income the people receive is spent and 25% is saved. What is the total amount of spending generated in the long run by a $10 billion tax rebate which is given to the country s citizens to stimulate the economy if saving habits do not change? Include the government rebate as part of the total spending. Solution: The question is asking us to determine what is spent from now to the end of time (assuming the pattern holds). Since we are including the government rebate as part of the spending, we see at time n 0, $10 billion is spent. But, according to what they tell us, the citizens then spend 75% of the $10 billion. So at time n 1, $10(.75) billion is spent. At time n 2, the citizens spend $10(.75)(.75) $ 10(.75) 2 billion and we continue on in this way. We assume the pattern holds indefinitely. Our goal is to find the total amount spent (measured in billions), which is the sum of all that is spent over time n 0, 1, 2,... This is described by the summation 10 n0 + 10(.75) + 10(.75) 2 + n2 n0 10(.75) n 10 n0 (.75) n. Because n 0 and.75 < 1, we can apply our formula for the geometric series to determine that the total amount spent (in billions) is

3 AN UNOFFICIAL GUIDE TO MATH FALL 2018 ( ) ( ) (.75) n (4) 40 billion n0 Example. How much should you invest today at an annual interest rate of 4% compounded continuously so that in years from today, you can make annual withdrawals of $2000 in perpetuity? Round your answer to the nearest cent. Solution: The question is asking: what do we need to invest today so that every year, we have $2000 in the bank. The formula for continuously compounded annual interest is A P e rt where r is the interest rate, t is time in years, A is the amount we have in the bank after t years, and P is the investment we make today. Let P be the amount we invest today so that in years, we have $2000. Then, at the interest rate we are given, 2000 P e.04() P 2000e.04(). Let P 4 be the amount we invest today so that in 4 years, we have $2000. Write 2000 P 4 e.04(4) P e.04(4). Similarly, for any year n > we can let P n be the amount we invest today so that after n years, we have $2000. Then 2000 P n e.04(n) P n 2000e.04(n). Where does this leave us? Well, the sum of all these P n gives the total amount we need to invest today so that we will always have $2000 in the bank each year beginning years from now. So Total P + P 4 + P e.04(n). To determine how much we need to invest today, we need to find the value of 2000e.04(n). We will need to use the formula for the geometric series but our n series is not in the correct form. So 2000e.04(n) n n 2000 ( e.04) n n 2000 ( e.04) n+ n ( e.04) ( ) e.04 n n0 ( 2000e ).04() e.04 n. n0

4 4 ELLEN WELD Now that this is in the correct form and e.04 < 1, we can apply the geometric formula. Our total is ( 2000e ) ( ).04() e.04 n 2000e.04() 1 $45, e.04 n0 Example people are sent to a colony on Mars and each subsequent year 500 more people are added to the population of the colony. The annual death proportion is 5%. Find the eventual population of the Mars colony after many years have passed, just before a new group of 500 people arrive. Solution: Let P k be the population of the colony on Mars at the start of year k. Then P because 500 people were sent to Mars initially. Moreover, Similarly, P (P 0.05P 0 ). people sent population already to Mars on Mars P (P 1.05P 1 ) people sent population already to Mars on Mars and we continue on in this pattern. But we want a nicer way to write this. Try and P (P 0.05P 0 ) P ( 500 ) P 0 P (P 1.05P 1 ) P ( (500) P 1 ) (500)+(.95) 2 (500). So our pattern is given by 500(.95) n. n0 This is in the correct form to apply the geometric series formula. So we can write 500(.95) n ,000. n0 We aren t quite done though. We were asked to find the population just before a new group of 500 people arrive. So we need to subtract 500. Thus, our answer is 9, Additional Examples 1. In a right triangle, a series of perpendicular line segments are drawn starting with the altitude using the vertex of the right angle in the right triangle then subsequently continuing to draw altitudes from the right angles in the new right triangles created which always include the vertex from the smallest

5 AN UNOFFICIAL GUIDE TO MATH FALL angle in the original right triangle. The series of altitudes are drawn so they move closer and closer to the smallest angle in the original right triangle. Find the sum of all these perpendicular line segments if one of the angles of the triangle is 47 and the side of the triangle adjacent to this angle is 2.7. Round your answer to the nearest hundredth. Solution: This is the most difficult problem in Math The biggest challenge is that, if you compute using the numbers given, it s very easy to oversimplify which makes you miss the overarching pattern. Instead of using the numbers given, we will use variables and then substitute what we are given at the very end. The first challenge in this problem is interpreting what object they are describing. The vertex of a right triangle is the point where the smaller legs (by legs of a triangle, I mean the two shorter sides of a right triangle) meet to form the right angle: The altitude from the vertex of a right triangle is the line starting from the vertex that makes a right angle with the hypotenuse: Now, we are drawing a series of altitudes in our triangles which always contains the smallest angle, which we will call θ. Consider the following picture:

6 6 ELLEN WELD We let this right triangle have sides a, b, c where a is opposite θ and c is the hypotenuse. Interestingly, we have a very nice formula for the length of an altitude when compared to the sides of the triangle. For example, if d 1 is the length of the altitude in the triangle abc, then d 1 ab c. In general, the length of an altitude of this type is the product of the legs of the triangle divided by the hypotenuse. Observe that because cos θ b c, we may conclude that d 1 ab ( ) b c a a cos θ. c With this in mind, we look at the next altitude in our sequence: Note that since we are not looking at the original triangle anymore (because this new altitude isn t an altitude in the first triangle) we have to change the lengths of our sides. Here, our triangle has sides d 1, c 1, and b (which was also in the first triangle). Here, b is the hypotenuse and d 1, c 1 are the legs of the triangle. We know what d 1, b are but we need to find c 1. Notice that cos θ c 1 b b cos θ c 1.

7 AN UNOFFICIAL GUIDE TO MATH FALL So, d 2 d 1c 1 b d 1(b cos θ) b d 1 cos θ (a cos θ) d 1 cos θ a(cos θ) 2. Next, we consider In this triangle, our sides are d 2, c 1, and b 1 which we need to compute. Further, c 1 is now the hypotenuse. Since Then, we know that cos θ b 1 c 1 b 1 c 1 cos θ. d d 2b 1 c 1 d 2c 1 cos θ d 2 cos θ (a(cos θ) 2 ) cos θ a(cos θ). c 1 d 2 From this we can determine a pattern. We see that the sum of the lengths of these altitudes is given by a(cos θ) n a cos θ + a(cos θ) 2 + a(cos θ) +. We put this in the correct form to apply the geometric series formula: a(cos θ) n a(cos θ) n+1 n0 (a cos θ)(cos θ) n n0 a cos θ 1 cos θ. Now that we have the general formula, we need to input the numbers they have given us.

8 8 ELLEN WELD After carefully re-reading the problem, we see that a 2.7 and that θ Thus, the sum of the lengths of all the altitudes is 2.7 cos(4 ) 1 cos(4 ) 7.5. TL;DR: The geometric series describing this situation is a(cos x) n+1 a cos x + a(cos x) 2 + a(cos x) + where a is the length of the side they give you and x is 90 minus the angle they give you (so if they give you the angle 60, x 0 ). Note that x must be measured in degrees. The actual sum is a cos x 1 cos x.