# 8.E: Sequences and Series (Exercises)

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## 8.1: Sequences

1. Finding limits of convergent sequences can be a challenge. However, there is a useful tool we can adapt from our study of limits of continuous functions at infinity to use to find limits of sequences. We illustrate in this exercise with the example of the sequence ln(n) n .

(a) Calculate the first 10 terms of this sequence. Based on these calculations, do you think the sequence converges or diverges? Why?

(b) For this sequence, there is a corresponding continuous function f defined by f (x) = ln(x) x . Draw the graph of f (x) on the interval [0, 10] and then plot the entries of the sequence on the graph. What conclusion do you think we can draw about the sequence ( ln(n) n ) if limx→∞ f (x) = L? Explain.

(c) Note that f (x) has the indeterminate form ∞ ∞ as x goes to infinity. What idea from differential calculus can we use to calculate limx→∞ f (x)? Use this method to find limx→∞ f (x). What, then, is limn→∞ ln(n) n ?

2. We return to the example begun in Preview Activity 8.1 to see how to derive the formula for the amount of money in an account at a given time. We do this in a general setting. Suppose you invest P dollars (called the principal) in an account paying r% interest compounded monthly. In the first month you will receive r 12 (here r is in decimal form; e.g., if we have 8% interest, we write 0.08 12 ) of the principal P in interest, so you earn P r 12 8.1. SEQUENCES 453 dollars in interest. Assume that you reinvest all interest. Then at the end of the first month your account will contain the original principal P plus the interest, or a total of P1 = P + P r 12 = P 1 + r 12 dollars.

(a) Given that your principal is now P1 dollars, how much interest will you earn in the second month? If P2 is the total amount of money in your account at the end of the second month, explain why P2 = P1 1 + r 12 = P 1 + r 12 2 .

(b) Find a formula for P3, the total amount of money in the account at the end of the third month in terms of the original investment P.

(c) There is a pattern to these calculations. Let Pn the total amount of money in the account at the end of the third month in terms of the original investment P. Find a formula for Pn.

3. Sequences have many applications in mathematics and the sciences. In a recent paper3 the authors write The incretin hormone glucagon-like peptide-1 (GLP-1) is capable of ameliorating glucose-dependent insulin secretion in subjects with diabetes. However, its very short half-life (1.5-5 min) in plasma represents a major limitation for its use in the clinical setting. The half-life of GLP-1 is the time it takes for half of the hormone to decay in its medium. For this exercise, assume the half-life of GLP-1 is 5 minutes. So if A is the amount of GLP-1 in plasma at some time t, then only A 2 of the hormone will be present after t + 5 minutes. Suppose A0 = 100 grams of the hormone are initially present in plasma.

(a) Let A1 be the amount of GLP-1 present after 5 minutes. Find the value of A1.

(b) Let A2 be the amount of GLP-1 present after 10 minutes. Find the value of A2.

(c) Let A3 be the amount of GLP-1 present after 15 minutes. Find the value of A3.

(d) Let A4 be the amount of GLP-1 present after 20 minutes. Find the value of A4.

(e) Let An be the amount of GLP-1 present after 5n minutes. Find a formula for An.

(f) Does the sequence {An} converge or diverge? If the sequence converges, find its limit and explain why this value makes sense in the context of this problem. 3Hui H, Farilla L, Merkel P, Perfetti R. The short half-life of glucagon-like peptide-1 in plasma does not reflect its long-lasting beneficial effects, Eur J Endocrinol 2002 Jun;146(6):863-9. 454 8.1. SEQUENCES

(g) Determine the number of minutes it takes until the amount of GLP-1 in plasma is 1 gram.

4. Continuous data is the basis for analog information, like music stored on old cassette tapes or vinyl records. A digital signal like on a CD or MP3 file is obtained by sampling an analog signal at some regular time interval and storing that information. For example, the sampling rate of a compact disk is 44,100 samples per second. So a digital recording is only an approximation of the actual analog information. Digital information can be manipulated in many useful ways that allow for, among other things, noisy signals to be cleaned up and large collections of information to be compressed and stored in much smaller space. While we won’t investigate these techniques in this chapter, this exercise is intended to give an idea of the importance of discrete (digital) techniques. Let f be the continuous function defined by f (x) = sin(4x) on the interval [0, 10]. A graph of f is shown in Figure 8.2. We approximate f by sampling, that is by partitioning the interval [0, 10] into uniform subintervals and recording the values of f at the endpoints.

Figure 8.2: The graph of f (x) = sin(4x) on the interval [0, 10]

(a) Ineffective sampling can lead to several problems in reproducing the original signal. As an example, partition the interval [0, 10] into 8 equal length subintervals and create a list of points (the sample) using the endpoints of each subinterval. Plot your sample on graph of f in Figure Figure 8.2. What can you say about the period of your sample as compared to the period of the original function?

(b) The sampling rate is the number of samples of a signal taken per second. As part (a) illustrates, sampling at too small a rate can cause serious problems with reproducing the original signal (this problem of inefficient sampling leading to an inaccurate approximation is called aliasing). There is an elegant theorem 8.1. SEQUENCES 455 called the Nyquist-Shannon Sampling Theorem that says that human perception is limited, which allows that replacement of a continuous signal with a digital one without any perceived loss of information. This theorem also provides the lowest rate at which a signal can be sampled (called the Nyquist rate) without such a loss of information. The theorem states that we should sample at double the maximum desired frequency so that every cycle of the original signal will be sampled at at least two points. Recall that the frequency of a sinusoidal function is the reciprocal of the period. Identify the frequency of the function f and determine the number of partitions of the interval [0, 10] that give us the Nyquist rate.

(c) Humans cannot typically pick up signals above 20 kHz. Explain why, then, that information on a compact disk is sampled at 44,100 Hz.

## 8.2: Geometric Series

1. There is an old question that is often used to introduce the power of geometric growth. Here is one version. Suppose you are hired for a one month (30 days, working every day) job and are given two options to be paid. Option 1. You can be paid $500 per day or Option 2. You can be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, 8 cents the fourth day, and so on, doubling the amount you are paid each day.

(a) How much will you be paid for the job in total under Option 1?

(b) Complete Table 8.3 to determine the pay you will receive under Option 2 for the first 10 days.

Day Pay on this day Total amount paid to date 1 $0.01 $0.01 2 $0.02 $0.03 3 4 5 6 7 8 9 10

Table 8.3: Option 2 payments

8.2. GEOMETRIC SERIES 463

(c) Find a formula for the amount paid on day n, as well as for the total amount paid by day n. Use this formula to determine which option (1 or 2) you should take.

2. Suppose you drop a golf ball onto a hard surface from a height h. The collision with the ground causes the ball to lose energy and so it will not bounce back to its original height. The ball will then fall again to the ground, bounce back up, and continue. Assume that at each bounce the ball rises back to a height 3 4 of the height from which it dropped. Let hn be the height of the ball on the nth bounce, with h0 = h. In this exercise we will determine the distance traveled by the ball and the time it takes to travel that distance.

(a) Determine a formula for h1 in terms of h.

(b) Determine a formula for h2 in terms of h.

(c) Determine a formula for h3 in terms of h.

(d) Determine a formula for hn in terms of h.

(e) Write an infinite series that represents the total distance traveled by the ball. Then determine the sum of this series.

(f) Next, let’s determine the total amount of time the ball is in the air.

(i) When the ball is dropped from a height H, if we assume the only force acting on it is the acceleration due to gravity, then the height of the ball at time t is given by H − 1 2 gt 2 . Use this formula to determine the time it takes for the ball to hit the ground after being dropped from height H.

(ii) Use your work in the preceding item, along with that in (a)-(e) above to determine the total amount of time the ball is in the air.

3. Suppose you play a game with a friend that involves rolling a standard six-sided die. Before a player can participate in the game, he or she must roll a six with the die. Assume that you roll first and that you and your friend take alternate rolls. In this exercise we will determine the probability that you roll the first six.

(a) Explain why the probability of rolling a six on any single roll (including your first turn) is 1 6 .

(b) If you don’t roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second 464 8.2. GEOMETRIC SERIES turn. Explain why the probability of this event is 5 6 5 6 1 6 = 5 6 2 1 6 .

(c) Now suppose you fail to roll the first six on your second turn. Explain why the probability is 5 6 5 6 5 6 5 6 1 6 = 5 6 4 1 6 that you to roll the first six on your third turn.

(d) The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is 1 6 + 5 6 2 1 6 + 5 6 4 1 6 + · · · . Find the sum of this series and determine the probability that you roll the first six.

4. The goal of a federal government stimulus package is to positively affect the economy. Economists and politicians quote numbers like “k million jobs and a net stimulus to the economy of n billion of dollars.” Where do they get these numbers? Let’s consider one aspect of a stimulus package: tax cuts. Economists understand that tax cuts or rebates can result in long-term spending that is many times the amount of the rebate. For example, assume that for a typical person, 75% of her entire income is spent (that is, put back into the economy). Further, assume the government provides a tax cut or rebate that totals P dollars for each person.

(a) The tax cut of P dollars is income for its recipient. How much of this tax cut will be spent?

(b) In this simple model, we will say that the spent portion of the tax cut/rebate from part (a) then becomes income for another person who, in turn, spends 75% of this income. After this “second round" of spent income, how many total dollars have been added to the economy as a result of the original tax cut/rebate?

(c) This second round of spending becomes income for another group who spend 75% of this income, and so on. In economics this is called the multiplier effect. Explain why an original tax cut/rebate of P dollars will result in multiplied spending of 0.75P(1 + 0.75 + 0.752 + · · · ). dollars. 8.2. GEOMETRIC SERIES 465

(d) Based on these assumptions, how much stimulus will a 200 billion dollar tax cut/rebate to consumers add to the economy, assuming consumer spending remains consistent forever.

5. Like stimulus packages, home mortgages and foreclosures also impact the economy. A problem for many borrowers is the adjustable rate mortgage, in which the interest rate can change (and usually increases) over the duration of the loan, causing the monthly payments to increase beyond the ability of the borrower to pay. Most financial analysts recommend fixed rate loans, ones for which the monthly payments remain constant throughout the term of the loan. In this exercise we will analyze fixed rate loans. When most people buy a large ticket item like car or a house, they have to take out a loan to make the purchase. The loan is paid back in monthly installments until the entire amount of the loan, plus interest, is paid. With a loan, we borrow money, say P dollars (called the principal), and pay off the loan at an interest rate of r%. To pay back the loan we make regular monthly payments, some of which goes to pay off the principal and some of which is charged as interest. In most cases, the interest is computed based on the amount of principal that remains at the beginning of the month. We assume a fixed rate loan, that is one in which we make a constant monthly payment M on our loan, beginning in the original month of the loan. Suppose you want to buy a house. You have a certain amount of money saved to make a down payment, and you will borrow the rest to pay for the house. Of course, for the privilege of loaning you the money, the bank will charge you interest on this loan, so the amount you pay back to the bank is more than the amount you borrow. In fact, the amount you ultimately pay depends on three things: the amount you borrow (called the principal), the interest rate, and the length of time you have to pay off the loan plus interest (called the duration of the loan). For this example, we assume that the interest rate is fixed at r%. To pay off the loan, each month you make a payment of the same amount (called installments). Suppose we borrow P dollars (our principal) and pay off the loan at an interest rate of r% with regular monthly installment payments of M dollars. So in month 1 of the loan, before we make any payments, our principal is P dollars. Our goal in this exercise is to find a formula that relates these three parameters to the time duration of the loan. We are charged interest every month at an annual rate of r%, so each month we pay r 12% interest on the principal that remains. Given that the original principal is P dollars, we will pay 0.0r 12 P dollars in interest on our first payment. Since we paid M dollars in total for our first payment, the remainder of the payment (M − r 12 P) goes to pay down the principal. So the principal remaining after the first payment (let’s call it P1) is the original principal minus what we paid on the principal, or P1 = P − M − r 12 P = 1 + r 12 P − M. 466 8.2. GEOMETRIC SERIES As long as P1 is positive, we still have to keep making payments to pay off the loan.

(a) Recall that the amount of interest we pay each time depends on the principal that remains. How much interest, in terms of P1 and r, do we pay in the second installment?

(b) How much of our second monthly installment goes to pay off the principal? What is the principal P2, or the balance of the loan, that we still have to pay off after making the second installment of the loan? Write your response in the form P2 = ( )P1 − ( )M, where you fill in the parentheses.

(c) Show that P2 = 1 + r 12 2 P − 1 + 1 + r 12 M.

(d) Let P3 be the amount of principal that remains after the third installment. Show that P3 = 1 + r 12 3 P − " 1 + 1 + r 12 + 1 + r 12 2 # M.

(e) If we continue in the manner described in the problems above, then the remaining principal of our loan after n installments is Pn = 1 + r 12 n P − Xn−1 k=0 1 + r 12 k M. (8.7) This is a rather complicated formula and one that is difficult to use. However, we can simplify the sum if we recognize part of it as a partial sum of a geometric series. Find a formula for the sum Xn−1 k=0 1 + r 12 k . (8.8) and then a general formula for Pn that does not involve a sum.

(f) It is usually more convenient to write our formula for Pn in terms of years rather than months. Show that P(t), the principal remaining after t years, can be written as P(t) = P − 12M r ! 1 + r 12 12t + 12M r . (8.9)

(g) Now that we have analyzed the general loan situation, we apply formula (8.9) to an actual loan. Suppose we charge $1,000 on a credit card for holiday expenses. If our credit card charges 20% interest and we pay only the minimum payment of $25 each month, how long will it take us to pay off the $1,000 charge? How much in total will we have paid on this $1,000 charge? How much total interest will we pay on this loan? 8.2. GEOMETRIC SERIES 467

(h) Now we consider larger loans, e.g. automobile loans or mortgages, in which we borrow a specified amount of money over a specified period of time. In this situation, we need to determine the amount of the monthly payment we need to make to pay off the loan in the specified amount of time. In this situation, we need to find the monthly payment M that will take our outstanding principal to 0 in the specified amount of time. To do so, we want to know the value of M that makes P(t) = 0 in formula (8.9). If we set P(t) = 0 and solve for M, it follows that M = rP 1 + r 12 12t 12 1 + r 12 12t − 1 .

(i) Suppose we want to borrow $15,000 to buy a car. We take out a 5 year loan at 6.25%. What will our monthly payments be? How much in total will we have paid for this $15,000 car? How much total interest will we pay on this loan?

(ii) Suppose you charge your books for winter semester on your credit card. The total charge comes to $525. If your credit card has an interest rate of 18% and you pay $20 per month on the card, how long will it take before you pay off this debt? How much total interest will you pay?

(iii) Say you need to borrow $100,000 to buy a house. You have several options on the loan: – 30 years at 6.5% – 25 years at 7.5% – 15 years at 8.25%.

(a) What are the monthly payments for each loan?

(b) Which mortgage is ultimately the best deal (assuming you can afford the monthly payments)? In other words, for which loan do you pay the least amount of total interest?

## 8.3: Series of Real Numbers

### Exercises

1. In this exercise we investigate the sequence ( b n n! ) for any constant b.

(a) Use the Ratio Test to determine if the series P 10k k! converges or diverges.

(b) Now apply the Ratio Test to determine if the series P b k k! converges for any constant b. 8.3. SERIES OF REAL NUMBERS 483

(c) Use your result from (b) to decide whether the sequence ( b n n! ) converges or diverges. If the sequence ( b n n! ) converges, to what does it converge? Explain your reasoning.

2. There is a test for convergence similar to the Ratio Test called the Root Test. Suppose we have a series P ak of positive terms so that an → 0 as n → ∞.

(a) Assume n √ an → r as n goes to infinity. Explain why this tells us that an ≈ r n for large values of n.

(b) Using the result of part (a), explain why P ak looks like a geometric series when n is big. What is the ratio of the geometric series to which P ak is comparable?

(c) Use what we know about geometric series to determine that values of r so that P ak converges if n √ an → r as n → ∞.

3. The associative and distributive laws of addition allow us to add finite sums in any order we want. That is, if Pn k=0 ak and Pn k=0 bk are finite sums of real numbers, then Xn k=0 ak + Xn k=0 bk = Xn k=0 (ak + bk ). However, we do need to be careful extending rules like this to infinite series.

(a) Let an = 1 + 1 2 n and bn = −1 for each nonnegative integer n.

(i) Explain why the series P∞ k=0 ak and P∞ k=0 bk both diverge.

(ii) Explain why the series P∞ k=0 (ak + bk ) converges.

(iii) Explain why X∞ k=0 ak + X∞ k=0 bk , X∞ k=0 (ak + bk ). This shows that it is possible to have to two divergent series P∞ k=0 ak and P∞ k=0 bk but yet have the series P∞ k=0 (ak + bk ) converge.

(b) While part (a) shows that we cannot add series term by term in general, we can under reasonable conditions. The problem in part (a) is that we tried to add divergent series. In this exercise we will show that if P ak and P bk are convergent series, then P (ak + bk ) is a convergent series and X (ak + bk ) = X ak + X bk .

(i) Let An and Bn be the nth partial sums of the series P∞ k=1 ak and P∞ k=1 bk , respectively. Explain why An + Bn = Xn k=1 (ak + bk ).

(ii) Use the previous result and properties of limits to show that X∞ k=1 (ak + bk ) = X∞ k=1 ak + X∞ k=1 bk . (Note that the starting point of the sum is irrelevant in this problem, so it doesn’t matter where we begin the sum.)

(c) Use the prior result to find the sum of the series P∞ k=0 2 k+3 k 5 k .

4. In the Limit Comparison Test we compared the behavior of a series to one whose behavior we know. In that test we use the limit of the ratio of corresponding terms of the series to determine if the comparison is valid. In this exercise we see how we can compare two series directly, term by term, without using a limit of sequence. First we consider an example.

(a) Consider the series X 1 k 2 and X 1 k 2 + k . We know that the series P 1 k 2 is a p-series with p = 2 > 1 and so P 1 k 2 converges. In this part of the exercise we will see how to use information about P 1 k 2 to determine information about P 1 k 2+k . Let ak = 1 k 2 and bk = 1 k 2+k .

(i) Let Sn be the nth partial sum of P 1 k 2 and Tn the nth partial sum of P 1 k 2+k . Which is larger, S1 or T1? Why?

(ii) Recall that S2 = S1 + a2 and T2 = T1 + b2. Which is larger, a2 or b2? Based on that answer, which is larger, S2 or T2?

(iii) Recall that S3 = S2 + a3 and T3 = T2 + b3. Which is larger, a3 or b3? Based on that answer, which is larger, S3 or T3?

(iv) Which is larger, an or bn? Explain. Based on that answer, which is larger, Sn or Tn?

(v) Based on your response to the previous part of this exercise, what relationship do you expect there to be between P 1 k 2 and P 1 k 2+k ? Do you expect P 1 k 2+k to converge or diverge? Why?

(b) The example in the previous part of this exercise illustrates a more general result. Explain why the Direct Comparison Test, stated here, works. The Direct Comparison Test. If 0 ≤ bk ≤ ak for every k, then we must have 0 ≤ X bk ≤ X ak 1. If P ak converges, then P bk converges. 2. If P bk diverges, then P ak diverges. Important Note: This comparison test applies only to series with nonnegative terms.

(i) Use the Direct Comparison Test to determine the convergence or divergence of the series P 1 k−1 . Hint: Compare to the harmonic series.

(ii) Use the Direct Comparison Test to determine the convergence or divergence of the series P k k 3+1 .

## 8.4: Alternating Series

### Exercises

1. Conditionally convergent series converge very slowly. As an example, consider the famous formula6 π 4 = 1 − 1 3 + 1 5 − 1 7 + · · · = X∞ k=0 (−1) k 1 2k + 1 . (8.17) In theory, the partial sums of this series could be used to approximate π.

(a) Show that the series in (8.17) converges conditionally.

(b) Let Sn be the nth partial sum of the series in (8.17). Calculate the error in 6We will derive this formula in upcoming work. 498 8.4. ALTERNATING SERIES approximating π 4 with S100 and explain why this is not a very good approximation.

(c) Determine the number of terms it would take in the series (8.17) to approximate π 4 to 10 decimal places. (The fact that it takes such a large number of terms to obtain even a modest degree of accuracy is why we say that conditionally convergent series converge very slowly.)

2. We have shown that if P (−1) k+1ak is a convergent alternating series, then the sum S of the series lies between any two consecutive partial sums Sn. This suggests that the average Sn+Sn+1 2 is a better approximation to S than is Sn.

(a) Show that Sn+Sn+1 2 = Sn + 1 2 (−1) n+2an+1.

(b) Use this revised approximation in (a) with n = 20 to approximate ln(2) given that ln(2) = X∞ k=1 (−1) k+1 1 k . Compare this to the approximation using just S20. For your convenience, S20 = 155685007 232792560 .

3. In this exercise, we examine one of the conditions of the Alternating Series Test. Consider the alternating series 1 − 1 + 1 2 − 1 4 + 1 3 − 1 9 + 1 4 − 1 16 + · · · , where the terms are selected alternately from the sequences 1 n and ( − 1 n 2 ) .

(a) Explain why the nth term of the given series converges to 0 as n goes to infinity. (b) Rewrite the given series by grouping terms in the following manner: (1 − 1) + 1 2 − 1 4 + 1 3 − 1 9 + 1 4 − 1 16 + · · · . Use this regrouping to determine if the series converges or diverges.

(c) Explain why the condition that the sequence {an} decreases to a limit of 0 is included in the Alternating Series Test.

4. Conditionally convergent series exhibit interesting and unexpected behavior. In this exercise we examine the conditionally convergent alternating harmonic series X∞ k=1 (−1) k+1 k and discover that addition is not commutative for conditionally convergent series. We will also encounter Riemann’s Theorem concerning rearrangements of conditionally 8.4. ALTERNATING SERIES 499 convergent series. Before we begin, we remind ourselves that X∞ k=1 (−1) k+1 k = ln(2), a fact which will be verified in a later section.

(a) First we make a quick analysis of the positive and negative terms of the alternating harmonic series.

(i) Show that the series X∞ k=1 1 2k diverges.

(ii) Show that the series X∞ k=1 1 2k + 1 diverges.

(iii) Based on the results of the previous parts of this exercise, what can we say about the sums X∞ k=C 1 2k and X∞ k=C 1 2k + 1 for any positive integer C? Be specific in your explanation.

(b) Recall addition of real numbers is commutative; that is a + b = b + a for any real numbers a and b. This property is valid for any sum of finitely many terms, but does this property extend when we add infinitely many terms together? The answer is no, and something even more odd happens. Riemann’s Theorem (after the nineteenth-century mathematician Georg Friedrich Bernhard Riemann) states that a conditionally convergent series can be rearranged to converge to any prescribed sum. More specifically, this means that if we choose any real number S, we can rearrange the terms of the alternating harmonic series X∞ k=1 (−1) k+1 k so that the sum is S. To understand how Riemann’s Theorem works, let’s assume for the moment that the number S we want our rearrangement to converge to is positive. Our job is to find a way to order the sum of terms of the alternating harmonic series to converge to S.

(i) Explain how we know that, regardless of the value of S, we can find a partial sum P1 P1 = Xn1 k=1 1 2k + 1 = 1 + 1 3 + 1 5 + · · · + 1 2n1 + 1 of the positive terms of the alternating harmonic series that equals or 500 8.4. ALTERNATING SERIES exceeds S. Let S1 = P1.

(ii) Explain how we know that, regardless of the value of S1, we can find a partial sum N1 N1 = − Xm1 k=1 1 2k = − 1 2 − 1 4 − 1 6 − · · · − 1 2m1 so that S2 = S1 + N1 ≤ S.

(iii) Explain how we know that, regardless of the value of S2, we can find a partial sum P2 P2 = Xn2 k=n1+1 1 2k + 1 = 1 2(n1 + 1) + 1 + 1 2(n1 + 2) + 1 + · · · + 1 2n2 + 1 of the remaining positive terms of the alternating harmonic series so that S3 = S2 + P2 ≥ S.

(iv) Explain how we know that, regardless of the value of S3, we can find a partial sum N2 = − Xm2 k=m1+1 1 2k = − 1 2(m1 + 1) − 1 2(m1 + 2) − · · · − 1 2m2 of the remaining negative terms of the alternating harmonic series so that S4 = S3 + N2 ≤ S.

(v) Explain why we can continue this process indefinitely and find a sequence {Sn} whose terms are partial sums of a rearrangement of the terms in the alternating harmonic series so that limn→∞ Sn = S.

## 8.5: Taylor Polynomials and Taylor Series

### Exercises

1. In this exercise we investigation the Taylor series of polynomial functions.

(a) Find the 3rd order Taylor polynomial centered at a = 0 for f (x) = x 3 − 2x 2 + 3x − 1. Does your answer surprise you? Explain.

(b) Without doing any additional computation, find the 4th, 12th, and 100th order Taylor polynomials (centered at a = 0) for f (x) = x 3 − 2x 2 + 3x − 1. Why should you expect this?

(c) Now suppose f (x) is a degree m polynomial. Completely describe the nth order Taylor polynomial (centered at a = 0) for each n.

2. The examples we have considered in this section have all been for Taylor polynomials and series centered at 0, but Taylor polynomials and series can be centered at any value of a. We look at examples of such Taylor polynomials in this exercise.

(a) Let f (x) = sin(x). Find the Taylor polynomials up through order four of f centered at x = π 2 . Then find the Taylor series for f (x) centered at x = π 2 . Why should you have expected the result?

(b) Let f (x) = ln(x). Find the Taylor polynomials up through order four of f centered at x = 1. Then find the Taylor series for f (x) centered at x = 1.

(c) Use your result from (b) to determine which Taylor polynomial will approximate ln(2) to two decimal places. Explain in detail how you know you have the desired accuracy.

3. We can use known Taylor series to obtain other Taylor series, and we explore that idea in this exercise, as a preview of work in the following section.

(a) Calculate the first four derivatives of sin(x 2 ) and hence find the fourth order Taylor polynomial for sin(x 2 ) centered at a = 0.

(b) Part (a) demonstrates the brute force approach to computing Taylor polynomials and series. Now we find an easier method that utilizes a known Taylor series. Recall that the Taylor series centered at 0 for f (x) = sin(x) is X∞ k=0 (−1) k x 2k+1 (2k + 1)! . (8.24)

(i) Substitute x 2 for x in the Taylor series (8.24). Write out the first several terms and compare to your work in part (a). Explain why the substitution in this problem should give the Taylor series for sin(x 2 ) centered at 0.

(ii) What should we expect the interval of convergence of the series for sin(x 2 ) to be? Explain in detail.

4. Based on the examples we have seen, we might expect that the Taylor series for a function f always converges to the values f (x) on its interval of convergence. We explore that idea in more detail in this exercise. Let f (x) = e −1/x 2 if x , 0, 0 if x = 0.

(a) Show, using the definition of the derivative, that f 0 (0) = 0.

(b) It can be shown that f (n) (0) = 0 for all n ≥ 2. Assuming that this is true, find the Taylor series for f centered at 0.

(c) What is the interval of convergence of the Taylor series centered at 0 for f? Explain. For which values of x the interval of convergence of the Taylor series does the Taylor series converge to f (x)?

## 8.6: Power Series

### Exercises

1. In this exercise we investigation the Taylor series of polynomial functions.

(a) Find the 3rd order Taylor polynomial centered at a = 0 for f (x) = x 3 − 2x 2 + 3x − 1. Does your answer surprise you? Explain.

(b) Without doing any additional computation, find the 4th, 12th, and 100th order Taylor polynomials (centered at a = 0) for f (x) = x 3 − 2x 2 + 3x − 1. Why should you expect this?

(c) Now suppose f (x) is a degree m polynomial. Completely describe the nth order Taylor polynomial (centered at a = 0) for each n.

2. The examples we have considered in this section have all been for Taylor polynomials and series centered at 0, but Taylor polynomials and series can be centered at any value of a. We look at examples of such Taylor polynomials in this exercise.

(a) Let f (x) = sin(x). Find the Taylor polynomials up through order four of f centered at x = π 2 . Then find the Taylor series for f (x) centered at x = π 2 . Why should you have expected the result?

(b) Let f (x) = ln(x). Find the Taylor polynomials up through order four of f centered at x = 1. Then find the Taylor series for f (x) centered at x = 1.

(c) Use your result from (b) to determine which Taylor polynomial will approximate ln(2) to two decimal places. Explain in detail how you know you have the desired accuracy.

3. We can use known Taylor series to obtain other Taylor series, and we explore that idea in this exercise, as a preview of work in the following section.

(a) Calculate the first four derivatives of sin(x 2 ) and hence find the fourth order Taylor polynomial for sin(x 2 ) centered at a = 0.

(b) Part (a) demonstrates the brute force approach to computing Taylor polynomials and series. Now we find an easier method that utilizes a known Taylor series. Recall that the Taylor series centered at 0 for f (x) = sin(x) is X∞ k=0 (−1) k x 2k+1 (2k + 1)! . (8.24)

(i) Substitute x 2 for x in the Taylor series (8.24). Write out the first several terms and compare to your work in part (a). Explain why the substitution in this problem should give the Taylor series for sin(x 2 ) centered at 0.

(ii) What should we expect the interval of convergence of the series for sin(x 2 ) to be? Explain in detail.

4. Based on the examples we have seen, we might expect that the Taylor series for a function f always converges to the values f (x) on its interval of convergence. We explore that idea in more detail in this exercise. Let f (x) = e −1/x 2 if x , 0, 0 if x = 0.

(a) Show, using the definition of the derivative, that f 0 (0) = 0.

(b) It can be shown that f (n) (0) = 0 for all n ≥ 2. Assuming that this is true, find the Taylor series for f centered at 0.

(c) What is the interval of convergence of the Taylor series centered at 0 for f? Explain. For which values of x the interval of convergence of the Taylor series does the Taylor series converge to f (x)?