Class Summary Page

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MATH 162

Calculus with Applications, II

December, 1997

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  • Monday, December 1. We began the class today by observing that the geometric series that we summed before Thanksgiving should already be familiar to us, albeit in a disguised way. We thought about decimal representations of rational numbers and realized that repeating decimals really were convergent geometric series. We looked at a pair of examples, the second (and more elaborate one) was d = 12.3454545.... We wrote d = 12 + 3/10 + 45/1000 + 45/100000 + 45/10000000 + ... = 123/10 + (45/1000)*(1 + 1/100 + 1/10000 + ...) = 123/10 + (45/1000)*(1 + 1/100 + 1/100^2 + ...) = 123/10 + (45/1000)*Sum((1/100)^k,k=0..Infinity). We recognize this series to be geometric and convergent (since here |r| = 1/100 < 1) with the sum 1/(1-1/100)). Thus d = 123/10 + (45/1000)*(1/(1-1/100)) = 123/10 + 45/(1000-10) = 123/10 + 45/990 = (123*99+45)/990 = 12222/990 = 679/55. We then investigated what we could say about a decimal whose digits weren't repeating. In particular, once we wrote this decimal as an infinite series, how could we prove that it converged? (The example we considered was a "random decimal" where the digits were produced by rolling a "10-sided" die --- actually an icosahedron with 20 sides, but each of the digits 0,1,...,9 appears on two sides.) We needed to introduce the Bounded Monotone Sequence Property (BMSP) --- see either the handout from last Friday, or Theorem 10 on page 605 in the course text. (Note that we have taken the BMSP as an axiom of the real numbers --- it distinguishes them from the set of rational numbers --- while Stewart prefers to take as his axiom something called the Completeness Axiom and to prove the BMSP from the Completeness Axiom. It turns out that the two statements, the Completeness Axiom and the BMSP, are equivalent in the sense that if we assume either of them, we can use it to prove the other. For this reason, it makes better sense for us to take as our axiom the statement that is in the form that we will find to be the most useful, and that is the BMSP.) After we looked at the graphs of several sequences in order to better understand the terms bounded and monotonic (note that what Stewart calls "increasing" and "decreasing", respectively, I prefer to call "nondecreasing" and "nonincreasing"), we used the BMSP to prove that the partial sums for our random decimal had to be convergent because they were nondecreasing, bounded below (say, by 0), and bounded above (to get an upper bound, just truncate the decimal after a certain number of digits, and then round the last digit up). We ended the class by reworking this example in the much more general (and important) context of nonnegative series, Sum(a_k,k=0..Infinity). If a_k >= 0 for each k, then the partial sums Sum(a_k,k=0..n) are clearly nondecreasing, and bounded below. Regarding the issue of whether they are bounded above, there are two possibilities. First, suppose that the partial sums are bounded above; then the BMSP tells us that they must converge, i.e., that Sum(a_k,k=0..Infinity) converges. On the other hand, if the partial sums are not bounded above then it is pretty clear that the series Sum(a_k,k=0..Infinity) diverges to Infinity. Thus, when we look at nonnegative series, the BMSP gives us the following dichotomy: either the series is convergent or it must diverge in a special way --- by diverging to Infinity. (Note that for series without the special condition that all of the terms have the same sign, there is an additional way to diverge: by oscillation.) Because of the "bounded and convergent" versus "unbounded and divergent" dichotomy for nonnegative series we sometimes indicate convergence by simply writing Sum(a_k,k=0..Infinity) < Infinity. At the end of class, a review sheet to aid in preparation for the fourth exam was distributed.

  • Wednesday, December 3. We continued our discussion of the implications of the Bounded Monotone Sequence Property (BMSP) for nonnegative series. In particular, we studied how one could begin to apply these ideas to power series. The problem here is that for most power series Sum(c_k*(x-a)^k,k=0..Infinity) the individual terms c_k*(x-a)^k do not have the same sign for all values of k and x --- in particular, if k is odd and c_k is nonzero, say positive, then c_k*(x-a)^k is positive for x > a but negative for x < a. In order to obtain a nonnegative series (so that we could use the BMSP) from the general power series Sum(c_k*(x-a)^k,k=0..Infinity), we hit upon the idea of considering the series of absolute values: Sum(|c_k|*|x-a|^k,k=0..Infinity). More generally, if Sum(|a_k|,k=0..Infinity) converges, then we say that Sum(a_k,k=0..Infinity) is absolutely convergent. We then proved the Theorem on the Relation Between Absolute Convergence and Convergence of Series: if Sum(a_k,k=0..Infinity) is absolutely convergent, then it is also convergent. (For the record, there do exist series that are convergent without being absolutely convergent.) In discussing the intuition behind this result we observed that when Sum(a_k,k=0..Infinity) is absolutely convergent, we should have -Sum(|a_k|,k=0..Infinity) <= Sum(a_k,k=0..Infinity) <= Sum(|a_k|,k=0..Infinity). We ended the class by proving this theorem using the BMSP. A key step was showing that if b_k = a_k+|a_k|, then the partial sums {Sum(b_k,k=0..n)} were bounded above. We did this by arguing that for each n, Sum(b_k,k=0..n) <= 2*Sum(|a_k|,k=0..n) <= 2*Sum(|a_k|,k=0..Infinity), which is just some finite number (by the assumption that Sum(a_k,k=0..Infinity) is absolutely convergent). Next class we will repackage this argument as a convergence test for nonnegative series (the Comparison Test). By the way, the proof of the Theorem on the Relation Between Absolute Convergence and Convergence of Series is in the course text on page 635, except that since Stewart has already developed the Comparison Test at this point in the book, he uses that test directly instead of appealing to the ideas underlying it (as we did). A practice version of the fourth exam (and a solution set) was distributed at the end of class.

  • Thursday, December 4. A review session was held from 7-9 pm in ACD 314.

  • Friday, December 5. The fourth exam was given today.

  • Monday, December 8. The fourth exam was returned today together with the solution set. Also a handout listing the major results on power series that we would be covering this week was handed out. We began by proving the convergence part of the Comparison Test; that's the part of this test that will be of the most use to us. Then we turned to the issue of the general nature of power series, Sum(c_k*(x-a)^k,k=0..Infinity). We can think of this as adding up the term-by-term products of two sequences {c_k} and {(x-a)^k}. The second of these is a geometric sequence with r=x-a, so it converges to zero quickly enough so that the series converges if x is close to a, and it diverges (in magnitude) if x is far from a so that its series must then diverge. When we multiply the c_k's by (x-a)^k's and try to sum the resulting products, what happens depends on what kind of sequence {c_k} is. There turn out to be three possible cases:
    (i) If the sequence {c_k} is strongly divergent [i.e., if it goes to infinity faster than any (nonzero) geometric sequence can converge to 0] then the product c_k*(x-a)^k will diverge as k->Infinity for every x no matter how close x is to a --- unless x=a.
    (ii) On the other hand, if {c_k} converges strongly to 0 (i.e., if it approaches zero faster than any geometric sequence can diverge to infinity) then the product c_k*(x-a)^k will converge to 0 fast enough for Sum(c_k*(x-a)^k,k=0..Infinity) to converge no matter how far x is from a.
    (iii) In between these two cases we have the interesting situation where the sequence {c_k} is comparable to some geometric sequence in the sense that if x is sufficiently close to a, then the convergence of (x-a)^k dominates and Sum(c_k*(x-a)^k,k=0..Infinity) converges, while if x is far enough from a then the divergence of (x-a)^k dominates and Sum(c_k*(x-a)^k,k=0..Infinity) diverges.
    We then looked at what the Theorem on the General Nature of Convergence of Power Series (on the handout) told us was happening. It gave a precise characterization of the three cases outlined above. In particular in the third case it told us that there was a number R (with 0 < R < Infinity) such that Sum(c_k*(x-a)^k,k=0..Infinity) converges if |x-a| < R and diverges if |x-a| > R. We called this R the radius of convergence, and then proceeded to define the radius of convergence to be 0 in case (i) and Infinity in case (ii). The set of points x where the power series converges is called the interval of convergence of that power series. In case (i), the interval of convergence is just the single point a which can be thought of as a degenerate closed interval, [a,a]. In case (ii), the interval of convergence is entire real line, or (-Infinity,Infinity). In the last case, the interval of convergence could be any one of the following four possibilities: (a-R,a+R) or (a-R,a+R] or [a-R,a+R) or [a-R,a+R]. In order to decide which of these four cases actually holds, we need to use various convergence tests such as the Integral Test (Section 10.3), the Integral Comparison Test (Section 10.4), and the Alternating Series Test (Section 10.5); when you look at it this way, that's a lot of material that we would need to cover in order to learn very little, namely, what was happening at the endpoints of the interval of convergence. As the schedule this semester has shortened our class-time by nearly a full week, we won't be covering those tests, but we're really not missing very much. We spent the last part of the class using the MAPLE worksheet "Animated Taylor Polynomials" to examine the Taylor series of several functions to see if their intervals of convergence seemed to be as described by the Theorem on the General Nature of Convergence of Power Series. Finally, the course evaluation forms were distributed.

  • Wednesday, December 10. We spent the class working through four examples that demonstrated the three different possibilities for the radius of convergence of a power series: R=0, R=Infinity, and 0 < R < Infinity.
    1. We first examined Sum(k!*x^k, k=0..Infinity). We reasoned that by an argument similar to that used on the fourth practice exam (see the solution set for that exam) to show that lim_(n->Infinity) 2^n/n! = 0, we could argue that for every x>0, lim_(k->Infinity) (1/x)^k/k! = 0. Taking reciprocals in this last equation, and using the Quotient Law for Limits, we have that lim_(k->Infinity) k!/x^k = Infinity. (For x<0, the divergence is even worse since not only are the magnitudes diverging, but the sign is oscillating.) So if x in not equal to 0, then we can use the Divergence Theorem to argue that since lim_(k->Infinity) k!/x^k is not equal to 0, Sum(k!*x^k, k=0..Infinity) must diverge. (Of course, when x=0, Sum(k!*x^k, k=0..Infinity) = 0!*1 + 1!*0 + 2!*0^2 + 3!*0^3 + ... = 1 + 0 + 0 + 0 + ... = 1.)
    2. We then quickly looked at a translated version of the last series: Sum(k!*(x-4)^k, k=0..Infinity), which diverges for every x not equal to 4, and which converges (to 1) when x=4.
    3. We next considered Sum(x^k/k^k, k=1..Infinity). We first observed that for large values of k, k^k must be much larger than x^k. We turned that observation into a comparison argument by choosing an integer m with m>=2*|x|. We wrote Sum(x^k/k^k, k=1..Infinity) = Sum(x^k/k^k, k=1..m-1) + Sum(x^k/k^k, k=m..Infinity), where the only issue was to establish the converge of the last series: note that the first sum on the right-hand side is a finite sum, so there are is no question of convergence here. For k>=m, we have that (|x|/k)^k ² (|x|/m)^k ² (|x|/(2*|x|)^k = (1/2)^k, with the second inequality following from our choice of m. Since Sum((1.2)^k, k=m..Infinity) is the "tail" of a convergent geometric series, it must converge. Thus, by the Comparison Test, Sum(|x|^k/k^k, k=m..Infinity), hence Sum(|x|^k/k^k, k=1..Infinity), must converge. Therefore the original series Sum(x^k/k^k, k=1..Infinity) is absolutely convergent for every x (note that nowhere in our analysis did we make any assumptions on x). We had an interesting discussion following this example. In response to a student's question it was pointed out that we could make different choices of m, but only some of these turn out to be useful. If we take m>=b*x, then the comparison argument we made works as long as b > 1, because we end up with (|x|/k)^k ² (|x|/m)^k ² (|x|/(b*|x|)^k = (1/b)^k for k>=m, and Sum((1/b)^k,k=m..Infinity) is the tail of a convergent geometric series (since 1/b <1). Note however that we cannot use the same comparison argument if we only know that m=b*|x| with some b²1 ... even though the conclusion of the argument that we just did perform is that Sum((1/b)^k,k=m..Infinity) converges for every m>=0.
    4. Finally we looked at Sum(7^k*(x-2)^k, k=0..Infinity), which we recognized to be a geometric series that converges [absolutely, to 1/(1-7*(x-2))] when -1 < 7*(x-2) < 1 (i.e., when 2-1/7 < x < 2+1/7) and which otherwise diverges.
    One thing that you should take from today's class is that with the current set of mathematical tools that we possess, it is difficult to determine when a power series converges or diverges unless it happens to be a geometric series; we will remedy this next class by developing some powerful tools to help us detect whether a power series is converging or diverging like some geometric series.

  • Friday, December 12. (The last day of class.) We looked at a fragment of a geometric sequence ... 9, 81, 243, 729, ... and found two ways to recognize that the common ratio in this sequence was 3: one was by taking an appropriate root of each term (9^(1/2)=3, 81^(1/3)=3, 243^(1/4)=3, ...), and the other was by looking at the ratios of successive terms (81/9=3, 243/81=3, 729/243=3, ...). We decided that we could build a pair of "geometric behavior detectors" for sequences {a_k} that were not exactly geometric, but which were close to geometric by computing lim_(k->Infinity) |a_k|^(1/k) or lim_(k->Infinity) |a_(k+1)|/|a_k|. Either one of these should give us an "effective common ratio," r of some geometric series that has the same behavior as {|a_k|} out at infinity in the sense that if r<1, then both sequences are converging to zero quickly enough so that their series are convergent, and if r>1, then both sequences are diverging (hence their series diverge by the Divergence Theorem). We illustrated these concepts first by looking at the sequences {(3*k/(k+1))^k} and {3^k*(k/(k+1))}, both of which we were able to show (the first by using limits of roots, and the second by using limits of ratios) behaved asymptotically (i.e., "out at infinity" like the sequence {3^k}. The Root and Ratio Tests were formally stated (see the handout distributed in class on Monday), and then applied to the examples from Wednesday. We used both the Ratio and Root Tests to conclude that Sum(7^k*(x-2)^k, k=0..Infinity) was absolutely convergent when 7*|x-2| < 1 and divergent when 7*|x-2| > 1. (Note that our analysis from Wednesday told us a little more, namely that the series also diverges when 7*|x-2| = 1.) We used the Root Test to see that that Sum(x^k/k^k, k=1..Infinity) was absolutely convergent for every x, and we ended using the Ratio Test to show that Sum(k!*x^k, k=0..Infinity) is divergent for every x not equal to 4 (and convergent for x=4). A final review sheet was distributed at the end of class.

  • Tuesday, December 16. A Review Session will be held in Commons 207 from 4-7 pm.

  • Wednesday, December 17. The Final Exam will be given in ACD 304 from 9:45 - 11:45 am.

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    This page last modified 12/13/97