RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS

Transcription

1 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS THOMAS KRAINER Abstract. This article provides read-to-use supplementar material on recursive sequences for a second semester calculus class. It equips first-ear calculus students with a basic methodical procedure based on which the can conduct a rigorous convergence or divergence analsis of man simple recursive sequences on their own without the need to invoke inductive arguments as is tpicall required in calculus tetbooks. The sequences that are accessible to this kind of analsis are predominantl (eventuall) monotonic, but also certain recursive sequences that alternate around their limit point as the converge can be considered. 1. Introduction Numerical analsis of autonomous ordinar differential equations via time discretization and Runge-Kutta methods, as well as numerical root finding based on Newton s method, lead to recursive sequences. These methods are relevant to man fields outside of mathematics, which matters for teaching first-ear calculus because most students in these classes aim towards a degree in the sciences or in engineering, not necessaril in mathematics. Most standard calculus tetbooks discuss recursive sequences onl ver marginall as an illustration of the Monotonic Sequence Theorem. In the process to establish monotonicit and boundedness of a particular recursive sequence, an inductive argument is tpicall invoked that is based on algebraic manipulations of inequalities and the particular form of the recurrence relation. Following such an argument represents a challenge for man students as it is tpicall their ver first and in most cases it remains their sole eposure to mathematical induction. Even fewer calculus students will be able to work problems out on their own that require inductive arguments. This is b no means surprising because induction, or proof techniques in general, are not central topics in a tpical calculus class. While an attempt of an in-depth treatment of recursive sequences would quickl lead to a full course on dnamical sstems and is therefore unreasonable within the confines of calculus, the question remains what can be done in the calculus classroom to give students more eposure to this kind of sequence, and, in particular, to give them tools that are appropriate for calculus to work out convergence or divergence problems b themselves. In these notes I share some basic material that I developed for use in m own calculus classes to address this question. These notes represent read-to-use supplementar material for a second semester calculus class that is otherwise based on a standard tet such as [4, 7, 8] (our department uses [7]). The notes came about after I unsuccessfull tried to find such material for m classes some ears 010 Mathematics Subject Classification. 97I30. Ke words and phrases. Calculus, recursive sequences, instructional materials. 1

2 THOMAS KRAINER ago. What I could find, then, went in m opinion either too far into a course on dnamical sstems, or became too theoretical, or remained a case-b-case analsis that was lacking a methodical approach. Having used this material for a few ears now, I found that it enables m calculus students to comprehend recursive sequences better, and, most importantl, that it equips them with a basic methodical procedure based on which the can conduct a rigorous convergence or divergence analsis of man simple recursive sequences on their own without the need to invoke inductive arguments. The sequences that are accessible to this kind of analsis are predominantl (eventuall) monotonic, but also certain recursive sequences that alternate around their limit point as the converge, such as the sequence of ratios of subsequent Fibonacci numbers, can be considered. The structure of this article is as follows: Sections and 3 are written in a stle that is directl geared towards calculus students. Section 4 is a compilation and adaptation of homework problems of the kind used in calculus or entr-level dnamical sstems tetbooks as well as pertinent internet resources. Appendi A contains rigorous analtic proofs of the arguments on which the methodical approach presented in Section 3 is based. The appendi is accessible to students attending an introductor real analsis class. Finall, for enrichment or additional guided independent stud for the most enthusiastic calculus students, I should mention that there are man ecellent introductor tets on dnamical sstems that are geared towards undergraduates such as [1,, 3, 5, 6] that are suitable for these purposes.. Graphical Analsis Cobweb Diagrams Consider a generic recursive sequence { an+1 = f(a n ), n = 1,, 3,..., = Given initial value. A cobweb diagram is a visual tool to track the behavior of the sequence {a n }. B definition, each term of the sequence is obtained b evaluating the function f on the previous term. One iterative step of passing from a sequence term a n to the net term a n+1 can be visualized as shown in Figure 1: a n+1 = a n = = f() a n = f() a n+1 a n a n+1 a n+1 a n Figure 1. Two eamples for passing from a n to a n+1. (1) Graph the function = f() and the diagonal = in the same coordinate sstem.

3 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS 3 () Position a marker at the point (a n, a n ) on the diagonal. The orthogonal projection of this point to the - or -ais marks the location of the sequence term a n on the aes. (3) Draw a vertical line that connects the point (a n, a n ) on the diagonal with the function graph = f(), and from that intersection point draw a horizontal line that connects again with the diagonal. The so obtained point on the diagonal has coordinates (a n+1, a n+1 ), and thus the orthogonal projection to the aes ield the position of the net sequence term a n+1. If a n is a fied point for f, i.e. if f(a n ) = a n, then (a n+1, a n+1 ) = (a n, a n ) and the sequence has become stationar. Starting with the initial term, this process can be iterated in the same coordinate sstem and allows for a qualitative visual analsis of the first few sequence terms b following along the path of concatenated vertical and horizontal lines that alternate between the diagonal and the function graph, see Figure. Because the entire information is contained in that concatenated path, the sequence terms are tpicall not separatel tracked on the - and -aes. = = a 5 a 3 a 4 = f() a a a 4 a 3 a 5 = f() Figure. Cobweb diagrams with and without tracking on the aes. As indicated in the right diagram in Figure, cobweb diagrams help us to visualize nicel that the limit of a continuous recurrence must be a fied point of the function f. 3. Limits of recursive sequences Definition 1. A sequence {a n } n=1 is called increasing if a n < a n+1 for all n N, and it is called decreasing if a n > a n+1 for all n N. Sequences that are either increasing or decreasing are also called monotonic. The sequence {a n } is called eventuall increasing if a n < a n+1 for all n large enough, and eventuall decreasing if a n > a n+1 for all n large enough. A sequence is eventuall monotonic if it is either eventuall increasing or eventuall decreasing. Consider a recursive sequence { an+1 = f(a n ), n = 1,, 3,..., = Given initial value. We make the standing assumption that f is continuous and defined on a union of open intervals.

4 4 THOMAS KRAINER We begin b eamining conditions that will insure that the sequence {a n } is monotonic. To this end consider the cobweb diagrams shown in Figure 3. We restrict our attention to the interval (a, b): b = = f() b = f() = lim an = b lim an = a a a b a a b Figure 3. Monotonic recursive sequences. (1) No fied point of f is located between a and b. Consequentl, for a < < b, the graph of = f() lies either entirel above or below the diagonal = as shown. () The function f maps the interval (a, b) into itself. This can be seen in the figure from the fact that the graph of = f() when restricted to a < < b stas entirel inside the interior of the highlighted square with diagonall opposite vertices (a, a) and (b, b). (3) Based on (1) and (), we can conclude that for an choice of initial value (a, b) we obtain a monotonic sequence {a n } that stas inside (a, b). If the graph of = f() lies above the diagonal =, that sequence is increasing with lim a n = b, and if the graph of = f() lies below the diagonal = it is decreasing with lim a n = a. Note that Conclusion (3) remains unchanged under both Conditions (1) and () if a or b (or both) are not fied points of f themselves as depicted in Figure 3, and that a or b ma also take the values ±. Continuit of f on (a, b) is a crucial assumption in an case. Figure 4 shows that both Conditions (1) and () are generall crucial for the validit of the conclusion in (3): The graph of = f() in the left diagram in Figure 4 satisfies Condition () but not Condition (1), while on the right Condition (1) is satisfied but not Condition (). In both cases, the depicted cobweb diagrams each show a sequence that alternates indefinitel between two values, and consequentl that sequence is not monotonic and does not have a limit (if Condition () fails, a sequence with generic initial value in (a, b) will tpicall also leave the interval (a, b) as indicated in Figure 4, see also Figure 5).

5 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS 5 b = f() = b = = f() a a a b a a b a Figure 4. The sequences alternate between and a. These observations lead to the following procedure that allows us to determine whether certain recursive sequences are eventuall monotonic, and to find the limit: Analzing for monotonicit and finding the limit Step 1. Solve the fied point equation f() =. Step. If is itself a fied point, the sequence is constant with a n = for all n, thus lim a n =. Otherwise, use the solutions of the fied point equation in Step 1. and the domain intervals of f to identif an open interval (a, b) such that a and b are fied points or domain interval endpoints (which ma include ± ) such that (a, b), and such that f does not have an fied points in (a, b). Step 3. Check whether f maps the interval (a, b) from Step. into itself. Step 4. If Step 3. reveals that f maps (a, b) into itself : The sequence {a n } is monotonic, stas inside the interval (a, b), and either lim a n = b if the sequence increases, or lim a n = a if it decreases. To verif which case occurs, it is enough to check whether a > (increasing case) or a < (decreasing case). If Step 3. reveals that f does not map (a, b) into itself : If b > a > check whether f maps (, b) into itself, and if a < a < check whether f maps (a, ) into itself. If it does, the above conclusion about monotonicit and the limit remains valid. Otherwise track some of the sequence terms and consider a new recursive sequence that is obtained b choosing among these higher terms (and the same f), and restart the process at Step. If tracking reveals that the original sequence leaves the interval (a, b), choose to be the first term outside of (a, b). Note that the new sequence is obtained from the original one simpl b removing terms at the beginning, thus the limiting behavior and eventual monotonicit for the new sequence are the same as for the original sequence.

6 6 THOMAS KRAINER The cobweb diagrams shown in Figure 3 represent tpical eamples for cases when Step 3. determines that the function f maps the interval (a, b) into itself. Figure 5 shows eamples where this is not the case, but the above procedure still leads to the conclusion that the sequence is eventuall monotonic and convergent. However, as the depicted eample on the right in Figure 4 shows, one cannot epect that this process will generall lead to a successful conclusion. = = = f() = f() = = f() Figure 5. Eventuall monotonic recursive sequences. Eample 1. Consider the recursive sequence a n+1 = a an n, n = 1,, 3,..., = 1. Show that {a n } converges, and find lim a n. Solution: Consider f() =, > 0. Then a n+1 = f(a n ) for all n = 1,, 3,... Step 1. Find the fied points: = ln() = ln() ln()( 1) = 0 = 1. Step. Identif the location of the initial term relative to the fied points and the domain intervals of f:

7 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS We continue to analze the sequence further on (0, 1). Step 3. f maps (0, 1) into itself: We have 0 < f() = = e ln() < e 0 = 1 for 0 < < 1. Note that ln() < 0 for 0 < < 1, so ln() < 0 for 0 < < 1. Step 4. From the previous steps we can conclude that the sequence {a n } is monotonic and stas inside the interval (0, 1). Because a = 1/ > 1/ =, the sequence is increasing, and we have lim a n = 1. An important component of the above procedure to analze recursive sequences for eventual monotonicit is to verif whether a continuous function f maps an interval (a, b), a < b, into itself, see Steps 3. and 4. This reall is the question about how large or small the values f() can become as varies over (a, b). Note that if a and b are both finite and f etends continuousl to [a, b], then answering this question is just what the Closed Interval Method accomplishes. In the general case we can still proceed ver similar to the Closed Interval Method: If all the local minima and maima of f as varies over (a, b) fall inside the range (a, b), and if the limits of f towards both interval ends eist and neither overshoot the value b nor undershoot the value a, then f maps (a, b) into itself. This leads to the following: Strateg for checking whether f maps (a, b) into itself We assume that f is continuous on (a, b) and that both lim f() eist, where each is either finite or ±. b A. Find all critical points of f in (a, b). B. Find lim f() and lim f(). a + b lim f() and a + Note that if f etends continuousl to a or b (if finite), the corresponding endpoint limit will just be f(a) or f(b), respectivel. C. Tabulate the function values of f at the critical points, and both limits. If the values at all the critical points fall inside (a, b), and the limits are both a and b, then f maps (a, b) into itself. Otherwise it does not. A Special Case: If f is increasing on (a, b) and etends continuousl to the finite endpoints of (a, b) (if an) with f(a) a as well as f(b) b, respectivel, then f maps (a, b) into itself. Note that if a or b are fied points, then the inequalities pertaining to the function values are satisfied. Recall that to check whether f is increasing on (a, b) it is sufficient to verif that f () > 0 for all a < < b. Eample. Consider the recursive sequence a n+1 = 5, n = 1,, 3,..., 6 a n = 4. Show that {a n } converges, and find lim a n.

8 8 THOMAS KRAINER Solution: Consider the function f() = 5 6. Observe that a n+1 = f(a n ) for all n = 1,, 3,... Step 1. Solve the fied point equation: 5 6 = 6 = = 0 = 1 or = 5. Step. Identif the location of the initial term relative to the fied points and the domain intervals of f: We thus analze the sequence further on the interval (1, 5). Step 3. f maps the interval (1, 5) to itself: Because both 1 and 5 are fied points, it is enough to show that f is > 0 for all 6. Step 4. Based on the previous three steps, we conclude that the sequence {a n } is monotonic and stas in the interval (1, 5). Since a = 5 < 4 = the sequence is decreasing, and we have lim a n = 1. increasing on (1, 5). But this is obvious in view of f () = 5 (6 ) Eample 3. Consider the recursive sequence a n+1 = a n, n = 1,, 3,..., a n 3 = An value other than 3. Show that {a n } converges, and find lim a n. Solution: Consider the function f() = 3. Then a n+1 = f(a n ) for all n = 1,, 3,... Step 1. Find the fied points: 3 = = = 0 = 1 or =. Step. The further analsis will depend on the location of relative to the marked points on the real number line shown: 1 3 We proceed with the process, being mindful that we have to distinguish several cases. Step 3. We begin b analzing where f increases and decreases: f () = ( 3) ( ) ( 3) = ( 1)( ) ( 3) = ( 3). From the formula we deduce that f () > 0 for < < 1 and < <. Since both = 1 and = are fied points, we get that f maps (, 1) into itself, and that f maps (, ) into itself. Because f () < 0 for 1 < < 3 and 3 < <, we deduce that f is decreasing on [1, 3 ) and on ( 3, ]. Consequentl, 1 = f(1) > f() for all 1 < < 3, which shows that f maps (1, 3 ) into (, 1). Likewise,

9 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS 9 = f() < f() for all 3 < <, which shows that f maps ( 3, ) into (, ). Step 4. From the previous steps, we can infer the following: If < < 1, the sequence {a n } is monotonic and stas inside the interval (, 1). Because f(0) = /3 > 0 we have that f() > on (, 1), and consequentl the sequence {a n } is increasing with lim a n = 1. If < <, the sequence {a n } is monotonic and stas inside the inverval (, ). Because f(3) = 7 3 < 3 we have that f() < on (, ), and consequentl the sequence {a n } is decreasing with lim a n =. If = 1 then a n = 1 for all n, and consequentl lim a n = 1. Likewise, if = then a n = for all n, and thus lim a n =. If 1 < < 3, then a = f( ) will be in the interval (, 1). Consequentl, starting from the second sequence term a, the sequence will increase with lim a n = 1. If 3 < <, then a = f( ) will be in the interval (, ). Thus, starting from the second sequence term, the sequence decreases with lim a n =. In conclusion: lim a n = { 1 if < 3, if > 3. Eample 4. The Fibonacci numbers are the sequence {F n } n=0 defined as F n+1 = F n + F n 1, n = 1,, 3,..., F 0 = 0, F 1 = 1. The sequence {a n } n=1 of ratios of subsequent Fibonacci numbers is given b a n = F n+1, n = 1,, 3,.... F n (a) Show that {a n } n=1 is a recursive sequence. Determine eplicitl a function g such that a n+1 = g(a n ) for all n = 1,, 3,... (b) Find,..., 0. Is the sequence {a n } monotonic? What if we consider separatel, a 3, a 5, a 7, a 9 (just the odd terms) or a, a 4, a 6, a 8, 0 (just the even terms)? (c) Define c n = a n 1 and d n = a n, n = 1,, 3,... Then {c n } and {d n } are subsequences that consist of the odd- and even-indeed terms of {a n }, respectivel. Show that both {c n } and {d n } are recursive sequences b determining eplicitl a function f with c n+1 = f(c n ) and d n+1 = f(d n ) for all n = 1,, 3,... (d) Show that both {c n } and {d n } converge to the same limit ϕ. Since the odd- and even-indeed subsequences of {a n } both converge to the same limit ϕ, we conclude that {a n } also converges with lim a n = ϕ. The limit ϕ, which we will find in the solution to be 1+ 5, is known as the Golden Ratio or the Golden Mean.

10 10 THOMAS KRAINER Solution: (a) Using the recurrence relation for the Fibonacci numbers we obtain a n = F n+1 F n Note that = F F 1 = F n + F n 1 F n = F1+F0 F 1 = 1 + F n 1 = F F n = n a F n 1 n 1 = 1. This leads to a n+1 = a n, n = 1,, 3,..., = 1. The function g is thus given b g() = (b) The first 10 sequence terms are shown in the table below: n a n 1 1/1 = 1 /1 = 3 3/ = /3 = /5 = /8 = /13 = /1 = /34 = /55 = The sequence {a n } is not monotonic. Instead, it seems to alternate: The table suggests that the successor to an odd-indeed term ma be greater, while the successor to an even-indeed term ma be smaller. Moreover, from just tracking the even- and odd-indeed terms separatel, it appears that the evenindeed terms decrease, while the odd-indeed terms appear to increase. We shall see below that this is indeed the case. (c) We have: c 1 = d 1 = a = g( ) c = a 3 = g(a ) = (g g)(c 1 ) d = a 4 = g(a 3 ) = (g g)(d 1 ) c 3 = a 5 = g(a 4 ) = (g g)(c ) d 3 = a 6 = g(a 5 ) = (g g)(d ) c 4 = a 7 = (g g)(c 3 ) d 4 = a 8 = (g g)(d 3 ) c 5 = a 9 = (g g)(c 4 ) d 5 = 0 = (g g)(d 4 ) { { cn+1 = (g g)(c n ) dn+1 = (g g)(d n ) c 1 = d 1 = a We see that {c n } and {d n } are recursive sequences with the function g g. Observe that this is generall true and not limited to the particular sequence {a n } considered in this eample. In the specific case considered here we have (g g)() = = ,

11 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS 11 so f() = We have c n+1 = c n + 1, n = 1,, 3,..., c n + 1 c 1 = 1 d n+1 = d n + 1, n = 1,, 3,..., d n + 1 d 1 = (d) We analze {c n } and {d n } analogousl to the previous eamples. Step 1. Find the fied points of f: = + 1 = + 1 = 0 = 1 ± 5. Step. Locate c 1 and d 1 relative to the fied points and the domain intervals of f: c d 1 We consequentl proceed to analze {c n } further on ( 1 5, 1+ 5 ), and {d n } on ( 1+ 5, ). Step 3. f maps both intervals ( 1 5, 1+ 5 ) and ( 1+ 5, ) into themselves: Since both 1± 5 are fied points, it is enough to show that f is increasing on both intervals. But since f () = 1 (+1) > 0 for all 1 this is clear. Step 4. Based on the previous steps we conclude that {c n } is increasing, stas in the interval ( 1 5, 1+ 5 ), and lim c n = Likewise, {d n } is decreasing, stas in the interval ( 1+ 5, ), and lim d n = The following cobweb diagrams illustrate the situation. = {a n } = g() = = = f() = f() {c n } {d n }

12 1 THOMAS KRAINER 4. Problems Draw a cobweb diagram for the recursive sequence defined b a n+1 = f(a n ), where the graph of f is shown and is marked on the -ais. Show at least a,..., a 7. = f() 1. = f(). Show that the following sequences converge, and find the limit. a n+1 = 1 an ( ), n 1, = 4 a n+1 = a n a n +, n 1, 4. = 3 { an+1 = 3a n, n 1, 5. = 1 { an+1 = 4 a n, n 1, 6. = 11 { an+1 = a /3 n, n 1, 7. = 1 a n+1 = 1, n 1, 8. + a n = 5 a n+1 = 5a n, n 1, a n = Consider the recursive sequence a n+1 = 1 ( a n + 9 ), n 1, 10. a n = 15 { an+1 = 1 + ln(a n ), n 1, 11. = 5 a n+1 = a n + sin(a n), n 1, 1. e an = 3π a n+1 = tan(a n), n 1, 13. = π 4 a n+1 = a n tan(a n) tan (a n ), n 1, = π 3 ( 1 ) a n+1 = a n 15. ln(a n), n 1, = 1 a n+1 = 4 arctan(a n), n 1, π where is arbitrar. Show that lim a n eists and find it. 17. Consider the recursive sequence a n+1 = 3a n a 3 n, n 1,

13 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS 13 where < < is arbitrar. Show that lim a n eists and find it. Show that the function f maps the interval I into itself. 18. f() = 3.9(1 ), I = (0, 1). 19. f() = 3(1 ln()), I = (0, e). 4 Use the approach from Eample 4(c),(d) to show that the following sequences converge, and find the limit. 0. a n+1 = 18 a n, n 1, = 4 1. a n+1 = + 1, n 1, + a n = 5. Show that for ever the recursive sequence defined b a n+1 = cos(a n ), n 1, converges to the unique fied point of the cosine function. 3. Repelling fied points: Let f be differentiable, and let f(a) = a. (a) Let f (a) > 1. Show that there is h 0 > 0 such that f() > for a < < a + h 0, and f() < for a h 0 < < a. Deduce that if {a n } is recursive with a n+1 = f(a n ), and if lim a n = a, then necessaril a n = a for all n large enough. Hint: Consider g() = f(). Then g(a) = 0 and g (a) > 0 (wh?). Then use the limit definition of the derivative to derive the stated inequalities. Finall, note that if a < a n < a + h 0, then a n+1 > a n > a, and if a h 0 < a n < a then a n+1 < a n < a, so the sequence moves awa from a. (b) Now let f (a) < 1. Use the first part and the approach from Eample 4 to show that if {a n } is recursive with a n+1 = f(a n ), and if lim a n = a, then a n = a for all n large enough. 4. Let f be twice differentiable. Let f(a) = 0. Suppose that f () 0 for a < b, and f() f () > 0 for all a < < b. Consider the sequence {a n } defined b Newton s method a n+1 = a n f(a n) f (a n ), n 1, = An value in (a, b). Show that {a n } stas in (a, b), and is decreasing with lim a n = a.

14 14 THOMAS KRAINER Appendi A. Rigorous Proofs Theorem 1. Let I = (a, b) be an open interval, where a < b, and let f : I R be continuous such that f() for all I. Then either (A) f() > for all I, or (B) f() < for all I. Now suppose further that f(i) I, and consider the recursive sequence { an+1 = f(a n ), n = 1,,..., = arbitrar initial value in I. This sequence is well-defined and contained in the interval I, and in Case (A) it is increasing with lim a n = b, and in Case (B) it is decreasing with lim a n = a. Proof. Consider the function g : I R defined b g() = f(). Then g is continuous, and b assumption we have g() 0 for all I. Therefore, b the Intermediate Value Theorem, we have either g() > 0 for all I which leads to Case (A), or g() < 0 for all I which leads to Case (B). Net consider the sequence {a n } n=1: We will first show inductivel that a n I for all n N, and then that {a n } n=1 is either increasing or decreasing depending on whether Case (A) or Case (B) holds for the function f. B assumption, the initial value belongs to I. So suppose that we know that a n I for some n N. Because f(i) I we obtain that a n+1 = f(a n ) I. This completes the induction and shows that the sequence {a n } n=1 is well-defined and contained in I. Moreover, for all n N we have a n+1 = f(a n ) > a n in Case (A), and a n+1 = f(a n ) < a n in Case (B), proving the claim about the monotonicit of the sequence {a n } n=1. It remains to prove the statement about the limit. Suppose that Case (A) holds for the function f, so the sequence {a n } n=1 is increasing. B the Monotonic Sequence Theorem, we have that lim a n = L = sup{a n n N}, which is either finite or infinite. Because a n < b for all n N, we obtain that necessaril L b. The assumption that L < b leads to a contradiction, as follows: As L < b, we necessaril have L I, and b continuit of the function f at the point L I we get L = lim a n+1 = lim f(a n) = f(l). B our stated assumption f does not have an fied points in I, so this is impossible. Consequentl, b = L. The argument in Case (B) is the same, thus concluding the proof of the theorem. Remark 1. Note that neither the statement of Theorem 1 nor its proof make an assumptions nor conclusions about the monotonicit of the function f : I R itself. The monotonicit assertion of the theorem eclusivel pertains to the sequence {a n } n=1. To clarif this further b means of an eample consider the function f() = ( 1 sin ) on the interval I = (0, 1 ). f is continuous (even C ) on I. Let us verif that 0 < f() < for all I, i.e., the function f satisfies the assumptions of Theorem 1

15 RECURSIVE SEQUENCES IN FIRST-YEAR CALCULUS 15 on the interval I, and Case (B) of the theorem applies: Because 0 < < 1 we see that ( 1 f() = 0 sin = ) 1 > 1, and thus there are no solutions to the equation f() = 0 in I. As 1 π I with f( 1 π ) = 1 π > 0 we necessaril obtain that f() > 0 on I b the Intermediate Value Theorem. Analogousl, using again that 0 < < 1, we obtain that ( 1 f() = sin = ) 1 < 1, and consequentl the equation f() = has no solutions in I either. Because f( 1 π ) = 1 π < 1 π we get that f() < on I from the Intermediate Value Theorem. Theorem 1 now implies that ever recursive sequence a n+1 = f(a n ) with arbitrar initial value 0 < < 1 is decreasing with lim a n = 0. The function f itself, however, is not monotonic in an neighborhood of the origin. This follows because f () = 1 ( 1 ) ( 1 sin + cos ) changes sign infinitel man times on ever interval of the form (0, ε) for ever 0 < ε < 1 as can be seen, for eample, from the fact that f ( 1 kπ ) = 1 + ( 1)k for all k N. For an illustration of this eample see Figure 6. = = f() Figure 6. An illustration of the eample discussed in Remark 1. Theorem. Let (a, b) be an open interval, where a < b, and let f : (a, b) R be continuous. Let α < β. Then the following are equivalent: (1) f maps (a, b) into (α, β). () For all critical points ξ of f in (a, b) we have α < f(ξ) < β. Moreover, α lim sup a + lim inf f() β. b f() and lim inf a + f() β, as well as α lim sup f() and b Proof. If f maps (a, b) into (α, β), all the claimed properties about the function values at critical points and the limits towards the interval ends are trivial. Hence suppose that α < f(ξ) < β for all critical points ξ (a, b), and that α lim sup f() a +

16 16 THOMAS KRAINER and lim inf a + f() β, as well as α lim sup b f() and lim inf f() β, hold. We b have to show that f maps (a, b) into (α, β). Assume that this was not the case. Then there eists 0 (a, b) such that either f( 0 ) β or f( 0 ) α. Without loss of generalit, we consider onl the case that f( 0 ) β since the argument in the other situation is analogous. We claim that there must eist a < c < 0 and 0 < d < b such that f(c) < f( 0 ) and f(d) < f( 0 ). Once this is warranted we obtain, using the Etreme Value Theorem, that M = ma{f() c d} f( 0 ) β, and there must eist c < ξ 0 < d with f(ξ 0 ) = M. But this implies that f attains a local maimum at ξ 0, and consequentl ξ 0 must be a critical point for f. B our present assumption we therefore have f(ξ 0 ) = M < β, which contradicts with M β. We need to establish the claim about the eistence of c and d with the stated properties. If there eists a critical point of f in (a, 0 ), choose c to be such a critical point. B our present assumption on the function values at critical points, we then have f(c) < β f( 0 ) as desired. If there is no critical point of f in (a, 0 ), we have that f eists everwhere and vanishes nowhere on (a, 0 ). B the Darbou Theorem, f has the intermediate value propert, so either f > 0 or f < 0 everwhere on (a, 0 ). The case that f < 0 is impossible as this would impl that f strictl decreases on (a, 0 ], which would entail that lim f() eists a + with lim f() > f( 0 ) β contrar to our present assumption on that limit. a + Consequentl, f > 0 on (a, 0 ), which shows that f is strictl increasing on (a, 0 ]. Hence we can pick c arbitraril with a < c < 0 and have f(c) < f( 0 ) as desired. The eistence of the point d with 0 < d < b and f(d) < f( 0 ) follows in the same manner. References [1] R. Devane, An Introduction to Chaotic Dnamical Sstems, Westview Press, 003. [] R. Holmgren, A first course in discrete dnamical sstems, nd Edition, Universitet, Springer, 000. [3] Y. Pesin and V. Climenhaga, Lectures on Fractal Geometr and Dnamical Sstems, Student Mathematical Librar, vol. 5, American Mathematical Societ, 009. [4] J. Rogawski, Calculus: Earl Transcendentals, Second Edition, W. H. Freeman, 011. [5] E. Scheinerman, Invitation to Dnamical Sstems, Dover Publications, 01. [6] S. Sternberg, Dnamical Sstems, Dover Publications, 010. [7] J. Stewart, Calculus: Earl Transcendentals, Seventh Edition, Cengage Learning, 010. [8] G.B. Thomas Jr., M.D. Weir, and J.R. Hass, Thomas Calculus, Single Variable, 1th Edition, Pearson, 009. Penn State Altoona, 3000 Ivside Park, Altoona, PA address: krainer@psu.edu