Additional Material On Recursive Sequences

Penn State Altoona MATH 141 Additional Material On Recursive Sequences 1. 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. () Position a marker at the point (a n, a n ) on the diagonal. The orthogonal projection of this point to either the - or -ais marks the location of the sequence term a n on the real number line. (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 on the real number line. 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. 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. 1

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.. 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. 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.

3 (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). 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:

4 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. 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.

5 = = = 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: 0 1 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.

6 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. 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 = 5 6 + 5 = 0 = 1 or = 5.

7 Step. Identif the location of the initial term relative to the fied points and the domain intervals of f: 1 5 6 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 = = 3 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) = 6 + 4 ( 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, = f() < f() for all 3 < <, which shows that f maps ( 3, ) into (, ).

8 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.

9 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 = 1 + 1 F F n = 1 + 1. n a F n 1 n 1 = 1. This leads to a n+1 = 1 + 1 a n, n = 1,, 3,..., = 1. The function g is thus given b g() = 1 + 1. (b) The first 10 sequence terms are shown in the table below: n a n 1 1/1 = 1 /1 = 3 3/ = 1.5 4 5/3 = 1.6 5 8/5 = 1.6 6 13/8 = 1.65 7 1/13 = 1.615384 8 34/1 = 1.619047 9 55/34 = 1.617647058835941 10 89/55 = 1.618 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 might 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)() = 1 + 1 1 + 1 = + 1 + 1,

10 so f() = +1 +1. 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 = + 1 = + 1 = 0 = 1 ± 5. Step. Locate c 1 and d 1 relative to the fied points and the domain intervals of f: 1 1 5 c 1 1+ 5 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 = 1+ 5. Likewise, {d n } is decreasing, stas in the interval ( 1+ 5, ), and lim d n = 1+ 5. The following cobweb diagrams illustrate the situation. = {a n } = g() = = = f() = f() {c n } {d n }

11 3. 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 + 8 3. ( ), n 1, a n+1 = 1 ( a n + 9 ), n 1, 10. a n = 4 = 15 a n+1 = a { n a n +, n 1, an+1 = 1 + ln(a n ), n 1, 4. = 3 11. = 5 { an+1 = a 3a n, n 1, n+1 = a n + sin(a n), n 1, 5. 1. e an = 1 { an+1 = = 3π 4 a n, n 1, 6. a = 11 n+1 = tan(a n), n 1, { 13. an+1 = a /3 n, n 1, = π 7. = 1 4 a n+1 = 1 a n+1 = a n tan(a n) 1, n 1, 14. 1 + tan (a n ), n 1, 8. + a n a = 5 1 = π a n+1 = 5a 3 ( n 1 ), n 1, a n+1 = a n 9. 3 + a n 15. ln(a n), n 1, = 1 = 1 16. Consider the recursive sequence a n+1 = 4 arctan(a n), n 1, π where is arbitrar. Show that lim a n eists and find it.

1 17. Consider the recursive sequence a n+1 = 3a n a 3 n, n 1, 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. a n+1 = 18 a n, n 1, a n+1 = + 1, n 1, 0. 1. + a n = 4 = 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.

13 4. Solutions = = = f() 1.. = f() 3. f() = 1 ( + 8). Step 1. f has onl one fied point: = 8. Step. Since = 4 we analze f further on (, 8). Step 3. We have f () = 1 > 0 with f(8) = 8. Thus f maps (, 8) into itself. Step 4. We have a = 6 > 4 =. Thus {a n } is increasing with lim a n = 8. 4. f() = +. Step 1. Fied points of f: + = if and onl if = 1 or =. Step. Since = 1.5 we consider f further on (1, ). Step 3. We have f () = = ( 1) > 0 on (1, ) with f(1) = 1 and f() =. Thus f maps (1, ) into itself. Step 4. We have a = 1.5 < 1.5 =, so {a n } is decreasing with lim a n = 1. 5. f() = 3, > 0. Step 1. Fied points of f (for > 0): Onl = 3. Step. Since = 1 consider f further on (0, 3). Step 3. We have f () = 3 > 0 on (0, 3) with f(0) = 0 and f(3) = 3. Thus f maps (0, 3) into itself. Step 4. We have a = 3 > 1 =, so {a n } is increasing with lim a n = 3. 6. f() = 4, < < 1. Step 1. f has onl one fied point: = 6. Step. Since = 11 we consider f further on ( 6, 1). Step 3. We have f 1 () = 4 > 0 on ( 6, 1), f( 6) = 6, and f(1) = 0 < 1. Thus f maps ( 6, 1) into itself. Step 4. We have a = < 11 =, so {a n } decreases with lim a n = 6. 7. f() = /3. Step 1. Fied points of f: = 0 and = 8. Step. Since = 1 we consider f further on (0, 8). Step 3. We have f () = 4 3 1/3 > 0 on (0, 8) with f(0) = 0 and f(8) = 8. Thus f maps (0, 8) into itself. Step 4. We have a = > 1 =, so {a n } is increasing with lim 8. f() = 1 +,. Step 1. Fied points of f: = ± 3. a n = 8.

14 Step. Since = 5 we consider f further on ( 3, ). Step 3. We have f 1 () = (+) > 0 on ( 3, ) with f( 3) = 3. Thus f maps ( 3, ) into itself. Step 4. We have a = 13 7 < 5 =, so {a n } is decreasing with lim a n = 3. 9. f() = 5 3+, 3. Step 1. Fied points of f: = 0 and =. Step. Since = 1 we consider f further on (0, ). Step 3. We have f () = 15 (3+) > 0 on (0, ) with f(0) = 0 and f() =. Thus f maps (0, ) into itself. Step 4. We have a = 5 4 > 1 =, so {a n } is increasing with lim a n =. 10. f() = ( 1 + 9 ), 0. Step 1. Fied points of f: = 3 and = 3. Step. Since = 15 we consider f further on (3, ). Step 3. We have f () = 9 > 0 on (3, ) with f(3) = 3. Thus f maps (3, ) into itself. Step 4. We have a = 39 5 < 15 =, so {a n } is decreasing with lim a n = 3. 11. f() = 1 + ln(), > 0. Step 1. Fied points of f: Onl = 1. Note: = 1 obviousl is a fied point. Since d d (f() ) = 1 1 0 for 1 there are no other solutions to f() = 0. Step. Since = 5 we consider f further on (1, ). Step 3. We have f () = 1 > 0 on (1, ) with f(1) = 1. Thus f maps (1, ) into itself. Step 4. We have a = 1 + ln(5) < 5 =, so {a n } decreases with lim a n = 1. 1. f() = + sin() e. Step 1. Fied points of f: = kπ, k Z. Step. Since = 3π we consider f further on (π, π). Step 3. We have f () = 1 + cos() sin() 1 + = e > 0 for > ln(), so e in particular on (π, π), and f(π) = π and f(π) = π. Thus f maps (π, π) into itself. Step 4. We have a = 3π 1 e 3π/ < 3π =, so {a n } decreases with lim a n = π. 13. f() = tan(), π < < π. Step 1. Fied points of f in ( π, π ): =, = 0, = for a unique π 4 < < π d. Note: Since d ( f()) = 1 1 sec () = 0 onl for = ± π 4 in ( π, π ) there are at most three solutions to f() = 0 in that interval b Rolle s Theorem. Note that = 0 is a solution. There cannot be an other solution in 0 < π 4, and at most one solution in π 4 < < π. Because π 4 f(π/4) > 0 and lim (π/) ( f()) = there is guaranteed one solution with π 4 < < π. Since the function f() is odd, is the third solution. Step. Since = π 4 we consider f further on (0, ). Step 3. We have f () = sec () > 0 on (0, ) with f(0) = 0 and f( ) =. Thus f maps (0, ) into itself. Step 4. We have a = 1 < π 4 =, so {a n } decreases with lim a n = 0. e e

14. f() = tan() 1 1+tan (), π < < π. Step 1. Fied points of f in ( π, π ): Onl = π 4. Step. Since = π 3 we consider f further on ( π 4, π ). Step 3. We have f () = tan()(tan() 1) sec () and lim (π/) f() = π. Thus f maps ( π 4, π > 0 on ( π 4, π ). Moreover, f( π 4 ) into itself. 15 ) = π 4 Step 4. We have a = π 3 3 1 4 < π 3 =. Consequentl, {a n } is decreasing with lim a n = π 4. 15. f() = ( 1 ln()), > 0. Step 1. Fied points of f: Onl = e 1/. Step. Since = 1 we consider f further on (e 1/, ). Step 3. We have f () = 1 ln() > 0 on (0, e 1/ ), and f () < 0 for > e 1/. f does not map the interval (e 1/, ) into itself because lim f() =. In fact, = 1 but a = 1 (0, e 1/ ). But f does map the interval (0, e 1/ ) into itself because f(e 1/ ) = e 1/ and lim f() = 0. 0 + Step 4. We have a 3 = ( 1 1 + ln()) > 1 = a. Consequentl, beginning with the second term, {a n } increases with lim a n = e 1/. 16. f() = 4 arctan() π. Step 1. Fied points: = 1, = 0, and = 1. Step. Depending on, we consider f on (, 1), ( 1, 0), (0, 1), or (1, ). Step 3. We have f () = 4 π(1+ ) > 0 for all. Thus f maps all intervals (, 1), ( 1, 0), (0, 1), and (1, ) into themselves. Step 4. We have f() > on (, 1) (0, 1) and f() < on ( 1, 0) (1, ). Thus lim a n = 1 if > 0, 0 if = 0, 1 if < 0. 17. f() = 3 3. Step 1. Fied points: = 1, = 0, and = 1. Step. Depending on location of we consider f on (, 1), ( 1, 0), (0, 1), and (1, ). Step 3. We have f () = 3 3. f () > 0 on ( 1, 0) and (0, 1) and 0, ±1 are fied points, so f maps each ( 1, 0) and (0, 1) into itself. f () < 0 on (, 1), so f decreases on that interval. We have f( ) = 1 and f( 1) = 1, so f decreases from = 1 to = 1 as sweeps out (, 1). Since = 3 ields f( 3) = 0 we conclude that f maps (, 3) into (0, 1), and f maps ( 3, 1) into ( 1, 0). f () < 0 on (1, ), so f decreases there. Since f(1) = 1 and f() = 1, f decreases from = 1 to = 1 as sweeps out

16 (1, ). For = 3 we have f( 3) = 0. Thus f maps (1, 3) into (0, 1), and f maps ( 3, ) into ( 1, 0). Step 4. We have f() > on (0, 1) and f() < on ( 1, 0). We obtain 1 if (, 3) (0, 3), lim a n = 0 if = 3, 0, 3, 1 if ( 3, 0) ( 3, ). The following cobweb diagrams illustrate some of these cases: = = = f() = f() = = = f() = f() 18. We have f () = 3.9 7.8, so f () = 0 for = 1. We have f(1/) = 3.9/4, so 0 < f(1/) < 1, and f(0) = 0 and f(1) = 0. Thus f maps (0, 1) into itself. 19. We have f () = 3 4 (1 + ln()). So f () = 0 for = e 1/. We have f(e 1/ ) = 3 e 1/, so 0 < f(e 1/ ) < e. We have f( e) = 0 and lim f() = 0 + 0. Thus f maps (0, e) into itself. 0. Let g() = 18. Then a n+1 = g(a n ). We have = 4 and a = 7. Let c n = a n 1 and d n = a n be the subsequences of odd- and even-indeed terms. Let Then { cn+1 = f(c n ), n 1, c 1 = 4, f() = (g g)() = 18 +. 4 { dn+1 = f(d n ), n 1, d 1 = 7. We proceed to analze both sequences {c n } and {d n }. Step 1. Fied points of f: = 6. Step. c 1 = 4 and d 1 = 7, so we consider f separatel on (, 6) (for {c n }) and on (6, ) (for {d n }). Step 3. We have f () = 1 4 > 0, and f(6) = 6. Thus f maps each of the intervals (, 6) and (6, ) into itself.

17 Step 4. We have c = 5.5 > 4 = c 1 and d = 6.5 < 7 = d 1. Thus {c n } increases and {d n } decreases, and we have lim c n = lim d n = 6. We obtain that {a n } converges with lim a n = 6. 1. Let g() = + 1 +. Then a n+1 = g(a n ). We have = 5 and a = 15 7. Let c n = a n 1 and d n = a n be the subsequences of odd- and even-indeed terms. Let f() = (g g)() = 0 + 9 9 + 4. Then { cn+1 = f(c n ), n 1, c 1 = 5, d n+1 = f(d n ), n 1, d 1 = 15 7. We proceed to analze both sequences {c n } and {d n }. Step 1. Fied points of f: = 5 and = 5. Step. c 1 = 5 and d 1 = 15 7, so we consider f separatel on ( 5, 5) (for {d n }) and on ( 5, ) (for {c n }). > 0, and f(± 5) = ± 5. Thus f maps each Step 3. We have f () = 1 (9+4) of the intervals ( 5, 5) and ( 5, ) into itself. Step 4. We have c = 65 9 < 5 = c 1 and d = 75 13 > 15 7 = d 1. Thus {c n } decreases and {d n } increases, and we have lim c n = lim d n = 5. We obtain that {a n } converges with lim a n = 5.. First observe that 1 cos() 1 for all, and 0 < cos() 1 for all 1 1, and finall 0 < cos() < 1 for all 0 < 1. Thus, for whatever value is, we have 1 a 1, 0 < a 3 1, and 0 < a 4 < 1. Note that the cosine function maps (0, 1) into itself, so we guaranteed have 0 < a n < 1 for all n 4. For the purposes of analzing convergence, we therefore ma assume for the further discussion without loss of generalit that 0 < < 1. Let c n = a n 1 and d n = a n be the subsequences of odd- and even-indeed terms. Let f() = cos(cos()). Then c n+1 = f(c n ) and d n+1 = f(d n ) for all n 1. f has onl one fied point, which is also the unique fied point of the cosine function. Note that d d ( f()) = 1 sin(cos ) sin > 0 for all, so there is at most one solution to the equation f() = 0, and we know that the fied point of the cosine function is a solution. f maps both intervals (0, ) and (, 1) into itself. This follows because f () = sin(cos ) sin > 0 for 0 < < 1, and because f( ) =, and 0 < f(0) < as well as < f(1) < 1. We also have f() > on (0, ) and f() < on (, 1). Thus, whatever the values of c 1 and d 1 inside (0, 1) are, we obtain that {c n } and {d n } converge with lim c n = lim d n =. Consequentl, also lim a n =. The cobweb diagrams below illustrate the situation (shown is a case where falls outside of (0, 1)):

18 {a n } = = cos() = = cos(cos ) = = cos(cos ) {c n } {d n } 3. (a) Let g() = f(). Then g(a) = f(a) a = 0, and g (a) = f (a) 1 > 0 b assumption. We have g g(a + h) g(a) g(a + h) (a) = lim = lim > 0. h 0 h h 0 h Thus there eists h 0 > 0 such that g(a+h) h > 0 for all h 0 < h < h 0, h 0. But this implies that g(a + h) > 0 for 0 < h < h 0 and g(a + h) < 0 for h 0 < h < 0. Setting = a+h this entails that g() > 0 for 0 < a < h 0, as well as g() < 0 for h 0 < a < 0. In view of the definition of g this just sas that f() > for a < < a + h 0 as well as f() < for a h 0 < < a which is the first claim. Now let {a n } be recursive with a n+1 = f(a n ) with lim a n = a. Then necessaril a h 0 < a n < a + h 0 for all n that are large enough. We claim that for all these n we must have a n = a. If that were not the case, then for the smallest n such the sequence falls into the interval (a h 0, a + h 0 ) and such that a n a we would either have a < a n < a + h 0 or a h 0 < a n < a. In the first case, we would get that a < a n < a n+1 < a n+ <... < a + h 0 because f() > for a < < a+h 0, and in the second case correspondingl a h 0 <... < a n+ < a n+1 < a n < a because f() < for a h 0 < < a. But this literall means that the sequence moves awa from a and not towards a in contradiction to lim a n = a. Thus a n = a for all n sufficientl large as claimed. (b) Let c n = a n 1 be the subsequence of odd-indeed terms, and d n = a n be the subsequence even-indeed terms. Then c n+1 = g(c n ) and d n+1 = g(d n ) with g() = (f f)(). We have g(a) = a and g (a) = f (f(a))f (a) = (f (a)) > 1. If lim a n = a, then also lim c n = a and lim d n = a. B the first part, we must have c n = a and d n = a for all sufficientl large n. Consequentl, also a n = a for all sufficientl large n. 4. Let g() = f() f (). Then a n+1 = g(a n ). We have g(a) = a. We will proceed to show that a < g() < (< b) for a < < b. Then, if is an arbitrar value in

19 (a, b), the recursive sequence {a n } decreases, stas inside (a, b), and lim a n = a as claimed. We first show that g() < for a < < b, and in particular that a is the onl fied point of g in a < b. B definition of g, g() < is equivalent to f() f () > 0 for a < < b. We have f(a) = 0 and f () 0 for a < b, so either f () > 0 for a < b or f () < 0 for a < b. In the first case, f is increasing and therefore f() > f(a) = 0 for a < < b, and in the second case f is decreasing and thus f() < f(a) = 0 for a < < b. In an case, f and f have the same sign on a < < b which shows that indeed f() f () > 0 for a < < b. We show net that a < g() for a < < b. Because g(a) = a this will follow if g () > 0 for a < < b, since then g is increasing and so g() > g(a) = a for a < < b. Indeed, g () = 1 (f ()) f()f () (f ()) = f()f () (f ()) > 0 for a < < b b our assumption that f() f () > 0 on a < < b.