ESSAY 1 : EXPLORATIONS WITH SOME SEQUENCES

Transcription

1 ESSAY : EXPLORATIONS WITH SOME SEQUENCES In this essay, in Part-I, I want to study the behavior of a recursively defined sequence f n =3.2 f n [ f n ] for various initial values f. I will also use a spreadsheet to display a table and graph for a sufficient range of n. In particular, I will investigate the behavior of f n for f in the range of f =.6875 In Part-II, I will investigate a slightly different sequence which is also recursively defined: f n =3.83 f n [ f n ]. We will see the results of this slight difference in comparing these two sequences. Finally in Part-III, I will investigate a sequence with a varying constant defined as: f n = A f n 2, with A 2 and f PART ONE Let's first change the notation a little bit: =3.2 Now this has a more compact form. This is one recursively defined sequence, therefore, each and every term of the sequence depends on the initial value a. Before attempting any theoretical work, it would be good to look at the behavior of the sequence for different a with a = n

2 with a = n with a = n

3 with a = n Generation of for different a whispers us some obvious results: () When a is negative, the sequence is divergent, therfore, not of our interest. (2) When a is greater than (without equality), the sequence is divergent again. (3) When a =, we have a constant sequence =. (4) When a =.6875, once again we have a constant sequence =.6875 (5) When a =, all = except for n=. Therefore, the sequence can be considered as a (almost) constant sequence. (6) When a, {.6875}, turns into an alternating sequence: For large n, the subsequence a 2 n has limit.53 whereas the subsequence a 2 n has limit.7994 These results may be summarized in the following table: a Convergent/Divergent Constant/Alternating Negative Not exist Divergent Neither > Not exist Divergent Neither Zero Zero Convergent Constant One Zero Convergent (Almost) Constant Convergent Constant a, {.6875} Exists Divergent Alternating

4 THEORETICAL WORK: In this paragraph, I will assume the more interesting case a, for the recursively defined sequence: =3.2, n N. It would be productive if I write each term in order as follows: a =3.2 a a, a 2 =3.2 a a, a 3 =3.2 a 2 a 2, a 4 =3.2 a 3 a 3, a 5 =3.2 a 4 a 4, =3.2. If I multiply the n equations side by side, and cancel the same terms, I can write the general term of the sequence in terms of every other term of the sequence as follows: n =3.2 n a a i. i= In this form, however, we might be mislead by the thought that the term 3.2 n is divergent. In fact, each term in the product sign, a i, has absolute value less than, therefore, the product n i= a i goes to zero as n goes to. Therefore, the fact that 3.2 n is divergent should not mislead us. There are some limit theorems which I will use below to explain the situation here. Now if we insert the term 3.2 n inside the product symbol, we realize that the general term of the sequence can be written as =a 3.2 a i. i= Now it is more meaningful to examine the term 3.2 a i rather than a i. First of all, for such a product to make sense, we would expect that it converges. And this will happen if and only if the absolute value of each term in the product symbol is less than. In other words, we require that: 3.2 a i. If we take one more step and divide each side by 3.2, we obtain: a i 3.2. In other words, a i.325. Surprisingly now, this becomes an inequality problem which the writer of this essay was not expecting at all. Since the absolute value of a i is less than.325, in other words, its distance form is less than.325, the term itself must lie somewhere in between n

5 -.325 and.325. Using a compound inequality, this can be written as:.325 a i.325. We are trying to solve for a i. Therefore, we may first multiply everything by -. We must of course keep in mind that the inequality symbols are reversed by multiplication by egative number. Namely:.325 a i.325. Finally, if we add to everything, we get:.325 a i.325. Equivalently:.6875 a i.325. Using interval notation, the solution set can be written as:.6875,.325. However, when we look at the results generated by spreadsheet above, we realize that the solution set here is not suitable for some of the terms. For instance, almost every term of the subsequence a 2 n lies in this interval, however, only few of the terms of the subsequence a 2 n are in this interval. Therefore, we realize that an analysis of this particular recursively defined sequence's general term in the way we described above yileds irrelevant results. This, I believe, is due to the fact that the general term of this particular sequence is not only recursively defined, but depends on the PRODUCT of the two most previous terms. We could, however, apply the strategy we used above for an ordinary sequence if want to prove or disprove its convergence. For example, if we define an ordinary sequence as a = 2, a 2 = 2 3, a 3 = 3 4,a 4 = 4 5,..., = n n, we realize that the absolute value of each term is less than. We also realize that it is convergent and has limit. Now, we may use this result to define another sequence as follows: b = 2,b 2=b 2 3,b 3=b 2 3 4,...,b n=b n n n. Observe that the sequence we defined here is recursive. Hence in order for b n to converge, we require that the absolute value of the general term b i be less than for each i. In fact, if we list some of the terms of this sequence, we realize that it is convergent and its limit is :

6 b n ={, 2, 2 3, 3 4, 4 5, 5 6,...}. Moreover, let us consider the defining equation b n =b n n n. Now, if the sequence has a limit, then b n and b n must have the same limit as n. Now since n n as n, b n has a limit. If we also remember from the paragraph above that = n n our results as follows: b n HAS A LIMIT BECAUSE = AS n. n n The following picture summarizes this part: a 2n.53 a 2n.53, we can summarize.6875 a 2n a 2n The following figures are MATLAB generated plots for different values of a =,.,.2,...,.9,

7 PART TWO Now we have =3.83 This is similar to the sequence defined in part one above. Once again each and every term of the sequence depends on the initial value a. Before attempting any theoretical work, let us once again see how the sequence behaves for different a with a = n

8 with a = n

9 with a = n with a =-/3.83=283/383= n

10 Imitating what we did in part one, we can say that generating for different a yields the following: () When a is negative, the sequence is divergent, therfore, not of our interest. (2) When a is greater than (without equality), the sequence is divergent again. (3) When a =, we have a constant sequence =. (4) When a = , once again we have a constant sequence = (5) When a =, all = except for n=. Therefore, the sequence can be considered as a (almost) constant sequence. (6) When a, { }, turns into an alternating sequence: For large n, the subsequence a 3n has limit whereas the subsequence a 3 n has limit and the subsequence a 3n 2 has limit Once again these results may be summarized as we did in part one: a Convergent/Divergent Constant/Alternating Negative Not exist Divergent Neither > Not exist Divergent Neither Zero Zero Convergent Constant One Zero Convergent (Almost) Constant Convergent Constant a, { } Exists Divergent Alternating THEORETICAL WORK: In this paragraph, I will assume the more interesting case a, for the recursively defined sequence: =3.83, n N. Multiplying the n equations side by side, and canceling the same terms, we can write the general term of the sequence in terms of every other term of the sequence as: n =3.83 n a a i. i= Inserting 3.83 n inside the product symbol, the general term of the sequence can be written as n =a i= 3.83 a i. Now it is more meaningful to examine

11 the term 3.2 a i rather than a i. If we require convergence, we have a similar inequality as obtained in part one: 3.83 a i. Solving this inequality problem, we get: 3.83 a i Equivalently: a i Using interval notation, the solution set can be written as: , Now almost every term of the subsequence a 3n 2 lies in this interval, however, only few of the terms of the subsequences a 3 n and a 3 n are in this interval. This remark once again backs up our conclusion from part one that an analysis of this particular recursively defined sequence's general term in the way we described above yileds irrelevant results perhaps because the general term of this particular sequence is not only recursively defined, but depends on the PRODUCT of the two most previous terms. The following picture summarizes this part: a 3n.56 a 3n.56 a 3n a 3n+.54 a 3n a 3n+2.957

12 The following figures are MATLAB generated plots for 5 different values of a =,.2,.4,...,.98,

13 PART THREE Now we investigate a sequence with a varying constant defined as: = A 2, with A 2 and a Now, I will follow a different approach. Let's not worry about a for the moment. Instead, let's do something else. 2, Let's assume that the sequence = A is well defined and has a limit, say, x. Then we must be able to argue that as n goes to, then both and go to. Therefore, we can write an equation for x: x= A x 2. The solutions of this equation are some questions: ± 4 A 2 A This brings Q. Are the solutions well defined? A. The term in the square root sign is positive for all A (remember <A<2) therefore the solutions are well-defined. Q2. How do they look like? A2. Please go to the site to download the following GSP animation file: Animate Parameter A between -.25 and 2 When we animate parameter A, point F traces trajectory of (A, f(a)) point G traces trajectory of (A, g(a)) 4 2 F A =.4 f(a)= g(a)= A =.77 2 A A = A -5 5 Question: Why to animate A in the interval (-.25, 2) only? -2 Answer: The functions f and g are not defined for A < G Q3. Is there an interesting value of A (remember < A < 2) for which what we choose as for the initial value a (remember -< a <) does not matter at all? A3. What do you expect? What is your guess? The following explains this in details.

14 When we look at the solutions of the equation x= A x 2, we realize that the square root term vanishes for A=.25. In that case, we have a double root and it is We might expect that in that special case, the initial value does not affect the limit of the sequence. Is that true? a = a =.5 a = a =.5 a = It looks like when A=.25, it does not matter what we choose for a all cases, the limit is 2 2. for in Since -< a <, let use look at some graphs for different A values:

15 Example: a =, with 9 different values of A=,.25,.5,.75,,.25,.5,.75, The first (A=) and the last (A=2) are not very interesting. These are (almost) constant sequences. When A=.25, (second figure on the first row) the sequence has a limit f.25 = , when A=.5 (third figure on the first row) the sequence has a limit f(.5)=.73, and when A=.75 (first figure on the second row) the sequence has a limit f(.75)=.67. In fact, in all these cases, the limit of the sequence is given by f A = 4 A 2 A The first figure on the third row is interesting (A=.5), though. The sequence does not converge and for big n, it visits different values as I understand form the spreadsheet. The second figure on the third row (A=.75) is also interesting, where in that case the sequence does not converge and it alternates between two different values for big n. I want to do a similar analysis for the other a o values. Please go to the site to download the MATLAB function file that I used to generate these plots.

16 Example: a =.5, with 2 different values of A=,.,.2,...,.9, It loks like when A is small, the sequence converges quickly. But when we look at the third figure on the third row (A=.8), the sequence alternates between two values.9 and.345. When A increases, the sequence alternates among 3 or even 4 different values for big n. (See figures on the sixth row). This happens to be the case until the last figure for which A=2 (the sequence is almost constant) I understand that the pattern is independent of the choice of initial a value. Therefore, the question is how the graph of the sequence depends on the parameter A? Is it enough to plot graphs for 2 values of A, or we should look at more tha 2 values? The following example answers these questions.

17 Example: a =, with 4 different values of A=,.25,.5,.75,...,.975, I would say, no matter how many times we partition the interval (,2), it looks like we get the same pattern: When A= and A=2, the sequence is not very interesting: constant sequence. For A values less than.75, the sequence seems to converge quickly, and has a limit which is given by f(a) For A values between.8 and.5, the sequence seems to alternate between 2 different values and is divergent. For A values between.5 and 2, the sequence seems to alternate even among 3 or 4 different values and is divergent. I want to study two more examples with more A values and complete this part with a conclusion.

18 Example: a =.5, with 5 different values of A=,.2,.4,.6,...,.98,

19 Last Example: a =, with different values of A=,.,.2,.3,...,.99,

20 CONCLUDING PART III. The behavior of the sequence is independent of the choice of the initial value a, 2. The behavior of the sequence is independent of the number of times we partition the interval A,2 3. When A= and A=2, the sequence is not very interesting: constant sequence. 4. For all A values less than a value close to.75 (I could not determine it exactly), the sequence converges quickly, and the limit is given by f A = 4 A 2 A 5. For all A values between.8 and.5 (I could not determine these values exactly), the sequence alternates between 2 different values and is divergent. 6. For A values between.5 and 2, the sequence alternates among 3 or 4 different values and is divergent.