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Given the sequence defined by the recursive relation a[n+1] = r*a[n](1-a[n])
You need to use the procedure iterate.
Throughout this problem you should choose initial values in the interval 0<a0<1.
(a) Let r=3/2. Calculate a moderate number of terms in the sequence (between 10 and 20). Does the sequence appear to be converging? If so to what value? Does the limit depend upon your choice of initial value? Plot the terms you have calculated
(b) Let r=2.8. Calculate a moderate number of terms in the sequence (between 10 and 20). Does the sequence appear to be converging? If so to what value? Does the limit
depend upon your choice of initial value? Plot the terms you have calculated How does this sequence differ from that in part (a).
(c) Let r=3.2. Calculate a moderate number of terms in the sequence (between 10 and 20). Show that the sequence does not appear to converging. Plot the terms you have calculated and describe how the sequence behaves in this case.
(d) Consider intermediate values between 2.8 and 3.2 to determine more precisely where the transition in behaviour takes place. Provide a few plots (no more than 4) showing the values you have investigated.
(e) Consider the values of r in the range 3.43<r<3.46. Determine as accurately as you can the value of r for which the period of oscillation doubles.
(f) As r increase further period doubling occurs. Try to find the when the sequence appears to oscillate between 8 values.
(g) Let r =3.65 and calculate a considerable number of terms (at least a few hundred) and plot your values.
(h) For r=3.65 choose a0=0.3 and then a0=0.301. Find and plot some terms in the sequence for each initial value. Determine how long the terms in the two sequences remain close together and when they begin to depart significantly from each other.

Dear all;

I need you help for solving this problem, and thanks in advantage for your help.

I have a polynom like  P =x^6-4*x^3+x-2;  and i would like to find an approximate value of the roots in some interval [a,b] =[-2,2] using sturm sequence.

The method is based on:

1) first construct the sturm sequence:

For given polynom P =x^6-4*x^3+x-2;

Let S0=P;


let   s:=quo(S0,S1,x);

.... S[k+1-rem(S[k-1],S[k]);


S[k] is the sturm sequence.

2) let f(a)= number of change of sign in the sturm sequence and f(b) the same . so f(b)-f(a) give the number of roots in the interval [a,b].

3) If f(b)-f(a) =0 so there are no roots

and if f(a)-f(b)=1 one can find the root

4) if f(a) -f(b) >2  :

given toterance tol=0.001; for example

if the abs(a-b)<2*epsilon we display a message that there are k roots at (b+a)/2

with our error tolerance

5) otherwise if c=(b+a)/2 is not a root of P_k(x)  for any k, ( where p_k is an element of the sturm sequence ) 

we divide the interval into equal halves [a,c] and [x,b] and we run step 2 on each interval

else if c is a root of one of these p_k(x) add any time account to c so that c lies close the middle of [a,b] and not a root

6) Give all the roots ( approximate the rrots with small error epsilon).


I kindly  appreciate your help




I need to find all combinations of a binary list of length 3.  For example: [0,0,0] , [1,0,0] , [0,1,0] , [0,0,1] , [1,1,0] , [1,0,1] , [0,1,1], [1,1,1].  Does anyone have a procedure or can write a loop that does this?

i recall having asked this question before and having received answers which unfortunately I cannot locate.

I wish to generate a sequence like
    a[1], b[1], a[2], b[2], a[3], b[3]

The following does not work for the obvious reason:

    seq(a[i], b[i], i=1..3);

What is the right way?




Bellow I have an example of my current problem:

I want to put a sequence of numbers in the place of Tau in the same command, so that it returns the answers in the same line divided by commas. On the picture I've done the closest I could to what I want, but if you know how to solve this, I think it would be cleaner to use an alternative.

Screenshot of the problem

Hello maple prime users

I have used the sequence command and series command in Maple before, but seem to have hit a little hurdle with a particular problem.

I want to run the expression:


   Where F:=(-1)^n*binomial(c,n)*N(a+n):

   Where N(a):=(-1)^n*[evalf(ln(2)) - 1 + 0.5 -1/3 + 1/4 +... + (-1)^a/a]:

I ran a test case using the following code:


for n from 0 to c do
      Seq2:=convert(Seq, `+`);




This works for the case where a =1, but will not work for a>1 as in the N(a+n) sequence it misses out the terms in the series for a-1, a-2 etc.... So when a = 2 it misses out the -1 of the sequence and when a = 3 it misses out the -1 and 1/2. Are there any tips on how to overcome this?


Best regards


- Yeti

Can Maple look for the limit of a sequence  f(n)  for n=1,2,3, ... ? Of course, if there is a limit of the function of continuous argument, the limit of the sequence is the same number. But it is easy to give examples of when there is the limit of the sequence, and the limit of the function does not exist:

assume(n, posint):

 limit(sin((n^2+1)/n*Pi), n=infinity);  # Obviously the correct answer is  0

                            -1 .. 1


I am trying to plot sequence like:



data:=[seq([n,arrayZ[n]], n=1..3000)]:
t1:=plot(data, style=point, symbol = solidcircle, color= red):

but get incorrect first argument error. How to fix this?

PS. In original sequence instead of null I have constant value, but I don't want it in my plot.

(a) Show that if {an} ∞ n=1 is Cauchy then {a 2 n} ∞ n=1 is also Cauchy. (b) Give an example of a Cauchy sequence {a 2 n} ∞ n=1 such that {an} ∞ n=1 is not Cauchy

how can we write a same result n-times near each other

for i from 1 to 4 do       # here the 4 can be a variable and can change...

I need the result to come in that form  :    3  ,  3  ,  3  ,  3
and because the number of times can change i can't just easily write x,x,x,x;

thanks in advance

Determine wether the sequence below converges or diverges, and if it converges, find the limit.





Hi all,


min/max command in Maple can return the minimum/maximum of a sequence or array.

 In my case, I want to find not only the minimum/maximum, but also where are them. How can I do?

For example, there is a squence [1,2,3,7,6,5,4].

Through max([1,2,3,7,6,5,4]), we can get 7.

But I still want to get "4" which is the index value of "7". 


Thank you.


I am required to generate a list containing the square of numbers 1 through k where k is an arbitrary int,defined from 1 to n. To do this, I've currently got the following commands:

local k, mylist:=[];

for k from 1 to n do


end do;

where sumsquare() is a procedure I defined to compute the sum of the squares of 1 through a number passed as an argument

At present this gives me an out of bounds error. 

How can I initialize mylist to be of size n, like in other languages such as C++?


let m3 = [[0; 1; 0]; [1; 0; 1]; [0; 1; 0]]

1. Firstly, express this matrix into sequence function expression

2. how to express this matrix in terms of forloop code

3. for complicated case such as 1 is not in easy pattern, can it intelligently express the matrix in terms of for loop code


is there exist extra tools to express matrix in terms of for loop code or sequence function code?

  • here is an exercise I got from a text book                                                                                                              calculate the first 10 terms of the following sequence :                                                                                              

u[0]=1                                                                                                                                                             u[n+1]=1/2(u[n]+2/u[n]) n>=0                                                                                                                          

  • estimate the differences u[3]-sqrt(2) , u[4]-sqrt(2), u[5]-sqrt(2), and u[6]-sqrt(2) with a precision of 50 numbers                                    
  • what can we conjecture about the sequence ?
  • how to prove that conjecture with MAPLE ?


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