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Is there a Maple function that given a set of substitutions in form object=set of substitute objects produces a sequence of sets, each a product of substitution from the next, remove repetitions. An example:

 

#Substitutions
a = {b, c, d}, b = {a, b, e}, c = {a, d, f};
#Result
a = ({e, f}, {b, c, d}), b = ({f}, {c, d}, {a, b, e}), c = ({e}, {b}, {a, d, f});

I need to do this with a sequence of a large number of elements but here is my question with a sequence of fewer elements.

a:={10,11,-13,9,20,74,10}:

I want to check each element of the sequence if it is negative. And if any element is negative I want to replace with 0 (or do something else). I tried the following code but it didn't work. 

for k from 1 to 7 do
  if (a[k]<0) then subs(a[k]=0, a) end if:
end do;

Please help me to figure it out. 

Hello,

Paraphrasing:

https://en.m.wikipedia.org/wiki/Langford_pairing

given a sequence of 2n numbers 1,1,2,2,3,3,...n,n in which the two 1's are 1 slot apart, the two 2's are 2 slots apart, the two 3's are 3 slots apart, -> generally the two n's are n slots apart.

for n=3, a possible sequence is [2,3,1,2,1,3]. 

There is sequences for n=4,7,8...

i havent a clue how to write a maple procedure to generate a sequence for given n, but I'm hoping an expert here can.

(Some) prime reciprocals have an interesting property. the repeating sequence has length p-1.

eg 7, has repeats after 6.

1/7=0.142857142857......

17 has repeats after 16

1/17=0.05882352941176470588235294117647059.........

of course some primes don''t have this property....

So what I need the experts here is for some code

myproc(17)=16, (and the sequence) 0588235294117647

my go:pinched from

http://199.71.183.11/questions/39621-Pattern-Matching-In-A-Sequence-Of-Digits

PriDigits := "" || (op(1, evalf(1/17, 50)));

reps := StringTools[Repeats](PriDigits);

lngth := seq(op(3, A), A in [reps]);

the leading 0 (which is part of the sequence)  is a problem.....

 

 

hi every one, i would like to know if it is possible to implement the following recursive sequence in Maple :

q(1)       = q(0)        +q_dot(0)
q_dot(1)  = q_dot(0)  +q_ddot(0)
q_ddot(1) = q_ddot(0)+q_dddot

 

q(2)       = q(1)        +q_dot(1)   = (q(0)+q_dot(0)) + (q_dot(0)+q_ddot(0)) = q(0)+2q_dot(0)+q_ddot(0)
q_dot(2)  = q_dot(1)  +q_ddot(1)
q_ddot(2) = q_ddot(1)+q_dddot

 

q(3)      = q(2)        +q_dot(2)=(q(0)+2q_dot(0)+q_ddot(0))+(q_dot(1)+q_ddot(1))= q(0)+3q_dot(0)+q_ddot(0)+q_dddot
q_dot(3) = q_dot(2)  +q_ddot(2)
q_ddot(3)= q_ddot(2)+q_dddot

I would like every q(n) to be expressed as following : q(n)=q(0)+n_1*q_dot(0)+n_2*q_ddot(0)+n_3*q_dddot.

By computing manually some sequences i am now sure that :
n_1=n
n_2=(n²-n)/2
n_3=?? (I would like to find this one)

The sequence can be written in a more general way as following :
q(n)        = q(n-1) + q_dot(n-1)
q_dot(n)   = q_dot(n-1) + q_ddot(n-1)
q_ddot(n) =q_ddot(n-1)+q_dddot

Can you guys please show me how this can be implementes in Maple ?

Thanks in advance

I have the following question:

Illustrate how the sequence N->R de fined by n ->n^2/n^2 + 31n + 228 can be shown to be
within a given epsilon > 0 of its limiting value x0.
(a) use an appropriate conditional statement to find N such that abs (xn -􀀀 x0) < epsilon for every n>=N
and produce an appropriate list of the data points (n,xn) to illustrate
this

I found N but without using any CONDITIONAL STATEMENT.Can you help me find N using IF FOR WHILE?

Hello everyone.  Let n be a positive number; I seek how to write the matrix Z\in M_{2^n,n} that is defined as follows:

for every i,j, Z_{ij}\in \{-1,1\}.

When I consider the 2^n rows of Z, I want to find ALL the possible sequences of length n with the entries +-1. A toy example, when n=2, is Z=\begin{pmatrix}1&1\\1&-1\\-1&1\\-1&-1\end{pmatrix}.

Thanks in advance;

How can I have a sequence as follow by using "seq" command?

x1,x2,...x200.

Please note that I need "xi" not "x_i".

I have the following sequence: n2/n2+31n+228.The question asks me to use an appropriate conditional statement to find N such that absolute (xn-x0)<epsilon for every n>=N. I found N but i don't know how to find it using Maple17.Can you please help me?

 

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;

S1=diff(p,x);

let   s:=quo(S0,S1,x);
       S2:=-rem(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?

 

Hello.

 

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:

h(a,b,c):=b!*(a+c)!*sum(F,n=0..c)

   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:

restart:
a:=1:
b:=2:
c:=3:
tot:=0:
F:=((-1)^(a+n)/(a+n)):

for n from 0 to c do
      Seq_F2:=(-1)^(a+n):
   
      Seq:=[seq(F,n=0..n)];
      Seq2:=convert(Seq, `+`);

      final:=Seq_F2*(evalf(ln(2))+Seq2);

      SS:=(-1)^n*binomial(c,n)*final;

      S:=(b!*(a+c)!)*SS;
      tot:=tot+S;
od:
tot;

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

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