Matlab limit help

Question

% (3) Your turn: generate the first 20 sequence terms

% in the recursively defined sequence

% X1 = 0.6, X_n = 1.5*X_(n-1).*(1-X_(n-1))

% Call the sequence X (like F above).

X(1)= [0.6]; % the initial seed

for n = 2:20

X(n) = 1.5*X(n-1).*(1-X(n-1));

end

disp(X)

% Follow-up: this sequence appears to have a limit!

% Provide it's exact form below, call it XLimit:

XLimit = 3/2

disp(rats(XLimit))

the sequence for X code is correct. what is the Limit of that sequence?

Answer 1

let me say this about this

x_n+1=bx_n(1-x_n) ,,,,,,,,0<x_n<1

If the coefficient that multiplies x_n=4, then the resulting sequence

goes into random chaos, with values anywhere between 0 and 1 as it is iterated,,, as long as x₁ not equal to stability values (1/4/,1/2,3/4). It is really neat to see.

In fact if the coefficient is 3.57 or greater then the recurrence equation does not

converge. At 3.57 the values biifurcate, then as it goes from 3.57 to 4, it bifurcates

more until at 4 the values are random. Reference: logistic map, population studies.

I agree with the method for determining the value of convergence

provided above by William, when for at least this recurrence equation

the coefficient is less than 3.5 (actually at 3.5 the convergence depends

on whether x₁ is less than or greater than .6). Chaos is found in these

iterative recurrence equations and the topic is complex.

If the coefficient>4, then encounter negative x_n+1

Answer 2

I don't have access to MATLAB, but I put

X₁ = 0.6

Xₙ = 1.5Xₙ₋₁(1 – Xₙ₋₁)

into an Excel spreadsheet and generated the first 20 sequence terms.

It's a decreasing sequence where's there is clearly a limit.

The Limit of Xₙ = 1.5Xₙ₋₁(1 – Xₙ₋₁) if X₁ = 0.6

When the sequence converges Xₙ → Xₙ₋₁

So, if you substitute Xₙ = Xₙ₋₁ into the recursive formula you can find that limit.

The limit is the same as one of the numbers you come up with by solving the following equation for x:

x = ³⁄₂x(1 –x) = ³⁄₂x – ³⁄₂x²

Subtracting x from both sides give

½x – ³⁄₂x² = 0 = ½x(1 – 3x)

x = 0 is one solution, but it's not the limit or the sequence (unless you start with X₁ = 0)

Solve 1 – 3x = 0 and you'll have the same number as the limit of your sequence.

Note

The sequence converges and approaches the same limit for any X₁ in the range 0 < X₁ < 1.

Answer

X_Limit = ³⁄₂x(1 –x)

where x is the number you get if you solve x = ³⁄₂x(1 –x) as described above.

Hope this helps.