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The logistic map is typically defined as x{n+1} = rx{n}(1-x{n})
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Let's call the value it converges to x*; the critical point of the
logistic map
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When
x* =
x* x*(1

x{n} = x* and x{n+1} = x*, we have
rx*(1 - x*)
rx*(1 - x*) = 0
- r(1 - x*)) = 0

Thus, the non-trivial critical point occurs when 1 - r(1 - x*) = 0,
or x = 1 - 1/r
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However, to
analyze the conditions for stability regarding a linearized logistic map, we
employ a Taylor approximation around our critical point x*
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Disregarding higher-order terms, we have
f(x) ≈ f(x*) + f'(x*)(x - x*)
Since f(x*) = rx*(1 - x*), then f'(x*) = r(1 - x*) - rx* = r(1 - 2x*)
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If we plug in our known values, we have
f(x) ≈ x* + r(1 - 2x*)(x - x*)
If the rule of a sequence dictates how each term changes one after the other,
we can think of this "change" transforming x{n} into x{n+1} as a function f
where f(x{n}) = x{n+1}
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x{n+1}
≈ x* + r(1 - 2x*)(x{n} - x*)
≈ x* + r(1 - 2x*)x{n} - rx*(1 - 2x*)
≈ r(1 - 2x*)x{n} + x* - rx*(1 - 2x*)
Let A = r(1 - 2x*) and B = x* - rx*(1 - 2x*)
Then, we have x{n+1} = Ax{n} + B
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x{n+1} = kA^(n+1) + B(A^n + A^(n-1) +
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+ 1 can be evaluated to the geometric
series (1 - A^(n+1))/(1 - A)
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Evaluating the limit provides us with
lim{n → ∞} x{n+1} = lim{n → ∞} kA^(n+1) + lim{n → ∞} B(1 - A^(n+1))/(1 - A)
= lim{n → ∞} kA^(n+1) + B/(1 - A) · lim{n → ∞} (1 - A^(n+1))
= 0 + B(1 - A)·(1)
= B/(1 - A)
Thus, the condition for stability of a linearized logistic map is for |A| < 1
and x{0} = k
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Since A = r(1 - 2x*) and x* = 1 - 1/r, then
|r(1 - 2(1 - 1/r))| < 1
-1 < r - 2r + (2/r)r < 1
-1 < -r + 2 < 1
-3 < -r < -1
1 < r < 3
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If |A| < 1, then 1 < r < 3, so 1 > 1/r > 1/3, and 0 < 1 - 1/r < 2/3
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This shows that x* is positive when |A| < 1
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Then, since x* is positive when |A| < 1, 0 < x*(1 - A) < 2x*, or 0 < B < 2x*
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Suppose we are near the critical point x* at some n of x{n}
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Thus, the only thing that
separates x{n} from x* is an additional ε
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Thus, we
have:
[1] x{n} = x* + ε{n}
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To do
this,
we must determine the rate at which each successive perturbation changes
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[2] x{n+1} = Ax{n} + B
If x{n} = x* + ε{n}, then x{n+1} = x* + ε{n+1}
From [2], we have
x* + ε{n+1} = A(x* + ε{n}) + B
Since B = x* - rx*(1 - 2x*) = x* - x*A,
x* + ε{n+1} = A(x* + ε{n}) + x* - x*A
ε{n+1} = Ax* + Aε{n} - Ax*
ε{n+1} = Aε{n}
[3] A = ε{n+1}/ε{n}
From [3], we see that the relative change of each successive perturbation from
the last is a constant A
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If we wish to impose a particular tolerance such that the difference between
x{n} and x* be sufficiently small, we can find an index n such that
[4] |x{n} - x*| < ζ
where ζ is an arbitrary tolerance
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From [1], x{n} = x* + ε{n} ⇒ ε{n} = x{n} - x*
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For n = 0, ε{1} = Aε{0}
n = 1, ε{2} = Aε{1} = A(Aε{0}) = A²ε{0}
n = 2, ε{3} = Aε{2} = A(A²ε{0}) = A³ε{0}

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From [1], |x{n} - x*| = |ε{n}| = |A^(n)ε{0}|
Thus, we have that
[6] |A^(n)ε{0}| < ζ
To acquire the number of iterations required for x{n} - ε{n} → x*, we
parameterize
our inequality for n
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When |A| < 1, ln(|A|) < 0, so
[7] n > ceiling[ln(ζ/|ε{0}|)/ln(|A|)]
when n ∈ ℕ{0}
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Alternatively, using [6] provides another restriction on A such that
|A| < (ζ/|ε{0}|)^(1/n)
As n → ∞, (ζ/|ε{0}|)^(1/n) → 1, re-establishing our necessary condition that
|A| < 1 for all n to converge
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To determine these effects and document these changes, we study the local
effects
A and B have on x*
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In terms of A and B, the critical point x* can be represented as
[8] x* = B/(1 - A)
Next, we find ∂x* from [8] with respect to A and B
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These results demonstrate the sensitivity the value x* has relative to changes
in A and B
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A relative change can be represented by the formula Δx/x, where x is an
independent variable
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Thus, we have
lim{Δx → 0} Δx/x = dx/x = ∂x/x
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For example, if A = 0
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1/0
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̅1
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1, there is a 0
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However, if A = 0
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99/0
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Thus, for every 1% increase in A when
A = 0
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Our findings show that as A → 0, changes in A have minimal effect to changes
in x*
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However, changes in B have a constant minimal effect on x*, with every 1% change
in B yielding a 1% change in x*
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