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Discrete Dynamical Systems
Discrete Dynamical Systems 1
J. F. Rabajante
IMSP, UPLB
1st Sem, AY 2012-2013
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Bank Account
Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?
Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelationSolution is x(k) = (1 + r)kx0.
Compounded m-times in a year:x(k + 1) = (1 + r
m )mx(k)Solution is x(k) = (1 + r
m )mkx0.
Compounded continuously (not anymore discrete):x(t) =
(limm→∞(1 + r
m )m)t x0 or dxdt = rx
Solution is x(t) = x0ert .
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Bank Account
Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?
Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelation
Solution is x(k) = (1 + r)kx0.
Compounded m-times in a year:x(k + 1) = (1 + r
m )mx(k)Solution is x(k) = (1 + r
m )mkx0.
Compounded continuously (not anymore discrete):x(t) =
(limm→∞(1 + r
m )m)t x0 or dxdt = rx
Solution is x(t) = x0ert .
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Bank Account
Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?
Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelationSolution is x(k) = (1 + r)kx0.
Compounded m-times in a year:x(k + 1) = (1 + r
m )mx(k)Solution is x(k) = (1 + r
m )mkx0.
Compounded continuously (not anymore discrete):x(t) =
(limm→∞(1 + r
m )m)t x0 or dxdt = rx
Solution is x(t) = x0ert .
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Bank Account
Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?
Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelationSolution is x(k) = (1 + r)kx0.
Compounded m-times in a year:x(k + 1) = (1 + r
m )mx(k)
Solution is x(k) = (1 + rm )mkx0.
Compounded continuously (not anymore discrete):x(t) =
(limm→∞(1 + r
m )m)t x0 or dxdt = rx
Solution is x(t) = x0ert .
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Bank Account
Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?
Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelationSolution is x(k) = (1 + r)kx0.
Compounded m-times in a year:x(k + 1) = (1 + r
m )mx(k)Solution is x(k) = (1 + r
m )mkx0.
Compounded continuously (not anymore discrete):x(t) =
(limm→∞(1 + r
m )m)t x0 or dxdt = rx
Solution is x(t) = x0ert .
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Bank Account
Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?
Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelationSolution is x(k) = (1 + r)kx0.
Compounded m-times in a year:x(k + 1) = (1 + r
m )mx(k)Solution is x(k) = (1 + r
m )mkx0.
Compounded continuously (not anymore discrete):x(t) =
(limm→∞(1 + r
m )m)t x0 or dxdt = rx
Solution is x(t) = x0ert .
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Bank Account
Suppose you have an initial deposit of x(0) = x0. The annualinterest rate is r . What is the corresponding mathematicalmodel?
Compounded annualy:x(k + 1) = (1 + r)x(k) — this is called a recursive/iterativerelationSolution is x(k) = (1 + r)kx0.
Compounded m-times in a year:x(k + 1) = (1 + r
m )mx(k)Solution is x(k) = (1 + r
m )mkx0.
Compounded continuously (not anymore discrete):x(t) =
(limm→∞(1 + r
m )m)t x0 or dxdt = rx
Solution is x(t) = x0ert .Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Biology
How many rabbits? (by Leonardo of Pisa)
Fibonacci sequence 1,1,2,3,5,8,13,...
x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1
We will show later that Fibonacci sequence is related to thegolden/divine ratio φ = 1.618....
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Biology
How many rabbits? (by Leonardo of Pisa)
Fibonacci sequence 1,1,2,3,5,8,13,...
x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1
We will show later that Fibonacci sequence is related to thegolden/divine ratio φ = 1.618....
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Biology
How many rabbits? (by Leonardo of Pisa)
Fibonacci sequence 1,1,2,3,5,8,13,...
x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1
We will show later that Fibonacci sequence is related to thegolden/divine ratio φ = 1.618....
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Biology
How many rabbits? (by Leonardo of Pisa)
Fibonacci sequence 1,1,2,3,5,8,13,...
x(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1
We will show later that Fibonacci sequence is related to thegolden/divine ratio φ = 1.618....
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Difference equations∆x = x(k + 1)− x(k) ;x(k + m) = f (x(k + m − 1), x(k + m − 2), ..., x(k))
Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous- autonomous and nonautonomous- order of a discrete dynamical system
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Difference equations∆x = x(k + 1)− x(k) ;x(k + m) = f (x(k + m − 1), x(k + m − 2), ..., x(k))
Let’s discuss- linear and nonlinear
- homogeneous and nonhomogeneous- autonomous and nonautonomous- order of a discrete dynamical system
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Difference equations∆x = x(k + 1)− x(k) ;x(k + m) = f (x(k + m − 1), x(k + m − 2), ..., x(k))
Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous
- autonomous and nonautonomous- order of a discrete dynamical system
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Difference equations∆x = x(k + 1)− x(k) ;x(k + m) = f (x(k + m − 1), x(k + m − 2), ..., x(k))
Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous- autonomous and nonautonomous
- order of a discrete dynamical system
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Difference equations∆x = x(k + 1)− x(k) ;x(k + m) = f (x(k + m − 1), x(k + m − 2), ..., x(k))
Let’s discuss- linear and nonlinear- homogeneous and nonhomogeneous- autonomous and nonautonomous- order of a discrete dynamical system
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b
= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b
= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b
= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b
= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b
= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b
= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b
= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Let us find the analytic solution to the IVPx(k + 1) = ax(k) + b, x(0) = x0.
x(1) = ax(0) + b
x(2) = ax(1) + b= a(ax(0) + b) + b= a2x(0) + ab + b= a2x(0) + b(a + 1)
x(3) = ax(2) + b= a(a2x(0) + b(a + 1)) + b= a3x(0) + ab(a + 1) + b= a3x(0) + b(a2 + a) + b= a3x(0) + b(a2 + a + 1)
In general, x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Continuation... x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
But note that an−1 + an−2 + ...+ a + 1 is a geometric series withratio a.
So an−1 + an−2 + ...+ a + 1 = an−1a−1 if a 6= 1, and
an−1 + an−2 + ...+ a + 1 = n if a = 1.
Therefore, we have the following theorem.
Theorem 1 The analytic solution to x(k + 1) = ax(k) + b,x(0) = x0 is
x(k) = akx0 + b ak−1a−1 if a 6= 1 (geometric series)
x(k) = akx0 + nb if a = 1 (arithmetic series).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Continuation... x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
But note that an−1 + an−2 + ...+ a + 1 is a geometric series withratio a.
So an−1 + an−2 + ...+ a + 1 = an−1a−1 if a 6= 1, and
an−1 + an−2 + ...+ a + 1 = n if a = 1.
Therefore, we have the following theorem.
Theorem 1 The analytic solution to x(k + 1) = ax(k) + b,x(0) = x0 is
x(k) = akx0 + b ak−1a−1 if a 6= 1 (geometric series)
x(k) = akx0 + nb if a = 1 (arithmetic series).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Continuation... x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
But note that an−1 + an−2 + ...+ a + 1 is a geometric series withratio a.
So an−1 + an−2 + ...+ a + 1 = an−1a−1 if a 6= 1, and
an−1 + an−2 + ...+ a + 1 = n if a = 1.
Therefore, we have the following theorem.
Theorem 1 The analytic solution to x(k + 1) = ax(k) + b,x(0) = x0 is
x(k) = akx0 + b ak−1a−1 if a 6= 1 (geometric series)
x(k) = akx0 + nb if a = 1 (arithmetic series).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Continuation... x(n) = anx(0) + b(an−1 + an−2 + ...+ a + 1).
But note that an−1 + an−2 + ...+ a + 1 is a geometric series withratio a.
So an−1 + an−2 + ...+ a + 1 = an−1a−1 if a 6= 1, and
an−1 + an−2 + ...+ a + 1 = n if a = 1.
Therefore, we have the following theorem.
Theorem 1 The analytic solution to x(k + 1) = ax(k) + b,x(0) = x0 is
x(k) = akx0 + b ak−1a−1 if a 6= 1 (geometric series)
x(k) = akx0 + nb if a = 1 (arithmetic series).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Amortization
The discrete IVP is x(k + 1) = (1 + r)x(k)− p, x(0) = L.
To do: Find the analytic solution. Then solve for p such thatx(n) = 0 where n is the last payment time.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Amortization
The discrete IVP is x(k + 1) = (1 + r)x(k)− p, x(0) = L.
To do: Find the analytic solution. Then solve for p such thatx(n) = 0 where n is the last payment time.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Amortization
The discrete IVP is x(k + 1) = (1 + r)x(k)− p, x(0) = L.
To do: Find the analytic solution. Then solve for p such thatx(n) = 0 where n is the last payment time.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Now, consider the following ax(k + 2) + bx(k + 1) + cx(k) = 0,a 6= 0.
Let x(k) = λk 6= 0.So we have aλk+2 + bλk+1 + cλk = 0.Factoring out, λk (aλ2 + bλ+ c) = 0.
The equation aλ2 + bλ+ c = 0 is a characteristic equation andthe solutions to the characteristic equation are calledeigenvalues.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Now, consider the following ax(k + 2) + bx(k + 1) + cx(k) = 0,a 6= 0.
Let x(k) = λk 6= 0.
So we have aλk+2 + bλk+1 + cλk = 0.Factoring out, λk (aλ2 + bλ+ c) = 0.
The equation aλ2 + bλ+ c = 0 is a characteristic equation andthe solutions to the characteristic equation are calledeigenvalues.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Now, consider the following ax(k + 2) + bx(k + 1) + cx(k) = 0,a 6= 0.
Let x(k) = λk 6= 0.So we have aλk+2 + bλk+1 + cλk = 0.
Factoring out, λk (aλ2 + bλ+ c) = 0.
The equation aλ2 + bλ+ c = 0 is a characteristic equation andthe solutions to the characteristic equation are calledeigenvalues.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Now, consider the following ax(k + 2) + bx(k + 1) + cx(k) = 0,a 6= 0.
Let x(k) = λk 6= 0.So we have aλk+2 + bλk+1 + cλk = 0.Factoring out, λk (aλ2 + bλ+ c) = 0.
The equation aλ2 + bλ+ c = 0 is a characteristic equation andthe solutions to the characteristic equation are calledeigenvalues.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Now, consider the following ax(k + 2) + bx(k + 1) + cx(k) = 0,a 6= 0.
Let x(k) = λk 6= 0.So we have aλk+2 + bλk+1 + cλk = 0.Factoring out, λk (aλ2 + bλ+ c) = 0.
The equation aλ2 + bλ+ c = 0 is a characteristic equation andthe solutions to the characteristic equation are calledeigenvalues.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Now, consider the following ax(k + 2) + bx(k + 1) + cx(k) = 0,a 6= 0.
Let x(k) = λk 6= 0.So we have aλk+2 + bλk+1 + cλk = 0.Factoring out, λk (aλ2 + bλ+ c) = 0.
The equation aλ2 + bλ+ c = 0 is a characteristic equation andthe solutions to the characteristic equation are calledeigenvalues.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Theorem 2 The general solution toax(k + 2) + bx(k + 1) + cx(k) = 0, a 6= 0 is
For real, distinct eigenvalues λ1 and λ2:x(k) = d1λ
k1 + d2λ
k2.
For repeated real eigenvalue λ: x(k) = d1λk + d2kλk .
For complex eigenvalues λ1,λ2= re±iθ:x(k) = r k (d1 cos(kθ) + d2 sin(kθ)).
Recall: α + βi = r cos(θ) + ir sin(θ) = reiθ.r =
√α2 + β2, θ = tan−1
(βα
).
Example: Try to graph the solution of2x(k + 2)− 2x(k + 1) + x(k) = 0.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Theorem 2 The general solution toax(k + 2) + bx(k + 1) + cx(k) = 0, a 6= 0 is
For real, distinct eigenvalues λ1 and λ2:x(k) = d1λ
k1 + d2λ
k2.
For repeated real eigenvalue λ: x(k) = d1λk + d2kλk .
For complex eigenvalues λ1,λ2= re±iθ:x(k) = r k (d1 cos(kθ) + d2 sin(kθ)).
Recall: α + βi = r cos(θ) + ir sin(θ) = reiθ.r =
√α2 + β2, θ = tan−1
(βα
).
Example: Try to graph the solution of2x(k + 2)− 2x(k + 1) + x(k) = 0.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Theorem 2 The general solution toax(k + 2) + bx(k + 1) + cx(k) = 0, a 6= 0 is
For real, distinct eigenvalues λ1 and λ2:x(k) = d1λ
k1 + d2λ
k2.
For repeated real eigenvalue λ: x(k) = d1λk + d2kλk .
For complex eigenvalues λ1,λ2= re±iθ:x(k) = r k (d1 cos(kθ) + d2 sin(kθ)).
Recall: α + βi = r cos(θ) + ir sin(θ) = reiθ.r =
√α2 + β2, θ = tan−1
(βα
).
Example: Try to graph the solution of2x(k + 2)− 2x(k + 1) + x(k) = 0.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Theorem 2 The general solution toax(k + 2) + bx(k + 1) + cx(k) = 0, a 6= 0 is
For real, distinct eigenvalues λ1 and λ2:x(k) = d1λ
k1 + d2λ
k2.
For repeated real eigenvalue λ: x(k) = d1λk + d2kλk .
For complex eigenvalues λ1,λ2= re±iθ:x(k) = r k (d1 cos(kθ) + d2 sin(kθ)).
Recall: α + βi = r cos(θ) + ir sin(θ) = reiθ.r =
√α2 + β2, θ = tan−1
(βα
).
Example: Try to graph the solution of2x(k + 2)− 2x(k + 1) + x(k) = 0.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Theorem 2 The general solution toax(k + 2) + bx(k + 1) + cx(k) = 0, a 6= 0 is
For real, distinct eigenvalues λ1 and λ2:x(k) = d1λ
k1 + d2λ
k2.
For repeated real eigenvalue λ: x(k) = d1λk + d2kλk .
For complex eigenvalues λ1,λ2= re±iθ:x(k) = r k (d1 cos(kθ) + d2 sin(kθ)).
Recall: α + βi = r cos(θ) + ir sin(θ) = reiθ.r =
√α2 + β2, θ = tan−1
(βα
).
Example: Try to graph the solution of2x(k + 2)− 2x(k + 1) + x(k) = 0.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Theorem 2 The general solution toax(k + 2) + bx(k + 1) + cx(k) = 0, a 6= 0 is
For real, distinct eigenvalues λ1 and λ2:x(k) = d1λ
k1 + d2λ
k2.
For repeated real eigenvalue λ: x(k) = d1λk + d2kλk .
For complex eigenvalues λ1,λ2= re±iθ:x(k) = r k (d1 cos(kθ) + d2 sin(kθ)).
Recall: α + βi = r cos(θ) + ir sin(θ) = reiθ.r =
√α2 + β2, θ = tan−1
(βα
).
Example: Try to graph the solution of2x(k + 2)− 2x(k + 1) + x(k) = 0.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.
x(k + 2)− x(k + 1)− x(k) = 0.
Eigenvalues: 1±√
52
General solution is: x(k) = d1
(1+√
52
)k+ d2
(1−√
52
)k.
Specific solution is: x(k) = 5+√
510
(1+√
52
)k+ 5−
√5
10
(1−√
52
)k.
Furthermore, get the ratio limk→∞x(k+1)
x(k) , the ratio is equal to Φ
(golden ratio).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.x(k + 2)− x(k + 1)− x(k) = 0.
Eigenvalues: 1±√
52
General solution is: x(k) = d1
(1+√
52
)k+ d2
(1−√
52
)k.
Specific solution is: x(k) = 5+√
510
(1+√
52
)k+ 5−
√5
10
(1−√
52
)k.
Furthermore, get the ratio limk→∞x(k+1)
x(k) , the ratio is equal to Φ
(golden ratio).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.x(k + 2)− x(k + 1)− x(k) = 0.
Eigenvalues: 1±√
52
General solution is: x(k) = d1
(1+√
52
)k+ d2
(1−√
52
)k.
Specific solution is: x(k) = 5+√
510
(1+√
52
)k+ 5−
√5
10
(1−√
52
)k.
Furthermore, get the ratio limk→∞x(k+1)
x(k) , the ratio is equal to Φ
(golden ratio).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.x(k + 2)− x(k + 1)− x(k) = 0.
Eigenvalues: 1±√
52
General solution is: x(k) = d1
(1+√
52
)k+ d2
(1−√
52
)k.
Specific solution is: x(k) = 5+√
510
(1+√
52
)k+ 5−
√5
10
(1−√
52
)k.
Furthermore, get the ratio limk→∞x(k+1)
x(k) , the ratio is equal to Φ
(golden ratio).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.x(k + 2)− x(k + 1)− x(k) = 0.
Eigenvalues: 1±√
52
General solution is: x(k) = d1
(1+√
52
)k+ d2
(1−√
52
)k.
Specific solution is: x(k) = 5+√
510
(1+√
52
)k+ 5−
√5
10
(1−√
52
)k.
Furthermore, get the ratio limk→∞x(k+1)
x(k) , the ratio is equal to Φ
(golden ratio).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.x(k + 2)− x(k + 1)− x(k) = 0.
Eigenvalues: 1±√
52
General solution is: x(k) = d1
(1+√
52
)k+ d2
(1−√
52
)k.
Specific solution is: x(k) = 5+√
510
(1+√
52
)k+ 5−
√5
10
(1−√
52
)k.
Furthermore, get the ratio limk→∞x(k+1)
x(k) ,
the ratio is equal to Φ
(golden ratio).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Example: Recall the Fibonacci sequencex(k + 2) = x(k + 1) + x(k), where x(0) = 1, x(1) = 1.x(k + 2)− x(k + 1)− x(k) = 0.
Eigenvalues: 1±√
52
General solution is: x(k) = d1
(1+√
52
)k+ d2
(1−√
52
)k.
Specific solution is: x(k) = 5+√
510
(1+√
52
)k+ 5−
√5
10
(1−√
52
)k.
Furthermore, get the ratio limk→∞x(k+1)
x(k) , the ratio is equal to Φ
(golden ratio).
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Exercise: Prove that the golden mean satisfies Φ = 1 + 1Φ . Also
prove that the golden ratio can also be rewritten as a continuedfraction: 1 + 1
1+ 11+...
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Equilibria and Stability: First Discussion
Consider x(k + 1) = f (x(k)). Applying the function f m-times,x(k + 1) = f (f (f (...f (x(0))))) = f n(x(0)).
Definition 1 An equilibrium or fixed point is a value x∗ for whichx∗ = f (x∗).
Example: Find the fixed point of x(k + 1) = ax(k) + b.Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel. Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Equilibria and Stability: First Discussion
Consider x(k + 1) = f (x(k)). Applying the function f m-times,x(k + 1) = f (f (f (...f (x(0))))) = f n(x(0)).
Definition 1 An equilibrium or fixed point is a value x∗ for whichx∗ = f (x∗).
Example: Find the fixed point of x(k + 1) = ax(k) + b.Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel. Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Equilibria and Stability: First Discussion
Consider x(k + 1) = f (x(k)). Applying the function f m-times,x(k + 1) = f (f (f (...f (x(0))))) = f n(x(0)).
Definition 1 An equilibrium or fixed point is a value x∗ for whichx∗ = f (x∗).
Example: Find the fixed point of x(k + 1) = ax(k) + b.Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel. Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Equilibria and Stability: First Discussion
Consider x(k + 1) = f (x(k)). Applying the function f m-times,x(k + 1) = f (f (f (...f (x(0))))) = f n(x(0)).
Definition 1 An equilibrium or fixed point is a value x∗ for whichx∗ = f (x∗).
Example: Find the fixed point of x(k + 1) = ax(k) + b.
Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel. Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Equilibria and Stability: First Discussion
Consider x(k + 1) = f (x(k)). Applying the function f m-times,x(k + 1) = f (f (f (...f (x(0))))) = f n(x(0)).
Definition 1 An equilibrium or fixed point is a value x∗ for whichx∗ = f (x∗).
Example: Find the fixed point of x(k + 1) = ax(k) + b.Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel.
Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
Equilibria and Stability: First Discussion
Consider x(k + 1) = f (x(k)). Applying the function f m-times,x(k + 1) = f (f (f (...f (x(0))))) = f n(x(0)).
Definition 1 An equilibrium or fixed point is a value x∗ for whichx∗ = f (x∗).
Example: Find the fixed point of x(k + 1) = ax(k) + b.Determine if the fixed point of x(k + 1) = 0.8x(k) + 2, x(0) = 1attract or repel. Determine if the fixed point ofx(k + 1) = −1.04x(k), x(0) = 1 attract or repel.
Rabajante MATH 191: Special Topics
Discrete Dynamical Systems
What is the fixed point of ax(k + 2) + bx(k + 1) + cx(k) = 0?
Rabajante MATH 191: Special Topics