SYST 201 Class Notes
SYST 201 Class Notes
Lecture 7
Example: Fibonacci numbers
Fibonacci series: 1, 1, 2, 3, 5, 8, 13, 21, 34 (the next number in the sequence is the sum of the previous two)
Formulate as a dynamic system:
[pic]
• Number of inner and outer rings in pinecones (and cauliflower!) are successive Fibonacci numbers.
[pic] [pic]
Solution to Fibonacci series: First find roots of characteristic equation:
[pic] or [pic]
[pic].
Thus, the general solution is:
[pic].
To find the particular solution, use initial conditions:
[pic]
[pic]
From the first equation, [pic]. Plugging into the second equation gives:
[pic]
[pic]
[pic]
[pic]
Thus, particular solution is:
[pic]
Amazingly, [pic] is always an integer!
Example: Imaginary Roots
[pic]
[pic]
First, evaluate iteratively:
[pic]
[pic]
[pic]
[pic]
[pic]
…
Solution: First find roots of characteristic equation:
[pic]
[pic]
Thus, the general solution is:
[pic].
To find the particular solution, use initial conditions:
[pic]
[pic]
From the first equation, [pic]. Plugging into the second equation gives:
[pic]
[pic]
[pic]
Note: When the roots of the characteristic equation are complex, the constants C1 and C2 are complex conjugates (assuming real coefficients in the characteristic equation). Therefore, you only need to find one of the constants (and let the other be its complex conjugate). But it is good practice to find both as a way to double-check your work.
Thus, particular solution is:
[pic]
Amazingly, [pic] is always real! We can check the particular solution:
[pic]
[pic]
[pic]
Same as derived previously.
Example: Double Root
[pic]
[pic]
Evaluate iteratively:
[pic]
[pic]
[pic]
…
Solution: First find roots of characteristic equation:
[pic] or [pic]
[pic]
[pic]
Since we have a double root, the general solution is:
[pic].
To find the particular solution, use initial conditions:
[pic]
[pic]
From the first equation, [pic]. Plugging into the second equation gives:
[pic]
Thus, particular solution is:
[pic]
Check the solution:
[pic]
[pic]
[pic]
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