Math Problem Statement
Solution
The given recurrence relation is:
for . We need to find the general form of .
Step 1: Solve the characteristic equation
This is a linear recurrence relation with constant coefficients, so we begin by solving the associated characteristic equation. The recurrence relation can be transformed into the characteristic equation by assuming a solution of the form .
Substitute , , and into the recurrence relation:
Divide through by :
Rearrange this to form a quadratic equation:
Step 2: Solve the quadratic equation
Now solve the quadratic equation using the quadratic formula:
So, the two roots of the characteristic equation are:
Step 3: General solution
Since the characteristic equation has distinct real roots, the general solution to the recurrence relation is of the form:
where and are constants determined by the initial conditions.
If you have specific initial values for and , we can proceed to find and .
Would you like to provide the initial conditions, or should I continue with the general solution?
Feel free to ask if you need more details on any step. Here are some related questions:
- What happens if the characteristic equation has complex roots?
- How would you find the constants and ?
- How can recurrence relations be solved using matrix methods?
- Can you solve recurrence relations with variable coefficients similarly?
- What are the applications of solving such recurrence relations in computer science?
Tip: Always check the characteristic equation first when solving linear recurrence relations with constant coefficients.
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Math Problem Analysis
Mathematical Concepts
Recurrence Relations
Algebra
Characteristic Equations
Formulas
Recurrence relation: an = 2an-1 + 4an-2
Characteristic equation: r^2 = 2r + 4
Quadratic equation: r^2 - 2r - 4 = 0
Quadratic formula: r = (-b ± √(b^2 - 4ac)) / 2a
Theorems
Quadratic Formula
Suitable Grade Level
Grades 11-12, College Level
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