116 lines
3.4 KiB
Markdown
116 lines
3.4 KiB
Markdown
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type: theoretical
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---
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A recurrence relation is an equation that defines a sequence based on its earlier terms, along with initial values.
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- The recurrence relation $a_n = a_{n-1} + 4$ with initial condition $a_1 = 3$ defines the sequence: $3, 7, 11, 15, \ldots$.
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### Techniques for Finding Explicit Formulas
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1. ***Backtracking*** involves repeatedly substituting the recurrence relation into itself until a pattern emerges.
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- For the recurrence relation $a_n = a_{n-1} + 4$, we repeatedly substitute:
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- $a_n = a_{n-1} + 4$
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- $a_n = (a_{n-2} + 4) + 4 = a_{n-2} + 2 \cdot 4$
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- $a_n = ((a_{n-3} + 4) + 4) + 4 = a_{n-3} + 3 \cdot 4$
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- $\ldots$
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- $a_n = a_{n-(n-1)} + (n-1) \cdot 4 = a_1 + (n-1) \cdot 4 = 3 + (n-1) \cdot 4$
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- So, the explicit formula for the sequence is:
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$$
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a_n = 3 + (n-1) \cdot 4
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$$
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2. ***Characteristic Equation*** applies to linear homogeneous recurrence relations.
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- A **LHR** relation of degree $k$ is of the form:
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$$
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s_n = a_1 s_{n-1} + a_2 s_{n-2} + \ldots + a_k s_{n-k},
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$$
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where $a_i \in \mathbb{R}$ are constants. [^1]
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- The ***characteristic equation*** is:
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$$
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x^k - a_1 x^{k-1} - a_2 x^{k-2} - \ldots - a_k = 0
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$$
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- The roots of the characteristic equation determine the explicit formula for the sequence. The sources focus on degree-2 relations, but the method generalizes to any degree.
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---
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### Solving Linear Homogeneous Recurrence Relations of Degree 2
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For $s_n = a s_{n-1} + b s_{n-2}$, the characteristic equation is $x^2 - ax - b = 0$. Let $r_1$ and $r_2$ be the roots:
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1. **In case of distinct roots** ($r_1 \neq r_2$):
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- The general solution is:
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$$
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s_n = c_1 r_1^n + c_2 r_2^n,
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$$
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where $c_1$ and $c_2$ are constants determined by initial conditions.
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2. **In case of repeated roots** ($r_1 = r_2 = r$):
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- The general solution is:
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$$
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s_n = r^n (c_1 + c_2 n),
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$$
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where $c_1$ and $c_2$ are constants determined by initial conditions. [^2]
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---
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## Example - Fibonacci Sequence
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The Fibonacci sequence is defined as:
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$$
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f_n =
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\begin{cases}
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0, & \text{if } n = 0, \\
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1, & \text{if } n = 1, \\
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f_{n-1} + f_{n-2}, & \text{if } n \geq 2.
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\end{cases}
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$$
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- The characteristic equation is:
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$$
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x^2 - x - 1 = 0
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$$
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- The roots are:
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$$
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r_1 = \frac{1 + \sqrt{5}}{2}, \quad r_2 = \frac{1 - \sqrt{5}}{2}.
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$$
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- Since $r_1 \neq r_2$, the explicit formula is:
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$$
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f_n = c_1 r_1^n + c_2 r_2^n.
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$$
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- Using the initial conditions:
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- $f_0 = 0 = c_1 + c_2$
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- $f_1 = 1 = c_1 \left(\frac{1 + \sqrt{5}}{2}\right) + c_2 \left(\frac{1 - \sqrt{5}}{2}\right)$
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- Solving this system, we get:
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$$
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c_1 = \frac{1}{\sqrt{5}}, \quad c_2 = -\frac{1}{\sqrt{5}}.
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$$
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- Therefore, the explicit formula for the Fibonacci sequence is:
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$$
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f_n = \frac{1}{\sqrt{5}} \left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}} \left(\frac{1 - \sqrt{5}}{2}\right)^n.
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$$
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---
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## Verifying Explicit Formulas
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The correctness of an explicit formula for a recurrence relation can be proven using ***strong mathematical induction***. For example, the explicit Fibonacci formula is verified by [Induction](Mathematical%20Proofs%20(Induction).md).
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## Footnotes
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[^1]: Linear homogeneous recurrence relations are equations where each term is a combination of earlier terms, with no added constants.
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[^2]: Repeated roots in a characteristic equation require modifying the solution to include a term that grows linearly with $n$.
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