Question

If \(S(n)=a S(n-1)+g(n)\) and \(g(n)

Short answer

\(S(n)\) grows as \(\Theta(a^n)\).

Step-by-step solution

Understand the Recurrence Relation

We are given the recurrence relation \(S(n) = a S(n-1) + g(n)\). Here, \(a\) is a constant and \(g(n)\) is a function where \(g(n) < c^n\) with \(1 \leq c < a\). This setup resembles a non-homogeneous linear recurrence relation where an additional term \(g(n)\) is present.

Analyze the Homogeneous Solution

First, consider the homogeneous part of the recurrence: \(S(n) = a S(n-1)\). This tells us that if \(g(n)\) were zero, \(S(n)\) would grow as \(S(n) = S(0) a^n\). Therefore, the homogeneous solution suggests that \(S(n) = C a^n\) for some constant \(C\).

Analyze the Particular Solution

For the particular solution, consider the effect of \(g(n)\). Since \(g(n) < c^n\) and \(c < a\), the solution growing induced by \(g(n)\) needs to be non-dominant compared to the homogeneous solution. Thus, the influence of \(g(n)\) is asymptotically negligible compared to the \(a^n\) growth, meaning the particular solution grows slower than the homogeneous solution.

Determine the Overall Growth

Combining both insights, the overall growth of \(S(n)\) is dominated by the \(a^n\) term, because \(g(n)\), growing as \(c^n\), remains asymptotically negligible. Therefore, we determine the growth rate of \(S(n)\) in big \(\Theta\) notation as \(\Theta(a^n)\).

Key concepts

Homogeneous Solutions

In the context of recurrence relations, a homogeneous solution addresses the scenario where there are no additional terms introduced, other than the terms dependent solely on the prior solutions. In the given exercise, this refers to the part of the recurrence relation where we have:

• The expression is simplified to the form: \(S(n) = a S(n-1)\)
• This implies that each term is calculated from its predecessor, scaled by the factor \(a\).

Consequently, if we assume no external influence from any additional function \(g(n)\), this part of the relation is solved as a geometric progression. That is, starting from some initial condition, say \(S(0)\), the solution forms a sequence given by: \(S(n) = S(0) a^n\). The term "homogeneous" refers to this consistent proportional growth, patterned only by the influence of the preceding term's value, leading to predictable exponential growth.
Understanding homogeneous solutions is crucial as it sets the backbone upon which other influences are analyzed, especially when non-homogeneous terms like \(g(n)\) are introduced.

Particular Solutions

When dealing with recurrence relations, particularly those that are non-homogeneous, a particular solution helps in understanding the additional effects that an external forcing function \(g(n)\) has on the overall solution. In this exercise, the particular solution pertains to how \(g(n)\), although present, behaves relative to the primary homogeneous term:

• The relation is of the form \(S(n) = a S(n-1) + g(n)\).
• Here, \(g(n)\) is a non-zero term but is constrained by the condition \(g(n) < c^n\), where \(c < a\).

The particular solution is crafted by assessing the long-term or asymptotic behavior of \(g(n)\). Given that \(c < a\), as \(n\) becomes larger, the influence of \(g(n)\) diminishes in comparison to the dominant term \(a^n\). Hence, any solution component arising from \(g(n)\) is gradually overshadowed, making its contribution negligible for large \(n\). This forms an important perspective in determining how an additional term briefly impacted the growth in the sequence before being eclipsed by the actual homogeneous solution.

Asymptotic Analysis

Asymptotic analysis is a mathematical approach to understanding the behavior of functions as inputs become large. In solving recurrence relations, it plays a vital role in determining which parts of the solution grow at certain rates and contribute significantly to the overall solution:

• The main goal is to identify the term that predominantly drives the growth of \(S(n)\).
• For the exercise, it compares \(a^n\) and \(g(n)\), assessing their growth rates relative to one another.

In our specific case, asymptotic analysis shows that the term involving \(g(n)\) grows as \(c^n\), which is slower than \(a^n\) due to the condition \(1 \leq c < a\). Thus, the asymptotic analysis leads us to conclude that the function \(S(n)\) is primarily determined by \(a^n\), supporting the conclusion of the growth rate being \(\Theta(a^n)\). This strategy confirms the decisions made by observing exponential dominance through comparison of terms, excluding trivial components that diminish with increasing \(n\). Asymptotic analysis thus provides clarity on the often complex relationships within recurrence solutions.