Question

Solve these recurrences using back substitution. Verify the solutions are correct by induction. (a) \(\quad a_{n}=\left\\{\begin{array}{ll}3 a_{n / 2}+4 & \text { for } k>0, n=2^{k} \\ 7 & \text { for } n=1\end{array}\right.\) (b) \(a_{n}=\left\\{\begin{array}{ll}5 a_{n / 5}+7 & \text { for } k>0, n=5^{k} \\\ 12 & \text { for } n=1\end{array}\right.\) (c) \(a_{n}=\left\\{\begin{array}{ll}2 a_{n / 3}+5 & \text { for } k>0, n=3^{k} \\\ 7 & \text { for } n=1\end{array}\right.\) (d) \(a_{n}=\left\\{\begin{array}{ll}7 a_{n / 4}+3 & \text { for } k>0, n=4^{k} \\\ 1 & \text { for } n=1\end{array}\right.\)

Short answer

Solutions: (a) \(a_n = 3^k \cdot 7 + 2(3^k - 1)\); (b) \(a_n = 5^k \cdot 12 + \frac{7(5^k - 1)}{4}\); (c) \(a_n = 2^k \cdot 7 + 5(2^k - 1)\); (d) \(a_n = 7^k + \frac{3(7^k - 1)}{6}\).

Step-by-step solution

Solving by Back Substitution (a)

Given: \(a_n = 3a_{n/2} + 4\) for \(n = 2^k\) and \(a_1 = 7\). Find a pattern by substituting back into the recurrence:- \(a_n = 3a_{n/2} + 4\)- \(a_{n/2} = 3a_{n/4} + 4\)- Substituting back: \[a_n = 3(3a_{n/4} + 4) + 4 = 9a_{n/4} + 12 + 4\]- Continue substitution: \[= 9(3a_{n/8} + 4) + 16 = 27a_{n/8} + 36 + 16 = 27a_{n/8} + 52\]- General form: \(a_n = 3^ka_{n/n} + 4(3^{k-1} + \ldots + 1)\)Use the geometric series formula: \(4(3^k - 1)/2\) when summing \(3^{k-1} + \cdots + 1\). So:\[a_n = 3^k \cdot a_1 + \frac{4(3^k-1)}{2} = 3^k \cdot 7 + 2(3^k - 1)\]\[= 3^k \cdot 7 + 2 \times 3^k - 2 = (3^k+2)(7) - 2\]

Verify by Induction (a)

Base Case: For \(n=1\), \(a_1 = 7\), matches the initial condition.Inductive Step: Assume \(a_{n/2} = 7(3^{k-1}) + (2 \cdot (3^{k-1}) - 2)\) holds, prove \(a_n = 7(3^k) + (2 \cdot (3^k) - 2)\):\[a_n = 3(7(3^{k-1}) + 2 \cdot (3^{k-1}) - 2) + 4\]\[= 7(3^k) + 2(3^k) - 6 + 4 = 7(3^k) + 2(3^k) - 2\]Hence, verified.

Solving by Back Substitution (b)

Given: \(a_n = 5a_{n/5} + 7\) for \(n = 5^k\) and \(a_1 = 12\):- Substitute in:- \(a_{n/5} = 5a_{n/25} + 7\)- Substituting back: \[a_n = 5(5a_{n/25} + 7) + 7 = 25a_{n/25} + 35 + 7\]- General form: \(a_n = 5^ka_{n/n} + 7(5^{k-1} + \ldots + 1)\)Geometric series: \(7(5^k - 1)/4\), hence:\[a_n = 5^k \cdot 12 + \frac{7(5^k-1)}{4}\]

Verify by Induction (b)

Base Case: For \(n=1\), \(a_1 = 12\), matches the initial condition.Inductive Step: Assume \(a_{n/5} = 5^{k-1}\cdot 12 + \frac{7(5^{k-1} - 1)}{4}\),\[a_n = 5\left(5^{k-1}\cdot 12 + \frac{7(5^{k-1}-1)}{4}\right) + 7\]Simplified, matches to \[5^k \cdot 12 + \frac{7(5^k - 1)}{4}\].

Solving by Back Substitution (c)

Given: \(a_n = 2a_{n/3} + 5\) for \(n = 3^k\) and \(a_1 = 7\).- Back substitute: \[a_{n/3} = 2a_{n/9} + 5\]- Continue back substitution: \[a_n = 2(2a_{n/9} + 5) + 5 = 4a_{n/9} + 10 + 5\]- General form is \(a_n = 2^ka_1 + 5(2^{k-1} + \ldots + 1)\)Apply geometric series sum: \(5(2^k - 1)\), hence:\[ a_n = 2^k \cdot 7 + 5(2^k - 1)\]

Verify by Induction (c)

Base Case: For \(n=1\), \(a_1 = 7\), matches the initial condition.Inductive Step: Assume \(a_{n/3} = 2^{k-1}\cdot 7 + 5(2^{k-1} - 1)\), prove for \(a_n\):\[a_n = 2(2^{k-1}\cdot 7 + 5(2^{k-1} - 1)) + 5\]Simplifies to match \[2^k \cdot 7 + 5(2^k - 1)\].

Solving by Back Substitution (d)

Given: \(a_n = 7a_{n/4} + 3\) for \(n=4^k\) and \(a_1 = 1\).- Back substitute: \[a_{n/4} = 7a_{n/16} + 3\]- Continue substitution to form: \[a_n = 7^k a_1 + 3(7^{k-1} + \ldots + 1)\]Sum the geometric series: \(3(7^k - 1)/6\), hence:\[a_n = 7^k + 3\cdot \frac{7^k - 1}{6}\].

Verify by Induction (d)

Base Case: For \(n=1\), \(a_1 = 1\), matches the initial condition.Inductive Step: Assume \(a_{n/4} = 7^{k-1} + \frac{3(7^{k-1} - 1)}{6}\),\[a_n = 7(7^{k-1} + \frac{3(7^{k-1} - 1)}{6}) + 3\], simplifying matches \[7^k + \frac{3(7^k - 1)}{6}\].

Key concepts

Back Substitution

Back substitution is a powerful method used to solve recurrence relations, which often appear in problems dealing with sequences. To apply back substitution, you repeatedly substitute the recurrence relation into itself to find a general pattern or formula.

This technique starts with expressing the given term in terms of its preceding terms, gradually breaking it down into simpler parts. For example, if you're given a relation like \( a_n = 3a_{n/2} + 4 \), you would substitute \( a_{n/2} \) with its own formula derived from the recurrence equation. This step-by-step process continues until a clear pattern emerges.

The goal is to express the nth term in terms of known initial conditions, allowing you to derive a closed formula.

• This method helps when sequences are defined with mutual dependencies, as typical in recursive algorithms.
• It’s crucial in computer science and discrete mathematics for automating complex tasks.

Recognizing patterns during substitution can reveal forms similar to geometric series, simplifying the final formula.

Mathematical Induction

Mathematical induction is a technique used to prove the correctness of statements or formulas, especially those derived from recurrence relations like those solved using back substitution. It consists of two key steps: the base case and the inductive step.

The base case involves showing that the statement holds true for the initial term of the sequence (e.g., \( n = 1 \)). This establishes the foundational truth of the statement.

The inductive step involves assuming that the statement is true for some arbitrary term \( n = k \) (the inductive hypothesis) and then proving it for the next term \( n = k+1 \). This step requires showing that if the statement holds for one case, it logically holds for the next.

• Induction helps ensure the pattern or formula derived through back substitution is indeed valid for all terms of the sequence.
• This method is essential in discrete mathematics and computer science to guarantee algorithms perform as expected.

By combining induction with back substitution, one can confidently apply derived formulas to solve problems across various fields.

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. It plays a crucial role in solving recurrence relations solved by back substitution.

In many recurrence problems, you may encounter terms that form a geometric series. For example, when you substitute back repeatedly in a recurrence relation, the additive components often follow a geometric progression.

An example of a geometric series is \( 1 + x + x^2 + \, \cdots + x^n \), whose sum can be calculated using the formula: \( \frac{x^{n+1} - 1}{x - 1} \). This sum formula is invaluable when simplifying expressions found in recurrence relations after substitution.

• The ability to recognize and sum these series simplifies solving complex recurrence relations, reducing complexity in computation.
• Knowledge of geometric series is vital for efficiently solving many problems in discrete mathematics and related areas.

Once detected, applying the geometric series sum formula allows quicker simplification of otherwise elaborate expressions.

Discrete Mathematics

Discrete mathematics involves the study of countable, distinct elements and is the mathematical underpinning of computer science. It's particularly relevant when discussing recurrence relations, as these are a fundamental concept in this field.

In discrete mathematics, recurrence relations describe processes where the next event or item depends on previous outcomes. This is perfectly aligned with the kinds of algorithms and data structures often explored in computer programming, where recursion and iteration are key.

Key areas where recurrence relations and discrete mathematics intersect include algorithm analysis, combinatorics, graph theory, and cryptography.

• Understanding recurrence and solving them using techniques like back substitution and induction are central skills in discrete mathematics.
• These solutions help in establishing the efficiency and feasibility of algorithms used in computational tasks.

Discrete mathematics provides the tools needed to formalize and solve problems found in computer science, significantly impacting technology advancements.