Question

Solve the recurrence \(T(n)=2 T(n-1)+n 2^{n}\), with the initial condition \(T(0)=1\).

Short answer

The solution to the recurrence is \( T(n) = \left(\frac{1}{2}n + 1\right)2^n \).

Step-by-step solution

Understand the Recurrence

Our task is to solve the recurrence relation given by \( T(n) = 2T(n-1) + n2^n \) with the initial condition \( T(0) = 1 \). This is a non-homogeneous recurrence relation due to the \( n2^n \) term.

Solve the Homogeneous Equation

First, solve the corresponding homogeneous equation \( T_h(n) = 2T_h(n-1) \). This can be solved by assuming a solution of the form \( T_h(n) = c \times 2^n \). The equation is satisfied for any constant \( c \).

Find a Particular Solution

Consider the non-homogeneous part, \( n2^n \). We propose a particular solution of the form \( T_p(n) = An2^n + B2^n \). Substituting into \( T(n) = 2T(n-1) + n2^n \) and matching coefficients can help us determine \( A \) and \( B \).

Substitute and Match Coefficients

Substitute \( T_p(n) = An2^n + B2^n \) and check:\[ T_p(n-1) = A(n-1)2^{n-1} + B2^{n-1} \]\[ T(n) = 2(A(n-1)2^{n-1} + B2^{n-1}) + n2^n = (2An-2A+2B+n)2^n \]Equate coefficients from \( T(n) = T_p(n) + T_h(n) = (An+B-n)2^n + n2^n \) to find values for \( A \) and \( B \).

Determine the Coefficients

From equating like terms in \[ (2An - 2A + 2B + n) = An + n \] We get two equations: 1. \( 2A = 1 \)2. \( -2A + 2B = 1 \)Solving these equations, \( A = \frac{1}{2} \) and \( B = 1 \).

Find the General Solution

The general solution of the recurrence \( T(n) \) combines the homogeneous and particular solutions:\[ T(n) = C2^n + \left(\frac{1}{2}n + 1\right)2^n = \left(C + \frac{1}{2}n + 1\right)2^n \]

Use Initial Condition to Solve for C

Substitute \( n = 0 \) and use \( T(0) = 1 \):\[ 1 = C \Rightarrow C = 0 \]Therefore, the solution that satisfies the initial condition is \[ T(n) = \left(\frac{1}{2}n + 1\right)2^n \].

Key concepts

Non-Homogeneous Recurrence

A non-homogeneous recurrence relation is distinct because it includes an additional function or sequence in its formula, unlike homogeneous recurrence relations. These types of recurrences are characterized by terms that do not solely depend on the previous terms in the sequence. Instead, an extra term, often called the non-homogeneous part, is present.

In the exercise, the recurrence relation is given by \(T(n) = 2T(n-1) + n2^n\), where the term \(n2^n\) is the non-homogeneous component. This term makes the equation more complex than a simple homogeneous recurrence relation because it introduces additional variability.

When dealing with such recurrences, the solution typically involves two parts: the solution to the associated homogeneous equation and a particular solution that accounts for the non-homogeneous part. Both of these solutions contribute to forming the general solution.

Homogeneous Solutions

To solve a non-homogeneous recurrence relation, the first step is often to find the solution to its homogeneous counterpart. This involves ignoring the non-homogeneous term and solving for the remaining relation.

In our exercise, the homogeneous part of the recurrence is \(T_h(n) = 2T_h(n-1)\). Solving this by assuming \(T_h(n) = c \, 2^n\) leads to a general solution that holds for any constant \(c\).

This solution essentially characterizes the behavior of the sequence without the influence of the non-homogeneous component. The form \(c \, 2^n\) emerges because it satisfies the equation \(T_h(n) = 2T_h(n-1)\) naturally, due to the exponential growth inherent in the functions involved.

Particular Solutions

Finding a particular solution involves determining a solution that specifically addresses the non-homogeneous component of the recurrence relation.

In the given problem, a proposed form for the particular solution is \(T_p(n) = An2^n + B2^n\). This form is chosen because it is capable of incorporating the non-homogeneous part \(n2^n\) effectively. By substituting \(T_p(n)\) into the original equation, you can solve for the constants \(A\) and \(B\).

Through algebraic manipulation and matching coefficients, you find \(A = \frac{1}{2}\) and \(B = 1\). These values allow the particular solution to cancel out the non-homogeneity introduced by \(n2^n\), thereby satisfying the entire equation when combined with the homogeneous solution.

Discrete Mathematics

Recurrence relations, such as the one addressed in this exercise, are a fundamental concept within discrete mathematics. Discrete mathematics deals with discrete elements and covers topics aligning with sequences, logic, and combinatorics.

Understanding recurrence relations is critical for fields like algorithm design, where these relations often describe the runtime complexity. They also appear in areas such as number theory and computer science.

Solving these relations involves identifying structures or patterns within sequences, as well as employing strategies to manage both homogeneous and non-homogeneous elements. By mastering such techniques, students can better understand various mathematical models used in theoretical and applied contexts.