Solve the following recurrence relations by using the method of generating functions as describe

Solve the following recurrence relations by using the method...

Key Concepts

Initial Conditions

Initial conditions provide the necessary starting values of the sequence and are essential for determining the unique solution of a recurrence relation. When solving via generating functions, these conditions are used to resolve any constants or unknown coefficients that appear in the resulting algebraic expression. Without them, the solution to the recurrence would remain ambiguous.

Algebraic Manipulation and Partial Fractions

Once the generating function is set up, algebraic manipulation is used to isolate and solve for it. This process often involves rewriting the generating function in a form that is suitable for partial fraction decomposition, a technique that breaks down a complex rational function into simpler fractions. These simpler fractions have known power series expansions, which can then be inverted term-by-term to provide the closed-form solution for the original recurrence.

Generating Functions

A generating function is a formal power series where the coefficient of each power of the variable represents a term in the sequence defined by the recurrence relation. Converting a recurrence into its generating function transforms the problem from one of recursion to one of algebra, making it easier to manipulate and solve. This method leverages power series techniques to obtain expressions that can ultimately be reversed to yield explicit formulas for the sequence.

Recurrence Relations

Recurrence relations are equations that define a sequence recursively by expressing each term as a function of preceding terms. In the context of solving these problems, the recurrences are linear and involve constant coefficients, and are often “homogeneous” meaning that every term is generated solely by its predecessors without an additional external term. The goal is usually to find a closed-form expression that satisfies both the recurrence and the provided initial conditions.

Power Series and Index Shifting

Using power series involves representing the sequence as a series expansion where the coefficients correspond to the terms of the sequence. Index shifting is a crucial step in aligning the terms when expressing the recurrence in the generating function form; this ensures that like powers of the variable are correctly combined. It enables the translation of recursion relations into algebraic equations involving the generating function.

Transcript

00:01 We are given recurrence relations with initial conditions when we are asked to solve.

00:12 In part a, we're given the recurrence relation, a .n equals 2, a .n minus 1, for n greater than are equal to 1, and a0 is 3.

00:36 We have that the characteristic equation is r squared minus 2r equals 0, or we have that r is equal to 0 or 2.

00:58 So we have distinct roots, and therefore, it follows that the solution is of the form an equals alpha 1, 0 to the n, plus alpha 2, 2 to the n.

01:27 And substituting, we have that a0, which is 3, is equal to, let's see, this is going to be 0 plus alpha 2 times.

01:47 1 and so we get that alpha 2 is equal to 3 and we have that it follows in general an is going to be 3 times 2 to the n.

02:14 In part b we are given the linear homogenous recurrence relation, an equals an minus 1 for n greater than or equal to 1 and a0 equals 2.

02:37 This is actually easy to solve if we notice simply a sequence of twos.

02:50 However, to use a different method, we have the characteristic equation is r squared minus r equals 0, which has roots r equals 0 and r equals 1, which are distinct.

03:06 And so it follows the general form of the solution for this relation is a n equals alpha 1 times 0 to the n plus alpha 2 times 1 to the n, which is really the same as alpha 2.

03:26 And we have from our initial condition that 2, which is a 0, is the same as alpha 2.

03:34 So it follows that in general solution is in part c we're given their occurrence relation an equals 5, a n minus 1 minus 6, a n minus 2, where n is great.

04:09 Greater than equal to 2, and we have the initial conditions, a 0 equals 1, and a 1 equals 0.

04:29 The characteristic equation for this linear, homogenous occurrence relation is r squared minus 5r plus 6 equals 0.

04:43 And this can be factored as r minus 2 times r minus 3 equals 0, so that r is equal to 2 or 3, these values are distinct, so it follows that the general solution is ann equals alpha 1, 2 to the n, plus alpha 2, 3 to the n.

05:08 And using our initial conditions, we have that 1, which is a0, is equal to a1 times 1 plus a2 times 1.

05:23 And we have that 0, which is a1, is 2 alpha 1 plus 3 alpha 2.

05:35 And so if we subtract twice equation 1 from equation 2, we get negative 2 is equal to alpha 2.

05:50 And we have that by back substitution, alpha 1 is 1 minus alpha 2, which is 1 minus negative 2 or 3.

06:01 And so the general form is a .n equals 3 times 2 to the n minus 2 times 3 to the n.

06:28 In part d, we are given the linear homogenous recurrence relation, a .n equals 4, a .n minus 1, minus 4, a .n minus 2.

06:49 4n greater than equal to 2.

06:51 And we're given the initial conditions, a0 equals 6, and a1.

06:58 Equals 8.

07:14 The characteristic equation is r squared minus 4r plus 4 equals 0, which can be factored as r minus 2 squared equals 0.

07:26 And so the characteristic roots are r equals 2 with a multiplicity of 2.

07:36 And so it follows that the general solution is an equals alpha 1 times 2 to the n plus alpha 2 times n 2 to the n.

07:56 And from our initial conditions, we have that 6, which is a0, this is equal to alpha 1, and 8, which is a1, is equal to 2 alpha 1 plus 2 alpha 2.

08:23 And then using back substitution, we have that alpha 2 is 1 half of 8 minus 2 times alpha 1, which is 6.

08:41 So we get 1 half of 8 minus 12 is 1 half of negative 4 is negative 2.

08:52 And so the specific solution is a .n equals 6 times 2 to the n plus negative 2 or minus.

09:09 2 n times 2 to the n.

09:26 In part e, we're given the relation, a .n equals negative 4, a .n minus 1, minus 4, a .n minus 2, where n is greater than are equal to 2...

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