a) Find all solutions of the recurrence relation an=2 an-1+2 n^2 . b) Find the solution of the r

A) Find all solutions of the recurrence relation an=2 an-1+2...

Key Concepts

Method of Undetermined Coefficients

The method of undetermined coefficients is a systematic approach used to find a particular solution to a non-homogeneous linear recurrence with constant coefficients. By guessing a form that matches the type of the non-homogeneous term and then determining the coefficients by equating the terms, this method simplifies the process of solving the recurrence.

Initial Conditions

Initial conditions provide specific starting values for the recurrence relation, which are essential for determining the unique sequence corresponding to the general solution. After obtaining the general solution comprising the homogeneous and particular parts, initial conditions are applied to solve for any arbitrary constants, thereby zeroing in on the solution that meets the original problem's criteria.

Particular Solution

A particular solution is a specific solution that satisfies the entire non-homogeneous recurrence relation including the added external function. Its form is often guessed based on the type of the non-homogeneous term. For the quadratic function in the recurrence, one typically assumes a polynomial form of similar degree and determines the coefficients by substituting it back into the original equation.

Recurrence Relation

A recurrence relation defines each term in a sequence as a function of one or more of the previous terms. In the context of the given problem, it specifically sets a term equal to a linear combination of earlier terms plus an additional function (in this case, a quadratic term), thereby forming a structured process to generate the sequence.

Linear Recurrence with Constant Coefficients

This type of recurrence relation is characterized by the fact that each term is expressed as a linear combination of previous terms with constant coefficients. Understanding this class of recurrences is important because it allows the use of systematic methods (like finding characteristic equations) to solve the homogeneous part of the recurrence.

Homogeneous Solution

The homogeneous solution refers to the solution of the part of the recurrence relation that involves only the terms from previous sequence values without any external (non-homogeneous) function. This solution is obtained by setting the non-homogeneous part to zero, and solving the resulting characteristic equation yields the general form of the solution for that part.

Transcript

00:01 In part a, we're given a recurrence relation, and we're asked to find all solutions.

00:07 Recurrence relation is an equals 2an minus 1 plus 2n squared.

00:13 This is a linear non -homogeneous recurrence relation, and so the associated linear homogeneous recurrence relation is an equals 2a .n.

00:28 Minus 1.

00:28 We have that the characteristic equation is r minus 2 equals 0, so the characteristic root is 2, and therefore we have that the general form for the solution to the associated linear homogenous recurrence relation is alpha times 2 to the n, where alpha is some constant, and and we have that in the non -homogeneous equation, f of n is equal to n squared.

01:24 We have this is the same as n squared times 1 to the n.

01:29 One is not a characteristic root.

01:32 And we have the n squared as a polynomial of degree 2.

01:37 So it follows that the general form for a solution, which is particular, is going to be p2 n squared plus p1n plus p0 times 1 to the n which is just 1 and we have that all solutions are given by the sum of the particular solution and the solution to the associated homogenous linear occurrence relation which is going to be my mistake.

02:38 This should be plus 2n squared.

02:41 So f of n is not n squared, it's 2n squared.

03:12 So to determine the coefficients on the particular solution, we'll plug it back into our non -hymogynous equation.

03:25 So we have that p2n squared plus p1n plus p0 is equal to 2 times and this is going to be p2 times n minus 1 squared plus p1 times n minus 1 plus p 0 and all of this plus 2n squared we can write this all terms on one side so we have p2 n squared plus p1n plus p0 is equal to 2 p2 times n squared and then minus 2n plus 1 plus p1 times n minus 1 plus p0 plus p0 plus 2n squared from which we obtain p2 n squared plus p 1n plus p 0 is equal to 2 p2 n squared but this is 2 p2 n squared plus 2 p2 n squared plus 2 n squared and then for n we have the coefficient negative 4 p2 and we have minus 2 p1 times n and for the constant term we have 2 p2 minus p1 i mean and plus 2 p0 and so we have that 0 is equal to p2 plus 2 n squared plus this is negative 4p2 minus 3p1 times n and then plus 2 p2 minus 2 p1 and plus p0 and it follows that the coefficients of the polynomial are all 0 so we have that p2 is equal to negative 2, we have that p1 is going to be equal to 4p2 over negative 3, so negative 4 thirds, p2, which is equal to negative 4 thirds of negative 2, which is 8 thirds, and p2 is equal to 1⁄2 of this is negative, not p2 though, i'm solving for p0.

07:23 My mistake, p0 is equal to negative 2p2 plus 2p1, which we have is equal to negative 2 times negative 2, which is 4, plus 2 times 8 thirds, which is 16 thirds.

07:43 So this is going to be 12 plus 16 is 28 thirds...

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