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Solve the recurrence relation \(a_{n}=2 a_{n-1}-a_{n-2}\) (a) What is the solution if the initial terms are \(a_{0}=1\) and \(a_{1}=2 ?\) (b) What do the initial terms need to be in order for \(a_{9}=30 ?\) (c) For which \(x\) are there initial terms which make \(a_{9}=x ?\)

Short Answer

Expert verified
Part (a): The solution with initial terms \(a_0=1\) and \(a_1=2\) is \(a_n=1+n\). Part (b): The initial terms needed for \(a_9=30\) are \(a_0=1\) and \(a_1=1+29/9\). Part (c): Any real number x can be the 9th term with appropriate initial terms.

Step by step solution

01

Identify the characteristic equation

For the recurrence relation \(a_n=2a_{n-1}-a_{n-2}\), we will find the characteristic equation. Assume the solution has the form \(a_n=r^n\). Substituting in the recurrence gives \(r^n = 2r^{n-1} - r^{n-2}\). Divide every term by \(r^{n-2}\) to obtain \(r^2 = 2r - 1\).
02

Solve the characteristic equation

We can solve the characteristic equation found in Step 1: \(r^2 - 2r + 1 = 0\). The solution is \((r-1)^2 = 0\), which has a double root at \(r=1\). This gives us a general solution of the form \(a_n = A\bullet1^n + B\bullet n\bullet1^n\), where A and B will be determined using initial conditions.
03

Determine A and B using initial conditions (Part a)

Use the initial terms \(a_0=1\) and \(a_1=2\) to find constants A and B. Plugging in, we get \(a_0 = A\bullet1^0 + B\bullet0\bullet1^0 = A = 1\) and \(a_1 = A\bullet1^1 + B\bullet1\bullet1^1 = A + B = 2\). Since we found A to be 1, B must be 1 as well. Therefore, the specific solution for part (a) is \(a_n = 1 + n\).
04

Use solution to find initial conditions for part (b)

Given \(a_9=30\), we use the specific solution \(a_n = 1 + n\) to find \(a_9 = 1 + 9 = 10\). To find a different solution that satisfies \(a_9=30\), we look for initial conditions that give us a sequence that has a different A and B. If the 9th term is 30, our equation from Step 3 becomes \(30 = A + 9B\). We can use one of the initial conditions, say A (since A must remain 1 to satisfy \(a_0=1\)), and adjust B to satisfy the equation \(30 = 1 + 9B\), which gives us \(B = 29 / 9\). Therefore, the sequence that satisfies the recurrence relation and has the 9th term of 30 is generated by the initial conditions \(a_0=1\) and \(a_1=1+29/9\).
05

Determine possible values for x in part (c)

Since we have established that \(a_n=A+nB\), and as long as A is fixed at 1 to meet the condition that \(a_0 = 1\), then B can be any real number. Thus for any 9th term \(a_9=x\), we can solve for B in the equation \(x = A + 9B\). So there will always exist initial terms \(a_0=1\) and \(a_1=1+B\) that allow for \(a_9=x\) for any real number value of x.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation
In studying recurrence relations, the characteristic equation plays a pivotal role. It is essentially a tool that translates a problem about sequences into a problem about polynomials, which are generally easier to solve. When faced with a linear homogeneous recurrence relation like \( a_n = 2a_{n-1} - a_{n-2} \), we seek solutions that follow a particular pattern, often something of the form \( a_n = r^n \).

By substituting this assumed form into the given relation and simplifying, we end up with a polynomial equation in terms of \( r \), namely the characteristic equation. For the recurrence relation provided, the characteristic equation is \( r^2 = 2r - 1 \), which can be rearranged to \( r^2 - 2r + 1 = 0 \). It's more than just finding the roots; it's about understanding that the solutions to this equation are the very building blocks of our sequence, guiding us to the general form of the sequence's terms.
Initial Conditions
When unraveling the mysteries of a sequence governed by a recurrence relation, initial conditions serve as the much-needed compass. They act as the starting points, setting the stage for the entire sequence. Specifically, they allow us to determine the specific constants that will satisfy the relation for a particular sequence.

For example, with the initial conditions \( a_0=1 \) and \( a_1=2 \) and the general solution to the recurrence relation \( a_n = A + Bn \), these conditions help us find the unique values of A and B. By substituting these initial terms into the general solution, we can solve for A and B, giving us the explicit formula for the sequence that traces back to those initial conditions. Likewise, through adjusting these constants, we can architect sequences fulfilling any desired properties, like a specified term further down the sequence.
Sequence and Series
A sequence is essentially a list of numbers usually generated based on some underlying rule, which, in our case, is the recurrence relation. It is the narrative of a sequence that tells us the progression of terms, one following another, creating a path of numerical breadcrumbs. A series, on the other hand, is the summation of the terms of a sequence, a collective total that accumulates value with each successive term.

In the context of recurrence relations and their solutions, understanding sequences and their behavior is indispensable. The explicit formula derived from the initial conditions and the characteristic equation isn't just an abstract mathematical expression; it concretizes the very nature of the sequence’s unfolding saga. Moreover, comprehending the explicit formula also opens up possibilities for exploring series, where we can venture into intriguing questions of convergence and the sums of infinite terms.

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Most popular questions from this chapter

Consider the three sequences below. For each, find a recursive definition. How are these sequences related? (a) \(2,4,6,10,16,26,42, \ldots .\) (b) \(5,6,11,17,28,45,73, \ldots\) (c) \(0,0,0,0,0,0,0, \ldots\)

Consider the sequence \(5,9,13,17,21, \ldots\) with \(a_{1}=5\) (a) Give a recursive definition for the sequence. (b) Give a closed formula for the \(n\) th term of the sequence. (c) Is 2013 a term in the sequence? Explain. (d) How many terms does the sequence \(5,9,13,17,21, \ldots, 533\) have? (e) Find the sum: \(5+9+13+17+21+\cdots+533\). Show your work. (f) Use what you found above to find \(b_{n},\) the \(n^{t h}\) term of \(1,6,15,28,45, \ldots,\) where \(b_{0}=1\)

You will prove that the Fibonacci numbers satisfy the identity \(F_{n}^{2}+\) \(F_{n+1}^{2}=F_{2 n+1} .\) One way to do this is to prove the more general identity, $$ F_{m} F_{n}+F_{m+1} F_{n+1}=F_{m+n+1} $$ and realize that when \(m=n\) we get our desired result. Note that we now have two variables, so we want to prove this for all \(m \geq 0\) and all \(n \geq 0\) at the same time. For each such pair \((m, n),\) let \(P(m, n)\) be the statement \(F_{m} F_{n}+F_{m+1} F_{n+1}=F_{m+n+1}\) (a) First fix \(m=0\) and give a proof by mathematical induction that \(P(0, n)\) holds for all \(n \geq 0 .\) Note this proof will be very easy. (b) Now fix an arbitrary \(n\) and give a proof by strong mathematical induction that \(P(m, n)\) holds for all \(m \geq 0\). (c) You can now conclude that \(P(m, n)\) holds for all \(m, n \geq 0 .\) Do you believe that? Explain why this sort of induction is valid. For example, why do your proofs above guarantee that \(P(2,3)\) is true?

Consider the sequence \(2,7,15,26,40,57, \ldots\) (with \(a_{0}=2\) ). By looking at the differences between terms, express the sequence as a sequence of partial sums. Then find a closed formula for the sequence by computing the \(n\) th partial sum.

Consider the sequence \(5,11,19,29,41,55, \ldots\) Assume \(a_{1}=5\) (a) Find a closed formula for \(a_{n},\) the \(n\) th term of the sequence, by writing each term as a sum of a sequence. Hint: first find \(a_{0},\) but ignore it when collapsing the sum. (b) Find a closed formula again, this time using either polynomial fitting or the characteristic root technique (whichever is appropriate). Show your work. (c) Find a closed formula once again, this time by recognizing the sequence as a modification to some well known sequence(s). Explain.

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