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Chapter 7: Problem 37

Solve the recurrence relation $$a_{n}=\sqrt{\frac{a_{n-2}}{a_{n-1}}}$$ with initial conditions \(a_{0}=8, a_{1}=1 /(2 \sqrt{2})\) by taking the logarithm of both sides and making the substitution \(b_{n}=\lg a_{n}\)

Short Answer

Expert verified
The solution to the recurrence relation is \(a_{n} = 2^{3 - 5n}\).

Step by step solution

01

Apply the substitution

The problem suggests to make the substitution \(b_{n} = \lg a_{n}\). By substituting this into the given recurrence relation, we have \(\lg a_{n} = \lg \left(\sqrt{\frac{a_{n-2}}{a_{n-1}}}\right)\). Using the properties of logarithms, we can rewrite the right side as \(\frac{1}{2} (\lg a_{n-2} - \lg a_{n-1})\). This means the original recurrence relation is equivalent to \(b_{n} = \frac{1}{2} (b_{n-2} - b_{n-1})\).
02

Rewrite the equation

We can further rewrite the equation from Step 1 as \(b_{n} - \frac{1}{2}b_{n-1} = \frac{1}{2}b_{n-2}\). By multiplying all terms by 2, we have \(2b_{n} - b_{n-1} = b_{n-2}\), which is a standard form of a linear homogeneous recurrence relation with constant coefficients.
03

Solve the resulting equation

To solve the resulting equation: \(2b_{n} - b_{n-1} = b_{n-2}\), we can rearrange to get \(b_{n-2} - 2b_{n-1} = -b_{n}\). The characteristic equation of this is \(r^2 - 2r + 1 = 0\). Solving this gives us \(r = 1\) (with multiplicity 2) as roots. Therefore, the general solution to this recurrence relation is \(b_{n} = A + nB\), where A and B are constants to be determined by the given initial conditions.
04

Apply Initial Conditions

We apply the initial conditions given in the problem. They were given for \(a_{n}\), but we can assume \(b_{0} = \lg a_{0} = \lg 8 = 3\) and \(b_{1} = \lg a_{1} = \lg (1 / 2\sqrt{2}) = -2\). By substituting \(n = 0\) into the general solution \(b_{n} = A + nB\), we get \(A = 3\). Substituting \(n = 1\), we get \(-2 = 3 + B\), leading to \(B = -5\). Thus \(b_{n} = 3 - 5n\). Finally, since \(b_{n} = \lg a_{n}\), we can find \(a_{n} = 2^{b_{n}} = 2^{3 - 5n}\), which is the solution for the original recurrence relation.

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Key Concepts

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

Homogeneous Recurrence Relation
A homogeneous recurrence relation is a way to express each term in a sequence as a function of previous terms without any addition of other terms or constants. This type of relation is defined entirely by its earlier terms, making it a "pure" pattern without external influences. In the exercise, after transforming the original recurrence relation into a log-based form, we obtain a linear homogeneous recurrence relation described by:
  • Equation: \( 2b_{n} - b_{n-1} = b_{n-2} \)
  • Form: The form \( b_{n} = \) solely relies on terms \( b_{n-1} \) and \( b_{n-2} \), illustrating the homogeneity.
These relations are characterized by constant coefficients, which means the factors multiplying each term in the recurrence are constants.
Characteristic Equation
When faced with recurrence relations, the characteristic equation comes in handy for finding solutions. It's a polynomial equation derived from a recurrence relation that helps us pinpoint the general solution or patterns of the sequence. For our exercise, the transition from the recurrence \( 2b_{n} - b_{n-1} = b_{n-2} \) forms the characteristic equation:
  • Equation: \( r^2 - 2r + 1 = 0 \)
  • Roots: Solving gives the root \( r = 1 \), with a multiplicity of 2.
The multiplicity indicates how many times a root appears as a solution of the characteristic polynomial. Since we have a repeated root, the solution takes the form \( b_{n} = A + nB \), revealing a predictable structure in our sequence.
Logarithmic Substitution
In this exercise, logarithmic substitution is an essential technique used to simplify and solve the recurrence relation. By substituting \( b_{n} = \lg a_{n} \), we transform a complex equation into a more manageable form:
  • Original equation: \( a_{n} = \sqrt{\frac{a_{n-2}}{a_{n-1}}} \)
  • Translated using logs: \( b_{n} = \frac{1}{2}(b_{n-2} - b_{n-1}) \)
Logarithms help to linearize multiplicative relationships, turning products into sums (or differences), and thus making them easier to tackle mathematically. Substitution aids in unveiling latent structures and patterns within a recurrence relation, smoothing the path to its solution.
Initial Conditions
Initial conditions are pivotal as they allow us to tailor the general solution of a recurrence to fit a specific sequence. In the provided solution, they specify the starting point of the sequence:
  • \( a_0 = 8 \)
  • \( a_1 = \frac{1}{2\sqrt{2}} \)
However, because we worked with \( b_{n} \), it translated to:
  • \( b_0 = \lg 8 = 3 \)
  • \( b_1 = \lg \left(\frac{1}{2\sqrt{2}}\right) = -2 \)
These conditions are fundamental in determining constants \( A \) and \( B \) for the equation \( b_{n} = 3 - 5n \). They ensure that the solution aligns accurately with the initial terms of the sequence, refining the solution to address specific initial setups.

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Most popular questions from this chapter

In Exercises \(47-52,\) Ackermann's original fiuncrion \(A O\) is defined by $$ \begin{array}{cc} A O(0, y, z)=z+1 & y, z \geq 0 \\ A O(1, y, z)=y+z & y, z \geq 0 \\ A O(2, y, z)=y z & y, z \geq 0 \\ A O(x+3, y, 0)=1 & x \geq 0, y \geq 1, \\ A O(x+3, y, z+1)= & \\ A O(x+2, y, A O(x+3, y, z)) & x \geq 0, y \geq 1, z \geq 0 . \end{array} $$ Exercises \(47-52\) refer to the function \(A O\) and to Ackermann's function \(A\). $$ \text { Show that } A(x, y)=A O(x, 2, y+3)-3 \text { for } x, y \geq 0 \text { . } $$

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