Discovering the General Term for a Recurrence Relation

Beyond the straightforward calculation of the first few terms in a recurrence relation, the challenge lies in determining the general term. This problem, similar to the one discussed in the referenced post, has garnered interest for its intriguing structure and the insights it provides into advanced mathematical techniques. In this article, we explore a more accessible approach suitable for high school students.

Introduction to the Problem

We are given a recurrence relation that generates a sequence of fractions. Specifically:

x1 1/2
xn 1 xn/2 2n/2n 1 - 1

When we calculate the first few terms, we get:

x1 0.5, x2 0.8, x3 10/11, x4 22/23, x5 46/47, x6 94/95.

From Fractions to Linear Transformations

One of the key insights is to view the problem through the lens of projective geometry. Let us define a new representation for the sequence. Representing x as a ratio x a:b allows us to rewrite the recurrence relation in a linear form. This approach simplifies the complex transformation into a more manageable process.

Projective Line and Linear Transformation

In projective geometry, a pair of numbers (a:b) represents the ratio a/b, where 1:0 is considered infinity, i.e., 1/0. The transformation that maps xn to xn 1 can be expressed as:

(a:b) → (2b:3b-a)

This linear transformation can be represented by a matrix:

(begin{pmatrix}0 2 -1 3end{pmatrix} )

Iterating this transformation is equivalent to raising this matrix to the power of n. To find the explicit form, we need to diagonalize the matrix. The eigenvalues and eigenvectors of this matrix provide the key to solving the problem.

Diagonalization and Eigenvalues

The matrix (begin{pmatrix}0 2 -1 3end{pmatrix}) has eigenvalues 2 and 1. The corresponding eigenvectors are (begin{pmatrix}1 1end{pmatrix}) and (begin{pmatrix}2 1end{pmatrix}). Using these, we decompose the matrix:

(begin{pmatrix}0 2 -1 3end{pmatrix} begin{pmatrix}1 2 1 1end{pmatrix}begin{pmatrix}2 0 0 1end{pmatrix}begin{pmatrix}-1 -2 -1 -1end{pmatrix})

We then apply this to our initial value, x0 -1, to find:

(begin{pmatrix}x_nend{pmatrix} begin{pmatrix}1 2 1 1end{pmatrix}begin{pmatrix}2^n 0 0 1end{pmatrix}begin{pmatrix}-1 -2 -1 -1end{pmatrix}begin{pmatrix}-1 1end{pmatrix})

Multiplying this out and taking the ratio, we obtain the explicit form:

(x_n 1 - frac{2}{3 cdot 2^n - 2})

Or equivalently:

(x_n frac{2^n cdot 2^{n-1} - 2}{2^n cdot 2^{n-1} - 1})

Conclusion

By employing the concepts of projective geometry and linear transformations, we manage to find a general term for the given recurrence relation. This method not only provides a clear formula for the sequence but also highlights the power of advanced mathematical techniques in simplifying complex problems.