1. Background Knowledge#
1. Recurrence Relation and General Term Formula#
In mathematics, a sequence is an ordered collection of numbers, typically represented as $a_1, a_2, a_3, \ldots$. A sequence can be defined by a recurrence relation, such as $a_{n+1} = pa_n + q^n$. The recurrence relation provides the relationship between adjacent terms in the sequence. The general term formula is a direct expression for any term in the sequence, usually in the form of $a_n$.
2. Preliminary Understanding of Matrices#
A matrix is a tabular data structure commonly used to represent linear transformations. The basic operations of matrices include addition, scalar multiplication, and matrix multiplication. Matrix addition involves adding the corresponding elements of two matrices of the same size; scalar multiplication involves multiplying each element of a matrix by the same scalar; matrix multiplication involves multiplying two matrices to obtain a new matrix. For sequence problems, we can convert coefficients into matrices and simplify the recurrence relations through matrix multiplication, thereby deriving the general term formula.
2. Introduction to the Problem#
Suppose we have a sequence with the recurrence relation $a_{n+1} = pa_n + q^n$ and the initial condition $a_1$. We want to find the general term formula for the sequence.
Conventional Method#
We use the method of undetermined coefficients to solve:
Given the recurrence relation $a_{n+1} = pa_n + q^n$
Our goal is to find a form $a_n + xq^{n+1} = p(a_n + xq^n)$, so that we can determine $x$ by comparing the forms on both sides, and thus find the general term formula for the sequence.
First, we rewrite the original recurrence as $a_{n+1} + xq^{n+1} = pa_n + q^n + xq^{n+1}$
Then, we want the form on the left to match the form on the right, i.e., $a_{n+1} + xq^{n+1} = p(a_n + xq^n)$
This means $pa_n + q^n + xq^{n+1} = pa_n + pxq^n$
By comparing the terms on both sides, we can derive $q^n + xq^{n+1} = pxq^n$
$q^n(1 + xq) = pxq^n$
Thus, we have $1 + xq = px$
$1 = px - xq$
$1 = x(p - q)$
$x = \frac{1}{p - q}$
So, if we find such an $x$, we can construct a new sequence $b_n = a_n + xq^n$, which will satisfy a simpler recurrence relation $b_{n+1} = pb_n$.
Finally, we can solve for the general term formula of the new sequence $b_n$, and then find the general term formula for the original sequence $a_n$ using $a_n = b_n - xq^n$.
Specifically, we have $b_n = a_n + \frac{q^n}{p-q}$
Since $b_{n+1} = pb_n$, this is a geometric sequence with the general term formula $b_n = b_1 \cdot p^{n-1}$
where $b_1 = a_1 + \frac{q}{p-q}$
Thus, $b_n = (a_1 + \frac{q}{p-q}) \cdot p^{n-1}$
So $a_n = (a_1 + \frac{q}{p-q}) \cdot p^{n-1} - \frac{q^n}{p-q}$
This is the general term formula for the sequence $a_n$: $a_n = (a_1 + \frac{q}{p-q}) \cdot p^{n-1} - \frac{q^n}{p-q}$
3. Provoking Thoughts#
In the solving process, we are always transforming the coefficients. Is this transformation related to some mathematical tool?
In fact, the linear operations of complex numbers and vectors also have similar properties. For example, complex numbers can be represented as two-dimensional vectors, and operations such as vector addition and scalar multiplication can be represented using matrices.
Linear Operations of Complex Numbers and Vectors#
Complex numbers can be viewed as a special case of two-dimensional vectors, where the real part corresponds to the x-component of the vector, and the imaginary part corresponds to the y-component. The addition and multiplication of complex numbers are similar to linear combinations in vector spaces.
Linear operations in vector spaces include vector addition, scalar multiplication, and linear combinations of vectors. These operations can be represented using matrices, and matrix multiplication provides a method to implement these transformations.
Example#
Suppose we have two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, where $a, b, c, d$ are all real numbers.
Addition#
The addition of complex numbers can be expressed as $z_1 + z_2 = (a + bi) + (c + di) = (a + c) + (b + d)i$
We can represent this operation using matrices: $\begin{pmatrix} a & b \ -b & a \end{pmatrix} + \begin{pmatrix} c & d \ -d & c \end{pmatrix} = \begin{pmatrix} a+c & b+d \ -(b+d) & a+c \end{pmatrix}$
Subtraction#
The subtraction of complex numbers can be expressed as $z_1 - z_2 = (a + bi) - (c + di) = (a - c) + (b - d)i$
Similarly, we can represent this with matrices: $\begin{pmatrix} a & b \ -b & a \end{pmatrix} - \begin{pmatrix} c & d \ -d & c \end{pmatrix} = \begin{pmatrix} a-c & b-d \ -(b-d) & a-c \end{pmatrix}$
4. Solving the Initial Problem with Matrices#
We will again use the matrix method to solve the initial problem:
Define the state vector $\mathbf{s}n = \begin{pmatrix} a_n \ q^n \end{pmatrix}$, and construct the matrix $A = \begin{pmatrix} p & 1 \ 0 & q \end{pmatrix}$, such that $\mathbf{s}{n+1} = A \mathbf{s}_n$
Calculate the power of the matrix $A^{n-1}$
$A^{n-1} = \begin{pmatrix} p & 1 \ 0 & q \end{pmatrix}^{n-1}$
Since $A$ is an upper triangular matrix, its powers remain upper triangular, with diagonal elements being the powers of $p$ and $q$: $A^{n-1} = \begin{pmatrix} p^{n-1} & * \ 0 & q^{n-1} \end{pmatrix}$
To find the upper right element, we can use mathematical induction: $A^2 = \begin{pmatrix} p^2 & p + q \ 0 & q^2 \end{pmatrix}$
$A^3 = \begin{pmatrix} p^3 & p^2 + pq + q \ 0 & q^3 \end{pmatrix}$
In general, $A^{n-1} = \begin{pmatrix} p^{n-1} & \sum_{k=0}^{n-2} p^kq^{n-2-k} \ 0 & q^{n-1} \end{pmatrix}$
Finally, $\mathbf{s}_n = A^{n-1} \mathbf{s}_1$
By calculating $\mathbf{s}_n$, we can obtain the expression for $a_n$.
Thus, the matrix method provides a new perspective for solving recurrence sequences.
This article is synchronized by Mix Space to xLog. The original link is https://blog.qwq.my/posts/math/reflections-on-the-method-of-undetermined-coefficients-for-finding-general-term-formulas