Find an explicit formula for a quadratic sequence using systems of equations
Use matrix systems to derive a general expression for the constants of an explicit formula for a quadratic sequence
What is a Quadratic Sequence?
A quadratic sequence is a sequence defined by a quadratic function.
For instance, consider the sequence:
It has a unique characteristic where the second order difference of each consecutive term is a constant value.
This means that by taking the difference between two consecutive terms, we get:
Notice how the differences isn’t constant, but they resemble the terms of an arithmetic sequence.
This means that the difference of the differences of each term should be constant.
Definition Quadratic Sequences
Given constants , , and , a quadratic sequence is a sequence defined by the explicit formula in the form of:
In a quadratic sequence, the sequence of the differences of two consecutive terms is an arithmetic sequence.
Therefore, the second order difference of a quadratic sequence is constant.
Finding Explicit Formulas for a Quadratic Sequence
One challenge when dealing with quadratic sequences is finding an explicit formula that describes the sequence.
This can be done by setting up using a system of linear equations.
Worked Example Finding an Explicit Formula for a Quadratic Sequence
Given the terms of sequence below, find the explicit formula that describes the terms of the sequence .
Solution
First, we identify if it is a quadratic sequence by verifying if the first-order difference is an arithmetic sequence.
Since the difference between differences is , we know that it is a quadratic sequence.
Knowing that is a quadratic sequence, we can say that:
By taking different values of , we can produce a system of linear equations.
Let :
Let :
Let :
Therefore, by combining these three equations we get the following system:
By solving this system of linear equations, we get the constants , , and
Finding an expression for using , we get:
Substituting on equation :
We set the last equation as .
Substituting on equation :
We set the last equation as .
Combining equations and , we get a new system of equations:
Finding an expression for using :
Substituting on equation :
Substituting on expression for on , we get:
Substituting and on our original expression for on , we get:
Therefore, the explicit formula for the quadratic sequence is:
Explicit Formula of a Quadratic Sequence from its Terms
We can generalize the process from the moment we identified that a sequence is a quadratic sequence.
To formulate our system of equations, we need three terms on our sequence.
For ease of generalization, we pick the terms , , and .
Using the general form of a quadratic sequence, we can substitute our terms to form our system of three equations.
Using ideas from matrix algebra, we can turn this system of equations as linear matric equation in the form
Here, is our coefficient matrix, constitutes our variable matrix, or the values for , , and , and matrix represents the terms on our other side of equation
In short, the equation might look like this:
By multiplying an inverse of the coefficient matrix on both sides of the equation, we get a matrix that represents the values of , , and .
We can find an inverse of the coefficient matrix by setting an augmented matrix with the identity matrix on one side.
We then use row operations to turn the coefficient matrix into an identity matrix.
The identity matrix will then become the inverse of the identity matrix
Proof Explicit Formula of a Quadratic Sequence from its Terms
Suppose that is a quadratic sequence in the form:
where , , and are constants.
By using three terms in our sequence , , and , we can form a system of three equations based on the constants , , and .
We can then represent them as the matrix equation:
where represents the coefficient matrix, represents the matrix for our terms, and represents the matrix for the missing variables, the variable matrix.
Using matrix algebra, by finding an inverse of the coefficient matrix , we can find a matrix equivalent to the variable matrix and therefore, we can find an expression for , , and .
Expressing our system as a matrix equation, we get:
Now, we use an augmented matrix and row operations to find an inverse of the coefficient matrix.
Therefore,
Multiplying to both sides of the matrix equation Be mindful of the noncommutativity of matrix multiplication.
By equating each entry in both matrices, we get an expression of the constants , , and in terms of the first, second and third terms of the quadratic sequence.
Therefore, we can say that:
Definition Explicit Formula of a Quadratic Sequence from First Three Terms
Given that is a quadratic sequence where , , and are the first three terms of the sequence, then the explicit formula of is
provided that , , and are:
We can also use the fact that the second order difference of a quadratic sequence (which is constant) is always .
By utilizing this, we can simplify the process by only solving for the constants and instead.
This leads to an alternate way for the explicit formula of a quadratic sequence.
Proof Explicit Formula of a Quadratic Sequence from its Second-Order Difference and its Terms
Suppose that is a quadratic sequence with a second order difference , which means that for all :
where is a real number constant.
Since , then:
Therefore, by substituting , and by using the first two terms of the sequence and , we can form a system of two equations.
Simplifying the system so all constants are on the right, we get:
Expressing our system as a matrix equation, we get.
Next, we find an inverse matrix for our coefficient matrix .
We can do this via an augmented matrix and row operations.
Therefore
Multiplying on both sides of the equation yields:
By equating each entry in both matrices, we get an expression for constants and in terms of the terms , , and the second-order difference .
Therefore, we can finally say that:
Definition Explicit Formula of a Quadratic Sequence from Second-Order Difference and First Two Terms
Given a quadratic sequence with a second-order difference of , its explicit formula can be expressed as: