[Linear Algebra] Closed formula for recursive sequence

In summary, the conversation involves finding a 3x3 matrix A that satisfies certain equations involving the elements of A and the values of x. The specific equations mentioned are eq. 1: A_{1,1} x_k + A_{1,2} x_{k+1} + A_{1,3} x_{k+2} = x_{k+1}, eq. 2: A_{2,1} x_k + A_{2,2} x_{k+1} + A_{2,3} x_{k+2} = x_{k+2}, and eq. 3: A_{3,1} x_k + A_{3,2} x_{
  • #1
reminiscent
131
2

Homework Statement


qsEKf8z.png


Homework Equations


a) the one given
b) det(A-λI) = 0
find λ values using A
c)use λ values to find eigenvectors

The Attempt at a Solution


This wasn't explained well enough so I can understand it in class. So far, I made the matrix being multiplied to A have the following elements from top to bottom: x0 = 1, x1 = 1, x2 = 2. The one on the right has the following elements: x1 = 1, x2= 2, x3=7. Is this correct thinking? Do I just find a matrix A that maps the vector on the left to the vector on the right?
I know how to do b) and c), I just don't know how to get A correctly.
 
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  • #2
reminiscent said:
So far, I made the matrix being multiplied to A have the following elements from top to bottom: x0 = 1, x1 = 1, x2 = 2

The usual way to describe an element of a matrix is to give it two indices, such as A[1][1] , A[1][2] or ## A_{1,1}, A_{1,2} ## etc. Which elements are you trying to describe with the phrase "from top to bottom" ?

You need a matrix A such that:

eq. 1: [itex] A_{1,1} x_k + A_{1,2} x_{k+1} + A_{1,3} x_{k+2} = x_{k+1} [/itex]
eq 2: [itex] A_{2,1} x_k + A_{2,2} x_{k+1} + A_{2,3} x_{k+2} = x_{k+2} [/itex]
eq 3: [itex] A_{3,1} x_k + A_{3,2} x_{k+1} + A_{3,3} x_{k+2} = x_{k+3} [/itex]

An solution to eq. 1 is [itex] A_{1,1} = 0, A_{1,2} = 1, A_{1,3} = 0 [/itex].
 
  • #3
Stephen Tashi said:
The usual way to describe an element of a matrix is to give it two indices, such as A[1][1] , A[1][2] or ## A_{1,1}, A_{1,2} ## etc. Which elements are you trying to describe with the phrase "from top to bottom" ?

You need a matrix A such that:

eq. 1: [itex] A_{1,1} x_k + A_{1,2} x_{k+1} + A_{1,3} x_{k+2} = x_{k+1} [/itex]
eq 2: [itex] A_{2,1} x_k + A_{2,2} x_{k+1} + A_{2,3} x_{k+2} = x_{k+2} [/itex]
eq 3: [itex] A_{3,1} x_k + A_{3,2} x_{k+1} + A_{3,3} x_{k+2} = x_{k+3} [/itex]

An solution to eq. 1 is [itex] A_{1,1} = 0, A_{1,2} = 1, A_{1,3} = 0 [/itex].
Sorry, I completely forgot that A was a 3x3 matrix.
Anyways, I see how you are getting the solutions.
For equation 2, I found these solutions:
A21 = 0 A22 =0 A23= 1
For equation 3, I found:
A31 = 0 A32 = 1 A33 =3

Are those correct? Does what they give for xk+3 have anything to do with the last equation?
 
  • #4
Stephen Tashi said:
The usual way to describe an element of a matrix is to give it two indices, such as A[1][1] , A[1][2] or ## A_{1,1}, A_{1,2} ## etc. Which elements are you trying to describe with the phrase "from top to bottom" ?

You need a matrix A such that:

eq. 1: [itex] A_{1,1} x_k + A_{1,2} x_{k+1} + A_{1,3} x_{k+2} = x_{k+1} [/itex]
eq 2: [itex] A_{2,1} x_k + A_{2,2} x_{k+1} + A_{2,3} x_{k+2} = x_{k+2} [/itex]
eq 3: [itex] A_{3,1} x_k + A_{3,2} x_{k+1} + A_{3,3} x_{k+2} = x_{k+3} [/itex]

An solution to eq. 1 is [itex] A_{1,1} = 0, A_{1,2} = 1, A_{1,3} = 0 [/itex].
So A can be more than 1 matrix?
 
  • #5
reminiscent said:
Sorry, I completely forgot that A was a 3x3 matrix.
Anyways, I see how you are getting the solutions.
For equation 2, I found these solutions:
A21 = 0 A22 =0 A23= 1
For equation 3, I found:
A31 = 0 A32 = 1 A33 =3

Are those correct? Does what they give for xk+3 have anything to do with the last equation?

You can (and should) check that for yourself. When you substitute in your values for ##A_{31}, A_{32}## and ##A_{33}## do you get the correct recursion?

And, for your next question in post #4, why do you think there is more than one matrix? The matrix ##A## has 9 elements called ##A_{11}, A_{12} \ldots, A_{33}##.
 
  • #6
Ray Vickson said:
You can (and should) check that for yourself. When you substitute in your values for ##A_{31}, A_{32}## and ##A_{33}## do you get the correct recursion?

And, for your next question in post #4, why do you think there is more than one matrix? The matrix ##A## has 9 elements called ##A_{11}, A_{12} \ldots, A_{33}##.
Yes, I do. I just changed the bottom row of A to be the constants of the recursive formula given, but I kept the first 2 rows I stated the same. I verified and got the eigenvalues in b.
I thought there was more than one matrix because in the equations, there are a combination of values for each element in A that can produce the desired elements. But it seems like there will be only one because the eigenvalues have to be verified, correct?
 
  • #7
reminiscent said:
Yes, I do. I just changed the bottom row of A to be the constants of the recursive formula given, but I kept the first 2 rows I stated the same. I verified and got the eigenvalues in b.
I thought there was more than one matrix because in the equations, there are a combination of values for each element in A that can produce the desired elements. But it seems like there will be only one because the eigenvalues have to be verified, correct?

I have no idea what you are saying.

While it is the case that IF you apply the matrix to some smal number of initial values like ##x_0=1,x_1=, x_2 =2## (and a few more obtained from the recursion) you will get numbers on the left and numbers on the right, so will have equations like
$$ A_{31}\: \text{number one} + A_{32}\: \text{number two} + A_{33} \:\text{number three} = \text{number four} $$
with actual numbers. Of course you can have many possible values of ##A_{31}, A_{32}, A_{33}## that "work". However, that is irrelevant: you are supposed to find a matrix that works for all values of ##x_k, x_{k+1}, x_{k+2}, x_{k+3}## that you will ever encounter in the calculations (say for ##x_{27}, x_{28}, x_{29}, x_{30}## or for ##x_{1000}, x_{1001}, x_{1002}, x_{1003}##, etc.) and when you use that fact there can be only one possible choice for the three numbers ##A_{31}, A_{32}, A_{33}##. There is only one matrix ##A##.
 
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  • #8
reminiscent said:
Are those correct?

You solution for eq. 3 doesn't work. Your solution gives [itex] x_{k+2} = x_{k+3} [/itex].

Does what they give for xk+3 have anything to do with the last equation?
Yes !
 
  • #9
Stephen Tashi said:
You solution for eq. 3 doesn't work. Your solution gives [itex] x_{k+2} = x_{k+3} [/itex].Yes ! (I see by post #6 that you figured it out.)
 

Related to [Linear Algebra] Closed formula for recursive sequence

1. What is a recursive sequence in linear algebra?

A recursive sequence in linear algebra is a sequence of numbers where the next term is calculated by using the previous terms in the sequence.

2. Why is a closed formula useful for recursive sequences?

A closed formula for a recursive sequence allows us to find any term in the sequence without having to calculate all the previous terms. This makes it easier and more efficient to work with the sequence.

3. How do you find a closed formula for a recursive sequence?

To find a closed formula for a recursive sequence, we first need to identify the pattern or rule that governs the sequence. This can be done by looking for relationships between the terms and using algebraic techniques to manipulate the terms and identify a pattern.

4. Can any recursive sequence have a closed formula?

No, not all recursive sequences have a closed formula. Some sequences may have a very complex or unpredictable pattern, making it difficult to find a closed formula.

5. How can a closed formula be used to solve problems in linear algebra?

A closed formula for a recursive sequence can be used to solve problems involving the sequence, such as finding a specific term or calculating the sum of a certain number of terms. It can also be used to analyze the behavior of the sequence and make predictions about its future terms.

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