Prove $x_n = c_1lambda_1^n + c_1lambda_2^n + dots + c_nlambda_n^n$
We have the recursive relation $x_n = a_{n-1}x_{n-1} + a_{n-2}x_{n-2} + dots + a_1x_1 + a_0x_0$. Prove that if the polynomial $t^n - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - dots - a_1t - a_0$ has distinct roots $lambda_1, lambda_2, dots, lambda_n$ then $$x_n = c_1lambda_1^n + c_1lambda_2^n + dots + c_nlambda_n^n$$
Here, coefficients $c_1, c_2, dots, c_n$ may be calculated from initial values $x_0, x_1, dots, x_n$.
I'm thinking of finding some matrix $A$ then looking for eigenvalues and eigenvectors so that we can do $A^n$ but I'm stuck.
linear-algebra recurrence-relations
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
asked Nov 25 at 7:32
sedrick
43912
add a comment |
We have the recursive relation $x_n = a_{n-1}x_{n-1} + a_{n-2}x_{n-2} + dots + a_1x_1 + a_0x_0$. Prove that if the polynomial $t^n - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - dots - a_1t - a_0$ has distinct roots $lambda_1, lambda_2, dots, lambda_n$ then $$x_n = c_1lambda_1^n + c_1lambda_2^n + dots + c_nlambda_n^n$$
Here, coefficients $c_1, c_2, dots, c_n$ may be calculated from initial values $x_0, x_1, dots, x_n$.
I'm thinking of finding some matrix $A$ then looking for eigenvalues and eigenvectors so that we can do $A^n$ but I'm stuck.
linear-algebra recurrence-relations
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
asked Nov 25 at 7:32
sedrick
43912
This is a basic result from the theory of linear homogeneous recurrence relations. As $lambda_i$ is a zero of that polynomial $x_n=lambda_i^n$ is a solution of the recurrence. Because the $lambda_i$s are distinct, those $n$ solutions are linearly independent (Vandermonde). Therefore you can write any initial segment $(x_0,x_1,ldots,x_{n-1})$ as a linear combination. After that we are basically done.
– Jyrki Lahtonen
Nov 25 at 8:03
add a comment |
We have the recursive relation $x_n = a_{n-1}x_{n-1} + a_{n-2}x_{n-2} + dots + a_1x_1 + a_0x_0$. Prove that if the polynomial $t^n - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - dots - a_1t - a_0$ has distinct roots $lambda_1, lambda_2, dots, lambda_n$ then $$x_n = c_1lambda_1^n + c_1lambda_2^n + dots + c_nlambda_n^n$$
Here, coefficients $c_1, c_2, dots, c_n$ may be calculated from initial values $x_0, x_1, dots, x_n$.
I'm thinking of finding some matrix $A$ then looking for eigenvalues and eigenvectors so that we can do $A^n$ but I'm stuck.
linear-algebra recurrence-relations
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
asked Nov 25 at 7:32
sedrick
43912
We have the recursive relation $x_n = a_{n-1}x_{n-1} + a_{n-2}x_{n-2} + dots + a_1x_1 + a_0x_0$. Prove that if the polynomial $t^n - a_{n-1}t^{n-1} - a_{n-2}t^{n-2} - dots - a_1t - a_0$ has distinct roots $lambda_1, lambda_2, dots, lambda_n$ then $$x_n = c_1lambda_1^n + c_1lambda_2^n + dots + c_nlambda_n^n$$
Here, coefficients $c_1, c_2, dots, c_n$ may be calculated from initial values $x_0, x_1, dots, x_n$.
I'm thinking of finding some matrix $A$ then looking for eigenvalues and eigenvectors so that we can do $A^n$ but I'm stuck.
linear-algebra recurrence-relations
linear-algebra recurrence-relations
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
asked Nov 25 at 7:32
sedrick
43912
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
asked Nov 25 at 7:32
sedrick
43912
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
edited Nov 25 at 7:59
Jyrki Lahtonen
107k12166365
107k12166365
asked Nov 25 at 7:32
sedrick
43912
asked Nov 25 at 7:32
sedrick
43912
asked Nov 25 at 7:32
sedrick
43912
43912
This is a basic result from the theory of linear homogeneous recurrence relations. As $lambda_i$ is a zero of that polynomial $x_n=lambda_i^n$ is a solution of the recurrence. Because the $lambda_i$s are distinct, those $n$ solutions are linearly independent (Vandermonde). Therefore you can write any initial segment $(x_0,x_1,ldots,x_{n-1})$ as a linear combination. After that we are basically done.
– Jyrki Lahtonen
Nov 25 at 8:03
add a comment |
This is a basic result from the theory of linear homogeneous recurrence relations. As $lambda_i$ is a zero of that polynomial $x_n=lambda_i^n$ is a solution of the recurrence. Because the $lambda_i$s are distinct, those $n$ solutions are linearly independent (Vandermonde). Therefore you can write any initial segment $(x_0,x_1,ldots,x_{n-1})$ as a linear combination. After that we are basically done.
– Jyrki Lahtonen
Nov 25 at 8:03
This is a basic result from the theory of linear homogeneous recurrence relations. As $lambda_i$ is a zero of that polynomial $x_n=lambda_i^n$ is a solution of the recurrence. Because the $lambda_i$s are distinct, those $n$ solutions are linearly independent (Vandermonde). Therefore you can write any initial segment $(x_0,x_1,ldots,x_{n-1})$ as a linear combination. After that we are basically done.
– Jyrki Lahtonen
Nov 25 at 8:03
This is a basic result from the theory of linear homogeneous recurrence relations. As $lambda_i$ is a zero of that polynomial $x_n=lambda_i^n$ is a solution of the recurrence. Because the $lambda_i$s are distinct, those $n$ solutions are linearly independent (Vandermonde). Therefore you can write any initial segment $(x_0,x_1,ldots,x_{n-1})$ as a linear combination. After that we are basically done.
– Jyrki Lahtonen
Nov 25 at 8:03
add a comment |
1 Answer
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One of the easiest approach: observe that the sequence space $$S ={(x_n)_{ngeq 0};|;x_{k+n} = a_{n-1}x_{n+k-1} + cdots + a_0 x_k, ;forall kgeq 0}$$ is of dimension $n$ since the whole sequence is determined by $n$-data, $x_0,ldots x_{n-1}$. And show that $lambda_j^n, ngeq 0$ forms a linearly independent subset of $S$, and thus forms a basis.
Matrix approach is also valid. Let $y_k = (x_k, ldots,x_{k-n+1})'$ for $kgeq n-1$. Then for companion matrix $C = (c_{ij})$ s.t. $c_{1j} = a_{n-j}$ and $c_{ij} = 1$ if $i-1 = j$ and $0$ otherwise, for $igeq 2$, it holds that $$y_{k+1} = C y_k.$$ Then the set of eigenvalues of $C$ is exactly ${lambda_j, 1leq j leq n}$ and $C$ is diagonalizable. That is, there is a basis consisting of its eigenvectors. You can see that $y_{n+k-1} = sum_j c_j lambda_j^k v_j$ where $v_j$'s are eigenvectors forming basis corresponding to $lambda_j$ and $y_{n-1}$ is represented as $sum_j c_j v_j$.
answered Nov 25 at 8:01
Song
3,985316
add a comment |
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1 Answer
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One of the easiest approach: observe that the sequence space $$S ={(x_n)_{ngeq 0};|;x_{k+n} = a_{n-1}x_{n+k-1} + cdots + a_0 x_k, ;forall kgeq 0}$$ is of dimension $n$ since the whole sequence is determined by $n$-data, $x_0,ldots x_{n-1}$. And show that $lambda_j^n, ngeq 0$ forms a linearly independent subset of $S$, and thus forms a basis.
Matrix approach is also valid. Let $y_k = (x_k, ldots,x_{k-n+1})'$ for $kgeq n-1$. Then for companion matrix $C = (c_{ij})$ s.t. $c_{1j} = a_{n-j}$ and $c_{ij} = 1$ if $i-1 = j$ and $0$ otherwise, for $igeq 2$, it holds that $$y_{k+1} = C y_k.$$ Then the set of eigenvalues of $C$ is exactly ${lambda_j, 1leq j leq n}$ and $C$ is diagonalizable. That is, there is a basis consisting of its eigenvectors. You can see that $y_{n+k-1} = sum_j c_j lambda_j^k v_j$ where $v_j$'s are eigenvectors forming basis corresponding to $lambda_j$ and $y_{n-1}$ is represented as $sum_j c_j v_j$.
answered Nov 25 at 8:01
Song
3,985316
add a comment |
One of the easiest approach: observe that the sequence space $$S ={(x_n)_{ngeq 0};|;x_{k+n} = a_{n-1}x_{n+k-1} + cdots + a_0 x_k, ;forall kgeq 0}$$ is of dimension $n$ since the whole sequence is determined by $n$-data, $x_0,ldots x_{n-1}$. And show that $lambda_j^n, ngeq 0$ forms a linearly independent subset of $S$, and thus forms a basis.
Matrix approach is also valid. Let $y_k = (x_k, ldots,x_{k-n+1})'$ for $kgeq n-1$. Then for companion matrix $C = (c_{ij})$ s.t. $c_{1j} = a_{n-j}$ and $c_{ij} = 1$ if $i-1 = j$ and $0$ otherwise, for $igeq 2$, it holds that $$y_{k+1} = C y_k.$$ Then the set of eigenvalues of $C$ is exactly ${lambda_j, 1leq j leq n}$ and $C$ is diagonalizable. That is, there is a basis consisting of its eigenvectors. You can see that $y_{n+k-1} = sum_j c_j lambda_j^k v_j$ where $v_j$'s are eigenvectors forming basis corresponding to $lambda_j$ and $y_{n-1}$ is represented as $sum_j c_j v_j$.
answered Nov 25 at 8:01
Song
3,985316
add a comment |
One of the easiest approach: observe that the sequence space $$S ={(x_n)_{ngeq 0};|;x_{k+n} = a_{n-1}x_{n+k-1} + cdots + a_0 x_k, ;forall kgeq 0}$$ is of dimension $n$ since the whole sequence is determined by $n$-data, $x_0,ldots x_{n-1}$. And show that $lambda_j^n, ngeq 0$ forms a linearly independent subset of $S$, and thus forms a basis.
Matrix approach is also valid. Let $y_k = (x_k, ldots,x_{k-n+1})'$ for $kgeq n-1$. Then for companion matrix $C = (c_{ij})$ s.t. $c_{1j} = a_{n-j}$ and $c_{ij} = 1$ if $i-1 = j$ and $0$ otherwise, for $igeq 2$, it holds that $$y_{k+1} = C y_k.$$ Then the set of eigenvalues of $C$ is exactly ${lambda_j, 1leq j leq n}$ and $C$ is diagonalizable. That is, there is a basis consisting of its eigenvectors. You can see that $y_{n+k-1} = sum_j c_j lambda_j^k v_j$ where $v_j$'s are eigenvectors forming basis corresponding to $lambda_j$ and $y_{n-1}$ is represented as $sum_j c_j v_j$.
answered Nov 25 at 8:01
Song
3,985316
One of the easiest approach: observe that the sequence space $$S ={(x_n)_{ngeq 0};|;x_{k+n} = a_{n-1}x_{n+k-1} + cdots + a_0 x_k, ;forall kgeq 0}$$ is of dimension $n$ since the whole sequence is determined by $n$-data, $x_0,ldots x_{n-1}$. And show that $lambda_j^n, ngeq 0$ forms a linearly independent subset of $S$, and thus forms a basis.
Matrix approach is also valid. Let $y_k = (x_k, ldots,x_{k-n+1})'$ for $kgeq n-1$. Then for companion matrix $C = (c_{ij})$ s.t. $c_{1j} = a_{n-j}$ and $c_{ij} = 1$ if $i-1 = j$ and $0$ otherwise, for $igeq 2$, it holds that $$y_{k+1} = C y_k.$$ Then the set of eigenvalues of $C$ is exactly ${lambda_j, 1leq j leq n}$ and $C$ is diagonalizable. That is, there is a basis consisting of its eigenvectors. You can see that $y_{n+k-1} = sum_j c_j lambda_j^k v_j$ where $v_j$'s are eigenvectors forming basis corresponding to $lambda_j$ and $y_{n-1}$ is represented as $sum_j c_j v_j$.
answered Nov 25 at 8:01
Song
3,985316
edited Nov 25 at 10:25
answered Nov 25 at 8:01
Song
3,985316
answered Nov 25 at 8:01
Song
3,985316
answered Nov 25 at 8:01
Song
3,985316
3,985316
add a comment |
add a comment |
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This is a basic result from the theory of linear homogeneous recurrence relations. As $lambda_i$ is a zero of that polynomial $x_n=lambda_i^n$ is a solution of the recurrence. Because the $lambda_i$s are distinct, those $n$ solutions are linearly independent (Vandermonde). Therefore you can write any initial segment $(x_0,x_1,ldots,x_{n-1})$ as a linear combination. After that we are basically done.
– Jyrki Lahtonen
Nov 25 at 8:03