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How to find the closed form of the recursive equation, if 2 of the 3 roots of the characteristic equation are...
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The given recursive equation is $$f(n) = f(n-1) + 3f(n-3) + 2n$$
The characteristic equation for $$f(n) = f(n-1) - 3f(n-3)$$ is $r^3 - r^2 - 3 = 0$, which has $2$ imaginary roots and $1$ real root whose approximate value is $1.8637$
The linear part is $2n$ and $f(n)$ can be transformed into a linear equation using the form $f(n) = cn + d$, where $c$ and $d$ are constants.
I am not sure about how to handle the cubic characteristic equation with $1$ real root. According to me, the solution to recursive part should be of the form $$(a_1 + a_2n + a_3n^2)times(1.8637)^n$$
$a_1, a_2, a_3$ can be found out using the initial conditions of the recursion. But this gives me a wrong answer.
$1.8637$ is the real root of the characteristic equation.
closed-form recursion
edited Jan 18 at 10:49
lioness99a
3,8362727
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
$endgroup$
add a comment |
$begingroup$
The given recursive equation is $$f(n) = f(n-1) + 3f(n-3) + 2n$$
The characteristic equation for $$f(n) = f(n-1) - 3f(n-3)$$ is $r^3 - r^2 - 3 = 0$, which has $2$ imaginary roots and $1$ real root whose approximate value is $1.8637$
The linear part is $2n$ and $f(n)$ can be transformed into a linear equation using the form $f(n) = cn + d$, where $c$ and $d$ are constants.
I am not sure about how to handle the cubic characteristic equation with $1$ real root. According to me, the solution to recursive part should be of the form $$(a_1 + a_2n + a_3n^2)times(1.8637)^n$$
$a_1, a_2, a_3$ can be found out using the initial conditions of the recursion. But this gives me a wrong answer.
$1.8637$ is the real root of the characteristic equation.
closed-form recursion
edited Jan 18 at 10:49
lioness99a
3,8362727
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
$endgroup$
add a comment |
$begingroup$
The given recursive equation is $$f(n) = f(n-1) + 3f(n-3) + 2n$$
The characteristic equation for $$f(n) = f(n-1) - 3f(n-3)$$ is $r^3 - r^2 - 3 = 0$, which has $2$ imaginary roots and $1$ real root whose approximate value is $1.8637$
The linear part is $2n$ and $f(n)$ can be transformed into a linear equation using the form $f(n) = cn + d$, where $c$ and $d$ are constants.
I am not sure about how to handle the cubic characteristic equation with $1$ real root. According to me, the solution to recursive part should be of the form $$(a_1 + a_2n + a_3n^2)times(1.8637)^n$$
$a_1, a_2, a_3$ can be found out using the initial conditions of the recursion. But this gives me a wrong answer.
$1.8637$ is the real root of the characteristic equation.
closed-form recursion
edited Jan 18 at 10:49
lioness99a
3,8362727
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
$endgroup$
The given recursive equation is $$f(n) = f(n-1) + 3f(n-3) + 2n$$
The characteristic equation for $$f(n) = f(n-1) - 3f(n-3)$$ is $r^3 - r^2 - 3 = 0$, which has $2$ imaginary roots and $1$ real root whose approximate value is $1.8637$
The linear part is $2n$ and $f(n)$ can be transformed into a linear equation using the form $f(n) = cn + d$, where $c$ and $d$ are constants.
I am not sure about how to handle the cubic characteristic equation with $1$ real root. According to me, the solution to recursive part should be of the form $$(a_1 + a_2n + a_3n^2)times(1.8637)^n$$
$a_1, a_2, a_3$ can be found out using the initial conditions of the recursion. But this gives me a wrong answer.
$1.8637$ is the real root of the characteristic equation.
closed-form recursion
closed-form recursion
edited Jan 18 at 10:49
lioness99a
3,8362727
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
edited Jan 18 at 10:49
lioness99a
3,8362727
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
edited Jan 18 at 10:49
lioness99a
3,8362727
edited Jan 18 at 10:49
lioness99a
3,8362727
edited Jan 18 at 10:49
lioness99a
3,8362727
3,8362727
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
asked Jan 18 at 9:51
Arnav KaushikArnav Kaushik
62
62
add a comment |
add a comment |
1 Answer
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$begingroup$
You can't ignore the complex roots. Take, for instance, $g(n) = -g(n-2)$, with the characteristic equation $r^2 + 1 = 0$. The general solution to this recursion is $g(n) = a_1i^n + a_2(-i)^n$. If $g(0)$ and $g(1)$ happen to be real (so that the entire sequence is real), then that just forces $a_1$ and $a_2$ to be just so that the imaginary part of $a_1i^n$ and the imaginary part of $a_2(-i)^n$ cancel. This $(a_1 + a_2n + a_3n^2)$ business is what you do if you have a multiple root.
With that in mind, the general solution to (the homogenous part of) your recursion is
$$
f(n) = a_1r_1^n + a_2r_2^n + a_3r_3^n
$$
where $r_1, r_2, r_3$ are the three distinct roots to the characteristic equation.
answered Jan 18 at 9:59
ArthurArthur
116k7116199
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1 Answer
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1 Answer
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$begingroup$
You can't ignore the complex roots. Take, for instance, $g(n) = -g(n-2)$, with the characteristic equation $r^2 + 1 = 0$. The general solution to this recursion is $g(n) = a_1i^n + a_2(-i)^n$. If $g(0)$ and $g(1)$ happen to be real (so that the entire sequence is real), then that just forces $a_1$ and $a_2$ to be just so that the imaginary part of $a_1i^n$ and the imaginary part of $a_2(-i)^n$ cancel. This $(a_1 + a_2n + a_3n^2)$ business is what you do if you have a multiple root.
With that in mind, the general solution to (the homogenous part of) your recursion is
$$
f(n) = a_1r_1^n + a_2r_2^n + a_3r_3^n
$$
where $r_1, r_2, r_3$ are the three distinct roots to the characteristic equation.
answered Jan 18 at 9:59
ArthurArthur
116k7116199
$endgroup$
add a comment |
$begingroup$
You can't ignore the complex roots. Take, for instance, $g(n) = -g(n-2)$, with the characteristic equation $r^2 + 1 = 0$. The general solution to this recursion is $g(n) = a_1i^n + a_2(-i)^n$. If $g(0)$ and $g(1)$ happen to be real (so that the entire sequence is real), then that just forces $a_1$ and $a_2$ to be just so that the imaginary part of $a_1i^n$ and the imaginary part of $a_2(-i)^n$ cancel. This $(a_1 + a_2n + a_3n^2)$ business is what you do if you have a multiple root.
With that in mind, the general solution to (the homogenous part of) your recursion is
$$
f(n) = a_1r_1^n + a_2r_2^n + a_3r_3^n
$$
where $r_1, r_2, r_3$ are the three distinct roots to the characteristic equation.
answered Jan 18 at 9:59
ArthurArthur
116k7116199
$endgroup$
add a comment |
$begingroup$
You can't ignore the complex roots. Take, for instance, $g(n) = -g(n-2)$, with the characteristic equation $r^2 + 1 = 0$. The general solution to this recursion is $g(n) = a_1i^n + a_2(-i)^n$. If $g(0)$ and $g(1)$ happen to be real (so that the entire sequence is real), then that just forces $a_1$ and $a_2$ to be just so that the imaginary part of $a_1i^n$ and the imaginary part of $a_2(-i)^n$ cancel. This $(a_1 + a_2n + a_3n^2)$ business is what you do if you have a multiple root.
With that in mind, the general solution to (the homogenous part of) your recursion is
$$
f(n) = a_1r_1^n + a_2r_2^n + a_3r_3^n
$$
where $r_1, r_2, r_3$ are the three distinct roots to the characteristic equation.
answered Jan 18 at 9:59
ArthurArthur
116k7116199
$endgroup$
You can't ignore the complex roots. Take, for instance, $g(n) = -g(n-2)$, with the characteristic equation $r^2 + 1 = 0$. The general solution to this recursion is $g(n) = a_1i^n + a_2(-i)^n$. If $g(0)$ and $g(1)$ happen to be real (so that the entire sequence is real), then that just forces $a_1$ and $a_2$ to be just so that the imaginary part of $a_1i^n$ and the imaginary part of $a_2(-i)^n$ cancel. This $(a_1 + a_2n + a_3n^2)$ business is what you do if you have a multiple root.
With that in mind, the general solution to (the homogenous part of) your recursion is
$$
f(n) = a_1r_1^n + a_2r_2^n + a_3r_3^n
$$
where $r_1, r_2, r_3$ are the three distinct roots to the characteristic equation.
answered Jan 18 at 9:59
ArthurArthur
116k7116199
answered Jan 18 at 9:59
ArthurArthur
116k7116199
answered Jan 18 at 9:59
ArthurArthur
116k7116199
answered Jan 18 at 9:59
ArthurArthur
116k7116199
116k7116199
add a comment |
add a comment |
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