Mixing
Solve the recurrence relation $a_n=a_{n-1}+2n+3a_{n-3}?$
$begingroup$
I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.
recurrence-relations homogeneous-equation
edited Jan 17 at 14:48
Adrian Keister
5,26571933
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
$endgroup$
|
show 1 more comment
$begingroup$
I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.
recurrence-relations homogeneous-equation
edited Jan 17 at 14:48
Adrian Keister
5,26571933
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
$endgroup$
$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41
$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43
$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49
$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51
1
$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16
|
show 1 more comment
$begingroup$
I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.
recurrence-relations homogeneous-equation
edited Jan 17 at 14:48
Adrian Keister
5,26571933
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
$endgroup$
I am currently solving a recurrence relation but I got stuck since I am not even able to find the basic root of the auxiliary equations.
recurrence-relations homogeneous-equation
recurrence-relations homogeneous-equation
edited Jan 17 at 14:48
Adrian Keister
5,26571933
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
edited Jan 17 at 14:48
Adrian Keister
5,26571933
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
edited Jan 17 at 14:48
Adrian Keister
5,26571933
edited Jan 17 at 14:48
Adrian Keister
5,26571933
edited Jan 17 at 14:48
Adrian Keister
5,26571933
5,26571933
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
asked Jan 17 at 14:36
Sagar SharmaSagar Sharma
143
143
$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41
$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43
$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49
$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51
1
$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16
|
show 1 more comment
$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41
$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43
$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49
$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51
1
$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16
$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41
$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41
$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43
$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43
$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49
$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49
$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51
$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51
1
1
$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16
$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let's rewrite this as
begin{align*}
a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
end{align*}
The generating function for this sequence is
begin{align*}
frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
end{align*}
Solving for $f(x)$ gives us
begin{align*}
f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
end{align*}
Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
end{align*}
Similarly, the red portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
end{align*}
The green portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
end{align*}
Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.
Finally, your sequence can be expressed as
begin{align*}
a_n = p_n + q_n - r_n - s_n
end{align*}
There is some heavy simplification you can do. For example, you can express
begin{align*}
p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
end{align*}
For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Let's rewrite this as
begin{align*}
a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
end{align*}
The generating function for this sequence is
begin{align*}
frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
end{align*}
Solving for $f(x)$ gives us
begin{align*}
f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
end{align*}
Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
end{align*}
Similarly, the red portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
end{align*}
The green portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
end{align*}
Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.
Finally, your sequence can be expressed as
begin{align*}
a_n = p_n + q_n - r_n - s_n
end{align*}
There is some heavy simplification you can do. For example, you can express
begin{align*}
p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
end{align*}
For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
$endgroup$
add a comment |
$begingroup$
Let's rewrite this as
begin{align*}
a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
end{align*}
The generating function for this sequence is
begin{align*}
frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
end{align*}
Solving for $f(x)$ gives us
begin{align*}
f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
end{align*}
Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
end{align*}
Similarly, the red portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
end{align*}
The green portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
end{align*}
Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.
Finally, your sequence can be expressed as
begin{align*}
a_n = p_n + q_n - r_n - s_n
end{align*}
There is some heavy simplification you can do. For example, you can express
begin{align*}
p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
end{align*}
For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
$endgroup$
add a comment |
$begingroup$
Let's rewrite this as
begin{align*}
a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
end{align*}
The generating function for this sequence is
begin{align*}
frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
end{align*}
Solving for $f(x)$ gives us
begin{align*}
f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
end{align*}
Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
end{align*}
Similarly, the red portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
end{align*}
The green portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
end{align*}
Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.
Finally, your sequence can be expressed as
begin{align*}
a_n = p_n + q_n - r_n - s_n
end{align*}
There is some heavy simplification you can do. For example, you can express
begin{align*}
p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
end{align*}
For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
$endgroup$
Let's rewrite this as
begin{align*}
a_{n+3} = a_{n+2} + 2(n+3) + 3a_n
end{align*}
The generating function for this sequence is
begin{align*}
frac{f(x) - a_0 - a_1x - a_2x^2}{x^3} = frac{f(x) - a_0 - a_1x}{x^2} + frac{2(3 - 2x)}{(1-x)^2} + 3f(x)
end{align*}
Solving for $f(x)$ gives us
begin{align*}
f(x) = color{blue}{frac{a_0(x-1)}{(3x^3+x-1)}} + color{red}{frac{a_1 x(x-1)}{(3x^3+x-1)}} - color{green}{frac{a_2x^2}{(3x^3+x-1)}} - color{orange}{frac{2(3-2x)x^3}{(x-1)^2(3x^3+x-1)}}
end{align*}
Let $r_1, r_2, r_3$ be the roots of $3x^3 + x - 1=0$. Then the series expansion of the blue term is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_0}{3}left(frac{1 - r_1}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{1 - r_2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{1 - r_3}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{p_n}x^n
end{align*}
Similarly, the red portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_1}{3}left(frac{r_1(1 - r_1)}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2(1 - r_2)}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3(1 - r_3)}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{q_n}x^n
end{align*}
The green portion is
begin{align*}
sum_{n=0}^{infty}underbrace{left{frac{a_2}{3}left(frac{r_1^2}{r_1^{n+1}(r_1 - r_2)(r_1 - r_3)} + frac{r_2^2}{r_2^{n+1}(r_2 - r_1)(r_2 - r_3)} + frac{r_3^2}{r_3^{n+1}(r_3 - r_1)(r_3 - r_2)}right)right}}_{r_n}x^n
end{align*}
Exercise Find the series expansion for the orange term, and call the coefficients as $s_n$.
Finally, your sequence can be expressed as
begin{align*}
a_n = p_n + q_n - r_n - s_n
end{align*}
There is some heavy simplification you can do. For example, you can express
begin{align*}
p_n + q_n - r_n=sum_{t=1}^{3}frac{1}{r_t^{n+1}}g_t(a_0, a_1, a_2, r_1, r_2, r_3)
end{align*}
For functions $g_1, g_2, g_3$, which you can pre-compute in a computer implementation.
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
edited Jan 17 at 16:38
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
answered Jan 17 at 16:30
Tom ChenTom Chen
1,568715
1,568715
add a comment |
add a comment |
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$begingroup$
Do you mean $a_n = a_{n-1} + 2n + 3a_{n-3}$ or $a_n = a_{n-2} + 2 a_{n-2} + 3 a_{n-3}$ (or something else entirely)?
$endgroup$
– Connor Harris
Jan 17 at 14:41
$begingroup$
First one!!!!!!
$endgroup$
– Sagar Sharma
Jan 17 at 14:43
$begingroup$
Do you have initial conditions?
$endgroup$
– Adrian Keister
Jan 17 at 14:49
$begingroup$
Initial conditions are varying.
$endgroup$
– Sagar Sharma
Jan 17 at 14:51
1
$begingroup$
Let $a_n=b_n+p+qn$ to eliminate $n$
$endgroup$
– lab bhattacharjee
Jan 17 at 15:16