Mixing
Solve recurrence relation $a_n = 3 a_{n - 3} + a_{n-1} + 2 n$
$begingroup$
I am not able to solve the above recurrence relation which involves $n$. The initial conditions $a_0 = 1, a_1 = 2, a_2 = 3$, for $n geq 0$. Please try to give closed form in terms of arbitrary $a_i geq 1 $ for $i geq 0$
recurrence-relations
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
$endgroup$
add a comment |
$begingroup$
I am not able to solve the above recurrence relation which involves $n$. The initial conditions $a_0 = 1, a_1 = 2, a_2 = 3$, for $n geq 0$. Please try to give closed form in terms of arbitrary $a_i geq 1 $ for $i geq 0$
recurrence-relations
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
$endgroup$
$begingroup$
Seems to be a duplicate of math.stackexchange.com/questions/3077056/…
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:27
add a comment |
$begingroup$
I am not able to solve the above recurrence relation which involves $n$. The initial conditions $a_0 = 1, a_1 = 2, a_2 = 3$, for $n geq 0$. Please try to give closed form in terms of arbitrary $a_i geq 1 $ for $i geq 0$
recurrence-relations
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
$endgroup$
I am not able to solve the above recurrence relation which involves $n$. The initial conditions $a_0 = 1, a_1 = 2, a_2 = 3$, for $n geq 0$. Please try to give closed form in terms of arbitrary $a_i geq 1 $ for $i geq 0$
recurrence-relations
recurrence-relations
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 17 at 16:20
Brij Raj KishoreBrij Raj Kishore
215112
215112
$begingroup$
Seems to be a duplicate of math.stackexchange.com/questions/3077056/…
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:27
add a comment |
$begingroup$
Seems to be a duplicate of math.stackexchange.com/questions/3077056/…
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:27
$begingroup$
Seems to be a duplicate of math.stackexchange.com/questions/3077056/…
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:27
$begingroup$
Seems to be a duplicate of math.stackexchange.com/questions/3077056/…
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is what you can do:
First write $a_{n+1}=3a_{n-2}+a_n+2n+2$ (I inserted $n=n+1$ to the equation) and subtract the original equation to obtain $a_{n+1}-a_n=3a_{n-2}-3a_{n-3}+a_n-a_{n-1}+2$ which is the same as $$a_{n+1}=2a_n-a_{n-1}+3a_{n-2}-3a_{n-3}+2$$
Now do the same trick again: $a_{n+2}=2a_{n+1}-a_n+3a_{n-1}-3a_{n-2}+2$ and subtract the above equation to obtain: $$a_{n+2}=3a_{n+1}-3a_n+4a_{n-1}-6a_{n-2}+3a_{n-3}$$ Now this is a homogenous recurrence relation. Since we increased the degree by $2$, we also need to compute $a_4$ and $a_5$ and then solve the equation via classical techniques.
Edit : To solve the relation completely, you first insert $x^n$ for $a_n$ to obtain the degree $5$ polynomial $x^5-3x^4+3x^3-4x^2+6x-3=0$. Then compute all the roots of this polynomial, say $x_1,dots,x_5$. If they are all distinct, general solution of the relation is of the form $$c_1cdot x_1^n+dots+c_5cdot x_5^n$$ for $c_iinmathbb{C}$ and you can compute $c_i$ by using the initial conditions. If there is a double root, say $x_1$, (in this case $1$ is a double root and there are $3$ distinct roots) then a general solution is of the form $$c_1cdot x_1^n+c_2cdot n x_1^n+c_3cdot x_2^n+dots+c_5cdot x_4^n$$ and again, you can compute $c_i$ using the initial conditions.
answered Jan 17 at 16:27
LeventLevent
2,729925
$endgroup$
$begingroup$
Since this eqn will become of degree 5, then how can we solve it? Can you give hint or link.
$endgroup$
– Brij Raj Kishore
Jan 17 at 16:36
$begingroup$
@BrijRajKishore Let me edit
$endgroup$
– Levent
Jan 17 at 16:37
$begingroup$
@BrijRajKishore Agree with you. It is even more difficult now!
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:47
$begingroup$
@mathcounterexamples.net well, the classical way to solve these kind of non homogeneous equations require some techniques (like separating a solution into two parts, one solution for the homogenous part and one solution for the non homogeneous part etc.). For me, I like to reduce the case to the homogenous case and then solve it using the basic ideas. It of course increases the degree, but I don't really think it makes it harder. But probably one needs a computer to compute the roots and the coefficients.
$endgroup$
– Levent
Jan 17 at 16:50
$begingroup$
Well you have an easier way to reduce it to an homogeneous equation. Namely changing $a_n = b_n + pn+q$ with $p,q$ well chosen. Then you stay with a degree 3 equation.
$endgroup$
– mathcounterexamples.net
Jan 17 at 17:03
|
show 1 more comment
$begingroup$
Another solution
Take $a_n= b_n -frac{2n}{3} -frac{20}{9}$ and plug $a_n$ in the initial difference equation.
You find that $b_n = 3 b_{n - 3} + b_{n-1}$ which is a linear difference equation of order $3$.
So now to find the solution you still have to find the root of $p(x) = x^3-x^2-3$ if we want to use the "classical method".
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is what you can do:
First write $a_{n+1}=3a_{n-2}+a_n+2n+2$ (I inserted $n=n+1$ to the equation) and subtract the original equation to obtain $a_{n+1}-a_n=3a_{n-2}-3a_{n-3}+a_n-a_{n-1}+2$ which is the same as $$a_{n+1}=2a_n-a_{n-1}+3a_{n-2}-3a_{n-3}+2$$
Now do the same trick again: $a_{n+2}=2a_{n+1}-a_n+3a_{n-1}-3a_{n-2}+2$ and subtract the above equation to obtain: $$a_{n+2}=3a_{n+1}-3a_n+4a_{n-1}-6a_{n-2}+3a_{n-3}$$ Now this is a homogenous recurrence relation. Since we increased the degree by $2$, we also need to compute $a_4$ and $a_5$ and then solve the equation via classical techniques.
Edit : To solve the relation completely, you first insert $x^n$ for $a_n$ to obtain the degree $5$ polynomial $x^5-3x^4+3x^3-4x^2+6x-3=0$. Then compute all the roots of this polynomial, say $x_1,dots,x_5$. If they are all distinct, general solution of the relation is of the form $$c_1cdot x_1^n+dots+c_5cdot x_5^n$$ for $c_iinmathbb{C}$ and you can compute $c_i$ by using the initial conditions. If there is a double root, say $x_1$, (in this case $1$ is a double root and there are $3$ distinct roots) then a general solution is of the form $$c_1cdot x_1^n+c_2cdot n x_1^n+c_3cdot x_2^n+dots+c_5cdot x_4^n$$ and again, you can compute $c_i$ using the initial conditions.
answered Jan 17 at 16:27
LeventLevent
2,729925
$endgroup$
$begingroup$
Since this eqn will become of degree 5, then how can we solve it? Can you give hint or link.
$endgroup$
– Brij Raj Kishore
Jan 17 at 16:36
$begingroup$
@BrijRajKishore Let me edit
$endgroup$
– Levent
Jan 17 at 16:37
$begingroup$
@BrijRajKishore Agree with you. It is even more difficult now!
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:47
$begingroup$
@mathcounterexamples.net well, the classical way to solve these kind of non homogeneous equations require some techniques (like separating a solution into two parts, one solution for the homogenous part and one solution for the non homogeneous part etc.). For me, I like to reduce the case to the homogenous case and then solve it using the basic ideas. It of course increases the degree, but I don't really think it makes it harder. But probably one needs a computer to compute the roots and the coefficients.
$endgroup$
– Levent
Jan 17 at 16:50
$begingroup$
Well you have an easier way to reduce it to an homogeneous equation. Namely changing $a_n = b_n + pn+q$ with $p,q$ well chosen. Then you stay with a degree 3 equation.
$endgroup$
– mathcounterexamples.net
Jan 17 at 17:03
|
show 1 more comment
$begingroup$
Here is what you can do:
First write $a_{n+1}=3a_{n-2}+a_n+2n+2$ (I inserted $n=n+1$ to the equation) and subtract the original equation to obtain $a_{n+1}-a_n=3a_{n-2}-3a_{n-3}+a_n-a_{n-1}+2$ which is the same as $$a_{n+1}=2a_n-a_{n-1}+3a_{n-2}-3a_{n-3}+2$$
Now do the same trick again: $a_{n+2}=2a_{n+1}-a_n+3a_{n-1}-3a_{n-2}+2$ and subtract the above equation to obtain: $$a_{n+2}=3a_{n+1}-3a_n+4a_{n-1}-6a_{n-2}+3a_{n-3}$$ Now this is a homogenous recurrence relation. Since we increased the degree by $2$, we also need to compute $a_4$ and $a_5$ and then solve the equation via classical techniques.
Edit : To solve the relation completely, you first insert $x^n$ for $a_n$ to obtain the degree $5$ polynomial $x^5-3x^4+3x^3-4x^2+6x-3=0$. Then compute all the roots of this polynomial, say $x_1,dots,x_5$. If they are all distinct, general solution of the relation is of the form $$c_1cdot x_1^n+dots+c_5cdot x_5^n$$ for $c_iinmathbb{C}$ and you can compute $c_i$ by using the initial conditions. If there is a double root, say $x_1$, (in this case $1$ is a double root and there are $3$ distinct roots) then a general solution is of the form $$c_1cdot x_1^n+c_2cdot n x_1^n+c_3cdot x_2^n+dots+c_5cdot x_4^n$$ and again, you can compute $c_i$ using the initial conditions.
answered Jan 17 at 16:27
LeventLevent
2,729925
$endgroup$
$begingroup$
Since this eqn will become of degree 5, then how can we solve it? Can you give hint or link.
$endgroup$
– Brij Raj Kishore
Jan 17 at 16:36
$begingroup$
@BrijRajKishore Let me edit
$endgroup$
– Levent
Jan 17 at 16:37
$begingroup$
@BrijRajKishore Agree with you. It is even more difficult now!
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:47
$begingroup$
@mathcounterexamples.net well, the classical way to solve these kind of non homogeneous equations require some techniques (like separating a solution into two parts, one solution for the homogenous part and one solution for the non homogeneous part etc.). For me, I like to reduce the case to the homogenous case and then solve it using the basic ideas. It of course increases the degree, but I don't really think it makes it harder. But probably one needs a computer to compute the roots and the coefficients.
$endgroup$
– Levent
Jan 17 at 16:50
$begingroup$
Well you have an easier way to reduce it to an homogeneous equation. Namely changing $a_n = b_n + pn+q$ with $p,q$ well chosen. Then you stay with a degree 3 equation.
$endgroup$
– mathcounterexamples.net
Jan 17 at 17:03
|
show 1 more comment
$begingroup$
Here is what you can do:
First write $a_{n+1}=3a_{n-2}+a_n+2n+2$ (I inserted $n=n+1$ to the equation) and subtract the original equation to obtain $a_{n+1}-a_n=3a_{n-2}-3a_{n-3}+a_n-a_{n-1}+2$ which is the same as $$a_{n+1}=2a_n-a_{n-1}+3a_{n-2}-3a_{n-3}+2$$
Now do the same trick again: $a_{n+2}=2a_{n+1}-a_n+3a_{n-1}-3a_{n-2}+2$ and subtract the above equation to obtain: $$a_{n+2}=3a_{n+1}-3a_n+4a_{n-1}-6a_{n-2}+3a_{n-3}$$ Now this is a homogenous recurrence relation. Since we increased the degree by $2$, we also need to compute $a_4$ and $a_5$ and then solve the equation via classical techniques.
Edit : To solve the relation completely, you first insert $x^n$ for $a_n$ to obtain the degree $5$ polynomial $x^5-3x^4+3x^3-4x^2+6x-3=0$. Then compute all the roots of this polynomial, say $x_1,dots,x_5$. If they are all distinct, general solution of the relation is of the form $$c_1cdot x_1^n+dots+c_5cdot x_5^n$$ for $c_iinmathbb{C}$ and you can compute $c_i$ by using the initial conditions. If there is a double root, say $x_1$, (in this case $1$ is a double root and there are $3$ distinct roots) then a general solution is of the form $$c_1cdot x_1^n+c_2cdot n x_1^n+c_3cdot x_2^n+dots+c_5cdot x_4^n$$ and again, you can compute $c_i$ using the initial conditions.
answered Jan 17 at 16:27
LeventLevent
2,729925
$endgroup$
Here is what you can do:
First write $a_{n+1}=3a_{n-2}+a_n+2n+2$ (I inserted $n=n+1$ to the equation) and subtract the original equation to obtain $a_{n+1}-a_n=3a_{n-2}-3a_{n-3}+a_n-a_{n-1}+2$ which is the same as $$a_{n+1}=2a_n-a_{n-1}+3a_{n-2}-3a_{n-3}+2$$
Now do the same trick again: $a_{n+2}=2a_{n+1}-a_n+3a_{n-1}-3a_{n-2}+2$ and subtract the above equation to obtain: $$a_{n+2}=3a_{n+1}-3a_n+4a_{n-1}-6a_{n-2}+3a_{n-3}$$ Now this is a homogenous recurrence relation. Since we increased the degree by $2$, we also need to compute $a_4$ and $a_5$ and then solve the equation via classical techniques.
Edit : To solve the relation completely, you first insert $x^n$ for $a_n$ to obtain the degree $5$ polynomial $x^5-3x^4+3x^3-4x^2+6x-3=0$. Then compute all the roots of this polynomial, say $x_1,dots,x_5$. If they are all distinct, general solution of the relation is of the form $$c_1cdot x_1^n+dots+c_5cdot x_5^n$$ for $c_iinmathbb{C}$ and you can compute $c_i$ by using the initial conditions. If there is a double root, say $x_1$, (in this case $1$ is a double root and there are $3$ distinct roots) then a general solution is of the form $$c_1cdot x_1^n+c_2cdot n x_1^n+c_3cdot x_2^n+dots+c_5cdot x_4^n$$ and again, you can compute $c_i$ using the initial conditions.
answered Jan 17 at 16:27
LeventLevent
2,729925
edited Jan 17 at 16:47
answered Jan 17 at 16:27
LeventLevent
2,729925
answered Jan 17 at 16:27
LeventLevent
2,729925
answered Jan 17 at 16:27
LeventLevent
2,729925
2,729925
$begingroup$
Since this eqn will become of degree 5, then how can we solve it? Can you give hint or link.
$endgroup$
– Brij Raj Kishore
Jan 17 at 16:36
$begingroup$
@BrijRajKishore Let me edit
$endgroup$
– Levent
Jan 17 at 16:37
$begingroup$
@BrijRajKishore Agree with you. It is even more difficult now!
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:47
$begingroup$
@mathcounterexamples.net well, the classical way to solve these kind of non homogeneous equations require some techniques (like separating a solution into two parts, one solution for the homogenous part and one solution for the non homogeneous part etc.). For me, I like to reduce the case to the homogenous case and then solve it using the basic ideas. It of course increases the degree, but I don't really think it makes it harder. But probably one needs a computer to compute the roots and the coefficients.
$endgroup$
– Levent
Jan 17 at 16:50
$begingroup$
Well you have an easier way to reduce it to an homogeneous equation. Namely changing $a_n = b_n + pn+q$ with $p,q$ well chosen. Then you stay with a degree 3 equation.
$endgroup$
– mathcounterexamples.net
Jan 17 at 17:03
|
show 1 more comment
$begingroup$
Since this eqn will become of degree 5, then how can we solve it? Can you give hint or link.
$endgroup$
– Brij Raj Kishore
Jan 17 at 16:36
$begingroup$
@BrijRajKishore Let me edit
$endgroup$
– Levent
Jan 17 at 16:37
$begingroup$
@BrijRajKishore Agree with you. It is even more difficult now!
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:47
$begingroup$
@mathcounterexamples.net well, the classical way to solve these kind of non homogeneous equations require some techniques (like separating a solution into two parts, one solution for the homogenous part and one solution for the non homogeneous part etc.). For me, I like to reduce the case to the homogenous case and then solve it using the basic ideas. It of course increases the degree, but I don't really think it makes it harder. But probably one needs a computer to compute the roots and the coefficients.
$endgroup$
– Levent
Jan 17 at 16:50
$begingroup$
Well you have an easier way to reduce it to an homogeneous equation. Namely changing $a_n = b_n + pn+q$ with $p,q$ well chosen. Then you stay with a degree 3 equation.
$endgroup$
– mathcounterexamples.net
Jan 17 at 17:03
$begingroup$
Since this eqn will become of degree 5, then how can we solve it? Can you give hint or link.
$endgroup$
– Brij Raj Kishore
Jan 17 at 16:36
$begingroup$
Since this eqn will become of degree 5, then how can we solve it? Can you give hint or link.
$endgroup$
– Brij Raj Kishore
Jan 17 at 16:36
$begingroup$
@BrijRajKishore Let me edit
$endgroup$
– Levent
Jan 17 at 16:37
$begingroup$
@BrijRajKishore Let me edit
$endgroup$
– Levent
Jan 17 at 16:37
$begingroup$
@BrijRajKishore Agree with you. It is even more difficult now!
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:47
$begingroup$
@BrijRajKishore Agree with you. It is even more difficult now!
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:47
$begingroup$
@mathcounterexamples.net well, the classical way to solve these kind of non homogeneous equations require some techniques (like separating a solution into two parts, one solution for the homogenous part and one solution for the non homogeneous part etc.). For me, I like to reduce the case to the homogenous case and then solve it using the basic ideas. It of course increases the degree, but I don't really think it makes it harder. But probably one needs a computer to compute the roots and the coefficients.
$endgroup$
– Levent
Jan 17 at 16:50
$begingroup$
@mathcounterexamples.net well, the classical way to solve these kind of non homogeneous equations require some techniques (like separating a solution into two parts, one solution for the homogenous part and one solution for the non homogeneous part etc.). For me, I like to reduce the case to the homogenous case and then solve it using the basic ideas. It of course increases the degree, but I don't really think it makes it harder. But probably one needs a computer to compute the roots and the coefficients.
$endgroup$
– Levent
Jan 17 at 16:50
$begingroup$
Well you have an easier way to reduce it to an homogeneous equation. Namely changing $a_n = b_n + pn+q$ with $p,q$ well chosen. Then you stay with a degree 3 equation.
$endgroup$
– mathcounterexamples.net
Jan 17 at 17:03
$begingroup$
Well you have an easier way to reduce it to an homogeneous equation. Namely changing $a_n = b_n + pn+q$ with $p,q$ well chosen. Then you stay with a degree 3 equation.
$endgroup$
– mathcounterexamples.net
Jan 17 at 17:03
|
show 1 more comment
$begingroup$
Another solution
Take $a_n= b_n -frac{2n}{3} -frac{20}{9}$ and plug $a_n$ in the initial difference equation.
You find that $b_n = 3 b_{n - 3} + b_{n-1}$ which is a linear difference equation of order $3$.
So now to find the solution you still have to find the root of $p(x) = x^3-x^2-3$ if we want to use the "classical method".
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
$begingroup$
Another solution
Take $a_n= b_n -frac{2n}{3} -frac{20}{9}$ and plug $a_n$ in the initial difference equation.
You find that $b_n = 3 b_{n - 3} + b_{n-1}$ which is a linear difference equation of order $3$.
So now to find the solution you still have to find the root of $p(x) = x^3-x^2-3$ if we want to use the "classical method".
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
$begingroup$
Another solution
Take $a_n= b_n -frac{2n}{3} -frac{20}{9}$ and plug $a_n$ in the initial difference equation.
You find that $b_n = 3 b_{n - 3} + b_{n-1}$ which is a linear difference equation of order $3$.
So now to find the solution you still have to find the root of $p(x) = x^3-x^2-3$ if we want to use the "classical method".
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
Another solution
Take $a_n= b_n -frac{2n}{3} -frac{20}{9}$ and plug $a_n$ in the initial difference equation.
You find that $b_n = 3 b_{n - 3} + b_{n-1}$ which is a linear difference equation of order $3$.
So now to find the solution you still have to find the root of $p(x) = x^3-x^2-3$ if we want to use the "classical method".
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 17 at 17:21
mathcounterexamples.netmathcounterexamples.net
26.9k22157
26.9k22157
add a comment |
add a comment |
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$begingroup$
Seems to be a duplicate of math.stackexchange.com/questions/3077056/…
$endgroup$
– mathcounterexamples.net
Jan 17 at 16:27