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finding a solution for this recurrence relation
Why does this recurrence relation generate a sinusoidal curve?Recurrence relation rabbit populationHow do I solve the recurrence relation without manually counting?Wolfram alpha giving wrong result on recurrence?Particular solution of a recurrence relationHow to find an explicit formula for sequence defined by recurrence relation?Modeling problems with recurrence relationHow to find the initial terms of the recurrence relation if you know the nth term?Recurrence Relation, Compound AnnuallySolving a nonhomogeneous recurrence relation?
$begingroup$
Find the sequence satisfying the recurrence relation $$n a_n+1 = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.
I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$fraca_n+1n+1 =fraca_nn +1.$$
After setting $b_k=a_n/n$, the resulting relation is $b_k+1 = b_k +1$.
$b_k+2 = b_k+1 +1$
by subtracting this relation from the one before it, I got :
$b_k+2 -2*b_k+1 +b_k = 0$
after that I replace $b_k = lambda^k$
I got :
$lambda^k+2 -2*lambda^k+1+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$
and the solution for the last relation is $lambda = 1$
but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.
can someone show me a proper way or a proper solution for this recurrence relation?
recurrence-relations recursion initial-value-problems
edited 4 hours ago
Abdalrohman Alsalkhadi
52
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Find the sequence satisfying the recurrence relation $$n a_n+1 = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.
I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$fraca_n+1n+1 =fraca_nn +1.$$
After setting $b_k=a_n/n$, the resulting relation is $b_k+1 = b_k +1$.
$b_k+2 = b_k+1 +1$
by subtracting this relation from the one before it, I got :
$b_k+2 -2*b_k+1 +b_k = 0$
after that I replace $b_k = lambda^k$
I got :
$lambda^k+2 -2*lambda^k+1+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$
and the solution for the last relation is $lambda = 1$
but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.
can someone show me a proper way or a proper solution for this recurrence relation?
recurrence-relations recursion initial-value-problems
edited 4 hours ago
Abdalrohman Alsalkhadi
52
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
$b_k+1 = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
48 mins ago
add a comment |
$begingroup$
Find the sequence satisfying the recurrence relation $$n a_n+1 = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.
I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$fraca_n+1n+1 =fraca_nn +1.$$
After setting $b_k=a_n/n$, the resulting relation is $b_k+1 = b_k +1$.
$b_k+2 = b_k+1 +1$
by subtracting this relation from the one before it, I got :
$b_k+2 -2*b_k+1 +b_k = 0$
after that I replace $b_k = lambda^k$
I got :
$lambda^k+2 -2*lambda^k+1+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$
and the solution for the last relation is $lambda = 1$
but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.
can someone show me a proper way or a proper solution for this recurrence relation?
recurrence-relations recursion initial-value-problems
edited 4 hours ago
Abdalrohman Alsalkhadi
52
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Find the sequence satisfying the recurrence relation $$n a_n+1 = (n+1) a_n+n(n+1)$$ with the initial condition $a_0=0$.
I'm trying to find a solution for this recurrence relation. After dividing both sides by $n(n+1)$, we get:
$$fraca_n+1n+1 =fraca_nn +1.$$
After setting $b_k=a_n/n$, the resulting relation is $b_k+1 = b_k +1$.
$b_k+2 = b_k+1 +1$
by subtracting this relation from the one before it, I got :
$b_k+2 -2*b_k+1 +b_k = 0$
after that I replace $b_k = lambda^k$
I got :
$lambda^k+2 -2*lambda^k+1+b_k$ and after dividing the relation by $lambda^k$ the result is $lambda^2 + lambda +1 = 0$
and the solution for the last relation is $lambda = 1$
but I stopped there because the initial Condition is
$a_0=0$ and I could't find the value of $a_1$.
can someone show me a proper way or a proper solution for this recurrence relation?
recurrence-relations recursion initial-value-problems
recurrence-relations recursion initial-value-problems
edited 4 hours ago
Abdalrohman Alsalkhadi
52
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Abdalrohman Alsalkhadi
52
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Abdalrohman Alsalkhadi
52
edited 4 hours ago
Abdalrohman Alsalkhadi
52
edited 4 hours ago
Abdalrohman Alsalkhadi
52
52
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
asked 5 hours ago
Ahmad RmmoAhmad Rmmo
212
212
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ahmad Rmmo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
$b_k+1 = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
48 mins ago
add a comment |
$begingroup$
$b_k+1 = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
48 mins ago
$begingroup$
$b_k+1 = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
48 mins ago
$begingroup$
$b_k+1 = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
48 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
$$
0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
$$
and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.
answered 5 hours ago
gt6989bgt6989b
36.3k22557
$endgroup$
add a comment |
$begingroup$
Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
$$n a_n+1 = (n+1) a_n+n(n+1)tag1$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$
answered 1 hour ago
SomosSomos
15.6k11437
$endgroup$
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
$begingroup$
This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
$$
0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
$$
and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.
answered 5 hours ago
gt6989bgt6989b
36.3k22557
$endgroup$
add a comment |
$begingroup$
This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
$$
0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
$$
and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.
answered 5 hours ago
gt6989bgt6989b
36.3k22557
$endgroup$
add a comment |
$begingroup$
This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
$$
0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
$$
and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.
answered 5 hours ago
gt6989bgt6989b
36.3k22557
$endgroup$
This is not a well-posed problem. In other words, given $a_0=0$, the value of $a_1$ can be any real number whatsoever since it must satisfy
$$
0 cdot a_1 = 1 cdot a_0 + 0 cdot 1,
$$
and both sides are $0$ for any value of $a_1$, so the equality holds. In other words, the specified conditions do not uniquely specify a sequence.
answered 5 hours ago
gt6989bgt6989b
36.3k22557
answered 5 hours ago
gt6989bgt6989b
36.3k22557
answered 5 hours ago
gt6989bgt6989b
36.3k22557
answered 5 hours ago
gt6989bgt6989b
36.3k22557
36.3k22557
add a comment |
add a comment |
$begingroup$
Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
$$n a_n+1 = (n+1) a_n+n(n+1)tag1$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$
answered 1 hour ago
SomosSomos
15.6k11437
$endgroup$
add a comment |
$begingroup$
Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
$$n a_n+1 = (n+1) a_n+n(n+1)tag1$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$
answered 1 hour ago
SomosSomos
15.6k11437
$endgroup$
add a comment |
$begingroup$
Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
$$n a_n+1 = (n+1) a_n+n(n+1)tag1$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$
answered 1 hour ago
SomosSomos
15.6k11437
$endgroup$
Given the value of $a_1$ you have $a_n = n(n-1) + na_1$ for all $n$. The recursion equation
$$n a_n+1 = (n+1) a_n+n(n+1)tag1$$ for $n=0$ is just $a_0=0.$ In other words, you should start with initial condition for $a_1$ and then all other values of $a_n$ for $n>1$ are determined, and also $a_0=0.$
answered 1 hour ago
SomosSomos
15.6k11437
answered 1 hour ago
SomosSomos
15.6k11437
answered 1 hour ago
SomosSomos
15.6k11437
answered 1 hour ago
SomosSomos
15.6k11437
15.6k11437
add a comment |
add a comment |
Ahmad Rmmo is a new contributor. Be nice, and check out our Code of Conduct.
Ahmad Rmmo is a new contributor. Be nice, and check out our Code of Conduct.
Ahmad Rmmo is a new contributor. Be nice, and check out our Code of Conduct.
Ahmad Rmmo is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
$b_k+1 = b_k +1$ gives $b_k = k -1 + b_1$.
$endgroup$
– lhf
48 mins ago