Linear recurrence solving for tighest possible big O bounds
$begingroup$
I am dealing with the following linear recurrence:
X0 = 1
X1 = 2
Xn = 3Xn-1 + 2Xn-2
I have proven that this has an upper bound of O(4n)
However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?
recurrence-relations
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
$endgroup$
add a comment |
$begingroup$
I am dealing with the following linear recurrence:
X0 = 1
X1 = 2
Xn = 3Xn-1 + 2Xn-2
I have proven that this has an upper bound of O(4n)
However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?
recurrence-relations
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
$endgroup$
$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13
$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28
$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13
add a comment |
$begingroup$
I am dealing with the following linear recurrence:
X0 = 1
X1 = 2
Xn = 3Xn-1 + 2Xn-2
I have proven that this has an upper bound of O(4n)
However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?
recurrence-relations
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
$endgroup$
I am dealing with the following linear recurrence:
X0 = 1
X1 = 2
Xn = 3Xn-1 + 2Xn-2
I have proven that this has an upper bound of O(4n)
However, I have been asked to come up with tighter bounds for this linear recurrence, but I dont know how to begin this. Does this involve solving the recurrence?
recurrence-relations
recurrence-relations
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
edited Dec 18 '18 at 13:13
Joel Reyes Noche
5,65733149
5,65733149
asked Dec 15 '18 at 22:09
user626552
$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13
$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28
$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13
add a comment |
$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13
$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28
$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13
$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13
$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13
$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28
$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28
$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13
$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write the solution as
$$
x_n = a lambda^n
$$
If you replace that in your original expression you get
$$
a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
$$
Which reduces to
$$
lambda^2 = 3 lambda + 2
$$
Solutions are
$$
lambda = frac{3}{2} pm frac{sqrt{13}}{2}
$$
So the solution is
$$
x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
$$
The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write the solution as
$$
x_n = a lambda^n
$$
If you replace that in your original expression you get
$$
a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
$$
Which reduces to
$$
lambda^2 = 3 lambda + 2
$$
Solutions are
$$
lambda = frac{3}{2} pm frac{sqrt{13}}{2}
$$
So the solution is
$$
x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
$$
The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Write the solution as
$$
x_n = a lambda^n
$$
If you replace that in your original expression you get
$$
a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
$$
Which reduces to
$$
lambda^2 = 3 lambda + 2
$$
Solutions are
$$
lambda = frac{3}{2} pm frac{sqrt{13}}{2}
$$
So the solution is
$$
x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
$$
The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
Write the solution as
$$
x_n = a lambda^n
$$
If you replace that in your original expression you get
$$
a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
$$
Which reduces to
$$
lambda^2 = 3 lambda + 2
$$
Solutions are
$$
lambda = frac{3}{2} pm frac{sqrt{13}}{2}
$$
So the solution is
$$
x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
$$
The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
$endgroup$
Write the solution as
$$
x_n = a lambda^n
$$
If you replace that in your original expression you get
$$
a lambda^n = 3 a lambda^{n - 1} + 2 alambda^{n-2}
$$
Which reduces to
$$
lambda^2 = 3 lambda + 2
$$
Solutions are
$$
lambda = frac{3}{2} pm frac{sqrt{13}}{2}
$$
So the solution is
$$
x_n = a left( frac{3}{2} + frac{sqrt{13}}{2} right)^n + bleft( frac{3}{2} - frac{sqrt{13}}{2} right)^n
$$
The constants $a$ and $b$ you can determine by setting $n=0$ and $n=1$ and using the conditions $x_0 = 1$ and $x_1 = 2$
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
answered Dec 15 '18 at 22:28
caveraccaverac
14.8k31130
14.8k31130
add a comment |
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$begingroup$
Do you know how to solve recurrence of this type? It is fairly straightforward
$endgroup$
– Shubham Johri
Dec 15 '18 at 22:13
$begingroup$
I think the best constant is $frac{3+sqrt{17}}{2} approx 3.55$ instead of $4$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:28
$begingroup$
Do not change your question after it has been answered.
$endgroup$
– Joel Reyes Noche
Dec 18 '18 at 13:13