How to solve this nonlinear recurrence relations?
$begingroup$
I know how to solve linear homogeneous and non-homogeneous recurrence relations. But today I found this problem.
$$F_1=1$$
$$F_2=7$$
$$F_3=13$$
$$F_n = 3F_{n-1} + k(F_{n-2}F_{n-3})+10$$
Where $k$ is a given number.
A friend of mine had this problem on a programming homework. And I wanted to impress him and told him there is a formula for this problem. But when I looked it more closely I realized it is not like other such problems that I have solved in discrete mathematics course, where we solved linear recurrence relations. So my question is could I find a formula for this recurrence relation and if I can how?
discrete-mathematics recurrence-relations recursion
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
$endgroup$
|
show 1 more comment
$begingroup$
I know how to solve linear homogeneous and non-homogeneous recurrence relations. But today I found this problem.
$$F_1=1$$
$$F_2=7$$
$$F_3=13$$
$$F_n = 3F_{n-1} + k(F_{n-2}F_{n-3})+10$$
Where $k$ is a given number.
A friend of mine had this problem on a programming homework. And I wanted to impress him and told him there is a formula for this problem. But when I looked it more closely I realized it is not like other such problems that I have solved in discrete mathematics course, where we solved linear recurrence relations. So my question is could I find a formula for this recurrence relation and if I can how?
discrete-mathematics recurrence-relations recursion
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
$endgroup$
$begingroup$
@Did that's the Laplace Transform, right? My guess, the product $F_{n-2}times F_{n-3}$ spoils the fun.
$endgroup$
– tpb261
Dec 18 '18 at 16:56
$begingroup$
Hmmm... Somehow, I missed the product. Sorry.
$endgroup$
– Did
Dec 18 '18 at 16:58
2
$begingroup$
Doubt there is a simple formula here (as the growth is huge; and non-linear equations rarely have closed forms). As $ntoinfty$ we have $F_n sim kF_{n-2}F_{n-3}$ so if $F_n sim e^{g(n)}$ then $g(n) sim g(n-2) + g(n-3) + log(k)$ which has solutions on the form $g(n) sim r^n$ for some $r approx 1.32$ so asymptotically we will have $F_n sim e^{r^n}$ (this is all very rough).
$endgroup$
– Winther
Dec 18 '18 at 17:18
1
$begingroup$
One can probably prove that $lim_{ntoinfty}frac{log(log(F_n))}{n} = log(r)$ where $r$ is the real root of $r^3 - r - 1 = 0$
$endgroup$
– Winther
Dec 18 '18 at 17:21
1
$begingroup$
Linear recurrence problems necessarily have a closed form (if I am remembering correctly) and there are few others that we get lucky and we can solve as well... but generally speaking the non-linear cases won't have simple formulas. Just to echo what Winther said above...
$endgroup$
– Mason
Dec 18 '18 at 20:50
|
show 1 more comment
$begingroup$
I know how to solve linear homogeneous and non-homogeneous recurrence relations. But today I found this problem.
$$F_1=1$$
$$F_2=7$$
$$F_3=13$$
$$F_n = 3F_{n-1} + k(F_{n-2}F_{n-3})+10$$
Where $k$ is a given number.
A friend of mine had this problem on a programming homework. And I wanted to impress him and told him there is a formula for this problem. But when I looked it more closely I realized it is not like other such problems that I have solved in discrete mathematics course, where we solved linear recurrence relations. So my question is could I find a formula for this recurrence relation and if I can how?
discrete-mathematics recurrence-relations recursion
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
$endgroup$
I know how to solve linear homogeneous and non-homogeneous recurrence relations. But today I found this problem.
$$F_1=1$$
$$F_2=7$$
$$F_3=13$$
$$F_n = 3F_{n-1} + k(F_{n-2}F_{n-3})+10$$
Where $k$ is a given number.
A friend of mine had this problem on a programming homework. And I wanted to impress him and told him there is a formula for this problem. But when I looked it more closely I realized it is not like other such problems that I have solved in discrete mathematics course, where we solved linear recurrence relations. So my question is could I find a formula for this recurrence relation and if I can how?
discrete-mathematics recurrence-relations recursion
discrete-mathematics recurrence-relations recursion
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
edited Dec 18 '18 at 17:02
rafa11111
1,1952417
1,1952417
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
asked Dec 18 '18 at 16:34
D.DimovD.Dimov
61
61
$begingroup$
@Did that's the Laplace Transform, right? My guess, the product $F_{n-2}times F_{n-3}$ spoils the fun.
$endgroup$
– tpb261
Dec 18 '18 at 16:56
$begingroup$
Hmmm... Somehow, I missed the product. Sorry.
$endgroup$
– Did
Dec 18 '18 at 16:58
2
$begingroup$
Doubt there is a simple formula here (as the growth is huge; and non-linear equations rarely have closed forms). As $ntoinfty$ we have $F_n sim kF_{n-2}F_{n-3}$ so if $F_n sim e^{g(n)}$ then $g(n) sim g(n-2) + g(n-3) + log(k)$ which has solutions on the form $g(n) sim r^n$ for some $r approx 1.32$ so asymptotically we will have $F_n sim e^{r^n}$ (this is all very rough).
$endgroup$
– Winther
Dec 18 '18 at 17:18
1
$begingroup$
One can probably prove that $lim_{ntoinfty}frac{log(log(F_n))}{n} = log(r)$ where $r$ is the real root of $r^3 - r - 1 = 0$
$endgroup$
– Winther
Dec 18 '18 at 17:21
1
$begingroup$
Linear recurrence problems necessarily have a closed form (if I am remembering correctly) and there are few others that we get lucky and we can solve as well... but generally speaking the non-linear cases won't have simple formulas. Just to echo what Winther said above...
$endgroup$
– Mason
Dec 18 '18 at 20:50
|
show 1 more comment
$begingroup$
@Did that's the Laplace Transform, right? My guess, the product $F_{n-2}times F_{n-3}$ spoils the fun.
$endgroup$
– tpb261
Dec 18 '18 at 16:56
$begingroup$
Hmmm... Somehow, I missed the product. Sorry.
$endgroup$
– Did
Dec 18 '18 at 16:58
2
$begingroup$
Doubt there is a simple formula here (as the growth is huge; and non-linear equations rarely have closed forms). As $ntoinfty$ we have $F_n sim kF_{n-2}F_{n-3}$ so if $F_n sim e^{g(n)}$ then $g(n) sim g(n-2) + g(n-3) + log(k)$ which has solutions on the form $g(n) sim r^n$ for some $r approx 1.32$ so asymptotically we will have $F_n sim e^{r^n}$ (this is all very rough).
$endgroup$
– Winther
Dec 18 '18 at 17:18
1
$begingroup$
One can probably prove that $lim_{ntoinfty}frac{log(log(F_n))}{n} = log(r)$ where $r$ is the real root of $r^3 - r - 1 = 0$
$endgroup$
– Winther
Dec 18 '18 at 17:21
1
$begingroup$
Linear recurrence problems necessarily have a closed form (if I am remembering correctly) and there are few others that we get lucky and we can solve as well... but generally speaking the non-linear cases won't have simple formulas. Just to echo what Winther said above...
$endgroup$
– Mason
Dec 18 '18 at 20:50
$begingroup$
@Did that's the Laplace Transform, right? My guess, the product $F_{n-2}times F_{n-3}$ spoils the fun.
$endgroup$
– tpb261
Dec 18 '18 at 16:56
$begingroup$
@Did that's the Laplace Transform, right? My guess, the product $F_{n-2}times F_{n-3}$ spoils the fun.
$endgroup$
– tpb261
Dec 18 '18 at 16:56
$begingroup$
Hmmm... Somehow, I missed the product. Sorry.
$endgroup$
– Did
Dec 18 '18 at 16:58
$begingroup$
Hmmm... Somehow, I missed the product. Sorry.
$endgroup$
– Did
Dec 18 '18 at 16:58
2
2
$begingroup$
Doubt there is a simple formula here (as the growth is huge; and non-linear equations rarely have closed forms). As $ntoinfty$ we have $F_n sim kF_{n-2}F_{n-3}$ so if $F_n sim e^{g(n)}$ then $g(n) sim g(n-2) + g(n-3) + log(k)$ which has solutions on the form $g(n) sim r^n$ for some $r approx 1.32$ so asymptotically we will have $F_n sim e^{r^n}$ (this is all very rough).
$endgroup$
– Winther
Dec 18 '18 at 17:18
$begingroup$
Doubt there is a simple formula here (as the growth is huge; and non-linear equations rarely have closed forms). As $ntoinfty$ we have $F_n sim kF_{n-2}F_{n-3}$ so if $F_n sim e^{g(n)}$ then $g(n) sim g(n-2) + g(n-3) + log(k)$ which has solutions on the form $g(n) sim r^n$ for some $r approx 1.32$ so asymptotically we will have $F_n sim e^{r^n}$ (this is all very rough).
$endgroup$
– Winther
Dec 18 '18 at 17:18
1
1
$begingroup$
One can probably prove that $lim_{ntoinfty}frac{log(log(F_n))}{n} = log(r)$ where $r$ is the real root of $r^3 - r - 1 = 0$
$endgroup$
– Winther
Dec 18 '18 at 17:21
$begingroup$
One can probably prove that $lim_{ntoinfty}frac{log(log(F_n))}{n} = log(r)$ where $r$ is the real root of $r^3 - r - 1 = 0$
$endgroup$
– Winther
Dec 18 '18 at 17:21
1
1
$begingroup$
Linear recurrence problems necessarily have a closed form (if I am remembering correctly) and there are few others that we get lucky and we can solve as well... but generally speaking the non-linear cases won't have simple formulas. Just to echo what Winther said above...
$endgroup$
– Mason
Dec 18 '18 at 20:50
$begingroup$
Linear recurrence problems necessarily have a closed form (if I am remembering correctly) and there are few others that we get lucky and we can solve as well... but generally speaking the non-linear cases won't have simple formulas. Just to echo what Winther said above...
$endgroup$
– Mason
Dec 18 '18 at 20:50
|
show 1 more comment
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$begingroup$
@Did that's the Laplace Transform, right? My guess, the product $F_{n-2}times F_{n-3}$ spoils the fun.
$endgroup$
– tpb261
Dec 18 '18 at 16:56
$begingroup$
Hmmm... Somehow, I missed the product. Sorry.
$endgroup$
– Did
Dec 18 '18 at 16:58
2
$begingroup$
Doubt there is a simple formula here (as the growth is huge; and non-linear equations rarely have closed forms). As $ntoinfty$ we have $F_n sim kF_{n-2}F_{n-3}$ so if $F_n sim e^{g(n)}$ then $g(n) sim g(n-2) + g(n-3) + log(k)$ which has solutions on the form $g(n) sim r^n$ for some $r approx 1.32$ so asymptotically we will have $F_n sim e^{r^n}$ (this is all very rough).
$endgroup$
– Winther
Dec 18 '18 at 17:18
1
$begingroup$
One can probably prove that $lim_{ntoinfty}frac{log(log(F_n))}{n} = log(r)$ where $r$ is the real root of $r^3 - r - 1 = 0$
$endgroup$
– Winther
Dec 18 '18 at 17:21
1
$begingroup$
Linear recurrence problems necessarily have a closed form (if I am remembering correctly) and there are few others that we get lucky and we can solve as well... but generally speaking the non-linear cases won't have simple formulas. Just to echo what Winther said above...
$endgroup$
– Mason
Dec 18 '18 at 20:50