Asymptotic for general Fibonacci sequence.
$begingroup$
Consider the sequence
$$f(n,k) = sum_{i=1}^{k}f(n-i,k)$$
with the following initial conditions $f(n,k) = 0$ for $n<0$ and $f(0,k)=1.$ I noticed that
$$lim_{nto infty} frac{f(n+1,k)}{f(n,k)}=phi(k)$$
and that
$$lim_{ktoinfty}phi(k)=2.$$
Based on the first limit I am guessing that
$$f(n,k)sim cphi(k)^{n}$$
where $c$ is some constant. But I am not sure how to determine $c$ and prove this asymptotic result. I think that it has something to do with this:
$$f(n,k)=frac{f(n,k)}{f(n-1,k)}cdot frac{f(n-1,k)}{f(n-2,k)}cdot frac{f(n-2,k)}{f(n-3,k)}cdots frac{f(1,k)}{f(0,k)}$$
but I am not able to write this rigorously. Perhaphs someone can give an argument that is rigorous enough?
recurrence-relations asymptotics fibonacci-numbers
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
$endgroup$
add a comment |
$begingroup$
Consider the sequence
$$f(n,k) = sum_{i=1}^{k}f(n-i,k)$$
with the following initial conditions $f(n,k) = 0$ for $n<0$ and $f(0,k)=1.$ I noticed that
$$lim_{nto infty} frac{f(n+1,k)}{f(n,k)}=phi(k)$$
and that
$$lim_{ktoinfty}phi(k)=2.$$
Based on the first limit I am guessing that
$$f(n,k)sim cphi(k)^{n}$$
where $c$ is some constant. But I am not sure how to determine $c$ and prove this asymptotic result. I think that it has something to do with this:
$$f(n,k)=frac{f(n,k)}{f(n-1,k)}cdot frac{f(n-1,k)}{f(n-2,k)}cdot frac{f(n-2,k)}{f(n-3,k)}cdots frac{f(1,k)}{f(0,k)}$$
but I am not able to write this rigorously. Perhaphs someone can give an argument that is rigorous enough?
recurrence-relations asymptotics fibonacci-numbers
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
$endgroup$
add a comment |
$begingroup$
Consider the sequence
$$f(n,k) = sum_{i=1}^{k}f(n-i,k)$$
with the following initial conditions $f(n,k) = 0$ for $n<0$ and $f(0,k)=1.$ I noticed that
$$lim_{nto infty} frac{f(n+1,k)}{f(n,k)}=phi(k)$$
and that
$$lim_{ktoinfty}phi(k)=2.$$
Based on the first limit I am guessing that
$$f(n,k)sim cphi(k)^{n}$$
where $c$ is some constant. But I am not sure how to determine $c$ and prove this asymptotic result. I think that it has something to do with this:
$$f(n,k)=frac{f(n,k)}{f(n-1,k)}cdot frac{f(n-1,k)}{f(n-2,k)}cdot frac{f(n-2,k)}{f(n-3,k)}cdots frac{f(1,k)}{f(0,k)}$$
but I am not able to write this rigorously. Perhaphs someone can give an argument that is rigorous enough?
recurrence-relations asymptotics fibonacci-numbers
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
$endgroup$
Consider the sequence
$$f(n,k) = sum_{i=1}^{k}f(n-i,k)$$
with the following initial conditions $f(n,k) = 0$ for $n<0$ and $f(0,k)=1.$ I noticed that
$$lim_{nto infty} frac{f(n+1,k)}{f(n,k)}=phi(k)$$
and that
$$lim_{ktoinfty}phi(k)=2.$$
Based on the first limit I am guessing that
$$f(n,k)sim cphi(k)^{n}$$
where $c$ is some constant. But I am not sure how to determine $c$ and prove this asymptotic result. I think that it has something to do with this:
$$f(n,k)=frac{f(n,k)}{f(n-1,k)}cdot frac{f(n-1,k)}{f(n-2,k)}cdot frac{f(n-2,k)}{f(n-3,k)}cdots frac{f(1,k)}{f(0,k)}$$
but I am not able to write this rigorously. Perhaphs someone can give an argument that is rigorous enough?
recurrence-relations asymptotics fibonacci-numbers
recurrence-relations asymptotics fibonacci-numbers
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
asked Jan 16 at 23:01
model_checkermodel_checker
4,45521931
4,45521931
add a comment |
add a comment |
1 Answer
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$begingroup$
First of all, the closed form of the sequence is
$$f(n, k) = sum_{i=0}^k c_ip_i^n$$
where $p_i$ are the roots of the equation
$$x^k = sum_{m=0}^{k - 1}x^m$$
and $c_i$ are constants to satisfy the initial conditions. It is a standard result which is easy to obtain looking at the generating function.
It is more or less obvious that asymptotic behavior is governed by the root with the largest absolute value. So your guess is correct. $f(n, k) sim phi(k)^n$ indeed, and $phi(k)$ is the largest root of the equation above.
The RHS of the equation is a geometric progression, therefore $x^k = dfrac{x^k - 1}{x-1}$ or $x^{k+1} - 2x^k + 1 = 0$. Now try to prove that as $k rightarrow infty$ the largest root approaches $2$ (easy to see that some root approaches $2$, a bit harder is to prove that it is indeed largest).
I am absolutely clueless on how to figure out $c$.
answered Jan 17 at 0:23
user58697user58697
1,851612
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
First of all, the closed form of the sequence is
$$f(n, k) = sum_{i=0}^k c_ip_i^n$$
where $p_i$ are the roots of the equation
$$x^k = sum_{m=0}^{k - 1}x^m$$
and $c_i$ are constants to satisfy the initial conditions. It is a standard result which is easy to obtain looking at the generating function.
It is more or less obvious that asymptotic behavior is governed by the root with the largest absolute value. So your guess is correct. $f(n, k) sim phi(k)^n$ indeed, and $phi(k)$ is the largest root of the equation above.
The RHS of the equation is a geometric progression, therefore $x^k = dfrac{x^k - 1}{x-1}$ or $x^{k+1} - 2x^k + 1 = 0$. Now try to prove that as $k rightarrow infty$ the largest root approaches $2$ (easy to see that some root approaches $2$, a bit harder is to prove that it is indeed largest).
I am absolutely clueless on how to figure out $c$.
answered Jan 17 at 0:23
user58697user58697
1,851612
$endgroup$
add a comment |
$begingroup$
First of all, the closed form of the sequence is
$$f(n, k) = sum_{i=0}^k c_ip_i^n$$
where $p_i$ are the roots of the equation
$$x^k = sum_{m=0}^{k - 1}x^m$$
and $c_i$ are constants to satisfy the initial conditions. It is a standard result which is easy to obtain looking at the generating function.
It is more or less obvious that asymptotic behavior is governed by the root with the largest absolute value. So your guess is correct. $f(n, k) sim phi(k)^n$ indeed, and $phi(k)$ is the largest root of the equation above.
The RHS of the equation is a geometric progression, therefore $x^k = dfrac{x^k - 1}{x-1}$ or $x^{k+1} - 2x^k + 1 = 0$. Now try to prove that as $k rightarrow infty$ the largest root approaches $2$ (easy to see that some root approaches $2$, a bit harder is to prove that it is indeed largest).
I am absolutely clueless on how to figure out $c$.
answered Jan 17 at 0:23
user58697user58697
1,851612
$endgroup$
add a comment |
$begingroup$
First of all, the closed form of the sequence is
$$f(n, k) = sum_{i=0}^k c_ip_i^n$$
where $p_i$ are the roots of the equation
$$x^k = sum_{m=0}^{k - 1}x^m$$
and $c_i$ are constants to satisfy the initial conditions. It is a standard result which is easy to obtain looking at the generating function.
It is more or less obvious that asymptotic behavior is governed by the root with the largest absolute value. So your guess is correct. $f(n, k) sim phi(k)^n$ indeed, and $phi(k)$ is the largest root of the equation above.
The RHS of the equation is a geometric progression, therefore $x^k = dfrac{x^k - 1}{x-1}$ or $x^{k+1} - 2x^k + 1 = 0$. Now try to prove that as $k rightarrow infty$ the largest root approaches $2$ (easy to see that some root approaches $2$, a bit harder is to prove that it is indeed largest).
I am absolutely clueless on how to figure out $c$.
answered Jan 17 at 0:23
user58697user58697
1,851612
$endgroup$
First of all, the closed form of the sequence is
$$f(n, k) = sum_{i=0}^k c_ip_i^n$$
where $p_i$ are the roots of the equation
$$x^k = sum_{m=0}^{k - 1}x^m$$
and $c_i$ are constants to satisfy the initial conditions. It is a standard result which is easy to obtain looking at the generating function.
It is more or less obvious that asymptotic behavior is governed by the root with the largest absolute value. So your guess is correct. $f(n, k) sim phi(k)^n$ indeed, and $phi(k)$ is the largest root of the equation above.
The RHS of the equation is a geometric progression, therefore $x^k = dfrac{x^k - 1}{x-1}$ or $x^{k+1} - 2x^k + 1 = 0$. Now try to prove that as $k rightarrow infty$ the largest root approaches $2$ (easy to see that some root approaches $2$, a bit harder is to prove that it is indeed largest).
I am absolutely clueless on how to figure out $c$.
answered Jan 17 at 0:23
user58697user58697
1,851612
answered Jan 17 at 0:23
user58697user58697
1,851612
answered Jan 17 at 0:23
user58697user58697
1,851612
answered Jan 17 at 0:23
user58697user58697
1,851612
1,851612
add a comment |
add a comment |
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