There is a straight solution for difference equation $f(x+1)=beta(x)f(x)$ such that $beta(1)=a$ and $x>1$?
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Recently, I stock in find a general or special solution of a difference equation $f(x+1)=beta(x)f(x)$ such that $beta(1)=a$ and $x>1$. I do not know much about this group of equations so any name, link, paper,... can helpful to me. I want to know more about this equations.
functions numerical-methods recurrence-relations
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
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add a comment |
$begingroup$
Recently, I stock in find a general or special solution of a difference equation $f(x+1)=beta(x)f(x)$ such that $beta(1)=a$ and $x>1$. I do not know much about this group of equations so any name, link, paper,... can helpful to me. I want to know more about this equations.
functions numerical-methods recurrence-relations
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
$endgroup$
$begingroup$
Do you assume anything about the solutions? Because as written, it is easy to construct solution by defining it piecewise on intervals $(n,n+1]$.
$endgroup$
– Wojowu
Jan 15 at 10:24
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You have $f(n)=a f(1)prodlimits_{i=2}^{n-1} beta(i)$
$endgroup$
– Henry
Jan 15 at 10:27
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I need a continuous function solution, so can I interpolate this?
$endgroup$
– Jamal Farokhi
Jan 15 at 10:38
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There are infinitely many such functions. Simply consider $f(x)=cos(2pi x)g(x)$ for example.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:53
add a comment |
$begingroup$
Recently, I stock in find a general or special solution of a difference equation $f(x+1)=beta(x)f(x)$ such that $beta(1)=a$ and $x>1$. I do not know much about this group of equations so any name, link, paper,... can helpful to me. I want to know more about this equations.
functions numerical-methods recurrence-relations
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
$endgroup$
Recently, I stock in find a general or special solution of a difference equation $f(x+1)=beta(x)f(x)$ such that $beta(1)=a$ and $x>1$. I do not know much about this group of equations so any name, link, paper,... can helpful to me. I want to know more about this equations.
functions numerical-methods recurrence-relations
functions numerical-methods recurrence-relations
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
asked Jan 15 at 10:21
Jamal FarokhiJamal Farokhi
454210
454210
$begingroup$
Do you assume anything about the solutions? Because as written, it is easy to construct solution by defining it piecewise on intervals $(n,n+1]$.
$endgroup$
– Wojowu
Jan 15 at 10:24
$begingroup$
You have $f(n)=a f(1)prodlimits_{i=2}^{n-1} beta(i)$
$endgroup$
– Henry
Jan 15 at 10:27
$begingroup$
I need a continuous function solution, so can I interpolate this?
$endgroup$
– Jamal Farokhi
Jan 15 at 10:38
$begingroup$
There are infinitely many such functions. Simply consider $f(x)=cos(2pi x)g(x)$ for example.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:53
add a comment |
$begingroup$
Do you assume anything about the solutions? Because as written, it is easy to construct solution by defining it piecewise on intervals $(n,n+1]$.
$endgroup$
– Wojowu
Jan 15 at 10:24
$begingroup$
You have $f(n)=a f(1)prodlimits_{i=2}^{n-1} beta(i)$
$endgroup$
– Henry
Jan 15 at 10:27
$begingroup$
I need a continuous function solution, so can I interpolate this?
$endgroup$
– Jamal Farokhi
Jan 15 at 10:38
$begingroup$
There are infinitely many such functions. Simply consider $f(x)=cos(2pi x)g(x)$ for example.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:53
$begingroup$
Do you assume anything about the solutions? Because as written, it is easy to construct solution by defining it piecewise on intervals $(n,n+1]$.
$endgroup$
– Wojowu
Jan 15 at 10:24
$begingroup$
Do you assume anything about the solutions? Because as written, it is easy to construct solution by defining it piecewise on intervals $(n,n+1]$.
$endgroup$
– Wojowu
Jan 15 at 10:24
$begingroup$
You have $f(n)=a f(1)prodlimits_{i=2}^{n-1} beta(i)$
$endgroup$
– Henry
Jan 15 at 10:27
$begingroup$
You have $f(n)=a f(1)prodlimits_{i=2}^{n-1} beta(i)$
$endgroup$
– Henry
Jan 15 at 10:27
$begingroup$
I need a continuous function solution, so can I interpolate this?
$endgroup$
– Jamal Farokhi
Jan 15 at 10:38
$begingroup$
I need a continuous function solution, so can I interpolate this?
$endgroup$
– Jamal Farokhi
Jan 15 at 10:38
$begingroup$
There are infinitely many such functions. Simply consider $f(x)=cos(2pi x)g(x)$ for example.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:53
$begingroup$
There are infinitely many such functions. Simply consider $f(x)=cos(2pi x)g(x)$ for example.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:53
add a comment |
1 Answer
1
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oldest
votes
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If you know nothing about $beta$, then $f(n)$ can be just about any sequence you want, so long as $f(2)=acdot f(1)$.
More strictly, the following is true:
Let $f_n$ be a piecewise constant nonzero function on all intervals $[n,n+1)$, that is, let $a_n$ be a sequence of real numbers, and let $f(x)$ equal $a_n$ on $[n,n+1)$. Let $a_2=acdot a_1$. Then, there exists a function $beta$ such that $f(x+1)=beta(x)cdot f(x)$ for all $x$.
You can easily see this is true by simply defining $beta(x)$ to equal $frac{a_{n+1}}{a_n}$ on $[n,n+1)$
answered Jan 15 at 10:28
5xum5xum
91.8k394161
$endgroup$
$begingroup$
$beta(x)$ can be any function and in a special case, for instance, if $beta(x)=s$, so $f(x)=Gamma(x)$. I want to find welldefind solutions.
$endgroup$
– Jamal Farokhi
Jan 15 at 10:40
$begingroup$
@JamalFarokhi If $beta$ can be any function, then $f$ can also be any function.
$endgroup$
– 5xum
Jan 15 at 10:48
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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$begingroup$
If you know nothing about $beta$, then $f(n)$ can be just about any sequence you want, so long as $f(2)=acdot f(1)$.
More strictly, the following is true:
Let $f_n$ be a piecewise constant nonzero function on all intervals $[n,n+1)$, that is, let $a_n$ be a sequence of real numbers, and let $f(x)$ equal $a_n$ on $[n,n+1)$. Let $a_2=acdot a_1$. Then, there exists a function $beta$ such that $f(x+1)=beta(x)cdot f(x)$ for all $x$.
You can easily see this is true by simply defining $beta(x)$ to equal $frac{a_{n+1}}{a_n}$ on $[n,n+1)$
answered Jan 15 at 10:28
5xum5xum
91.8k394161
$endgroup$
$begingroup$
$beta(x)$ can be any function and in a special case, for instance, if $beta(x)=s$, so $f(x)=Gamma(x)$. I want to find welldefind solutions.
$endgroup$
– Jamal Farokhi
Jan 15 at 10:40
$begingroup$
@JamalFarokhi If $beta$ can be any function, then $f$ can also be any function.
$endgroup$
– 5xum
Jan 15 at 10:48
add a comment |
$begingroup$
If you know nothing about $beta$, then $f(n)$ can be just about any sequence you want, so long as $f(2)=acdot f(1)$.
More strictly, the following is true:
Let $f_n$ be a piecewise constant nonzero function on all intervals $[n,n+1)$, that is, let $a_n$ be a sequence of real numbers, and let $f(x)$ equal $a_n$ on $[n,n+1)$. Let $a_2=acdot a_1$. Then, there exists a function $beta$ such that $f(x+1)=beta(x)cdot f(x)$ for all $x$.
You can easily see this is true by simply defining $beta(x)$ to equal $frac{a_{n+1}}{a_n}$ on $[n,n+1)$
answered Jan 15 at 10:28
5xum5xum
91.8k394161
$endgroup$
$begingroup$
$beta(x)$ can be any function and in a special case, for instance, if $beta(x)=s$, so $f(x)=Gamma(x)$. I want to find welldefind solutions.
$endgroup$
– Jamal Farokhi
Jan 15 at 10:40
$begingroup$
@JamalFarokhi If $beta$ can be any function, then $f$ can also be any function.
$endgroup$
– 5xum
Jan 15 at 10:48
add a comment |
$begingroup$
If you know nothing about $beta$, then $f(n)$ can be just about any sequence you want, so long as $f(2)=acdot f(1)$.
More strictly, the following is true:
Let $f_n$ be a piecewise constant nonzero function on all intervals $[n,n+1)$, that is, let $a_n$ be a sequence of real numbers, and let $f(x)$ equal $a_n$ on $[n,n+1)$. Let $a_2=acdot a_1$. Then, there exists a function $beta$ such that $f(x+1)=beta(x)cdot f(x)$ for all $x$.
You can easily see this is true by simply defining $beta(x)$ to equal $frac{a_{n+1}}{a_n}$ on $[n,n+1)$
answered Jan 15 at 10:28
5xum5xum
91.8k394161
$endgroup$
If you know nothing about $beta$, then $f(n)$ can be just about any sequence you want, so long as $f(2)=acdot f(1)$.
More strictly, the following is true:
Let $f_n$ be a piecewise constant nonzero function on all intervals $[n,n+1)$, that is, let $a_n$ be a sequence of real numbers, and let $f(x)$ equal $a_n$ on $[n,n+1)$. Let $a_2=acdot a_1$. Then, there exists a function $beta$ such that $f(x+1)=beta(x)cdot f(x)$ for all $x$.
You can easily see this is true by simply defining $beta(x)$ to equal $frac{a_{n+1}}{a_n}$ on $[n,n+1)$
answered Jan 15 at 10:28
5xum5xum
91.8k394161
answered Jan 15 at 10:28
5xum5xum
91.8k394161
answered Jan 15 at 10:28
5xum5xum
91.8k394161
answered Jan 15 at 10:28
5xum5xum
91.8k394161
91.8k394161
$begingroup$
$beta(x)$ can be any function and in a special case, for instance, if $beta(x)=s$, so $f(x)=Gamma(x)$. I want to find welldefind solutions.
$endgroup$
– Jamal Farokhi
Jan 15 at 10:40
$begingroup$
@JamalFarokhi If $beta$ can be any function, then $f$ can also be any function.
$endgroup$
– 5xum
Jan 15 at 10:48
add a comment |
$begingroup$
$beta(x)$ can be any function and in a special case, for instance, if $beta(x)=s$, so $f(x)=Gamma(x)$. I want to find welldefind solutions.
$endgroup$
– Jamal Farokhi
Jan 15 at 10:40
$begingroup$
@JamalFarokhi If $beta$ can be any function, then $f$ can also be any function.
$endgroup$
– 5xum
Jan 15 at 10:48
$begingroup$
$beta(x)$ can be any function and in a special case, for instance, if $beta(x)=s$, so $f(x)=Gamma(x)$. I want to find welldefind solutions.
$endgroup$
– Jamal Farokhi
Jan 15 at 10:40
$begingroup$
$beta(x)$ can be any function and in a special case, for instance, if $beta(x)=s$, so $f(x)=Gamma(x)$. I want to find welldefind solutions.
$endgroup$
– Jamal Farokhi
Jan 15 at 10:40
$begingroup$
@JamalFarokhi If $beta$ can be any function, then $f$ can also be any function.
$endgroup$
– 5xum
Jan 15 at 10:48
$begingroup$
@JamalFarokhi If $beta$ can be any function, then $f$ can also be any function.
$endgroup$
– 5xum
Jan 15 at 10:48
add a comment |
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$begingroup$
Do you assume anything about the solutions? Because as written, it is easy to construct solution by defining it piecewise on intervals $(n,n+1]$.
$endgroup$
– Wojowu
Jan 15 at 10:24
$begingroup$
You have $f(n)=a f(1)prodlimits_{i=2}^{n-1} beta(i)$
$endgroup$
– Henry
Jan 15 at 10:27
$begingroup$
I need a continuous function solution, so can I interpolate this?
$endgroup$
– Jamal Farokhi
Jan 15 at 10:38
$begingroup$
There are infinitely many such functions. Simply consider $f(x)=cos(2pi x)g(x)$ for example.
$endgroup$
– Simply Beautiful Art
Jan 15 at 14:53