There is a straight solution for difference equation $f(x+1)=beta(x)f(x)$ such that $beta(1)=a$ and $x>1$?












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

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0












$begingroup$

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 $[n,n+1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $beta(x)$ 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












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $[n,n+1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $beta(x)$ 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
















0












$begingroup$

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 $[n,n+1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $beta(x)$ 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














0












0








0





$begingroup$

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 $[n,n+1)$






share|cite|improve this answer









$endgroup$



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 $[n,n+1)$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    $beta(x)$ 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


















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