How to find $x_n$ from $x_{n+1} = frac{x_n}{1-a+a x_n}$?












3












$begingroup$


For $ngeq 0$ let $x_{n+1} = frac{x_n}{1-a+a x_n}$, where $ain (0,1)$.



I would like to know if it is possible to express $x_n$ as a function of $a$ and $x_0$ for all $ngeq0$.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Add more details...what you tried?
    $endgroup$
    – Cloud JR
    Jan 29 at 18:23










  • $begingroup$
    What is p0? Explain
    $endgroup$
    – Cloud JR
    Jan 29 at 18:24










  • $begingroup$
    The RSolve[x[n+1]==x[n]/(1-a+a x[n]),x[n],n] command in Mathematica yields $$x_n=-frac{a-1}{1-(1-a)^n-a+a(1-a)^n+C(1-a)^n},$$ where $C$ is a constant.
    $endgroup$
    – Adrian Keister
    Jan 29 at 18:25






  • 1




    $begingroup$
    Welcome to MES! I changed your title so that it better reflects your question. It is always better to provide a more self-contained one. Also, it is even better (and is highly recommend) to provide your own ideas, attempts or contexts. Finally, have you ever heard of fractional linear transformation?
    $endgroup$
    – Sangchul Lee
    Jan 29 at 18:33








  • 1




    $begingroup$
    @AdrianKeister Not sure a naked formula helps much the OP's mathematical understanding... Of course, one could reverse engineer the formula to understand where it comes from, but this requires more mathematical maturity than what we can guess the OP possesses.
    $endgroup$
    – Did
    Jan 29 at 18:45


















3












$begingroup$


For $ngeq 0$ let $x_{n+1} = frac{x_n}{1-a+a x_n}$, where $ain (0,1)$.



I would like to know if it is possible to express $x_n$ as a function of $a$ and $x_0$ for all $ngeq0$.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Add more details...what you tried?
    $endgroup$
    – Cloud JR
    Jan 29 at 18:23










  • $begingroup$
    What is p0? Explain
    $endgroup$
    – Cloud JR
    Jan 29 at 18:24










  • $begingroup$
    The RSolve[x[n+1]==x[n]/(1-a+a x[n]),x[n],n] command in Mathematica yields $$x_n=-frac{a-1}{1-(1-a)^n-a+a(1-a)^n+C(1-a)^n},$$ where $C$ is a constant.
    $endgroup$
    – Adrian Keister
    Jan 29 at 18:25






  • 1




    $begingroup$
    Welcome to MES! I changed your title so that it better reflects your question. It is always better to provide a more self-contained one. Also, it is even better (and is highly recommend) to provide your own ideas, attempts or contexts. Finally, have you ever heard of fractional linear transformation?
    $endgroup$
    – Sangchul Lee
    Jan 29 at 18:33








  • 1




    $begingroup$
    @AdrianKeister Not sure a naked formula helps much the OP's mathematical understanding... Of course, one could reverse engineer the formula to understand where it comes from, but this requires more mathematical maturity than what we can guess the OP possesses.
    $endgroup$
    – Did
    Jan 29 at 18:45
















3












3








3





$begingroup$


For $ngeq 0$ let $x_{n+1} = frac{x_n}{1-a+a x_n}$, where $ain (0,1)$.



I would like to know if it is possible to express $x_n$ as a function of $a$ and $x_0$ for all $ngeq0$.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




For $ngeq 0$ let $x_{n+1} = frac{x_n}{1-a+a x_n}$, where $ain (0,1)$.



I would like to know if it is possible to express $x_n$ as a function of $a$ and $x_0$ for all $ngeq0$.






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 10:32







furyo

















asked Jan 29 at 18:19









furyofuryo

313




313












  • $begingroup$
    Add more details...what you tried?
    $endgroup$
    – Cloud JR
    Jan 29 at 18:23










  • $begingroup$
    What is p0? Explain
    $endgroup$
    – Cloud JR
    Jan 29 at 18:24










  • $begingroup$
    The RSolve[x[n+1]==x[n]/(1-a+a x[n]),x[n],n] command in Mathematica yields $$x_n=-frac{a-1}{1-(1-a)^n-a+a(1-a)^n+C(1-a)^n},$$ where $C$ is a constant.
    $endgroup$
    – Adrian Keister
    Jan 29 at 18:25






  • 1




    $begingroup$
    Welcome to MES! I changed your title so that it better reflects your question. It is always better to provide a more self-contained one. Also, it is even better (and is highly recommend) to provide your own ideas, attempts or contexts. Finally, have you ever heard of fractional linear transformation?
    $endgroup$
    – Sangchul Lee
    Jan 29 at 18:33








  • 1




    $begingroup$
    @AdrianKeister Not sure a naked formula helps much the OP's mathematical understanding... Of course, one could reverse engineer the formula to understand where it comes from, but this requires more mathematical maturity than what we can guess the OP possesses.
    $endgroup$
    – Did
    Jan 29 at 18:45




















  • $begingroup$
    Add more details...what you tried?
    $endgroup$
    – Cloud JR
    Jan 29 at 18:23










  • $begingroup$
    What is p0? Explain
    $endgroup$
    – Cloud JR
    Jan 29 at 18:24










  • $begingroup$
    The RSolve[x[n+1]==x[n]/(1-a+a x[n]),x[n],n] command in Mathematica yields $$x_n=-frac{a-1}{1-(1-a)^n-a+a(1-a)^n+C(1-a)^n},$$ where $C$ is a constant.
    $endgroup$
    – Adrian Keister
    Jan 29 at 18:25






  • 1




    $begingroup$
    Welcome to MES! I changed your title so that it better reflects your question. It is always better to provide a more self-contained one. Also, it is even better (and is highly recommend) to provide your own ideas, attempts or contexts. Finally, have you ever heard of fractional linear transformation?
    $endgroup$
    – Sangchul Lee
    Jan 29 at 18:33








  • 1




    $begingroup$
    @AdrianKeister Not sure a naked formula helps much the OP's mathematical understanding... Of course, one could reverse engineer the formula to understand where it comes from, but this requires more mathematical maturity than what we can guess the OP possesses.
    $endgroup$
    – Did
    Jan 29 at 18:45


















$begingroup$
Add more details...what you tried?
$endgroup$
– Cloud JR
Jan 29 at 18:23




$begingroup$
Add more details...what you tried?
$endgroup$
– Cloud JR
Jan 29 at 18:23












$begingroup$
What is p0? Explain
$endgroup$
– Cloud JR
Jan 29 at 18:24




$begingroup$
What is p0? Explain
$endgroup$
– Cloud JR
Jan 29 at 18:24












$begingroup$
The RSolve[x[n+1]==x[n]/(1-a+a x[n]),x[n],n] command in Mathematica yields $$x_n=-frac{a-1}{1-(1-a)^n-a+a(1-a)^n+C(1-a)^n},$$ where $C$ is a constant.
$endgroup$
– Adrian Keister
Jan 29 at 18:25




$begingroup$
The RSolve[x[n+1]==x[n]/(1-a+a x[n]),x[n],n] command in Mathematica yields $$x_n=-frac{a-1}{1-(1-a)^n-a+a(1-a)^n+C(1-a)^n},$$ where $C$ is a constant.
$endgroup$
– Adrian Keister
Jan 29 at 18:25




1




1




$begingroup$
Welcome to MES! I changed your title so that it better reflects your question. It is always better to provide a more self-contained one. Also, it is even better (and is highly recommend) to provide your own ideas, attempts or contexts. Finally, have you ever heard of fractional linear transformation?
$endgroup$
– Sangchul Lee
Jan 29 at 18:33






$begingroup$
Welcome to MES! I changed your title so that it better reflects your question. It is always better to provide a more self-contained one. Also, it is even better (and is highly recommend) to provide your own ideas, attempts or contexts. Finally, have you ever heard of fractional linear transformation?
$endgroup$
– Sangchul Lee
Jan 29 at 18:33






1




1




$begingroup$
@AdrianKeister Not sure a naked formula helps much the OP's mathematical understanding... Of course, one could reverse engineer the formula to understand where it comes from, but this requires more mathematical maturity than what we can guess the OP possesses.
$endgroup$
– Did
Jan 29 at 18:45






$begingroup$
@AdrianKeister Not sure a naked formula helps much the OP's mathematical understanding... Of course, one could reverse engineer the formula to understand where it comes from, but this requires more mathematical maturity than what we can guess the OP possesses.
$endgroup$
– Did
Jan 29 at 18:45












1 Answer
1






active

oldest

votes


















6












$begingroup$

Note that $$frac1{x_{n+1}}=frac{1-a}{x_n}+a$$ which implies that $$frac1{x_{n+1}}-1=(1-a)left(frac1{x_n}-1right)$$ hence $$frac1{x_n}-1=(1-a)^nleft(frac1{x_0}-1right)$$ from which an explicit formula for $x_n$ in terms of $n$, $a$ and $x_0$ follows.



(Of course, there is some well known theory behind all this...)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, could you please tell me the name of this famous theory?
    $endgroup$
    – furyo
    Jan 30 at 10:39










  • $begingroup$
    The study of homographies.
    $endgroup$
    – Did
    Jan 30 at 10:43














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Note that $$frac1{x_{n+1}}=frac{1-a}{x_n}+a$$ which implies that $$frac1{x_{n+1}}-1=(1-a)left(frac1{x_n}-1right)$$ hence $$frac1{x_n}-1=(1-a)^nleft(frac1{x_0}-1right)$$ from which an explicit formula for $x_n$ in terms of $n$, $a$ and $x_0$ follows.



(Of course, there is some well known theory behind all this...)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, could you please tell me the name of this famous theory?
    $endgroup$
    – furyo
    Jan 30 at 10:39










  • $begingroup$
    The study of homographies.
    $endgroup$
    – Did
    Jan 30 at 10:43


















6












$begingroup$

Note that $$frac1{x_{n+1}}=frac{1-a}{x_n}+a$$ which implies that $$frac1{x_{n+1}}-1=(1-a)left(frac1{x_n}-1right)$$ hence $$frac1{x_n}-1=(1-a)^nleft(frac1{x_0}-1right)$$ from which an explicit formula for $x_n$ in terms of $n$, $a$ and $x_0$ follows.



(Of course, there is some well known theory behind all this...)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer, could you please tell me the name of this famous theory?
    $endgroup$
    – furyo
    Jan 30 at 10:39










  • $begingroup$
    The study of homographies.
    $endgroup$
    – Did
    Jan 30 at 10:43
















6












6








6





$begingroup$

Note that $$frac1{x_{n+1}}=frac{1-a}{x_n}+a$$ which implies that $$frac1{x_{n+1}}-1=(1-a)left(frac1{x_n}-1right)$$ hence $$frac1{x_n}-1=(1-a)^nleft(frac1{x_0}-1right)$$ from which an explicit formula for $x_n$ in terms of $n$, $a$ and $x_0$ follows.



(Of course, there is some well known theory behind all this...)






share|cite|improve this answer









$endgroup$



Note that $$frac1{x_{n+1}}=frac{1-a}{x_n}+a$$ which implies that $$frac1{x_{n+1}}-1=(1-a)left(frac1{x_n}-1right)$$ hence $$frac1{x_n}-1=(1-a)^nleft(frac1{x_0}-1right)$$ from which an explicit formula for $x_n$ in terms of $n$, $a$ and $x_0$ follows.



(Of course, there is some well known theory behind all this...)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 29 at 18:39









DidDid

248k23226466




248k23226466












  • $begingroup$
    Thank you for your answer, could you please tell me the name of this famous theory?
    $endgroup$
    – furyo
    Jan 30 at 10:39










  • $begingroup$
    The study of homographies.
    $endgroup$
    – Did
    Jan 30 at 10:43




















  • $begingroup$
    Thank you for your answer, could you please tell me the name of this famous theory?
    $endgroup$
    – furyo
    Jan 30 at 10:39










  • $begingroup$
    The study of homographies.
    $endgroup$
    – Did
    Jan 30 at 10:43


















$begingroup$
Thank you for your answer, could you please tell me the name of this famous theory?
$endgroup$
– furyo
Jan 30 at 10:39




$begingroup$
Thank you for your answer, could you please tell me the name of this famous theory?
$endgroup$
– furyo
Jan 30 at 10:39












$begingroup$
The study of homographies.
$endgroup$
– Did
Jan 30 at 10:43






$begingroup$
The study of homographies.
$endgroup$
– Did
Jan 30 at 10:43




















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