Mixing
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
asked Jan 29 at 18:19
furyofuryo
313
$endgroup$
|
show 4 more comments
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
asked Jan 29 at 18:19
furyofuryo
313
$endgroup$
$begingroup$
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
$begingroup$
TheRSolve[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
|
show 4 more comments
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
asked Jan 29 at 18:19
furyofuryo
313
$endgroup$
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
asked Jan 29 at 18:19
furyofuryo
313
asked Jan 29 at 18:19
furyofuryo
313
edited Jan 30 at 10:32
furyo
asked Jan 29 at 18:19
furyofuryo
313
asked Jan 29 at 18:19
furyofuryo
313
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
$begingroup$
TheRSolve[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
|
show 4 more comments
$begingroup$
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
$begingroup$
TheRSolve[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
|
show 4 more comments
1 Answer
1
active
oldest
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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...)
answered Jan 29 at 18:39
DidDid
248k23226466
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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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...)
answered Jan 29 at 18:39
DidDid
248k23226466
$endgroup$
$begingroup$
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
add a comment |
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...)
answered Jan 29 at 18:39
DidDid
248k23226466
$endgroup$
$begingroup$
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
add a comment |
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...)
answered Jan 29 at 18:39
DidDid
248k23226466
$endgroup$
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...)
answered Jan 29 at 18:39
DidDid
248k23226466
answered Jan 29 at 18:39
DidDid
248k23226466
answered Jan 29 at 18:39
DidDid
248k23226466
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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