$alpha_n x_n = 1-p + p alpha_n x_{n-1}$
$begingroup$
I have the equation $$alpha_n x_n = 1-p + p alpha_n x_{n-1},$$ with $alpha_n, x_n,p in (0,1)$ for all $ngeq0$.
I would like to express $x_n$ as a function of $(alpha_n)$ and $p$ for every $n$. I know a formula when $(alpha_n)$ is constant but not in general.
real-analysis sequences-and-series
asked Jan 24 at 15:59
furyofuryo
313
$endgroup$
add a comment |
$begingroup$
I have the equation $$alpha_n x_n = 1-p + p alpha_n x_{n-1},$$ with $alpha_n, x_n,p in (0,1)$ for all $ngeq0$.
I would like to express $x_n$ as a function of $(alpha_n)$ and $p$ for every $n$. I know a formula when $(alpha_n)$ is constant but not in general.
real-analysis sequences-and-series
asked Jan 24 at 15:59
furyofuryo
313
$endgroup$
1
$begingroup$
$x_n = (1 - p) / alpha_n + p x_{n-1}$. Then you should be able to find the general solution via iteration.
$endgroup$
– user295959
Jan 24 at 16:03
add a comment |
$begingroup$
I have the equation $$alpha_n x_n = 1-p + p alpha_n x_{n-1},$$ with $alpha_n, x_n,p in (0,1)$ for all $ngeq0$.
I would like to express $x_n$ as a function of $(alpha_n)$ and $p$ for every $n$. I know a formula when $(alpha_n)$ is constant but not in general.
real-analysis sequences-and-series
asked Jan 24 at 15:59
furyofuryo
313
$endgroup$
I have the equation $$alpha_n x_n = 1-p + p alpha_n x_{n-1},$$ with $alpha_n, x_n,p in (0,1)$ for all $ngeq0$.
I would like to express $x_n$ as a function of $(alpha_n)$ and $p$ for every $n$. I know a formula when $(alpha_n)$ is constant but not in general.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 24 at 15:59
furyofuryo
313
asked Jan 24 at 15:59
furyofuryo
313
asked Jan 24 at 15:59
furyofuryo
313
asked Jan 24 at 15:59
furyofuryo
313
asked Jan 24 at 15:59
furyofuryo
313
313
1
$begingroup$
$x_n = (1 - p) / alpha_n + p x_{n-1}$. Then you should be able to find the general solution via iteration.
$endgroup$
– user295959
Jan 24 at 16:03
add a comment |
1
$begingroup$
$x_n = (1 - p) / alpha_n + p x_{n-1}$. Then you should be able to find the general solution via iteration.
$endgroup$
– user295959
Jan 24 at 16:03
1
1
$begingroup$
$x_n = (1 - p) / alpha_n + p x_{n-1}$. Then you should be able to find the general solution via iteration.
$endgroup$
– user295959
Jan 24 at 16:03
$begingroup$
$x_n = (1 - p) / alpha_n + p x_{n-1}$. Then you should be able to find the general solution via iteration.
$endgroup$
– user295959
Jan 24 at 16:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can observe that
$x_n=px_{n-1}+frac{1-p}{alpha_n}$
so you can deduce the following formula
$x_n=p^n x_0+(1-p)sum_{k=1}^nfrac{p^{n-k}}{alpha_k}$
The case $n=1$ is trivially true so you can hypothesize that it is true for $n$ and you can prove that it is true also for $n+1$.
$x_{n+1}= px_{n}+frac{1-p}{alpha_{n+1}}=$
$= p^{n+1} x_0+(1-p)sum_{k=1}^nfrac{p^{(n+1)-k}}{alpha_k} +frac{1-p}{alpha_{n+1}} =$
$= p^{n+1} x_0+(1-p)sum_{k=1}^{n+1}frac{p^{(n+1)-k}}{alpha_k} $
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You can observe that
$x_n=px_{n-1}+frac{1-p}{alpha_n}$
so you can deduce the following formula
$x_n=p^n x_0+(1-p)sum_{k=1}^nfrac{p^{n-k}}{alpha_k}$
The case $n=1$ is trivially true so you can hypothesize that it is true for $n$ and you can prove that it is true also for $n+1$.
$x_{n+1}= px_{n}+frac{1-p}{alpha_{n+1}}=$
$= p^{n+1} x_0+(1-p)sum_{k=1}^nfrac{p^{(n+1)-k}}{alpha_k} +frac{1-p}{alpha_{n+1}} =$
$= p^{n+1} x_0+(1-p)sum_{k=1}^{n+1}frac{p^{(n+1)-k}}{alpha_k} $
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
$endgroup$
add a comment |
$begingroup$
You can observe that
$x_n=px_{n-1}+frac{1-p}{alpha_n}$
so you can deduce the following formula
$x_n=p^n x_0+(1-p)sum_{k=1}^nfrac{p^{n-k}}{alpha_k}$
The case $n=1$ is trivially true so you can hypothesize that it is true for $n$ and you can prove that it is true also for $n+1$.
$x_{n+1}= px_{n}+frac{1-p}{alpha_{n+1}}=$
$= p^{n+1} x_0+(1-p)sum_{k=1}^nfrac{p^{(n+1)-k}}{alpha_k} +frac{1-p}{alpha_{n+1}} =$
$= p^{n+1} x_0+(1-p)sum_{k=1}^{n+1}frac{p^{(n+1)-k}}{alpha_k} $
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
$endgroup$
add a comment |
$begingroup$
You can observe that
$x_n=px_{n-1}+frac{1-p}{alpha_n}$
so you can deduce the following formula
$x_n=p^n x_0+(1-p)sum_{k=1}^nfrac{p^{n-k}}{alpha_k}$
The case $n=1$ is trivially true so you can hypothesize that it is true for $n$ and you can prove that it is true also for $n+1$.
$x_{n+1}= px_{n}+frac{1-p}{alpha_{n+1}}=$
$= p^{n+1} x_0+(1-p)sum_{k=1}^nfrac{p^{(n+1)-k}}{alpha_k} +frac{1-p}{alpha_{n+1}} =$
$= p^{n+1} x_0+(1-p)sum_{k=1}^{n+1}frac{p^{(n+1)-k}}{alpha_k} $
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
$endgroup$
You can observe that
$x_n=px_{n-1}+frac{1-p}{alpha_n}$
so you can deduce the following formula
$x_n=p^n x_0+(1-p)sum_{k=1}^nfrac{p^{n-k}}{alpha_k}$
The case $n=1$ is trivially true so you can hypothesize that it is true for $n$ and you can prove that it is true also for $n+1$.
$x_{n+1}= px_{n}+frac{1-p}{alpha_{n+1}}=$
$= p^{n+1} x_0+(1-p)sum_{k=1}^nfrac{p^{(n+1)-k}}{alpha_k} +frac{1-p}{alpha_{n+1}} =$
$= p^{n+1} x_0+(1-p)sum_{k=1}^{n+1}frac{p^{(n+1)-k}}{alpha_k} $
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
edited Jan 24 at 16:37
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
answered Jan 24 at 16:29
Federico FalluccaFederico Fallucca
2,235210
2,235210
add a comment |
add a comment |
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$begingroup$
$x_n = (1 - p) / alpha_n + p x_{n-1}$. Then you should be able to find the general solution via iteration.
$endgroup$
– user295959
Jan 24 at 16:03