Mixing
Discrete mathematics solving recurrence relation with param
0
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I was given this equation to solve and I'm without any idea how to work the parameter which is given as p in the equation and therefore I can't figure out what the correct solution is. Any help would be very much appreciated.
$$a_{n+2}=pa_{n+1}+(1-p)a_n$$
where $pin(0,1)setminus{1/2}$ is a given parameter.
discrete-mathematics recurrence-relations
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
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add a comment |
0
$begingroup$
I was given this equation to solve and I'm without any idea how to work the parameter which is given as p in the equation and therefore I can't figure out what the correct solution is. Any help would be very much appreciated.
$$a_{n+2}=pa_{n+1}+(1-p)a_n$$
where $pin(0,1)setminus{1/2}$ is a given parameter.
discrete-mathematics recurrence-relations
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
$endgroup$
$begingroup$
If $p$ were a given number, say $0.75$, what would you do to solve the equation then? Can you then see how you might generalise to an arbitrary $p$?
$endgroup$
– postmortes
Jan 16 at 10:24
$begingroup$
@postmortes as the p is there it's very much confusing for me as I've never came across similar equations. I really have 0 ideas on how to solve this.
$endgroup$
– lostandconfused
Jan 16 at 10:52
$begingroup$
So I probably figured out the general solution as $$a_n=C_1 + C_2 (p - 1)^n$$
$endgroup$
– lostandconfused
Jan 16 at 11:25
$begingroup$
You should write your own answer to your question then, so people can see how you got there :)
$endgroup$
– postmortes
Jan 16 at 11:26
add a comment |
0
0
0
$begingroup$
I was given this equation to solve and I'm without any idea how to work the parameter which is given as p in the equation and therefore I can't figure out what the correct solution is. Any help would be very much appreciated.
$$a_{n+2}=pa_{n+1}+(1-p)a_n$$
where $pin(0,1)setminus{1/2}$ is a given parameter.
discrete-mathematics recurrence-relations
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
$endgroup$
I was given this equation to solve and I'm without any idea how to work the parameter which is given as p in the equation and therefore I can't figure out what the correct solution is. Any help would be very much appreciated.
$$a_{n+2}=pa_{n+1}+(1-p)a_n$$
where $pin(0,1)setminus{1/2}$ is a given parameter.
discrete-mathematics recurrence-relations
discrete-mathematics recurrence-relations
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
asked Jan 16 at 10:18
lostandconfusedlostandconfused
1
1
$begingroup$
If $p$ were a given number, say $0.75$, what would you do to solve the equation then? Can you then see how you might generalise to an arbitrary $p$?
$endgroup$
– postmortes
Jan 16 at 10:24
$begingroup$
@postmortes as the p is there it's very much confusing for me as I've never came across similar equations. I really have 0 ideas on how to solve this.
$endgroup$
– lostandconfused
Jan 16 at 10:52
$begingroup$
So I probably figured out the general solution as $$a_n=C_1 + C_2 (p - 1)^n$$
$endgroup$
– lostandconfused
Jan 16 at 11:25
$begingroup$
You should write your own answer to your question then, so people can see how you got there :)
$endgroup$
– postmortes
Jan 16 at 11:26
add a comment |
$begingroup$
If $p$ were a given number, say $0.75$, what would you do to solve the equation then? Can you then see how you might generalise to an arbitrary $p$?
$endgroup$
– postmortes
Jan 16 at 10:24
$begingroup$
@postmortes as the p is there it's very much confusing for me as I've never came across similar equations. I really have 0 ideas on how to solve this.
$endgroup$
– lostandconfused
Jan 16 at 10:52
$begingroup$
So I probably figured out the general solution as $$a_n=C_1 + C_2 (p - 1)^n$$
$endgroup$
– lostandconfused
Jan 16 at 11:25
$begingroup$
You should write your own answer to your question then, so people can see how you got there :)
$endgroup$
– postmortes
Jan 16 at 11:26
$begingroup$
If $p$ were a given number, say $0.75$, what would you do to solve the equation then? Can you then see how you might generalise to an arbitrary $p$?
$endgroup$
– postmortes
Jan 16 at 10:24
$begingroup$
If $p$ were a given number, say $0.75$, what would you do to solve the equation then? Can you then see how you might generalise to an arbitrary $p$?
$endgroup$
– postmortes
Jan 16 at 10:24
$begingroup$
@postmortes as the p is there it's very much confusing for me as I've never came across similar equations. I really have 0 ideas on how to solve this.
$endgroup$
– lostandconfused
Jan 16 at 10:52
$begingroup$
@postmortes as the p is there it's very much confusing for me as I've never came across similar equations. I really have 0 ideas on how to solve this.
$endgroup$
– lostandconfused
Jan 16 at 10:52
$begingroup$
So I probably figured out the general solution as $$a_n=C_1 + C_2 (p - 1)^n$$
$endgroup$
– lostandconfused
Jan 16 at 11:25
$begingroup$
So I probably figured out the general solution as $$a_n=C_1 + C_2 (p - 1)^n$$
$endgroup$
– lostandconfused
Jan 16 at 11:25
$begingroup$
You should write your own answer to your question then, so people can see how you got there :)
$endgroup$
– postmortes
Jan 16 at 11:26
$begingroup$
You should write your own answer to your question then, so people can see how you got there :)
$endgroup$
– postmortes
Jan 16 at 11:26
add a comment |
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$begingroup$
If $p$ were a given number, say $0.75$, what would you do to solve the equation then? Can you then see how you might generalise to an arbitrary $p$?
$endgroup$
– postmortes
Jan 16 at 10:24
$begingroup$
@postmortes as the p is there it's very much confusing for me as I've never came across similar equations. I really have 0 ideas on how to solve this.
$endgroup$
– lostandconfused
Jan 16 at 10:52
$begingroup$
So I probably figured out the general solution as $$a_n=C_1 + C_2 (p - 1)^n$$
$endgroup$
– lostandconfused
Jan 16 at 11:25
$begingroup$
You should write your own answer to your question then, so people can see how you got there :)
$endgroup$
– postmortes
Jan 16 at 11:26