Mixing
Simplify recurrence relation made of 3 series
$begingroup$
Whats the best way to approach simplifying these three dependent relations:
$a_n = b_n + 2c_n$
$b_n = 2c_{n-1} $
$c_n = 2c_{n-1} + 2b_{n-1}$
(where $a_1=5, a_2=16$)
I tried to plug $b_n$ to $c_n$, and then simplifying $a_n$, but got stuck in the process. I know that the solution should be: $a_n = 2a_{n-1} + 4a_{n-2}$
recurrence-relations
asked Feb 2 at 20:08
BBLNBBLN
1194
$endgroup$
add a comment |
$begingroup$
Whats the best way to approach simplifying these three dependent relations:
$a_n = b_n + 2c_n$
$b_n = 2c_{n-1} $
$c_n = 2c_{n-1} + 2b_{n-1}$
(where $a_1=5, a_2=16$)
I tried to plug $b_n$ to $c_n$, and then simplifying $a_n$, but got stuck in the process. I know that the solution should be: $a_n = 2a_{n-1} + 4a_{n-2}$
recurrence-relations
asked Feb 2 at 20:08
BBLNBBLN
1194
$endgroup$
add a comment |
$begingroup$
Whats the best way to approach simplifying these three dependent relations:
$a_n = b_n + 2c_n$
$b_n = 2c_{n-1} $
$c_n = 2c_{n-1} + 2b_{n-1}$
(where $a_1=5, a_2=16$)
I tried to plug $b_n$ to $c_n$, and then simplifying $a_n$, but got stuck in the process. I know that the solution should be: $a_n = 2a_{n-1} + 4a_{n-2}$
recurrence-relations
asked Feb 2 at 20:08
BBLNBBLN
1194
$endgroup$
Whats the best way to approach simplifying these three dependent relations:
$a_n = b_n + 2c_n$
$b_n = 2c_{n-1} $
$c_n = 2c_{n-1} + 2b_{n-1}$
(where $a_1=5, a_2=16$)
I tried to plug $b_n$ to $c_n$, and then simplifying $a_n$, but got stuck in the process. I know that the solution should be: $a_n = 2a_{n-1} + 4a_{n-2}$
recurrence-relations
recurrence-relations
asked Feb 2 at 20:08
BBLNBBLN
1194
asked Feb 2 at 20:08
BBLNBBLN
1194
asked Feb 2 at 20:08
BBLNBBLN
1194
asked Feb 2 at 20:08
BBLNBBLN
1194
asked Feb 2 at 20:08
BBLNBBLN
1194
1194
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
if you take
$$Y_n =
left(
begin{array}{c}
b_n \
c_n
end{array}
right)
$$
then
$$ Y_n = M Y_{n-1}, $$
where
$$M =
left(
begin{array}{cc}
0&2 \
2&2
end{array}
right)
$$
I prefer writing
$$ Y_{n+1} = M Y_{n}. $$
From the trace $2$ and determinant $-4,$ we find $M^2 - 2M-4I = 0,$ or
$M^2 = 2M + 4I.$ This is Cayley-Hamilton. We point out that
$$ Y_{n+2} = M^2 Y_{n}. $$ So
$$ Y_{n+2} = M^2 Y_{n} = (2M+4I)Y_n = 2MY_n + 4 Y_n = 2 Y_{n+1} + 4 Y_n. $$
Or
$$ Y_{n+2} = 2 Y_{n+1} + 4 Y_n. $$
Therefore
$$ b_{n+2} = 2 b_{n+1} + 4 b_n $$ and
$$ c_{n+2} = 2 c_{n+1} + 4 c_n. $$
What does this say about $a_n ; ?$
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
This is introduction class so I'm afraid I cant understand this solution yet :)
$endgroup$
– BBLN
Feb 2 at 20:40
add a comment |
$begingroup$
The first equation gives
begin{align}
a_n & = b_n+2c_n\
-2a_{n-1} & = -2b_{n-1}-4c_{n-1}\
-4a_{n-2} & = -4b_{n-2}-8c_{n-2}.
end{align}
Now the last two equations give $c_n=2c_{n-1}+4c_{n-2}$, by replacing $b_{n-1}$ in the last equation by using the second equation. Similarly we obtain $b_n=2b_{n-1}+4b_{n-2}$. Now summing up the three equations give the claim.
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
$begingroup$
I see that this solution is right, but how would you start without knowing/guessing the simplified form? Whats the best way to approach this without knowing how many elements there would be in the final form
$endgroup$
– BBLN
Feb 2 at 20:43
$begingroup$
The best way would be to start with two equations only, say the last two ones: We have $c_n=2c_{n-1}+2b_{n-1}$. Oh, let's get rid of the $2b_{n-1}$. By the second equation we can replace it by $4c_{n-2}$. We get $c_n=2c_{n-1}+4c_{n-2}$. This is already the "simplified" form. Now we just need to see how this also works for the $b$ and $a$ sequence.
$endgroup$
– Dietrich Burde
Feb 2 at 20:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
$begingroup$
if you take
$$Y_n =
left(
begin{array}{c}
b_n \
c_n
end{array}
right)
$$
then
$$ Y_n = M Y_{n-1}, $$
where
$$M =
left(
begin{array}{cc}
0&2 \
2&2
end{array}
right)
$$
I prefer writing
$$ Y_{n+1} = M Y_{n}. $$
From the trace $2$ and determinant $-4,$ we find $M^2 - 2M-4I = 0,$ or
$M^2 = 2M + 4I.$ This is Cayley-Hamilton. We point out that
$$ Y_{n+2} = M^2 Y_{n}. $$ So
$$ Y_{n+2} = M^2 Y_{n} = (2M+4I)Y_n = 2MY_n + 4 Y_n = 2 Y_{n+1} + 4 Y_n. $$
Or
$$ Y_{n+2} = 2 Y_{n+1} + 4 Y_n. $$
Therefore
$$ b_{n+2} = 2 b_{n+1} + 4 b_n $$ and
$$ c_{n+2} = 2 c_{n+1} + 4 c_n. $$
What does this say about $a_n ; ?$
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
This is introduction class so I'm afraid I cant understand this solution yet :)
$endgroup$
– BBLN
Feb 2 at 20:40
add a comment |
$begingroup$
if you take
$$Y_n =
left(
begin{array}{c}
b_n \
c_n
end{array}
right)
$$
then
$$ Y_n = M Y_{n-1}, $$
where
$$M =
left(
begin{array}{cc}
0&2 \
2&2
end{array}
right)
$$
I prefer writing
$$ Y_{n+1} = M Y_{n}. $$
From the trace $2$ and determinant $-4,$ we find $M^2 - 2M-4I = 0,$ or
$M^2 = 2M + 4I.$ This is Cayley-Hamilton. We point out that
$$ Y_{n+2} = M^2 Y_{n}. $$ So
$$ Y_{n+2} = M^2 Y_{n} = (2M+4I)Y_n = 2MY_n + 4 Y_n = 2 Y_{n+1} + 4 Y_n. $$
Or
$$ Y_{n+2} = 2 Y_{n+1} + 4 Y_n. $$
Therefore
$$ b_{n+2} = 2 b_{n+1} + 4 b_n $$ and
$$ c_{n+2} = 2 c_{n+1} + 4 c_n. $$
What does this say about $a_n ; ?$
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
This is introduction class so I'm afraid I cant understand this solution yet :)
$endgroup$
– BBLN
Feb 2 at 20:40
add a comment |
$begingroup$
if you take
$$Y_n =
left(
begin{array}{c}
b_n \
c_n
end{array}
right)
$$
then
$$ Y_n = M Y_{n-1}, $$
where
$$M =
left(
begin{array}{cc}
0&2 \
2&2
end{array}
right)
$$
I prefer writing
$$ Y_{n+1} = M Y_{n}. $$
From the trace $2$ and determinant $-4,$ we find $M^2 - 2M-4I = 0,$ or
$M^2 = 2M + 4I.$ This is Cayley-Hamilton. We point out that
$$ Y_{n+2} = M^2 Y_{n}. $$ So
$$ Y_{n+2} = M^2 Y_{n} = (2M+4I)Y_n = 2MY_n + 4 Y_n = 2 Y_{n+1} + 4 Y_n. $$
Or
$$ Y_{n+2} = 2 Y_{n+1} + 4 Y_n. $$
Therefore
$$ b_{n+2} = 2 b_{n+1} + 4 b_n $$ and
$$ c_{n+2} = 2 c_{n+1} + 4 c_n. $$
What does this say about $a_n ; ?$
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
$endgroup$
if you take
$$Y_n =
left(
begin{array}{c}
b_n \
c_n
end{array}
right)
$$
then
$$ Y_n = M Y_{n-1}, $$
where
$$M =
left(
begin{array}{cc}
0&2 \
2&2
end{array}
right)
$$
I prefer writing
$$ Y_{n+1} = M Y_{n}. $$
From the trace $2$ and determinant $-4,$ we find $M^2 - 2M-4I = 0,$ or
$M^2 = 2M + 4I.$ This is Cayley-Hamilton. We point out that
$$ Y_{n+2} = M^2 Y_{n}. $$ So
$$ Y_{n+2} = M^2 Y_{n} = (2M+4I)Y_n = 2MY_n + 4 Y_n = 2 Y_{n+1} + 4 Y_n. $$
Or
$$ Y_{n+2} = 2 Y_{n+1} + 4 Y_n. $$
Therefore
$$ b_{n+2} = 2 b_{n+1} + 4 b_n $$ and
$$ c_{n+2} = 2 c_{n+1} + 4 c_n. $$
What does this say about $a_n ; ?$
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
answered Feb 2 at 20:26
Will JagyWill Jagy
104k5103202
104k5103202
$begingroup$
This is introduction class so I'm afraid I cant understand this solution yet :)
$endgroup$
– BBLN
Feb 2 at 20:40
add a comment |
$begingroup$
This is introduction class so I'm afraid I cant understand this solution yet :)
$endgroup$
– BBLN
Feb 2 at 20:40
$begingroup$
This is introduction class so I'm afraid I cant understand this solution yet :)
$endgroup$
– BBLN
Feb 2 at 20:40
$begingroup$
This is introduction class so I'm afraid I cant understand this solution yet :)
$endgroup$
– BBLN
Feb 2 at 20:40
add a comment |
$begingroup$
The first equation gives
begin{align}
a_n & = b_n+2c_n\
-2a_{n-1} & = -2b_{n-1}-4c_{n-1}\
-4a_{n-2} & = -4b_{n-2}-8c_{n-2}.
end{align}
Now the last two equations give $c_n=2c_{n-1}+4c_{n-2}$, by replacing $b_{n-1}$ in the last equation by using the second equation. Similarly we obtain $b_n=2b_{n-1}+4b_{n-2}$. Now summing up the three equations give the claim.
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
$begingroup$
I see that this solution is right, but how would you start without knowing/guessing the simplified form? Whats the best way to approach this without knowing how many elements there would be in the final form
$endgroup$
– BBLN
Feb 2 at 20:43
$begingroup$
The best way would be to start with two equations only, say the last two ones: We have $c_n=2c_{n-1}+2b_{n-1}$. Oh, let's get rid of the $2b_{n-1}$. By the second equation we can replace it by $4c_{n-2}$. We get $c_n=2c_{n-1}+4c_{n-2}$. This is already the "simplified" form. Now we just need to see how this also works for the $b$ and $a$ sequence.
$endgroup$
– Dietrich Burde
Feb 2 at 20:45
add a comment |
$begingroup$
The first equation gives
begin{align}
a_n & = b_n+2c_n\
-2a_{n-1} & = -2b_{n-1}-4c_{n-1}\
-4a_{n-2} & = -4b_{n-2}-8c_{n-2}.
end{align}
Now the last two equations give $c_n=2c_{n-1}+4c_{n-2}$, by replacing $b_{n-1}$ in the last equation by using the second equation. Similarly we obtain $b_n=2b_{n-1}+4b_{n-2}$. Now summing up the three equations give the claim.
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
$begingroup$
I see that this solution is right, but how would you start without knowing/guessing the simplified form? Whats the best way to approach this without knowing how many elements there would be in the final form
$endgroup$
– BBLN
Feb 2 at 20:43
$begingroup$
The best way would be to start with two equations only, say the last two ones: We have $c_n=2c_{n-1}+2b_{n-1}$. Oh, let's get rid of the $2b_{n-1}$. By the second equation we can replace it by $4c_{n-2}$. We get $c_n=2c_{n-1}+4c_{n-2}$. This is already the "simplified" form. Now we just need to see how this also works for the $b$ and $a$ sequence.
$endgroup$
– Dietrich Burde
Feb 2 at 20:45
add a comment |
$begingroup$
The first equation gives
begin{align}
a_n & = b_n+2c_n\
-2a_{n-1} & = -2b_{n-1}-4c_{n-1}\
-4a_{n-2} & = -4b_{n-2}-8c_{n-2}.
end{align}
Now the last two equations give $c_n=2c_{n-1}+4c_{n-2}$, by replacing $b_{n-1}$ in the last equation by using the second equation. Similarly we obtain $b_n=2b_{n-1}+4b_{n-2}$. Now summing up the three equations give the claim.
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
The first equation gives
begin{align}
a_n & = b_n+2c_n\
-2a_{n-1} & = -2b_{n-1}-4c_{n-1}\
-4a_{n-2} & = -4b_{n-2}-8c_{n-2}.
end{align}
Now the last two equations give $c_n=2c_{n-1}+4c_{n-2}$, by replacing $b_{n-1}$ in the last equation by using the second equation. Similarly we obtain $b_n=2b_{n-1}+4b_{n-2}$. Now summing up the three equations give the claim.
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
edited Feb 2 at 20:43
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
answered Feb 2 at 20:26
Dietrich BurdeDietrich Burde
82k649107
82k649107
$begingroup$
I see that this solution is right, but how would you start without knowing/guessing the simplified form? Whats the best way to approach this without knowing how many elements there would be in the final form
$endgroup$
– BBLN
Feb 2 at 20:43
$begingroup$
The best way would be to start with two equations only, say the last two ones: We have $c_n=2c_{n-1}+2b_{n-1}$. Oh, let's get rid of the $2b_{n-1}$. By the second equation we can replace it by $4c_{n-2}$. We get $c_n=2c_{n-1}+4c_{n-2}$. This is already the "simplified" form. Now we just need to see how this also works for the $b$ and $a$ sequence.
$endgroup$
– Dietrich Burde
Feb 2 at 20:45
add a comment |
$begingroup$
I see that this solution is right, but how would you start without knowing/guessing the simplified form? Whats the best way to approach this without knowing how many elements there would be in the final form
$endgroup$
– BBLN
Feb 2 at 20:43
$begingroup$
The best way would be to start with two equations only, say the last two ones: We have $c_n=2c_{n-1}+2b_{n-1}$. Oh, let's get rid of the $2b_{n-1}$. By the second equation we can replace it by $4c_{n-2}$. We get $c_n=2c_{n-1}+4c_{n-2}$. This is already the "simplified" form. Now we just need to see how this also works for the $b$ and $a$ sequence.
$endgroup$
– Dietrich Burde
Feb 2 at 20:45
$begingroup$
I see that this solution is right, but how would you start without knowing/guessing the simplified form? Whats the best way to approach this without knowing how many elements there would be in the final form
$endgroup$
– BBLN
Feb 2 at 20:43
$begingroup$
I see that this solution is right, but how would you start without knowing/guessing the simplified form? Whats the best way to approach this without knowing how many elements there would be in the final form
$endgroup$
– BBLN
Feb 2 at 20:43
$begingroup$
The best way would be to start with two equations only, say the last two ones: We have $c_n=2c_{n-1}+2b_{n-1}$. Oh, let's get rid of the $2b_{n-1}$. By the second equation we can replace it by $4c_{n-2}$. We get $c_n=2c_{n-1}+4c_{n-2}$. This is already the "simplified" form. Now we just need to see how this also works for the $b$ and $a$ sequence.
$endgroup$
– Dietrich Burde
Feb 2 at 20:45
$begingroup$
The best way would be to start with two equations only, say the last two ones: We have $c_n=2c_{n-1}+2b_{n-1}$. Oh, let's get rid of the $2b_{n-1}$. By the second equation we can replace it by $4c_{n-2}$. We get $c_n=2c_{n-1}+4c_{n-2}$. This is already the "simplified" form. Now we just need to see how this also works for the $b$ and $a$ sequence.
$endgroup$
– Dietrich Burde
Feb 2 at 20:45
add a comment |
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