Finding general formula for a recursively defined series [closed]
$begingroup$
Can someone please show me how to solve the following question please, I am very lost. Suppose that the sequence $x_0, x_1, x_2...$ is defined by $x_0 = 6$, $x_1 = 2$, and $x_{k+2} = −6x_{k+1}−8x_k$ for k≥0. Find a general formula for $x_k$.
sequences-and-series
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
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closed as off-topic by amWhy, rschwieb, Nosrati, Brian Borchers, RRL Dec 9 '18 at 6:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, rschwieb, Nosrati, Brian Borchers, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Can someone please show me how to solve the following question please, I am very lost. Suppose that the sequence $x_0, x_1, x_2...$ is defined by $x_0 = 6$, $x_1 = 2$, and $x_{k+2} = −6x_{k+1}−8x_k$ for k≥0. Find a general formula for $x_k$.
sequences-and-series
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
$endgroup$
closed as off-topic by amWhy, rschwieb, Nosrati, Brian Borchers, RRL Dec 9 '18 at 6:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, rschwieb, Nosrati, Brian Borchers, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Try listing out the first few terms i.e $x_2, x_3, x_4, x_5,...$. Do you notice a pattern?
$endgroup$
– Aniruddh Venkatesan
Dec 9 '18 at 2:29
add a comment |
$begingroup$
Can someone please show me how to solve the following question please, I am very lost. Suppose that the sequence $x_0, x_1, x_2...$ is defined by $x_0 = 6$, $x_1 = 2$, and $x_{k+2} = −6x_{k+1}−8x_k$ for k≥0. Find a general formula for $x_k$.
sequences-and-series
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
$endgroup$
Can someone please show me how to solve the following question please, I am very lost. Suppose that the sequence $x_0, x_1, x_2...$ is defined by $x_0 = 6$, $x_1 = 2$, and $x_{k+2} = −6x_{k+1}−8x_k$ for k≥0. Find a general formula for $x_k$.
sequences-and-series
sequences-and-series
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
edited Dec 9 '18 at 3:00
Aniruddh Venkatesan
144112
144112
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
asked Dec 9 '18 at 1:48
Mike1029Mike1029
62
62
closed as off-topic by amWhy, rschwieb, Nosrati, Brian Borchers, RRL Dec 9 '18 at 6:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, rschwieb, Nosrati, Brian Borchers, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, rschwieb, Nosrati, Brian Borchers, RRL Dec 9 '18 at 6:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, rschwieb, Nosrati, Brian Borchers, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Try listing out the first few terms i.e $x_2, x_3, x_4, x_5,...$. Do you notice a pattern?
$endgroup$
– Aniruddh Venkatesan
Dec 9 '18 at 2:29
add a comment |
$begingroup$
Try listing out the first few terms i.e $x_2, x_3, x_4, x_5,...$. Do you notice a pattern?
$endgroup$
– Aniruddh Venkatesan
Dec 9 '18 at 2:29
$begingroup$
Try listing out the first few terms i.e $x_2, x_3, x_4, x_5,...$. Do you notice a pattern?
$endgroup$
– Aniruddh Venkatesan
Dec 9 '18 at 2:29
$begingroup$
Try listing out the first few terms i.e $x_2, x_3, x_4, x_5,...$. Do you notice a pattern?
$endgroup$
– Aniruddh Venkatesan
Dec 9 '18 at 2:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is similar to finding the general and particular solution to a differential equation $y''+ay'+b = 0$ using undetermined coefficients, but instead of our answers being in terms of $e^x$, they will be in terms of $a^x$.
To find the general solution of this recursion, we need a characteristic equation. Let $u^k = x_k$ so the characteristic equation is $$u^2 + 6u + 8 = 0$$
This is an easy factorization, so we get $$(u+4)(u+2)=0 Rightarrow u = -2, -4$$
Our general equation will look like $$x_k = A(-4)^k + B(-2)^k$$ but we need to find the coefficients A and B.
For $x_0=6$, $$A(-4)^0+B(-2)^0 = 6 Rightarrow A+B = 6$$
For $x_1=2$, $$A(-4)^1+B(-2)^1 = 2 Rightarrow -4A-2B = 2$$
Now we must find $A text { and } B$, and we do so by solving the system of equations $$begin {matrix} A + B = 6 \ -4A - 2B = 2 end {matrix}$$
and doing so we get $A = -7$ and $B = 13.$ Thus the general equation for $x_k$ is $$x_k = -7(-4)^k + 13 (-2)^k$$
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is similar to finding the general and particular solution to a differential equation $y''+ay'+b = 0$ using undetermined coefficients, but instead of our answers being in terms of $e^x$, they will be in terms of $a^x$.
To find the general solution of this recursion, we need a characteristic equation. Let $u^k = x_k$ so the characteristic equation is $$u^2 + 6u + 8 = 0$$
This is an easy factorization, so we get $$(u+4)(u+2)=0 Rightarrow u = -2, -4$$
Our general equation will look like $$x_k = A(-4)^k + B(-2)^k$$ but we need to find the coefficients A and B.
For $x_0=6$, $$A(-4)^0+B(-2)^0 = 6 Rightarrow A+B = 6$$
For $x_1=2$, $$A(-4)^1+B(-2)^1 = 2 Rightarrow -4A-2B = 2$$
Now we must find $A text { and } B$, and we do so by solving the system of equations $$begin {matrix} A + B = 6 \ -4A - 2B = 2 end {matrix}$$
and doing so we get $A = -7$ and $B = 13.$ Thus the general equation for $x_k$ is $$x_k = -7(-4)^k + 13 (-2)^k$$
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
$endgroup$
add a comment |
$begingroup$
This is similar to finding the general and particular solution to a differential equation $y''+ay'+b = 0$ using undetermined coefficients, but instead of our answers being in terms of $e^x$, they will be in terms of $a^x$.
To find the general solution of this recursion, we need a characteristic equation. Let $u^k = x_k$ so the characteristic equation is $$u^2 + 6u + 8 = 0$$
This is an easy factorization, so we get $$(u+4)(u+2)=0 Rightarrow u = -2, -4$$
Our general equation will look like $$x_k = A(-4)^k + B(-2)^k$$ but we need to find the coefficients A and B.
For $x_0=6$, $$A(-4)^0+B(-2)^0 = 6 Rightarrow A+B = 6$$
For $x_1=2$, $$A(-4)^1+B(-2)^1 = 2 Rightarrow -4A-2B = 2$$
Now we must find $A text { and } B$, and we do so by solving the system of equations $$begin {matrix} A + B = 6 \ -4A - 2B = 2 end {matrix}$$
and doing so we get $A = -7$ and $B = 13.$ Thus the general equation for $x_k$ is $$x_k = -7(-4)^k + 13 (-2)^k$$
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
$endgroup$
add a comment |
$begingroup$
This is similar to finding the general and particular solution to a differential equation $y''+ay'+b = 0$ using undetermined coefficients, but instead of our answers being in terms of $e^x$, they will be in terms of $a^x$.
To find the general solution of this recursion, we need a characteristic equation. Let $u^k = x_k$ so the characteristic equation is $$u^2 + 6u + 8 = 0$$
This is an easy factorization, so we get $$(u+4)(u+2)=0 Rightarrow u = -2, -4$$
Our general equation will look like $$x_k = A(-4)^k + B(-2)^k$$ but we need to find the coefficients A and B.
For $x_0=6$, $$A(-4)^0+B(-2)^0 = 6 Rightarrow A+B = 6$$
For $x_1=2$, $$A(-4)^1+B(-2)^1 = 2 Rightarrow -4A-2B = 2$$
Now we must find $A text { and } B$, and we do so by solving the system of equations $$begin {matrix} A + B = 6 \ -4A - 2B = 2 end {matrix}$$
and doing so we get $A = -7$ and $B = 13.$ Thus the general equation for $x_k$ is $$x_k = -7(-4)^k + 13 (-2)^k$$
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
$endgroup$
This is similar to finding the general and particular solution to a differential equation $y''+ay'+b = 0$ using undetermined coefficients, but instead of our answers being in terms of $e^x$, they will be in terms of $a^x$.
To find the general solution of this recursion, we need a characteristic equation. Let $u^k = x_k$ so the characteristic equation is $$u^2 + 6u + 8 = 0$$
This is an easy factorization, so we get $$(u+4)(u+2)=0 Rightarrow u = -2, -4$$
Our general equation will look like $$x_k = A(-4)^k + B(-2)^k$$ but we need to find the coefficients A and B.
For $x_0=6$, $$A(-4)^0+B(-2)^0 = 6 Rightarrow A+B = 6$$
For $x_1=2$, $$A(-4)^1+B(-2)^1 = 2 Rightarrow -4A-2B = 2$$
Now we must find $A text { and } B$, and we do so by solving the system of equations $$begin {matrix} A + B = 6 \ -4A - 2B = 2 end {matrix}$$
and doing so we get $A = -7$ and $B = 13.$ Thus the general equation for $x_k$ is $$x_k = -7(-4)^k + 13 (-2)^k$$
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
edited Dec 9 '18 at 20:35
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
answered Dec 9 '18 at 2:45
bjcolby15bjcolby15
1,20811016
1,20811016
add a comment |
add a comment |
$begingroup$
Try listing out the first few terms i.e $x_2, x_3, x_4, x_5,...$. Do you notice a pattern?
$endgroup$
– Aniruddh Venkatesan
Dec 9 '18 at 2:29