How to find unique solutions to recurrence relation $a_n =10a_n-1-32a_n-2+32a_n-3$ with $a_0=5$, $a_1=18$, $a_2=76$ The Next CEO of Stack OverflowHow do I resolve a recurrence relation when the characteristic equation has fewer roots than terms?Solving for the closed term solution of a third order recurrence relation with real constant coefficientsFinding the closed form solution of a third order recurrence relation with constant coefficientsHow to find recurrence relation of a given solution.Imaginary solutions of a recurrence relationRecurrence relation with characteristic equation that has only 1 root and complex rootsHelp with solving this Recurrence RelationSolve recurrence relation: $a_n=3a_n−1+5a_n−2−7a_n−3$Solve the following recurrence relation: $a_n=10a_n-2$How to find the initial terms of the recurrence relation if you know the nth term?
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How to find unique solutions to recurrence relation $a_n =10a_n-1-32a_n-2+32a_n-3$ with $a_0=5$, $a_1=18$, $a_2=76$
The Next CEO of Stack OverflowHow do I resolve a recurrence relation when the characteristic equation has fewer roots than terms?Solving for the closed term solution of a third order recurrence relation with real constant coefficientsFinding the closed form solution of a third order recurrence relation with constant coefficientsHow to find recurrence relation of a given solution.Imaginary solutions of a recurrence relationRecurrence relation with characteristic equation that has only 1 root and complex rootsHelp with solving this Recurrence RelationSolve recurrence relation: $a_n=3a_n−1+5a_n−2−7a_n−3$Solve the following recurrence relation: $a_n=10a_n-2$How to find the initial terms of the recurrence relation if you know the nth term?
$begingroup$
How to find the unique solutions to the recurrence relation given initial conditions and using the characteristic root technique?
$a_n =10a_n-1-32a_n-2+32a_n-3$ with $a_0=5$, $a_1=18$, $a_2=76$
Using the characteristic root technique, I create the characteristic root equation $x^3 -10x^2+32x-32=0$, which reduces to $(x-4)^2(x-2)=0$, so the characteristic roots are x=4 and x=2
I am confused here about how to set up my system of equations because for differing roots we use $a_n=ar_1^n + br_2^n$ where $r$ are the roots. While for repeated roots we use $a_n=ar^n + bnr^n$. How do I set up the system of equations to solve when an equation has repeated and differing roots?
recurrence-relations
asked Mar 19 at 11:57
SamSam
45118
$endgroup$
add a comment |
$begingroup$
How to find the unique solutions to the recurrence relation given initial conditions and using the characteristic root technique?
$a_n =10a_n-1-32a_n-2+32a_n-3$ with $a_0=5$, $a_1=18$, $a_2=76$
Using the characteristic root technique, I create the characteristic root equation $x^3 -10x^2+32x-32=0$, which reduces to $(x-4)^2(x-2)=0$, so the characteristic roots are x=4 and x=2
I am confused here about how to set up my system of equations because for differing roots we use $a_n=ar_1^n + br_2^n$ where $r$ are the roots. While for repeated roots we use $a_n=ar^n + bnr^n$. How do I set up the system of equations to solve when an equation has repeated and differing roots?
recurrence-relations
asked Mar 19 at 11:57
SamSam
45118
$endgroup$
add a comment |
$begingroup$
How to find the unique solutions to the recurrence relation given initial conditions and using the characteristic root technique?
$a_n =10a_n-1-32a_n-2+32a_n-3$ with $a_0=5$, $a_1=18$, $a_2=76$
Using the characteristic root technique, I create the characteristic root equation $x^3 -10x^2+32x-32=0$, which reduces to $(x-4)^2(x-2)=0$, so the characteristic roots are x=4 and x=2
I am confused here about how to set up my system of equations because for differing roots we use $a_n=ar_1^n + br_2^n$ where $r$ are the roots. While for repeated roots we use $a_n=ar^n + bnr^n$. How do I set up the system of equations to solve when an equation has repeated and differing roots?
recurrence-relations
asked Mar 19 at 11:57
SamSam
45118
$endgroup$
How to find the unique solutions to the recurrence relation given initial conditions and using the characteristic root technique?
$a_n =10a_n-1-32a_n-2+32a_n-3$ with $a_0=5$, $a_1=18$, $a_2=76$
Using the characteristic root technique, I create the characteristic root equation $x^3 -10x^2+32x-32=0$, which reduces to $(x-4)^2(x-2)=0$, so the characteristic roots are x=4 and x=2
I am confused here about how to set up my system of equations because for differing roots we use $a_n=ar_1^n + br_2^n$ where $r$ are the roots. While for repeated roots we use $a_n=ar^n + bnr^n$. How do I set up the system of equations to solve when an equation has repeated and differing roots?
recurrence-relations
recurrence-relations
asked Mar 19 at 11:57
SamSam
45118
asked Mar 19 at 11:57
SamSam
45118
asked Mar 19 at 11:57
SamSam
45118
asked Mar 19 at 11:57
SamSam
45118
asked Mar 19 at 11:57
SamSam
45118
45118
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint
The treatment with the repeated roots is as follows:
if a root $r$ was repeated of order, say $k$ , then the corresponding terms in the general recurrence will become$$t_n=P_k(n)cdot r^n$$where $P_k(n)$ is a polynomial of degree $k-1$.
In our case, it will be:$$a_n=(a+bn)cdot 4^n+ccdot 2^n$$
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint
The treatment with the repeated roots is as follows:
if a root $r$ was repeated of order, say $k$ , then the corresponding terms in the general recurrence will become$$t_n=P_k(n)cdot r^n$$where $P_k(n)$ is a polynomial of degree $k-1$.
In our case, it will be:$$a_n=(a+bn)cdot 4^n+ccdot 2^n$$
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
The treatment with the repeated roots is as follows:
if a root $r$ was repeated of order, say $k$ , then the corresponding terms in the general recurrence will become$$t_n=P_k(n)cdot r^n$$where $P_k(n)$ is a polynomial of degree $k-1$.
In our case, it will be:$$a_n=(a+bn)cdot 4^n+ccdot 2^n$$
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
The treatment with the repeated roots is as follows:
if a root $r$ was repeated of order, say $k$ , then the corresponding terms in the general recurrence will become$$t_n=P_k(n)cdot r^n$$where $P_k(n)$ is a polynomial of degree $k-1$.
In our case, it will be:$$a_n=(a+bn)cdot 4^n+ccdot 2^n$$
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
Hint
The treatment with the repeated roots is as follows:
if a root $r$ was repeated of order, say $k$ , then the corresponding terms in the general recurrence will become$$t_n=P_k(n)cdot r^n$$where $P_k(n)$ is a polynomial of degree $k-1$.
In our case, it will be:$$a_n=(a+bn)cdot 4^n+ccdot 2^n$$
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 19 at 12:02
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
add a comment |
add a comment |
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