Solving non-homogeneous recurrence relationships with exponents
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Find all solutions of the recurrence relation $a_n = 2a_{n-1} + 2n^2$.
I am having trouble with coming up with a guess for $2n^2$. Would it just be $an^2 + bn + c$?
I feel like there's an easier way based on ways to solve other recurrances.
recurrence-relations
asked Nov 25 at 21:24
Jersey Fonseca
356
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up vote
0
down vote
favorite
Find all solutions of the recurrence relation $a_n = 2a_{n-1} + 2n^2$.
I am having trouble with coming up with a guess for $2n^2$. Would it just be $an^2 + bn + c$?
I feel like there's an easier way based on ways to solve other recurrances.
recurrence-relations
asked Nov 25 at 21:24
Jersey Fonseca
356
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find all solutions of the recurrence relation $a_n = 2a_{n-1} + 2n^2$.
I am having trouble with coming up with a guess for $2n^2$. Would it just be $an^2 + bn + c$?
I feel like there's an easier way based on ways to solve other recurrances.
recurrence-relations
asked Nov 25 at 21:24
Jersey Fonseca
356
Find all solutions of the recurrence relation $a_n = 2a_{n-1} + 2n^2$.
I am having trouble with coming up with a guess for $2n^2$. Would it just be $an^2 + bn + c$?
I feel like there's an easier way based on ways to solve other recurrances.
recurrence-relations
recurrence-relations
asked Nov 25 at 21:24
Jersey Fonseca
356
asked Nov 25 at 21:24
Jersey Fonseca
356
asked Nov 25 at 21:24
Jersey Fonseca
356
asked Nov 25 at 21:24
Jersey Fonseca
356
asked Nov 25 at 21:24
Jersey Fonseca
356
356
add a comment |
add a comment |
1 Answer
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If you're familiar with the $Z$-transform you can find a systematic way to solve these kind of relations. Otherwise, a guess of the form
$$
a_n = alpha + beta n + gamma n^2 + delta 2^{n}
$$
The actual solution is
$$
a_n = left(frac{a_1}{2} + 11right) 2^n - 2 n^2 - 8 n -12
$$
answered Nov 25 at 22:51
caverac
12.5k21027
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you're familiar with the $Z$-transform you can find a systematic way to solve these kind of relations. Otherwise, a guess of the form
$$
a_n = alpha + beta n + gamma n^2 + delta 2^{n}
$$
The actual solution is
$$
a_n = left(frac{a_1}{2} + 11right) 2^n - 2 n^2 - 8 n -12
$$
answered Nov 25 at 22:51
caverac
12.5k21027
add a comment |
up vote
0
down vote
If you're familiar with the $Z$-transform you can find a systematic way to solve these kind of relations. Otherwise, a guess of the form
$$
a_n = alpha + beta n + gamma n^2 + delta 2^{n}
$$
The actual solution is
$$
a_n = left(frac{a_1}{2} + 11right) 2^n - 2 n^2 - 8 n -12
$$
answered Nov 25 at 22:51
caverac
12.5k21027
add a comment |
up vote
0
down vote
up vote
0
down vote
If you're familiar with the $Z$-transform you can find a systematic way to solve these kind of relations. Otherwise, a guess of the form
$$
a_n = alpha + beta n + gamma n^2 + delta 2^{n}
$$
The actual solution is
$$
a_n = left(frac{a_1}{2} + 11right) 2^n - 2 n^2 - 8 n -12
$$
answered Nov 25 at 22:51
caverac
12.5k21027
If you're familiar with the $Z$-transform you can find a systematic way to solve these kind of relations. Otherwise, a guess of the form
$$
a_n = alpha + beta n + gamma n^2 + delta 2^{n}
$$
The actual solution is
$$
a_n = left(frac{a_1}{2} + 11right) 2^n - 2 n^2 - 8 n -12
$$
answered Nov 25 at 22:51
caverac
12.5k21027
answered Nov 25 at 22:51
caverac
12.5k21027
answered Nov 25 at 22:51
caverac
12.5k21027
answered Nov 25 at 22:51
caverac
12.5k21027
12.5k21027
add a comment |
add a comment |
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