Verifying correctness of a solution to a recurrence relation
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I have been trying to solve a recurrence relation:
$a_n = 6_{a_{n-1}} - 9_{a_{n-2}}$
As it is a 2nd order linear, homogeneous recurrence, I solved it using the substitution method to get the general solution:
$a_n = A3^n + Bn3^n$
With the initial conditions $a_0 = -1$ and $a_1 = 1$, I got the particular solution:
$a_n = -3^n + frac{4}{3}n3^n$
It satisfies the two initial conditions but now I must substitute it into the recurrence relation to prov that it satisfies the recurrence relation.
I understand that I have to substitute it in to get:
$a_n = 6(-3^{n-1} + frac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + frac{4}{3}(n-2)
; 3^{n-2})$
How do I go to solve this to prove that it does in fact satisfy the recurrence relation?
recurrence-relations
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
$endgroup$
add a comment |
$begingroup$
I have been trying to solve a recurrence relation:
$a_n = 6_{a_{n-1}} - 9_{a_{n-2}}$
As it is a 2nd order linear, homogeneous recurrence, I solved it using the substitution method to get the general solution:
$a_n = A3^n + Bn3^n$
With the initial conditions $a_0 = -1$ and $a_1 = 1$, I got the particular solution:
$a_n = -3^n + frac{4}{3}n3^n$
It satisfies the two initial conditions but now I must substitute it into the recurrence relation to prov that it satisfies the recurrence relation.
I understand that I have to substitute it in to get:
$a_n = 6(-3^{n-1} + frac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + frac{4}{3}(n-2)
; 3^{n-2})$
How do I go to solve this to prove that it does in fact satisfy the recurrence relation?
recurrence-relations
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
$endgroup$
1
$begingroup$
Do you mean something like $$ a_n = 6 a_{n-1} - 9 a_{n-2} $$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 11:53
$begingroup$
Indeed you have to prove that $$6(-3^{n-1} + tfrac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + tfrac{4}{3}(n-2) ; 3^{n-2})$$ equals $$-3^n+tfrac43n3^n$$ (but watch out your parentheses, the correct version is in this comment...). This should not be that difficult...
$endgroup$
– Did
Dec 14 '18 at 12:23
$begingroup$
You are missing parentheses: you should have $6(-3^{n-1}+ frac{4}{3}(n- 1)3^{n-1})- 9(-3^{n-2}+ frac{4}{3}(n-2)3^{n-2})$
$endgroup$
– user247327
Dec 14 '18 at 12:24
$begingroup$
Yes, but I don't understand how to prove it.
$endgroup$
– Eddy99
Dec 14 '18 at 15:30
add a comment |
$begingroup$
I have been trying to solve a recurrence relation:
$a_n = 6_{a_{n-1}} - 9_{a_{n-2}}$
As it is a 2nd order linear, homogeneous recurrence, I solved it using the substitution method to get the general solution:
$a_n = A3^n + Bn3^n$
With the initial conditions $a_0 = -1$ and $a_1 = 1$, I got the particular solution:
$a_n = -3^n + frac{4}{3}n3^n$
It satisfies the two initial conditions but now I must substitute it into the recurrence relation to prov that it satisfies the recurrence relation.
I understand that I have to substitute it in to get:
$a_n = 6(-3^{n-1} + frac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + frac{4}{3}(n-2)
; 3^{n-2})$
How do I go to solve this to prove that it does in fact satisfy the recurrence relation?
recurrence-relations
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
$endgroup$
I have been trying to solve a recurrence relation:
$a_n = 6_{a_{n-1}} - 9_{a_{n-2}}$
As it is a 2nd order linear, homogeneous recurrence, I solved it using the substitution method to get the general solution:
$a_n = A3^n + Bn3^n$
With the initial conditions $a_0 = -1$ and $a_1 = 1$, I got the particular solution:
$a_n = -3^n + frac{4}{3}n3^n$
It satisfies the two initial conditions but now I must substitute it into the recurrence relation to prov that it satisfies the recurrence relation.
I understand that I have to substitute it in to get:
$a_n = 6(-3^{n-1} + frac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + frac{4}{3}(n-2)
; 3^{n-2})$
How do I go to solve this to prove that it does in fact satisfy the recurrence relation?
recurrence-relations
recurrence-relations
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
edited Dec 14 '18 at 12:56
Eddy99
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
asked Dec 14 '18 at 11:44
Eddy99Eddy99
61
61
1
$begingroup$
Do you mean something like $$ a_n = 6 a_{n-1} - 9 a_{n-2} $$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 11:53
$begingroup$
Indeed you have to prove that $$6(-3^{n-1} + tfrac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + tfrac{4}{3}(n-2) ; 3^{n-2})$$ equals $$-3^n+tfrac43n3^n$$ (but watch out your parentheses, the correct version is in this comment...). This should not be that difficult...
$endgroup$
– Did
Dec 14 '18 at 12:23
$begingroup$
You are missing parentheses: you should have $6(-3^{n-1}+ frac{4}{3}(n- 1)3^{n-1})- 9(-3^{n-2}+ frac{4}{3}(n-2)3^{n-2})$
$endgroup$
– user247327
Dec 14 '18 at 12:24
$begingroup$
Yes, but I don't understand how to prove it.
$endgroup$
– Eddy99
Dec 14 '18 at 15:30
add a comment |
1
$begingroup$
Do you mean something like $$ a_n = 6 a_{n-1} - 9 a_{n-2} $$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 11:53
$begingroup$
Indeed you have to prove that $$6(-3^{n-1} + tfrac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + tfrac{4}{3}(n-2) ; 3^{n-2})$$ equals $$-3^n+tfrac43n3^n$$ (but watch out your parentheses, the correct version is in this comment...). This should not be that difficult...
$endgroup$
– Did
Dec 14 '18 at 12:23
$begingroup$
You are missing parentheses: you should have $6(-3^{n-1}+ frac{4}{3}(n- 1)3^{n-1})- 9(-3^{n-2}+ frac{4}{3}(n-2)3^{n-2})$
$endgroup$
– user247327
Dec 14 '18 at 12:24
$begingroup$
Yes, but I don't understand how to prove it.
$endgroup$
– Eddy99
Dec 14 '18 at 15:30
1
1
$begingroup$
Do you mean something like $$ a_n = 6 a_{n-1} - 9 a_{n-2} $$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 11:53
$begingroup$
Do you mean something like $$ a_n = 6 a_{n-1} - 9 a_{n-2} $$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 11:53
$begingroup$
Indeed you have to prove that $$6(-3^{n-1} + tfrac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + tfrac{4}{3}(n-2) ; 3^{n-2})$$ equals $$-3^n+tfrac43n3^n$$ (but watch out your parentheses, the correct version is in this comment...). This should not be that difficult...
$endgroup$
– Did
Dec 14 '18 at 12:23
$begingroup$
Indeed you have to prove that $$6(-3^{n-1} + tfrac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + tfrac{4}{3}(n-2) ; 3^{n-2})$$ equals $$-3^n+tfrac43n3^n$$ (but watch out your parentheses, the correct version is in this comment...). This should not be that difficult...
$endgroup$
– Did
Dec 14 '18 at 12:23
$begingroup$
You are missing parentheses: you should have $6(-3^{n-1}+ frac{4}{3}(n- 1)3^{n-1})- 9(-3^{n-2}+ frac{4}{3}(n-2)3^{n-2})$
$endgroup$
– user247327
Dec 14 '18 at 12:24
$begingroup$
You are missing parentheses: you should have $6(-3^{n-1}+ frac{4}{3}(n- 1)3^{n-1})- 9(-3^{n-2}+ frac{4}{3}(n-2)3^{n-2})$
$endgroup$
– user247327
Dec 14 '18 at 12:24
$begingroup$
Yes, but I don't understand how to prove it.
$endgroup$
– Eddy99
Dec 14 '18 at 15:30
$begingroup$
Yes, but I don't understand how to prove it.
$endgroup$
– Eddy99
Dec 14 '18 at 15:30
add a comment |
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1
$begingroup$
Do you mean something like $$ a_n = 6 a_{n-1} - 9 a_{n-2} $$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 11:53
$begingroup$
Indeed you have to prove that $$6(-3^{n-1} + tfrac{4}{3}(n-1); 3^{n-1}) - 9(-3^{n-2} + tfrac{4}{3}(n-2) ; 3^{n-2})$$ equals $$-3^n+tfrac43n3^n$$ (but watch out your parentheses, the correct version is in this comment...). This should not be that difficult...
$endgroup$
– Did
Dec 14 '18 at 12:23
$begingroup$
You are missing parentheses: you should have $6(-3^{n-1}+ frac{4}{3}(n- 1)3^{n-1})- 9(-3^{n-2}+ frac{4}{3}(n-2)3^{n-2})$
$endgroup$
– user247327
Dec 14 '18 at 12:24
$begingroup$
Yes, but I don't understand how to prove it.
$endgroup$
– Eddy99
Dec 14 '18 at 15:30