Mixing
Problem with solving the recurrence relation $a_n=a_{n-1}+6a_{n-2}+30$ for $ngeq2$, $a_0=0$, $a_1=-10$
$begingroup$
My task:
$a_n=a_{n-1}+6a_{n-2}+30$ for $ngeq2$, $a_0=0$, $a_1=-10$
My solution
$x^{2}-x-6$
$Delta=25$
$x1=-2 $
$x2=3$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-2)^{n}+b*3^{n}$
$a_0=0=a+b$
$a_1=-10=-2a+3b$
$b=-2$
$a=2$
$a_n=2*(-2)^{n}-2*3^{n}+30$
I calculate $a_2$
$a_2=-10+0+30=20$
Which is correct with above formula
$a_2=2*(-2)^{2}-2*3^{2}+30=20$
but for $a_3$ and above results are not matching, am I doing something wrong here?
discrete-mathematics
asked Jan 27 at 16:17
GorossoGorosso
315
$endgroup$
add a comment |
$begingroup$
My task:
$a_n=a_{n-1}+6a_{n-2}+30$ for $ngeq2$, $a_0=0$, $a_1=-10$
My solution
$x^{2}-x-6$
$Delta=25$
$x1=-2 $
$x2=3$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-2)^{n}+b*3^{n}$
$a_0=0=a+b$
$a_1=-10=-2a+3b$
$b=-2$
$a=2$
$a_n=2*(-2)^{n}-2*3^{n}+30$
I calculate $a_2$
$a_2=-10+0+30=20$
Which is correct with above formula
$a_2=2*(-2)^{2}-2*3^{2}+30=20$
but for $a_3$ and above results are not matching, am I doing something wrong here?
discrete-mathematics
asked Jan 27 at 16:17
GorossoGorosso
315
$endgroup$
add a comment |
$begingroup$
My task:
$a_n=a_{n-1}+6a_{n-2}+30$ for $ngeq2$, $a_0=0$, $a_1=-10$
My solution
$x^{2}-x-6$
$Delta=25$
$x1=-2 $
$x2=3$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-2)^{n}+b*3^{n}$
$a_0=0=a+b$
$a_1=-10=-2a+3b$
$b=-2$
$a=2$
$a_n=2*(-2)^{n}-2*3^{n}+30$
I calculate $a_2$
$a_2=-10+0+30=20$
Which is correct with above formula
$a_2=2*(-2)^{2}-2*3^{2}+30=20$
but for $a_3$ and above results are not matching, am I doing something wrong here?
discrete-mathematics
asked Jan 27 at 16:17
GorossoGorosso
315
$endgroup$
My task:
$a_n=a_{n-1}+6a_{n-2}+30$ for $ngeq2$, $a_0=0$, $a_1=-10$
My solution
$x^{2}-x-6$
$Delta=25$
$x1=-2 $
$x2=3$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-2)^{n}+b*3^{n}$
$a_0=0=a+b$
$a_1=-10=-2a+3b$
$b=-2$
$a=2$
$a_n=2*(-2)^{n}-2*3^{n}+30$
I calculate $a_2$
$a_2=-10+0+30=20$
Which is correct with above formula
$a_2=2*(-2)^{2}-2*3^{2}+30=20$
but for $a_3$ and above results are not matching, am I doing something wrong here?
discrete-mathematics
discrete-mathematics
asked Jan 27 at 16:17
GorossoGorosso
315
asked Jan 27 at 16:17
GorossoGorosso
315
edited Jan 27 at 16:23
Gorosso
asked Jan 27 at 16:17
GorossoGorosso
315
asked Jan 27 at 16:17
GorossoGorosso
315
asked Jan 27 at 16:17
GorossoGorosso
315
315
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The $+30$ is the problem. Find $c$ such that $b_n:=a_n+cimplies b_n=b_{n-1}+6b_{n-2}$, then use your usual techniques.
Edit to add detail:
$$b_n=b_{n-1}-c+6(b_{n-2}-c)+30+c=b_{n-1}+6b_{n-2}+30-6c$$ Set $c=5$ so $$b_n=b_{n-1}+6b_{n-2}impliesexists a,,b:,b_n=a(-2)^n+b3^n.$$From $b_0=5,,b_1=-5$, you can find $a,,b$. Then $a_n=b_n-5$.
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
$endgroup$
$begingroup$
I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it.
$endgroup$
– Gorosso
Jan 27 at 16:41
$begingroup$
@Gorosso See my edit.
$endgroup$
– J.G.
Jan 27 at 17:15
$begingroup$
Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$?
$endgroup$
– Gorosso
Jan 27 at 18:00
$begingroup$
@Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$.
$endgroup$
– J.G.
Jan 27 at 18:14
$begingroup$
Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/…
$endgroup$
– Gorosso
Jan 27 at 19:22
add a comment |
$begingroup$
Since for all $n$ we have $$a_n-a_{n-1}-6a_{n-2}=30$$ we have also
$$a_{n-1}-a_{n-2}-6a_{n-3}=30$$
so $$a_n-a_{n-1}-6a_{n-2}=a_{n-1}-a_{n-2}-6a_{n-3}$$
and thus we get l.r. :
$$a_n-2a_{n-1}-5a_{n-2}+6a_{n-3}=0$$
and so on...
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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active
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votes
$begingroup$
The $+30$ is the problem. Find $c$ such that $b_n:=a_n+cimplies b_n=b_{n-1}+6b_{n-2}$, then use your usual techniques.
Edit to add detail:
$$b_n=b_{n-1}-c+6(b_{n-2}-c)+30+c=b_{n-1}+6b_{n-2}+30-6c$$ Set $c=5$ so $$b_n=b_{n-1}+6b_{n-2}impliesexists a,,b:,b_n=a(-2)^n+b3^n.$$From $b_0=5,,b_1=-5$, you can find $a,,b$. Then $a_n=b_n-5$.
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
$endgroup$
$begingroup$
I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it.
$endgroup$
– Gorosso
Jan 27 at 16:41
$begingroup$
@Gorosso See my edit.
$endgroup$
– J.G.
Jan 27 at 17:15
$begingroup$
Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$?
$endgroup$
– Gorosso
Jan 27 at 18:00
$begingroup$
@Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$.
$endgroup$
– J.G.
Jan 27 at 18:14
$begingroup$
Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/…
$endgroup$
– Gorosso
Jan 27 at 19:22
add a comment |
$begingroup$
The $+30$ is the problem. Find $c$ such that $b_n:=a_n+cimplies b_n=b_{n-1}+6b_{n-2}$, then use your usual techniques.
Edit to add detail:
$$b_n=b_{n-1}-c+6(b_{n-2}-c)+30+c=b_{n-1}+6b_{n-2}+30-6c$$ Set $c=5$ so $$b_n=b_{n-1}+6b_{n-2}impliesexists a,,b:,b_n=a(-2)^n+b3^n.$$From $b_0=5,,b_1=-5$, you can find $a,,b$. Then $a_n=b_n-5$.
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
$endgroup$
$begingroup$
I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it.
$endgroup$
– Gorosso
Jan 27 at 16:41
$begingroup$
@Gorosso See my edit.
$endgroup$
– J.G.
Jan 27 at 17:15
$begingroup$
Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$?
$endgroup$
– Gorosso
Jan 27 at 18:00
$begingroup$
@Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$.
$endgroup$
– J.G.
Jan 27 at 18:14
$begingroup$
Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/…
$endgroup$
– Gorosso
Jan 27 at 19:22
add a comment |
$begingroup$
The $+30$ is the problem. Find $c$ such that $b_n:=a_n+cimplies b_n=b_{n-1}+6b_{n-2}$, then use your usual techniques.
Edit to add detail:
$$b_n=b_{n-1}-c+6(b_{n-2}-c)+30+c=b_{n-1}+6b_{n-2}+30-6c$$ Set $c=5$ so $$b_n=b_{n-1}+6b_{n-2}impliesexists a,,b:,b_n=a(-2)^n+b3^n.$$From $b_0=5,,b_1=-5$, you can find $a,,b$. Then $a_n=b_n-5$.
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
$endgroup$
The $+30$ is the problem. Find $c$ such that $b_n:=a_n+cimplies b_n=b_{n-1}+6b_{n-2}$, then use your usual techniques.
Edit to add detail:
$$b_n=b_{n-1}-c+6(b_{n-2}-c)+30+c=b_{n-1}+6b_{n-2}+30-6c$$ Set $c=5$ so $$b_n=b_{n-1}+6b_{n-2}impliesexists a,,b:,b_n=a(-2)^n+b3^n.$$From $b_0=5,,b_1=-5$, you can find $a,,b$. Then $a_n=b_n-5$.
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
edited Jan 27 at 17:15
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
answered Jan 27 at 16:21
J.G.J.G.
31.9k23250
31.9k23250
$begingroup$
I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it.
$endgroup$
– Gorosso
Jan 27 at 16:41
$begingroup$
@Gorosso See my edit.
$endgroup$
– J.G.
Jan 27 at 17:15
$begingroup$
Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$?
$endgroup$
– Gorosso
Jan 27 at 18:00
$begingroup$
@Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$.
$endgroup$
– J.G.
Jan 27 at 18:14
$begingroup$
Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/…
$endgroup$
– Gorosso
Jan 27 at 19:22
add a comment |
$begingroup$
I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it.
$endgroup$
– Gorosso
Jan 27 at 16:41
$begingroup$
@Gorosso See my edit.
$endgroup$
– J.G.
Jan 27 at 17:15
$begingroup$
Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$?
$endgroup$
– Gorosso
Jan 27 at 18:00
$begingroup$
@Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$.
$endgroup$
– J.G.
Jan 27 at 18:14
$begingroup$
Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/…
$endgroup$
– Gorosso
Jan 27 at 19:22
$begingroup$
I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it.
$endgroup$
– Gorosso
Jan 27 at 16:41
$begingroup$
I don't really get it, could you show me how its supposed to look in my case? That's the best way for me to understand it.
$endgroup$
– Gorosso
Jan 27 at 16:41
$begingroup$
@Gorosso See my edit.
$endgroup$
– J.G.
Jan 27 at 17:15
$begingroup$
@Gorosso See my edit.
$endgroup$
– J.G.
Jan 27 at 17:15
$begingroup$
Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$?
$endgroup$
– Gorosso
Jan 27 at 18:00
$begingroup$
Where does c=5 come from? Also what I am supposed to do with $a_n=b_n-5$?
$endgroup$
– Gorosso
Jan 27 at 18:00
$begingroup$
@Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$.
$endgroup$
– J.G.
Jan 27 at 18:14
$begingroup$
@Gorosso The technique you used originally only works when there isn't a constant in the recursion relation, so I chose $c$ to delete it. Once you have a formula for $b_n$, you have one for $a_n$.
$endgroup$
– J.G.
Jan 27 at 18:14
$begingroup$
Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/…
$endgroup$
– Gorosso
Jan 27 at 19:22
$begingroup$
Now results are matching, so I guess I did everything according to your tips. But I am having some issue in similar reccurence using your method, could you check it out? math.stackexchange.com/questions/3090006/…
$endgroup$
– Gorosso
Jan 27 at 19:22
add a comment |
$begingroup$
Since for all $n$ we have $$a_n-a_{n-1}-6a_{n-2}=30$$ we have also
$$a_{n-1}-a_{n-2}-6a_{n-3}=30$$
so $$a_n-a_{n-1}-6a_{n-2}=a_{n-1}-a_{n-2}-6a_{n-3}$$
and thus we get l.r. :
$$a_n-2a_{n-1}-5a_{n-2}+6a_{n-3}=0$$
and so on...
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
$endgroup$
add a comment |
$begingroup$
Since for all $n$ we have $$a_n-a_{n-1}-6a_{n-2}=30$$ we have also
$$a_{n-1}-a_{n-2}-6a_{n-3}=30$$
so $$a_n-a_{n-1}-6a_{n-2}=a_{n-1}-a_{n-2}-6a_{n-3}$$
and thus we get l.r. :
$$a_n-2a_{n-1}-5a_{n-2}+6a_{n-3}=0$$
and so on...
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
$endgroup$
add a comment |
$begingroup$
Since for all $n$ we have $$a_n-a_{n-1}-6a_{n-2}=30$$ we have also
$$a_{n-1}-a_{n-2}-6a_{n-3}=30$$
so $$a_n-a_{n-1}-6a_{n-2}=a_{n-1}-a_{n-2}-6a_{n-3}$$
and thus we get l.r. :
$$a_n-2a_{n-1}-5a_{n-2}+6a_{n-3}=0$$
and so on...
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
$endgroup$
Since for all $n$ we have $$a_n-a_{n-1}-6a_{n-2}=30$$ we have also
$$a_{n-1}-a_{n-2}-6a_{n-3}=30$$
so $$a_n-a_{n-1}-6a_{n-2}=a_{n-1}-a_{n-2}-6a_{n-3}$$
and thus we get l.r. :
$$a_n-2a_{n-1}-5a_{n-2}+6a_{n-3}=0$$
and so on...
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
answered Jan 27 at 17:23
Maria MazurMaria Mazur
48.4k1260121
48.4k1260121
add a comment |
add a comment |
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