Mixing
Solve the following recurrence $b(n)=0.9*b(n-1)+60, b(1)=300$
$begingroup$
How to solve the following recurrence: $$b(n)=0.9*b(n-1)+60, b(1)=300$$
I do it with telescoping, like this:
$$b(n)=0.9*(0.9*b(n-2)+60)+60$$
$$b(n)=0.9*(0.9*(0.9*b(n-3)+60)+60)+60$$
Which means that for $n=4$ we get $$b(n)=0.9^{(n-1)}*300+(60)^n-1$$
But the correct solution is $b(n)=-300*(0.9^n)+600$
Can someone please explain where is my mistake?
discrete-mathematics recurrence-relations
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
asked Jan 18 at 11:58
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
How to solve the following recurrence: $$b(n)=0.9*b(n-1)+60, b(1)=300$$
I do it with telescoping, like this:
$$b(n)=0.9*(0.9*b(n-2)+60)+60$$
$$b(n)=0.9*(0.9*(0.9*b(n-3)+60)+60)+60$$
Which means that for $n=4$ we get $$b(n)=0.9^{(n-1)}*300+(60)^n-1$$
But the correct solution is $b(n)=-300*(0.9^n)+600$
Can someone please explain where is my mistake?
discrete-mathematics recurrence-relations
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
asked Jan 18 at 11:58
ponikoliponikoli
416
$endgroup$
$begingroup$
Looks like a geometric sum to me.
$endgroup$
– Matti P.
Jan 18 at 12:02
add a comment |
$begingroup$
How to solve the following recurrence: $$b(n)=0.9*b(n-1)+60, b(1)=300$$
I do it with telescoping, like this:
$$b(n)=0.9*(0.9*b(n-2)+60)+60$$
$$b(n)=0.9*(0.9*(0.9*b(n-3)+60)+60)+60$$
Which means that for $n=4$ we get $$b(n)=0.9^{(n-1)}*300+(60)^n-1$$
But the correct solution is $b(n)=-300*(0.9^n)+600$
Can someone please explain where is my mistake?
discrete-mathematics recurrence-relations
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
asked Jan 18 at 11:58
ponikoliponikoli
416
$endgroup$
How to solve the following recurrence: $$b(n)=0.9*b(n-1)+60, b(1)=300$$
I do it with telescoping, like this:
$$b(n)=0.9*(0.9*b(n-2)+60)+60$$
$$b(n)=0.9*(0.9*(0.9*b(n-3)+60)+60)+60$$
Which means that for $n=4$ we get $$b(n)=0.9^{(n-1)}*300+(60)^n-1$$
But the correct solution is $b(n)=-300*(0.9^n)+600$
Can someone please explain where is my mistake?
discrete-mathematics recurrence-relations
discrete-mathematics recurrence-relations
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
asked Jan 18 at 11:58
ponikoliponikoli
416
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
asked Jan 18 at 11:58
ponikoliponikoli
416
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
edited Jan 18 at 12:06
Thomas Shelby
3,7092525
3,7092525
asked Jan 18 at 11:58
ponikoliponikoli
416
asked Jan 18 at 11:58
ponikoliponikoli
416
asked Jan 18 at 11:58
ponikoliponikoli
416
416
$begingroup$
Looks like a geometric sum to me.
$endgroup$
– Matti P.
Jan 18 at 12:02
add a comment |
$begingroup$
Looks like a geometric sum to me.
$endgroup$
– Matti P.
Jan 18 at 12:02
$begingroup$
Looks like a geometric sum to me.
$endgroup$
– Matti P.
Jan 18 at 12:02
$begingroup$
Looks like a geometric sum to me.
$endgroup$
– Matti P.
Jan 18 at 12:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $b_n = ab_{n-1} + c, b_1 = d$. Then by expanding, one gets: $b_n = a(ab_{n-2} + c) + c = a^{n-1}b_1 + sum_{i=0}^{n-1}{a^ic}$. Finally $b_n =a^{n-1}d + cfrac{a^n-1}{a-1}$.
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
This answer is a better one since it does not depend on any choice to nullify a part of the expression.
$endgroup$
– Martigan
Jan 18 at 12:28
add a comment |
$begingroup$
Let
$$b_{n}=a_n+c$$
So we have that
$$a_n+c=0.9(a_{n-1}+c)+60$$
$$a_n=0.9a_{n-1}+60-0.1c$$
So if we let $60-0.1c=0$, we get that $c=600$. But this means that
$$a_n=0.9a_{n-1}$$
Which I think you can solve. From this, we will get that
$$b_n=a_n+600$$
answered Jan 18 at 12:12
BotondBotond
5,8802832
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $b_n = ab_{n-1} + c, b_1 = d$. Then by expanding, one gets: $b_n = a(ab_{n-2} + c) + c = a^{n-1}b_1 + sum_{i=0}^{n-1}{a^ic}$. Finally $b_n =a^{n-1}d + cfrac{a^n-1}{a-1}$.
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
This answer is a better one since it does not depend on any choice to nullify a part of the expression.
$endgroup$
– Martigan
Jan 18 at 12:28
add a comment |
$begingroup$
Let $b_n = ab_{n-1} + c, b_1 = d$. Then by expanding, one gets: $b_n = a(ab_{n-2} + c) + c = a^{n-1}b_1 + sum_{i=0}^{n-1}{a^ic}$. Finally $b_n =a^{n-1}d + cfrac{a^n-1}{a-1}$.
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
$endgroup$
$begingroup$
This answer is a better one since it does not depend on any choice to nullify a part of the expression.
$endgroup$
– Martigan
Jan 18 at 12:28
add a comment |
$begingroup$
Let $b_n = ab_{n-1} + c, b_1 = d$. Then by expanding, one gets: $b_n = a(ab_{n-2} + c) + c = a^{n-1}b_1 + sum_{i=0}^{n-1}{a^ic}$. Finally $b_n =a^{n-1}d + cfrac{a^n-1}{a-1}$.
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
$endgroup$
Let $b_n = ab_{n-1} + c, b_1 = d$. Then by expanding, one gets: $b_n = a(ab_{n-2} + c) + c = a^{n-1}b_1 + sum_{i=0}^{n-1}{a^ic}$. Finally $b_n =a^{n-1}d + cfrac{a^n-1}{a-1}$.
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
answered Jan 18 at 12:10
lightxbulblightxbulb
945311
945311
$begingroup$
This answer is a better one since it does not depend on any choice to nullify a part of the expression.
$endgroup$
– Martigan
Jan 18 at 12:28
add a comment |
$begingroup$
This answer is a better one since it does not depend on any choice to nullify a part of the expression.
$endgroup$
– Martigan
Jan 18 at 12:28
$begingroup$
This answer is a better one since it does not depend on any choice to nullify a part of the expression.
$endgroup$
– Martigan
Jan 18 at 12:28
$begingroup$
This answer is a better one since it does not depend on any choice to nullify a part of the expression.
$endgroup$
– Martigan
Jan 18 at 12:28
add a comment |
$begingroup$
Let
$$b_{n}=a_n+c$$
So we have that
$$a_n+c=0.9(a_{n-1}+c)+60$$
$$a_n=0.9a_{n-1}+60-0.1c$$
So if we let $60-0.1c=0$, we get that $c=600$. But this means that
$$a_n=0.9a_{n-1}$$
Which I think you can solve. From this, we will get that
$$b_n=a_n+600$$
answered Jan 18 at 12:12
BotondBotond
5,8802832
$endgroup$
add a comment |
$begingroup$
Let
$$b_{n}=a_n+c$$
So we have that
$$a_n+c=0.9(a_{n-1}+c)+60$$
$$a_n=0.9a_{n-1}+60-0.1c$$
So if we let $60-0.1c=0$, we get that $c=600$. But this means that
$$a_n=0.9a_{n-1}$$
Which I think you can solve. From this, we will get that
$$b_n=a_n+600$$
answered Jan 18 at 12:12
BotondBotond
5,8802832
$endgroup$
add a comment |
$begingroup$
Let
$$b_{n}=a_n+c$$
So we have that
$$a_n+c=0.9(a_{n-1}+c)+60$$
$$a_n=0.9a_{n-1}+60-0.1c$$
So if we let $60-0.1c=0$, we get that $c=600$. But this means that
$$a_n=0.9a_{n-1}$$
Which I think you can solve. From this, we will get that
$$b_n=a_n+600$$
answered Jan 18 at 12:12
BotondBotond
5,8802832
$endgroup$
Let
$$b_{n}=a_n+c$$
So we have that
$$a_n+c=0.9(a_{n-1}+c)+60$$
$$a_n=0.9a_{n-1}+60-0.1c$$
So if we let $60-0.1c=0$, we get that $c=600$. But this means that
$$a_n=0.9a_{n-1}$$
Which I think you can solve. From this, we will get that
$$b_n=a_n+600$$
answered Jan 18 at 12:12
BotondBotond
5,8802832
answered Jan 18 at 12:12
BotondBotond
5,8802832
answered Jan 18 at 12:12
BotondBotond
5,8802832
answered Jan 18 at 12:12
BotondBotond
5,8802832
5,8802832
add a comment |
add a comment |
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$begingroup$
Looks like a geometric sum to me.
$endgroup$
– Matti P.
Jan 18 at 12:02