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How to write recursive formula as explicit (specifically, exponential)?
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I recall learning in school how to convert arithmetic and geometric sequence formulas between recursive and explicit, but I don't remember learning a systematic method to approach it. For example, let $a_n=2a_n-1+2$ where $a_0=5$. How would I convert this to an explicit form defined in terms of $n$? Beyond this specific example, is there a generalized method to do so?
sequences-and-series functions exponential-function recursive-algorithms
asked Aug 6 at 0:51
AARON TSAI
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I recall learning in school how to convert arithmetic and geometric sequence formulas between recursive and explicit, but I don't remember learning a systematic method to approach it. For example, let $a_n=2a_n-1+2$ where $a_0=5$. How would I convert this to an explicit form defined in terms of $n$? Beyond this specific example, is there a generalized method to do so?
sequences-and-series functions exponential-function recursive-algorithms
asked Aug 6 at 0:51
AARON TSAI
163
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I recall learning in school how to convert arithmetic and geometric sequence formulas between recursive and explicit, but I don't remember learning a systematic method to approach it. For example, let $a_n=2a_n-1+2$ where $a_0=5$. How would I convert this to an explicit form defined in terms of $n$? Beyond this specific example, is there a generalized method to do so?
sequences-and-series functions exponential-function recursive-algorithms
asked Aug 6 at 0:51
AARON TSAI
163
I recall learning in school how to convert arithmetic and geometric sequence formulas between recursive and explicit, but I don't remember learning a systematic method to approach it. For example, let $a_n=2a_n-1+2$ where $a_0=5$. How would I convert this to an explicit form defined in terms of $n$? Beyond this specific example, is there a generalized method to do so?
sequences-and-series functions exponential-function recursive-algorithms
asked Aug 6 at 0:51
AARON TSAI
163
asked Aug 6 at 0:51
AARON TSAI
163
asked Aug 6 at 0:51
AARON TSAI
163
asked Aug 6 at 0:51
AARON TSAI
163
163
add a comment |Â
add a comment |Â
3 Answers
3
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1
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There is a systematic way of solving something like $a_n=2a_n-1+2a_n-2$, which is called a linear recurrence relation. For example this pdf shows you how to do it.
But for your specific example you have a constant $2$, so it's not a linear recurrence. However you can rewrite it as $a_n+2=2(a_n-1+2),a_0+2=7$ where it's obvious that $a_n+2=7cdot 2^n$ so $a_n=7cdot 2^n-2$.
For a completely general recurrence there isn't really a systematic way to convert it to a formula, but you can always try tricks like the above (linear recurrence, substitution) to see if they work.
answered Aug 6 at 2:09
Akababa
2,557922
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It's equivalent to $left(~mboxwith a_0 = 5~right)$:
beginalign
a_n + 2 & = 2left(a_n - 1 + 2right) = 2^2left(a_n - 2 + 2right) =
2^3left(a_n - 3 + 2right) = cdots = 2^nleft(a_0 + 2right) = 7 times 2^n
\[5mm] implies & bbox[15px,border:1px solid navy]a_n = 7 times 2^n - 2
endalign
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
add a comment |Â
up vote
1
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Leu us consider the general case of $$a_n=alpha, a_n-1+beta$$ Let first $a_n=b_n+gamma$ which makes
$$b_n+gamma=alpha,(b_n-1+gamma)+betaimplies b_n=alpha,b_n-1+(alphagamma+beta-gamma)$$ To come back to something simple, let (assuming $alphaneq 1$)
$$alphagamma+beta-gamma=0 implies gamma=frac beta 1-alpha$$ So, we are left with
$$ b_n=alpha,b_n-1 implies b_n=c ,alpha^n-1implies a_n=c, alpha^n-1+frac beta 1-alpha$$
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There is a systematic way of solving something like $a_n=2a_n-1+2a_n-2$, which is called a linear recurrence relation. For example this pdf shows you how to do it.
But for your specific example you have a constant $2$, so it's not a linear recurrence. However you can rewrite it as $a_n+2=2(a_n-1+2),a_0+2=7$ where it's obvious that $a_n+2=7cdot 2^n$ so $a_n=7cdot 2^n-2$.
For a completely general recurrence there isn't really a systematic way to convert it to a formula, but you can always try tricks like the above (linear recurrence, substitution) to see if they work.
answered Aug 6 at 2:09
Akababa
2,557922
add a comment |Â
up vote
1
down vote
There is a systematic way of solving something like $a_n=2a_n-1+2a_n-2$, which is called a linear recurrence relation. For example this pdf shows you how to do it.
But for your specific example you have a constant $2$, so it's not a linear recurrence. However you can rewrite it as $a_n+2=2(a_n-1+2),a_0+2=7$ where it's obvious that $a_n+2=7cdot 2^n$ so $a_n=7cdot 2^n-2$.
For a completely general recurrence there isn't really a systematic way to convert it to a formula, but you can always try tricks like the above (linear recurrence, substitution) to see if they work.
answered Aug 6 at 2:09
Akababa
2,557922
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There is a systematic way of solving something like $a_n=2a_n-1+2a_n-2$, which is called a linear recurrence relation. For example this pdf shows you how to do it.
But for your specific example you have a constant $2$, so it's not a linear recurrence. However you can rewrite it as $a_n+2=2(a_n-1+2),a_0+2=7$ where it's obvious that $a_n+2=7cdot 2^n$ so $a_n=7cdot 2^n-2$.
For a completely general recurrence there isn't really a systematic way to convert it to a formula, but you can always try tricks like the above (linear recurrence, substitution) to see if they work.
answered Aug 6 at 2:09
Akababa
2,557922
There is a systematic way of solving something like $a_n=2a_n-1+2a_n-2$, which is called a linear recurrence relation. For example this pdf shows you how to do it.
But for your specific example you have a constant $2$, so it's not a linear recurrence. However you can rewrite it as $a_n+2=2(a_n-1+2),a_0+2=7$ where it's obvious that $a_n+2=7cdot 2^n$ so $a_n=7cdot 2^n-2$.
For a completely general recurrence there isn't really a systematic way to convert it to a formula, but you can always try tricks like the above (linear recurrence, substitution) to see if they work.
answered Aug 6 at 2:09
Akababa
2,557922
answered Aug 6 at 2:09
Akababa
2,557922
answered Aug 6 at 2:09
Akababa
2,557922
answered Aug 6 at 2:09
Akababa
2,557922
2,557922
add a comment |Â
add a comment |Â
up vote
1
down vote
It's equivalent to $left(~mboxwith a_0 = 5~right)$:
beginalign
a_n + 2 & = 2left(a_n - 1 + 2right) = 2^2left(a_n - 2 + 2right) =
2^3left(a_n - 3 + 2right) = cdots = 2^nleft(a_0 + 2right) = 7 times 2^n
\[5mm] implies & bbox[15px,border:1px solid navy]a_n = 7 times 2^n - 2
endalign
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
add a comment |Â
up vote
1
down vote
It's equivalent to $left(~mboxwith a_0 = 5~right)$:
beginalign
a_n + 2 & = 2left(a_n - 1 + 2right) = 2^2left(a_n - 2 + 2right) =
2^3left(a_n - 3 + 2right) = cdots = 2^nleft(a_0 + 2right) = 7 times 2^n
\[5mm] implies & bbox[15px,border:1px solid navy]a_n = 7 times 2^n - 2
endalign
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's equivalent to $left(~mboxwith a_0 = 5~right)$:
beginalign
a_n + 2 & = 2left(a_n - 1 + 2right) = 2^2left(a_n - 2 + 2right) =
2^3left(a_n - 3 + 2right) = cdots = 2^nleft(a_0 + 2right) = 7 times 2^n
\[5mm] implies & bbox[15px,border:1px solid navy]a_n = 7 times 2^n - 2
endalign
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
It's equivalent to $left(~mboxwith a_0 = 5~right)$:
beginalign
a_n + 2 & = 2left(a_n - 1 + 2right) = 2^2left(a_n - 2 + 2right) =
2^3left(a_n - 3 + 2right) = cdots = 2^nleft(a_0 + 2right) = 7 times 2^n
\[5mm] implies & bbox[15px,border:1px solid navy]a_n = 7 times 2^n - 2
endalign
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
answered Aug 6 at 2:20
Felix Marin
65.6k7105136
65.6k7105136
add a comment |Â
add a comment |Â
up vote
1
down vote
Leu us consider the general case of $$a_n=alpha, a_n-1+beta$$ Let first $a_n=b_n+gamma$ which makes
$$b_n+gamma=alpha,(b_n-1+gamma)+betaimplies b_n=alpha,b_n-1+(alphagamma+beta-gamma)$$ To come back to something simple, let (assuming $alphaneq 1$)
$$alphagamma+beta-gamma=0 implies gamma=frac beta 1-alpha$$ So, we are left with
$$ b_n=alpha,b_n-1 implies b_n=c ,alpha^n-1implies a_n=c, alpha^n-1+frac beta 1-alpha$$
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
add a comment |Â
up vote
1
down vote
Leu us consider the general case of $$a_n=alpha, a_n-1+beta$$ Let first $a_n=b_n+gamma$ which makes
$$b_n+gamma=alpha,(b_n-1+gamma)+betaimplies b_n=alpha,b_n-1+(alphagamma+beta-gamma)$$ To come back to something simple, let (assuming $alphaneq 1$)
$$alphagamma+beta-gamma=0 implies gamma=frac beta 1-alpha$$ So, we are left with
$$ b_n=alpha,b_n-1 implies b_n=c ,alpha^n-1implies a_n=c, alpha^n-1+frac beta 1-alpha$$
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Leu us consider the general case of $$a_n=alpha, a_n-1+beta$$ Let first $a_n=b_n+gamma$ which makes
$$b_n+gamma=alpha,(b_n-1+gamma)+betaimplies b_n=alpha,b_n-1+(alphagamma+beta-gamma)$$ To come back to something simple, let (assuming $alphaneq 1$)
$$alphagamma+beta-gamma=0 implies gamma=frac beta 1-alpha$$ So, we are left with
$$ b_n=alpha,b_n-1 implies b_n=c ,alpha^n-1implies a_n=c, alpha^n-1+frac beta 1-alpha$$
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
Leu us consider the general case of $$a_n=alpha, a_n-1+beta$$ Let first $a_n=b_n+gamma$ which makes
$$b_n+gamma=alpha,(b_n-1+gamma)+betaimplies b_n=alpha,b_n-1+(alphagamma+beta-gamma)$$ To come back to something simple, let (assuming $alphaneq 1$)
$$alphagamma+beta-gamma=0 implies gamma=frac beta 1-alpha$$ So, we are left with
$$ b_n=alpha,b_n-1 implies b_n=c ,alpha^n-1implies a_n=c, alpha^n-1+frac beta 1-alpha$$
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
answered Aug 6 at 3:36
Claude Leibovici
112k1055126
112k1055126
add a comment |Â
add a comment |Â
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