Can $a(n) = frac{n}{n+1}$ be written recursively?
$begingroup$
Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$
Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$
Can you write this as a recursive function as well?
A pattern I have noticed:
- Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.
I am currently in Algebra II Honors and learning sequences
sequences-and-series number-theory recursion
edited 7 mins ago
user1952500
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Levi KLevi K
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$begingroup$
Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$
Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$
Can you write this as a recursive function as well?
A pattern I have noticed:
- Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.
I am currently in Algebra II Honors and learning sequences
sequences-and-series number-theory recursion
edited 7 mins ago
user1952500
829712
asked 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$
Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$
Can you write this as a recursive function as well?
A pattern I have noticed:
- Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.
I am currently in Algebra II Honors and learning sequences
sequences-and-series number-theory recursion
edited 7 mins ago
user1952500
829712
asked 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Take the sequence $$frac{1}{2}, frac{2}{3}, frac{3}{4}, frac{4}{5}, frac{5}{6}, frac{6}{7}, dots$$
Algebraically it can be written as $$a(n) = frac{n}{n + 1}$$
Can you write this as a recursive function as well?
A pattern I have noticed:
- Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.
I am currently in Algebra II Honors and learning sequences
sequences-and-series number-theory recursion
sequences-and-series number-theory recursion
edited 7 mins ago
user1952500
829712
asked 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 mins ago
user1952500
829712
asked 1 hour ago
Levi KLevi K
262
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Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 7 mins ago
user1952500
829712
edited 7 mins ago
user1952500
829712
edited 7 mins ago
user1952500
829712
829712
asked 1 hour ago
Levi KLevi K
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asked 1 hour ago
Levi KLevi K
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asked 1 hour ago
Levi KLevi K
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262
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3 Answers
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$begingroup$
After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$
answered 1 hour ago
Levi KLevi K
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$begingroup$
begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
$endgroup$
add a comment |
$begingroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = frac{na_{n-1} + 1}{n+1}$$
with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
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active
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votes
$begingroup$
After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$
answered 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$
answered 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$
answered 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = ({2 - A_{n}})^{-1}$$ where $${A_1 = 1/2}$$
answered 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Levi KLevi K
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Levi KLevi K
262
answered 1 hour ago
Levi KLevi K
262
262
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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add a comment |
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$begingroup$
begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
$endgroup$
add a comment |
$begingroup$
begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
$endgroup$
add a comment |
$begingroup$
begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
$endgroup$
begin{align*}
a_{n+1} &= frac{n+1}{n+2} \
&= frac{n+2-1}{n+2} \
&= 1 - frac{1}{n+2} text{, so } \
1 - a_{n+1} &= frac{1}{n+2} text{, } \
frac{1}{1 - a_{n+1}} &= n+2 &[text{and so } frac{1}{1 - a_n} = n+1]\
&= n+1+1 \
&= frac{1}{1- a_n} +1 \
&= frac{1}{1- a_n} + frac{1-a_n}{1-a_n} \
&= frac{2-a_n}{1- a_n} text{, then } \
1 - a_{n+1} &= frac{1-a_n}{2- a_n} text{, and finally } \
a_{n+1} &= 1 - frac{1-a_n}{2- a_n} \
&= frac{2-a_n}{2- a_n} - frac{1-a_n}{2- a_n} \
&= frac{1}{2- a_n} text{.}
end{align*}
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
answered 41 mins ago
Eric TowersEric Towers
33.5k22370
33.5k22370
add a comment |
add a comment |
$begingroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = frac{na_{n-1} + 1}{n+1}$$
with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
$endgroup$
add a comment |
$begingroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = frac{na_{n-1} + 1}{n+1}$$
with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
$endgroup$
add a comment |
$begingroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = frac{na_{n-1} + 1}{n+1}$$
with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
$endgroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = frac{na_{n-1} + 1}{n+1}$$
with $a_1 = 1/2$. I mostly derived this by noticing developing this would be easier if I negated the denominator of the previous term (thus multiplying by $n$), adding $1$ to the result (giving the increase in $1$ in the denominator), and then dividing again by the desired denominator ($n+1$).
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
answered 1 hour ago
Eevee TrainerEevee Trainer
9,91631740
9,91631740
add a comment |
add a comment |
Levi K is a new contributor. Be nice, and check out our Code of Conduct.
Levi K is a new contributor. Be nice, and check out our Code of Conduct.
Levi K is a new contributor. Be nice, and check out our Code of Conduct.
Levi K is a new contributor. Be nice, and check out our Code of Conduct.
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