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 recursion
edited Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
asked Apr 6 at 3:25
Levi KLevi K
435
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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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$endgroup$
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 recursion
edited Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
asked Apr 6 at 3:25
Levi KLevi K
435
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 recursion
edited Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
asked Apr 6 at 3:25
Levi KLevi K
435
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 recursion
sequences-and-series recursion
edited Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
asked Apr 6 at 3:25
Levi KLevi K
435
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 Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
asked Apr 6 at 3:25
Levi KLevi K
435
New contributor
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
edited Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
edited Apr 6 at 4:39
Jyrki Lahtonen
110k13172390
110k13172390
asked Apr 6 at 3:25
Levi KLevi K
435
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.
asked Apr 6 at 3:25
Levi KLevi K
435
asked Apr 6 at 3:25
Levi KLevi K
435
435
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add a comment |
5 Answers
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oldest
votes
$begingroup$
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 Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
$endgroup$
$begingroup$
Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
$endgroup$
– Levi K
2 days ago
add a comment |
$begingroup$
After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$
edited Apr 6 at 4:36
user1952500
917712
answered Apr 6 at 3:33
Levi KLevi K
435
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 |
$begingroup$
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 Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
$endgroup$
add a comment |
$begingroup$
Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$
answered 2 days ago
robjohn♦robjohn
270k27313642
$endgroup$
$begingroup$
I like the way you thought about it!
$endgroup$
– Levi K
yesterday
add a comment |
$begingroup$
You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$
or
$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$
etc.
equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$
and equation 2, simply notes:
$$n=na_{n-1}+1$$
etc.
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
$endgroup$
$begingroup$
Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
$endgroup$
– Levi K
2 days ago
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 Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
$endgroup$
$begingroup$
Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
$endgroup$
– Levi K
2 days ago
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 Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
$endgroup$
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 Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
answered Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
answered Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
answered Apr 6 at 3:52
Eric TowersEric Towers
33.6k22370
33.6k22370
$begingroup$
Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
$endgroup$
– Levi K
2 days ago
add a comment |
$begingroup$
Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
$endgroup$
– Levi K
2 days ago
$begingroup$
Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
$endgroup$
– Levi K
2 days ago
$begingroup$
Brilliant! I got the same answer, but I appreciate all the work you put in showing how to get there. I just found it by noticing a pattern with 2's, but your solution is a great mathematical approach. Thank you!
$endgroup$
– Levi K
2 days ago
add a comment |
$begingroup$
After some further solving, I was able to come up with an answer
It can be written $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$
edited Apr 6 at 4:36
user1952500
917712
answered Apr 6 at 3:33
Levi KLevi K
435
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 $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$
edited Apr 6 at 4:36
user1952500
917712
answered Apr 6 at 3:33
Levi KLevi K
435
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 $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$
edited Apr 6 at 4:36
user1952500
917712
answered Apr 6 at 3:33
Levi KLevi K
435
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 $${A_{n + 1}} = frac{1}{2 - A_{n}}$$ where $$A_1 = frac{1}{2}$$
edited Apr 6 at 4:36
user1952500
917712
answered Apr 6 at 3:33
Levi KLevi K
435
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 Apr 6 at 4:36
user1952500
917712
edited Apr 6 at 4:36
user1952500
917712
edited Apr 6 at 4:36
user1952500
917712
917712
answered Apr 6 at 3:33
Levi KLevi K
435
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 Apr 6 at 3:33
Levi KLevi K
435
answered Apr 6 at 3:33
Levi KLevi K
435
435
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.
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.
Levi K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
$endgroup$
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 Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
$endgroup$
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 Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
$endgroup$
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 Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
answered Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
answered Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
answered Apr 6 at 3:32
Eevee TrainerEevee Trainer
10.2k31742
10.2k31742
add a comment |
add a comment |
$begingroup$
Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$
answered 2 days ago
robjohn♦robjohn
270k27313642
$endgroup$
$begingroup$
I like the way you thought about it!
$endgroup$
– Levi K
yesterday
add a comment |
$begingroup$
Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$
answered 2 days ago
robjohn♦robjohn
270k27313642
$endgroup$
$begingroup$
I like the way you thought about it!
$endgroup$
– Levi K
yesterday
add a comment |
$begingroup$
Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$
answered 2 days ago
robjohn♦robjohn
270k27313642
$endgroup$
Perhaps a bit simpler is to note that
$$
overbrace{ frac1{1-a_n} }^{n+1}+1=overbrace{frac1{1-a_{n+1}}}^{n+2}\
$$
solving for $a_{n+1}$ yields
$$
frac1{2-a_n}=a_{n+1}\
$$
answered 2 days ago
robjohn♦robjohn
270k27313642
answered 2 days ago
robjohn♦robjohn
270k27313642
answered 2 days ago
robjohn♦robjohn
270k27313642
answered 2 days ago
robjohn♦robjohn
270k27313642
270k27313642
$begingroup$
I like the way you thought about it!
$endgroup$
– Levi K
yesterday
add a comment |
$begingroup$
I like the way you thought about it!
$endgroup$
– Levi K
yesterday
$begingroup$
I like the way you thought about it!
$endgroup$
– Levi K
yesterday
$begingroup$
I like the way you thought about it!
$endgroup$
– Levi K
yesterday
add a comment |
$begingroup$
You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$
or
$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$
etc.
equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$
and equation 2, simply notes:
$$n=na_{n-1}+1$$
etc.
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
$endgroup$
add a comment |
$begingroup$
You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$
or
$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$
etc.
equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$
and equation 2, simply notes:
$$n=na_{n-1}+1$$
etc.
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
$endgroup$
add a comment |
$begingroup$
You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$
or
$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$
etc.
equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$
and equation 2, simply notes:
$$n=na_{n-1}+1$$
etc.
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
$endgroup$
You can, and in multiple ways. Such as:$$a_n=frac{1}{a_{n-1}+frac{2}{n}}tag{1}$$
or
$$a_n=frac{na_{n-1}+1}{n+1}tag{2}$$
etc.
equation 1, is simply noting: $$frac{n-1}{n}+frac{2}{n}=frac{n+1}{n}=frac{1}{a_n}$$ Where the first fraction in the sum is $a_{n-1}$
and equation 2, simply notes:
$$n=na_{n-1}+1$$
etc.
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
answered 2 days ago
Roddy MacPheeRoddy MacPhee
742118
742118
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Levi K is a new contributor. Be nice, and check out our Code of Conduct.
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