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 yesterday
Jyrki Lahtonen
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asked yesterday
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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 recursion
edited yesterday
Jyrki Lahtonen
110k13172390
asked yesterday
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 yesterday
Jyrki Lahtonen
110k13172390
asked yesterday
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 yesterday
Jyrki Lahtonen
110k13172390
asked yesterday
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 yesterday
Jyrki Lahtonen
110k13172390
asked yesterday
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 yesterday
Jyrki Lahtonen
110k13172390
edited yesterday
Jyrki Lahtonen
110k13172390
edited yesterday
Jyrki Lahtonen
110k13172390
110k13172390
asked yesterday
Levi KLevi K
435
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asked yesterday
Levi KLevi K
435
asked yesterday
Levi KLevi K
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435
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5 Answers
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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 yesterday
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
15 hours 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 yesterday
user1952500
833712
answered yesterday
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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$endgroup$
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 yesterday
Eevee TrainerEevee Trainer
10k31742
$endgroup$
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 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
$endgroup$
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 4 hours ago
robjohn♦robjohn
270k27313642
$endgroup$
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 yesterday
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
15 hours 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 yesterday
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
15 hours 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 yesterday
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 yesterday
Eric TowersEric Towers
33.6k22370
answered yesterday
Eric TowersEric Towers
33.6k22370
answered yesterday
Eric TowersEric Towers
33.6k22370
answered yesterday
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
15 hours 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
15 hours 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
15 hours 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
15 hours 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 yesterday
user1952500
833712
answered yesterday
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 yesterday
user1952500
833712
answered yesterday
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 yesterday
user1952500
833712
answered yesterday
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 yesterday
user1952500
833712
answered yesterday
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 yesterday
user1952500
833712
edited yesterday
user1952500
833712
edited yesterday
user1952500
833712
833712
answered yesterday
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 yesterday
Levi KLevi K
435
answered yesterday
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 yesterday
Eevee TrainerEevee Trainer
10k31742
$endgroup$
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 yesterday
Eevee TrainerEevee Trainer
10k31742
$endgroup$
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 yesterday
Eevee TrainerEevee Trainer
10k31742
$endgroup$
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 yesterday
Eevee TrainerEevee Trainer
10k31742
answered yesterday
Eevee TrainerEevee Trainer
10k31742
answered yesterday
Eevee TrainerEevee Trainer
10k31742
answered yesterday
Eevee TrainerEevee Trainer
10k31742
10k31742
add a comment |
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 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
$endgroup$
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 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
$endgroup$
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 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
$endgroup$
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 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
answered 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
answered 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
answered 5 hours ago
Roddy MacPheeRoddy MacPhee
720118
720118
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 4 hours ago
robjohn♦robjohn
270k27313642
$endgroup$
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 4 hours ago
robjohn♦robjohn
270k27313642
$endgroup$
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 4 hours 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 4 hours ago
robjohn♦robjohn
270k27313642
answered 4 hours ago
robjohn♦robjohn
270k27313642
answered 4 hours ago
robjohn♦robjohn
270k27313642
answered 4 hours ago
robjohn♦robjohn
270k27313642
270k27313642
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.
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