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Can $a(n) = fracnn+1$ be written recursively?
Formula for a sequenceUse of Recursively Defined FunctionsRecursive Sequence from Finite SequencesFinding the lowest common value in repeating sequencesUnexplanied pattern from increasing rational sequencesWhat would describe the following basic sequence?Understanding sub-sequencesCan the Fibonacci sequence be written as an explicit rule?Turning a recursively defined sequence into an explicit formulaWhat's the formula for producing these series?
$begingroup$
Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$
Algebraically it can be written as $$a(n) = fracnn + 1$$
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
New contributor
$endgroup$
add a comment |
$begingroup$
Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$
Algebraically it can be written as $$a(n) = fracnn + 1$$
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
New contributor
$endgroup$
add a comment |
$begingroup$
Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$
Algebraically it can be written as $$a(n) = fracnn + 1$$
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
New contributor
$endgroup$
Take the sequence $$frac12, frac23, frac34, frac45, frac56, frac67, dots$$
Algebraically it can be written as $$a(n) = fracnn + 1$$
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
New contributor
New contributor
edited 6 mins ago
user1952500
829712
829712
New contributor
asked 1 hour ago
Levi KLevi K
262
262
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New contributor
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add a comment |
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 $$A_1 = 1/2$$
New contributor
$endgroup$
add a comment |
$begingroup$
beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*
$endgroup$
add a comment |
$begingroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = fracna_n-1 + 1n+1$$
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 ($n+1$).
$endgroup$
add a comment |
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3 Answers
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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 $$A_1 = 1/2$$
New contributor
$endgroup$
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 $$A_1 = 1/2$$
New contributor
$endgroup$
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 $$A_1 = 1/2$$
New contributor
$endgroup$
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 $$A_1 = 1/2$$
New contributor
New contributor
answered 59 mins ago
Levi KLevi K
262
262
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add a comment |
add a comment |
$begingroup$
beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*
$endgroup$
add a comment |
$begingroup$
beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*
$endgroup$
add a comment |
$begingroup$
beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*
$endgroup$
beginalign*
a_n+1 &= fracn+1n+2 \
&= fracn+2-1n+2 \
&= 1 - frac1n+2 text, so \
1 - a_n+1 &= frac1n+2 text, \
frac11 - a_n+1 &= n+2 &[textand so frac11 - a_n = n+1]\
&= n+1+1 \
&= frac11- a_n +1 \
&= frac11- a_n + frac1-a_n1-a_n \
&= frac2-a_n1- a_n text, then \
1 - a_n+1 &= frac1-a_n2- a_n text, and finally \
a_n+1 &= 1 - frac1-a_n2- a_n \
&= frac2-a_n2- a_n - frac1-a_n2- a_n \
&= frac12- a_n text.
endalign*
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
$$a_n = fracna_n-1 + 1n+1$$
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 ($n+1$).
$endgroup$
add a comment |
$begingroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = fracna_n-1 + 1n+1$$
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 ($n+1$).
$endgroup$
add a comment |
$begingroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = fracna_n-1 + 1n+1$$
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 ($n+1$).
$endgroup$
Just by playing around with some numbers, I determined a recursive relation to be
$$a_n = fracna_n-1 + 1n+1$$
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
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.
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