Can a(n) = n/n+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?
Fully-Firstable Anagram Sets
Character reincarnated...as a snail
Can I ask the recruiters in my resume to put the reason why I am rejected?
LaTeX: Why are digits allowed in environments, but forbidden in commands?
Codimension of non-flat locus
RSA: Danger of using p to create q
Important Resources for Dark Age Civilizations?
Intersection point of 2 lines defined by 2 points each
Is it legal for company to use my work email to pretend I still work there?
Malcev's paper "On a class of homogeneous spaces" in English
Alternative to sending password over mail?
Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?
How is it possible to have an ability score that is less than 3?
strTok function (thread safe, supports empty tokens, doesn't change string)
I'm flying to France today and my passport expires in less than 2 months
Paid for article while in US on F-1 visa?
Why do I get two different answers for this counting problem?
What is the word for reserving something for yourself before others do?
When a company launches a new product do they "come out" with a new product or do they "come up" with a new product?
dbcc cleantable batch size explanation
Could an aircraft fly or hover using only jets of compressed air?
Cross compiling for RPi - error while loading shared libraries
What is a clear way to write a bar that has an extra beat?
Client team has low performances and low technical skills: we always fix their work and now they stop collaborate with us. How to solve?
Can a(n) = n/n+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 1/2, 2/3, 3/4, 4/5, 5/6, 6/7, ...
Algebraically it can be written as a(n) = 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
New contributor
$endgroup$
add a comment |
$begingroup$
Take the sequence 1/2, 2/3, 3/4, 4/5, 5/6, 6/7, ...
Algebraically it can be written as a(n) = 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
New contributor
$endgroup$
add a comment |
$begingroup$
Take the sequence 1/2, 2/3, 3/4, 4/5, 5/6, 6/7, ...
Algebraically it can be written as a(n) = 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
New contributor
$endgroup$
Take the sequence 1/2, 2/3, 3/4, 4/5, 5/6, 6/7, ...
Algebraically it can be written as a(n) = 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
New contributor
New contributor
New contributor
asked 1 hour ago
Levi KLevi K
262
262
New contributor
New contributor
add a comment |
add a comment |
3 Answers
3
active
oldest
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 $$A_1 = 1/2$$
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 |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Levi K is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3176633%2fcan-an-n-n1-be-written-recursively%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
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 $$A_1 = 1/2$$
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 52 mins ago
Levi KLevi K
262
262
New contributor
New contributor
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 34 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 54 mins 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.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3176633%2fcan-an-n-n1-be-written-recursively%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown