Finding the least upper bound for a sequence Announcing the arrival of Valued Associate #679:...
How were pictures turned from film to a big picture in a picture frame before digital scanning?
If Windows 7 doesn't support WSL, then what is "Subsystem for UNIX-based Applications"?
Why is a lens darker than other ones when applying the same settings?
Found this skink in my tomato plant bucket. Is he trapped? Or could he leave if he wanted?
How to ask rejected full-time candidates to apply to teach individual courses?
A term for a woman complaining about things/begging in a cute/childish way
Weaponising the Grasp-at-a-Distance spell
After Sam didn't return home in the end, were he and Al still friends?
Why not use the yoke to control yaw, as well as pitch and roll?
How can a team of shapeshifters communicate?
How does light 'choose' between wave and particle behaviour?
How to write capital alpha?
How to change the tick of the color bar legend to black
Is openssl rand command cryptographically secure?
Project Euler #1 in C++
What are the main differences between the original Stargate SG-1 and the Final Cut edition?
Nose gear failure in single prop aircraft: belly landing or nose-gear up landing?
What does 丫 mean? 丫是什么意思？
Rationale for describing kurtosis as "peakedness"?
Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?
How can I save and copy a screenhot at the same time?
What is the origin of 落第?
What does it mean that physics no longer uses mechanical models to describe phenomena?
One-one communication
Finding the least upper bound for a sequence
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Upper bound a binomial-like summationWhat is the value of a limit of sums with $sin k$Confusion with $O$ functionFinding a general term for the sequenceEquality of Floors of some Partial SumsEquation form for a seriesFinding the nth element of a sequenceGeometric Sequence with formula for kth term.Upper bound on a series: $sum_{k = M}^{infty} (a/sqrt(k))^k$Finding an upper bound on composite $C^{1}$ functions
$begingroup$
Define $f(1) = 10$ and $f(n) = f(lfloor n/2rfloor) + f(lfloor n/3rfloor) + 17n$.
Find the smallest value $k$ so that $f(n) leq kn$ for all $n$.
I plugged in $f(1)$ into the function, which lead me to $2 cdot f(0) + 17 = 10,$ so $f(0) = -7/2$.
Then I calculated $f(2) = f(1) + f(0) + 34 = 10 -7/2 + 34 =40.5$
Also $f(3) = f(1) + f(1) + 17(3) = 71$ and $f(4) = f(2) + f(1) + 17(4) = 118.5$
I don't see a clear pattern. I think it oscillates back and forth, but I can't get anything.
Any help is appreciated.
sequences-and-series algebra-precalculus functions upper-lower-bounds
$endgroup$
|
show 1 more comment
$begingroup$
Define $f(1) = 10$ and $f(n) = f(lfloor n/2rfloor) + f(lfloor n/3rfloor) + 17n$.
Find the smallest value $k$ so that $f(n) leq kn$ for all $n$.
I plugged in $f(1)$ into the function, which lead me to $2 cdot f(0) + 17 = 10,$ so $f(0) = -7/2$.
Then I calculated $f(2) = f(1) + f(0) + 34 = 10 -7/2 + 34 =40.5$
Also $f(3) = f(1) + f(1) + 17(3) = 71$ and $f(4) = f(2) + f(1) + 17(4) = 118.5$
I don't see a clear pattern. I think it oscillates back and forth, but I can't get anything.
Any help is appreciated.
sequences-and-series algebra-precalculus functions upper-lower-bounds
$endgroup$
$begingroup$
In $fleft(3right)=fleft(1right)+fleft(1right)+17left(3right)$, you forgot to add $17left(3right)$.
$endgroup$
– Jake
Mar 26 at 0:55
$begingroup$
i fixed the mistakes.
$endgroup$
– user651921
Mar 26 at 0:56
$begingroup$
There is still a mistake. You wrote $17$ rather than $17left(3right)$. $fleft(3right)ne37$.
$endgroup$
– Jake
Mar 26 at 0:57
1
$begingroup$
It looks as if $f$ is increasing with respect to $n$, so if $mle n$, then $fleft(mright)le fleft(nright)$. So, $$fleft(nright)=fleft(leftlfloorfrac{n}{2}rightrfloorright)+fleft(leftlfloorfrac{n}{3}rightrfloorright)+17nle fleft(frac{n}{2}right)+fleft(frac{n}{3}right)+17n.$$ You can use this to find an upper-bound for the right-hand side of that inequality.
$endgroup$
– Jake
Mar 26 at 1:11
$begingroup$
$f(6n)=f(3n)+f(2n)+102n$. I wrote a short program to compute $lfloor f(n)/nrfloor$ and found that between $n=5times10^9$ and $n=1times 10^{10}$ the value hangs at $101$ making the $102n$ look very interesting.
$endgroup$
– John Wayland Bales
Mar 26 at 3:35
|
show 1 more comment
$begingroup$
Define $f(1) = 10$ and $f(n) = f(lfloor n/2rfloor) + f(lfloor n/3rfloor) + 17n$.
Find the smallest value $k$ so that $f(n) leq kn$ for all $n$.
I plugged in $f(1)$ into the function, which lead me to $2 cdot f(0) + 17 = 10,$ so $f(0) = -7/2$.
Then I calculated $f(2) = f(1) + f(0) + 34 = 10 -7/2 + 34 =40.5$
Also $f(3) = f(1) + f(1) + 17(3) = 71$ and $f(4) = f(2) + f(1) + 17(4) = 118.5$
I don't see a clear pattern. I think it oscillates back and forth, but I can't get anything.
Any help is appreciated.
sequences-and-series algebra-precalculus functions upper-lower-bounds
$endgroup$
Define $f(1) = 10$ and $f(n) = f(lfloor n/2rfloor) + f(lfloor n/3rfloor) + 17n$.
Find the smallest value $k$ so that $f(n) leq kn$ for all $n$.
I plugged in $f(1)$ into the function, which lead me to $2 cdot f(0) + 17 = 10,$ so $f(0) = -7/2$.
Then I calculated $f(2) = f(1) + f(0) + 34 = 10 -7/2 + 34 =40.5$
Also $f(3) = f(1) + f(1) + 17(3) = 71$ and $f(4) = f(2) + f(1) + 17(4) = 118.5$
I don't see a clear pattern. I think it oscillates back and forth, but I can't get anything.
Any help is appreciated.
sequences-and-series algebra-precalculus functions upper-lower-bounds
sequences-and-series algebra-precalculus functions upper-lower-bounds
edited Mar 26 at 0:57
asked Mar 26 at 0:51
user651921
$begingroup$
In $fleft(3right)=fleft(1right)+fleft(1right)+17left(3right)$, you forgot to add $17left(3right)$.
$endgroup$
– Jake
Mar 26 at 0:55
$begingroup$
i fixed the mistakes.
$endgroup$
– user651921
Mar 26 at 0:56
$begingroup$
There is still a mistake. You wrote $17$ rather than $17left(3right)$. $fleft(3right)ne37$.
$endgroup$
– Jake
Mar 26 at 0:57
1
$begingroup$
It looks as if $f$ is increasing with respect to $n$, so if $mle n$, then $fleft(mright)le fleft(nright)$. So, $$fleft(nright)=fleft(leftlfloorfrac{n}{2}rightrfloorright)+fleft(leftlfloorfrac{n}{3}rightrfloorright)+17nle fleft(frac{n}{2}right)+fleft(frac{n}{3}right)+17n.$$ You can use this to find an upper-bound for the right-hand side of that inequality.
$endgroup$
– Jake
Mar 26 at 1:11
$begingroup$
$f(6n)=f(3n)+f(2n)+102n$. I wrote a short program to compute $lfloor f(n)/nrfloor$ and found that between $n=5times10^9$ and $n=1times 10^{10}$ the value hangs at $101$ making the $102n$ look very interesting.
$endgroup$
– John Wayland Bales
Mar 26 at 3:35
|
show 1 more comment
$begingroup$
In $fleft(3right)=fleft(1right)+fleft(1right)+17left(3right)$, you forgot to add $17left(3right)$.
$endgroup$
– Jake
Mar 26 at 0:55
$begingroup$
i fixed the mistakes.
$endgroup$
– user651921
Mar 26 at 0:56
$begingroup$
There is still a mistake. You wrote $17$ rather than $17left(3right)$. $fleft(3right)ne37$.
$endgroup$
– Jake
Mar 26 at 0:57
1
$begingroup$
It looks as if $f$ is increasing with respect to $n$, so if $mle n$, then $fleft(mright)le fleft(nright)$. So, $$fleft(nright)=fleft(leftlfloorfrac{n}{2}rightrfloorright)+fleft(leftlfloorfrac{n}{3}rightrfloorright)+17nle fleft(frac{n}{2}right)+fleft(frac{n}{3}right)+17n.$$ You can use this to find an upper-bound for the right-hand side of that inequality.
$endgroup$
– Jake
Mar 26 at 1:11
$begingroup$
$f(6n)=f(3n)+f(2n)+102n$. I wrote a short program to compute $lfloor f(n)/nrfloor$ and found that between $n=5times10^9$ and $n=1times 10^{10}$ the value hangs at $101$ making the $102n$ look very interesting.
$endgroup$
– John Wayland Bales
Mar 26 at 3:35
$begingroup$
In $fleft(3right)=fleft(1right)+fleft(1right)+17left(3right)$, you forgot to add $17left(3right)$.
$endgroup$
– Jake
Mar 26 at 0:55
$begingroup$
In $fleft(3right)=fleft(1right)+fleft(1right)+17left(3right)$, you forgot to add $17left(3right)$.
$endgroup$
– Jake
Mar 26 at 0:55
$begingroup$
i fixed the mistakes.
$endgroup$
– user651921
Mar 26 at 0:56
$begingroup$
i fixed the mistakes.
$endgroup$
– user651921
Mar 26 at 0:56
$begingroup$
There is still a mistake. You wrote $17$ rather than $17left(3right)$. $fleft(3right)ne37$.
$endgroup$
– Jake
Mar 26 at 0:57
$begingroup$
There is still a mistake. You wrote $17$ rather than $17left(3right)$. $fleft(3right)ne37$.
$endgroup$
– Jake
Mar 26 at 0:57
1
1
$begingroup$
It looks as if $f$ is increasing with respect to $n$, so if $mle n$, then $fleft(mright)le fleft(nright)$. So, $$fleft(nright)=fleft(leftlfloorfrac{n}{2}rightrfloorright)+fleft(leftlfloorfrac{n}{3}rightrfloorright)+17nle fleft(frac{n}{2}right)+fleft(frac{n}{3}right)+17n.$$ You can use this to find an upper-bound for the right-hand side of that inequality.
$endgroup$
– Jake
Mar 26 at 1:11
$begingroup$
It looks as if $f$ is increasing with respect to $n$, so if $mle n$, then $fleft(mright)le fleft(nright)$. So, $$fleft(nright)=fleft(leftlfloorfrac{n}{2}rightrfloorright)+fleft(leftlfloorfrac{n}{3}rightrfloorright)+17nle fleft(frac{n}{2}right)+fleft(frac{n}{3}right)+17n.$$ You can use this to find an upper-bound for the right-hand side of that inequality.
$endgroup$
– Jake
Mar 26 at 1:11
$begingroup$
$f(6n)=f(3n)+f(2n)+102n$. I wrote a short program to compute $lfloor f(n)/nrfloor$ and found that between $n=5times10^9$ and $n=1times 10^{10}$ the value hangs at $101$ making the $102n$ look very interesting.
$endgroup$
– John Wayland Bales
Mar 26 at 3:35
$begingroup$
$f(6n)=f(3n)+f(2n)+102n$. I wrote a short program to compute $lfloor f(n)/nrfloor$ and found that between $n=5times10^9$ and $n=1times 10^{10}$ the value hangs at $101$ making the $102n$ look very interesting.
$endgroup$
– John Wayland Bales
Mar 26 at 3:35
|
show 1 more comment
0
active
oldest
votes
Your Answer
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
});
}
});
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%2f3162539%2ffinding-the-least-upper-bound-for-a-sequence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3162539%2ffinding-the-least-upper-bound-for-a-sequence%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
$begingroup$
In $fleft(3right)=fleft(1right)+fleft(1right)+17left(3right)$, you forgot to add $17left(3right)$.
$endgroup$
– Jake
Mar 26 at 0:55
$begingroup$
i fixed the mistakes.
$endgroup$
– user651921
Mar 26 at 0:56
$begingroup$
There is still a mistake. You wrote $17$ rather than $17left(3right)$. $fleft(3right)ne37$.
$endgroup$
– Jake
Mar 26 at 0:57
1
$begingroup$
It looks as if $f$ is increasing with respect to $n$, so if $mle n$, then $fleft(mright)le fleft(nright)$. So, $$fleft(nright)=fleft(leftlfloorfrac{n}{2}rightrfloorright)+fleft(leftlfloorfrac{n}{3}rightrfloorright)+17nle fleft(frac{n}{2}right)+fleft(frac{n}{3}right)+17n.$$ You can use this to find an upper-bound for the right-hand side of that inequality.
$endgroup$
– Jake
Mar 26 at 1:11
$begingroup$
$f(6n)=f(3n)+f(2n)+102n$. I wrote a short program to compute $lfloor f(n)/nrfloor$ and found that between $n=5times10^9$ and $n=1times 10^{10}$ the value hangs at $101$ making the $102n$ look very interesting.
$endgroup$
– John Wayland Bales
Mar 26 at 3:35