What is the Big-Ω of the following function?












2












$begingroup$


For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
leq sum_{n=1}^{2n}x^2+x^2
$$

$$
= sum_{n=1}^{2n}2x^2
$$

$$
leq 2n ( 2(2n)^2)
$$

$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.










share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago
















2












$begingroup$


For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
leq sum_{n=1}^{2n}x^2+x^2
$$

$$
= sum_{n=1}^{2n}2x^2
$$

$$
leq 2n ( 2(2n)^2)
$$

$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.










share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago














2












2








2





$begingroup$


For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
leq sum_{n=1}^{2n}x^2+x^2
$$

$$
= sum_{n=1}^{2n}2x^2
$$

$$
leq 2n ( 2(2n)^2)
$$

$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.










share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




For the following function:



$$
sum_{n=1}^{2n}x+x^2
$$



It is easy to see the (tightest) Big-Oh is $O(n^3)$, but I am not so sure about the Big-Omega. Here is my attempt:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
geq sum_{n=1}^{2n}x^2
$$



But not sure how to continue. Also, does tightest Big-Ω always equal tightest Big-Oh?



EDIT: The way I worked out the Big-Oh is shown below:



$$
sum_{n=1}^{2n}x+x^2
$$

$$
leq sum_{n=1}^{2n}x^2+x^2
$$

$$
= sum_{n=1}^{2n}2x^2
$$

$$
leq 2n ( 2(2n)^2)
$$

$$
= 8n^3
$$



Therefore the Big-Oh is $O(n^3)$.







time-complexity asymptotics runtime-analysis






share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







TigerHix













New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









TigerHixTigerHix

134




134




New contributor




TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






TigerHix is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago














  • 1




    $begingroup$
    Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
    $endgroup$
    – Apass.Jack
    4 hours ago










  • $begingroup$
    @Apass.Jack Updated the question.
    $endgroup$
    – TigerHix
    2 hours ago








1




1




$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
4 hours ago




$begingroup$
Please show in the question how "to see the (tightest) Big-$O$ is $O(n^3)$".
$endgroup$
– Apass.Jack
4 hours ago












$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
2 hours ago




$begingroup$
@Apass.Jack Updated the question.
$endgroup$
– TigerHix
2 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.





Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    20 mins ago













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: "419"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});






TigerHix is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f102874%2fwhat-is-the-big-%25ce%25a9-of-the-following-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.





Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    20 mins ago


















2












$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.





Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    20 mins ago
















2












2








2





$begingroup$

$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.





Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$






share|cite|improve this answer











$endgroup$



$$begin{align}
sum_{x=1}^{2n}(x+x^2) gt sum_{x=1}^{2n}x^2 gt sum_{x=n+1}^{2n}x^2gtsum_{x=n+1}^{2n}n^2= n^3
end{align}
$$



So the function is $Omega(n^3)$. You have shown it is $O(n^3)$. So the function is $Theta(n^3)$.



Although $n^3$, the asymptotic lower bound ignoring a constant factor is the same as the asymptotic upper bound ignoring a constant factor in the current case, it is not always true because some functions fluctuate. For example,
$$f(n)=begin{cases}1 quad text {when }ntext{ is odd,}\ n quad text{when }ntext{ is even.}end{cases}$$ $f(n)=O(n)$ and $f(n)=Omega(1)$. Both bounds are tightest.





Exercise. Show that $$sum_{x=1}^{m}(x+x^2)=frac{m(m+1)(m+2)}{3}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









Apass.JackApass.Jack

8,1121633




8,1121633












  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    20 mins ago




















  • $begingroup$
    Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
    $endgroup$
    – TigerHix
    2 hours ago










  • $begingroup$
    That clears it up, thanks!
    $endgroup$
    – TigerHix
    1 hour ago






  • 1




    $begingroup$
    @TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
    $endgroup$
    – Draconis
    20 mins ago


















$begingroup$
Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
$endgroup$
– TigerHix
2 hours ago




$begingroup$
Thank you for the detailed answer! Could you also answer the other question: does the tightest Big-Ω always equal the tightest Big-Oh?
$endgroup$
– TigerHix
2 hours ago












$begingroup$
That clears it up, thanks!
$endgroup$
– TigerHix
1 hour ago




$begingroup$
That clears it up, thanks!
$endgroup$
– TigerHix
1 hour ago




1




1




$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
20 mins ago






$begingroup$
@TigerHix The fact that $O$ and $Omega$ can be different, btw, is why $Theta$ exists.
$endgroup$
– Draconis
20 mins ago












TigerHix is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















TigerHix is a new contributor. Be nice, and check out our Code of Conduct.













TigerHix is a new contributor. Be nice, and check out our Code of Conduct.












TigerHix is a new contributor. Be nice, and check out our Code of Conduct.
















Thanks for contributing an answer to Computer Science 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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f102874%2fwhat-is-the-big-%25ce%25a9-of-the-following-function%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Knooppunt Holsloot

Altaar (religie)

Gregoriusmis