Mixing
Proving $3n^2 - 4n in Omega(n^2)$
$begingroup$
Proving $3n^2 - 4n in Omega(n^2)$
Attempt:
$3n^2 - 4n geq cn^2$
$n(3n-4) geq cn^2$
$(3n-4) geq cn$
$3n - 4 - cn geq 0$
$n(3-c) geq 4$
$n geq frac{4}{3-c}$
would do I choose for $n_0$ and $c$ to satisfy this proof?
asymptotics computer-science
edited Jan 24 at 14:01
Namaste
1
asked Jan 24 at 6:26
Bas basBas bas
49012
$endgroup$
add a comment |
$begingroup$
Proving $3n^2 - 4n in Omega(n^2)$
Attempt:
$3n^2 - 4n geq cn^2$
$n(3n-4) geq cn^2$
$(3n-4) geq cn$
$3n - 4 - cn geq 0$
$n(3-c) geq 4$
$n geq frac{4}{3-c}$
would do I choose for $n_0$ and $c$ to satisfy this proof?
asymptotics computer-science
edited Jan 24 at 14:01
Namaste
1
asked Jan 24 at 6:26
Bas basBas bas
49012
$endgroup$
add a comment |
$begingroup$
Proving $3n^2 - 4n in Omega(n^2)$
Attempt:
$3n^2 - 4n geq cn^2$
$n(3n-4) geq cn^2$
$(3n-4) geq cn$
$3n - 4 - cn geq 0$
$n(3-c) geq 4$
$n geq frac{4}{3-c}$
would do I choose for $n_0$ and $c$ to satisfy this proof?
asymptotics computer-science
edited Jan 24 at 14:01
Namaste
1
asked Jan 24 at 6:26
Bas basBas bas
49012
$endgroup$
Proving $3n^2 - 4n in Omega(n^2)$
Attempt:
$3n^2 - 4n geq cn^2$
$n(3n-4) geq cn^2$
$(3n-4) geq cn$
$3n - 4 - cn geq 0$
$n(3-c) geq 4$
$n geq frac{4}{3-c}$
would do I choose for $n_0$ and $c$ to satisfy this proof?
asymptotics computer-science
asymptotics computer-science
edited Jan 24 at 14:01
Namaste
1
asked Jan 24 at 6:26
Bas basBas bas
49012
edited Jan 24 at 14:01
Namaste
1
asked Jan 24 at 6:26
Bas basBas bas
49012
edited Jan 24 at 14:01
Namaste
1
edited Jan 24 at 14:01
Namaste
1
edited Jan 24 at 14:01
Namaste
1
1
asked Jan 24 at 6:26
Bas basBas bas
49012
asked Jan 24 at 6:26
Bas basBas bas
49012
asked Jan 24 at 6:26
Bas basBas bas
49012
49012
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that the final inequality is equivalent to the first one if $c<3$ (you divided by $(3-c)$). Then, according to your work, by taking for example $c=1$ and $n_0=4/(3-c)=2$, it follows that
$$3n^2 - 4n geq cn^2 quadforall ngeq n_0$$
Hence we may conclude that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
$endgroup$
add a comment |
$begingroup$
$3n^2 - 4n in Omega(n^2)$ means:
there are $c>0$ and $n_0 in mathbb N$ such that $3n^2-4n ge cn^2$ for $n ge n_0$.
Since $frac{3n^2-4n}{n^2}= 3-frac{4}{n} to 3$ as $n to infty$, there is $n_0$ such that $frac{3n^2-4n}{n^2} ge 2$ for $n ge n_0$.
Hence $3n^2-4n ge 2n^2$ for $n ge n_0$.
This shows that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
FredFred
48.3k1849
$endgroup$
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
});
}
});
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%2f3085545%2fproving-3n2-4n-in-omegan2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that the final inequality is equivalent to the first one if $c<3$ (you divided by $(3-c)$). Then, according to your work, by taking for example $c=1$ and $n_0=4/(3-c)=2$, it follows that
$$3n^2 - 4n geq cn^2 quadforall ngeq n_0$$
Hence we may conclude that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
$endgroup$
add a comment |
$begingroup$
Note that the final inequality is equivalent to the first one if $c<3$ (you divided by $(3-c)$). Then, according to your work, by taking for example $c=1$ and $n_0=4/(3-c)=2$, it follows that
$$3n^2 - 4n geq cn^2 quadforall ngeq n_0$$
Hence we may conclude that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
$endgroup$
add a comment |
$begingroup$
Note that the final inequality is equivalent to the first one if $c<3$ (you divided by $(3-c)$). Then, according to your work, by taking for example $c=1$ and $n_0=4/(3-c)=2$, it follows that
$$3n^2 - 4n geq cn^2 quadforall ngeq n_0$$
Hence we may conclude that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
$endgroup$
Note that the final inequality is equivalent to the first one if $c<3$ (you divided by $(3-c)$). Then, according to your work, by taking for example $c=1$ and $n_0=4/(3-c)=2$, it follows that
$$3n^2 - 4n geq cn^2 quadforall ngeq n_0$$
Hence we may conclude that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
answered Jan 24 at 6:38
Robert ZRobert Z
101k1069142
101k1069142
add a comment |
add a comment |
$begingroup$
$3n^2 - 4n in Omega(n^2)$ means:
there are $c>0$ and $n_0 in mathbb N$ such that $3n^2-4n ge cn^2$ for $n ge n_0$.
Since $frac{3n^2-4n}{n^2}= 3-frac{4}{n} to 3$ as $n to infty$, there is $n_0$ such that $frac{3n^2-4n}{n^2} ge 2$ for $n ge n_0$.
Hence $3n^2-4n ge 2n^2$ for $n ge n_0$.
This shows that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
FredFred
48.3k1849
$endgroup$
add a comment |
$begingroup$
$3n^2 - 4n in Omega(n^2)$ means:
there are $c>0$ and $n_0 in mathbb N$ such that $3n^2-4n ge cn^2$ for $n ge n_0$.
Since $frac{3n^2-4n}{n^2}= 3-frac{4}{n} to 3$ as $n to infty$, there is $n_0$ such that $frac{3n^2-4n}{n^2} ge 2$ for $n ge n_0$.
Hence $3n^2-4n ge 2n^2$ for $n ge n_0$.
This shows that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
FredFred
48.3k1849
$endgroup$
add a comment |
$begingroup$
$3n^2 - 4n in Omega(n^2)$ means:
there are $c>0$ and $n_0 in mathbb N$ such that $3n^2-4n ge cn^2$ for $n ge n_0$.
Since $frac{3n^2-4n}{n^2}= 3-frac{4}{n} to 3$ as $n to infty$, there is $n_0$ such that $frac{3n^2-4n}{n^2} ge 2$ for $n ge n_0$.
Hence $3n^2-4n ge 2n^2$ for $n ge n_0$.
This shows that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
FredFred
48.3k1849
$endgroup$
$3n^2 - 4n in Omega(n^2)$ means:
there are $c>0$ and $n_0 in mathbb N$ such that $3n^2-4n ge cn^2$ for $n ge n_0$.
Since $frac{3n^2-4n}{n^2}= 3-frac{4}{n} to 3$ as $n to infty$, there is $n_0$ such that $frac{3n^2-4n}{n^2} ge 2$ for $n ge n_0$.
Hence $3n^2-4n ge 2n^2$ for $n ge n_0$.
This shows that $3n^2 - 4n in Omega(n^2)$.
answered Jan 24 at 6:38
FredFred
48.3k1849
answered Jan 24 at 6:38
FredFred
48.3k1849
answered Jan 24 at 6:38
FredFred
48.3k1849
answered Jan 24 at 6:38
FredFred
48.3k1849
48.3k1849
add a comment |
add a comment |
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%2f3085545%2fproving-3n2-4n-in-omegan2%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
