Mixing
Is n O(n)? Is n Ω(n)?
$begingroup$
I have a homework assignment (though this isn't part of it!) which I want to be sure on. This may be a stupid question.
The functions in question are $f(n) = 2^n$ and $g(n) = 3^n$. I'm pretty sure about the following:
$f$ is $O(g)$ as $2^n leq 3^n forall n in mathbb{N}$, using $c = 1$.
$f$ is also $Omega(g)$. Proof:
$f$ being $Omega(g)$ means that for some $c > 0$ we have that $c cdot 2^n geq 3^n$ for sufficiently large $n$.
Taking $log_3$ of both sides gives us $log_3(c cdot 2^n) geq n$.
We can use change-of-base to get: $frac{log_2(c cdot 2^n)}{log_2(3)} geq n$.
Log rules give us $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$.
This shows that $f$ is $Omega(g)$ if $n$ is $Omega(n)$, with some $c$ equal to $frac{1}{log_2(3)}$.
If there are any problems, please let me know.
Thanks!
logarithms asymptotics
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
$endgroup$
|
show 5 more comments
$begingroup$
I have a homework assignment (though this isn't part of it!) which I want to be sure on. This may be a stupid question.
The functions in question are $f(n) = 2^n$ and $g(n) = 3^n$. I'm pretty sure about the following:
$f$ is $O(g)$ as $2^n leq 3^n forall n in mathbb{N}$, using $c = 1$.
$f$ is also $Omega(g)$. Proof:
$f$ being $Omega(g)$ means that for some $c > 0$ we have that $c cdot 2^n geq 3^n$ for sufficiently large $n$.
Taking $log_3$ of both sides gives us $log_3(c cdot 2^n) geq n$.
We can use change-of-base to get: $frac{log_2(c cdot 2^n)}{log_2(3)} geq n$.
Log rules give us $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$.
This shows that $f$ is $Omega(g)$ if $n$ is $Omega(n)$, with some $c$ equal to $frac{1}{log_2(3)}$.
If there are any problems, please let me know.
Thanks!
logarithms asymptotics
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
$endgroup$
2
$begingroup$
You don't switch the inequality sign when taking logs....
$endgroup$
– mathworker21
Jan 10 at 4:03
$begingroup$
@mathworker21 you're right - changed
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
Your last inequality implies this is only true for $nlefrac{log_2c}{log_23-1}=k$, so whatever value of $c$ you chose, it will not be true for $n>k$
$endgroup$
– Shubham Johri
Jan 10 at 4:09
$begingroup$
@ShubhamJohri could you explain that further, please?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:11
$begingroup$
@mathworker21 Thanks. Realized that that wasn't the main problem (and deleted the comment). But now that is the problem.
$endgroup$
– DirkGently
Jan 10 at 4:13
|
show 5 more comments
$begingroup$
I have a homework assignment (though this isn't part of it!) which I want to be sure on. This may be a stupid question.
The functions in question are $f(n) = 2^n$ and $g(n) = 3^n$. I'm pretty sure about the following:
$f$ is $O(g)$ as $2^n leq 3^n forall n in mathbb{N}$, using $c = 1$.
$f$ is also $Omega(g)$. Proof:
$f$ being $Omega(g)$ means that for some $c > 0$ we have that $c cdot 2^n geq 3^n$ for sufficiently large $n$.
Taking $log_3$ of both sides gives us $log_3(c cdot 2^n) geq n$.
We can use change-of-base to get: $frac{log_2(c cdot 2^n)}{log_2(3)} geq n$.
Log rules give us $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$.
This shows that $f$ is $Omega(g)$ if $n$ is $Omega(n)$, with some $c$ equal to $frac{1}{log_2(3)}$.
If there are any problems, please let me know.
Thanks!
logarithms asymptotics
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
$endgroup$
I have a homework assignment (though this isn't part of it!) which I want to be sure on. This may be a stupid question.
The functions in question are $f(n) = 2^n$ and $g(n) = 3^n$. I'm pretty sure about the following:
$f$ is $O(g)$ as $2^n leq 3^n forall n in mathbb{N}$, using $c = 1$.
$f$ is also $Omega(g)$. Proof:
$f$ being $Omega(g)$ means that for some $c > 0$ we have that $c cdot 2^n geq 3^n$ for sufficiently large $n$.
Taking $log_3$ of both sides gives us $log_3(c cdot 2^n) geq n$.
We can use change-of-base to get: $frac{log_2(c cdot 2^n)}{log_2(3)} geq n$.
Log rules give us $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$.
This shows that $f$ is $Omega(g)$ if $n$ is $Omega(n)$, with some $c$ equal to $frac{1}{log_2(3)}$.
If there are any problems, please let me know.
Thanks!
logarithms asymptotics
logarithms asymptotics
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
edited Jan 10 at 4:07
FibroMyAlgebra
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
asked Jan 10 at 3:59
FibroMyAlgebraFibroMyAlgebra
184
184
2
$begingroup$
You don't switch the inequality sign when taking logs....
$endgroup$
– mathworker21
Jan 10 at 4:03
$begingroup$
@mathworker21 you're right - changed
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
Your last inequality implies this is only true for $nlefrac{log_2c}{log_23-1}=k$, so whatever value of $c$ you chose, it will not be true for $n>k$
$endgroup$
– Shubham Johri
Jan 10 at 4:09
$begingroup$
@ShubhamJohri could you explain that further, please?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:11
$begingroup$
@mathworker21 Thanks. Realized that that wasn't the main problem (and deleted the comment). But now that is the problem.
$endgroup$
– DirkGently
Jan 10 at 4:13
|
show 5 more comments
2
$begingroup$
You don't switch the inequality sign when taking logs....
$endgroup$
– mathworker21
Jan 10 at 4:03
$begingroup$
@mathworker21 you're right - changed
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
Your last inequality implies this is only true for $nlefrac{log_2c}{log_23-1}=k$, so whatever value of $c$ you chose, it will not be true for $n>k$
$endgroup$
– Shubham Johri
Jan 10 at 4:09
$begingroup$
@ShubhamJohri could you explain that further, please?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:11
$begingroup$
@mathworker21 Thanks. Realized that that wasn't the main problem (and deleted the comment). But now that is the problem.
$endgroup$
– DirkGently
Jan 10 at 4:13
2
2
$begingroup$
You don't switch the inequality sign when taking logs....
$endgroup$
– mathworker21
Jan 10 at 4:03
$begingroup$
You don't switch the inequality sign when taking logs....
$endgroup$
– mathworker21
Jan 10 at 4:03
$begingroup$
@mathworker21 you're right - changed
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
@mathworker21 you're right - changed
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
Your last inequality implies this is only true for $nlefrac{log_2c}{log_23-1}=k$, so whatever value of $c$ you chose, it will not be true for $n>k$
$endgroup$
– Shubham Johri
Jan 10 at 4:09
$begingroup$
Your last inequality implies this is only true for $nlefrac{log_2c}{log_23-1}=k$, so whatever value of $c$ you chose, it will not be true for $n>k$
$endgroup$
– Shubham Johri
Jan 10 at 4:09
$begingroup$
@ShubhamJohri could you explain that further, please?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:11
$begingroup$
@ShubhamJohri could you explain that further, please?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:11
$begingroup$
@mathworker21 Thanks. Realized that that wasn't the main problem (and deleted the comment). But now that is the problem.
$endgroup$
– DirkGently
Jan 10 at 4:13
$begingroup$
@mathworker21 Thanks. Realized that that wasn't the main problem (and deleted the comment). But now that is the problem.
$endgroup$
– DirkGently
Jan 10 at 4:13
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
You have correctly (as far as I can see from a quick read) rewritten the condition to
$$frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$$
But then you need to argue that you can make this true for all sufficiently large $n$ just by choosing $c$ right -- and that is not the case.
The factor $frac{1}{log_2 3}$ is less than $1$, so the difference between the two terms involving $n$ gets ever larger the larger $n$ is. Therefore, no matter what you take $c$ to be, this difference will eventually be more than the constant $frac{log_2c}{log_23}$, and therefore your rewritten inequality does not hold for all large enough $n$ -- also no matter what you take "large enough" to mean.
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
There we go. Thanks! I really appreciate the help. I'm pretty new to the wonderful world of asymptotics.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
Simpler approach: $$frac{c 2^n}{3^n} to 0$$ no matter what $c$ is, so $2^n$ is not $Omega(3^n)$.
Using your approach:
Because $log_2(3) > 1$ we will always have $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} le n$ for all sufficiently large $n$, no matter what $c$ is.
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
he's probably asking what the flaw in his logic is
$endgroup$
– mathworker21
Jan 10 at 4:04
$begingroup$
Yes. Could you explain further?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
@angryavian Sorry, did I mess up my ineq signs or did you mess up yours? I think you mean $geq$ there as $f = Ω(g)$ means: $exists c > 0$, $n_0 geq 0$ such that $g(n) leq cf(n)$ for all $n geq n_0$.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:10
$begingroup$
@FibroMyAlgebra From your work, you showed that you want to find a $c$ such that $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} ge n$ for all large $n$. In my answer I note that this is impossible, since the left-hand side is a line (as a function of $n$ with a smaller slope than the slope of the right-hand side.
$endgroup$
– angryavian
Jan 10 at 4:14
$begingroup$
Ah, I see. @henningmakholm also answered this. Thanks!
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
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%2f3068218%2fis-n-on-is-n-%25ce%25a9n%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
$begingroup$
You have correctly (as far as I can see from a quick read) rewritten the condition to
$$frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$$
But then you need to argue that you can make this true for all sufficiently large $n$ just by choosing $c$ right -- and that is not the case.
The factor $frac{1}{log_2 3}$ is less than $1$, so the difference between the two terms involving $n$ gets ever larger the larger $n$ is. Therefore, no matter what you take $c$ to be, this difference will eventually be more than the constant $frac{log_2c}{log_23}$, and therefore your rewritten inequality does not hold for all large enough $n$ -- also no matter what you take "large enough" to mean.
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
There we go. Thanks! I really appreciate the help. I'm pretty new to the wonderful world of asymptotics.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
You have correctly (as far as I can see from a quick read) rewritten the condition to
$$frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$$
But then you need to argue that you can make this true for all sufficiently large $n$ just by choosing $c$ right -- and that is not the case.
The factor $frac{1}{log_2 3}$ is less than $1$, so the difference between the two terms involving $n$ gets ever larger the larger $n$ is. Therefore, no matter what you take $c$ to be, this difference will eventually be more than the constant $frac{log_2c}{log_23}$, and therefore your rewritten inequality does not hold for all large enough $n$ -- also no matter what you take "large enough" to mean.
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
There we go. Thanks! I really appreciate the help. I'm pretty new to the wonderful world of asymptotics.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
You have correctly (as far as I can see from a quick read) rewritten the condition to
$$frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$$
But then you need to argue that you can make this true for all sufficiently large $n$ just by choosing $c$ right -- and that is not the case.
The factor $frac{1}{log_2 3}$ is less than $1$, so the difference between the two terms involving $n$ gets ever larger the larger $n$ is. Therefore, no matter what you take $c$ to be, this difference will eventually be more than the constant $frac{log_2c}{log_23}$, and therefore your rewritten inequality does not hold for all large enough $n$ -- also no matter what you take "large enough" to mean.
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
$endgroup$
You have correctly (as far as I can see from a quick read) rewritten the condition to
$$frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} geq n$$
But then you need to argue that you can make this true for all sufficiently large $n$ just by choosing $c$ right -- and that is not the case.
The factor $frac{1}{log_2 3}$ is less than $1$, so the difference between the two terms involving $n$ gets ever larger the larger $n$ is. Therefore, no matter what you take $c$ to be, this difference will eventually be more than the constant $frac{log_2c}{log_23}$, and therefore your rewritten inequality does not hold for all large enough $n$ -- also no matter what you take "large enough" to mean.
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
edited Jan 10 at 4:17
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
answered Jan 10 at 4:13
Henning MakholmHenning Makholm
240k17305541
240k17305541
$begingroup$
There we go. Thanks! I really appreciate the help. I'm pretty new to the wonderful world of asymptotics.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
There we go. Thanks! I really appreciate the help. I'm pretty new to the wonderful world of asymptotics.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
$begingroup$
There we go. Thanks! I really appreciate the help. I'm pretty new to the wonderful world of asymptotics.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
$begingroup$
There we go. Thanks! I really appreciate the help. I'm pretty new to the wonderful world of asymptotics.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
Simpler approach: $$frac{c 2^n}{3^n} to 0$$ no matter what $c$ is, so $2^n$ is not $Omega(3^n)$.
Using your approach:
Because $log_2(3) > 1$ we will always have $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} le n$ for all sufficiently large $n$, no matter what $c$ is.
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
he's probably asking what the flaw in his logic is
$endgroup$
– mathworker21
Jan 10 at 4:04
$begingroup$
Yes. Could you explain further?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
@angryavian Sorry, did I mess up my ineq signs or did you mess up yours? I think you mean $geq$ there as $f = Ω(g)$ means: $exists c > 0$, $n_0 geq 0$ such that $g(n) leq cf(n)$ for all $n geq n_0$.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:10
$begingroup$
@FibroMyAlgebra From your work, you showed that you want to find a $c$ such that $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} ge n$ for all large $n$. In my answer I note that this is impossible, since the left-hand side is a line (as a function of $n$ with a smaller slope than the slope of the right-hand side.
$endgroup$
– angryavian
Jan 10 at 4:14
$begingroup$
Ah, I see. @henningmakholm also answered this. Thanks!
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
Simpler approach: $$frac{c 2^n}{3^n} to 0$$ no matter what $c$ is, so $2^n$ is not $Omega(3^n)$.
Using your approach:
Because $log_2(3) > 1$ we will always have $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} le n$ for all sufficiently large $n$, no matter what $c$ is.
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
he's probably asking what the flaw in his logic is
$endgroup$
– mathworker21
Jan 10 at 4:04
$begingroup$
Yes. Could you explain further?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
@angryavian Sorry, did I mess up my ineq signs or did you mess up yours? I think you mean $geq$ there as $f = Ω(g)$ means: $exists c > 0$, $n_0 geq 0$ such that $g(n) leq cf(n)$ for all $n geq n_0$.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:10
$begingroup$
@FibroMyAlgebra From your work, you showed that you want to find a $c$ such that $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} ge n$ for all large $n$. In my answer I note that this is impossible, since the left-hand side is a line (as a function of $n$ with a smaller slope than the slope of the right-hand side.
$endgroup$
– angryavian
Jan 10 at 4:14
$begingroup$
Ah, I see. @henningmakholm also answered this. Thanks!
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
Simpler approach: $$frac{c 2^n}{3^n} to 0$$ no matter what $c$ is, so $2^n$ is not $Omega(3^n)$.
Using your approach:
Because $log_2(3) > 1$ we will always have $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} le n$ for all sufficiently large $n$, no matter what $c$ is.
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
$endgroup$
Simpler approach: $$frac{c 2^n}{3^n} to 0$$ no matter what $c$ is, so $2^n$ is not $Omega(3^n)$.
Using your approach:
Because $log_2(3) > 1$ we will always have $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} le n$ for all sufficiently large $n$, no matter what $c$ is.
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
edited Jan 10 at 4:06
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
answered Jan 10 at 4:04
angryavianangryavian
40.7k23380
40.7k23380
$begingroup$
he's probably asking what the flaw in his logic is
$endgroup$
– mathworker21
Jan 10 at 4:04
$begingroup$
Yes. Could you explain further?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
@angryavian Sorry, did I mess up my ineq signs or did you mess up yours? I think you mean $geq$ there as $f = Ω(g)$ means: $exists c > 0$, $n_0 geq 0$ such that $g(n) leq cf(n)$ for all $n geq n_0$.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:10
$begingroup$
@FibroMyAlgebra From your work, you showed that you want to find a $c$ such that $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} ge n$ for all large $n$. In my answer I note that this is impossible, since the left-hand side is a line (as a function of $n$ with a smaller slope than the slope of the right-hand side.
$endgroup$
– angryavian
Jan 10 at 4:14
$begingroup$
Ah, I see. @henningmakholm also answered this. Thanks!
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
add a comment |
$begingroup$
he's probably asking what the flaw in his logic is
$endgroup$
– mathworker21
Jan 10 at 4:04
$begingroup$
Yes. Could you explain further?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
@angryavian Sorry, did I mess up my ineq signs or did you mess up yours? I think you mean $geq$ there as $f = Ω(g)$ means: $exists c > 0$, $n_0 geq 0$ such that $g(n) leq cf(n)$ for all $n geq n_0$.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:10
$begingroup$
@FibroMyAlgebra From your work, you showed that you want to find a $c$ such that $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} ge n$ for all large $n$. In my answer I note that this is impossible, since the left-hand side is a line (as a function of $n$ with a smaller slope than the slope of the right-hand side.
$endgroup$
– angryavian
Jan 10 at 4:14
$begingroup$
Ah, I see. @henningmakholm also answered this. Thanks!
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
$begingroup$
he's probably asking what the flaw in his logic is
$endgroup$
– mathworker21
Jan 10 at 4:04
$begingroup$
he's probably asking what the flaw in his logic is
$endgroup$
– mathworker21
Jan 10 at 4:04
$begingroup$
Yes. Could you explain further?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
Yes. Could you explain further?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
@angryavian Sorry, did I mess up my ineq signs or did you mess up yours? I think you mean $geq$ there as $f = Ω(g)$ means: $exists c > 0$, $n_0 geq 0$ such that $g(n) leq cf(n)$ for all $n geq n_0$.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:10
$begingroup$
@angryavian Sorry, did I mess up my ineq signs or did you mess up yours? I think you mean $geq$ there as $f = Ω(g)$ means: $exists c > 0$, $n_0 geq 0$ such that $g(n) leq cf(n)$ for all $n geq n_0$.
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:10
$begingroup$
@FibroMyAlgebra From your work, you showed that you want to find a $c$ such that $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} ge n$ for all large $n$. In my answer I note that this is impossible, since the left-hand side is a line (as a function of $n$ with a smaller slope than the slope of the right-hand side.
$endgroup$
– angryavian
Jan 10 at 4:14
$begingroup$
@FibroMyAlgebra From your work, you showed that you want to find a $c$ such that $frac{log_2(c)}{log_2(3)} + frac{n}{log_2(3)} ge n$ for all large $n$. In my answer I note that this is impossible, since the left-hand side is a line (as a function of $n$ with a smaller slope than the slope of the right-hand side.
$endgroup$
– angryavian
Jan 10 at 4:14
$begingroup$
Ah, I see. @henningmakholm also answered this. Thanks!
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
$begingroup$
Ah, I see. @henningmakholm also answered this. Thanks!
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:15
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%2f3068218%2fis-n-on-is-n-%25ce%25a9n%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
2
$begingroup$
You don't switch the inequality sign when taking logs....
$endgroup$
– mathworker21
Jan 10 at 4:03
$begingroup$
@mathworker21 you're right - changed
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:05
$begingroup$
Your last inequality implies this is only true for $nlefrac{log_2c}{log_23-1}=k$, so whatever value of $c$ you chose, it will not be true for $n>k$
$endgroup$
– Shubham Johri
Jan 10 at 4:09
$begingroup$
@ShubhamJohri could you explain that further, please?
$endgroup$
– FibroMyAlgebra
Jan 10 at 4:11
$begingroup$
@mathworker21 Thanks. Realized that that wasn't the main problem (and deleted the comment). But now that is the problem.
$endgroup$
– DirkGently
Jan 10 at 4:13