$\ln(x)$ and Big O notation

Question

I have tried to assert that $\ln(x)=O(x^0)$ a few times, but it seems fairly obvious that this statement should be false, and so I've been faced with some rightful speculation.

My reason is that

$$\ln(x)=\lim_{k\to0}\frac{x^k-1}{k}=O(x^0)?$$

But if not, then I ask

What is the smallest value of $k$ for $\ln(x)=O(x^k)?$ A proof would be nice, and possibly some cleanuping if my notation is used incorrectly.

I couldn't figure out if there was a tag for O notation. If there is one, please add it. — Simply Beautiful Art, May 22 '16 at 12:51
$\ln(x)$ is not $O(1)$ because it diverges. However it is $O(x)$, so there's your answer. However, this is not strict; $\ln(x)$ is called a sub-polynomial function because its asymptotic behavior is strictly slower than any non-constant polynomial — Alex G., May 22 '16 at 12:57
$x^0=1$ so claiming something to be $O(x^0)$ is to claim that it's bounded by a constant, $\ln x$ clearly isn't. — Henrik supports the community, May 22 '16 at 12:57
No, that doesn't make sense. If anything, that would be saying $\ln(x)$ is $O(1)$, which it isn't — Alex G., May 22 '16 at 12:59
@AlexG. I've also tried looking up sub-polynomial function, but it turns up pretty much nothing. — Simply Beautiful Art, May 22 '16 at 13:03
1....$x^0=1$.....2.Are you asking about the asymptotic behavior of $\ln x$ as $x\to 0$ or as $x\to \infty$? — DanielWainfleet, May 22 '16 at 13:03
@user254665 Both. There is Clement's answer and another, $O(x^k)$ as $x\to\infty$ and $k\to0$. — Simply Beautiful Art, May 22 '16 at 13:28
For some context, see http://math.stackexchange.com/q/1793655/. Let us hope that users on this page have more success getting the message through, than two users on the other page. — Did, May 22 '16 at 13:31

score 3 · Accepted Answer · answered May 22 '16 at 13:00

3

The first point is: when you use asymptotic notations, you need to specify with regard to what point the asymptotics is taken. Here, it looks like this is when $x\to 0^+$; note that this could have equally been $x\to\infty$, so specifying it is required.

Now, you have that for any fixed $\alpha > 0$, $$ \ln x = O(x^{-\alpha}) $$ when $x\to 0^+$, since $\left\lvert x^\alpha \ln x\right\rvert\xrightarrow[x\to0^+]{} 0$. Yet, this does not "go to the limit": there is no minimum $\alpha\geq 0$ for which this holds.

Parenthesis: you have $\ln x = x^{\frac{\ln\ln x}{\ln x}}$, so you could technically write $\ln x = \Theta(x^{\frac{\ln\ln x}{\ln x}})$... but note that this means "taking $\alpha = \alpha(x) = \frac{\ln\ln x}{\ln x}$."

answered May 22 '16 at 13:00

Clement C.

67,323

What about $O(x^\alpha)$ as $x\to\infty$ and $\alpha\to0?$ – Simply Beautiful Art May 22 '16 at 13:30
@Clement For some context: http://math.stackexchange.com/q/1793655/. Good luck... :-) – Did May 22 '16 at 13:31
You can do the same sort of think when x tends to infinity. But if you want asymptotics with regard to TWO parameters at the same time, you have to start defining what you mean by the O () notation then. – Clement C. May 22 '16 at 13:32
@Did Oh. ${}{}{} $ – Clement C. May 22 '16 at 13:34
1

@ClementC. "Oh", in Bachmann-Landau's sense, I guess... – Did May 22 '16 at 13:35
@Did my comment for the O was for the OP. I am not sure i am (nor that he is) clear on taking the asymptotics according to two parameters at the same time, here. – Clement C. May 22 '16 at 13:37
@ClementC. That is ok, your answer was insightful nonetheless. Can we compare them separately at least? – Simply Beautiful Art May 22 '16 at 13:40
Well, for fixed x, when alpha goes to zero you get constant vs. constant. The log is a constant wrt alpha, and x^alpha goes to 1. – Clement C. May 22 '16 at 13:43
1

@Did also, just got the joke. – Clement C. May 22 '16 at 16:45

score 1 · Answer 2 · answered May 22 '16 at 15:08

1

For $r>0$ take $M>1$ such that $1/M<r.$ For $y>M$ we have $$0<\ln y=\int_1^y (1/z)\;dz= \int_1^M(1/z)\;dz+\int_M^y(1/z)\;dz=$$ $$=\ln M+\int_M^y(1/z)\;dz<\ln M+\int_M^y(1/M)\;dz=\ln M+(y-M)/M.$$ $$\text { So, }\quad 0<(\ln y)/y<(\ln M)/y+(1-M/y)/M.$$ Therefore $$0\leq \sup_{x\geq y}\;(\ln x)/x\leq \sup_{x\geq y}(\;(\ln M)/x+(1-M/y)/M\;)=$$ $$=(\ln M)/y+(1-M/y)/M.$$ $$\text {Therefore, }\quad 0\leq \lim_{y\to \infty}\sup_{x\geq y}\;(\ln x)/x\leq$$ $$\leq \lim_{y\to \infty}\;(\ln M)/y+(1-M/y)/M=1/M<r.$$ Since $r$ can be as small as desired, we have therefore $\lim_{y\to \infty}(\ln y)/y=0.$

For positive $a,b$ we have $$(\ln y)^a/y^b=((\ln y)/y^{b/a})^a=(((a/b)\ln y^{b/a})/y^{b/a})^a.$$ Let $z=y^{b/a}.$ Then $z\to \infty$ as $y\to \infty$ and we have $$(\ln y)^b/y^a=((a/b)^a\cdot ((\ln z)/z))^a\to 0$$ as $y$ (and $z$) go to $\infty.$

In particular with $a=1$ and $b>0$ we have $\lim_{y\to \infty}(\ln y)/y^b=0.$ So as $y\to \infty$ we have $\ln y=O(y^b)$ for any $b>0. $

Actually we have $\ln y=o(y^b)$ as $y\to \infty$ for any $b>0$ because $\lim_{y\to \infty}(\ln y)/y^b=0.$

For $0<y<1$and $y\to 0$ let $x=1/y.$ Then $|\ln y|/y=x\ln x$ and use the above inequalities.

answered May 22 '16 at 15:08

DanielWainfleet

57,985

:/ I feel like you avoided the question on whether or not I can say $\ln y=\lim_{b\to0}O(y^b)$ as $y\to\infty$. But thank you on the proofs. – Simply Beautiful Art May 22 '16 at 15:12
1

I dk what is meant by $\lim_{b\to 0} O(y^b)$. It does not appear in your Q. – DanielWainfleet May 23 '16 at 16:02
It is implied by the equations in my Q, for if not $\ln x=O(x^0)$, then at least my Q implies $\ln(x)=\lim_{k\to0}O(x^k)$ – Simply Beautiful Art May 23 '16 at 20:07
@SimpleArt The OP is rightfully asking you the meaning of the (stunning) assertion: $\ln x=\lim_{k\to0}O(x^k)$. Why not provide this meaning? (Personally I couldn't, but surely you have a precise idea in mind when you continue to write these horrors after several people mentioned they are problematic...) – Did May 23 '16 at 20:27
@Did Probably because I am no longer sure how to put it in any meaningful or "correct" way, if one were to exist. – Simply Beautiful Art May 23 '16 at 20:29
@SimpleArt But 28 minutes ago, you stated as a matter of fact that your question implies this statement... Are you a troll? – Did May 23 '16 at 20:37
@Did One does not take math as seriously as I and be a troll. I am a high school student who hasn't even taken Calculus, (and if you search hard, this statement has been repeated) to be honest, I just lack the experience you have. I'm just... unsure of how to approach the community sometimes, and this is a question I'd like to have a solid and satisfactory answer on, that's all. – Simply Beautiful Art May 23 '16 at 20:42
@SimpleArt Then repeating ad nauseam the absurd statement that $\ln x=\lim_{k\to0}O(x^k)$ is not the way to go. Listening to what people are explaining to you might be more fruitful. – Did May 23 '16 at 20:44
@SimpleArt "Let us continue this discussion in chat." Are you kidding me? No thanks. – Did May 23 '16 at 20:45
@Did Oops, my bad. And don't worry, I don't plan on making that statement anymore. At most I'll probably say $\ln(x)=O(x^k)$ for $k>0$ – Simply Beautiful Art May 23 '16 at 20:45
@Did http://math.stackexchange.com/q/1795502 A troll does not spend 3 hours on one question so that the OP understands it. And I'm sorry, but I don't feel this is no longer comment appropriate. – Simply Beautiful Art May 23 '16 at 20:47
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My difficulty with your notation is that $O(x^k)$ is not a function or even a set of functions, but an abbreviation for a condition.... $\ln x=O(x^k)$ as $x\to \infty$ for any $k>0,$ but $\ln x$ is unbounded as $x\to \infty$.... And $\ln y=O(y^{-k})$ as $y\to 0$ for any $k>0$ but $\ln y$ is unbounded as $y\to 0$. – DanielWainfleet May 24 '16 at 01:24

$\ln(x)$ and Big O notation

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