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Does $x \sim y \implies \ln x \sim \ln y$?

I would have thought not, but the following has almost persuaded me otherwise:

Assume $x \sim y.$ Does this imply that $$\tag{1}I = \frac{\int_{1}^{x}\frac{dt}{t}}{\int_{1}^{y}\frac{dt}{t}}\sim 1?$$

WLOG let $(y - x) = d, d > 0.$ At worst the difference of the integrals will be less than $(1/x)(y-x).$

Then $$ I = \frac{\ln x}{\ln x + (y-x)(\frac{1}{x})} = \frac{\ln x }{\ln x + \frac{y}{x} -1}. $$

Since $\frac{y}{x}\sim 1,$ $$ \lim_{x \to \infty} I = 1.$$

Is this correct (and if so is there a better proof)? Thanks.

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Define $\sim$ in this context. –  Thomas Andrews Jan 25 '13 at 17:11
    
$x \sim y$ means $\lim x/y = 1 $ as x gets big. –  daniel Jan 25 '13 at 17:12
    
@daniel Note that for big $x$, $\log 2x \sim \log x$ even though $2x \not\sim x$. –  Erick Wong Jan 25 '13 at 17:13
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@daniel My point is that $\log x \sim \log y$ is much weaker than $x \sim y$ when $x,y$ are large. Logarithms turn bounded relative differences into bounded absolute differences, so your intuition should be that $x \sim y$ does imply $\log x \sim \log y$. –  Erick Wong Jan 25 '13 at 17:46
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Ah. Good point and intuition is the crux of the question. Thanks. –  daniel Jan 25 '13 at 17:56
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3 Answers

up vote 1 down vote accepted

By the definition, $y/x\to 1$ as $x\to\infty$, you get that for any $\epsilon>0$ there is an $N$ such that if $x>N$ $y/x\in (1-\epsilon,1+\epsilon)$, or $y\in ((1-\epsilon)x,(1+\epsilon)x)$.

Taking the log of both sides gives you:

$$\log (1-\epsilon) + \log x < \log y < \log(1+\epsilon) +\log x$$

Dividing by $\log x$, you get, for $x>N$:

$$1+\frac{\log (1-\epsilon)}{\log x} < \frac {\log y}{\log x} < 1 + \frac{\log(1+\epsilon)}{\log x}$$

Now show that this implies your condition.

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Amongst other things, you introduce $d$ in your argument, then never use it. I also don't see how you get absolute equality in $$I = \frac{\ln x}{\ln x + (y-x)(\frac{1}{x})} = \frac{\ln x }{\ln x + \frac{y}{x} -1}$$ Neither equality looks true. –  Thomas Andrews Jan 25 '13 at 17:32
    
But you don't know $y-x>0$. $x\sim y$ does not imply $y>x$. –  Thomas Andrews Jan 25 '13 at 17:38
    
But it is not always true one way the other - for one value of $x$ $y>x$ and another value $y<x$. It's even possible, of course, for $y=x$. –  Thomas Andrews Jan 25 '13 at 17:39
    
OK thanks, I will think about this and appreciate the comments. –  daniel Jan 25 '13 at 17:40
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That's true if $y\to \infty$ (or, more precisely, if $\log(y)$ does not tend to zero). Under this assumption (that you implicitely used in your last formula), we get:

$$\frac{x}{y} \to 1 \implies \log\left(\frac{x}{y}\right)=\log(x)-\log(y)\to 0 \implies \frac{\log(x)}{\log(y)}\to 1$$

BTW, the last implication is not reversible, and the converse is not true.

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I like the economy of this and noting of tacit assumptions. –  daniel Jan 26 '13 at 14:27
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I took the question to mean: if $x(t)/y(t) \to 1$ as $t \to \infty$, does it follow that $\log(x(t))/\log(y(t))\to 1$ also? Then, as leonbloy hints, we can get a counterexample, $$ x(t) = 1+\frac{1}{t},\qquad y(t) = 1+\frac{1}{t^2} . $$ As the other answers show, there is no such counterexample if also $x(t) \to \infty$.

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Appreciate the counter-example. –  daniel Jan 26 '13 at 14:30
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