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Let $\sim$ mean if $a \sim b$ then $\lim_{x \to \infty} \frac{a}{b} =1.$

The following is a threshold question. It seems that $x \sim y \implies \pi(x) \sim \pi(y).$

Pf. $\pi(x) \sim \frac{x}{\log x}, \pi(y)\sim \frac{y}{\log y}.$ So $\log y\cdot \pi(y)\sim y , \log x\cdot \pi(x) \sim x,$ and so $\log y \cdot \pi(y)\sim \log x\cdot \pi(x)$ or $\frac{\log y}{\log x} \sim \frac{\pi(x)}{\pi(y)}.$ Since $x\sim y,$ we know that $\log x \sim \log y.$ So $\frac{\log y}{\log x} \sim 1 \sim \frac{\pi(x)}{\pi(y)}$ and finally $\pi(x) \sim \pi(y). $

If this much is true here is the question.

We have that $\pi(x) \sim \frac{x}{\log x}$ or $ \pi(x)\cdot \log x \sim x.$ Then it must be true that

$$\pi[ \pi(x)\cdot \log x] \sim \pi[x] \rightarrow \frac{\pi(x) \log x }{\log(\pi(x) \log x)} \sim \frac{x}{\log x}. $$

and then

$$\pi[ \frac{\pi(x) (\log x)^2 }{\log(\pi(x) \log x)} ] \sim \pi(x).... $$

...and so on?

When we plot some of these the left hand side is an increasingly distant cousin of the right. So maybe the error of the PNT is a factor but does it affect the validity of the process as we compound this process ad infinitum?

Thanks for any insight.

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you say that $a\sim b$ if $\lim_{x\rightarrow\infty} \frac{a}{b} = 1$, are $a$ and $b$ functions of $x$ here? –  Guest 86 Jan 25 '13 at 18:43
    
I think that would be the usual definition. But if we say the $x\sim y$ implies $\log x \sim \log y$ do we use that? –  daniel Jan 25 '13 at 18:46

1 Answer 1

up vote 1 down vote accepted

The asymptotic behavior is enough to guarantee it for a finite amount of repetitions but not for an infinite amount.

I cannot say for $\Pi$ but say you have some other function

$$\mu(x) = x \left(1+\epsilon(x)\right)$$

where $\lim_{x\rightarrow\infty} \epsilon(x) = 0$. It's easy to see that $\mu(x) \sim x$ but:

Applying your process works like $$ \mu(...\mu(\mu(\mu(x)))...) = \mu(...\mu(\mu(x(1+\epsilon(x))))...) \approx_{x\gg 1} \mu(...\mu(x(1+\epsilon(x))^2)...)$$

So that after $k$ applications you get

$$\mu^k(x) \approx_{x\gg 1} x(1+\epsilon(x))^k$$

Clearly for any finite $k$, $\mu^k(x) \sim x$. But as $k\rightarrow\infty$, how fast $\epsilon(x)$ goes to zero becomes an issue.

So that the relation you wrote may hold for a finite number of applications but it has to be justified properly if you wish to make the claim for an infinite number.

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This seems a counter-example to the general case. Does it disprove the case $\mu(x) = \pi(x)?$ –  daniel Jan 25 '13 at 19:08
    
The idea is correct but the implementation was wrong here, sorry. Having a dispersive dependency rather than simply $\Pi(x) = x / \log(x)$, I would assume the accumulated mistake may ruin the relation. I only briefly encountered $\Pi$, I cannot say much about it. But my point is that the requirement of proof is on you, if you're making the claim you did - The move from finite to infinite needs justification. –  Guest 86 Jan 25 '13 at 19:14
    
Yes I think I agree with your last statement. I'm definitely not making the claim! I think yours is a good example of why the claim is suspicious. –  daniel Jan 25 '13 at 19:17
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I think the point is much stronger now. I think error for $\pi(x)$ is $O(x/(\log x)^2))$ so it may be possible to prove/disprove.+1 –  daniel Jan 25 '13 at 19:57
1  
In such a case, after a finite amount of steps $k$ the accumulated error seems to be $O(k x/\log^2 x)$ resulting in the inability to take the limit of $k\rightarrow\infty$ as the rate of growth compared to that of $x$ seems to matter. –  Guest 86 Jan 25 '13 at 22:17

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