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I want to show that the limit of the geometric mean of primes less than or equal to $x$ is $e$ as $x \to \infty$. Is this correct?

Using the product law of logarithms we have $$\ln \prod\limits_{p \leqslant x} p = \sum\limits_{p \leqslant x} {\ln (p)}$$ but $$\sum\limits_{p \leqslant x} {\ln (p)} = \vartheta (x) \sim x$$ hence $$\prod\limits_{p \leqslant x} p \sim {e^x}$$ and the geometric mean is $${\left( {\prod\limits_{p \leqslant x} p } \right)^{1/x}} \sim e.$$

Edit: I realized that this product does not represent the geometric mean. What does it represent?

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You are not calculating the geometric mean. The geometric mean would involve taking the $1/\pi(x)$ power, not the $1/x$ power. That is, the number of primes would need to be involved. The geometric mean of these primes is certainly an increasing function of $x$. –  Matthew Conroy Nov 6 '12 at 19:40
    
@MatthewConroy I am not sure what this is called, but it looks similar to the geometric mean. –  glebovg Nov 6 '12 at 19:46
    
Which step(s) in your argument are you unsure of? –  Matthew Conroy Nov 6 '12 at 19:52
    
Are my arguments valid and what does ${\left( {\prod\nolimits_{p \leqslant x} p } \right)^{1/x}}$ represent? –  glebovg Nov 6 '12 at 21:18
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2 Answers 2

up vote 4 down vote accepted

The geometric mean of the collection of primes numbers less than $x$ can be written:

$$G(x)=\left( \prod_{p \le x} p\right)^{\frac{1}{\pi(x)}}$$ $$\log(G(x))=\frac{1}{\pi(x)}\sum_{p \le x}\log(p)=\frac{\vartheta(x)}{\pi(x)}$$

For large $x$, we can use the PNT and an asymptotic for the first Chebyshev function given here:

$$\log(G(x)) \sim \frac{\log(x)}{x} \cdot x= \log(x)$$

Thus, $G(x) \sim x$. Here's a plot:

Plot of geometric mean of primes less than $x$ for $0 \le x \le 10,000$.

A best fit line reveals $G(x) \approx .31473x$. So I think the more interesting question to ask is what is $\displaystyle \lim_{x \rightarrow \infty} \frac{G(x)}{x}$? Can it be expressed in terms of known constants?

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This is fascinating because the arithmetic mean of all primes is asymptotically $x/2$, so the geometric mean is asymptotically $e$ to the power of the arithmetic mean. Truth amazing. Thanks. –  glebovg Nov 6 '12 at 21:15
    
What does ${\left( {\prod\nolimits_{p \leqslant x} p } \right)^{1/x}}$ represent? –  glebovg Nov 6 '12 at 21:17
    
"Amazing"...and truly wrong. Because $e^{x/2}$ exceeds $x$ for all positive $x$, this answer claims that an average of values all of which are less than $x$ actually exceeds $x$! (The source of the error appears to be failure to divide $s(n)$ by $n$ in the linked answer.) –  whuber Nov 6 '12 at 22:09
    
I used the wrong asymptotic. I'm summing the $\log$ of primes up to $x$, not the sum of primes up to $x$. If you can provide an appropriate asymptotic for the sum of the log of primes then this should be solvable. –  Jackson Walters Nov 6 '12 at 23:23
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$\log G(x)\sim\log x$ does not imply $G(x)\sim x$ but only $G(x)=x^{1+o(1)}.$ –  Charles Apr 24 '13 at 19:15
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I am surprised the answer to this query is no where on the web?

Taking logs, you want to take $S(N)=\sum_{p\le N}\log p$ [the product] and then quotient this by $T(N)=\sum_{p\le N} 1$ [the number of primes] before reexponentiating for the final result. By partial summation, we have

$$S(N)=T(N)\log N-\int_2^N T(u){du\over u}$$

and the integral is $N/\log(N)$ to first order, for $T(u)\sim u/\log(u)$ by PNT. So ${S(N)\over T(N)}$ is $(\log N)-1$, and exponentiating gets you $N/e$ as the answer.

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