Estimating a product over primes (brainfreeze) - Page 2

Zeta-Flux · 2009-06-11, 19:44

Quote:

I found the explicit bounds (thanks again!), but I didn't see that.

That O-term is what you get from the explicit bounds.

CRGreathouse · 2009-06-11, 19:54

I see what you mean. I was mistranslating your statement as
¡ $\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log^3 x}\right)\right)$ !
rather than
$\frac{e^{-\gamma}}{\log x}+O\left(\frac{1}{\log^3 x}\right).$

R.D. Silverman · 2009-06-12, 12:33

Quote:

Originally Posted by CRGreathouse

I see what you mean. I was mistranslating your statement as
¡ $\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log^3 x}\right)\right)$ !
rather than
$\frac{e^{-\gamma}}{\log x}+O\left(\frac{1}{\log^3 x}\right).$

But these are the same! Only the implied constants in the big O
estimates differ, and they are not known anyway.

CRGreathouse · 2009-06-12, 13:47

Quote:

Originally Posted by R.D. Silverman

Quote:

Originally Posted by CRGreathouse

I see what you mean. I was mistranslating your statement as
¡ $\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log^3 x}\right)\right)$ !
rather than
$\frac{e^{-\gamma}}{\log x}+O\left(\frac{1}{\log^3 x}\right).$

But these are the same! Only the implied constants in the big O
estimates differ, and they are not known anyway.

$\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log^3 x}\right)\right)=\frac{e^{-\gamma}}{\log x}+O\left(\frac{1}{\log^4 x}\right).$

Zeta-Flux · 2009-06-12, 14:55

Quote:

Originally Posted by R.D. Silverman

But these are the same! Only the implied constants in the big O
estimates differ, and they are not known anyway.

Look at the parentheses. I think you are making the same mistake CRGreathouse initially made.

R.D. Silverman · 2009-06-12, 18:27

Quote:

Originally Posted by CRGreathouse

$\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log^3 x}\right)\right)=\frac{e^{-\gamma}}{\log x}+O\left(\frac{1}{\log^4 x}\right).$

Ooops. I missed the exponent '3'

CRGreathouse · 2010-08-27, 14:31

Dusart 2010 has
$\frac{e^{-\gamma}}{\log x}\left(1-\frac{0.2}{\log^2x}\right)<\prod_{p\le x}1-\frac1p\ <\ \frac{e^{-\gamma}}{\log x}\left(1+\frac{0.2}{\log^2x}\right)$
for x >= 2973.

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2009-06-11, 19:54	#13
CRGreathouse Aug 2006 3·1,993 Posts	I see what you mean. I was mistranslating your statement as ¡ $\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log^3 x}\right)\right)$ ! rather than $\frac{e^{-\gamma}}{\log x}+O\left(\frac{1}{\log^3 x}\right).$

2010-08-27, 14:31	#18
CRGreathouse Aug 2006 1011101011011₂ Posts	Dusart 2010 has $\frac{e^{-\gamma}}{\log x}\left(1-\frac{0.2}{\log^2x}\right)<\prod_{p\le x}1-\frac1p\ <\ \frac{e^{-\gamma}}{\log x}\left(1+\frac{0.2}{\log^2x}\right)$ for x >= 2973.