euler's constant

qweretyq · Riemann Hypothesis Offline Joined: 06 Jul 2004 Posts: 273

euler's constant

**Posted:** Thu Mar 17, 2005 1:03 am

nr1337

There's two of them:

[URL=http://mathworld.wolfram.com/Euler-MascheroniConstant.html]The Euler-Mascheroni Constant :approx: 0.5772...[/URL]
e :approx: 2.718281828459045

**Posted:** Thu Mar 17, 2005 1:11 am

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#3

Euler has the distinction of being (maybe?) the only person with two constants named after him. The first, $e$ , is
$e=\lim_{x\to\infty}\left(1+1/x\right)^x\approx 2.718281828$

The second, also known as the 'Euler-Mascheroni Constant' and commonly symbolized by $\gamma$ is
$\gamma=\lim_{x\to\infty}\left(\sum_{k=1}^x {\frac{1}k} - \ln x\right)\approx0.577215665$

**Posted:** Thu Mar 17, 2005 1:16 am

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Nomendil · New Member Offline Joined: 10 Mar 2005 Posts: 11

#4

In my experience, $e$ is generally referred to as the "euler number" rather than as his constant.

**Posted:** Thu Mar 17, 2005 12:06 pm

Magnara

#5

Mascheroni sounds more like a noodle than a constant.

And I've usually seen $e$ referred to as the exponential constant. That's what the $e$ stands for, not "Euler."

**Posted:** Thu Mar 17, 2005 11:57 pm

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Celeborn · Hodge Conjecture Offline Joined: 09 Apr 2004 Posts: 55

#6

**Posted:** Mon Mar 21, 2005 2:32 am

billzhao

#7

Actually, $e$ is more often known as the Napier's constant.

**Posted:** Mon Mar 21, 2005 3:11 am

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ComplexZeta

#8

I hadn't heard $e$ referred to as Napier's constant before, but if Napier deserves his name on any constant, it would be $1/e$ rather than $e$ itself. (His logarithms were essentially in base $1/e$ , except he used 10 million for $\infty$ .)

**Posted:** Mon Mar 21, 2005 8:02 am

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problem_solver · P versus NP Offline Joined: 18 Jan 2007 Posts: 35

#9

**Posted:** Thu Mar 01, 2007 12:31 am

Treething

#10

well...

$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\log(n)$

is really the same as

$(\frac12-\log\frac21)+(\frac13-\log\frac32)+...+(\frac1n-\log\frac{n}{n-1})$

just speculating, the mathworld link probably has more in-depth and more correct information but i'm too lazy to decipher math language into english

**Posted:** Thu Mar 01, 2007 5:01 am

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paladin8

#11

$\sum_{k=1}^{x}\frac{1}{k}< 1+\int_{1}^{x}\frac{dt}{t}$ so we know the limit of the difference is less than $1$ (and also that it is positive by a similar argument).

**Posted:** Thu Mar 01, 2007 7:59 am

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