Gamma Function and the Euler-Mascheroni Constant

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SUMMARY

The discussion centers on the relationship between the gamma function and the Euler-Mascheroni constant, specifically through the product expansion of the multiplicative inverse. The user derived a unique equation involving the digamma function and its logarithmic derivative, leading to the conclusion that the limit as s approaches zero of the logarithm of the gamma function relates directly to the Euler-Mascheroni constant. The conversation also touches on the concept of extending products over continuous intervals, referencing Bigeometric Calculus and the work of Volterra.

PREREQUISITES
  • Understanding of the gamma function and its properties
  • Familiarity with the digamma function and its logarithmic derivative
  • Knowledge of limits and l'Hôpital's rule
  • Basic concepts of Bigeometric Calculus and product integrals
NEXT STEPS
  • Explore the properties of the digamma function and its applications
  • Research the derivation and implications of the Euler-Mascheroni constant
  • Study Bigeometric Calculus and its historical context in mathematics
  • Investigate product integrals and their geometric interpretations
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Mathematicians, students of real analysis, and anyone interested in advanced calculus concepts, particularly those exploring the gamma function and its applications in mathematical analysis.

epr1990
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I was taking a break from studying from my real analysis, electrodynamics, and nuclear physics exams this week, and, being a math-phile, I decided to play around with the gamma-function for some reason. Anyway, I used the common product expansion of the multiplicative inverse, and I arrived at a very simple but interesting relation to the exact value of the Euler-Mascheroni Constant. I searched the internet and couldn't seem to find anything similar to what I did. So, I was wondering if anyone else has seen anything like the result that I have attached as a .pdf file.


Thanks in advance!
 

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Now that I think about it, doesn't it basically just follow from l'Hospitals rule and

\Gamma^\prime(1) = \Psi(1) = \gamma
 
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I didn't even think to look at digamma, but it seems that as usual, you are indeed correct. It follows straight from a series representation for it. Basically, the last line in my analysis is almost exactly this

ψ(x+1)=-γ+∑(1/k-1/(x+k))
 
Actually, it follows immediately if you change the LHS of the last line to the equivalent

-(log(gamma(1+s))-log(gamma(1))/s

since log(gamma(1))=log(1)=0 and the digamma function ψ is defined to be the logarithmic derivative of the gamma function, so that under the limit as s ->0 this is precisely -ψ(1). Thanks again!
 
Still, it's a pretty neat equation that I haven't seen before. I'm happy you've made this thread!
 
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Yea I think so too. This is a little bit off topic, but, it was also my reason for looking at the properties of the gamma function. Have you seen any theory on extending a product over a continuous interval, as is done with the sum to create the integral?

I have tried to develop an approach analogous to that of the limiting case of a Riemann sum to no avail. I have no idea what this would describe geometrically, but that's just another obstacle. Anyway, I have never seen or heard of anyone doing anything like that, but I would think that you should be able to construct such a thing for any binary operation defined on a complete and compact set. Any guidance would be greatly appreciated.
 
Just in case anyone wanted to know, I found a lot about it. It's called Bigeometric (or Multiplicative) Calculus.
 
The so called product integral was developed in 1887 by Volterra according to http://en.wikipedia.org/wiki/Product_integral

And here is the page for multiplicative calculus, just in case anyone is interested:
http://en.wikipedia.org/wiki/Multiplicative_calculus
 

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