How was this derived? I know why it is equal to (n-1)! and why it satisfies the condition x * G(x) = G(x+1), but I don't know what process was used to discover the function in the first place. How do you start out trying a curve to fit the factorial function and end up here?
>>8180085
Iva also wanted to know this for a while
pick a small n (0,1,2)
integrate it by hand
see the pattern?
>yes
you are done
>no
increase n by one, integrate again
>>8180085
Yea I'm pretty curious about this as well
Bumping for science
>>8180085
if you try to find a function satisfying [math]f(n+1)=nf(n)[\math] then you'll get to the gamma function very smoothly... it is natural to search for something written in the form exp times polynomial, so it is easy to integrate by parts and takes the value 0 at both 0 and infinity). then all you have to do is tune the parameters.
the real "genius trick" was to search for the functional equation [math]f(n+1)=nf(n)[\math] satisfied by the factorial and NOTdirectly for the factorial itself.
>>8180239
sry for bad editing, here's the same with tex
if you try to find a function satisfying [math]f(n+1)=nf(n)[/math] then you'll get to the gamma function very smoothly... it is natural to search for something written in the form exp times polynomial, so it is easy to integrate by parts and takes the value 0 at both 0 and infinity). then all you have to do is tune the parameters.
the real "genius trick" was to search for the functional equation [math]f(n+1)=nf(n)[/math] satisfied by the factorial and NOTdirectly for the factorial itself.
>>8180240
I imagine they just discovered it by coincidence to be honest.
It's a pretty simple integral. It must have come up early in the study of calculus.
You basically just had to be Euler, honestly (this is the case for a lot of Gauss and Euler's work), at least to put it all together from scratch. Anyone can show it works, but coming up with it is harder.
>>8180661
>Anyone can show it works,
I can't.
>>8180880
Then you're no-one.
