April 28 Information

Let ${\displaystyle \Gamma _{a}(x+a)=x\cdot \Gamma _{a}(x).}$ Then for ${\displaystyle a=1}$ we have ${\displaystyle \Gamma _{_{1}}(x)=\Gamma (x),}$ whose integral expression is ${\displaystyle \int _{0}^{\infty }{\frac {t^{x-1}}{e^{t}}}dt.}$ I was wondering whether such expressions also exist for this generalized version of the ${\displaystyle \Gamma }$ function. — 79.113.194.139 ( talk) 04:30, 28 April 2014 (UTC) reply

Yes, it's just ${\displaystyle \Gamma _{a}(x)=\int _{0}^{\infty }{\frac {t^{x-a}}{e^{t}}}dt.}$ Widener ( talk) 08:27, 28 April 2014 (UTC) reply

I think you answered the question that was literally asked, but perhaps not the one that was intended. My guess is that the OP meant to write

${\displaystyle \Gamma _{a}(x+a)=x\cdot \Gamma _{a}(x)}$

How about it, 79.113? Did I guess right? -- Trovatore ( talk) 09:15, 28 April 2014 (UTC) reply

Yes. Sorry. — 79.113.194.139 ( talk) 09:44, 28 April 2014 (UTC) reply

Then that recursive relationship doesn't uniquely define ${\displaystyle \Gamma _{a}}$. What's the boundary condition? The gamma function was motivated to coincide with the factorial. What is the motivation behind ${\displaystyle \Gamma _{a}}$? 203.45.159.248 ( talk) 09:59, 28 April 2014 (UTC) reply

Assuming you also want ${\displaystyle \Gamma _{a}(0)=1}$, then ${\displaystyle \Gamma _{a}(x)=a^{x/a}\Gamma (x/a)}$. Sławomir Biały ( talk) 11:31, 28 April 2014 (UTC) reply

Wow! Unbelievable as always, Slawomir! :-) If you could you also please explain the logic and/or intuition which helped you arrive at this expression ? Thanks! — 86.125.209.133 ( talk) 17:56, 28 April 2014 (UTC) reply

From the equation ${\displaystyle \Gamma _{a}(x+a)=x\Gamma _{a}(x)}$, I see that the solution should be something like ${\displaystyle \Gamma (x/a)}$. This doesn't quite work, so I try to write ${\displaystyle \Gamma _{a}(x)=f(x)\Gamma (x/a)}$. Imposing the functional equation again gives ${\displaystyle f(x+a)=af(x)}$ which has ${\displaystyle f(x)=a^{x/a}}$ as a solution. (Clearly there will be many functions solving this; I'm not sure what conditions are needed to ensure uniqueness.) Sławomir Biały ( talk) 22:02, 28 April 2014 (UTC) reply

Resolved

The Cube is a Space-filling Solid...

The Dodecahedron is NOT!!!

Four Dodecahedrons have a empty Corner: 0xyz with 1,5^o

Could it be a SPACE with Space-filling Dodecahedrons???...

Could it be a SPACE where the Icosahedron = 20 Tetrahedrons???...

I can Imagine them BUT could they be Calculated???...

Are these "Mistakes" say something about our Universe???...

THANK you VERY-VERY much!!!...

"Have a nice Day/Night..."

SPYROY Kosta - Greece - Honeycomp ( talk) 14:36, 28 April 2014 (UTC) reply

In hyperbolic space, you can tile appropriately scaled dodecahedra. Their dihedral angles vary according to their size, so you can make them have dihedral angles of precisely 90° ( order-4 dodecahedral honeycomb), 72° ( order-5 dodecahedral honeycomb), 60° ( order-6 dodecahedral honeycomb), ... , 0° ( infinite-order dodecahedral honeycomb). In elliptic space, you can also tile appropriately scaled dodecahedra 3 ({5,3,3}, 120-cell) or 2 ({5,3,2}, dodecahedral dichoron, each dodecahedron takes up a 3-hemisphere) at a corner. Double sharp ( talk) 14:47, 28 April 2014 (UTC) reply

Though if the order is more than 6 the vertices stick out beyond infinity. And as to {5,3,2}, how do you distinguish it from a great sphere? — Tamfang ( talk) 03:13, 29 April 2014 (UTC) reply

It's analogous to the pentagonal dihedron {5,2}: the faces of the dodecahedra in {5,3,2} tile a great sphere, just as the sides of the pentagons in {5,2} tile a great circle. Double sharp ( talk) 15:24, 29 April 2014 (UTC) reply

I don't think this says anything significant about our universe - Euclidean 3-space is only a local approximation of its actual geometry. There's only one space-filling fully regular honeycomb in 3-space - the cube, which has three sets of parallel faces. So if you want to consider the philosophical impact of this, I guess the question is "Why cubes?" AlexTiefling ( talk) 15:57, 28 April 2014 (UTC) reply

You are probably thinking of the rhombic dodecahedron. This is well-known to fill 3-space, and can be thought of as the 3-space analogy of the regular hexagon (which fills (tessellates) 2-space), though it is not regular. RomanSpa ( talk) 19:11, 28 April 2014 (UTC) reply

It's sufficiently clear, in my humble opinion, that that's not what he has in mind. — Tamfang ( talk) 03:13, 29 April 2014 (UTC) reply

Here are two views of a curved space filled with regular dodecahedra [1] [2] — and two views of a space where 20 tetrahedra form an icosahedron [3] [4] — Tamfang ( talk) 06:08, 29 April 2014 (UTC) reply