5 typo (op-correction); added 4 characters in body

[Background:] Looking at the powerseries for the gamma-function

$\Gamma(1+x) = 1 + a_1 x + a_2 x^2 - a_3 * x^3 + ...$

then we can arrive at a decomposition

$\Gamma(1+x) = r(x) + g(x)$

where g(x) is constructed by the sum of the (taylor-expansions of) geometric series

          1      1      1   1     1   1
g(x) = ---  - --- +  ---*--- - ---*--- + ...
1+x    2+x    3+x  2!   4+x  3!

[edit:corrected a sign-error]


and from this the powerseries for r(x)

$r(x) = \Gamma(1+x) - g(x)$

[end background]

That function r(x) begins with

$r(x) = 1/e + 0.21938 * x + 0.09784 *x^2 + \ldots$

The function has then some nice properties. By heuristics and inspection of its powerseries it seems for instance, that

• apparently it is entire, has no zero except that $lim_{x-> \infty} r(-x) = 0$

• $r(0) = 1/e$ where $e = \exp(1)$

• $r(k) = r(k-1)*k + 1/e$

Just today I found, that in fact this is the incomplete gamma- function as defined/implemented in mathematica as "gamma(1+x,1)". But this may not be of concern here, because I want to understand how to think the other way round:

• Question 1:

Assume we had only the functional relation and the initial value

$r(x) = r(x-1)*x +1/e$
$r(0) = 1/e$

what else would we need to make r(x) unique and arrive at the solution

$r(x) = \Gamma(1+x)-g(x)$

?

• Question 2:

Is there any way to generalize that construction scheme to get some function f(x) where the functional equation depends on a constant parameter c =/= 0

$f(x) = f(x-1)*x + c$

For instance let $c=1/2$ . What would a -for instance convex - function $f(x)$ look like?

[update]: Question 2 seems to be easy - at the integer x f(x) is simply a scaling of r(x) by c and e: $f(x)= r(x)*c*e$ so I'd assume the same can be assumed for fractional x.

4 sign-error corrected

[Background:] Looking at the powerseries for the gamma-function

$\Gamma(1+x) = 1 + a_1 x + a_2 x^2 - a_3 * x^3 + ...$

then we can arrive at a decomposition

$\Gamma(1+x) = r(x) + g(x)$

where g(x) is constructed by the sum of the (taylor-expansions of) geometric series

          1      1      1   1     1   1
g(x) = ---  + --- - ---*--- -- +  ---*--- - ---*--- + ...
1+x    2+x    3+x  2!   4+x  3!

[edit:corrected a sign-error]


and from this the powerseries for r(x)

$r(x) = \Gamma(1+x) - g(x)$

[end background]

That function r(x) begins with

$r(x) = 1/e + 0.21938 * x + 0.09784 *x^2 + \ldots$

The function has then some nice properties. By heuristics and inspection of its powerseries it seems for instance, that

• apparently it is entire, has no zero except that $lim_{x-> \infty} r(-x) = 0$

• $r(0) = 1/e$ where $e = \exp(1)$

• $r(k) = r(k-1)*k + 1/e$

Just today I found, that in fact this is the incomplete gamma- function as defined/implemented in mathematica as "gamma(1+x,1)". But this may not be of concern here, because I want to understand how to think the other way round:

• Question 1:

Assume we had only the functional relation and the initial value

$r(x) = r(x-1)*x +1/e$
$r(0) = 1/e$

what else would we need to make r(x) unique and arrive at the solution

$r(x) = \Gamma(1+x)-g(x)$

?

• Question 2:

Is there any way to generalize that construction scheme to get some function f(x) where the functional equation depends on a constant parameter c =/= 0

$f(x) = f(x-1)*x + c$

For instance let $c=1/2$ . What would a -for instance convex - function $f(x)$ look like?

3 minor edits

Background:

[Background:] Looking at the powerseries for the gamma-function

$\Gamma(1+x) = 1 + a_1 x + a_2 x^2 - a_3 * x^3 + ...$

then we can arrive at a decomposition

$\Gamma(1+x) = r(x) + g(x)$

where g(x) is constructed by the sum of the (taylor-expansions of) geometric series

          1      1      1   1     1   1
g(x) = ---  + --- -  ---*--- + ---*--- - ...
1+x    2+x    3+x  2!   4+x  3!


and from this the powerseries for r(x)

$r(x) = \Gamma(1+x) - g(x)$

[end background]

That function r(x) begins with

$r(x) = 1/e + 0.21938 * x + 0.09784 *x^2 + \ldots$

The function has then some nice properties. By heuristics and inspection of its powerseries it seems for instance, that

• apparently it is entire, has no zero except that $lim_{x-> \infty} r(-x) = 0$

• $r(0) = 1/e$ where $e = \exp(1)$

• $r(k) = r(k-1)*k + 1/e$

Just today I found, that in fact this is the incomplete gamma- function as defined/implemented in mathematica as "gamma(x,1)". gamma(1+x,1)". But this may not be of concern here, because I want to understand , how to think the other way round:

• Question 1:

Assume we had only the functional relation and the initial value

$r(x) = r(x-1)*x +1/e$
$r(0) = 1/e$

what else would we need to make r(x) unique and arrive at the solution

$r(x) = \Gamma(1+x)-g(x)$

?

• Question 2:

Is there any way to generalize that construction scheme to get some function f(x) where the functional equation depends on a constant parameter c =/= 0

$f(x) = f(x-1)*x + c$

For instance let $c=1/2$ . What would a -for instance convex - function $f(x)$ look like?

2 made property of function r(x) more precise
1