9

The Question

I noticed that the Gamma function has been expanded to the complex numbers.. and that an expansion to the dual numbers requires very little effort.. but now, I'm curios.. how about the split-complex numbers?

About The Gamma Function

To those unaware, the Gamma function, $Γ(x)$, extends the factorial function to the real numbers, maintaining the property of $Γ(x)/x = Γ(x-1)$.

About The Split-Complex Numbers

The split-complex numbers are essentially the "opposite" of the complex numbers, as in, they introduce a constant, $j$, where $j^2 = 1$ but $j \neq \pm1$. The square of the absolute value can be found via $|a+bj|^2 = a^2 - b^2$, and through this, any number with an absolute value of $0$, is a zero divisor, meaning dividing by it essentially acts like dividing by zero. $1+j$ is one of those zero divisors. The version of Euler's formula for these numbers acts as follows: $e^{nj} = \cosh(n) + j\sinh(n)$. Some split-complex numbers also square to themselves, complicating things just a bit.

The Goal

Create a generalized formula for the $Γ(x)$ extending it to the split-complex numbers, with special cases where needed. It is preferable that $Γ(a+0j)$ remains equal to $Γ(a)$ in nearly every scenario, unless branches form that introduce other values.

WawaWeegee
  • 129
  • 8

2 Answers2

14

Yes. Any real function can be extended to split-complex numbers. This is because split-complex numbers are isomorphic to $\mathbb{R}^2$, in other words, pairs of real numbers.

In other words,

$a+bj \cong (a-b,a+b)$

So, for a real function $f(x)%$, its extension to split-complex numbers is as follows: $f(a+bj)\cong (f(a-b),f(a+b))\cong \frac12 (f(a+b)+f(a-b))+\frac{j}2 (f(a+b)-f(a-b))$.

So, for Gamma function (like any other) we have:

$\Gamma(a+bj)=\frac12 (\Gamma(a+b)+\Gamma(a-b))+\frac{j}2 (\Gamma(a+b)-\Gamma(a-b))$

Anixx
  • 10,161
  • 3
    Is this consistent with just putting a split complex number directly into the taylor series definition of the gamma function? – Sidharth Ghoshal Jun 29 '25 at 20:53
  • In this sense any real function can be extended to the complex numbers as well? – Jannik Pitt Jun 30 '25 at 09:14
  • @JannikPitt extending to complex numbers is much more difficult and complicated. A function can gain non-trivial properties, like Zeta function. – Anixx Jun 30 '25 at 09:26
  • @JannikPitt to split-complex numbers we can extent any real function, not necessarily analytic or differentiable, but for complex numbers there is no obvious way to extend non-analytic functions. For comparison, to extend a funtion to dual numbers it should be differentiable, so it is somewhat in between complex and split-complex numbers in terms of complicatedness of extending functions. – Anixx Jun 30 '25 at 10:18
  • 2
    @SidharthGhoshal The isomorphism mentioned in this answer is a ring isomorphism, so it is indeed consistent. It's markedly different from $\mathbb C$, which is isomorphic to $\mathbb R^2$ only as an additive group and has a different multiplicative structure. – Carmeister Jun 30 '25 at 13:02
  • 1
    The isomorphism with $\mathbb R^2$ is given by $(1,0)=\tfrac{1-j}2,;(0,1)=\tfrac{1+j}2$. After this change of basis, we have simply $\Gamma((x,y))=(\Gamma(x),\Gamma(y))$. – mr_e_man Jun 30 '25 at 19:29
11

$$\Gamma(a+bj)=\int_{0}^{\infty}t^{a-1+bj}e^{-t}\,\mathrm dt=\int_{0}^{\infty}t^{a-1}e^{-t}e^{bj\ln t}\,\mathrm dt,$$ $$e^{bj\ln t}=\sum_{k=0}^{\infty}\frac{(b\ln t)^k j^k}{k!}=\sum_{k=0}^{\infty}\frac{(b\ln t)^{2k+1} j^{2k+1}}{(2k+1)!}+\frac{(b\ln t)^{2k} j^{2k}}{(2k)!}\\=\sum_{k=0}^{\infty}\frac{(b\ln t)^{2k+1} j}{(2k+1)!}+\frac{(b\ln t)^{2k}}{(2k)!}=\cosh(b\ln t)+j\sinh(b\ln t)$$ $$\implies \Gamma(a+bj)=\int_{0}^{\infty}t^{a-1}e^{-t}\left(\cosh(b\ln t)+j\sinh(b\ln t)\right)\,\mathrm dt$$

Edit: Using the formula $$\cosh x=\frac{e^x+e^{-x}}{2},\sinh x=\frac{e^x-e^{-x}}{2}$$ $$\implies \Gamma(a+bj)=\frac{1}{2}\int_{0}^{\infty}\left(t^{a+b-1}+t^{a-b-1}\right)e^{-t}+j\left(t^{a+b-1}-t^{a+b-1}\right)\,\mathrm dt\\=\frac{1}{2}(\Gamma(a+b)+\Gamma(a-b))+\frac{j}{2}(\Gamma(a+b)-\Gamma(a-b))$$ Which matches @Anixx's answer

Thinh Dinh
  • 8,233