0
$\begingroup$

I was wondering what the expansion series of the function

$$ f(x) = -\frac{1}{x^3} \cdot \frac{1}{\Gamma(x) \cdot \Gamma(-(\exp(\frac{2}{3}\pi\cdot i))x) \cdot \Gamma(-(\exp(\frac{4}{3}\pi \cdot i))x)} $$ is, at $x = 0$. I'm also interested in the method behind computing the series expansion.

You might think: "why don't you just paste this equation in wolframalpha and then see what the expansion series is?" Well, I did exactly that, but wolframalpha couldn't compute it! Does that mean the expansion series doesn't exist or is it more likely that it's just too 'hard' for the computational knowledge engine to find the series?

Thanks,

Max

EDIT For more information: this function is function (27) at this page, when n=3. I offer my apologies for not stating the question well at first.

  • 0
    The Maclaurin series for the reciprocal gamma function is given [here](http://dlmf.nist.gov/5.7); *Higher Transcendental Functions* by Erdelyi *et al.* should have details on the derivation.2010-11-01
  • 0
    @ J.M. thanks. Please notice I made a mistake in the question and it's different now. I hope you can answer that one, too.2010-11-01
  • 0
    @Max: Do you really want the expansion at x = 0 ? To use (27) to evaluate the infinite products one employs the expansion at x = 1.2010-11-01
  • 0
    @ Mister Dubuque: I think I want the expansion at x = 0. In any case, I don't wish to re-evaluate the results found by Prudnikov et al. . I would like to use the series expansion to find closed form expressions of some infinite (sum) series.2010-11-01
  • 1
    @Max: It would be helpful tell us more about the series you are trying to evaluate.2010-11-01
  • 0
    @ Bill Dubuque: I will tell you all in the next question.2010-11-01

2 Answers 2

2

Mathematica can handle (rather easily) that function. For example, the expansion at x=0 to order 5.

$f(x) = -1-2 \gamma x-2 \gamma ^2 x^2+\displaystyle \frac{1}{6} x^3 \left(-8 \gamma ^3-\psi ^{(2)}(1)\right)-$

$-\displaystyle\frac{1}{3} x^4 \left(\gamma \left(2 \gamma ^3+\psi ^{(2)}(1)\right)\right)+\frac{1}{60} x^5 \left(-16 \gamma ^5-20 \gamma ^2 \psi ^{(2)}(1)+\psi ^{(4)}(1)\right)+O\left(x^6\right)$

(same notation than the one used in Mariano's post).

  • 0
    No problem, Max :-)2010-11-01
2

Up to a constant and a power of $x$, your function is $\frac1{x\Gamma(x)}$. Mathematica tells me that the Taylor series at zero is $$ 1+\gamma x+\left(\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) x^2+\frac{1}{12} x^3 \left(2 \gamma ^3-\gamma \pi ^2-2 \psi ^{(2)}(1)\right)+\frac{x^4 \left(60 \gamma ^4-60 \gamma ^2 \pi ^2+\pi ^4-240 \gamma \psi ^{(2)}(1)\right)}{1440}+\dots$$

($\gamma$ is Euler's constant, and the $\psi$'s are poly-gamma functions.)

  • 0
    @ Mariano Suarez-Alvarez: Thanks a lot, but wolframlpha could tell me this, too! I'm very interested in how to find the series expansion of the particular function I stated, and how it's found.2010-11-01
  • 2
    Well, you clearly stated that W|A couldn't compute the series...2010-11-01
  • 0
    yes, it couldn't compute the series I mentioned, but it can compute f(x)=1/(x*Gamma(x)) . There's a difference. You mention that "up to a constant and a power of x, your function is..." That constant and that power of x is important, because because it tells me exactly what term I should use to solve a particular problem and what the exact value of that term is.2010-11-01
  • 0
    The powers of $x$ are *not* important: they "shift" the series. And the constant, well, it is a constant! It just gets multiplied to each coefficient of my function to get those of yours.2010-11-01
  • 1
    But when alpha (or Mathematica) can't do something, it often helps to take it apart. The last two gamma functions in the denominator are constants and alpha can do them separately. The extra terms of x in the denominator can be divided out afterward.2010-11-01
  • 0
    To be more explicit... you divide the series for the reciprocal gamma by an appropriate power of x. Since the reciprocal gamma is entire, the series for this divided by an integer power of x will necessarily have $x^{-j}$ terms.2010-11-01
  • 0
    @ Ross Millikan: No they are not constants (any more...). My mistake.2010-11-01
  • 0
    @ Ross Millikan and JM: Thanks, I will keep that in mind.2010-11-01
  • 0
    @ Mariano Suarez-Alvarez: Ok.. so the powers of x 'shift'. Does this mean that $\gamma$x becomes $\gamma$x^4 in 'my' function ?2010-11-01