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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.

Max Lonysa Muller
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2 Answers2

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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.)

Mariano Suárez-Álvarez
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  • @ 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. – Max Lonysa Muller Nov 01 '10 at 14:37
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    Well, you clearly stated that W|A couldn't compute the series... – Mariano Suárez-Álvarez Nov 01 '10 at 14:45
  • 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. – Max Lonysa Muller Nov 01 '10 at 14:52
  • 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. – Mariano Suárez-Álvarez Nov 01 '10 at 14:55
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    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. – Ross Millikan Nov 01 '10 at 14:57
  • 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. – J. M. ain't a mathematician Nov 01 '10 at 14:58
  • @ Ross Millikan: No they are not constants (any more...). My mistake. – Max Lonysa Muller Nov 01 '10 at 15:08
  • @ Ross Millikan and JM: Thanks, I will keep that in mind. – Max Lonysa Muller Nov 01 '10 at 15:30
  • @ Mariano Suarez-Alvarez: Ok.. so the powers of x 'shift'. Does this mean that $\gamma$x becomes $\gamma$x^4 in 'my' function ? – Max Lonysa Muller Nov 01 '10 at 15:50
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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).

r_31415
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