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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.

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The Maclaurin series for the reciprocal gamma function is given here; Higher Transcendental Functions by Erdelyi et al. should have details on the derivation. –  J. M. Nov 1 '10 at 14:54
    
@ 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. –  Max Muller Nov 1 '10 at 16:00
    
@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. –  Bill Dubuque Nov 1 '10 at 16:41
    
@ 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. –  Max Muller Nov 1 '10 at 16:59
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@Max: It would be helpful tell us more about the series you are trying to evaluate. –  Bill Dubuque Nov 1 '10 at 17:18
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2 Answers 2

up vote 2 down vote accepted

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).

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Thanks a lot, Robert. –  Max Muller Nov 1 '10 at 18:10
    
No problem, Max :-) –  Robert Smith Nov 1 '10 at 18:27
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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.)

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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 Muller Nov 1 '10 at 14:37
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Well, you clearly stated that W|A couldn't compute the series... –  Mariano Suárez-Alvarez Nov 1 '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 Muller Nov 1 '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-Alvarez Nov 1 '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 1 '10 at 14:57
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