3
$\begingroup$

$$\frac14+\frac{x-4}{2!x^2}-\frac{(x-4)(2x-4)(3x-4)}{4!x^4}+\frac{(x-4)(2x-4)(3x-4)(4x-4)(5x-4)}{6!x^6}\mp\ldots$$

Can anyone deduce the sum of this series?

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    shouldn't the first term be 4, so one can have a uniform pattern?2012-10-26
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    @F'OlaYinka: I think it's OK as it is.2012-10-26
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    It's equal to $4^{\frac{1}{x}-1} \cos\big[\frac{\pi}{x}\big]$2012-10-26
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    is the series well defined this way? $\displaystyle \cfrac 14 + \sum_{k\ge1} (-1)^{k+1}\cfrac { \prod_{n=1}^{2k-1} (nx -4)} {(2k)! \space x^{2k}}$2012-10-26
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    @F'OlaYinka $\prod_{n=1}^{2k-1}(nx-4)$ :)2012-10-26
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    @J.G. Thanks! fixed it.2012-10-26
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    Ok Alex is right. How did you know Alex?2012-10-26
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    My dear friend Mathematica knew it :)2012-10-26
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    Maple doesn't seem to :(2012-10-26
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    Hehe you cheated lol. Stupid Wolfram didn't though surprisingly. But does that mean it's not original?2012-10-26

1 Answers 1

8

This is

$$ -\frac1x\sum_{k=0}^\infty\binom{2k-\frac4x}{2k}\frac1{2k-\frac4x}(-1)^k\;. $$

With

$$ s(q,n):=\sum_{k=0}^\infty\binom{k+n}k\frac{q^k}{k+n}=\frac{(1-q)^{-n}}n\;, $$

we have

$$ \begin{align} -\frac1x\sum_{k=0}^\infty\binom{2k-\frac4x}{2k}\frac1{2k-\frac4x}(-1)^k &=-\frac1{2x}\left(s\left(\mathrm i,-\frac4x\right)+s\left(-\mathrm i,-\frac4x\right)\right) \\ &=-\frac1{2x}\left(\frac{(1-\mathrm i)^{4/x}}{-4/x}+\frac{(1+\mathrm i)^{4/x}}{-4/x}\right) \\ &=\frac18\left((1+\mathrm i)^{4/x}+(1-\mathrm i)^{4/x}\right) \\ &=\frac182^{2/x}\left(\mathrm e^{\mathrm i\pi/x}+\mathrm e^{-\mathrm i\pi/x}\right) \\ &=4^{1/x-1}\cos\frac\pi x\;,\end{align} $$

as Alex rightly stated in a comment.

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    Isn't it $\cos[\frac{\pi}{x}]$?2012-10-26
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    @Alex: Sorry, yes, I lost the division somewhere along the way -- thanks, fixed.2012-10-26
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    Wow you guys are good. Thanks for that. So it's not original then?2012-10-26
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    @KingChem: I'm not sure exactly what you mean by that. If you mean whether anyone ever wrote down that series for that function before, then it may very well be original, since probably noone ever had occasion to do so. The fact that I was able to reconstruct the closed form from the series doesn't mean I knew the series beforehand. But whether it's relevant to anything is another question entirely...2012-10-26
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    @joriki: Yeah that's what I meant. I have infact found a use for it; φ/2^(3/5) =1+4/(5^2 2!)-(4×6×11)/(5^4 4!)+(4×6×11×16×21)/(5^6 6!)-(4×6×11×16×21×26×31)/(5^8 8!)+⋯2012-10-27
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    @KingChem: What's $\phi$?2012-10-27
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    @joriki: The Golden Ratio.2012-10-28