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

Can anyone deduce the sum of this series? The reason I ask is because I made it and showed it to my professor who was impressed but was concerned as to whether it is original. I figured that a good test would be to see if anyone is able to say how it is derived.

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shouldn't the first term be 4, so one can have a uniform pattern? –  user31280 Oct 26 '12 at 22:34
    
@F'OlaYinka: I think it's OK as it is. –  joriki Oct 26 '12 at 22:34
    
It's equal to $4^{\frac{1}{x}-1} \cos\big[\frac{\pi}{x}\big]$ –  Alex Oct 26 '12 at 22:35
    
is the series well defined this way? $\displaystyle \cfrac 14 + \sum_{k\ge1} (-1)^{k+1}\cfrac { \prod_{n=1}^{2k-1} (nx -4)} {(2k)! \space x^{2k}}$ –  user31280 Oct 26 '12 at 22:40
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My dear friend Mathematica knew it :) –  Alex Oct 26 '12 at 22:46

1 Answer 1

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 $\cos[\frac{\pi}{x}]$? –  Alex Oct 26 '12 at 22:52
    
@Alex: Sorry, yes, I lost the division somewhere along the way -- thanks, fixed. –  joriki Oct 26 '12 at 22:53
    
Wow you guys are good. Thanks for that. So it's not original then? –  KingChem Oct 26 '12 at 23:03
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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... –  joriki Oct 26 '12 at 23:07
    
@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!)+⋯ –  KingChem Oct 27 '12 at 11:31

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