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Application to Laurent series |
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:<math>\cos z = \prod_{k=0}^{\infty} \left(1 - \frac{z^2}{(k + \frac{1}{2})^2\pi^2}\right).</math>
==Application to Laurent series==
The partial fraction expansion for a function can also be used to find a Laurent series for it by simply replacing the rational functions in the sum with their Laurent series, which are often not difficult to write in closed form. This can also lead to interesting identities if a Laurent series is already known.
Recall that
:<math>\tan(z) = \sum_{k=0}^{\infty} \frac{-2z}{z^2 - (k + \frac{1}{2})^2\pi^2} = \sum_{k=0}^{\infty} \frac{-8z}{4z^2 - (2k + 1)^2\pi^2}</math>
We can expand the summand using a geometric series:
:<math>\frac{-8z}{4z^2 - (2k + 1)^2\pi^2} = \frac{8z}{(2k + 1)^2\pi^2} \frac{1}{1 - (\frac{2z}{(2k + 1)\pi})^2} = \frac{8}{(2k + 1)^2\pi^2}\sum_{n=0}^{\infty} \frac{2^{2n}}{(2k + 1)^{2n}\pi^{2n}} z^{2n + 1}</math>
Substituting back,
:<math>\tan(z) = 2\sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \frac{2^{2n+2}}{(2k + 1)^{2n+2}\pi^{2n+2}} z^{2n + 1},</math>
which shows that the coefficients ''a<sub>n</sub>'' in the Laurent (Taylor) series of ''tan(z)'' about ''z'' = 0 are
:<math>a_{2n+1} = \frac{T_{2n+1}}{(2n+1)!} = \frac{2^{2n+3}}{\pi^{2n+2}} \sum_{k=0}^{\infty} \frac{1}{(2k + 1)^{2n+2}}</math>
:<math>a_{2n} = \frac{T_{2n}}{(2n)!} = 0,</math>
where ''T<sub>n</sub>'' are the [[tangent numbers]].
Conversely, we can compare this formula to the Taylor expansion for tan(''z'') about z = 0 to calculate the infinite sums:
:<math>\tan(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 + \cdots</math>
:<math>\sum_{k=0}^{\infty} \frac{1}{(2k + 1)^2} = \frac{\pi^2}{2^3} = \frac{\pi^2}{8}</math>
:<math>\sum_{k=0}^{\infty} \frac{1}{(2k + 1)^4} = \frac{1}{3} \frac{\pi^4}{2^5} = \frac{\pi^4}{96}</math>
==See also==
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