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The following mathematical equation was shown during the television show Fringe which aired on Friday, April 22nd. Any idea what it is?

Fringe formula

(edit by J.M.: for reference, this was Sam Weiss scribbling formulae in his notebook.)

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    @BBischof, ditto.2012-03-30

3 Answers 3

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The last formula seems to be an integral expression for the Dirichlet $\eta$ function:

$\eta(s)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^s}\quad \Re(s) > 0$

Using the relationship with the usual Riemann $\zeta$ function

$\eta(s)=(1-2^{1-s})\zeta(s)$

and this integral expression, you get that integral in the notebook:

$\eta(s) = \frac1{\Gamma(s)} \int_0^{\infty}\frac{x^{s-1}}{e^x+1}\mathrm dx$

There is also this closely related integral (which Arturo mentions in his answer).

The (first part of the) second line looks to be the chain of relations relating Riemann $\zeta$, Dirichlet $\eta$, and Dirichlet $\lambda$:

$\frac{\zeta(s)}{2^s}=\frac{\lambda(s)}{2^s-1}=\frac{\eta(s)}{2^s-2}$

In the second part, the expressions look to be the differentiation of Dirichlet $\eta$, but the screenshot is fuzzy around that region...

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    The succeeding scene had Sam writing a 0 after the =, so I suppose yes.2011-06-02
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By sheer coincidence, this equation occurs in P. Mark Kayll's paper Integrals don't have anything to do with discrete Math, do they? (Mathematics Magazine 84 no. 2, April 2011, pages 108-119, doi:10.4169/math.mag.84.2.108, which I was reading through today. It is equation (6) on page 111.

$\zeta(s)$ is the Riemann zeta function, $\zeta(s) = \sum_{k=1}^{\infty}\frac{1}{k^s}.$

$\Gamma(s)$ is the Gamma function, $\Gamma(s) = \int_0^{\infty} t^{s-1}e^{-t}\,dt.$

$R(s)\gt \frac{1}{2}$ says that the real part of $s$ is greater than $\frac{1}{2}$.

The final equation is just a recasting of the equality $\zeta(x)\Gamma(x) = \int_0^{\infty}\frac{t^{x-1}}{e^t-1}\,dt$ which holds whenever $\Gamma(x)$ is finite.

Nothing to get too excited about... unless you don't know what any of the symbols mean, which I suspect holds for a very large portion of the viewership of Fringe.

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    @J.M. I did double check; this is the DOI listed on the bottom of the first page of the article.2011-04-28
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It's Riemann's zeta-function as a Mellin transform. See http://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin_transform

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    Oh, it's $+1$, not $-1$. And it's $\eta$, not $\zeta$...2011-04-28