The following mathematical equation was shown during the television show Fringe which aired on Friday, April 22nd. Any idea what it is?
(edit by J.M.: for reference, this was Sam Weiss scribbling formulae in his notebook.)
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$\begingroup$ I haven't watched: was that Walternate writing? $\endgroup$ – J. M. is a poor mathematician Apr 28 '11 at 2:43
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3$\begingroup$ I appreciate this question. I think it's fun and lively - what is Fringe? $\endgroup$ – davidlowryduda♦ Apr 28 '11 at 4:14
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$\begingroup$ @mix: This TV show. $\endgroup$ – J. M. is a poor mathematician Apr 28 '11 at 4:30
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3$\begingroup$ @J.M. : No, it was that bowling guy. He was observing a vortex and taking notes. $\endgroup$ – Gunnar Þór Magnússon Apr 28 '11 at 11:30
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2$\begingroup$ adds popular-math to followed tags... +1 $\endgroup$ – BBischof Apr 28 '11 at 14:13
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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...
answered Apr 28 '11 at 2:34
J. M. is a poor mathematicianJ. M. is a poor mathematician
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$\begingroup$ Ok. Now that I look closely, it is a $e^x +1$, not a $e^x-1$. +1. $\endgroup$ – Aryabhata Apr 28 '11 at 2:37
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$\begingroup$ A confession: I was actually looking at the left hand side, and it sure looked like an $\eta$ to me... as for the equations on the second line, check out formula 3 in the MathWorld link I gave. $\endgroup$ – J. M. is a poor mathematician Apr 28 '11 at 2:41
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$\begingroup$ It is quite hard to read the second line, so I actually ignored those. Now that you point out, it does look like Eqn (3) :-) $\endgroup$ – Aryabhata Apr 28 '11 at 2:44
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$\begingroup$ Given the R(s) > 1/2 above that last integral, I wonder if the show was trying to show him solving (or attempting to solve) the Riemann Hypothesis... $\endgroup$ – Steven Stadnicki May 23 '11 at 19:09
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$\begingroup$ The succeeding scene had Sam writing a 0 after the =, so I suppose yes. $\endgroup$ – J. M. is a poor mathematician Jun 2 '11 at 6:13
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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.
answered Apr 28 '11 at 2:34
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$\begingroup$ Actually it is $e^x + 1$, not $e^x -1$ (sure looked like that to me too!), so J.M's answer is more accurate. (And the symbol and $Re(s) \gt 1/2$ fit better for that I suppose). $\endgroup$ – Aryabhata Apr 28 '11 at 2:38
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$\begingroup$ Arturo, the DOI doesn't seem to work. Maybe link to JSTOR instead? $\endgroup$ – J. M. is a poor mathematician Apr 28 '11 at 2:38
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1$\begingroup$ @J.M., also available at umt.edu/math/reports/kayll/Karply-MMag-submit09.pdf $\endgroup$ – lhf Apr 28 '11 at 2:44
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$\begingroup$ @lhf: Thanks! I shall read it now... $\endgroup$ – J. M. is a poor mathematician Apr 28 '11 at 2:47
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1$\begingroup$ @J.M. I suspect the doi has not been created yet; the issue just came out in the past couple of weeks. $\endgroup$ – Arturo Magidin Apr 28 '11 at 3:22
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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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$\begingroup$ Oh, it's $+1$, not $-1$. And it's $\eta$, not $\zeta$... $\endgroup$ – lhf Apr 28 '11 at 2:47
protected by Zev Chonoles Jul 18 '15 at 7:41
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