I think we should have a journal dedicated to scientific fiction. Occasionally, I stumble on the arxiv across a paper that actually wanted to to be a science fiction story. It is scientifically accurate but somewhat far fetched - and begging for a narrative.
A good example is the Learned et al's recent paper on The Cepheid Galactic Internet, arguing that aliens would use modulation of Cepheid variable stars to encode messages to other civilizations. The Dyson Sphere is an older example. A more recent one, Hsu and Zee's suggestion that the CMB contains a messages. Various papers on warp-drives, wormholes and time-travel would also qualify. What do you think?
A good example is the Learned et al's recent paper on The Cepheid Galactic Internet, arguing that aliens would use modulation of Cepheid variable stars to encode messages to other civilizations. The Dyson Sphere is an older example. A more recent one, Hsu and Zee's suggestion that the CMB contains a messages. Various papers on warp-drives, wormholes and time-travel would also qualify. What do you think?
"You do not really understand something unless you can explain it to your grandmother." ~ Albert Einstein
from a large population of “real-world” objects. Examples include 
which 
arising from different independent factors); and 
, the proportion of 
is approximately 
. Thus, for instance, 
about 
of the time, but a first digit of 
only about 
of the time. 
largest value of 
for the first few 
and some parameters 
. In many cases, 
is close to 
. 
digits (before the decimal point), where 
is above the median number of digits, should obey an approximate exponential law, i.e. be approximately of the form 
for some 
. Again, in many cases 
is close to 
.
, which is what we are most familiar with, but the laws hold for any base (after replacing all the occurrences of 
in the above laws with the new base, of course). The laws tend to break down if the hypotheses (i)-(iv) are dropped. For instance, if the statistic 
) 
) 
) 
)
) 
)
) 
)
) 
)
) 
)
) 
)
) 
)
) 
) 
, with the parameters 
and 
(selected by log-linear regression), again rounding to three significant figures:
.
, then country populations range in a decent number of scales (the majority of countries have population between 
and 
), and we begin to see the law emerge, where 
and 
:
binary digit populations 
, on the average. This approximate doubling of countries with each new scale begins to falter at about the population 
(i.e. at around 
million), for the simple reason that one has begun to run out of countries. (Note that the median-population country in this set, Singapore, has a population with 
binary digits.)
of the countries have population with first digit 
have population with first digit 
. (This makes the somewhat implausible assumption that growth rates are uniform across all countries; I will talk about what happens when one omits this hypothesis later.) To further simplify the experiment, suppose that no countries are created or dissolved during this time period. What happens to Benford’s law when passing from 
?
. If, for instance, the first digit of 
is either 
or 
; conversely, if the first digit of 
. This is consistent with Benford’s law holding for both &bg=ffffff&fg=000000&s=0 "\displaystyle \log_{10} \frac{2}{1} = \log_{10} \frac{3}{2} + \log_{10} \frac{4}{3} ( = \log_{10} \frac{4}{2} )")
after rounding). Indeed one can check the other digit ranges also and that conclude that Benford’s law for 
is the first digit of 
totally fails to be preserved under doubling.+=+%5Clog_%7B10%7D+%5Cfrac%7B%5Cbeta%7D%7B%5Calpha%7D+%5C+%5C+%5C+%5C+%5C+(1)&bg=ffffff&fg=000000&s=0 "\displaystyle {\Bbb P}( \alpha 10^n \leq X < \beta 10^n \hbox{ for some integer } n ) = \log_{10} \frac{\beta}{\alpha} \ \ \ \ \ (1)")
, where the left-hand side denotes the proportion of data for which 
and 
for some integer 
. Suppose now that we generalise Benford’s law to the continuous Benford’s law, which asserts that 
. Then it is not hard to show that a statistic 
does, and similarly with 
is uniformly distributed.) In fact, the continuous Benford law is the only distribution for the quantities on the left-hand side of +=+c+10%5E%7B-m/%5Calpha%7D+%5C+%5C+%5C+%5C+%5C+(2)&bg=ffffff&fg=000000&s=0 "\displaystyle {\Bbb P}( X \geq 10^m ) = c 10^{-m/\alpha} \ \ \ \ \ (2)")
, just as with Benford’s law. Once one does that, one can phrase the Pareto distribution law independently of any base as +=+c+x%5E%7B-1/%5Calpha%7D+%5C+%5C+%5C+%5C+%5C+(3)&bg=ffffff&fg=000000&s=0 "\displaystyle {\Bbb P}( X \geq x ) = c x^{-1/\alpha} \ \ \ \ \ (3)")
much larger than the median value of 
half the time and 
is assumed to be just as likely to be high-growth as one whose population begins with 
.
begins with either 
of being high-growth regardless of the first digit of their population, we conclude the identity +=+%5Cfrac%7B1%7D%7B2%7D+%7B%5CBbb+P%7D(+X+%5Chbox%7B+has+first+digit+%7D+2,+3+)+%5C+%5C+%5C+%5C+%5C+(4)&bg=ffffff&fg=000000&s=0 "\displaystyle {\Bbb P}( \tilde X \hbox{ has first digit } 2, 3 ) = \frac{1}{2} {\Bbb P}( X \hbox{ has first digit } 2, 3 ) \ \ \ \ \ (4)")
&bg=ffffff&fg=000000&s=0 "\displaystyle + \frac{1}{2} {\Bbb P}( X \hbox{ has first digit } 1 )")
which is independent of the first digit of 
), one obtains another quantity 
which also obeys the continuous Benford’s law. (Indeed, we have already seen this to be the case when 
is a deterministic constant, and the case when 
is random then follows simply by conditioning 
also obeys Benford’s law – even if 
, the uniform distribution from 
does not obey Benford’s law; for instance, if one picks a random number from 
then each digit from 
also obeys Benford’s law (ignoring for now the distinction between continuous and discrete distributions), as it can be viewed as the product of 
and 
, thus +%5Cleq+%5Clog_%7B10%7D+%5Cfrac%7B%5Cbeta%7D%7B%5Calpha%7D+++%5Cvarepsilon+%5C+%5C+%5C+%5C+%5C+(5)&bg=ffffff&fg=000000&s=0 "\displaystyle \log_{10} \frac{\beta}{\alpha} - \varepsilon \leq {\Bbb P}( \alpha 10^n \leq X < \beta 10^n \hbox{ for some integer } n ) \leq \log_{10} \frac{\beta}{\alpha} + \varepsilon \ \ \ \ \ (5)")
, or more generally 
. But now suppose one uses a variable multiplier; for instance, suppose one uses the model discussed earlier in which 
half the time and %7D&bg=ffffff&fg=000000&s=0 "{{\Bbb P}( X \hbox{ has first digit } 2, 3 )}")
and %7D&bg=ffffff&fg=000000&s=0 "{{\Bbb P}( X \hbox{ has first digit } 1 )}")
differs from the Benford’s law predictions of 
and 
respectively by at most 
. Since the left-hand side of 
. But the averaging opens up an opportunity for cancelling; for instance, an overestimate of 
for %7D&bg=ffffff&fg=000000&s=0 "{{\Bbb P}( X \hbox{ has first digit } 2, 3 )}")
could cancel an underestimate of 
for %7D&bg=ffffff&fg=000000&s=0 "{{\Bbb P}( X \hbox{ has first digit } 1 )}")
to produce a spot-on prediction for 
, then there would be a noticeable deviation from Benford’s law, particularly in digits 
and 
, due to this growth bottleneck. But this is not a particularly plausible scenario (being somewhat analogous to 
is not fixed, but varies with 
, then the effect of these sorts of multiplicative changes is to blur and average together the various values of 
equal to the 
of +=+c+X_n%5E%7B-1/%5Calpha%7D&bg=ffffff&fg=000000&s=0 "\displaystyle \frac{n}{N} = {\Bbb P}( X \geq X_n ) = c X_n^{-1/\alpha}")
with %5E%5Calpha%7D&bg=ffffff&fg=000000&s=0 "{C := (Nc)^\alpha}")
. Conversely one can deduce the Pareto distribution from Zipf’s law. These deductions are only formal in nature, because the Pareto distribution can only hold exactly for continuous distributions, whereas Zipf’s law only makes sense for discrete distributions, but one can generate more rigorous variants of these deductions without much difficulty. 
of the sample space), whereas the Pareto distribution in regions closer to the bulk (e.g. between the top 
and and top 
). But this is mostly a difference of degree rather than of kind, though in some cases (such as with the example of the 2007 country populations data set) the exponent 
and 
, and those 
and 
, it seems that the former is larger by a factor of 
, which upon summing by 
is locally uniform at unit scales, while the Pareto distribution (or Zipf’s law) asserts that the tail distribution of 
of winning a game on your serve and a probability 
of winning it on your opponent’s serve, then over the next two games you have a probability 
of winning both, +q(1-p)&bg=ffffff&fg=000000&s=0 "p(1-q)+q(1-p)")
of winning one, and (1-q)&bg=ffffff&fg=000000&s=0 "(1-p)(1-q)")
of losing both, and the order the games are played in makes no difference.
that A will win and a probability 
that B will win, whereas if player B is serving then these probabilities are 
. Moreover, the outcomes of all the points are independent. 
if you serve near the T, but if you serve out wide every time, then your opponent can take advantage of this by standing wider, at which point the probability of winning the point will no longer be 
. 
and you'll serve a fault with probability 
. (That ignores the possibility that your opponent might occasionally be able to get your serve back, but let us indeed ignore that.) And let's suppose that if you go for a serve that is merely very hard to return, then you'll succeed with probability 
, and furthermore that if you succeed then your opponent will give you the opportunity to smash, and that your smash will win the point with probability 
(which is very close to 
) and will go out with probability 
. Then if you go for an ace, you win with probability 
. 
, and rather boringly I don't have any idea. By roughly how much do professional tennis players increase their first-serve percentage if they make their serves very slightly slower, or very slightly less close to the lines? And how much can they afford to do that before their serves become easy to return? Obviously it depends a lot on who is playing, so a different question might be whether it is in fact the case that there are professional tennis players who are trying for aces when it is clearly not the best strategy. (Of course, the mixing-it-up principle could complicate this question and lead to the conclusion that one should sometimes try for an ace and sometimes go for a good first serve that leaves one in a strong position in the ensuing rally.)&bg=ffffff&fg=000000&s=0 "0.95\times 0.75+0.25(0.95\times 0.75)")
, which is &bg=ffffff&fg=000000&s=0 "0.95\times(0.75\times 1.25)")
, which is 19/20 times 15/16, which is certainly at least 7/8. That gives your opponent almost no chance of winning a game. &bg=ffffff&fg=000000&s=0 "q(p)")
, which tells me the probability that I will win the point if I do a probability-=1&bg=ffffff&fg=000000&s=0 "q(p)=1")
, whereas if I go for a safer first serve then &bg=ffffff&fg=000000&s=0 "q(p)")
correspondingly lower.&bg=ffffff&fg=000000&s=0 "pq(p)")
, and we are assuming that its maximal value is +(1-p)r&bg=ffffff&fg=000000&s=0 "pq(p)+(1-p)r")
. What difference does that make? Well, we are assuming that 
will decrease &bg=ffffff&fg=000000&s=0 "o(\delta)")
(since the derivative is zero) but will increase r&bg=ffffff&fg=000000&s=0 "(1-p)r")
by 
. Therefore, the maximum of +(1-p)r&bg=ffffff&fg=000000&s=0 "pq(p)+(1-p)r")
will occur at a higher value of 
that says "If I hit a probability-
shot then I will win the rally with probability &bg=ffffff&fg=000000&s=0 "r(p')")
." But things are more complicated, because &bg=ffffff&fg=000000&s=0 "pq(p)")
, but this can be broken down further as a wish to maximize )&bg=ffffff&fg=000000&s=0 "p(1-\max p'r(p,p'))")
.
measure the riskiness of the shots and let &bg=ffffff&fg=000000&s=0 "r(p_1,\dots,p_k)")
denote my probability of winning if I chose to play a 
-shot, you chose to play a 
-shot, and so on. (This is a slight change because it focuses on my probability of winning and saves me having to write "
".) Then it looks as though my best strategy is to maximize &bg=ffffff&fg=000000&s=0 "p_1r(p_1)")
, which is the same as maximizing &bg=ffffff&fg=000000&s=0 "p_1\min_{p_2} p_2r(p_1,p_2)")
, which is the same as maximizing &bg=ffffff&fg=000000&s=0 "p_1\min_{p_2}(p_2\max_{p_3}r(p_1,p_2,p_3)")
, and so on. Let me just write out the 
th version of this expression in full gory detail. My probability of winning appears to be (I haven't checked this carefully) &bg=ffffff&fg=000000&s=0 "\max_{p_1}p_1\min_{p_2}p_2\max_{p_3}p_3\dots\min_kp_kr(p_1,\dots,p_k)")
if 
is even and a similar expression ending with a max if 
isn't bounded, so it seems as though we need to take some kind of limit as 
looks pretty unpleasant. So are there simplifying assumptions that would at least allow one to justify some of the advice given to players about when to attack, when to go for winners, and so on.
where 
is a nonsquare positive integer. The usual method is to compute the continued fraction
.
by
etcetera.
tends to be very small and, if you compute long enough, for some 
you will have 
. +=+t%5E%7B2g+2%7D+++d_%7B2g+1%7D+t%5E%7B2g+1%7D+++%5Ccdots+++d_1+t+++d_0&bg=ffffff&fg=000000&s=0 "D(t) = t^{2g+2} + d_{2g+1} t^{2g+1} + \cdots + d_1 t + d_0")
. Before I get into details, I want to tell you about something gorgeous that I won’t explain at all. Using the methods in their paper, van der Poorten and Trap can discover identities like%7D.&bg=ffffff&fg=000000&s=0 "\displaystyle{ \int \frac{3 x dx}{\sqrt{x^4+2x}} = \log \left( x^3+1+x \sqrt{x^4+2x} \right)}.")
%7D&bg=ffffff&fg=000000&s=0 "\sqrt{D(t)}")
is actually much prettier than for integers. Everything should be understood in terms of the curve 
cut out by &bg=ffffff&fg=000000&s=0 "y^2 = D(t)")
. This is a curve of genus 
, with two points at infinity. (One of these points is the limit of %7D)&bg=ffffff&fg=000000&s=0 "(t, \sqrt{D(t)})")
and the other is the limit of %7D)&bg=ffffff&fg=000000&s=0 "(t, -\sqrt{D(t)})")
.) I’ll call these two points 
and 
. The theory is controlled by the line bundles &bg=ffffff&fg=000000&s=0 "\mathcal{O}(k \infty_+ + \ell \infty_-)")
. In particular, there are nontrivial solutions to %5E2+-+D(t)+y(t)%5E2+=1&bg=ffffff&fg=000000&s=0 "x(t)^2 - D(t) y(t)^2 =1")
if and only if the continued fraction is periodic, if and only if +=+%5Cmathcal%7BO%7D(k+%5Cinfty_-)&bg=ffffff&fg=000000&s=0 "\mathcal{O}(k \infty_+) = \mathcal{O}(k \infty_-)")
for some 
. +=+Y_k+t%5Ek+++Y_%7Bk-1%7D+t%5E%7Bk-1%7D+++%5Ccdots&bg=ffffff&fg=000000&s=0 "Y(t) = Y_k t^k + Y_{k-1} t^{k-1} + \cdots")
in 
, we define the continued fraction of &bg=ffffff&fg=000000&s=0 "Y(t)")
.![[Y(t)] := \sum_{i=0}^k Y_i t^i](http://s2.wordpress.com/latex.php?latex=%5BY(t)%5D+:=+%5Csum_%7Bi=0%7D%5Ek+Y_i+t%5Ei&bg=ffffff&fg=000000&s=0 "[Y(t)] := \sum_{i=0}^k Y_i t^i")
. Set ![a_0 = [Y]](http://s3.wordpress.com/latex.php?latex=a_0+=+%5BY%5D&bg=ffffff&fg=000000&s=0 "a_0 = [Y]")
and define 
by 
. Then set ![a_1 = [F_1]](http://s3.wordpress.com/latex.php?latex=a_1+=+%5BF_1%5D&bg=ffffff&fg=000000&s=0 "a_1 = [F_1]")
and 
. Continuing in this way, we get a sequence 
of polynomials, a sequence 
of power series, and a continued fraction
.
as before; they do converge to 
in the sense that each ratio 
agrees with 
to a higher order than the ratio does. &bg=ffffff&fg=000000&s=0 "D(t)")
is a polynomial of the form 
.%7D%7D+=+t%5E%7Bg+1%7D+++(1/2)+d_%7B2g+1%7D+t%5Eg+++%5Ccdots.&bg=ffffff&fg=000000&s=0 "\displaystyle{\sqrt{D(t)}} = t^{g+1} + (1/2) d_{2g+1} t^g + \cdots.")
&bg=ffffff&fg=000000&s=0 "a_i(t)")
, &bg=ffffff&fg=000000&s=0 "F_i(t)")
, &bg=ffffff&fg=000000&s=0 "x_i(t)")
and &bg=ffffff&fg=000000&s=0 "y_i(t)")
from above.&bg=ffffff&fg=000000&s=0 "a_i(t)")
is a polynomial in 
, its only poles are at 
, and it has a pole of the same order at both 
’s. So, other than 
, the function +-+a_i(t)&bg=ffffff&fg=000000&s=0 "F_i(t) - a_i(t)")
has the same poles as 
. Then &bg=ffffff&fg=000000&s=0 "F_{i+1} = 1/(F_i - a_i)")
has zeroes at the poles of 
. 
. Suppose that &bg=ffffff&fg=000000&s=0 "F_i(t)")
has a pole of order 
at 
at 
. Then 
’s. The difference +-+a_i(t)&bg=ffffff&fg=000000&s=0 "F_i(t) - a_i(t)")
has a zero of order 
at 
and a pole of order 
has a pole of order 
at 
and a zero of order 
.
and let the zeroes be 
. Then the poles of 
are 
, for some 
and some 
and the zeroes are 
. (Here 
, 
and 
are supported away from 
and a sequence of divisors 
such that the poles of 
while the zeroes are 
.
is the degree of 
. Note also that 
in the Picard group, so+-+p_%7Bi%7D+%5Cinfty_%7B+%7D&bg=ffffff&fg=000000&s=0 "P_i \equiv P_0 + p_0 \infty_{-} + \sum p_j (\infty_{-} - \infty_{+}) - p_{i} \infty_{+}")
.
has degree 
and all the other 
has degree 
(exercise!), so all of +%5Cinfty_%7B-%7D+-+i+%5Cinfty_%7B+%7D&bg=ffffff&fg=000000&s=0 "P_i \equiv (g+i) \infty_{-} - i \infty_{+}")
in the Picard group. You may remember that a generic divisor of degree 
has a unique effective representative in PIcard; 
is that unique representative. So, we have just found an explicit way to write down an arithmetic progression in &bg=ffffff&fg=000000&s=0 "Pic^g(C)")
, where 
is a hyperelliptic curve.
is torsion in the Picard group or, in other words, when there is a unit +++y(t)+%5Csqrt%7BD(t)%7D&bg=ffffff&fg=000000&s=0 "x(t) + y(t) \sqrt{D(t)}")
in the coordinate ring of 
. Then, eventually, the sequence in Picard will repeat. It turns out, when this happens, the corresponding approximation /y_i(t)&bg=ffffff&fg=000000&s=0 "x_i(t)/y_i(t)")
gives your unit!
? The /B_i&bg=ffffff&fg=000000&s=0 "(A_i + \sqrt{D})/B_i")
: how do we relate the polynomials 
and 
to the divisors 
“. Specifically I discussed the Borel-Weil theorem, Ginzburg’s construction using Springer fibres, and the geometric Satake correspondence. I focused on 
to keep the root system combinatorics and the geometry as elementary as possible.
&bg=ffffff&fg=000000&s=0 "f(R)")
gravity. The other mechanism is called Vainshtein effect, which is incorporated in DGP and some massive gravity models. The scalar degree of freedom becomes strongly coupling and frozen because of the derivative self-interactions. The theory effectively becomes GR. ++%7D++%5Csqrt%7B+-+%5Cbar%7B%5Cnabla%7D%5E2++%7D+C++++++=++%5Cfrac%7B3%7D%7B2%7D+%5Cfrac%7B%5Ceta++-1++%7D%7B%5Ceta%7D+%5Cdelta+++&bg=ffffff&fg=000000&s=0 "\bar{\nabla}^2 \phi - \frac{1}{\eta} \sqrt{ - \bar{\nabla}^2 } \phi + \frac{1}{2 \eta} \bar{\nabla}^2 C + \frac{ 3 \eta^2 - 5 \eta + 1 }{2 \eta^2 (2 \eta -1) } \sqrt{ - \bar{\nabla}^2 } C = \frac{3}{2} \frac{\eta -1 }{\eta} \delta")
%5E2+++++++++%5Calpha++%5Cbar%7B%5Cnabla%7D%5E2+C+++-+(%5Cbar%7B+%5Cnabla%7D_%7Bij%7D+C)%5E2++++++%5Cfrac%7B+3+%5Cbeta+(%5Ceta+-1)+%7D%7B2+%5Ceta-1+%7D++++%5Csqrt%7B+-+%5Cbar%7B%5Cnabla%7D%5E2++%7D+C+++++=+%5Cfrac%7B+3(+%5Ceta+-1+)+%7D+%7B%5Ceta+%7D+(+1-+%5Cbeta+%5Cbar%7B%5Cnabla%7D%5E%7B-1%7D+)+%5Cdelta,+++++++++++&bg=ffffff&fg=000000&s=0 "(\bar{\nabla}^2 C)^2 + \alpha \bar{\nabla}^2 C - (\bar{ \nabla}_{ij} C)^2 + \frac{ 3 \beta (\eta -1) }{2 \eta-1 } \sqrt{ - \bar{\nabla}^2 } C = \frac{ 3( \eta -1 ) } {\eta } ( 1- \beta \bar{\nabla}^{-1} ) \delta,")
can be easily handled in the Fourier space. The real headache comes from the nonlinear derivative terms %5E2+&bg=ffffff&fg=000000&s=0 "(\bar{\nabla}^2 C)^2")
and %5E2+&bg=ffffff&fg=000000&s=0 "(\bar{ \nabla}_{ij} C)^2")
. One of the major achievement in this paper is that we developed a convergent method to solve this set of equations consistently. It involves alternate use of relaxation and Fast Fourier transform (so we call it FFT-relaxation method). Although that is a main result of the paper, I am not going to talk about it in details so as not to get too technical and dry. But interested readers are welcome to read the original paper. 
and 
in the lower figure. In the large scales (small k), the full nonlinear DGP model reduces to the linear one. More interestingly in the nonlinear regime (large k limit), the fully nonlinear DGP model approaches the GR with the same expansion history. This demonstrates that Vainshtein effect drives the model towards GR limit in the large k regime. The transition is broad and the limit is not yet fully attained in the range shown here. 
is about 1 h/Mpc. For more details, please refer to our original paper 
