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# Documents  20G15 | enregistrements trouvés : 5

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## Post-edited  Descent in Bruhat-Tits theory Prasad, Gopal (Auteur de la Conférence) | CIRM (Editeur )

Bruhat-Tits theory applies to a semisimple group G, defined over an henselian discretly valued field K, such that G admits a Borel K-subgroup after an extension of K. The construction of the theory goes then by a deep Galois descent argument for the building and also for the parahoric group scheme. In the case of unramified extension, that descent has been achieved by Bruhat-Tits at the end of [BT2]. The tamely ramified case is due to G. Rousseau [R]. Recently, G. Prasad found a new way to investigate the descent part of the theory. This is available in the preprints [Pr1, Pr2] dealing respectively with the unramified case and the tamely ramified case. It is much shorter and the method is based more on fine geometry of the building (e.g. galleries) than algebraic groups techniques.
Bruhat-Tits theory applies to a semisimple group G, defined over an henselian discretly valued field K, such that G admits a Borel K-subgroup after an extension of K. The construction of the theory goes then by a deep Galois descent argument for the building and also for the parahoric group scheme. In the case of unramified extension, that descent has been achieved by Bruhat-Tits at the end of [BT2]. The tamely ramified case is due to G. ...

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## Post-edited  Triality Elduque, Alberto (Auteur de la Conférence) | CIRM (Editeur )

Duality in projective geometry is a well-known phenomenon in any dimension. On the other hand, geometric triality deals with points and spaces of two different kinds in a sevendimensional projective space. It goes back to Study (1913) and Cartan (1925), and was soon realizedthat this phenomenon is tightly related to the algebra of octonions, and the order 3 outer automorphisms of the spin group in dimension 8.
Tits observed, in 1959, the existence of two different types of geometric triality. One of them is related to the octonions, but the other one is better explained in terms of a class of nonunital composition algebras discovered by the physicist Okubo (1978) inside 3x3-matrices, and which has led to the definition of the so called symmetric composition algebras.
This talk will review the history, classification, and their connections with the phenomenon of triality, of the symmetric composition algebras.
Duality in projective geometry is a well-known phenomenon in any dimension. On the other hand, geometric triality deals with points and spaces of two different kinds in a sevendimensional projective space. It goes back to Study (1913) and Cartan (1925), and was soon realizedthat this phenomenon is tightly related to the algebra of octonions, and the order 3 outer automorphisms of the spin group in dimension 8.
Tits observed, in 1959, the ...

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## Multi angle  The rationality problem for forms of moduli spaces of stable marked curves Reichstein, Zinovy (Auteur de la Conférence) | CIRM (Editeur )

Let $\overline{M_{g,n}}$ be the moduli space of stable curves of genus $g$ with $n$ marked points. It is a classical problem in algebraic geometry to determine which of these spaces are rational over $\mathbb{C}$. In this talk, based on joint work with Mathieu Florence, I will address the rationality problem for twisted forms of $\overline{M_{g,n}}$ . Twisted forms of $\overline{M_{g,n}}$ are of interest because they shed light on the arithmetic geometry of $\overline{M_{g,n}}$, and because they are coarse moduli spaces for natural moduli problems in their own right. A classical result of Yu. I. Manin and P. Swinnerton-Dyer asserts that every form of $\overline{M_{0,5}}$ is rational. (Recall that the $F$-forms $\overline{M_{0,5}}$ are precisely the del Pezzo surfaces of degree 5 defined over $F$.) Mathieu Florence and I have proved the following generalization of this result.
Let $n\geq 5$ is an integer, and $F$ is an infinite field of characteristic $\neq$ 2.
(a) If $n$ is odd, then every twisted $F$-form of $\overline{M_{0,n}}$ is rational over $F$.
(b) If $n$ is even, there exists a field extension $F/k$ and a twisted $F$-form of $\overline{M_{0,n}}$ which is unirational but not retract rational over $F$.
We also have similar results for forms of $\overline{M_{g,n}}$ , where $g \leq 5$ (for small $n$ ). In the talk, I will survey the geometric results we need about $\overline{M_{g,n}}$ , explain how our problem reduces to the Noether problem for certain twisted goups, and how this Noether problem can (sometimes) be solved.

Keywords: rationality - moduli spaces of marked curves - Galois cohomology - Noether's problem
Let $\overline{M_{g,n}}$ be the moduli space of stable curves of genus $g$ with $n$ marked points. It is a classical problem in algebraic geometry to determine which of these spaces are rational over $\mathbb{C}$. In this talk, based on joint work with Mathieu Florence, I will address the rationality problem for twisted forms of $\overline{M_{g,n}}$ . Twisted forms of $\overline{M_{g,n}}$ are of interest because they shed light on the ...

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## Multi angle  Quasisemisimple classes Michel, Jean (Auteur de la Conférence) | CIRM (Editeur )

This is a report on joint work with François Digne. Quasisemisimple elements are a generalisation of semisimple elements to disconnected reductive groups (or equivalently, to algebraic automorphisms of reductive groups). In the setting of reductive groups over an algebraically closed field, we discuss the classification of quasisemisimple classes, including isolated and quasi-isolated ones. The talk will start with the basic theory of non-connected reductive groups.
This is a report on joint work with François Digne. Quasisemisimple elements are a generalisation of semisimple elements to disconnected reductive groups (or equivalently, to algebraic automorphisms of reductive groups). In the setting of reductive groups over an algebraically closed field, we discuss the classification of quasisemisimple classes, including isolated and quasi-isolated ones. The talk will start with the basic theory of n...

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## Multi angle  Bruhat-Tits theory of quasi-split groups Rémy, Bertrand (Auteur de la Conférence) | CIRM (Editeur )

The goal of this lecture is to present the construction of the Bruhat-Tits buildings attached to a quasi-split (that is admitting a Borel subgroup) semisimple group G defined over an henselian discretly valued field K and also the construction of the parahoric group schemes parametrized by the points of the buildings. The building part is [BT1] and the group scheme part corresponds to the four first sections of [BT2] but could also be treated by Yu's method [Y] namely by using Raynaud's theory of group schemes [BLR].
The goal of this lecture is to present the construction of the Bruhat-Tits buildings attached to a quasi-split (that is admitting a Borel subgroup) semisimple group G defined over an henselian discretly valued field K and also the construction of the parahoric group schemes parametrized by the points of the buildings. The building part is [BT1] and the group scheme part corresponds to the four first sections of [BT2] but could also be treated by ...

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# Referencing figures doesn't work [duplicate]

Possible Duplicate:
Where to put the \label on a figure environment?

I'm having a problem with referencing figures in LaTeX. I have the following text:

\begin{figure}[H]
\center
\label{fig:enn1015}
\includegraphics[scale=0.40]{imagens/outputs/ENN_10_15.eps}
\caption{ENN unbalanced 10\%}
\end{figure}

The figure \ref{fig:enn1015} (...)

For some reason, the figure caption is:

Figure 3.5 Enn unbalanced 10%

But the text is:

The figure 3.2.1 (...)

The figure 3.5

It is printing the chapter.section.subsection the figure is in, and not the figure. Can anybody help me?

-

## migrated from stackoverflow.comDec 7 '11 at 15:08

This question came from our site for professional and enthusiast programmers.

## marked as duplicate by Werner, Torbjørn T., lockstep, Joseph Wright♦Dec 9 '11 at 10:08

Try putting your \label{} statement immediately after your \caption{} statement.
Edit: I'm not a total expert on how the \label command works, but I think it assigns a label to the last-seen "labelable" thing, so you want it right after your caption, which is "labelable" (my terminology). Currently, it is labeling whatever the last "labelable" thing was before your figure, probably a section heading or something (hence the 3.2.1).

format: plain


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abs_normal: Solve a Quadratic Program With Box Constraints

Syntax
ok = qp_box(      level, a, b, c, C, g, G, epsilon, maxitr, xin, xout )

Prototype

template <class Vector>
bool qp_box(
size_t        level   ,
const Vector& a       ,
const Vector& b       ,
const Vector& c       ,
const Vector& C       ,
const Vector& g       ,
const Vector& G       ,
double        epsilon ,
size_t        maxitr  ,
const Vector& xin     ,
Vector&       xout    )

Source
This following is a link to the source code for this example: qp_box.hpp .

Purpose
This routine could be used to create a version of abs_min_linear that solved quadratic programs (instead of linear programs).

Problem
We are given $a \in \B{R}^n$, $b \in \B{R}^n$, $c \in \B{R}^m$, $C \in \B{R}^{m \times n}$, $g \in \B{R}^n$, $G \in \B{R}^{n \times n}$, where $G$ is positive semi-definite. This routine solves the problem $$\begin{array}{rl} \R{minimize} & \frac{1}{2} x^T G x + g^T x \; \R{w.r.t} \; x \in \B{R}^n \\ \R{subject \; to} & C x + c \leq 0 \; \R{and} \; a \leq x \leq b \end{array}$$ The matrix $G + C^T C$ must be positive definite on components of the vector $x$ where the lower limit minus infinity and the upper limit is plus infinity; see a and b below.

Vector
The type Vector is a simple vector with elements of type double.

level
This value is less that or equal two. If level == 0 , no tracing is printed. If level >= 1 , a trace of the qp_box operations is printed. If level == 2 , a trace of the qp_interior sub-problem is printed.

a
This is the vector of lower limits for $x$ in the problem. If a[j] is minus infinity, there is no lower limit for $x_j$.

b
This is the vector of upper limits for $x$ in the problem. If a[j] is plus infinity, there is no upper limit for $x_j$.

c
This is the value of the inequality constraint function at $x = 0$.

C
This is a row-major representation of thee the inequality constraint matrix $C$.

g
This is the gradient of the objective function.

G
This is a row-major representation of the Hessian of the objective function. For $j = 0 , \ldots , n-1$, $- \infty < a_j$ or $b_j < + \infty$ or $G_{j,j} > 0.0$.

epsilon
This argument is the convergence criteria; see KKT conditions below. It must be greater than zero.

maxitr
This is the maximum number of qp_interior iterations to try before giving up on convergence.

xin
This argument has size n and is the initial point for the algorithm. It must strictly satisfy the constraints; i.e.,       a < xin,  xin < b,  C * xin - c < 0
xout
This argument has size is n and the input value of its elements does no matter. Upon return it is the primal variables $x$ corresponding to the problem solution.

ok
If the return value ok is true, convergence is obtained; i.e., $$| F ( x , y_a, s_a, y_b, s_b, y_c, s_c ) |_\infty < \varepsilon$$ where $|v|_\infty$ is the infinity norm of the vector $v$, $\varepsilon$ is epsilon , $x$ is equal to xout , $y_a, s_a \in \B{R}_+^n$, $y_b, s_b \in \B{R}_+^n$ and $y_c, s_c \in \B{R}_+^m$.

KKT Conditions
Give a vector $v \in \B{R}^m$ we define $D(v) \in \B{R}^{m \times m}$ as the corresponding diagonal matrix. We also define $1_m \in \B{R}^m$ as the vector of ones. We define $$F ( x , y_a, s_a, y_b, s_b, y_c, s_c ) = \left( \begin{array}{c} g + G x - y_a + y_b + y_c^T C \\ a + s_a - x \\ x + s_b - b \\ C x + c + s_c \\ D(s_a) D(y_a) 1_m \\ D(s_b) D(y_b) 1_m \\ D(s_c) D(y_c) 1_m \end{array} \right)$$ where $x \in \B{R}^n$, $y_a, s_a \in \B{R}_+^n$, $y_b, s_b \in \B{R}_+^n$ and $y_c, s_c \in \B{R}_+^m$. The KKT conditions for a solution of this problem is $$F ( x , y_a, s_a, y_b, s_b, y_c, s_c ) = 0$$

Example
The file qp_box.cpp contains an example and test of qp_box. It returns true if the test passes and false otherwise.
Input File: example/abs_normal/qp_box.hpp


text:
## Tuesday, February 23, 2016

### Book Review: Rock Breaks Scissors by Poundstone

http://www.amazon.com/Rock-Breaks-Scissors-Outguessing-Outwitting/dp/1491531185

This is the third book I have read recently that is about forecasting (see here and here for the previous two).

Poundstone did a pretty good job overall. The first chapter of the book is a general discussion on the human psychology of perceiving randomness. The following one half gives specific examples of how to outperform in various more light-hearted and trivial prediction exercises (e.g. multiple choice tests, rock-paper-scissors, passwords etc.), and the remaining one half does the same thing but with more serious topics (e.g. sports betting, web big data, stock market etc.).

Once again I'll try to distill the essence of the book here:
1. The human mind is handicapped in creating and detecting randomness
2. We can take advantage of the above in many situations for a statistically significant gain
3. In situations where randomness rules, the real pros do not care so much about views; they care more about odds

## Sunday, February 7, 2016

### Book Review: Superforecasting by Tetlock and Gardner

The theme of this book is very similar to Nate Silver's The Signal and the Noise. The first few chapters are worth reading, while as it drags on the final one-third of the book is less appealing, as it indulges in corny talks of team works etc.. The new ideas I got from this book, in addition to those already listed in the review above, include:
1. The use of Brier score for rating forecast, because 'good' forecast should be in probabilistic terms
2. As a corollary, one has to make enough forecasts in order to have a statistically significant score (=/= good score)
3. Update predictions as frequently as appropriate
4. There is a vicious cycle going on in the 'Mainstream' of our society: people are not aware of the importance/option to think probabilistic -> they crave for binary, determined answers -> the media are aware of this demand -> the media cultivate the guru/pundit culture -> the people get more addicted to this way of thinking
All in all, it is not as readable a book as Silver's, but I would still recommend it to anyone who is serious about improving and understanding the art of forecasting.

## Thursday, March 19, 2015

### Book Review: The Signal and the Noise by Nate Silver

An anecdotal account without much technical materials, nonetheless an enjoyable read. Through the case studies (weather forecast, earthquake prediction, economic forecast, sports betting, poker, etc.) the readers are gradually introduced to what quantitative prediction is/should be.

If I am to distill three most important points made in the book, they would be
1. Think probability; or even better, think CONDITIONAL probability, aka Bayesian.
2. Giving your prediction a context, or "having a story," would result in forecasts that are much better than blindly dive into datamining.
3. The evaluation of the performance of predictions should not be based (solely) on whether they got it right. Maintaining best practices is more important.
There may not be much originality in the materials, but the writing is decent and there are valuable lessons to take home. Recommended especially for people who work with quantitative data manipulation.

## Tuesday, July 1, 2014

They introduce the $-Game, and explain why it (instead of the Minority Game) is a better model for the financial market. Yeung et al, 2008, "Models of financial markets with extensive participation incentives" They introduce the Wealth Game, and define the temperature of the grand canonical ensemble. More interestingly, they connect this theoretical framework to the real market by calibrating the model to market data, thus identifying where we are on the "phase diagram." De Martino et al, "From perceived risk to collective behavior in minority games" ## Monday, June 30, 2014 ### Volatility Smile and Volatility Models: A Short Note Local Volatility (LV) Model 1. LV preserves market completeness (is it a good/must-have thing?) 2. LV gives forward smiles that are too flat compared to market 3. In fact, LV doesn't really qualify as a model: it fits implied vol perfectly but trivially, with no capability/intention to consider the underlying dynamics 4. LV also produces poor smile dynamics prediction More General Notes 1. The most sought-after quality for a model should be accurate hedging, not mathematical tractability etc. 2. In fact, there is no good reason to limit ourselves to vanillas only when calibrating a vol model; calibrating only to vanillas has been common practice, which is unfortunate 3. The information carried by vanillas is the unconditional (risk-neutral) probabilities that can be recovered by taking second derivatives w.r.t. strikes (i.e. the stock price is here at t=0, what is the probability that it will end up there at t=T?) 4. Conditional probabilities, on the other hand, is not contained in vanillas (i.e. conditional on the stock price being here at t=s 5. Another way of saying this is that vanillas are not sensitive to unconditional distributions 6. Therefore, one way to proceed is to include some exotics in the calibration: forward starting options are sensitive to conditional probabilities; while American digitals (or barriers) are sensitive to jumps 7. As a rule of thumb, jumps are "responsible" for short-dated smile, and stochastic volatility is "responsible" for longer-dated smile Reference: Ayache 2004, Can anyone solve the smile problem? Ayache 2005, Can anyone solve the smile problem? A Post-Scriptum Also, Rebonato's Volatility and Correlation: The Perfect Hedger and the Fox is relevant, and it's a great read, but that would be another episode. ## Saturday, March 1, 2014 ### How do you model the dynamics of asset prices? A self-reflection (Epilogue of 3) As the title suggests, this was supposed to be a three part series. But I feel like I have left out so much that I wanted to discuss, so although the first 3 parts have already provided an overview on what I would call the "classical" quant modeling approach (loosely speaking, one that focuses on describing the asset price time series with a random process), I would take this opportunity to talk about what lies beyond such approach. We are interested particularly in capturing the tails of the distribution. What about putting the tail on the center stage? That is the spirit of extreme value theory. It has had wide application in natural disaster forecast. The basic idea is to posit a probability distribution that specifically fits the tail of the distribution. Contrast this with, say, the stochastic volatility approach in which we tried to fatten the tail, while still keeping our focus on the 'body' of the return distribution. Now we are directly and un-apologetically modeling the tail itself. The advantage is obvious: we have more freedom in fitting the tails closely by dedicating the calibration to it. The short coming is more subtle. In order to perform extreme value analysis, we have to specify the threshold at where the tail starts, so that we can attach a distribution to it. If you define daily return of -20% as the threshold, then you can use some distribution to catch the sub-sample of daily return less than -20%. But how do you choose the threshold? Choose it too deep into the body and you miss the point of using extreme value models. After all, we are trying to focus on the tail by not worrying about modeling the boring part of the distribution. On the other hand, choose it too far off into the tail and you suffer by not having enough historical data points for an accurate calibration. The other issue is that extreme value analysis would not solve the "you don't know what you don't know" problem. You pick a threshold, you do the fitting, and chances are sooner or later a much worse draw down would take you by surprise, proving to you that the tail is even fatter that you have assumed. Now you might say, "Hey, isn't that a necessary evil of any perceivable model in finance? That one has to make assumptions and wait to be surprised?" And I tend to agree to the sad fact. Perhaps except for if we follow a completely different paradigm. Recently some researchers having been looking at crash forecasting using a wide range of unconventional tools. They range from fractal analysis that studies the (change in) scaling of time series; agent-based models that take a bottom-up approach trying to produce asset price dynamics by considering (grossly simplified) behaviors of traders; and critical point/phase change models that are inspired by Ising model in solid state physics. The common theme is the acknowledgement of market non-linearity and the lack of a single, forever stable market state. Unfortunately there is no straight forward way to connect these models to option prices, but with the advancements of numerical methods we may some day see wider applications of them. ## Friday, February 28, 2014 ### Looping the Strings Problem This one is a classic that I believe is found in one of the popular interview books. You start with N pieces of strings that are mixed into a single pile. You start randomly choosing two free ends at a time and joining them together. After N rounds, you will be done as there will be no more free ends left. If the number of closed loops is k, what is F = E[k]? If you go down the combinatorial path, you will soon see how hopeless it is to try figuring out the many different intermediate states. So perhaps combinatorial is not the right tool here. In fact, this problem can be readily solved with a closed-form solution using induction and recurrence. Before arriving at that, let's get some sense and inspiration by stating with a small N. N = 1 is trivial and we see that k = 1 surely, so F = 1. Consider N = 2. During the first round of joining, when one free end is picked, it can be joined to any of the remaining 3 free ends. In one out of these three cases, we end up with k = 2 (namely, if the free end is joined to another end of itself); while in the remaining two cases, we end up with k = 1 (namely, if the free end is joined to any end of the other string). So$ F_2 = \frac 1 3 (1+F_1) + \frac 2 3 (F_1)=\frac 4 3 $. More generally, it's easy to show that$ F_N = \frac 1 {2N-1} (1+F_{N-1}) + \frac {2N-2} {2N-1} (F_{N-1})$This recurring relation can be rearranged to give the difference equation$ F_N = F_{N-1} + \frac 1 {2N-1} $. With the initial condition of$ F_1 = 1$, we can alternatively express it as a finite sum. Note how slow it grows with N (~ log(N)). ## Thursday, January 30, 2014 ### How do you model the dynamics of asset prices? A self-reflection (3 of 3) Last time we added jumps to the recipe to enrich the original random walk dynamics. Jump diffusion model is able to fit to market implied vol much better than the diffusion only counterpart. Before moving on, just two reminders: 1. The numerical pricing procedure becomes more involved, especially for grid/finite differencing methods (because we end up with an integro-differential equation, as opposed to a PDE) 2. Although we usually speak of "the" jump diffusion model, we are free in specifying the probability distribution of the jump amplitude How about good old Gaussian? So is jump diffusion the ultimate holy grail? And is pure diffusion model done with? For the first question, we will see later in the series that there are many other approaches that supersede jump diffusion, but for now let's take a step back and address the second question. It turns out that the use of mixtures is able to tackle at least some of the shortcomings of the original random walk model. A mixture, not surprisingly, is formed by combining more than one 'sub-distributions.' (usually, but not necessarily, of the same kind) Here we focus on Gaussian mixture. There are a number of possibilities regarding how the sub-distributions are combined, the use of Markov chain being a prominent one. Essentially, every time we need a new random variable to be generated, we first toss a coin to determine which sub-distribution out of the mixture to draw from. In the case of a mixture of two Gaussians, if one sub-distribution has a larger variance than the other, then the mixture would have fatter tail, although the individual sub-distributions do not. Candidate: Mixture • Markovian • Most likely not tractable • Excess kurtosis • Market incomplete The use of mixture models in option pricing is rare, possibly because not everyone likes the idea of the hidden Markov chain that comes with the same package. Justifying and interpreting it is easier in the context of econometric modeling that deals with longer time frames, in which case the state changes can be perceived as a change in regimes. For something like intra-day option pricing, the model may require that we draw from alternate sub-distributions every 5 minutes or so, and it's not trivial to make sense of what that means. (other than the fact that it fits the data well) Another French Mathematician If producing a fatter tail is all we care about, then what about substituting Gaussian process with something with excess kurtosis? The resulting model would likely not have analytic solution, but we can fall back on numerical methods as our last resort. For example, ask the computer to generate 10000 random numbers from a Student's t distribution and use those in a Monte Carlo simulation. The idea sounds neat, but one has to be cautious about it. We may not be aware of it, but when we were modeling asset prices with Gaussian or Poisson processes, both of the processes are closed under linear transformation. In other words, aggregating multiple Gaussian (or Poisson) random variables would result in yet another Gaussian (or Poisson) variable. That means when we use Gaussian process to model asset prices, even though it may not be a very faithful description of reality, at least we are being consistent and we know what we are getting ourselves into: namely, if the one-day distribution is an N(0.01, 0.05), then the one-month distribution would be an N(0.25,0.25), etc. Other distributions in general do not have this property. If in our hypothetical Monte Carlo simulation we draw from a t distribution with a degree of freedom of 10 for daily return, how would the monthly return be distributed? How about quarterly? These problems can be avoided by going for non-Gaussian processes while staying within the territory of Levy processes, which is a broad family of processes that possess the nice property of being closed under linear transformation. (obviously both Gaussian and Poisson processes are members of this family) ## Tuesday, January 28, 2014 ### Other Interview Questions Q: It's 12:00 now. What time would it be when the minute hand crosses the hour hand again? A: Between 1200 and 2400, this will happen for 11 times. So divide 12 hr by 11 and add to 1200, i.e. 13:05:27 Q: A cell can do 4 things: 1) Dies; 2) Nothing; 3) Duplicate; or 4) Triplicate. What is P(colony dies out) if we start with one single cell? A: Solve the polynomial$ p = \frac {1} {4} + \frac {1}{4} p + \frac {1}{4} p^2 + \frac {1}{4} p^3$Q: A 6x6 matrix has entries of either +1 or -1. How many unique combinations are there such that all column products and row products are +1? A: Consider the top left 5x5 sub-matrix. It can take any permutation (as long as each row/column contains no more than 3 same number) because we can rely on the rest of the 6x6 matrix (an "L" shape) to "fix" it. So the answer is$ 2^{25} $Q: There is an integer X. The product of the digits is 96. What is the largest/smallest possible value of X? A: smallest = 268, largest = 322222 Q: What is the sum of all digits of all the numbers between 1 and 1000000? A: Convince yourself that all digits appear for an equal number of times. The answer is 27000001 Q: A bus makes a stop, and 3/4 of all passengers get off, while 7 get on. This repeats for two more times. What is the smallest number of initial passengers? A: 52 Q: Paint the outside of a cube, then dice it into 27 equal smaller cubes. Pick one at random and roll it. What is P(no paint visible after the roll)? A:$ 6/27 \times 1/6 + 1/27 $Q: Two equally talented teams are playing a series of games in which the first team to win 4 games wins the series. What is P(7th game has to be played)? A:$ \frac {6!}{3!3!} \frac {1}{2^6} = \frac {5}{16}$Q: I roll a die, spin a roulette and draw a card from a deck of 52. What is the probability that all three give the same number? A:$ 5 \times \frac 1 6 \times \frac 1 {13} \times 1 {38} $Q: What is the last digit of 3 to the power 33? A: Some experimentation would indicate that$3^0=1$,$3^1=3$,$3^2=9$,$3^3=27$,$3^4=81$, ..., hence$3^33  mod  10 = 3$Q: N couples shake hand with others. It is known that 1) They don't shake hand with their own partners; and 2) The first (2N - 1) people surveyed all have different number of handshakes. What is the number of handshake the remaining person made? A: N - 1 (Can draw graphs to deduce; any analytic solution?) Q: What is P(3 random points on a circle are within an 180 degree arc)? A: There are two approaches: 1) Area on an abstract x-y plane, x and y are distances from the first point pinned down (call it z, which is arbitrary); or 2)$ N \times \frac {1}{2^{N-1}} $, with$ N=3$. Both give the answer of 3/4. Q: How do you simulate a 6-sided die using a fair coin? What's the expected number of toss to produce the number? A: Use a binary scheme. With 3 tosses, there are 8 possible outcomes, which is more than enough to simulate 6 sides. The trick is to know which 2 to exclude. {HHH,TTT} would be a bad choice because you always have to wait until the 3rd toss to know if you failed. So we should choose something like {HHH,HHT} or {THH,THT}. The expected number of toss is given by solving$E[N]=\frac 6 8 \times 3 + \frac 2 8 (2+E[N])$Q: Place 3 points randomly on the perimeter of a square. What is P(they don't form a triangle) and P(all lie on different edges)? A:$P(they_don't_form_a_triangle) = 4/{4^3} = 1/16$;$P(all_lie_on_different_edges) =\frac {4 \times _3P_3}{4^3} = 3/8\$

Q: 10 light bulbs are lined up in a row, and it is known that no two of the adjacent light bulbs can be both on. What is the possible number of arrangements?
A: Fibonacci sequence, 144

Q: You have a large number of red and blue balls (equally many) in a large container. What is P(odd number of red) if you draw a) 3 balls from the container? b) 10 balls from the container?
A: 0.5 regardless of how many you draw. For example if 3 are drawn, the binomial distribution is 1(odd) 3(even) 3(odd) 1(even); if 4 are drawn, the binomial distribution is 1(even) 4(odd) 6(even) 4(odd) 1(even).

Q: Transport as many apples as possible to a destination 1000 km away starting with 3000 apples. For each km of transportation, you lose 1 apple.
A:  The best tactic is to always start with full load so as to maximize number of "apple km" traveled per apple lost. So the 1st stop is optimally at 3000 - 3X = 2000 <=> X = 333; the 2nd stop is optimally at 2000 - 2X = 1000 <=> X = 500. After 833 km, we are left with 1000 apples. Hence we will have 833 apples at the destination.

Q: You are playing against two players A and B individually in a tournament, with A being the better player. If you have to win at least 2 consecutive games out of 3 in order to be the champion, would you prefer the order ABA or BAB?
A: Out of the 8 outcomes {LLL,LLW,LWL,LWW,WLL,WLW,WWL,WWW}, only {WWW,WWL,LWW} are desirable. If q = P(winning against A) and p = P(winning against B) so that p > q,
P(champion with the order ABA) = pqp + pq(1-p) + (1-p)qp = 2pq - ppq
P(champion with the order BAB) = qpq + qp(1-q) + (1-q)pq = 2pq - pqq
So  P(champion with the order BAB) > P(champion with the order ABA)

## Friday, January 17, 2014

### Short Notes on HMM (Hidden Markov Model)

1. Regarding calibration
Numerical optimization is computatinoal intensive and prone to local max, but it is able to estimate any general model specification. On the other hand, EM algorithms such as Baum-Welch are more reliable in seeking global max, but they are only capable of backing out the parameters of models with distributions that are fully described by moments (Gaussian, Poisson, ...)
2. Problem with HMM, and extensions
As Soltan discusses in his paper, ordinary HMM is by definition prone to instability due to the exponentiated matrix structure. The instability dominates in the long run, meaning that state occupancy duration estimation of ordinary HMM is usually unreliable. There are two ways to go: 1) HSMM (a.k.a. explicit duration HMM, inhomogeneous HMM, etc.), or 2) Treat duration as an extra degree of state space (see Soltan).
References:


text:
# If the solutions of $X'=AX$ have a constant norm then $A$ is skew symmetric.

Mathematics Asked by As soon as possible on January 7, 2022

I have a differential equation $X’=AX$ where $Ainmathcal M_n(Bbb R)$. The question is to prove that if all the solutions have a constant norm then $A$ is skew-symmetric matrix.

What I have tried so far: let $varphi(t)=Vert X(t)Vert^2$ where $X$ is a solution. Then
$$0=varphi'(t)=2langle X'(t),X(t)rangle=2langle AX(t),X(t)rangle=2langle X(t),A^TX(t)rangle$$
How can I complete my answer? Any other suggestion?

## 3 Answers

The desired result is a consequence of the following

Observation: A symmetric matrix

$$B in mathcal M_n(Bbb R) tag 1$$

such that

$$langle y, By rangle = 0 tag 2$$

for all vectors $$y$$ vanishes, i.e.,

$$B = 0.tag 3$$

Proof of Observation:

Suppose that

$$B in mathcal M_n(Bbb R) tag{4}$$

is symmetric,

$$B^T = B; tag{5}$$

then there exists an orthogonal matrix

$$O in mathcal M_n(Bbb R), tag{6}$$

that is,

$$O^TO = OO^T = I, tag{7}$$

which diagonalizes $$B$$:

$$OBO^T = text{diag}(mu_1, mu_2, ldots, mu_n), tag{8}$$

where the $$mu_i$$, $$1 le i le n = text{size}(A)$$ are the (real) eigenvalues of $$B$$. Now consider the $$n times 1$$ (column) vectors

$$e_i = (delta_{ij})_{j = 1}^n, tag{9}$$

that is, $$i$$-th row entry of $$e_i$$ is $$1$$ and all other entries are $$0$$; we easily see that (8) implies

$$(OBO^T)e_i = mu_i e_i, tag{10}$$

and thus,

$$langle O^Te_i, BO^Te_i rangle = langle e_i, OBO^Te_i rangle = langle e_i, mu_i e_i rangle = mu_i langle e_i, e_i rangle = mu_i; tag{11}$$

setting

$$y_i = O^T e_i, tag{12}$$

we have

$$langle y_i, By_i rangle = mu_i; tag{13}$$

now if for all vectors $$y$$,

$$langle y, By rangle = 0, tag{14}$$

then in accord with (13),

$$mu_i = 0, ; 1 le i le n, tag{15}$$

and thus, via (8)

$$B = O^T text{diag}(mu_1, mu_2, ldots, mu_n) O = 0. tag{16}$$

Thus a symmetric matrix such that

$$langle y, By rangle = 0 tag{17}$$

for all vectors $$y$$ must itself be the $$0$$ matrix. End: Proof of Observation.

We apply this Observation to the problem at hand as follows:

with

$$langle x, x rangle = text{constant}, tag{18}$$

we have

$$langle x, x rangle' = 0, tag{19}$$

whence
$$langle dot x, x rangle + langle x, dot x rangle = langle x, x rangle' = 0; tag{20}$$

given that

$$dot x = Ax, tag{21}$$

we write (20) in the form

$$langle Ax, x rangle + langle x, Ax rangle = 0; tag{22}$$

whence

$$langle x, A^Tx rangle + langle x, Ax rangle = 0, tag{23}$$

or

$$langle x, A^Tx + Ax rangle = 0, tag{24}$$

or

$$langle x, (A^T + A)x rangle = 0; tag{25}$$

now $$A + A^T$$ is symmetric:

$$(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T, tag{26}$$

and since $$x$$ may be any solution to (21), it may be assumed that $$x(t)$$ may be any vector at time $$t$$, and hence (25) yields

$$langle x(t), (A^T(t) + A(t))x(t) rangle = 0; tag{27}$$

we now invoke the above Observation and conclude that

$$A^T(t) + A(t) = 0, tag{28}$$

that is,

$$A^T(t) = -A(t) tag{29}$$

for all values of $$t$$.

Answered by Robert Lewis on January 7, 2022

Think that

$$dot X = X wedge vec v = A X mbox{ with } A mbox{skew symmetric}$$

then

$$< X, dot X > = < X, X wedge vec v > = 0 Rightarrow ||X|| = C_0$$

NOTE

The external product can be generalized to $n$ dimensions.

Answered by Cesareo on January 7, 2022

Now conclude from $x^TAx=0$ for all $x$ that $A_{ii}=0$ for all $i$ by setting $x=e_i$ and then $A_{ij}+A_{ji}=0$ by setting $x=e_i+e_j$.

Answered by Lutz Lehmann on January 7, 2022

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text:
Homework Help: Escape velocity of a spoked wheel.

1. Feb 4, 2010

yuiop

Consider a spoked wheel of radius r.
The instantaneous tangential velocity of a point on the rim is v.
A test particle in the form of a small bead that can freely slide along a spoke when released, is placed as close to the axis as possible so that for the sake of argument its radial displacent is aproximately zero. If the bead is released what would the terminal velocity of the bead be when it it hit the rim of the wheel, in terms of r and v?

Please ignore any slowdown of the wheel due to consevation of angular momentum, as the test particle is considered to have negligable mass relative to the mass of the large wheel.

2. Feb 4, 2010

cesiumfrog

Have you yet attempted to integrate the centrifugal force, and change frame?

3. Feb 5, 2010

yuiop

This is my attempt to integrate the centrifugal acceleration (g) expressed as:

$$g = r \omega ^2$$

I have used angular velocity (omega) rather than tangential velocity because omega is independent of radius.

$$v_{e} = \int_0^r \, {r \omega^2} \, dr = \frac{r^2 \omega^2}{2}$$

Substituting the tangential velocity expression $v = r \omega$ back in gives:

$$v_{e}= \frac{v^2}{2}$$

However this answer is not entirely satisfactory to me because of something for reasons to do with the EP principle in relativity and secondly I have to question whether the sliding bead really leaves the end of the spoke at a velocity that is the square of the rim velocity? That is before I add in the component of velocity tangential to the spoke that the bead will aquire. Any thoughts?

Last edited: Feb 5, 2010
4. Feb 5, 2010

cesiumfrog

Since when does acceleration multiplied by distance give velocity? Dimension check fail.

Hint: force x distance gives?

Last edited: Feb 5, 2010
5. Feb 5, 2010

yuiop

OK, fair comment. I'll try again.

$$KE = \int_0^r \, {mr \omega^2} \, dr = \frac{mr^2 \omega^2}{2}$$

Substituting the tangential velocity expression $v = r \omega$ back in gives:

$$KE= \frac{mv^2}{2}$$

OK, it would appear that the kinetic energy aquired by the sliding test particle as it reaches the end of the spoke is the same as the kinetic energy of a similar test particle at rest on the rim of the wheel. Does this mean that the potential energy of a particle on the rim has the same value and this represents the work required to "raise" the particle back up to the axis of the wheel again?

Also, does this mean the final outward velocity of the sliding test particle in the frame of the rotating wheel is the same as the tangential rim velocity of a similar test particle in the non rotating frame outside the wheel?

Basically, I am trying to figure out the analogues for potential energy and "escape velocity" in a gravitational field are, in the context of a rotating disc. This is relevant to a discussion about the Equivalence Principle in another forum.

6. Feb 6, 2010

cesiumfrog

The work done on the particle in the rotating frame tells you its final velocity (which clearly is in the radial direction) in the rotating frame. This leaves one of the simplest exercises in trigonometry to calculate the final velocity in the inertial frame, and hence the multiple by which this bead (released from the axis) would be flung faster (presuming its path is not blocked at the rim) than it would if it were instead just released directly from the rim.

Huh? What form of potential? You do realise energy is frame dependent?

7. Feb 6, 2010

yuiop

The combined radial and tangential velocity is $\sqrt{(2v^2)}$ so "the multiple by which this bead (released from the axis) would be flung faster (presuming its path is not blocked at the rim) than it would if it were instead just released directly from the rim" is simply $\sqrt{2}$ but that is not really what I was trying to find out. I am more interested in the purely radial velocity as that is all that would be measured by an observer at rest with the rotating wheel frame, while the bead remains on the spoke.

I mean the equivalent of gravitational potential bu as applied to a rotating disc in flat space. The equivalence principle of General Relativity implies that there is an equivalence between a rotating frame and a gravitational frame and it is that relationship that I am trying to explore.

Most of the time I spend in physicsforums is in the Special & General relativity section, so yes, I am vaguely aware that some quantities are frame dependent.


text:
# Solve for x to the nearest hundredth. 3^x = 1.295

## Question:

Solve for {eq}x {/eq} to the nearest hundredth.

{eq}3^x = 1.295 {/eq}

## Solving for a variable

We have to solve the given equation for x. However, x is an exponent and, as the base on both sides of the equation can not be made, the only way of bringing it down is by using the natural log.

## Answer and Explanation:

The equation is solved as follows by taking the natural log of both sides.

\begin{align} 3^x& = 1.295\\ x\ln 3&=\ln 1.295&&&&\text{Taking the natural log of both the sides}\\ x&=\frac{\ln 1.295}{\ln 3}\\ \therefore x&=0.2353 \end{align}

#### Learn more about this topic:

How to Solve Logarithmic Equations

from

Chapter 10 / Lesson 8
42K


text:
# Y4 Maths Fractions Part 2

Question:
What is 36% as a decimal?
Question:
What is 60% plus $\frac{1}{5}$ as a decimal ?


text:
1st Example

Imagine we have three web pages $$A$$, $$B$$ and $$C$$ such there is only one link from page $$A$$ pointing to page $$B$$ and another link from page $$B$$ pointing to page $$C$$. Using the PageRank formula with $$p=0,85$$, we find

$$\begin{array}{ccl} PR\left(A\right) & = & 0,15\\ PR\left(B\right) & = & 0,15+0,85\times PR\left(A\right)=0,15+0,85\times0,15=0,2775\\ PR\left(C\right) & = & 0,15+0,85\times PR\left(B\right)=0,15+0,85\times0,2775=0,385875 \end{array}$$

We were able to easily find the exact PageRank values for each web page.

Note that $$PR\left(A\right)<PR\left(B\right)<PR\left(C\right)$$.

2nd Example

we have now three pages $$A$$, $$B$$ and $$C$$ such that one link pointing from $$A$$ to $$B$$, one link pointing from $$B$$ to $$C$$ and one link pointing from $$C$$ to $$A$$. Using the PageRank formula with $$p=0,85$$, we find

$PR\left(A\right)=0,15+0,85\times PR\left(C\right)$ $PR\left(B\right)=0,15+0,85\times PR\left(A\right)$ $PR\left(C\right)=0,15+0,85\times PR\left(B\right)$

Now we can not immediately find the exact values for the PageRank for each page. We need to solve the simultaneous equations with respect to $$PR\left(A\right), PR\left(B\right)$$ and $$PR\left(C\right)$$. It is easily verified that the solution is $PR\left(A\right)=PR\left(B\right)=PR\left(C\right)=1.$

As expected, all the three pages have equal PageRank value. You may try other examples using this applet, where you may play with at most $$6$$ pages ($$A$$, $$B$$, $$C$$, $$D$$, $$E$$ and $$F$$), and define how they are linked to each other.

In general , to compute the PageRank value for a set of $$n$$ web pages linked to each other, we will need to simultaneously solve $$n$$ equations (the equations for the PageRank for each page) with $$n$$ unknowns (the PageRank values for reach page). When $$n$$ is small, this is not a problem. However, the Google database includes several millions of web pages, and this raises difficulties on finding the exact solution for the simultaneous equations. In such case, the approach consists on defining an iterative procedure to find approximate values for the PageRank, the approximation becoming progressively better each time we perform a new iteration.

Let us look at an example considering the three web pages $$A$$, $$B$$ and $$C$$, and links as described in the 2nd example, taking as initial approximations the value $$1$$ for the three PageRank values. Using the PageRank formula for each page, we find

$PR\left(A\right)=0,15+0,85\times PR\left(C\right)=0,15+0,85\times1=1$ $PR\left(B\right)=0,15+0,85\times PR\left(A\right)=0,15+0,85\times1=1$ $PR\left(C\right)=0,15+0,85\times PR\left(B\right)=0,15+0,85\times1=1.$

That is, we found the same PageRank values, all remaining equal to $$1$$. This happened because we chose the initial values equal to the exact PageRank value for each page that we had been able to compute previously.

What happens if our initial guesses for the PageRank values were different from the exact ones? To have an idea of what will happen, choose our initial guesses all equal to $$0,5$$. Using the PageRank formula, we find

$PR\left(A\right)=0,15+0,85\times PR\left(C\right)=0,15+0,85\times0,5=0,575$ $PR\left(B\right)=0,15+0,85\times PR\left(A\right)=0,15+0,85\times0,5=0,575$ $PR\left(C\right)=0,15+0,85\times PR\left(B\right)=0,15+0,85\times0,5=0,575$.

Using the PageRank formula again with the values obtain above, we now find

$PR\left(A\right)=0,15+0,85\times PR\left(C\right)=0,15+0,85\times0,575=0,63875$ $PR\left(B\right)=0,15+0,85\times PR\left(A\right)=0,15+0,85\times0,575=0,63875$ $PR\left(C\right)=0,15+0,85\times PR\left(B\right)=0,15+0,85\times0,575=0,63875.$

Now we may repeat this procedure as long we like. After $$10$$ repetitions, or iterations, of this procedure, we find

$PR\left(A\right)=PR\left(B\right)=PR\left(C\right)=0,90156...,$

while, after $$67$$ iterations, we find the approximations

$PR\left(A\right)=PR\left(B\right)=PR\left(C\right)=0,99999....$

We see that the computed approximations for the PageRank values are getting closer to the exact ones. This will happen regardless of our initial guess for the PageRank for each web page (for this example with three web pages, if we start with an initial guess equal to $$1000$$ for each page, after performing $$100$$ iterations the approximation obtained is $$1,00044$$ for each page). Moreover, this behaviour for the approximations constructed will happen for any other set of web pages and links among them. It would be nice to have an explanation (and a proof) for this observation.

Remark: The iterative procedure described may be interpreted as an application of an iterative algorithm to solve simultaneous equations, known as the Jacobi method. There is an alternative iterative method, known as the Gauss-Seidel method, differing from the described one by immediately using the newly calculated approximations in the subsequent calculations. To illustrate how this Gauss-Seidel method works, look again at the second usage of the iterative procedure described above. After choosing the initial guess equal to $$0,5$$ for the PageRank for each page, we would find for the first iteration

$$\begin{array}{ccl} PR\left(A\right) & = & 0,15+0,85\times PR\left(C\right)=0,15+0,85\times0,5=0,575\\ PR\left(B\right) & = & 0,15+0,85\times PR\left(A\right)=0,15+0,85\times0,575=0,63875\\ & & (\mbox{using }PR\left(A\right)=0,575\mbox{ instead of }PR\left(A\right)=0,5)\\ PR\left(C\right) & = & 0,15+0,85\times PR\left(B\right)=0,15+0,85\times0,63875=0,69293.\\ & & (\mbox{using } PR\left(B\right)=0,63875 \mbox{ instead of }PR\left(B\right)=0,5) \end{array}$$

In general, the Gauss-Seidel method produces approximations that get closer to the exact values faster than the approximations constructed by the Jacobi method.


text:
• ### Gaia Data Release 2: The astrometric solution(1804.09366)

April 25, 2018 astro-ph.IM
Gaia Data Release 2 (Gaia DR2) contains results for 1693 million sources in the magnitude range 3 to 21 based on observations collected by the European Space Agency Gaia satellite during the first 22 months of its operational phase. We describe the input data, models, and processing used for the astrometric content of Gaia DR2, and the validation of these results performed within the astrometry task. Some 320 billion centroid positions from the pre-processed astrometric CCD observations were used to estimate the five astrometric parameters (positions, parallaxes, and proper motions) for 1332 million sources, and approximate positions at the reference epoch J2015.5 for an additional 361 million mostly faint sources. Special validation solutions were used to characterise the random and systematic errors in parallax and proper motion. For the sources with five-parameter astrometric solutions, the median uncertainty in parallax and position at the reference epoch J2015.5 is about 0.04 mas for bright (G<14 mag) sources, 0.1 mas at G=17 mag, and 0.7 mas at G=20 mag. In the proper motion components the corresponding uncertainties are 0.05, 0.2, and 1.2 mas/yr, respectively. The optical reference frame defined by Gaia DR2 is aligned with ICRS and is non-rotating with respect to the quasars to within 0.15 mas/yr. From the quasars and validation solutions we estimate that systematics in the parallaxes depending on position, magnitude, and colour are generally below 0.1 mas, but the parallaxes are on the whole too small by about 0.03 mas. Significant spatial correlations of up to 0.04 mas in parallax and 0.07 mas/yr in proper motion are seen on small (<1 deg) and intermediate (20 deg) angular scales. Important statistics and information for the users of the Gaia DR2 astrometry are given in the appendices.
• Parallaxes for 331 classical Cepheids, 31 Type II Cepheids and 364 RR Lyrae stars in common between Gaia and the Hipparcos and Tycho-2 catalogues are published in Gaia Data Release 1 (DR1) as part of the Tycho-Gaia Astrometric Solution (TGAS). In order to test these first parallax measurements of the primary standard candles of the cosmological distance ladder, that involve astrometry collected by Gaia during the initial 14 months of science operation, we compared them with literature estimates and derived new period-luminosity ($PL$), period-Wesenheit ($PW$) relations for classical and Type II Cepheids and infrared $PL$, $PL$-metallicity ($PLZ$) and optical luminosity-metallicity ($M_V$-[Fe/H]) relations for the RR Lyrae stars, with zero points based on TGAS. The new relations were computed using multi-band ($V,I,J,K_{\mathrm{s}},W_{1}$) photometry and spectroscopic metal abundances available in the literature, and applying three alternative approaches: (i) by linear least squares fitting the absolute magnitudes inferred from direct transformation of the TGAS parallaxes, (ii) by adopting astrometric-based luminosities, and (iii) using a Bayesian fitting approach. TGAS parallaxes bring a significant added value to the previous Hipparcos estimates. The relations presented in this paper represent first Gaia-calibrated relations and form a "work-in-progress" milestone report in the wait for Gaia-only parallaxes of which a first solution will become available with Gaia's Data Release 2 (DR2) in 2018.
• Context. The first Gaia Data Release contains the Tycho-Gaia Astrometric Solution (TGAS). This is a subset of about 2 million stars for which, besides the position and photometry, the proper motion and parallax are calculated using Hipparcos and Tycho-2 positions in 1991.25 as prior information. Aims. We investigate the scientific potential and limitations of the TGAS component by means of the astrometric data for open clusters. Methods. Mean cluster parallax and proper motion values are derived taking into account the error correlations within the astrometric solutions for individual stars, an estimate of the internal velocity dispersion in the cluster, and, where relevant, the effects of the depth of the cluster along the line of sight. Internal consistency of the TGAS data is assessed. Results. Values given for standard uncertainties are still inaccurate and may lead to unrealistic unit-weight standard deviations of least squares solutions for cluster parameters. Reconstructed mean cluster parallax and proper motion values are generally in very good agreement with earlier Hipparcos-based determination, although the Gaia mean parallax for the Pleiades is a significant exception. We have no current explanation for that discrepancy. Most clusters are observed to extend to nearly 15 pc from the cluster centre, and it will be up to future Gaia releases to establish whether those potential cluster-member stars are still dynamically bound to the clusters. Conclusions. The Gaia DR1 provides the means to examine open clusters far beyond their more easily visible cores, and can provide membership assessments based on proper motions and parallaxes. A combined HR diagram shows the same features as observed before using the Hipparcos data, with clearly increased luminosities for older A and F dwarfs.
• ### Gaia data release 1, the photometric data(1612.02952)

Dec. 9, 2016 astro-ph.IM
Context. This paper presents an overview of the photometric data that are part of the first Gaia data release. Aims. The principles of the processing and the main characteristics of the Gaia photometric data are presented. Methods. The calibration strategy is outlined briefly and the main properties of the resulting photometry are presented. Results. Relations with other broadband photometric systems are provided. The overall precision for the Gaia photometry is shown to be at the milli-magnitude level and has a clear potential to improve further in future releases.
• ### Multi-scale and multi-domain computational astrophysics(1402.4821)

Feb. 19, 2014 astro-ph.IM
Astronomical phenomena are governed by processes on all spatial and temporal scales, ranging from days to the age of the Universe (13.8,Gyr) as well as from km size up to the size of the Universe. This enormous range in scales is contrived, but as long as there is a physical connection between the smallest and largest scales it is important to be able to resolve them all, and for the study of many astronomical phenomena this governance is present. Although covering all these scales is a challenge for numerical modelers, the most challenging aspect is the equally broad and complex range in physics, and the way in which these processes propagate through all scales. In our recent effort to cover all scales and all relevant physical processes on these scales we have designed the Astrophysics Multipurpose Software Environment (AMUSE). AMUSE is a Python-based framework with production quality community codes and provides a specialized environment to connect this plethora of solvers to a homogeneous problem solving environment.
• ### The Astrophysical Multipurpose Software Environment(1307.3016)

July 11, 2013 astro-ph.IM
We present the open source Astrophysical Multi-purpose Software Environment (AMUSE, www.amusecode.org), a component library for performing astrophysical simulations involving different physical domains and scales. It couples existing codes within a Python framework based on a communication layer using MPI. The interfaces are standardized for each domain and their implementation based on MPI guarantees that the whole framework is well-suited for distributed computation. It includes facilities for unit handling and data storage. Currently it includes codes for gravitational dynamics, stellar evolution, hydrodynamics and radiative transfer. Within each domain the interfaces to the codes are as similar as possible. We describe the design and implementation of AMUSE, as well as the main components and community codes currently supported and we discuss the code interactions facilitated by the framework. Additionally, we demonstrate how AMUSE can be used to resolve complex astrophysical problems by presenting example applications.


text:
# Why the 1700 Japanese could have discovered calculus and modern science by 2200 or 2700

Originally published at 狗和留美者不得入内. You can comment here or there.

Some thoughts on Greek, Persian, Arab, Indian, and Chinese science

In the above piece, I omitted the accomplishments of Japanese mathematics based on the soropan (a Japanese version of the abacus imported from China in the 14th century via the Korean peninsula) and a Chinese 13th century mathematics text by Zhu Shijie [3]. These are well documented at this Japanese website [8] and in a book [4] by early 20th century math historian Yoshio Mikami. I have also uploaded here a PDF of that book: historyofjapanes00smitiala_bw.

I had known about Seki Takakazu [1] before, who was the first Japanese mathematician to do original work that was history making. There were Japanese mathematicians before him, but they mostly wrote books pertaining to the soropan as well as regurgitation of some results from 13th century Chinese math. I had long read that he discovered determinants and even Bernoulli numbers, and in addition to that, computed pi to 10 decimal places through a method rediscovered in the West in 20th century.  Today, I was shocked to learn that a Japanese mathematician 20 some years younger than Seki by the name of Takebe Kenkō [6] actually figured the infinite series for $\arcsin^2 x$. I skimmed thru the part on that in Mikami’s book, and it appears that he did in a computationally observational way. He also computed pi to 41 digits using some other mathematical method. What is of course more impressive of the two is how he managed to figure an infinite series for a trigonometric function, though of course it would have been much more so if he did so in deductive way, as Madhava in 14th century India perhaps or likely did. Mikami in his book notes that Seki never left, so it is unlikely that he got ideas from the West. And while the West and Arab world was leagues ahead at the theoretical level, China was ahead at the computational level, with relates to how Seki figured out determinants and resultants before the Western mathematicians did, not to mention Bernoulli numbers around the same time.

Before I easily noticed a certain relative absence of Chinese in mathematics, with an emphasis on lack of rigorous proof or axiomatic method. But in high school, I already knew of the likes of Liu Hui, Zu Chongzhi, Yang Hui (Pascal’s triangle in China is named after him), etc. Zu Chongzhi’s book wherein he obtained $3.1415926 < \pi < 3.1415927, \pi \approx \frac{355}{113}$ unfortunately was lost by the 12th century (they say the later students found it too difficult). Nonetheless, the result was recorded by Li Chunfeng and thus still preserved. The Chinese with the Confucian tradition, the imperial exam system well established around 700, and a large nation to govern later deemphasized mathematics. On the other hand, Japan, have gotten its culture, including its writing, much from China, were not burdened by these. So when the mathematics knowledge from China entered Japan, the tradition of wasan established itself quite quickly and solidly, and very quickly, the Japanese vastly exceeded what the Chinese had done in mathematics.

Given that in addition to $\arcsin^2$ series result, a different Japanese mathematician by name of Toshikiyo KAMATA [2] figured out the series for $\sin$ as well. So one can only imagine, given that the wasan tradition was already deeply embedded by then, that the Japanese, assuming left alone, would have by 2000 AD figured out the series of more trigonometric function. And given that those wasan mathematicians in Japan also figured out much later, presumably devoid of direct Western influence, a technique for finding minimums and maximums of polynomials that Fermat figured out in 17th century, it’s reasonable to expect that they would have discovered calculus at the level of Leibniz and Newton by 2500 AD at the latest.

It’s interesting how much Japan, after learning what it could from China, complemented China in an area where China was relatively weak in, relative to its size especially. Again, Japan was not burdened by imperial exams, exclusive use of logographic characters for writing, study of the classics, and an gigantic land and population to govern. As for the problem of Chinese characters, the Koreans introduced an alphabet in the 15th century, Hangul, and given that the first book printed with metal movable type happened in Korea, one can imagine that eventually, there would have been a printing explosion in Korea as well, notwithstanding some opposition to Hangul by the elite, which was revived in the late 16th century. There was significant interaction between Japan and Korea, especially in the late 16th and early 17th century when Japan tried and failed to conquer Korea, which according to Wikipedia due to military pressure had developed cannons more advanced than Chinese ones.

There’s a lot of anti-Japanese sentiment in China (much for Japanese conquests in China in modern times), much of which comes across as rather infantile. Some Chinese still make fun of Japan for having gotten its writing from China, viewing Japan as very derivative in general. However, the Japanese elite, despite their much lower start, noticed the fun and potential importance of mathematical research much more than did the Chinese elite did and made some serious progress in that regard. There are some Westerners who believe that without the axiomatic system of the Greeks and with Confucianism and an imperial exam system inhibiting innovation in science and technology, especially theoretical science, East Asians would have never gotten anywhere serious. I used to believe this too, but after seeing what the Japanese did, no longer so. It is true of course that the East Asian way of going about mathematics was a high M lower V type, while the Greek way, so abstract and theoretical, was the high V lower M type, which is consistent with psychometric results. Interestingly though, in the modern era after intellectual contact with the West, the Japanese have excelled more in algebra than in analysis.

Seeing how the Japanese after taking in some Chinese mathematics soon made the Chinese in mathematics look like idiots, my opinion of the Han Chinese can’t help but go down further. Relatedly, Japanese used the earlier Mongols and Manchus conquests of China to justify their conquest of China in the modern era as well. Speaking of which, during the Qing Dynasty, the only original discovery made in mathematics in China was actually done by a Mongol born in Inner Mongolia, who also helped the government with surveying in Xinjiang. In detail, after seeing that French Jesuit brought to China series for sine and cosine but not their derivations, the Mongol Ming Antu set out to find them and succeeded, discovering Calatan numbers and some other series or identities in the process. Of course, unlike with Seki, if not for direct Western influence, he almost certainly would not have worked on this problem.

In any case, seeing what the Japanese accomplished in mathematics in only  a few centuries drawing up only the work of earlier Chinese mathematicians, I believe that East Asians could have developed modern science by themselves. It might have taken a while, but I believe in Japan, it would have happened eventually. Given military competition between Japan and Korea, I can imagine that sooner or later, wasan would have been applied towards military matters as well, with hints towards physics, not to mention that according to [8], it had proliferated into the real world as well. Seki Takakazu was a contemporary of Newton, and although he achieved nowhere near as math, I believe he was just as talented. After all, unlike Newton, he and his Japanese colleagues had access to neither movable type printing on an alphabet nor all that scientific literature of the Greeks and Arabs, not to mention the Western European scientific community post-Renaissance. So in this regard, Seki really was a genius of first order, who did so much heaving lifting himself, and not only that, he founded the self-sustaining tradition of wasan in Japan. The giant whose shoulders he stood on directly was Zhu Shijie, whose works were translated and disseminated by a few of his Japanese predecessors.

Though Japan was not a civilization of its own, its strong discernment and interest in science, technology, and military, as well as ambitious and revolutionary improvement based upon what it took in from the outside world much contrasts it from China. In other words, the Japanese were quick to realize that knowledge and technology from the outside world were superior to their own, but without willing kowtow to the outsiders politically or intellectually, they single-mindedly and courageously developed their own school to eventually exceed those outsiders. One saw the same happening in Japanese theoretical physics and mathematics in the 20th century. The Japanese did not have the smug, self-centered, and narrow-minded attitude that the Chinese had with respect to the outside world, who even engaged in much self-delusion upon visibly superior European mathematics, astronomy, and military technology. There are too many spineless opportunist Chinese using American name schools and companies to engage in political, academic, and business comprador-ship and shenanigans in China; in general, I don’t think that Chinese actually inherently respect knowledge all that much, instead treating academics more as a way to gain status, climb up the ladder, and feed their narcissism and engaging in much intellectual posturing. If Chinese want to improve their culture and their country, they need to actively oust those people, and instead actively promote the likes of Seki Takakazu in their home country.

References

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• #### 中国姓氏的南北分布

有意思的是分布是很不均匀的，也就是说好多姓是南方比例远高于北方比例或相反。从这一点，你如果知道一个人的父母的姓氏，你就可以对他是北方人还是南方人有个有点统计显著的估计。 读者可以在包含大姓分布图的…

• #### Why I think Stanford professor James Landay's piece on computing in China in 2011 exposed his idiocy

I dittoed his blog post on my LiveJournal too, see https://gmachine1729.livejournal.com/175023.html. For the original, see…

• #### China Will Overtake the US in Computing…Maybe, Someday… by Stanford professor James Landay in 2011

Original at https://dubfuture.blogspot.com/2011/12/china-will-overtake-us-in.html. [note: the following is a rough draft — I appreciate…

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# If an account that earns interest compounded continuously takes 12 years to do in value, how long will it take to triple in value?

Dec 8, 2016

We first need the growth factor $g$, the number that we use to multiply the amount with every year.

#### Explanation:

Example: if the yearly interest is 5% then the amount is multiplied by (100%+5%)/(100%)=1,05 every year, so after three years it is multiplied three times, or by
$1.05 \times 1.05 \times 1.05 = {1.05}^{3} = 1.157625$

We need to find $g$ from ${g}^{12} = 2$
$g = \sqrt[12]{2} = {2}^{\frac{1}{12}} = 1.05946 \ldots$

Next we need to solve ${\left(1.05946 \ldots\right)}^{t} = 3$ for $t$ (in years).

Method 1 - Using logs:
$\log {\left(1.05946 \ldots\right)}^{t} = \log 3 \to$
$t \cdot \log 1.05946 \ldots = \log 3 \to$
$t = \frac{\log 3}{\log 1.05946 \ldots} = 19.02 \ldots \to$

Answer : $t = 19.02$ years.

Method 2 - Using the GC:
Set ${Y}_{1} = {1.05946}^{X}$
Set ${Y}_{2} = 3$
$X : 0 \to 24$ (twice the time for doubling)
$Y : 0 \to 6$ (so the horizontal line will be in the middle)


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Question

# f(t)=3e^{2t} Determine L[f] Let f be a function defined on an interval [0,infty) The Laplace transform of f is the function F(s) defined by F(s) =int_0^infty e^{-st}f(t)dt provided that the improper integral converges. We will usually denote the Laplace transform of f by L[f].

Laplace transform
ANSWERED
asked 2021-01-08
$$f(t)=3e^{2t}$$
Determine L[f]
Let f be a function defined on an interval $$[0,\infty)$$
The Laplace transform of f is the function F(s) defined by
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$
provided that the improper integral converges. We will usually denote the Laplace transform of f by L[f].

## Answers (1)

2021-01-09
Step 1
we have to find the laplace of $$f(t)=3e^{2t}$$ by using the definition of the laplace transform.
as we know that the laplace transform of f is the function F(s) defined by
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$
therefore,
$$L[f]=\int_0^\infty e^{-st}f(t)dt$$
$$L[3e^{2t}]=\int_0^\infty e^{-st}(3e^{2t})dt$$
$$=3\int_0^\infty e^{-st}(e^{2t})dt$$
$$=3\int_0^\infty e^{-st+2t}dt$$
$$=3\int_0^\infty e^{-(s-2)t}dt$$
Step 2
now as we know that $$\int e^{bx}dx=\frac{e^{bx}}{b}$$
therefore,
$$L[3e^{2t}]=3\left(\frac{e^{-(s-2)t}}{-(s-2)}\right)_0^\infty$$
$$=-\frac{3}{(s-2)}\left(e^{-(s-2)t}\right)_0^\infty$$
$$=-\frac{3}{(s-2)}\left(e^{-(s-2)\infty}-e^{-(s-2)(0)}\right)$$
$$=-\frac{3}{(s-2)}(e^{\infty}-e^0)$$
$$=-\frac{3}{(s-2)}(0-1)$$
$$=\frac{3}{(s-2)}$$
Step 3
therefore the laplace transform of $$f(t)=3e^{2t} \text{ is } \frac{3}{(s-2)}$$

expert advice

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## This Paladin PC sounds too OP. If they are, how can I fix it?

I just constructed a Level 5 Human Paladin character for one of my players and this is what I got.

His base stats are:

• STR 15
• DEX 14
• CON 18
• INT 14
• WIS 17
• CHR 12

This was done by using the 4d6 minus the lowest number method.

He has an Armor Class of 18 because he has Chain mail armor and a shield.

He chose Oath of The Ancients and thus has ensnaring strike, speak with animals, moonbeam and misty step as well as choosing bless and shield of faith. He also talks celestial.

His flaw is that he puts others before himself and will always jump in front of danger to save one of his companions but even with that, having 65 hit points (obtained through rolling) and an armor class of 18 while having the strength and proficiency to not be over encumbered.

The part the really confuses me is that he has 30 speed, while wearing heavy armor and carrying a large shield; all because of his strength and proficiency’s.

This all sounds way too powerful for Level 5; am I wrong? And if I’m not, how could I balance this, preferably without telling the player to change his character (like fighting enemies that exploit a certain loophole in the character’s build)?

## Does the target choose which side they move to for Wall of Ice?

The description for Wall of Ice states

“If the wall cuts through a creature’s space when it appears, the creature within its area is pushed to one side of the wall and must make a Dexterity saving throw. On a failed save, the creature takes 10d6 cold damage, or half as much damage on a successful save.”

For both Wall of Stone and Wall of Force the spells specificity it is “Your choice” which side they arrive on, with Wall of Stone triggering a Dex save to see if they can avoid it. Wall of Ice’s save only state doing damage, but nothing about getting a chance to escape.

Does that lack of specifying the caster’s choice mean that it is the target’s choice by default? Do creatures surrounded by the walls not get a chance to escape as long as they are untouched?

My current reading is that if a target is hit by the wall they can escape freely, but if you can enclose the targets without the wall touching them they would be trapped without a save to escape. Would that be correct?

## Does the creature in room 27 of White Plume Mountain use the magic items they have?

Room 27 contains a creature who has some additional items listed as treasure:

Are those items active on the creature (changing some of their stats and giving some additional options)? Or are those not meant to be active/used and to be given as treasure?

If they do use them, does that adjust their CR?

## Assign resources that each have a certain amount of work they can provide to tasks that require a certain amount of work

I’m attempting to do a hobby automation project and have come to a roadblock. I have a certain problem I need to solve, but can’t think of the solution nor what the name for the problem would be.

Say we have n tasks where each task requires $$x_i$$ amount of work to be done and m resources where each resource can provide $$y_j$$ work. The total amount of work required will equal the the total amount of work the resources can provide, i.e. $$\sum_{i=1}^{n} y_i = \sum_{j=1}^{m}x_j$$. For all j from 1 to m, $$y_j \in \left \{1, 2, 6, 12, 24\right \}$$ and each $$x_i = a*1 +b *2 + c*6+d*12+e*24$$. I was looking at task assignment problems, but those seemed to be a bit overkill since they were concerned with optimization where I’m just simply trying to slot the correct blocks in the right place so that I don’t have tasks that are given too few or too many resources.

My current guess is that you can iterate over each task, and give it the largest resource available that doesn’t go over the amount of work that is left for the task to be completed. It’s almost analagous to filling a jar with different sized rocks; the best way is to start with the largest rocks and then go down in size from there, so that the smaller rocks fill in the space between the larger ones. Am I not taking something into consideration that complicates this problem further? I’m sorry if this is an obvious one, but I’m a hobbyist programmer and couldn’t think of the name of the problem or of a good set of keywords to google. The closest I could find is task assignment so far.

## Are Half-Elves supposed to have a slender build like Elves, or are they supposed to have a build that’s intermediate between Humans and Elves? [5e]

The Player’s Handbook’s weight descriptions for Half-Elves are inconsistent. The Half-Elf section in Chapter 2 (pg. 38) says:

They range from under 5 feet to about 6 feet tall, and from 100 to 180 pounds, with men only slightly taller and heavier than women.

However, the “Height and Weight Range” table in Chapter 4 (pg. 121) gives a weight formula for Half-Elves of 110 + (2d8) x (2d4), which is 114 to 238 pounds.

This is very different.

Do we have any reason to label one description Correct and the other A Mistake? The 2018 PHB Errata are silent on this point.

Looking at the rest of the table’s formulas, Humans are 114 to 270 lbs, whereas Wood Elves are 102 to 180 lbs and High Elves are 92 to 170 lbs. (I’ve excluded the Drow because they’re significantly shorter.)

So are Half-Elves supposed to have a slender build like elves (100-180 lbs), or are they supposed to have a build that’s intermediate between Human and Elf (114-238 lbs)?

[As a side note, changing the Half-Elves’ weight modifier in the table from its current 2d4 (which is like a Human’s) to only 1d4 (like the Wood and High Elves’) would yield a calculated range of 112 to 176 pounds.]

## Players get frustrated when they couldn’t solve a hard diplomatic problem, how to get them to think out of the box

Problem:

Players got themselves into a diplomatic problem that they know is probably above their pay grade in terms of difficulty. They spent a session trying to figure out this problem with talking to people, and rolling different charisma checks in order to try to persuade people they probably had no business persuading (rolls were average, arguments weren’t extremely compelling). The party didn’t plan any grand schemes, any extraordinary strategies, no clever ideas on the spot, but rather tried very basic head first dialogue.

This has happened in the past in regards to combat, and the party has with a recent deadly encounter had to think out of the box more (one player even said: “guys we need to plan more and think less about just hacking and slashing sometimes”). Now it’s a more diplomatic problem that doesn’t seem as easy as rolling a single charisma check and hoping it works out.

In the end, the party did not manage to solve the diplomatic problem (although there is room in the future for them to try again with the upper hand), and one of the players said they did not enjoy the session. Player enjoyment is my top priority. But I also think dnd is best when there’s risk, when you can fail rolls, when the PCs don’t always win (not that I actively seek this out though).

How can I get the party to perform less linearly in dialogue-related problems?

An example problem:

P: If trying to out smart a bad person with a lot of influence in the town

A: There are options for framing the person, bribing people, seeking dirt on this person to find their weakness, tarnishing their reputation, trying to prove their wrong doing by seeking out evidence, and a bunch of other possibilities.

I’ve tried to have a brief session-0 talk again about if they want dialogue-related problems handicapped, and they didn’t seem to take to that, but rather felt like they tried everything and didn’t know what else to do. I also did a postmortem on this problem and tried to give different options they could have tried, but I get a feeling the players feel like they still tried everything and failed and the session was “a waste” (even though they still got exp, still got some loot, and got some more plot).

Kind of at a loss of how to tackle this issue that isn’t just: “Go watch some dnd podcasts to get ideas, or go read X, Y, and Z resource on the subject”.

## Is it possible to use a whip as an instrument because it creates a “sonic boom?” Furthermore, can they attack while “performing?”

I noticed that a whip when used creates a sonic boom. The definition of a musical instrument is as follows

A musical instrument is a device created or adapted to make musical sounds. In principle, any object that produces sound can be considered a musical instrument—it is through purpose that the object becomes a musical instrument.

Your Charisma (Performance) check determines how well you can delight an audience with music, dance, acting, storytelling, or some other form of entertainment.

Because of this, I was wondering if I could do something akin to a bladedancer with whips for dancing and use my charisma modifier to attack as my character danced around the battlefield. As far as I know, that’s not an option in 5e (at least not with officially available material, but material allowed under unearthed arcana would work.

If an altered bladedancer is not an option, I was thinking maybe bard/ranger (bard 1st) with war caster feat and dual wielding. Would this work?

## Can a creature take turns as normal if they are inside an Antimagic Field while another creature casts Time Stop?

Time stop is a spell that stops time for other creatures, allowing one creature to take multiple turns in a row. It says:

You briefly stop the flow of time for everyone but yourself. No time passes for other creatures, while you take 1d4 + 1 turns in a row, during which you can use actions and move as normal.

Certainly, this is a magical effect. The spell causes the flow of time to stop for other creatures, and while you are taking multiple turns, no time flows for them. I imagine antimagic field can defeat it. The relevant text says:

Spells. Any active spell or other magical effect on a creature or an object in the sphere is suppressed while the creature or object is in it.

So, imagine combat between Annie, Tim, and Charlie. Ordinarily, initiative might look something like this:

Annie → Charlie → Tim

Suppose Annie casts time stop and rolled a 1 on their d4. Thus, they take 2 turns in a row. Initiative would look like this:

Annie → Annie → Charlie → Tim

Now imagine that Tim cast antimagic field, and following their turn, Annie casts time stop. Suppose they rolled a 1 on their d4 so that they can take 2 turns in a row. What would the initiative order look like?

Here are some possible resolutions I can think of, but none satisfy me totally:

1. Time stop defeats antimagic field. The initiative order is: Annie → Annie → Charlie → Tim. The reason this is unsatisfactory is that time stop shouldn’t seem to prevail because it’s a spell, and antimagic field defeats spells.

2. Time stop cannot be cast while there is an active antimagic field, because there exists some creatures you can’t stop time for. The reason this is unsatisfactory is there is no rule that prevents these two spells from being active at the same time. Also, since things like beholders exist, it’s not unreasonable to say there is almost always an active area of antimagic somewhere in the world, and that means time stop can almost never be used.

3. The caster of time stop and antimagic field take their turns as normal while everyone else is frozen in time. Thematically and narratively, this seems the most logical. So we go through the turn order, treating every turn Alice would have taken as one full round. For this scenario, since Alice takes two turns in a row, then we can imagine two rounds going by. Ordinarily, everyone but Alice takes a turn, but now we unfreeze anyone inside an area of antimagic. So initiative would be: Annie (time stop starts) → Charlie (frozen in time) → Tim (unfrozen) → Annie (time stop ends) → Charlie → Tim. The reason this is unsatisfactory is because we’re advancing the “round count” now, which feasibly triggers things like lair actions that activate on a certain initiative count. It does have the side effect of allowing Tim to act normally though, affecting other creatures if he wants, because he isn’t bound by time stop and Annie isn’t the one doing the violations of the rules of the spell.

4. There is no answer to this question, and this is solidly in the zone of DM adjudication. This is unsatisfactory because, well, all questions answered that way tend to be unsatisfactory.

So which is it? Or is it an option I haven’t listed here? Can Tim take turns as normal while inside an antimagic field if Annie casts time stop?


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# Aristarchus of Samos

Aristarchus of Samos (/ˌærəˈstɑːrkəs/; Greek: Ἀρίσταρχος ὁ Σάμιος, Aristarkhos ho Samios; c. 310 – c. 230 BC) was an ancient Greek astronomer and mathematician who presented the first known heliocentric model that placed the Sun at the center of the known universe with the Earth revolving around it. He was influenced by Philolaus of Croton, but Aristarchus identified the "central fire" with the Sun, and he put the other planets in their correct order of distance around the Sun.[2] Like Anaxagoras before him, he suspected that the stars were just other bodies like the Sun, albeit further away from Earth. His astronomical ideas were often rejected in favor of the geocentric theories of Aristotle and Ptolemy. Nicolaus Copernicus attributed the heliocentric theory to Aristarchus.[3]

Aristarchus of Samos
Statue of Aristarchus of Samos at the Aristotle University of Thessaloniki
Bornc. 310 BC
Diedc. 230 BC (age c. 80)
NationalityGreek
Occupation
• Scholar
• Mathematician
• Astronomer

## Heliocentrism

The original text has been lost, but a reference in Archimedes' book The Sand Reckoner (Archimedis Syracusani Arenarius & Dimensio Circuli) describes a work by Aristarchus in which he advanced the heliocentric model as an alternative hypothesis to geocentrism:

You are now aware ['you' being King Gelon] that the "universe" is the name given by most astronomers to the sphere the centre of which is the centre of the earth, while its radius is equal to the straight line between the centre of the sun and the centre of the earth. This is the common account (τά γραφόμενα) as you have heard from astronomers. But Aristarchus has brought out a book consisting of certain hypotheses, wherein it appears, as a consequence of the assumptions made, that the universe is many times greater than the "universe" just mentioned. His hypotheses are that the fixed stars and the sun remain unmoved, that the earth revolves about the sun on the circumference of a circle, the sun lying in the middle of the orbit, and that the sphere of the fixed stars, situated about the same centre as the sun, is so great that the circle in which he supposes the earth to revolve bears such a proportion to the distance of the fixed stars as the centre of the sphere bears to its surface.[4]

Aristarchus suspected the stars were other suns[5] that are very far away, and that in consequence there was no observable parallax, that is, a movement of the stars relative to each other as the Earth moves around the Sun. Since stellar parallax is only detectable with telescopes, his accurate speculation was unprovable at the time.

It is a common misconception that the heliocentric view was held as sacrilegious by the contemporaries of Aristarchus.[6] Lucio Russo traces this to Gilles Ménage's printing of a passage from Plutarch's On the Apparent Face in the Orb of the Moon, in which Aristarchus jokes with Cleanthes, who is head of the Stoics, a sun worshipper, and opposed to heliocentrism.[6] In the manuscript of Plutarch's text, Aristarchus says Cleanthes should be charged with impiety.[6] Ménage's version, published shortly after the trials of Galileo and Giordano Bruno, transposes an accusative and nominative so that it is Aristarchus who is purported to be impious.[6] The resulting misconception of an isolated and persecuted Aristarchus is still transmitted today.[6][7]

According to Plutarch, while Aristarchus postulated heliocentrism only as a hypothesis, Seleucus of Seleucia, a Hellenistic astronomer who lived a century after Aristarchus, maintained it as a definite opinion and gave a demonstration of it[8] but no full record has been found. In his Naturalis Historia, Pliny the Elder later wondered whether errors in the predictions about the heavens could be attributed to a displacement of the Earth from its central position.[9] Pliny[10] and Seneca[11] referred to the retrograde motion of some planets as an apparent (and not real) phenomenon, which is an implication of heliocentrism rather than geocentrism. Still, no stellar parallax was observed, and Plato, Aristotle, and Ptolemy preferred the geocentric model, which was held as true throughout the Middle Ages.

The heliocentric theory was revived by Copernicus,[12] after which Johannes Kepler described planetary motions with greater accuracy with his three laws. Isaac Newton later gave a theoretical explanation based on laws of gravitational attraction and dynamics.

## Distance to the Sun (lunar dichotomy)

Aristarchus's 3rd-century BC calculations on the relative sizes of (from left) the Sun, Earth and Moon, from a 10th-century AD Greek copy

The only known surviving work usually attributed to Aristarchus, On the Sizes and Distances of the Sun and Moon, is based on a geocentric world view. It has historically been read as stating that the angle subtended by the Sun's diameter is two degrees, but Archimedes states in The Sand Reckoner that Aristarchus had a value of ½ degree, which is much closer to the actual average value of 32' or 0.53 degrees. The discrepancy may come from a misinterpretation of what unit of measure was meant by a certain Greek term in the text of Aristarchus.[13]

Aristarchus claimed that at half moon (first or last quarter moon), the angle between the Sun and Moon was 87°.[14] He might have proposed 87° as a lower bound, since gauging the lunar terminator's deviation from linearity to one degree of accuracy is beyond the unaided human ocular limit (with that limit being about three degrees of accuracy). Aristarchus is known to have also studied light and vision.[15]

Using correct geometry, but the insufficiently accurate 87° datum, Aristarchus concluded that the Sun was between 18 and 20 times farther away than the Moon.[16] (The true value of this angle is close to 89° 50', and the Sun's distance is actually about 400 times that of the Moon.) The implicit false solar parallax of slightly under three degrees was used by astronomers up to and including Tycho Brahe, c. AD 1600. Aristarchus pointed out that the Moon and Sun have nearly equal apparent angular sizes, and therefore their diameters must be in proportion to their distances from Earth; thus, the diameter of the Sun was calculated to be between 18 and 20 times the diameter of the Moon.[17]

## Notes

1. ^ "Aristarchus of Samos: Mathematician and astronomer". World History. 8 September 2015. Archived from the original on 7 May 2018. Retrieved 29 November 2018.
2. ^ Draper, John William (2007) [1874]. "History of the Conflict Between Religion and Science". In Joshi, S. T. (ed.). The Agnostic Reader. Prometheus. pp. 172–173. ISBN 978-1-59102-533-7.
3. ^ George Kish (1978). A Source Book in Geography. Harvard University Press. p. 51. ISBN 978-0-674-82270-2.
4. ^ Heath, Thomas (1913), p. 302. The italics and parenthetical comments are as they appear in Heath's original.
5. ^ Louis Strous. "Who discovered that the Sun was a star?". solar-center.stanford.edu. Retrieved 2014-07-13.
6. Russo, Lucio (2013-12-01). The Forgotten Revolution: How Science Was Born in 300 BC and Why it Had to Be Reborn. Translated by Levy, Silvio. Springer Science & Business Media. p. 82, fn.106. ISBN 9783642189043. Retrieved 13 June 2017.; Russo, Lucio; Medaglia, Silvio M. (1996). "Sulla presunta accusa di empietà ad Aristarco di Samo". Quaderni Urbinati di Cultura Classica (in Italian). Fabrizio Serra Editore. New Series, Vol. 53 (2): 113–121. JSTOR 20547344.
7. ^ Plutarch. "De facie quae in orbe lunae apparet, Section 6". Perseus Digital Library. Tufts University. Retrieved 13 June 2017.
8. ^ Plutarch, Platonicae quaestiones, VIII, i
9. ^ Neugebauer, O. (1975). A History of Ancient Mathematical Astronomy. Studies in the History of Mathematics and Physical Sciences. 1. Springer-Verlag. pp. 697–698.
10. ^ Naturalis historia, II, 70
11. ^ Naturales quaestiones, VII, xxv, 6–7
12. ^ Joseph A. Angelo (14 May 2014). Encyclopedia of Space and Astronomy. Infobase Publishing. p. 153. ISBN 978-1-4381-1018-9.
13. ^ http://www.dioi.org/vols/we0.pdf
14. ^ Greek Mathematical Works, Loeb Classical Library, Harvard University, 1939–1941, edited by Ivor Thomas, volume 2 (1941), pages 6–7
15. ^ Heath, 1913, pp. 299–300; Thomas, 1942, pp. 2–3.
16. ^ A video on reconstruction of Aristarchus' method, in Turkish without subtitles.
17. ^ Kragh, Helge (2007). Conceptions of cosmos: from myths to the accelerating universe: a history of cosmology. Oxford University Press. p. 26. ISBN 0-19-920916-2.