TITLE: What are the alternative theories of dark energy? ($w \neq -1$)
QUESTION [37 upvotes]: There has been a lot of related questions about dark energy around here but these are usually 2-4 years old and the closest question to mine hasn't really been answered, so I am going to proceed. Experts may skip to the last paragraph, but I am going to give a short account for others.

Standard Cosmology run-through
As I understand the matter, assuming Einstein equations and an almost homogeneous and isotropic universe, the current standard theory of cosmology states from multiple kinds of observations (the cosmic microwave background, supernovae, cosmic shear...) that there have to be missing terms on the right hand side of the so-called first Friedmann equation
$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3} (\rho_{radiation} + \rho_{matter} + ???) $$
This equation basically states: the rate of expansion of the universe is proportional to density of matter and radiation ($\rho_{matter},\,\rho_{radiation}$). Even if we account for matter we detect only indirectly (i.e. dark matter), there still is another $\rho$ which gives nowadays as much as three times a bigger contribution than all the other densities added up. We call this density usually "dark energy density" $\rho_\Lambda$, but it is really just a label for the need a of a formal energy density instead of $???$ in the equation.
Normal energy densities however dilute with $a(t)$ which is a measure of expansion/contraction of any length in the universe. Every kind of energy/matter density gets diluted at a different rate. It follows that some types of densities such as radiation density $\rho_{radiation}$ got already diluted and nowadays plays basically a negligible role in the expansion. On the other hand, matter gets diluted slower and manages to persist at least as a minor element that determines the expansion.
We describe the rate of dilution by a parameter $w$, which is different for every kind of density. It dilutes as:
$\rho \propto a^{-3(1+w)}$. Since we assume a perfect fluid, the specific form of this equation is given by
$$w = \frac{\rho}{P}$$
where $P$ is the pressure in the fluid. I, however, consider talking about "pressure of dark energy" as very far-fetched and confusing when just discussing its formal need in cosmology, since we cannot just suddenly blurp out it is a perfect fluid.
For dark energy, we vary the parameter $w$ to get different cosmological models (we also  have to vary the actual density $\rho_\Lambda$, we deduce the value only for certain points in our cosmic history) and compare the results with actual observations of the universe. The best fits have a value of $w \approx -1$. I.e., dark energy seems almost not to dilute - $\rho \propto a^{\approx 0}$.
In the case of completely constant, undiluting $\rho_\Lambda$, dark energy is more of a property of space, fixed on a unit of volume, rather than a density of any kind of a dynamical particle. Then, the best choice to describe it is the cosmological constant, a modification of gravity, or vacuum energy a result of quantum theory (in a theory of everything, these could turn out to be basically the same thing).

THE QUESTION :
Now to the question - observational cosmology tries hard to assess the parameter $w$ to a higher degree of precision by multiple methods of observation - just to be really sure. Some of the astronomers I have met personally seem to be thrilled about the possibility that $w\neq -1$.
But my question is - is there a viable theory which would be able to give us the microscopical model of a density with $w\neq -1$ (but $w\approx -1$)? If yes, how does it relate to current standard and beyond-standard theories? And the last, more technical question, what would then be the "loose ends" of the theory such as free parameters?
(You may obviously also describe relevant theories which would predict inconsistent $w$ obtained by different observational/fitting methods. But again, consider that all methods have shown consistent results within observational uncertainty.)

EDIT (a review of Dark energy theories):
I have found a very nice comprehensive review of explanations of dark energy here. It basically assesses that Dark energy can be explained by modifying the "right hand side" of Einstein's equations (sources of gravity), "left hand side" (effective laws of gravity) or by modifying the assumptions of the FLRW equations (i.e. no homogeneity on large scales).
The left hand side is either modified by a self-interaction dominated scalar field (quintessence), a scalar field with a non-quadratic kinetic energy (k-essence) and combinations of these. Coupling with dark matter can be introduced, or even the single scalar field can be used to explain both dark energy and dark matter. Sources with postulated equations of state such as Chaplygin gas can always be introduced.
The right hand side can be modified to a broad class of gravitational theories called $f(R)$ theories, which introduce higher orders of curvature $R$ into the GTR action $S \sim \int R$. These can be effective terms arising from vacuum polarization or can be fundamental modifications of gravity. It can be shown that an $f(R)$ theory can be conformally transformed into Einstein gravity with a scalar field source. As to my observation, the main advantage of $f(R)$ theories is the fact they are not so arbitrary as quintessence as much more observational constraints are placed on them due to the fact of being a universal gravitational theory.
The other gravity modifications result from the usual portrayal of our universe as a 4-dimensional brane with at least one "large" extra dimension. The Dvali, Gabadadze and Porrati model described in the linked article accounts for the observation of current acceleration as a transition from a high-matter-density limiting case, where Friedmann equations apply, to a low-density case, where an effective source term of the modified gravitational interaction becomes non-negligible.
The last explanation in the article portrays our observed accelerated expansion as an effect arising from being in a very large under-dense region of the universe. The so-called Lemaitre-Tolman-Bondi models are plausible as to explaining the standard candle observations, the location of the first acoustic peak, but have to assume we are very close to the center of the under-dense region (as long as we keep observing isotropy e.g. in the CMB).
Nevertheless, the cited article is two years old, most notably not aware of data from the Planck mission. It would be great to have a commentary on the observational plausibility on the mentioned (or new) theories from an expert.

REPLY [30 votes]: Measuring $w$ is actually what I do for a living. The current best measurements put $w$ at $-1$ but with an uncertainty of $5\%$, so there's a little room for $w \ne -1$ models, but it's not big and getting smaller all the time. Indeed, we'd all be thrilled if, as measurements got more precise, $w \ne -1$ turns out to be the case because the $\Lambda$CDM model (where Dark Energy is just a scalar correction to General Relativity) is just horribly arbitrary and we'd much rather get evidence for something new and exciting.
You ask if models can reproduce a $w \approx -1$ and then ask about free parameters in such models. You're on the right track with asking about free parameters. There are some ways in which you can explain that $\Omega_\Lambda$ and $\Omega_m$ are fairly close to each other today (quintessence models do this with tracker models, inhomogeneity inducing corrections to the Friedmann equations can do this because the inhomogeneities get stronger as structure grows) while the $w \approx -1$ value is not a particularly unnatural value for many models (it is for some, and those are already ruled out). So you're in an odd situation where you could certainly imagine, say, a quintessence model with say $w \approx -1$, but you're not too tempted to because $\Lambda$CDM is just much simpler. To get out of this impasse, you're on the right track when you ask about free parameters (or loose ends as you call them). Quintessence for example would have a total of $4$, which represents the number of free parameters in your quintessence field that have an impact on the rate of evolution, whereas $\Lambda$CDM has only one. This is why, for the time being, we go with $\Lambda$CDM : our current probes can only constrain a single parameter, namely the current effective equation of state, and it is consistent with $\Lambda$CDM, which then plays the role of a sort of "placeholder" model given that it requires the least number of degrees of freedom. Current constraints on the evolution of Dark Energy properties are a joke, and you might as well consider them non existent. The image below is a confidence contour in the $w$ VS $w_a$ plane given the latest CMB, BAO, and supernova data. $w_a$ is a somewhat arbitrarily chosen parameter ($w(a) = w_0 + (1 - a) w_a$ where $a$ is the scale factor) but it gives you an idea of your probe's sensitivity to the evolution of $w$.

Future probes may be able to meaningfully constrain $w_a$, and if it turns out to be different than $0$, then that'll rule out $\Lambda$CDM and open up a whole new ball game. Remember, $\Lambda$CDM corresponds to a static $w_a = 0$. It'll be hard, since we'll need to go to much higher redshifts while maintaining similar levels of accuracy in our measurements, if not improvements because given its current properties (and whatever its evolution) Dark Energy is weaker in the past ($\Omega \propto a^{-3(1+w)}$, so while a $w \approx -1$ Dark Energy maintains a similar density, matter density, which corresponds to $w = 0$, increases as you look back in time). Being able to better constrain this evolution is the name of the game if you're interested in ruling out the current standard model.
The classification of possible explanations that your review suggests is indeed the best way to broadly categorize the proposed solutions.
A popular model which modifies the right hand side is quintessence. Quintessence  employs a common solution (or common attempt at a solution) to many problems : When in doubt, throw in a scalar field. No known field satisfied the required properties, so it would most certainly lie outside the Standard Model of particle physics. In general, one can consider the impact on expansion of the addition of any fundamental field to the Lagrangian of particle physics. The only condition this field would have to satisfy for this field to be a candidate for dark energy is $w < -1/3$. For a quintessence field $\phi$ with a potential field $V(\phi)$, $p$ and $\rho$ are such that $w$ is given by :
\begin{eqnarray}
  w_\phi &=& \frac{p}{\rho} & \mbox{ (by definition)} \\
 &=& \frac{\frac{1}{2} \dot{\phi}^2 - V(\phi) - \frac{1}{6a^2} (\nabla \phi)^2}{\frac{1}{2} \dot{\phi}^2 + V(\phi) + \frac{1}{2a^2} (\nabla \phi)^2} & \mbox{ ($p$ and $\rho$ for quintessence fields)} \\
 &=& \frac{\frac{1}{2} \dot{\phi}^2 - V(\phi)}{\frac{1}{2} \dot{\phi}^2 + V(\phi)} & \mbox{ (use spatial homogeneity to get rid of $\nabla \phi$ terms)} \\
 &\stackrel{\text{$\dot{\phi} \rightarrow 0$}}{\approx}& -1 & \mbox{ (for a slow rolling field)}
\end{eqnarray}
The couple of $\nabla \phi$ terms in there disappear when you invoke a homogeneous universe (using the Comological Principle). Hence, many models with a $V(\phi)$ that yields a $w_\phi < -1/3$ have been considered simply by constructing the proper $V(\phi)$, and it is not hard nor particularly unnatural to construct models where $w \approx -1$ today, simply by having a slow rolling field where $\dot{\phi}$ is small and therefore $w_\phi \approx -1$. The difficulty lies in making such models distinguishable from the standard case of a $w = -1$ that is static in time, given our experimental insensitivity to the time evolution of the field. The field hasn't necessarily been in slow roll for the history of the universe. On the whole this is actually a whole class of models, and many models and couplings with matter can be invoked to explain all sorts of things about Dark Energy (like the rough similarity of $\Omega_\phi$ and $\Omega_m$ today, and why this happens when the field is slowly rolling etc etc). Again though, we're not yet sensitive to any of the observational implications of any of this. In the near future, we may be able to constrain 2 parameters of freedom with regards to the evolution of expansion, and not the 4 that are required to fully characterize a quintessence model, meaning that even then we will only be able to, at best, favor quintessence models over others, but it'll still be a stretch before we can definitively prove them.
Another right hand side explanation is k-essence, but this is essentially a modified version of quintessence where the kinetic energy term is not the canonical one, and rather than play around with $V(\phi)$ it is thought that it is the properties of this kinetic term that could lead to the required properties. Recall that our own kinetic term was $\frac{1}{2} \dot{\phi}^2 - \frac{1}{6a^2} (\nabla \phi)^2$, and that we got rid of it entirely by assuming $\nabla \phi$ homogeneity and that $\dot{\phi}$ was small. The idea for k-essence is not to get rid of these terms, but to come up with a new form for the kinetic energy such that the k-essence field will have to desired properties. The novelty here (relative to quintessence) is that we don't need a very low upper bound on the mass of field to make it slow rolling. But observationaly, this remains just as hard to prove because the properties of the field (your review gives a very comprehensive overview of this) leave open a number of degrees of freedom as well.
$f(R)$ models modify the left hand side. Here, again, it's best to think of this as a class of models. As the name indicates, $f(R)$ gravity works by generalizing the Einstein-Hilbert action :
$$
S[g] = \int \frac{1}{16 \pi G c^{-4}} R \sqrt{-|g_{\mu \nu}|} d^4x \rightarrow S[g] = \int \frac{1}{16 \pi G c^{-4}} f(R) \sqrt{-|g_{\mu \nu}|} d^4x
$$
Thus the action now depends on an unspecified function of the Ricci scalar. The case $f(R) = R$ represents classical General Relativity. It is this function that we tune in these models to get the desired properties. To test the general case rather than thest each $f(R)$ individually, you can taylor expand $f(R) = c_0 + c_1 R + c_2 R^2 + ...$ and constrain your expansion to however many degrees of freedom you think your observations can constrain. Just having $c_0$ obviously corresponds to a cosmological constant, and $c_1 = 1$ is, as we have said, the usual GR. Note however, that because this modifies gravity as a whole, and not just cosmology, other observations could be brought into the fold to constrain more parameters, and we are not limited here by the number of degrees of freedom cosmological probes can constrain. If cosmological tests of GR agree with smaller scale tests of GR in a manner consistent with an $f(R)$ modified gravity, these models could be proven to be onto something. Here of course we must look at all the tests of GR to get a fuller picture, but that is another review in and of itself. Here we'll simply note that more generally, the challenge when modifying the left hand side is to only change cosmology, but remain consistent with the smaller scale tests. The theory of general relativity is well tested on a number of scales. The accelerated expansion of the universe is the only potential anomaly when comparing general relativity to observations. To modify it such that the Friedmann equations gain an acceleration term while all other tests remain consistent is tricky business for theorists working in the field. The $\Lambda$ correction works because it essentially maintains the same theory except for adding a uniform scalar field in the theory. Hence, while its impact on the global metric is significant, it has no impact on local tests of gravity because it doesn't introduce a gradient to the energy density field. Other theories run into tuning problems precisely because they cannot naturally mimic this behavior. Chameleon theories get around this problem by adding a field that becomes extremely massive (i.e. an extremely short scale force) when surrounded by matter. Hence, its local impact on gravity is negligible in all the contexts in which general relativity has been tested. Ideally, your model would remain consistent with pre-existing tests, but still offer hope for some testable differences (ideally with a rigorous prediction set by the scale of acceleration) beyond cosmological acceleration so as to make your model more believable.
Finally, we can consider if the assumptions that went into the Friedmann equations are correct. One approach (not the one you mentioned) is to consider the impact of inhomogeneities. While it was always obvious that homogeneity did not hold below certain scales, the traditional approach has been to simply conclude that while the Friedmann equations will therefore not accurately describe the "local" scale factor, it would describe the "average" scale factor so long as we only considered it to be accurate above the inhomogeneous scales. In other words, they describe the expansion of the "average" scale factor. Mathematically speaking, this is not the rigorous way to get around this problem.
Ideally, what one would do instead, is to calculate the average of the expansion of the scale factor. Some theorists have proposed methods of doing this in ways that would add an acceleration term to the Friedmann equations. For a matter dominated universe these become :
\begin{eqnarray}
\left( \frac{\dot{a}}{a} \right)^2 &=& \frac{8 \pi G}{3} \left( \left\langle \rho \right\rangle - \frac{k}{a^2} \right) - \frac{1}{2} Q(t) \\
\frac{\ddot{a}}{a} &=& \frac{- 4 \pi G}{3} \left\langle \rho \right\rangle + Q(t) \\
\mbox{Where } Q(t) &=& \frac{2}{3} \left( \left\langle a^2 \right\rangle - \left\langle a \right\rangle^2 \right) - 2 \left\langle \sigma^2 \right\rangle
\end{eqnarray}
Where $\sigma$ depends on the level of structure formation of the universe. Because this model builds on the effect of inhomogeneities, it could explain why acceleration is a late time effect. Earlier in the universe's history, structure fromation was in its infancy and could not provide the necessary level of inhomogeneity. As the universe grows, structure formation kicks in, and so too does this effect become more significant. The biggest challenge for these models is providing a definite value for $Q$. This has proven to be a difficult task because estimating $\left\langle a^2 \right\rangle - \left\langle a \right\rangle^2$ analytically is extremely difficult, and simulating this sort of GR on such a large scale presents problems of its own.
Another potential effect of these inhomogeneities is the loss of energy as light enters large potential wells, only to exit them after they have significantly expanded. In this view, the observations would be explained by what happens to the photons along the line of sight, and no actual acceleration would take place. In some ways, this is equivalent to a sort of physically motivated tired light model. So far, these models have not been able to explain the observations assuming only a matter dominated universe. It is, however, entirely possible that a small portion of the observed effect is due to this. Keep in mind that this explanation would only affect supernovae measurements, and hence could not explain the now independent confirmation of other probes regarding acceleration. At best, they could one day reduce tension between supernovae and other probes.
The Lemaitre-Tolman-Bondi model assumes that homoegeneity is just not true at all, and that we live inside a gravitationaly collapsing system, near its center. It's a little out there, but the thing about cosmology is that it's rather easy to make far fetched models and go "hey, you never know", because hey, we don't :P It's a big universe out there and we only have access to a tiny portion of it, so why not consider I guess. The key to disproving this one will be to increase the agreement between the various probes. The metric it leads to doesn't have the same impact on all the different distance probes in the same way that $\Lambda$CDM does. So it's disprovable, and for the time being can be tuned to explain the observations, but a little far fetched to begin with. I'd guess that you can change the toy model they use to model the inhomogeneous shape of the universe and almost always get something that'll fit the data, but the more you do that the less convincing your theory becomes.
As a side note, though this isn't your main question, it's not odd that we're using an effective fluid model. It's not counting on Dark Energy having properties similar to a fluid in the traditional sense, it's just telling you that you're using certain symmetries to simplify you $T_{\mu \nu}$ tensor when solving for the Friedmann equations. Essentially, if you're in the reference frame where the momenta of the universe sum to $0$ then $T_{\mu \nu} = diag(\rho, p_x, p_y, p_z) = diag(\rho, p, p, p)$, and $w$ is just the ratio of of $\rho$ to $p$. It's just a way to simplify equations and clarify your degrees of freedom.


TITLE: If $x^{-} = \textrm{argmin}_x f(x) $, $x^{+} = \textrm{argmin}_x g(x) $, and $x^{-} \neq x^{+} $, then can we state the following:
QUESTION [0 upvotes]: If $x^{-} = \operatorname{argmin}_x f(x) $, $x^{+} = \operatorname{argmin}_x g(x) $, and $x^{-} \neq x^{+} $, then can we state that $x^* \neq x^-, x^+$ where $x^{*} = \operatorname{argmin}_x f(x) + g(x)$.
If the optimal point for $f(x)$ and $g(x)$ are different. Then can say that the optimal point for their summation is none of their optimal points, assuming that both functions are non-zero.
If this does not hold, are there conditions under which they hold?
As suggested in the comments, $f$ and $g$ have unique minimizers.

REPLY [1 votes]: The answer is no. For instance, $f(x)=|x|$ has a minimizer at $0$ and $g(x)=0.5|x-1|$ has a minimizer at $1$; however $f(x)+g(x)$ has a minimizer at $0$.
Look at this graph for a demonstration. Note that both $f$ and $g$ are convex in this example.


TITLE: Proving a Set is not a Vector Space (Using a Nonstandard Definition of Scalar Multiplication)
QUESTION [0 upvotes]: I have a question about proving that a set is not a vector space. In my Linear Algebra textbook, we have this example:
"Let V be the set of all ordered pairs of real numbers, with the standard operation of addition and the nonstandard definition of scalar multiplication listed below."
$c( x_1 , x_2) = ( cx_1, 0)$
The book says that the first 9 axioms pass, but that the 10th axiom (Multiplicative Identity Property) fails because:
$1(1,1) = (1,1) \neq (1,0)$
But I am not certain as to why... Is it because $1(x_1, x_2) = (1x_1,1x_2)$ which simplifies to $(x_1,x_2)$, thus resulting in $(1,1)$ as opposed to $(1,0)$?
Though, I have doubts, because at the same time I can see 1 being c in the way scalar multiplication is defined, and thus $1(x_1, x_2) = (x_1, 0)$.

REPLY [0 votes]: According to the axiom that we mentioned, you should have $1(1,1)=(1,1)$. But what we actually have for this scalar product is $1(1,1)=(1,0)$. Since $(1,1)\neq(1,0)$, that axiom doesn't hold and therefore we don't have a vector space.


TITLE: how to prove using elementary comparisons that $x \log x \leq \log(2x^2-3x+2)$?
QUESTION [3 upvotes]: I am trying to show that $x \log x \leq \log(2x^2-3x+2)$ on $[1,2]$. The things is that these two functions are pretty close to each other:

so I have a hard time finding a simple function which trivially lies between the two.

REPLY [1 votes]: We have to compare $(1+x)\log(1+x)$ and $\log(1+x+2x^2)$ on $I=[0,1]$.
That is the same as comparing:
$$ f(x)=(1+x)\log(1+x)-2x\log(2),\quad g(x)=\log(1+x+2x^2)-2x\log(2)\tag{1} $$
that are both vanishing functions on the endpoint of $I$. $f(x)$ is a convex function, hence $f(x)\leq 0$ on $I$. Since $g\left(\frac{1}{2}\right)=0$ but $g$ is positive on $\left(\frac{1}{2},0\right)$, it is enough to prove $f(x)\leq g(x)$ on $J=\left[0,\frac{1}{2}\right]$. We have $f(x)\leq (1-2\log 2)x+\frac{x^2}{2}$ over such interval, hence it is enough to prove:
$$\forall x\in\left[0,\frac{1}{2}\right],\quad \log(1+x+2x^2)\geq x+\frac{x^2}{2}.\tag{2}$$
The last inequality is easy to prove if we notice that $\frac{\log(1+x+2x^2)}{x}$ is a concave function on $J$, or that $x+\frac{3x^2}{2}-\frac{5x^3}{3}$ is between the two terms given by $(2)$.


TITLE: Let $f$ be a differentiable function defined for all real $x$, where $f(x)\ge 0$ for all $x\in[0,a]$
QUESTION [2 upvotes]: Let $f$ to be a differentiable function defined for all real $x$, where $f(x)\ge 0$ for all $x\in[0,a]$.If $$\int_0^a f(x)\,dx = a, $$ then $$2\int_0^{5a}\left(f\left( \frac x 5 \right) + 3 \right)\,dx$$ is equal to
Can anyone help me with this question. Don't know what to do.
Thanks.

REPLY [0 votes]: We can use substitution and properties of integrals to simplify this problem. We will get something easier to deal with: $2\int_0^{5a}(f(x/5)+3)dx$ = $2$ $[$ $\int_0^{5a}$ $f(x/5)$$dx$ + $\int_0^{5a}$$3$$dx$$ ].$ Let  u = $x/5$. Then, $du = dx/5 $. So, we get: $2$ $[$ $\int_0^{5a}$ $f(u)$$5du$ + $\int_0^{5a}$$3$$dx$$ ].$ Can you finish it from here?


TITLE: Ranks composition of linear maps
QUESTION [0 upvotes]: We have $a: A \to B$ and $b:B \to C$, where $a$ and $b$ are linear maps of vector spaces. Show that rank$(a) \geq $ rank $ (b \circ a)$ with equality if (not iff) $b$ is injective.

Can anybody give me a hint on how to solve this? I'm not exactly sure about the rank of a composition of two linear maps.

REPLY [1 votes]: Hint:
$\operatorname{rank}a=\dim(\operatorname{Im}a)$ and, if $E\subset E'$ are subspaces, $\;\dim E\le\dim E'$.


TITLE: Elaborating $a$ in $\sum_{i=0}^{k} (a_i 10^i)$
QUESTION [0 upvotes]: I found this summation in https://oeis.org/A067030, but I don't understand the basic claim that:

Any $(k+1)$-digit number $N$ can be represented as:

a.
$$N = \sum_{i=0}^{k} (a_i 10^i)$$
At the end it is stated that:
b.
$$N+Rev(N) = \sum_{i=0}^{k}a_i(10^i+10^{k-i})$$
How would you get for example integer $110 =37+73$ from the above summation? I understand $k$ being the number of digits. But does it refer to the result $(110 \rightarrow 3)$ or the input $(37,73 \rightarrow 2)$ in equation b.?

REPLY [1 votes]: With $N=37$, we kave $k=1$, so the claim is that $73=\sum_{i=0}^ka_i10^i$ for suitable $a_i$. Of course, this just says $37=a_0+10a_1$ and we can (in fact, must) pick $a_0=7$ and $a_1=3$.
This makes $\operatorname{rev}(N)=\sum_{i=0}^ka_i10^{k-i}=10a_0+a_1=70+3=73$
and $N+\operatorname{rev}(N)=\sum_{i=0}^ka_i(10^i+10^{k-i})=a_0\cdot 11+a_1\cdot 11=110.$


text:
\begin{document}

\title[New Solutions for The weighted Fermat-Torricelli problem]{Two new analytical solutions and two new geometrical solutions for the weighted Fermat-Torricelli problem in the Euclidean Plane}
\author{Anastasios N. Zachos}
\address{University of Patras, Department of Mathematics, GR-26500 Rion, Greece}
\email{azachos@gmail.com} \keywords{weighted Fermat-Torricelli
problem, weighted Fermat-Torricelli point} \subjclass{51M04,
51E10}

\begin{abstract}
We obtain two analytic solutions for the weighted
Fermat-Torricelli problem in the Euclidean Plane which states
that: Given three points in the Euclidean plane and a positive
real number (weight) which correspond to each point, find the
point such that the sum of the weighted distances to these three
points is minimized.
 Furthermore, we give two new geometrical solutions  for the
the weighted Fermat-Torricelli problem (weighted Fermat-Torricelli
point), by using the floating equilibrium condition of the
weighted Fermat-Torricelli problem (first geometric solution) and
a generalization of Hofmann's rotation proof under the condition
of equality of two given weights (second geometric solution).
\end{abstract}\maketitle

\section{Introduction}

We state the weighted Fermat-Torricelli problem in
$\mathbb{R}^{2}:$

\begin{problem}
Given a triangle $\triangle A_{1}A_{2}A_{3}$ with vertices
$A_{1}=(x_{1},y_{1}),$ $A_{2}=(x_{2},y_{2}),$
$A_{3}=(x_{3},y_{3}),$ find a fourth point $A_{F}=(x_{F},y_{F})$
which minimizes the objective function

\begin{equation}\label{obj1}
f(x,y)=\sum_{i=1}^{3}B_{i}\sqrt{(x-x_{j})^{2}+(y-y_{j})^{2}}
\end{equation}

where $B_{i}$ is a positive real number (weight) which corresponds
to $A_{i}.$

\end{problem}

By replacing $B_{1}=B_{2}=B_{3}$ in (\ref{obj1}), we obtain the
(unweighted) Fermat-Torricelli problem which was first stated by
Pierre de Fermat (1643).

The solution of the weighted Fermat-Torricelli problem (Problem~1)
is called the weighted Fermat-Torricelli point $A_{F}.$

The existence and uniqueness of the weighted Fermat-Torricelli
point and a complete characterization of the "floating case" and
"absorbed case" has been established by Y. S Kupitz and H. Martini
(see \cite{Kup/Mar:97}, theorem 1.1, reformulation 1.2 page 58,
theorem 8.5 page 76, 77). A particular case of this result for
three non-collinear points in $\mathbb{R}^{2},$ is given by the
following theorem:

\begin{theorem}{\cite{BolMa/So:99},\cite{Kup/Mar:97}}\label{theor1}
Let there be given a triangle $\triangle A_{1}A_{2}A_{3},$ $A_{1},
A_{2}, A_{3}\in\mathbb{R}^{2}$ with corresponding positive
weights $B_{1}, B_{2}, B_{3}.$ \\
(a) The weighted Fermat-Torricelli point $A_{F}$ exists and is
unique. \\
(b) If for each point $A_{i}\in\{A_{1},A_{2},A_{3}\}$

\begin{equation}\label{floatingcase}
\|{\sum_{j=1, i\ne j}^{3}B_{j}\vec u(A_i,A_j)}\|>B_i,
\end{equation}

 for $i,j=1,2,3$  holds,
 then \\
 ($b_{1}$) the weighted Fermat-Torricelli point $A_{F}$ (weighted floating equilibrium point) does not belong to $\{A_{1},A_{2},A_{3}\}$
 and \\
 ($b_{2}$)

\begin{equation}\label{floatingequlcond}
 \sum_{i=1}^{3}B_{i}\vec u(A_F,A_i)=\vec 0,
\end{equation}
where $\vec u(A_{k} ,A_{l})$ is the unit vector from $A_{k}$ to
$A_{l},$ for $k,l\in\{0,1,2,3\}$
 (Weighted Floating Case).\\
 (c) If there is a point $A_{i}\in\{A_{1},A_{2},A_{3}\}$
 satisfying
 \begin{equation}
 \|{\sum_{j=1,i\ne j}^{3}B_{j}\vec u(A_i,A_j)}\|\le B_i,
\end{equation}
then the weighted Fermat-Torricelli point $A_{F}$ (weighted
absorbed point) coincides with the point $A_{i}$ (Weighted
Absorbed Case).
\end{theorem}

A direct consequence of the weighted floating case and the
weighted absorbed case of theorem~\ref{theor1} gives
Corollary~\ref{cor1} (Torricelli's theorem) and
Corollary~\ref{cor2} (Cavalieri's alternative) (see in
\cite{BolMa/So:99}, p~236).
\begin{corollary}\label{cor1}
If $B_{1}=B_{2}=B_{3}$ and all three angles of the triangle
$\triangle A_{1}A_{2}A_{3}$  are less than $120^{\circ},$ then
$A_{F}$ is the isogonal point (interior point) of $\triangle
A_{1}A_{2}A_{3}$ whose sight angle to every side of
$A_{1}A_{2}A_{3}$ is $120^{\circ}$ (Torricelli's theorem).
\end{corollary}\label{cor2}
\begin{corollary}If $B_{1}=B_{2}=B_{3}$ and one of the angles of
the triangle $\triangle A_{1}A_{2}A_{3}$ is equal or greater than
$120^{\circ},$ then $A_{F}$ is the vertex of the obtuse angle of
$\triangle A_{1}A_{2}A_{3}$ (Cavalieri's alternative).
\end{corollary}

Concerning the solution of the weighted Fermat-Torricelli problem
with the use of analytic geometry and trigonometry, we mention the
works of \cite{GreenbergRobertello:65}, \cite{Jones:88},
\cite{Pesamosca:91} \cite{MHAjja:94}, \cite{Gue/Tes:02},
\cite{Zachos/Zou:08} and \cite{Zachos:13}.

Recently, an analytic solution, which express explicitly the
coordinates of the weighted Fermat-Torricelli point with respect
to the coordinates of the three points $A_{i}$ and the three
weights $B_{1},$ $B_{2}$ $B_{3}$ for the weighted
Fermat-Torricelli problem with respect to the weighted floating
case of Theorem~1 has been derived in \cite{Uteshev:12} and for
the case $B_{1}=B_{2}=B_{3}=1$ has been derived in
\cite{Roussos:12}.

In this paper, we present two new analytical solutions for the
weighted Fermat-Torricelli problem in $\mathbb{R}^{2}$ in the
weighted floating case of Theorem~\ref{theor1}. The first
analytical solution gives the coordinates of the weighted
Fermat-Torricelli point as a function of the coordinates of the
three non collinear points and the three given weights (real
positive numbers) in a different way from \cite{Uteshev:12} and
\cite{Roussos:12} (Theorem~\ref{analytic1}. Section~2).

The second analytical solution gives the location of the weighted
Fermat-Torricelli point as a function of two inscribed angles of
the circumscribed circle which passes form the three non collinear
points and the three given weights by applying a coordinate
independent approach given in \cite{Zachos:13}
(Theorem~\ref{angular}, Section~3).

The first geometrical solution of the weighted Fermat-Torricelli
point with ruler and compass focuses on constructing the
intersection of two Simpson lines (weighted case) by applying the
duality of the weighted Fermat-Torricelli problem which was
introduced in \cite{Uteshev:12} (Problem~\ref{firstprob},
Section~4).

Finally, the second geometric solution of the weighted
Fermat-Torricelli point focuses on finding the angle of rotation
of the three non collinear points about one of them and
generalizes Hofmann's rotation proof (\cite{BolMa/So:99},
\cite{Hofmann:27}, \cite{Tikh:86}) regarding the equality of the
given weights (Problem~\ref{secgeomprob}, Corollary~\ref{corhof},
Section~4).

\section{Analytical solution of the weighted Fermat-Torricelli problem}

We obtain an analytic solution for the floating case of Theorem~1,
i.e the weighted Fermat-Torricelli point $A_{F}$ is an interior
point of $\triangle A_{1}A_{2}A_{3},$ such that the coordinates
$x_{F}$ and $y_{F}$ of $A_{F}$ are expressed explicitly as a
function of $x_{i},y_{i}$ and $B_{i},$ for $i=1,2,3,$ by using
analytic geometry in $\mathbb{R}^{2}.$

We denote by $a_{ij}$ the length of the linear segment $A_iA_j$
and $\alpha_{ikj}$ the angle $\angle A_{i}A_{k}A_{j}$ for
$i,j,k=0,1,2,2',3,3', i\neq j\neq k$ (See fig.~\ref{fig1}).

Without loss of generality, we set $A_{1}=(0,0),$
$A_{2}=(a_{12},0),$ $A_{3}=(x_{3},y_{3}).$

We need the following two lemmata:

\begin{lemma}{\cite{Gue/Tes:02},\cite{Zachos/Zou:08}}\label{lem1}
Under the condition (\ref{floatingcase}) and the weighted floating
equilibrium condition (\ref{floatingequlcond}) the following
equation is satisfied:

\begin{equation}\label{sinlawweights}
\frac{B_{i}}{\sin\alpha_{203}}=\frac{B_{2}}{\sin\alpha_{103}}=\frac{B_{3}}{\sin\alpha_{102}}=C,
\end{equation}

where
$C=\frac{2B_{1}B_{2}B_{3}}{\sqrt{(B_{1}+B_{2}+B_{3})(B_{2}+B_{3}-B_{1})(B_{1}+B_{3}-B_{2})(B_{1}+B_{2}-B_{3})}}$

\end{lemma}

\begin{lemma}{\cite{BolMa/So:99},\cite{MHAjja:94},\cite{Gue/Tes:02},\cite{Zachos/Zou:08}}\label{lem2}
Under the condition (\ref{floatingcase}) and the weighted floating
equilibrium condition (\ref{floatingequlcond}) the angle
$\alpha_{i0j}$ is expressed as a function of $B_{1}, B_{2}$ and
$B_{3}:$

\begin{equation}\label{alphai0j}
\alpha_{i0j}=\arccos\left(\frac{B_{k}^2-B_{i}^2-B_{j}^2}{2B_{i}B_{j}}\right)
\end{equation}
for $i,j,k=1,2,3,$ and $i\ne j\ne k.$

\end{lemma}

\begin{theorem}\label{analytic1}
Under the condition (\ref{floatingcase}) of the weighted floating
case, the coordinates of the weighted Fermat-Torricelli point
$A_{F}$ $(x_{F},y_{F})$ are given by the following relations:

\begin{equation}\label{xanalytic}
x_{F}=-\frac{(a_{12}-x_{2}')(x_{3}'y_{3}-x_{3}y_{3}')+d_{3}(x_{3}-x_{3}')}{(a_{12}-x_{2}')(y_{3}'-y_{3})-(y_{3}-x_{3}')y_{2}'},
\end{equation}

\begin{equation}\label{yanalytic}
y_{F}=\frac{y_{2}'(a_{12}y_{3}-x_{3}'y_{3}-a_{12}y_{3}'+x_{3}y_{3}')}{x_{3}y_{2}'-x_{3}y_{2}'+a_{12}y_{3}-x_{2}'y_{3}-a_{12}y_{3}'+x_{2}'y_{3}'}
\end{equation}

where

\begin{equation}\label{x2prime}
x_{2}'=-\frac{B_{3}}{B_{2}}\left(x_{3}\frac{B_{1}^2-B_{2}^2-B_{3}^2}{2B_{2}B_{3}}+y_{3}\sqrt{1-\left(\frac{B_{1}^2-B_{2}^2-B_{3}^2}{2B_{2}B_{3}}\right)^{2}},
\right)
\end{equation}

\begin{equation}\label{y2prime}
y_{2}'=\frac{B_{3}}{B_{2}}\left(x_{3}\sqrt{1-\left(\frac{B_{1}^2-B_{2}^2-B_{3}^2}{2B_{2}B_{3}}\right)^{2}}-y_{3}\frac{B_{1}^2-B_{2}^2-B_{3}^2}{2B_{2}B_{3}}\right),
\end{equation}

\begin{equation}\label{x3prime}
x_{3}'=-a_{12}\frac{B_{2}}{B_{3}}\left(\frac{B_{1}^2-B_{2}^2-B_{3}^2}{2B_{2}B_{3}}\right)
\end{equation}
and
\begin{equation}\label{y3prime}
y_{3}'=-a_{12}\frac{B_{2}}{B_{3}}\left(\sqrt{1-\left(\frac{B_{1}^2-B_{2}^2-B_{3}^2}{2B_{2}B_{3}}\right)^{2}}\right).
\end{equation}

\end{theorem}

\begin{proof}[Proof of Theorem~\ref{analytic1}:]

\begin{figure}
\centering
\includegraphics[scale=0.8]{analyticsolution2d2new}
\caption{}\label{fig1}
\end{figure}

We apply the weighted Torricelli configuration which is similar to
the configuration used in \cite{ENGELBRECHt:1877} and we construct
two similar triangles $\triangle A_{1}A_{2}A_{3'}$ and $\triangle
A_{1}A_{3}A_{2'},$ such that:
\begin{equation}\label{alpha13p2132p}
\alpha_{13^{\prime}2}=\alpha_{132^{\prime}}=\pi-\alpha_{102},
\end{equation}
\begin{equation}\label{alpha13p2132p}
\alpha_{213^{\prime}}=\alpha_{312^{\prime}}=\pi-\alpha_{203},
\end{equation}
and
\begin{equation}\label{alpha13p2132p}
\alpha_{123^{\prime}}=\alpha_{12^{\prime}3}=\pi-\alpha_{103}.
\end{equation}

From (\ref{alpha13p2132p}), (\ref{alpha13p2132p}) and
(\ref{alpha13p2132p}), we derive that the point of intersection of
the two circles which pass from $A_{1}, A_{3^{\prime}}, A_{2}$ and
$A_{1}, A_{2^{\prime}}, A_{3},$ respectively, is the weighted
Fermat-Torricelli point $A_{F}$ (fig.~\ref{fig1}). Therefore, we
obtain that $A_{F}$ is the intersection point of the lines
(weighted Simpson lines) defined by $A_{2}A_{2^{\prime}}$ and
$A_{3}A_{3^{\prime}}.$

Thus, we have:

\begin{equation}\label{eqanal1}
x_{2}'=a_{12'}\cos(\alpha_{213}+\pi-\alpha_{203}),
\end{equation}

\begin{equation}\label{eqanal2}
y_{2}'=a_{12'}\sin(\alpha_{213}+\pi-\alpha_{203}),
\end{equation}

\begin{equation}\label{eqanal3}
x_{3}'=a_{13'}\cos(\pi-\alpha_{203}),
\end{equation}

and

\begin{equation}\label{eqanal4}
y_{3}'=-a_{13'}\sin(\pi-\alpha_{203}).
\end{equation}

By applying the sine law in $\triangle A_{1}A_{2}A_{3'}$ and
$\triangle A_{1}A_{3}A_{2'},$ we get, respectively,:

\begin{equation}\label{eqanal5}
a_{13'}=a_{12}\frac{B_{2}}{B_{3}}.
\end{equation}

and

\begin{equation}\label{eqanal6}
a_{12'}=a_{13}\frac{B_{3}}{B_{2}}.
\end{equation}

By replacing (\ref{eqanal5}),(\ref{eqanal6}) and (\ref{alphai0j})
from lemma~\ref{lem2} in (\ref{eqanal1}), (\ref{eqanal2}),
(\ref{eqanal3}) and (\ref{eqanal4}), we obtain (\ref{x2prime}),
(\ref{y2prime}), (\ref{x3prime}) and (\ref{y3prime}).

The equations of the lines defined by $A_{2}A_{2'}$ and
$A_{3}A_{3'},$ respectively, are as follows:

\begin{equation}\label{eqanal7}
\frac{y_{2}'}{x_{2}'-a_{12}}=\frac{y}{x-a_{12}}
\end{equation}
and
\begin{equation}\label{eqanal8}
\frac{y_{3}-y_{3}'}{x_{3}-x_{3}'}=\frac{y_{3}-y}{x_{3}-x}
\end{equation}

Solving (\ref{eqanal7}) and (\ref{eqanal8}) with respect to
$(x,y)$ we derive the point of intersection $A_{F}=(x_{F},y_{F}),$
and the coordinates $x_{F}$ and $y_{F}$ are given by
(\ref{xanalytic}) and (\ref{yanalytic}), respectively.

\end{proof}

\section{An explicit angular solution of the weighted Fermat-Torricelli problem}

It is well known that the barycenter $A_{m}$ of $\triangle
A_{1}A_{2}A_{3}$ is constructed by the relation
$a_{im}=\frac{2}{3}a_{i,jk}=\frac{1}{3}\sqrt{2a_{ij}^2+2a_{ik}^2-a_{jk}^2},$
where $a_{i,jk}$ is the length of the midline that connects the
vertex $A_{i}$ with the midpoint of the line segment $A_{j}A_{k}$
for $i,j,k=1,2,3,$ $i\ne j\ne k$ and the median minimizes the
objective function $a_{m1}^{2}+a_{m2}^{2}+a_{m3}^{2}.$

A natural question to ask is if the location of the weighted
Fermat-Torricelli problem could be given with respect to the
lengths of the sides of $\triangle A_{1}A_{2}A_{3}$ and the
constant positive weights $B_{1}, B_{2},B_{3}.$

A positive answer to this question is given by the following lemma
(\cite[Corollary~2]{Zachos:13}):

\begin{lemma}{(\cite[Corollary~2]{Zachos:13})}\label{lem3}The explicit solution of the weighted Fermat-Torricelli problem in $\mathbb{R}^{2}$, under the condition
(\ref{floatingcase}) (weighted floating case)  is given by:
\begin{equation}\label{mainres}
\alpha_{013}=\operatorname{arccot}\left(\frac{\sin(\alpha_{213})-\cos(\alpha_{213})
\cot(\arccos{\frac{B_{3}^{2}-B_{1}^2-B_{2}^2}{2B_{1}B_{2}}})-
\frac{a_{13}}{a_{12}
}\cot(\arccos{\frac{B_{2}^{2}-B_{1}^2-B_{3}^2}{2B_{1}B_{3}}})}
{-\cos(\alpha_{213})-\sin(\alpha_{213})
\cot(\arccos{\frac{B_{3}^{2}-B_{1}^2-B_{2}^2}{2B_{1}B_{2}}})+
\frac{a_{13}}{a_{12} }}\right)
\end{equation}
and
\begin{equation}\label{mainresb}
a_{10}=\frac{\sin\left(\alpha_{013}+\arccos{\frac{B_{2}^{2}-B_{1}^2-B_{3}^2}{2B_{1}B_{3}}}\right)a_{13}}{\sin\left(\arccos{\frac{B_{2}^{2}-B_{1}^2-B_{3}^2}{2B_{1}B_{3}}}\right)}.
\end{equation}
where

\begin{equation}\label{mainresbr2bb}
\alpha_{213}=\arccos\left(\frac{a_{12}^2+a_{13}^2-a_{23}^2}{2a_{12}a_{13}}\right)
\end{equation}

and $\alpha_{013}$ and $a_{10}$ depend on $B_{1}, B_{2}, B_{3},$
$a_{13},$ $a_{12}$ and $a_{23}.$

\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lem3}:]
We assume that the weighted floating case occurs (see
theorem~\ref{theor1}, Case~b), in order to locate it in the
interior of the $\triangle A_1A_2A_3.$

From the cosine law in $\triangle A_{1}A_{0}A_{2},$ and $\triangle
A_{1}A_{0}A_{3}$ we get, respectively:
\begin{equation}\label{i23cos}
a_{02}^{2}=a_{01}^{2}+a_{12}^{2}-2a_{01}a_{12}\cos(\alpha_{213}-\alpha_{013})
\end{equation}
and
\begin{equation}\label{i23cos3}
a_{03}^{2}=a_{01}^{2}+a_{13}^{2}-2a_{01}a_{13}\cos(\alpha_{013}).
\end{equation}
From (\ref{i23cos}) and (\ref{i23cos3}), $a_{02}$ and $a_{03}$ are
expressed with respect to the two variables $a_{01}$ and
$\alpha_{013}:$
\[a_{0i}=a_{0i}(a_1,\alpha_{013}),\]

for $i=2,3.$ By differentiating (\ref{obj1}) with respect to
$a_{01}$ and $\alpha_{013},$ respectively, we get:
\begin{equation}\label{der1}
B_1+B_2\frac{\partial a_{02}}{\partial a_{01}}+B_3\frac{\partial
a_{03}}{\partial a_{01}}=0,
\end{equation}
\begin{equation}\label{der2}
B_2\frac{\partial a_{02}}{\partial \alpha_{013}}+B_3\frac{\partial
a_{03}}{\partial \alpha_{013}}=0.
\end{equation}

From Appendix~A, by replacing (\ref{dera2a1}) and (\ref{dera3a1})
in (\ref{der1}), we obtain:

\begin{equation} \label{eqcos}
B_2\cos(\alpha_{102})+B_3\cos(\alpha_{103})=-B_{1}
\end{equation}

By replacing (\ref{dera2a013}) and (\ref{dera3a013}) in
(\ref{der2}), we obtain:
\begin{equation} \label{eqsin}
-B_2\sin(\alpha_{102})+B_3\sin(\alpha_{103})=0
\end{equation}

By squaring both parts of (\ref{eqcos}) and (\ref{eqsin}) and by
adding the two derived equations, we get:

\begin{equation}\label{cos203}
\cos(\alpha_{203})=\frac{B_{1}^2-B_{2}^2-B_{3}^2}{2B_{2}B_{3}}.
\end{equation}

Similarly by expressing the objective function with respect to the
two variables $a_{2},$ $\alpha_{023},$ and with respect to the two
variables $a_{3},$ $\alpha_{031},$ we derive, respectively:

\begin{equation}\label{cos103}
\cos(\alpha_{103})=\frac{B_{2}^2-B_{1}^2-B_{3}^2}{2B_{1}B_{3}}
\end{equation}
and
\begin{equation}\label{cos102}
\cos(\alpha_{102})=\frac{B_{3}^2-B_{1}^2-B_{2}^2}{2B_{1}B_{2}}.
\end{equation}

From the sine law in $\triangle A_1A_0A_2$, $\triangle A_1A_0A_3,$
we get, respectively:

\begin{equation}\label{sinl12}
\frac{a_{12}}{\sin(\alpha_{102})}=\frac{a_{01}}{\sin(\alpha_{213}-\alpha_{013}+\alpha_{102})}
\end{equation}
and
\begin{equation}\label{sinl13}
\frac{a_{13}}{\sin(\alpha_{103})}=\frac{a_{01}}{\sin(\alpha_{013}+\alpha_{103})}
\end{equation}

By eliminating $a_{01}$ from (\ref{sinl12}) and (\ref{sinl13}), we
obtain:
\begin{equation} \label{result}
\alpha_{013}=arccot(\frac{\sin(\alpha_{213})-\cos(\alpha_{213})
\cot(\alpha_{102})- \frac{a_{31}}{a_{12} }\cot(\alpha_{103})}
{-\cos(\alpha_{213})-\sin(\alpha_{213}) \cot(\alpha_{102})+
\frac{a_{31}}{a_{12} }})
\end{equation}
By replacing (\ref{cos102}) and (\ref{cos103}) in (\ref{result}),
we obtain (\ref{mainres}). From the sine law in $\triangle
A_{1}A_{0}A_{3},$ we derive (\ref{mainresb}).

The values of $a_{01}$ and $\alpha_{013}$ give the location of the
weighted Fermat-Torricelli point $A_{F}.$
\end{proof}

\begin{remark}
The explicit solution of the weighted Fermat-Torricelli problem is
similar with the definition of a complex number in a polar form:\\
$z=r\exp(i(\alpha_{213}-\alpha_{013})),$ where the absolute value
of z is $r=a_{1}$ and the argument of z is $\arg
z=\alpha_{213}-\alpha_{013}.$
\end{remark}

Let $C(Q,R)$ be the inscribed circle with center $Q$ and radius
$R$ which passes from the vertex $A_{i},$ for $i=1,2,3.$

Each of the three central angles is given by the relation:
\begin{equation}\label{centralij}
c_{iQj}=2\alpha_{imj},
\end{equation}
such that:
\[c_{1Q2}+c_{2Q3}+c_{1Q3}=2\pi\]
or
\begin{equation}\label{central1Q2}
c_{1Q2}=2\pi-c_{1Q3}-c_{2Q3},
\end{equation}
 for $i\ne m \ne j,$ $i,m,j=1,2,3.$ From the sine law in
$\triangle A_{1}A_{2}A_{3}$ and taking into account
(\ref{centralij}), we get:
\begin{equation}\label{centralijsine}
\frac{a_{13}}{\sin(\frac{c_{1Q3}}{2})}=\frac{a_{12}}{\sin(\frac{c_{1Q2}}{2})}=2R.
\end{equation}
By replacing (\ref{centralij}), (\ref{central1Q2})
(\ref{centralijsine}) in (\ref{mainres}) and (\ref{mainresb}) of
lemma~\ref{lem3}, we derive the following result:
\begin{theorem}\label{angular}
An explicit angular solution of the weighted Fermat-Torricelli
problem in $\mathbb{R}^{2},$ under the condition
(\ref{floatingcase}) is given by:
\begin{equation}\label{mainresang}
\cot\alpha_{013}=\frac{\sin(\frac{c_{2Q3}}{2})-\cos(\frac{c_{2Q3}}{2})
\cot(\arccos{\frac{B_{3}^{2}-B_{1}^2-B_{2}^2}{2B_{1}B_{2}}})-
\frac{\sin(\frac{c_{1Q3}}{2})}{\sin(\frac{c_{1Q3}+c_{2Q3}}{2})
}\cot(\arccos{\frac{B_{2}^{2}-B_{1}^2-B_{3}^2}{2B_{1}B_{3}}})}
{-\cos(\frac{c_{2Q3}}{2})-\sin(\frac{c_{2Q3}}{2})
\cot(\arccos{\frac{B_{3}^{2}-B_{1}^2-B_{2}^2}{2B_{1}B_{2}}})+
\frac{\sin(\frac{c_{1Q3}}{2})}{\sin(\frac{c_{1Q3}+c_{2Q3}}{2}) }}
\end{equation}
and
\begin{equation}\label{mainresbang}
a_{10}=2R\frac{\sin(\alpha_{013}+\arccos{\frac{B_{2}^{2}-B_{1}^2-B_{3}^2}{2B_{1}B_{3}}})
\sin(\frac{c_{1Q3}}{2})}{\sin(\arccos{\frac{B_{2}^{2}-B_{1}^2-B_{3}^2}{2B_{1}B_{3}}})},
\end{equation}
where  $\alpha_{013}$ and $a_{10}$ depend on $B_{1}, B_{2},
B_{3},$ $c_{1Q3},$ $c_{2Q3},$ and $R.$

\end{theorem}

\begin{remark} We conclude that by setting $R=1$ (unit radius circumscribed circle) in
(\ref{mainresbang}), the explicit solution depends only on five
given elements: $B_{1}, B_{2}, B_{3},$ $c_{1Q3}$ and $c_{2Q3}.$
This unique result holds only if the inequalities
(\ref{floatingcase}) of the weighted floating case of Theorem~1,
Case (b) are satisfied.
\end{remark}

\begin{remark}
We note that lemma~\ref{lem3} and theorem~\ref{angular} provide an
analytic solution for the weighted Fermat-Torricelli problem in
$\mathbb{R}^{2}$ without using the coordinates of the points
$A_{i},$ for $i=1,2,3$ (Coordinate independent approach) taking
into account only given Euclidean elements (lengths and angles).
\end{remark}

\section{Two new geometrical solutions of the weighted Fermat-Torricelli point in the Euclidean plane}

We present two new geometrical solutions to find the weighted
Fermat-Torricelli point in the weighted floating case.

The first solution deals with the position of $A_{1'}$ and
$A_{3'}$ which shall give the position of the two Simpson lines
defined by $A_{2}$ and $A_{2'}$ and $A_{3},$ $A_{3'}$
(fig.~\ref{fig2})

The second solution deals with the generalization of Hofmann's
rotation proof (\cite{Tikh:86},\cite{Hofmann:27}) for
$B_{1}=B_{2},$ such that $B_{1}$

\begin{problem}\label{firstprob}Construct the solution of Problem~1 (Weighted Fermat-Torricelli problem) using ruler and compass, under the condition (\ref{floatingcase}).
\end{problem}

\begin{figure}
\centering
\includegraphics[scale=0.8]{analyticsolution2d2newd}
\caption{}\label{fig2}
\end{figure}

\begin{proof}[Solution of Problem~\ref{firstprob}:]

We need to construct the vertices $A_{3'}$ and $A_{2'}$ (see
fig.~\ref{fig2}).

First, we construct the vertex $A_{3'}.$ We select a point $K$
which belongs to the linear segment $A_{1}A_{2},$ such that
$\|A_{1}K\|=B_{3}$ and we construct a triangle $\triangle A_{1}KL$
with the other two sides $\|A_{1}L\|=B_{2}$ and $\|KL\|=B_{3}.$

Thus, lemma~\ref{lem1} yields $\angle
A_{2}A_{1}A_{3'}=\alpha_{213'}=\pi-\alpha_{203}.$

We need to calculate $a_{13'},$ in order to find the location of
$A_{3}'.$

Take a point $G$ to the line defined by $A_{1}A_{2},$ such that
$\|A_{1}A_{2}\|=a_{12},$ $\|A_{2}G\|=B_{3}$ where
$\|A_{1}G\|=a_{12}+B_{3}$ and construct with a ruler and compass
the perpendicular linear segment at the point $G$ to the line
defined by  $A_{1}A_{2}$ and take a point $H$ such that
$\\|GH\|=B_{2}.$ We denote by $I$ the point of intersection of the
line defined by $A_{2}H$ and the perpendicular line at the point
$A_{1}$ with respect to the line defined by $A_{1}A_{2}$
(fig.~\ref{fig3}). Taking into account the similar triangles
$\triangle A_{1}A_{2}I$ and  $\triangle A_{2}GH,$ we get:

\begin{equation}\label{eqnalc1}
a_{13'}=a_{12}\frac{B_{2}}{B_{3}}.
\end{equation}

\begin{figure}
\centering
\includegraphics[scale=0.8]{analyticsolution2d2newd1}
\caption{}\label{fig3}
\end{figure}

Similarly, we construct the vertex $A_{2'}.$ We select a point $M$
which belongs to the linear segment $A_{1}A_{3},$ such that
$\|A_{1}M\|=B_{2}$ and we construct a triangle $\triangle A_{1}MN$
with the other two sides $\|A_{1}N\|=B_{3}$ and $\|MN\|=B_{1}.$

Thus, lemma~\ref{lem1} yields $\angle
A_{3}A_{1}A_{2'}=\alpha_{312'}=\pi-\alpha_{203}.$

We need to calculate $a_{12'},$ in order to find the location of
$A_{2}'.$

Take a point $Q$ to the line defined by $A_{1}A_{3},$ such that
$\|A_{1}A_{3}\|=a_{13},$ $\|A_{2}G\|=B_{3}$ where
$\|A_{1}Q\|=a_{13}+B_{3}$ and construct with a ruler and compass
the perpendicular linear segment at the point $Q$ to the line
defined by  $A_{1}A_{3}$ and take a point $R$ such that
$\\|RQ\|=B_{3}.$ We denote by $P$ the point of intersection of the
line defined by $A_{3}R$ and the perpendicular line at the point
$A_{1}$ with respect to the line defined by $A_{1}A_{3}$
(fig.~\ref{fig4}). Taking into account the similar triangle
$\triangle A_{1}A_{3}P$ and  $\triangle A_{3}QR,$ we get:

\begin{equation}\label{eqnalc1}
a_{12'}=a_{13}\frac{B_{3}}{B_{2}}.
\end{equation}

\begin{figure}
\centering
\includegraphics[scale=0.8]{analyticsolution2d2newd2}
\caption{}\label{fig4}
\end{figure}

\end{proof}

\begin{problem}\label{secgeomprob}Solve Problem~1 (Weighted Fermat-Torricelli problem)  by generalizing Hofmann's rotation, under the condition
(\ref{floatingcase}) and  $B_{1}=B_{2},$ such that
$B_{1}+B_{2}+B_{3}=1,$ $B_{1}>\frac{1}{4}$ and
$\alpha_{132}>\pi-\arccos\left(-1+\frac{(1-2B_{1})^{2}}{2B_{1}^{2}}\right).$
\end{problem}

\begin{proof}[Solution of Problem~\ref{secgeomprob}:]
\begin{figure}
\centering
\includegraphics[scale=0.8]{constructsymhof}
\caption{}\label{fig5}
\end{figure}

We consider a weight $B_{i}$ which corresponds to the vertex
$A_{i}$ in $\mathbb{R}^{2},$ for $i=1,2,3.$

By replacing $B_{1}=B_{2}$ in (\ref{alphai0j}) of
lemma~\ref{lem2}, we derive that:

\begin{equation}\label{alpha203103}
\alpha_{203}=\alpha_{103}=\arccos\left(1-\frac{1}{2B_{1}}\right)
\end{equation}

and

\begin{equation}\label{alpha102000}
\alpha_{102}=\arccos\left(-1+\frac{(1-2B_{1})^{2}}{2B_{1}^{2}}\right)
\end{equation}
which yields $B_{1}>\frac{1}{4}.$

Taking into account (\ref{alpha203103}) and (\ref{alpha102000}),
we rotate the triangle $\triangle A_{1}A_{2}A_{3}$ about $A_{3}$
through $\pi-\alpha_{102}=2\alpha_{103}-\pi rad$ and we obtain the
triangle $\triangle A_{3}A_{1}^{\prime}A_{2}^{\prime}.$ Let
$A_{F}^{\prime}$ be the corresponding weighted Fermat-Torricelli
point of $\triangle A_{1}^{\prime}A_{2}^{\prime}A_{3},$ for
$B_{1}^{\prime}=B_{1}$ and $B_{2}=B_{2}^{\prime}$
(fig.~\ref{fig5}). Thus, the points $A_{2},$ $A_{F},$
$A_{F}^{\prime}$ and $A_{3}^{\prime}$ are collinear
(fig.~\ref{fig5}), because $\triangle A_{F}A_{F}^{\prime}A_{3}$ is
an isosceles triangle and \[\angle A_{F}A_{F}^{\prime}A_{3}=\angle
A_{F}^{\prime}A_{F}A_{3}=\pi-\alpha_{103}.\]

\end{proof}

\begin{corollary}{\cite{Hofmann:27}, \cite{Tikh:86},
\cite{BolMa/So:99}}\label{corhof} For $B_{1}=B_{2}=B_{3}$ the
solution of Problem~\ref{secgeomprob} is given by rotating the
$\triangle A_{1}A_{2}A_{3}$ about $A_{3}$ through $60^{\circ}.$
\end{corollary}

\begin{proof}
By replacing $B_{1}=B_{2}=B_{3}=1$ in the solution of
Problem~\ref{secgeomprob}, we deduce that the rotation about
$A_{3}$ need to be $\pi-120^{\circ}=60^{\circ}<\alpha_{132}$
(Hofmann's rotation).
\end{proof}

The author is sincerely grateful to Professor Dr. Vassilios G.
Papageorgiou for his very valuable comments, fruitful discussions
and for his permanent attention to this work.

\appendix

\section{}
We mention two methods of the length of a linear segment with
respect to (I) a variable length and (II) a variable angle, which
have been used, in order to find the weighted Fermat-Torricelli
point.

I. A method of differentiating the length of a linear segment with
respect to the length of a variable linear segment is given first
in \cite[Proposition~2.6, (b), Remark~2.4,
Corollary~3.3]{Zachos/Zou:08}, \cite[formula~(4),
p.~413]{Zachos/Zou:08b} and has been explained in detail in
\cite{Cots/Zach:11}, \cite[Corollary~2]{Zachos/Cots:10}.
Specifically, by differentiating (\ref{i23cos}) with respect to
$a_{1},$ and by replacing in the derived equation
$\cos(\alpha_{213}-\alpha_{013})$ taken from (\ref{i23cos}) , we
obtain:
\begin{equation}\label{dera2a1}
\frac{\partial a_2}{\partial a_{1}}=\cos(\alpha_{102}).
\end{equation}
Similarly, by differentiating (\ref{i23cos3}) with respect to
$a_{1},$ and by replacing in the derived equation
$\cos(\alpha_{013})$ taken from (\ref{i23cos3}) , we obtain:
\begin{equation}\label{dera3a1}
\frac{\partial a_3}{\partial a_{1}}=\cos(\alpha_{103}).
\end{equation}
We mention a method of differentiating the length of a linear
segment with respect to a variable angle, which have been used, in
order to find the weighted Fermat-Torricelli point
(\cite[Proposition~2.6 (b)]{Zachos/Zou:08}) in $\mathbb{R}^{2}.$
By mentioning this technique of differentiation, we correct some
typographical errors which appear in \cite{Zachos/Zou:08}.
Specifically, by differentiating (\ref{i23cos}) with respect to
$\alpha_{013},$ we get:

\begin{equation}\label{dera2013}
\frac{\partial a_{02}}{\partial
\alpha_{013}}=-a_{01}\frac{a_{12}}{a_{02}}\sin(\alpha_{213}-\alpha_{013})
\end{equation}
From the sine law in $\triangle A_{1}A_{0}A_{2},$ we get:

\begin{equation}\label{sin102}
\frac{a_{12}}{\sin(\alpha_{102})}=\frac{a_{02}}{\sin(\alpha_{213}-\alpha_{013})}
\end{equation}

By replacing (\ref{sin102}) in (\ref{dera2013}), we obtain:
\begin{equation}\label{dera2a013}
\frac{\partial a_{02}}{\partial
\alpha_{013}}=-a_{01}\sin(\alpha_{102}).
\end{equation}
Similarly, by differentiating (\ref{i23cos3}) with respect to
$\alpha_{013},$ we obtain:
\begin{equation}\label{dera3a013}
\frac{\partial a_{03}}{\partial
\alpha_{013}}=a_{01}\sin(\alpha_{103}).
\end{equation}

config: arxiv


TITLE: if $M_1,M_2$ are two structures for FOL $L$ that differ in interpretations of signs not in statement $A$, then $M_1 \vdash A$ iff $M_2 \vdash A$
QUESTION [0 upvotes]: Let $L$ be a first order language. Let $A$ be statement, and let $M_1,M_2$ be two structures for $L$.
I want to prove that if $M_1$ and $M_2$ differ in interpretation only of signs that aren't in $A$ then: $M_1 \vdash A$ iff $M_2 \vdash A$
Naturally I would think to prove it by induction on the structure of $A$. However, in first order logic, formulas are created recursively, and not statements. So I can't do that.
Then I thought about using induction on the structure of formulas and the induction claim will be: for every formula without free variables. (because a statement is a formula without free variables).
But here I also have a problem in that step: $A = \exists_x B$, because $x$ might be free in $B$ and thus, I won't be able to use the induction claim.
So my real problem is here is to determine what exactly will be the induction claim.
Help would be appreciated.

REPLY [0 votes]: As you suspected you have to proceed by induction over the formulas (not necessarily closed).
The trick is to consider the semantics of first-order logic as an assignment of truth functions to open formulas: if $M$ is a $L$-structure and $\varphi(x_1,\dots,x_n)$ is a $L$-formula (with free variables $x_1,\dots,x_n$) then its truth-function is the
$$
F_\varphi^M \colon M^n \longrightarrow \{0,1\}
$$
where $F_\varphi^M(a_1,\dots,a_n)=1$ if and only if $M \models\varphi(a_1,\dots,a_n)$,
or equivalently if $M,a_1,\dots,a_n \models \varphi(x_1,\dots,x_n)$, where $M,a_1,\dots,a_n$ is the $L \cup \{x_1,\dots,x_n\}$-structure that interprets the symbols of $L$ as in $M$ and the constants $x_1,\dots,x_n$ with $a_1,\dots,a_n$ respectively
What you need to show is that for every formula $\varphi$ you have that $F_\varphi^{M_1}$ and $F_\varphi^{M_2}$ are equal.
This is the statement to be proven by induction on the complexity of the formulas.
Of course in the special case where $\varphi$ is a closed formula, and so the relative functions $F_\varphi^{M_1}$ and $F_{\varphi}^{M_2}$ are $0$-ary functions (that is they are constants), the theorem states that the formula has the same truth value in the two models.
Hope this helps.


TITLE: Prove perpendicular bisectors of non-parallel lines intersect
QUESTION [0 upvotes]: Suppose that $A$, $B$ and $C$ are points and that $AB$ and $BC$ are not parallel. Show that the perpendicular bisector of $AB$, $l$, and the perpendicular bisector of $BC$, $l'$, are not parallel and so
intersect.

REPLY [0 votes]: Assume otherwise. Then $AB$ is perpendicular to both $l$ and to its parallel $l'$. As $BC$ is also perpendicular to $l'$, we conclude that $AB$ and $BC$ are parallel.


TITLE: Partial derivative with respect to a matrix
QUESTION [0 upvotes]: I am struggling on the derivative of a function with respect to a matrix. Let$f(B)=a^\prime_{1\times n} B_{n\times k}\left( B^{T}A_{n\times n}B\right) ^{-1}B^{T}a_{n\times 1}$, and $a$ is a $n\times 1$ vector of constant, $A_{n\times n}$ is a $n\times n$ full rank constant matrix, and $B_{n\times k}$ is a $n\times k$ matrix with rank $k$ (i.e., I assume $k<n$). What's partial derivative of $f(B)$ with respetive to $B$, i.e., $\frac{\partial f(B)}{\partial B}$? Thanks a lot!

REPLY [1 votes]: Define a new matrix variable and calculate the various differentials of it
$$\eqalign{
 M &= B^TAB \cr
dM &= dB^T\,AB+B^TA\,dB \cr
dM^{-1} &= -M^{-1}\,dM\,M^{-1} \cr
}$$
Now write down the function and find the differential of it
$$\eqalign{
 f &= BM^{-1}B^T:aa^T \cr\cr
df
 &= (dB\,M^{-1}B^T+B\,dM^{-1}\,B^T+BM^{-1}\,dB^T):aa^T \cr
 &= dB:aa^TBM^{-T} -BM^{-1}\,dM\,M^{-1}B^T:aa^T+aa^TBM^{-1}:dB \cr
&= \Big(aa^TB(M^{-1}+M^{-T})-ABM^{-1}B^Taa^TBM^{-1}-A^TBM^{-T}B^Taa^TBM^{-T}\Big):dB \cr\cr
\frac{\partial f}{\partial B}
 &= aa^TB(M^{-1}+M^{-T})-ABM^{-1}B^Taa^TBM^{-1}-A^TBM^{-T}B^Taa^TBM^{-T} \cr\cr
}$$
In the above, colons are used to denote the trace/Frobenius product
$$\eqalign{ A:B = {\rm tr}(A^TB) }$$


TITLE: Minimal right ideals
QUESTION [0 upvotes]: Let $I$ be a minimal right ideal of a ring $R$ with $1$. If $r\in R$, could we say that $rI$ is zero or a minimal right ideal?
I assumed a right ideal $J$ in $rI$ and intersecting it with $I$ got a right ideal in $I$ which is equal to zero or equal to $I$ (due to minimality). If the intersection equals $I$, we get $I⊆rI$. But, I could not proceed any more, and would thank for any contribution!

REPLY [1 votes]: Intuitively, $rI$ might be very different from $I$ in terms of its members, so intersecting its subset with $I$ might not be useful.
If $J$ is a right ideal in $rI$, then, being a subset of $rI$, it has to arise as $rX$ for some $X\subset I$. Can you prove that $X$ is a right ideal? You might need to tighten the specification of $X$ a little to make it work.


TITLE: When $\left(2+\frac{x}{y}+\frac{y}{x}\right)^x$ is larger that $(1+x)^{1+x}$?
QUESTION [0 upvotes]: I need to find when:
$$\left(2+\frac{x}{y}+\frac{y}{x}\right)^x>(1+x)^{1+x}$$
given that $0<y<x<1$. I cannot manage to find a necessary and sufficient condition.

REPLY [1 votes]: $$2+\frac{x}{y}+\frac{y}{x}\ge2+2=4$$ with equality iff $x=y$.
$(1+x)^{1+\frac{1}{x}}$ is an increasing function for $x\ge-1$, since its derivative is $$(1+x)^{1+1/x}(1-\ln(1+x)/x)/x\ge0.$$ (Note $\ln(1+x)\le x$.) So on the given range it takes its maximum value at $x=1$ of $2^2$. Hence the two expressions meet at the value of 4 when $x=1=y$.


TITLE: Calculate most parallel vector
QUESTION [1 upvotes]: Let's suppose I have a vector $\vec{a}$ and arbitrary many vectors (in my example two: $\vec{b}$ and $\vec{c}$), which all have a common starting point P. Now I want to find out which of these vectors is most "aligned" to $\vec{a}$, so which vector is most parallel to it (On my image, it would be $\vec{b}$).

I know that I could use the dot product to calculate the angle between a and the other vectors and take the one with the smallest angle.
But the challenge here is that I must not to use functions/operations which yield irrational output. So when I take for example the dot product, I need to calculate the unit vector, which requires the square root. But a square root produces an irrational number. So is it possible to solve this problem just with "rational operations"?

REPLY [1 votes]: You want $\left( \dfrac{|\vec{a} \cdot \vec{b}|}{||\vec{a}||\, ||\vec{b}||} \right) \gt  \left(\dfrac{|\vec{a} \cdot \vec{c}|}{||\vec{a}||\, ||\vec{c}||} \right)$, i.e. $\dfrac{|\vec{a} \cdot \vec{b}|}{ ||\vec{b}||}  \gt \dfrac{|\vec{a} \cdot \vec{c}|}{ ||\vec{c}||} $
This is equivalent to  $\dfrac{(\vec{a} \cdot \vec{b})^2}{\vec{b} \cdot \vec{b}} \gt \dfrac{(\vec{a} \cdot \vec{c})^2}{\vec{c} \cdot \vec{c}} $, which may count as just using rational operations.
Just find the equivalent for multiple vectors and choose the maximum.


TITLE: Calculating $\lim_{n\to\infty}\left(\frac{\sin(2\sqrt 1)}{n\sqrt 1\cos\sqrt 1} +\cdots+\frac{\sin(2\sqrt n)}{n\sqrt n\cos\sqrt n}\right)$
QUESTION [3 upvotes]: Using the trigonometric identity of $\sin 2\alpha = 2\sin \alpha \cos \alpha$, I rewrote the expression to:
$$\lim_{n\to\infty}\left(\frac{\sin(2\sqrt 1)}{n\sqrt 1\cos\sqrt 1} + \cdots+\frac{\sin(2\sqrt n)}{n\sqrt n\cos\sqrt n}\right) = \lim_{n\to\infty}\left(\frac{2\sin(\sqrt 1)}{n\sqrt 1} +\cdots+\frac{2\sin(\sqrt n)}{n\sqrt n}\right)$$
I then tried using the squeeze theorem to get the limit, but I can't get an upper sequence that converges to zero, and the best I could get is a sequence that converges to two:
$$\frac{2\sin(\sqrt 1)}{n\sqrt 1} + \cdots+\frac{2\sin(\sqrt n)}{n\sqrt n} \leq \frac{2}{n\sqrt 1} + \cdots+\frac{2}{n\sqrt n} \leq \frac{2}{n}\cdot n \longrightarrow 2$$
What am I missing?

REPLY [0 votes]: You are considering
$$S_n=\frac 1n \sum_{k=1}^n \frac{\sin(2\sqrt k)}{\sqrt k\cos(\sqrt k)}=\frac 2n \sum_{k=1}^n \frac{\sin(\sqrt k)}{\sqrt k}$$
Using the integral test
$$\int \frac{\sin(\sqrt k)}{\sqrt k}\,dk=2\int \sin(t) dt=-2\cos(t)$$
$$\int_1^n \frac{\sin(\sqrt k)}{\sqrt k}\,dk=2 \left(\cos (1)-\cos \left(\sqrt{n}\right)\right)$$
$$\frac 2n \int_1^n \frac{\sin(\sqrt k)}{\sqrt k}\,dk=\frac{4 \left(\cos (1)-\cos \left(\sqrt{n}\right)\right)} n$$ and you can conclude.


TITLE: Why can we use flabby sheaves to define cohomology?
QUESTION [8 upvotes]: In my algebraic geometry class, we defined sheaf cohomology using flabby sheaves, and the functor on the category of sheaves on a space $X$:
$$
D: \mathcal F \mapsto D\mathcal F
$$
where
$$
D\mathcal F(U)=\left\{s:U\to \bigsqcup_{p\in X}\mathcal F_p\middle| s(p)\in\mathcal F_p\right\}
$$
$D\mathcal F$ is then a flabby sheaf containing $\mathcal F$.
But when I did some further research into derived functor cohomology, it became clear that the important property you need is that sheaf categories 'have enough injectives' - so every sheaf is a subsheaf of some injective sheaf.  At first I thought that flabby sheaves and injective sheaves were the same thing, but then I found out that in fact, being injective is a stronger property: every injective sheaf is flabby, but not every flabby sheaf is injective.
How then, were we able to develop the cohomology theory using flabby sheaves, when one in general needs to consider injective objects?

REPLY [12 votes]: At first, the problem to solve is of purely categorical nature and does involve neither injective nor flabby sheaves: Given any left-exact functor ${\mathcal F}:{\mathscr A}\to{\mathscr B}$ between abelian categories, define a measure of failure of its right exactness.
One attempt to formalize this is provided by the notion of $\delta$-functors: A $\delta$-functor is an ${\mathbb N}_{\geq 0}$-indexed family of additive functors ${\mathcal T}^i: {\mathscr A}\to{\mathscr B}$ with ${\mathcal T}^0\cong{\mathcal F}$ together with, for any short exact sequence ${\mathscr E}: 0\to X\to Y\to Z\to 0$ in ${\mathcal A}$ and any $i\geq 0$, a connecting morphism $\delta^i({\mathscr E}): {\mathcal T}^i(Z)\to{\mathcal T}^{i+1}(X)$ such that the following long sequence in ${\mathscr B}$ is exact:
$$...\to {\mathcal T}^{i-1}(Z)\to{\mathcal T}^i(X)\to{\mathcal T}^i(Y)\to{\mathcal T}^i(Z)\to{\mathcal T}^{i+1}(X)\to ...$$
Then the ${\mathcal T}^i$ with $i>0$ can be thought of as compensating for the failure of right-exactness of ${\mathcal T}^0$.
The questions of existence and uniqueness arise.
For uniqueness, one has the notion of universality: A $\delta$-functor ${\mathcal T}^{\ast}$ is universal if for any other $\delta$-functor ${\mathcal S}^i$ any natural transformation ${\mathcal T}^0\to{\mathcal S}^0$ can uniquely be extended to a morphism of $\delta$-functors ${\mathcal T}\to{\mathcal S}$. Fixing ${\mathcal T}^0$, universal $\delta$-functors are unique up to unique isomorphism, the most important criterion for universality is then the following: If ${\mathcal T}$ is effecable in the sense that for any $X\in{\mathscr A}$ there exists a monomorphism $X\hookrightarrow I$ such that ${\mathcal T}^i(I)=0$ for all $i>0$, then ${\mathcal T}^{\ast}$ is universal.
What about existence of an effecable $\delta$-functor?
Firstly think about which objects $I\in{\mathscr A}$ are candidates for the required ones satisfying ${\mathcal T}^i(I)=0$ for all $i>0$. Consider ${\mathcal T}^1$ first: If $0\to I\to A\to B\to 0$ is a short exact sequence, then ${\mathcal T}^1(I)$ should measure the failure of right-exactness of $0\to {\mathcal T}^0 I\to {\mathcal T}^0 A\to {\mathcal T}^0 B\to 0$ - in particular, it might/should vanish whenever this sequence is already exact? So say that $I$ is ${\mathcal T}^0$-preacyclic (not standard terminology, but I need to distinguish it from the standard notion of ${\mathcal T}^0$-acyclicity which is stronger) if any short exact sequence $0\to I\to A\to B\to 0$ stays exact under application of ${\mathcal T}^0$. Any ${\mathscr A}$-injective object is ${\mathcal T}^0$-preacyclic: any short exact sequence $0\to I\to A\to B\to 0$ with $I$ injective is split exact, and split exactness is preserved under any additive functor. However, particular functors may admit more preacyclic objects - for example, flabby sheaves are preacyclic for the global sections functor.
The ${\mathcal T}^0$-preacyclic objects are those you intuitively would like the functors ${\mathcal T}^i, i>0$ to vanish at (this will only be true for the stronger notion of ${\mathcal T}^0$-acyclic objects in the end, but at the moment, that's the idea). Interestingly, it turns out that this already forces the definition of ${\mathcal T}^i$ completely, and the argument for this is usually called dimension shifting: If you already defined ${\mathcal T}^0,...,{\mathcal T}^n$, and if $X\in{\mathscr A}$ is given, pick any $0\to X\to I\to X^{\prime}\to 0$ with $I$ a preacyclic object. Then the long exact sequences yields the isomorphism ${\mathcal T}^{n+1} X\cong\text{coker}\left({\mathcal T}^n I\to {\mathcal T}^n X^{\prime}\right)$, so provided you already defined ${\mathcal T}^n$ and you know which objects are to be preacyclic, the definition of ${\mathcal T}^{n+1}$ is forced.
All this was a bit sloppy but hopefully motivates, in the end, the idea of defining ${\mathcal T}^{\ast}$ through the choice of ${\mathcal T}^0$-preacyclic resolutions. It turns out however that in order to make this work one cannot work with arbitrary ${\mathcal T}^0$-preacyclic resolutions but instead has to use ${\mathscr S}$-resolutions, where ${\mathscr S}$ is any class of ${\mathcal T}^0$-preacyclic objects which additionally has the following properties:

Any object admits an embedding into an object in ${\mathscr S}$.
For $0\to X\to Y\to Z\to 0$ with $X,Y\in{\mathscr S}$, also $Z\in{\mathscr S}$.
${\mathscr S}$ is closed under summands.

If ${\mathscr A}$ has enough injectives, you can always choose ${\mathscr S}$ to be the class of injective objects, but in case of ${\mathcal T}^0 = \Gamma$ the global sections functor, you can also take for ${\mathscr S}$ the class of flabby sheaves.
A posteriori, you learn that there is even a largest class ${\mathscr S}$ with the above properties, namely the class of those $I\in{\mathscr A}$ satisfying ${\mathcal T}^i(I)=0$ for all $i>0$, and are called ${\mathcal T}^0$-acyclic. However, even without knowing ${\mathcal T}^{\ast}$ in advance you can prove acyclicity of objects by showing that they belong to a class ${\mathscr S}$ with the properties above.
From my experience, it is common to define ${\mathcal T}^i$ through injective resolutions and to show then that it can be computed using ${\mathscr S}$-resolutions as well. This has the advantage of making it simpler to check the independence of the choice of resolution, since injective resolutions are unique up to homotopy.


TITLE: What is the connection between density and distribution function?
QUESTION [2 upvotes]: X is a r.v. with distribution function_
$$F_X(x)=\begin{cases} 0 & x< 0\\ \frac{x}{2} & 0\leq x<1 \\\frac{1}{2} & 1\leq
 x<2 \\ \frac{3}{4} & 2\leq x <3 \\ 1 & x\geq 3 \end{cases}$$
It has density
$$f(x)=\begin{cases} \frac{1}{2} & 0\leq x <1 \\ \frac{1}{4} & x=2,3 \end{cases}$$
Im just wondering how they got the result of the density.
Im kinda confused on the relation between density and distribution function. There is a definition stating that if:
$$F_X(x)=\int_{-\infty}^x f_x(u) d\mu(u) $$ Then $f_x$ is the density of X. Did they use this definition to get the density functions in those two cases, or is it something trivial that im missing?

REPLY [0 votes]: Densities are linked with underlying measures.
Is the function $f$ mentioned in your question a density of the distribution function mentioned in your question?
Yes, but this only if the underlying measure is a very peculiar one.
An example (there are more) is the measure $$\mu:=\lambda+\delta_2+\delta_3\tag1$$ where $\lambda$ denotes the Lebesgue measure and for constant $c\in\mathbb R$ measure $\delta_c$ is prescribed by $A\mapsto1_A(c)$.
Then $$\begin{aligned}\int_{-\infty}^{x}f(u)\mu(du) & =\int_{-\infty}^{x}f(u)(\lambda+\delta_{2}+\delta_{3})(du)\\
 & =\int_{-\infty}^{x}f(u)\lambda(du)+\int_{-\infty}^{x}f(u)\delta_{2}(du)+\int_{-\infty}^{x}f(u)\delta_{3}(du)\\
 & =\int_{-\infty}^{x}f(u)\lambda(du)+f(2)1_{(-\infty,x]}(2)+f(3)1_{(-\infty,x]}(3)\\
 & =\int_{-\infty}^{x}f(u)\lambda(du)+\frac{1}{4}1_{(-\infty,x]}(2)+\frac{1}{4}1_{(-\infty,x]}(3)\\
 & =\int_{-\infty}^{x}1_{\left[0.1\right)}\left(u\right)+1_{\left\{ 2,3\right\} }\left(u\right)\lambda(du)+\frac{1}{4}1_{(-\infty,x]}(2)+\frac{1}{4}1_{(-\infty,x]}(3)\\
 & =\int_{-\infty}^{x}1_{\left[0.1\right)}\left(u\right)\lambda(du)+\frac{1}{4}1_{(-\infty,x]}(2)+\frac{1}{4}1_{(-\infty,x]}(3)\\
 & =F_{X}\left(x\right)
\end{aligned}
$$
If some function gets the predicate of density of distribution then - unless stated otherwise - it will be a density with respect to the Lebesgue measure.
Using peculiar measures like $(1)$ in most cases makes no sense and causes confusion. You are the living proof of that.


TITLE: Factor a rank one operator on a Hilbert C*-module.
QUESTION [2 upvotes]: Let $A \subset \mathcal{L}(\mathcal{H}_0)$ be a concrete $C^*$-algebra and $X$ be a right Hilbert A-module. For each $x,y \in X$ we have rank one operators
\begin{align}
\theta_{x,y}: X & \to X \\
 z & \mapsto x\langle y, z\rangle_A
\end{align}
The set $\mathcal{K}_A(X):=\overline{\mathrm{span}\{\theta_{x,y} : x,y \in X\}}$ is well known to be a $C^*$ algebra.

Question: Can I always find a representation of $\mathcal{K}_A(X)$ on some Hilbert space $\mathcal{H}_1$ such that for any $x \in X$ the map $\theta_{x,x}$, regarded as an element of $ \mathcal{L}(\mathcal{H}_1)$, "factors trhough $\mathcal{H}_0$"? To be precise, can I always find an element $b_x \in \mathcal{L}(\mathcal{H}_1, \mathcal{H}_0)$ such that $\theta_{x,x}=b_x^* b_x$?

I can't seem to find this in any of my usual references for Hilbert modules. It seems to be true for all the examples I've tried. For instance, for a usual Hilbert space regarded as a right Hilbert $\mathbb{C}$-module, we have $b_x=\langle x|$ so that $b_x^*$ is multiplication by $x$.
However, I can't quite prove it in general nor find a counterexample. I don't mind assuming that $X$ is full, but I am not sure this is needed. Also, it might be that this only works for a carefully chosen $\mathcal{H}_1$, but this is fine as I care for a "concrete" result rather than a necessarily abstract one.
Any reference or help to prove/disprove the claim will be much appreciated.

REPLY [1 votes]: Given $\mathcal{H}_0$ define $\mathcal{H}_1 := X \otimes_{A} \mathcal{H}_0$, which is a Hilbert space carrying a representation of $\mathcal{K}(X)$ given by $T(x \otimes h) = (Tx) \otimes h$. Define $b_x$ to be the "creation" operator $b_x \in \mathcal{L}(\mathcal{H}_0,\mathcal{H}_1)$ satisfying $b_x(h) = x \otimes h$ and then $b_x^*$ is the corresponding "annihilation" operator  $b_x^*(y \otimes h)  = \pi((x \mid y)_A) h$, where $\pi$ is the representation of $A$ on $\mathcal{H}_0$. Then
$$
b_x b_x^* (y \otimes h) = x \otimes \pi((x \mid y)_A)h = x(x \mid y)_A \otimes h = \Theta_{x,x}(y) \otimes h.
$$
Of course there's some checking to be done to make sure everything is well-defined.
I should also mention that the representation of $\mathcal{K}(X)$ on $\mathcal{H}_1$ is faithful since $\pi$ is faithful.


TITLE: Curvature via hessian in Taylor expansion
QUESTION [2 upvotes]: In the case of a univariate function, the smaller the second derivative in its Taylor expansion, the smaller is the curvature of the univariate function.
Now, how is the curvature of the function measured if it is a multivariate function, via its Hessian in the Taylor expansion? Is it some norm of the Hessian matrix or does it have something to do with the spectral properties of this matrix? What does the analogous of 'curvature' mean- in this case, and how is it measured/represented?

REPLY [4 votes]: "Curvature of a function" is not a standard mathematical term as far as I know, though it may be something found in a calculus textbook. If you mean the curvature of the graph of a function $y=f(x)$, then it involves the first derivative as well as the second. One can make the first derivative go away by changing the system of coordinates so that one of the axis is tangent to the curve at the point of interest. Then indeed, the second derivative $f''$ gives the curvature  of the graph.
For the surface  $z=f(x,y)$, assuming $\nabla f$ vanishes at the point of interest, we get the principal curvatures (plural) as the eigenvalues of the Hessian. The determinant of the Hessian gives Gaussian curvature, while the trace (that is, Laplacian $\Delta f$) gives mean curvature times two.
The relation is more complicated at the points where $\nabla f$ is nonzero, and the entire concept of curvature gets much more complex when we move from surfaces to higher-dimensional manifolds. Then the curvature cannot be adequately measured by a single number. An accessible reference is Curvatures of Hypersurfaces.


TITLE: Find $\lim_{x \to 0} \frac{e^{2x}-1}{\tan x}$
QUESTION [1 upvotes]: I'd like help finding
$$\lim_{x \to 0} \frac{e^{2x}-1}{\tan x}$$
without the use of L'Hôpital's rule.
So far I did this:
$$\lim_{x \to 0} \frac{e^{2x}-1}{\tan x}$$
$$=\lim_{x \to 0} \frac{\cos x(e^{2x}-1)}{\sin x}$$
$$=\lim_{x \to 0} \frac{x\cos x(e^{2x}-1)}{x\sin x}$$
$$=\lim_{x \to 0} \frac{\cos x(e^{2x}-1)}{x}$$
Basically I got nowhere. Any hints or partial solutions to help me?
By the way, the is the 1969 AP BC Multiple Choice #28.

REPLY [0 votes]: Here's where the logic for most of the answers comes from, and how we avoid the use L'Hopital's Rule.
First, let's observe a very important limit, namely $\displaystyle \lim_{h\to0}{e^h-1\over h}$. We know that $$e = \lim_{x\to\infty}\left(1+{1\over x}\right)^{x} = \lim_{y\to0}\left(1 + y\right)^{1/y}.$$ So, using some creative algebra we have $$\begin{align}\lim_{h\to0}{e^h - 1 \over h} &= \lim_{y\to0}{y\over\ln(y+1)} \tag{Let $y = e^h - 1 \iff h = \ln(y+1)$}\\ &=\lim_{y\to0}\left({1\over{1\over y}\ln(y+1)}\right) \\ &=\lim_{y\to0}{1\over\ln[(1+y)^{1/y}]}\\&={1\over\ln(e)}\\&={1\over1} \\&= 1.\end{align}$$ Notice how the limit changes from $h\to0$ to $y\to0$ because as $x\to 0$, $y = e^h - 1 \to e^0 - 1 = 1-1=0$, hence $y\to 0$. Now, we begin the important part of proving the limit.
Using this result, it is very easy to prove that $$\begin{align}\lim_{x\to0}{e^{2x} - 1\over x} &= \lim_{x\to0}\left({2\over2}\right)\left({e^{2x}-1\over x}\right)\\&=2\lim_{x\to0}{e^{2x}-1\over2x}\\&=2(1)\\&=2.\end{align}$$
Therefore, we have that $$\begin{align}\lim_{x\to0}{e^{2x}-1\over\tan x} &=\lim_{x\to0}[(e^{2x} - 1)\cot(x)] \\ &=\lim_{x\to0}\left((e^{2x}-1)\cdot{\cos(x)\over \sin(x)}\right) \\ &=\lim_{x\to0}\left((e^{2x}-1)\cdot{2x\over2x}\cdot{\cos(x)\over \sin(x)}\right)\\&=2\lim_{x\to0}\left[{e^{2x}-1\over2x}\cdot{x\over\sin(x)}\cdot\cos(x)\right]\\&= 2\lim_{x\to0}\left[{e^{2x}-1\over2x}\cdot{1\over{\sin(x)\over x}}\cdot\cos(x)\right]\\&=2\left(1\cdot{1\over1}\cdot1\right)\\&=2.\end{align}$$


TITLE: An urn starts with $5$ white and $7$ red balls.
QUESTION [0 upvotes]: I have a question about the following problem:
Consider a sample of size 3 drawn in the following manner: We start with an
urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the following probabilities: 0 white balls, 1 white ball, 3 white balls, and 2 white balls.
I believe the probabilities for white and red are 5/12 and 7/12. If there were 3 balls drawn, then that would mean that three additional balls for the same color are drawn as well, as noted above. So for instance, 0 white balls would mean RRRRRR = 5/12 * 5/12 * 5/12? Is that correct? I guess I'm confused on how to account for the additional ball of the same color being added.
Thanks in advance.

REPLY [0 votes]: a)$\frac{7}{12}\cdot\frac{8}{13}\cdot\frac{9}{14}=\frac{3}{13}$
b)$\frac{3\cdot(5\cdot6\cdot7)}{12\cdot13\cdot14}=\frac{5}{13}$
c)$\frac{5\cdot6\cdot7}{12\cdot13\cdot14}=\frac{5}{52}$
d)$\frac{3\cdot(5\cdot6\cdot7)}{12\cdot13\cdot14}=$


TITLE: Finding a min on $\sqrt{f(x)}$ is equal to min on $f(x)$
QUESTION [1 upvotes]: During a discussion about RMS, one said that finding the min of a function or of it square root is the same because square root is monotonic increasing.
Is this make any sense?

REPLY [0 votes]: Finding a minimum of a function is equivalent to finding the point at which the derivative of the function is equal to zero on a given interval. If we can show that the sign of the derivative of any $f(x)$ is equal to the sign of the derivative of any $\sqrt{f(x)}$ then we can show that the min or minimums for $\sqrt{f(x)}$ exist for exactly each minimum of $f(x)$.
Case 1:
Assume $f'(x) > 0$.
$\frac{d}{dx}\sqrt{f(x)}$ by the chain rule is equal to $\frac{1}{2\sqrt{f(x)}} * f'(x)$.We know the first term must be greater than zero since the square root cannot be negative. Thus $\frac{d}{dx}\sqrt{f(x)} > 0$
Case 2:
Assume $f'(x) < 0$
$\frac{d}{dx}\sqrt{f(x)} = \frac{1}{2\sqrt{f(x)}} * f'(x)$ Again, the first term is positive, and the second term is assumed to be negative, yielding a negative derivative.
$\frac{d}{dx}\sqrt{f(x)} < 0$
Case 3:
Assume $f'(x) = 0$
$\frac{d}{dx}\sqrt{f(x)} = \frac{1}{2\sqrt{f(x)}} * f'(x)$
Thus  $\frac{d}{dx}\sqrt{f(x)} = 0$
For the last case, and the case we are after, if f'(x) = 0, the location of all possible minimums, $\frac{d}{dx}\sqrt{f(x)}$ also equals zero. Technically, we have shown that the critical points must exist at the same location,  but since we have shown also that they increase and decrease over the exact same intervals, the nature of the critical points must also match. Thus we have shown that if a function is continuous, differentiable, and strictly positive, the minimums of its square root match the minimums of the function.