TITLE: Why do some people write the gravitational force as proportional to $\mathbf{r}/|\mathbf{r}|^3$?
QUESTION [12 upvotes]: I'm reading Mathematical Aspects of Classical and Celestial Mechanics, Second Edition by Arnold, Kozlov, and Neishtadt. It occurred to me that many people like to use third power when mention the law of universal gravitation.
For example,

$$F_{ij}=\frac{\gamma m_i m_j}{|r_{ij}|^3}r_{ij}$$ where $r_{ij}=r_i-r_j, \gamma =\text{const}>0$

Is this just some conversion/tradition/habit, or does it carry any significance? i.e. the absolute value changed sign and derivative?

REPLY [5 votes]: It is a matter of convenience:
Let $\vec{r_{ij}}$ be a distance vector with magnitude $r_{ij}$ along the line connecting the masses $m_i$ and $m_j$. Then:
(I): $\vec{r_{ij}}$ squared is a scalar whose value equals its magnitude squared. Proof: $\vec{r_{ij}}^2=\vec{r_{ij}} \cdot \vec{r_{ij}}=r_{ij} r_{ij} \cos{0}=r_{ij}^2$.
(II): $\vec{r_{ij}}$ can be written as $\vec{r_{ij}}=r_{ij} \hat{r_{ij}}$, with $\hat{r_{ij}}$ being a unit vector having the same direction as $\vec{r_{ij}}$. Proof: This goes as an axiom ;-)
If ones writes Newton’s law of gravity as
$$F_{ij}=\frac{γm_im_j}{\vec{r_{ij}}^2}=\frac{γm_im_j}{r_{ij}^2} \tag{1}$$
Then (1) is an incomplete description of the gravitational force. Equation (1) only represents the magnitude of the gravitational force, as can be noticed by the fact that the right side of it is just a scalar.
It is a well-established observation that gravity is a force whose direction is along the line connecting the masses $m_i$ and $m_j$. As a force, gravity must be written as a vector, and therefore the right form would be
$$\vec{F_{ij}}=\frac{γm_im_j}{r_{ij}^2} \hat{r_{ij}} \tag{2}$$
From (II), equation (2) can be written as
$$\vec{F_{ij}}=\frac{γm_im_j}{r_{ij}^2} \frac{\vec{r_{ij}}}{r_{ij}}=\frac{γm_im_j}{r_{ij}^3} \vec{r_{ij}} \tag{3}$$
Typing vectors and unit vectors with arrows and “hats” atop (as in current case) is a little cumbersome. To solve for typing vectors with arrows, one can select other options as boldfacing them for example. To avoid unit vectors one uses the convention $\frac{\vec{r}}{r}$, i.e., the choice of (3) is a convenient form of writing these sort of equations, especially for works containing a huge amount of text.
It is not only easier (if one uses a WYSIWYG text editor of typing machine) to type equation (3) using these conventions, but visually cleaner as well. Just check:
\begin{equation}
\mathbf{F}_{ij} \boldsymbol{=}\mathrm γ\dfrac{m_i m_j}{\mathbf{r}_{ij}^3 } \mathbf{r}_{ij}
\tag{4}
\end{equation}
As was pointed out by @AndrewSteane, there are even mathematical advantages by using this notation, hence, it is just more convenient to use the $\frac{\bf r}{r^3}$ notation.
The problem with this, however, is that people (almost) always is introduced to Newton’s law of gravity in its non-vector form (equation 1) and is expecting to find the inverse square law in a first glance, therefore, it may look odd for beginners until one gets used to it some point in future ;-)


TITLE: What has general relativity got to do with special relativity?
QUESTION [3 upvotes]: What has general relativity got to do with special relativity? Arguably, special relativity is encompassed in the requirement that the dynamical laws of physics be Poincare invariant (as opposed to Galileo, or something else).
General relativity, on the other hand, can be characterised first off by the claim that free-fall motion is geodesic motion, and secondly that the connection determining the geodesics is curved in the presence of matter.
At least on these two readings, the theories are independent of each other in their basic assumptions. Indeed, gravity in Newton-Cartan theory arises from a curved connection while assuming a Newtonian absolute time.
Yes, in both SR and GR there is a Lorentzian metric. But I'm taking that to be a non-essential feature of GR, the essential one being that free-fall worldlines are geodesics.
Historically, it appears that some (Einstein I believe) thought of the two subjects as linked in their foundations, and somewhat interdependent. If this was believed at some point, what was the link supposed to be? Why are they both 'relativity' and not the theory of relativity (SR), and the theory of spacetime curvature (GR)?
Edit:
It seems the question has been misunderstood. Poincare invariance and geometrised gravity are independent (just recall the Newton-Cartan theory a.k.a. geometrised Newtonian gravity). My question is why historically geometrised gravity came up in the context of Lorentzian spacetime, not, for instance, a classical (Newtonian) spacetime.

REPLY [2 votes]: Brief introduction
In Newtonian mechanics, the correlation between dynamical quantities and geometry was not that eminent. Sure, kinematics used geometry, but dynamics was good enough explained in terms of “forces”. The profound dependence of dynamical quantities with the underlying geometry was not that obvious. Just have a look into Galilean transformation for example
$$
\begin{cases}
t=t'\\
x=x'+Vt'\\
y=y'\\
z=z'
\end{cases}
\tag{1}
$$
It would be interesting if one wrote this in the form
$$ \left(
\begin{array}
&t\\
x\\
y\\
z\\
\end{array}
\right)=
\left[
\begin{matrix}
1&0&0&0\\
V&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{matrix}
\right]
\left(
\begin{array}
&t'\\
x'\\
y'\\
z'\\
\end{array}
\right)
\tag{2}
$$
But there is no much symmetry between space and time, and by those times there was no reason to believe that there was anything else there, especially because the absoluteness of time was completely established.
3 years after Einstein’s SR, with its radical changes in the concepts of space and time, Minkowski presented a reformulation of SR in terms of geometric concepts. In fact, most physical quantities seemed to be easily adapted to SR if they were written in the form of a 4-vector, including dynamical ones. An impressive symmetry between space and time transform, made people to believe that there was more in that than a simple coincidence.
Answering to your question
Have a look into the homogeneous Lorentz group:
$$ \left(
\begin{array}
&t\\
x\\
y\\
z\\
\end{array}
\right)=
\left[
\begin{matrix}
-1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{matrix}
\right]
\left(
\begin{array}
&t'\\
x'\\
y'\\
z'\\
\end{array}
\right)
\tag{3}
$$
There is an astonishing symmetry between time and space here. More importantly, we have already accepted the idea that time was not that special and could change due to speed (position derivative) effects, but not with position itself, hence why not try to generalize this to include positional effects along with speed effects. The natural way to generalize it is by writing
$$ \left(
\begin{array}
&t\\
x\\
y\\
z\\
\end{array}
\right)=
\left[
\begin{matrix}
F_1& F_2& F_3& F_4\\
F_5& F_6& F_7& F_8\\
F_9& F_{10}& F_{11}& F_{12}\\
F_{13}& F_{14}& F_{15}& F_{16}\\
\end{matrix}
\right]
\left(
\begin{array}
&t'\\
x'\\
y'\\
z'\\
\end{array}
\right)
\tag{4}
$$
With $F_i$ being functions of position, time and dynamical quantities plus the condition that (4) should become (3) for small space-time volumes. This is the connection between GR and SR. It generalizes the Lorentz homogeneous group.
Now, it should be noted that Newtonian Gravity was not Lorentz invariant and thus was out of the scope of SR. A Scar in the theory that needed to be addressed and Einstein saw the opportunity to link the form of those $F_i$ by linking then to gravity.
As a final thought, one could consider SR to encompass the changes in physics due to speed (position derivative) and GR as including both changes due to position and its derivative speed.
PS: I used the notation $c=1$ and the metric signature $(-,+,+,+)$.


TITLE: How to interpret tensor form PDE in terms of matrix algebra
QUESTION [3 upvotes]: From this mathwork page "c for system", the usual second order PDE is written in tensor form:
$$ -\nabla\cdot(\mathbf{c} \otimes \nabla \mathbf{u})+\mathbf{a}\mathbf{u}=\mathbf{f}
$$
and $\mathbf{c}$ is claimed to be a $N$-by-$N$-by-$2$-by-$2$ tensor. Each side of the equation should correspond to a $N$-by-$1$ vector in this case, and now I am confused... I want to express this equation in matrix form, since $\mathbf{c}$ is a 4th rank tensor while $\nabla{u}$ is 2nd order tensor($2$-by-$N$ matrix?), how to define the tensor operation $\otimes$ then? Seems wiki doesn't give the answer.
Cand anybody elaborate the rule here? (Also how to cast the gradient operator in this case?) Any physical insight is honestly appreciated!

REPLY [1 votes]: Using abstract index notation:
There are two types of indices here: I will use $i,j,\ldots$ for indices relative to the domain (which appears to be $\mathbb{R}^2$) and $A,B,\ldots$ for indices relative to the co-domain (basically from the number of equations) which is $\mathbb{R}^N$. We can write $\mathbf{c}$ as the rank 4 tensor in index notation
$$ c^{A;ij}_{B} $$
(I included a semicolon just for clarity to separate the upper case and lower case indices.)
and adopting the Einstein summation convention your equation is
$$ - \nabla_i ( c^{A;ij}_{B} \nabla_j u^B) + a^A_B u^B = f^A $$

Without using indices:
Your tensor $\mathbf{c}$ can be thought of a $(N\times 2)\times (N\times 2)$ matrix, that is a square matrix with $N\times 2$ rows and columns. You can flatten $\nabla \mathbf{u}$ to an $N\times 2$ column vector. Then $\mathbf{c}$ acts on $\nabla \mathbf{u}$ by matrix multiplication.
The key, of course, is to flatten things the same way from $\mathbf{c}$ and $\mathbf{u}$.


text:
\section{Conclusions}

We have presented new computational tools for particle tracking, based on Belief
Propagation and Loop Series and compared them to the Markov Chain Monte Carlo scheme
for the estimation of the permanent~\cite{JSV:poly_perm}. We have specifically
considered tracking of passive particles in fluid dynamics experiments. The methods
are quite general though and applications to other tracking problems, e.g. to
self-propelling biological objects, are possible and will be pursued. Our long-term
goal is to develop computational tools effective enough to permit on-the-fly
processing of particle tracking frames. Algorithms presented here ensure an
excellent accuracy and the BP-improved scheme is already orders of magnitudes faster
than the MCMC scheme.

config: arxiv


TITLE: What causes tangential velocity of a planet to change in an elliptical orbit?
QUESTION [3 upvotes]: I am confused about this concept of central force motion. We have read that object undergoing a change in velocity must possess acceleration. However, in elliptical orbit motion gravitational force is directed towards the radial direction, then what causes tangential velocity to change? I have read somewhere that there a component of gravitational force in the tangential direction as well, but I don't understand how?
Can anyone please help me with this?

REPLY [0 votes]: The main point is that the Earth is round. As the body moves in its vicinity, the direction of anything tangential continually changes. Without gravity (and another force) the body continues along direct line with constant velocity - even in this case the tangential velocity (speed) changes!


TITLE: How is computer science and physics related?
QUESTION [11 upvotes]: Not sure if this is a 'real' question, but what is the relationship between physics and computer science? A lot of physicists are also computer scientists and vice versa. My professor has a PhD in Physics but is a Computer Scienctist. What's the relation? Why is this so common?

REPLY [6 votes]: I think the main reason why this is so common is that many people who are of the tenured professor age now (50-60) were in graduate school before most colleges offered a Ph.D. in computer science.  So back then, people who were interested in theoretical computer science got their doctorate in Mathematics, and people who were interested in applied computer science got their degrees in a field of the natural sciences.  Since physics offers the most readily answerable questions computationally, particularly with the computational power available in the 1950s and 1960s, many applied computer scientists ended up in physics programs.  When schools began to offer full bachelor's, master's, and doctoral programs in computer science, these people returned to their first love.  (I gathered most of this from talking to a CS professor I had in undergrad who followed exactly the same path you describe.)  I'm pretty sure there are a lot more physics Ph.D.s in computer science professorships than vice versa (even though I'm sure there are exceptions, and I'll get several comments pointing this out).  I realize there is a lot of overlap between the two fields, but if you want to know the real reason what there are so many professors in CS from physics, this is why.


TITLE: question related to Bayes' rule and Bays' risk.
QUESTION [0 upvotes]: Let $X_1, X_2, X_3, \ldots, X_n$ be a random sample for $N(e,1)$. Let the prior p.d.f. of $e$ be $N(0,\sigma^2)$ under the square error loss function $L(e,d)={(d-e)}^2$. Find the Bayes' decision rule and the Bayes' risk of the Bayes' rule.
EDIT: Thanks for the help but I am just not able to the Bayes' risk by myself and the instructor is being very supportive for this problem. This is a simple homework problem which he gave casually in the class. But it has become a pain in my a**. Please help me solve the Bayes' risk as well. Thank you.

REPLY [0 votes]: This may be somewhat backwards, but I would actually find the Bayes' rule first and then derive the risk afterwards.
Rewriting your $e$ as $\mu$ to follow convention, note that under squared error loss, the Bayes decision rule is the mean of the posterior distribution of $\mu$.
$$X_1, \ldots, X_n \sim N(\mu, 1) \quad \textrm{and} \quad \mu \sim N(0, \sigma^2)$$
To find the posterior distribution, we only concern ourselves with the kernel of the density (i.e. only with parts of the density that involve $\mu$). This will simplify things.
$$\begin{align*}
\pi(\mu \,|\, x) &\propto \exp \left\{ -\frac{1}{2} \left( \frac{\sum(x_i - \mu)^2}{1} + \frac{\mu^2}{\sigma^2} \right) \right\}\\
&\propto \exp \left\{ -\frac{1}{2\sigma^2} \left( \sigma^2 \sum(x_i - \mu)^2 + \mu^2 \right) \right\}
\end{align*}$$
Next, write $\sum(x_i - \mu)^2$ as $\sum(x_i - \overline{x} + \overline{x} + \mu)^2$ to see that it simplifies to $\sum(x_i - \overline{x})^2 + n(\overline{x} - \mu)^2$. The first term does not involve $\mu$ so we drop it to eliminate the summation, and expand the second term to get:
$$\begin{align*}
\pi(\mu \,|\, x) &\propto \exp \left\{ -\frac{1}{2\sigma^2} \left( \sigma^2 (n\mu^2 - 2 n \overline{x} \mu) + \mu^2 \right) \right\}\\
&\propto \exp \left\{ -\frac{1}{2\sigma^2} \left( \mu^2(\sigma^2 n + 1) - 2\mu(\sigma^2 n \overline{x})  \right) \right\}\\
&\propto \exp \left\{ -\frac{1}{2\frac{\sigma^2}{\sigma^2 n + 1}} \left( \mu^2 - 2\mu \frac{\sigma^2 n \overline{x}}{\sigma^2 n + 1}  \right) \right\}\\
&\propto \exp \left\{ -\frac{1}{2\frac{\sigma^2}{\sigma^2 n + 1}} \left(\mu - \frac{\sigma^2 n \overline{x}}{\sigma^2 n + 1} \right)^2 \right\}
\end{align*}$$
This is the kernel of a normal distribution with mean $\frac{\sigma^2 n}{\sigma^2 n + 1} \overline{x}$. So that is your Bayes' decision rule.
You can see why this makes sense. Normally, without prior information, we would just estimate the mean by $\overline{x}$. But since we know that $\mu$ is centered at 0, we take that information into account by shrinking the naive estimator $\overline{x}$ towards 0 by the factor $\frac{\sigma^2 n}{\sigma^2 n + 1}$. Furthermore, if we have lots of data (i.e. $n$ is large) or if the variance of $\mu$ is known to be large, then the shrinkage factor is closer to 1, meaning we give less weight to our prior knowledge.
To find the Bayes' risk, just apply the bias-variance decomposition to the Bayes' rule to find the "ordinary" risk, and apply the definition of Bayes' risk. Then simply evaluate the integral to get the Bayes' risk.


TITLE: Limits $\lim_{x\rightarrow 0}\int_x^{2x}\frac{e^t}{t}dt$ and $\lim_{x\rightarrow 0}\frac{1}{x}\int_x^{2x}\frac{e^t}{t}dt$.
QUESTION [1 upvotes]: Let us consider the following two limits:
$$1) \lim_{x\rightarrow 0}\int_x^{2x}\frac{e^t}{t}dt$$
$$2) \lim_{x\rightarrow 0}\frac{1}{x}\int_x^{2x}\frac{e^t}{t}dt$$
For the first, since
$$e^{x} \int_x^{2x}\frac{1}{t}dt\leq\int_x^{2x}\frac{e^t}{t}dt\leq e^{2x} \int_x^{2x}\frac{1}{t}dt$$
and $\int_x^{2x}\frac{1}{t}dt=\ln 2$, for $x\rightarrow 0$ we have (by comparison theorem) that
$$\lim_{x\rightarrow 0}\int_x^{2x}\frac{e^t}{t}dt=\ln 2$$
For the second, by denoting $F(x)=\int_0^{x}\frac{e^t}{t}dt$ and by applying L'Hôpital's rule:
$$\lim_{x\rightarrow 0}\frac{1}{x}\int_x^{2x}\frac{e^t}{t}dt=\left[\int_x^{2x}\frac{e^t}{t}dt\right]'=\lim_{x\rightarrow 0} \left[F(2x)-F(x)\right]'=\lim_{x\rightarrow 0} \left(2F'(2x)-F'(x)\right)$$
The result is $1$.
Now we come to my question. If first limit is $=\ln 2$, I would expect that the second $\rightarrow \infty$. Why this did not happen?

REPLY [2 votes]: I think your second calculation is wrong: firstly, since you discovered that the limit of the numerator exists and is nonzero, you don't have an indeterminate form and cannot apply L'Hôpital's Rule.
To compute the limit, you can use the same technique you used in the first bit, to see it blows up. Or note that the integral is the average value of the function on $[x,2x]$, and hence must be between the maximum and minimum of the function:
$$ \frac{e^{2x}}{2x} \leqslant \frac{1}{x}\int_x^{2x} \frac{e^t}{t} \, dt \leqslant \frac{e^x}{x}, $$
for sufficiently small $x$, when the function is decreasing in $x$. Hence it is bounded below by $e^x/x \to \infty$, and hence does diverge as you suppose.


TITLE: Check whether a system {$v_1,...,v_m$} of vectors in $\mathbb R^n$ (in $\mathbb R[x]$) is linearly independent.
QUESTION [0 upvotes]: Check whether a system {$v_1,...,v_m$} of vectors in $\mathbb R^n$ (in $\mathbb R[x]$) is linearly independent.
These are my thoughts:
For {$v_1,...,v_m$} to be linearly independent, prove that:
$\lambda_1v_1 + \lambda_2v_2 + ... + \lambda_mv_m = \theta$ where $\lambda_1, ..., \lambda_m \in F$ and $\lambda_1 +... + \lambda_m = 0$ and $\theta$ represents the null vector ($\underline 0$).
So I'm assuming that each vector is in $\mathbb R^n$ but there are $m$ vectors in this system so I wrote them as a linear combination with scalars. Is this right?
So now how is the best way to do this proof. Can you do it in $\mathbb R$ and $\mathbb R^2$ then generalize or is that normally not allowed?
Will you need to use mathematical induction to generalize? I would just like some advice before I waste my time.

REPLY [1 votes]: Easy way to do so is to build a matrix which $ {v}_{1}, {v}_{2}, ... $ are its rows or columns.
Calculate the Determinant of this matrix.
If the determinant is different from zero, the vectors are independent.
Let $ M \in {\mathbb{R}}^{n \times m} $ such that $ M = \begin{bmatrix}
\vdots & \vdots & \vdots & \vdots \\
{v}_{1} & {v}_{2} & \cdots & {v}_{m} \\
\vdots & \vdots & \vdots & \vdots
\end{bmatrix} $.
Then the following holds, $ \det \left( M \right ) = 0 \Leftrightarrow \left \{ {v}_{1}, {v}_{2}, \cdots {v}_{m} \right \}, \; Linear \, Dependent $.
The intuition behind it comes from the meaning of the determinant, where you can read about at the Wikipedia Article - The Determinant.
This works for $ m = n $.
For the case $ m > n $ they must be dependent.
For the case $ m < n $ you can use the SVD.
If the number of singular values is lower than $ m $ they are dependent.
By the way, this is equivalent of the question whether the equation $ M x = 0 $ has a solution or not.


text:
\chapter{Derivations on Hahn Fields}\label{derha}

\noindent
In this chapter $\smallk$ is a {\em differential\/} field with derivation $\der$ (possibly trivial), and $\Gamma$ is an ordered abelian group. Let $\alpha,\beta, \gamma$ range over $\Gamma$, and fix an additive map $c:\Gamma \rightarrow \smallk$.

In the first section we consider how the map $c$ gives rise to a strong derivation on the Hahn field $\smallk((t^\Gamma))$ extending the given derivation $\der$ of $\smallk$.

In the second section we
consider how to solve differential equations $y-ay'=f$ for $a,f\in \smallk((t^\Gamma))$ with $a\prec 1$. The main result in this chapter is Theorem~\ref{thmA}, which is about preserving truncation closedness under adjoining solutions of such equations to
truncation closed differential subfields of $\smallk((t^\Gamma))$.
This will be important in the next chapter.

In the third section we summarize a general construction from
\cite{DMM} that involves (partial) exponential maps on Hahn fields.
This will also be needed in the next chapter.

\section{Extending the Derivation on the Coefficient Field to the Hahn Field}\label{extderco}

\noindent
The map $c$ allows us to extend $\der$ to a derivation on the field $\smallk((t^\Gamma))$, also denoted by $\der$, by declaring for $f=\sum_{\gamma}f_\gamma t^\gamma\in \smallk((t^\Gamma))$ that
\[\der \left(f\right)\ :=\  \sum_{\gamma} \Big(\der (f_\gamma) +f_\gamma c(\gamma)\Big)t^\gamma.\]
Thus $(t^\gamma)'=c(\gamma)t^\gamma$.
For $f\in \smallk((t^\Gamma))$ we have $\supp f'\subseteq \supp f$, so $f'\preceq f$. In particular, $\der$ is small, and the natural field isomorphism from
$\smallk$ onto the (differential) residue field $\bigO/\smallo$ of $\smallk((t^\Gamma))$ is an isomorphism of differential fields.
 What are the constants of $\smallk((t^\Gamma))$? In the  extreme case
$c(\Gamma)=\{0\}$, the constant field of $\smallk((t^\Gamma))$ is
$C_{\smallk}((t^\Gamma))$. For us the opposite extreme is more relevant:

\begin{lemma}\label{cgood}
The following are equivalent:
\begin{enumerate}
	\item[\rm{(i)}] the constant field of $\smallk((t^\Gamma))$ is the same as the constant field of $\smallk$;
    \item[\rm{(ii)}] $c$ is injective and $c(\Gamma) \cap \smallk^\dagger = \{0\}$;
    \item[\rm{(iii)}] for all $f\in \smallk((t^\Gamma))^\times$ with
$f\not\asymp 1$ we have $\der(f)\asymp f$, hence
$f^\dagger \asymp 1$.
\end{enumerate}
\end{lemma}

\noindent
Thus the valued differential field $\smallk((t^\Gamma))$ is asymptotic under the conditions of Lemma~\ref{cgood}.

\medskip\noindent
Sometimes it is more natural to consider a Hahn field $\smallk[[\fM]]$, where $\fM$ is not given in the form $t^\Gamma$, and then
a map $c: \fM\to \smallk$ is said to be \textbf{additive} if $c(\fm\fn)=c(\fm) + c(\fn)$
for all $\fm, \fn\in \fM$. Again, such $c$ allows us to extend $\der$ to a derivation $\der$ on the field $\smallk[[\fM]]$ by
\[\der \left(f\right)\ :=\  \sum_{\fm} \Big(\der (f_\fm) +f_\fm c(\fm)\Big)\fm.\]
This operator $\der$ on $\smallk[[\fM]]$ has support $\{1\}$, and so is strong.

 \medskip\noindent
{\bf Examples.} For $\fM=x^{\mathbb{Z}}$ with $x\succ 1$ we have the usual derivation
$\frac{d}{dx}$ with
respect to $x$ on the field of Laurent series
$\smallk[[x^{\mathbb{Z}}]]=\smallk((t^{\mathbb{Z}}))$ (with $t=x^{-1}$), but it is not of the form considered above.
The derivation $x\frac{d}{dx}$, however, does have the form above, with the
trivial derivation on $\smallk$ and $c(x^k)=k\cdot 1\in \smallk$ for
$k\in \mathbb{Z}$. Likewise for $\fM=x^{\mathbb{Q}}$ ($x\succ 1$),
and $\operatorname{char}(\smallk)=0$: then  $x\frac{d}{dx}$ has the above
form, with the trivial derivation on $\smallk$ and
$c(x^q)=q\cdot 1\in \smallk$ for
$q\in \mathbb{Q}$.

\bigskip\noindent
We now return to the setting of  $\smallk((t^\Gamma))$, and observe that
$\der(f|_\gamma) = \der(f)|_\gamma$ for $f\in \smallk((t^\Gamma))$. Thus by Proposition \ref{D}:

\begin{corollary}\label{lem1}
If $R$ is a truncation closed subring of $\smallk((t^\Gamma))$ and $f\in R$ is such that $\der(g)\in R$ for every proper truncation $g$ of $f$,
then all proper truncations of $\der(f)$ lie in $R$ and thus
$R[\der(f)]$ is truncation closed.
\end{corollary}

\begin{lemma}\label{DifFieldGenTrunc}
Let $R$ be a truncation closed subring of $\smallk((t^\Gamma))$.
Then the differential subring of $\smallk((t^\Gamma))$ generated by $R$ is
truncation closed.
\end{lemma}

\begin{proof}
Let $R_0 = R$ and $R_{n+1} = R_n[\der(f): f\in R_n]$.
Assume inductively that $R_n$ is a truncation closed subring of
$\smallk((t^\Gamma))$ .
For $f\in R_n$ and $\gamma\in \Gamma$ we have $\der(f)|_\gamma = \der(f|_\gamma)$, so all truncations of $\der(f)$ lie in $R_{n+1}$, and thus $R_{n+1}$ is truncation closed by Proposition~\ref{D}(ii).
Since the differential ring generated by $R$ is $\bigcup_{n} R_n$, and a union of truncation closed subsets of $\smallk((t^\Gamma))$ is truncation closed, we conclude that the differential subring of $\smallk((t^\Gamma))$ generated by $R$ is truncation closed.
\end{proof}

\noindent
Note that Lemma~\ref{DifFieldGenTrunc} goes through with ``subfield'' instead of
``subring''.

\section{Adjoining Solutions of $y-ay'=f$}\label{adj}

\noindent
We wish to preserve truncation closedness under
adjoining solutions of differential equations
\[y-ay'\ =\ f \qquad(a,f \in \smallk((t^\Gamma)),\ a\prec 1).\]
This differential equation is expressed more suggestively as
$(I-a\der)(y)=f$, where $I$ is the identity operator on
$\smallk((t^\Gamma))$ and $a\der$ is considered as a (strong)
operator on $\smallk((t^\Gamma))$ in the usual way. Note that $a\der$
is even small as defined earlier, hence
$I-a\der$ is bijective, with inverse

\[(I-a\der)\inv\ =\ \sum_n (a\der)^n\]
Thus the above differential equation has a unique solution
$y=(I-a\der)\inv(f)$ in $\smallk((t^\Gamma))$. This is why we now turn our attention to the operator $(I-a\der)\inv$.

\medskip\noindent
For $n\geq 1$, $0\leq m\leq n$ we define $G^n_m(X)\in \mathbb{Z}\{X\}$ recursively as follows:
\begin{itemize}
\item $G^n_0 = 0$,
\item $G^n_n = X^n$,
\item $G^{n+1}_m = X(\der(G^n_m) +G^{n}_{m-1})$ for $1\leq m\leq n$.
\end{itemize}
This recursion easily gives
\[(a\der)^n\ =\ \sum_{m=1}^n G^n_m(a)\der^m \qquad(n\ge 1),\]
hence
\begin{equation}
(I-a\der)\inv = I + \sum^{\infty}_{n=1} \sum_{m=1}^n G^n_m(a)\der^m.
\end{equation}

\noindent Since $G^n_m(X)$ is homogeneous of degree $n$, we have
$G^n_m(a)\in \smallk t^{n\alpha}$ for $a\in \smallk t^\alpha$.

\begin{lemma}\label{OpLemma1}
Suppose $R$ is a truncation closed differential subring of
$\smallk((t^\Gamma))$, $a\in R\cap\smallk t^\Gamma$, $a\prec 1$, $f\in R$, and  $(I-a\der)\inv(g)\in R$
for all $g\truncof f$. Then all proper truncations of $(I-a\der)\inv (f)$ lie in $R$.
\end{lemma}

\begin{proof} We have
$(I-a\der)\inv(f) = f+  \sum^{\infty}_{n=1} \sum_{m=1}^n G^n_m(a)\der^m(f)$.
Also $a=a_{\alpha}t^\alpha$ with $\alpha>0$, hence $\supp  \left(G^n_m(a)\der^m(f)\right)\subseteq n\alpha+\supp f$ for
$1\le m\le n$. Consider a proper truncation $(I-a\der)\inv(f)|_{\gamma}$
of $(I-a\der)\inv(f)$; we have to show that this truncation lies in $R$.
The truncation being proper gives $N\in \mathbb{N}^{\ge 1}$ and $\beta\in \supp f$ with
$\gamma\le N\alpha +\beta$.
Let $f_1:= f|_\beta$ and $f_2:=f-f_1$. Then $f_1, f_2\in R$ and
\[(I-a\der)\inv(f)\ =\ (I-a\der)\inv(f_1) + f_2 + \sum^{\infty}_{n=1} \sum_{m=1}^n G^n_m(a)\der^m(f_2).\]
Using $\supp f_2\ge \beta$ and truncating at $\gamma$ gives
\[(I-a\der)\inv(f)|_{\gamma}\ =\ (I-a\der)\inv(f_1)|_\gamma + f_2|_\gamma + \left.\left(\sum^{N-1}_{n=1} \sum_{m=1}^n G^n_m(a)\der^m(f_2)\right)\right|_\gamma,\]
which lies in $R$, since $f_1\truncof f$ and thus $(I-a\der)\inv(f_1)\in R$.
\end{proof}

\noindent
The key inductive step is provided by the next lemma.

\begin{lemma}\label{OperatorLemma} Let $R$ be a truncation closed differential subring of $\smallk((t^\Gamma))$. Let $a,f\in R$ be such that $a\prec 1$ and for all $b,g\in R$,
\begin{enumerate}
\item[\rm{(i)}] $g\truncof f \Rightarrow (I-a\der)\inv (g)\in R$,
\item[\rm{(ii)}] $b\truncof a \Rightarrow (I-b\der)\inv(g)\in R$.
\end{enumerate}
Then all proper truncations of $(I-a\der)^{-1}(f)$ lie in $R$.
\end{lemma}

\begin{proof} Assume $(I-a\der)^{-1}(f)|_\gamma\truncof (I-a\der)^{-1}(f)$. Then we have
$\alpha\in \supp a$ and $\beta\in \supp f$, and $N\in \mathbb{N}^{\ge 1}$ such that
 $\gamma\leq N\alpha + \beta$.
Put $a_1:=a|_{\alpha}$, $a_2:= a-a_1$, $f_1:=f|_\beta$, and $f_2:=f-f_1$.
Set $P=a_1\der$, $Q=a_2\der$, so $a\der=P+Q$, $P,Q$ are small, and
\begin{align*} (I-a\der)\inv(f)\ &=\ (I-a\der)\inv(f_1) + (I-a\der)\inv(f_2)\\
 &=\ (I-a\der)\inv(f_1) + \big(I-(P+Q)\big)\inv(f_2).\end{align*}
Now $Q(I-P)\inv=\sum_n QP^n$ is small, and we have the identity
\begin{align*} I-(P+Q)\ &=\ \big(I-Q(I-P)\inv\big)(I-P),\ \text{ so}\\
\big(I-(P+Q)\big)\inv\ &=\ (I-P)\inv \big(I-Q(I-P)\inv\big)\inv\\
&=\ \sum_{m=0}^\infty (I-P)\inv\big(Q(I-P)\inv\big)^m, \text{ hence}\\
(I-a\der)\inv(f)\ &=\ (I-a\der)\inv(f_1) + \sum_{m=0}^\infty (I-P)\inv \big(Q(I-P)\inv\big)^m(f_2).
\end{align*}
For $m\ge N$ we have $\supp \big(Q(I-P)\inv\big)^m(f_2)\ge \gamma$, so truncating at $\gamma$ yields
\[(I-a\der)\inv(f)|_{\gamma}\ =\ (I-a\der)\inv(f_1)|_{\gamma} + \left.\left(\sum_{m=0}^{N-1}(I-P)\inv \big(Q(I-P)\inv\big)^m(f_2)\right)\right|_\gamma.\]
Since $f_1\truncof f$ the first summand of the right hand side lies in $R$. Since $a_1\truncof a$, we have $(I-P)\inv(h)\in R$ for all $h\in R$, and thus, using $a_2, f_2\in R$,
\[\sum_{m=0}^{N-1}(I-P)\inv (Q(I-P)\inv)^m (f_2)\in R.\]
Therefore $(I-a\der)^{-1}(f)|_\gamma\in R$.
\end{proof}

\begin{theorem}\label{thmA} Let $E$ be a truncation closed differential subfield
of $\smallk((t^\Gamma))$. Let $\widehat{E}$ be the smallest differential subfield of
$\smallk((t^\Gamma))$ that contains $E$ and is closed under $(I-a\der)\inv$ for all $a\in \widehat{E}^{\prec 1}$. Then $\widehat{E}$ is truncation closed.
\end{theorem}

\begin{proof}
Let $F$ be a maximal truncation closed differential subfield of $\widehat{E}$
containing $E$. (Such $F$ exists by Zorn's Lemma.) It suffices to show that
$F= \widehat{E}$. Assume $F\ne \widehat{E}$. Then there exist  $a\in F^{\prec 1}$ and $f\in F$ with
$(I-a\der)\inv(f) \notin F$. Take such $a$ and $f$ for which
$(\alpha, \beta)$ is minimal for lexicographically ordered pairs of ordinals,
where $\alpha$ is the order type of the support of $a$ and $\beta$ that of
 $f$. By minimality of $(\alpha,\beta)$ we can apply Lemma \ref{OperatorLemma} to $F,a$, and $f$ to get that $F\big((I-a\der)\inv(f)\big)$ is
truncation closed,
using also parts (i) and (ii) of Proposition~\ref{D}. Since $F\big((I-a\der)\inv(f)\big)$ is a
differential subfield of $\widehat{E}$, this contradicts the maximality of $F$.
\end{proof}

\subsection*{Complementing the previous results}
Suppose $\smallk_1$ is a differential
subfield of $\smallk$ and $E$ is a truncation closed
subfield of
$\smallk((t^\Gamma))$ such that $a\in \smallk_1$ and
$c(\gamma)\in \smallk_1$ whenever $at^\gamma\in E$, $a\in \smallk^\times$, $\gamma\in \Gamma$. Let $\Gamma_1$ be the subgroup of $\Gamma$ generated by the
$\gamma\in \Gamma$ with $at^\gamma\in E$ for some $a\in \smallk^\times$.
In connection with Lemma~\ref{DifFieldGenTrunc} we note:

\medskip\noindent
{\em The differential subfield of $\smallk((t^\Gamma))$ generated by $E$ is
contained in  $\smallk_1((t^{\Gamma_1}))$.}

\medskip\noindent
This is because $E\subseteq \smallk_1((t^{\Gamma_1}))$ and $\smallk_1((t^{\Gamma_1}))$ is a differential subfield of
$\smallk((t^\Gamma))$. Likewise we can complement Theorem~\ref{thmA}:

\medskip\noindent
{\em If
$\der E\subseteq E$ and $\widehat{E}$ is the smallest
differential subfield of $\smallk((t^\Gamma))$ that contains $E$ and is closed under
$(I-a\der)\inv$ for all $a\in \widehat{E}^{\prec 1}$,
then $\widehat{E}\subseteq \smallk_1((t^{\Gamma_1}))$.}

\medskip\noindent
This is because $\smallk_1((t^{\Gamma_1}))$ is closed under
$(I-a\der)\inv$ for all $a\in\smallk_1((t^{\Gamma_1}))^{\prec 1}$.

\section{Exponentiation}

\noindent
Our aim is to apply the material above
to the differential field $\T_{\exp}$ of purely exponential transseries.
The construction of $\T_{\exp}$ involves an
iterated formation of Hahn fields, where at each step we apply
 the following general procedure stemming from \cite{DMM} (but copied from \cite[Appendix A]{ADH}).

\medskip\noindent
Define a {\bf pre-exponential ordered field}
to be a tuple~$(E, A, B, \exp)$ such that
\begin{enumerate}
\item $E$ is an ordered field;
\item $A$ and $B$ are additive subgroups of $E$ with $E=A\oplus B$ and $B$
convex in $E$;
\item $\exp\colon B\to E^{\times}$ is a strictly increasing group morphism (so $\exp(B)\subseteq E^>$).
\end{enumerate}
Let $(E,A,B,\exp)$ be a pre-exponential ordered field. We view $A$ as the part of $E$ where exponentiation is not yet defined, and accordingly we introduce
a ``bigger'' pre-exponential ordered field $(E^*,A^*,B^*,\exp^*)$ as follows:
Take a {\em multiplicative\/} copy $\exp^*(A)$ of the ordered additive
group $A$ with order-preser\-ving isomorphism
$\exp^*\colon A\to\exp^*(A)$,
and put $E^*\ :=\ E[[\exp^*(A)]]$. Viewing $E^*$ as an ordered Hahn field over
the ordered coefficient field $E$, we set
\[ A^*\ :=\ E[[\exp^*(A)^{\succ 1}]],\qquad B^*\ :=\ (E^*)^{\preceq 1}\ =\ E\oplus
(E^*)^{\prec 1}\ =\ A \oplus B \oplus (E^*)^{\prec 1}.\]
Note that $\exp^*(A)^{\succ 1}=\exp^*(A^{>})$. Next we extend
$\exp^*$ to $\exp^*\colon B^*\to (E^*)^{\times}$ by
\[\exp^*(a+b+\varepsilon)\ :=\ \exp^*(a)\cdot \exp(b)\cdot \sum_{n=0}^\infty
\frac{\varepsilon^n}{n!} \qquad(a\in A,\ b\in B,\ \varepsilon\in (E^*)^{\prec 1}).\]
Then $E\subseteq B^*=\operatorname{domain}(\exp^*)$, and $\exp^*$ extends $\exp$. Note that $E<(A^*)^>$ (but $\exp^*(E)$ is cofinal in $E^*$ if $A\ne \{0\}$). In particular, for $a\in A^{>}$, we have
\[\exp^*(a)\in \exp^*(A^{>})\ \subseteq\ (A^*)^>,\ \text{  so $\exp^*(a)\ >\ E$.}\]

\medskip\noindent
Assume also that a derivation $\der$ on the field $E$ is given that respects
exponentiation, that is, $\der(\exp(b))=\der(b)\exp(b)$ for all
$b\in B$. Then we extend $\der$ uniquely to a
derivation on
the field $E^*$, also denoted by $\der$, by requiring
that $\der$ is strong as an operator on the Hahn field $E^*$
over $E$, and that $\der(\exp^*(a))=\der(a)\exp^*(a)$
for all $a\in A$. This falls under the general construction at the beginning of this section with $\smallk=E$ and $\fM= \exp^*(A)$ by taking the additive function
$c: \exp^*(A) \to E$ given by $c(\exp^*(a))=\der(a)$.
This extended derivation on $E^*$ again respects
exponentiation in the sense that
$\der(\exp^*(b))=\der(b)\exp^*(b)$ for all
$b\in B^*$.

\subsection*{Exponential ordered fields} For the sake of completeness and because we use this notion in the next chapters
we define an {\bf exponential ordered field\/} to be an ordered field $E$ equipped with a strictly increasing group morphism
$\exp: E \to E^{\times}$ from the additive group of $E$ into its multiplicative group. Note that then $\exp(E)\subseteq E^{>}$, and
that this is basically the case of a pre-exponential ordered field
$(E, A, B, \exp)$ with $A=\{0\}$ and $B=E$.


TITLE: Combinatorics problem involving multinomial theorem
QUESTION [1 upvotes]: here is my problem:

4 digits are picked randomly (from 0 to 9), in how many combinations can you have only 2 different digits? Meaning that I would have let us say 2,2,4,4 or 5,5,5,8.

So the way I wanted to do this was imagining that the digits were actually like 4 oranges going to 10 children and how many possibilities are there for only 2 children getting oranges. So I figured this would be separating the groups first so
10 choose 8 choose 1 choose 1 x 4 choose 2 choose 2

for the possibilities where both children get 2 oranges. And now for the possibilities where 1 of the children gets 3 and the other gets 1, would be
10 choose 8 choose 1 choose 1 x 4 choose 3 choose 1.

Unfortunately this gives me $$\frac{10!}{8!}\times \frac{4!}{2!2!}+\frac{10!}{8!}\frac{4!}{3!} = 900$$ possibilities whereas the right answer is $630.
Can anyone help me out please! and with this type of solving the problem preferably as I am trying to figure out the subtleties of the multinomial theorem!

REPLY [1 votes]: As you point out, we can have a $2$-$2$ split or a $3$-$1$ split.
For $2$-$2$ splits, we choose $2$ numbers from $10$. There are $\binom{10}{2}=45$ ways to do this.  For each choice, there are $\binom{4}{2}$ ways of arranging them in a row, total $270$.
For $3$-$1$ splits, the number we have $3$ of can be chosen in $10$ ways, and the number we have $1$ of can be chosen in $9$ ways.  Then the digits can be arranged in $4$ ways, total $360$.
Remark: If we are not interested in counting ordered arrangements, then the total is $135$. To produce $630$, we need to take order into account.


TITLE: Semi-free resolutions
QUESTION [6 upvotes]: Let $\mathscr{C}$ be a DG category (not much will be lost if you assume that $\mathscr{C}$ has one object, i.e. is a DG algebra). One way to construct the unbounded derived category of $\mathscr{C}$-modules is by using semi-free resolutions. Recall that a DG module $F$ is free if it is a sum of shifts of corepresentable modules, or semi-free if it has an exhaustive filtration $0 = F_0 \subset F_1 \subset \cdots \subset F$ such that the subquotients $F_{n+1}/F_n$ are free.
I've seen the existence of semi-free resolutions stated as follows: for any DG module $M$, there is a semi-free module $F$ and a surjective quasi-isomorphism $F \to M$. But how does one actually construct $F$? I'm especially confused about how to make $F \to M$ surjective, since generators of free modules are cycles.
Edit: I found a construction of $F \to M$ in Drinfeld's paper "DG quotients of DG categories." Start by choosing a free module $F_1$ and a morphism $F_1 \to M$ which is surjective on cohomology. Now induct: given $F_n \to M$, let $C_n = \text{Cone}(F_n \to M)$ then find a free module $P_n$ and a morphism $P_n \to C_n$ which is surjective on cohomology. Using the map $C_n \to F_n[1]$ we get $P_n \to F_n[1]$, and Drinfeld takes $F_{n+1} = \text{Cone}(P_n[-1] \to F_n)$. Of course $F$ is the direct limit of the $F_n$.
This is nice, but I still don't understand how to make $F \to M$ surjective. Does this happen automatically for the construction given above, or do we have to do something extra?

REPLY [6 votes]: There is a recent paper of Tobi Barthel, Emily Riehl, and myself
that answers this question in a model categorical framework.  We
were lazy and only considered the one object case, although we
believe our work generalizes to DG categories.  In another respect we
were not lazy: we work over a general commutative ring $R$, not just
a field, and this introduces interesting subtleties.  The reference is
https://arxiv.org/abs/1310.1159
The standard model structure on (unbounded) DG modules over a DG
$R$-algebra $A$ takes quasi-isomorphisms as weak equivalences.
Cofibrant approximations are more general than semi-free resolutions,
but they are retracts of semi-free resolutions given by model theoretic
cellular DG modules.  There is an early construction by my adviser, John Moore,
in the 1959-60 Cartan seminar, which we modernize.  This uses bicomplexes,
as usual in differential homological algebra.  There is another construction,
originally due to Gugenheim and myself, dating from 1974, which we also
modernize.  Qiaochu, you will be interested that we drop surjectivity in that
construction, meaning that we do not have model theoretic fibrations.  We use
multicomplexes, which are bigraded but have differentials that are sums of
pieces that mimic differentials in spectral sequences.   The drop of
surjectivity and the use of multicomplexes gives a great gain of
computability, as we illustrate by modernizing 1974 applications to the
computation of the cohomology of many homogeneous spaces.
We also explain the role of the bar construction.  When R is a field it
constructs cofibrant approximations in the standard model structure.  In
general, it constructs cofibrant approximations in a relative model
structure for which the weak equivalences are the maps of DG $A$-modules
which are chain homotopy equivalences of underlying $R$-modules, and in
that generality it is not semi-free.


TITLE: Triangle formed by circumcenter, orthocenter and incenter
QUESTION [1 upvotes]: I'm trying to solve the following question from my analytical geometry course:

In a right-angled triangle $T$ the circumcenter $O$, the orthocenter $H$ and the incenter $I$ are the vertex of the triangle $T'$. The sides sizes of $T$ are $a$, $b$ and $c$ and its centroid is $G$.

What relation must a, b and c satisfy so that the segment $I$ $G$ is height of the triangle $T'$?

What relation must a, b and c satisfy so that the segment $I$ $G$ is angle bissector of the triangle $T'$?

I've found the Euler Line (the line that contains $O$, $G$ and $C$) equation of $T$,
$$y = x\cdot-\frac bc + \frac{b(b+c)}{a+b+c}$$
and the equation of the segment $I$ $G$ line,
$$y = x\cdot \frac{c(a-2b+c)}{b(a+b-2c)} + \frac{c(b-c)}{a+b-2c}$$
but then I can't finish the exercise. Can someone help me?

REPLY [0 votes]: Spoiler: the answer to both questions is: there is no such triangle.

Using usual notation for the right-angled $\triangle ABC$
with side lengths $a,b,c:\ c^2=a^2+b^2$,
semiperimeter $\rho=\tfrac12(a+b+c)$, inradius $r=\tfrac12(a+b-c)$, circumradius $R=2c$,
incenter $I$, centroid $G$, circumcenter $O$ and orthocenter $H=C$.
We can use known relations:
\begin{align}
|GI|^2&=\tfrac19\,(\rho^2+5r^2-16rR)
\tag{1}\label{1}
,\\
|GO|^2&=R^2-\tfrac29\,(\rho^2-r^2-4rR)
\tag{2}\label{2}
,\\
|OI|^2&=R\,(R-2r)
\tag{3}\label{3}
,\\
|IH|^2&=(r+2R)^2+2r^2-\rho^2
\tag{4}\label{4}
,\\
|CG|&=2|GO|
\tag{5}\label{5}
.
\end{align}
For the first question we must have
\begin{align}
|GI|^2+|GO|^2-|OI|^2&=0
\tag{6}\label{6}
,\\
\text{or }\quad
\rho&=\sqrt{7r^2+10rR}
\tag{7}\label{7}
.
\end{align}
For the second question we must have
\begin{align}
|IH|-2\,|OI|&=0
\tag{8}\label{8}
,\\
\text{or }\quad
|IH|^2-4\,|OI|^2&=0
\tag{9}\label{9}
,\\
\text{or }\quad
\rho&=\sqrt{3r^2+12rR}
\tag{10}\label{10}
.
\end{align}
To check if
the condition \eqref{7} or \eqref{10}
is achievable for a valid right-angled triangle,
let's consider
a map of all possible shapes of triangles in terms of
two dimensionless parameters: $u=\rho/R$ and $v=r/R$.
It is known, that for a valid triangle $v\in(0,\tfrac12)$,
and $u(v)\in\Big(u_{\min}(v),u_{\max}(v)\Big)$, where
\begin{align}
u_{\min}(v)&=
\sqrt{27-(5-v)^2-2\sqrt{(1-2v)^3}}
\tag{11}\label{11}
,\\
u_{\max}(v)&=
\sqrt{27-(5-v)^2+2\sqrt{(1-2v)^3}}
\tag{12}\label{12}
.
\end{align}
Moreover, any point $(v,u)$ of the area $\mathcal{T}$, bounded by
\eqref{11}, \eqref{12} and the vertical line $v=0$,
corresponds to a unique valid triangle with $R=1,\ \rho=u,\ r=v$.
Boundary curves \eqref{11}, \eqref{12} correspond to isosceles shapes,
and the point $(\tfrac12,\tfrac{3\sqrt3}2)$ corresponds
to the equilateral shape.
The condition for the right-angled triangles is
given by
\begin{align}
u(v)&=v+2
\tag{13}\label{13}
,
\end{align}
this is a straight line, crossing the
area of all valid triangular shapes $\mathcal{T}$.
Conditions \eqref{7} and \eqref{10}
combined with \eqref{13}
result in two values of $v$,
\begin{align}
v_1&=\tfrac{\sqrt{33}}6-\tfrac12
\tag{14}\label{14}
\\
\text{and }\quad
v_2&=\sqrt6-2
,\quad\text{ respectively}
\tag{15}\label{15}
,
\end{align}
with corresponding $u_1=v_1+2$ and $u_2=v_2+2$.
Both points $(v_1,u_1),\ (v_2,u_2)$
are located outside
the area of valid triangular shapes,
below the lower boundary \eqref{11}.
Thus the answer to both questions is:
there is no any valid triangle with given properties.
The sides $a,b,c$ of the triangle with $R=1$
and given pair $(v,u)$ can be found as
the roots of cubic equation
\begin{align}
x^3-2u\,x^2+(u^2+v^2+4v)\,x-4\,uv
&=0
\tag{16}\label{16}
,
\end{align}
and for all points $(v,u)$ from the validity region $\mathcal{T}$,
all the roots are positive and correspond to a valid triangle.
It's easy to check that both solutions of \eqref{16}
with given pairs $(v_1,u_1),\ (v_2,u_2)$
result one side length equal $2$ (the hypotenuse),
and the other two roots are complex.
This is the illustration of the map, where
the blue and red are the boundary curves,
the black line correspond to the right triangles,
and green and orange lines correspond to
the conditions of the first and the second question,
respectively.


TITLE: Let $f'(x)=e^{x^2}$; compute $\lim_{x\to0}\frac{f(2)-f(x+2)}x$
QUESTION [3 upvotes]: Let $f : [0,\infty] \to \mathbb{R}$ be a function such that its derivative $f'(x)=e^{x^2}$.
Compute $$\lim_{x\to 0} \frac{f(2)-f(x+2)}{x}$$
The definition of the derivative is
$$f'(a)= \lim_{x\to 0} \frac{f(a+x)-f(a)}{x}$$
so if $f'(a)=e^{a^2}$ then
$$e^{a^2}=\lim_{x\to 0} \frac{f(a+x)-f(a)}{x}$$
if $a=2$ then
$$e^{a^2}=\lim_{x\to 0} \frac{f(2+x)-f(2)}{x}$$
I'm a little bit stumped here, my naive approach would be to multiply both sides by $(-1)$ and get that the answer is $-e^{2^2}$ but I don't think I can do that.

REPLY [3 votes]: By an elementary property of the limit, and by the definition of the derivative,
$$\lim_{x\to 0} \frac{f(2)-f(x+2)}{x}=-\lim_{x\to 0} \frac{f(x+2)-f(x)}{x}=-f'(2).$$

REPLY [3 votes]: We have
$$f'(a)= \lim_{x\to 0}\frac{f(a+x)-f(a)}x$$
but you made a mistake in replacing the LHS with $e^{x^2}$. It should be $e^{a^2}$:
$$e^{a^2}=\lim_{x\to 0}\frac{f(a+x)-f(a)}x$$
Set $a=2$ and negate:
$$-e^{2^2}=\lim_{x\to0}\frac{f(2)-f(x+2)}x=-e^4$$


TITLE: Show that function is convex if other function is convex
QUESTION [0 upvotes]: Let C be a convex subset of $\mathbb{R}^{n}$. Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$. For any two points x and y in C, define the function $\phi: [\ 0,1 ]\ \rightarrow \mathbb{R}$ as $\phi(\lambda) = f(\lambda x + (1-\lambda)y)$. I have to show that if $\phi$ is convex for any pair of x and y points in C, then f is convex.
I began by assuming $\phi$ is convex and taking $\lambda_{1}$ and $\lambda_{2}$, such that $\forall \mu \in (0,1)$
\begin{equation}
\phi(\mu \lambda_{1} + (1-\mu)\lambda_{2}) \leq \mu\phi(\lambda_{1}) + (1-\mu)\phi(\lambda_{2})
\end{equation}
Let $\alpha = \mu \lambda_{1} + (1-\mu)\lambda_{2}$, then
\begin{equation}
f(\alpha x + (1-\alpha)y) \leq \mu f(\lambda_{1} x + (1-\lambda_{1})y) + (1-\mu)f(\lambda_{2} x + (1-\lambda_{2})y)
\end{equation}
From here on I don't know how to proceed
PS: I think the statement is true, if it isn't then a counterexample will suffice
Thanks for the help in advance

REPLY [0 votes]: $$\begin{aligned}
f(\lambda x+(1-\lambda)y)&=\phi(\lambda)=\phi(\lambda\cdot1+(1-\lambda)\cdot0)\\
&\leq\lambda(\phi(1)+(1-\lambda)\phi(0) \\
&=\lambda f(x)+(1-\lambda)(f(y)
\end{aligned}$$