TITLE: Need introduction to coordinate-free geometry
QUESTION [1 upvotes]: Forgive my whiny tone -- I'm driving myself nuts.  These questions have
been driving me crazy and I'm losing objectivity.
I need a serious introduction to coordinate-free geometry.
Let me ask more than one question, not because I deserve multiple
answers but because I want to show exactly how I am confused and
thus get a better reference.  I'm after a reference, not answers.
I've searched Amazon's book section and it returns
books that are pretty obviously the stuff I studied back in
the 70s.  A google search for coordinate-free geometry tutorial
tends to return stuff that breezes past all the elementary stuff
and launches into differential geometry and so on.  I'm not ready
for that.
This was provoked by chapter 1 of Thorne and Blandford's book,
wherein the dot product is defined by
$$A \cdot B = \frac{1}{4}[(A+B)^2 - (A-B)^2]$$
Now, I can expand the rhs with the best of them.
$$A \cdot B = \frac{1}{2}[AB + BA]$$
But, what, precisely, is $AB$ in the above?
This is the kind of stuff I'm having trouble with.
Where can I get a real introduction to coord-free geometry that
will show me the algebra?
Allow me to drone on a bit.  Since I'm having trouble getting
graphics into my latex doc let me play a little fast and loose with
my example.  Consider the unit circle at the origin.  There's a tangent
line to that circle which is horizontal and runs through (0,1).
That's what I think of when I hear the word tangent.
Aside:  if I want to bend a sheet of aluminum I can take a
knife and a straightedge and score a line on the aluminum
and then bend it.  OTOH, as Lincoln said, ``four score and
seven years ago....''  Now, that's one word, score, used for
two completely separate things.
Back to my question:  If I try to read a book on manifolds
I tend to run into tangent space.  I can't get anyone to say
that tangent space doesn't mean  tangent at all.
If we "reserve" tangent to mean straight lines sitting
on curves, we don't "lose" anything by calling tangent space
something like George space.  That is, there's nothing tangent
about tangent space.  Right?
From the descriptions of tangent space it appears
that what it really boils down to is the set of all vectors using
the particular point as the origination point.  That is, that's
what I'm getting out of it.  Ah, here's another way to get
at my issue:  I'm worried that there's something I'm missing in
the descriptions of tangent space that somehow includes
some facet of tangent.
(Sigh.)  One more thing occurs to me:  Two points, $P$ and $Q$.
Deliberately ignoring/omitting the vectors between $P$ and $Q$,
assuming I'm right about what a tangent space is, that means
that a vector originating at $Q$ is not part of $P$'s tangent space.
Am I right?
(Deeper sigh.)  Also, I need a proper definition of this operator:  $\otimes$.

REPLY [0 votes]: Your product AB is called the geometric product, and it's what defines vectors in the first place. A dot product is the symmetric part of that basic product.


TITLE: Normal space and regular space
QUESTION [0 upvotes]: 1) Let $X$ be a regular space with a countable base. Let $U$ be an open set in $X$.
Show that $U$ equals a countable union of closed sets of $X$.
Please help me !

REPLY [0 votes]: Let $B_n, n \in \Bbb N$ be a countable base for $X$.
Then by regularity $$U=\bigcup \{\overline{B_n}: \overline{B_n} \subseteq U\}$$
so $U$ is an at most countable union of closed sets.
(If $x \in U$, for some $O$ open we have $x \in O \subseteq \overline{O} \subseteq U$ by regularity of $X$. Then for some $n$: $x \in B_n \subseteq O$ as the $B_n$ form a base. It follows that $x \in \overline{B_n}$ which is part of the right hand union. If OTOH $x$ is in the right hand union, it's in some $\overline{B_n}$ that by definition (!) is a subset of $U$ so that $x \in U$ as well. Two inclusions, hence equality of sets.)

REPLY [0 votes]: Let $(B_i)$ be a countable basis. Let $x \in U$. Then (by regularity)  there exist an open set $V_x$ such that $x \in V_x $ and $\overline {V_x} \subseteq U$. There exist $i_x$ such that $x \in B_{i_x} \subseteq  V_x$. Now verify that $U$ is the union of the closed sets $\overline {B_{i_x}}$.


TITLE: Induction Proof on String
QUESTION [2 upvotes]: Formally prove the correctness of the union construction as follows. Let

$M_1$ and $M_2$ be the two $\lambda$-NFA's constructed for $R_1$ and $R_2$ and let
$N$ be the $\lambda$-NFA constructed so that $L(N) = R_1 + R_2$. Let
$w$ be a string such that ${\Delta_N}^{*}(q,w,f)$, where
$q$ is the start state and
$f$ is the final state.

Prove that either ${\Delta_{M_1}}^{*}(q_1, w, f_1)$ or ${\Delta_{M_2}}^{*}(q_2, w, f_2)$. Use induction on all strings $w$.

I've never done induction on strings before, so I'm a bit confused. I was thinking about using the empty string as the base case, but I'm not sure what I can even say about that. If there's a path from $q$ to $f$ by way of $\lambda$, then obviously this must be the case for one of the two $M$ NFAs... but why? How can I explain this?
Getting past the base case, I'm imagining the proof showing all types of strings $w$: $\text{'a'}$, $\text{'ab'}$, $\text{'a + b'}$, and $\text{'a*'}$ succeeding in the same way as the base case sufficing to prove this for all strings? Thanks for any help, I'm reading and re-reading this section in the book and I'm getting nothing out of it.

REPLY [0 votes]: Hint. Very often, induction on strings means induction on the length of a string.
Here is a general scheme to prove that some property $P$ holds for all words of $A^*$. As I said, you prove the result by induction on the length $n$ of a word $u$. You first prove $P$ for $n = 0$ (which means that $u$ is the empty word, right?). Then you suppose that $P$ holds for all words of length $\leqslant n$ and you prove $P$ for a word $u$ of length $n+1$. Very often$^{(1)}$, you start by observing that $u = pa$, where $p$ is the prefix of length $n$ of $u$ and $a$ is a letter and you apply the induction step on $p$.
You may now try this method on your exercise.
(1) In some cases, you may have to write $u=as$, where $s$ is the suffix of length $n$ of $u$ and $a$ is a letter and apply the induction step on $s$.


TITLE: Find a Linear transformation $ T:\mathbb{R}^3\rightarrow \mathbb{R}^3 $
QUESTION [2 upvotes]: Find a Linear transformation $ T:\mathbb{R}^3\rightarrow \mathbb{R}^3 $ with
(a) Range($T$) =span{(1,1,1)}
(b) Range($T$) =span{(1,2,3),(1,3,2)}
I have done like that. Since Range($T$) =span{(1,1,1)} let us take $T(1,0,0)=(1,1,1)$ and $T(0,1,0)=(0,0,0)$,$\ $  $T(0,0,1)=(0,0,0)$.   And then we can find the transformation. But this transformation is not unique.
Am I right?

REPLY [1 votes]: It looks like you've got the right idea. Here's an alternate perspective.
Recall that an $n\times m$ matrix $A$ defines a linear map $T:\Bbb R^m\to\Bbb R^n$ by $T(x)=Ax$. In this situation, the column-space of $A$ is equal to the range of $T$
$$
\DeclareMathOperator{Col}{Col}
\DeclareMathOperator{Range}{Range}
\Col(A)=\Range(T)
$$
Since $\Col(A)$ is the span of the columns of $A$ we see that every $3\times 3$ matrix whose columns are scalar multiples of $\langle 1,1,1\rangle$ give an example for (a). Of course, it follows that a linear map with this property is not unique.
Can you construct a matrix that gives an example for (b)?


TITLE: Determine the interval on which f(x) is decreasing, increasing and find points of inflection.
QUESTION [0 upvotes]: $$f(x) =\frac{1}{1+e^{-2x}}$$
Find all asymptotes of the graph of f(x).
Determine the interval on which f(x) is decreasing.
Determine the interval on which f(x) is increasing.
Determine all the points of inflection of f(x)
The asymptotes are vertical none , horizontal 0 and 1
then to find the interval decreasing , increasing you will have to get first derivative but I do use the chain rule?

REPLY [1 votes]: Let's simplify $f(x)$ a bit.
$$f(x) =\frac{1}{1+e^{-2x}} = \frac{1}{\left(1+\frac{1}{e^{2x}}\right)}$$
$$ = \frac{1}{\frac{\left(e^{2x}+1\right)}{e^{2x}}}$$
$$ = \frac{e^{2x}}{e^{2x}+1}$$
Now this is something that you can differentiate using chain rule/quotient rule.
$$f'(x) = \frac{e^{2x}\left(2\right)\left(e^{2x}+1\right)-e^{2x}\left(e^{2x}\right)\left(2\right)}{\left(e^{2x}+1\right)^2} $$
$$ f'(x) = \frac{2e^{2x}\left(e^{2x}+1-e^{2x}\right)}{\left(e^{2x}+1\right)^2}$$
$$ f'(x) = \frac{2e^{2x}}{\left(e^{2x}+1\right)^2}$$
What we notice here is that $f'(x)$ is never zero, and thus it has $no$ critical points. So, if you plug in any random value, you will see that the second deriative is $always$ positive, and thus always $increasing$.

Then you find $f''(x)$ for inflection points.


TITLE: Understanding when a vector is parallel vs. perpendicular?
QUESTION [2 upvotes]: I am having some difficulty with this concept. I am studying a problem:

Find an equation for the plane that passes through the point $P = (1, 2, 3)$
  and contains the line $L$ given by the parametric equation
  $x(t) = 1 − 3t,\, y(t) = 3$, and $z(t) = 6 + 2t$.

The solution states that $v = \left< 3, 0, −2 \right>$ is parallel to $L$ and hence the plane. $Q = (1, 3, 6)$ is in the plane and
hence $PQ = \left< 0, 1, 3 \right>$ is parallel to the plane.
$n = v \times PQ = \left< −2, 9, −3 \right>$ is perpendicular to the plane.
Therefore, the plane is $−2x + 9y − 3z = −2 \cdot 1 + 9 \cdot 2 − 3 \cdot 3 = 7$.
Could someone help me with this reasoning? I understand $v$ is parallel to $L$ because it is a scaled version of the normal to $L$, where the normal is $\left< -3, 0, 2 \right>$. I understand why $Q$ is in the plane.
Why is $PQ$ parallel to the plane? Why do we know that $v \times PQ$ will be perpendicular to the plane? Is this because the cross product of two vectors parallel to a plane will be perpendicular to the plane? Thank you for any clarification or help in advance! Just trying to get a better hold on this concept.

REPLY [1 votes]: To begin with, $v$ isn’t normal to $L$. That’s pretty much the opposite of being parallel. In vector form, the given parameterization of the line is $\langle1,3,6\rangle+t\langle-3,0,2\rangle$ from which it should be obvious that $Q=(1,3,6)$ is a point on the line (for $t=0$) and $v=\langle-3,0,2\rangle$ is the line’s direction vector, so is by definition parallel to $L$.
The points $P$ and $Q$ are on the plane, so the line segment $\overline{PQ}$ is also contained in the plane. The vector $Q-P$ (what you’re calling $PQ$) is parallel to this line segment, which means that it is also parallel to any plane that contains the line segment.
Finally, the cross product of two vectors is perpendicular to them both. Since both of the vectors are parallel to the plane you’re constructing, their cross product will also be perpendicular to that plane.


TITLE: Using the Normal to approximate the Binomial
QUESTION [1 upvotes]: A vote on whether to allow the use of medical marijuana is being held. A polling company will survey $200$ individuals to measure support for the new law. If in fact $53\%$ of the population oppose the new law, use the normal approximation to the binomial, with a continuity correction, to approximate the probability that the poll will show a majority in favor?
Attempt
Let $X$ be the number of people who favor the law. The sample size is $n=200$. The probability of success $p =  0.47 $. Since $np = 94 >5$ with can approximate $X$ with a normal distribution. Notice, $\mu = np = 94$ is the number of people who favor the law. Thus, we want
\begin{align} P(X > 94) &\implies P(X \geq 94.5 ) \implies \\[2ex] &P\left( Z \geq \frac{94.5-\mu}{\sigma} \right) = P(Z \geq 0.0708) = 1 - \Phi (0.0708) = 0.4721\end{align}
But, this is far off from the answer which is $\boxed{0.1788}$. Is my approach correct? OR am I missing something?

REPLY [2 votes]: You want the probability that the poll will show a majority in favor, which happens if $X>100$. Otherwise, your solution appears to be correct.
Approximately we have that $X \sim \mathcal N (np,np(1-p))$ with $n=200$ and $p=0.47$, so with the continuity correction
$$ P(X>100.5) = P\left(Z>\frac{100.5-np}{\sqrt{np(1-p)}}\right)\approx P\left(Z>0.9209\right)\approx 0.1785,$$
where $Z \sim \mathcal N (0,1)$.


TITLE: The proof of Banach-Tarski paradox on Wikipedia
QUESTION [0 upvotes]: https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox#Step_3
In the part of the proof that deals with the fixed points on $S^2$, the page says "Let $\lambda$ be some line through the origin that does not intersect any point in $D$. This is possible since $D$ is countable." where $D$ is the set of the fixed points. Could you formally explain why we can find such line that does not intersect any point in $D$? My understanding is that since $S^2$ is a uncountable set, there are infinitely many points on $S^2$ that are not in $D$. But I don't know how to explain such line through origin can intersect with $S^2$ on "two" points that are not in %D%.
The next line of the proof is "Let $J$ be the set of angles, $\alpha$, such that for some natural number $n$, and some $P$ in $D$, $r(n\alpha)P$ is also in $D$, where $r(n\alpha)$ is a rotation about $\lambda$ of $n\alpha$. Then $J$ is countable." Exactly why is $J$ countable?
Thank you in advance!

REPLY [1 votes]: For the first part, I will repeat my answer from comments above - the point is that there are also uncountably many possible lines going through (at least) one point not in D, since (informally) "removing countably many points from an uncountable set leaves uncountably many points remaining". So if every such line also went through a point in D, this would imply that D is uncountable.
For the second part, we first fix a point $P$ of $D$ (if we can show the set of rotations about $P$ taking $P$ to another point of $D$ is countable, since $D$ is countable we will be done). Denote by $J_P$ the subset of $J$ of the angles $\alpha$ corresponding to rotations from $P$. Well, we know there are only countably many points in $D$, so in particular there are only countably many angles $\beta$ which we can rotate $P$ by to obtain another point of $D$. Now, each $\beta$ is in $J_P$, but we also have $$\beta = 2 \cdot (\beta/2) = 3 \cdot (\beta/3) = \ ... \ ,$$ so in fact each $\alpha_{\beta,n} := \beta/n$ is in $J_P$ too. But for each $\beta$, the set of $\alpha_{\beta,n}$ is countable, and I claim that $J_P = \{ \alpha_{\beta,n} \}$; indeed, if $\alpha$ is a rotation in $J_P$, then $n \alpha = \beta$ for some $n$ and angle $\beta$ as above. Thus $J_P$ is in particular countable (we can index this set by $(\beta,n) \in \{\text{these countably many angles} \ \beta \} \times \mathbb{Z}$, a countable set).
Since $J_P$ is countable for each $P$, and $D$ is countable, we deduce that $J$ is countable. More formally, we can index $D$ as $D = \{P_i \}_{i} = \{P_1, P_2, \ ... \}$. Then $J = J_{P_1} \cup J_{P_2} \cup \ ...$ is countable, since (assuming the axiom of choice) the countable union of countable sets is countable (see: https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable).
Hope this makes sense! Let me know if you have any questions.


TITLE: Residue of a complex function with essential singularity
QUESTION [0 upvotes]: Compute the residue of the function:
$$
\frac{e^{1/z}}{1-z}
$$
when $z_0 = 0$
I don't know how to proceed with this exercise. I can't find the Laurent Series of the function around $z_0 = 0$. Could someone give me a hint?
Thanks in advance

REPLY [2 votes]: You have:
$$\begin{split} e^{1/z} &= \sum_{n=0}^\infty \frac{1}{n!}\ z^{-n} \\ \frac{1}{1 - z} &= \sum_{n=0}^\infty z^n \end{split}$$
hence you can write:
$$\begin{split} e^{1/z} &= \sum_{n=-\infty}^\infty a_n\ z^n \\ \frac{1}{1 - z} &= \sum_{n=-\infty}^\infty b_n z^n \end{split}$$
where:
$$\begin{split} a_n &= \begin{cases} \frac{1}{(-n)!} &\text{, if } n\leq 0 \\ 0 &\text{, if } n>0 \end{cases} \\ b_n &= \begin{cases} 0 &\text{, se } n<0 \\ 1 &\text{, if } n \geq 0 \end{cases} \end{split}\; .$$
Therefore the Laurent series of your function is given by:
$$\frac{e^{1/z}}{1 - z} = \left( \sum_{n=-\infty}^\infty a_n\ z^n \right) \cdot \left( \sum_{n=-\infty}^\infty b_n z^n \right) = \sum_{n=-\infty}^\infty \underbrace{\left( \sum_{h+k=n} a_hb_k \right)}_{=:c_n}\ z^n\; .$$
Now, you're interested only in $c_{-1} = \operatorname{Res}(f;0)$ and this coefficient evaluates into:
$$\begin{split} c_{-1} &= \sum_{h+k=-1} a_hb_k \\ &= \sum_{k=0}^\infty a_{-1-k} b_k \\ &= \sum_{k=0}^\infty \frac{1}{(k+1)!} \\ &\stackrel{i=k+1}{=} \sum_{i=1}^\infty \frac{1}{i!} \\ &= e-1\; .\end{split}$$


TITLE: Rayleigh Scattering, Atmosphere-Dependant
QUESTION [0 upvotes]: I've come across the Wikipedia article on Rayleigh Scattering and it quotes that

In detail, the intensity $I$ of light scattered by any one of the small spheres of diameter $d$ and refractive index $n$ from a beam of unpolarized light of wavelength $λ$ and intensity $I_0$ is given by
$$I = I_o \left(\frac{1 + cos^2\theta}{2R^2}\right)\left(\frac{2\pi}{\lambda}\right)^4\left(\frac{n^2-1}{n^2+2}\right)^2\left(\frac{d}{2}\right)^6$$

In another article, it says that a hydrogen-dominated atmosphere exhibits stronger Rayleigh scattering effect as compared to an atmosphere which is oxygen-dominated. From the formula quoted above, I can't see how is this so. In any way, the "particle size" of a hydrogen molecule is probably less than that of an oxygen molecule. If $d$ was the factor, it would have said that the intensity of the scattered light should have been lower, contradicting the statement above.
Is there another way to think about this problem and possibly explaining the difference in the scattering effect in two different atmospheres?

REPLY [0 votes]: In line with the general guidance (not to put answers in comments), I'll offer this bare-bones answer. However, I'm sure that real experts can give much more comprehensive ones: I am not an expert on planetary atmospheres! You may wish to wait for those answers.
It looks like the source of the claim

a hydrogen-dominated atmosphere exhibits stronger Rayleigh scattering effect as compared to an atmosphere which is oxygen-dominated

was this web page on exoplanet atmospheres. So "stronger Rayleigh scattering effect" needs to be set in that context. Integrating the given formula over angles gives the scattering cross section per particle. Then the total amount of scattering (as a function of wavelength) will involve integrating along the path of the light (in fact, over all the parallel paths which pass through the atmosphere at various heights): the extent of the atmosphere as well as the density of molecules in the atmosphere will enter the formula. What is left (the spectrum of unscattered light) is what we observe.
So the bottom line is that the quantities in that equation
are not the only ones affecting the overall Rayleigh scattering of light passing through planetary atmospheres.
As I said, I am sure that more detail can be given.


TITLE: Reasoning about an [M/G]/1 queue
QUESTION [1 upvotes]: I am considering a queue where 1 - 4 requests can be serviced at once (min 1, max 4), but where requests can only be admitted in serial order. Is it reasonable to consider this as though it was an M/G/1 queue (assuming the M/G bit is correct)?
My reasoning runs that if all I am concerned about is how long requests have to wait to begin service then the fact that four are being serviced in parallel is irrelevant if they can only access the service in serial order. As we can simply consider the four being serviced in parallel as though the distribution of service times was other than it is in reality (ie I don't care what request it is that completes to allow my request in, I only care about how long I have to wait for that to happen).

REPLY [0 votes]: I do not think you can model this as an M/G/1 queue. First, in your system, must the server process 4 requests at once? In other words, if the server is idle and a request arrives, does it process the request right away, or wait until 3 more arrive?
If the server must process 4 requests at once, then a request that arrives to an empty system would have to wait for more arrivals, meaning it will wait longer than it would in an M/G/1 queue.
If the server does not need to wait for 4 requests, then suppose the server finishes a batch, and there are 4 requests waiting in the queue. Jobs #2-4 in the queue will get processed right away (along with job #1), but in an M/G/1 queue those jobs would have to wait their turn.
So, I do not think this is equivalent to an M/G/1 queue, even if the only performance measure you care about is wait time in the queue.


TITLE: What is the process transforming $K_L$ mesons into $K_S$ mesons?
QUESTION [1 upvotes]: First of all, I could not find any good book/webpage where I can read about K mesons, can somebody point me to a good reading?
The second point is about $K_L$ and $K_S$ mesons. As I can read, they are eigenstates of $K^0$ with opposite strangeness. In principle they should decay into pions, but I was wondering if a $K_L$ can spontaneously change into a $K_S$ instead of decay into pions or there would be an interaction with protons/nuclei needed for this transformation to occur.

REPLY [2 votes]: $K_L$s cannot spontaneously transform to $K_S$s in free space. They are eigenstates of the free hamiltonian and maintain their identity. Their mass difference is $\Delta m \sim 10^{-14} m$.  To effect a
$K_L\to K_S$ transformation ("regeneration") they must go through matter and interact coherently and strongly with its nuclei: absolutely nothing spontaneous about it. I would strongly recommend changing your title, as the present one is odd.

Let's ignore CP violation to avoid mission creep and muddying the waters in the process, so just call
$$  |K_{L}\rangle \sim \dfrac{1}{\sqrt2}\left( |K^{0}\rangle - |\bar{K^{0}}\rangle \right)  \\   |K_{S}\rangle \sim\dfrac{1}{\sqrt2}\left(|K^{0}\rangle + |\bar{K^{0}}\rangle \right).$$
The $|K^{0}\rangle ,  |\bar{K^{0}}\rangle $ states are production states, that is they have well-defined strangeness, but they do not propagate or decay. The $|K^{0}\rangle\sim \bar{s} d$ is the isopartner of $K^+$, with S=1; while  the  $|\bar{K^{0}}\rangle\sim \bar{d}s$ is the isopartner of $K^-$, with S=-1.
They differ in strangeness by two, so only a doubly weak interaction mixes them, the celebrated box diagram. When this is  taken into account, the free eigenstates are $K_L, K_S$ with very different lifetimes, so a beam of neutral kaons will quickly purify to a beam of surviving $K_L$, which are longer lived. In vacuum, no conversion to $K_S$s.
As this beam enters a nuclear target,
the two "strongly distinct"  states (since strangeness is preserved in the strong interactions) are treated very very differently. The $K^0$ scatters its quarks by the exchange of gluons, pions, whatever, and exchanges quarks with the nucleons and back: elastic & charge-exchange scattering. You may write 137000 Feynman diagrams. Coherently, and then goes out, slightly diminished.
The $\bar{K^{0}}$ is different. It may do all of the above, but it also has additional channels: Its s quark may replace a nucleon quark to form a hyperon, Λ or Σ, so it is absorbed and gone. Whatever the hyperon does (decay), it does not return to the original state, so the absorption crudely degrades the $\bar{K^{0}}$ beam more, and increases strangeness, if you must think of it that way. (People would frown, with justice).
The crucial thing is that, since the two components of the $K_L$ wave function were absorbed and phase-shifted differently, many times (coherent scattering), the outgoing wave function is not $\dfrac{1}{\sqrt2}\left( |K^{0}\rangle - |\bar{K^{0}}\rangle \right) $ anymore, but a slightly distorted version thereof. When resolved to free propagation eigenstates, the distortion amounts to a component of $K_S$ now present: $K_S$ has been regenerated after strong interaction with matter.
The standard textbook by D. Perkins should answer any questions you might have.


TITLE: solving $1+\frac{1}{x} \gt 0$
QUESTION [7 upvotes]: In solving a larger problem, I ran into the following inequality which I must solve:
$$ 1+\frac{1}{x} \gt 0.$$
Looking at it for a while, I found that $x\gt 0$ and $x\lt -1$ are solutions. Please how do I formally show that these are indeed the solutions.

REPLY [5 votes]: After reading your comment here is how to go about it when you multiply by $x^2$. So we need to solve
$$x^2 + x > 0.$$
That can be written as
$$x(x + 1) > 0.$$
Now the product of two numbers is positive when both numbers are positive or when both are negative. Therefore we have two cases to solve:
Case 1:
$x >0$ and $x + 1> 0$
Both of these inequalities are satisfied when $x > 0$.
Case 2:
$x < 0$ and $x + 1 < 0$
Both of these inequalities are satisfied when $x < -1$.
Hope that helps!

REPLY [3 votes]: note that $\frac{x+1}{x}>0$ and $(x+1)x>0$ have the same solution(or, as you mentioned, multiply them both with $x^2$)
generally speaking, bringing the denominator over the bar doesn't change the answer, but be careful that the denominator can't be zero.


TITLE: $R^nf_*\mathbb{Z}$ trivial for a morphism with hypersurface fibers.
QUESTION [3 upvotes]: I have some questions on local systems. If $f:X\to Y$ is a morphism of projective complex algebraic varieties, $Y$ being a curve, I want to prove that if the fibers of $f$ are smooth hypersurfaces in $\mathbb P^n$ then $R^nf_*\mathbb{Z}$ is a trivial localy system, isomorphic to the constant sheaf $\mathbb Z$. But I have no clue why such an assertion might be true.
The only thing I can think of would be to base change to $U$, an open set in the curve, such that U is in the smooth locus of f, there I could use that if I take U sufficiently small, the f is a fibration, and compute the cohomology $H^n(U, \mathbb Z)$. But firstly I dont see how that helps. And in anycase I miss the place where $f$ is not smooth.
Could somebody help me?

REPLY [0 votes]: I assume $n\geq 3$ below.
The morphism $f:X\to Y$ in your question is a smooth projective family of $n-1$-dimensional hypersurfaces in $\mathbb P^n$. (You say in your first paragraph that the fibers are smooth hypersurfaces.)
The cohomology of a single hypersurface of dimension $n-1$ in $\mathbb P^n$ in degree $\neq n$ is quite simple to compute by the Lefschetz hyperplane theorem. Indeed, if $F$ is a smooth $n-1$-dimensional hypersurface in $\mathbb P^n$ (with $n\geq 3$) then the following holds.
If $i$ is even and $i\neq n-1$, then $H^i(X,\mathbb Z) = \mathbb Z$.
If $i$ is odd and $i\neq n-1$, then $H^i(X,\mathbb Z) = 0$.
So if $n$ is even, then the cohomology of the fibers of $f$ is $\mathbb Z$. This then implies that $R^n f_\ast \mathbb Z = \mathbb Z$ (by some arguments I omit for now).
If $n$ is odd, then the cohomology of the fibers of $f$ is $0$. Therefore, $R^n f_\ast \mathbb Z = 0$.


TITLE: Prove that $a^2(p-q)(p-r)+ b^2(q-r)(q-p)+ c^2(r-p)(r-q) =4(\delta)^2$
QUESTION [2 upvotes]: If $p$,$q$,$r$ are the perpendiculars drawn from the vertices of a triangle ABC upon any straight line meeting the sides externally in D,E,F.

where a,b,c are the sides opposite to angles A,B,C in Triangle ABC and delta represents the area of the triangle ABC.
Also prove that EF= $(2pq)/((p-q)(p-r))$
I have tried this problem in this way but couldn't solve the problem, can anyone help me from here??

REPLY [0 votes]: The following pictures explain how we found out EF using similar triangles approach..we also used the fact that the area of our triangle + a trapezium is equal to the area of 2 trapeziums[see[][]2


TITLE: Proof closure of $\mathbb{Z}$ under addition, especially that $m+(-n) \in \mathbb{Z}$ with $m\in\mathbb{N}, n\in\mathbb{N}$
QUESTION [0 upvotes]: I need help with the following task:
Prove that $\mathbb{Z}$ is closed under addition
We define the set of integers as $\mathbb{Z}$= {${n:n\in\mathbb{N}}$} $\cup$ {$0$} $\cup$ {$-n:n\in\mathbb{N}$}. Also $0$ $\notin$ $\mathbb{N}$
The additon of integers, defined as the sum of two integers (or the result we get on adding two or more integers), has the following properties:
A1) $x+(y+z)=(x+y)+z$ $\forall x,y,z \in \mathbb{Z}$
A2) $x+y=y+x$ $\forall x,y,z \in \mathbb{Z}$
A3) There exists exactly one element $0$ $\in \mathbb{Z}$ with the property $x+0=x$ $\forall x\in \mathbb{Z}$
A4) For each $x\in\mathbb{Z}$ there exists exactly one element $-x\in\mathbb{Z}$ with the property $x+(-x)=0$
It seems to be a very easy task. Unfortunately I don't have a clue how to prove the case that $$m+(-n) \in \mathbb{Z}   ,( m\in\mathbb{N}, n\in\mathbb{N})$$
I found posts here were people simply answered, yeah m+(-n) = m-n and the difference of two natural numbers is an integer. But without a proof, this is not sufficient and unfortunately I didn't found any proof for this. We obviousle have to differentiate between thre cases: m>n, m=n, m<n. I wanted to prove this task with help of induction, since it helped me for my proof that $\mathbb{N}$ is closed under addition. But in this case, it gets complicated.
Anyone an idea, how to prove this easily? I am thankful for any advice.

REPLY [0 votes]: I think at the very least you should assume this property from the naturals, this is the one derived directly from the successors definition.
$$m\ge n\implies \exists a\in\mathbb N\mid m=n+a$$
Then you can proceed from the axioms you gave:
$m+(-n)=(n+a)+(-n)\,\stackrel{A_1}{=}\,(n+(-n))+a\,\stackrel{A_4}{=}\,0+a\,\stackrel{A_2}{=}\,a+0\,\stackrel{A_3}{=}\,a\in\mathbb Z$
For the case $m\le n\ $ or $\,n=m+a$ you have to prove first that: $$-(m+a)=(-m)+(-a)$$
$(m+a)+(-m)+(-a)=(m+(-m))+(a+(-a))=0+0=0$
And then proceed the same way as before:
$m+(-n)=m+(-m)+(-a)=0+(-a)=(-a)+0=(-a)\in\mathbb Z$


TITLE: Can a point and a compact set in a Tychonoff space be separated by a continuous function into an arbitrary finite dimension Lie group?
QUESTION [2 upvotes]: Given a topological space $X$ which is Tychonoff (i.e., completely regular and Hausdorff), we know that given a compact set $K\subseteq X$ and a point $p \in X$ with $p\not\in K$, we can construct a continuous function $f:X\to \mathbb{R}$ such that $f(K)=0$ and $f(\{p\})=1$.
If, instead of $\mathbb{R}$, I take an arbitrary finite dimensional Lie group $M$, can I still construct a continuous function $f':X\to M$ that will separate $K$ and $p$ for two different arbitrary points in $M$?
Is it enough to assume $X$ to be Tychonoff or do I need further or different assumptions about $X$? For example, is $X$ being metric sufficient?

REPLY [0 votes]: As far as I remember, each Lie group $M$ is locally Euclidean, so there is a homeomorphic embedding $i:\Bbb R\to M$, which yields the required separating map $f'=i\circ f$.


text:
\chapter{Undecidability Results for Hahn Fields with Truncation} \label{TU}

\noindent
In this chapter we consider some model theoretic properties of (valued) Hahn fields with truncation. We show (Corollary \ref{KGamvTUnd})  that such  structures
are very wild in the sense that they can even interpret the theory of $(\N;+,\times)$ via an interpretation of $(\N, \mathcal{P}(\N); +,\in)$. In particular, they
are undecidable; this answers a question posed informally by van den Dries some years ago. As an application we show in Section 3.3 that the exponential field $\T$ of transseries with its canonical monomial group $G^{\operatorname{LE}}$ is undecidable.
In Section 3.4 we indicate for  Hahn fields with truncation a definable binary relation with ``bad'' properties such as the strict order property and the tree property of the second kind. In Section 3.5 we record an undecidability result for valued Hahn fields $C((t^{\Q}))$ with the derivation $d/dt$; this doesn't involve truncation.

In connection with Theorem~\ref{NinK} the author would like to thank Philipp Hieronymi and Erik Walsberg for bringing monadic second-order logic to his attention.

\section{The Set of Natural Numbers in Monadic Second-Order Logic}
\noindent
We start with the following well-known result.
\begin{theorem}\label{Nund}
The theory of $(\N;+,\times)$ is undecidable.
\end{theorem}

\subsection*{Monadic second-order logic} Given a structure $\cM=(M;\ldots)$, \textbf{monadic second-order logic of} $\cM$ extends first-order logic over $\cM$ by allowing quantification over subsets of $M$.
More precisely it amounts to considering the two-sorted structure $(M,\cP(M);\ldots,\in)$, where the usual interpretation is given to $\in \subseteq M\times \cP(M)$.
The following positive result and its proof appear in \cite{G}, and are in contrast with the negative Corollary \ref{MOSNplus}.
\begin{theorem}
The theory of $(\N,\cP(\N);\in)$ is decidable.
\end{theorem}

\begin{lemma}\label{timesinMOSNplus}
Multiplication on $\N$ is definable in $(\N,\mathcal{P}(\N);+, \in)$.
\end{lemma}
\begin{proof}
If the multiplication of consecutive numbers is defined, then general multiplication of two natural numbers can be defined in terms of addition:
\[n=mk \iff (m+k)(m+k+1) = m(m+1) + k(k+1) + n + n.\]

\noindent
If divisibility is defined, then multiplication of consecutive numbers is defined by
\[n=m(m+1)\iff \forall k(\in \N) (n|k \leftrightarrow [m|k \wedge (m+1)|k]).\]

\noindent
Divisibility can be defined using addition by
\[m|n \iff \forall S \big(0\in S \wedge \forall x (x\in S \rightarrow x+m\in S) \rightarrow n\in S\big)\,\]
where the variable $S$ ranges over $\cP(\N)$ and the variable $x$ ranges over $\N$.
Since addition is a primitive, multiplication is defined in $(\N,\mathcal{P}(\N); +, \in)$.
\end{proof}

\begin{corollary}\label{MOSNplus}
The theory of $(\N,\mathcal{P}(\N); +, \in)$ is undecidable.
\end{corollary}
\begin{proof}
This follows from Lemma \ref{timesinMOSNplus} and Theorem \ref{Nund}.
\end{proof}

\section{Hahn Fields with Truncation}\label{HFWT}
\noindent
Let $K=\smallk((t^\Gamma))$ be a Hahn field with nontrivial value group $\Gamma$. We consider $K$ below  as an $L_{\bigO, \fM, \cU}$-structure where
\[L_{\bigO, \fM, \cU}\ :=\ \{0,1,+,\times,\bigO, \fM, \cU\},\] and the unary predicate symbols $\bigO, \fM,$ and $\cU$ are interpreted respectively as the valuation ring, the canonical monomial group $t^\Gamma$, and the canonical additive complement to $\bigO$. For $\gamma \in \Gamma$ and $\fm = t^\gamma$ we set $f|_\fm : = f|_\gamma$. Then we have the equivalence (for $f$, $u \in  K$)

\[f|_1 = u \iff u\in \cU\ \&\ \exists g\in \bigO (f=u+g),\]
showing that truncation at $1$ is definable in the $L_{\bigO, \fM, \cU}$-structure $K$.
For $\fm \in t^\Gamma$ and $f\in K$ we have
\[f|_\fm =  g \iff (\fm^{-1}f)|_0=\fm^{-1}g,\]
showing that the operation $(f,\fm)\mapsto f|_\fm : K\times t^\Gamma \rightarrow K$ is definable in the $L_{\bigO, \fM, \cU}$-structure $K$.

For convenience of notation we introduce the asymptotic relations $\preceq,\prec ,$ and $\asymp$ on $K$ as follows.
For $f,g\in K$, $f\preceq g$ if and only if there is $h\in \bigO$ such that $f=gh$, likewise $f\prec g$ if and only if $f\preceq g$ and $g\not\preceq f$, and  $f\asymp g$ if and only if $f\preceq g$ and $g\preceq f$.
Let $R := \{(\fm,f)\in t^\Gamma\times K: \fm\in t^{\supp(f)}\}$. Then $R$ is definable in the $L_{\bigO, \fM, \cU}$-structure $K$ since for $a,b\in K$
\[(a,b)\in R\iff  a\in t^\Gamma \text{ and } b-b|_a \asymp a.\]

\begin{theorem}\label{NinK}
The $L_{\bigO, \fM, \cU}$-structure $K$ interprets $(\N,\mathcal{P}(\N); +, \in)$.
\end{theorem}
\begin{proof}
Let $\approx$ be the definable equivalence relation on $K$ such that $f\approx g$, for $f,g\in K$, if and only if $\supp(f) = \supp(g)$.
Take $\fn\in t^\Gamma$ such that $\fn\prec 1$. Consider the element $f= \sum_n \fn^n\in K$, and the set $S= \{g\in K: \supp(g) \subseteq \supp(f)\}$.
Let $E\subseteq t^{\supp(f)}\times (S/\hspace{-1mm}\approx)$  be given by
\[(\fm,g/\hspace{-1mm}\approx)\in E :\iff \fm \in t^{\supp(g)},\]
and note that $E$ is definable in the $L_{\bigO, \fM, \cU}$-structure $K$ since $R$ is. Define $\iota: \N \rightarrow t^{\supp(f)}$ by $\iota(n) = \fm^n$, and note that $\iota$ induces an isomorphism $(\N, \cP(\N);\in) \stackrel{\sim}{\longrightarrow} (t^{\supp(f)},S/\hspace{-1mm}\approx;E)$, such that $\iota(m+n) = \iota(m)\iota(n)$.
\end{proof}

\begin{corollary} \label{UndL}
The theory of the $L_{\bigO, \fM, \cU}$-structure $K$ is undecidable.
\end{corollary}
\begin{proof}
This follows easily from Theorem \ref{NinK} and Theorem \ref{MOSNplus}
\end{proof}

\noindent
In the next subsections we consider cases where we can replace $L_{\bigO, \fM, \cU}$ by another language.

\subsection*{The case of a real closed coefficient field.}
In this subsection we assume that the coefficient field $\smallk$ of $K$ is real closed. Then we have a field ordering $\leq$ on $K$ that extends the ordering of $\smallk$; it is given by: $f>0$ if and only if the leading coefficient of $f$ is $>0$. Note that then $f\in \bigO$ if and only if $|f|<u$ for all positive $u\in \cU$. Equipped with this ordering we now consider $K$ as an $L_{\le, \fM, \cU}$-structure where $L_{\le, \fM,\cU}:=\{0,1,+,\times,\le, \fM, \cU\}$.  We just observed that
$\bigO$ is $0$-definable in this $L_{\fM,\cU}$-structure. Thus:

\begin{corollary} The theory of $K$ as an $L_{\le, \fM,\cU}$-structure is undecidable.
\end{corollary}

\begin{corollary}  Suppose $\Gamma$ is divisible.  Then the theory of $K$ in the language $\{0,1,+,\times,\fM,\cU\}$ is undecidable.
\end{corollary}

\begin{proof}
The divisibility of  $\Gamma$ yields that $K$ is real closed, so the above ordering is the only field ordering on $K$ and is definable in the field $K$.
\end{proof}

\subsection*{Defining the coefficient field $\smallk$} We now consider $K=\smallk((t^\Gamma))$ as an $L_{\bigO, \cU}$-structure, where
\[L_{\bigO, \cU}\  :=\ \{0,1,+,\times, \bigO,\cU\}.\]
Note that for $f\in \bigO$ we have
\[f\cU\subseteq \cU\  \iff\  f\in \smallk,\]
where we identify $\smallk$ with $\smallk t^0$. Thus we can define the coefficient field $\smallk$ in the $L_{\bigO, \cU}$-structure $K$.

\medskip\noindent
\textbf{Question:} Is it possible to define the monomial group $t^\Gamma$ in the $L_{\bigO,\cU}$-structure $K$?

\subsection*{An approach without the monomial group} Alternatively we may work in the setting of the two-sorted structure $(K, \Gamma; v, T)$ where $K$ denotes the underlying field, $\Gamma$ is the ordered value group, $v$ is the valuation, and $T:K\times \Gamma \rightarrow K$ is such that $T(f,\gamma) = f|_\gamma$. Then we can define the binary relation $R\subseteq \Gamma\times K$ by
\[(\gamma,f) \in R :\iff v(f-T(f,\gamma)) = \gamma .\]
We then obtain the following;
\begin{theorem}
The two-sorted structure $(K,\Gamma;v,T)$ interprets $(\N,\mathcal{P}(\N); +,\in)$.
\end{theorem}
The proof is similar to the proof of Theorem \ref{NinK}.
\begin{corollary}\label{KGamvTUnd}
The theory of the two-sorted structure $(K, \Gamma; v, T)$ is undecidable.
\end{corollary}

\section[Undecidability of Transeries with Exponential function and Monomials]{Undecidability of the Exponential Field of Transseries with a Predicate for its Multiplicative Group of Transmonomials}
\noindent
The field $\T$ of transseries is a real closed field extension of $\R$ with a canonical exponential operation \[\exp: \T\to  \T\] that extends the usual exponentiation
with base $e$ on $\R$. The elements of $\T$ are called \textbf{transseries} or  \textbf{logarithmic-exponential series}.  See \cite[Appendix A]{ADH} for a detailed
construction of $\T$. Below in Chapter 8 we repeat some key features
of this construction in connection with the canonical derivation of $\T$, but in this section the derivation plays no role, and we only  list here
some general facts about $\T$ that are relevant in connection with the undecidability results below.

First, it is known from  \cite{DMM1} that $\T$ as an exponential field is an elementary extension of the real exponential field $\R_{\exp}$.  The field $\T$ also has a natural valuation on it given by the valuation ring
\[\{f\in \T:\ |f|\le n \text{ for some }n\},\]
which makes $\T$ a valued field with its subfield $\R$ as a field of representatives for the residue field.
The theory of $\T$ as a {\em valued exponential field\/} has also been analyzed, leading to model-completeness in a natural language, with various
 tameness properties such as weak o-minimality (and thus NIP) as a consequence; this is a special case of results in \cite{DII} and \cite{DL}.

As a valued field, $\T$ has a distinguished monomial group $G^{\operatorname{LE}}$ whose elements are called \textbf{transmonomials} or \textbf{logarithmic-exponential monomials}. In fact,
$\T$ is construed as a truncation closed subfield of the Hahn field  $\R[[G^{\operatorname{LE}}]]$, with $\R(G^{\operatorname{LE}})\subseteq \T$.  As a valued field with monomial group, the theory of $\T$ is also well-understood as a special case of the AKE-theorems, with AKE standing for Ax, Kochen, and Ersov. In particular, this complete theory is axiomatized by the
axioms for real closed fields with a nontrivial convex valuation and a monomial group of positive elements.

\medskip\noindent
We now show that these positive results do not go through if we equip the valued field $\T$ simultaneously with its canonical exponentiation and  its distinguished monomial group $G^{\operatorname{LE}}$. In fact, the theory of this structure is undecidable in a rather strong way, and it turns out that we don't even need the valuation. The basic reason is that  $G^{\operatorname{LE}}=\exp \cU$, where $\cU=\{f\in \T:\ \supp f \succ 1\}$, in other words, $\cU$ is the natural infinite part of $\T$, that is, $\cU$ is the intersection of $\T$ with the canonical additive complement to the valuation ring of $\R[[G^{\operatorname{LE}}]]$.  In particular, $\cU$ is definable in the  exponential field $\T$ equipped with $G^{\operatorname{LE}}$.

Using this, we can define truncation on $\T$ (with respect to $\cU$) in the same way we defined truncation for $\smallk((t^\Gamma))$. The same proof of Theorem \ref{NinK} goes through to show:
\begin{theorem} The valued exponential field $\T$ equipped with its monomial group $G^{\operatorname{LE}}$ interprets  the structure $(\N,\cP(\N);+,\in)$
\end{theorem}
\begin{corollary}
The theory of the valued exponential field  $\T$ equipped with its monomial group $G^{\operatorname{LE}}$ is undecidable.
\end{corollary}

\medskip\noindent
Note that the valuation ring of $\T$ is convex. Thus we can define the valuation ring from the infinite part $\cU$ and the ordering, since
\[\bigO\  =\  \{f\in K:\  |f| \leq u\text{ for all positive } u\in \cU\}.\]
Furthermore, the ordering is definable from the field structure, so we obtain:

\begin{corollary}
The theory of the exponential field $\T$ with monomial group $G^{\operatorname{LE}}$ is undecidable.
\end{corollary}

\medskip\noindent

\section{Witnessing the Strict Order Property and a Tree Property}

\noindent
We have already shown how $(\N;+,\times)$ can be interpreted in the $L$-structure $K$ and thus we know that it has the strict order property and the tree property of the second kind among others. In this section we make explicit a binary relation that witnesses these properties inside $K$.
\subsection*{The independence property}
Let $L$ be a language and $\cM= (M;\ldots)$ an $L$-structure. We say that an $L$-formula $\phi(x;y)$ \textbf{shatters} a set $A\subseteq M^x$ if for every subset $S$ of $A$ there is $b_S\in M^y$ such that for every $a\in A$ we have that $M\models \phi(a;b_S)$ if and only if $a\in S$. Let $T$ be an $L$-theory. We say that $\phi(x;y)$ has the \textbf{independence property with respect to }$T$, or \textbf{IP} for short, if there is a model $M$ of $T$, such that $\phi(x;y)$ shatters an infinite subset of $M^x$.

For a partitioned formula $\phi(x;y)$ we let $\phi^{opp}(y;x) = \phi(x;y)$, that is, $\phi^{opp}$ is the same formula $\phi$ but where the role of the parameter variables and type variables is exchanged.

\begin{lemma}\label{IPopp}
A formula $\phi(x;y)$ has $\IP$ if $\phi^{opp}$ has $\IP$.
\end{lemma}
\begin{proof}
By compactness the formula $\phi(x;y)$ shatters some set $\{a_J: J\in \cP(\N)\}$. Let the shattering be witnessed by $\{b_I:I\subseteq \cP(\N)\}$. Let $B= \{b_{I_i}: i\in\N\}$ be such that $I_i = \{Y\subseteq \N: i\in Y\}$. Then we have
\[\models\phi(a_J, b_{I_i}) \iff i \in J,\]
and thus $\phi^{opp}$ shatters $B$.
\end{proof}

\subsection*{The Strict Order Property}
We say that a formula $\phi(x;y)$ has the Strict Order Property, or $\SOP$ for short, if there are $b_i\in M^y $, for $i\in \N$, such that $\phi(M^x,b_i) \subset \phi(M^x,b_j)$ whenever $i<j$.

\begin{proposition}
The formula $\varphi(x;y)$, defining the relation $R$ as in section \ref{HFWT}, has $\SOP$.
\end{proposition}

\begin{proof}
Let $\Theta= \{\theta_i: i\in \N\}$ be any subset of $\Gamma$ such that $\theta_i <\theta_j$ for $i<j$, and consider the set $\{f_n = \sum_{i=0}^n t^{\theta_i}: i \in \N\}$. Note that $\varphi(K,f_m)\subset \varphi(K,f_n)$ for $m<n$.
\end{proof}

\subsection*{The tree property of the second kind}
We say that a formula $\phi(x;y)$ has the \textbf{tree property of the second kind}, or $\TP$ for short, if there are tuples $b_j^i\in M^y$, for $i,j\in \N$, such that for any $\sigma:\N\rightarrow \N$ the set $\{\phi(x;b^i_{\sigma(i)}): i\in \N\}$ is consistent and for any $i$ and $j\neq k$  we have $\{\phi(x;b^i_j),\phi(x;b^i_k)\}$ is inconsistent.

\begin{lemma}\label{TP2IP}
If $\phi(x;y)$ has $\TP$ then $\phi$ has $\IP$.
\end{lemma}
\begin{proof}
Let $\{\phi(x,b^i_j)\}_{i,j\in \N}$ witness $\TP$ for $\phi(x;y)$. Fix $j$. Without loss of generality we will assume that $j=0$. Consider the set $\{b^i_0\}$. Let $I\subseteq \N$. By $\TP$ there is $a_I \in M^x$ such that \[\cM\models  \phi(a_I;b^i_j) \iff (i\in I \text{ and }j=0, \text{ or } i\notin I \text{ and } j=1).\]
Thus by Lemma \ref{IPopp} $\phi(x;y)$ has IP.
\end{proof}

\begin{lemma}\label{TP2Char}
Let $A=\{a_i:i\in \N\}\subseteq M^x$ and $B=\{b_I:I\in \mathcal{P}(\N)\}\subseteq M^y$. Assume that there is $\phi(x;y)$ such that for any fixed $b_I\in B$
\[\models \phi(a;b_I)\iff \text{ there is }  i\in I \text{ such that }a=a_i .\]
Then $\phi$ has $\TP$.
\end{lemma}

\begin{proof}
Let $\phi$, $A$, and $B$ be as in the hypothesis of the Lemma. Let $P=\{p_i \in \N\}$ be the set of primes where $p_i\neq p_j$ for $i\neq j$. We construct $A^i_j\subseteq \N$ recursively as follows:
\begin{itemize}
	\item $A^0_j:= \{p_j^{n_0}:n_0>0\}$
    \item $A^i_j:= \{p_{n_i}^m: n_i\in \N,\ m\in A^{i-1}_{j}\}.$
\end{itemize}
So for example $A^1_2 =\{p_{n_1}^{p_2^{n_0}}: n_1,n_0>0\}$. \\
\textbf{Claim 1} For $\alpha \in \N^{n}$ we have that $\bigcap_{i<n} A^i_{\alpha(i)}\neq \emptyset$.\\
It is not hard to check that
\[p_{\alpha(0)}^{\text{\reflectbox{$\ddots$}} ^{p_{\alpha(n-1)}}}\in \bigcap_{i<n} A^i_{\alpha(i)}.\]
\textbf{Claim 2:} For fixed $i$, and $j\neq k$ we have $A^i_j\cap A^i_k = \emptyset$. \\
For simplicity in notation we prove the case where $i=1$. Let $m\in A^1_j\cap A^1_k$. Then $m = p_{m_1}^{p_j^{m_0}} = p_{n_1}^{p_k^{n_0}}$. Since $p_j^{m_0}$ and $p_k^{n_0}$ are nonzero, we have that $p_{m_1} = p_{n_1}$, and thus $p_j^{m_0} = p_k^{n_0}$. Similarly, since $m_0$ and $n_0$ are nonzero we conclude that $p_j=p_k$, and thus $j=k$.

Now let $b^i_j = b_{A^i_j}$. By compactness, together with Claim 1, we get that the set $\{\phi(x;b^i_{\sigma(i)}): i\in \N\}$ is consistent.
By the hypothesis of the Lemma, together with claim 2, we get that for any $i$ and $j\neq k$  we have $\{\phi(x;b^i_j),\phi(x;b^i_k)\}$ is inconsistent.
\end{proof}
 If $\phi(x;y)$ and $A$ are as in the lemma, we say that $\phi(x;y)$ and $B$ \textbf{only shatter} $A$ \textbf{in} $M$. Note that in this case $A$ is in fact a definable set.

\begin{proposition}\label{TP2forK}
	The formula $\varphi(x;y)$, defining the relation $R$ as in section \ref{HFWT}, has $\TP$.
\end{proposition}
\begin{proof}
Let $\Theta$ be a well-ordered subset of $\Gamma$ and consider the sets \[t^\Theta=\{t^\theta:\theta \in \Theta\}, \text{ and } B= \left\{\sum_{\delta\in \Delta} t^\delta: \Delta \subseteq \Theta\right\}.\]
It is clear then that $\varphi(x;y)$ and $B$ only shatter $t^\Theta$, and thus by Lemma \ref{TP2Char} the formula $\varphi(x;y)$ has $\TP$.
\end{proof}

\begin{corollary}
The formula $\varphi(x;y)$, defining the relation $R$ as in section \ref{HFWT}, has $\IP$.
\end{corollary}
\begin{proof}
The result follows directly from proposition \ref{TP2forK} and lemma \ref{TP2IP}.
\end{proof}

\section{Undecidability for some Differential Hahn Fields}
\noindent
In this section we consider the Hahn field $K=C((t^\Gamma))$ where $C$ is a field of characteristic zero, and $\Gamma$ is either $\Z$ or $\Q$. We equip $K$ with the derivation $t\frac{d}{dt}$ that is trivial on $C$ and sends $t^\gamma$ to $\gamma t^\gamma$. We prove below that the theory of the valued differential field $K$ is undecidable. This result is folklore for $\Gamma= \Z$ but we include it here since it is not easily found in the literature.

\begin{lemma}
$\{c\in C:\ y^\dagger = cf^\dagger \text{ for some } y,f\in K^\times, f\not\asymp 1\}=\Q$
\end{lemma}
\begin{proof}
Let $y= at^r(1+\delta)$ and $f= bt^s(1+\epsilon)$ in $K^\times$ be such that $y^\dagger = cf^\dagger$ with $a,b,c\in C$, $r,s\in \Gamma$, $s\neq0$, and $\delta, \epsilon\in K^{\prec 1}$. Then $r +\frac{\delta'}{1+\delta} = c(s+ \frac{\epsilon'}{1+\epsilon})$, so $r= cs$ and thus $c=\frac{r}{s}\in \mathbb{Q}$.
For the other direction it is clear that if $c =\frac{p}{q}\in \Q$, then $y= t^p$ and $f= t^q$ satisfy $y^\dagger = cf^\dagger$.
\end{proof}

\begin{theorem}\label{undCtGam}
The theory of the valued differential field $K$ is undecidable.
\end{theorem}
\begin{proof}
By the Lemma above the subring $\Q$ of $C$ is (existentially) definable in the valued differential field $C((t^\Gamma))$. Applying Julia Robinson's result from \cite{RJ} we obtain that the theory is undecidable.
\end{proof}
\noindent
Using the derivation $t\frac{d}{dt}$ is only done to simplify certain expressions: it is clear that Theorem \ref{undCtGam} goes through with $\frac{d}{dt}$ as the derivation instead of $t\frac{d}{dt}$.

\medskip\noindent
The valuation ring of the Hahn field $C((t^\Z))$ is definable in the field $C((t^\Gamma))$ by a theorem of Ax, (see for example \cite{FJ}). Therefore:
\begin{corollary} \label{undCtZ}
The theory of the differential field $C((t^\Z))$ is undecidable.
\end{corollary}

config: arxiv


text:
\subsection{Proof of Universality}
\label{sec:universality-proof}

\begin{theorem}[Involutive generative models are universal]
\label{maintheorem}
Let $n\in \Nats$, and let random variables
$\pi \sim N(0^{n+6},\Id_{n+6})$
and $\pi' \sim N(0,1)$.

For all compact sets
$\Omega \subseteq \Reals^n$,
and all continuous functions
$T \colon \Reals^n \times \Reals \to \Omega$
    there exists a sequence $\{\widehat{\cI}_m\}_{m\in \Nats}$ of involutive neural networks
that induce continuous functions $\Reals^{2n+6} \to \Reals^{2n+6}$
such that for all
$\phi \in \Omega$
the random variables
${\hcI_{m}}[\phi] \defas {\hcI_m(\phi \conc \pi})_{1..n}$ converge in distribution to
$T[\phi] \defas T(\phi \conc \pi')$, as $m\to\infty$.
\end{theorem}

We begin with an outline of the proof technique.
For each $m\in\Nats$, we aim to define a involutive neural network $\hcI_{m}$.
We begin by exhibiting an involutive function parametrized by a positive real $\fcnerr<1$ depending on $m$, such that when the auxiliary variable is treated as a random variable, the involutive function matches the cumulative distribution function of the desired state transition arbitrarily well as $\fcnerr \to 0$. Then we show that this involutive function can be uniformly approximated arbitrarily well by an involutive neural network parameterized by some other positive real $\neterr<1$ depending on both $\fcnerr$ and $m$, as $\neterr \to 0$.
(In the main proof of Theorem~\ref{maintheorem} below, we will impose tighter constraints on $\fcnerr$ and $\neterr$.)

We will use Hornik's Universal Approximation Theorem to obtain such a uniform approximation.

\begin{theorem}[{\citet[Thm.~3]{hornik1991approximation}}]
    \label{thm:Hornik-UAT}
    Let $a, b$ be positive integers,
$F \colon \Reals^a \to \Reals^b$ be a continuous function, and $\psi \colon \Reals \to \Reals$ be a continuous bounded nonconstant activation function.
    For any compact set $\Omega \subseteq \Reals^a$ and $\delta > 0$ there is a neural network consisting of a single hidden layer with activation $\psi$ that induces a continuous function
    $\widehat{F}_\delta \colon \Reals^a \to \Reals^b$ satisfying
    \[\max_{x \in \Omega} \bigl|F(x)-\widehat{F}_\delta(x)\bigr|_1 \leq \delta.\]
\end{theorem}

We now define and prove several facts about some objects that will be useful in the proof.

Define $\F \colon \Reals^{n+3} \to \Reals^{n+3}$ by
    \[
        \F(x) \defas
\begin{cases}
      0^{n+3} & \IF q \leq -\frac{1}{2}\\
      (q+\frac{1}{2})  (T(\phi,\pi)-\phi) \conc 0^3 & \IF -\frac{1}{2} < q < \frac{1}{2} \\
      (T(\phi,\pi)-\phi) \conc 0^3 & \IF \frac{1}{2} \leq q,
\end{cases}
\]
where $\phi \defas x_{1..n}$,  $\pi \defas \frac{x_{n+1}}{\fcnerr}$, and $q \defas x_{n+3} - x_{n+2}$.

Define $\G \defas \Reals^{n+3} \to \Reals^{n+3}$ by
\[
\G(x) \defas
\begin{cases}
    0^{n+3} & \IF q \leq -\frac{1}{2} \\
      (q+\frac{1}{2}) \phi' \conc 0^3 & \IF -\frac{1}{2} < q < \frac{1}{2} \\
      \phi' \conc 0^3& \IF \frac{1}{2} \leq q,
\end{cases}
\]
where $\phi' \defas x_{1..n}$ and $q \defas x_{n+3} - x_{n+2}$.

Define $\g \colon \Reals^{2n+6} \to \Reals^{2n+6}$ by
    \[
        \g(x)
        \defas
        x \odot (1^n \conc \fcnerr 1^{n+6}) +(0^{n+2} \conc 1 \conc 0^{n+2} \conc 1),
    \]
where $\odot$ denotes pointwise multiplication. Note that its inverse satisfies
    \[
        \textstyle
        \g^{-1}(x)=(x-(0^{n+2} \conc 1 \conc 0^{n+2} \conc 1)) \odot (1^n \conc \frac{1}{\fcnerr} 1^{n+6}).
    \]

Define $\h \colon \Reals^{2n+6} \to \Reals^{2n+6}$ by
    \[
        \h \defas \, \bigl(x_{1..n+3} +\G(v)\bigr) \conc v
\]
where $v \defas x_{n+4..2n+6} + \F(x_{1..n+3})$.

Note that its inverse satisfies
\[
    \h^{-1}(x) = w
    \conc \bigl(x_{n+4..2n+6} - \F(w)\bigr)
\]
where $w \defas x_{1..n+3} - \G(x_{n+4..2n+6})$.

Let $\sigma$ be the permutation of $\{1, \ldots, 2n+6\}$ that transposes $n+2$ with $n+3$ and transposes $2n+5$ with $2n+6$ (and leaves all other elements fixed).

Now consider the involutive function $\INV \colon \Reals^{2n+6} \to \Reals^{2n+6}$ defined by $\INV \defas \g^{-1} \circ \h^{-1} \circ I_P^{2n+6,\sigma} \circ \h \circ \g$.

Let $\A$ denote the event that $\PI_3 - \PI_2 > -\frac{1}{2 \fcnerr}$ and $\PI_{n+6} - \PI_{n+5} > -\frac{1}{2 \fcnerr}$ and $|\PI|<\frac{1}{\fcnerr}$ all hold.

Note that
\[
    \label{eq:neterr-assumption}
    \tag{$\dagger$}
    \lim_{\fcnerr \to 0} P(\A) = 1.
\]

\begin{lemma}
    \label{lem:conditioned}
    Conditioned on the event $\A$, we have
    \[
        \INV(\phi \conc \PI)_{1..n} = T(\phi,\PI_1)+\fcnerr \PI_{4..n+3}.
    \]
\end{lemma}
\begin{proof}
The event $\A$ fully determines the branches of $\F$ and $\G$ that are taken in the evaluation of $\INV(\phi \conc \PI)$, which enables us to simplify the expression $\INV(\phi \conc \PI)_{1..n}$ to the stated form.
\end{proof}

For $\phi \in \Omega$, define the random variable $\INVr[\phi] \defas \INV(\phi \conc \PI)_{1..n}$.
The following lemma is immediate.
\begin{lemma}
    \label{lem:rv-convergence-in-dist}
Conditioned on the event $\A$, the random variable $\INV[\phi]$
    converges in distribution to $T[\phi]$ as $\fcnerr \to 0$.
    \qed
\end{lemma}

We will show that conditioned on the event $\A$, and for appropriately small $\fcnerr$ and $\neterr$, we can approximate $\INV$ arbitrarily well with an involutive neural network $\aINVNET$.

Since $\Omega$ is a compact subset of $\Reals^n$, it is bounded, and hence contained in a ball of finite radius $r \in \Reals$. Let $\Omega^+ \subseteq \Reals^{n+3}$ be the closure of the ball of radius $7r+10$, and let $\Omega^{++} \subseteq \Reals^{n+3}$ be the closure of the ball of radius $14r+21$. The sets $\Omega^+$ and $\Omega^{++}$ are closed and bounded, hence compact subsets of $\Reals^{n+3}$.

Since $\F$ and $\G$ are continuous, by Theorem~\ref{thm:Hornik-UAT},
for any $\neterr > 0$ there are neural networks each with a single hidden layer and sigmoid activation that induce continuous functions
$\aF \colon \Reals^{n+3} \to \Reals^{n+3}$ and $\aG \colon \Reals^{n+3} \to \Reals^{n+3}$,
 satisfying
\[\max_{x \in \Omega^{++} } \bigl|\F(x)-\aF(x)\bigr|_1 \leq \neterr\]
and
\[\max_{x \in \Omega^{++} } \bigl|\G(x)-\aG(x)\bigr|_1 \leq \neterr.\]
That is, $\aF$ and $\aG$ converge uniformly to $\F$ and $\G$, respectively, on $\Omega^{++}$ as $\neterr \to 0$.

Define $\ah \colon \Reals^{2n+6} \to \Reals^{2n+6}$ by
    \[
        \ah \defas \, \bigl(x_{1..n+3} +\aG(v')\bigr) \conc v'
\]
where $v' \defas x_{n+4..2n+6} + \aF(x_{1..n+3})$.

Note that its inverse satisfies
\[
    \ah^{-1}(x) = w'
    \conc \bigl(x_{n+4..2n+6} - \aF(w')\bigr)
\]
where $w' \defas x_{1..n+3} - \aG(x_{n+4..2n+6})$.

Now form the involutive neural network that induces a function $\aINVNET \colon \Reals^{2n+6} \to \Reals^{2n+6}$ defined as follows,
\[
    \aINVNET \defas \g^{-1} \circ \ah^{-1} \circ I_P^{2n+6,\sigma} \circ \ah \circ \g,
\]
by composing the layers or neural nets corresponding to each term in the function definition.

\begin{lemma}
    \label{lem:hhat-convergence}
    As $\neterr \to 0$,
the function
    $\ah$ converges uniformly to $\h$ on $\Omega \times B_3(0^{n+6})$ and
its inverse $\ah^{-1}$ converges to $\h^{-1}$ on $\Omega^+ \times \Omega^{+}$.
\end{lemma}

\begin{proof}
First observe that $|\G(x)| \leq |x|$ and $|\F(x)| \leq 2r$, so that $|\h(x)| \leq 3|x|+4r$.

For $x \in \Omega \times B_3(0^{n+6})$, we have $|x_{n+4..2n+6}+\aF(x_{1..n+3})| < |\aF(x_{1..n+3})| + 3 < |\F(x_{1..n+3})| + 4 < 2r+4$, and hence we have $x_{n+4..2n+6}+\aF(x_{1..n+3}) \in \Omega^{++}$. Thus for $x \in \Omega \times B_3(0^{n+6})$, all applications of $\aF$ and $\aG$ in $\ah(x)$ are on points in $\Omega^{++}$ (where $\aF$ and $\aG$ converge uniformly to $\F$ and $\G$), and so $\ah$ converges to $\h$ on $\Omega \times B_3(0^{n+6})$ as $\neterr \to 0$. Furthermore, the convergence is uniform, since it is formed from sums and compositions of uniformly converging functions.

For $x \in \Omega^+ \times \Omega^{+}$ we have $|x_{1..n+3}-\aG(x_{n+4..2+6})| < |\aG(x_{n+4..2+6})| + 7r+10 < |\G(x_{n+4..2+6})| + 7r+11 < 14r+21$, and hence we have $x_{n+4..2n+6}-\aG(x_{1..n+3}) \in \Omega^{++}$. Thus for $x \in \Omega^+ \times \Omega^{+}$, all applications of $\aF$ and $\aG$ in $\ah^{-1}(x)$ are on points in $\Omega^{++}$ (where $\aF$ and $\aG$ converge to $\F$ and $\G$) and so $\ah^{-1}$ converges to $\h^{-1}$ on $\Omega^+ \times \Omega^{+}$ as $\neterr \to 0$.
\end{proof}

    Note that by Lemma~\ref{lem:hhat-convergence}, we have
\[
    \label{eq:first-fcnerr-assumption}
    \tag{$\ddagger$}
    \max_{x \in \Omega \times B_{1/\fcnerr}(0^{n+6})}\Bigabs{\ah(g(x))-\h(g(x))}<1
\]
    for sufficiently small $\neterr$.

\begin{lemma}
    \label{lem:pointwise-convergence}
Consider the random variable $x \defas \phi \conc \PI$.
Conditioned on the event $\A$, the function $\aINVNET(x)$ converges pointwise to $\INV(x)$, as $\neterr \to 0$.
\end{lemma}
\begin{proof}
Condition on $\A$ and assume $\neterr$ is sufficiently small that
\eqref{eq:first-fcnerr-assumption} holds.
    Then notice that $x \in \Omega \times B_{1/\fcnerr}(0^{n+6})$, so that $g(x) \in \Omega \times B_3(0^{n+6})$, and hence $\bigl|I_P^{2n+6,\sigma}(\ah(g(x)))\bigr|=\bigl|\ah(g(x))\bigr|<|\h(g(x))|+1 < 3|g(x)|+4r+1 < 7r+10$. Thus $I_P^{2n+6,\sigma}\bigl(\ah(g(x))\bigr) \in \Omega^+ \times \Omega^+$. Hence when evaluating $\aINVNET(x)$, the inputs to both $\ah$ and $\ah^{-1}$ are in the domains where their respective convergence properties stated in Lemma~\ref{lem:hhat-convergence} hold.
Therefore each function occurring in the definition of $\aINVNET$ convergence pointwise to the corresponding function in the definition of $\INV$. Further, all such functions are continuous, and so the result holds.
\end{proof}

For $\phi \in \Omega$, define the random variable $\aINVNETr[\phi] \defas \aINVNET(\phi \conc \pi)_{1..n}$.
Let $\xi_m \sim N(0^n,\fcnerr \Id_n)$ be an independently chosen Gaussian.

The following result is immediate from Lemma~\ref{lem:pointwise-convergence}.
\begin{lemma}
    \label{lem:rv-pointwise-convergence}
Conditioned on the event $\A$, the random variable $\aINVNETr[\phi]$
    converges in distribution to $\INV[\phi]$ as $\neterr \to 0$.
    \qed
\end{lemma}

We now prove Theorem~\ref{maintheorem}.

\begin{proof}[Proof of Theorem~\ref{maintheorem}]
    Fix $m\in\Nats$; we will define $\hcI_{m}$ such that the sequence $\bigset{\hcI_m}_{m\in \Nats}$ has the desired convergence property using Lemmas~\ref{lem:rv-convergence-in-dist} and~\ref{lem:rv-pointwise-convergence}.

We may decompose the CDF of $\aINVNETr[\phi]$ in terms of $\A$ and $\barA$:
\begin{align*}
    F_{\aINVNETr[\phi]}(\phi') =\,& P(\A)  \cdot F_{\aINVNETr[\phi] | \A}(\phi') \: \\
&+ (1-P(\A)) \cdot F_{\aINVNETr[\phi] | \barA}(\phi').
\end{align*}

Now we form a sequence $\bigset{{{\hcI_{m}}}[\phi]}_{m\in\Nats}$ and demonstrate that $\lim_{m \to \infty} F_{{{\hcI_{m}}}[\phi]}(\phi') = F_{T[\phi]}(\phi')$ at all points of continuity $\phi'$ of $T[\phi]$.

By Lemma~\ref{lem:rv-convergence-in-dist}
choose $\fcnerr$ such that
    \[
        |F_{T[\phi]}(\phi') - (F_{\INV[\phi]|\A})(\phi')|<\tfrac{1}{3m}
    \]
    holds for all points of continuity $\phi'$ of $T[\phi]$ and
    $P(\A) < \frac{1}{3m}$ holds, which is possible by \eqref{eq:neterr-assumption}.

By Lemma~\ref{lem:rv-pointwise-convergence}
choose $\neterr$ so that
\[
    \bigl|(F_{\INV[\phi]|\A})(\phi') - F_{\aINVNETr[\phi] | \A}(\phi')\bigr|  <  \tfrac{1}{3m}
\]
    holds for all points of continuity $\phi'$ of $T[\phi]$
and \eqref{eq:first-fcnerr-assumption} holds.
    This is possible because conditioned on $\A$, the random variable $\aINVNETr[\phi]$ converges in distribution to $\INV[\phi]$ and every point of continuity of $T[\phi]$ is also a point of continuity of $\INV[\phi]$.

Now define $\hcI_{m} \defas \aINVNETr$. Observe that for any point of continuity $\phi'$ of $T[\phi]$, we have
\begin{align*}
    \bigabs{F_{{{\hcI_{m}}}[\phi]}&(\phi') - F_{T[\phi]}(\phi')} \\
    &= \bigabs{F_{\aINVNETr[\phi]}(\phi') - F_{T[\phi]}(\phi')}\\
    &< \bigabs{F_{\aINVNETr[\phi] | \A}(\phi') - F_{T[\phi]}(\phi')} + \tfrac{1}{3m}\\
    &< \bigabs{F_{\INV[\phi]|\A}(\phi') - F_{T[\phi]}(\phi')} + \tfrac{2}{3m}\\
    &< \tfrac{1}{m}.
\end{align*}

Finally, considering this fact for all $m\in\Nats$, we see that ${{\hcI_{m}}}[\phi]$ converges in distribution to $T[\phi]$.
\end{proof}


TITLE: The double integral $\int_{1}^{\infty}\int_{1}^{\infty} A(x+y)^2 e^{-(x+y)}\, dx \,dy$
QUESTION [0 upvotes]: I met this double integral while reviewing calculus,
the equation is like:
$$\int_{1}^{\infty}\int_{1}^{\infty} A(x+y)^2 e^{-(x+y)} dxdy=1$$
and we should find the value of $A$ by doing the integral,
I tried to change coordinate into polar, namely, let $x=r\cos{\theta}$ and $y=r\sin{\theta}$, but we can see the upper limit of $x$ is infinity, so there is definitely no such $r$ that satisfied this condition.
Then I tried to first integrate over variable $x$ (treat $y$ as constant first), but also find this method not very convenient,
so any idea how to do this integral? Thanks in advance.

REPLY [1 votes]: Define
$$c_n:=\int_1^\infty x^n e^{-x}\>dx\qquad(n\geq 0)\ .$$
Then $c_0={1\over e}$, and
$$c_n=-x^n e^{-x}\biggr|_1^\infty+n\int_1^\infty x^{n-1}e^{-x}\>dx={1\over e}+ n c_{n-1}\qquad(n\geq1)\ .$$
We shall only need $c_1={2\over e}$ and $c_2={5\over e}$. Now your integral
$$J:=\int_1^\infty\int_1^\infty(x+y)^2 e^{-(x+y)}\>dy\>dx$$
can be decomposed into
$$J=c_2c_0+2c_1^2 +c_0c_2={18\over e^2}\ ,$$
because
$$\int_1^\infty\int_1^\infty x^p e^{-x}\> y^q e^{-y}\>dy\>dx=c_p\>c_q\ .$$
The solution to your original problem is then $$A={e^2\over 18}\ .$$


text:
/-
Copyright (c) 2021 OpenAI. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kunhao Zheng, Stanislas Polu, David Renshaw, OpenAI GPT-f
-/
import mathzoo.imports.miniF2F

open_locale nat rat real big_operators topological_space

theorem mathd_algebra_160
  (n x : ℝ)
  (h₀ : n + x = 97)
  (h₁ : n + 5 * x = 265) :
  n + 2 * x = 139 :=
begin
  linarith,
end

subset_name: curated