TITLE: WolframAlpha blows simple substitution?
QUESTION [2 upvotes]: I am to find a formula for the n-th derivative of $$\frac{1+x}{1-x}$$ I came up with $$\frac{(-1)^{n+1} \times 2 \times n!}{(1-x)^{n+1}}$$
This seems right, but I noticed that WolframAlpha somehow manages to eat the sign for any n.
For example, for $n=2$ I have: $\frac{-4}{(1-x)^{3}}$, but WolframAlpha gives $\frac{4}{(1-x)^3}$. I am not particularly good at calculating, but I would like to think that I have mastered the numbers below $10$ so I'm pretty sure I'm right here and WolframAlpha did something very weird.
EDIT: Yep, my derivative is wrong. BUT this still doesn't answer the question why WolframAlpha responds with a positive fraction all the time - I didn't ask it to calculate a derivate, I just entered my formula.
PS: Who removed my WolframAlpha input? That was sort of essential to the question, I was never that interested in the derivative itself, I just wanted to know why WA responded that way.
(Exact input: (-1)^(n+1)*2*n!/((1-x)^(n+1)), n=2 )

REPLY [4 votes]: For your Wolfram Alpha input "(-1)^(n+1)*2*n!/((1-x)^(n+1))", I get the output $4\over (x-1)^3$ (not what you have written, I surmise you just misread it).
Note that
$$
{4\over(x-1)^3}  ={4\over \bigl(-(1-x)\bigr)^3}={4\over (-1)^3(1-x)^3}={-4\over(1-x)^3};
$$
which is exactly what your formula gave.


TITLE: For any integer $a$, $\gcd(11a+5,2a+1)=1$.
QUESTION [1 upvotes]: How would I go by proving this statement?
What I did was I tried using Proposition GCD Of One, so that
$(11a+5)x + (2a+1)y = 1$, and
$(11x+2y)a + (5x+y) = 1$.
But I have no idea what to do from here.
What can I do?

REPLY [4 votes]: Hint: Can you justify the following?
\begin{align*}
\gcd(11a + 5, 2a + 1) &= \gcd\Big((11a + 5) - 5(2a + 1), 2a + 1\Big) \\
&= \gcd(a, 2a + 1)
\end{align*}
Can you finish it from here?

REPLY [2 votes]: Hint $\ $ If $\,d\,$ is a common factor then $\,{\rm mod}\ d\!:\ 2a+1\equiv 0\equiv 11a+5.\,$ Now  eliminate $\,a\,$ by cross multiplying  $\ 0\equiv \color{#0a0}{11}(2a+1)-\color{#c00}2(11a+5) \equiv \color{#0a0}{11}-\color{#c00}{10}\equiv 1\ $ i.e. $\,0\equiv 1,\,$ so $\, d\mid 1\!-\!0,\,$ so $\,d = 1.$
Remark $\ $ More conceptually, put the fractions for $\,-a\,$ over the common denominator $\,22$
$$ {\rm mod}\ d\!:\ \  \frac{1}{2} \equiv -a\,\equiv \frac{5}{11}\ \Rightarrow\ \frac{\color{#0a0}{11}}{22}\,\equiv\, \frac{\color{#c00}{10}}{22}\qquad$$
Note $ $ The fractions uniquely exist mod $\,d\,$ since $\,2,11\,$ are invertible, being coprime to $\,d\,$ (else $2\mid d\mid 2a+1\,\Rightarrow 2\mid 1,\,$ and $\,11\mid d\mid 11a+5\,\Rightarrow\, 11\mid 5\,)$.


TITLE: How to calculate air flow from a pressurized tank?
QUESTION [0 upvotes]: I am writing an app that deals with simulated air brakes, and I need to simulate the release of air from one container of a given pressure into another container of a lower pressure. It doesn't have to be perfect, as this is just a rail simulator, but I need the air pressure to seem realistic. In reading about how to simulate air flow from an orifice, I was a little overwhelmed. Is there a way to get it "close enough" for my needs?

REPLY [0 votes]: For air, when the downstream absolute pressure is 52.8% or less of the upstream absolute pressure, the flow is choked (limited by the sonic velocity, mass flow rate independent of the sink pressure).  Your comment above says the source pressure is about 130 psi, so depending on the hose  (sink) pressure the flow may very well be choked.
You can find relationships for choked (and non-choked) flow in textbooks and articles on fluid flow.  For example, see articles on the web, or a text that discusses fluid flow, such as Elements of Thermodynamics and Heat Transfer by Obert and Young, or Flow of Fluids through Valves, Fittings, and Pipe, Crane Technical Paper 410.  The Crane reference provides equations, and graphs of discharge coefficients and other parameters in the equations.


TITLE: Conditions for Fubini's theorem in simple terms
QUESTION [3 upvotes]: Could someone explain the conditions required to satisfy Fubini's theorem in layman's terms, without going into measure theory? I'm a high school student trying to gain a simple understanding of when Fubini's theorem doesn't apply, but everything I've found online is in terms of measure spaces. Would it be possible to explain the conditions in terms of continuity and boundedness over the region of integration?

REPLY [1 votes]: Another version of Fubini's theorem states that if $A$ and $B$ are $\sigma$-finite measure spaces, not necessarily complete, and if either
$$
\int_{A}\left(\int_{B}|f(x, y)| \mathrm{d} y\right) \mathrm{d} x<\infty \quad \text { or } \quad \int_{B}\left(\int_{A}|f(x, y)| \mathrm{d} x\right) \mathrm{d} y<\infty
$$
then
$$
\int_{A \times B}|f(x, y)| \mathrm{d}(x, y)<\infty
$$
and
$$
\int_{A}\left(\int_{B} f(x, y) \mathrm{d} y\right) \mathrm{d} x=\int_{B}\left(\int_{A} f(x, y) \mathrm{d} x\right) \mathrm{d} y=\int_{A \times B} f(x, y) \mathrm{d}(x, y)
$$
In this version the condition that the measures are $\sigma$-finite is necessary.


TITLE: Why does everyone say that the faster you move through space, the slower you move through time, when that's not the case?
QUESTION [17 upvotes]: Suppose, I'm on earth and my brother is moving away from earth at a constant speed, $v=0.8c$. Now, if 5 seconds $(t_0)$ pass for me, the amount of time that will pass for my brother according to me will be $t$:
$$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$t=\frac{5}{\sqrt{1-(0.8)^2}}$$
$$t=8.33s$$
So, if $5$ decades pass for me, $8.33$ decades will pass for my brother. He will experience rapid aging according to me. So, why is everyone saying time will go slower for him when the case is the exact opposite?

REPLY [1 votes]: None of the answers point to the fact that time dilation is not an effect that happens to the clocks themselves. To explain it correctly you need to first identify what are the two spacetime event points between which you are calculating the time interval. If the two events are on the world-line of the brother on board the spaceship, then his time should be considered as the proper time and the earth-bound brother's clock will show the dilated time. For example if the spaceship-brother goes from earth to some distant star then his starting point and finishing point, both are on his worldliness. Starting the journey and finishing it are the two events. If by the spaceship clock the journey took say 5 yrs, the earth clock will show the corresponding time interval dilated by some amount depending on the velocity of the ship. It is 5 yrs times the Lorentz factor if you ignore the initial acceleration and final deceleration of the ship. If you care to include these two non-uniform motions , no big deal, you just need to use the time dilation formula with integration. Look up the 1st chapter of Landau Lifshitz Vol 3.
You may look up a few videos I made on this, here are the links:
https://youtu.be/KEeEhP89SZs
https://youtu.be/8iybUbKDTLs


text:
\subsection{More Results for Model Variants}\label{sec:experiment-additional-results}

Table~\ref{tbl:node_classification_on_citation_network} shows the result of the prediction accuracies of models that use MLPs with various layer size $L$ as transformation functions $\mathcal{B}^{(t)}$.
It also shows the results for the input injection model (GB-GNN-II) we have introduced in Section~\ref{sec:more-model-variants} and the SAMME.R model (GB-GNN-SAMME.R) in Section~\ref{sec:model-details}.
Figures~\ref{fig:train-loss-all-cora}--\ref{fig:train-loss-all-pubmed} show the transition of the training loss for $L=0, \ldots, 4$ (Figure~\ref{fig:train-loss-all-cora}: Cora, Figure \ref{fig:train-loss-all-citeseer}: CiteSeer, Figure~\ref{fig:train-loss-all-pubmed}: PubMed).
Figures~\ref{fig:test-loss-all-cora}--\ref{fig:test-loss-all-pubmed} show the transition of the training loss for $L=0, \ldots, 4$ (Figure \ref{fig:test-loss-all-cora}: Cora, Figure \ref{fig:test-loss-all-citeseer}: CiteSeer, Figure~\ref{fig:test-loss-all-pubmed}: PubMed).
Figures~\ref{fig:cosine-all-cora}--\ref{fig:cosine-all-pubmed} show the transition of the cosine values between the negative gradient $-\nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})$ and the weak learner $f^{(t)}$ at the $t$-th iteration for models that has $L=0$ to $4$ layers (Figure~\ref{fig:cosine-all-cora}: Cora, Figure~\ref{fig:cosine-all-citeseer}: CiteSeer, Figure~\ref{fig:cosine-all-pubmed}: PubMed).

\input{image/train_loss_all}
\input{image/test_loss_all}
\input{image/cosine_all}
\input{table/citation_for_deep_mlps}

config: arxiv


TITLE: Combinatorial sequences whose ratios $a_{n+1}/a_{n}$ are integers
QUESTION [21 upvotes]: I have a proof technique in search of examples. I'm looking for combinatorially meaningful sequences $\{a_n\}$ so that $a_{n+1}/a_n$ is known or conjectured to be an integer, such that there is a relation between the $n$th case and $n+1$st, but not an obvious $a_{n+1}/a_n\to 1$ map. This means $a_n$ is the $n$th partial product of an infinite sequence of integers, but there isn't an obvious product structure.

The prototype was an enumeration of
domino tilings of an Aztec
diamond of order $n$, $a_n =
   2^{n(n+1)/2}$, so $a_{n+1}/a_n =
   2^{n+1}$. (There is a nice $2^{n+1}$
to 1 map unrelated to my technique,
but it isn't obvious.)
Another application was a proof that
$\det \{B_{i+j}\}_{i,j=0}^n =
   \prod_{i=1}^n i! $  where $B_n$ is
the $n$th Bell number, equation
25 in the linked page.
The counts of alternating sign matrices 1, 2, 7, 42, ... are not an example, since
$ASM(n+1)/ASM(n) = \frac{ (3n+1)!n!}{2n! (2n+1)!}$ which is not always an integer, e.g,     7/2 is not.

What are some other interesting combinatorial families whose ratios $a_{n+1}/a_n$ are known or (preferably) conjectured to be integers?

Thanks.

REPLY [1 votes]: The number of acyclic orientations on the complete graph on $n$ vertices is $n!$, and the number of acyclic orientation on a unit-interval graph on $n$ vertices is given by a product with $n$ factors, so you can easily construct many sequences from this.


TITLE: Show that $p\{\limsup_{n\to \infty }A_n\geq M\}\geq \limsup_{n\to \infty }p\{A_n\geq A\}$
QUESTION [0 upvotes]: Let $M\in\mathbb R$. How can I show that $$p\{\limsup_{n\to \infty }A_n\geq M\}\geq \limsup_{n\to \infty }p\{A_n\geq M\}\ \ \ ?$$
What I tried is $$p\{\limsup_{n\to\infty }A_n\geq M\}=p\left\{\bigcap_{n\in\mathbb N}\bigcup_{k=n}^\infty A_n\geq M\right\}=\lim_{n\to \infty }p\left\{\bigcup_{k=n}^\infty A_k\geq M\right\}$$
but $\bigcup_{k=n}^\infty A_n\geq M$ and $\bigcap_{n\in\mathbb N}\bigcup_{k=n}^\infty A_n\geq M$ doesn't has a sense... (they are probably not sets as I could consider).
Any idea ?

REPLY [3 votes]: By the reverse Fatou's lemma, we have
$$ \limsup_{n\to\infty}\Bbb{P}(A_n \geq M)
\leq \Bbb{E}( \limsup_{n\to\infty} \mathbf{1}_{\{ A_n \geq M \}} )
= \Bbb{P}( A_n \geq M \text{ i.o.} ) $$
But on the event $\{ A_n \geq M \text{ i.o.} \}$ we have $\limsup A_n \geq M$. Therefore the inequality follows.