TITLE: An alternating series convergence
QUESTION [0 upvotes]: I was wondering whether the following series converges or diverges,
$$\sum_{n=1}^\infty (-1)^n \sqrt[n]{a}$$ $$\forall a>0,  a\ne1$$
The divergence test cannot be applied, since the sequence does not have a limit.

REPLY [4 votes]: Since $$\lim_{n \to \infty} (-1)^n \sqrt[n]{a} \neq 0$$
the series is not convergent.

REPLY [2 votes]: Note that $\sqrt[n]a \to 1$, hence $(-1)^n \sqrt[n]a \not\to 0$, so the series diverges.


TITLE: Proof of mathematic induction via the property of $\mathbb{N}$
QUESTION [1 upvotes]: Assuming that any nonempty set of positive integers has a minimal element, prove:
If the proposition $P$ is such that
(1) $P(m_0)$ is true,
(2) $P(m)$ is true whenever $P(a)$ is true for all $a$ such that $m_0\leqslant a<m$,
then $P(n)$ is true for all $n\geqslant m_0$.
My solution: Let $A$ be the following set $$A=\{n\geqslant m_0: P(n)\  \text{is false}\}$$ Since $m_0\in A$ ($\Rightarrow A\neq \varnothing$) and $A\subset \mathbb{N}$ then $A$ has the smallest element, let's call it $m$. Then $m\in A, \ m\geqslant m_0$ and $P(m)$ is false. Since $P(m_0)$ is true $\Rightarrow$ $P(m_0+1)$ also true $\Rightarrow$$...$$\Rightarrow$ $P(m)$ is true. This is contradiction and hence $A=\varnothing$. Thus $P(m)$ is true for all $m\geqslant m_0$
EDIT: Let's define the following set $$A=\{k\geqslant m_0: P(k) \ \text{is false}\}$$ Suppose that this set is non-empty. Then being the subset of $\mathbb{N}$ it has minimal element $m$. Using minimality of $m$ we have that for $a$ such that $m_0\leqslant a<m$ we have that $P(a)$ is true. Then by 2) $P(m)$ is true. But since $m\in A$ we have $P(m)$ is false. This is the contradiction. It follows that the set $A$ is empty meaning that $P(k)$ is true for all $K\geqslant m_0$.

REPLY [1 votes]: Let $B = \{k\ge m_0| P(k) \text{ is false}\}\subset \mathbb N$.
Either $B$ is empty; or $B$ has a minimum element.
If $B$ has a minimum element $m = \min B$, then what can you say about $a: m_0 \le a < m$?[$*$]  Is $P(a)$ true or false or can you not determine.
And what does condition 2) tell you about whether $P(m)$ is true or not? And what does $m = \min B$ tell you about whether $P(m)$ is true or not?
Is a minimum element of $B$ possible?  What does it mean if it isn't?
[$*$]  Warning:  Do any $m_0 \le a < m$ exist?  Do we know that $m > m_0$.


TITLE: Greatest elements in crystallographic root systems
QUESTION [2 upvotes]: I have a question regarding a remark in the book "Reflection Groups And Coxeter Groups" by James E. Humphreys (unfortunately the book is not to be found as a whole on google books or such).
In chapter 2 we encounter the definition of a crystallographic root system ($\varPhi$ crystallographic iff  $\frac{2(a,b)}{(b,b)} \in \mathbb{Z}\forall a,b \in \varPhi$) and its associated Weyl group $W$ (the group generated by all reflections belonging to elements of $\varPhi$).
After the definition Humphreys follows up with some facts about crystallographic root systems. It is quite easy to see that in the case where $\varPhi$ is irreducible (i.e. the Coxeter graph is connected) there can occur at most two squared length of roots (the roots are then called long and short roots respectively) and $W$ acts transitively on the long and the short roots. After fixing a simple system $\Delta \subset \varPhi$ a partial ordering is introduced on $\varPhi$ via $a \leq b$ iff $b-a$ is a non-negative linear combination of $\Delta$ and the author writes that there is a unique greatest element of $\varPhi$ with respect to this ordering which is a long root.
I can show (using transitivity and a fundamental domain given earlier in the book) that there can be at most two greatest elements one being a long root and one a short root. However I am unable to see why the long root should be greater than the short root. Is there a proof for this fact (preferably one that does not use Lie theory) that does not use the classification and construction of all possible root systems (this is at the moment the only way I could find)?
Any help would be appreciated. Thanks in advance.
Edit: As requested the definition of a simple system: A subset $\Delta \subset \varPhi$ of a rootsystem $\varPhi$ is called simple if $\Delta$ is a basis for the span of $\varPhi$ which satisfies the additional condition that any element of $\varPhi$ is either a non-negative or a non-positive linear combination of $\Delta$.

REPLY [5 votes]: As promised here the complete proof: There are two cases we need two consider:
Case A: One of the labels of the edges of the coxeter graph is "4". Then the ratio of squared lengths of long roots and short roots is $2$. Wlog we will assume that $(a,a)=2$ for long roots and $(b,b)=1$ for short roots. Let $b$ be a positive short root. By transitivity there exists a long root $a$ such that $(a,b)=-1$ and we have $a+b=s_a(b)$ is a short root orthogonal to $b$. Hence we have two subcases.
subcase A.1: $a+b$ is negative. Then $-a-b$ is positive and we can conclude that $-a=-a-b+b$ is a root (since $a$ was a root) which is greater in our partial ordering than $b$.
subcase A.2: $a+b$ is positive. Then $a+b+b=a+2b=s_b(a)$ is in fact a root and in fact again greater than $b$.
Case B: One of the labels of the edges of the coxeter graph is "6". Then the ratio of squared lengths of long roots and short roots is $3$. Wlog we will assume that $(a,a)=3$ for long roots and $(b,b)=1$ for short roots. Let $b$ be a positive short root. By transitivity there exists a long root $a$ such that $(a,b)=-\frac 3 2$. We have $a+b=s_a(b)$ is a short root and $(a+b,b)=-\frac 1 2$. The same subcases arise:
subcase B.1: b+a is negative. Then $-b-a$ is positive and $-a=-b-a+b$ is a root greater than $b$.
subcase B.2: b+a is positive. Then $2b+a=b+a+b=s_{a+b}(b)$ is a root which is greater than $b$.
Therefore we find an even greater root for every positive short root $b$ in our irreducible system $\varPhi$ and a greatest element with respect to our partial ordering has to be a long root. (This long root $a$ is uniquely determined since $W$ acts transitively on the long roots, $(a,c) \geq 0 \forall c \in \Delta$ and the fact that the subset of our vector space defined by this condition forms a fundamental domain for the action of $W$ on the vector space.)


TITLE: complex series summation up to infinity
QUESTION [1 upvotes]: Sum  of series
  $$S(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots , |x|<1$$

What I tried.
Let $$A=\ln\bigg[(1+x)(1+x^2)(1+x^3)(1+x^4)\cdots \cdots \bigg]$$
$$A = \ln\bigg[(1+x+x^2+2x^3+2x^4+3x^5+\cdots+ )\bigg]$$
How do is solve it help me please

REPLY [1 votes]: Regarding the relationship with the $A$ term
$$
S_n(x) = \sum_{k=1}^n\frac{g_k'(x)}{g_k(x)} = \frac{d}{dx}\sum_{k=1}^n\ln g_k(x) = \frac{d}{dx}\ln\prod_{k=1}^ng_k(x) = \frac{d}{dx}A
$$
here $g_k(x) = 1+x^k$.  Also $\prod_{k=1}^n g_k(x) = \frac 12(-1;x)_{n+1}$  (see) and
$$
S_n(x) = \frac{d}{dx}\ln\left(\frac 12(-1;x)_{n+1}\right)
$$


TITLE: How to integrate an equation with multiple non-independent variables
QUESTION [0 upvotes]: I'm a little lost with this particular equation, I have three variables which need to be integrated and can't quite wrap my mind to get the correct result.
I have this:
$$
\frac{dH}{dt}=8\pi V_{0}-3H^{2}(t)
$$
Can I solve this for H(t)?
Also, H can be written as: $H=\frac{da}{adt}$, where $a$ is a function of time giving me:
$$
\frac{d}{dt}\left(\frac{da}{adt}\right)=8\pi V_{0}-3\left(\frac{da}{adt}\right)^{2}
$$
I'm just lost at this point, anyone that can keep a clear head and do this I'd be very appreciative if you can show me an applicable method...
Many thanks

REPLY [1 votes]: HINT :
The ODE is of separable kind :
$$dt=\frac{dH}{8\pi V_0-3H^2}$$
$$t=\int \frac{dH}{8\pi V_0-3H^2}= \frac{ 1 }{2\sqrt{6\pi V_0 }}\tanh^{-1}\left(\frac{\sqrt{3}}{2\sqrt{\pi V_0}} H\right)+\text{constant}$$
This gives $t$ as a function of $H$. Inverse this function to find $H(t)$.
Note :
If you don't understand the integral, this could help you :
$\int \frac{dH}{8\pi V_0-3H^2}= \int \frac{\frac{dH(t)}{dt}}{8\pi V_0-3H(t)^2}dt \quad$ and apply the chain rule for integration.


TITLE: Trigonometric proof? Product to Sum.
QUESTION [0 upvotes]: I'm having trouble with proving these:
$$\sin^2(70^\circ)\sin^2(50^\circ)\sin^2(10^\circ)=\frac{1}{64}$$
$$16\sin(10^\circ)\sin(30^\circ)\sin(50^\circ)\sin(70^\circ)\sin(90^\circ)=1$$
I don't know, but I think we have to use
$$\cos(140^\circ)=-\sin(50^\circ).$$

REPLY [2 votes]: Another way:
As $\sin(3\cdot10^\circ)=\dfrac12,$
$\sin(3\cdot50^\circ)=\sin(180^\circ-30^\circ)=\cdots=\dfrac12,$
and $\sin(3\cdot70^\circ)=\sin(180^\circ+30^\circ)=\cdots=-\dfrac12,$
So, $\sin10^\circ,\sin50^\circ,\sin(-70^\circ)=-\sin70^\circ$ are the roots of $$3\sin x-4\sin^3x=\dfrac12\iff4\sin^3x-3\sin x+\dfrac12=0$$
Using Vieta's formula,
$$\sin10^\circ\cdot\sin50^\circ\cdot(-\sin70^\circ)=-\dfrac{\dfrac12}4$$
Can you take it from here?


TITLE: Does $f'''$ exist or not?
QUESTION [1 upvotes]: Does there exist  a  function  $f : \mathbb{R} \rightarrow \mathbb{R}$  such  that  the  derivatives  $f' , f''$  exist  but  not $f'''$?

I  thinks this exists.  Take  $f(x) = |x|^3$ . Then $f'$ and $f''$ exist everywhere, but $f'''$  does not exist at $0$. Is this true?

REPLY [2 votes]: In fact there exists a twice continuously differentiable function $f$ on $\mathbb R$ such that $f'''(x)$ exists nowhere. Proof: Let $W$ be a continuous function on $\mathbb R$ that is nowhere differentiable (Weierstrass). Then set
$$f(y)=\int_0^y\int_0^x W(t)\,dt.$$
Then $f''(y)=W(y)$ everywhere. But since $W$ is not differentiable anywhere, $f'''(y)$ exists nowhere.


text:
\subsubsection{Figures for advection tests}
\label{A:AdvTestPlots}

\begin{figure}[b]
\begin{subfigure}{0.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_PA_1_cosine_bell_1_limiter_no_ICON_FFSL_2D.png}
  \caption{}
  \label{fig:PACosineExact}
\end{subfigure}
\begin{subfigure}{0.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_cosine_bell_1_limiter_4_ICON_FFSL_2D.png}
  \caption{}
  \label{fig:PACosineICON}
\end{subfigure}
\begin{subfigure}{0.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_cosine_bell_1_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:PACosineAdjoint}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_cosine_bell_1_limiter_4_SOURCE_2D.png}
  \caption{}
  \label{fig:PACosineSource}
\end{subfigure}
\caption{Solid body rotation, cosine bell, a)-d) - contour plots with color bar:
                                           a) exact solution
                                           b) ICON-FFSL, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                           c) standard adjoint
                                           d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                           }
\label{fig:CosineBellContours}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_cosine_bell_1_EXACT_ADJOINT_SOURCE_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACosineAdjointSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_cosine_bell_1_EXACT_ADJOINT_SOURCE_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACosineAdjointSourceLim4}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_cosine_bell_1_ADJOINT_SOURCE_DIFF_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACosineAdjointSourceDiffLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_cosine_bell_1_ADJOINT_SOURCE_DIFF_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACosineAdjointSourceDiffLim4}
\end{subfigure}
\caption{Solid body rotation, cosine bell, a)-d) - along the curve exact vs standard adjoint vs art. source adjoint :
                                           a) solutions, art. source with limiter ~\cite{zalesak1979fully}, \cite{schar1996synchronous}
                                           b) solutions, art. source with limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                           c) errors, art. source with limiter ~\cite{zalesak1979fully}, \cite{schar1996synchronous}
                                           d) errors, art. source with limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                           }
\label{fig:CosineBellCurve}
\end{figure}

\begin{figure}
\begin{subfigure}{0.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_PA_1_slotted_cylinder_2_limiter_no_ICON_FFSL_2D.png}
  \caption{}
  \label{fig:PACylinderExact}
\end{subfigure}
\begin{subfigure}{0.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_slotted_cylinder_2_limiter_3_ICON_FFSL_2D.png}
  \caption{}
  \label{fig:PACylinderIcon}
\end{subfigure}
\begin{subfigure}{0.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_slotted_cylinder_2_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:PACylinderAdjoint}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_slotted_cylinder_2_limiter_3_SOURCE_2D.png}
  \caption{}
  \label{fig:PACylinderSource}
\end{subfigure}
\caption{Solid body rotation, slotted cylinder, a)-d) - contour plots with color bar:
                                           a) exact solution
                                           b) ICON-FFSL, limiter ~\cite{zalesak1979fully}, \cite{schar1996synchronous}
                                           c) standard adjoint
                                           d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{schar1996synchronous}
                                           }
\label{fig:SlottedCylinderContours}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_slotted_cylinder_2_EXACT_ADJOINT_SOURCE_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACylinderAdjointSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_slotted_cylinder_2_EXACT_ADJOINT_SOURCE_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACylinderAdjointSourceLim4}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_slotted_cylinder_2_ADJOINT_SOURCE_DIFF_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACylinderAdjointSourceDiffLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{PA_1_slotted_cylinder_2_ADJOINT_SOURCE_DIFF_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:PACylinderAdjointSourceDiffLim4}
\end{subfigure}
\caption{Solid body rotation, slotted cylinder, a)-d) - along the curve exact vs standard adjoint vs art. source adjoint:
                                                 a) solutions, art. source with limiter ~\cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                 b) solutions, art. source with limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                                 c) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                 d) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                 }
\label{fig:SlottedCylinderCurve}
\end{figure}

\begin{figure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_DF1_2_cosine_bells_4_limiter_no_ICON_FFSL_2D.png}
  \caption{}
  \label{fig:DF1CosineExact}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF1_2_cosine_bells_4_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:DF1CosineAdjoint}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF1_2_cosine_bells_4_limiter_3_SOURCE_2D.png}
  \caption{}
  \label{fig:DF1CosineSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF1_2_cosine_bells_4_limiter_4_SOURCE_2D.png}
  \caption{}
  \label{fig:DF1CosineSourceLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}=0$, cosine bells, a)-d) - contour plots with color bar:
                                          a) exact solution
                                          b) standard adjoint,
                                          c) art. source adjoint, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                          d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                          }
\label{fig:DefFl1:2CosBellContour}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_cosine_bells_4_EXACT_ADJOINT_SOURCE_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CosineAdjointSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_cosine_bells_4_EXACT_ADJOINT_SOURCE_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CosineAdjointSourceLim4}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_cosine_bells_4_ADJOINT_SOURCE_DIFF_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CosineAdjointSourceDiffLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_cosine_bells_4_ADJOINT_SOURCE_DIFF_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CosineAdjointSourceLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}=0$, cosine bells, a)-d) - along the curve  exact vs standard adjoint vs art. source adjoint:
                                                                       a) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                                       b) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                                       c) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                                       d) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                                       }
\label{fig:DefFl1:2CosBellCurve}
\end{figure}

\begin{figure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_DF1_2_slotted_cylinders_5_limiter_no_ICON_FFSL_2D.png}
  \caption{}
  \label{fig:DF1CylinderExact}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF1_2_slotted_cylinders_5_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:DF1CylinderAdjoint}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF1_2_slotted_cylinders_5_limiter_3_SOURCE_2D.png}
  \caption{}
  \label{fig:DF1CylinderSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF1_2_slotted_cylinders_5_limiter_4_SOURCE_2D.png}
  \caption{}
  \label{fig:DF1CylinderLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}=0$, slotted cylinders, a)-d) - contour plots with color bar:
                                          a) exact solution
                                          b) standard adjoint,
                                          c) art. source adjoint, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                          d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                          }
\label{fig:DefFl1:2SlottedCylinderContour}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_slotted_cylinders_5_EXACT_ADJOINT_SOURCE_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CylinderAdjoinrSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_slotted_cylinders_5_EXACT_ADJOINT_SOURCE_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CylinderAdjointSourceLim4}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_slotted_cylinders_5_ADJOINT_SOURCE_DIFF_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CylinderAdjointSourceDiffLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF1_2_slotted_cylinders_5_ADJOINT_SOURCE_DIFF_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF1CylinerAdjointSourceDiffLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}=0$, slotted cylinders, a)-d) - along the curve exact vs standard adjoint vs art. source adjoint:
                                                                       a) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                                       b) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                                       c) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                                       d) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                                       }
\label{fig:DefFl1:2SlottedCylinderCurve}
\end{figure}

\begin{figure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_DF3_3_cosine_bells_4_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:DF3CosineExact}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF3_3_cosine_bells_4_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:DF3CosineAdjoint}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF3_3_cosine_bells_4_limiter_3_SOURCE_2D.png}
  \caption{}
  \label{fig:DF3CosineSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF3_3_cosine_bells_4_limiter_4_SOURCE_2D.png}
  \caption{}
  \label{fig:DF3CosineSourceLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}\neq0$, cosine bells, a)-d) - contour plots with color bar:
                                          a) exact solution
                                          b) standard adjoint,
                                          c) art. source adjoint, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                          d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                          }
\label{fig:DefFl2:2CosBellContour}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_cosine_bells_4_EXACT_ADJOINT_SOURCE_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CosineAdjointSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_cosine_bells_4_EXACT_ADJOINT_SOURCE_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CosineAdjointSourceLim4}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_cosine_bells_4_ADJOINT_SOURCE_DIFF_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CosineAdjointSourceDiffLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_cosine_bells_4_ADJOINT_SOURCE_DIFF_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CosineAdjointSourceDiffLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}\neq0$, cosine bells, a)-d) - along the curve exact vs standard adjoint vs art. source adjoint:
                                                                       a) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                                       b) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                                       c) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                                       d) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                                       }
\label{fig:DefFl2:2CosBellCurve}
\end{figure}

\begin{figure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_DF3_3_slotted_cylinders_5_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:DF3CylinderExact}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF3_3_slotted_cylinders_5_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:DF3CylinderAdjoint}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF3_3_slotted_cylinders_5_limiter_3_SOURCE_2D.png}
  \caption{}
  \label{fig:DF3CylinderSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_DF3_3_slotted_cylinders_5_limiter_4_SOURCE_2D.png}
  \caption{}
  \label{fig:DF3CylinderLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}\neq0$, slotted cylinders, a)-d) - contour plots with color bar:
                                          a) exact solution
                                          b) standard adjoint,
                                          c) art. source adjoint, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                          d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                          }
\label{fig:DefFl2:2SlottedCylinderContour}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_slotted_cylinders_5_EXACT_ADJOINT_SOURCE_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CylinderAdjointSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_slotted_cylinders_5_EXACT_ADJOINT_SOURCE_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CylinderAdjointSourceLim4}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_slotted_cylinders_5_ADJOINT_SOURCE_DIFF_LIM3_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CylinderAdjointSourceDiffLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{DF3_3_slotted_cylinders_5_ADJOINT_SOURCE_DIFF_LIM4_2D_CURVE_v3.png}
  \caption{}
  \label{fig:DF3CylinderAdjointSourceDiffLim4}
\end{subfigure}
\caption{Deformational flow, $\nabla \vec{v}\neq0$, slotted cylinders, a)-d) - along the curve  exact vs standard adjoint vs art. source adjoint :
                                                          a) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                          b) solutions, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                          c) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                                          d) errors, art. source with limiter ~ \cite{zalesak1979fully}, \cite{harris2011flux}
                                                          }
\label{fig:DefFl2:2SlottedCylinderCurve}
\end{figure}

\begin{figure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_MV_4_vortex_3_limiter_no_ICON_FFSL_2D.png}
  \caption{}
  \label{fig:MVExact}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_MV_4_vortex_3_limiter_no_ADJOINT_2D.png}
  \caption{}
  \label{fig:MVAdjoint}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_MV_4_vortex_3_limiter_3_SOURCE_2D.png}
  \caption{}
  \label{fig:MVSourceLim3}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_MV_4_vortex_3_limiter_4_SOURCE_2D.png}
  \caption{}
  \label{fig:MVSourceLim4}
\end{subfigure}
\caption{Moving vortices, a)-d) - contour plots with color bar:
                                          a) exact solution
                                          b) standard adjoint,
                                          c) art. source adjoint, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                          d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                          }
\label{fig:MovingVortexContour}
\end{figure}

\begin{figure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{exact_MV_4_vortex_3_limiter_no_ICON_FFSL_3D_LONLAT.png}
  \caption{}
  \label{fig:MVExact3D}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_MV_4_vortex_3_limiter_no_ADJOINT_3D_LONLAT.png}
  \caption{}
  \label{fig:MVAdjoint3D}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_MV_4_vortex_3_limiter_3_SOURCE_3D_LONLAT.png}
  \caption{}
  \label{fig:MVSourceLim33D}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_MV_4_vortex_3_limiter_4_SOURCE_3D_LONLAT.png}
  \caption{}
  \label{fig:MVSourceLim43D}
\end{subfigure}
\caption{Moving vortices, a)-d) - unrotated grid $(\lambda, \theta)\in[-\pi; \pi]\times[-\pi/2; \pi/2]$, plot viewing angle $(90^\circ, 0^\circ)$:
                                   a) exact solution
                                   b) standart adjoint
                                   c) art. source adjoint, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                   d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                   }
\label{fig:MovingVortexAtT}
\end{figure}

\begin{figure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_vortex_3_limiter_3_ICON_FFSL_3D.png}
  \caption{}
  \label{fig:PAMV}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_vortex_3_limiter_no_ADJOINT_3D.png}
  \caption{}
  \label{fig:PAMVAdjoint3D}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_vortex_3_limiter_3_SOURCE_3D.png}
  \caption{}
  \label{fig:PAMVSourceLim33D}
\end{subfigure}
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=\textwidth]{numerical_PA_1_vortex_3_limiter_4_SOURCE_3D.png}
  \caption{}
  \label{fig:PAMVSourceLim43D}
\end{subfigure}
\caption{Solid body rotation wind field, moving vortex initial scalar field,a)-d) - plot viewing angle $(90^\circ, 0^\circ)$, $t=T/2$:
                                   a) ICON FFSL, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                   b) standart adjoint
                                   c) art. source adjoint, limiter ~ \cite{zalesak1979fully}, \cite{schar1996synchronous}
                                   d) art. source adjoint, limiter ~\cite{zalesak1979fully}, \cite{harris2011flux}
                                   }
\label{fig:MovingVortexAtT/2}
\end{figure}

config: arxiv


text:
\section{Introduction}

The Law of Import-Export
denotes the principle that a right-nested conditional of the form $A\to (B \to C)$ is logically equivalent to the simple conditional $(A\wedge B) \to C$ where both antecedents are united by conjunction. The Law holds in classical logic for material implication, and if there is a

logic for the indicative conditional of ordinary language, it appears Import-Export ought to be a part of it. For instance, to use an example from \citep[300]{cooper1968propositional}, the sentences ``If Smith attends and Jones attends then a quorum will be present'', and ``if Smith attends, then if Jones attends, a quorum will be present'' appear to convey the same hypothetical information. The same appears to hold more generally, at least when $A$, $B$ and $C$ themselves are non-conditional sentences, and the equivalence has been described as ``a fact of English usage'' \citep{mcgee1989conditional}.\footnote{Import-Export has been challenged on linguistic grounds, see for instance \cite{khoo2019triviality}, drawing on examples from Fitelson. The alleged counterexamples are subtle, however, and even Khoo and Mandelkern accept a version of the law. See also Appendix \ref{sec:IE}.}

In a celebrated paper, however, Allan \citet{gibbard1980two} showed that a binary conditional connective `$\to$' collapses to the material conditional of classical logic `$\supset$' if the following conditions hold: (i) the conditional connective satisfies Import-Export, (ii) it is at least as strong as the material conditional

($A\to C \models_{L} A\supset C$), where $\models_{L}$ is the consequence relation of the target logic of conditionals, (iii) it is supraclassical in the sense that it reproduces the valid inferences of classical logic in conditional form

($\models_{L} A\to C$ whenever $A\models_{\sf CL} C$). From (i)--(iii) and some natural background assumptions, Gibbard infers $A\supset C \models_{L} A\to C$. Given (ii), $\to$ and $\supset$ are thus logically equivalent,

according to the logic of conditionals ($\models_{L}$) under consideration. Prima facie, the conditional then needs to support all inference schemes validated by the material conditional in classical logic. However, inferences such as $\neg A \models A \to C$ (one of the paradoxes of material implication) enjoy little plausibility in ordinary reasoning with conditionals.

Gibbard's result poses a challenge for theories that compete with material implication as an adequate analysis of the indicative conditional.\footnote{Notable defenders of the material implication analysis are \citet{Lewis1976}, \citet{Jackson1979} and \citet{Grice1989}.} For example, Stalnaker's logic {\sf C2} \citep{stalnaker1968} and Lewis's logic {\sf VC} \citep{lewis1973} are both supraclassical and make the conditional stronger than the material conditional, but they invalidate Import-Export for that matter.

Not all theories make that choice, however. All of the above logics operate in a \textit{bivalent} logical setting,

thus limiting their options. In this paper, we explore how certain \textit{trivalent} logics of conditionals address Gibbard's challenge. These logics, which retain truth-functionality, analyze an indicative conditional of the form ``if $A$ then $C$'' as a \textit{conditional assertion} that is \emph{void} if the antecedent turns out to be false, and that takes the truth value of the consequent $C$ if $A$ is true \citep{reichenbach1935wahr,definetti1936logique,quine1950methods,belnap1970conditional}. This analysis assigns a third truth value (``neither true nor false'') to such ``void'' assertions, and gives rise to various logics that combine a truth-functional conditional connective with existing frameworks for trivalent logics \citep[e.g.,][]{cooper1968propositional,farrell1979material, milne1997bruno,cantwell2008logic,baratgin2013uncertainty,EgreRossiSprenger2020a,EgreRossiSprenger2020b}.

This chapter clarifies the scope and significance of Gibbardian collapse results with specific attention to such trivalent logics, in which the conditional is undefined when its antecedent is false. We begin with a precise explication of Gibbard's result, including a more formal version of his original proof sketch (Section \ref{sec:Gibbard}). Then we present two trivalent logics of indicative conditionals, paired with Strong Kleene semantics for conjunction and negation, and we examine how they deal with  Gibbardian collapse (Section \ref{sec:Tri} and \ref{sec:TriSK}). We then turn to trivalent logics that replace Strong Kleene operators with \citeauthor{cooper1968propositional}'s quasi-connectives where the conjunction of the True and the third truth value is the True (Section \ref{sec:TriQ}). Specifically, we show why rejecting superclassicality---and retaining both Import-Export and a stronger-than-material conditional---is a viable way of avoiding Gibbardian collapse.

In the second part of the paper, we consider a recent strengthening of Gibbard's result

due to Branden Fitelson and apply it to the above trivalent logics (Section \ref{sec:FitRes}, \ref{sec:FitInt} and \ref{sec:Block}). From this analysis it emerges that Gibbard's result may be better described as a \textit{uniqueness result}: we cannot have two conditional connectives that satisfy Import-Export as well as Conjunction Elimination, where one is strictly stronger than the other, and where the weaker (already) satisfies Modus Ponens.
We also provide three appendices:

Appendix \ref{sec:IE}  rebuts a recent attempt at a \textit{reductio} of Import-Export, Appendix \ref{sec:technical} provides the proofs of various lemmata stated in the paper, and Appendix \ref{sec:khoo} gives a more constrained derivation of Gibbardian collapse than his original proof, of particular relevance for the first trivalent system we discuss. For more in-depth treatment of trivalent logics of conditionals, we refer the reader to our comprehensive survey and analysis in \citet{EgreRossiSprenger2020a,EgreRossiSprenger2020b}.


TITLE: What is the name of the solid created by joining two tetrahedra face-to-face?
QUESTION [1 upvotes]: When I connect the bases of two pyramids, I will receive an octahedron. I have a question about an analogical situation with two tetrahedra. What is the name of a geometrical figure, which is a combination of two tetrahedra?

REPLY [2 votes]: It's a triangular dipyramid, which is one of the $92$ Johnson solids (solids all of whose faces are regular but where the vertices do not all look the same).

REPLY [2 votes]: When joining two regular tetrahedra, the result is one of the Johnson solids, the triangular bipyramid.

Two general tetrahedra joined face-to-face can also be called triangular bipyramids, but these joins are not Johnson solids.


TITLE: Calculating combined errors
QUESTION [0 upvotes]: I am trying to learn to combine errors, but I am confused on how to calculate it once I string it all together into my equation.
The equation is given by
$$
y = A + (n - n_1) \cdot \frac{B - A}{n_2 - n_1}
$$
and the errors are
\begin{align}
\sigma_{n} &= 0.143 \\
\sigma_{n_1} &= 0.209 \\
\sigma_{n_2} &=1.27
\end{align}
How would I combine all these errors to work out the total errors for $y$?
My attempt was:
$$
\textrm{Total error} = \sqrt{ \sqrt{n^2 + n_1^2} + \frac{1}{\sqrt{n_2^2 + n_1^2}} }
$$
This yields the $1.01$.
But from my actual data there is no way this is correct, that would be many orders of magnitude off from my actual measurements, so it can't be the correct way to calculate the error.

REPLY [5 votes]: Your function is
$$f(n,n_1,n_2)=A+(B-A)\frac{n-n_1}{n_2-n_1}$$
The error propagation formula states that the error $\sigma_f$ on this function is given by:
$$\sigma_f=\sqrt{\left(\frac{\partial f}{\partial n}\sigma_n\right)^2+\left(\frac{\partial f}{\partial n_1}\sigma_{n_1}\right)^2+\left(\frac{\partial f}{\partial n_2}\sigma_{n_2}\right)^2}$$
Substituting into this formula will give you the correct error estimate.
EDIT:
If you can't take partial derivatives for some reason, then in this particular case, since only addition, subtraction, multiplication and division are involved, you only need to know two rules:

For adding and subtracting quantities, the absolute error on the sum/difference is equal to the absolute errors added in quadrature.
For multiplying and dividing quantities, the relative error on the product/quotient is equal to the relative errors (e.g. $\sigma_a/a$) added in quadrature.

So, first compute the error on the differences $d_1=n-n_1$ and $d_2=n_2-n_1$. Then use those errors $\sigma_{d_1}$ and $\sigma_{d_2}$ to compute the error on the quotient $\frac{d_1}{d_2}$. The final error will be equal to $(B-A)$ times the error on $\frac{d_1}{d_2}$.


TITLE: Why does the progression of values of $\sin(x)$ seem so arbitrary?
QUESTION [2 upvotes]: I don't know how to ask this in a way that doesn't sound a little stupid, but what I mean is, the progression of values of say, sin(x) seems really arbitrary. Is there an intuitive way to understand why, for example:
   x            y
-------   -------------
sin(45) = 0.85090352453
sin(46) = 0.90178834764

What is determining the amounts that Y goes up or down by as X changes? And even better, is there something (fairly accessible to non-mathematicians) I can read that can explain this sort of thing in a very intuitive way? I don't want to prove that it works, just understand how come. Thanks!

REPLY [1 votes]: It seems that you are using radians rather than degrees.  Degrees are most common in day to day use (a full circle is $360$).  Serious maths usually uses radians (a full circle is $2 \pi$).  Calculators that support sine usually offer both but the default varies.  If you did not intend to use radians then check the instructions for a degree mode.  Computing languages also vary but radians are common.  An option for degrees is less common but it can be handled.  If $\theta$ is your angle in degrees then use $\sin(\theta  \pi / 180)$.
As Doug M explains , $1$radian is a quite large and fairly arbitrary amount, around $57.3$ degrees.  He explains the consequences well.
If you switch to degrees then it will look quite different.  $45$ is quite special in degrees and its $\sin$ is $\frac{\sqrt(2)}{2}$ which is approximately $0.7071$.  $1$ degree is fairly small.  If you calculate $\sin$ for $45, 46, 47, ...$ then the pattern should be clearer.  It will grow approximately at a constant rate but gradually slowing until it reaches $90$ when it will reach its maximum of $1$ and then it will start to fall.
Alternatively, go back to radians but use a smaller step e.g. $0.01, 0.02, 0.03, ...$.  Again, you will see a pattern.


TITLE: Sum of infinitely many i.i.d. random variables is infinite with probability 1
QUESTION [3 upvotes]: How do I solve this? I'm really confused.

If $X_1,X_2,\ldots$ are non-negative independently and identically distributed random variables with $P(X_i>0)>0$, show that $\displaystyle P\left(\sum_{i=1}^\infty X_i=\infty\right)=1$.

REPLY [2 votes]: Hint: It suffices to show for each $n$ that $$P\bigg\{\omega:\sum_{i=1}^\infty X_i(\omega)>n\bigg\}=1,$$ which is the case iff for all $0<\delta<1$
$$P\bigg\{\omega:\sum_{i=1}^\infty X_i(\omega)>n\bigg\}>\delta.$$ There is $\epsilon>0$ such that $P(X_i>\epsilon)>0$.
Now for each $m$ we have $$P\bigg\{\omega:\sum_{i=1}^m X_i(\omega)>n\bigg\}\leq P\bigg\{\omega:\sum_{i=1}^\infty X_i(\omega)>n\bigg\},$$ so it suffices to show, using the $\epsilon$, that for some $m$ large enough, $$P\bigg\{\omega:\sum_{i=1}^m X_i(\omega)>n\bigg\}>\delta,$$
which is an elementary argument. Try it.


TITLE: $v^2 = 2ax$ or $v^2 = ax$?
QUESTION [4 upvotes]: As far as I am aware, $v^2 = 2ax$ is the formula to find the velocity in various questions.
If kinetic energy = work,
$$\frac{1}{2}mv^2=Fx$$
$$mv^2=2max$$
$$v^2=2ax$$
We use this formula to solve some questions in school.
But when i just fiddle around with basic formulas i get this.
$$x/v = t$$
$$v/a = t$$
$$t = t$$
$$x/v = v/a$$
$$v^2 = ax$$
And this just confuses me. I assume that $x/v=t$ and $v/a=t$ is actually kind of simplified or else I cannot see why $v^2$ equals $ax$ on one and $2ax$ on the other.
Can someone explain to me what am I doing wrong?

REPLY [6 votes]: In your second derivation, the correct formulas are
$$\begin{align}
\frac{\Delta x}{v} &\approx \Delta t &
\frac{\Delta v}{a} &\approx \Delta t
\end{align}$$
I'm sure you can easily find some examples to show you why $x/v = t$ and $v/a = t$ don't make any sense. Anyway, when you put these together, you get $v\Delta v \approx a\Delta x$, with the approximation becoming more accurate the smaller the $\Delta$s are.
Note that if you take the limit as $\Delta v$ and $\Delta x$ go to zero, then integrate, you get
$$\begin{align}
\int_{v_i}^{v_f} v\,\mathrm{d}v &= \int_{x_i}^{x_f} a\,\mathrm{d}x \\
\frac{v_f^2 - v_i^2}{2} &= a(x_f - x_i) \\
v_f^2 &= v_i^2 + 2a\Delta x
\end{align}$$
which is exactly the correct formula.


TITLE: What's the efficient method to find the farthest vertex from centroid
QUESTION [0 upvotes]: Say I have a arbitrary convex polygon, what I wonder is the longest length from its centroid to its vertex, and which vertex it is.
I've looked it up on Wikipedia finding that I have to calculate its area firstly, then there's a formula determine the coordinates of centroid. I can then calculate all the lengths. Is there any other way faster?

REPLY [0 votes]: Actually the area of a convex polygon is computationally quite easy: it is the sum of the (oriented) areas of all the triangles formed by the origin of the coordinate system and two adjacent vertices (in a fixed order around the circumference of the polygon).
The double of the oriented area of such an individual triangle is $A_i=v_{i,x}v_{i+1,y}-v_{i,y}v_{i+1,x}.$ You will also need the individual $A_i,$ not just their sum, in your formula for the centroid.
The idea is that you can apply the "subdivision" from Wikipedia using any point of the plane as the third vertex of your triangles, even if it lies outside the polygon (as the origin of the coordinate system might), as long as you use oriented areas. Remember that oriented areas may be negative.
The centroid of an individual triangle is also easily computed as (two thirds of) $v_i+v_{i+1}.$
This means that your computation has to loop through the vertices once to calculate the $x$ and $y$ coordinates of the centroid of the polygon, and one more time to find the vertex (or vertices) with minimal distance.


TITLE: Proving that a graph is connected?
QUESTION [3 upvotes]: I'm trying to prove that this graph is connected given the provided information.
Let $G$ be a simple undirected graph with $n \geq 2$ vertices. Prove that if $δ(G) \geq \frac{n}{2}$, then $G$ is connected.
I can see from testing a few examples that it's definitely true. As for the actual proof, I'm stuck:
If we have $n$ vertices, then we have at most $\frac{n(n-1)}{2}$ edges. However, I'm still not seeing what the minimum vertex degree has to do with the number of edges at any other vertices, and especially in showing that we have enough edges to guarantee connectedness.
I don't feel like I'm approaching this problem the right way. If anyone could help, that would be much appreciated.

REPLY [2 votes]: HINT: Let $u$ and $v$ be any two vertices. Let $N(u)$ be the vertex $u$ together with its neighbors, and define $N(v)$ similarly. Is it possible that $N(u)\cap N(v)=\varnothing$?

REPLY [1 votes]: By contradiction: suppose the graph is not connected, then it has at the very least $2$ connected components, so the size of the smallest component is at most $\frac{n}{2}$ Meaning the degree of a vertex in that component is at most $\frac{n}{2}-1$ meaning the minimum degree is at most $\frac{n}{2}-1$. Thus a graph with minimum degree at least $\frac{n}{2}$ must be connected.


TITLE: Solving Equation of form: *num* = $x^3 + ax^2 + bx + c$.
QUESTION [1 upvotes]: The solution to the following:
$$
45113 = x^3 + ax^2 + bx + c
$$
is:
$$
45113 = 31^3 + 15 · 31^2 + 29 · 31 + 8
$$
My question is how does one go about solving this set of equation, I'd
like to write a program that solves this but need to know what kind of equation I'm looking at and what steps are involved at getting to the
solution. I'm familiar with Linear Diophantine and Chinese Remainder Theorem solving but not sure where, to begin with, this one. I just need to find a guide or know how to step by step solve this equation, but google searches aren't coming up with anything.
Any papers or guides on how to do this would be helpful, thank you.
Ok, here is how to solve, choose a value below the cube root of 45113. Then perform the following steps:
If there are restrictions on x apply them in selecting the integer x. For example select a positive integer less than the cube root of 45113, ie ≤ 35, for x.

Subtract x³ from 45113
Divide the result by x². The integer result is the coefficient “a” .
Subtract ax² from the result of step 1.
Divide this amount by x. The integer amount is the coefficient “b” .
Subtract bx from the result of step 3. The new result is “c” .
This will work, no matter the integer value of x.

REPLY [4 votes]: You have some degrees of freedom here. Assuming you want a solution or some solutions and not all solutions, I would do something like this based on Vieta's formulas
$$\begin{align*} 0 &= x^3+ax^2+bx+(c-n) \\
\\
&= x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_2r_3+r_1r_3)x-r_1r_2r_3
\end{align*}$$

Choose a $c$. $ (n-c)=r_1r_2r_3$ is the product of the roots of the cubic equation.
Factor an integer factor out of $n-c$, and call that root $r_1$.
Choose the 2 other roots, $r_2$ and $r_3$, such that $r_2r_3 =\dfrac{n-c}{r_1}$.  They need not be integers, and they can be complex conjugates if you want.  They should sum to an integer, and things are easier to manage if the real parts of their individual products with $r_1$ are also integers.
Set the values of $a = -(r_1+r_2+r_3)$, and $b=(r_1r_2+r_2r_3+r_1r_3)$ from Vieta's formulas.

Now you should have integer $c, r_1, a, b$ so you can write
$$n=r_1^3+ar_1^2+br_1+c$$


TITLE: Proving zero identities in vector calculus with simple arguments involving index counting or symmetry?
QUESTION [1 upvotes]: Consider the following table describing four second derivative operators.
             L   R
curl grad    2   1   vanishes identically
curl curl    3   1
div curl     3   0   vanishes identically
div grad     2   0

The idea of the $L$ and $R$ numbers is that they're counts of the number of indices you would expect to see on the left and right sides of an expression involving this operator, if it was expressed in index notation, with the following rules:

Because indices can be contracted, all numbers are to be interpreted modulo 2, but in the above table I've just counted the raw number of indices without contraction.
We ignore the fact that a curl requires a Levi-Civita symbol with three additional indices.

For example, if we take the equation $\operatorname{div}\operatorname{grad}f = g$ and translate it into index notation, we have
$$\nabla_i \nabla^i f = g.$$
There are two indices on the left, so $L=2$, and no indices on the right, $R=0$.
So in these four examples, I make the following observation: the derivative operator vanishes identically iff $L\ne R\ (\text{mod}\ 2)$. This rule would seem to make a lot of sense, and would probably convince me as a proof of the two zero identities, if the curl really was a tensor. But in reality the curl is not a tensor with one index, it's a tensor contracted with a Levi-Civita symbol, and because the Levi-Civita symbol has an odd number of indices in three dimensions, this messes up the parity of $L$.
I have two questions:
(1) Can this argument be patched up somehow?
(2) Can this type of argument be converted into a more elementary argument about symmetry, which could be presented in a way that was understandable to sophomore-level students who didn't know about tensors?

REPLY [0 votes]: This is not really a complete answer to my own question, only to part 2 for one particular zero identity, but it's too long for a comment, so I thought I'd post it here and see if it attracts any comments. I'm interested in whether it seems correct, and whether it can be simplified, clarified, or generalized to apply to more than one zero identity.
As a possible step toward the desired generalization, it's interesting to note that $\operatorname{curl}\operatorname{grad}$ and $\operatorname{div}\operatorname{curl}$ are identical when written in index notation. They differ only in that in $\operatorname{div}\operatorname{curl}$ we apply it to a field by contracting one index.
Argument that the divergence of a curl is zero
The kind of field that looks like it would have a nonvanishing value of this operation is something like $\textbf{F}=zx\hat{\textbf{y}}$, the idea being that we take a field like $x\hat{\textbf{y}}$ that has a curl in the $z$ direction, and then we give it some $z$ dependence so that there could be a divergence. In fact, any field that is differentiable near the origin can have its components approximated in that neighborhood by a function of this general form (a second-order mixed polynomial), so if we can prove that $\operatorname{div}(\operatorname{curl}\textbf{F})=0$ for this particular $\textbf{F}$, it follows that the div of a curl is zero for any $\textbf{F}$.
Now we show based on symmetry that $\operatorname{div}(\operatorname{curl}\textbf{F})=0$ for this $\textbf{F}$. Suppose we rotate our coordinate system by 180 degrees about the $y$ axis. This doesn't change the field or its components, but the curl lies in the $x$-$z$ plane, so the output of the curl operation has its components sign-flipped because of the new coordinate system used to describe it. This is an over-all reversal of the curl's direction, and since the div is linear, the effect is to sign-flip $\operatorname{div}(\operatorname{curl}\textbf{F})$.
But the divergence is a scalar, so the final result of taking $\operatorname{div}(\operatorname{curl}\textbf{F})$ cannot change just because we change our coordinates.
We have proved that $\operatorname{div}(\operatorname{curl}\textbf{F})$ flips its sign, but also that it doesn't change its sign. This is only possible if it is zero, as claimed.


TITLE: Verification of the integral: $\int \frac{dx}{a\cos^2x+b\sin(2x)+c\sin^2x}$
QUESTION [2 upvotes]: I have solved several integrals (of the form below) with given coefficients for $a, b, c$ and then though whether I could generalize that or not. I would like to ask for verification. Problem statement:

Evaluate:
  $$
\int \frac{dx}{a\cos^2x+b\sin(2x)+c\sin^2x}
$$

Below is what I got so far. I've started by factoring out $1\over \cos^2x$ to obtain:
$$
I = \int \frac{1}{\cos^2x}\cdot\frac{dx}{a + 2b\tan x + c\tan^2x}
$$
Now substitute $t = \tan x$:
$$
I = \int \frac{dt}{a + 2bt + ct^2}
$$
Now depending on the coefficients, we might have three cases: either there are 2 distinct real roots, both roots are equal, or there are no roots in $\Bbb R$.
If there are no roots, then we could complete the square and consider a standard integral. Suppose:
$$
ct^2 + 2bt + a = c(t-h)^2 + k \\
ct^2 + 2bt + a = ct^2 + 2ch t+ch^2 + k
$$
After solving a linear system of equations one may obtain:
$$
h = {b\over c}\\
k = {ac + b^2\over c}
$$
Hence the integral becomes:
$$
I = {1\over c}\int \frac{dt}{\left(t - {b\over c}\right)^2 + {ac + b^2 \over c^2}}
$$
Which is a standard integral:
$$
\boxed{I = {1\over \sqrt{ac + b^2}}\arctan\left(\frac{c\tan x + b}{\sqrt{ac-b^2}}\right) + Const}
$$

As for the second case, namely when there are two distinct roots. After solving  a quadratic equation:
$$
t_{1,2} = \frac{-b \pm \sqrt{b^2 - ac}}{c}
$$
Let:
$$
R_1 = \frac{-b + \sqrt{b^2 - ac}}{c}\\
R_2 = \frac{-b - \sqrt{b^2 - ac}}{c}
$$
So the integral becomes:
$$
I = \int \frac{dt}{(t-R_1)(t-R_2)}
$$
Using partial fraction decomposition:
$$
\frac{1}{ct^2 + 2bt + a} = \frac{A}{t - R_1} + \frac{B}{t-R_2} \\
At - AR_2 + Bt - BR_1 = 1\\
\begin{cases}
A + B = 0\\
-AR_2-BR_1 = 1
\end{cases}\\
A = -B\\
B = {1\over (R_2-R_1)}
$$
Therefore:
$$
I = \int \frac{1}{(R_1 - R_2)(t - R_1)}dt - \int \frac{1}{(R_1-R_2)(t-R_2)}dt
$$
Which yields:
$$
\begin{align}
I &= {1\over R_1 - R_2}\ln|t - R_1| - {1\over R_1 - R_2}\ln|t - R_2| \\
&={1\over R_1 - R_2}\ln|\tan x - R_1| - {1\over R_1 - R_2}\ln|\tan x - R_2| \\
&=\boxed{{1\over R_1 - R_2}\ln\left|{\tan x - R_1\over \tan x - R_2}\right| + Const}
\end{align}
$$

When there are two equal roots, then let the root be denoted by $R_1 = R_2 = R$:
$$
I = \int {dt\over (t - R)^2} = -\frac{1}{t - R}
$$
Or simply:
$$
\boxed{I = -\frac{1}{\tan x - R} = \frac{1}{R-\tan x} + Const}
$$

I'm wondering whether I've done this right, could someone please verify the above? Thank you!

REPLY [3 votes]: Taking into account the small corrections in the comments, this is correct for the cases with real roots. The case with no real roots has a subtle error that often gets glossed over in these sorts of problems.
The issue is that when there are no real roots, $[a \cos(x)^2 + b\sin(2x) + c \sin(x)^2]^{-1}$ is continuous everywhere. This means its antiderivative must be everywhere differentiable, but what you have for its antiderivative is not, since $\tan^{-1}$ jumps down by $\pi$ every time its argument switches from $+\infty$ to $-\infty$ (or up if it's switching the other way because $c < 0$). To restore continuity, we need to add (or subtract, if $c < 0$) $\pi$ to it every time $x$ passes an odd multiple of $\pi/2$. This can be done with the floor function, giving
$$
I = \frac{1}{\sqrt{ac - b^2}}\left(\tan^{-1}\left[\frac{c\tan(x)+b}{\sqrt{ac-b^2}}\right] +\pi\left\lfloor\frac{x}{\pi}+\frac{1}{2}\right\rfloor\operatorname{sgn}[c]\right) + Const,
$$
where $\operatorname{sgn}$ is the sign function. As you can see in this plot for the $a=2,\,b=c=1$ case, the extra term restores continuity of the antiderivative.


TITLE: Maximal torus of linear algebraic group over a ring
QUESTION [4 upvotes]: Let $G$ be a linear algebraic group over a $k$-algebra $A$, where $k$ is an algebraically closed field.
Consider the structure morphism $G\rightarrow U={\rm Spec}(A)$.
Assume that for every $k$-point of $A$, the algebraic group $G_k$ has a maximal $r$-dimensional $k$-torus.

Question. Does it follow that $G$ has a maximal $r$-dimensional $A^\times$-torus?

REPLY [5 votes]: The following is a counterexample. Let $S = \operatorname{Spec} k[[t]]$ and $G \to S$ an affine group scheme with special fiber $\mathbf{G}_a$ and generic fiber $\mathbf{G}_m$. The special fiber contains a maximal torus (namely the zero-dimensional torus). But now a zero-dimensional torus in $G$ cannot possibly be maximal because over the generic fiber it is strictly contained in $\mathbf{G}_m$.
Remark: Sean Cotner has written up a nice article that gives a description of such $G \to S$ by taking centralizers of certain elements in $\text{SL}_{2,S}$.


TITLE: Parities of a permutation
QUESTION [0 upvotes]: Is there in general any difference between a cyclic and an even permutation? Or can they be used interchangeably. Is an acyclic permutation the same as an odd permutation?

REPLY [0 votes]: Is there in general any difference between a cyclic and an even permutation?

Yes.
The permutation $(12)(34)$ is even but not cyclic. The permutation $(1234)$ is cyclic but odd.


TITLE: Why didn't heavier elements settle at the core of the solar system?
QUESTION [1 upvotes]: As the solar system formed, why didn't all of the heavier elements such as iron, collect where the sun is leaving the lighter elements in the outer solar system?

REPLY [3 votes]: Most of the iron and heavy elements in the solar system are in the Sun.
They aren't all in the Sun, because the material from which the solar system formed was well-mixed and turbulent.
The reason that the concentration of heavier elements is higher in the small, rocky, inner planets (the gas giants have an abundance mixture similar to the Sun), is that firstly, only heavier, "refractory" elements could condense into solids in the hotter inner parts of the protosolar nebula; and secondly, once these had clumped to form protoplanets, they were unable to capture and hold onto the vast amounts of hydrogen and helium in the early protoplanetary nebula, because of their low gravities and high temperatures.
The sinking/diffusion you refer to requires, stratified, stable fluids. This condition was not satisfied in the proto-solar-system, but was satisfied in the early history of the planets; hence the ample evidence for chemical differentiation in most of them, with heavier elements sinking towards the middle.


TITLE: Find $\Pi^{7}_{k=1}(z-e^{-k\pi i})$
QUESTION [1 upvotes]: Find $$\Pi^{7}_{k=1}(z-e^{-k\pi i})$$
Calculating by hands seems possible but it requires a lot of work. Is there any tricks? Periodicity helps?

REPLY [4 votes]: As you noted, periodicity is very helpful. $e^{-k\pi i} = e^{\pi i} = -1$ for odd $k$ and $e^{-k\pi i} = e^{2\pi i} = 1$ for even $k$. Your expression then simplifies to
$$(z+1)^4(z-1)^3 = (z^2 - 1)^3(z+1)$$

REPLY [3 votes]: Using Euler's formula, $e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1,$ so $(e^{i\pi})^{2r+1}=-1$ and $(e^{i\pi})^{2r}=1$
$1\le k\le 2n+1 $ contains $n+1$ odd ,$n$ even values of $k$.
So, $$\Pi^{2n+1}_{k=1}(z-e^{-k\pi i})=\{z-(-1)\}^{n+1}(z-1)^n=(z+1)(z^2-1)^n$$
Here $n=3$.
$1\le k\le 2n $ contains $n$ odd ,$n$ even values of $k$.
So, $$\Pi^{2n}_{k=1}(z-e^{-k\pi i})=\{z-(-1)\}^n(z-1)^n=(z^2-1)^n$$