text:
\section{Physics context and summary of main results}
\label{sec:introduction}

\subsection{Feynman graphs and integrals}

Quantum Field Theory is among the main frameworks of the physics of our time, and the backbone of all computational techniques to compare theory and experiments in high-energy physics. The interactions among the quantum fields and states are encoded by correlation functions and on-shell scattering amplitudes. While it is not  in general  possible to compute these quantities exactly,  they can be expanded into a perturbative series in cases where the theory contains a small parameter.

The perturbative series can be neatly organised in terms of Feynman graphs, and the $L$-th order in the perturbative expansion receives contributions from Feynman graphs with $L$ loops. The precise definition of a Feynman graph is not important for the purposes of this paper (see, for example,~\cite{Brown:2015fyf}). Here it suffices to say that to each Feynman graph one can associate a \emph{Feynman integral}, which depends on the dimension $d$ of space-time and is a function of the external kinematic data (e.g., the masses and momenta of all external particles).

Whenever it converges, a Feynman integral
defines a family of periods depending on kinematic parameters that is a generalisation of the notion of period in the sense of Kontsevich and Zagier~\cite{PeriodsZagier}. For algebraic values of the masses and momenta it is exactly a period in their  sense.  One can show that Feynman integrals can be promoted (at least when the masses and momenta are generic and $d$ is even) to `motivic periods' of the cohomology of a family of algebraic varieties  \cite{Brown:2015fyf}. The same is almost certainly true in all cases.

\begin{rem}
Feynman integrals are often divergent and need to be  regularised. While various different regularisations exist, the most commonly used regularisation in physics is \emph{Dimensional Regularisation}~\cite{tHooft:1972tcz,Bollini:1972ui,Cicuta:1972jf}. Loosely speaking, it consists in replacing the space-time dimension $d$ by $D=d-2\eps$, where $\eps$ is a variable taking values in $\mathbb{C}$. One obtains in this way a meromorphic function of $\eps$~\cite{speer},
which admits a Laurent expansion around $\eps=0$. The objects of interest are the Laurent coefficients, which are then also families of  periods in the spirit of Kontsevich and Zagier~\cite{Bogner:2007mn}. In applications one is only interested in  the first few terms in the Laurent expansion, because only a finite number of Laurent coefficients contribute to the physical observable of interest (which must be finite and independent of the chosen regularisation).
\end{rem}

\subsection{Iterated integrals and multiple polylogarithms}
It is known that large classes of Feynman integrals can be evaluated in terms of iterated integrals. Let  $X$ be a smooth $m$-dimensional complex manifold. Let $\gamma:[0,1]\to X$ be a piecewise smooth path on $X$, and let $\omega_1,\ldots,\omega_n$ be smooth  one-forms on $X$. We denote the pullback of $\omega_i$ to the interval $[0,1]$ by $dt\,f_i(t) = \gamma^\ast\omega_i$. The iterated integral of the forms $\omega_1,\ldots,\omega_n$ along $\gamma$ is defined as
\beq\bsp
\int_{\gamma}&\,\omega_1\cdots\omega_n :=\int_{0\le t_1\le t_2\le \cdots\le t_n\le 1}\gamma^{\ast}\omega_1\wedge\ldots\wedge \gamma^{\ast}\omega_n\\
&= \int_{0\le t_1\le t_2\le \cdots\le t_n\le 1}dt_1\,f_1(t_1)\cdots dt_n\,f_n(t_n)\\
&=\int_0^1dt_n\,f_n(t_n)\int_0^{t_n}dt_{n-1}\cdots\int_0^{t_2}dt_1\,f_1(t_1)\,.
\esp\eeq
More generally, an iterated integral is any linear combination of such integrals.  The
empty iterated integral (when $n = 0$) is defined to be the constant function 1.
We will only be interested in homotopy-invariant iterated integrals, i.e., linear combinations that only depend on the homotopy class of the path $\gamma$ in $X$. All iterated integrals that appear in the computation of Feynman integrals are of this type, where $X$ denotes the complex points of a smooth algebraic variety over $\overline{\Q}$,  and all iterated integrals are $\overline{\Q}$-linear combinations of integrals of  forms $\omega_i$ which are globally defined logarithmic forms on $X$ which are defined over $\overline{\Q}$.
To spell this out in more detail, Feynman integrals  typically give rise to homotopy-invariant iterated integrals of forms $\omega_i $ which are holomorphic or of the form $d\log f$ where $f$ is a rational function on $X$. Frequently,  $X$ is an open subset of (a finite covering of)  an affine  space with coordinates $x_1,\ldots, x_n$, and one  can write
$\omega_i = d\log R_i(x_1,\ldots,x_m)$, where $R_i(x_1,\ldots,x_m)$ is a rational (or algebraic) function.

A particularly important representative of iterated integrals of $d\log$-forms are \emph{multiple polylogarithms} (also known as hyperlogarithms), which were first introduced in the works of Poincar\'e, Kummer and Lappo-Danilevsky~\cite{Kummer,Lappo:1927} and have recently reappeared in both mathematics~\cite{GoncharovMixedTate,Goncharov:1998kja,Brown:2011ik} and physics~\cite{Remiddi:1999ew,Gehrmann:2000zt,Ablinger:2011te}. Multiple polylogarithms can be defined as
\beq\label{eq:G_def}
I(a_1,\ldots,a_n;z) = \int_0^z\frac{dt}{t-a_1}\,I(a_2,\ldots,a_n;t)\,,
\eeq
where $a_i$, $z\in\mathbb{C}$, and the recursion starts at $I(;z)=1$. If $a_n=0$ the integral in~\eqref{eq:G_def} diverges, and we define instead
\beq
I(\underbrace{0,\ldots,0}_{n\textrm{ times}};z) = \frac{1}{n!}\log^nz\,.
\eeq
Multiple polylogarithms are well-studied in mathematics and in physics. In particular, it is well understood how to perform algebraic manipulations of multiple polylogarithms.
In addition, there are several fast numerical implementations of these functions that can be used for their evaluation at high precision~\cite{Gehrmann:2001pz,Gehrmann:2001jv,Vollinga:2004sn,Buehler:2011ev,Frellesvig:2016ske,Ablinger:2018sat,Naterop:2019xaf}. Given the good algebraic and numerical control one has over multiple polylogarithms, it is often desirable to express Feynman integrals, scattering amplitudes and correlation functions in terms of multiple polylogarithms whenever possible.

If an iterated integral is homotopy-invariant and the functions $R_i(x_1,\ldots,x_m)$ are rational functions of the variables $x_i$ (with coefficients in $\overline{\mathbb{Q}}$ say), and if the base point of the integration path is algebraic, then one can always write the integral in terms of multiple polylogarithms evaluated at algebraic arguments. Indeed, we can use homotopy-invariance and replace the path of integration by a homotopic path along the edges of a hypercube where all but one of the variables are constant.
However, if the $R_i(x_1,\ldots,x_m)$ are not rational, no such algorithm exists. Nonetheless, many examples of iterated integrals of $d\log$-forms with non-rational arguments that have appeared in physics can be evaluated in terms of multiple polylogarithms (see, e.g.,~\cite{Heller:2019gkq}). This has led to folklore conjectures in the physics community that \emph{every} (iterated) integral of $d\log$-forms with algebraic arguments can be expressed in terms of multiple polylogarithms evaluated at algebraic points (at least in principle). The purpose of this paper is to show that this is false by  providing an explicit example of a double iterated integral of $d\log$-forms which cannot be expressed in terms of any linear combination of multiple polylogarithms evaluated at algebraic arguments (assuming the standard period conjecture). We highlight the implications for quantum field theory below, after a brief technical summary of our results.

\subsection{Summary of results}
Let $\rho = -e^{2\pi i/3} = e^{-i\pi/3}$, and $\bar{\rho}$ denote its complex conjugate.
In \S\ref{sec:two_periods} we consider the iterated integrals:
\begin{align}
\label{eq:IE_def}
I_{\E} & = 2 \, \mathrm{Re} \int_{-1\leq x_1 \leq x_2\leq \infty}  \frac{1}{\rho- \overline{\rho}}   \,  d\log \left(\frac{x_1-\rho }{x_1- \overline{\rho}
  } \right)  \wedge  d\log\left(\frac{\sqrt{1+x_2^3}+1}{\sqrt{1+x_2^3}-1}\right)\,,\\
\label{eq:I_def}
I&=\int_{2\le x_1\le x_2\le\infty}\frac{1}{\rho-\bar{\rho}}\,d\log\left(\frac{x_1-\rho}{x_1-\bar{\rho}}\right)\wedge d\log\left(\frac{\sqrt{1+x_2^3}+1}{\sqrt{1+x_2^3}-1}\right)\,.
\end{align}

In order to interpret these integrals geometrically, consider the algebraic curve in $\mathbb{P}^2$ defined by the equation
\beq\label{eq:E_equation}
zy^2 = z^3+x^3\ .
\eeq
It defines an elliptic curve $\E$, so in particular  it is not possible to find any change of variables such that the argument of the logarithm in the integrands in~\eqref{eq:IE_def} and~\eqref{eq:I_def} becomes rational.  By constructing the underlying `motives' of these integrals (we shall use the word `motive' loosely to mean an object in a category of realisations which arises from the cohomology of an algebraic variety) and proving that they contain a non-trivial mixed elliptic extension, we prove in Corollary~\ref{cor:main_1} and~\ref{cor:main_2}  that the motivic versions $I_{\E}^{\mathfrak{m}}$ and $I^{\mathfrak{m}}$ of these integrals are algebraically independent from all motivic polylogarithms at algebraic points. It then follows from a version of  Grothendieck's period conjecture that $I_{\E}=\per(I_{\E}^{\mathfrak{m}})$ and $I=\per(I^{\mathfrak{m}})$ cannot be expressed in terms of multiple polylogarithms evaluated at any algebraic argument.

The obstruction to being polylogarithmic is the same for both $I_{\E}^{\mathfrak{m}}$ and $I^{\mathfrak{m}}$. More precisely, we show in \S\ref{sec:I_pol} that there is a linear combination $I^{\mm}_{\mathrm{Pol}}$ of $I_{\E}^{\mathfrak{m}}$ and $I^{\mathfrak{m}}$:
\beq\label{eq:IPol}
I^{\mm} = \frac{1}{6}\, I^{\mm}_{\mathcal{E}} +   I^{\mm}_{\mathrm{Pol}}\,,
\eeq
where $I^{\mm}_{\mathrm{Pol}}$ is a motivic period of a mixed Artin-Tate object, which numerically evaluates to a linear combination of  dilogarithms and logarithms. The obstruction itself is an extension of $H^1(\mathcal{E})$ by a certain Dirichlet motive $\Q_{\chi}(-2)$. The non-triviality of this extension is precisely detected  by the non-vanishing of the integral $I_{\E}$.   Furthermore, Beilinson's conjecture~\cite{Bec,Be} then predicts that $I_{\E}$, which is essentially the regulator, is proportional to the (non-critical) value at $2$ of the $L$ function of $\mathcal{E}$. Indeed, this is what we find numerically, and  could almost certainly be proven rigorously using the theory of iterated integrals of Eisenstein series (i.e., multiple modular values) as we now explain.

 A key point is that $\E$ defined by~\eqref{eq:E_equation} admits a modular parametrisation. Let $\Gamma(N)\subset \SL_2(\mathbb{Z})$ be the principal congruence subgroup of level $N$, $\mathfrak{H} = \{\tau\in\mathbb{C} : \textrm{Im }\tau>0\}$ the complex upper half-plane, and $Y(N)$  the modular curve obtained by taking the (orbifold) quotient of $\mathfrak{H}$ by the usual action of $\Gamma(N)$ via M\"obius transformations. There is an isomorphism $\varphi: Y(6) \to \E\backslash\mathcal{C}$, where $\mathcal{C}$ denotes a finite set of points.  The pullback $\varphi^* \omega$ of a logarithmic differential form $\omega$  on $\E$ with poles along $\mathcal{C}$ can be identified with a modular form of weight two for $\Gamma(6)$. The  holomorphic differential on $\E$ pulls back to the unique (normalised) cusp form of weight two, whereas $d\log$-forms pull back to linear combinations of Eisenstein series and this cusp form. As a result the integral $I$ in~\eqref{eq:I_def} can be expressed  (\S\ref{sec:modular}) as  a   double iterated integral
\beq\label{eq:I_Eisenstein}
I=\frac{1}{\rho-\bar{\rho}}\int_{0\le t_1\le t_2\le\infty}\frac{dt_1\wedge dt_2}{(2\pi)^2}\,E_1(it_1)\,E_2(it_2)\,,
\eeq
where $E_1(\tau)$ and $E_2(\tau)$ are certain  Eisenstein series of weight two for $\Gamma(6)$. In~\cite{Brown:mmv} it was shown that double iterated integrals of Eisenstein series of small weight for the full modular group $\Gamma(1)$ evaluate to multiple zeta values,  and periods of simple extensions of motives of cusp forms for $\Gamma(1)$. The latter include  non-critical  $L$-values  of cusp forms (amongst other quantities) and  first   appear when the sum of the modular weights of the two Eisenstein series is twelve, because the first cusp form for $\Gamma(1)$ has weight twelve. Since $\Gamma(6)$ has genus one, the first cusp form already appears in weight two:
\beq
f(\tau) = \eta(\tau)^4\,,
\eeq
where $\eta(\tau)$ is the Dedekind $\eta$ function.
 By the general theory, we therefore expect the double iterated integral in~\eqref{eq:I_Eisenstein} to evaluate to a linear combination of multiple polylogarithms evaluated at sixth roots of unity and the value at $2$ of the completed $L$ function of the cusp form $f$:
 \beq
\Lambda(f,2) = \int_{i\infty}^0d\tau\,f(\tau)\,\tau = 0.85718907492991773071685111\ldots\,
\eeq
Using the PSLQ algorithm, we find (with $I_{\mathrm{Pol}} = \per(I^{\mm}_{\mathrm{Pol}})$):
\begin{align}
I_{\E} &= -4 \, \pi \sqrt{3}\,  \Lambda(f,2)\,,\\
I_{\mathrm{Pol}} &= \frac{5}{\sqrt{3}}\,\textrm{Cl}_2\left(\frac{\pi}{3}\right)\,,
\end{align}
where $\textrm{Cl}_2\left(\frac{\pi}{3}\right) = \textrm{Im }\textrm{Li}_2(e^{i\pi/3})$. These evaluations could be proven rigorously with a more detailed analysis: the first should follow from an application of a version of the Rankin-Selberg method to double Eisenstein integrals, as was done in \cite{Brown:mmv2} in level $1$. The second could be proven using, for example,  the theory of multiple elliptic polylogarithms and unipotent completion of the motivic fundamental group of the universal elliptic curve. Since these computations are quite lengthy and technical and are not required for the main  point of this paper, they are not presented here.

\subsection{Implications for Quantum Field Theory}
The integrals $I_{\E}$ and $I$ are explicit examples of integrals of $d\log$-forms that cannot be evaluated in terms of multiple polylogarithms at algebraic points. Here we discuss some implications  for perturbative Quantum Field Theory, because these integrals are concrete counter-examples to certain folklore  beliefs in the community stating that  all integrals of $d\log$-forms evaluate to polylogarithms.

{\bf The role of the integration cycle.} An integrand that can be written in $d\log$-form with algebraic arguments is  insufficient for an integral to evaluate to multiple polylogarithms. Whether an integral evaluates to polylogarithms is not determined by the integrand alone, but also the integration cycle, which plays an important role. To illustrate this point, let us return to~\eqref{eq:IPol}:  whilst $I^{\mm}_{\mathcal{E}}$ and $I^{\mm}$ are examples of motivic periods  which  are algebraically independent from motivic polylogarithms, the period $\mathrm{per}\,(I^{\mm}_{\mathrm{Pol}})$ does evaluate to multiple polylogarithms, even though the periods of all three objects are integrals involving the same
$d\log$-form in the integrand.\footnote{The fact that we only consider the real part of $I_{\mathcal{E}}$ immaterial here. See \S\ref{sec:Sec4} for the precise definition of the integration cycles used to compute $I_{\mathcal{E}}$ and $I$.} Thus, looking at the integrand alone is insufficient to decide if an integral can be evaluated in terms of multiple polylogarithms, and the choice of the integration cycle is  important.

{\bf Canonical differential equations in \emph{d}log-form.} It follows from our examples that it is not clear that families of (dimensionally-regularised) Feynman integrals that satisfy a system of linear differential equations in `canonical $d\log$-form' (cf.~\cite{Henn:2013pwa}) can always be evaluated in terms of polylogarithms.  Instead, other classes of iterated integrals -- such as iterated integrals of modular forms or multiple elliptic polylogarithms -- may also show up even in the case of a differential equation in `canonical $d\log$-form'. This can happen whenever the $d\log$-forms involve algebraic arguments that cannot be rationalised via a suitable parametrisation of the external kinematic data (see, e.g.,~\cite{Besier:2018jen,Besier:2019kco} for a review). This situation is known to occur for example in Feynman integrals contributing to two-loop QED corrections to Bhabbha scattering~\cite{Henn:2013woa,Festi:2018qip} as well as for the two-loop mixed QCD-QED corrections to the Drell-Yan process~\cite{Bonciani:2016ypc,Besier:2019hqd}. In~\cite{Heller:2019gkq} it was shown that in four space-time dimensions these results can be expressed in terms of multiple polylogarithms depending on complicated algebraic functions of the external kinematic data. Our examples from the previous section show that this is not the rule, and there is no reason why the same should be true for other integrals that involve $d\log$-forms depending on square roots that cannot be rationalised, nor is there a reason why it should be true for the higher orders in the Laurent expansion in the dimensional regulator $\eps$ for the integrals considered in~\cite{Henn:2013woa,Bonciani:2016ypc,Heller:2019gkq}.

{\bf Planar $\mathcal{N}$ = 4 super Yang--Mills theory.} The examples of the previous section may also have implications on conjectures about the analytic structure of certain special Quantum Field Theories, like for example the planar $\cN=4$ super Yang-Mills theory. In~\cite{Arkani-Hamed:2016byb} it was argued that scattering amplitudes in this theory for certain assignments of the quantum numbers of the external states -- the so-called \emph{maximally-helicity-violating (MHV)} and \emph{next-to-MHV (NMHV)} amplitudes -- can be expressed in terms of polylogarithms for any number of loops and external particles. A central point in the argument is a (conjectural) procedure to write the loop integrand of these amplitudes in a form which only involves $d\log$-forms. For up to seven external particles, the arguments of these $d\log$-forms are rational functions obtained from cluster algebras (of finite type) associated to certain Grassmannian spaces, see, e.g.,~\cite{Golden:2013xva,Golden:2014xqa,Drummond:2017ssj,Drummond:2018dfd,Drummond:2019cxm,Golden:2019kks,Mago:2019waa,Arkani-Hamed:2019rds,Henke:2019hve,Arkani-Hamed:2020cig}.
It is also known that two-loop MHV amplitudes for any number of external particles can be expressed in terms of polylogarithms~\cite{CaronHuot:2010ek,Golden:2014xqa}.
Starting from eight external particles, however, two-loop NMHV and three-loop MHV amplitudes involve $d\log$-forms with algebraic arguments~\cite{Prlina:2017azl,Prlina:2017tvx}. It is currently not known if one can parametrise the external kinematic data in a way which would rationalise these algebraic arguments (though it is known that in some cases all contributions from these algebraic arguments cancel in the full amplitude~\cite{Bourjaily:2019igt}). As a consequence, conjectures stating  that  MHV and NMHV amplitudes in planar $\cN=4$ Super Yang-Mills are always expressible in terms of polylogarithms should be taken with a pinch of salt: there is no firm supporting evidence, nor a counterexample, for this conjecture beyond seven particles.

\subsubsection*{Acknowledgements} The authors  thank Matija Tapu\v{s}kovi\'{c} for discussions and Johannes Broedel, Andrew McLeod and Lorenzo Tancredi for comments on the manuscript. The authors would also thank the organisers of the `Mathemamplitudes' workshop in Padova, where the ideas presented in this paper were first discussed, for the invitation to the workshop.
This project has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement nos. 724638 and 637019).


TITLE: How is mass conserved when a muon decays?
QUESTION [2 upvotes]: A small disclaimer: I am a layperson and not a formal student of physics so forgive any glaring stupidity betrayed in the nature of the question.
A muon is supposed to always decay into an electron and two different neutrinos,
However, the mass of a muon is 207 times greater than an electron, a disparity that is seemingly too great for the addition of two insignificant neutrinos.
How is conservation of mass conserved here?

REPLY [4 votes]: I'm going to expand on dmckee's answer because this used to puzzle me in my younger days, and I think it's a fascinating part of modern physics.
It's tempting to think of a particle like a muon as a chunk of matter whizzing around - something like a tiny billiard ball. However the physicist's description of a particle is much, much stranger. Quantum field theory describes the particle as an excitation in a quantum field.
For every type of particle there is an associated quantum field that fills all of spacetime. When you add energy to this field it appears as a particle i.e. you can create a new particle by adding energy to the field. Likewise, removing energy from the field destroys a particle. When you add energy to a quantum field some of the energy goes to the rest mass energy of the particle and some to the particle's energy of motion. dmckee has given the equation for the total energy:
$$ E = \sqrt{(pc)^2 + (mc^2)^2} $$
and roughly speaking in this equation the $mc^2$ bit is the contribution from the rest mass (yes, as in the famous equation $E = mc^2$) and the $pc$ is from the motion.
But let's get back to the specific case you asked about of a muon decaying to an electron, an electron antineutrino and a muon neutrino:
$$ \mu \rightarrow e + \bar{\nu}_e + \nu_\mu $$
In the decay:

an energy $E_\mu$ is removed from the muon field destroying a muon
an energy $E_e$ is added to the electron field creating an electron
an energy $E_{\bar{\nu}_e}$ is added to the electron antineutrino field creating an electron antineutrino
an energy $E_{\nu_\mu}$ is added to the muon neutrino field creating a muon neutrino

The exchange of energy destroys one particle and creates three new ones. Energy is always conserved so we can say with confidence that the energy before equals the energy after:
$$ E_\mu = E_e + E_{\bar{\nu}_e} + E_{\nu_\mu} $$
But as dmckee points out, mass is not conserved. For example when we add an energy $E_e$ to the electron field some of this energy goes into the electron rest mass and some goes into the kinetic energy of the electron. Likewise for the neutrinos. In effect the missing mass gets turned into energy of motion of the three product particles.


TITLE: Inductively Defined Family of Sets
QUESTION [1 upvotes]: Suppose I am given the following inductive definition$^1$: Let $T$ be the smallest family of sets $\{T_1,T_2,T_3,\ldots\}$ such that

$0 \leq k < n \Rightarrow k \in T_n$
$t_1 \in T_n \wedge n > 0 \Rightarrow \lambda.t_1 \in T_{n - 1}$
$t_1 \in T_n \wedge t_2 \in T_n \Rightarrow (t_1 ~ t_2) \in T_n$

The elements of $T_n$ are the nameless $\lambda$-terms (with de Bruijn indices) with at most $n$ free variables numbered between 0 and $n - 1$. I want to express $T$ inductively, but am having trouble. In particular, I'm not sure how to express (2) and (3). For instance, suppose I try to define the set $T_{n + 1}$ in terms of $T_n$ for (2): in order to define $T_{n + 1}$, I need to know which elements occur in $T_{n + 2}$. For (3), in order to define $T_{n + 1}$, I need to know which elements occur in $T_{n + 1}$. I am being led in circles!
I don't understand the underlying principle which would permit me to define the family of sets $T$, because it's not a straightforward inductive definition. I assume the answer lies in defining mutually recursive sets, then finding their union. However, I'm stuck. Can anyone enlighten me?

Pierce, Types and Programming Languages, 6.1.2.

REPLY [2 votes]: First note that this problem is given in the context of some larger set $S$ that already satisfies the required properties.  In this case, the "smallest" $T$ (should it exist) would mean a subset $T$ of $S$ satisfying these properties $\mathscr{P}$ such that every set $X \subseteq S$ having $\mathcal{P}$ has $T \subseteq S$.  But the reason such a $T$ will exist is because we are implicitly given a well-defined procedure without choices of how to build up our $T$.  Because of this fact, you can express the properties in a form that's equivalent and adheres to the more formal definition of induction, but this is unnecessary and can inhibit understanding.
To the more practical side of the matter, definitions like these fall into the "closure category" and often involve iterating steps infinitely many times starting with the emptyset and repeating iteratively until you've met all of the conditions (which is usually after an infinite number of steps).  The intuition here is that if you've started with the empyset and added elements only as required by the definitions ("closing under the required properties"), then any elements that you added must always be contained in the "smallest" set.  Many times, stepping through all of the Natural numbers will be sufficient.
Let's begin the procedure to get an idea.  Here, you will be building up each of the $T_n$'s simultaneously as required by the conditions:
Start with $T_n = \emptyset$ for all $n \in \mathbb{N}$.
Step (1): After condition (1), $T_n = \{0, 1, \ldots, n-1\}$ for all $n$.
After condition (2), $T_n = \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\}$.
After condition (3), $T_n = \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \cup$
$\{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \times \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\}$ (union of result from condition (2) and all possible ordered pairs from condition (2)).
Now start with result from Step (1), condition (3): $T_n = \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \cup$
$\{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\} \times \{0, 1, \ldots, n-1, \lambda.0, \lambda.1, \ldots \lambda.n\}$ and repeat conditions (1) - (3) with this set instead of the emptyset for Step (2).  Then plug in the final result from Step (2), and repeat conditions (1) - (3) with this set for Step (3).  Keep repeating for each Natural number step.  Of course condition (1) doesn't need to be repeated in this process because it won't add anything new.
If you continue this process of what is known as "closing under the operations" ad infinitum (all Natural numbers), you will wind up with the desired set.  The reason for this is intuitively simple.  Any terms $t_1$ and $t_2$ in the final set were added by some Natural number steps $m$ and $n$, say $n > m$ here.  But then in step $n+1$, you make sure you've satisfied conditions (2) and (3) for any elements in the set so far constructed, including the elements $t_1$ and $t_2$.


text:
\section{Cosine of Complex Number}
Tags: Cosine Function, Complex Numbers, Cosine of Complex Number

\begin{theorem}
Let $a$ and $b$ be [[Definition:Real Number|real numbers]].
Let $i$ be the [[Definition:Imaginary Unit|imaginary unit]].
Then:
:$\cos \left({a + b i}\right) = \cos a \cosh b - i \sin a \sinh b$
where:
:$\cos$ denotes the [[Definition:Cosine Function|cosine function]] ([[Definition:Real Cosine Function|real]] and [[Definition:Complex Cosine Function|complex]])
:$\sin$ denotes the [[Definition:Real Sine Function|real sine function]]
:$\sinh$ denotes the [[Definition:Hyperbolic Sine|hyperbolic sine function]]
:$\cosh$ denotes the [[Definition:Hyperbolic Cosine|hyperbolic cosine function]]
\end{theorem}

\begin{proof}
{{begin-eqn}}
{{eqn | l = \cos \paren {a + b i}
      | r = \cos a \cos \paren {b i} - \sin a \sin \paren {b i}
      | c = [[Cosine of Sum]]
}}
{{eqn | r = \cos a \cosh b - \sin a \sin \paren {b i}
      | c = [[Hyperbolic Cosine in terms of Cosine]]
}}
{{eqn | r = \cos a \cosh b - i \sin a \sinh b
      | c = [[Hyperbolic Sine in terms of Sine]]
}}
{{end-eqn}}
{{qed}}
\end{proof}

\begin{proof}
{{begin-eqn}}
{{eqn | l = \cos a \cosh b - i \sin a \sinh b
      | r = \frac {e^{i a} + e^{-i a} } 2 \frac {e^b + e^{-b} } 2 - i \frac {e^{i a} - e^{-i a} } {2 i} \frac {e^b - e^{-b} } 2
      | c = [[Cosine Exponential Formulation]], {{Defof|Hyperbolic Cosine}}, [[Sine Exponential Formulation]], {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac {e^{b + i a} + e^{-b + i a} + e^{b - i a} + e^{-b - i a} - e^{b + i a} + e^{-b + i a} + e^{b - i a} - e^{-b - i a} } 4
      | c = simplifying
}}
{{eqn | r = \frac {e^{-b + i a} + e^{b - i a} } 2
      | c = simplifying
}}
{{eqn | r = \frac {e^{i \paren {a + b i} } + e^{-i \paren {a + b i} } } 2
      | c =
}}
{{eqn | r = \cos \paren {a + b i}
      | c = [[Cosine Exponential Formulation]]
}}
{{end-eqn}}
{{qed}}
\end{proof}

\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \cosh {a + b i}
      | r = \cosh a \map \cosh {b i} + \sinh a \map \sinh {b i}
      | c = [[Hyperbolic Cosine of Sum]]
}}
{{eqn | r = \cosh a \cos b + \sinh a \map \sinh {b i}
      | c = [[Cosine in terms of Hyperbolic Cosine]]
}}
{{eqn | r = \cosh a \cos b + i \sinh a \sin b
      | c = [[Sine in terms of Hyperbolic Sine]]
}}
{{end-eqn}}
{{qed}}
\end{proof}

config: wiki


TITLE: product between symmetric matrix and transpose matrix
QUESTION [1 upvotes]: Suppose that B is a symmetric matrix and A generic matrix with $A^T$ its transpose, is it true that:
$B*A = B*A^T $ ?
i think that $B*A = (B^T)*(A^T) = B*(A^T) $ so is true
And if A,B were 2 tensors?

REPLY [2 votes]: No it is false because $B=I $ is a symmetric matrix but for all non-symmetric matrix $A$ you have that
$A=I*A\neq I* A^t$


TITLE: Find Probablity Mass Function of Opening Boxes
QUESTION [0 upvotes]: There are 5 closed boxes on the table. 3 of the boxes have prizes, the other 2 do not. You open boxes one at a time until you find a prize. Let $X$ be the number of boxes you open.
Find the probability mass function of $X$.

$X\in\{1, 2, 3\}$
$P(X=1)=\frac{3}{5}$
$P(X=2)=\frac{2}{5}\cdot\frac{3}{4}$
$P(X=3)=\frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{3}$
$P(X=x)=$ ?

REPLY [1 votes]: You are done.
$$P(X=x) = \begin{cases}\frac{3}{5}, & ,x=1 \\
\frac{2}{5}\cdot\frac{3}{4}, & ,x=2\\
\frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{3}  & ,x=3\\
0, & \text{,Otherwise}\end{cases}$$


TITLE: Is there a principal difference in acceleration from an impulse and the free fall of a body in the orbit around a gravitational mass?
QUESTION [1 upvotes]: Especially taking in account that acceleration need a force which leads to some conversion of energy. This is not the case for orbital motion, such a acceleration does not convert energy. That seems to be a good point why we talk in General Relativity in first line about curvature of space and time and not more about gravitational force.
So the question is, is there a principal difference in acceleration from an impulse and the free fall of a body in the orbit around a gravitational mass? This question is related to this about Can gravity accelerate light?

REPLY [0 votes]: we talk in General Relativity [...] about curvature of space and time

At least (without implying strict separability) about curvature of spacetime; right.
In this context it is however possible to distinguish "free (inertial, falling) motion" from "any other motion", The latter is accordingly said to be "due to (external) influence(s)", and its ("first order") deviation from "free motion" is quantified as a non-zero value of (instantaneous) acceleration.

Is there a principal difference [...] ?

The difference is most plainly and strikingly that the acceleration value is either zero, or non-zero.
A general prescription how to evaluate acceleration (from given more basic geometric notions, namely given values of spacetime interval ratios) is sketched here (PSE/a/146397).
Note that this mentioned prescription is based on non-zero spacetime intervals; it is indeed intended to be evaluated for ratios of certain time-like intervals. It is not even defined for a set of events which are all (pairwise) separated by null intervals. However, it may be argued that it is therefore consistent to attribute to the propagation of a signal front strictly an acceleration value of zero; calling it strictly "free" with respect to the possibly curved spacetime geometry of the experimental region under consideration.


TITLE: Proofreading there is degree of divisor polynomial is less or equal to $n/2$
QUESTION [2 upvotes]: Given a degree $n$, reducible polynomial $f(x)$ over a field $\mathbb{K}$, prove there is less or equal $n\over 2$ degree divisor polynomial,
My current proof:
$$\text{Assume the statement is false, imply all the degree of divisor polynomials are greater than n/2}$$
$$\text{Let all the divisors of the polynomial are }P_{1}(x)P_{2}(x)\ldots P_{k}(x)=f(x)$$
$$\Rightarrow \deg(P_{1}(x)) + \deg(P_{2}(x)) + \ldots + \deg(P_{k}(x)) > \deg(f(x))=n$$
$$\text{This contradicts }$$
$$\text{if} \quad g(x)h(x) = f(x)\text{ then }\quad \deg(g(x))+\deg(h(x)) = \deg(f(x))$$

REPLY [0 votes]: Yes, it is correct.
Since you don't need to divide the polynomial to irreducible factors, it is enough to say that

Since $f$ is reducible, we can write $f=gh$. By the indirect assumption, we would have $\deg(g),\,\deg(h)>n/2$ which already contradicts
  $$n=\deg(f)=\deg(g)+\deg(h)\,.$$