TITLE: The equivalence between Cauchy integral and Riemann integral for bounded functions
QUESTION [18 upvotes]: Definitions
Suppose $P\colon a=x_0<x_1<\dotsb<x_n=b$ is a partition of $[a,b]$. Let $\Delta x_k=x_k-x_{k-1}$ and $\lVert P\rVert$ denotes $\max_{0<k\le n}\Delta x_k$.
The Cauchy integral of a function $f$ on closed interval $[a,b]$ equals to $I$ if and only if for each $\epsilon>0$, there's some $\delta>0$, for each partition $P$ of $[a,b]$ such that $\lVert P\rVert<\delta$, we have $\left\lvert\sum_{k=1}^nf(x_k)\Delta x_k-I\right\rvert<\epsilon$.
Problem
If $f$ is bounded on $[a,b]$ whose Cauchy integral equals to $I$, then $f$ is Riemann-integrable and $\int_a^bf=I$.
Background
It's an exercise from our calculus(analysis) problemset book, and there's a hint: consider the partitions whose $x_k-x_{k-1}$ is a constant for different $k$'s, and try to estimate the Riemann sum for each of these partitions through the Cauchy integral.
I have no idea about such estimation. After drawing some pictures, I discouraged. I googled on the Internet and found an article. I realized that it's a quite different approach and with some advanced techniques (such as the analysis of a positive measure set -- discontinuities). I hope there will be some simpler approachers, just as the hint says. I need a more detailed hint, or a solution. Can anybody help me? Thanks!

REPLY [1 votes]: I think for a calculus class/book, this might be overcomplicated. Though, before having a proper measure theory course, I was taught that a function is integrable if it has a limited number of discontinuities (or class I and II discontinuities), apart from being bounded.
$$\sum_{k=1}^{n}\left | f(t_{k})\cdot \bigtriangleup x_{k} - I \right | = \sum_{k=1}^{n}\left | f(t_{k})\cdot \bigtriangleup x_{k} - f(x_{k})\cdot \bigtriangleup x_{k} + f(x_{k})\cdot \bigtriangleup x_{k} - I \right | \leq \sum_{k=1}^{n} \left \{ \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} + \left | f(x_{k})\cdot \bigtriangleup x_{k} - I \right | \right \} \leq \varepsilon + \sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} $$
Now, if $f(x)$ is continuous, then it is uniform continuous on $[a,b]$ so that we can pick up $P$ such that $max \left | f(t_{k}) - f(x_{k}) \right | \leq \frac{\varepsilon }{b-a}$ and then:
$$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} \leq \frac{\varepsilon }{b-a} \sum_{k=1}^{n} \bigtriangleup x_{k} = \varepsilon$$
If $f(x)$ is bounded and has a limited number of discontinuities, then we can split $[a,b]$ into sub-segments and "make" $f(x)$ continuous on those sub-segments. It should be enough to prove this for the case with one discontinuity, and use induction after that.
However, the necessary and sufficient condition states that the set of points of discontinuity should have measure of zero. If it is so, then:
$$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k}$$ can be split into:
$$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} \leq M \cdot \mu(D_{n-m}) + \sum_{i=1}^{m} \left | f(t_{k_{i}}) - f(x_{k_{i}}) \right |\cdot \bigtriangleup x_{k_{i}} $$ where summation is on the continuous part of the $[a,b]$, $M=max|f(x)|$, $D_{n-m}$ is the remaining part of $[a,b]$ containing the set of points of discontinuity and $\mu(D_{n-m})$ is its measure which goes to zero. So, this is the biggest question; how, using the fact that Cauchy integral is equal to $I$, to prove that the set of points of discontinuity has the measure of zero? I tend to agree with @Tony Piccolo answer ...


TITLE: Is it necessary to prove the existence of an operator representing symmetry on Hilbert space?
QUESTION [1 upvotes]: Is there any need to prove the existence of an operator $U$ which represents the action of symmetry transformation on rays in Hilbert space? Or is it enough just to prove that it is unitary and linear or antiunitary and antilinear?
I do not see how we should go about finding this operator. I got the answer that this operator is given, but I don't see how. The only thing that is given is the symmetry transformation which takes the whole ray $R_1$ into, let's say, $R_2$. But what then is the action of the operator on this ray? Which vector in $R_1$ will be taken into which in $R_2$?
My guess is that we can show that coefficients of the old and a new transformed vector are the same, so we can just say that in this way the transformation is translated into an operator. Every orthonormal vector of the old basis has the same coefficient as the one in the new basis.

REPLY [0 votes]: Source of confusion was obviously that I did not understand what it means to prove the existance of such a thing as an operator. Details of the proof aside, in order to prove this, you simply construct the operator and in proofs of Wigners theorem that is exactly what is done. On the basis of the fundamental assumption about our transformation which is probability conservation, we made a connection betwen the old and a new set of vectors by defining a legal maping that is, defining a maping that associates exatly one vector from the new set to the vectors from the old one. After all, new vectors in a given ray differ from the old ones by a multiplicative constant in the first place, so just by seeing that we can choose a phase of the new vectors to be this or that, ve can construct the operator. Of course there is much more to this story but I believe, in the end, that this was my confusion. To add, the operator is purely mathematical object as opposed to transformation. So we have a certain freedom in its construction.


TITLE: How to prove that no two-to-one function can be continuous?
QUESTION [3 upvotes]: This is question

I have almost done this question, but why can we only assume $a<c<d\leq b$ and why not $a\leq c<d<b$ and also the case when $a=c$ and $b=d$. That is my professor asked me, could you explain for me?

REPLY [2 votes]: The point is that since at least two of $c, d, e, f$ are in the interior of $[a, b]$, we must have:

EITHER at least one of $c, d$ is in the interior of $[a, b]$,
OR at least one of $e, f$ is in the interior of $[a, b]$.

(Of course, perhaps both possibilities hold.)
Each possibility further refines into two more possibilities: e.g., if we have that at least one of $c, d$ is in the interior of $[a, b]$, then

EITHER $c$ is in the interior of $[a, b]$ - in which case $a<c<d\le b$ -
OR $d$ is in the interior of $[a, b]$ - in which case $a\le c<d<b$.

(Again, perhaps both possibilities hold.)
So we know that (at least) one of the following four cases holds:

$a<c<d\le b$,
$a\le c<d<b$,
$a<e<f\le b$, or
$a\le e<f<b$.

The point is that each of these cases is handled in the same way, so we only have to explicitly write out one of them.
That said, you should understand why they are each handled in the same way - look at the proof of the "$a<c<d\le b$" case, and see why $(i)$ nothing special about "maxima" (vs. "minima") is used, and $(ii)$ why nothing special about the left extremum being in the interior (vs. the right extremum being in the interior) is used.


TITLE: Vector analysis: understanding formulas for normal and tangent
QUESTION [2 upvotes]: I'm studying a chapter on vector analysis and I don't really understand why the following equations are switched for the two situations. (My course is not in English, so I apologize if I translated something wrong.)
Equations for the tangent plane and normal in a point on a surface

The equation for a tangent plane in $p(x_0, y_0, z_0)$ on a surface $\varphi(x,y,z) = 0$:

$$\left(\frac{\partial \varphi}{\partial x}\right)_p (x-x_0) +
\left(\frac{\partial \varphi}{\partial y}\right)_p (y-y_0) +
\left(\frac{\partial \varphi}{\partial z}\right)_p (z-z_0) = 0$$

The equation for the normal in $p(x_0, y_0, z_0)$ on a surface $\varphi(x,y,z) = 0$:

$$\frac{x-x_0}{\left(\frac{\partial \varphi}{\partial y}\right)_p} =
\frac{y-y_0}{\left(\frac{\partial \varphi}{\partial y}\right)_p} =
\frac{z-z_0}{\left(\frac{\partial \varphi}{\partial z}\right)_p}
$$
Equations for the tangent and normal plane in a point on a curve
For $p(t=t_0) = \{f_1(t_0), f_2(t_0), f_3(t_0)\}$ with $t \in R$

The equation for the tangent in $p$:

$$\frac{x-x_0}{f'_1(t_0)} = \frac{y-y_0}{f'_2(t_0)} = \frac{z-z_0}{f'_3(t_0)}
$$

The equation for the normal plane in $p$:

$$f'_1(t_0) (x-x_0) +
f'_2(t_0)(y-y_0) +
f'_3(t_0)(z-z_0) = 0$$
I tried to find a graphical explanation for this to understand how the equations were derived, but without much success.

REPLY [2 votes]: The equation
$$ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 $$
describes an affine plane passing through the point $(x_0,y_0,z_0)$ whose normal vector is $(a,b,c)$ (so $(a,b,c)$ should be non-zero). The equations
$$ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} $$
describe an affine line passing through $(x_0, y_0, z_0)$ whose direction vector is $(a,b,c)$ (this representation is possible only if $a,b$ and $c$ are non-zero).
In the first case, your surface $S$ is described as a level set of a single function $\varphi \colon \mathbb{R}^3 \rightarrow \mathbb{R}$ (so $S = \{ (x,y,z) \, | \, \varphi(x,y,z) = 0\}$). If $p = (x_0,y_0,z_0)$ is a point on the surface, then the gradient $\nabla \varphi|_p$ is perpendicular to the tangent plane of $S$ at $p$. Hence, the gradient is a normal vector for the tangent plane at $p$ and you get the resulting equation for the tangent plane. Similarly, the normal line is a line passing through $(x_0,y_0,z_0)$ in the direction of a normal to the tangent plane (which is $\nabla \varphi|_p$), hence the resulting equation for the normal line.
In the second case, the curve $\gamma$ is described as the image of a parametrization $\mathbf{f} \colon \mathbb{R} \rightarrow \mathbb{R}^3$ (so $\gamma  = \mathbf{f}(\mathbb{R})$). The tangent line to $\gamma$ at $\mathbf{f} (t_0)$ is the line passing through $p = (f_1(t_0), f_2(t_0), f_3(t_0))$ in the direction of the velocity vector $\mathbf{f}'(t_0)$, hence the resulting equation for the tangent line. The normal plane to $\gamma$ at $p$ is the plane passing through $p$ whose normal vector is the velocity vector $\mathbf{f}'(t_0)$ (because the normal plane should be perpendicular to the velocity vector), hence the resulting equation for the plane.


text:
\begin{document}
\maketitle
\begin{abstract}
The anticipated increase in the number of plug-in electric vehicles (EV) will put
additional strain on electrical distribution circuits.  Many control schemes have been
proposed to control  EV charging.  Here, we develop control algorithms based on
randomized EV charging start times and simple one-way broadcast communication allowing for a time delay between communication events.  Using arguments from queuing theory and
statistical analysis, we seek to maximize the utilization of excess distribution circuit capacity while keeping the probability of a circuit overload negligible.

{\it  Key Words:} Power Distribution, Battery Chargers, Queuing Theory

\end{abstract}

\section{Introduction}
\label{sec:intro}

If electric vehicles (EV) make up a significant fraction of the future vehicle fleet,
there will be significant impacts on electrical generation and transmission due to the
increase in electricity consumption.  Some researchers have viewed the potential
increase in EVs as an opportunity to utilize the on-board battery storage in an
interactive way to provide two-way energy flows to buffer time-variable renewables\cite{09CF,97KL,05KT} or to provide ancillary services such as (short-term) frequency regulation\cite{02V2G}.  We also view
these potential applications as mutually beneficial to consumers, utilities, and
generators, however, the first potential impact of EVs,
independent of mass deployment of renewable generation, is filling in the nighttime
load-curve minimum typical of most regions.\cite{07PNNL} The electrical system as it is currently configured would benefit in several ways.  The primary benefit is a higher
utilization of currently deployed assets resulting in faster paybacks on generation,
transmission, and distribution investments. Second, in some regions, the locational
marginal prices for wholesale energy become negative during the nighttime hours.  If
properly controlled, EVs will provide a reliable increased nighttime load making the
operation of baseload generation more profitable.  Modifying the load curve in this way
will also reduce the relative impact of demand charges relative to energy charges for
many distribution utilities.

To serve this new load, distribution utilities must determine how many EVs can be
reliably served on a given distribution circuit without substantially increasing the
probability of an overload or other system upset. A likely overload scenario is the
synchronization of EV charging start times that would occur as their owners begin EV
charging immediately after they arrive home from work during the weekday evening hours.  We seek to avoid this load synchronization by controlling EV charging start times.

Although sophisticated control schemes relying on high-speed two-way communication
between all EV loads and centralized controllers could allow distribution circuits to run at nearly full capacity, the reliability of electrical service would then
compromised due to real-time reliance upon a sophisticated communications network.  Instead, we seek a distributed control method that relies upon no or minimal communication, where the
decision of when to begin charging is made local to each EV charger.  We propose to
regulate EV charging start times by modulating a single, circuit-wide EV connection rate (or arrival rate) that determines, on average, how many EV chargers can commence charging per unit time. The circuit-wide rate may be sent to each charger via one-way, broadcast
communication and updated periodically at a frequency that suits a particular utility's or
circuit's needs.  As a limit of this scheme, we also consider a simpler version where a
fixed arrival rate is preprogrammed into the charger by the utility.

For this initial work, we consider a few simplifying assumptions.  We envision that each utility or circuit will have a predetermined EV-charging `window' of length $\Delta T$
with start and stop times that are correlated with the minimum in the load curve for
that utility or circuit, schematically shown in Fig~\ref{fig:schematic}a.  We also
assume that, during $\Delta T$, the other loads are known well enough that an available
excess capacity $P_{excess}$ can be reliably determined and that all EV chargers consume the same amount of power $P_{con}$ so that the capacity for EV charging is expressed in terms of the maximum number of EVs charging without an overload, i.e. $N=P_{excess}/P_{con}$.  Each of these assumption can relaxed to allow for charging at
any point in the load curve,  fluctuations in uncontrolled, non-EV loads on the circuit, and multiple classes of EV charging powers.

Schematically shown in Fig~\ref{fig:schematic}b, the window $\Delta T$ is split into
many smaller time intervals of length $\tau$ with beginning and end times $t_i$.  At
each $t_i$, the number $n_i$ of EVs charging is measured{\footnote {To develop intuition and to keep the computations simple in this initial work, we consider the situation
where $n_i$ is available to the entity controlling the system.  Although this may
require two-way communication, this can easily be relaxed by instead measuring the real
or apparent power drawn by the circuit using a current transformer at the substation and controlling EV charging based the apparent power available below the circuit capacity.}} and $n_i$ is used to determine the connection (or arrival) rate of EV chargers
$\lambda(n_i)$ that will apply until the end of the internal at $t_{i+1}$, i.e. for a
time $\tau$.  In this context, $\tau$ is the time interval between utility broadcast
communication events.  During this interval, EVs are connecting at a rate $\lambda(n_i)$ but are also leaving at an average rate $\mu n(t)$ where $\mu=\Delta E/P_{con}$ is the
inverse of the average EV charging time and $\Delta E$ is the average amount of energy
required to reach full EV battery charge.  Although the individual decisions to start
and stop charging are made locally, control of the charging load is exercised through
broadcasting a single parameter $\lambda(n_i)$ which amounts to feedback control because it depends on the number of EV charging at time $t_i$.

The EV service concept proposed above is extremely robust.  The connection of EV
chargers is inherently a random process naturally avoiding any of the dangerous
synchronization that can occur in EV charging.  The reliance on the communication network is minimal since decisions to start or stop charging are distributed and based upon a single number $\lambda (n_i)$ received via rate-limited, one-way, broadcast communication. In addition, misoperation by one or a handful of EV chargers will not greatly affect the overall load.  However, the randomness of the EV connection process will cause the
aggregate EV load fluctuate (in addition to the fluctuations in the underlying load). Excluding the possibility of real-time intervention
during the time interval $\tau$, a utility must carefully choose $\lambda(n_i)$ to
maximize EV service and asset utilization without overloading or damaging the circuit.
The remainder of this manuscript analyzes this problem and seeks to make good choices for $\lambda(n_i)$.
\begin{figure}
\centering \caption{a) Example schematic load curve showing a load valley lasting for time $\Delta T$ during the nighttime hours where excess capacity on the circuit can accommodate a maximum of $N$ EV chargers without overloading where each EV charger consumes $P_{con}$ real power. b) Schematic representation of control scheme.  The open circles represent the sparse measurements of $n_i$ at the beginning of a time interval $\tau$.  See main text for details.
} \includegraphics[width=0.55\textwidth]{schematic}
\label{fig:schematic}
\end{figure}

\section{Calculations}
\label{sec:setup}
Within any period $\tau$, the EV charging process described above closely resembles the
birth-death queuing models typically used in telecommunications where both the arrival
and completion of calls is often assumed to be a random process with an average arrival rate $\lambda$ and departure (or end of service) rate $\mu$.  For our EV-charging
problem, one implication of adopting this model is that the time between the start of
EV-charing events is exponentially distributed.  The same holds for the time between
completion of EV-charging events.  In practice, the arrival and departure-time
distributions will certainly be cut off by $\Delta T$ and $\Delta E_0/P_{con}$,
respectively, where $\Delta E_0$ is the EV battery capacity.  However, for a large
number of EVs in the system, the effect of eliminating these tails should be minimal.

We begin with a simplified `constant-rate' version of the
problem letting $\tau \rightarrow \infty$ corresponding to the case where
$\lambda=\lambda_0$ is preprogrammed and the problem is essentially one of design and
not control.  In this limit, the problem reduces to an $M/M/\infty$ queuing
problem\cite{kleinrock} where the probability of finding $n$ EVs charging is given by
\begin{equation}\label{pk}
\pi_n=e^{-\bar{n}} \bar{n}^n/n!,
\end{equation}
and the probability of a circuit overload, i.e. $n\geq N$, is given by
\begin{equation}\label{overload1}
P_N=\sum_{m=N}^\infty \pi_m=1-\Gamma(N,\bar{n})/\Gamma(N).
\end{equation}
Here, $\bar{n}=\sum_n n \pi_n=\lambda_0/\mu$ is the average number of EVs charging and
$\Gamma (\cdot)$ and $\Gamma (\cdot,\cdot)$ are the Gamma and incomplete Gamma
functions.

Next, we consider a `variable-rate' version of the problem where we control EV charging
by allowing $\lambda$ to depend on $n_i$.  The goal of this control protocol is to
maximize the utilization of excess circuit capacity while keeping the probability of
overload $P_N$ acceptably low. For a given control protocol $\lambda(n)$, the probability distribution $\pi_n^i$ of the number $n_i=n$ of EVs charging at time $t_i$, obeys a simple Markov chain equation:
\begin{equation}
 \pi_n^i = \sum_m W_{nm}(\lambda(m)\tau) \pi_m^{i-1}
\end{equation}
where we have introduced $W_{mn}(\lambda\tau)$ - the transition kernel of the
$M/M/\infty$ process with the rates $\lambda$ and $\mu$.  For $n>m$, the transition kernel can be expressed as the following sum \cite{schoutens2000stochastic}:
\begin{equation}
   W_{nm}(\lambda\tau) = \exp(-q)\sum_{k=0}^n \frac{n! q^{n-k} \epsilon^{m-k}(1-\epsilon)^k}{k!(n-k)!(m-k)!},
\end{equation}
where $\epsilon = 1 - \exp(-\mu \tau)$ and $q = \lambda \epsilon/\mu$.
Assuming that the system was not overloaded at time $t_{i-1}$, the probability of an overload at time $t_i$ (i.e. $n_i \geq N$) is given by $P_N =
\sum_{m\geq N} \pi_m^i$ and can be written as
\begin{equation} \label{PNineq}
 P_N = \sum_{n<N,m\geq N} W_{mn}(\lambda(n)\tau) \pi_n^{i-1} \leq \max_{n<N} \sum_{m\geq
 N}W_{mn}(\lambda(n)\tau)
\end{equation}
The last inequality, which is based on the normalization of $\pi_n^{i-1}$ and that $\pi_n^{i-1}=0$ for $n \geq N$,  forms the foundation for constructing our variable-rate protocol. We define the control protocol $\lambda(n)$ as a solution of
\begin{equation}\label{lambdaeq}
\sum_{m\geq N} W_{mn}(\lambda(n)\tau)= P_N^* \ll 1, \quad n<N
\end{equation}
and set $\lambda(n)=0$ for $n\geq N$. For a given value of $P_N^*$ this protocol is
guaranteed to have an overload probability bounded from above: $P_N < P_N^*$. We note
that this control scheme might be suboptimal because we impose more strict constraints on the overload probability than required; instead of requiring the total overload probability to be less then $P_N^*$, we look for a solution where the probability of overload conditioned on the starting with $n_{i-1}$ EVs connected is bounded by $P_N^*$ for {\it every} $n_{i-1}$.
However, as we will show, the performance of this control scheme is almost perfect, so
further optimization can result only in diminishingly small improvements.  The result also indicates that one starting condition $n_{i-1}$ dominates the transition to the overloaded state.

As a first measure of the control system's performance, we consider the time response of the control or equivalently the time to relax to a steady state.  In the constant-rate scheme, this time is estimated as the product of the average number of EVs charging in equilibrium and the typical inter-arrival time $\bar{n}/\lambda_0$, which according to the discussion below Eq.~(\ref{overload1}) is $\bar{n}/\lambda_0=1/\mu$.  Control in this case can be slow because the relaxation time is the typical time required to charge an EV, which can be several hours for present EV battery and charging technology\cite{07PNNL,10SAE}.  In the variable-rate case, the relaxation time can be estimated as the number of time intervals $\tau$ it takes, starting with $n_0=0$, to reach $n\sim N$. To leading order, the solution of (\ref{lambdaeq}) can be estimated as
$\lambda(n)\tau\approx N-n - \Delta_n$, where $\Delta_n \sim -\log P_N \sqrt{N-n}$ for
$N-n\gg 1$. Thus, in the limit $N\gg 1, \mu\tau \ll 1$ and $\mu N \tau \lesssim 1$,
there will be $n\approx N+\log P_N \sqrt{N}$ cars in the system after one interval $\tau$.  After
the second interval $\tau$, $n\approx N+ \log P_N \sqrt{-\log P_n \sqrt{P_N}}$. Thus, the total time
required to reach stationary state can be estimated as $\tau \log (-N \log P_N)$.
For realistic values of $N$ and $P_N$, the response time of the variable-rate protocol is now under the control of the utility which can set $\tau$ based on technological or economic considerations.

A second measure of system's performance, which we address in the remainder of this
manuscript, is the average capacity utilization defined as $\bar{n}/N$ ranging from $0$
with no EVs charging to $1$ when the excess circuit capacity is fully utilized.  Our
control schemes seek to maximize $\bar{n}/N$ while keeping $P_N$ acceptably small.  The
proximity of $\bar{n}/N$ to one for a given $P_N$ is a measure of the quality of the
control scheme.  Exact calculation of this quantity is a computationally challenging
problem that is beyond the scope of this work. Instead, here we obtain the lower bound
for the capacity by exploiting the convexity of the function $\lambda(n)$ derived from Eq.~(\ref{lambdaeq}). We first note, that the stability of the queue (i.e.
existence of the stationary distribution) implies that the following relation holds:
\begin{equation}\label{conslaw}
 \overline{\lambda(n)\tau} = \mu\tau \bar{n}
\end{equation}
It is not possible to extract the value of $\bar{n}$ from Eq.~(\ref{conslaw}) directly because $\lambda(n)$ is a nonlinear function of $n$. However, for convex and monotonically
decreasing functions $\lambda(n)$ that solve the equation (\ref{lambdaeq}) one can use
the relation (\ref{conslaw}) to find the lower bound for $\bar{n}$. It follows from
Jensen's inequality that $\mu \bar{n}-\overline{\lambda(n)} \geq \mu
\bar{n}-\lambda(\bar{n})$ and from motonicity of $\mu n - \lambda(n)$ we conclude that
$\bar{n}\geq n^*$ with $n^*$ being the solution of the equation
\begin{equation}\label{nbareq}
 \lambda(n^*) =\mu n^*,
\end{equation}
where we have assumed some convex and monotonic continuation of the discrete function
$\lambda(n)$.

Equations~(\ref{lambdaeq}) and (\ref{nbareq}) have been solved numerically with
Newton's method with the help of the approximate expressions for the transition kernel
$\sum_{m\geq N}W_{mn}(\lambda \tau)$. For a given value of $P_N^*$, a table of
$\lambda(n)$ was found for all $n=0\dots N-1$ which was then continued to
real values of $n$ with second-order spline interpolation. Equation~(\ref{nbareq})
was then solved to find the lower bound on $\bar{n}$, i.e. $n^*$, as a function of the
overload probability bound $P_N^*$.

\section{Results and Discussion}
\label{sec:results}
To understand the differences between the constant-rate and variable-rate protocols, Fig.~\ref{fig:protocols} displays $\lambda(n)$ versus $n$ for $N=100$ and $P_N=10^{-10}$.  For the variable-rate case, the condition on the overload probability in Eq.~(\ref{lambdaeq}) suppresses $\lambda(n)$ as $n$ approaches $N$ while no such condition exists in the constant-rate case.  The resetting of $\lambda(n)$ at the beginning of each time interval $\tau$ provides a mechanism for controlling the fluctuations in $n(t)$ due to the random EV connection process.  For example, if fluctuations in the rate of EVs connections between $t_{i-1}$ and $t_i$ generated a few more EV connections that was expected on average, i.e. more than $\tau \lambda(n_{i-1})$, then average connection rate in the subsequent period is decreased.  The opposite control response occurs if the average number of connections falls short of the expected average.

We find that the particular shape of $\lambda(n)$ as $n$ approaches $N$ is important.  In Fig.~\ref{fig:protocols}, $N=100$ and $\lambda(n)$ could be closely approximated by a linear function up to about $n=60$.  Extending that line to $\lambda=0$ near $n=70$ would provide a very familiar type of control, i.e. proportional feedback control with a deadband between $n=70$ and $n=100$.  However, our analysis of this modified control protocol finds this linearized $\lambda(n)$ shows little improvement in average capacity utilization over the constant-rate case demonstrating that the details of $\lambda(n)$ for $n\sim N$ are crucial to optimal control performance.
\begin{figure}
\centering \caption{EV connection rate $\lambda(n)$ versus $n$ for the constant-rate (red, dashed line) and variable-rate cases (blue, solid curve) for $N=100$ and $P_N = 10^{-10}$.  For the variable-rate case, $\mu \tau=0.001$.
} \includegraphics[width=0.49\textwidth]{protocols}
\label{fig:protocols}
\end{figure}

Figure~\ref{fig:plt100} shows the average capacity utilization for both the constant and variable-rate cases.  The ability of the variable-rate control to essentially clip fluctuations in $n$ by modifying $\lambda$ is clearly evident in the the higher values of $\bar{n}/N$ achieved.  Figure~\ref{fig:plt100} also demonstrates the incremental value of broadcast-communication control over the case of no communication.  Considering low overload probabilities, e.g. $P_N=10^{-10}$, Fig.~\ref{fig:plt100} shows even when the optimal value of $\lambda_0$ is used, the best excess capacity utilization in the constant-rate scheme is only $\approx 30\%$.  Simply adding one-way broadcast communication to allow the utility to update the connection rate every $\tau=0.01/\mu$, the utilization jumps to $\approx 80\%$.  To provide a sense of the communication time scale, a typical charging time for an EV using today's technology\cite{10SAE} is approximately 5 hours.  Therefore, to achieve $80\%$ utilization at $P_N=10^{-10}$ would require a communication rate of once every 3 minutes, which is achievable even with slow communication methods such as power line carrier (PLC).  Figure~\ref{fig:plt100} shows that increasing the communication rate by a factor of ten (about once every 20 seconds) only boosts the average capacity utilization from $80\%$ to $90\%$.   From this analysis, we conclude that the presence of relatively slow, one-way, broadcast communication is sufficient to allow utilities to effectively manage EV charging.
\begin{figure}
\centering \caption{Average capacity utilization for the constant-rate case (red, dash-dot curve) and the variable-rate case with $N=100$ and $\mu \tau = 0.01$ (blue, dashed curve) and $\mu \tau = 0.001$ (blue, solid curve).  The horizontal axis is the common log of the inverse of the probability of overload. Intuitively for all cases, the average capacity utilization falls as the probability of overload become smaller.  In addition, the utilization also falls as $\tau$ increases in the variable-rate case.
} \includegraphics[width=0.49\textwidth]{plt100}
\label{fig:plt100}
\end{figure}

Figure~\ref{fig:plt1000} provides the same analysis as Fig.~\ref{fig:plt100} except the available excess capacity $N$ has been increased from 100 to 1000 EVs.  The performance of both control methods improves as $N$ increases.  In each case, the better performance is due to fractionally smaller variance in the equilibrium distribution $\pi_n$ for the constant-rate case and in the number of EV connections during $\tau$ for the variable-rate.  As $N$ is increased, the value of even slow communication is still apparent due to the increase in excess capacity utilization from $70\%$ to about $95\%$. However, increasing the speed of communication becomes less and less important.

\begin{figure}
\centering \caption{Same as Fig.~\ref{fig:plt100} except here $N=1000$.  For larger $N$ using variable-rate control, the capacity utilization approaches 1 and becomes less sensitive to $\tau$.
} \includegraphics[width=0.49\textwidth]{plt1000}
\label{fig:plt1000}
\end{figure}

\section{Conclusion and Future Directions}
\label{sec:conclusions}
Avoiding synchronization of EV-charging start times is crucial to avoiding circuit overloads as the penetration of EVs on a distribution circuit increases.  Randomization of start times is one method that avoids this synchronization and also allows the properties of the EV charging load to be analyzed from the standpoint of queuing  theory.  Our initial analysis shows that even relatively slow, one-way, broadcast communication such as PLC can be used to modulate the random EV connection process and effectively control the EV-charging load.  When there is a large amount of excess circuit capacity for EV charging (e.g. 1000 EV connected at a time), our control algorithm easily achieves $95\%$ utilization of the capacity with a minimal probability of overloading the circuit.  Our analysis also shows that compared to no communication, even relatively slow, one-way, broadcast communication boosts the utilization of circuit capacity dramatically.  However, as the speed of that communication increases, there are diminishing returns for EV charging.

Our initial work lays the foundation for further study into this EV-charging control method that could also be applied to other forms of demand response if the consumer finds randomization of load start times acceptable.  Our work can be expanded on in several ways:
\begin{itemize}
\item{To avoid having to define an accurate excess circuit capacity for every distribution circuit, a utility may choose to use measurements of the current injected into the circuit at the substation to determine excess capacity in real time.  These measurements would also capture the statistical fluctuations in the non-EV-charging loads.  Our analysis should be performed in this setting to account for time-dependent average excess capacity and its fluctuations.}

\item{Our analysis assumed that the population of EVs connected to their chargers but not presently charging was relatively large so that the number of unconnected EVs had little dependence on the number of connected EVs.  During daylight hours when EVs are driven to the workplace, the number of EVs on a residential distribution circuit may be greatly reduced and also fluctuate greatly as they are used for day trips.  Our analysis should be extended to account for these effects.}
\item{Our analysis assumes that once an EV begins charging, it stays connected until its battery is fully charged.  A more realistic setting would include the possibility of EVs disconnecting before reaching full charge either due to the requirements of the owner or perhaps due to a broadcast control signal from the utility.}
\item{The constraint in our model is based only on circuit capacity and does not include any effects of power flow.  Including the effects of reactive power would make for a much richer problem because EVs at different locations on the circuit would contribute different amounts of current draw at the substation.}
\item{Our model was confined to a single distribution circuit.  Natural extensions would consider both capacity and power flow effects at substation or even local transmission scales.}
\item{Finally, a more general model would also incorporate the transportation grid and may account for delays in transportation flows,  EVs changing stations (e.g. from home to work and back) and possibly even cross-correlations between irregularities in traffic flows and EV charging.}
\end{itemize}

\section*{Acknowledgment}

We are thankful to all the participants of the ``Optimization and Control for Smart
Grids" LDRD DR project at Los Alamos
and Smart Grid Seminar Series at CNLS/LANL for multiple fruitful discussions.  Research
at LANL was carried out under the auspices of the National Nuclear Security
Administration of the U.S. Department of Energy at Los Alamos National Laboratory under
Contract No. DE C52-06NA25396.

\bibliographystyle{IEEEtran}
\bibliography{SmartGrid}

\end{document}

config: arxiv


TITLE: how to check if a singularity is isolated?
QUESTION [1 upvotes]: I have a function $1/(\sin(1/z))$ and I must show if the singularities are isolated or not.
Is taking the limit of the number a little to the right and a little to the left enough?
If not, how can I prove it is isolated or note?

REPLY [2 votes]: Find all singularities. In your case, $z=0$ and all points $z$ such that $\sin \frac1z =0$, i.e. $\frac1z = n\pi$ where $n\in \mathbb{Z}$.
Since $0 = \lim_{n\to\infty}\frac1{n\pi}$, $0$ is not an isolated singularity, (though the other ones are).


TITLE: The product of two distributions is not a distribution
QUESTION [1 upvotes]: Suppose $u(x)$ is the Heaviside function, which takes value $1$ when $x\ge0$ and takes value $0$ when $x<0$. Then the derivative of $u$ is a delta function $\delta_0$, or a distribution.
Now my question is can we properly define the integral:
$$\int_{-1}^1x^2|u'(x)|^2dx.$$
If there is no square inside the integral, i.e., $\int_{-1}^1xu'(x)dx$ is easy to decide.
But for the square case, I'm more inclined to the negative answer, and typically in some Functional Analysis textbooks, it is always remarked that the product of two distributions are generally not a distribution, but none of them provide a proof.
So here I post this question, hope those have a profound understanding of functional analysis could help me.
Any advice is appreciated.

REPLY [1 votes]: Rather than asserting that the "product of two distributions is generally not a distribution" the textbook I'm learning from (Hunter & Nachtergaele - Applied Analysis (2001)) says it is not possible to define a product with the same algebraic properties as the pointwise product of functions (pg 295). They do not prove this, but rather state the problem precisely as an exercise (11.7) and provide a hint which I found very enlightening: compute the product $x \cdot \delta(x) \cdot \text{p.v.}(1/x)$
where $x$ is interpreted as a distribution, and $\text{p.v.}(1/x)$ is the distribution defined by
$$ \text{p.v.}(1/x)(\varphi) = \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{1}{x} \varphi(x) \ dx.$$
The counter example amounts to
\begin{align*}
\delta \cdot (x \cdot \text{p.v.}(1/x)) &= 1 \cdot \delta = 1 \\
(\delta \cdot x) \cdot \text{p.v.}(1/x) &= 0 \cdot \text{p.v.}(1/x) = 0
\end{align*}
where every multiplication is a multiplication of a distribution by a smooth function of polynomial growth which is reasonable: if $f$ is smooth and of polynomial growth, and $T$ is a distribution, define the product as the distribution
$$
(fT)(\phi) = T(f \cdot \phi)
$$
where $\cdot$ is pointwise multiplication and is well defined for smooth functions.
In effect, we cannot create a product operator that is consistent with pointwise multiplication for smooth functions, and has the usual properties of a multiplication operator (specifically, associativity in this counter example).


TITLE: An equation that is undefined at every value except one (plotting a single point)
QUESTION [1 upvotes]: Suppose I wanted to plot an equation that would only place a single point at a specific (x,y) value, and nothing anywhere else, that is, the equation, y=g(x), would be undefined for all other x values except the x from that one (x,y) point. What might that look like?
I toyed around with dividing by zero in the equation (to make all x-values undefined and therefor not plotted), but I wasn’t sure how to not divide by zero IFF x is the x-value I want plotted. I know Kronecker Delta function could be used here but afaik it only has a definition as a piecewise function and obviously that defeats the entire purpose of what I’m trying to do. I suspect the answer will involve limits. Any ideas?
I also have an idea of plotting the equation for a circle with its center at the (x,y) point and the radius approaching zero, but I don’t know enough about limits to know if that will suffice?

REPLY [3 votes]: You can just write $(x-x_0)^2+(y-y_0)^2=0$.  This equation has only the point $(x_0,y_0)$ as solution.


TITLE: Combinatorics: Distinct Award Distribution Problem
QUESTION [0 upvotes]: Five separate awards (best scholarship, best leadership qualities, and so on) are to be presented to selected students from a class of $30$. How many different outcomes are
  possible if:
a) a student can receive any number of awards?
b) each student can receive at most one award?

The common answer seems to be $30^5$ for part a, and $30.29.28.27.26$ for part b. What I really have a hard time understanding is since the awards are distinct they also impact the way the combinations are computed. In other words if one of the students if named "Joe Schmo", Joe Schmo with best leadership award is different than Joe Schmo with best scholarship and we cannot just multiply away all the students $5$ times.
Also, the answers $30^5$ and the other ones were obtained from the official solution manual to a certain book, so this also adds to my confusion.

REPLY [0 votes]: 1) The awards are distinguishable (let them have numbers $1,2,3,4,5$).
a) (a student can receive any number of awards)
For award $1$ there are $30$ possibilities.
The same is true again for award $2$ so there are $30^2$ possibilities for the awards $1,2$.
This goes on, and there are $30^5$ possibilities for $5$ awards.
b) (each student can receive at most one award)
For award $1$ there are $30$ possibilities.
The person who received award $1$ cannot receive award $2$ so there $29$ possibilities for award $2$, so that there are $30\cdot29$ possibilities for award $1$ and $2$.
This goes on resulting in $30\cdot29\cdot28\cdot27\cdot26$ possibilities for $5$ awards.

2) The awards are not distinguishable.
a) (a student can receive any number of awards)
If for $i=1,2,\dots,30$ pupil $i$ receives $a_i$ awards then the $a_i$ are nonnegative integers with: $$a_1+\cdots+a_{30}=5$$So to be found is the number of such sums. This can be done with stars and bars, leading to $\binom{5+29}{29}$ possibilities.
b) (each student can receive at most one award)
$5$ different pupils must be selected for receiving the awards. That gives $\binom{30}{5}$ possibilities in total.


TITLE: Unique solution to a matrix equations
QUESTION [0 upvotes]: Given any $n \times k$ real matrix $M$, where $n<k$ and $rank(M)=n$, I consider the following equation (where $M'$ is the transpose of $M$):
$$
MM' = MAM'
$$
Then clearly, $A = \mathbb{1}_k $, the $k\times k$ identity matrix, is a possible solution.
Is that unique, or there are other possible solutions?

REPLY [2 votes]: Here is a counterexample.
I will use MATLAB notation for matrices, so ; ends each row of the matrix.
Let M be the 2 by 3 matrix M = [1 2 3;4 5 6].
Then A = [ 1.25 0 0.25;0 0 0;0.25 0 1.25].
MM' = MAM'= [14 32;32 77].


TITLE: If $B$ is the set of condensation points of a set $A$, show that every point of $B$ is a condensation point of $B$.
QUESTION [1 upvotes]: This problem comes from Dieudonné's Foundations of Modern Analysis:

Let $E$ be a separable metric space. A condensation point $x$ of a subset $A$ of $E$ is a point $x \in E$ such that in every neighborhood of $x$, there is a nondenumerable set of points of $A$. Show that: (a) If $A$ has no condensation point, it is denumerable (consider the intersections of $A$ with the sets of a basis for the open sets of $E$).
(b) If $B$ is the set of condensation points of a set $A$, show that every point of $B$ is a condensation point of $B$, and that $A \cap (\complement B)$ is at most denumerable. (Observe that $B$ is closed, and use (a).)

I've managed to solve (a) in the following way:
Let $x$ be any point of $A$. There exists a neighbourhood $V_x$ of $x$, such that it contains at most $\aleph_0$ elements from $A$ (since $A$ has no condensation points). Since $E$ is separable, there exists a denumerable basis $(G_\lambda)$ for all open sets of $E$, and there exists an element of the basis $G_x$ such that $x \in G_x \subset V$. Now we have $A \subset \bigcup_{x \in A} G_x$. But since every set $G_x$ contains at most $\aleph_0$ elements of $A$ (as it is a subset of $V_x$) and the sum $\bigcup_{x \in A} G_x$ is a sum of at most $\aleph_0$ distinct sets, $\bigcup_{x \in A} G_x$ contains itself at most $\aleph_0$ elements of $A$, and thus $|A| \leq \aleph_0$.
However, I can't tackle (b), even with the hint. I don't really feel how closeness of $B$ is supposed to help us here.

REPLY [1 votes]: For part (a): Your first part may not be correct. Since you don't apriori know about countability (or uncountability of $A$), you don't know apriori whether  $ \bigcup_{x \in A} G_x$ is a countable union or uncountable union. (Note that countable union of countable sets is necessarily countable).
Therefore, you may proceed as follows for this part:
Let $(G_n)$ be the countable base of $E$.
Since $A$ does not have any condensation points, if $A$ is empty, we are done. If $A$ is non empty, then let $N=\{k\in \mathbb N: G_{n_k}\cap A \text{ is countable }\}$.
Define $S=\cup_{k\in N} G_{n_k}$.  Let $E\setminus S\ne\emptyset $. Then for any $x$ in $E\setminus S$, if $B(x,\delta)$ is any arbitrary open ball around it then there exists a $G_{n_{k'}}\in (G_n)$ such that $x\in G_{n_{k'}}\subset B(x,\delta)$ and clearly
$n_{k'}\notin N$. It follows that $x$ is a condensation point of $A$, which is a contradiction. Hence $E\setminus S=\emptyset$. It follows that $E=S\implies A$ is countable.
For part $(b)$: First of all, note that $B$ is closed. If not, let $cl(B)-B\ne \emptyset$.
Let $x\in cl (B)-B$. It follows that $x$ is not a condensation point of $A$ and hence $\exists \delta\gt 0$ such that $B(x,\delta)\cap A$ is countable.
Since $x$ is a limit point of $B, B(x,\delta)\cap B$ is infinite. Choose $y\in B(x,\delta)\cap B$ such that $y\ne x$ and take $r=\frac 12\min\{d(x,y),\delta -d(x,y)\}$ so that $B(y,r)\subset B(x,\delta) \implies B(y,r)$ contains countably many points of $A$ which is a contradiction as $y$ is a condensation point of $A$.  Therefore, $cl (B)=B\iff B$ is closed. It follows that $B^c$ is open.
Now let $C= \text{set of condensation points of }A\cap B^c$.
Claim: $C$ is empty.
Proof: If not, suppose that $C\ne \emptyset$. Let $p\in C$ be any point in $C$.  For any $\epsilon \gt 0, B(p,\epsilon)\cap (A\cap B^c)\subset B(x,\epsilon)\cap A\implies B(p,\epsilon)\cap (A\cap B^c)$ is countable which is a contradiction as $p$ is a condensation point of $A\cap B^c$. Hence it follows by contradiction that $C=\emptyset\implies A\cap B^c$ is countable.


text:
\begin{document}

\title[Tannakian Formalism and Langlands Conjectures]
{The Tannakian Formalism and the Langlands Conjectures}

\author{David Kazhdan}
\email{kazhdan@math.huji.ac.il}
\address{Einstein Institute of Mathematics\\
    Hebrew University \\
    Givat Ram, Jerusalem 91904\\
    Israel}

\author{Michael Larsen}
\email{mjlarsen@indiana.edu}
\address{Department of Mathematics\\
    Indiana University \\
    Bloomington, IN 47405\\
    U.S.A.}

\author{Yakov Varshavsky}
\email{vyakov@math.huji.ac.il}
\address{Einstein Institute of Mathematics\\
    Hebrew University \\
    Givat Ram, Jerusalem 91904\\
    Israel}

\thanks{DK was partially supported by ISF grant 1438/06.
ML was partially supported by NSF Grant DMS-0800705.
YV was partially supported by ISF grant 598/09.}

\begin{abstract}
Let $H$ be a connected reductive group over an algebraically closed field of characteristic zero, and let $\Gm$ be an abstract group.
In this note we show that  every homomorphism of  Grothendieck semirings $\phi:K_0^+[\Gm]\to K_0^+[H]$, which maps irreducible
representations to irreducible, comes from a group homomorphism  $\rho:\Gm\to H(K)$. We also connect this result with the Langlands conjectures.

\end{abstract}

\maketitle

\section*{Introduction}
Let $F$ be a global function field, $\Gm_F$ the absolute Galois group of $F$,  $G$ a split connected reductive group over $F$,   $l$  a prime number different from the characteristic of $F$, and  $\hat G={}^LG^0$ the connected Langlands dual group over $\Qlb$.

Recall that a weak  Langlands conjecture asserts that for every $(\pi,\om)$, where $\pi$ is
an automorphic representation $\pi$ of $G$, whose central character is of finite order, and $\om$ is a representation of $\hat G$, there exists  a unique semisimple $\ell$-adic representation $\rho_{\pi,\om}$ of $\Gm_F$, whose
$L^S$-function is equal to the $L^S$-function of $(\pi,\om)$.

 Moreover, a strong Langlands conjecture asserts that there exists a $\hat G$-valued $\ell$-adic  representation
 $\rho_{\pi}:\Gm_F\to \hat G(\Qlb)$ (not unique in general) such that the composition $\om\circ \rho_{\pi}$ is isomorphic to $\rho_{\pi,\om}$ for each representation
 $\om$.

The main result of this note implies that in some cases the strong Langlands conjecture follows from the weak one. More specifically we show the existence
 of $\rho_{\pi}$  in the case when $\rho_{\pi,\om}$ is irreducible for each irreducible representation $\om$. Moreover, in this case, $\rho_{\pi}$ is unique up to conjugation, and the Zariski closure of its image contains the derived group of $G$.

Our result is a corollary of the following variant of the Tannakian formalism. Let $H$ be a connected reductive group over an algebraically closed field $K$ of characteristic zero, and let $\Gm$ be an abstract group. Then every homomorphism  of groups $\rho:\Gm\to H(K)$ induces a homomorphism of  Grothendieck
semirings $\rho^*:K_0^+[\Gm]\to K_0^+[H]$.  In this note we show a partial converse of this assertion. Namely, we show that every homomorphism of  Grothendieck
semirings $\phi:K_0^+[\Gm]\to K_0^+[H]$, which maps irreducible representations to irreducibles, comes from a group homomorphism  $\rho:\Gm\to H(K)$.
In particular, we show that a connected reductive group is determined by its Grothendieck semiring.

This note was inspired by a combination of a work in progress \cite{KV}, where it is shown that a weak Langlands conjecture holds in many cases, and a work
\cite{LP}, which indicates that one does not need the full Tannakian structure in order to reconstruct a connected reductive group.

We would like to acknowledge the contribution of Pavel Etingof, whose suggestions led to substantial simplification and conceptual clarification of this paper.

Part of the work was done while DK and YV visited the University of Chicago.
The rest of the work was done while YV visited Indiana University.
We thank both these institutions for stimulating atmosphere and financial support.
We would also like to thank Michael Mueger for calling our attention to two previous proofs of Theorem~\ref{kgroups} below.

\section{Main results}
Let $K$ be an algebraically closed field of  characteristic zero.

\begin{Emp}
(a) For every  algebraic group $G$ over $K$, we denote by $K_0^+[G]$ the Grothendieck semiring of the category of rational representations
of $G$.

In other words, $K_0^+[G]$ is the set of equivalence classes of finite-dimensional semisimple representations of $G$. For every representation
$\om$ of $G$, we denote by $[\om]$ its class (or more precisely, the class of its semisimplification) in $K_0^+[G]$.  For every pair of semisimple representations $\om_1$ and $\om_2$ of $G$, we have
$[\om_1]+[\om_2]=[\om_1\oplus\om_2]$ and $[\om_1]\cdot[\om_2]=[\om_1\otimes\om_2]$.

(b) Note that a representation $\om$ of $G$ is irreducible, if and only if its class $[\om]\in K_0^+[G]$ is irreducible, that is, it cannot be realized as a nontrivial sum
$[\om_1]+[\om_2]$ of elements of  $K_0^+[G]$.

(c) Every homomorphism $\rho:G\to H$ of algebraic group over $K$ gives rise to the  homomorphism
$\rho^*:K_0^+[H]\to K_0^+[G]$ of semirings, where $\rho^*([\om]):=[\om\circ\rho]$.

\end{Emp}
The following result asserts that each connected reductive group is determined by its Grothendieck semiring.
\begin{thm}
\label{kgroups}
Let $G$ and $H$ be two connected reductive groups over $K$, and let $\phi:K_0^+[G]\isom K_0^+[H]$ be an isomorphism of semirings.

Then there exists an isomorphism $\rho\colon H \isom G$ such that $\rho^*=\phi$. Moreover, $\rho$ is unique up to conjugation.
\end{thm}

\begin{Emp}
{\bf Remark.} Note, by comparison, that if $G$ is connected, semisimple, and simply connected, then the Grothendieck \emph{ring} $K_0[G]$ is isomorphic to
$\Z[x_1,\ldots,x_r]$, where $r$ is the rank of $G$.  Thus, for such groups, $K_0[G]$ encodes only the rank.
\end{Emp}

\begin{Emp}
Now let  $\Gm$ be an abstract group, and let $K_0^+[\Gm]$ be the Grothendieck semiring of the category of finite-dimensional representations
of $\Gm$ over $K$. Every group homomorphism $\rho:\Gm\to G(K)$ gives rise to the homomorphism
 $\rho^*:K_0^+[G]\to K_0^+[\Gm]$ of Grothendieck semirings.
 \end{Emp}

We have  the following version of the Tannakian formalism.

\begin{thm}
\label{main}
Let $\phi:K_0^+[G]\to K_0^+[\Gm]$ be a homomorphism of semirings which maps irreducible elements to irreducibles.

Then  there exists a  homomorphism
$\rho\colon \Gamma \to G(K)$ such that  $\rho^*=\phi$.
Moreover,  the Zariski closure of the image of each such $\rho$ contains $G^{\der}$, and $\rho$ is unique up to conjugation.
\end{thm}
\begin{Emp}
\label{remmain}
{\bf Remarks.}

(a) Conversely, let  $\rho\colon \Gamma \to G(K)$ be a homomorphism such that  the Zariski closure of $\rho(\Gamma)$ contains $G^{\der}$.
Then the homomorphism $\rho^*:K_0^+[G]\to K_0^+[\Gm]$ maps irreducible elements to irreducibles.

(b) The result fails completely if one does not assume that $\phi$ maps irreducible elements to irreducible.

Indeed, let $G$ be $\SL_2$, and let $\Gm$ be the group with one element. In this case,  for each integer
$k\geq 2$ there exists a (unique) homomorphism of semirings

$$\phi_k: K_0^+[\SL_2]\to K_0^+[\Gm]=\Z_{\geq 0},$$

which maps the standard representation of
$\SL_2$ to $k\in\Z_{\geq 0}$.  Only $\phi_2$ corresponds to a (unique) homomorphism $\Gm\to \SL_2(K)$.
 \end{Emp}

\begin{Emp}
\label{cg}
{\bf Chevalley space.}
(a) Let $c_G:=\Spec K[G]^G$ be the  Chevalley space of $G$. For every representation $\om$ of $G$, its trace $\Tr_{\om}\in K[G]^G\subset K[G]$ is a
regular function on $c_{G}$.

(b) Let $\chi_G\colon G\to c_G$ be the canonical projection,
induced by the embedding $K[c_G]=K[G]^G\hookrightarrow K[G]$. Then  for each $g\in G$
and each representation $\om$ of $G$, we have an equality $\Tr_{\om}(\chi_G(g))=\Tr_{\om}(g)$.
\end{Emp}

The following result is a more explicit formulation of Theorem~\ref{main}.

\begin{cor}
\label{cmain}
Let $f:\Gm\to c_G(K)$ be a map of sets.

Suppose that  for every irreducible algebraic representation  $\om$ of $G$, there exists an
irreducible finite dimensional representation $\rho_\om$ of $\Gm$ over $K$ such that

\begin{equation}
\label{trace-condition}
\Tr_{\rho_\om}(\gm) = \Tr_\om(f(\gm))  \text{ for all }\gm\in\Gm.
\end{equation}

Then there exists a  homomorphism $\rho\colon \Gamma \to G(K)$ such that

\begin{equation}
\label{lift}
\chi_G(\rho(\gm)) = f(\gm ) \text{ for all }\gm\in\Gm.
\end{equation}

Moreover, the Zariski closure of $\rho(\Gamma)$ contains $G^{\der}$, and $\rho$ is unique up to conjugation.
\end{cor}
\begin{Emp}
{\bf Remark.}
Conversely, assume that there exists a homomorphism   $\rho\colon \Gamma \to G(K)$ satisfying (\ref{lift}) and such that
 the Zariski closure of $\rho(\Gamma)$ contains $G^{\der}$. Then for every irreducible representation $\om\colon G\to \GL_{n}$
 the representation

 $$\rho_{\om}:=\om\circ\rho:\Gm\to \GL_{n}(K)$$

 is irreducible  and satisfies  (\ref{trace-condition})
 (use \ref{cg} (b)).
 \end{Emp}

\begin{Emp}
{\bf Application to the Langlands conjectures.}

Let $F$ be a global field, $F^{\mathrm{sep}}$ a separable closure, $\Gamma_F = \mathrm{Gal}(F^{\mathrm{sep}}/F)$ the absolute Galois group, and
$\ell$ a prime number, different from the characteristic of $F$. Let $\hat G$ be a connected reductive group over $\Qlb$.

By an {\em $\ell$-adic} (resp. {\em $\hat G$-valued $\ell$-adic}) representation of $\Gm_F$, we mean a continuous homomorphism $\rho\colon \Gamma_F\to \GL_n(\Qlb)$ (resp. $\rho\colon \Gamma_F\to \hat G(\Qlb)$) which is unramified for almost all places of $F$.

There is a well-defined trace
$\Tr_{\rho}(\Frob_v)$ (resp.  $\Tr_{\om\circ\rho}(\Frob_v)$) for almost all places  $v$ of $F$ (resp. and all representations $\om$ of $\hat G$).
\end{Emp}

The following analogue of Corollary~\ref{cmain} has applications to Langlands conjectures.

\begin{cor}
\label{galrep}
Let $\hat G$ be a reductive group over $\Qlb$,  $\Sigma$ a cofinite subset of the places of $F$,  and $f\colon \Sigma\to c_{\hat G}(\Qlb)$  any map of sets.

Assume that for every irreducible algebraic representation $\om$ of $G$, there exists an irreducible
$\ell$-adic  representation $\rho_\om$ of $\Gamma_F$ such that

\begin{equation}
\label{frob-condition}
\Tr_{\rho_\om}(\Frob_v) = \Tr_{\om}(f(v))\text { for almost all } v\in \Sigma.
\end{equation}

Then there exists a $\hat G$-valued $\ell$-adic representation $\rho\colon \Gamma_K\to \hat G(\Qlb)$ such that

\begin{equation}
\label{lift1}
\chi_{\hat G}(\rho(\Frob_v)) = f(v)\text{  for almost all }v\in \Sigma.
\end{equation}

Moreover, the Zariski closure of $\rho(\Gamma_F)$ contains $\hat G^{\der}$,  and $\rho$ is unique up to conjugation.
\end{cor}

\section{Determining connected reductive group by its Grothendieck semiring}
In this section, we are going to prove Theorem~\ref{kgroups}.
Michael Mueger called our attention to the fact that at least two proofs of this theorem already exist in the literature: \cite{Mc} and \cite{Ha}.  Nevertheless, we feel that this new proof has
merits (including brevity) which justify presenting it.

Let $G$ be a connected reductive group. We will fix a Borel subgroup $B\subset G$ and a maximal torus $T\subset B$.
Let $\al_1,\ldots,\al_r$ be the simple roots of $G$ with respect to $(B,T)$, and let $W$ be the Weyl group of $(G,T)$.

\begin{Emp}
\label{notation-weights}

(a) For each subset $X\subset X^*(T)$ we denote by $\Conv(X)$ the convex hull of
$X\subset X^*(T)\otimes\R$.

(b) For each dominant weight $\nu$ of $G$ we denote by $V_{\nu}$ the irreducible representation of $G$ with highest weight $\nu$.

(c) We define a partial order on $X^*(T)$ by the rule $\mu\leq\la$ if and only if $\la=\mu+\sum_{i=1}^r x_i\al_i$ and $x_i\geq 0$ for all $i$.

\end{Emp}

\begin{prop}
\label{order}
Let $\mu$ and $\la$ be two dominant weights of $G$. The following conditions are equivalent:

(a) $\mu\leq \la$;

(b) $\Conv(W\mu)\subset \Conv(W\lambda)$;

 (c) There exists a finite dimensional representation $V'$ of $G$ such that
for every $n$, every irreducible factor of $V_\mu^{\otimes n}$ is a factor of
$V_\lambda^{\otimes n}\otimes V'$.
\end{prop}

\begin{proof}

$(a)\implies (b)$.
Notice that since $\mu$ is dominant, we have $w\mu\leq\mu$ for all $w\in W$.
Therefore our assumption $\mu\leq\la$ implies that $w\mu\leq\la$ for all $w\in W$.
Thus our assertion follows from the following lemma.

\begin{lemma}
\label{worbit}
Let $\mu$ and $\la$ be two weights of $G$ such that  $w\mu\leq\la$ for all $w\in W$.
Then $\Conv(W\mu)\subset \Conv(W\lambda)$.
\end{lemma}

\begin{proof}
Suppose $\Conv(W\mu)$ is not contained in $\Conv(W\lambda)$.
Then there exists $w\in W$ such that $w\mu\not\in \Conv(W\lambda)$. As
$\Conv(W\lambda)$ is $W$-stable, it follows that $\mu\not\in \Conv(W\lambda)$, and hence also
$w\mu\notin \Conv(W\lambda)$  for all $w\in W$.

By the separation lemma, there exists  $\theta\in V^*$ such that
$\theta(\mu) > \theta(w\la)$ for all $w\in W$.
This is an open condition, so we may choose $\theta$ such that $\theta(\alpha_i)\neq 0$
for each $i=1,\ldots,r$. Replacing $\theta$ by $w\theta$ and $\mu$ by $w\mu$ for some
$w\in W$, we may assume in addition that $\theta(\alpha_i)>0$ for each $i=1,\ldots,r$.

By our assumption, $\mu=\la-\sum_{i=1}^r x_i\al_i$ with each $x_i\geq 0$. Therefore

$$\theta(\mu)=\theta(\la)-\sum_{i=1}^r x_i\theta(\al_i)\leq\theta(\la),$$

contradicting our assumption
$\theta(\mu) > \theta(\la)$.
\end{proof}

$(b)\implies (c)$.
We start with the following lemma.

\begin{lemma}
\label{convex}
Let $X$ be a finite subset of a finite dimensional Euclidean space $E$. Then there exists a compact subset $Y$ of $E$ such that

\begin{equation}
\label{quantize}
\Conv(nX) \subset Y+\underbrace{X+X+\cdots+X}_n
\end{equation}

for all positive integers $n$.
\end{lemma}

\begin{proof}
Let $m:= |X|$, and let $Y$ denote the ball of vectors of norm at most
$R := 2m \max_{x\in X} \Vert x\Vert.$
We claim that inclusion (\ref{quantize}) holds for this $Y$.

Let $X$ be the set $\{x_1,\ldots,x_m\}$. Then every vector in $\Conv(nX)$ is of the form

$$v := a_1 n x_1 + \cdots + a_m n x_m,$$

where the $a_i$ are non-negative and sum to $1$.  Let $b_i := \lfloor na_i\rfloor$ for $i\ge 2$
and $b_1 = n-(b_2+\cdots+b_m)$.  As $|b_i - a_i n| < 1$ for $i>1$, we have

$$|b_1 - a_1 n| = |n-(b_2+\cdots+b_m) + (a_2 n+\cdots+a_m n-n)| < m-1.$$

Thus,

$$\Vert(b_1 x_1 + \cdots + b_m x_m) - v \Vert \le \sum_{i=1}^m |b_i - a_i n| \Vert x_i\Vert
< R,$$

and of course $b_1 x_1+\cdots+b_m x_m$ belongs to the $n$-fold iterated sum of $X$.
\end{proof}

Now we return to the proof of the proposition. We assume that $\Conv(W \mu)\subset \Conv(W \lambda)$, let $X = W\lambda$, and fix a compact set $Y$ satisfying (\ref{quantize}).  Denote by $V'$  the direct sum of all representations $V_\nu$ where $\nu$ ranges over the dominant weights in $W Y$.

If $n$ is a positive integer, the highest weight $\chi$ of any irreducible factor of $V_\mu^{\otimes n}$ is a weight of
$V_\mu^{\otimes n}$. Therefore $\chi\leq n\mu$, hence by the implication $(a)\implies(b)$ shown above, $\chi$ is an element
of
\[
\Conv(Wn\mu)= n\Conv(W\mu)\subset n\Conv(W\la).
\]
By (\ref{quantize}), $\chi$ can be written as a sum of $n$ elements of $W\la$ and an element of $W Y$,
which is necessarily in the weight group.
Thus $\chi$ has the form $\sum_{i=1}^n w_i\la+w'\nu$ for some $w_1,\ldots,w_n,w'\in W$ and some highest weight $\nu$ of $V'$.

Using the conjecture of Parthasarthy, Ranga-Rao, and Varadarajan, proven by Kumar \cite{Ku}, we conclude that $V_{\chi}$
is an irreducible factor of $V_\lambda^{\otimes n}\otimes V_{\nu}$, hence also an irreducible factor of $V_\lambda^{\otimes n}\otimes V'$.

$(c)\implies(a)$.
Now suppose that there exists a finite dimensional representation $V'$ of $G$ such that
for every $n$, every irreducible factor of $V_\mu^{\otimes n}$ must be a factor of
$V_\lambda^{\otimes n}\otimes V'$ as well.  Then every weight of $V_\mu^{\otimes n}$
must be a weight of $V_\lambda^{\otimes n}\otimes V'$, and in particular, this is true for the weight
$n\mu$. Thus $n\mu=\la_n+\nu_n$ for some weights $\la_n$ of  $V_\lambda^{\otimes n}$ and $\nu_n\in V'$.

Note that $\la_n=n\la-\sum_{i=1}^r n_i\al_i$ for some $n_i\in\Z_{\geq 0}$. Therefore $n\mu$ is equal to
$n\la-\sum_{i=1}^r n_i\al_i+\nu_n$, hence for each $n\in\N$ we have an equality
\[
\la-\mu=\sum_{i=1}^r \frac{n_i}{n}\al_i-\frac{1}{n}\nu_n.
\]
Next we recall that the set of weights  of $V'$ is finite, so the expression $\frac{1}{n}\nu_n\in X^*(T)\otimes\R$
tends to zero when $n$ tends to infinity. Hence the difference $\la-\mu$ equals $\sum_{i=1}^r x_i\al_i$, where each $x_i=\lim_{n\to\infty}\frac{n_i}{n}$
is non-negative. This shows that $\mu\leq\la$.
\end{proof}

\begin{cor}
\label{root datum}
The root datum of $G$ can be reconstructed from the semiring $K_0^+[G]$.
\end{cor}

\begin{proof}
We divide our construction into steps as follows.

{\bf Step 1.} First we claim that a partially ordered set of dominant weights of $G$ can be reconstructed from the semiring $K_0^+[G]$.

For this we note that the map $\mu\mapsto [V_{\mu}]$
gives a bijection between the set of dominant weights of $G$  and the set of irreducible objects of of $K_0^+[G]$.

Proposition~\ref{order} implies that for two dominant weights $\mu,\la$ of $G$  we have $\mu\le \lambda$ if and only if there exists $\theta\in K_0^+[G]$ such that for all $n\in \N$ and all irreducible elements $[V_\nu]\in K_0^+[G]$, we have

$$[V_\mu]^n - [V_\nu] \in K_0^+[G]\Rightarrow [V_\lambda]^n\theta-[V_\nu] \in K_0^+[G].$$

{\bf Step 2.}
For every triple $\la,\mu,\nu$ of dominant weights  of $G$,
we have $\nu=\la+\mu$
if and only if $\la$ is the largest dominant weight such that $V_{\la}$ is an irreducible factor of $V_{\mu}\otimes V_{\nu}$. Therefore
Proposition~\ref{order} implies that the semigroup of dominant weights of $G$ can be reconstructed  from  the semiring $K_0^+[G]$.

{\bf Step 3.} The group of weights $X^*(T)$ of $G$ is the group completion of the semigroup of dominant weights.  The group of coweights of $G$, $X_*(T)$,  is given as the group of homomorphisms

$$X_*(T)=\Hom(X^*(T),\Z).$$

Note that there is a canonical isomorphism between $\Aut(X^*(T))$ and $\Aut(X_*(T))$.

{\bf Step 4.} We claim that $\al\in X^*(T)$ is a simple root if and only if it is a minimal weight of $T$ for which there exists a dominant weight $\lambda\in X^*(T)$ such that   $V_{2\lambda-\al}$  is an irreducible factor of $V_{\lambda}^{\otimes 2}$.

Indeed, for every simple root $\al$, we choose any dominant weight $\la$ such that
$\langle\check{\al},\la\rangle=1$. Then it follows from Kumar's theorem \cite{Ku} that
$2\lambda-\alpha = \lambda + s_{\alpha}(\lambda)$ is the highest weight of a factor of $V_\lambda^{\otimes 2}$. Conversely, if $V_{2\lambda-\beta}$ is an irreducible factor of $V_\lambda^{\otimes 2}$, then $2\lambda-\beta$ is a weight of $V_\lambda^{\otimes 2}$, thus
$\beta$ is of the form $a_1\alpha_1+\cdots+a_r\alpha_r$ with all $a_i$ non-negative integers. Hence any minimal such $\beta$ has to be a simple root.

By the Step 1, the set of simple roots can  therefore be reconstructed from the semiring $K_0^+[G]$.

{\bf Step 5.} For each simple root $\al$ of $G$ the corresponding simple coroot $\check{\al}\in X_*$ can be characterized by the following condition:
for every dominant weight $\mu$ the pairing $\langle\check{\al},\mu\rangle$ is the unique element  $m\in \Z_{\geq 0}$ such that
$2\mu-m\al$ is dominant, but $2\mu-(m+1)\al$ is not dominant.  Indeed,
\[
\langle\check{\al},2\mu-m\al\rangle=2\langle\check{\al},\mu\rangle-m\langle \check{\al},\al\rangle= 2\langle\check{\al},\mu\rangle-2m
\]
is non-negative if and only if $m\leq \langle\check{\al},\mu\rangle$, while for every other simple root $\al'\neq\al$ of $G$ with a corresponding
simple coroot $\check{\al}'$, we have
\[
\langle\check{\al}',2\mu-m\al\rangle=2\langle\check{\al}',\mu\rangle-m\langle \check{\al}',\al\rangle\geq 2\langle\check{\al}',\mu\rangle\geq 0
\]
for all $m\geq 0$. Thus the set of simple coroots can also
be reconstructed from $K_0^+[G]$.

{\bf Step 6.} After having reconstructed  all simple coroots $\check{\al}$, we reconstruct all simple reflections $s_{\al}\in \Aut(X_*(T))$, hence the Weyl group $W\subset \Aut(X_*(T))$, as the subgroup generated by simple reflections. Next we reconstruct the set of all roots of $G$, as images of the simple roots under $W$,
and likewise for the coroots of $G$.
This completes the reconstruction of the whole root datum of $G$.
\end{proof}

\begin{Emp}
\begin{proof}[Proof of Theorem~\ref{kgroups}]
An isomorphism of semirings $\phi:K_0^+[G]\isom K_0^+[H]$ induces  a bijection between irreducible objects, hence a bijection between  dominant weights of $G$ and $H$, which we  denote by $\wt{\phi}$.

The proof of  Corollary~\ref{root datum} shows that $\wt{\phi}$ extends to the isomorphism between the root data of $G$ and $H$, hence it comes from an isomorphism of algebraic groups $\rho:H\isom G$.

We claim that $\rho^*:K_0^+[G]\isom K_0^+[H]$ is equal to $\phi$. It is enough to show that for each dominant weight $\la$ of $G$, we have $\phi([V_{\la}])=\rho^*([V_{\la}])$.  Both expressions, however, are equal to $[V_{\wt{\phi}(\la)}]$.

Conversely, if $\rho:H\to G$ is an isomorphism such that $\rho^*=\phi$, then for each dominant weight $\mu$ of $G$ we have $\rho^*([V_{\la}])=\phi([V_{\la}])=[V_{\wt{\phi}(\la)}]$, so $\rho$  induces the isomorphism $\wt{\phi}$ between the root data, hence $\rho$ is unique up to conjugation.

\end{proof}
\end{Emp}

\section{The Tannakian formalism}
In this section we are going to prove Theorem~\ref{main}. Throughout the section we will assume that  the hypotheses of Theorem~\ref{main} hold. For each irreducible representation $\om$ of $G$ we choose an irreducible representation $\rho_{\om}$
of $\Gm$ such that $[\rho_\om]=\phi([\om])$.

\begin{lemma}
\label{isom}
(a) Let $\om'$ and $\om''$ be two irreducible representations of $G$, and
let $\om'\otimes\om''\cong \oplus\om_i$ be a decomposition of their tensor product
into irreducibles. Then $\rho_{\om'}\otimes\rho_{\om''}\cong \oplus\rho_{\om_i}$.

(b)  If $\om$ is a trivial (one-dimensional) representation $\1$ of $G$, then $\rho_{\om}$ is a trivial representation of $\Gm$.

(c) The representation $\om$ is one-dimensional if and only if $\rho_{\om}$ is one-dimensional.

(d) For each irreducible representation $\om$ of $G$, we have $\rho_{\om^*}\cong(\rho_{\om})^*$.

(e)  Let $\om'$ and $\om''$ be two irreducible representations
of $G$ such that $\rho_{\om'}\cong\rho_{\om''}$. Then restrictions $\om'|_{G^{\der}}$ and $\om''|_{G^{\der}}$ are isomorphic.
\end{lemma}

\begin{proof}
(a) By hypothesis, we have $[\om']\cdot[\om'']=\sum_i [\om_i]$. Since $\phi$ is a homomorphism of semirings we conclude that
\[
[\rho_{\om'}\otimes\rho_{\om''}]=\phi([\om'])\cdot\phi([\om''])=\sum_i \phi([\om_i])=[\oplus\rho_{\om_i}].
\]
Since   $\rho_{\om'}$ and $\rho_{\om''}$ are irreducible, their tensor product  $\rho_{\om'}\otimes\rho_{\om''}$ is semisimple (see \cite[p.~88]{Ch}).
Therefore $\rho_{\om'}\otimes\rho_{\om''}\cong \oplus\rho_{\om_i}$.

(b) This follows from the observation that $\om=\1$ if and only if $\om\otimes\om\cong\om$.

(c) This follows from the observation that $\om$ is one-dimensional if and only if $\om\otimes\om$ is irreducible.

(d) Note that the representation $\om\otimes \om^*$ has a trivial subrepresentation $\1$. Therefore by (a) and (b), the representation
 $\rho_{\om^*}\otimes\rho_{\om}$ has a subrepresentation $\rho_{\1}\cong \1$. Since $\rho_{\om}$ and
 $\rho_{\om^*}$ are irreducible, this implies that $\rho_{\om^*}\cong(\rho_{\om})^*$.

 (e) If $\rho_{\om'}\cong\rho_{\om''}$, then   the tensor product $\rho_{\om'}\otimes(\rho_{\om''})^*\cong \rho_{\om'}\otimes\rho_{\om''^*}$ contains a
 subrepresentation $\1$. Using  (a) and (c), we conclude that the tensor product
 $\om'\otimes\om''^*$ has a one-dimensional subrepresentation $\xi$. Since $\om'$ and $\om''$ are irreducible, we conclude that
 $\om'\cong\om''\otimes\xi$, thus the restrictions $\om'|_{G^{\der}}$ and $\om''|_{G^{\der}}$ are isomorphic.
\end{proof}

\begin{Emp}
For every irreducible representation $\om$ of $G$ we denote by $z_{\om}$ its central character.
Let $Z$ be the center of $G$, and denote by  $\iota$ the embedding $\Gm\isom\Gm\times\{1\}\hookrightarrow\Gm\times Z(K)$.
\end{Emp}

\begin{lemma}
\label{density}
(a) There exists a unique homomorphism of semirings
$\widetilde{\phi}:K_0^+[G]\to K_0^+[\Gm\times Z(K)]$  such that

\begin{equation}
\label{center}
\widetilde{\phi}([\om])=[\rho_{\om}\boxtimes z_{\om}] \text{  for each irreducible } \om.
\end{equation}

(b) Moreover, $\widetilde{\phi}$ is injective, maps irreducibles to irreducibles and satisfies $\iota^*\circ \widetilde{\phi}=\phi$.

(c) Assume that there exists a homomorphism $\rho:\Gm\to G(K)$ such that $\rho^*=\phi$, and let
$\wt{\rho}:\Gm\times Z(K)\to G(K)$ be a homomorphism defined by $\wt{\rho}(\gm,z):=\rho(\gm)\cdot z$. Then $\wt{\rho}^*=\wt{\phi}$.
\end{lemma}

\begin{proof}
(a) Since the additive Grothendieck semigroup $K_0^+[G]$ is freely generated by irreducible elements $[\om]$, there exists a unique homomorphism of semigroups
$\widetilde{\phi}:K_0^+[G]\to K_0^+[\Gm\times Z(K)]$ which satisfies (\ref{center}). It remains to show  that for every two representations $\om',\om''$ of $G$ we have an equality

\begin{equation}
\label{mult}
\widetilde{\phi}([\om']\cdot [\om''])=\widetilde{\phi}([\om'])\cdot\widetilde{\phi}([\om'']).
\end{equation}

By the additivity of $\widetilde{\phi}$, we may assume that $\om'$ and $\om''$ are irreducible.
Let $\om'\otimes\om''\cong \oplus\om_i$ be a decomposition of their tensor product
into irreducibles. Then $[\om']\cdot [\om'']=\sum_i[\om_i]$, hence the left hand side of (\ref{mult}) is equal to
\[
\widetilde{\phi}(\sum_i[\om_i])=\sum_i\widetilde{\phi}([\om_i])=\sum_i[\rho_{\om_i}\boxtimes z_{\om_i}],
\]
while the right hand side of (\ref{mult}) is equal to
\[
[\rho_{\om'}\boxtimes z_{\om'}]\cdot[\rho_{\om''}\boxtimes z_{\om''}]=[(\rho_{\om}\otimes\rho_{\om''})\boxtimes z_{\om'} z_{\om''}].
\]
Since the central character of each $\om_i$ is equal to $z_{\om'}z_{\om''}$, equality (\ref{mult}) follows from Lemma~\ref{isom} (a).

(b) By construction, for each irreducible element $[\om]$, the element   $\widetilde{\phi}([\om])=[\rho_{\om}\boxtimes z_{\om}]$ is irreducible, and

$$\iota^*\widetilde{\phi}([\om])=\iota^*([\rho_{\om}\boxtimes z_{\om}])=[\rho_{\om}]=\phi([\om]).$$

This  implies that $\wt{\phi}$ maps irreducibles to irreducibles
and satisfies   $\iota^*\circ \widetilde{\phi}=\phi$.

Finally, since as additive semigroups,  $K_0^+[G]$ and $K_0^+[\Gm]$ are freely generated by irreducibles, in order to show that $\wt{\phi}$ is injective, it is enough to show that it is injective on irreducibles.

Let $\om'$ and $\om''$ are two irreducible representations of $G$ such that $\wt{\phi}([\om'])=\wt{\phi}([\om''])$.
Then $\rho_{\om'}\cong \rho_{\om''}$ and $z_{\om'}=z_{\om''}$. Using Lemma~\ref{isom} (e), we conclude that  $\om'|_{G^{\der}}\cong \om''|_{G^{\der}}$ and
$\om'|_{Z}=\om''|_Z$. Hence $\om'\cong\om''$, implying the injectivity.

(c) It is enough to show that $\wt{\rho}^*([\om])=\wt{\phi}([\om])$ when $[\om]$ is irreducible. Both expressions, however, are equal to $[\rho_{\om}\boxtimes z_{\om}]$.

\end{proof}

\begin{Emp}

\begin{proof}[Proof of  Theorem~\ref{main}]
First we will show the existence of $\rho$ under the assumption that  $\phi:K_0^+[G]\to K_0^+[\Gm]$ is injective.

Let $\cC$ be the full subcategory of $\Rep\Gm$ consisting of semisimple representations
$\tau\in \Rep\Gm$ such that $[\tau]=\phi([\om])$ for some $[\om]\in K_0^+[G]$. Since $\phi([\om])$ is irreducible for each irreducible $[\om]$,
$\cC$  is a semisimple abelian subcategory.  Since $\phi$ is a homomorphism of semirings, $\cC$ is a rigid tensor subcategory  of $\Rep\Gm$
(use Lemma~\ref{isom} (a)--(d)), hence a Tannakian category. Let $f:\cC\to\Vect$ be the forgetful functor, and let $H:=\Aut^{\otimes}(f)$  be the group of tensor automorphisms of $f$.

By the Tannakian formalism (see, for example, \cite[Thm 2.11]{DM})  $H$ is an affine group scheme, and $f$
induces an equivalence of tensor categories $\cC\isom\Rep H$.  Since $G$ is an algebraic group, the category $\Rep G$ has a tensor generator $\om$.
Then an element $\rho_{\om}\in\Rep\Gm$ such that $[\rho_{\om}]=\phi([\om])$ must be a tensor generator of $\cC\cong\Rep H$. This implies that
$H$ is an algebraic group (see \cite[Prop 2.20]{DM}).  Moreover, since $\cC\cong\Rep H$ is semisimple, the group $H$ is reductive
(see \cite[Prop 2.23]{DM}).

Every element of $\gm\in\Gm$ defines a tensor automorphism of $f$ over $K$.
Hence we get a group homomorphism $\pi:\Gm\to H(K)$ such that
$\pi^*:\Rep H\to\Rep\Gm$ is the inverse of the equivalence $f:\cC\isom\Rep H$.

By construction, the homomorphism $\phi:K_0^+[G]\to K_0^+[\Gm]$ decomposes as
$K_0^+[G]\overset{\phi'}{\longrightarrow}K_0^+[H] \overset{\pi^*}{\longrightarrow}K_0^+[\Gm]$, and the homomorphism $\phi'$ is surjective.
By our assumption, $\phi'$ is also injective, hence it is an isomorphism.  Since $G$ is connected, we conclude that $H$ is connected as well (use, for example,
\cite[Cor 2.22]{DM}). Therefore by Theorem~\ref{kgroups} there exists an isomorphism $\rho':H\isom G$ such that $\phi'=\rho'^*$.
Then the composition $\rho:=\rho'\circ\pi:\Gm\to G(K)$ satisfies $\rho^*=\phi'\circ\pi^*=\phi$.

To show the existence of $\rho$ in general, we consider the homomorphism of Grothendieck semirings $\wt{\phi}:K_0^+[G]\to K_0^+[\Gm\times Z(K)]$, considered in
Lemma~\ref{center}.

Then $\wt{\phi}$ is injective, so by the particular case  shown above, there exists a homomorphism
$\wt{\rho}:\Gm\times Z(K)\to G(K)$  such that $\wt{\rho}^*=\wt{\phi}$. Then the composition
$\rho:=\wt{\rho}\circ\iota:\Gm\to G(K)$ satisfies
$\rho^*=\iota^*\circ\wt{\phi}=\phi$.

Conversely, let $\rho:\Gm\to G(K)$ be a homomorphism such that $\rho^*=\phi$. To show that the Zariski closure of $\rho(\Gm)$ contains $G^{\der}$, it
suffice to show that the homomorphism $\wt{\rho}:\Gm\times Z(K)\to G(K)$ from Lemma~\ref{center} (c) has a Zariski closed image.

Let $H\subset G$ be the
Zariski closure of the image of $\wt{\rho}$, and denote by $i$ the inclusion $H\hookrightarrow G$. Then
$\wt{\rho}^*=\wt{\phi}:K_0^+[G]\to K_0^+[\Gm\times Z(K)]$ factors through $i^*: K_0^+[G]\to K_0^+[H]$.
In particular, $i^*$ is injective, and maps irreducibles to irreducibles. Then using Chevalley's theorem (\cite[Th.~5.1]{Bo}) or (\cite[Prop. 2.21]{DM}),
$i$ has to be an isomorphism.

Finally, to show that $\rho$ is unique up to  conjugation, it suffice to show that $\wt{\rho}:\Gm\times Z(K)\to G(K)$
is unique up to conjugation.  Thus we can replace $\rho$ by $\wt{\rho}$, and $\phi$ by $\wt{\phi}$,  thereby assuming that $\phi$ is injective.

Then, using  the notation of the existence part, the tensor functor $\rho^*:\Rep G\to\Rep\Gm$ decomposes as a composition
$\Rep G\overset{\psi}{\longrightarrow}\Rep H \overset{\pi^*}{\longrightarrow}\Rep \Gm$ of tensor functors. By the Tannakian formalism,
there exists a homomorphism $\rho':H\to G$ such that $\rho'^*=\psi$. Then $\rho$ is conjugate to the composition $\rho'\circ \pi$,
so it remains to show that the conjugacy class of $\rho'$ is uniquely defined.

We have seen that $\phi$ decomposes as  $K_0^+[G]\overset{\phi'}{\longrightarrow}K_0^+[H] \overset{\pi^*}{\longrightarrow}K_0^+[\Gm]$, therefore
$\rho'^*:K_0^+[G]\isom K_0^+[H]$ coincides with $\phi'$. Hence the uniqueness assertion for $\rho'$  follows from Theorem~\ref{kgroups}.
\end{proof}
\end{Emp}

\section{Two Corollaries}
In this section we are going to prove Corollaries ~\ref{cmain} and \ref{galrep}.

\begin{lemma}
\label{equiv}
 Assume that  the hypotheses of Corollary~\ref{cmain} hold.

(a)   There exists a unique
homomorphism of semirings $\phi:K_0^+[G]\to K_0^+[\Gm]$ such that $\phi([\om])=[\rho_{\om}]$ for each irreducible $\om$.

(b) Let $\rho:\Gm\to G(K)$ be a group homomorphism. Then equality (\ref{lift}) holds for $\rho$ if and only if
$\om\circ\rho\cong\rho_{\om}$ for all irreducible $\om$.
\end{lemma}

\begin{proof}
(a)  Since the semigroup $K_0^+[G]$ is freely generated by irreducible elements,
there exists a unique homomorphism of semigroups $\phi:K_0^+[G]\to K_0^+[\Gm]$ such that $\phi([\om])=[\rho_{\om}]$ for each irreducible $\om$.

It remains to show that  for every two representations $\om',\om''$ of $G$ we have an equality

$$\phi([\om']\cdot[\om''])=\phi([\om'])\cdot\phi([\om'']).$$

Since a semisimple representation is determined by its trace,
it is enough to show that

\begin{equation}
\label{tr2}
\Tr_{\phi([\om']\cdot[\om''])}(\gm) = \Tr_{\phi([\om'])}(\gm)\cdot \Tr_{\phi([\om''])}(\gm)
\end{equation}

for all $\gm\in\Gm$. First we observe that for all $\gm\in\Gm$ and  all $[\om]\in K_0^+[G]$ we have an equality

\begin{equation}
\label{tr}
\Tr_{\phi([\om])}(\gm) = \Tr_{[\om]}(f(\gm)).
\end{equation}

Indeed, by the additivity, it is enough to show (\ref{tr}) for $\om$  irreducible. In this case the assertion follows from equalities  $\phi([\om])=[\rho_{\om}]$ and (\ref{trace-condition}).

Using  (\ref{tr}), our desired equality (\ref{tr2}) can be written in the form

$$\Tr_{[\om']\cdot[\om'']}(f(\gm))=\Tr_{[\om']}(f(\gm))\cdot \Tr_{[\om'']}(f(\gm)).$$

Therefore it follows from the multiplicativity of the trace map
$\Tr:K_0^+[G]\to K[G]$.

(b) Since functions $\Tr_{\om}$ with $\om$ irreducible generate $K[c_G]$ as a $K$-vector space (see \cite[Theorem 6.1(a)]{St}), equality (\ref{lift}) is equivalent to the equality

\begin{equation}
\label{lift'}
\Tr_{\om}(\chi_G(\rho(\gm)))=\Tr_{\om}(f(\gm))
\end{equation}

for all $\gm\in\Gm$ and all irreducible $\om$.
Since the left side of (\ref{lift'}) equals $\Tr_{\om\circ\rho}(\gm)$ (see section \ref{cg} (b)), while the right hand side of (\ref{lift'})
equals $\Tr_{\rho_{\om}}(\gm)$ by (\ref{trace-condition}), equality (\ref{lift'}) is equivalent to the equality $\Tr_{\om\circ\rho}=\Tr_{\rho_{\om}}$
for all irreducible $\om$. But this is equivalent to the desired isomorphism $\om\circ\rho\cong\rho_{\om}$.
\end{proof}

\begin{Emp}
\begin{proof}[Proof of Corollary~\ref{cmain}]
By Lemma~\ref{equiv}  (a), there exists a unique homomorphism of semirings $\phi:K_0^+[G]\to K_0^+[\Gm]$
such that $\phi([\om])=[\rho_{\om}]$ for each irreducible $\om$. Then by Theorem~\ref{main} there exists a
homomorphism $\rho:\Gm\to G(K)$ such that $\rho^*=\phi$. In particular, we have $[\om\circ\rho]$ is equal
to $\phi([\om])=[\rho_{\om}]$ for each irreducible $\om$. Then by Lemma~\ref{equiv}  (b),  the equality (\ref{lift})
holds for $\rho$.

Conversely, let $\rho:\Gm\to G(K)$ be a homomorphism, satisfying (\ref{lift}). Then  by Lemma~\ref{equiv}  (b),
$\rho^*([\om])=[\om\circ\rho]$ is equal to $\phi([\om])=[\rho_{\om}]$ for each irreducible $\om$. Thus
$\rho^*:K_0^+[G]\to K_0^+[\Gm]$ is equal to $\phi$.
Therefore it follows from Theorem~\ref{main} that $\rho$ is unique up to conjugation,
and that the Zariski closure of $\rho(\Gm)$ contains $G^{\der}$.
\end{proof}
\end{Emp}

\begin{Emp}
\begin{proof}[Proof of Corollary~\ref{galrep}]
The argument is almost identical to that of Corollary~\ref{cmain}.

As in Lemma~\ref{equiv} (a), there exists a unique homomorphism of semirings $\phi:K_0^+[\hat G]\to K_0^+[\Gm_F]$ such that $\phi([\om])=[\rho_{\om}]$
for each irreducible $\om$. Indeed, arguing as in Lemma~\ref{equiv} (a) word for word, we reduce ourselves to the  equality (\ref{tr2}).  Moreover, by the Chebotarev density theorem, it is enough to show equality (\ref{tr2}) when $\gm=\Frob_v$ for almost all $v\in\Sigma$.

Then we reduce the problem to showing that

$$\Tr_{\phi([\om])}(\Frob_v)=\Tr_{[\om]}(f(v))$$

for all irreducible $[\om]$ and almost all $v\in\Sigma$. But the latter equality follows from equalities
$\phi([\om])=[\rho_{\om}]$ and (\ref{frob-condition}).

By Theorem~\ref{main}, there now exists a homomorphism $\rho:\Gm_F\to \hat G(\Qlb)$ such that  $\rho^*=\phi$.

We claim that for every representation $\om$ of $G$, the composition $\om\circ\rho$ is a semisimple $\ell$-adic representation.
By additivity, it is enough to show in the case when $\om$ is irreducible. However, in this case,

$$[\om\circ\rho]=\rho^*([\om])=[\rho_{\om}]$$

is irreducible,
hence $\om\circ\rho\cong \rho_{\om}$ is an irreducible $\ell$-adic representation.

Choosing $\om$ to be a faithful representation of $\hat G$, we conclude that $\rho$ is continuous and unramified almost everywhere.

Finally, arguing exactly as in Lemma~\ref{equiv} (b) (and using the isomorphisms $\om\circ\rho\cong \rho_{\om}$)
we conclude that $\rho$ satisfies the equality (\ref{lift1}).

Conversely, let  $\rho:\Gm_F\to \hat G(\Qlb)$ be a $\hat G$-valued $\ell$-adic homomorphism satisfying
(\ref{lift1}). Again arguing exactly as in Lemma~\ref{equiv} (b) and using the Chebotarev density theorem, we conclude that
$\rho^*([\om])=[\om\circ\rho]$ is equal to $\phi([\om])=[\rho_{\om}]$ for each irreducible $\om$. Thus
$\rho^*:K_0^+[\hat G]\to K_0^+[\Gm_F]$ is equal to $\phi$.

Therefore it follows from Theorem~\ref{main} that $\rho$ is unique up to conjugation,
and that the Zariski closure of $\rho(\Gm_F)$ contains $\hat G^{\der}$.
\end{proof}
\end{Emp}


TITLE: Evaluate the integrals in $\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$
QUESTION [1 upvotes]: Note that when $0\le x \le 1$ we have $$\frac 12 \le \frac 1 {1+x^2} \le 1.$$
Hence, $$\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$$
Evaluate the integrals above to show that $$\frac1{1260}\le\frac {22}7-\pi\le\frac1{630}.$$
All the integrands include the expression $x^4(1-x)^4$ but I'm not sure how to continue.

REPLY [1 votes]: The first and third integrals can be solved using the substitution $x=\sin^2(\theta)$. The integral will take the form $2\displaystyle\int_0^{\pi/2}\sin^{2a-1}(\theta)\cos^{2b-1}(\theta)d\theta$, which is equal to the beta function $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$. The middle integral can be split into fractions according to Travis's answer.


TITLE: Find the volume of the solid formed by rotating the
QUESTION [0 upvotes]: Find the volume of the solid formed by rotating the region in the 1st quadrant enclosed by $$y = \frac{1}{11} x\ \ \ \text{and} \ \ \ y = x^{1/3}$$ about the y-axis.

REPLY [0 votes]: For the sake of closure on the result, I'm working out the numerical values by both the "disk method", which Semsem set up, and by the "shell method".
The "bounding functions" can be expressed as $ \ x = 11y \ $ and $ \ x = y^3 \ $ for applying "disks" (integration in the $ y-$ direction).  The intersection points of the curves are given by
$$ \frac{1}{11}x \ - \ x^{1/3} \ = \ 0 \ \ \Rightarrow \ \ x^{1/3} \ (\frac{1}{11}x^{2/3} \ - \ 1 ) \ = \ 0 \ , $$
so we have either $ \ x = 0 \ \Rightarrow \ y = 0 \ \ $ or $  \ \ x^{2/3} = 11 \ \Rightarrow \ x = 11^{3/2} \ \Rightarrow \ y = 11^{1/2} \ . $  The volume integral Semsem shows is then evaluated as
$$ \pi \ \left[ \ \frac{11^2}{3} y^3 \ - \ \frac{1}{7}y^7 \ \right]_0^{11^{1/2}} \ = \  \pi \ \left[ \ \frac{11^2}{3} \cdot (11^{1/2})^3 \ - \ \frac{1}{7} \cdot (11^{1/2})^7 \ - \ 0\ \right]  $$
$$ = \  \pi \ \left[ \ \frac{11^2}{3} \cdot (11^{3/2}) \ - \ \frac{1}{7} \cdot (11^{7/2})  \right] \ = \ \pi \  (11^{7/2})\  \left(  \frac{1}{3} \ - \ \frac{1}{7}   \right) \ = \ \frac{4 \pi}{21} \cdot  (11^{7/2}) \ ,  $$
which may also be written as $ \ \frac{4 \ \cdot \ 11^3 \ \cdot \ \pi}{21} \cdot  (11^{1/2}) \ $ or $ \ \frac{5324 \  \pi}{21} \sqrt{11} \ . $
Applying the "shell method" instead, which involves an integration in the $ x-$ direction, the "radius arm" extending from the axis of rotation, the $ \ y-$ axis, is $ \ r = x \ , $ and the height of the shells is given by $ \ x^{1/3} - \frac{1}{11}x \ . $  Here, the volume integral is
$$ 2 \pi \ \int_0^{11^{3/2}}  \ x \ (x^{1/3} \ - \ \frac{1}{11}x) \ \ dx \ = \ 2 \pi \ \int_0^{11^{3/2}}  \ x^{4/3} \ - \ \frac{1}{11}x^2 \ \ dx   $$
$$ = \ 2 \pi \ \left[ \ \frac{3}{7} y^{7/3} \ - \ \frac{1}{33}y^3 \ \right]_0^{11^{1/2}} \ = \  2 \pi \ \left[ \ \frac{3}{7} \cdot (11^{3/2})^{7/3} \ - \ \frac{1}{33} \cdot (11^{3/2})^3 \ - \ 0\ \right]  $$
$$ = \  2 \pi \ \left[ \ \frac{3}{7} \cdot (11^{7/2}) \ - \ \frac{1}{3 \cdot 11} \cdot (11^{9/2})  \right] \ = \ 2 \pi \  (11^{7/2})\  \left(  \frac{3}{7} \ - \ \frac{1}{3}   \right) \ = \ \frac{4 \pi}{21} \cdot  (11^{7/2})   $$
once again; the decimal expression is indeed $ \ \approx \ 2641.587 \ . $  So the result Andy shows in his comment looks to be correct; it is unclear from what is said here why he believed it to be otherwise.  (Perhaps this was a case of a textbook answer written in a different but equivalent form, or a computer-based system rejecting the way the answer was typed...)


TITLE: Is my proof of the continuity of $\frac{1}{x}$ correct?
QUESTION [3 upvotes]: I need to prove that $\frac{1}{x}$ is continuous, but my proof, as far as I did now, only applies to the interval $x\ge1$ wich is a progress. So, there's how I did:
$$\frac{|x-a|}{|x|}<|x-a|<\delta \tag{for $x>1$}$$
Then, by choosing $\delta=\epsilon|a|$ we have:
$$\frac{|x-a|}{|x|}<\epsilon|a|\implies\frac{|x-a|}{|x||a|}<\epsilon\implies\frac{|x-a|}{|xa|}<\epsilon \implies \frac{|a-x|}{|xa|}<\epsilon \implies \left|\frac{1}{x} - \frac{1}{a}\right|<\epsilon$$
This, implies that $\lim \limits_{x\to a}\frac{1}{x} = \frac{1}{a}$ for $x>1$. I did not think so much about the $x\le1$ case, but I would like if this proof is correct.

REPLY [3 votes]: There are a few not very big things wrong with the proof, and a major problem: you do not explain the logic of the argument.
Let $a\ne 0$, and let $f(x)=\frac{1}{x}$. We want to show that $f$  is continuous at $x=a$.  In what follows, we will assume that $a\gt 0$. By putting in the appropriate absolute value signs, one can transform the proof into one that works for all $a\ne 0$.
We want to show that for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-a|\lt \delta$, then $|f(x)-f(a)\lt \epsilon$.
If following the argument poses difficulty, let $a=0.4$.  Now let $\epsilon \gt 0$ be given.
It is easy to see that for any $x\ne 0$,  we have $|f(x)-f(a)|=\frac{|x-a|}{a|x|}$.
We have control over $|x-a|$, since we have great freedom in selecting $\delta$. The number $a$ is fixed. We need to control the size of $x$, since if $x$ is close to $0$, the $x$ in the denominator may make $|f(x)-f(a)|$ big.
Now comes the first key step. We will make sure ultimately that $\delta\le \frac{a}{2}$. Why? Well, then if $|x-a|\le \delta$, then we will have $\frac{a}{2}\lt x\lt \frac{3a}{2}$. The important part is that we will have $x\gt \frac{a}{2}$, so $x$ will stay nicely away from  $0$.
Thus if $\delta\le \frac{a}{2}$, and $|x-a|\lt \delta$, we have
$$|f(x)-f(a)|\lt \frac{2}{a^2}|x-a|.$$
Now to make sure that $|f(x)-f(a)|\lt \epsilon$, it is enough to make sure, in addition to the $\delta \le \frac{a}{2}$, that $\delta\le \frac{a^2\epsilon}{2}$.
Finally, we choose our $\delta$: it is $\min\left(\frac{a}{2},\frac{a^2\epsilon}{2}\right)$. By our arguments above, if $|x-a|\lt \delta$, then $|f(x)-f(a)|\lt \epsilon$.
Remark: The restriction to $x\ge 1$ in the argument was an attempt to avoid small denominators, so you were aware of the small denominators problem. In your proof, we can make sure that $x\ge 1$ if $a\gt 1$ by picking $\delta$ small enough, for example $\delta\le \frac{a-1}{2}$. But you did not do that.
There is a problem with the assertion that if $\delta=\epsilon |a|$ then everything is OK. Imagine, to be  silly, that $a=2$ and $\epsilon=47$. (This is the silly part.) Then $x$ being within $\epsilon |a|$ of $a$ guarantees nothing,  $x$ could even be equal to $0$.
You may think the argument of the answer tedious, and overkill. True, and true. But I wanted to describe the issues in detail, since the basic pattern is fairly standard.


TITLE: Integral of $\sin{x} (1 + 2 \cos{x})^4$
QUESTION [1 upvotes]: I'm looking for a way to solve

$$ \int{ \sin{x} \, ( 1 + 2 \cos{x})^4 }. $$

I think there are some theorems which help me but right now I don't know how to do it. I know that $$ 1 = \cos^2{x} + \sin^2{x} $$  but I don't know how to apply it. I think this must be another tricky one.

REPLY [2 votes]: HINT:
Substitute  $1+2\cos x=u$

REPLY [2 votes]: Hint: $$\int{ \sin x ( 1 + 2 \cos{x})^4 } dx = -\int{ (1 + 2 \cos{x})^4 } d(\cos x).$$

REPLY [2 votes]: Hint
$$ \int f^n(x) f'(x)dx  =  \frac{f^{n+1}(x)}{n+1} +c $$


TITLE: What is the problem here (all integers are irrational proof...I think so)?
QUESTION [8 upvotes]: Let us assume $a$ is an integer which is rational which implies $a=p/q$ (where $p$ and $q$ are integers and $q$ not equal to $0$). If $p$ and $q$ are not coprime, let us simplify the fraction so this it is (I don't know how to talk like mathematicians).
Which implies, $$a=b/c$$ (where $b$ and $c$ are coprime integers). Squaring on both sides,
\begin{align}
a^2&=b^2/c^2\\
a^2c^2&=b^2
\end{align}
So $a^2$ is a factor of $b^2$, and also of $b$, due to the uniqueness of the fundamental theorem of arithmetic.
So,
\begin{align}
b &=a^{2}d   \tag{where $d$ is an integer}\\
b^2 &= a^{4}d^{2}
\end{align}
But $b^2=a^2c^2$ So,
\begin{align}
a^2c^2 &= a^4d^2\\
c^2 &= a^2d^2
\end{align}
So, $a^2$ is a factor of $c^2$ and $c$ due to the fundamental theorem of arithmetic. So $b$ and $c$ have $a^2$ as a common factor. But this contradicts the fact that $b$ and $c$ are coprime. This is because we have taken $a$ as a rational integer, so $a$ cannot be a rational integer.
What's wrong here (genuinely asking)?

REPLY [2 votes]: Basic facts missing  $ac=b$ is a lot easier to use.  $a^2$ does not need to divide $b$. A fraction sharing  no common factor other than 1, between the number on top ( numerator), and the number on the bottom ( denominator), is said to be in lowest terms .
Anyways starting from $a={b\over c}$ we get $ac=b$ showing c divides b, sharing no factor other than 1, and therefore, $c=1$, implying $a=b$ so $a={a\over 1}$ it Also can be used to show :$a={-a\over -1}$