TITLE: Why matrices are multiplied the way they are multiplied?
QUESTION [4 upvotes]: I couldn't see a specific reason for multiplying every row of A with every column of B. Is this an arbitrary property of multiplication function of matrices?
Instead, why don't we simply multiply row#1 of matrix A with row#1 of matrix B, which would help us to do multiplication easier without all that confusion of what we are gonna multiply with what? In this case, product of A(mxn) and B(kxn) would be P(mxk).
I know a lot of things would change in today's mathematics if we define multiplication this way. But would that approach cause any problem in the future?

REPLY [3 votes]: If we consider matrices simply as tables of numbers than we can define many possible different binary operations that we can call '' multiplications'', simply using this name to distinguish this operation from the addition (defined as the sum of corresponding elements). Obviously different definitions give different properties of the ''multiplication'' and someone can be useful in some contest, but not in other.
As an example the Hadamard product of two matrix (defined as the product of the corresponding elements) is associative, distributive and also commutative, but can be defined only for matrices that have the same dimension, and ( as far as I know) is used in computer graphic.
The Kroneker product is another possible kind of multiplication, that has usefull properties and has important applications being related to the tensor product of linear transformations.
The usual row-column product  has the advantage that it can represent the action of linear transformations between vector spaces, and capture all properties of these transformations (linearity, associativity, non commutativity, existence of a neutral element and of not invertible elements). There is some amount of convection in the definition,  in the sense that we can chose the row to the left and column to the right ( as usual) or vice versa, but really these two possible alternative give isomorphic structures.


TITLE: Can there be any doubt that atmospheric "backradiation" is absorbed at the surface?
QUESTION [0 upvotes]: I am deliberately asking this question in Physics rather than Earth Sciences because I think it does come down to specific physical processes.
The background is that I recently had a discussion where someone who claimed to be an experienced physicist claimed that the greenhouse effect cannot work.
This was not for the usual reasons that involve incorrect invocation of the 2nd Law of Thermodynamics or whatever, but for the apparently straightforward reason that IR at the frequency re-emitted as "backradiation" by CO2 is not absorbed by most materials except at extremely low temperatures. The upshot is that the IR is scattered at the surface and ends up going out to space anyway.
I am not a physicist. I only have a layman's interest and I had previously worked on a hand waving assumption that upward heat loss minus downwelling IR essentially sums to reduce the rate of heat lost to space. However, I can see that if the IR is not absorbed at the surface then it does appear to argue against my very naïve view of it.
As I say, I am not a physicist and I did not have an immediate answer to what this person is saying. I am sceptical, if only because it seems to be a simple argument and, if there were any truth in it, it would surely have gained some traction by now.
My actual question is in a few parts:

Is this nonsense, as I suspect? What uncontroversial physics (ideally something
relatively simple) refutes the claim?
Is there any possible grain of truth at all? The problem I had was that I could not
even get to the bottom of what physics they were invoking so that I could look into it
myself.
I'm now intrigued about what actually is happening at the level of IR photons
hitting the surface. Assuming that what this person claims is wrong and that the
backradiation does slow heat loss, what is the actual process happening at an atomic
level that results in this? I assume there has to be absorption for there to be an
effect.

REPLY [1 votes]: There are two major problems with what they say: First, that IR (like any light) is straightforwardly either absorbed or reflected, depending on the colour of the material, the Earth's surface is not particularly reflective at IR wavelengths (easily measured), and temperature makes little difference. The second problem is that the greenhouse effect doesn't work that way, anyway.
I have heard this claim before, where it was then justified (supposedly) on the basis of the 2nd law of thermodynamics forbidding heat from spontaneously flowing from cold objects to hot ones. The idea was that a cold object radiating at long wavelengths (like the sky, at an effective radiative temperature of about -20 C) couldn't warm a hotter object (like the surface of the Earth at + 15 C), because that would involve heat flowing the 'wrong way'. This is a misunderstanding - the 2nd Law applies to the net heat flow. Because the hot object radiates more heat towards the cold object than vice versa, the net heat flow is still hot to cold, even though there is some heat flowing cold to hot.
The most everyday-familiar refutation of this I know is a microwave oven. The wavelength of microwaves is even longer than IR, corresponding to black body radiation from an object pretty close to absolute zero. (You can think about the cosmic microwave background from the big bang being about 2.7 C above absolute zero.) Neverthless, your dinner can absorb it, and get hot.
The second problem is that the greenhouse effect in a convective atmosphere does not work by backradiation from the atmosphere keeping the surface warm. (If that was actually the mechanism, the oceans would boil, since liquid water is an even more powerful greenhouse agent than water vapour. See below.)
To understand the greenhouse effect, you first have to understand that gases have the property that they warm up if they are compressed, and cool down if they are allowed to expand. As air circulates convectively in the atmosphere, it is constantly rising and falling, and as it rises it cools, and as it descends again it warms. This sets up a fixed temperature gradient in the atmosphere, called the adiabatic lapse rate. Taking into account the latent heat carried by water vapour, the result is that the atmosphere tends to settle with a lapse rate of 6.5 C per km altitude. For every 100 metres you climb, you find the air temperature will drop about 0.65 C. If it changes at a rate any steeper than that, the warm air near the surface rises convectively, cools, but only at 6.5 C/km so it remains warmer than the surroundings it moves into. This mixes in and warms the air aloft, eliminating the excess lapse rate. It is a bit like the way that a pan of boiling water is held fixed at 100 C. If you turn the heat up, the water just boils more vigorously, carrying heat away faster by convection/evaporation, and stays at the same temperature. The atmosphere stops the lapse rate exceeding the adiabatic limit in the same way, by convecting surface heat upwards faster until the limit is met again.
Now consider the Earth from space. Heat can only enter or leave the system by radiation. The Earth receives heat (mostly at visible wavelengths) from the sun, and emits heat (in the infrared) to outer space, and the two must balance almost exactly, or the Earth would warm up at an enormous rate. (A cumulative 1 Watt per square metre difference would dump 31 MJ/year cumulatively into every square metre of land and ocean. That's the same as 31 kettles running at 1 kW for about twenty minutes.) The Earth's temperature adjusts so that it emits black body radiation at precisely the rate needed to balance the inward flow. Given the intensity of light from the sun, and the fraction the Earth absorbs rather than reflecting back into space, this temperature turns out to be about -20 C.
However, the surface that has to approach this temperature is the visible surface of the planet at infrared wavelengths, and the atmosphere is opaque in the infrared, because of greenhouse gases and clouds. The dense, moist part of the atmosphere is about 10 km thick, and IR is radiated to space across the whole span, so the average altitude of emission of radiation to space is about 5 km. This is the part of the atmosphere that has to be at -20 C to maintain the energy balance.
If the atmosphere 5 km up is held at -20 C, then the surface will be held at roughly $-20 + 5\times 6.5=12.5$ C by the adiabatic lapse rate. Air descending from this altitude is warmed by compression. This warming is the greenhouse effect.
Backradiation does occur, is absorbed, and is required for the energy flows to balance. However, it has no effect on the surface temperature, because any excess heat it might supply is quickly convected away by the processes establishing the adiabatic lapse rate. The factors that affect greenhouse warming are the energy absorbed from sunlight, the lapse rate itself (which varies with the moisture content of the air), and the average altitude of IR emission to space (which is affected by greenhouse gases and the altitude of clouds).
I will give some references, since this being a controversial subject, I want to make clear that my version above is mainstream atmospheric physics.
Soden and Held 2000 'Water Vapour Feedback and Global Warming', Annu. Rev. Energy Environ. 2000. 25:441–75. See figure 1 and the discussion around it in particular.
Manabe and Strickler 1964 'Thermal Equilibrium of the Atmosphere with a Convective Adjustment' Journal of the Atmospheric Sciences 21 (1964): 361-385.
Sagan 1960, 'The radiation balance of Venus. Technical Report No. 32-34', Jet Propulsion Laboratory, NASA Contract No. NASW-6 (1960). In this paper, the famous Carl Sagan estimated the surface temperature of Venus using the discovery that the cloud tops were at least 30 km above the surface, and the adiabatic lapse rate for the Venusian atmosphere was about 10 C/km (page 7), predicting a massive 300 C greenhouse effect! The adiabatic lapse rate formula I use here is given as eqn 7 on page 5. The clouds are actually about 50-80 km high, and the adiabatic lapse rate about 8 C/km, giving about 500-600 C of greenhouse warming.
I have heard that Schwarzchild (of black hole fame) developed the principle in his work on stellar atmospheres in the 1930s, and it was carried across to atmospheric physics in the late 1950s and early 1960s. I don't have any references to support that, though.
--
I mentioned above that the greenhouse effect applies in liquid water, too. Liquid water is opaque to IR after a few millimetres, which means we can divide the body of the water up into horizontal slices 2 mm thick, each of which radiates black body radiation both upwards and downwards into the next layer. Sunlight shines down into the water at visible wavelengths, at which water is transparent, and is absorbed at depth. This same amount of energy has to be radiated to space from the topmost surface, for energy balance. Black body radiation is isotropic - the same in all directions - so it also radiates this same amount downwards into the layer below as backradiation. The layer below therefore has to radiate twice the amount both upwards and downwards, so that we get the required net upward flow. The layer below that has to radiate three times the same amount, and so on. The radiated power of the backradiation being emitted within the body of the water therefore increases linearly with depth, and by the Stefan-Botzman law, the temperature needed to generate this additional backradiation rises with the fourth root of the depth. One metre of water contains 500 layers, so the absolute temperature increase has to vary by a factor of 4.73 between top and bottom. If it's anywhere close to ambient temperature at the top, it has to be boiling at the bottom.
This does not actually happen, because as soon as the temperature gradient exceeds the adiabatic lapse rate for water (which is tiny, because water is nearly incompressible), the water starts convecting, carrying the heat up to the surface. Radiation is extremely inefficient for transporting heat through water - convection works a lot faster, and prevents any heat building up below the surface. Despite liquid water being a greenhouse agent 20,000 times more powerful than water vapour in the air (mainly because it's denser), we do not get heat buiding up in the depths.
If you suppress convection, by dissolving salt in the water to change its density with depth, the bottom of the pool does heat up. This is the basis of the solar pond, and can achieve temperatures of 90 C at the bottom with only a few metres of water.
Likewise, actual greenhouses (the buildings made out of glass) also work by suppressing convection. As can be demonstrated by using a transparent pane of rock salt for a roof, which is transparent to IR, and so emits no backradiation, but still keeps the contents warm.


TITLE: Randomly Assigning a Bill
QUESTION [0 upvotes]: Let's say there are $n$ people at the restaurant table. After finishing their meal they would like to decide randomly who will pay the whole bill. Each person should have a $1/n$ probability of paying the bill. Is there a good way of arranging this lottery at the restaurant table?
Can you do this game with a fair coin? For example, $n=2$ is trivial, but if $n=3$ how should you throw the coin in order for everyone to have a $1/n$ chance?

REPLY [2 votes]: We shall only consider the case that $n$ is not a power of $2$.
Find the binary expansions of $\frac1n, \frac2n, \ldots, \frac{n-1}n$ (they are non-terminating eventually periodic). Now repeatedly toss a fair coin and write $0$ for heads, $1$ for tail, thus obtaining the binary expansion of a number $x$ between $0$ and $1$. You may stop as soon as there is only one $k\in\{1,\ldots,n\}$ with $\frac {k-1}n\le x<\frac{k}n$. Then "player" number $k$ has "won".

Example: There are $n=5$ people, thus you compute
$$\begin{align}
\frac15&=0_2001100110011\ldots\\
\frac25&=0_2011001100110\ldots\\
\frac35&=0_2100110011001\ldots\\
\frac45&=0_2110011001100\ldots\\
\end{align} $$
My random generator just produced $HTHHHTHTH\ldots$, which translates to $0_2010001010\ldots$. In real life, I would already have stopped after three coin tosses $HTH$, because any number starting $0_2010\ldots$ must definitely be bigger than $0_2001\ldots=\frac15$ and smaller than $0_011\ldots=\frac25$. Thus player $2$ "wins". But I could not have stopped already after $HT$ because the number might still grow bigger than $\frac25$ (e.g. $0_201110$).


TITLE: Fewest number of multiplications to reach x^2000
QUESTION [1 upvotes]: What is the fewest number of multiplications required to reach $x^{2000}$ from $x$, using only previously generated powers of $x$? For example,
$$x \rightarrow x^2 \rightarrow x^4 \rightarrow x^8\rightarrow x^{16}\rightarrow x^{32}\rightarrow x^{64}\rightarrow x^{128}\rightarrow x^{256}\rightarrow x^{512}\rightarrow \\
x^{1024}\rightarrow x^{1536}\rightarrow x^{1792}\rightarrow x^{1920}\rightarrow x^{1984}\rightarrow x^{2000},$$
uses $15$ multiplications.

REPLY [5 votes]: Your method with $15$ multiplications is not the most efficient way to compute $x^{2000}.$ There is a scheme with $13$ multiplications, derived from the optimal addition chain for $n=2000$, see e.g.
http://wwwhomes.uni-bielefeld.de/achim/addition_chain.html
 0            x^1
 1( 0)        x^2
 2( 1)        x^4
 3( 2, 0)     x^5
 4( 3)        x^10
 5( 4, 3)     x^15
 6( 5)        x^30
 7( 6)        x^60
 8( 7)        x^120
 9( 8, 3)     x^125
10( 9)        x^250
11(10)        x^500
12(11)        x^1000
13(12)        x^2000

The number $13$ is confirmed in the table from https://oeis.org/A003313.


TITLE: How to transform a rotation matrix so it rotates from on its right corner vertex rather than its left corner vertx
QUESTION [1 upvotes]: I am trying to figure out what rotation matrix will allow me to rotate a picture in 2d from its right corner point.
Below you can see an image, on the left the picture is rotating from its left corner vertex and on the right it rotates on its right corner vertex.

What do i plug into the rotation matrix in order to get this to happen?
$$ R(\theta)=\left[
    \begin{array}{}
      \cos \theta&-\sin\theta\\
      \sin\theta&\cos\theta
    \end{array}
\right] $$
I have tried to use rotation then mirroring and all other such variations, but none of them satisfy the question, as for the picture on the right, the house starts with its left corner point at (2,2)

REPLY [0 votes]: To rotate the plane about any given point:

Translate the plane so that the point moves to the origin
Rotate the plane about the origin
Translate the plane so that the origin moves back to the given point

The translations in 1 and 3 are inverses of each other.
In terms of complex numbers, you could use the map
$$f_{w,\theta}(z)=w +e^{i\theta}(z-w)$$
Here $w$ is the center of rotation and $\theta$ is the angle of rotation in radians measured counterclockwise.
You should be able to convert this to matrix notation.


TITLE: Prove $C^n(\mathbb{R})\cong C(\mathbb{R})$ (linear isomorphism)
QUESTION [1 upvotes]: I need to prove that
$C^n(\mathbb{R})\cong C(\mathbb{R})$,
that is, there exists a bijective linear function from $C^n(\mathbb{R})$ onto  $C(\mathbb{R})$.

REPLY [0 votes]: It's enough to show they have the same dimension (here the dimension is a certain infinite cardinal, namely the cardinality of a basis).
Since $C^n\subset C$ it's clear that $\dim(C^n)\le\dim(C)$. For the other inequality:

If $f\in C$ there exists $g\in C^1$ with $g'=f$.

Proof: $g(x)=\int_0^x f(t)\,dt$.
By induction:

If $f\in C$ there exists $g\in C^n$ with $g^{(n)}=f$.

So if we define $T:C^n\to C$ by $Tf=f^{(n)}$ then $T$ is surjective; hence $\dim(C^n)\ge \dim(C)$.
Here we used a basic fact about dimension:

If $T:X\to Y$ is linear then $\dim(TX)\le\dim(X)$.

Proof: Say $B$ is  a basis for $X$. Then $T(B)$ spans $TX$, so $T(B)$ contains a basis for  $TX$; hence $\dim(TX)\le\text{card}(T(B))\le\text{card}(B)=\dim(X)$.


TITLE: Morphisms of a category with one object, which is a group
QUESTION [2 upvotes]: I've read two articles(1,2), but still have a question about structure of group, say $G$, as an one-object category. Let's call this category $C$.

I understand that morphisms of $G$, which is denoted by $Mor(G,G)$, forms a monoid, which is explained in the first article(1). However, how can we say that $Mor(G,G)$ is a group? We just said $C$ has an object which is a group, not its morphism also should satisfy group's definition. How can we induce group properties directly from $C$'s definition?
Also, I think it is natural to regard $Aut(G) \subseteq Mor(G,G)$, since every automorphism satisfy definition of a morphism. However, if we admit $Mor(G,G) \cong G$, then $Aut(G) \subseteq G$ can be nonsense, since we know the case that $G \cong Inn(G) \subset Aut(G) $, which implies $|G| < |Aut(G)|$.

What is wrong with my idea?

REPLY [4 votes]: The answer to your first question is no: as you've observed, we only need that $\operatorname{Hom}_{C}(G,G)$ is a monoid. You could take it to be the free monoid on one generator, which is just $\mathbb{N}$. (This is sort of a general model of a discrete dynamical system.)
I think the confusion in your second question arises from a type error. You're letting $G$ name the single object in your category, but you're also letting $G$ name the abstract group which is that single object's automorphism group.
Every automorphism is a morphism. In fact, in any category $\mathbf{C}$, $\operatorname{Aut}_{\mathbf{C}}(X) \subseteq \operatorname{Hom}_{\mathbf{C}}(X)$ for any object $X$ of $\mathbf{C}$ as monoids.
(Also, note that if $G$ is abelian, it has only one inner automorphism.)


text:
\begin{document}

\title{On Studying the Phase Behavior of the Riemann Zeta Function Along the Critical Line}
\author{Henrik Stenlund}
\date{June 1st, 2018}
\maketitle

\begin{abstract}
The critical line of the Riemann zeta function is studied from a new viewpoint. It is found that the ratio between the zeta function at any zero and the corresponding one at a conjugate point has a certain phase and its absolute value is unity. This fact is valid along the whole critical line and only there. The common  functional equation is used with the aid of the function ratio between any zero and its negative side pair, a complex conjugate. As a result, an equation is obtained for solving the phase along the critical line.
\footnote{Visilab Report \#2018-06}
\footnote{The author is with Visilab Signal Technologies Oy, Finland. The author is grateful for full support from the organization}
\subsection{Keywords}
Riemann zeta function, zeros of zeta function, recursion relation of zeta function, functional equation of zeta function
\subsection{Mathematical Classification}
MSC: 11M26
\end{abstract}

\section{Introduction}

\subsection{Motivation}
Developing methods for studying of the nature of the nontrivial zeros on the critical line of the Riemann zeta function are very important. One is attempting in this paper to find a way of eliminating the zeta function out of the functional equations allowing possibly new expressions to be developed in terms of more elementary functions. We are attempting to find an expression for the phase of the zeta function along the critical line.

\subsection{Preliminaries}
As is well known, the Riemann zeta function has a number of zeros, along the negative real axis at even integer points and along the critical line $x=\frac{1}{2}$. These latter appear at points whose exact positions are not known beforehand but require a lot of numerical work, the points being of a transcendental nature. Every zero in the upper half of the complex plane has a pair mirroring on the negative half plane. This work is now focusing in these nontrivial zeros. The following treatment is relying on the validity of the Riemann hypothesis \cite{Titchmarsh1999}. It requires all nontrivial zeros to reside on the critical line $x=\frac{1}{2}$ and none outside it. We are living the times of seeing the hypothesis finally proven, if not already done.

Now what will happen if a zeta function approaching a zero is divided by another zeta approaching a corresponding pairing zero on the negative half plane? As both of them go to zero one would expect the ratio to become singular. While investigating numerically the behavior of the Riemann zeta function approaching some of its zeros on the critical line, it was noted that the phase behavior of the function has a particular feature. When the argument $s$ approaches any zero $s_0$ of the zeta function $\zeta(s)$. The ratio seems to be
\begin{equation}
\lim_{s\rightarrow{s_0}}\frac{\zeta(s)}{\zeta(\bar{s})}=e^{i\theta}  \label{eqn2}
\end{equation}
with $s=\frac{1}{2}+i{t}$. One can evaluate this ratio while approaching the zero at $s_0$ from any direction and it appears that the ratio will not be singular. It can be proven as done in the following section. This leads one to think that there could exist another way to study the zeros of the zeta function.

\section{Elimination of the Zeta Function}
Since the zeta function is real on the real axis and meromorphic elsewhere, one has according to Schwarz's reflection principle
\begin{equation}
\zeta(\bar{s})=\bar{\zeta(s)}  \label{eqn30}
\end{equation}
The function has a general form
\begin{equation}
\zeta(s)=e^{i\phi}\rho  \label{eqn40}
\end{equation}
over the complex plane (with $\phi,\rho \in \textbf{\large{R}}$). Especially one does have along the critical line
\begin{equation}
\zeta(s_0)=e^{i\phi_0}\rho_0  \label{eqn42}
\end{equation}
and then one has the following
\begin{equation}
\bar{\zeta(s_0)}={{\rho}_0}e^{-i\phi_0}  \label{eqn52}
\end{equation}
But according to equation (\ref{eqn30}) this is equal to
\begin{equation}
\zeta(\bar{s_0})={{\rho}_0}e^{-i\phi_0}  \label{eqn56}
\end{equation}
Therefore,
\begin{equation}
\frac{\zeta(s_0)}{\zeta(\bar{s_0})}=e^{2i\phi_0}  \label{eqn60}
\end{equation}
proving the assertion and verifying the original numerical observation. This holds along the entire critical line. The function ratio (\ref{eqn2}) can be processed as follows at or very near a zero $s_0$
\begin{equation}
\frac{\zeta(s_0)}{\zeta(\bar{s_0})}=\frac{\zeta(s_0)}{\zeta(1-s_0)}  \label{eqn20}
\end{equation}
Approaching the zero can be done along the critical line from either direction. Then the ratio's amplitude remains at unity and the phase angle alone is varying. It can be done from a close distance from other directions on the complex plane while the amplitude is not unity and the phase angle is varying. The correct value is finally obtained at the zero and the amplitude reaches unity.

The functional equation of the Riemann zeta function with the argument $s \in \textbf{\large{C}}$ is well known \cite{Riemann1858}, \cite{Siegel1932}, \cite{Edwards2001}, \cite{Titchmarsh1999}. One applies it in the following form.
\begin{equation}
\frac{\zeta(s)}{\zeta(1-s)}=\frac{(2\pi)^s}{2cos(\frac{\pi{s}}{2})\Gamma(s)}  \label{eqn71}
\end{equation}
From this it follows that at any nontrivial zero along the critical line
\begin{equation}
\frac{\zeta(s_0)}{\zeta(\bar{s_0})}=\frac{(2\pi)^{s_0}}{2cos(\frac{\pi{s_0}}{2})\Gamma(s_0)}=e^{2i\phi_0+2\pi{iN}}  \label{eqn100}
\end{equation}
This is true for any point $s_0$ on the complex plane and the right side equality is to be used in the following. One has eliminated the zeta function from this equation. One has implemented the $Mod(2\pi{i}), N\in\textbf{\large{N}}$ to be carried further on the way to the final expressions. The Weierstrass formula for the Gamma function $\Gamma(s)$ is valid for $s \in{\textbf{\large{C}}}$
\begin{equation}
\frac{1}{\Gamma(s)}={s}e^{\gamma{s}}\prod_{k=1}{(1+\frac{s}{k})e^{-\frac{s}{k}}}  \label{eqn120}
\end{equation}
and one can substitute it getting
\begin{equation}
e^{2i\phi_0+2\pi{iN}}=\frac{{(2\pi)}^{s_0}{s_0}e^{\gamma{s_0}}\prod_{k=1}{(1+\frac{s_0}{k})e^{-\frac{s_0}{k}}}}{2cos(\frac{\pi{s_0}}{2})}  \label{eqn140}
\end{equation}
Taking the logarithm of the equation above one will obtain
\begin{equation}
2i\phi_0+2\pi{iN}={s_0}(\gamma+ln(2\pi))+ln(\frac{s_0}{2})-ln(cos(\frac{\pi{s_0}}{2}))+\sum_{k=1}{(-\frac{s_0}{k}+ln(1+\frac{s_0}{k}))}  \label{eqn160}
\end{equation}
This equation with a complex variable $s_0$ is valid for all nontrivial zeros.

Breaking down the $s_0$ as $s_0=\frac{1}{2}+i{t}$ where $t \in \textbf{\large{R}}$, and substituting it will lead to
\begin{equation}
2i\phi_0+2\pi{iN}=({\frac{1}{2}+i{t}})(\gamma+ln(2\pi))+ln(\frac{\frac{1}{2}+i{t}}{2})-ln(cos(\frac{\pi({\frac{1}{2}+i{t}})}{2}))  \nonumber
\end{equation}
\begin{equation}
+\sum_{k=1}{[-\frac{\frac{1}{2}+i{t}}{k}+ln(1+\frac{\frac{1}{2}+i{t}}{k})]}  \label{eqn200}
\end{equation}
The complex logarithms and the cosine term can further be broken down to real and imaginary parts as well and then one will get
\begin{equation}
2i\phi_0+2\pi{iN}=({\frac{1}{2}+i{t}})(\gamma+ln(2\pi))+ln(\frac{\sqrt{1+4{t^2}}}{4})-ln(\sqrt{\frac{cosh^2(\frac{\pi{t}}{2})+sinh^2(\frac{\pi{t}}{2})}{2}})  \nonumber
\end{equation}
\begin{equation}
+i\cdot{artan(2{t})}+i\pi{M}-i\cdot{artan(-tanh(\frac{\pi{t}}{2}))}+i\pi{L}  \nonumber
\end{equation}
\begin{equation}
+\sum_{k=1}{[-\frac{1}{2k}-\frac{it}{k}+ln(\sqrt{(1+\frac{1}{2k})^2+\frac{t^2}{k^2}})+i\cdot{artan(\frac{t}{k+\frac{1}{2}})}]}+i\pi{K}  \label{eqn220}
\end{equation}
Obviously, the real and imaginary parts of each term succeeded to get a linear form allowing a trivial separation.

\subsection{The Real Part}
For studying the zeros, one would be interested in the real part, equation (\ref{eqn220})
\begin{equation}
0=\frac{1}{2}(\gamma+ln(2\pi))+ln(\frac{\sqrt{1+4{t^2}}}{4})+  \nonumber
\end{equation}
\begin{equation}
-ln\sqrt{\frac{cosh^2(\frac{\pi{t}}{2})+sinh^2(\frac{\pi{t}}{2})}{2}}+  \nonumber
\end{equation}
\begin{equation}
+\sum_{k=1}{[-\frac{1}{2k}+\frac{1}{2}ln(1+\frac{1}{k}+\frac{1}{4k^2}+\frac{t^2}{k^2})]}  \label{eqn250}
\end{equation}
This can be developed further simplifying to
\begin{equation}
\gamma+ln(\frac{\pi}{4})=ln(\frac{cosh(\pi{t})}{1+4{t^2}})  \nonumber
\end{equation}
\begin{equation}
-\sum_{k=1}{ln[e^{-\frac{1}{k}}({1+\frac{1}{k}+\frac{1}{4k^2}+\frac{t^2}{k^2}})]}  \label{eqn270}
\end{equation}
In spite of its apparent complexity, this equation is an identity which is valid for all $t$.

\subsection{Solving the Phases}
The imaginary part of the equation (\ref{eqn220}) can be used for getting a general expression for the limit phase $\phi_0$ at a point along the critical line
\begin{equation}
\phi_0+\frac{\pi{M}}{2}=\frac{t}{2}(\gamma+ln(2\pi))+\frac{1}{2}artan(2{t})-\frac{1}{2}{artan(-tanh(\frac{\pi{t}}{2}))}  \nonumber
\end{equation}
\begin{equation}
+\frac{1}{2}\sum_{k=1}{[\frac{-t}{k}+artan(\frac{t}{k+\frac{1}{2}})]}  \label{eqn320}
\end{equation}
The result has become $Mod({\frac{\pi}{2}}), M\in\textbf{\large{N}}$. The resulting function is odd with respect to $t$.

\subsection{Numerical Evaluation of the Phase}
A crude calculation of the ratio from equation (\ref{eqn60}) shows a value of -80.95 degrees $\pm{90}$ degrees for the first nontrivial zero and for the second one 77.36 degrees. One can plot the curve as a function of $t$ along the axis $x=0.5$ as in Figure  \ref{Fig.1}.
\begin{figure}[htbp]
	\centering
		\includegraphics[width=1.100\textwidth]{zeros.jpg}
	\caption{Phase in degrees along $x=0.5$}
	\label{Fig.1}
\end{figure}
This is the angle $\phi_0$ as such. The $Mod({\frac{\pi}{2}})$ of the angle is more natural and produces Figure \ref{Fig.2}.
\begin{figure}[htbp]
	\centering
		\includegraphics[width=1.100\textwidth]{zerosmod90.jpg}
	\caption{$Mod({\frac{\pi}{2}})$ of the phase in degrees along $x=0.5$}
	\label{Fig.2}
\end{figure}
The negative side curve is of opposite sign as the function is odd.

\section{Discussion}
A few interesting results have been obtained. The first is that according to equation (\ref{eqn60}) the ratio between the zeta function at any zero and its conjugate is not singular, but always with unity absolute value and with a particular phase. This fact holds along the whole critical line ($x=\frac{1}{2}$) but fails immediately outside of it. The failure is not dramatic but will cause an error which increases while the point of focus is moving farther from the critical line. The second is that the equation (\ref{eqn100}) has only elementary functions left for studying the zeros of the zeta function. The third interesting finding is ensuing from the previous two, equation (\ref{eqn320}) presenting a simplified expression for calculation of phase at a zero. These equations form the main results of this work.

config: arxiv


TITLE: questions about proof of reverse formula of Fourier transform
QUESTION [2 upvotes]: We had the following theorem in class for a fourier transform $\widehat f$:

Let $\widehat f$ be the restriction of a $\mathbb C$ definied meromorphic function $F$. Let $F$ have a finite number of poles and let $z\cdot F(z)$ be bounded for big $z$. Then for $t>0$\begin{align*}
f(t)=i\cdot\sum_{\Im(z)>0}\textrm{res}_z\left(F(z)e^{i zt}\right)\end{align*}

My proof:
Consider $\gamma_r(s):=re^{i s}$:
The residue theorem gives us
for big $r$\begin{align*}
\int_{\gamma_r}F(z)e^{i zt}\,d z+\int_{-r}^r\widehat f(\omega)e^{i \omega t}\,d \omega=2\pi i\cdot\sum_{\Im(z)>0}\textrm{res}_z\left(F(z)e^{i zt}\right)
\tag{$\ast$}\end{align*}
Now $|z\cdot F(z)|\leq M$ for $z\geq R$, so for $r\geq R$\begin{align*}
\left|\int_{\gamma_r}F(z)e^{i zt}\,d z\right|&=\left|\int_0^\pi F(re^{i s})i re^{i s}\cdot e^{i \gamma_r(s)t}\,d s\right|\\
&\leq \int_0^\pi r\left|F\left(re^{i s}\right)\right|e^{-rt\sin s}\,d s\\
&\leq M\int_0^\pi e^{-rt\sin s}\,d s
\end{align*}
For $0\leq s\leq \frac{\pi}{2}$ we have $\sin s\geq\frac{2s}{\pi}$, and so
\begin{align*}
2M\int_0^{\frac{\pi}{2}}e^{-rt\sin s}\,d s&\leq2M\int_0^{\frac{\pi}{2}}e^{-rt\frac{2s}{\pi}}\,d s\\
&=2M\left[\left(\frac{-\pi}{2rt}\right)\cdot e^{-rt\cdot\frac{2s}{\pi}}\right]_0^{\frac{\pi}{2}}\\
&=\frac{M\pi}{rt}\cdot\left(1-e^{-rt}\right)\\
&\xrightarrow[]{r\rightarrow\infty}0
\end{align*}
So we get\begin{align*}
f(t)=\frac1{2\pi}\lim_{r\rightarrow\infty}\int_{-r}^r\widehat f(\omega)e^{i\omega t}\,d\omega=i\cdot\sum_{\Im(z)>0}\textrm{res}_z\left(F(z)e^{i zt}\right)
\end{align*}

So we had ($\ast$) be given in class, but I don't understand why this equation holds. How does it follow from the residue theorem?
And second why is $|e^{i\gamma_r(s)t}|=e^{-rt\sin s}$? (I did it with wolfram alpha because I didn't get any nice result)
Thanks.

REPLY [0 votes]: 1) It seems like you have assumed that $0 \leq s \leq \pi$ for $\gamma_r(s):=re^{i s}$, such that it is the arc of an upper half-circle. Together with the line $-r$ to $r$ this gives you a half-circle like in this picture, and for large $r$, all poles in the upper half plane will be inside it (since there are finitely many). Then it is just the residue theorem.
2) Remember that $\gamma_r(s):=re^{i s} = r \cos(s) + ir \sin(s)$, and that $|e^{ik}|$ is one when $k$ is real.


TITLE: Doubt about proof of For every real $x > 0$ and every integer $n > 0$ there is one and only one positive real $y$ such that $y^n = x$
QUESTION [1 upvotes]: This is the Theorem 1.21 of Walter Rudin Principles of Mathematical Analysis. I follow the same reasoning of Walter Rudin except that I did take another path. Then I would like to know if my proof is ok like this. I will answer my own question with the proof. However, I used the same reasoning that Walter Rudin, but, I wonder if there are a proof more intuitive than this, because, his proof seems very tricky and mine too (In the contradiction part). Thanks in advance

REPLY [1 votes]: Proof
Uniqueness
Since $y_1 > y_2 > 0 \implies y_1^n > y_2^n > 0$ then if $y_1 \neq y_2 \implies y_1^n \neq y_2^n$ we proved that if there exists a $y$ such that $y^n = x$ it has to be unique
Existence
Let $B = \{ t \in \mathbb{R} : x < t^n \}$
B is bounded below, since $t > 0$, $0$ is a below bound. B is nonempty, because, by Archimedian property there exists $my> x$ with $m \in \mathbb{N}, y \in \mathbb{R}, x \in \mathbb{R} $ If we put $y = 1$ we get $m > x$ if $x \geq 1$ then $m^n > x$ if $0<x < 1$ we can find that $1^n > x$ then we proved that  is nonempty, since at least $M \in B$
By Completness property, B have a infimun, let's denote it by $z = \inf B$
To prove that $z^n = x$ we wil procede by contradition
Suppose that $z^n > x$
Note that if $0 < a < b$
$b^n - a^n = (b -a)(b^{n-1} + b^{n-2}a + ... + a^{n-1})$ as $a < b$ then
$ b^n - a^n < (b-a)(b^{n-1} + b^{n-2}b + ... b^{n-1})$
$ \implies b^n - a^n < (b-a)(nb^{n-1})$ $(*)$
Let $0 < h < \frac{z^n - x}{n(z+1)^{n-1}}$
$\implies hn(z+1)^{n-1} < z^n - x $ $(**)$
Since $n(z+1)^{n-1} \neq 0$
Using $(*)$ with $a = z - h, b = z$
$z^n - (z -h)^n < hnz^{n-1}$
With this and $(**)$
$z^n - (z - h)^n < hnz^{n-1} < hn(z+1)^{n-1} < z^n - x$
By transition $z^n - (z - h)^n <  z^n - x$
$\implies x < (z - h)^n$ this means that $(z-h) \in B$ but $z - h < z$ and $ z$ is the infimum of $B$. This lead us to a contradiction.
Suppose $z^n < x$
same reasoning
Let $0 < h < 1$ and $h < \frac{x - z^n}{n(z+1)^{n-1}}$ $(***)$
And using $(*)$ with $a = z, b = (z +h)$
$(z + h)^n - z^n < hn(z+h)^{n-1}$
Combining this with $(***)$
$(z + h)^n - z^n < hn(z+h)^{n-1} < hn(z +1)^{n-1} < x - z^n$
$\implies (z + h)^n < x$ that means that $(z + h) \notin B$ but $z + h > z$ and since $B$ isn't bounded above, this lead us to a contradicion
Hence $z^n = x$


TITLE: If I have an equation of degree 2 or 3 and one of the coefficients is unknown, how to find that coefficient giving that one of the poles is given.
QUESTION [0 upvotes]: For example, $S^3 + 3S^2 + 18S + K$
Given that one of the roots is $-10$. How would you find K that satisfies this condition?

REPLY [0 votes]: It worked!
K = 880.
I tried solving it with my calculator and it gave me the roots which one of them is -10.


TITLE: order of an integral on the real line
QUESTION [1 upvotes]: How to prove that, the following integral, $d \in \mathbb{N}$, $d > 1$,
$$
\int_{n^{1/4}}^{\infty} x^{d-1} \exp(-x^2) \, dx
$$
is of order $O(n^{\frac{d-1}{4}} \exp(-n^{1/2}) ) $ ?

REPLY [1 votes]: Integration by parts after substituting $x\mapsto x^{1/2}$
$$
\begin{align}
\int_{n^{1/4}}^\infty x^{d-1}e^{-x^2}\,\mathrm{d}x
&=\frac12\int_{n^{1/2}}^\infty x^{d/2-1}e^{-x}\,\mathrm{d}x\\
&=\frac12\left(n^{d/4-1/2}e^{-n^{1/2}}-\left(\frac d2-1\right)\int_{n^{1/2}}^\infty x^{d/2-2}e^{-x}\,\mathrm{d}x\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac12n^{d/4-1/2}e^{-n^{1/2}}\left(1+O\left(n^{-1/2}\right)\right)}
\end{align}
$$
since
$$
\int_{n^{1/2}}^\infty x^{d/2-2}e^{-x}\,\mathrm{d}x
\le n^{-1/2}\int_{n^{1/2}}^\infty x^{d/2-1}e^{-x}\,\mathrm{d}x
$$
Furthermore
$$
\frac12n^{d/4-1/2}e^{-n^{1/2}}=O\left(n^{d/4-1/4}e^{-n^{1/2}}\right)
$$


TITLE: Let $x\in\Bbb R$. Then there exists a unique $w\in\Bbb R$ such that $x+w=0$
QUESTION [1 upvotes]: Let $x,y\in\Bbb R$. We define addition operation $(+)$ on $\Bbb R$ by $$x+ y:=\inf\{r+ s\mid r,s\in\Bbb Q \text{ and } x<r  \text{ and } y<s\}$$
Then there exists a unique $w\in\Bbb R$ such that $x+w=0$.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

Lemma:

Let $x,y,z\in\Bbb R$.

$x+y=y+x$
$(x+y)+z=x+(y+z)$
$x+0=x$

If $r\in\Bbb Q$ and $r>0$, there exists $n\in\Bbb N\setminus\{0\}$ such that $\dfrac{1}{n}<r$.
For every $x\in \Bbb R$ and  $n\in\Bbb N\setminus\{0\}$, there exist $r,s\in\Bbb Q$ such that $r<x\le s$ and $s-r\le\dfrac{1}{n}$.

It is obvious that $x=\inf \{r \mid r\in\Bbb Q,x<r\}$.
Existence
Let $w=\inf\{-s\mid s\in\Bbb Q,s<x\}$.
First, we prove that $x+w=\inf\{r-s \mid r,s\in\Bbb Q,s<x<r\}$.

By definition, $x+w=\inf\{r+p\mid r,p\in\Bbb Q,x<r,w<p\}$.
Substituting $-p$ for $p$, we get $x+w=\inf\{r-p\mid r,-p\in\Bbb Q,x<r,w<-p\}=$ $\inf\{r-p\mid r,p\in\Bbb Q,x<r,w<-p\}$.
We have $p\in\Bbb Q$ and $w<-p \iff p\in\Bbb Q$ and $-p>-s$ for some $s\in\Bbb Q$ such that $s<x$ $\iff p\in\Bbb Q$ and $p<s$ for some $s\in\Bbb Q$ such that $s<x$ $\iff p=s$ for some $s\in\Bbb Q$ such that $s<x$.
Hence $x+w=\inf\{r-s\mid r,s\in\Bbb Q,x<r,s<x\}=\inf\{r-s\mid r,s\in\Bbb Q,s<x<r\}$.

Second, we prove that $x+w=0$.

We have $s<x<r \implies r-s>0 \implies x+w=\inf\{r-s\mid r,s\in\Bbb Q,s<x<r\} \ge 0$.
Assume the contrary that $x+w>0$. Since $\Bbb Q$ is dense in $\Bbb R$, there exists $p\in\Bbb Q$ such that $0<p<x+w$. By Lemma 2, there exists $n\in\Bbb N\setminus\{0\}$ such that $\dfrac{1}{n}<p$. By Lemma 3, there exist $r_0,s_0\in\Bbb Q$ such that $r_0<x< s_0$ and $s_0-r_0\le\dfrac{1}{n}<p<x+w$. This is clearly a contradiction. Hence $x+w=0$.

Uniqueness
Assume that $x+w=0$ and $x+w'=0$. By Lemma 1, $w=w+0=w+(x+w')=w+(w'+x)=(w+w')+x=(w'+w)+x=w'+(w+x)=w'+(x+w)=w'+0=w'.$
Hence such $w$ is unique.

REPLY [0 votes]: By a similar reasoning, I would like to present a proof of the below theorem.

Let $\Bbb R^+=\{x\in\Bbb R \mid x>0\}$ and $x,y\in\Bbb R^+$. We define multiplication operation $(\cdot)$ on $\Bbb R^+$ by $$x\cdot y:=\inf\{r\cdot s\mid r,s\in\Bbb Q \text{ and } x<r  \text{ and } y<s\}$$
Then there exists a unique $w\in\Bbb R^+$ such that $x\cdot w=1$

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

Lemma:

Let $x,y,z\in\Bbb R^+$.

$x\cdot y=y\cdot x$
$x\cdot (y+z)=x\cdot y+x\cdot z$
$(x\cdot y)\cdot z=x\cdot (y\cdot z)$
$x\cdot 1=x$

For every $x,k\in\Bbb R$ such that $0<x,1<k$. There exists $s,r\in\Bbb Q$ such that $0<s<x<r$ and $r/s<k$.

Since $\Bbb Q$ is dense in $\Bbb R$, $x=\inf \{r \mid r\in\Bbb Q,x<r\}$.
Existence
Since $\Bbb Q$ is dense in $\Bbb R$, $x=\inf \{r \mid r\in\Bbb Q,x<r\}$. Let $w=\inf\{1/s\mid s\in\Bbb Q,0<s<x\}$.
We prove that $x\cdot w=\inf\{r/s\mid r,s\in\Bbb Q,0<s<x<r\}$.

By definition, $x\cdot w=\inf\{r\cdot p\mid r,p\in\Bbb Q,x<r,w<p\}$.
We substitute $1/p$ for $p$ and get $x\cdot w=\inf\{r\cdot (1/p)\mid r,1/p\in\Bbb Q,x<r,w<1/p\}$ $=\inf\{r/p\mid r,p\in\Bbb Q,x<r,w<1/p\}$.
We have $p\in\Bbb Q$ and $w<1/p$ $\iff p\in\Bbb Q$ and $1/p>1/s$ for some $s\in\Bbb Q$ such that $0<s<x$ $\iff p\in\Bbb Q$ and $0<p<s$ for some $s\in\Bbb Q$ such that $0<s<x$ $\iff p=s$ for some $s\in\Bbb Q$ such that $0<s<x$.
Hence $x\cdot w=\inf\{r/s\mid r,s\in\Bbb Q,x<r,0<s<x\}=\inf\{r/s\mid r,s\in\Bbb Q,0<s<x<r\}$.

Second, we prove $x\cdot w=1$.

We have $0<s<r \implies 1<r/s \implies x\cdot w=\inf\{r/s\mid r,s\in\Bbb Q,0<s<x<r\} \ge 1$.
Assume the contrary that $x\cdot w>1$. Since $\Bbb Q$ is dense in $\Bbb R$, there exists $p\in\Bbb Q$ such that $1<p<x\cdot w$. By Lemma 2, there exist $s,r\in\Bbb Q$ such that $0<s<x<r$ and $r/s<p<x\cdot w$. This is clearly a contradiction. Hence $x\cdot w=1$.

Uniqueness
Assume that $x\cdot w=x\cdot w'=1$. By Lemma 1, $w=w\cdot 1=w\cdot (x\cdot w')=w\cdot (w'\cdot x)=$ $(w\cdot w')\cdot x=(w'\cdot w)\cdot x=w'\cdot (w\cdot x)=w'\cdot (x\cdot w)=w'\cdot 1=w'.$ Hence such $w$ is unique.


TITLE: How to maximise stomach drop when driving through a dip
QUESTION [4 upvotes]: I have a couple of young kids, and on our school commute each morning we drive through a dip in the road. The dip is probably about 5 meters deep (15 feet) and the speed limit in the area is 60 km/h (40 m/h). We find it great fun to pretend its a dip on a rollercoaster, and I've been trying to work out the most efficient pattern of accelerating / decelerating through the dip to maximise the "stomach-drop" effect. But I'd like to get science on my side!
How do I maximise the stomach-drop? When should I be accelerating, coasting or braking?

Above are some sample segments of the dip, if the recommended accelerating/coasting/braking phases do not match up please specify better segments
Right now my hunch is to accelerate to point B (as the dip starts), then coast / brake through to point C, then accelerate again to point E, then coast again. But if I can eke out even more stomach drop, I cannot wait to test it out on my kids!

REPLY [0 votes]: After giving this way too much thought..
I believe a solution of the problem would be as follows. Consider a slightly updated "dip-chart":

Assuming constant acceleration going forwards, at points A, C & E the force acting on our stomach would just be gravity pulling it downwards at 9.8m/s:

As we start going into the dip (point A), the car starts getting an additional downward acceleration. Our stomach, being a bag of liquid suspended inside our body, is similar to a mass held in place by a pair of springs. As the car accelerates downwards, the stomach lifts "up". The maximum amount of "upward" lift is at point B:

As we continue on the curve toward point C, the downward acceleration of the car tapers off and we reach point C and our original balanced stomach:

Now as we climb up out of the dip, the acceleration of the car gains an upward direction too, and so the "slow to react" stomach is pulled downwards on its springs, reaching its maximum at point D:

Finally, we exit the dip at point E and once again have a resting stomach:

Now for maximizing the "stomach drop feeling" (via acceleration / coasting / braking)
If we acceleration between point A & point B, we add an additional sideways force onto our stomach spring. So not only does it "move upward", it will always move "backward" a little:

We then coast / maintain speed from point B to point C, and then from point C to point D we decelerate / brake to make our stomach get a horizontal forward movement:

That gives a final set of instructions as follows:

Up to point A, maintain constant speed
Point A -> B, accelerate
Point B -> C, maintain constant speed
Point C -> D, decelerate
Point D -> onwards, accelerate back up to speed


TITLE: What is the first $w$ such that a rectangle, $R_{w\times w-1}$ is minimally-square-partitioned by less than $w$ squares.
QUESTION [2 upvotes]: Motivated by:
Tiling an orthogonal polygon with squares,
How to prove that the minimum square partition of a 3X2 rectangle has 3 squares,
Minimum square partitions for 4x3 and 5x4 rectangles,
What is the minimum square partition of an almost-square rectangle?.

Let:

An almost-square-rectangle be a rectangle that has a width  $w$ and height $h=w-1$.
A square partitioning be a covering by non-overlapping squares; the entire rectangle must be covered, all the squares must be disjoint.
A minimum-square-partitioning be a square partitioning, for which is no square partitioning that is made of a lesser number of squares.

The $R_{w\times w-1}$ rectangles illustrated below are minimally-square-partitioned by $w$ squares. Proofs here: How to prove that the minimum square partition of a 3X2 rectangle has 3 squares and Minimum square partitions for 4x3 and 5x4 rectangles).

However, in general, a $R_{w\times w-1}$ rectangle can be partitioned into $\mathcal O(\log w)$ squares (see What is the minimum square partition of an almost-square rectangle?, and Minimum tiling of a rectangle by squares, which refer to a paper called Tiling a Rectangle with the Fewest Squares by Richard Kenyon.
My question:

What is the minimum $w$, such that a $w\times w-1$ rectangle, $R_{w\times w-1}$, can be minimally-square-partitioned into $f(w)$ squares, and $f(w)<w$.

REPLY [1 votes]: By the proofs cited above, $w\geq 6$, and by Yuval's answer cited above, $w\leq 7$. So the only question is now whether a $6 \times 5$ rectangle can be covered with 5 squares. I think the answer is yes: it can be covered by three $2 \times 2$ squares and two $3 \times 3$ squares. So $w=6$.


TITLE: What is the set $\mathcal{D} = \{x| f(x) \leq 0\} \cap \{x| x\geq 0\}$? where $f$ is convex
QUESTION [0 upvotes]: Let $f:\mathbb{R}\to \mathbb{R}$ be continuous and convex. Assume that $f(0) \leq 0$. What is the set $\mathcal{D} = \{x| f(x) \leq 0\} \cap \{x| x\geq 0\}$? Assume that $\mathcal{D}$ is bounded.

I think $D = [0, \alpha]$, where $\alpha = \max\{x| f(x) = 0\}$. Any idea or proof?

REPLY [0 votes]: $f$ is convex so is it's sublevel (Just apply the definition of convexity of $f$) . Thus the set $\{x ~ | f(x) \leq 0 \}$ is convex (closed) subset of $\Bbb R$ so it is a closed interval.  Therefore
$$D = \text{closed interval} ~ \cap ~[0 , + \infty)  = [0 , \alpha] $$
Note that $\alpha  < + \infty$ since we assumed $D$ is bounded.


text:
(*  Title:      RSAPSS/Productdivides.thy
    Author:     Christina Lindenberg, Kai Wirt, Technische Universität Darmstadt
    Copyright:  2005 - Technische Universität Darmstadt
*)

section "Lemmata for modular arithmetic with primes"

theory Productdivides
imports Pdifference
begin

lemma productdivides: "\<lbrakk>x mod a = (0::nat); x mod b = 0; prime a; prime b; a \<noteq> b\<rbrakk> \<Longrightarrow> x mod (a*b) = 0"
  by (simp add: mod_eq_0_iff_dvd primes_coprime divides_mult)

lemma specializedtoprimes1:
  fixes p::nat
  shows "\<lbrakk>prime p; prime q; p \<noteq> q; a mod p = b mod p ; a mod q = b mod q\<rbrakk>
         \<Longrightarrow> a mod (p*q) = b mod (p*q)"
by (metis equalmodstrick1 equalmodstrick2 productdivides)

lemma specializedtoprimes1a:
  fixes p::nat
  shows "\<lbrakk>prime p; prime q; p \<noteq> q; a mod p = b mod p; a mod q = b mod q; b < p*q \<rbrakk>
    \<Longrightarrow> a mod (p*q) = b"
  by (simp add: specializedtoprimes1)

end

subset_name: curated


TITLE: If $\int_a^b g(x)dx = 0$, then $g(x) = 0$ for a.e. $x$
QUESTION [1 upvotes]: Let $g\in L^2(\mathbb{R})$. Prove that if for $\forall a,b\in\mathbb{R}$ s.t. $a<b$, $\int_a^b g(x)dx = 0$, then $g = 0$ for a.e. $x$.
I want to prove by contradiction. Suppose $g > 0$ on a set of positive measure, say $E$, then $\exists \epsilon >0$ s.t. $g(x) > \epsilon$ for $\forall x\in E$.
Take an open set $O \supseteq E$ s.t. $m(O) < m(E) + \delta$ for some $\delta > 0$, and $m$ is the Lebesgue measure. And $O = \cup_{n\in \mathbb{N}} (a_n,b_n)$(union of disjoint open intervals), as O is open.
Then $\int_O g(x)dx = \sum_n \int_{a_n}^{b_n} g(x)dx = 0$, but $$\int_O g(x)dx = \int_{O-E} g(x)dx + \int_E g(x)dx > \int_{O-E} g(x)dx + \epsilon m(E)$$
I want make the last term postive so that I can derive a contradiction. But I don't know how to control $g$ on $O$-$E$. Can someone help me with this? Or is there a better way to prove the statement?

REPLY [3 votes]: Pick $a<b$, then $g \in L^2[a,b] \subset L^1[a,b] $.
Then $\phi(t) = \int_{a}^t g(x) dx $, then the Lebesgue differentiation theorem tells us that $\phi'(t) = g(t)$ for ae. $x$. Since $\phi' = 0$ we see that $g = 0$ ae. $x \in [a,b]$. Since $a,b$ are arbitrary, we have $g=0$ ae.


TITLE: Weak convergence of unit mass
QUESTION [0 upvotes]: I have found in my notes the statement bellow:

$\mu_n\Rightarrow\mu$ as $n\rightarrow \infty$, if and only if $x_n\rightarrow x$ as $n\rightarrow \infty$ where $\mu$ is the unit mass at $x$ and $\mu_n$ is a unit mass at $x_n$ for $n=1,2,...$.

I do not understand this statement clearly as why this holds.
Any help on this would be appreciated!

REPLY [0 votes]: I am going to assume $\mu_n$ and $\mu$ are probability measures on $\mathbb{R}$
The definition of $\mu_n \Rightarrow \mu$ is that $\int f \,d\mu_n \to \int f \,d\mu$ for all bounded continuous functions $f$.  If $\mu_n$ is a point mass at $x_n$ and $\mu$ is a point mass at $x$, then the above condition says that $f(x_n) \to f(x)$ for all bounded continuous functions $x$.  So you need to prove that $f(x_n) \to f(x)$ for all bounded continuous $f$ if and only if $x_n \to x$.
In one direction, if $x_n \to x$, then it simply follows from (one of) the definition(s) of continuity that $f(x_n) \to f(x)$ whenever $f$ is continuous.  The other direction is a bit trickier.  Suppose that $x_n$ does not converge to $x$.  Then there is a positive distance $\epsilon$ such that $|x_n - x| > \epsilon$ for infinitely many $n$.  Now let $f$ be a continuous function such that $f(x) = 1$ but $f(y) = 0$ whenever $|y-x| > \epsilon$ (I will leave it to you to figure out what such a function looks like).  Then $f(x) = 1$ but $f(x_n) = 0$ for infinitely many $n$, so $f(x_n)$ does not converge to $f(x)$.


TITLE: Finding point on a line a given distance from another point on that line, 3D
QUESTION [0 upvotes]: I have two coordinates: $A(0.25, 0.0337, 0.5)$ and $B(0.3912, 0.1558, 0.3796)$
I would like to find the coordinates of a point $C$ along that line that is 0.1014 away from point $B$, but not between $A$ and $B$.
The equation for that line between $A$ and $B$ (going from A to B) is $r(t) = <0.25, 0.0337, 0.5> + t<0.1412, 0.1221, -0.1204>$
Or in parametric form:
$x(t) = 0.25 + 0.1412t$
$y(t) = 0.0337 + 0.1221t$
$z(t) = 0.5 - 0.1204t$
Using the distance formula, I would have:
$0.1014^2 = (x - 0.3912)^2 + (y - 0.1558)^2 + (z - 0.3796)^2$
I'm assuming I need two more sets of equations to solve for x, y, and z (one of which could be the line equation above), and that I'll get two sets of solutions; one giving the coordinates between $A$ and $B$, and the other giving a point outside of $A$ and $B$.
Unsure how to use the parametric equations, and what other equation I may need. Thanks for any assistance!

REPLY [0 votes]: As long as your parametrized equations for $x,y,z$ along the line in terms of $t$ are correct, and it appears that $B$ lies on the line for some $t > 0$, while $A$ is obtained with $t = 0$, then all you have to do is plug in the equations $x(t),y(t),z(t)$ into your formula for distance from $B$, and set it equal to the distance you want, and this will give you a quadratic in $t$ that has two solutions, and you want the solution that has the maximum value of $t$. Then plugging this value of $t$ into $x(t),y(t),z(t)$ gives you your point.


text:
\section{Product of Complex Conjugates/Examples/3 Arguments}
Tags: Product of Complex Conjugates

\begin{theorem}
Let $z_1, z_2, z_3 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\overline z$ denote the [[Definition:Complex Conjugate|complex conjugate]] of the [[Definition:Complex Number|complex number]] $z$.
Then:
:$\overline {z_1 z_2 z_3} = \overline {z_1} \cdot \overline {z_2} \cdot \overline {z_3}$
\end{theorem}

config: wiki


TITLE: $\frac{\pi}{2}  =\tan^{-1}(\infty)$
QUESTION [3 upvotes]: Using the result, $\tan^2{\alpha} - A \tan{\alpha} + 1 = 0~$, where A is a constant, prove that the two solutions to this equation (such that $0 \leq \alpha \leq \frac{\pi}{2}$) are complementary (i.e. $\alpha_1 + \alpha_2=\large \frac{\pi}{2}~$)

To solve this equation, suppose the two roots are $\alpha_1$ and $\alpha_2$. We can take the product of roots $\Pi\tan{\alpha}: \tan{\alpha_1}\tan{\alpha_2}=1~$, but noting that $\tan(\alpha_1 + \alpha_2) = \large \frac{\tan{\alpha_1}+\tan{\alpha_2}}{1 - \tan{\alpha_1}\tan{\alpha_2}}~$ from the $\tan$ expansion.
Substituting $\tan{\alpha_1}\tan{\alpha_2}=1~$ into that expansion:
$$\tan(\alpha_1 + \alpha_2) = \frac{\tan{\alpha_1}+\tan{\alpha_2}}{1 - 1}=\frac{\tan{\alpha_1}+\tan{\alpha_2}}{0}=\infty~$$
Therefore, following on from this, $\alpha_1 + \alpha_2 = \tan^{-1}(\infty) = \large \frac{\pi}{2}$
However, is this proof flawed, especially in equating $\large \frac{\tan{\alpha_1}+\tan{\alpha_2}}{0}$ with $\infty~$, and $\tan^{-1}(\infty)$ with $\large \frac{\pi}{2}~$? Would limits be required for a more proper treatment?

REPLY [7 votes]: We need to make the question more precise.  After all, for any real number $u$, there are infinitely many $\alpha$ such that $\tan \alpha=u$.  So we will restate the problem.
Suppose that $A>0$ and $A^2-4 \ge 0$.  Then the equation
$$x^2 -Ax+1=0$$
has two (possibly equal) positive solutions, say $x_1$ and $x_2$.
Let $\alpha_i$ be the angle (number) in the interval $(0,\pi/2)$ such that $\tan\alpha_i=x_i$.  Show that $\alpha_1 +\alpha_2=\pi/2$.
The addition formula for $\tan$ seems to me not the best way to handle the problem, even though, as Gerry Myerson points out, there is a precise way of interpreting the situation when the denominator is $0$.
It seems to me simpler to note that if $\alpha_i$ are angles in the interval $(0,\pi/2)$ we have
$$\alpha_1+\alpha_2=\frac{\pi}{2} \qquad \text{iff}\qquad \tan\alpha_2=\frac{1}{\tan\alpha_1}$$
This follows from primitive properties of right triangles.
So the only thing to verify is that the product of the roots of $x^2-Ax+1=0$ is $1$, and this is clear from the shape of the equation.
Added comment: If one takes great care, or if one has very good intuition, it is possible to handle "$\infty$" without making errors.  Euler (mostly) did it, but we are not all Euler. Improper handling of "$\infty$" is an all too frequent source of student mistakes.  So it is best to do "defensive thinking," and avoid trying to handle "$\infty$" as if it were a number. After all, it isn't.


TITLE: Show that continuous functions preserve sequence convergence in topological spaces
QUESTION [2 upvotes]: Let $(X, \mathcal{T}) and (Y, \mathcal{U})$ be topological spaces and let $f : X \rightarrow Y$.
Suppose $f$ is continuous, and $\{x_n\}_{n=1}^{\infty}$ is a sequence in $X$ converging to a point $x$.
I need to show that $\{f(x_n)\}_{n=1}^{\infty}$ converges to $f(x)$.
This is what I have so far:

Since $f$ is continuous, we have that $f(\overline A) =$
  $\overline{f(A)}$. We see that $\overline{\{x_n\}_{n=1}^{\infty}} =$
  $\{x_n\}_{n=1}^{\infty} \cup \{x\}$ and so we have
  $f(\{x_n\}_{n=1}^{\infty} \cup \{x\}) =$
  $\overline{\{f(x_n)\}_{n=1}^{\infty}} = \{f(x_n)\}_{n=1}^{\infty} \cup$
  $f(x)$. Therefore, there exists a sequence in
  $\{f(x_n)\}_{n=1}^{\infty}$ that converges to $f(x)$.

But now I'm stuck because I don't know how to show that this sequence is $\{f(x_n)\}_{n=1}^{\infty}$ and not just some subsequence of it?

REPLY [5 votes]: For any neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood of $x$, so there exists some $N\in \mathbb{N}$ such that $x_n\in f^{-1}(V) \forall n>N,$ $i.e. f(x_n)\in V\forall n>\mathbb{N}$.


text:
\begin{definition}[Definition:Set of Integer Multiples]
The [[Definition:Set|set]] $n \Z$ is defined as:
:$\set {x \in \Z: n \divides x}$
for some $n \in \Z_{>0}$.
That is, it is the [[Definition:Set|set]] of all [[Definition:Integer|integers]] which are [[Definition:Divisor of Integer|divisible]] by $n$, that is, the '''set of integer multiples of $n$'''.
Thus we have:
:$n \Z = \set {\ldots, -3 n, -2 n, -n, 0, n, 2 n, 3 n, \ldots}$
\end{definition}


TITLE: Möbius transformation image
QUESTION [3 upvotes]: Let $f(z)=\frac{az+b}{z+d}$, when $d\in\mathbb{R}$, $d\not=0$ $a,b\in\mathbb{C}$ and $f$ is not constant.
I want to find the image of the real and imaginary axes under $f$. I've found that the image of the real axis is $\{z:z=\frac{a+b}{1+d}+\frac{ad-b}{d(1+d)}u,u\in\mathbb{R}\}$. The pole of f is not on the imaginary axis therefore his image is a circle. From the angle preservation principle the diameter of the circle is on the image of the real axis. I would like a hint.

REPLY [2 votes]: Your Möbius transformation goes from the $z$-plane to the $w$-plane, where
$$w = \frac{az+b}{z+d}$$
You need to rearrange this equation to make $z$ the subject. Let's say you have
$$z=\frac{pw+q}{rw+s}$$
An equation for the imaginary axis in the $z$-plane is $|z-1|=|z+1|$. (The imaginary axis is the perpendicular bisector of the chord joining $-1$ and $1$.) This equation then becomes
$$\left|\frac{pw+q}{rw+s}-1\right| = \left|\frac{pw+q}{rw+s}+1\right|$$
The solution set of this equation is something in the $w$-plane. It is the image of the imaginary axis.
Make the substitution $w=u+\mathrm{i}v$, apply the definition of modulus, simplify, complete the square on $u$ and $v$ and then read off the answer. See my example below:
Example
Consider the Mobius transformation from the $z$-plane to the $w$-plane where
$$w=\frac{z+1}{z+2} \implies z=\frac{1-2w}{w-1}$$
Now we substitute this expression for $z$ into $|z-1|=|z+1|$ and then substitute $w=u+\mathrm{i}v$
$$\begin{eqnarray*}
|z-1| &=& |z+1| \\ \\
\left|\frac{1-2w}{w-1}-1\right| &=& \left|\frac{1-2w}{w-1}+1\right| \\ \\
\left|\frac{2-3w}{w-1}\right| &=& \left|\frac{-w}{w-1}\right| \\ \\
\frac{|2-3w|}{|w-1|} &=& \frac{|-w|}{|w-1|} \\ \\
|3w-2| &=& |w| \\ \\
|(3u-2)+3\mathrm{i}v| &=& |u+\mathrm{i}v| \\ \\
(3u-2)^2 + 9v^2 &=& u^2 + v^2 \\ \\
9u^2 - 12u + 4 + 9v^2 &=& u^2 + v^2 \\ \\
8u^2 - 12u +8v^2 + 4 &=& 0 \\ \\
8\left(u-\frac{3}{4}\right)^2+8v^2 &=& \frac{1}{2} \\ \\
\left(u-\frac{3}{4}\right)^2+v^2 &=& \frac{1}{16}
\end{eqnarray*}$$
Finally, we see that the image of the imaginary axis is a circle, centre $w=\frac{3}{4}$ with radius $\frac{1}{4}$.


TITLE: $x^{\frac{2}{3}}$ is not locally lipschitz at $x=0$
QUESTION [1 upvotes]: In order to show $x^{\frac{2}{3}}$ is not globally Lipschitz I can say that
$\frac{u^{\frac{2}{3}}-0}{u-0} \rightarrow\infty$  as $u \rightarrow 0$
However, why does this tell me that $x^{\frac{2}{3}}$ is not locally lipschitz at $x=0$?

REPLY [1 votes]: If the function $f(u)=|u|^{2/3}$ were locally Lipschitz at $u=0$, then there would have to exist some $ϵ>0$ and some constant $C>0$ so that, among other inequalities,
$$
||u|^{2/3}−0|≤C|u−0|
$$
for all $u\in(-ϵ,ϵ)$.
Clearly, your computations show that this is not possible, however small you choose $ϵ$, the singularity at $u=0$ is still contained in $(-ϵ,ϵ)$ and thus the difference quotients are unbounded.


TITLE: If $a_n \geq 0$ $\forall n \in \mathbb{N}$ and $\sum_{n=1}^ \infty a_n < \infty$, what can be said about $\lim_{k\to\infty}\sum_{n=k}^\infty a_n$?
QUESTION [1 upvotes]: If $a_n \geq 0$ $\forall n \in \mathbb{N}$ and $\sum_{n=1}^ \infty a_n < \infty$, what can be said about $\lim_{k\to\infty}\sum_{n=k}^\infty a_n$?
My claim is that $\lim_{k\to\infty}\sum_{n=k}^\infty a_n = 0$. To see this, first note that for all $k\in \mathbb{N}$, the series $\sum_{n=k}^\infty a_n$ converges. Write $L_k = \sum_{n=k}^\infty a_n$ for each $k \in \mathbb{N}$. Then we have
$$
L_{k+1} = \sum_{n=k+1}^\infty a_n \leq a_k + \sum_{n=k+1}^\infty a_n = \sum_{n=k}^\infty a_n = L_k.
$$
So the sequence $\{L_k\}$ is monotone decreasing. Moreover, $0 \leq L_k$ for all $k\in \mathbb{N}$. From these two facts, we can apply the monotone convergence theorem for sequences which tells us that
$$
\lim_{k \to \infty} L_k \to \inf \{\sum_{n=k}^\infty a_n : k \in \mathbb{N}\}.
$$
So it suffices to show that
$$
\inf \{\sum_{n=k}^\infty a_n : k \in \mathbb{N}\} =0.
$$
Let $\epsilon >0$. Since $\sum_{n=1}^\infty a_n$ converges, we can choose $N \in \mathbb{N}$ such that for all $j > i \geq N$ we have
$$
\sum_{n=i}^j a_n = \left| \sum_{n=i}^j a_n \right| < \epsilon.
$$
Hence, for this $N$ we have
$$
\sum_{n=N}^\infty a_n = \lim_{j \to \infty}\sum_{n=N}^j a_n < 0 + \epsilon.
$$
This shows that
$$
\inf \{\sum_{n=k}^\infty a_n : k \in \mathbb{N}\} =0.
$$
Can anyone critique my proof?

REPLY [0 votes]: Your proof is too lengthy.
Note that $$\sum _1^\infty a_n =\sum _1^{k-1} a_n+\sum _{k}^\infty a_n$$
Let $k\to \infty $ and you get $$ \lim _{k\to \infty}\sum _k^\infty a_ n = \sum _1^\infty a_n-\sum _1^\infty a_n=0$$


TITLE: Prove an inner product space
QUESTION [0 upvotes]: Prove $(L^2[0,1],\|·\|_g)$ is an inner product space, where the norm of $\|·\|_g$ is given by $\|f\|_g=(\int_0^1 g(x)|f(x)|^2dx)^{\frac{1}{2}}$. I know that I need to prove this through proving it satisfies the four properties of inner product space, I just don't know how to express the inner product from the norm. (Consider it in the real space.)
Is $\langle f,g \rangle=\int_0^1 g(x)|f(x)||g(x)|dx$ the corresponding inner product?

REPLY [1 votes]: You shouldn't use $g$ as the symbol for two functions, one the inner product's weight, the other an argument of the inner product.
Instead we want $\langle f,\,h\rangle=\int_0^1g(x)f^\ast(x)h(x)dx$ in the complex case, or $\langle f,\,h\rangle=\int_0^1g(x)f(x)h(x)dx$ in the real case.
This recovers the desired norm as a special case. Further, it's unique by the polarization identity, which @copperhat already gave in the real case. (In the complex case, it's $4\langle f,\,h\rangle=\sum_{n=0}^3i^{-n}\Vert f+i^nh\Vert_g$.)
You can now verify the inner product axioms require $g>0$ almost everywhere in $[0,\,1]$.  (Merely $g\ge0$ isn't enough, because e.g. if $g=0$ on an interval of non-zero width we only have positive semidefiniteness.)