TITLE: Inequality involving AM-GM but its wierd
QUESTION [1 upvotes]: Let a, b, c be positive real numbers. Prove that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$
Ohk now i know using AM-GM that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3] {\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \cdot 1=3 $
Now if i could have shown that the other term $\geq 1$. I would be done.
But the problem is that (again using AM-GM)
$\sqrt[3]{abc} \leq \frac{a+b+c}{3}
\Rightarrow 3 \cdot \sqrt[3]{abc} \leq a+b+c \Rightarrow \frac{3 \cdot \sqrt[3]{abc}}{a+b+c} \leq 1$
So if first part is $\geq 3$ and second part is $\leq 1$, How will i show that it is greater than $4$?
Is my approach correct? Or is there something wrong with the question?
Thanks.
(Source: https://web.williams.edu/Mathematics/sjmiller/public_html/161/articles/Riasat_BasicsOlympiadInequalities.pdf ,pg-$13$ exercise $1.3.4.a$ )

REPLY [2 votes]: From
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{(a+b+c)^2}{ab+bc+ca},$$
and $ \sqrt[3]{abc} \geqslant \frac{3abc}{ab+bc+ca},$ we need to prove
$$\frac{(a+b+c)^2}{ab+bc+ca} + \frac{9abc}{(ab+bc+ca)(a+b+c)} \geqslant 4,$$
equivalent to
$$ (a+b+c)^2 + \frac{9abc}{a+b+c} \geqslant 4(ab+bc+ca),$$
or
$$a^2+b^2+c^2 + \frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$
Which is Schur is inequality.


TITLE: Abelian varieties as analytic manifolds
QUESTION [3 upvotes]: Assume we have an Abelian varieties over the p-adic numbers, namely $
k=\mathbb{Q}_p$. Then the question is whether $A(k)$, the rational points over $k$, will form a p-adic analytic manifold.
I am reading a Book by Serre "Lie Algebras and Lie groups". i took the definition of Analytic manifolds from this text, I guess it should be an standard definition.

REPLY [7 votes]: Yes, this is true. For any smooth finite type $k$-scheme $X$, $k$ a non-Archimedean field, $X(k)$ has a canonical structure of $k$-analytic manifold in the sense of Serre's book (or Bourbaki). Usually nowadays one calls these locally $k$-analytic manifolds. I think this can be proved using "standard smooth affine opens" of $X$, or the Jacobian criterion for smoothness if you like, plus the inverse function theorem (proved in Serre's book).


TITLE: Quantum Mechanics- Quantum Zeno Paradox
QUESTION [2 upvotes]: I'm struggling with showing both results in the question and help will be very much appreciated.
Thanks.

REPLY [2 votes]: It looks like there are two eigenstates $\chi_{\pm}$ of $Q$,  and a measurement yielding $Q=\pm1$ leaves the system in the eigenstate $\chi_+$ for $Q=+1$.  The time evolution operator being
$e^{-iHt/\hbar}$, after time $T/n$ where $n$ is large this state becomes
$e^{-iHT/(n \hbar)} \chi_+ = \left(I - \frac{i H T}{n\hbar}\right) \chi_+ + {\cal O}(1/n^2)$.  So apparently (and this will be the result of whatever "the previous example" was), we must have
$$ \langle \chi_+,   H  \chi_+ \rangle= \dfrac{E_1 + E_2}{2}$$
If $A_n$ is as given, show that $|A_n|^2 = 1 + {\cal O}(1/n^2)$.  What does this say about
$\log(|A_n|^2)$ and then about $\log(|A_n|^{2n}) = n \log(|A_n|^2)$ and $|A_n|^{2n}$?


TITLE: Optimization Calculus.. a box/shelter with sides missing..
QUESTION [0 upvotes]: I'm solving a problem involving calculus optimization. The problem is the following:
"We plan to build a boxshaped shelter with no floor and one side open. (Hence we need a roof and three sides). The base of the shelter is a square. The volume of the shelter is 144 cubic meters. Find the dimensions of the shelter that minimizes the material, i.e. the surface area. Use differentiation."
Everything I say from here on out are my conclusions:
I'm assuming that the shelter is a rectangular prism. Since the base of the shelter is a square, I assume the roof is as well.
So since the roof is a square, and a square's sides are all of equal length, that means that the width/length of the roof is really the same variable. So let x = length = width. Let y denote the height.
We want to minimize the surface area of the rectangle. The surface area of a rectangle is as follows:
$$SA = 2LW + 2HW + 2LH$$
However, since we are missing the bottom of the rectangle, $L*W$, and one of the sides, either $L*W$ or $H*W$, our SA formula becomes:
$$SA = x^2 + yx + 2xy$$ or $$SA = x^2 + 3xy$$
Remember that $x = length = width$ and $y = height$.
We have more than one variable in that equation, let us connect the two. We're given that the volume of the shelter is 144. So:
$$144 = x^2 * y$$
$$y = \frac{144}{x^2}$$
And this means that:
$$SA = x^2 + \frac{432}{x}$$
Now we differentiate the function and attempt to find the critical points (where $f'(x) = 0$ or $f'(x)$ is undefined. I.e. we want the local minimum..
Differentiating SA, it becomes:
$$SA' = 2x - 432x^{-2}$$
Is this really correct or did I go very wrong somewhere along the way? If by some miracle this is correct, how would I go about finding the local minimum?

REPLY [1 votes]: This looks about right to me. You would need to set the 1st derivative you found to zero, and find the critical value. To make sure it is a minimum, you would then calculate the 2nd derivative. If the 2nd derivative with the critical value put in is positive, then you have a minimum.


TITLE: breaking up fractions
QUESTION [0 upvotes]: I have these two fractions
${11 \over 31 }+{-11 \over 61}$
Adding them gives
$330 \over 1891$
But how do I go back to the two fractions, once I've added them?
I can get the denominators just by factoring 1891, but how do I find the numerators without brute force?

REPLY [0 votes]: I eventually got around to writing a library to do this.
Link to github.


TITLE: norm induced by inner product and triangle inequality
QUESTION [1 upvotes]: Let $\langle\cdot,\cdot\rangle$ be a scalar product on a space $X$, and let $\lVert \cdot\rVert$ denote the norm induced by this scalar product. I need to show that for $x,y\in X$, $\lVert x+y\rVert=\lVert x\rVert+\lVert y\rVert $ holds $\Longleftrightarrow$ $x$ and $y$ are linear transformations of each other, i.e. $\exists \alpha\geq 0$ with $x = \alpha y$. While one direction is clear ($\Longleftarrow$) I am struggling with $\Longrightarrow$. How can I construct such an $\alpha$?

REPLY [5 votes]: Do you know that the Cauchy-Schwarz inequality
$$\lvert \langle x, y \rangle \rvert \leq \lVert x \rVert \lVert y \rVert$$
is an equality if and only if $x = \alpha y$ for some $\alpha$? (If not, here's a proof.)
With that, you can prove the other direction by noting that
$$\lVert x \rVert^2 +\lVert y \rVert^2 + 2\lVert x \rVert \lVert y \rVert = (\lVert x \rVert + \lVert y \rVert)^2 = \lVert x + y \rVert^2 =\langle x+y , x+y \rangle \\
= \langle x , x \rangle + \langle y , y \rangle + 2\langle x, y \rangle = \lVert x \rVert^2 + \lVert y \rVert^2 + 2 \langle x , y \rangle.$$
This implies that $\langle x , y \rangle = \lVert x \rVert \lVert y \rVert$, which by the first claim means that $x = \alpha y$ for some $\alpha$.


TITLE: rank of $(A+A^{T})$ when $A$ is skew symmetric
QUESTION [0 upvotes]: If $A$ is skew-symmetric and rank of $A$ is $2$. Then what can we say about $\operatorname{rank}\left(A+A^{T}\right)$? I don't find any theorem giving me a concise idea about this. (Maybe I am missing something).

REPLY [2 votes]: Any vector in the image of $A+A^T$ is clearly the sum of a vector in the image of $A$ and a vector in the image of $A^T$ (up to here, this works with any $B$ instead of $A^T$). Consequently,
$$\operatorname{rank}(A+A^T)\le \operatorname{rank}(A)+\operatorname{rank}(A^T) =2\operatorname{rank}(A)$$
Note that for any $k$ with $2k\le n$, we find an example
$$ A=\begin{pmatrix}0_{k\times (n-k)}&I_k\\0_{(n-k)\times(n-k)}&0_{(n-k)\times k}\end{pmatrix}$$where the inequality is sharp.
On the other hand, even for $k=n-1$, we find an example
$$ A=\begin{pmatrix}0&1&0&\cdots\\-1&0&1&0&\cdots\\0&-1&0&1&\\\vdots&0&-1&0&\ddots\\
&\vdots&&\ddots&\ddots\end{pmatrix}$$
where the resulting rank is $0$.


TITLE: For $n\in\mathbb{N}$ is the $n$th power of a hermitian operator always hermitian?
QUESTION [1 upvotes]: It would seem to me that $\hat{Q}^n$ being a hermitian operator when $\hat{Q}$ is one, with $n$ being a positive integer, is not true in general. After all: if $n$ is uneven then it is impossible to split the operator into the product of two of the same operators raised to the same power, like so:
$$\hat{Q}^n = \hat{Q}^{n/2}\hat{Q}^{n/2}$$
And since two different hermitian operators will not necessarily commute they might not be hermitian in that case.
Yet I've found some suggestion online that a hermitian operator raised to a power $n$ is always hermitian? Is this true, or am I misunderstanding something?

REPLY [3 votes]: And since two different hermitian operators will not necessarily commute they might not be hermitian in that case.

That may be the case, but an operator commutes with all its powers, i.e. $[A^m,A^n] = 0$ for any operator $A$ and any $m,n\in\mathbb{N}$. This follows simply from $[A,A] = 0$ and writing out the definition of $A^n$.


TITLE: Tensor is not simple
QUESTION [1 upvotes]: I'm trying to solve the following:
Let $e_1,e_2 \in V$ linearly independent. I want to show that $e_1 \otimes e_2 +e_2 \otimes e _1 \in V \otimes V$ is not simple. $V$ is a finite dimensional vector space.
By contradiction, I assume that there exist $p,q \in V$ such that
$$p \otimes q = e_1 \otimes e_2 +e_2 \otimes e _1 \tag{1}$$
If now $\{ e_1, e_2, e_3, ..., e_n\}$ represents the basis of $V$ I can express $p$ as
$$p=\lambda_1 e_1 + \lambda_2 e_2 + \sum_{i=3}^{n} \lambda_i e_i   \tag{2}$$
and
$$q=\tilde{\lambda_1} e_1 + \tilde{\lambda_2} e_2 + \sum_{i=3}^{n} \tilde{\lambda_i} e_i \tag{3}$$
With a naive approach I thought about plugging in (2) and (3) in (1), expanding it and compare the coefficients which leads to
$e_1 \otimes e_2 +e_2 \otimes e_1  \stackrel{!}{=}   p \otimes q = \lambda_1 \tilde\lambda_2 \otimes (e_1,e_1) + \lambda_1 \tilde{\lambda}_2 \otimes (e_1,e_2) + \lambda_1 \sum_{i=3}^n \tilde{\lambda}_i \otimes (e_1,e_i)+\lambda_2\tilde{\lambda}_1 \otimes (e_2,e_1) + \lambda_2 \tilde{\lambda}_2 \otimes(e_2,e_2)+ ...$
By comparing the coefficients the following equations must hold
$\lambda_1 \tilde{\lambda}_2=1$
$\lambda_2 \tilde{\lambda}_1=1$
$\lambda_1 \tilde{\lambda_1}=0$
$\lambda_2 \tilde{\lambda_2}=0$
Obviously, this leads to a contradiction, i.e. $e_1 \otimes e_2 +e_2 \otimes e _1 \in V \otimes V$ is not simple. I this approach correct? I'm very new to the topic of tensors so I'm just wondering if it's really that easy.

REPLY [1 votes]: There is a slight problem with notation, in that you are writing $\lambda_1\tilde{\lambda_2}\otimes (e_1,e_2)$ when you should be writing $\lambda_1\tilde{\lambda_2}(e_1\otimes e_2)$, etc.  However, this is easy to fix without fundamentally changing the proof.
Here is a higher level approach that relies on properties of tensor products, and has some connections with some ideas you will see later.
Suppose that $e_1\otimes e_2 + e_2\otimes e_1=v\otimes w$ for some vectors $v,w\in V$.  Then for any bilinear map $\psi$, we must have $\psi(v,w)=\psi(e_1,e_2)+\psi(e_2,e_1)$.  So all we must do is find the right bilinear maps to get a contradictory set of conditions on $v$ and $w$.
Fix some projection map $P:V\to \mathbb F^2$ with $P(e_1)=e_1, P(e_2)=e_2$. Then $P\otimes P:V\otimes V\to \mathbb F^2\otimes \mathbb F^2$, and so $P(v)\otimes  P(w)=e_1\otimes e_2 + e_2\otimes e_1$.  Therefore, we can reduce to the case that $V$ is $\mathbb F^2$.
We could use bilinear maps to pick off the coefficients and obtain a contradiction that way, mimicking the original proof.  However, for variety's sake, here is another more geometric approach.
Consider the bilinear map $(v,w)\mapsto \det(v|w)$ where $(v|w)$ is the matrix with columns $v, w$.  Then since $\det(e_1|e_2)=-\det(e_2|e_1)$, we must have $\det(v|w)=0$, and so $v, w$ are linearly dependent, so $w=\alpha v$ for some $\alpha \in \mathbb F^{\times}$.  In higher level language, what we have shown is a shadow of the fact that since $e_1\wedge e_2+e_2\wedge e_1=0$, then $v\wedge w=0$, and so $v$ and $w$ are linearly dependent.  This more general fact doesn't require using the projection to two dimensions.
Define the map $S_c(e_1)=c e_2, S_c(e_2)=e_1$ for $c\in \mathbb F$.  By taking $\phi(v,w)=\det(v|S(w))$ and noting that $\det(e_1|e_1)+\det(e_2|c e_2)=0$, the exact same argument as above shows that $v$ and $S_c(w)$ are linearly dependent.  This means that $w$ and $S_c(w)$ are also dependent (since $v\neq 0$), so $w$ is an eigenvector of $S_c$ for every $c$.  But this is impossible.


TITLE: What has been measured at the Higgs experiment and what do we know now?
QUESTION [5 upvotes]: Explained at the level of a 5$^{\text {th}}$ semester physics student (i.e. pre QFT, but far beyond the level of a news article for non-physicists, which avoids all details and only deals in analogies) ...

What has been measured at CERN some days ago?
What are the essential ingredients of the theory necessary to interpret said measurement? And so how do we deduce from the results that there is a new field/particle observed?
How to read the most relevant graphs in the presentation of the results?

REPLY [5 votes]: What has been measured at CERN some days ago?

Once you have an event (as @user1504 describes in his answer), i.e. a proton proton interaction, and have the four vectors of all the interaction products, i.e. jets, photons, leptons, you can generate invariant masses of the interaction products. The Higgs particle has been predicted  theoretically and is the last piece in the jig saw puzzle that the Standard Model has put together in one picture. The invariant mass of two photons shows an enhancement consistent with the Higgs properties. Also the invariant masses of Z Z to a lesser extent. The combined statistics of all possible decay channels of the Higgs seen give a 5sigma certainty that the resonance is there and has the expected decay behavior within statistics of the individual decay channels.

What are the essential ingredients of the theory necessary to interpret said measurement? And so how do we deduce from the results that there is a new field/particle observed?

The theoretical part is a different question and should be posed independently: why and how the standard model theory accommodates all particle data.
That there is a new particle observed with 5 sigma certainty is what the data tells us, and it is new because we have never observed a 125GeV resonance before the present experiments.
To gauge whether it is  the expected from the SM theory Higgs particle one needs good statistics in all the decay channels plus angular distributions that will establish the spin parity. Th theoretical Higgs has 0 spin and positive parity.

How to read the most relevant graphs in the presentation of the results?

The resonance is clear in fig 3. The sigma of this channel is its statistical significance for it alone. Similar for the ZZ channel, fig 4.
Fig 5 gives the probability,

The observed probability (local p-value) that the background-only hypothesis would yield the same or more events as are seen in the CMS data, as a function of the SM Higgs boson mass for the five channels considered. The solid black line shows the combined local p-value for all channels.


TITLE: Notation issue - Asymptotic behaviour: is $\sim$ too restrictive?
QUESTION [1 upvotes]: As a student I am completely unable to understand unambiguously what is meant by a notation such as $$f \sim g $$ when in Physics the behaviour of two functions at infinity is evaluated.
I found a definition according to which the notation is equivalent to stating that $$ \lim_{x \to \infty} \frac {f}{g} = 1$$ but is this not too restrictive?
This then means is not correct to say that $$x^2$$ and $$\frac{1}{2} x^2$$ have  the same asymptotic behaviour at infinity. I find it puzzling.
Thanks
PS: The question is admittedly trivial but an extensive search left it unanswered, so I appeal to your patience...

REPLY [1 votes]: In some italian textbooks we use two different symbols: as $x \to x_0$

$f\sim g$ to mean that $\lim_{x \to x_0} \frac{f(x)}{g(x)}=\ell \in \mathbb{R} \setminus \{0\}$, and
$f \asymp g$ to mean that $\lim_{x \to x_0} \frac{f(x)}{g(x)}=1$.

This notation is not really popular, but it helps you to understand where a multiplicative constant is not enough.


text:
\begin{document}

\title[Double difference property for H\"{o}lder continuous functions]{The double difference property for the class of locally H\"{o}lder continuous functions}

\author{Rashid A. Aliev}

\address{Institute of Mathematics and Mechanics, NAS of Azerbaijan, Baku, Azerbaijan \newline Baku State University, Baku, Azerbaijan}

\email{aliyevrashid@mail.ru}

\author{Aysel A. Asgarova}

\address{Azerbaijan University of Languages, Baku, Azerbaijan}

\email{asgarova2016@mail.ru}

\author{Vugar E. Ismailov}

\address{Institute of Mathematics and Mechanics, NAS of Azerbaijan, Baku, Azerbaijan}

\email{vugaris@mail.ru}

\subjclass[2000]{26B05, 39A70, 39B22, 47B39}

\keywords{Cauchy functional equation; additive function; difference
property; double difference property; modulus of continuity; H\"{o}lder
continuity}

\begin{abstract}
In this paper, we show that the pair of classes of locally H\"{o}lder continuous functions
(considered on $\mathbb{R}$ and $\mathbb{R}^{2}$, respectively) has the double difference property.
\end{abstract}

\maketitle

\section{Introduction}

The notions \textit{difference property} and \textit{double difference property}
are due to de Bruijn \cite{1} and Laczkovich \cite{5},
respectively. These properties for various classes of real functions were
investigated by many authors. We refer the reader to Laczkovich's survey
paper \cite{6} for a detailed source of information on this topic.

For a fixed function $g:\mathbb{R}\rightarrow \mathbb{R}$ and any $h\in
\mathbb{R}$ we define the difference function $\Delta _{h}g:\mathbb{R}
\rightarrow \mathbb{R}$ by
\begin{equation*}
\Delta _{h}g(x)=g(x+h)-g(x)
\end{equation*}
and the double difference function $Dg:$ $\mathbb{R}^{2}\rightarrow \mathbb{R
}$ by
\begin{equation*}
Dg(x,y)=g(x+y)-g(x)-g(y).
\end{equation*}

Let $\mathcal{F}$ be a class of functions defined on $\mathbb{R}$ and $
\mathcal{F}_{2}$ be a class of functions defined on $\mathbb{R}^{2}$. The
class $\mathcal{F}$ is said to have the difference property if every
function $g:\mathbb{R}\rightarrow \mathbb{R}$, for which $\Delta _{h}g\in
\mathcal{F}$ for each $h\in \mathbb{R}$, is of the form $g=f+A$, where $f\in
\mathcal{F}$ and $A$ is an additive function (see \cite{1}). A function $A$
is called additive if it satisfies the Cauchy functional equation $
A(x+y)=A(x)+A(y)$. The pair $\left( \mathcal{F},\mathcal{F}_{2}\right) $ is
said to have the double difference property if whenever $Dg\in \mathcal{F}
_{2}$ holds for a function $g:\mathbb{R}\rightarrow \mathbb{R}$, then $g$ is
of the form $g=f+A$, where $f\in \mathcal{F}$ and $A$ is additive (see \cite
{5}).

de Bruijn \cite{1} was the first who showed that the class of continuous
functions has the difference property and thus resolved Erd\"{o}s's famous
conjecture. He also proved that the difference property holds for a large
number of essential function classes (see \cite{1,2}). Some of these classes
are

\smallskip

1) $C^{k}(\mathbb{R)}$, functions with continuous derivatives up to order $k$
;

2) $C^{\infty }(\mathbb{R)}$, infinitely differentiable functions;

3) analytic functions;

4) functions which are absolutely continuous on any finite interval;

5) functions having bounded variation over any finite interval;

6) algebraic polynomials;

7) trigonometric polynomials;

8) Riemann integrable functions.

\smallskip

However, the class $\mathcal{L}$ of Lebesgue measurable functions fails to
have this property if we assume the continuum hypothesis (see \cite{1,5}).
It was conjectured by Erd\"{o}s that every function $g:\mathbb{R}\rightarrow
\mathbb{R}$ for which $\Delta _{h}g(x)$ is measurable for each $h$, is of
the form $g=f+A+S$, where $f$ is measurable, $A$ is additive and $S$ has the
property that $\Delta _{h}S(x)=0$ for almost all $x$. Laczkovich \cite{5}
solved this conjecture affirmatively and moreover proved that the pair $(
\mathcal{L},\mathcal{L}_{2})$ has the double difference property, where $
\mathcal{L}_{2}$ denotes the class of Lebesgue measurable functions defined
on $\mathbb{R}^{2}$. It was also proved in \cite{5} that the double
difference property holds for Baire $\alpha $ functions. Later Tabor and
Tabor \cite{7} proved that the class $C^{n}(X,Y)$ of $n$-times continuously
differentiable functions defined on a real normed space $X$ and taking
values in a real Banach space $Y$ has the double difference property.
Kotlicka \cite{3} showed that several pairs of classes of functions have the
double difference property. Among them there are approximately continuous
functions, pointwise continuous functions, essentially continuous functions
(considered on $\mathbb{R}$ and $\mathbb{R}^{2}$, respectively) and $L_{p}$
-classes for $0<p<\infty $ (considered on $\mathbb{T}$ and $\mathbb{T}^{2}$,
respectively, where $\mathbb{T}$ is the torus).

In \cite{T1}, Tabor proved that the pair of classes of Lipschitz functions
defined on a metric semigroup $G$ and $G\times G$, respectively, with values
in a reflexive Banach space $E$ has the double difference property. For
finite dimensional Banach spaces $X$ and $Y$, Tabor and Tabor \cite{T2}
showed that the double difference property holds for the pair of classes of $
Y$-valued Lipschitz functions defined on a convex set $K\subset X$ such that
$0\in K$ $\ $and on the set
\begin{equation*}
C(K)=\left\{ (x,y)\in X\times X:x\in K,y\in K,x+y\in K\right\} ,
\end{equation*}
respectively. Consequently, the double difference property holds for the
pair of classes of real Lipschitz functions defined on an interval $I$
containing zero and on the set $C(I)$, respectively.

In this paper, we prove that for any $\alpha \in (0,1]$ the pair of classes
of locally H\"{o}lder real continuous functions (considered on $\mathbb{R}$
and $\mathbb{R}^{2}$, respectively) with exponent $\alpha $ has the double
difference property.

\bigskip

\section{Main result}

We start this section with the definition of modulus of continuity of a
multivariate function and some notation. Let $f(\mathbf{x}
)=f(x_{1},...,x_{s})$, $s\geq 1,$ be any $s$-variable function defined on a
set $\Omega \subset \mathbb{R}^{s}$. The function
\begin{equation*}
\omega (f;\delta ;\Omega )=\sup \left\{ \left\vert f(\mathbf{x})-f(\mathbf{y}
)\right\vert :\mathbf{x},\mathbf{y}\in \Omega ,\text{ }\left\vert \mathbf{x}-
\mathbf{y}\right\vert \leq \delta \right\} ,\text{ }0\leq \delta \leq
diam\Omega ,
\end{equation*}
is called the modulus of continuity of $f$ on $\Omega $. We will also use
the notation $\omega _{\mathbb{Q}}(f;\delta ;\Omega )$, which stands for the
function $\omega (f;\delta ;\Omega \cap \mathbb{Q}^{s})$. Here $\mathbb{Q}
^{s}$ denotes the space of $s$-dimensional vectors with rationale
coordinates. Clearly, $\omega _{\mathbb{Q}}(f;\delta ;\Omega )$ makes sense
if the set $\Omega \cap \mathbb{Q}^{s}$ is not empty. Note that we always
have the inequality $\omega _{\mathbb{Q}}(f;\delta ;\Omega )\leq \omega
(f;\delta ;\Omega )$ and the strong equality $\omega _{\mathbb{Q}}(f;\delta
;\Omega )=\omega (f;\delta ;\Omega )$ holds for continuous $f$ and certain
sets $\Omega $. For example, this holds if for any $\mathbf{x},\mathbf{y}\in
\Omega $ with $\left\vert \mathbf{x}-\mathbf{y}\right\vert \leq \delta $
there exist sequences $\left\{ \mathbf{x}_{n}\right\} ,\left\{ \mathbf{y}
_{n}\right\} \subset \Omega \cap \mathbb{Q}^{s}$ such that $\mathbf{x}
_{n}\rightarrow \mathbf{x}$, $\mathbf{y}_{n}\rightarrow \mathbf{y}$ and $
\left\vert \mathbf{x}_{n}-\mathbf{y}_{n}\right\vert \leq \delta ,$ for all $
n $. There are many sets $\Omega $, which satisfy this property.

The class $H_{\alpha }^{\left( loc\right) }\left( \mathbb{R}^{s}\right) $ of
locally H\"{o}lder continuous functions with exponent $\alpha $ is defined
as the class of functions $f$ for which $\omega (f;\delta ;\Omega )\leq
K\delta ^{\alpha }$ for any compact set $\Omega \subset \mathbb{R}^{s}$.
Here $K$ depends on $\Omega $.

\bigskip

Our main result is the following theorem.

\bigskip

\textbf{Theorem 2.1.} \textit{Assume a function $g:\mathbb{R}\rightarrow
\mathbb{R}$ is such that the bivariate function $\mathit{g(x+y)}-g(x)-g(y)$
is locally H\"{o}lder continuous with exponent $\alpha$. Then there exist a
function $f\in H_{\alpha }^{\left( loc\right) }\left( \mathbb{R}\right) $
and an additive function $A$ such that $g=f+A$.}

\bigskip

To prove this theorem we need the following auxiliary lemma.

\bigskip

\textbf{Lemma 2.1.} \textit{Assume a function $F\in C(\mathbb{R}^{2})$ has
the form}
\begin{equation}
F(x,y)=g(x+y)-g(x)-g(y),  \label{3.1}
\end{equation}
\textit{where $g$ is an arbitrarily behaved function. Then the following
inequality holds}

\begin{equation}
\omega _{\mathbb{Q}}(g;\delta ;[-M,M])\leq 2\delta \left\vert
g(1)-g(0)\right\vert +3\omega \left( F;\delta ;[-M,M]^{2}\right) ,
\label{3.2}
\end{equation}
\textit{where $\delta \in \left( 0,\frac{1}{2}\right) \cap \mathbb{Q}$ and $
M\geq 1$.}

\bigskip

\begin{proof} Consider the function $h(t)=g(t)-g(0)$ and write \eqref{3.1}
in the form

\begin{equation}
G(x,y)=h(x+y)-h(x)-h(y),  \label{3.3}
\end{equation}
where

\begin{equation*}
G(x,y)=F(x,y)+g(0).
\end{equation*}
Note that the functions $g$ and $h$, as well as the functions $F$ and $G,$
have the common modulus of continuity. Thus we prove the lemma if we prove
it for the pair $\left\langle G,h\right\rangle .$

\bigskip Since $h(0)=0,$ it follows from \eqref{3.3} that

\begin{equation}
G(x,0)=G(0,y)=0.  \label{3.4}
\end{equation}
Obviously, for any real number $x,$
\begin{eqnarray*}
G(x,x) &=&h(2x)-2h(x); \\
G(x,2x) &=&h(3x)-h(x)-h(2x); \\
&&\cdot \cdot \cdot \\
G(x,(k-1)x) &=&h(kx)-h(x)-h((k-1)x).
\end{eqnarray*}

We obtain from the above equalities that

\begin{eqnarray*}
h(2x) &=&2h(x)+G(x,x), \\
h(3x) &=&3h(x)+G(x,x)+G(x,2x), \\
&&\cdot \cdot \cdot \\
h(kx) &=&kh(x)+G(x,x)+G(x,2x)+\cdot \cdot \cdot +G(x,(k-1)x).
\end{eqnarray*}
Thus for any nonnegative integer $k$,

\begin{equation}
h(x)=\frac{1}{k}h(kx)-\frac{1}{k}\left[ G(x,x)+G(x,2x)+\cdot \cdot \cdot
+G(x,(k-1)x)\right] .  \label{3.5}
\end{equation}

Consider now the simple fraction $\frac{p}{n}\in (0,\frac{1}{2})$ and set $
m_{0}=\left[ \frac{n}{p}\right] .$ Here $[r]$ denotes the whole number part
of $r$. Clearly, $m_{0}\geq 2$ and the remainder $p_{1}=n-m_{0}p<p.$ Taking $
x=\frac{p}{n}$ and $k=m_{0}$ in \eqref{3.5} gives us the following equality

\begin{equation*}
h\left( \frac{p}{n}\right) =\frac{1}{m_{0}}h\left( 1-\frac{p_{1}}{n}\right)
\end{equation*}
\begin{equation}
-\frac{1}{m_{0}}\left[ G\left( \frac{p}{n},\frac{p}{n}\right) +G\left( \frac{
p}{n},\frac{2p}{n}\right) +\cdot \cdot \cdot +G\left( \frac{p}{n},(m_{0}-1)
\frac{p}{n}\right) \right] .  \label{3.6}
\end{equation}
On the other hand, since

\begin{equation*}
G\left( \frac{p_{1}}{n},1-\frac{p_{1}}{n}\right) =h(1)-h\left( \frac{p_{1}}{n
}\right) -h\left( 1-\frac{p_{1}}{n}\right) ,
\end{equation*}
it follows from \eqref{3.6} that

\begin{equation*}
h\left( \frac{p}{n}\right) =\frac{h(1)}{m_{0}}-\frac{1}{m_{0}}\left[ G\left(
\frac{p}{n},\frac{p}{n}\right) +\cdot \cdot \cdot +G\left( \frac{p}{n}
,(m_{0}-1)\frac{p}{n}\right) +G\left( \frac{p_{1}}{n},1-\frac{p_{1}}{n}
\right) \right]
\end{equation*}
\begin{equation}
-\frac{1}{m_{0}}h\left( \frac{p_{1}}{n}\right) .  \label{3.7}
\end{equation}

Put $m_{1}=\left[ \frac{n}{p_{1}}\right] $, $p_{2}=n-m_{1}p_{1}.$ Clearly, $
0\leq p_{2}<p_{1}$. Similar to \eqref{3.7}, we can write that

\begin{equation*}
h\left( \frac{p_{1}}{n}\right) =\frac{h(1)}{m_{1}}-\frac{1}{m_{1}}\left[
G\left( \frac{p_{1}}{n},\frac{p_{1}}{n}\right) +\cdot \cdot \cdot +G\left(
\frac{p_{1}}{n},(m_{1}-1)\frac{p_{1}}{n}\right) +G\left( \frac{p_{2}}{n},1-
\frac{p_{2}}{n}\right) \right]
\end{equation*}
\begin{equation}
-\frac{1}{m_{1}}h\left( \frac{p_{2}}{n}\right) .  \label{3.8}
\end{equation}

Let us make a convention that \eqref{3.7} is the $1$-st and \eqref{3.8} is
the $2$-nd formula. One can continue this process by defining the chain of
pairs $(m_{2},p_{3}),$ $(m_{3},p_{4})$ until the pair $(m_{k-1},p_{k})$ with
$p_{k}=0$ and writing out the corresponding formulas for each pair. For
example, the last $k$-th formula will be of the form
\begin{equation*}
h\left( \frac{p_{k-1}}{n}\right) =\frac{h(1)}{m_{k-1}}
\end{equation*}
\begin{equation*}
-\frac{1}{m_{k-1}}\left[ G\left( \frac{p_{k-1}}{n},\frac{p_{k-1}}{n}\right)
+\cdot \cdot \cdot +G\left( \frac{p_{k-1}}{n},(m_{k-1}-1)\frac{p_{k-1}}{n}
\right) +G\left( \frac{p_{k}}{n},1-\frac{p_{k}}{n}\right) \right]
\end{equation*}
\begin{equation}
-\frac{1}{m_{k-1}}h\left( \frac{p_{k}}{n}\right) .  \label{3.9}
\end{equation}
Note that in \eqref{3.9}, $h\left( \frac{p_{k}}{n}\right) =0$ and $G\left(
\frac{p_{k}}{n},1-\frac{p_{k}}{n}\right) =0$. Considering now the $k$-th
formula in the $(k-1)$-th formula, then the obtained formula in the $(k-2)$
-th formula, and so forth, we will finally arrive at the equality

\begin{equation*}
h\left( \frac{p}{n}\right) =h(1)\left[ \frac{1}{m_{0}}-\frac{1}{m_{0}m_{1}}
+\cdot \cdot \cdot +\frac{(-1)^{k-1}}{m_{0}m_{1}\cdot \cdot \cdot m_{k-1}}
\right]
\end{equation*}

\begin{equation*}
-\frac{1}{m_{0}}\left[ G\left( \frac{p}{n},\frac{p}{n}\right) +\cdot \cdot
\cdot +G\left( \frac{p}{n},(m_{0}-1)\frac{p}{n}\right) +G\left( \frac{p_{1}}{
n},1-\frac{p_{1}}{n}\right) \right]
\end{equation*}

\begin{equation*}
+\frac{1}{m_{0}m_{1}}\left[ G\left( \frac{p_{1}}{n},\frac{p_{1}}{n}\right)
+\cdot \cdot \cdot +G\left( \frac{p_{1}}{n},(m_{1}-1)\frac{p_{1}}{n}\right)
+G\left( \frac{p_{2}}{n},1-\frac{p_{2}}{n}\right) \right]
\end{equation*}

\begin{equation*}
+\cdot \cdot \cdot +
\end{equation*}
\begin{equation}
\frac{(-1)^{k}}{m_{0}m_{1}\cdot \cdot \cdot m_{k-1}}\left[ G\left( \frac{
p_{k-1}}{n},\frac{p_{k-1}}{n}\right) +\cdot \cdot \cdot +G\left( \frac{
p_{k-1}}{n},(m_{k-1}-1)\frac{p_{k-1}}{n}\right) \right] .  \label{3.10}
\end{equation}
Taking into account \eqref{3.4}, we obtain from \eqref{3.10} that

\begin{equation*}
\left\vert h\left( \frac{p}{n}\right) \right\vert \leq \left[ \frac{1}{m_{0}}
-\frac{1}{m_{0}m_{1}}+\cdot \cdot \cdot +\frac{(-1)^{k-1}}{m_{0}m_{1}\cdot
\cdot \cdot m_{k-1}}\right] \left\vert h(1)\right\vert
\end{equation*}

\begin{equation}
+\left[ 1+\frac{1}{m_{0}}+\cdot \cdot \cdot +\frac{1}{m_{0}\cdot \cdot \cdot
m_{k-2}}\right] \omega \left( G;\frac{p}{n};[0,1]^{2}\right).  \label{3.11}
\end{equation}
Since $m_{0}\leq m_{1}\leq \cdot \cdot \cdot \leq m_{k-1},$ it is not
difficult to see that in \eqref{3.11}

\begin{equation*}
\frac{1}{m_{0}}-\frac{1}{m_{0}m_{1}}+\cdot \cdot \cdot +\frac{(-1)^{k-1}}{
m_{0}m_{1}\cdot \cdot \cdot m_{k-1}}\leq \frac{1}{m_{0}}
\end{equation*}
and

\begin{equation*}
1+\frac{1}{m_{0}}+\cdot \cdot \cdot +\frac{1}{m_{0}\cdot \cdot \cdot m_{k-2}}
\leq \frac{m_{0}}{m_{0}-1}.
\end{equation*}
Considering the above two inequalities in \eqref{3.11} we obtain that

\begin{equation}
\left\vert h\left( \frac{p}{n}\right) \right\vert \leq \frac{\left\vert
h(1)\right\vert }{m_{0}}+\frac{m_{0}}{m_{0}-1}\omega \left( G;\frac{p}{n}
;[0,1]^{2}\right) .  \label{3.12}
\end{equation}
Since $m_{0}=\left[ \frac{n}{p}\right] \geq 2,$ it follows from \eqref{3.12}
that

\begin{equation}
\left\vert h\left( \frac{p}{n}\right) \right\vert \leq \frac{2p\left\vert
h(1)\right\vert }{n}+2\omega \left( G;\frac{p}{n};[0,1]^{2}\right) .
\label{3.13}
\end{equation}

Let now $\delta \in \left( 0,\frac{1}{2}\right) \cap \mathbb{Q}$ be a
rational increment, $M\geq 1$ and $x,x+\delta $ be two points in $\left[ -M,M
\right] \cap \mathbb{Q}.$ By \eqref{3.3}, \eqref{3.4} and \eqref{3.13} we
can write that

\begin{equation}
\left\vert h(x+\delta )-h(x)\right\vert \leq \left\vert h(\delta
)\right\vert +\left\vert G(x,\delta )\right\vert \leq 2\delta \left\vert
h(1)\right\vert +3\omega \left( G;\delta ;[-M,M]^{2}\right) .  \label{3.14}
\end{equation}
Now \eqref{3.2} follows from \eqref{3.14} and the definitions of $h$ and $G$.
\end{proof}

\bigskip

\textit{Remark 1.} Under the assumptions of Lemma 2.1, the restriction of $g$
to the set of rational numbers is uniformly continuous on any interval $
[-M,M]\cap \mathbb{Q}$ and hence continuous on $\mathbb{Q}$.

\bigskip

Now we are ready to prove Theorem 2.1.

\begin{proof} Let us put

\begin{equation}
F(x,y)=g(x+y)-g(x)-g(y)  \label{3.00}
\end{equation}
and consider the function

\begin{equation*}
u(t)=g(t)-\left[ g(1)-g(0)\right] t.
\end{equation*}
Obviously, $u(1)=u(0)$ and

\begin{equation}
F(x,y)=u(x+y)-u(x)-u(y).  \label{3.17}
\end{equation}
By Lemma 2.1, the restriction of $u$ to $\mathbb{Q}$ is continuous and
uniformly continuous on every interval $[-M,M]\cap \mathbb{Q}$. Denote this
restriction by $v$.

Let $y$ be any real number and $\{y_{n}\}_{n=1}^{\infty }$ be any sequence
of rational numbers converging to $y$. We can choose $M>0$ so that $y_{n}\in
\lbrack -M,M]$ for any $n\in \mathbb{N}$. It follows from the uniform
continuity of $v$ on $[-M,M]\cap \mathbb{Q}$ that the sequence $
\{v(y_{n})\}_{n=1}^{\infty }$ is Cauchy. Thus there exits a finite limit $
\lim_{n\rightarrow \infty }v(y_{n})$. It is not difficult to see that this
limit does not depend on the choice of $\{y_{n}\}_{n=1}^{\infty }$.

Let $f$ denote the following extension of $v$ to the set of real numbers.

\begin{equation*}
f(y)=\left\{
\begin{array}{c}
v(y),\text{ if }y\in \mathbb{Q}\text{;} \\
\lim_{n\rightarrow \infty }v(y_{n}),\text{ if }y\in \mathbb{R}\backslash
\mathbb{Q}\text{ and }\{y_{n}\}\text{ is a sequence in }\mathbb{Q}\text{
tending to }y.
\end{array}
\right.
\end{equation*}
In view of the above arguments, $f$ is well defined on the whole real line.
Let us prove that $f$ is the function we seek.

Consider an arbitrary point $(x,y)\in \mathbb{R}^{2}$ and a sequence of
points $\{(x_{n},y_{n})\}_{n=1}^{\infty }$ with rationale coordinates
tending to $(x,y)$. Taking into account \eqref{3.17}, we can write that

\begin{equation}
F(x_{n},y_{n})=v(x_{n}+y_{n})-v(x_{n})-v(y_{n}),\text{ for all }n=1,2,...,
\label{3.18}
\end{equation}
since $v$ is the restriction of $u$ to $\mathbb{Q}$. Tending $n\rightarrow
\infty $ in both sides of \eqref{3.18} we obtain that

\begin{equation}
F(x,y)=f(x+y)-f(x)-f(y).  \label{3.0}
\end{equation}

Set $A=g-f$. It follows from \eqref{3.00} and \eqref{3.0} that $A$ is
additive. Let us now prove that $f\in H_{\alpha }^{\left( loc\right) }\left(
\mathbb{R}\right) $. Since $v(1)=v(0)$ we obtain from \eqref{3.17} and
\eqref{3.2} that for $\delta \in \left( 0,\frac{1}{2}\right) \cap \mathbb{Q}$
, $M\geq 1$ and any numbers $a,b\in \lbrack -M,M]\cap \mathbb{Q}$, $
\left\vert a-b\right\vert \leq \delta ,$ the following inequality holds

\begin{equation}
\left\vert v(a)-v(b)\right\vert \leq 3\omega \left( F;\delta
;[-M,M]^{2}\right) .  \label{3.19}
\end{equation}
Consider now any real numbers $r_{1}$ and $r_{2}$ satisfying $r_{1},r_{2}\in
\lbrack -M,M]$, $\left\vert r_{1}-r_{2}\right\vert \leq \delta $ and take
sequences $\{a_{n}\}_{n=1}^{\infty }\subset \lbrack -M,M]\cap \mathbb{Q}$, $
\{b_{n}\}_{n=1}^{\infty }\subset \lbrack -M,M]\cap \mathbb{Q}$ with the
property $\left\vert a_{n}-b_{n}\right\vert \leq \delta ,$ $n=1,2,...,$ and
tending to $r_{1}$ and $r_{2}$, respectively. By \eqref{3.19},
\begin{equation*}
\left\vert v(a_{n})-v(b_{n})\right\vert \leq 3\omega \left( F;\delta
;[-M,M]^{2}\right) .
\end{equation*}
If we take limits\ on both sides of the above inequality, we obtain that

\begin{equation}
\left\vert f(r_{1})-f(r_{2})\right\vert \leq 3\omega \left( F;\delta
;[-M,M]^{2}\right) ,  \label{3.20}
\end{equation}
which means that $f$ is uniformly continuous on $[-M,M]$ and hence it is
continuous on the whole real line.

It follows from \eqref{3.20} that

\begin{equation}
\omega \left( f;\delta ;[-M,M]\right) \leq 3\omega \left( F;\delta
;[-M,M]^{2}\right) .  \label{3.21}
\end{equation}
Note that in \eqref{3.21} $\delta $ is a rational number from the interval $
\left( 0,\frac{1}{2}\right) $. Since the modulus of continuity of a
continuous function is continuous from the right (see \cite{Kol}), it
follows that, in fact, \eqref{3.21} is valid for all $\delta \in \left[ 0,
\frac{1}{2}\right) $. Since $F\in H_{\alpha }^{\left( loc\right) }\left(
\mathbb{R}^{2}\right) $, we obtain from \eqref{3.21} that

\begin{equation}
\omega \left( f;\delta ;[-M,M]\right) \leq K\delta ^{\alpha },\text{ where }
0\leq \delta <\frac{1}{2}\text{.}  \label{3.22}
\end{equation}

Let now $\frac{1}{2}\leq \delta \leq 2M$. We can write that

\begin{equation}
\omega \left( f;\delta ;[-M,M]\right) \leq 2\left\Vert f\right\Vert
_{C([-M,M])}\leq 2^{1+\alpha }\left\Vert f\right\Vert _{C([-M,M])}\delta
^{\alpha },\text{ where }\frac{1}{2}\leq \delta \leq 2M\text{.}  \label{3.23}
\end{equation}

The inequalities \eqref{3.22} and \eqref{3.23} show that $f$ is H\"{o}lder
continuous on $[-M,M]$ with exponent $\alpha $. Since $M$ is an arbitrary
number not less than $1$ and any compact $\Omega $ is contained in a closed
interval of the form $[-M,M]$, we obtain that $f\in H_{\alpha }^{\left(
loc\right) }\left( \mathbb{R}\right) $.
\end{proof}

\textit{Remark 2.} The above proof shows that for any compact set $\Omega
\subset $ $\mathbb{R}$ the pair of H\"{o}lder continuous function classes $
H_{\alpha }\left( \Omega \right) $ and $H_{\alpha }\left( \Omega \times
\Omega \right) $ has the double difference property. This holds, in
particular, for the pair of classes of Lipschitz functions defined on $
\Omega $ and $\Omega \times \Omega $, respectively. The last assertion
complements the corresponding result of J. Tabor and J. Tabor \cite{T2} in
the real space setting (see Introduction).

\bigskip

\textit{Remark 3.} Theorem 2.1 is not only an existence result. It's proof
gives a recipe for constructing the function $f$. It also allows us to
estimate the modulus of continuity of $f$ in terms of the modulus of
continuity of $g(x+y)-g(x)-g(y)$ (see (2.21)).

\bigskip

config: arxiv


TITLE: How to explain some gaps in understanding of the following interpretation of a given proof?
QUESTION [1 upvotes]: I've been teaching myself linear algebra, and have come to the point where I'm studying the theorem which states that RREF matrices are always unique.
I found this proof, but I'm not sure if my understanding is entirely complete.
I'll start with rank, taken from the proof:
"
First, having a matrix with two reduced row echelon forms means there are two reduced row echelon form matrices $E$ and $E'$ and an inversible matrix $P$ such that $E'=PE$. The goal is then to prove $E=E'$ (of size $m\times n$).
Now, first remark they must have the same rank $r$, which we can assume at least one or the result is obvious, so we have $r\geqslant1$."
My understanding is that a matrix's rank is essentially the amount of its rows (or columns) which are linearly independent.
With a rank of 1, we have a matrix with at least one linear independent row; we'll call this $R_1$. Any other row $R_i$ can be derived from a linear combination of $\alpha R_1$, where $\alpha$ is some constant such that $\alpha \ne 0$; or is the result of $\alpha R_1 + R_i$ for $0 \le i, j \le m$, where $E$, $E^{\prime}$, and $P$ are $m \times n$ matrices.
Moving on,
"
Denote by $p_1,\dots,p_r$ the indices of the pivot columns in $E$, and likewise $p_1',\dots,p_r'$ for $E'$, and complete by $p_{r+1}=p_{r+1}'=n$ for convenience.
"
My assumption is that, because $r$ was referred to as the rank of our matrices, $r$'s value also applies in the subscript of these indices. Furthermore, the statement $p_{r+1} = p^{\prime}_{r+1} = n$ appears to imply that the very last column of $E$ is actually linear independent; otherwise, we would have $n = p_{r} = p^{\prime}_{r}$.
The author then continues with the following method:
"
Define for $j$ such that $0\leqslant j\leqslant r$ the assertion $H_j$ by the following three conditions:
1. $p_{j+1}=p_{j+1}'$
2. the columns 1 to $p_{j+1}-1$ are equal in $E$ and $E'$
3. the first $j$ columns of $P$ are like those of the identity matrix
It is clear that $H_r$ gives $E=E'$.
"
Defining $H_r$: according to the assertion, the first condition evaluates to

$p_{r+1} = p^{\prime}_{r+1} = n$

While the second just states that $E = E^{\prime}$.
The third, on the other hand, sets P to the identity of an $m \times n$ matrix.
If we apply this to the statement $E^{\prime} = PE$, it's not hard to verify its correctness for columns $p_r = p^{\prime}_r$. Rather,
if we consider for a moment that $\vec{u}$ is a column vector
in $\mathbb{R}^m$, and $\vec{u}$ is equivalent to the column $p_r$ of $E$, then
$$\vec{u}^{\prime} = P\vec{u} = \vec{u}$$
because P is the identity matrix.
Ok, so we've established that.
This is where things really begin to trip me up, though:
"
Let us start with $H_0$. If we multiply the relation $E'=PE$ with the matrix whose $p_1-1$ columns are $k_1,\dots,k_{p_1-1}$, we are in the part of $E$ where all columns are zero, and hence multiplying by $P$, we still get zero.
"
What I'm gathering from this about RREF is that, given a matrix $M$ in RREF form, there is no guarantee that the first pivot element for $M$ will be $[M]_{11}$.
This actually seems to somewhat contradict my understanding of the nature of a system of linear equations, because a system must have a first variable of some kind, right? In other words, if we have the following system:
$$
\begin{matrix}
a_{11}x_1 + \cdots + a_{1n}x_n = b_1 \\
\vdots \\
a_{m1}x_1 + \cdots + a_{mn}x_n = b_m\\
\end{matrix}
$$,
where $1 \le m, n$, there should be at least one equation with $a_{ij}x_j$ where $a_{ij} \ne 0$, and $1 \le j \le n$ for all $x_j$. In other words,
there must be at least one non-zero instance of each variable within the entire system.
If this is the case, then this implies that the following RREF matrix is impossible:
$$
\begin{bmatrix}
0 & a_{12} & 0 & 0 & \cdots & a_{1n}\\
0 & 0 & a_{23} & 0 & \cdots & a_{2n}\\
0 & 0 & 0 & a_{34} & \cdots & a_{3n}\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & a_{mn} \\
\end{bmatrix}
$$
because $x_1$ cannot be completely zero'd out.
The last quote insinuates this with its definition for $H_0$, in particular that there is a set of columns right before the first pivot column. These columns then must have all zeros, and are indexed by $k_1,\cdots, k_{p_1-1}$.
This is the main road block; everything else is for the most part clear.
Questions
1) why does $j$ begin at 0 when referring to $p$?
2) how does making $p_{r+1} = p^{\prime}_{r+1} = n$ add convenience?
3) It seems that there are situations where undefined values emerge. For example, $p_{1} - 1$ is stated for j = 0; how exactly is this guaranteed to exist? Isn't the first pivot column for matrix derived from a system of linear equations always the very first element of the matrix, thus applying it to be 0? Or, are there other valid reasons for this?

REPLY [0 votes]: I think what makes that proof hard to follow is that it uses matrix algebra exclusively. This is a lot simpler if you use the following - I am basically following a proof outlined in Cullen (Theorem 2.18):
If $A=PB$ and $P$ is invertible then the row space of $A$ and $B$ is the same. (you can try proving this? - the proof is also in the same textbook)
So now you can get that the non-zero rows of any RREF (say $E$) of $A$ will span the row space of $A$. In fact the non-zero rows of $E$ will be a basis for the row space of $A$, since the non-zero rows of $E$ are linearly independent. This comes down to: any two vectors with first non-zero entry in different component positions are linearly independent (again, you can prove it yourself if you want to).
What this gives us is that if the pivot columns are columns $k_1, k_2, \ldots, k_n$ then we cannot have a non-zero vector $\gamma$ in the row space of $A$ which has all zeros in components $k_1, k_2, \ldots, k_n$. Without loss of generality we can further assume that $k_1< k_2< \ldots< k_n$ as a direct consequence of the definition of RREF.
Suppose $E'$ is another RREF of $A$. Surely $E'$ must have the same pivot columns as $E$, else you will have a vector in $E'$ which is linearly independent from the vectors in $E$. Let $\rho_1$ be the first non-zero row of $E$ and $\gamma_1$ the first non-zero row of $E'$ (leading 1's in position $k_1$). Then $\rho_1-\gamma_1$ has all zeros in components $k_1, k_2, \ldots, k_n$, and since both are in the row space of $A$ we also have $\rho_1-\gamma_1$ is in the row space of $A$ (row space is a subspace and so is closed under vector addition and scalar multiplication). So in fact $\rho_1-\gamma_1=0$. We can repeat this argument for each of the remaining $k_i$ to show that all of the non-zero rows of $E$ and $E'$ are the same.


TITLE: Does there exist an unbiased estimator for the absolute value of the mean?
QUESTION [4 upvotes]: Given random variable $Z$, with samples $\{z_1,\dots,z_n\}$ is there an unbiased estimator for $|\mathbb{E}[Z]|$?
If $Z$ is positive (or negative) with probability 1 this is straightforward, as then the absolute value can be essentially ignored.
This is a followup to my previous question here. Based on the answers provided and after having thought about it for a while, I am intuitively convinced that no such estimator can exist (for general random variables). However, I am lost on how to prove it.

REPLY [7 votes]: Unbiased estimators of $|E(Z)|$ need not exist in general. In particular, here is an example of a scenario where there are no unbiased estimators: Let $Z \sim Normal(\mu, \sigma^2)$ and let $Z_1, ..., Z_n$ be a random sample of $Z$. Our parameter of interest is $\theta = |\mu|$.
This is perhaps a much stronger result than we need to show that there does not exist an unbiased estimator of $\theta$, but there is a paper by Hirano and Porter (see here) that deals with this situation. The main result is as follows: Let $Z \sim N(h, \sigma^2)$ with $h \in \mathbb{R}$ and $\sigma^2 > 0$, let $\kappa(h)$ be a function of $h$, and let $T(Z)$ be an estimator of $\kappa(h)$. If there exists a point $h_0$ in the parameter space such that $\kappa$ is not continuously differentiable at $h_0$, then $T$ cannot be unbiased. (Their result is actually more general and deals with multivariate normal distributions, but I've simplified it to the univariate case.)
The general proof is a bit complicated, but the idea should be straightforward: Suppose that $E(T(Z)) = \kappa(h)$. Using a bounding inequality for the exponential function, we get differentiability under the integral sign and so the derivative of $E(T(Z))$ with respect to $h$ is well-defined and exists everywhere. However, $\kappa(h)$ is not continuously differentiable at $h = h_0$, a contradiction.
In our case, $\kappa(h) = |h|$, which is not continuously differentiable at $h = 0$. As such, there cannot be an unbiased estimator $T$ of $\kappa(h)$.


TITLE: Conditional Expectation of Binomial Given $X \leq x$
QUESTION [1 upvotes]: Are there any neat formulas to reduce something like $\sum_{i=0}^{x} i \binom{n}{i} p^i (1-p)^{n-i}$ where $x<n$?
This would be proportional to $\mathbb{E}(X\leq x)$ where $X$~$\text{Bin}(n,p)$. We'd just have to divide by $\Pr(X\leq x)$. I realize there are a lot of asymptotic results, but I'd like to be able to consider small $n$.

REPLY [0 votes]: I don't know if you'd call it "neat", but it can be expressed using a hypergeometric function:
$$np - \left( x+1 \right) {n\choose x+1}{p}^{x+1} \left( 1-p \right) ^{n-
x-1}{\mbox{$_2$F$_1$}\left([1,1-n+x],\;[x+1],\,{\frac {-p}{1-p}}\right)}
$$


TITLE: $f_n(x) = n^{1/p}\chi(nx)$ in $L^p(\mathbb{R})$ weakly converges to 0 for $1 \leq p <\infty$
QUESTION [3 upvotes]: I am trying to show that $f_n(x) = n^{1/p}\chi(nx)$ in $L^p(\mathbb{R})$ weakly converges to 0 when $1<p<\infty$. Here I have let $\chi$ denote the characteristic function on $[0,1]$. So far I have noticed that for $g\in L^q(\mathbb{R})$, for $q$ the conjugate pair of $p$,
$$\int_\mathbb{R} f_n(x) g(x) d\mu(x) = \int_{[0,1/n]}n^{1/p}g(x) d\mu(x) = n^{1/p}\int_{[0,1/n]}g(x) d\mu(x)$$
From here, it seems like we should use some kind of change of variables like $x = u/n$. However, this does not seem to yield the final result. Also, why does the weak convergence fail when $p =1$? Thank you in advance!

REPLY [3 votes]: $L^{p}(\mathbb  R)$ always refers to $L^{p}$ space w.r.t. Lebesgue measure on $\mathbb R$.
By Holder's inequality we have  $$ n^{1/p}\int_0^{1/n}g(x) d\mu(x)$$ $$ \leq  n^{1/p}\Big(\int_0^{1/n} |g(x)|^{q}d\mu(x)\Big)^{1/q} \Big(\int_0^{1/n} 1^{p}\Big)^{1/p} $$ $$=\Big(\int_0^{1/n} |g(x)|^{q}d\mu(x)\Big)^{1/q} \to 0.$$
When $p=1$ just take $g \equiv 1$ to see that your integral does not tend to $0$.


TITLE: Finding lacunary points of a complex function
QUESTION [0 upvotes]: I understand that lacunary points are those which don't belong to the image of a function. For example, since $e^{z}$ can never assume value zero, it's lacunary point is $0$.  But I don't get why $i$ and $-i$ are lacunary points of $\tan z$ and $\cot z$.

REPLY [0 votes]: If $\, w := \exp(i z),\,$ then $\, \tan(z) = i (1-w^2)/(1+w^2)\,$ and  $\, \tan(z) = i\,$ implies $\, w = 0 \,$ but $\,\exp(z)\,$ is lacunary.


text:
\begin{document}
\title{Convergence of the homotopy analysis method}
\author{M. Turkyilmazoglu}
\maketitle
\begin {center}
{Mathematics Department, University of Hacettepe, 06532-Beytepe,
Ankara, Turkey, Tel: 009003122977850, Fax: 009003122972026, Email:
turkyilm@hotmail.com}
\end {center}

\renewcommand{\baselinestretch}{1.}
\begin{abstract}
The homotopy analysis method is studied in the present paper. The
question of convergence of the homotopy analysis method is
resolved. It is proven that under a special constraint the
homotopy analysis method does converge to the exact solution of
the sought solution of nonlinear ordinary or partial differential
equations. An optimal value of the convergence control parameter
is given through the square residual error. An error estimate is
also provided. Examples, including the Blasius flow, clearly
demonstrate why and on what interval the corresponding homotopy
series generated by the homotopy analysis method will converge to
the exact solution.
\end{abstract}
{\bf Key words:} Nonlinear equations, Approximate solution,
Homotopy analysis method, Convergence

\section{Introduction}
\label{introduction} The search for a better and easy to use tool
for the solution of nonlinear equations illuminating the nonlinear
phenomena of our life keeps continuing.

A variety of methods therefore were proposed to find approximate
solutions. One of the most recent popular technique is the
homotopy analysis method, which is a combination of the classical
perturbation technique and homotopy concept as used in topology.
In the homotopy analysis method, which requires neither a small
parameter nor a linear term in a differential equation, a homotopy
with an embedding parameter $p \in [0, 1]$ is constructed. In
\cite{Liao92a} a basic idea of homotopy analysis method for
solving nonlinear differential equations was presented. In this
method, the solution is considered as the sum of an infinite
series, which converges rapidly to accurate solutions. A numerous
nonlinear problems in science, finance and engineering were
recently treated by the method, see at least
\cite{Abbasbandy2006}, \cite{Zhu2006}, \cite{Hayat2008},
\cite{VanGorder2008}, \cite{Akyildiz2008}, \cite{Wu2009},
\cite{Liang2009},\cite{Molabahrami2009} and \cite{Zhao2009}.
Particularly, a few new solutions of some nonlinear problems were
found by means of the method \cite{Liao2007a}, which were
neglected by other analytic methods and even by numerical
techniques. All of these show the potential of the homotopy
analysis method for strongly nonlinear problems. Recently an
optimal homotopy analysis approach for strongly nonlinear
differential equations were proposed in \cite{Liao2010}. Even
though a great deal of equations were solved using the homotopy
analysis method, the question of convergence of the method is yet
to be answered.

We in the present paper investigate the homotopy analysis
technique from a mathematical point of view. The aim is to analyze
the method and to show that under certain circumstances the
homotopy analysis method converges to the exact solution desired,
without a prior knowledge of the exact solution. An optimal value
of the convergence control parameter is defined through the square
residual error concept. Our another emphasis is to address the
error estimate of the approximate solution. The given theorem is
justified exemplifying it by basic examples from ordinary and
partial differential equations from the literature. The presented
theory not only gives the convergence, but it also provides the
information about the interval of convergence of the homotopy
series.

In the rest of the paper, \S \ref{Method} lays the basis of
homotopy analysis method. A theory is outlined in \S \ref{Results}
for the convergence and error estimate. Illustrative examples in
\S \ref{Examp} are followed by the conclusions in \S
\ref{conclusions}.

\section{The Homotopy Analysis Method}
\label{Method} Liao \cite{Liao92a} described the early form of the
homotopy analysis method in 1992. The essential idea of this
method is to introduce a homotopy parameter, say $p$, which varies
from 0 to 1 and a nonzero auxiliary parameter so-called the
convergence control parameter $h$. At $p=0$, the system of
equations usually has been reduced to a simplified form which
normally admits a rather simple solution. As $p$ gradually
increases continuously toward 1, the system goes through a
sequence of deformations, and the solution at each stage is close
to that at the previous stage of the deformation. Eventually at
$p=1$, the system takes the original form of the equation and the
final stage of the deformation gives the desired solution. To
illustrate the basic ideas of this method, consider the nonlinear
boundary value problem
\begin{eqnarray}\label{eq1}
   N(u(r))=0;\;r\in \Omega, \quad B(u(r),{du\over dn}) = 0;\;r\in
   \Gamma
\end{eqnarray}
where $u(r)$ defined over the region $\Omega$ is the function to
be solved under the boundary constraints in $B$ defined over the
boundary $\Gamma$ of $\Omega$. The homotopy analysis technique
defines a homotopy $u(r,p): R\times [0,1]\rightarrow R$ so that
\begin{eqnarray}\label{eq2}
  H(u,p) &=& (1-p)[L(u)-L(u_0)]+p h N(u),
\end{eqnarray}where $L$ is a suitable auxiliary linear operator,
$u_0$ is an initial approximation of equation (\ref{eq1})
satisfying exactly the boundary conditions. It is obvious from
equation (\ref{eq2}) that
\begin{eqnarray}\label{eq3}

  H(u,0) &=& L(u)-L(u_0), \quad H(u,1)\; =\; N(u).
\end{eqnarray}As $p$ moves from 0 to 1, $u(r,p)$ moves from
$u_0(r)$ to $u(r)$. In topology, this called a deformation and
$L(u)-L(u_0)$ and $N(u)$ are said to be homotopic. Our basic
assumption is that the solution of equation (\ref{eq2}) when
equated to zero can be expressed as a power series in $p$
\begin{eqnarray}\label{eq4}
   u(r,p) &=& u_0(r)+pu_1(r)+p^2u_2(r)+\cdots =
   \sum_{k=0}^\infty u_k(r)p^k.
\end{eqnarray}
The appropriate solutions of the coefficients $u_k(r)$ in
(\ref{eq4}) can be found from the homotopy deformation equations,
see \cite{Liao2003book}. Hence, the approximate solution of
equation (\ref{eq1}) can be readily obtained as
\begin{eqnarray}\label{eq5}
   u(r)=\lim_{p\to 1}u(r,p) = \sum_{k=0}^\infty u_k(r).
\end{eqnarray}It was found that the auxiliary parameter $h$ can
adjust and control the convergence region and rate of homotopy
series solutions (\ref{eq4}). Whenever the series (\ref{eq4}) is
known to be convergent, then (\ref{eq5}) represents the exact
solution of (\ref{eq1}), as proved in \cite{Liao2003book}.

\section{A convergence Theorem and error estimate}\label{Results}
Using the methodology underlined above, the number of problems
treated by the homotopy analysis method approaches a couple of
hundreds now. However, the very basic question of why the series
obtained by setting $p=1$ in (\ref{eq4}) should be convergent
remains unanswered till today. To remedy this issue up to a point,
we provide the subsequent theorems here. It should be noted that
how to find a proper convergence control parameter $h$, or even
better, to get a faster convergent one, to be used in Theorem 1
will be clarified later in this section.

\vspace{0.2in}{\bf Theorem 1.} Suppose that $A\subset R$ be a
Banach space donated with a suitable norm $\|\|$ (depending on the
physical problem considered), over which the sequence $u_k(t)$ of
(\ref{eq4}) is defined for a prescribed value of $h$. Assume also
that the initial approximation $u_0(t)$ remains inside the ball of
the solution $u(t)$. Taking $r\in R$ be a constant, the following
statements hold true:

{\bf (i)} If $\|v_{k+1}(t)\|\le r \|v_k(t)\|$ for all $k$, given
some $0<r<1$, then the series solution $u(t,p) = \sum_{k=0}^\infty
u_k(t)p^k$, defined in (\ref{eq4}) converges absolutely at $p=1$
to (\ref{eq5}) over the domain of definition of $t$,

{\bf (ii)} If $\|v_{k+1}(t)\|\ge r \|v_k(t)\|$ for all $k$, given
some $r>1$, then the series solution $u(t,p) = \sum_{k=0}^\infty
u_k(t)p^k$, defined in (\ref{eq4}) diverges at $p=1$ over the
domain of definition of $t$.

\vspace{0.2in}{\bf Proof.} {\bf (i)} In compliance with the ratio
test for the power series in $p$, the proof is clear. However, in
order to give an estimate to the truncation error of homotopy
analysis method, we shortly give the whole proof here. If $S_n(t)$
denote the sequence of partial sum of the series (\ref{eq5}), we
need to show that $S_n(t)$ is a Cauchy sequence in $A$. For this
purpose, consider,
\begin{eqnarray}\label{eq6}

  \|S_{n+1}(t)-S_n(t)\| = \|u_{n+1}(t)\|\le r \|u_n(t)
  \|\le r^2 \|u_{n-1}(t)\|\le \cdots \le r^{n+1}\|u_0(t)\|.
\end{eqnarray}
It should be remarked that owing to (\ref{eq6}), all the
approximations produced by the homotopy method (\ref{eq2}) will
lie within the ball of $u(t)$. For every $m,n \in N$, $n\ge m$,
making use of (\ref{eq6}) and the triangle inequality
successively, we have,
\begin{eqnarray}\label{eq7}

  \|S_{n}(t)-S_m(t)\| = \|(S_n(t)-S_{n-1}(t))+(S_{n-1}(t)-S_{n-2}(t))+
  \cdots+(S_{m+1}(t)-S_m(t))\|\\\nonumber \le {1-r^{n-m}\over
  1-r}r^{m+1}\|u_0(t)\|.
\end{eqnarray}
Since $0<r<1$, we get from (\ref{eq7})
\begin{eqnarray}\label{eq8}

  \lim_{n,m\to\infty}\|S_{n}(t)-S_m(t)\| &=& 0.
\end{eqnarray}
Therefore, $S_n(t)$ is a Cauchy sequence in the Banach space $A$,
and this implies that the series solution (\ref{eq5}) is
convergent. This completes the proof {\bf (i)}. $\diamond$

The proof of {\bf (ii)} follows from the fact that under the
hypothesis supplied in {\bf (ii)}, there exist a number $l$,
$l>r>1$, so that the interval of convergence of the power series
(\ref{eq4}) is $|p|<1/l<1$, which obviously excludes the case of
$p=1$. $\diamond$

\vspace{0.2in}{\bf Remark 1.} Since the finite number of terms
does not affect the convergence, Theorem 1 is equally valid if the
inequalities stated in {\bf (i-ii)} are true for sufficiently
large $k'$s. Thus, it is sufficient to keep track of magnitudes of
the ratios $rat_k$ defined by
\begin{eqnarray}\label{rat}

  rat_k &=& {\|v_{k+1}(t)\|\over \|v_k(t)\|},
\end{eqnarray}and whether they remain less than unity.

\vspace{0.2in}{\bf Remark 2.} By enforcing the ratio in {\bf (i)}
to hold true in the infinite limit, the validity region of $t$ for
the series solution can also be constructed, see Example 1 below.

\vspace{0.2in}{\bf Theorem 2.} If the series solution defined in
(\ref{eq5}) is convergent, then it converges to an exact solution
of the nonlinear problem (\ref{eq1}).

\vspace{0.2in}{\bf Proof.} The proof can be found in
\cite{Liao2003book}.

\vspace{0.2in}{\bf Theorem 3.} Assume that the series solution
$\sum_{n=0}^\infty u_n(t)$, defined in (\ref{eq5}) is convergent
to the solution $u(t)$ for a prescribed value of $h$. If the
truncated series $\sum_{n=0}^Mu_n(t)$ is used as an approximation
to the solution $u(t)$ of problem (\ref{eq1}), then an upper bound
for the error, $E_M(t)$, is estimated as
\begin{eqnarray}\label{eq10}

  E_M(t) \le {r^{M+1}\over 1-r}\|u_0(t)\|.
\end{eqnarray}

\vspace{0.2in}{\bf Proof.} Making use of the inequality
(\ref{eq7}) of Theorem 1, we immediately obtain
\begin{eqnarray}\label{eq11}

  \|u(t)-S_M(t)\| \le {1-r^{n-M}\over
  1-r}r^{M+1}\|u_0(t)\|,
\end{eqnarray}
and taking into account $(1-r^{n-M})<1$, (\ref{eq11}) leads to the
desired formula (\ref{eq10}). This completes the proof. $\diamond$

\vspace{0.2in}{\bf Remark 3.} An optimal value of the convergence
control parameter $h$ can be found by means of the exact square
residual error integrated in the whole region of interest
$\Gamma$, at the order of approximation $M$, that is,
\begin{eqnarray}\label{eqres1}

  Res(h) &=& \int_\Gamma [N(\sum_{k=0}^Mu_k(r))]^2dr.
\end{eqnarray}
Obviously, the more quickly $Res(h)$ in (\ref{eqres1}) decreases
to zero, the faster the corresponding homotopy series solution
(\ref{eq5}) converges. So, at the given order of approximation
$M$, the corresponding optimal value of the convergence control
parameter $h$ is given by the minimum of $Res(h)$, corresponding
to a nonlinear algebraic equation of the form
\begin{eqnarray}\label{eqres2}

  {d Res \over d h} &=& 0.
\end{eqnarray}

Therefore, the convergence control parameter obtained via
(\ref{eqres1}-\ref{eqres2}) can be supplied into the Theorem 1.
However, it is unfortunate that the exact square residual error
$Res(h)$ defined by (\ref{eqres1}) needs too much CPU time to
calculate even if the order of approximation is not very high, and
thus is often useless in practice. To avoid the time-consuming
computation, Liao in \cite{Liao2003book} suggested to investigate
the convergence of some special quantities, which often have
important physical meanings. For example, one can consider the
convergence of $u'(0)$ and $u''(0)$ of a nonlinear differential
equation (\ref{eq1}), if they are unknown. It is found by the
homotopy analysis researchers that there often exists such a
region that certain values of $h$ give a convergent series
solution of such kind of quantities. Besides, such a region can be
found, although approximately, by plotting the curves of these
unknown quantities versus $h$. These curves are called {\em
$h$-curves} or {\em curves for convergence-control parameter},
which have been successfully applied in many nonlinear problems as
cited herein. This approach constitutes another way of finding
proper values of $h$ for the Theorem 1.

\vspace{0.2in}{\bf Remark 4.} Similar to the constant $h-$curves
idea of Liao, an approximate interval of convergence for $h$ can
be determined by application of Theorem 1 to some certain physical
quantities, see Example 3 below.

\section{Illustrative Examples}\label{Examp}
To illustrate the validity of the Theorems outlined, we take into
account the following examples taken from the homotopy analysis
studies in the literature.

\vspace{0.2in}{\bf Example 1.} Consider the first-order
nonlinear differential equation \cite{Liao2003book}
\begin{eqnarray}\label{eq12}
   u'+u^2 &=& 1,\quad u(0) = 0,
\end{eqnarray}that governs the steady free convection flow over
a vertical semi-infinite flat plate which is embedded in a fluid
saturated porous medium of ambient temperature \cite{Crane70} and
also the steady-state boundary-layer flows over a permeable
stretching sheet \cite{Keller2000}. In accordance with Theorem 1,
choosing $u_0(t)=1-e^{-2t}$ and $L={d\over dt}+2$, the homotopy
series solution via the homotopy approach (\ref{eq2}) can be
straightforwardly constructed. Employing the $L^2$ norm in $R$,
then Theorem 1 assures the convergency of (\ref{eq12}) provided
that
\begin{eqnarray}\label{eq13}

  {\|u_{n+1}(t)\| \over \| u_n(t)\|} <1.
\end{eqnarray}
Table \ref{Table 1} presents evolution of the ratio (\ref{eq13})
for a variety of the convergence control parameter $h$ for the
problem (\ref{eq12}). Additionally, Table \ref{Table 1} gives the
error, see the last column, defined by
\begin{eqnarray}\label{eq14}

  err &=& \int_0^\infty|u_e(t)-u(t)|dt,
\end{eqnarray}where $u_e(t)$ is the solution of (\ref{eq12})
calculated numerically. Data displayed in Table \ref{Table 1}
clearly explains why the homotopy analysis method generates
completely convergent series solution to the problem (\ref{eq12})
for the chosen parameters $h$.
\begin{table}[!htb]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
  \hline $h$ &$M=1$    & $M=10$  & $M=20$  & $M=30$  & $M=40$  & $M=50$  & $err$\\\hline
  2.0        & 0.44721 & 0.87228 & 0.93102 & 0.95275 & 0.96407 & 0.97044 & $9.609\times 10^{-5}$ \\
  1.5        & 0.15811 & 0.43509 & 0.46523 & 0.47625 & 0.48196 & 0.48517 & $3.768\times 10^{-20}$\\
  1.3        & 0.15811 & 0.31877 & 0.33354 & 0.33881 & 0.34152 & 0.34304 & $8.261\times 10^{-26}$\\
  1.0        & 0.31623 & 0.45743 & 0.47701 & 0.48425 & 0.48802 & 0.49015 & $8.384\times 10^{-18}$\\
  0.5        & 0.65192 & 0.69906 & 0.71905 & 0.72798 & 0.73295 & 0.73583 & $1.492\times 10^{-8}$ \\\hline
\end{tabular}
\end{center}
\caption{The evolution of the ratio (\ref{eq13}) for the equation
(\ref{eq12}). Last column corresponds to the error defined by
(\ref{eq14}).}\label{Table 1}
\end{table}

The optimal value of convergence control parameter $h$ is further
calculated from equations (\ref{eqres1}-\ref{eqres2}) only for
$M=7$ as $h=1.30405$, which strongly indicates the reason of the
faster converge of the homotopy series near this value, see Table
\ref{Table 1}. Moreover, the uniform validity region of the
homotopy series solution (\ref{eq5}) can also be analytically
evaluated for the special value of $h=1$. In this particular case,
the terms in the homotopy series (\ref{eq5}) satisfy the ratio
\begin{eqnarray}\label{eq15}

  u_{n+1}(t)\over u_n(t) &=& {1\over 2}(1-e^{-2t}).
\end{eqnarray}Hence, regarding (\ref{eq15}), Theorem 1 assures
the convergence of the corresponding homotopy series (\ref{eq5})
for all values of $t$ valid in the interval $t>-{\ln 3\over 2}$.

\vspace{0.2in}{\bf Example 2.} Consider now the second-order
nonlinear differential equation
\begin{eqnarray}\label{eq16}
   2u''+u-u^2 &=& 0,\quad u(0) = 0,\quad u(\infty)=1,
\end{eqnarray}that governs the steady mixed convection flow past a plane
of arbitrary shape under the boundary layer and Darcy-Boussinesq
approximations \cite{Keller2001}. With the choices of $L={d^2\over
dt^2}-1$ and $u_0(t)=1-e^{-t}$, the homotopy (\ref{eq2}) generates
a homotopy series (\ref{eq5}) whose ratios of successive terms are
tabulated in Table \ref{Table 2} at several order of
approximations for some selected values of $h$. As observed from
the Table, $h=1.2$ generates divergent homotopy solution, and also
the convergence of the homotopy series to the exact solution takes
place at a considerably slow rate near $h=1$ as compared to the
smaller values of $h$.
\begin{table}[!htb]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
  \hline $h$ &$M=1$    & $M=10$  & $M=20$  & $M=30$  & $M=40$  & $M=50$  & $err$\\\hline
  1.2        & 0.42070 & 1.24479 & 1.31373 & 1.34407 & 1.35657 & 1.36525 & $8828.16$              \\
  1.0        & 0.24684 & 0.88919 & 0.94119 & 0.96001 & 0.96970 & 0.97513 & $3.444\times 10^{-4}$\\
  0.8        & 0.21484 & 0.63937 & 0.68278 & 0.69872 & 0.70700 & 0.71167 & $1.332\times 10^{-9}$\\
  0.6        & 0.36425 & 0.69932 & 0.74524 & 0.76240 & 0.77137 & 0.77643 & $1.753\times 10^{-7}$\\\hline
\end{tabular}
\end{center}
\caption{The ratio of successive terms in the homotopy series
corresponding to equation (\ref{eq16}). Last column corresponds to
the error defined by (\ref{eq14}).}\label{Table 2}
\end{table}

The optimal value of convergence control parameter $h$ is
calculated from equations (\ref{eqres1}-\ref{eqres2}) only for
$M=7$ as $h=0.73258$, which explains why the homotopy series
(\ref{eq5}) should converge faster near this value to the solution
of (\ref{eq16}), see Table \ref{Table 2}.

\vspace{0.2in}{\bf Example 3.} Consider now the nonlinear partial
differential Burger's equation
\begin{eqnarray}\label{eq17}
   u_t+uu_x &=& u_{xx},\quad u(x,0) = 2x,
\end{eqnarray}
that has been found to describe various kind of phenomena, such as
a mathematical model of turbulence and the approximate theory of
the flow through a shock wave traveling in a viscous fluid
\cite{Benton66a}. Equation (\ref{eq17}) admits an exact solution
given by
\begin{eqnarray}\label{eq18}
   u(x,t) &=& \frac{2x}{1+2t}.
\end{eqnarray}

To approximate the exact solution (\ref{eq18}), we choose the
auxiliary parameters as $u_0(x,t)=2x$ and $L={\partial \over
\partial t}$. Then, the homotopy (\ref{eq2}) turns out to be
\begin{eqnarray}\label{eq19}
   \left(1+\frac{1-p}{h p}\right)u_t(x,t,p)+u(x,t,p)u_x(x,t,p)-
   u_{xx}(x,t,p) = 0,\; u(x,0,p)=2x.
\end{eqnarray}
Equation (\ref{eq19}) produces the below homotopy series for the
solution of (\ref{eq17})
\begin{eqnarray}\label{eq19a}
   u(x,t) =
   2 x - 4 h t x + 4 h t (-1 + h + 2 h t) x -
 4 h t (-1 + h + 2 h t)^2 x \\\nonumber + 4 h t (-1 + h + 2 h t)^3 x -
 4 h t (-1 + h + 2 h t)^4 x+\cdots,
\end{eqnarray}whose convergence to the exact solution (\ref{eq18})
takes place for the values satisfying
$\lim_{n\to\infty}{|u_{n+1}|\over |u_n|}=|1 - h (1 + 2 t)|<1$.
Thus, in accordance with Theorem 1, this holds exactly for the
values of $-\frac{1}{2}<t<\frac{2-h}{2 h}$ together with $0<h<2$.
It is easy to demonstrate that the case $h=1$ corresponds to the
traditional Taylor series expansion of the solution (\ref{eq18}),
which is only valid in the region $-\frac{1}{2}<t<\frac{1}{2}$.

\vspace{0.2in}{\bf Example 4.} Consider now the following
fourth-order parabolic partial differential equation arising in
the study of the transverse vibrations of a uniform flexible beam
\cite{Gorman75}
\begin{eqnarray}\label{eq20}
   u_{tt}+({y+z\over 2\cos x}-1)u_{xxxx}+({z+x\over 2\cos y}-1)u_{yyyy}
   +({x+y\over 2\cos z}-1)u_{zzzz}&=& 0,\\\nonumber
   u(x,y,z,0) =-u_t(x,y,z,0)= x+y+z-(\cos x+\cos y+\cos z),
\end{eqnarray}
whose exact solution is given by
\begin{eqnarray}\label{eq21}
   u(x,t) &=& (x+y+z-\cos x-\cos y-\cos z)e^{-t}.
\end{eqnarray}
To approximate the exact solution (\ref{eq21}), if we choose the
auxiliary parameters respectively, $u_0(x,t)=(x+y+z-\cos x-\cos
y-\cos z)(1-t)$ and $L={\partial^2 \over \partial t^2}$, the
homotopy (\ref{eq2}) for $h=1$ then generates the subsequent
homotopy series
\begin{eqnarray}\label{eq22}

  u(x,t) &=& \sum_{n=0}^\infty u_n=\sum_{n=0}^\infty(x+y+z-\cos x-\cos y-\cos z)
  \left(\frac{t^{2n}}{(2n)!}-\frac{t^{2n+1}}{(2n+1)!}\right)
\end{eqnarray}which leads to $\lim_{n\to \infty} {|u_{n+1}|\over |u_n|}=0$.
So, the homotopy series (\ref{eq22}) converges to the exact
solution of (\ref{eq20}) for all $t$.

\vspace{0.2in}{\bf Example 5.} Let us consider the linear partial
differential equation
\begin{eqnarray}\label{eq23}
   u_t+u_x-2u_{xxt}&=&0,\quad u(x,0) =e^{-x},
\end{eqnarray}
whose exact separable solution is given by
\begin{eqnarray}\label{eq24}
   u(x,t) &=& e^{-x-t}.
\end{eqnarray}
To approximate the exact solution (\ref{eq24}), if we choose the
auxiliary parameters $u_0(x,t)=e^{-x}$ and $L={\partial \over
\partial t}$ respectively, then the homotopy (\ref{eq2}) turns out to be
\begin{eqnarray}\label{eq25}
   (1-p)u_t(x,t,p)+ p h(u_t(x,t,p)+u_x(x,t,p)-2
   u_{xxt}(x,t,p)) = 0,\; u(x,0,p)=e^{-x}.
\end{eqnarray}
When $h=1$, it appears that the homotopy series solution is
convergent only for $t\leq 0$, that is out of physical interest.
On the other hand, for $h=-1$, the successive ratio yields
$$|{u_{n+1}(t)\over u_n(t)}|={|t|\over n+1},$$ whose limit gives
rise to zero. This explains why the homotopy series solution
(\ref{eq5}) in this case represents the real physical solution
over $t\geq 0$ for $h=-1$, as also explained in \cite{Liang2009}.

\vspace{0.2in}{\bf Example 6.} As a final example, we consider the
classical Blasius flat-plate flow problem of fluid mechanics
governed by the nonlinear initial-value of third-order
\cite{Liao99a}
\begin{eqnarray}\label{eq26}
   y'''+{yy''\over 2} &=& 0,\quad y(0)=y'(0)=y'(\infty)-1=0,
   \quad \eta \in [0,\infty).
\end{eqnarray}With the transformations $$y={u\over \lambda},\quad t=\lambda \eta,$$
where $\lambda$ is a scaling parameter taken here as 4, system
(\ref{eq26}) is converted into
\begin{eqnarray}\label{eq27}
   u'''+{uu''\over 2\lambda^2} &=& 0,\quad u(0)=u'(0)=u'(\infty)-1=0,
   \quad t\in [0,\infty).
\end{eqnarray}
Taking $L={d^3\over dt^3}+{d^2\over dt^2}$ with
$u_0(t)=-1+t+e^{-t}$, the values for the convergence control
parameters $h$ from the homotopy (\ref{eq2}) are shown in Table
\ref{Table 3} along with the corresponding residual errors. It is
seen that the optimal value for the physical problem considered is
$h=-3/2$. Figures \ref{fig1}(a-d) demonstrate the ratios
(\ref{rat}) in Theorem 1 which were computed from the homotopy
series (\ref{eq5}) for the values of $h$ given in Table \ref{Table
3}. The convergence is assured further by the values of the ratios
less than unity, suggesting that limit tends to 0.83, 0.77, 0.76
and 0.78, for the values of $h$ shown.
\begin{table}[!htb]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
  \hline $h$ & $M=1$    & $M=10$                & $M=20$               & $M=30$               \\\hline
  1.6        & 1.258621 & $1.201\times 10^{-3}$  & $2.463\times 10^{-7}$ & $1.032\times 10^{-6}$ \\
  1.5        & 1.027450 & $1.372\times 10^{-4}$  & $3.602\times 10^{-7}$ & $2.573\times 10^{-9}$ \\
  1.4        & 0.820771 & $1.399\times 10^{-4}$  & $7.621\times 10^{-7}$ & $5.454\times 10^{-9}$ \\
  1.0        & 0.238796 & $7.621\times 10^{-4}$  & $1.511\times 10^{-5}$ & $4.540\times 10^{-7}$ \\\hline
\end{tabular}
\end{center}
\caption{The square residual errors $Res(h)$ for the homotopy
series solutions corresponding to equation (\ref{eq27}) at several
values of convergence control parameters $h$.}\label{Table 3}
\end{table}

\begin{figure}[!htb]
\begin{center}
\mbox{ \subfigure[]{
\includegraphics[width=6cm,height=5cm]{graph1.eps}}
\subfigure[]{
\includegraphics[width=6cm,height=5cm]{graph2.eps}}}
\mbox{ \subfigure[]{
\includegraphics[width=6cm,height=5cm]{graph3.eps}}
\subfigure[]{
\includegraphics[width=6cm,height=5cm]{graph4.eps}}}
\caption[Basic flow]{A list plot of the ratios $rat$ from the
theorem to reveal the convergence of the HAM solutions for the
Blasius equation (\ref{eq27}) for different choices of $h$. (a)
$h=-1$, (b) $h=-1.4$, (c) $h=-1.5$ and (d) $h=-1.6$.} \label{fig1}
\end{center}
\end{figure}

\section{Concluding remarks}
\label{conclusions} In this paper, the homotopy analysis method
has been analyzed with an aim to investigate the conditions which
result in the convergence of the generated homotopy solutions of
the nonlinear ordinary and partial differential equations. The
theorems outlined in the paper have proved that if specific values
are assigned to the auxiliary parameters in the homotopy analysis
method, then the approximate homotopy results successfully
converge to the exact solution. An optimal value approach for the
convergence control parameter has also been given. Examples have
been provided to verify the theory. Via the theorems provided
here, not only the question of the convergence of the homotopy
series is answered, but also the region of validity of the space
variable ensuring the convergence is determined. The traditional
flat-plate boundary layer flow problem has been finally treated by
the convergence theorem.
\bibliographystyle{unsrt}
\bibliography{mybib}
\end{document}


TITLE: Solving $ y(x)-2-\int_0^x e^{y(t)-t} \, dt = 0 $
QUESTION [1 upvotes]: So, I'm trying to solve the following differential equation methodically:
$$ y(x)-2-\int_0^x e^{y(t)-t} \, dt = 0 $$
I rearranged the equation a bit and differentiated both sides and got:
$$ e^x \cdot y'(x)=e^{y(x)} $$
Now,  I immediately see that a possible solution is $ y(x) = x $, but how do I solve this more methodically?

REPLY [2 votes]: Separation of variables:
$$ e^x \cdot y'=e^{y} $$
$$ e^x \cdot \frac{dy}{dx}=e^{y} $$
$$ e^{-y} \cdot \frac{dy}{dx}=e^{-x} $$
$$\int e^{-y} \, dy=\int e^{-x} \, dx$$