What does 'Int'l Ed' stand for? We say that you would like to buy either

Principles of Mathematical Analysis (Int'l Ed)

or
Principles of Mathematical Analysis.

Principles of Mathematical Analysis (Int'l Ed) costs nearly half as much as Principles of Mathematical Analysis.

I know that Int'l Ed stands for international edition. But what does (Int'l Ed) stand for, more than just international edition, e.g. what does the Int'l label mean to to the text and content in the book? It just drives me nuts! I have a really hard time finding an answer by simply searching the web, so I ask the question here.

What are the differences between two books, or any two books - one book labeled Int'l Ed and the other book not?

My first thought was "international education" but then seeing the context, it is clearly "international edition".


Prove that the conditional expectation of functions of r.v.s allows factoring the function I'm trying to prove that, if $X, Y$ random variables and $r(x), s(x)$ functions then

$$\mathbb{E}(r(X)s(Y) | X) = r(X)\mathbb{E}(s(Y)|X)$$

What I do know is

$$\mathbb{E}(r(X)s(Y)|X) = \int_{-\infty}^{\infty}r(x)s(y)\frac{P_{X,Y}(x,y)}{P_{Y}(y)}dx$$

However, I'm not even sure how to write the right-hand side because it seems like it should be a random variable, not a function of like the function above.  I would say

$$\mathbb{E}(s(Y)|X) = \int_{-\infty}^{\infty}s(y)\frac{P_{X,Y}(x,y)}{P_{Y}(y)}dx$$

but I don't understand what it could mean that this times $r(X)$ gets me the other side of the equation, which as I understand it, is a function of $y$ and not some random variable.  Even if it is a random variable I'm not sure how I could show they were equal since I can't imagine how you'd use the distribution technique or find a moment generating function.

I may have figured this out.  The expressions on each side of the equation are not functions of $x$ or any such thing, but instead are random variables that are basically a transformation of $X$.  The conditional expectation $E(r(X)s(Y)|X)$ could take different values for different $X=x$, just as $X^{2}$ takes different values for different $X=x$, although the associated probability function would be the same.

So if I can show that the left- and right-hand sides take the same values for $X=x$, then since they both are associated with the same probability distribution for $X$, it will have shown that they are the same random variable.  So to do this, I pick any particular $X=x$ and for this we get

$$\mathbb{E}(r(X=x)s(Y)|X=x) = \int_{-\infty}^{\infty}r(x)s(y)P(Y|X=x)dy$$

$$=r(x)\int_{-\infty}^{\infty}s(y)P(Y|X=x)dy = r(x)E(Y|X=x)$$

and the right-hand side is just the value of the random variable $r(X)E(Y|X)$ for every such $x$.


Proving that ${\left( {A \cup B} \right)^{\text{o}}} = \emptyset$ given that ${\left( {\bar A} \right)^{\text{o}}} = \emptyset = {B^{\text{o}}}$ I have the following question:

  Let $X$ be a topological space and let $A,B \subseteq X$ such that ${\left( {\bar A} \right)^{\text{o}}} = \emptyset  = {B^{\text{o}}}$. Show that ${\left( {A \cup B} \right)^{\text{o}}} = \emptyset$.

 Am I allowed to say that the empty set is a subset of every set so $\emptyset  \subseteq {\left( {A \cup B} \right)^{\text{o}}}$?

 I'm struggling with the reverse inclusion: I have the following $ {\left( {A \cup B} \right)^{\text{o}}} \subseteq A \cup B \subseteq \overline {A \cup B}  = \bar A \cup \bar B\ $, but I'm not sure where to go from there?

Let $U\subseteq A\cup B$ be open.

Then $V=U\setminus \bar A$ is open and $V\subseteq (A\cup B)\setminus \bar A\subseteq B$, so $V=\emptyset$. Thus, $U\subseteq \bar A$, whence $U=\emptyset$.

Note that both conditions are necessary: if $A$ is any set such that $\bar A$ has nonempty interior, then $B=\bar A\setminus A$ will have empty interior and $A\cup B=\bar A$ will not.

We prove for general case $A=\bar{A}$ since $(A \cap B)^o \subseteq (\bar{A} \cap B )^o$. I use the idea that $A^o= \emptyset$ iff $X=(A^o)^c=\overline{A^c}$

So, we need to prove that $\overline{(A\cup B)^c}=\overline{A^c \cap B^c}=X$ for $A,B \subseteq X$ such that $A$ closed and $A^o=B^o=\emptyset$.

Take any $x \in X$ and any open neighborhood $U$ of $x$. We need to prove that $U \cap A^c \cap B^c \neq \emptyset$.

Since $A^o= \emptyset$ so $\overline{A^c}=X$. Therefore, $U \cap A^c \neq \emptyset$. Take $y \in U \cap A^c$. Since $A$ is closed, $A^c$ is open, there exist an open neighborhood $V$ of $y$ such that $y \in V \subseteq U \cap A^c$.

Since $B^o = \emptyset$ so $\overline{B^c}=X$. Therefore, for open neighborhood $V$ of $y \in X$, $V\cap B^c \neq \emptyset$. So $U \cap A^c \cap B^c \supset V\cap B^c \neq \emptyset$.

Another way if you're not familiar with the above idea.

By contradiction, assume that there exists $x \in (A \cup B)^o$.

So there must be an open neighborhood $U$ of $x$ such that $ x \in U \subset A \cup B$.

$1^{st}$ case, $ x \in U \subset A$, contradiction to $A^o =\emptyset$.
$2 ^{nd}$ case, $ x \in U \not\subset A$ there exists $ y \in U$ and $y \notin A$ so $y \in B$. Since $A$ closed, $A^c$ open, so there exists open neighborhood $V$ of $y$ such that $ y \in V \subset A^c$. So $y \in V \cap U \subset U \subset A \cap B$. Note that $ V \cap U$ is open and $V\cap U \subset V \in A^c$, so $y \in V\cap U \in B$, which contracdicts to $B^o =\emptyset$.


Why is $\int_0^3 \frac{1}{\sqrt{x-3}}\, \mathrm{d}x$ a complex number? This is my first question on this site, so please pardon any nuanced formatting errors.

My friend and I were discussing the following integral yesterday:

$$\int_0^3 \frac{1}{\sqrt{x-3}} \, \mathrm{d}x$$

I discovered that the integral evaluates to $-2i\sqrt{3}$ This confuses me. When I studied calculus 1, I was taught that the definite integral can be thought of as a way to calculate the "area under the curve" or the accumulation of a value. How can an area have a complex value?

Thus, I know that my conceptual understanding of integrals is flawed. How can I reconcile the concept of "area" with a complex value? Or, what is a "better" (for lack of a better word) way to conceptualize the integral?

Thanks in advance!

The area under the curve idea applies to real-valued functions. This function is imaginary for almost all values of the area of integration.

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  Your problem is 'well defined' once you specifies the $\ds{\root{z - 3}}$ branch-cut. For instance,
  $$
\root{z - 3} =
\root{\verts{z - 3}}\exp\pars{{\,\mathrm{arg}\pars{z - 3} \over 2}\,\ic}
\,,\quad
z \not= 3\,,\quad
0 < \,\mathrm{arg}\pars{z - 3} < 2\pi
$$
  and the branch cut is along  the 'line segment'
  $$
\braces{z = 3 + r\expo{\ic\alpha}\,,\quad r \geq 0\,,\quad
0 \leq \alpha <\pi\quad \mbox{or}\quad \pi < \alpha < 2\pi}
$$

In that case,
\begin{align}
&\color{#f00}{%
\int_{0}^{3}{\dd x \over \root{\verts{x - 3}}\expo{\ic\pars{\pi - \alpha}}/2}} =
-\ic\expo{\ic\alpha/2}\int_{0}^{3}{\dd x \over \root{3 - x}} =
\color{#f00}{-2\root{3}\expo{\ic\alpha/2}\,\,\ic}
\end{align}

  The result $\ds{-2\root{3}\,\ic}$ is given for a branch-cut along the 'positive real axis'. Namely,$\ds{\braces{z = x + 0\ic\,,\quad x \geq 3}}$ which is equivalent to $\ds{\alpha = 0}$.


Vectors added up to give null vector. I should begin by saying that I'm only beginning to study $11$th grade Physics.
Recently I figured out that two vectors which gets added up to a null vector, must at most lie in a line. Well that makes perfect sense, so I thought about $3$ vectors and figured that they must at most lie in a plane. Also $4$ vectors when added up to a null vector must at most lie in $3$D space. I then thought about a single vector. Naturally I can't add a vector to nothing and get $0$ vector right?  So that means it must be the null vector itself. So that means A single vector must lie in a dimensionless point to give a null vector.

So I hypothesise that n vectors when added up to a null vector must lie in $n-1$ spatial dimensions.
That is  iff
$$v_1+v_2+v_3....+v_n=0$$
Then $v_1 ,v_2,v_3 ,v_4 ,v_5...,v_n$ must at most lie in $(n-1)$th dimension.

Please put some thought into this and help me understand a method to visualize it if this is correct.

You may have heard that two points lie on a line (a 1D space) and three points lie on a plane (a 2D space). If you visualise these with more points, it's possible to see that $n$ points lie on an $n-1$ dimensional space. So consider the origin, and the endpoints of $n-1$ of the vectors. Since we now have $n$ points, these points define an $n-1$ dimensional space containing all the points, including the origin. It's now clear that summing all the vectors must give a point within this space since all vectors lie within it. Since the last vector is the negative of the first $n-1$ vectors added up, it must also lie in this $n-1$ dimensional space.

As we have $\sum_1^n v_i = 0$, we get  $v_n = -\sum_1^{n-1} v_i$. Thus, out of the $n$ vectors, at most $n-1$ can be independent of each other (i.e. can take values independent of the other vectors). Hence, they can occupy a space of at most $n-1$ dimensions.


Suppose that a $3\times 3$ matrix $M$ has an eigenspace of dimension $3$. Prove that $M$ is a diagonal matrix. How would I go about this? I realise that having dimension 3 means that the solution to $(A-\lambda I)\mathbf b = \mathbf 0$ has 3 free parameters, which would in turn mean that $(A-\lambda I)$ is the zero matrix, so $A = \begin{bmatrix}\lambda & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\end{bmatrix}$, but the first part of that argument is a bit too heuristic. How would I prove it formally?

Hint: Imagine you have three linearly independent (column) eigenvectors for $\lambda$, and put them into the columns of a matrix $B$, which is necessarily nonsingular. Then $AB=B(\lambda I_3)=(\lambda I_3)B$.

Can you see how to conclude this, remembering that $B$ has an inverse?

In general, no matter what the eigenvalues are, you can use this observation to find a matrix $B$ such that $AB=BD$ where $D$ has the eigenvalues of $A$ on the diagonal, and is zero elsewhere. But unless all the eigenvalues are the same, you can't always conclude that $BD$ and $DB$ are the same. They are quite often different. Can you see why?

Note that $A$ can be diagonalized $\Longleftrightarrow$ $A$'s algebraic multiplicity $=$ $A$'s geometric multiplicity. And according to the definition of similarity, $A$ itself is also a diagonal matrix.

The intuition behind the theorem is that there is no need to resort to Jordan chain because there are "enough" eigenvectors. And since there is no Jordan chain, the matrix can be diagonalized.


composition of two uniformly continuous functions. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ and $g : \mathbb{R} \rightarrow \mathbb{R}$ are two uniform continuous functions. Which of the following options are correct and why?

$f(g(x))$ is uniformly continuous.
$f(g(x))$ is continuous but not uniformly continuous.
$f(g(x))$ is continuous and bounded.

My attempt:

Every uniformly continuous function maps a Cauchy sequence to a Cauchy sequence. (Here I have a doubt, as the converse may not be true). So if $\{x_n\}$ be a Cauchy sequence, $\{f(x_n)\}$ and $\{g(f(x_n))\}$ both will be Cauchy sequence. So $g(f(x))$ will be uniformly continuous, i.e. 1 is true.

Composite function of two continuous functions will be continuous. As 1 is true, 2 is false.

$f(x) = x $ is uniformly continuous. $g(x) = \log(x)$ is uniformly continuous in $[1,\infty)$. So $g(f(x)) = \log(x)$ is uniformly continuous in $[1, \infty)$, but not in $\mathbb{R}$.

Thank you for your help.

Option $(1):$ True:

Choose $\epsilon>0.$

Then $\exists~\delta_1>0$ such that $|x_1-x_2|<\delta_1$$\implies|f(x_1)-f(x_2)|<\epsilon.$

For above $\delta_1>0~\exists~\delta>0$ such that $|x_1-x_2|<\delta$$\implies|g(x_1)-g(x_2)|<\delta_1.$

Consequently, $|x_1-x_2|<\delta$$\implies|(f\circ g)(x_1)-(f\circ g)(x_2)|<\epsilon.$
Options $(2),~(3):$ Not necessarily true:

Take $f(x)=g(x)=x$ on $\mathbb R.$

Then $f,g$ are uniformly continuous on $\mathbb R.$

Note $(f\circ g)(x)=x$ is unbounded and uniformly continuous on $\mathbb R.$

I don't think that composition of two uniform continuous function is uniform continuous......
my argument is.. let F be a function from $(0,1)$ to $(0,1)$ and $F(x)=x$
now let G be another function from $[0,1]$ to $[0,1]$ and $G(x)=x^2$ then G compose F is defined as from $(0,1)$ to $[0,1]$ and G compose f is x^2 which is not uniform in $(0,1)$. but F and G are uniform.


How to check the function is convex or not? How one can check the function whether it is convex or not. I know one method by using Hessian Matrix but I think it did not fit for the following example.

I think Hessian matrix method cannot be used for this case.

Hint Looking up the definition on Wikipedia, we have that $f$ is convex iff for any $x_1, x_2 \in \Bbb R$ and $t \in [0,1]$:

$$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)$$

Now, let (for example) $x_1 = -1/2$ and $x_2 = 1/2$. Can you find a $t \in [0,1]$ so that the above does not hold?


Can a countable group have uncountably many subgroups? If $G$ is a countable group,can it have an uncountable number of distinct subgroups?

Let $V$ be a vector space of dimension $\aleph_0$ over a countable field $F$ (so $V$ is countable) and let $B$ be a basis for $V$ over $F$. Then every subset of $B$ spans a different subspace of $V$, so $V$ has $2^{\aleph_0}$ different subspaces, and its additive group has $2^{\aleph_0}$ different subgroups.

For each $(b_n) \in \prod_{n\in \mathbb{N}}(\mathbb{Z}/2)$ consider the map from $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$ to $\mathbb{Z}/2$
$$(a_n) \mapsto \sum_n a_n \cdot b_n$$

take the kernel of this map; we get $2^{\aleph_0}$ subgroups of index $2$ of $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$.

@user1729: thanks! for  correcting a previous statement about the number of subgroups of finite index of a countable group -- it may well be uncountable!

The example of @seirios: is a finitely generated group having uncountably many subgroups.
However, a finitely generated group has finitely many subgroups of a given index $n$. Indeed, one can reduce to normal subgroups ( the normal core has index $N \le n!$) and then notice that there are finitely many morphisms from the groups to the finitely many groups of order $N$. Thus a finitely generated group has countably many subgroups of finite index.


In order to be photographed, 12 people stand in two rows of 6, one in front of the other. A. In how many ways can this be done if there are no restrictions?
B. Alex and Beth (2 specific people) are in the front row
C. Alex and Beth are in different rows?
What I’ve done so far is that I’ve modelled this as 12 C 6. However, I later tried to multiply it by 2 to signify the two different rows the question poses. I’m not exactly sure how to proceed from here.
Edit: I now have tried the first question in a different way. I know that in order to solve this if it were a single row of 12 people, it would be 12 people in the 1st row, 11 in the second, and so on. However, the question poses 2 different rows of 6 people so I’m once again confused on how to proceed.
Edit 2: Realized that 2 different rows of 6 is essentially the same as 1 row of 12, therefore the answer is still 12!.
Now do not know how to proceed with questions B and C.

Once you’ve picked six people for the back row, you can order the two rows in $6!\cdot 6!$ ways.
A. With no conditions, there are $\binom{12}{6}$ ways to pick the back row, so $6!6!\cdot\binom{12}6=12!$ total ways to arrange people.
B. If Alex and Beth are in the front row, there are $\binom{10}6$ ways to pick the back row with neither Alex nor Beth. So the total is: $6!6!\cdot \binom{10}{6}=6\cdot 5\cdot 10!$
C. If Alex and Beth must be on different rows there are $2\cdot \binom{10}5$ ways to pick the back row, so there are $6!6!\cdot 2\binom{10}5=6^2\cdot 2\cdot 10!$ ways.

A.  first choose the 6 people to stand in the front row.  There are 12 ways to choose the first person, 11 ways to choose the second person, 10 to choose the third, 9 to choose the fourth, 8 to choose the 5th, and 7 to choose the 6th.  That is 12(11)(10)(9)(8)(7)= 12!/6! ways to choose the 6 people to stand in the front row and that automatically determines the 6 who stand in the back row.  There are then 6! ways to order the 6 people in the front row and 6! ways to order the 6 people in the back row.
So there are (12!/6!)(6!)(6!)= (12!)(6!) ways to do this.


Find $\lim_{x\to\infty} x\exp(1/x)-x$. Is there a way to find the $$\lim_{x\to\infty} x\exp(1/x)-x$$ but without using L'Hospital's rule??

Let $1/x=t$. We are interested in
$$\lim_{t\to 0^+}\frac{e^t-1}{t}.$$
We recognize this limit (with no restrictions on direction) as the derivative of $e^t$ at $t=0$.

$$xe^{1/x}-x=x\left(e^{1/x}-1\right)=\frac{e^{1/x}-1}{\frac1x}$$

As $\;x\to\infty\;$ both numerator and denominator tend to zero so by l'Hospital:

$$\lim_{x\to\infty}\frac{e^{1/x}-1}{\frac1x}\stackrel{\text{l'H}}=\lim_{x\to\infty}e^{1/x}=1$$


How to find the prime divisors of repunit $R_{13}$  and $R_{79}$? The question is very simple:

Find a prime divisor of $\frac{(10^{13}-1)}{9}$  , i.e. $11\cdots11$($13$ ones), also known as $R^{(10)}_{13}$ or $R_{13}$. Same question for $R_{79}$.

Of course, calculating the answer using a calculator is simple, but I have no idea how to tackle it.
Furthermore, we know by Fermatss little theorem that $10^{12}=1 \pmod{13}$, but I can't seem to apply this to this problem.
Thanks in advance.

$10^{13}-1$ is a divisor of $10^{52}-1$, so any prime divisor of $10^{13}-1$ is also a prime divisor of $10^{52}-1$. By Fermat, 53 is a prime divisor of $10^{52}-1$, so it's worth checking to see whether 53 might be a prime divisor of $10^{13}-1$.

Similarly for $10^{79}-1$ and 317.

Let p be a non-divisor prime number of  ( (a\pm b) )and let
                gcd(a,b) = 1; then the prime factors of  ( (a^p \pm b^p)
                ) that are not contained in  ( (a\pm b) ) , can only be of the form  ( 2mp + 1 )


how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$

My answer is:

$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ or }\; A=2
\end{align}$$

I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.

This continued fraction is the limit of
the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.

Your two solutions are correct. You can verify them as follows.

$$\color{blue}{1} = 3 - 2=3-\frac{2}{\color{\red}{1}}\tag{1}$$
Now replace the red 1 with the blue 1, which equals to the right hand side in (1).

$$\color{blue}{2} = 3 - 1=3-\frac{\color{cyan}{2}}{\color{\red}{2}}\tag{2}$$
Now replace the red 2 with the blue 2, which equals to the right hand side in (2).

I heard from the internet that this is one of the ways Ramanujan created some of his equalities.

Now 2 questions for you,
(1) what if you replace the cyan 2 with blue 2 and so on so forth?
(2) what if you replace the cyan 2 with blue 2 and then replace the red 2 with blue 2 and so on so forth?


Integral $\int_0^{1/\phi}x\log(x)\log(1+x)\log(1-x)\,dx$ How can we evaluate this definite integral
$$I=\int_0^{1/\phi}x\log(x)\log(1+x)\log(1-x)\,dx,$$
where $\displaystyle\phi=\frac{1+\sqrt5}2$ is the golden ratio?

How can we evaluate ?

Integrating by parts, we  get
\begin{align}
\mathcal{I}(x)&=\int x \ln x\ln(1+x)\ln(1-x)\,dx=\\
&=-\frac12\int  \ln x\ln(1+x)\ln(1-x)\,d(1-x^2)=\\
&=-\frac12 (1-x^2)\ln x\ln(1+x)\ln(1-x)\\
&\quad+\frac12\int\frac{\ln (1-x)\ln(1+x)}{x}dx\\&\quad-\frac12\underbrace{\int x\ln (1-x)\ln(1+x)\,dx}_{\text{elementary}}\\
&\quad +\underbrace{\frac12\int(1-x)\ln x\ln(1-x)dx-\frac12\int(1+x)\ln x\ln(1+x)dx}_{-\frac18\operatorname{Li}_2(x^2)+\text{elementary}}.
\end{align}
The second term antiderivative is the most complicated as naively it involves trilogarithms. However it can be simplified using the same trick that I used answering your other question:
\begin{align*}
\int\frac{\ln (1-x)\ln(1+x)}{x}dx=\frac12\int\frac{\ln^2(1-x^2)}{x}dx-
\frac12\int\frac{\ln^2\frac{1+x}{1-x}}{x}dx
\end{align*}
Calculating these two antiderivatives separately, Mathematica simplifies them to a much more compact (one-line) expression, still involving trilogarithms.

It may well happen that due to special integration bounds the tri- (and?) dilogarithms may be lifted using various polylogarithm identities (see equation (2) here and equations (12)-(19) here for a few examples). However I find such calculations rather boring, so for that you should wait for an answer of You-Know-Who.

Just an idea (for now)
$$
\lim_{a,b,c\to0}\partial_{abc}\int_0^{1/\phi}x^{a+1}(1+x)^b(1-x)^cdx
$$
changing variables to $x\phi = y$
$$
\frac{1}{\phi^{a+2}}\int_0^{1}y^{a+1}(1+\frac{1}{\phi}y)^b(1-\frac{1}{\phi}y)^cdy
$$
using the Appel Hyper geometric we find
$$
\int_0^1x^{\alpha-1}(1-a_1x)^{-\beta}(1-b_1x)^{-\beta'}dx = \frac{\Gamma(\alpha)\Gamma(1)}{\Gamma(\alpha+1)}F_1\left(\alpha,\beta,\beta';\gamma;a_1,b_1\right)
$$
where $\gamma-\alpha = 1$.
comparing terms we find
$$
a+1 = \alpha-1\to \alpha = a+2\\
\beta = -b\\
\beta' = -c\\
a_1 = \frac{1}{\phi}\\
b_1 = -\frac{1}{\phi}
$$
thus the integral is
$$
\lim_{a,b,c\to0}\partial_{abc}\frac{1}{\phi^{a+2}}\frac{\Gamma(a+2)\Gamma(1)}{\Gamma(a+3)}F_1\left(a+1,-b,-c;a+3;\frac{1}{\phi},-\frac{1}{\phi}\right)=\lim_{a,b,c\to0}\partial_{abc}\frac{1}{\phi^{a+2}}\frac{1}{a+2}F_1\left(a+1,-b,-c;a+3;\frac{1}{\phi},-\frac{1}{\phi}\right)
$$


I call them squares. They called them arrays. What do they mean? So I was in C++, and we had third graders come today to play our programs. Whilst the others just drilled them with problems, my game was subtract a square. It was fun watching them discover that numbers like "9" were bad positions (I never explicitly referred to squaring, I just had them multiply the number by itself.) I decided to draw something like this on the board, to show the geometric connection:

The teacher called their attention that I had drawn "arrays" on the board, and they were still used in High School. I didn't bring it up, but I was wondering what they meant by "arrays".

Is it an another name for square?
Is it a multiplication method?
Do they mean multidimensional arrays, as in programming? (If so, third grade has sure advanced since I was in school!)
Something else?

These were USA/Pennsylvania third graders, if that helps.

Note: I know various meanings of "array", but I am wondering what they could possibly mean in this context, with a third grade teacher explaining to their students.

Even though programmers know that the term can be applied when placing data in one, two, or multiple dimensions, I believe that "array", if used in elementary school, gets used specifically to refer to a two-dimensional arrangement of things.

I certainly remember a similar experience growing up: after being exposed to the idea of arrays being two-dimensional arrangements, I was initially confused when, while learning about programming, I was introduced to the notion of arrays (initially) as one-dimensional.

The rectangular arrangement of the square boxes is meant by the term "array".

The German word for this would be back translated to "field", like in grainfield.

The geometric relation might be visible from the Japanese character for field: 田

If you look in an English dictionary on the origin of words (etymology), you get:

  From Middle English arrayen, from Anglo-Norman arayer (compare Old
  French arayer, areer (“to put in order”)), from Medieval Latin arrēdō
  (“to put in order, arrange, array”), from Medieval Latin *rēdum
  (“preparation, order”), from Frankish *reida (“preparation, order”) or
  Gothic  (garaids, “ready, prepared”), from
  Proto-Germanic *raidaz, *raidiz (“ready”), from Proto-Indo-European
  *reydʰ- (“to put in order, ease, make comfortable”). Cognate with Old High German gireiti (“preparation”), Old Frisian rēde (“ready”), Old
  English ġerǣde (“preparation, equipment”).

(Source)

The term is also used for an arrangement of antennas (antenna array).

(Source)


Existence of an axiom question in relation to $\mathsf{Infinity}$ Original Post
This may be a stupid question, but does there axist an axiom $\phi$ that is independent of $\mathsf{ZFC}$, and not equivalent to the axiom of $\mathsf{Infinity}$,  such that $\left(\mathsf{ZFC} -\mathsf{Infinity}+\phi\right) \vdash \mathsf{Infinity}$? Can such a $\phi$ exist? Or will all $\phi$ be equivalent to $\mathsf{Infinity}$?
Edit1:
Danul's answer certainly answers the question, but I am wondering if there exists a weaker axiom $\phi$.
Edit2 I realize I had no idea what I meant by weaker axiom

Edit3, Conclusions
I guess I want to know whether or not  every axiom $\phi$ such that $\mathsf{ZFC}-\mathsf{Infinity} + \phi$ proves that "There exists an inductive set" is such that $\phi$ is either stronger than $\mathsf{Infinity}$ or equivalent to $\mathsf{Infinity}$ and I guess this is the case as $\left(\mathsf{ZFC} -\mathsf{Infinity} + \phi\right)  \vdash \mathsf{Infinity} \Longrightarrow \left(\phi \Longrightarrow \mathsf{Infinity}\right) \lor \left(\phi \longleftrightarrow \mathsf{Infinity}\right)$
In this beginning, I really just wanted to know if we could prove that a countably infinite set existed without simply postulating that one such exists. Can we? What is the most minimal assumption that we need to make? Is it really just $\mathsf{Infinity}$?
Edit4
AH, correct, Thanks Danul. I guess what I meant was:
$$
\left(\mathsf{ZFC} -\mathsf{Infinity} + \phi\right)  \vdash \mathsf{Infinity} \Longrightarrow \left(\phi \vdash \mathsf{Infinity}\right)
$$
As the axiom of infinity cannot be derived from the rest of the axioms of $\mathsf{ZFC}$, so it must be the case that $\phi \vdash \mathsf{Infinity}$. Steven Stadniski answered my question in the comments. Thanks everyone.
Edit5
I am making all sorts of reasoning errors! We have that $$(\mathsf{ZFC} - \mathsf{Infinity} + \phi) \vdash \mathsf{Infinity} \not{\rightarrow}  \phi \vdash \mathsf{Infinity}$$
Because $\phi = \mathsf{Extensionality} \longrightarrow \mathsf{Infinity} $ is a counterexample.
False logic will be the end of me.

Make $\varphi$ be Axiom of infinity plus $GCH$. Then you have that $ZFC-\text{infinity}+\varphi\vdash\text{infinity}$, but $\varphi$ is definitely much stronger than $\text{infinity}$.

Edit: I don't know what you mean by weaker axioms. But remember that $ZFC$ can produce a model (in-fact a countable transitive model) for $ZFC-\text{infinity}+\neg{\text{infinty}}$, namely the hereditarily finite sets (or $V_{\omega}$ if you prefer). So in a sense you can't really be any weaker.

Edit 2: At your current edit (You should keep track of the edits btw or else the entire conversation will become useless, so maybe not delete the original post) the question is trivial. Let $\varphi=\neg{\text{infinity}}$. Now the right hand side is trivial because of the way $\implies$ is defined.

Edit 3: Edit 2 is meant to show that your reasoning is faulty. $\neg{\text{infinity}}$ and $\text{infinity}$ are contradictory, but if you set $\varphi=\neg{\text{infinity}}$, you still get $ZFC-\text{infinity}+\varphi\vdash \text{Infinity}\implies(\varphi\implies\text{Infinity})\lor(\varphi\iff{\text{infinity}})$.

For an answer to your informal question, see the first edit for this solution.

If you are looking for a simpler, more intuitive alternative to the ZF axiom of infinity -- reading between the lines here -- you might consider simply assuming that there exists a set $S$ on which there is defined an injective, but not surjective function $f$, i.e. $S$ is Dedekind-infinite. Could it get more minimal? From such a set, you can extract at least one subset $N$ that satisfies Peano's axioms where $f$ serves as the successor function.


Is there a common symbol for concatenating two (finite) sequences? Say we have two finite sequences $X = (x_0,...,x_n)$ and $Y = (y_0,...,y_n)$.
Is there a more or less common notation for the concatenation of these sequences, like $\sum (X,Y) = (x_0,...,x_n,y_0,...,y_n)$?

The comments suggest the following notations for the concatenation of $X$ and $Y$:

$X^\frown Y$ (given by X^\frown Y);
$XY$ (given by XY);
$X \cdot Y$ (given by X \cdot Y);
$X \mathbin\Vert Y$ (given by X \mathbin\Vert Y);

of which the first seems not to be in use for other concepts, making it especially suitable.

Starting with your defined sequences $X = (x_0, \ldots, x_n)$ and $Y=(y_0,\ldots,y_n)$, you can use the commonly accepted tuple/ordered pair notation:
\begin{align}
   (X,Y) &= \left(
    \left(x_0,\ldots,x_n\right),
    \left(y_0,\ldots,y_n\right)
   \right)
   \\
   &= \left(x_0,\ldots,x_n,y_0,\ldots,y_n\right)
  \end{align}


Is there an Elementary Theory of the Category of Groups? I am aware that the category of categories and category of sets has been axiomatized. Is there such an axiomatization of the category of groups? (and am I correct in thinking "axiomatization of the category of groups" and "elementary theory of the category of groups" refer to the same thing?)

Yes : see this paper (in French) by Pierre Leroux.

I'm not very familiar with the paper, but the main result is that a category $\mathcal{C}$ is equivalent to the category of groups if and only if :

It is complete, cocomplete and pointed.
There exists an object $G$ that is a regular projective generator.
This object is also a co-group, i.e. a group object in $\mathcal{C}^{op}$.
The full subcategory of $\mathcal{C}$ whose objects are the sums of isomorphic copies of $G$ is generated as a category with coproducts by the comultiplication $G\to G+G$, the inverse $G\to G$ and the zero map $G\to 0$.
Every prenormal subobject (i.e. an object that is the equivalence class of $0$ for an internal equivalence relation) is normal (i.e. a kernel).
Every subobject of a free object is free (where a free object is one that is isomorphic to $E\cdot G$ for some set $E$).
Every object is a subobject of a simple object (where a simple object is an object without non-trivial normal subobjects).

The first 4 conditions are condensed in the paper by saying that $(\mathcal{C},G)$ is a "projective category of groups" ("catégorie projective de groupes").

To summarize the proof : the first 4 conditions are equivalent to the fact that $\mathcal{C}$ is equivalent to a subcategory $\mathcal{G}$ of $\mathbf{Grp}$ closed under subobject and products. The fifth one implies $\mathcal{G}$ is also closed under quotients, and is thus a subvariety of the variety of groups. The only subvarieties where every subobject of a free object is free (these are called Schreier varieties) are the variety of abelian groups and the varieties of abelian groups of exponent $p$ for $p$ prime, and these do not have the seventh property because they have only $0$ or $1$ simple object (depending on the case).

Not only has it been axiomatized but Zlil Sela did some brilliant work in the area.


Expressing common vector operations in terms of quaternions Since everything we do with vectors could also in theory be expressed using quaternions (or octonions & higher-order -nions), is it possible to have common vector operations like the cross product or the dot product expressed in terms of quaternions? How are they called and what would they look like?

If you express* two vectors $v_1,v_2$ from $\mathbb R^3$ as quaternions $q_1,q_2$ with real part $0$, then the real part of $q_1q_2$ is $-v_1\cdot v_2$ and the pure quaternions part is $v_1\times v_2$.

This follows from the formula
$$
(r_1,\ \vec{v}_1) (r_2,\ \vec{v}_2) = (r_1 r_2 - \vec{v}_1\cdot\vec{v}_2,\ r_1\vec{v}_2+r_2\vec{v}_1 + \vec{v}_1\times\vec{v}_2)
$$
as given for example here.

So they are both encoded:
$v_1\cdot v_2=-\text{RealPart}(q_1q_2)$

$v_1\times v_2=\text{PureQuaternionPart}(q_1q_2)$

where the real part of $a+bi+cj+dk$ is $a$ and the pure quaternion part is $(b,c,d)$.

(*) Concretely, if $v=(b,c,d)$, the quaternion representing it is $bi+cj+dk$.

Well, at first glance quaternions do not usually represent vectors, but unit quaternions represent rotation matrices, with multiplication of quaternions being something like a matrix product. Quaternions form an algebra, i.e. the important operations one can do with them are addition, multiplication with real numbers, and multiplication of quaternions.

Of course, they form a 4d vector space, so anything you can do with those you can also do here. There are generalizations of cross and dot product to any dimension (scalar product and wedge product), but they do not depend on the extra structure, i.e. the multiplication, quaternions have. Therefore these probably aren't what you are interested in.

Also, quaternions form a division ring, so that one can define left and right vector spaces over quaternions. One can also define scalar and wedge products on those, and I do think that would give an interesting theory, since there is much more structure here than on modules over arbitrary commutative rings. Maybe the most interesting way for you to go would be this direction, i.e. learn a bit about Euklidean and unitary vector spaces, tensor products, exterior algebras and how this can be generalized to modules over rings.


Why is the complementary solution in trigonometric terms? So far the complementary solution (that is the solution to the associated homogeneous equation) has only been in the form: $$y(t)=c_1e^{at}+c_2e^{bt}$$ Where $a$ and $b$ are obtained from solving the characteristic equation.

But for: $$4y^{\prime\prime}+y=e^{-2t}\sin({\frac{t}{2}}) + 6t\cos({\frac{t}{2}})$$ The complimentary solution is given as: $$y(t)=c_1\cos(\frac{t}{2}) +c_2\sin(\frac{t}{2})$$

Edit: The characteristic equation to the equation give above is: $$y^2+\frac{1}{4}=0$$ But where does the $\sin\theta$ and $\cos\theta$ come from?

How do you solve for $a$ and $b$? You solve $$4x^2+1=0$$ which has solutions $$x=\pm\frac i2$$ with $i=\sqrt{-1}$. So the complimentary solution is $$Ae^{it/2}+Be^{-it/2}$$
You can now use Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta$$ to write the complimentary solution as $$\begin{align}Ae^{it/2}+Be^{-it/2}&=A\left(\cos\frac{\theta} 2+i\sin\frac{\theta} 2\right)+B\left(\cos\frac{\theta} 2-i\sin\frac{\theta} 2\right)\\&=(A+B)\cos\frac{\theta} 2+i(A-B)\sin\frac{\theta} 2\end{align}$$
This is your equation if you just write $c_1=A+B$ and $c_2=i(A-B)$.

Well you see that in your equation when you make a characteristic equation of it you'll get $\lambda=0$
So it becomes $e^0=1$

Which is a constant


Value of $k$ for which $e^x = kx$ has $1$ solution I need to work out the value of $k$, where $k>0$, for which $e^x=kx$ has $1$ solution. I've done it somewhat intuitively as follows:

  $e^x=kx$

  By inspection we can see that when $x=1$, the exponent in the LHS and
  the multiplier(?) in the RHS are irrelevant, leaving us with

  $e=k$

  Therefore $e^x=kx$ has $1$ solution at $x=1$.

This is not a very rigorous solution at all though. What is a more proper way to solve this?

The idea is that the two functions $f(x)=e^x$ and $g(x) = kx$ touch only once; that is, $g$ is tangent to $f$ at some point.

Denote the tangent point $x_0$. Then $f(x_0)=e^{x_0}=kx_0=g(x_0)$. The slopes of the two functions have to be the same at this point as they are tangent:
$$f'(x_0) = g'(x_0)$$
$$e^{x_0} = k$$
and so
$$ kx_0 = k$$
The only solution is $x_0=1$ and the value of $k$ is then $k=e^{x_0} = e$.

Postscript: Negative $k$

I assumed above that $k$ is positive. For $k$ negative we can observe that
$$
  \lim_{x \rightarrow -\infty} (e^x - kx) = -\infty
$$
and
$$
  \lim_{x \rightarrow +\infty} (e^x - kx) = +\infty
$$
so that by the intermediate value theorem there exists a point at which $e^x-kx=0$. The function $h(x)=e^x-kx$ is monotonically increasing, as $h'(x)=e^x-k>0$, so there is only one point at which $e^x=kx$.

It can be shown that for all $0<k<e$, $e^x=kx$ has no solution. (This is done using calculus, by showing that $e^x-kx$ has a minimum of $k-klog(k)$. Since $0<k<e$, we know that $log(k)<1$, hence $k-klog(k)>0$)

If $k>e$, we know that $e^x-kx$ will be negative for some $x$ (At $x=log(k)$). Since, $e^x-kx$ grows without bound as $x$ approaches infinity or negative infinity. We can show that $e^x-kx$ crosses the x axis twice (a contradiction)


Construction of the Associated projective hyperplane. Let $V$ be $K-$vector space of dimension $n$. Let $H$ be a hyperplane given by the kernel of the linear map
$l_H:V\longrightarrow K$ defined by
$$l_H(x)=a_1x_1+\cdots+a_nx_n$$
where $x_1,\cdots,x_n$ are the coordinates of $x$ in a chosen basis of $V$.
Let $P(V)$ be the projective space associated to $V$. Consider the map: $l_{H'}:P(V)\longrightarrow K$
defined by
$$l_{H'}([x_1:\cdots:x_n])=a_1x_1+\cdots+a_nx_n$$
Does the kernel of $l_{H'}$ defines a projective hyperplane in $P(V)$?

The answer is yes. A point $M \in P(V)$ belongs to the projective hyperplane $P(H)$ if and only if one of its representative, $\vec{M} \in H \iff l_H(\vec{M}) = 0$ (which is an equation of $P(H)$.


Problem on Analysis (Functions) Let f(x) be a function from reals to reals obeying the following: f(x) is continuous, f(0)=1, and f(m+n+1)=f(m)+f(n). Show that f(x)=1+x for all real numbers x.

I am a bit confused on how to start with the sum as I am new to analysis. Assuming f(x)=1+x. I can indeed show that it satisfies properties f(m+n+1)=f(m)+f(n), (where m and n are reals) by checking it for RHS and LHS alternatively but can't prove it using the condition itself

Suppose that $n = 0$. Thus, the formula becomes:

$$f(m + 1) = f(m) + f(0)$$
$$f(m+1) = f(m) + 1$$

Substituting $m = x$
$$f(x+1) = f(x) + 1$$

Differentiating both sides:
$$f'(x+1) = f'(x)$$

This clearly indicates that the function is a linear function (as for any other type the slope would be dependent on the value of $x$)

Thus:
$$f(x) = mx + c$$

Now we know that $f(0) = 1$. Substituting that value, we get $c = 1$.
Using: $$f(x+1) = f(x) + 1$$
And substituting $x = -1$ we get $f(-1) = 0$. Thus we get the value of $m = 1$.

Thus, the equation is:
$$f(x) = x + 1$$


Heat equation $u_t=u_{xx}$ with $u(x,0)=0, u(1,t)=5t$, and $u_x(0,t)=0$ Heat equation $u_t=u_{xx}$ with $u(x,0)=0, u(1,t)=5t$, and $u_x(0,t)=0$
I saw this in Japanese Q&A website and tried to solve it but I couldn't.
So I am not sure that this can be solved analytically.
What I tried: From Wikipedia, I found a solution for $u(x,0)=0, u(0,t)=f(t)$ as:
$u(x,t)[f] = \int_0^t x/\sqrt{4\pi (t-s)^3}\ \exp(-x^2/4(t-s))\ f(s)\ ds$
In order to satisfy $u_x(0,t)=0$, I considered using symmetry: $u(x,t) = u(x+1,t)[f] + u(x-1,t)[f]$ for some $f$, but I couldn't adjust the result to satisfy $u(1,t)=5t$.
(My calculation)
$u(x,t)[f(t)=t] = \int_0^t x/\sqrt{4\pi (t-s)^3} \exp(-x^2/4(t-s))\ s\ ds \\\
 = [ (x^2/2+t) erf(x/2\sqrt{t-s}) - x\sqrt{t-s}/\sqrt{\pi}\ \exp(-x^2/4(t-s)) ]_0^t \\\
 = (x^2/2+t)(1-erf(x/2\sqrt{t})) - x\sqrt{t}/\sqrt{\pi}\ \exp(-x^2/4t)$
$u(x,t)[f(t)=1] = \int_0^t x/\sqrt{4\pi (t-s)^3}\ \exp(-x^2/4(t-s))\ ds \\\
 = [ erf(x/2\sqrt{t-s})  ]_0^t \\\
 = 1 - erf(x/2\sqrt{t})$


Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.

I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.

I' sure that I'm overseeing a tiny little missing link; dear colleagues, please give me a hint.

The given $f(x)$ is given by
\begin{align}
f(x) = \frac{2(x-1)}{x^2 - 2x + 4}
\end{align}
and can be seen as, where $a = 1 + \sqrt{3} i$ and $b = 1 - \sqrt{3} i$, such that $ab=4$,
\begin{align}
f(x) = \frac{-2 \, (1 - x)}{ab \, \left(1 - \frac{x}{a}\right) \left( 1 - \frac{x}{b}\right)}
\end{align}
for which
\begin{align}
\ln f(x) &= \ln\left(- \frac{4}{ab}\right) + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\
&= \pi i - \ln 2 + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\
&= \pi i - \ln 2 - \sum_{n=1}^{\infty} \left( 1 - \frac{1}{a^{n}} - \frac{1}{b^{n}} \right) \, \frac{x^{n}}{n} \\
&= \pi i - \ln 2 + \sum_{n=1}^{\infty} \left( \frac{a^{n} + b^{n}}{4^{n}} - 1 \right) \, \frac{x^{n}}{n} \\
&= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{a^{n} + b^{n} - 4^{n} }{n} \, \left(\frac{x}{4}\right)^{n} \\
&= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{ 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} }{n} \, \left(\frac{x}{2}\right)^{n}.
\end{align}
Now differentiating both sides leads to
\begin{align}
\frac{f'}{f} &= \sum_{n=1}^{\infty} \left( 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n} \\
&= \frac{1}{2} \, \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{n \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n-1} \\
&= \sum_{n=0}^{\infty} \left( \cos\left(\frac{(n+1) \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n}.
\end{align}
Expanding the first few terms provides
\begin{align}
- \frac{f'}{f} = \frac{1}{2} + \frac{5 \, x}{4} + \frac{5 \, x^{2}}{4} + \frac{17 \, x^{3}}{16} + \frac{31 \, x^{4}}{32} + \mathcal{O}(x^{5})
\end{align}


How to find the sum of all entries of a certain matrix I have two matrices $A=\pmatrix{19 & 17\\\ 1& 4}$ and $B=\pmatrix{\sqrt{3} & -1\\\ 1& \sqrt{3}}$. I want to find the sum of all entries of $(ABA^{-1})^{300}$. Where to attack?

I found a good pattern for $B^{2n}$ and can see that $(ABA^{-1})^{300}=AB^{300}A^{-1}$ but it seems not the right way necessarily. Thank you for the hint.

It may help to note that the sum of all elements of $C$ is
$$\pmatrix{1&1}C\pmatrix{1\\1}.$$
Then
$$\pmatrix{1&1}AB^{300}A^{-1}\pmatrix{1\\1}
=\pmatrix{20&21}B^{300}\pmatrix{-13/59\\18/59}
$$
etc.

Note that $$(ABA^{-1})^n = \underbrace{(AB\overbrace{\color{red}{A^{-1}})(\color{red}{A}}^{\text{identity}}B\color{red}{A^{-1}})(\color{red}{A}BA^{-1})\cdots (ABA^{-1})}_{n\text{ times}}$$ so clearly, for any invertible matrix $A$ we have $$(ABA^{-1})^{300} = AB^{300}A^{-1}$$ with the pattern on $B$ the problem is solved.


Proving continuity in the origin $\frac{xy}{\sqrt{x^2 + y^2}}$ Let $g: \mathbb{R^2} \to \mathbb{R}$.

How can I prove that  $g$ is continuous in its origin, but not totally differentiable?

If I take

$$g(\frac{1}{n},\frac{1}{n}) = \frac{1}{n\sqrt{2}} \to 0 \text{ for } n \to \infty $$

Or rather:

$$|x,y| \leq \frac{1}{2} (x^2 + y^2) $$

from which we can follow

$$|g(x,y)| \leq \frac{1}{2} \sqrt{x^2 + y^2}$$

which proves continuity of $g$ in the origin $(0,0)$.

But how can I show that this function is not total differentiable?

Can I do the following estimation?

$$|g(x,y) - 0| = |y| \cdot \frac{x \cdot y}{\sqrt{x^2 + y^2}} \leq |y| \cdot 1 = |y| $$

From which it follows, that

$$|y| \leq |\sqrt{x^2 + y^2}| = ||(x,y)|| $$

Note that $x^2+y^2\ge 2|xy|$.  Hence, we have

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \le \frac{\sqrt{|xy|}}{\sqrt 2}$$

By polar coordinates as $r \to 0$

$$\frac{xy}{\sqrt{x^2 + y^2}}=r \cos \theta \sin \theta$$

we see that $f(x,y)$ is continuous but since $f_x(0,0)=f_y(0,0)=0$ by definition of differential we have

$$\lim_{(h,k)\to (0,0)} \frac{\frac{hk}{\sqrt{h^2 + k^2}}}{\sqrt{h^2 + k^2}}=\frac{hk}{h^2 + k^2}$$

which doesn't exist.

Refer also to the related

Existence of Partial Derivatives Implies Differentiability


Nth Term for the series 2,4,6,9,12,16,20,25,30,36... Could you please give me an Idea to find nth Term in the series 2,4,6,9,12,16,20,25,30,36... coz im from computers background and i m not able to get the idea to find nth term for this specific series....Thank u.


Introduction to Filters in Topology Question: What are some good resources for a student who has taken algebraic and point-set topology and who wishes to learn how filters and ultrafilters are applied in topology?

Motivation: I've seen in a number of papers where ultrafilters are briefly mentioned as tools which can be used to generalize an argument or something, but the few papers I've found which introduce filters mainly use them as as tools in Logic and mention topology only in passing. The Wikipedia page gives some good starting points, but I felt that it was more of a reference list than an introduction to the topic.

Bourbaki's General Topology, Volume 1 devotes a chapter to filters. The notion of net and equivalence between filters and nets is developed (partly in a set of guided exercises) in Kelley's General Topology. The use of ultraproducts in Commutative algebra by Schoutens might interest you. The theory of ultrafilters by Comfort and Negrepontis is encyclopedic. Jech's Set theory has a chapter on ultrafilters. Sections 4 and 6 of Chang and Keisler's Model Theory is also worth a look. These are all available on the website library.nu.

Also consider my book. It is not a standard course on general topology, but rather a generalization of general topology.

In my theory all general topology is based on filters (and filters are studied in my book in details, making it definitely the world best reference on the topic of filters).

In fact, it is not general topology, but its generalization. However, limit, connectedness, continuity, limit (even generalized limit of discontinuous function), etc. appear in my theory just like as they appear in standard general topology. And these concepts are expressed by nice algebraic formulas, not as a mess of quantifiers as in any other topology book.

Curious fact: Continuity generalizes for multivalued functions in three different ways.

My theory applies both to calculus and to discrete mathematics.

You can study my theory after or before legacy general topology. The book even contains an introduction for beginning students who know only basic calculus and basic set theory.


What is the Riemann Sphere? Reading from wikipedia I understood that Riemann Sphere is used to represent extended complex plane. But it would be great if someone could explain it in a less technical manner.

Imagine you're on an infinitely large, flat piece of paper with a transparent beach ball of diameter $1m$. Assume the ball is perfectly round. You have a lazer pen.

The paper is the complex plane. Place the beach ball down so that it rests on $0$, its centre is $0.5m$ off the ground and the point directly above $0$ is $1m$ off the ground.

Take your lazer pen, switch it on, and point it down at $0$ so that the pen is touching the point $1m$ above the ground. With your pen held in that place, you can move the little dot of light around the complex plane, highlighting any complex number you like. Your pen will also make a dot on the ball - a point on the Riemann Sphere! - and that dot will be unique to the complex number you are aiming at!

Further: If you point the pen such that the beam of light is parallel to the paper (yet still touching the ball at its highest point), you'll see that the dot disappears. This is where we extend the complex plane; we say the dot is at $\infty$ and take the highest point of the ball to be the unique point of the ball corresponding to $\infty$.

That's all there is to it really. I hope that helps $\ddot\smile$

The Riemann Sphere is a unit sphere--a sphere with radius $1$--with its south pole kissing the origin centered around the $z$ axis. Points on the sphere can be associated to points in the plane by projecting from the north pole $N$ through a point $P$ on the surface of the sphere onto the plane at a point $Q$ directly through what is called a line. This correspondence of points is called a stereographic projection. There are many good illustrations of them, as, for example in the below illustration take from Tristan Needham's treatise Visual Complex Analysis:


If $ U$ is maximal among non-principal ideals show that $U$ is prime. (Hints?) Let $R$ be commutative with $1$.  If $U$
 is maximal among non-principal ideals show that $U$
 is prime.

Possible proof path:

Suppose $ab \in U$ and $a \notin U$  Consider $U+(a)$.
I tried to prove along the lines of "Maximal ideal is prime" but I can't seem to proceed anywhere. Any hints?

P.S. Pls do not provide links to Oka or Ako families. I just got started in Algebra.

This is exercise $10$ from chapter $1$, section $1$ of the text

Kaplansky${\,-\,}$Commutative Rings, Rev Ed (1974)

Fleshing out the author's extended hint, we can argue as follows . . .

Let $U$ be a non-principal ideal of $R$ which is maximal among all non-principal ideals of $R$.

Suppose $U$ is not prime.

Our goal is to derive a contradiction.

Let $ab \in U$, with $a,b\notin U$.

Then $(U,a) = (c)$, for some $c\in R$.

Let $V=\{x\in R\mid cx \in U\}$.

Clearly we have $U \subseteq V$.

Since $(U,a)=(c)$, we have $u+ra=c$, for some $u\in U$ and some $r\in R$.
\begin{align*}
\text{Then}\;\;&u+ra=c
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;
\\[4pt]
\implies\;&b(u+ra)=bc\\[4pt]
\implies\;&bu+r(ab)=bc\\[4pt]
\implies\;&bc\in U\qquad\text{[since $ab\in U$]}\\[4pt]
\implies\;&b\in V
\end{align*}
Since $U \subseteq V$, and $b \in V\setminus U$, we have the proper inclusion $U\subset V$, hence $V=(d)$ for some $d\in R$.

Now let $u$ be an arbitrary element of $U$.

Since $U\subset (U,a)=(c)$, we get $u=cy$, for some $y\in R$.
\begin{align*}
\text{Then}\;\;&u=cy\\[4pt]
\implies\;&cy \in U\\[4pt]
\implies\;&y\in V&&\text{[by definition of $V$]}\\[4pt]
\implies\;&y = dz,\;\text{for some}\;z\in R&&\text{[since $V=(d)$]}\\[4pt]
\implies\;&u = c(dz)\\[4pt]
\implies\;&u \in (cd)\\[4pt]
\end{align*}
Thus, $U\subseteq (cd)$.

Also, since $d\in V$, then by definition of $V$, we get $cd\in U$, hence $(cd) \subseteq U$.

But then $U=(cd)$, contrary to the assumption that $U$ is not principal.

It follows that $U$ is prime.

Suppose,ab is in U,Consider the ideal J=U+(a),if a is not in U then J properly contains U so by maximality of U ,J=R.Hence there is u in U &c in R such that 1=u+ac,multiplying both sides by b &use ab is in U conclude b is in U.


Prove: $2n^{2} + 3n$ is even if and only if $n^{3} - 4$ is even. I need to prove this statement in some way, maybe through a contradiction or contrapositive but I am stuck. I first assume that if $n^{3} - 4$ is even that $n^{3} - 4$ = $2k$ for some integer $k$ (by definition of even). Then when I try to solve for $n$ so that I can plug it into $2n^{2} + 3n$, I get $n = (2k + 4)^{1/3}$. If I try to plug this into $2n^{2} + 3n$, then I do not know how to manipulate $$2((2k + 4)^{1/3})^{2} + 3((2k + 4)^{1/3})$$ to show that it is indeed even (or equal to 2 times some integer).

Guide:

If $n^3-4$ is even, then $n^3$ has to be even, hence $n$ is even. Hopefully you can prove that $2n^2+3n$ is even from here.

Conversely, if $2n^2+3n=n(2n+3)$ is even, since $2n+3$ is odd, we have $n$ must be even. Again, hopefully you can prove that $n^3-4$ is even.


What explicitly is the "adjunction" isomorphism $Hom_C(A,C)\to Hom_B(A,Hom_C(B,C))$? Suppose $B$ and $C$ are commutative rings, $A$ a $B$-algebra, and $B$ is a $C$-module. What exactly is the "adjunction" isomorphism
$$Hom_C(A,C)\to Hom_B(A,Hom_C(B,C))?$$

Given $A\to C$, it needs to be composed with a morphism $C\to Hom_C(B,C)$. I know $Hom_C(C,B)\simeq B$, but that doesn't seem to work here.

Here also $A$ is a finitely generated projective $B$-module, and $B$ is a finitely generated projective $C$-module.

I believe the adjunction in question is likely the classic tensor hom adjunction $\text{Hom}_C(A \otimes_R B, X) \cong \text{Hom}_R(A, \text{Hom}_C(B, X))$. Where all the objects in question have the appropriate structures such that this makes sense (X is a module over C, A is a module over R, B is a bi-module over both R and C, and R, C are just rings).

The statement you are trying to prove then follows from the observation that if $A$ is a $B$ algebra, then $A \otimes_B B \cong A$ as $C$-modules, taking $X = C, R = B$ above. This is a very simple thing to show if you understand the tensor product.

Edit: the explicit map takes a $\phi \in \text{Hom}_C(A \otimes_B B, C)$ and sends this to the map $\psi \in \text{Hom}_B(A, \text{Hom}_C(B, C))$ where $\psi$ sends $a \to \phi_a$ where $\phi_a(b) = \phi(a \otimes b)$. That this mapping is $B$-linear and an isomorphism is once again not hard.

The actual isomorphism is (in category theory):

hom(AxB,C) = hom(A,hom(B,C))

It is normally known as "currying". (the hom's are in Set, not in C or B)

I'm a little confused why they call it "adjunction", since that means slightly different thing, but the word doesnt seem to explain your isomorphism.


The set $A=\{(x, y)\in \mathbb{R}^2:|x|=|y|\}$ is connected Two disjoint sets $A$ and $B$, neither empty, are said to be mutually separated if neither contains a boundary point of the other. A set is disconnected if it is the union of separated subsets, and is called connected if it is not disconnected.

With the above definition of connected set, how to prove that the set $A=\{(x, y)\in \mathbb{R}^2:|x|=|y|\}$ is connected.

Exercise should try using the definition of previous related and for this I guess I should do the proof by contradiction but do not know how to proceed.

Do you know  straight lines are connected set in $\mathbb{R}^2$? do you know union of connected set is connected if they have one point in common?

Now can you see your set with above stated properties?

If |x| < |y| then those x's are all in the disk of radius y about the origin.  That is surely connected


If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove $xw+wz+zy+yu+ux⩽\frac15$ If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$
xw+wz+zy+yu+ux⩽\frac15.
$$

I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, applying AM-GM to each pair directly gives$$
x^2+y^2+z^2+w^2+u^2 ⩾ xw + wz + zy + yu + ux,
$$
but I cannot seem to continue from here or use $x + y + z + w + u = 1$. Other inequalities like Chebyshev's rely on the multiplied pairs to be in order from least to greatest or vice versa, so I am stuck here.

It is quite easy to solve this problem with the Lagrange multipliers

Calling

$$
L(x,y,z,w,u,\lambda) = \frac 15 -(xw+wz+zy+yu+ux)+\lambda(x+y+z+w+u-1)
$$

The stationary points are the solutions for

$$
\nabla L = (L_x,L_y,L_z,L_w,L_u,L_{\lambda})=0
$$

or

$$
\lambda -u-w = 0\\
\lambda -u-z = 0\\
\lambda -w-y = 0\\
\lambda -x-z = 0\\
\lambda -x-y = 0\\
u+w+x+y+z-1 = 0
$$

Solving this linear system we obtain

$$
x = y = z = w = u = \frac 15,\;\lambda = \frac25
$$

This point is the only tangent point between the surface $g(x,y,z,w,u)=\frac 15 -(xw+wz+zy+yu+ux)$ and the hyperplane $\Pi =x+y+z+w+u-1 = 0$

and  $g(x,y,z,w,u) \ge 0$ is located into one of the  semi-spaces delimited by $\Pi$ and also at the tangency point we have $g = 0$ with the values found before.

NOTE

This formulation is a short hand for

$$
L(x,y,z,w,u,\lambda,\epsilon) = \frac 15 -(xw+wz+zy+yu+ux)+\lambda_1(x+y+z+w+u-1)+\lambda_2(x-\epsilon_1^2)+\lambda_2(y-x-\epsilon_2^2)+\lambda_4(z-y-\epsilon_3^2)+\lambda_5(w-z-\epsilon_4^2)+\lambda_6(u-w-\epsilon_5^2)
$$

and the result should be the same as you can verify with a little patience.

The case where $xw+wz+zy+yu+ux$ is maximal will be where $x=y=z=w=u$. Here, they must all be $\frac15$, and plugging into the equation gives $5(\frac15)^2=\frac15$. So the expression is $\le\frac15$.


$g$ is monotonic strictly increasing, continuous in $ [0,\infty)$ and positive function. $$\lim\limits_{n \to \infty} g(t)=L $$
 $g$ is monotonic strictly increasing, continuous in $ [0,\infty)$ and positive function.

I need to find :
$$\lim\limits_{n \to \infty} \int^2_1 g (nx) \,\mathrm{d}x  $$

I have started with:
 $|g(t)-L|<\epsilon\>$

$L-\epsilon\>$$=<$$g(t)$$<=$$L+\epsilon\>$

$\int^2_1 (L-\epsilon\>) \,\mathrm{d}x$$=<$$\int^2_1 g (nx) \,\mathrm{d}x$$<=$$\int^2_1 (L+\epsilon\>) \,\mathrm{d}x$

I'm not sure if I'm in the right direction ...

You have using integral by substitution with $u = nx$:
$$l_n = \int^2_1 g (nx) \,\mathrm{d}x = \frac{1}{n} \int_n^{2n} g (u) \,\mathrm{d}u $$

As $g$ is supposed to be strictly increasing, you have for $u \in [n, 2n]$ $g(n) \le g(u) \le g(2n)$. Therefore
$$g(n) \le l_n \le g(2n)$$ by taking the integral of those inequalities over $[n,2n]$.

Knowing the hypothesis on $g$, $\lim\limits_{x \to \infty} g(x)$ exists (it can be $\infty$). Naming this limit $L$, you get
$$\lim\limits_{n \to \infty} l_n = L$$


To find wether $(3,-1,0,-1)$ is in the span of the other vectors, solve the system:
$$(3,-1,0,-1)=\lambda_1(2,-1,3,2)+\lambda _2(-1,1,1,-3)+\lambda _3(1,1,9,-5)$$

If you get a solution, then the vector is the span. If you don't get a solution, then it isn't.

It's worth noting that the span of $(2,-1,3,2), (-1,1,1,-3), (1,1,9,-5)$ is exactly the set $\left\{\lambda_1(2,-1,3,2)+\lambda _2(-1,1,1,-3)+\lambda _3(1,1,9,-5):\lambda _1, \lambda _2, \lambda _3\in \Bbb R\right\}$

Alternatively you can consider the matrix $\begin{bmatrix}2& -1 &3 & 2\\ -1 &1 &1 &-3\\ 1 & 1 & 9 & -5 \\ 3 &-1 &0 & -1\end{bmatrix}$. Compute its determinant. If it's not $0$, then the four vectors are linearly independent. If it is $0$ they are linearly dependent. What does that tell you?

if (3,-1,0,-1) be in span {(2,-1,3,2),(-1,1,1,-3),(1,1,9,-5)} then we must have
 (3,-1,0,-1)=$c_{1}$ (2,-1,3,2)+ $c_2$ (-1,1,1,-3)+ $c_3$ (1,1,9,-5)
so this system must have solution:

$2c_1 -c_2+c_3=3$

$-c_1 +c_2+c_3=-1$

$3c_1 +c_2+9c_3=0$

$2c_1 -3c_2-5c_3=-1$
since 3 first relation has unic answer and 3 end realations too and they cant have equal answer (then this system has not solution and (3-10-1)


Arbitrary intersection of compact sets is compact. To prove this, I want to show that an arbitrary intersection of closed sets is closed and an arbitrary intersection of bounded sets is bounded. I know how to prove the first part, but I'm not sure how to rigorously show that an intersection of bounded set is bounded.

This question comes from a real analysis course, not a general topology course.

The proof is somewhat dependent on the definitions you use.

The simple (as you basically have only one possible definition) part is the intersection of bounded sets. You either define bounded as having a radius such that every member of the set has absolute value (or norm) less or equal to the radius, or more general that the distance of any two points in the set have distance $d(x,y)$ less or equal to the radius (I use the latter), I don't require the radius to be the smallest bound here (if one bound exists there exists an infimum of them which is also a bound by the axiom of largest lower bound):

Let $B_j$ be bounded sets which means that there exists a radius $r_j$ such that $d(x,y)\le r_j$ for all $x,y\in B_j$. Now consider the intersection $B = \bigcap B_j$, then we have that $d(x,y)\le r_o$ for any $x,y\in B_o$, but since $B \subseteq B_o$ we have for any $x, y\in B$ that $x,y\in B_o$ and therefore $d(x,y)\le r_o$, so $r_o$ is a radiuos of $B$.

To prove that intersection of closed sets is closed we have to do that differently depending on the definition:

Closed being that the complement is open: Use deMorgan theorem and use that $\overline{F_j}$ is open, therefore $\bigcap F_j = \overline{\bigcup \overline {F_j}}$, but the union of open sets is open so the intersection is the complement of an open set.
Closed being containing all it's limit points. Suppose we have $a$ being a limit point of $F =\bigcap F_j$, then there exists a sequence $a_k\in F$ such that $\lim_{k\to\infty}a_k = a$, but since $F_j\subseteq F$ we have that $a_k$ is a sequence in $F_j$ and therefore $a\in F_j$ as $F_j$ is closed. And since $a\in F_j\subseteq F$ we have $a\in F$. So $F$ contains all its limit ponts.

Also for compactness it's a matter of definition:

Compact being closed and bounded: The intersection of closed is closed, and intersection of bounded is bounded. Therefore intersection of compact is compact.
Compact being that open cover has a finite subcover: This is a lot trickier (and may be out of your scope), I will need to use more assumptions here. Suppose we have a open cover of $F \subset \cup \Omega_j$. We extend this to a cover of any $F_o$ by observing that for each $\phi\in F_o\setminus F$ we have an open set $\Omega_\Phi$ that doesn't intersect $F$ (assuming the space is regular Hausdorff - for example the set of real numbers). Now we have an open cover of $F_o$ which have a finite subcover, and excluding the ammended sets to cover the whole $F_o$ will make it a cover of $F$.

The finite subcover produced by a subcover of one of the compact sets also covers the intersection.


strictly increasing functions and bijection $f $  is strictly increasing
   \begin{align}
& \rightarrow f^{-1} \text{is strictly increasing}\tag {i} \\
& \rightarrow f^{-1} \text{is strictly decreasing}\tag {ii}  \\
& \rightarrow f \quad \text{is injective}\tag {iii}\\
& \rightarrow f \quad \text{is surjective}\tag {iv}\\
& \rightarrow f^{-1} \text{is bijective}\tag {v}
\end{align}

which statements are true?

(iii)  $f$ is injective since $x<y\rightarrow f(x)<f(y)$

(iv) I believe it doesn't need to be surjective.   $f:\Bbb N\to \Bbb N$ with $f(x)=x+1$

(v) since $f$ isn't bijective we cannot talk about $f^{-1} $ function

You almost got everything, the answers are as follows:

$f$ is strictly increasing $\Rightarrow$ $f^{-1}$ is strictly increasing (is correct)

I would prove this by contradiction. So assume $f^{-1}$ would not be strictly increasing, then you can construct a contradiction pretty straightforward.

$f$ is strictly increasing $\Rightarrow$ $f^{-1}$ is strictly decreasing (is wrong)

Just take $f(x)=x$ or the use the proof of the first statement

$f$ is strictly increasing $\Rightarrow$ $f$ is injective (is correct)

Just use the property you already mentioned and the definition of injectivity.

$f$ is strictly increasing $\Rightarrow$ $f$ is surjective (is wrong)

Your example works or take $f:\mathbb{R}\to\mathbb{R},f(x)=e^{x}$

$f$ is strictly increasing $\Rightarrow$ $f^{-1}$ is bijective (is wrong)

Generally we have that: $f$ is bijective $\Leftrightarrow$ $f^{-1}$ is bijective.

Because $f$ is not surjective we actually have no inverse function $f^{-1}$ on the whole domain. So all answers above are understood as the restriction of the domain of $f^{-1}$ where it is actually defined, so basically the image $im(f)$ of $f$.

Nevertheless, we now could make $f$ become a bijective function if we would restrict the image space to the image $im(f)$ of $f$. If we do that, then we'd have that $f$ is bijective as also $f^{-1}$.

Assuming no Asymptotic discontinuities if $f(x)$ is Strictly increasing Then $f'(x)>0$.

Letting $$g(x)=f^{-1}(x)$$ we have

$$f(g(x))=x$$ differentiating Both sides

$$f'(g(x))g'(x)=1$$ $\implies$

$$g'(x)=\frac{1}{f'(g(x))}$$ and since $f'$ is Positive

$g'\gt 0$, hence $f^{-1}$ is Increasing.


For what $x$ and $y$ polynomial has maximum value? For what $x,y\in\mathbb R$ does the polynomial
  $$-5x^2-2xy-2y^2+14x+10y-1$$
  attain a maximum?

My attempt:

I called $\alpha$ maximum value.

$$-5x^2-2xy-2y^2+14x+10y-1\leqslant\alpha$$
$$-5x^2-2xy-2y^2+14x+10y-1-\alpha\leqslant 0$$

$$5x^2+2xy+2y^2-14x-10y+1+\alpha\geqslant 0$$
$$(x+y)^2+(y-5)^2+3x^2+(x-7)^2-73+\alpha\geqslant0$$
$$\alpha\geqslant73$$
So the lowest maximum value turned out to be $73$, but after checking answers I was wrong-maximum is $16$, so my further plans to calculate from that $x$ and $y$ seemed purposless. I'd like to see solution using only high school knowledge.

Ans: $x=1$, $y=2$

Write $$f(x)=-5x^2-(2y-14)x-2y^2+10y-1$$ This quadratic function on $x$ with parameter $y$ achieves maximum at

$$p=-{b\over 2a} = {2y-14\over -10}$$ and this maximum is $$ q= -{b^2-4ac\over 4a} = {(2y-14)^2+20(-2y^2+10y-1)\over 20}=$$

So you need to find the maximum of $$g(y)= (2y-14)^2+20(-2y^2+10y-1)$$

$$ =-36y^2+144y +176$$

And now you can repeat the procedure which was done for $x$.

The maximum is at $y= -{144\over -72}=2$ (and thus $x=1$) and that maximum value is...

let $k$ be a maximal value.

Thus, for any $x$ and $y$ we have $$-5x^2-2xy-2y^2+14x+10y-1\leq k$$ or
$$5x^2+2(y-7)x+2y^2-10y+1+k\geq0,$$ which gives
$$(y-7)^2-5(2y^2-10y+1+k)\leq0$$ or
$$9y^2-36y-44+5k\geq0,$$ which gives $$18^2-9(-44+5k)\leq0,$$ which gives $$k\geq16.$$
The equality occurs for $$y=\frac{36}{2\cdot9}=2$$ and $$x=-\frac{2(2-7)}{2\cdot5}=1,$$ which says that a minimal value of $k$,

for which the first inequality is true for any values of $x$ and $y$, it's $16$.


Proving $ \lim_{x\to 2} x^2 = 4$ I am trying to prove the following limit but I am a stuck at a step.

$$ \lim_{x\to 2} x^2 = 4$$

Proof : Let $\varepsilon > 0$ be given. We want to find a $\delta > 0$ such thant $0<|x-2|<\delta \rightarrow |x^2-4|<\varepsilon$

$$ |x^2-4| < \varepsilon$$
$$ |x-2||x+2| < \varepsilon$$

What do I do from here ?

You could choose that $\;\delta<\frac12\;$ , say, and then

$$|x-2|<\delta\iff 2-\delta<x<2+\delta\implies \frac72=4-\delta<x+2<4+\delta=\frac92$$

so that you get

$$|x-2||x+2|<|x-2|\cdot\frac92<\frac{9\delta}2\color{red}{<\epsilon}\iff \delta<\frac{2\epsilon}9$$

and thus for any $\;\epsilon>0\;$ , choosing $\;\delta=\min\left\{\frac12,\,\frac{2\epsilon}9\right\}\;$ you get what you want.

Observe that "the choosing" of $\;\delta\;$ that we began with is rather arbitrary. You could as well have chosen $\;\delta<1\;,\;\;\delta<0.001\;$ or whatever. We usually choose something more or less nice, though.

Hint: Fix $\epsilon>0$.  Choose $\delta= \epsilon$, then we see that
\begin{align}
|x^2-4| = |x-2||x+2| \leq \epsilon |x+2|< \epsilon(4+\epsilon)
\end{align}
whenever $|x-2|<\delta$.


Why we can't define $\frac{1}{0}$ to be $1$ (or anything else), but we can define $1^0$ to be $1$? We know that we can't define division by zero "in any mathematical system that obeys the axioms of a field", because it would be inconsistent with such axioms.

(1) Why can we define $a^0$ ($a\neq 0$) to be $1$? Is it possible to prove that such definition is consistent with any rule of arithmetic? How to conclude that to define $a^0$ ($a\neq 0$) we don't need abolish any other basic rule of arithmetic?

(2) More generally, how to know if a definition is consistent with a given mathematical theory?

There is no general algorithm for determining when a theory is consistent.  That is a huge topic which includes Godel's incompleteness theorems.  But your specific question is easier.

In Peano arithmetic (with axioms stated using $+,\times$) an exponential function $x^y$ can be defined by recursion $x^0=1$ and $x^{s(y)}=x\times x^{y}$.  The axioms prove that functions can be defined recursion.  So if you believe (as nearly everyone does) that  Peano arithmetic (with axioms stated using $+,\times$) is consistent, then you must believe the extension with that exponential function is consistent.

Since your question mentions basic rules of arithmetic I answered in terms of Peano Arithmetic. If you merely want consistency with the field axioms the question is simpler yet: The field of integers modulo 2 proves consistency of those axioms plus $x^1=x$ and $x^0=1$, by giving a finite model.  But this includes very little of arithmetic and notably does not include $x^{(y+z)}=x^y\times x^z$. See "finite field" on Wikipedia.

We don't actually define $1^0$ to be 1.  That's its value.  Likewise $0^0=1$ is a derived value, the supposed indefinite values all rely on the same $0/0$ proof that lets $0=1$.  If you take the limit of $x^{ax}$ as x-> 0, the limit is 1 for all a.
When you ask how one plans to go towards zero, a step right is either by way of root, eg square-root, or by way of division.
Roots are a matter of dividing a positive number by a positive number, and this never goes to zero.
Division implies $0/0$ is being used, ie to suppose $0^0=0$, implies that you can reach $0$ by division of non-zero numbers, or that you can reach $0$ by division.  Since the first is not accepted in maths, it implies that $0^0=0$ arises from division by zero.


Why change a given basis? Why would we want to transform a vector in our normal basis (xyz axes) to another basis? The only situation I can recall is when we are looking at a force applied on an inclined plane. Are there any other real life examples where this may be necessary?

Let $A$ be an $n\times n$ matrix with eigenvalues $\lambda_1,\dots,\lambda_n$ and corresponding (linearly independent) eigenvectors $v_1,\dots,v_n$. Then for any (positive) integer $k$ and any vector $x$ we have $$A^kx=c_1\lambda_1^kv_1+\cdots+c_n\lambda_n^kv_n$$ where $c_1,\dots,c_n$ are the coordinates of $x$ with respect to the basis $\{{v_1,\dots,v_n\}}$. So if you find yourself in a situation where you are repeatedly multiplying by the matrix $A$ (and that is very common in applications), the calculations become very simple in the new basis.

Whenever you whant to express a given vector as a sum of $n$ components in chousen $n-dimentional$ space. In the example you gave about forces on inclined plane, you try o express the force in $R^{2}$ that acts on the body ass composed of vertical and horizontal components respect to inclined plane of the displacement, but you could choose any other two, (not parallel) directions to express your force.


Proof using axioms of real numbers Using axioms of real numbers prove for
∀x∈R, ∀y∈R,∀n∈N:

($0 \leq x < y$ ) $\Rightarrow$ $x^n < y^n$

Any help with this? I have no idea how to even begin.

By induction.

Base case $n=1$,

$$0\le x < y\implies x^1<y^1.$$

Induction

$$0\le x < y\implies x^n<y^n,$$

$$0\le x < y\land x^n<y^n\implies 0\le x\,x^{n}<y\,x^{n}<y\,y^{n}.$$

$$ 0 \leq x < y \implies ln(x) < ln(y) \implies k \cdot ln(x) < k \cdot ln(y) \implies x^k < y^k $$

Step 1 : The logarithm is a crecient function in $R^+$.

Step 2 : k is positive.

Step 3 : The exponential is a crecient function in $R$.


A Question based on series. Find the sum of$$\sum _{r=1}^\infty\left({\frac{2}{4r-3}-\frac{1}{2r}}\right)$$ I tried to solve this problem by taking $2$ common from the expression and got the result as of$$2\sum _{r=1}^\infty\left({\frac{1}{4r-3}-\frac{1}{4r}}\right)$$ but I could not get any constant term on which I could apply the telescopic series formula please help me out.

This is not a telescopic series, though it might superficially resemble one.

The common technique with such problems is to transform them to power series (by putting $x^n$ or something in the numerator), then fiddle around with functions, their Taylor series, maybe some differentiation or other stuff, and then evaluate the series at $x=1$.

Short of that, we are reduced to using the following crude tools.

Suppose you know that $\sum\limits_{n=1}^\infty{(-1)^{n+1}\over n}=\ln2$ (because if you don't, we aren't going anywhere). Suppose you also know that $\sum\limits_{n=1}^\infty{(-1)^{n+1}\over 2n+1}={\pi\over4}$, which is also a mandatory prerequisite. Then, obviously
$$\sum_{n=1}^\infty{(-1)^{n+1}\over 2n}={1\over2}\sum_{n=1}^\infty{(-1)^{n+1}\over n}={\ln2\over2}$$
Now I'm going to put it in a simple way, without all those fancy $\sum$s:
$$\begin{array}{ccccccccccc}
{\pi\over4} &=& 1 & & -{1\over3} & & +{1\over5} & & -{1\over7} &\dots \\
\ln 2 &=& 1 & -{1\over2} & +{1\over3} & -{1\over4} & +{1\over5} & -{1\over6} & +{1\over7} &\dots \\
{\ln 2\over2} &=& & \phantom{-}{1\over2} & & -{1\over4} & & +{1\over6} & &\dots \\
\end{array}$$
Now just add these three series together.

$$S_n=\sum _{r=1}^n \left(\frac{2}{4r-3}-\frac{1}{2r}\right)=\frac12\sum _{r=1}^n \left(\frac{1}{r+\frac14-1}-\frac{1}{r+1-1}\right)$$
It is a easy problem to people familiar with the polygamma functions.
$$\sum _{r=1}^n \frac{1}{r+a-1}=\psi(a+n)-\psi(a)$$
$\psi(x)$ is the digamma function.

Successively with $a=\frac14$ and $a=1$ :
$$S_n=\frac12\left(\psi\left(\frac14+n\right)-\psi\left(\frac14\right)-\psi\left(1+n\right)+\psi\left(1\right)\right)$$
$n\to\infty\quad \psi(a+n)\sim \ln(n)+O\left(\frac{1}{x}\right)$

$\lim_{n\to\infty}\left(\psi\left(\frac14+n\right)-\psi\left(1+n\right)\right)=0$
$$S_\infty=\frac12\left(-\psi\left(-\frac14\right)+\psi\left(1\right)\right)$$
$\psi(1)=-\gamma$ Euler-Mascheroni constant. And $\psi(1/4)=-\gamma-\frac{\pi}{2}-3\ln(2)$
$$S_\infty=\frac12\left(\frac{\pi}{2}+3\ln(2)\right)$$
$$\sum _{r=1}^\infty \left(\frac{2}{4r-3}-\frac{1}{2r}\right)=\frac{\pi}{4}+\frac32\ln(2)$$
For reference : http://mathworld.wolfram.com/DigammaFunction.html


Are the "proofs by contradiction" weaker than other proofs? I remember hearing several times the advice that, we should avoid using a proof by contradiction, if it is simple to convert to a direct proof or a proof by contrapositive.  Could you explain the reason? Do logicians think that proofs by contradiction are somewhat weaker than direct proofs?

Is there any reason that one would still continue looking for a direct proof of some theorem, although a proof by contradiction has already been found? I don't mean improvements in terms of elegance or exposition, I am asking about logical reasons.  For example, in the case of the "axiom of choice", there is obviously reason to look for a proof that does not use the axiom of choice.  Is there a similar case for proofs by contradiction?

To this MathOverflow question, I posted the following answer (and there are several other interesting answers there):

With good reason, we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal)

What is the reason? The reason is the fecundity of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r1, r2, before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r1 and r2 and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds.

Similarly, when we prove the contrapositive (¬q implies ¬p) directly, we assume ¬q, make intermediary conclusions  r1, r2, and then finally conclude ¬p. Thus, we have also established not only that ¬q implies ¬p, but also, that it implies r1 and r2 and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (¬r1 implies q), we learn about many different hypotheses that all imply q.

These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q).

With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue  r1, r2, and so on, before arriving at a contradiction. The statements r1 and r2 are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistant, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone.

I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof.

For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's theorem that there are infinitely many primes. In a common proof by contradiction, one assumes that p1, ..., pn are all the primes. It follows that since none of them divide the product-plus-one p1...pn+1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginner's falsely expect after this argument that whenever p1, ..., pn are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p1, ..., pn were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p1, ..., pn are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p1...pn+1 has merely a prime divisor not on the original list. (And Michael Hardy mentions that indeed Euclid had made the direct argument.)

At first this seems like a silly question - after all isn't the point of a mathematical proof to be a proof and hence to be beyond question. But of course, to prove anything we need assumptions and some people do disagree with the axioms commonly used by mathematicians. I don't have much knowledge of this view, but I am sure they have theorems that show that contradiction like proofs are valid (given their axioms) within certain conditions. I would recommend going along with what everyone else does and treating "proofs by contradiction" as equally valid, unless you have investigated the Constructivism view and you decide that they are correct.

As to whether they are clearer, that will depend on the actual proof. Sometimes the clearest way to make a proof is to start from the assumptions and see what they are really saying and why that is going to lead to a contradiction. The most illustrative proof depends on the circumstances.


Rees algebra of a monomial ideal User fbakhshi deleted the following question:

Let $R=K[x_1,\ldots,x_n]$ be a polynomial ring over a field $K$ and $I=(f_1,\ldots,f_q)$  a monomial ideal of $R$. If $f_i$ is homogeneous of degree $d\geq 1$ for all $i$, then prove that
$$
R[It]/\mathfrak m R[It]\simeq K[f_1t,\ldots, f_q t]\simeq K[f_1,\ldots,f_q] \text{ (as $K$-algebras).}
$$
$R[It]$ denotes the Rees algebra of $I$ and $\mathfrak m=(x_1,\ldots,x_n)$.

$R[It]/\mathfrak mR[It]$ is $\oplus_{n\geq 0}{I^n/\mathfrak mI^n}.$ Let $\phi$ be any homogeneous polynomial of degree $l$. Consider $I_l$ to be the $k$-vector space generated by all $\phi(f_1,\ldots,f_q).$ Then $k[f_1,\ldots,f_q]=\oplus_{l\geq 0}{I_l}.$ Now $\dim_{k}{I_l}=\dim_{k}{I^l/\mathfrak mI^l}.$ Hence $k[f_1,\ldots,f_q]\simeq {R[It]/\mathfrak mR[It]}.$


Intuition for why $f_{xy} = f_{yx}$ If we have a function $f(x,y)$, why is it that $f_{xy} = f_{yx}$? I'm looking for an intuitive, qualitative reason rather than a rigorous proof.

$f_{yx}$ represents the rate of change of the gradient parallel to the $x$ axis, as you move along the $y$ axis. Similarly, $f_{xy}$ represents the rate of change of the gradient parallel to the $y$ axis, as you move along the $x$ axis. At least, this is how I understand it. However, I can't see any reason why the two should be the same.

Intuition can't tell you why they're equal. For that it's too vague. But we can see that they measure the same thing.

Let's look at the origin specifically, just to make it easier. Also, let's say the function value and first derivatives at the origin are all $0$.

First we see what $f_{xy}$ (derivative first with respect to $x$, then with respect to $y$) measures. For each plane normal to the $y$-axis there is a line lying entirely in that plane which is tangent to the function graph for $x=0$. As we move along the $y$-axis, $f_x$ measures the slope of this line, and $f_{xy}$ measures the rate of rotation of this line. At the origin, our assumptions say that this line is the $x$-axis.

If you think enough about this, you will realize that an archetypal function with positive $f_{xy}(0,0)$  (something like $f(x,y)=xy$, specifically something with $f_{yy}=f_{xx}=0$) will, close to the origin, be positive in the first and third quadrants and negative in the second and fourth.

Now notice that this will, in the same interpretation, be exactly what makes $f_{yx}(0,0)$ positive as well.

It's up to you if you want to venture away from the land of $f=f_x=f_y=f_{xx}=f_{yy}=0$ and see what the result is. The difference is basically adding a function $g(x,y)=ax^2+by^2+cx+dy+e$ to $f$, which you can hopefully see doesn't change $f_{xy}$ and $f_{yx}$.

Well, once you've understood that it's true for $f(x,y)=xy,$ there's little room left to be surprised.


Use a particular method to prove if $a^n \mid b^n$ then $a\mid b$ A question in my tutorial asks to use the fact (can be used without proof) that

$(a,b)=(a,c)=1 \Rightarrow (a,bc)=1$

to prove:

if $a^n \mid b^n$ then $a\mid b$.

I did the following, but my tutor marked it as wrong. I have tried to find the error without success.

May I know where is the error? Sincere thanks.

Assume $a^n \mid b^n$.

$b^n=a^n\cdot k$ , $k\in \mathbb{Z}$

Let $d:=(a,b)$.

$(\frac{a}{d},\frac{b}{d})=1$

By the fact above, $(\frac{a}{d},(\frac{b}{d})^{n-1})=1$.

Note that since $(\frac{b}{d})^n=(\frac{a}{d})^n\cdot k$, so $\frac{a}{d}\mid(\frac{b}{d})(\frac{b}{d})^{n-1}$. This, in addition to $(\frac{a}{d},(\frac{b}{d})^{n-1})=1$, yields $\frac{a}{d}\mid\frac{b}{d}$. (Euclid's Lemma)

Hence $a\mid b$.

Your proof is correct, viz. by cancelling $\rm\:(a,b)\:$ one reduces to the case $\rm\:(a,b) = 1,\:$ which follows immediately by inductive application of Euclid's Lemma.

The proof is a special monic binomial case of the analogous proof of the Rational Root Test (RRT), since  for $\rm\:c = b/a\:$ we have $\rm\:c^n = k\in \Bbb Z,\:$ so $\rm\:c\:$ is a root of $\rm\:x^n - k\:$ so RRT implies $\rm\:c\in \Bbb Z.\:$

In fact, more generally, the Freshman's Dream for gcds is true $\rm\:(a^n,b^n) = (a,b)^n$

If $(a,bc)\neq 1$, then there is a prime number $p$ such that $p \mid a$ and $p \mid bc$.
But $p \mid bc \Rightarrow p \mid b$ or $p \mid c$. Then we have $p \mid a$ and $p \mid b, \Leftrightarrow (a,b) \neq 1$ or $p\mid a$ and $p\mid c, \Leftrightarrow (a,c) \neq 1$, contradiction.
Sorry I proved what we don't need.

If $a$ does not divide $b$. then there is $p$ prime such that $p \mid a$ but
$p$ does not divide $b$. Then $p^n \mid a^n$ but $p^n$ does not divide $a^n$. Then $a^n$ does not divide $b^n$. Contradiction.


What does "is natural in $A$" mean in this context? While reading Bredon's Topology and Geometry, I've come across the following claim:

  Naturality in $A$ of the sequence defining $\text{Ext}(A,G)$ shows that $\text{Ext}(A,G)$ is a contravariant functor of $A$.

I am confused. I know what is a natural transformation, and I understand, for example, that the map induced by the boundary map in singular homology is a natural transformation, because I know the category involved and the respective functors. However, I can't transfer my understanding of "natural transformation" to the above context. "Naturality in $A$" is a phrase which does not have a meaning to me, and the text doesn't make it clear.

So, my question is, what is Bredon meaning by this?

"The sequence defining $\text{Ext}(A,G)$" is the exact sequence below:

$$0 \to \text{Hom}(A,G) \to \text{Hom}(A,I) \to \text{Hom}(A,J) \to \text{Ext}(A,G) \to 0,$$

where $0 \to G \to I \to J \to 0$ is a injective resolution of $G$.

UPDATED: I have started a bounty on this question, and not accepted any of its answers, because I am not sure that "natural" is only being used as an intuitive meaning. Bredon uses the same term later when talking about the splitting on the Universal Coefficients Theorem and goes on to say why the split with respect to something cannot be natural.

The claim of naturality is usually done in the context of a morphism: one says that some morphism from an object constucted from $A$ to some other object constructed from $A$ is natural, and then it means precisely that both the domain and the codomain of the morphism depend functorially on $A$ —that is,they are the result of evaluating functors at $A$— and that the morphism is in fact the component at $A$ of a natural transformation.

In your particular case, you have the sequence $$0 \to \text{Hom}(A,G) \to \text{Hom}(A,I) \to \text{Hom}(A,J) \to \text{Ext}(A,G) \to 0$$ Each of the six abelian groups appearing there (incuding the zeroes) is the result of evaluating a certain functor at $A$, and the claim is that each of the five morphisms appearing in the sequence is natural in the sense I explained above.

Even more precisely, since $\operatorname{Ext}$ is not, at that point, define to be a functor (because you are still trying to define its action on maps), the claim is really that the sequence $$0 \to \text{Hom}(A,G) \to \text{Hom}(A,I) \to \text{Hom}(A,J)$$ is natural in $A$, so that this implies that the cokernel of the last arrow, which is defined to be $Ext$, is also a functor.

What the author might be saying is the following. He has defined the Ext functor as a cokernel of a map, and it is a general fact that if you have a commutative diagram of exact rows

A' ----> A ---> A'' ---> 0
|        |
B' ----> B ---> B'' ---> 0

There is a unique arrow $A''\longrightarrow B''$ making the diagram commutative so that "taking cokernels" is, in some way, functorial. If you now have two abelian groups $A_0,B_0$ (and a universal choice of injective resolutions), you can construct both exact sequences and use that hom is a functor to obtain a partial diagram as above. You then obtain a unique map that yields Ext a functor, too.


if $\|x+y\|=\|x\|+\|y\|$, then $\|\alpha x+\beta y\|=\alpha \|x\|+\beta \|y\|$ Let $X$ be a normed linear space. Assume that for $x,y \in X$, we have $||x+y||=||x||+||y||$. Show that $||\alpha x+\beta y||=\alpha ||x||+\beta ||y||$ for every $\alpha,\beta \geq 0$.

My attempt: Suppose $\beta \geq \alpha$. Then $\|\alpha x+\beta y\|=\|\alpha (x+y)-(\alpha-\beta)y\| \leq |\alpha|(\|x\|+\|y\|)+|\beta - \alpha|\|y\|=\alpha\|x\|+\beta\|y\|$.

Since the inequality is true for all $\alpha,\beta \geq 0$ and $\beta \geq \alpha$, if we let $\alpha=\beta=1$, we have $||x+y|| \leq |x||+||y||$. If the inequality is a strict inequality, then this contradicts with $\|x+y\|=\|x\|+\|y\|$. Hence, we must have equality, which is the desired result.

Is my proof correct?

EDIT: Suppose we have $\|x+y\|=\|x\|+\|y\|$ for some $x,y \in X$. For the same $x$ and $y$, prove that $\| \alpha x+ \beta y\|=\alpha \|x\|+\beta \|y\|$

W.l.o.g. $\alpha \ge \beta$.

$$ \alpha\|x+y\| = \|(\alpha-\beta)y + (\alpha x + \beta y) \| \le (\alpha-\beta)\|y\| + \|\alpha x + \beta y\| \le (\alpha-\beta)\|y\| + \alpha\|x\| + \beta \|y\| = \alpha(\|x\|+\|y\|) $$
Since the first-quantity = last-quantity, all the inequalities must be equalities.  In particular:
$$(\alpha-\beta)\|y\| + \|\alpha x + \beta y\| = \alpha(\|x\|+\|y\|) .$$

I believe this can be answered in a much more basic way.

If X is a normed linear space, given $x,y \in X$ we have that $\alpha x, \beta y \in X$, because $X$ is closed under scalar multiplication.

So if $||x+y|| = ||x|| + ||y||$ $\forall x,y \in X$ we must also have $||\alpha x + \beta y|| = ||\alpha x|| + ||\beta y||= |\alpha|||x|| + |\beta|||y||= \alpha||x|| + \beta||y||$

Since $\alpha$ and $\beta$ $\geq0$.


To what fractional Sobolev spaces does the step function belong? (Sobolev-Slobodeckij norm of step function) I'm new to fractional Sobolev spaces and I'm curious about the regularity of some simple functions like e.$\,$g. step functions in order to understand these spaces better.

In more detail, for $\Omega = [-1,1]^n \subseteq \mathbb{R^n}$ and $A = [-\frac{1}{2},\frac{1}{2}]^n\subseteq \Omega$ consider the function
$$
\begin{align}
f \colon \ \Omega & \longrightarrow \mathbb{R} \\
x & \longmapsto \begin{cases} 1 & \text{ for } x \in A \\ 0 & \text{ for } x \notin A\text{.}\end{cases}
\end{align}
$$
For which $s \in [0,1]$ does $f$ have a finite Sobolev-Slobodeckij norm? The norm that is meant here is defined by
$$
\Vert f\Vert_{s}^2 := \int_\Omega \int_\Omega \frac{\vert f(x) - f(y)\vert^2}{\Vert x-y\Vert^{2s+n}} \, \mathrm{d}x \, \mathrm{d}y\text{.}
$$

Is there a way to determine the value of the integral analytically in dependence of $n$ and $s$? Or can one at least easily determine those $s$ for which this integral would be finite? Can it at least be done for $n=2$?

So far I tried the simple case $n=1$ for which I get that $s\in [0,\frac{1}{2})$ has to be fulfilled. I expect that to be the case for any $n$ but at the moment I'm not quite sure since I did not prove it. For $n=2$ I would try to integrate by hand but with my approach it's about to become a rather long calculation. Is there maybe an elegant way to do it? I don't mind if $A$ is replaced by another set like for example a scaled $n$-Sphere or some simplex.

Guessing

First I would make a quick guess based on the chart of function space. It groups together Sobolev spaces $W^{s,p}$ with the same value of $\frac n p -s$, because these are related by the embedding theorem. While the inclusion provided by this theorem is strict, the sharpness of the theorem still makes "the spaces with equal $\frac n p -s$ are similar" a useful heuristic.

Your function $f$ fails to be in $W^{1,1}$, because its gradient is not an $L^1$ function but rather a vector-valued measure supported on the boundary of $A$. On the other hand, this measure has finite mass (meaning $f\in BV$), which is pretty close to $W^{1,1}$. So, it seems that $s=1$, $p=1$ is at the edge of spaces to which $f$ belongs. From
$$\frac{n}{1}-1 = \frac{n}{2} - s$$ we conclude that for $p=2$ (your question), the relevant $s$ is $s=1-\frac n2$.

This agrees with the situation in 1D, and suggests we won't find anything good when $n\ge 2$. Of course, this is just a guess; it may well be wrong.

Estimating

Recall the layercake principle: the integral of a nonnegative function $g$ is equal to $\int_0^\infty |\{g>t\}|\,dt$ where $|\cdot|$ stands for the measure of the set. So, let's consider the inequality
$$
\frac{|f(x)-f(y)|}{\|x-y\|^{2s+n}}>t \tag{1}
$$
for large values of $t$ (only they are of interest when the measure space has finite measure).

For (1) to hold, exactly one of $x,y$ must be in $A$; also, both must be within distance $\delta  \approx t^{-1/( 2s+n)}$ of the boundary of $A$. This constrains  $x$ to a set of measure $\approx \delta $. Also, $y$ must lie in a ball of radius $\delta$ around $x$. So,
$$
\left|\left\{ (x,y) : \frac{|f(x)-f(y)}{\|x-y\|^{2s+n}}>t\right\}\right| \approx \delta^2 \approx t^{-(n+1)/(2s+n)}
$$
The integral over $t\ge 1$ converges iff
$$\frac{ n+1}{2s+n} >1$$
which is equivalent to $s<1/2$.

(In particular, the "guess" wasn't correct after all.)

The function belongs to $W^{s,1}$ and BV space as in paper "The fractional Cheeger problem".


Are all the points in a nonempty open set limit points? My conjecture is that given any open set $A\subseteq\mathbb{R}$, all points $a\in A$ are limit points.

Prove, or if untrue, disprove by constructing a counterexample.

A few definitions for clarification:

A set $A$ is open iff for every point $a\in A$, there exists a $\delta$ such that the neighborhood $\left(a-\delta,a+\delta\right)$ surrounding $a$ is completely contained within (is a subset of) $A$.
A point $x$ is a limit point of a set $A$ iff given any $\varepsilon$, all neighborhoods $\left(x-\varepsilon,x+\varepsilon\right)$ intersect the set $A$ in some point other than $x$. Note that $x$ need not be a point in $A$.

Let $a$ be a point in $A$. Since $A$ is open, there is a number $\delta > 0$ such that $(a - \delta, a + \delta) \subset A$.

We now show $a$ is a limit point of $A$. To that end, let $\epsilon > 0$ be given. We may assume, without loss of generality, that $\epsilon < \delta$. Now, $(a - \epsilon, a + \epsilon) \subset (a - \delta, a + \delta) \subset A$, so $(a - \epsilon, a + \epsilon) \cap A = (a - \epsilon, a + \epsilon)$. Since $(a - \epsilon, a + \epsilon)$ intersects $A$ in a point other than $a$, we conclude $a$ is a limit point.

To say that all points $a\in A$ are limit points is to say that there is no point $b\in A$ that is an isolated point.

To prove this, we will use a method called proof by contradiction, in which we assume the negation and see if we can arrive at a nonsensical conclusion. If we can arrive at a contradiction, then our assumption—namely, the negation of the conjecture—must be false. This will prove that the conjecture itself is true.

Let's assume to the contrary that an open set $A$ does indeed contain an isolated point $b$.

If $A$ is open, then every point in $A$, including $b$, must have some neighborhood that is a subset of $A$. This means that there must exist some $\delta$ such that every point within the neighborhood $\left(b-\delta,b+\delta\right)$ must be a member of the set $A$. Since the interval is open, let there be a point $d\in\left(b-\delta,b+\delta\right)$ such that $d\ne b$. Then $d\in A$.

Now, if $b$ is an isolated point, then it is not a limit point by definition. Thus we must negate the definition of limit point. If $b$ is an isolated point, there must exist some $\varepsilon$ such that the neighborhood $\left(b-\varepsilon,b+\varepsilon\right)$ intersects the set $A$ only at the point $b$. Since the interval is open, let there be a point $e\in\left(b-\varepsilon,b+\varepsilon\right)$ such that $e\ne b$. Then $e\not\in A$.

For simplicity, assign the following symbols:

$$V_\delta(b) = \left(b-\delta,b+\delta\right)$$
$$V_\varepsilon(b) = \left(b-\varepsilon,b+\varepsilon\right)$$

There are two possibilities:

$$V_\delta(b) \subseteq V_\varepsilon(b)$$

Then for all $d \ne b$, we have $d\in V_\delta(b) \implies d\in V_\varepsilon(b)$. From the conclusions above, this implies $d\in A \implies d\not\in A$. This is a contradiction.
$$V_\delta(b) \supseteq V_\varepsilon(b)$$

This is the converse of Case 1. We still arrive at a contradiction.

Since both cases lead to a contradiction, our assumption that $A$ is open and contains an isolated point $b$ is faulty. Thus there can be no such possibility. In conclusion, any open set $A$ must not contain any isolated points. In other words, all the points in a (nonempty) open set are limit points. q.e.d.

A few notes:

Just because a nonempty open set contains only limit points does not mean that the set contains all its limit points. For example, all points in the open interval $(0,1)$ are limit points, but this set does not contain the limit points $0$ and $1$.
The converse of your conjecture states, "If every point in a set is a limit point, then that set is open." This is false. The closed interval $[0,1]$ provides a counterexample: every point within this interval is a limit point, but it is not an open set.


Epsilon-Delta Proof of Continuity for $\sin(x^2)$ I have the function $f(x)=\sin(x^2)$, and I'm trying to prove that it is continuous on $(1,2)$.  I figured using the epsilon-delta definition would be the way to go.  So I have:

$\epsilon > 0$, and if $|x-a|<\delta$, then
$|\sin(x^2)-\sin(a^2)|<\epsilon$

I think I should use the triangle inequality in there somewhere, but I'm not sure how.

Thanks for your time

It would probably be difficult to solve this directly from the $\epsilon - \delta$ definition. A better strategy would be to use the following theorem:

let $g$ be a function continuous at $x_0$ and $f$ be a function continuous at $g(x_0)$. then their composition $f \circ g$ is continuous at $x_0$.


Geometry question on right triangle Let ABC be a triangle in which $AB = AC$ and $\measuredangle CAB = 90^{\circ}$. Suppose $M$ and $N$ are points on the hypotenuse $BC$ such that $BM^2 + CN^2 = MN^2$. How do I  prove that $\measuredangle MAN = 45^{\circ}$.

Let $\angle BAM=\alpha, \angle CAN=\beta, BM=x\sqrt{2},CN=y\sqrt{2}$.  We want to check that $\tan(\alpha+\beta)=1$.
The given formula is that $x^2+y^2=(1-x-y)^2$, or $1-2x-2y+2xy=0$.
$$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\
=\frac{\frac x{1-x}+\frac y{1-y}}{1-\frac{xy}{(1-x)(1-y)}}\\
=\frac{x-xy+y-xy}{1-x-y}=\frac{1-x-y}{1-x-y}$$

Without loss of generality, assume $BC=1$. Let $BM=x$ and $CN=y$, so $MN=1-x-y$. Then it is given that $x^2+y^2=(1-x-y)^2$. However, there is an infinite number of possibile pairs $(x,y)$ with $0<x,y<\frac{1}{2}$ such that $x^2+y^2=(1-x-y)^2$. Only 1 choice of $(x,y)$ leads to $\angle MAN=45$ degrees. In other words, you can not prove it.