TITLE: Solve system of diff equations using laplace transform and evaluate x(1)
QUESTION [1 upvotes]: I keep getting the wrong answer, and wolphram seems to back me up.
Here's the system of equations

The answer I get for $x(1)$ is 10492.1... The supposedly right answer is -1426.16
Can anyone try this problem and tell me what they find out?
Thanks a lot.
EDIT:
Using Cramer's Rule I got
$X(s)=\frac{10}{(s-9)(s-3)}-\frac{20}{s(s-9)(s-3)}$
Using partial fractions decomposition I found the inverse laplace transform and got:
$(10/6-20/54)e^{9t}+(-10/6+20/18)e^{3t}-20/27$

REPLY [1 votes]: You're right, the answer's wrong. Let's just solve for $Y(s)$ from the second equation, $$Y(s) = -\frac{X(s)}{s-2}.$$ Subbing this into the first equation gives, after rearranging,
$$
X(s) = \frac{10s-20}{s(s-3)(s-9)},
$$
the partial fraction decomposition of which is
$$
X(s) = \frac{35}{27(s-9)} - \frac{5}{9(s-3)} - \frac{20}{27s}.
$$
Note that we could sub this into our first line to get $Y(s)$. Inverting the transform,
$$
x(t) = \frac{35}{27}\mathrm{e}^{9t} - \frac{5}{9}\mathrm{e}^{3t} - \frac{20}{27},
$$
so $x(1) \approx 10492.098$.


TITLE: Distance between two ordered sets
QUESTION [3 upvotes]: Is there a way to measure the "distance" between two ordered sets?
Say i got two sequences of letters:
$$
S_1  \{A, B, C, D, E, F\}
$$
$$
S_2  \{B, C, D, A, F, E\}
$$
How could I find an "amount of difference" between $S_1$ and $S_2$?
Is there some generic metric on a space of ordered sets?

REPLY [2 votes]: There are different ways to do that. You could use euclidean distance (or any other metric applicable to vector spaces) if the alphabet is in a metric space. As a special case if you use $L^1$ distance and your alphabet is ${0, 1}$ with usual metric you will get the hamming distance.
Since you're talking permutation distance, then yes it's a metric too given that you use the least number of exchanges to go from one permutation to the next. In your example I'd guess that the distance would be $4$ (swap A-B, A-C, A-D and then E-F, my guess is that this is the least amount of swaps you could do to get to $S_2$). It's obvious that the axioms of metric is fulfilled:

Since both $a$ and $b$ can be expressed as a number of swaps from the identity map ($I$) and therefore theres a way from $I$ to $a$ and $I$ to $b$ (and reversing that makes a way from $b$ to $I$) you have a way from $b$ to $a$ via $I$ - at least one way makes for a shortest way. That is for each $b$ and $a$ theres a $d(b,a)$ being the lowest number of swaps to get from $b$ to $a$.
$d(a,b) \ge 0$ (you can't swap a negative number of times so the minimum of swaps has to be non-negative)
$d(a,b) = 0 \Rightarrow a=b$ (if you can get from $a$ to $b$ without swaps then they have to be the same)
$d(a,b) = d(b,a)$ (you can get from $b$ to $a$ with the same swaps in reverse order as those that take you from $a$ to $b$, that is the same number of steps - the minimum of those has to be the same)
$d(a,b) + d(b, c) \ge d(a,c)$ (you can get from $a$ to $c$ by using the $d(a,b)$ steps from $a$ to $b$ and the $d(b,c)$ steps from $b$ to $c$, the minimum number of steps from $a$ to $c$ can't be greater than that.


TITLE: Find a subgroup of order $12$
QUESTION [3 upvotes]: Suppose that $H$ is a normal subgroup of $G$. If $|H|=4$ and for $g \in G$, $gH$ has order $3$ in $G/H$, find a subgroup of order $12$ in $G$.
My Approach: $|g| \in \{3,6,12\}$.
This is only thing i have calculated

REPLY [1 votes]: A simple way to look at things you wrote is following.

Take subgroup $K$ generated by $H$ and $g$; this clearly makes sense.

This contains $H$, and $H$ is normal in it (why?).
Further, $K/H$ is generated by just $gH$, which has order $3$, and so $K/H$ is of order $3$.
From the order of normal subgroup and order of quotient group, you get order of $K$, what it is?


TITLE: Use the $\epsilon$-$\delta$ definition to prove the following limit
QUESTION [0 upvotes]: Here is the question I am trying to solve:
$\lim_{x\rightarrow0}\frac{x^2+1}{x+1}=1$
Here is what I have so far:
Proof
Let $\epsilon>0$. We will show given $\delta\geq0$, we have $0<|x|<\delta$ implies that
$$\lvert\frac{x^2+1}{x+1}-1\lvert  =  \lvert\frac{x^2+1-x-1}{x+1}\lvert  = \lvert\frac{x^2-x}{x+1}\lvert = \frac{|x||x-1|}{|x+1|}$$
But this is where I get stuck. I am unsure what to do to make sure the denominator is bounded away from $0$. Any help welcome!

REPLY [0 votes]: Let $\epsilon >0$; we are searching for a $δ>0$, such that $x\in (-δ,δ)\setminus \{0\}\Rightarrow |\frac {x^2+1}{x+1}-1|<\epsilon\iff |\frac {x^2-x}{x+1}|<\epsilon\iff \frac {|x||x-1|}{|x+1|}<\epsilon$ (1). But, for $δ\in (0,\frac 12]$, $x\in (-δ,δ)\setminus \{0\}$ gives $x\in (-\frac 12,\frac 12)\setminus \{0\}$, so $x-1\in (-\frac 32,-\frac 12)\setminus \{-1\}\Rightarrow \frac 12<|x-1|<\frac 32, x\neq 0$ and $x+1\in (\frac 12,\frac 32)\setminus \{1\}\Rightarrow \frac 12<|x+1|<\frac 32, x\neq 0$. Inspired from (1) we demand $\frac {|x||x-1|}{|x+1|}<\frac {δ\frac 32}{\frac 23}\leq\epsilon\Rightarrow 0<δ\leq \frac 49 \epsilon$. By choosing $0< δ\leq min\{\frac 12, \frac 49\epsilon\}$ we have $x\in (-δ,δ)\setminus \{0\}\Rightarrow |\frac {x^2+1}{x+1}-1|=\frac {|x||x-1|}{|x+1|}<\frac {δ\frac 32}{\frac 23}\leq \frac {\frac 49 \epsilon \frac 32}{\frac 23}=\epsilon$.   q.e.d.


TITLE: Determine which sequences are equicontinuous.
QUESTION [4 upvotes]: I'd appreciate if somebody could check if my proofs below are correct and also give me some hints on the equicontinuity of $(k_n)$ (part c)). $\mathbb{R}$ represents the real line with the standard topology.
Consider the following sequences of functions in $\mathcal{C}(\mathbb{R}, \mathbb{R})$, (the set of continuous functions $f : \mathbb{R} \to \mathbb{R}$):
a) $f_n (x) = n + \sin(x)$
b) $h_n (x) = |x|^{1/n}$
c) $k_n (x) = n\sin(x/n)$
where $n\ge 1$.
Question
Which sequences are equicontinuous / pointwise bounded?
Answer
a) $(f_n)$ is not pointwise bounded, since given any $x, |n + \sin(x)| \to \infty$. For equicontinuity at a point $x$, we need to show that given any $\epsilon > 0$, there is an open neighborhood $U$ of $x$ such that
$$ \forall n\ge 1, \forall x^* \in U: |f_n(x^*) - f_n(x)| < \epsilon $$
Since $x \mapsto \sin(x)$ is continuous, let $U$ be an open set containing $x$ such that
$$x^* \in U \implies |\sin(x^*) - \sin(x)| < \epsilon$$
Since $|f_n(x^*) - f_n(x)| = |\sin(x^*) - \sin(x)|$ for all $n$, we see that $U$ satisfies the required condition. Since $x$ was arbitrary, we conclude that the sequence $(f_n)$ is equicontinuous.

b) $(h_n)$ is indeed pointwise bounded, since for any given $x$, $|x|^{1/n} \to 1$. I claim that it is not equicontinuous, since it is not equicontinuous at $x=0$: Assume it is equi. cont. at $x=0$. Pick $\epsilon = 1/2$. Then there exists $(-\delta, \delta)$ for some $\delta > 0$ such that
$$ \forall x^* \in (-\delta, \delta) \land \forall n\ge 1: |x^*|^{1/n} < 1/2$$
But this is not possible, since $|x^*|^{1/n} \to 1$.

c) $(k_n)$ is clearly not pointwise bounded. For equicontinuity, I'm a bit lost. Any tips?

REPLY [2 votes]: I agree with your answers to 1) and 2).
Regarding 3), $k_n$ is pointwise bounded as for any $x \in \mathbb R$ you have
$$\left\lvert k_n(x) \right\rvert = \left\lvert n \sin \frac{x}{n} \right\rvert \le n  \left\lvert \frac{x}{n} \right\rvert = \lvert x \rvert.$$ Moreover, $k_n$ is equicontinuous as for any $x,y \in \mathbb R$ you have the inequality (that can be proven using mean value theorem)
$$\lvert \sin x - \sin y \rvert \le \lvert x - y \rvert$$ and therefore
$$\lvert k_n(x) - k_n(y) \rvert \le \lvert x- y \rvert.$$ So, $k_n$ is even uniformly equicontinuous.


TITLE: Sum of $1^2+3^2+\cdots+(2n+1)^2$ Have trouble with proof.
QUESTION [2 upvotes]: I've been working through a question on Courant's What is Mathematics?
This is the question: Prove $1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$. I called this $S_{(2n+1)^2}$.
What I've been given as a hint is that I can use the sum $1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ (I called this $S_{n^2}$) in some way.
So here's my line of thought:
What do I need to do to $S_{n^2}=1^2+2^2+3^2+\cdots+n^2$ to get $S_{(2n+1)^2}=1^2+3^2+\cdots+(2n+1)^2$?
When I compared them, I thought that $S_{n^2}-(2^2+4^2+\cdots+(2n)^2) = S_{(2n+1)^2}$.
I defined $2^2+4^2+\cdots+(2n)^2$ as $S_{(2n)^2}$.
After a bit of head-scratching I managed to figure out $S_{(2n)^2}=2^2(0)+2^2(1)^2+2^2(2)^2+\cdots+2^2(n)^2=4(1^2+2^2+3^2+\cdots+n^2)=4S_{n^2} =\frac{2}{3}n(n+1)(2n+1)$
Now is where my confusion starts. So from my logic above I went forward with $S_{n^2}-S_{(2n)^2}=S_{(2n+1)^2}$ which gave me $-\frac{1}{2}(n(n+1)(2n+1))$ which isn't $S_{(2n+1)^2}$.
I don't know where I really went wrong. One of my main suspicions is that $S_{n^2}-S_{(2n)^2}=S_{n^2}-4S_{n^2}=-3S_{n^2}\neq S_{(2n+1)^2}$. But I don't see why.

REPLY [0 votes]: Using your hint, here is one way to go:
$$1^2 + 3^2 +\cdots +(2n+1)^2 = (1^2 + 2^2 + 3^2 + 4^2+\cdots +(2n+1)^2) - (2^2 + 4^2 + 6^2+\cdots+ (2n)^2)\\
(1^2 + 2^2 + 3^2 + 4^2+\cdots+ (2n+1)^2) - 4(1^2 + 2^2 + 3^2+\cdots+ n^2)$$
$$\frac{(2n+1)(2n+2)(4n+3)}{6} - \frac{4(n)(n+1)(2n+1)}{6}$$
Alternatively, you could use a proof by induction.
Base case:
$$n = 1\\
1^2 + 3^2 = \frac {1\cdot2\cdot3\cdot5}{3} = 10$$
Suppose:
$$1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$$
We will show that:
$$1^2+3^2+\cdots+(2n+1)^2 + (2n+3)^2=\frac{(n+1)(n+2)(2n+3)(2n+5)}{3}$$
By our hypothesis:
$$\frac{n(n+1)(2n+1)(2n+3)}{3} + (2n+3)^2=\frac{(n+1)(n+2)(2n+3)(2n+5)}{3}$$
And simplify.


TITLE: The Direct Product of a finite number of cyclic groups, is again a cyclic group?
QUESTION [1 upvotes]: Can anyone please brief me on this question, do you prove or disprove? If it is to be disproved can you give a counterexample? if there is no generator, then what is an example of such a Direct Product of a finite number of cyclic groups?

REPLY [2 votes]: Recall the Klein 4-group: This is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$, the direct product of the cyclic group $\mathbb{Z}_2$ with itself.
If you recall, although this group is abelian, it is not cyclic: there is no one element that generates all of $\mathbb{Z}_2 \times \mathbb{Z}_2$. It has three proper subgroups of order $2$ (in which the single non-identity element is its own inverse) and the trivial group, which you know contains only the identity.
Any direct product of $\mathbb{Z}_m$ and $\mathbb{Z}_n$ is non-cyclic, UNLESS $m, n$ are coprime: UNLESS, $\gcd(m,n) = 1$.  When $\gcd(m, n) = 1$, then $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn}$, which IS cyclic.


TITLE: Help with interpreting the physical meaning of the gradient
QUESTION [0 upvotes]: I understand that the physical meaning of the gradient is a vector in the direction of maximum increase with the magnitude equivalent to the maximum rate of change. However, I'm simply not seeing it. Is there a good way of visualizing this?
Here is my thinking, which I am visualizing to be a counter example to the interpretation of the gradient. I'm not sure on how to graph this, so I'll explain in words. Assume a function T = x + y, where it is constant in the Z dimension (which allows me to put the T function on the z-axis). At the origin, the function increases equally and the most in both the x direction and the y direction, but along the line of y = x, the function decreases the most.
The gradient of this function at the origin, is the vector addition of the increase in the x direction and increase in the y direction. How then is the vector pointing in $T = <x, y>$ which is along the line of y = x, equivalent to the direction of maximum increase?

REPLY [0 votes]: Here's one way to look at it, if you accept directional derivatives (also has a nice illustration of the gradient).
Suppose we are interested in the direction of greatest increase of some $f(x,y)$. Then, by definition we should look at the rate of increase of $f$ in the direction $v$ via its directional derivative:
$$ R_f(v) = \nabla f\cdot \hat{v} $$
which is the rate of change of $f$ in the direction $v$. Intuitively, this is because you are projecting the total change (i.e. the gradient) of $f$ onto the direction vector $v$. It might be obvious that $R_f(v)$ is maximal when $v=\nabla f$, but you can easily see this by letting $$ \hat{v} = (\cos(\theta),\sin(\theta)) = \frac{(v_1,v_2)}{||(v_1,v_2)||_2} = \frac{v}{||v||_2}$$
so that (in 2D) $R_f(v) = \cos(\theta)\partial_xf + \sin(\theta)\partial_yf$. Notice that $\theta$ is simply the angle with the $x$ axis in the plane, so it determines the direction fully. Let's find the maximal $\theta$. Take $\partial_\theta R_f(v) = 0$ and you'll find $\theta=\arctan(\partial_y f/\partial_x f)$, which means $v = (v_1,v_2) = (\partial_x f, \partial_y f)=\nabla f$, as we expected. I.e. $ \max_v R_f(v) = (\nabla f\cdot \nabla f)/||\nabla f||_2 = ||\nabla f||_2 $. So the direction of maximal $R_f$ is $\nabla f$.

Ok, well for your specific example, $T(x,y)=x+y$, so $\nabla T = (1,1)$.
How can we see that walking along the gradient gives the greatest increase?
Let's just take a unit step (from the origin) in a few different directions to see why. Note that $T(0,0)=0$
First, in the $x$ direction, we take a step $s=(1,0)$, so $T(s)=1$. So we took a step of size $1$, and $T$ increased by 1.
Let's try in the $y$ direction: $s=(0,1)$, thus $T(s)=1$, so same as last time.
What about $s=(1,1)/\sqrt{2}$, which is a unit step in the gradient direction?
Well, $T(s)=\sqrt{2} > 1$. So even though we took a step of size $1$, our function grew about ${\sim}{1.4}$, rather than just by $1$, as it did in the other directions.
What about something in between? Say, $s=(0.921,0.395)$, so then $T(s)=1.316$. So, taking a step of size $1$ led to an increase of ${\sim}1.3$. This is between the other values, as we would kind of expect.


TITLE: Normal distribution - Probability
QUESTION [0 upvotes]: The post office has to transport a large number of letters and packets daily.
The number of letters is distributed normally with $\mu_1=7300$ ans $\sigma_1=500$.
For the independent number of packets we consider the normal distribution with $\mu_2=450$ and $\sigma_2=90$.
I want to calculate the probability that in a day at most $6500$ letters will be delivered.
I also want to calculate the probability that together more than $7500$ letters and packets have to be delivered in one day.
$$$$
I have done the following:
Let $X$ describe the number of letters and $Y$ the number of packets.
For the first one:
The probability is equal to $$P(X\leq 6500)=\Phi \left (\frac{6500-\mu_1}{\sigma_1}\right )=\Phi \left (\frac{6500-7300}{500}\right )=\Phi \left (\frac{-800}{500}\right )=\Phi \left (-1,6\right )=0.05480$$
Is this correct?
For the second one:
Do we want to calculate the probability $P(X+Y>7500)$ ? If yes, how could we do that?

REPLY [1 votes]: Since $X$ and $Y$ are independent normals, their sum is also normally distributed. In particular $X+Y$ is normal with mean
$$
E(X+Y)=EX+EY=7300+450=7750
$$
and variance
$$
\text{Var}(X+Y)=\text{Var}{X}+\text{Var}{Y}=500^2+90
^2$$
as $X$ and $Y$ are independent. With this knowledge you can do a similar computation to compute $P(X+Y>7500)$ as you did for the first question.
For a proof see this wikipedia article.


TITLE: Prove that if $G=L\rtimes_\theta H_1=L\rtimes_\theta H_2$ (inner semidirect products) then $H_1 \cong H_2$
QUESTION [1 upvotes]: Given the following equality between two inner semidrect products: $$G=L\rtimes_\theta H_1=L\rtimes_\theta H_2$$ ($L$ is normal in $G$ while $H$ is a subgroup in $G$)
I want to show that there is an isomorphism between $H_1$ and $H_2$. I'm not sure how to show this. I thought of mapping $1h \in H_1$ to $(1,h)$ and was hoping to justify why this can be mapped to $(1,h')$ (and then, in turn to $1h' \in H_2$), but I'm not seeing how to justify this step.
What am I missing?

REPLY [5 votes]: $G$ is the semidirect product of $L$ and $H_{i} $ iff   the ses $1\to L\to G\to H_{i}\to 1 $ splits, So, $H_{1} \cong G/L \cong H_{2}$


text:
\begin{document}

\title[Solution to a pair of matrix equations]{On new existence of a unique common solution to a pair of non-linear matrix equations}

\author[H.\ Garai, L.K. \ Dey, W. Sintunavarat,  S. \ Som, S. \ Raha]
{Hiranmoy Garai$^{1}$, Lakshmi Kanta Dey$^{2}$, Wutiphol Sintunavarat$^{3}$,  Sumit Som$^{4}$, Sayandeepa Raha$^{5}$}

\thanks{Corresponding author: W. Sintunavarat (E-mail: wutiphol@mathstat.sci.tu.ac.th)}
\address{{$^{1}$\,} Hiranmoy Garai,
                    Department of Mathematics,
                    National Institute of Technology
                    Durgapur, India.}
                    \email{hiran.garai24@gmail.com}
\address{{$^{2}$\,} Lakshmi Kanta Dey,
                    Department of Mathematics,
                    National Institute of Technology
                    Durgapur, India.}
                    \email{lakshmikdey@yahoo.co.in}
\address{{$^{3}$\,} Wutiphol Sintunavarat, Department of Mathematics and Statistics,    					        Faculty of Science and Technology,
                    Thammasat University Rangsit Center, Pathum Thani 12120, Thailand.}
                    \email{wutiphol@mathstat.sci.tu.ac.th}
\address{{$^{4}$\,} Sumit Som,
                    Department of Mathematics,
                    National Institute of Technology
                    Durgapur, India}
                    \email{somkakdwip@gmail.com}
\address{{$^{5}$\,} Sayandeepa Raha,
                    Department of Biotechnology,
                    National Institute of Technology
                    Durgapur, India}
                    \email{rahasayandeepa@gmail.com}
\subjclass{$47H10$, $54H25$.}
\keywords{Common fixed point, non-linear matrix equation, Thompson metric.}

\begin{abstract}
The main goal of this article is to study the existence of a unique positive definite common solution to a pair of matrix equations of the form \begin{eqnarray*}
X^r=Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i \mbox{ and } X^s=Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(X)A_i
\end{eqnarray*}
where $Q_1,Q_2\in P(n)$, $A_i\in M(n)$ and $F,G:P(n)\to P(n)$ are certain functions and $r,s>1$. In order to achieve our target, we take the help of elegant properties of Thompson metric on the set of all $n \times n$ Hermitian positive definite matrices. To proceed this, we first derive a common fixed point result for a pair of mappings utilizing a certain class of control functions in a metric space. Then, we obtain some sufficient conditions to assure a unique positive definite common solution to the said  equations. Finally, to validate  our results, we  provide a couple of numerical examples with diagrammatic representations of  the convergence behaviour of iterative sequences.
\end{abstract}

\maketitle

\setcounter{page}{1}

\section{Introduction}

The metrical fixed point theory is one of essential and important tools for solving various types equations arising in different mathematical problems.
Throughout the last hundred years, many researchers have studied metrical fixed point theory by investigating several interesting existence (and uniqueness) fixed point results for different types of contractions, and they have utilized those results for solving different equations arising in  optimization theory, approximation theory, learning theory,  variational inequality and many more.
Based on the  impact of several real-world problems, the metrical fixed point theory becomes one of the foremost subject in modern research.
The topic of common fixed points for a pair or a family of mappings in metric  spaces is of great interest and play important roles in many applications. Throughout the last decade, many mathematicians develop the study of common fixed points by establishing several sufficient conditions involving compatibility, weak compatibility, commutativity, continuity, the E.A. property, the common limit in the range property and numerous others for existence (and uniqueness) of common fixed point of pair and family of mappings. Motivated by some important and interesting fixed point and common fixed point results, we enrich the common fixed point theory in terms of a control function in Section 2 of this  work.

On the another side, there is a huge application of matrix equations in various types  of problems in engineering, computer sciences, stability analysis and many more,  (see \cite{W1,TW} and the references therein). Throughout the last few years, many researchers have achieved different kinds of sufficient conditions for ensuring general solution(s) of different types of matrix equations, see \cite {RR,ER,BS,  HBP, HBK, SS,SS1, SS2}. Besides these, very recently, Garai and Dey \cite{GD} noticed that there are some problems in non-interacting control theory, where two different matrix equations arise and these types of problems require common solution to certain pair of matrix equations.  Garai and Dey \cite{GD} were the first to handle this situation. More precisely, they obtained some adequate conditions to ensure the existence of unique common positive definite solution to the following equations
\begin{equation}\label{me1}
X=Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i,
\end{equation}
\begin{equation}\label{me01}
X=Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(X)A_i,
\end{equation}
where $Q_1\ Q_2\in P(n)$ (set of all $n\times n$ Hermitian positive definite matrices), $A_i\in M(n)$ (set of all $n \times n$ matrices), $A_i^*$ denotes the conjugate transpose of $A_i$, $F, G$ are mappings from $H(n)$ (set of all $n \times n$ Hermitian matrices) into itself. But the above type of equations are particular case of the general matrix equation
\begin{equation}\label{me001}
X^r=Q + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i,
\end{equation}
where $Q,\ A_i, F$ are as above and $r\geq 1$. But, in reality, most of the time we have to handle the general matrix equations.

Motivated by the above scenario, our major goal of the present paper is to  focus on two pairs of matrix equations. The first pair is as follows:
\begin{align}
X^s=Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i \label{add01},\\
X^s=Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(X)A_i \label{add02}
\end{align}
where $Q_1,~Q_2\in P(n)$, $A_i's$ are arbitrary nonsingular complex matrices of order $n$, $F,~G$ are two functions from $P(n)$ to $P(n)$ and $s>1$ is a real number, meanwhile, the second pair is as follows:
\begin{align}
X^r=\displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i, \label{add03}\\
X^s=\displaystyle \sum_{i=1}^{m} {A_i}^*G(X)A_i \label{add04}
\end{align}
where $F,G$ are same as the first system, $A_i$'s are arbitrary $n \times n$ unitary matrices, and $r,s>1$ are real numbers.

In order to solve the two mentioned systems, we first obtain a common fixed point result using  a special type of family of functions in complete metric spaces. Based on this common fixed point result, we present some adequate conditions for the existence of unique common positive definite solution of both the systems. Finally, we authenticate our obtained sufficient conditions by presenting some numerical examples along with clear diagrammatic representations of  the convergence behaviour of iterative sequences.

\section{The common fixed point results}
In this section, we present a new common fixed point result for a pair of mappings.  Before investigating such a common fixed point result, first, we introduce a collection of some control functions.

For a fixed real number $\alpha \in [0,1)$, we use the symbol $\Psi_{\alpha}$ to denote the collection of all mappings $\psi:\mathbb{R_+}^3 \to[0,\infty)$ satisfying the following conditions:
\begin{enumerate}
\item[$(\psi')$] $\psi$ is continuous;
\item[$(\psi'')$] if $b \leq \psi(a,a,b)$ or $b \leq \psi(b,a,a)$ or $b \leq \psi(a,b,a)$, then $b\leq \alpha a$.
\end{enumerate}
Some examples of mappings belonging to the class $\Psi_{\alpha}$ are given by the following:
\begin{enumerate}
\item[$(i)$] a mapping $\psi:\mathbb{R_+}^3 \to[0,\infty)$ which is defined by
$$\psi(a,b,c)=\alpha a$$
for all $a,,b,c\in \mathbb{R_+}$, where $\alpha \in [0,1)$;
\item[$(ii)$] a mapping $\psi:\mathbb{R_+}^3 \to[0,\infty)$ which is defined by
$$\psi(a,b,c)=Ma+Nb+Oc$$
for all $a,b,c \in \mathbb{R_+}$, where $M,N,O$ are nonnegative real numbers with $M+N+O<1$;
\item[$(iii)$] a mapping $\psi:\mathbb{R_+}^3 \to[0,\infty)$ which is defined by
$$\psi(a,b,c)=\alpha \max\{a,b,c\}$$
for all $a,b,c \in \mathbb{R_+}$, where $\alpha \in [0,1)$.
\end{enumerate}

\begin{theorem}\label{th1}
Let $(C,d)$ be a complete metric space and $T_1,T_2:C\to C$ be two mappings satisfying the following condition:
\begin{equation}\label{e1}
d(T_1(x),T_2(y)) \leq \psi\left(d(x,y),d(x,T_1(x)),d(y,T_2(y))\right)
\end{equation}
for all $x,y \in C$, where $\psi \in \Psi_{\alpha}$. Then the following assertions hold:
\begin{enumerate}
\item[$(i)$] $T_1$ and $T_2$ have a unique common fixed point;
\item[$(ii)$] for any $u_0\in C$, the sequence $\{u_n\}$ converges to that common fixed point, where $u_{2k}=T_2(u_{2k-1})$ and $u_{2k+1}=T_1(u_{2k})$ for all $k \in \mathbb{N},$ and the error estimation is given by
$$d(u_n,z)\leq \frac{\alpha^{n-1}}{1-\alpha}d(u_0,u_1),$$
for all $n \in \mathbb{N}$, where $z$ is the unique common fixed point of $T_1$ and $T_2$.
\end{enumerate}
\end{theorem}

\begin{proof}
Let $u_0\in C$ be arbitrary but fixed and consider the sequence $\{u_n\}$ which is defined by taking
\begin{center}
	$u_{2k}=T_2(u_{2k-1})$ and $u_{2k+1}=T_1(u_{2k})$
\end{center}
for all $k \in \mathbb{N}$. First, we assume that $n$ is even. Then putting $x=u_n$ and $y=u_{n+1}$ in  \eqref{e1}, we get
\begin{eqnarray}
	d(u_{n+1},u_{n+2})
		&=& d(T_1(u_n),T_2(u_{n+1})) \nonumber \\
		&\leq& \psi(d(u_n,u_{n+1}),d(u_n,T_1(u_n)),d(u_{n+1},T_2(u_{n+1}))) \nonumber \\
		&=& \psi(d(u_n,u_{n+1}),d(u_n,u_{n+1}),d(u_{n+1},u_{n+2})). \nonumber
\end{eqnarray}
By using the property of $\psi$, we get $$d(u_{n+1},u_{n+2})\leq \alpha d(u_n,u_{n+1}),$$
where $\alpha$ is a constant lying in $[0,1)$.
Next, we assume that $n$ is odd. Then the proceeding in a similar manner as above, we can show that $$d(u_{n+1},u_{n+2})\leq \alpha d(u_n,u_{n+1}).$$
Therefore, we have
\begin{eqnarray}
d(u_{n+1},u_{n+2})
&\leq& \alpha d(u_n,u_{n+1}) \nonumber \\
&\leq& \alpha^2 d(u_{n-1},u_{n}) \nonumber \\
& \vdots& \nonumber  \\
&\leq& \alpha^n d(u_{0},u_{1}) \nonumber
\end{eqnarray}
for any $n \in \mathbb{N}$.
Since $\alpha\in [0,1)$, it follows that the infinite series $\sum_{n=1}^{\infty} d(u_n,u_{n+1})$ is convergent. Thus, $\{u_n\}$ is a Cauchy sequence in $C$ and so there exists $z\in C$ such that $u_n\to z$ as $n\to \infty$. So the subsequences $\{u_{2k}\}$ and $\{u_{2k+1}\}$ converge to $z$.

Now, for any $k \in \mathbb{N}$, we have
\begin{eqnarray}
d(u_{2k+1},T_2(z))
&=& d(T_1(u_{2k}),T_2(z))	 \nonumber \\
&\leq &  \psi(d(u_{2k},z),d(u_{2k},T_1(u_{2k})),d(z,T_2(z))) \nonumber \\
&=& \psi(d(u_{2k},z),d(u_{2k},u_{2k+1}),d(z,T_2(z))). \nonumber \\
\end{eqnarray}
By taking the limit as $k \to \infty$ in the above inequality, we get
$$d(z,T_2(z)) \leq \psi(0,0,d(z,T_2(z)))$$
and so $d(z,T_2(z)) \leq \alpha \cdot 0$. This implies that $z=T_2(z)$. Similarly, we can show that $z=T_1(z)$. So $z$ is a common fixed point of $T_1$ and $T_2$.

Next, we will show the uniqueness of the common fixed point of $T_1$ and $T_2$. For this, let $z_1$ be another common fixed point of $T_1$ and $T_2$. Then putting $x=z$, $y=z_1$ in \eqref{e1}, we get
\begin{eqnarray}
	d(T_1(z),T_2(z_1))
	&\leq & \psi(d(z,z_1),d(z,T_1(z)),d(z_1,T_2(z_1))) \nonumber \\
	&=& \psi(d(z,z_1),0,0) \nonumber
\end{eqnarray}
and so $d(z,z_1) \leq \alpha \cdot 0$. It yields that $z=z_1$. Therefore, $z$ is the unique common fixed point of $T_1$ and $T_2$. This proves $(i)$.

Finally, we will show that the assertion $(i)$ holds.
Now for any $n,m \in \mathbb{N}$ with $n<m$, we have
\begin{eqnarray}
d(u_n,u_m)&\leq& d(u_n,u_{n+1})+d(u_{n+1}.u_{n+2})+ \hdots + d(u_{m-1},u_m)\nonumber\\
&\leq& (\alpha^{n-1}+\alpha^{n}+\hdots+\alpha^{m-2})d(u_0,u_1)\nonumber\\
&=&\frac{\alpha^{n-1}-\alpha^m}{1-\alpha} d(u_0,u_1). \nonumber
\end{eqnarray}
Keeping $n$ fixed and letting $m \to \infty$ in the above equation, we get
$$d(u_n,z)\leq \frac{\alpha^{n-1}}{1-\alpha} d(u_0,u_1)$$
for all $n\in \mathbb{N}$.
This proves $(ii)$.
\end{proof}
\section{Common solution to matrix equations}

Right through this section, we consider matrices over the set of complex numbers.  We use the notation $I_n$ as the $n\times n$ identity matrix. The Thompson metric $d$ on the set $P(n)$ is defined by

$$d(A,B)= \max\{\log W(A/B), \log W(B/A)\}$$
for all $A,B \in P(n)$, where $$W(A/B):=\inf\{\delta>0:A\leq \delta B\}=\lambda^{+}(B^{-\frac{1}{2}}AB^{-\frac{1}{2}})$$
is the maximum eigenvalue of the matrix $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$. It is well known that the set $P(n)$ is complete with respect to the Thompson metric $d$. The following properties involving  the Thompson metric are necessary to recall for the sake of our  developments.

\begin{lemma}[\cite{L,T}]
	Let $d$ be the Thompson metric on the set $P(n)$. Then
	\begin{itemize}
		\item[$(i)$] $d(A,B) = d(A^{-1},B^{-1})= d(MAM^{*},MBM^{*})$ for all $A,B \in P(n)$ and non-singular matrix $M$.
		\item[$(ii)$] $d(A^{r},B^{r})\leq \mid r\mid d(A,B)$ for all $A,B \in P(n)$ and $r \in [-1,1]$.
		\item[$(iii)$] $d(A+B,C+D)\leq \max\{d(A,C),d(B,D)\}$ for all $A,B,C,D \in P(n)$. In particular, $d(A+B,A+D)\leq d(B,D)$.
	\end{itemize}
\end{lemma}

\begin{theorem}\label{tt1}
Consider the  pair of  matrix equations  as follows:
\begin{align}
X^s=Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i \label{te1},\\
X^s=Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(X)A_i \label{te20}
\end{align}
where $Q_1,~Q_2$ are two $n\times n$ Hermitian positive definite matrices, $A_i's$ are arbitrary nonsingular complex matrices of order $n$, $F,~G$ are two functions from $P(n)$ to $P(n)$ and $s>1$, $s\in \mathbb{R}$. Assume that there is a  $a\in [0,\infty)$ such that for any $X\in P(n)$ with $d(X,I_n) \leq e^a$, where $d$ is the Thompson metric on $P(n)$, the following conditions hold:
\begin{enumerate}
\item [$(A)$] $\lambda_{max} \left(Q_1^{-\frac{1}{2}}Q_2Q_1^{-\frac{1}{2}}\right) \leq \lambda_{max} \left(F(X)^{-\frac{1}{2}}G(Y) F(X)^{-\frac{1}{2}}\right)$ and $\lambda_{max} \left(Q_2^{-\frac{1}{2}}Q_1Q_2^{-\frac{1}{2}}\right) \leq \lambda_{max} \left(G(Y)^{-\frac{1}{2}}F(X) G(Y)^{-\frac{1}{2}}\right)$ ;
\item [$(B)$]  $\lambda_{max} \left(F(X)^{-\frac{1}{2}}G(Y) F(X)^{-\frac{1}{2}}\right) \leq \left( \lambda_{max} \left(X^{-\frac{1}{2}}YX^{-\frac{1}{2}}\right)\right)^l$ and \\ $\lambda_{max} \left(G(Y)^{-\frac{1}{2}}F(X) G(Y)^{-\frac{1}{2}}\right) \leq  \left(\lambda_{max} \left(Y^{-\frac{1}{2}}X Y^{-\frac{1}{2}}\right)\right)^l$, where $l>0$ is a real number such that $l<s$;
\item [$(C)$]  $\lambda_{max} \left( I_n^{-\frac{1}{2}}\left(Q_1+ \sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^\frac{1}{s} I_n^{-\frac{1}{2}}\right),$ $\lambda_{max} \left( \left(\left(Q_1+ \sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^\frac{1}{s} \right)^{-\frac{1}{4}}\right),$  $\lambda_{max} \left( I_n^{-\frac{1}{2}}\left(Q_2+ \sum_{i=1}^{m} {A_i}^*G(X)A_i\right)^\frac{1}{s} I_n^{-\frac{1}{2}}\right)$, $\lambda_{max} \left( \left(\left(Q_2+ \sum_{i=1}^{m} {A_i}^*G(X)A_i\right)^\frac{1}{s} \right)^{-\frac{1}{4}}\right)$ $\leq e^a$.
\end{enumerate}
Then the following assertions hold:
\begin{enumerate}
\item[$(i)$] the pair of matrix equations \eqref{te1} and \eqref{te20} posses a unique common solution $\bar{X}$ such that $\bar{X}$ is positive definite and $d(\bar{X},I_n)\leq a$;
\item [$(ii)$] the unique solution is provided by the limit of the iterative sequence $\{X_k\}$, where
\begin{equation}
	X_k =\left\{
	\begin{tabular}{ccc}
		$\left(Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X_{k-1})A_i\right)^{\frac{1}{s}}$ & if & $k$ is odd, \\
		$\left(Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(X_{k-1})A_i\right)^{\frac{1}{s}}$ & if & $k$ is even,
	\end{tabular}
	\right.
\end{equation}
and
$X_0 \in P(n)$ is any element with $d(X_0,I_n) \leq a$, and the error estimation is given by $$d(X_k,\bar{X}) \leq \frac{\alpha^{k-1}}{1-\alpha}d(X_0,X_1)$$
for all $k\in \mathbb{N}$, where $\alpha:=\frac{l}{s}$.
\end{enumerate}
\end{theorem}

\begin{proof}
Consider the set $\mathcal{C}=\{X\in P(n):d(X,I_n)\leq a\}$
and the Thompson metric $d$ on $\mathcal{C}$.
Then $\mathcal{C}$ is complete with respect to the Thompson metric $d$.  Let us now consider a pair of mappings $T_1,T_2 :\mathcal{C} \to P(n)$ by
\begin{equation}\label{te3}
T_1(X)=\left(Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^{\frac{1}{s}}
\end{equation}
\begin{equation}\label{te4}
T_2(X)=\left(Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(X)A_i\right)^{\frac{1}{s}}
\end{equation}
for all $X \in \mathcal{C}$.
From the condition $(C)$, we get
$$d(T_1(X),I_n)=d\left(\left(Q_1+ \sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^\frac{1}{s},I_n\right)\leq a$$
and
$$d(T_2(X),I_n)=d\left(\left(Q_2+ \sum_{i=1}^{m} {A_i}^*G(X)A_i\right)^\frac{1}{s},I_n\right)\leq a.$$
This yields that $T_1,T_2$ are two self-mappings on $\mathcal{C}$.

Next, we will show that $T_1$ and $T_2$ satisfy the contractive condition in Theorem  \ref{th1}. Let $X, Y\in \mathcal{C}$ be arbitrary. Then we have
\begin{eqnarray}
d\left(T_1(X),T_2(Y)\right)
&=&d\left(\left(Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^{\frac{1}{s}},\left(Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(Y)A_i\right)^{\frac{1}{s}}\right)\nonumber\\
&\leq& \frac{1}{s} d\left(\left(Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i\right),\left(Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(Y)A_i\right)\right)\nonumber\\
&\leq& \frac{1}{s} \max \left\{d(Q_1,Q_2),d\left(\displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i,\sum_{i=1}^{m} {A_i}^*G(Y)A_i\right)\right\}. \nonumber \\
\label{te30}
\end{eqnarray}
Now, we have
\begin{align*}
&d\left(\displaystyle \sum_{i=m-1}^{m} {A_i}^*F(X)A_i,\sum_{i=m-1}^{m} {A_i}^*G(Y)A_i\right)\\
&\leq \max\left\{d\left(A_{m-1}^*F(X)A_{m-1},A_{m-1}^*G(Y)A_{m-1}\right),d\left(A_{m}^*F(X)A_{m},A_{m}^*G(Y)A_{m}\right)\right\}\\
&= \max\left\{d(F(X),G(Y)),d(F(X),G(Y))\right\}\\
&=d(F(X),G(Y)).
\end{align*}
Again, we have
\begin{align*}
&d\left(\displaystyle \sum_{i=m-2}^{m} {A_i}^*F(X)A_i,\sum_{i=m-2}^{m} {A_i}^*G(Y)A_i\right)\\
&\leq \max \left\{d\left(A_{m-2}^*F(X)A_{m-2},A_{m-2}^*G(Y)A_{m-2}\right),d\left(\displaystyle \sum_{i=m-1}^{m} {A_i}^*F(X)A_i,\sum_{i=m-1}^{m} {A_i}^*G(Y)A_i\right) \right\}\\
&\leq \max\left\{d(F(X),G(Y)),d(F(X),G(Y))\right\}\\
&=d(F(X),G(Y)).
\end{align*}
Continuing in this way, we can show that
\begin{equation}\label{te40}
d\left(\displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i,\sum_{i=1}^{m} {A_i}^*G(Y)A_i\right) \leq d(F(X),G(Y)).
\end{equation}
Using the equation \eqref{te40} in the equation \eqref{te30}, we get
\begin{equation}\label{te5}
d(T_1(X),T_2(Y))\leq \frac{1}{s} \max\left\{d(Q_1,Q_2),d(F(X),G(Y))\right\}.
\end{equation}
From the condition $(A)$, we get
$$\log W(Q_2/Q_1)\leq \log W(G(Y)/F(X))$$ and  $$\log W(Q_1/Q_2)\leq \log W(F(X)/G(Y)).$$
This implies that
\begin{align*}
\max \left\{\log W(Q_2/Q_1),\log W(Q_2/Q_1)\right\}&\leq  \max\left\{\log W(G(Y)/F(X)),\log W(G(Y)/F(X))\right\}\\
\Rightarrow d(Q_1,Q_2)&\leq d(F(X),G(Y)).
\end{align*}
Using this fact in the equation \eqref{te5}, we get
\begin{equation}\label{add 1}
	d(T_1(X),T_2(Y))\leq \frac{1}{s}d(F(X),G(Y)).
\end{equation}
It follows from the condition $(B)$ that $d(F(X),G(Y))\leq k d(X,Y)$. Using this fact with (\ref{add 1}), we get
\begin{equation}\label{te6}
d(T_1(X),T_2(Y))\leq \frac{k}{s}d(X,Y).
\end{equation}
Let us take $\psi(t_1,t_2,t_3)=\frac{k}{s}t_1$ for all $t_1,t_2,t_3 \in \mathbb{R}_+$. Then $\psi \in \Psi_{\frac{k}{s}}$. Utilizing the definitions of $\phi$  in the equation \eqref{te6}, we get
$$d(T_1(X),T_2(Y))\leq \psi(d(X,Y),d(X,T_1(X)),d(Y,T_2(Y))).$$
So all the hypotheses of Theorem  \ref{th1} hold, and so
 there exists $\bar{X}\in \mathcal{C}$ such that
$$T_1(\bar{X})=\bar{X}~\mbox{and}~ T_2(\bar{X})=\bar{X},$$ i.e.,
$$\bar{X}^s=Q_1 + \displaystyle \sum_{i=1}^{m} {A_i}^*F(\bar{X})A_i ~\mbox{and}~
\bar{X}^s=Q_2 + \displaystyle \sum_{i=1}^{m} {A_i}^*G(\bar{X})A_i. $$
Thus  $\bar{X}$ is a common solution of the pair of matrix equations \eqref{te1} and \eqref{te20}. Again since $\bar{X} \in \mathcal{C}$, we have $d(\bar{X},I_n) \leq a$. Moreover, since the common fixed point of $T_1$ and $T_2$ is unique, it follows that the common solution of equations \eqref{te1} and \eqref{te20} is also unique.

Finally, by using the final part of  Theorem \ref{th1}, the given sequence $\{X_k\}$ considered in the statement  converges to the common solution $\bar{X}$.
\end{proof}

\begin{theorem} \label{tt2}
Consider the  pair of  matrix equations  as follows:
\begin{align}
X^r=\displaystyle \sum_{i=1}^{m} {A_i}^*F(X)A_i, \label{te7}\\
X^s=\displaystyle \sum_{i=1}^{m} {A_i}^*G(X)A_i \label{te8}
\end{align}
where $F,G$  are same as theorem \ref{tt1}, $A_i$'s are abritrary $n \times n$ unitary matrices, and $r,s>1$ are real  numbers. Assume that there is a real number $a\geq 0$ such that for any $X\in P(n)$ with $d(X,I_n)\leq e^{ra}$, the following conditions hold:
\begin{enumerate}
\item[$(A)$] $\lambda_{max}  \left(F(X)\right)\leq \frac{e^{ra}}{m},$ $\lambda_{\max}\left((F(X))^{-1}\right)\leq m e^{ra}$ and $\lambda_{max}  \left(G(X)\right)\leq \frac{e^{ra}}{m},$ $\lambda_{\max}\left((G(X))^{-1}\right)\leq m e^{ra}$;
\item [$(B)$]  $\lambda_{\max}\left(F(X)\right)\leq \frac{1}{m2^r} \left(\lambda_{\max}\left(Y^{-\frac{1}{2}}XY^{-\frac{1}{2}}\right)\right)^l$, $\lambda_{\max}\left(G(X)\right) \leq \frac{1}{m2^s} \left(\lambda_{\max}\left(Y^{-\frac{1}{2}}XY^{-\frac{1}{2}}\right)\right)^l$ and $\lambda_{\max}\left((F(X))^{-1}\right)$, $\lambda_{\max}\left((G(X))^{-1}\right) \leq m \left(\lambda_{\max}\left(X^{-\frac{1}{2}}YX^{-\frac{1}{2}}\right)\right)^l$,
where $l>0$ is a real number such that $3l<\frac{rs}{r+s}.$
\end{enumerate}
Then the following assertions hold:
\begin{enumerate}
\item[$(i)$]the pair of matrix equations \eqref{te7} and \eqref{te8} posses a unique common solution $\bar{X}$ such that $\bar{X}$ is positive definite and $d(\bar{X},I_n)\leq ra$;
\item [$(ii)$] the unique solution is provided by the limit of the iterative sequence $\{X_k\}$, where
\begin{equation}
X_k =\left\{
\begin{tabular}{ccc}
$\left(\displaystyle \sum_{i=1}^{m} {A_i}^*F(X_{k-1})A_i\right)^{\frac{1}{r}}$ & if & $k$ is odd, \\
$\left(\displaystyle \sum_{i=1}^{m} {A_i}^*G(X_{k-1})A_i\right)^{\frac{1}{s}}$ & if & $k$ is even,
\end{tabular}
\right.
\end{equation}
and $X_0 \in P(n)$ is any element with $d(X_0,I_n) \leq ra$, and the error estimation is given by
$$d(X_k,\bar{X}) \leq \frac{\alpha^{k-1}}{1-\alpha}d(X_0,X_1),$$ where $\alpha:=3l\left(\frac{1}{r}+\frac{1}{s}\right)$.
\end{enumerate}
\end{theorem}

\begin{proof}
Consider the set $\mathcal{C}=\{X\in P(n):d(X,I_n)\leq ra\}$ and the Thompson metric $d$ on $\mathcal{C}$. Then $\mathcal{C}$ is complete with respect to the Thompson metric $d$.
We now consider a pair of mappings $T_1,T_2:\mathcal{C}\to P(n)$ by
\begin{align}
T_1(X)= \left(\sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^\frac{1}{r}, \label{te9}\\
T_2(X)= \left(\sum_{i=1}^{m} {A_i}^*G(X)A_i\right)^\frac{1}{s} \label{te10}
\end{align}
for all $X\in \mathcal{C}$.  For any $X\in \mathcal{C}$, we have
\begin{align*}
d(T_1(X),I_n) &= d\left(\left(\sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^\frac{1}{r},(I_n)^\frac{1}{r}\right)\\
&\leq \frac{1}{r} d\left(\left(\sum_{i=1}^{m} {A_i}^*F(X)A_i\right), I_n\right)\\
&=\frac{1}{r} d\left(\left(\sum_{i=1}^{m} {A_i}^*F(X)A_i\right), \sum_{i=1}^{m}A_i^*\left(\frac{1}{m}I_n\right){A_i}\right)\\
&\leq \frac{1}{r} d\left(F(X),\left(\frac{1}{m}I_n\right)\right).
\end{align*}
Using condition $(A)$, we get
\begin{align*}
\lambda_{\max}\left(\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}F(X)\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}\right)&=\lambda_{\max}\left(mF(X)\right)\\
&=m\lambda_{\max}\left(F(X)\right)\\
&\leq e^{ra}\\
\Rightarrow \log \left(\lambda_{\max}\left(\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}F(X)\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}\right)\right) &\leq ra\\
\Rightarrow \log \left(W\left(F(X) \Bigg{/}\frac{1}{m}I_n\right)\right)&\leq ra,
\end{align*}
and
\begin{align*}
\lambda_{\max}\left((F(X))^{-\frac{1}{2}}\frac{1}{m}I_n(F(X))^{-\frac{1}{2}}\right)&=\lambda_{\max}\left(\frac{1}{m}(F(X))^{-1}\right)\\
&=\frac{1}{m}\lambda_{\max}\left((F(X))^{-1}\right)\\
&\leq e^{ra}\\
\Rightarrow \log\left(\lambda_{\max}\left((F(X))^{-\frac{1}{2}}\frac{1}{m}I_n(F(X))^{-\frac{1}{2}}\right)\right)&\leq ra\\
\Rightarrow \log \left(W\left(\frac{1}{m}I_n \Bigg{/}F(X)\right)\right)&\leq ra.
\end{align*}
Therefore, $d(T_1(X),I_n)\leq ra$. Similarly, we have $d(T_2(X),I_n)\leq ra$. Thus $T_1,T_2$ are self-mappings on $\mathcal{C}$.

Next, we will show that $T_1$ and $T_2$ satisfies the contractive condition in Theorem  \ref{th1}. For  $X,Y \in \mathcal{C}$, we have
\begin{eqnarray}
	d(T_1(X),T_2(Y))
	&\leq& d\left(T_1(X),T_1(X)+\frac{1}{2}I_n\right)+d\left(T_1(X)+\frac{1}{2}I_n,T_1(X)+T_2(Y)\right)\nonumber\\
	&& +d\left(T_2(Y),T_2(Y)+\frac{1}{2}I_n\right)+d\left(T_2(Y)+\frac{1}{2}I_n,T_1(X)+T_2(Y)\right)\nonumber\\
	&\leq& d\left(T_1(X),I_n\right)+d\left(I_n,T_1(X)+\frac{1}{2}I_n\right)+d\left(T_2(Y),\frac{1}{2}I_n\right)\nonumber\\
	&&+d\left(T_2(Y),I_n\right)+d\left(I_n,T_2(Y)+\frac{1}{2}I_n\right)+d\left(T_1(X),\frac{1}{2}I_n\right)\nonumber \label{t11}\\
	&\leq& d\left(T_1(X),I_n\right)+d\left(T_2(Y),I_n\right)+2d\left(T_1(X),\frac{1}{2}I_n\right)+2d\left(T_2(Y),\frac{1}{2}I_n\right). \nonumber \\
\end{eqnarray}

Using condition $(B)$, we get
\begin{align*}
\lambda_{\max}\left(\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}F(X)\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}\right)&=\lambda_{\max}\left(mF(X)\right)\\
&=m\lambda_{\max}\left(F(X)\right)\\
&\leq m2^r\lambda_{\max}\left(F(X)\right)\\
&\leq \left(\lambda_{\max}\left(Y^{-\frac{1}{2}}XY^{-\frac{1}{2}}\right)\right)^k\\
\Rightarrow \log \left(\lambda_{\max}\left(\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}F(X)\left(\frac{1}{m}I_n\right)^{-\frac{1}{2}}\right)\right) &\leq k \log\left(\lambda_{\max}\left(Y^{-\frac{1}{2}}XY^{-\frac{1}{2}}\right)\right)\\
\Rightarrow \log \left(W\left(F(X) \Bigg{/}\frac{1}{m}I_n\right)\right)&\leq k\log \left(W\left(X \big{/}Y\right)\right)
\end{align*}
and
\begin{align*}
\lambda_{\max}\left((F(X))^{-\frac{1}{2}}\frac{1}{m}I_n(F(X))^{-\frac{1}{2}}\right)&=\lambda_{\max}\left(\frac{1}{m}(F(X))^{-1}\right)\\
&=\frac{1}{m}\lambda_{\max}\left((F(X))^{-1}\right)\\
&\leq \left(\lambda_{\max}\left(X^{-\frac{1}{2}}YX^{-\frac{1}{2}}\right)\right)^k\\
\Rightarrow \log\left(\lambda_{\max}\left((F(X))^{-\frac{1}{2}}\frac{1}{m}I_n(F(X))^{-\frac{1}{2}}\right)\right)&\leq k \log\left(\lambda_{\max}\left(X^{-\frac{1}{2}}YX^{-\frac{1}{2}}\right)\right)\\
\Rightarrow \log \left(W\left(\frac{1}{m}I_n \Bigg{/}F(X)\right)\right)&\leq k \log\left(W\left(Y \big{/}X\right)\right).
\end{align*}
Therefore,
{\small
\begin{align*}
\max\left\{\log \left(W\left(F(X) \Bigg{/}\frac{1}{m}I_n\right)\right),\log \left(W\left(\frac{1}{m}I_n \Bigg{/}F(X)\right)\right)\right\}&\leq k\max\left\{\log \left(W\left(X \big{/}Y\right)\right),\log \left(W\left(Y \big{/}X\right)\right)\right\}\\
\Rightarrow d\left(F(X),\frac{1}{m}I_n\right)&\leq k d(X,Y)\\
\Rightarrow d(T_1(X),I_n)&\leq \frac{k}{r}d(X,Y).
\end{align*}
}
Similarly, we can show that $d(T_2(Y),I_n)\leq \frac{k}{s}d(X,Y).$ Again
\begin{align*}
d\left(T_1(X),\frac{1}{2}I_n\right)&=d\left(\left(\sum_{i=1}^{m} {A_i}^*F(X)A_i\right)^\frac{1}{r},\left(\frac{1}{2^r}I_n\right)^\frac{1}{r}\right)\\
&\leq \frac{1}{r} d\left(\left(\sum_{i=1}^{m} {A_i}^*F(X)A_i\right),\left(\frac{1}{2^r}I_n\right)\right)\\
&=\frac{1}{r} d\left(\left(\sum_{i=1}^{m} {A_i}^*F(X)A_i\right),\left(\sum_{i=1}^{m}A_i^*\left(\frac{1}{m 2^r}I_n\right)A_i\right)\right)\\
&\leq \frac{1}{r} d\left(F(X),\frac{1}{m 2^r}I_n\right).
\end{align*}
Using the condition $(B)$,we get
\begin{align*}
\lambda_{\max}\left(\left(\frac{1}{m2^r}I_n\right)^{-\frac{1}{2}}F(X)\left(\frac{1}{m2^r}I_n\right)^{-\frac{1}{2}}\right)&=\lambda_{\max}\left(m2^rF(X)\right)\\
&=m2^r \lambda_{\max}(F(X))\\
&\leq \left(\lambda_{\max}\left(Y^{-\frac{1}{2}}XY^{-\frac{1}{2}}\right)\right)^k\\
\Longrightarrow \log \left(\lambda_{\max}\left(\left(\frac{1}{m2^r}I_n\right)^{-\frac{1}{2}}F(X)\left(\frac{1}{m2^r}I_n\right)^{-\frac{1}{2}}\right)\right) &\leq k \log\left(\lambda_{\max}\left(Y^{-\frac{1}{2}}XY^{-\frac{1}{2}}\right)\right)\\
\Longrightarrow \log \left(W\left(F(X) \Bigg{/}\frac{1}{m2^r}I_n\right)\right)&\leq k\log \left(W\left(X \big{/}Y\right)\right),
\end{align*}
and
\begin{align*}
\lambda_{\max}\left(\left(F(X)\right)^{-\frac{1}{2}}\frac{1}{m2^r}I_n\left(F(X)\right)^{-\frac{1}{2}}\right)&=\lambda_{\max}\left(\frac{1}{m2^r}\left(F(X)\right)^{-1}\right)\\
&=\frac{1}{m2^r}\lambda_{\max}\left(\left(F(X)\right)^{-1}\right)\\
&\leq \frac{1}{m}\lambda_{\max}\left(\left(F(X)\right)^{-1}\right)\\
&\leq \left(\lambda_{\max}\left(X^{-\frac{1}{2}}YX^{-\frac{1}{2}}\right)\right)^k\\
\Rightarrow \log\left(\lambda_{\max}\left(\left(F(X)\right)^{-\frac{1}{2}}\frac{1}{m2^r}I_n\left(F(X)\right)^{-\frac{1}{2}}\right)\right)&\leq k \log\left(\lambda_{\max}\left(X^{-\frac{1}{2}}YX^{-\frac{1}{2}}\right)\right)\\
\Rightarrow \log\left(W\left(\frac{1}{m2^r}I_n\Bigg{/}F(X)\right)\right)&\leq k \log\left(Y/X\right).
\end{align*}
Therefore,
{\small
\begin{align*}
\max\left\{\log \left(W\left(F(X) \Bigg{/}\frac{1}{m2^r}I_n\right)\right),\log\left(W\left(\frac{1}{m2^r}I_n\Bigg{/}F(X)\right)\right)\right\}&\leq k\max\left\{\log \left(W\left(X \big{/}Y\right)\right),\log \left(W\left(Y \big{/}X\right)\right)\right\}\\
\Longrightarrow d\left(F(X),\frac{1}{m2^r}I_n\right)&\leq kd(X,Y)\\
\Longrightarrow d\left(T_1(X),\frac{1}{2}I_n\right)&\leq \frac{k}{r}d(X,Y).
\end{align*}
}
Similarly, we can show that $d\left(T_2(Y),\frac{1}{2}I_n\right)\leq \frac{k}{s}d(X,Y).$  Utilizing all these calculations in equation \eqref{t11}, we deduce
$$d(T_1(X),T_2(Y))\leq \alpha d(X,Y),$$
where $\alpha=3k\left(\frac{1}{r}+\frac{1}{s}\right)<1$.

Let us consider $\psi(t_1,t_2,t_3)=\alpha t_1$ for all $t_1,t_2,t_3 \in \mathbb{R}_+$. Then $\psi\in \Psi_{\alpha}$ and
$$ d(T_1(X),T_2(Y)) \leq \psi\left(d(X,Y),d(X,T_1(X)),d(Y,T_2(Y))\right)$$ for all $X,Y \in \mathcal{C}$. The rest of the proof is similar to that of Theorem \ref{tt1} and so omitted.
\end{proof}
\section{Numerical examples}
In this section, we validate the results obtained in the previous section by some examples and showing
the numerical approximations of convergence of iterated sequences using  diagrams.
\begin{example}
Consider the system of matrix equations \eqref{te1} and \eqref{te20} for $s=2$, $m=1$, $F(X)=X^{\frac{1}{2}}$ and $G(X)=X^{\frac{1}{3}}$, that is,
\begin{align}
X^2=Q_1&+A_1^*X^{\frac{1}{2}}A_1 \label{ne1}\\
X^2=Q_2&+A_1^*X^{\frac{1}{3}}A_1 \label{ne2},
\end{align}
where we choose \( A_1=
  \begin{pmatrix}
    4+5i & 2+7i & -3+2i  \\
    5-6i & 4-3i & -i+8\\
    9-i & 5+2i & 6-2i
  \end{pmatrix},\
Q_1=
  \begin{pmatrix}
    2 & -1 & 0  \\
    - 1 & 2 & -1\\
    0 & -1 & 2
  \end{pmatrix} \mbox{and }
  \) \\
\(Q_2=
  \begin{pmatrix}
    6 & 0 & 0 \\
    0 & 6 & 0\\
    0 & 0 & 6
  \end{pmatrix}.
\)

Then if we choose $a=10$, then using MATLAB, we check that  for any $X\in P(3)$ with $d(X,I_3)\leq e^{10}$, the assumptions $(A)-(C)$ of Theorem \ref{tt1} are fulfilled. So by the consequences of this theorem, matrix equations \eqref{ne1} and \eqref{ne2} posses a unique common solution $\bar{X}$ such that $\bar{X}$ is positive definite and $d(\bar{X},I_3)\leq 10$.
 Indeed, we see that the unique positive definite solution is
\[ \bar{X}=
  \begin{pmatrix}
    5.6933  & 2.4413+1.3428i & 1.7040+0.5152i\\
 2.4413-1.3428i & 4.4438 &  0.7308+0.3634i\\
 1.7040-0.5152i & 0.7308-0.3634i & 5.0193
  \end{pmatrix}.
\]
Note that $d(\bar{X},I_3)=2.17614\leq 10$. The graphical representation of the convergence history  of the sequence $\{X_n\}$ considered in Theorem \ref{tt1} to the unique solution by taking two distinct initial values is shown in Figure 1.
\begin{figure}[h!]
\begin{center}
\textbf{Convergence behaviour}
\includegraphics[scale=1.00]{Image_1}
\textbf{Figure 1}

\end{center}
\end{figure}
\end{example}
\begin{example}
We consider the system of matrix equations \eqref{te7} and \eqref{te8} for $r=2$, $s=3$, $m=1$, $F(X)=X^{\frac{1}{2}}$ and $G(X)=X^{\frac{1}{4}}$, that is, we consider the following system of equations
\begin{align}
X^2=A_1^*X^{\frac{1}{2}}A_1\label{ne3}\\
X^3=A_1^*X^{\frac{1}{4}}A_1 \label{ne4},
\end{align}
where we choose
\[ A_1=
  \frac{1}{3}\begin{pmatrix}
    2 & -2 & 1  \\
    1 & 2 & 2\\
    2 & 1 & -2
  \end{pmatrix}.\]
 Then for $a=2$, using MATLAB, we check that for any $X\in P(3)$ with $d(X,I_3)\leq e^{2\cdot 2}=e^4$, the assumptions $(A),\ (B)$ of Theorem \ref{tt2} are fulfilled.  So by this theorem, matrix equations \eqref{ne3}and \eqref{ne4} posses a unique common solution $\bar{X}$ such that $\bar{X}$ is positive definite and $d(\bar{X},I_3)\leq 2$.  We obtain by calculating  that the unique common positive definite solution $\bar{X}$ is
\[ \bar{X}=
  \begin{pmatrix}
    1  & 0 & 0\\
 0 & 1 &  0\\
0 & 0 & 1
  \end{pmatrix}.
\]
Note that $d(\bar{X},I_3)=0\leq 2$. The graphical representation of the convergence history  of the sequence $\{X_n\}$ considered in Theorem \ref{tt2} to the unique solution by taking two distinct initial values is shown in Figure 2.
 \begin{figure}[h!]
\begin{center}
\textbf{Convergence behaviour}\\
\includegraphics[scale=2.50]{Image_2}\\
\textbf{Figure 2}

\end{center}
\end{figure}
\end{example}

\begin{Acknowledgement}
The first and second authors are  thankful to CSIR, Government of India for their financial support (Ref. No. 09/973 $(0018)/2017$-EMR-I and $25(0285)/18/$EMR-II).
\end{Acknowledgement}

config: arxiv


TITLE: $\cos x=12/13$, where $0 \lt x \lt 90$ degrees. What is the value of $\sin(2x)?$
QUESTION [2 upvotes]: apologies if this is an overly simplistic question to answer:
I have the value of $\cos x = \frac{12}{13}$, and I need to find the value of $\sin(2x)$, where $x$ is between $0$ and $90$ degrees (first quadrant).
I have $\sin(2x)=2\sin x(\frac{12}{13})$ but I am stumped trying to find the value of $\sin x$.
Any help would be greatly appreciated

REPLY [3 votes]: Because a picture can help:

The triangle is drawn so that $\cos x = \frac{12}{13}$.  Use the Pythagorean theorem to find $y$, and then $\sin x = \frac{y}{13}$.


TITLE: Evaluating $\lim_{n\to\infty}( f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0) )$
QUESTION [7 upvotes]: I'm doing an exercise in Real Analysis and I would like to have my proof check since I'm not 100% sure of it. The problem says to consider a function $f:\mathbb{R}\rightarrow\mathbb{R}$ differentiable at $x_0$ and with $f(x_0)>0$ (which I guess it's useless at this point but the exercise had other points in which it was necessary so I put it anyway). Then, we are asked to evaluate the following
$$\lim_{n\to\infty} \left(f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0)\right).$$
Here is how I would do it:

$f$ is differentiable and hence continuous at $x_0$;

We apply Taylor's Theorem at $x_0$, so that there is a neighborhood of $x_0$, $I=(x_0-\delta,x_0+\delta)$, and a function $h$ such that, for all $x\in I$,
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+h(x)(x-x_0)$$
and $h(x)\rightarrow 0$ as $x\rightarrow x_0$.

Now, if $n>\frac 1{\delta}$ then each of the $n$ points $x_0+\frac 1{n^2},\dots,x_0+\frac 1n$ are in $I$ so we can apply the previous point to get

$$f\left(x_0+\frac k{n^2}\right)=f(x_0)+f'(x_0)\frac k{n^2}+h\left(x_0+\frac k{n^2}\right)\frac k{n^2}.$$

Summing from $k=1$ to $k=n$ we get

$$f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)=nf(x_0)+\frac {f'(x_0)}{n^2}(1+\dots+n)+\frac 1{n^2}\left(h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right)$$

Using the fact that $1+\dots+n=\frac {n(n+1)}2$ our original expression simplifies to

$$f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0)=\frac {f'(x_0)}{n^2}\frac {n(n+1)}2+\frac 1{n^2}\left(h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right).$$

Fix $\epsilon>0$, since $\frac {n(n+1)}{2n^2}\rightarrow \frac 12$ there is an $\bar{n}$ such that $n\geq \bar{n}$ implies that $\left|\frac {n(n+1)}{2n^2}-\frac 12\right|<\epsilon$. On the other hand, there is a $\delta'$ such that $|h(x)|<\epsilon$ on $(x_0-\delta',x_0+\delta')$. Since $x_0+\frac k{n^2}\rightarrow x_0$ and $\frac k{n^2}\leq \frac 1n$ we have that for $n>\frac 1{\delta'}$

$$\left|h\left(x_0+\frac k{n^2}\right)\right|<\epsilon.$$

By picking $n\geq\max\left\{\frac 1{\delta},\bar{n},\frac 1{\delta'}\right\}$ we obtain that

$$\left|\frac {f'(x_0)}{n^2}\frac {n(n+1)}2+\frac 1{n^2}\left(h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right)-\frac {f'(x_0)}2\right|\leq$$
$$f'(x_0)\left|\frac {n(n+1)}{2n^2}-\frac 12\right|+\frac 1{n^2}\left|h\left(x_0+\frac 1{n^2}\right)+\dots+nh\left(x_0+\frac 1n\right)\right|<$$
$$f'(x_0)\epsilon+\frac 1{n^2}\left|h\left(x_0+\frac 1{n^2}\right)\right|+\dots+\frac n{n^2}\left|h\left(x_0+\frac 1n\right)\right|<f'(x_0)\epsilon+\epsilon(1+\dots+n)=$$
$$=f'(x_0)\epsilon+\frac {\epsilon}{n^2}\frac {n(n+1)}2<f'(x_0)\epsilon+\epsilon\left(\frac 12+\epsilon\right).$$

Since $\epsilon$ is arbitrary, we can conclude that

$$\lim_{n\to\infty} \left(f\left(x_0+\frac 1{n^2}\right)+\dots+ f\left(x_0+\frac n{n^2}\right)-nf(x_0)\right) = \frac {f'(x_0)}2.$$
Is my proof correct? Is there any smarter/shorter way to do that? Thanks is advance for any help!

REPLY [1 votes]: Something you could use is the fact that $\lim_{k\to 0}\frac{h(x_0+k)}{k}=0$ so there exists an $n_0$ such that $h(x_0+k)<k$ for $k<1/n_0$, since the limit is taken over $n$ and $\frac{k}{n^2}\leq\frac{1}{n}$ we have that there exists an $n_0$ such that $$\frac{1}{n^2}\sum_{i=1}^n ih\left(x_0+\frac{i}{n^2}\right)<\frac{1}{n^2}\sum_{i=1}^n\frac{i^2}{n^2}<\frac{1}{n^2}\sum_{i=1}^n1=\frac{1}{n}$$
For all $n\geq n_0$, that sequence converges to $0$, so $\lim_n\frac{1}{n^2}\sum_{i=1}^{n}ih(x_0+\frac{i}{n^2})=0$


TITLE: Example 2, Sec. 29, in Munkres' TOPOLOGY, 2nd ed: Local compactness of $\mathbb{R}^\omega$
QUESTION [1 upvotes]: In Example 2, Sec. 29, in the book Topology by James R. Munkres, 2nd edition, the author shows that $\mathbb{R}^\omega$ (in the product topology) is not locally compact (at any point). However, the argument he gives is not as detailed as I would like it to be. Accordingly, I would like to present the following proof:

By definition
$$
\mathbb{R}^\omega := \mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \cdots, \tag{Definition 0}
$$
that is, $\mathbb{R}^\omega$ is the set of all the (infinite) sequences of real numbers.

Here $\mathbb{R}^\omega$ has the product topology determined by the standard (or usual) topology on $\mathbb{R}$.

Let $\mathbf{x} := \left( x_1, x_2, x_3, \ldots \right)$ be any point of $\mathbb{R}^\omega$. [Refer to (Definition 0) above.] We show that $\mathbb{R}^\omega$ is not locally compact at $\mathbf{x}$.

Suppose, if possible, that $\mathbb{R}^\omega$ is locally compact at $\mathbf{x}$. Then by definition there exists a compact subspace $\mathbf{C}$ of $\mathbb{R}^\omega$ containing a neighborhood $\mathbf{U}$ of $\mathbf{x}$, that is, $\mathbf{U}$ is an open set of $\mathbb{R}^\omega$ and $\mathbf{C}$ is a compact subspace of $\mathbb{R}^\omega$ such that
$$
\mathbf{x} \in \mathbf{U} \subset \mathbf{C}. \tag{0}
$$

Now as $\mathbf{x} \in \mathbf{U}$ and $\mathbf{U}$ is an open set of the product space $\mathbb{R}^\omega$, so there exists a canonical basis set $\mathbf{B} := U_1 \times U_2 \times U_3 \times \cdots$ for the product topology on $\mathbb{R}^\omega$, where $U_{n_1}, \ldots, U_{n_k}$ are open sets of $\mathbb{R}$ for some finitely many natural numbers $n_1, \ldots, n_k$ and $U_n = \mathbb{R}$ for all the other natural numbers $n$, such that
$$
\mathbf{x} \in \mathbf{B} \subset \mathbf{U}. \tag{1}
$$

Now as $\mathbf{x} = \left( x_1, x_2, x_3, \ldots \right) \in \mathbf{B} = U_1 \times U_2 \times U_3 \times \cdots$, so for each $j = 1, \ldots, k$, we have $x_{n_j} \in U_{n_j}$, and since $U_{n_j}$ is an open set of $\mathbb{R}$, there exists an open interval $\left( a_{n_j}, b_{n_j} \right)$ such that
$$
x_{n_j} \in \left( a_{n_j}, b_{n_j} \right) \subset U_{n_j}. \tag{2}
$$
Let us now put
$$
\mathbf{B}^* := B_1 \times B_2 \times B_3 \times \cdots, \tag{Definition 1}
$$
where
$$
B_n := \begin{cases} \left( a_{n_j}, b_{n_j} \right) & \mbox{ if } n = n_j \mbox{ for some } j = 1, \ldots, k; \\ U_n = \mathbb{R} & \mbox{ otherwise}.  \end{cases} \tag{Definition 2}
$$
Then by (Definition 2), (Definition 1), (2), (1), and (0) above, we have
$$
\mathbf{x} \in \mathbf{B}^* \subset \mathbf{B} \subset \mathbf{U} \subset \mathbf{C},
$$
which implies that
$$
\mathbf{x} \in \mathbf{B}^* \subset \mathbf{C},
$$
and hence
$$
\overline{\mathbf{B}^*} \subset \overline{\mathbf{C}} = \mathbf{C},
$$
because $\mathbf{C}$, being a compact subspace of the (metrizable and hence) Hausdorff space $\mathbb{R}^\omega$, is also closed in $\mathbb{R}^\omega$. Now as $\overline{\mathbf{B}^*}$ is a closed subset of $\mathbb{R}^\omega$, so $\overline{\mathbf{B}^*} \cap \mathbf{C} = \overline{\mathbf{B}^*}$ is also closed in the subspace $\mathbf{C}$ of $\mathbb{R}^\omega$. Therefore $\overline{\mathbf{B}^*}$, being a closed subspace of the compact space $\mathbf{C}$, is also compact as a subspace of $\mathbf{C}$, and since $\mathbf{C}$ in turn is a subspace of $\mathbb{R}^\omega$, therefore we can conclude that $\overline{\mathbf{B}^*}$ is also compact as a subspace of $\mathbb{R}^\omega$.

But
$$
\overline{\mathbf{B}^*} = \overline{B_1} \times \overline{B_2} \times \overline{B_3} \times \cdots,
$$
where
$$
\overline{B_n} = \begin{cases} \left[ a_{n_j}, b_{n_j} \right] & \mbox{ if } n = n_j \mbox{ for some} j = 1, \ldots, k; \\ \mathbb{R} & \mbox{ otherwise}. \end{cases}
$$
Let $n_0$ be any natural number distinct from $n_1, \ldots, n_k$. Then
$$
\pi_{n_0} \left( \overline{\mathbf{B}^*}  \right) = \mathbb{R}.
$$
And, as the mapping $\pi_{n_0} \colon \mathbb{R}^\omega \longrightarrow \mathbb{R}$ is continuous and as $\overline{\mathbf{B}^*}$ is a compact subspace of $\mathbb{R}^\omega$, so $\pi_{n_0} \left( \overline{\mathbf{B}^*}  \right) = \mathbb{R}$ is also compact, which is a contradiction.

Therefore our supposition that $\mathbb{R}^\omega$ is locally compact at $\mathbf{x}$ is wrong.

Hence $\mathbb{R}^\omega$ is not locally compact at any point $\mathbf{x} \in \mathbb{R}^\omega$.

Is this proof correct and clear enough? If so, is my argument any clearer than the argument given by Munikres?
This Math Stack Exchange post of mine is also somewhat relevant.

REPLY [1 votes]: Your proof is correct. But to be honest, I do not think it is better than Munkres' proof. Perhaps Munkres is a bit short and leaves some details to the reader, and you filled in these details.
Munkres shows that none of the basic open sets having the form
$$B = (a_1,b_1) \times \ldots (a_n,b_n) \times \mathbb R \times \mathbb R \times  \ldots $$
has compact closure. Here he invokes Theoem 19.5.
Why does this suffice to prove that $\mathbb R^\omega$ is not locally compact? Given $x =(x_i) \in \mathbb R^\omega$, assume that there is an open $U \subset \mathbb R^\omega$ and a compact $C \subset \mathbb R^\omega$ such that $x \in U \subset C$. Take a basic open $V = \prod_{i=1}^\infty V_i$ (i.e. $V_i  =\mathbb R$ for almost all $i$) such that $x \in V \subset U$. Choose $n$ such that $V_i = \mathbb R$ for $i > n$. Then pick open intervals $(a_i,b_i)$ such that $x_i  \in (a_i,b_i) \subset V_i$ for $i \le n$. This gives you a basic open $B$ as considered by Munkres such that $x \in B \subset C$. But we have $\overline B \subset \overline C = C$, thus $\overline B$ must be compact since it is a closed subset of a compact set.
Here is one more alternative proof.
The above assumption leads to $x \in V \subset C$ with a basic open $V$. The projections $p_i : \mathbb R^\omega \to \mathbb R, p_i(x) = x_i$, are continuous, thus the $C_i = p_i(C)$ are compact. Clearly $V_i = p_i(V) \subset C_i$. This contradicts the fact that almost all $V_i = \mathbb R$.


TITLE: When is the Ito representation deterministic?
QUESTION [1 upvotes]: Consider a functional on Wiener space $F\in L^2(\Omega, \Bbb R)$. Then by Ito's representation theorem we have that $$F=E[F]+\int_0^T\phi(s)dB(s)$$ for some nice $\phi$. Question, when is $\phi$ deterministic? I know that if $\phi$ is deterministic then $F$ must be Gaussian as the Ito integral of a deterministic function is Gaussian.
However is the converse true? If $F\sim \mathcal N(\mu,\sigma^2)$ can we find a deterministic $\phi$ so that $$F=\mu+\int_0^T \phi(s) dB(s)$$
What about just letting $\phi(s)\equiv \sigma/\sqrt{T}$?

REPLY [1 votes]: Example: Take
$$
\phi(s) = \cases{ 1,&  $0\le s\le T/2$;\cr \xi,& $T/2<s\le T$.\cr}
$$
where $\xi$ is the sign of $B_{T/2}$. This $\phi$ is non-deterministic.  But, using the facts that $\xi$ is (i) $\mathcal F_{T/2}$ measurable and therefore (ii) independent of the post-$T/2$ increments of the Brownian motion,  one can check that $F=\int_0^T\phi(s)\,dB_s=B_{T/2}+\xi\cdot(B_T-B_{T/2})$ has the normal distribution with mean $0$ and variance $T$.


TITLE: A function $ f : [0,1] \to \Bbb{C} $ is differentiable with derivative 0, is constant.
QUESTION [0 upvotes]: In Conway's complex analysis while proving proposition 2.10
"If $G$ is an open connected set and  $ f : G \to \Bbb{C} $ such that $f'(z)=0 \ \forall z \in G$ then f is constant"
He fixes some $z_{0} \in G$ and defines the set $A=$ {$z \in G: f(z)=f(z_0)$} and shows that this set is both open as well as closed, since it is a non-empty subset of a connected set, we have, $A=G$ which would complete the proof.
I can't understand the proof to show it's open, where he defines $ g(t)=f(zt+(1-t)a) \ , t \in [0,1]$ for some fixed $a \in A $ and $\epsilon >0$ such that $z\in B(a,\epsilon)\subset G$ He wants to show that $B(a,\epsilon) \subset A.$
By proving
$g'(t)=0 \ $on  $ [0,1]$ he concludes that g is constant.
I thought this result was only true for real valued functions. I am clear about everything else, why the function has been constructed in this way and all. I am just not able to convince myself of the above conclusion.

REPLY [0 votes]: If $g:[0,1]\to \mathbb  C$ is differentiable and $g'=0$ for all $t$ then  its real and imaginary parts are constants (because there derivatives are also $0$) , right? That proves that $g$ is a complex constant.


TITLE: Equation for reversible adiabatic expansion
QUESTION [0 upvotes]: The heat capacity of one mole of an ideal gas is found to be $C_{v}=3R(1+aRT)/2$ where $a$ is a constant. The book provides the solution to this problem but there is a step in which $(1+aRT)=e^{aRt}$ How is this possible?
Please explain.

REPLY [2 votes]: If $aRT$ is extremely small then $$e^{aRT} \approx 1+aRT$$


TITLE: How can I determine whether balloons inside a balloon are filled with the same gas?
QUESTION [3 upvotes]: We recently awarded ourselves some helium balloons to celebrate something, and they came with about a dozen little balloons inside, filling it to just under half way.
I was discussing with a colleague whether it would be possible to determine whether the inner balloons were also helium-filled, without popping the large outer balloon.
Obviously, if the inner balloons were filled with something moderately lighter than helium, they would float, but they're huddled at the bottom, so it's either helium or plain air.
How about it?
p.s. you guys really need a 'balloon' tag. :)

REPLY [1 votes]: One indication whether the 'inner' balloons are filled with helium or something heavier is whether the 'outer' balloon still floats in air or not.
Whether it floats or not depends on whether its overall density is lower than air's density or not. Its overall density $d$ is simply given by:
$$d=\frac{\Sigma m}{V}$$
Where $\Sigma m$ is the sum of all the masses that make up the balloon and $V$ its total volume.
The sum of all masses means the sum of:

Mass of outer balloon material
Mass of helium
Masses of all inner balloons material
Masses of all gas contained in the inner balloons

If the inner balloons are filled with air and there are too many of them the overall density will exceed that of air and the outer balloon will not float.
But even if the inner balloons are all filled with helium, the outer balloon may still not float. That's because a larger number of inner balloons making up the same volume of a smaller number of inner balloons with create more mass because of the increased surface area and the inherent 'cost' in mass of larger number of inner balloons. For that reason:

Obviously, if the inner balloons were filled with something moderately lighter than helium, they would float, but they're huddled at the bottom, so it's either helium or plain air.

... is not necessarily correct.
If the mass of the outer balloon material, the total volume of the balloon, the mass of the inner balloon material and the volume of the inner balloons was known, then depending on whether the outer balloon still floats in air or not, the lower or lower limit of the density of the inner balloon gas could be calculated.
Call $m$, $d$ and $V$, with resp. suffixes $o$ and $i$ for outer and inner balloons, the mass of balloon material, the density of the filling gas and volume of the balloons. Let there be $n$ identical inner balloons. The overall density of the balloon is then:
$$d=\frac{m_o+d_o(V_o-nV_i)+nm_i+nd_iV_i}{V_o}$$
If the balloon just about floats in air, then:
$$d=d_{air}$$
From which we can deduce:
$$d_i=\frac{d_{air}V_o-m_o-d_o(V_o-nV_i)-nm_i}{nV_i}$$


TITLE: About the sheafification
QUESTION [1 upvotes]: Look at the following definition of the shefification (the source is the Stacks Project):

I don't understand what is the projection $\prod_{u\in U}\mathcal F_u\longrightarrow\prod_{v\in V}\mathcal F_v$. In particular, if $u\in U\setminus V$ what is the image of $s_u$ under this map?
Thanks in advance.

REPLY [4 votes]: The actual question turns out to not be particular to sheaves, so I will adjust my notation accordingly. The elements of a product $\prod_{a\in\mathcal A}X_a$ are tuples (well, choice functions) $(x_a)_{a\in\mathcal A}$. If $\mathcal B\subset\mathcal A$, then we can assign $$(x_a)_{a\in\mathcal A}\mapsto (x_b)_{b\in\mathcal B}.$$ This map $\prod_{a\in\mathcal A}X_a\to\prod_{b\in\mathcal B}X_b$ is what the author was referring to.
Also, note that you can use the universal property of products in categories to determine the corresponding projections in non-concrete categories.


TITLE: For formulas $\phi$, $\psi$ show (($\forall$$x_1$)($\phi$ -> $\psi$) ->(($\forall$$x_1$)$\phi$ -> ($\forall$$x_1$)$\psi$)) is logically valid.
QUESTION [1 upvotes]: What is the difference between saying
($\forall$$x_1$)($\phi$ -> $\psi$) and saying ($\forall$$x_1$)$\phi$ -> ($\forall$$x_1$)$\psi$
Thanks
EDIT
I am trying to prove that for formulas $\phi$ and $\psi$ the formula (($\forall$$x_1$)($\phi$ -> $\psi$) ->(($\forall$$x_1$)$\phi$ -> ($\forall$$x_1$)$\psi$)) is logically valid.
I am trying to do so using contradiction and showing that if $\theta$=(($\forall$$x_1$)($\phi$ -> $\psi$) ->(($\forall$$x_1$)$\phi$ -> ($\forall$$x_1$)$\psi$)) then v(\$theta$) cannot be F (hence the formula is logically valid).
if i need to show that v(((∀$x_1$)(ϕ -> ψ) ->((∀$x_1$)ϕ -> ($∀x_1$)ψ)))=T to show that it is logically valid, why is it wrong to try a contradiction and show that v((($∀x_1$)(ϕ -> ψ) ->(($∀x_1$)ϕ -> ($∀x_1$)ψ))) cannot be F, thus it must be T as required?

REPLY [1 votes]: For any formulas $\varphi, \psi$ show that $(∀x_1)(\varphi \rightarrow \psi) \rightarrow ((∀x_1)\varphi \rightarrow (∀x_1)\psi))$ is logically valid [from Alan Hamilton, Logic for mathematicians (1978), Exercise 19, page 70].

We can follow the hint of page 69 :

In general, to prove logical validity we must show that an arbitrary valuation in an arbitrary interpretation satisfies the wff concerned.

Let $\mathcal I$ be an interpretation with domain $D_\mathcal I$ and let $v$ be a valuation in $\mathcal L$.
If $v$ does not satisfy $(∀x_1)(\varphi \rightarrow \psi)$ then $v$ satisfy the above formula.
Consider now a $v$ which satisfy $(∀x_1)(\varphi \rightarrow \psi)$ and suppose that $v$ does not satisfy $(∀x_1)\varphi \rightarrow (∀x_1)\psi$.
This means that $v$ satisfy $(∀x_1)\varphi$ and $v$ does not satisfy $(∀x_1)\psi$.
We have to recall Definition 3.20, page 70 :

(iv) $v$ satisfies $(\forall x_i)\beta$ if every valuation $v'$ which is $i$-equivalent
  to $v$ satisfies $\beta$

and Definition 3.19, page 69 :

Two valuations $v$ and $v'$ are $i$-equivalent if $v(x_j) = v'(x_j)$ for every $j \ne i$.

Thus, from definitions, if $v$ does not satisfy $(∀x_1)\psi$, there is a valuation $v'$ which is $1$-equivalent to $v$ which does not satisfy $\psi$.
But $v$ satisfy $(∀x_1)\varphi$; this means that every valuation which is $1$-equivalent to $v$ satisfy $\varphi$.
In conclusion, we have a valuation $v'$ which satisfy $\varphi$ and not satisfy $\psi$; thus, $v'$ does not satisfy $\varphi \rightarrow \psi$.
But this contradicts the fact that $v$ satisfy $(∀x_1)(\varphi \rightarrow \psi)$, which means that every valuation $v'$ which is $1$-equivalent to $v$ satisfies $\varphi \rightarrow \psi$.


TITLE: Why does the following relationship hold?
QUESTION [0 upvotes]: My problem is the following:

During the solution the following relationship is found to be true (last two lines):

Namely, $E_{e}=|\overrightarrow{p_{e}}| \cdot \beta_{e} $, where $\beta_{e}$ is the velocity of the incoming electron.
Why is this the case?
Thanks,
T.D.

REPLY [0 votes]: Well you have if $c=1$:
$$E = m\gamma$$
$$\mathbf{p} = m\gamma\boldsymbol{\beta}$$
Taking the norm of the second one and dividing the first one by the second one you get:
$$\frac{E}{p} = \beta$$
This is the relation that is used in the exercise.


TITLE: Difference between spinor and vector field
QUESTION [9 upvotes]: How do we distinguish spinors and vector fields? I want to know it in terms of physics with mathematical argument.

REPLY [2 votes]: Some mathematical representation has been given by others. If you like some discussion on the physics difference between these two objects you can consider the following. The difference is in the rotational symmetry of these fields. The wave function of a vector particle  needs to be rotated once (360 degrees) in order to return to its original configuration (to be identical). The wave function of a spinor particle, must be rotated twice (720 degrees) to return to its original configuration. This is because of the spin difference of these two objects being $S = 1\hbar$ and $S = 1/2\hbar$ respectively. For a rotation by an arbitrary angle 'a' the rotation matrix operator is of the general form
$R(a)=\exp (iS.a/\hbar)$.
So you can see that if $S = 1\hbar$ then an angle $a = 2\pi$ returns the field to its original configuration, while if $S = (1/2)\hbar$ then you need $a = 4\pi$ to return the "object" to its original configuration. I hope this helps see the difference in the physics of these two fields.


TITLE: Prove that $\limsup _{n\to \infty}(a_n+b_n)\leq \limsup _{n\to \infty}a_n + \limsup _{n\to \infty}b_n$
QUESTION [3 upvotes]: Let $(a_n)$ and $b_n$ be bounded sequences of real numbers. Prove that $$\limsup _{n\to \infty}(a_n+b_n)\leq \limsup _{n\to \infty}a_n + \limsup _{n\to \infty}b_n$$
How can this be proved?
Using the definition of limit, can I use the fact that $\sup|a_n + b_n|\leq \sup|a_n| + \sup|b_n|$?

REPLY [5 votes]: Proposition: Let $\limsup\limits_{n\rightarrow \infty}{x_n}=x$ and let $x<\alpha$, then there is an integer $N$ such that $n\ge N$
  implies $x_n<\alpha$.

Proof: Suppose there is an $\alpha>x$ such that for every $n\ge N$ implies $x_n\ge \alpha$ for infinitely many values of $n$. Let $A$ be the set of all such $x_n$, if $A$ has an upper bound then by least upper bound property of $\mathbb{R}$, $A$ has a supremum, $\gamma\ge\alpha>x$. Otherwise there is a subsequence $x_{n_k}$ such that $x_{n_k}\rightarrow +\infty$. Both cases contradict the definition of $\limsup. \qquad \square$

Let $\limsup\limits_{n\rightarrow \infty}{a_n}=a$ and $\limsup\limits_{n\rightarrow \infty}{b_n=b}$ as it is given that the sequences $a_n$ and $b_n$ are bounded, hence $a$ and $b$ are real.
For arbitrary $\varepsilon>0$ there exits an $N$ such that $a_n<a+\frac{\varepsilon}{2}$ and $b_n<b+\frac{\varepsilon}{2}$. Hence $$\begin{array}{rcl}a_n+b_n&<&a+b+\varepsilon\implies\\\limsup\limits_{n\rightarrow \infty}{(a_n+b_n)}&\le& a+b+\varepsilon \implies \\ \limsup\limits_{n\rightarrow \infty}{(a_n+b_n)}&\le&\limsup\limits_{n\rightarrow\infty}{a_n}+\limsup\limits_{n\rightarrow\infty}{b_n} \end{array}$$

Note: The following is also true but with a weaker hypothesis.
For any two real sequences $\{a_n\}$ and $\{b_n\}$ $$\limsup\limits_{n\rightarrow \infty}{(a_n+b_n)}\le \limsup\limits_{n\rightarrow \infty}{a_n} +\limsup\limits_{n\rightarrow \infty}{b_n}$$
provided that the right hand side is not of the form $\infty-\infty$. You can see here for a proof.


TITLE: Pushforward of a log canonical pair
QUESTION [0 upvotes]: This question comes from learning the paper "Existence of minimal models for varieties of log general type", where they define the log terminal model (See Definition 3.6.7). However, the question itself does not need that definition.
Let $\pi: X \to U$ be a projective morphism of normal quasi-projective varieties. Suppose that $K_X + \Delta$ is log canonical and let $\phi: X -\to Y$ be a birational contraction of normal quasi-projective varieties over $U$, where $Y$ is projective over $U$ (here, $\phi$ is a "birational contraction" means $\phi^{-1}$ does not contract any divisor, for example, when $\phi$ is a morphism). Set $\Gamma=\phi_* \Delta$. Then is it true $$\phi_*(K_X + \Delta)=K_Y +\Gamma\quad?$$
I am equally satisfied with an answer in the simplified case as $U$ is a point, or $\phi$ is a morphism.

REPLY [2 votes]: Yes, since
$$\phi_*K_X=K_Y.$$
Here is the reason. We take a common resolution of $X$ and $Y$, say
$$
p:W\rightarrow X,  \\q: W \rightarrow Y.
$$
Then we can right
$$
K_W-p^*K_X=E,\\
K_W-q^*K_Y=F,
$$
where $E$ is exceptional over $X$ (and over $Y$ because "contraction") and $F$ is exceptional over $Y$. If apply $q_*$ to
$$
p^*K_X+E=q^*K_Y=F,
$$
we get the equality.


TITLE: $K$ is bounded in $(\mathbb R^n,||.||)$ iff $K$ is bounded in $(\mathbb R^n,||.||_2).$
QUESTION [2 upvotes]: Let $||.||$ be any norm on $\mathbb R^n$ and $K\subset\mathbb R^n.$ How to show that
Verify: $K$ is bounded in $(\mathbb R^n,||.||)$ iff $K$ is bounded in $(\mathbb R^n,||.||_2).$
I tried using the equivalence of two norms on a finite dimension NLS but couldn’t come to any conclusion.

REPLY [3 votes]: Hint: the equivalence of norms mean that there are positive constants
$a$ and $b$ such that for all $x\in {\mathbb R}^n$,
$$a \|x\|\le \|x\|_2\le b\|x\|$$


TITLE: If an object just spontaneously appeared, would it have gravitational potential energy?
QUESTION [0 upvotes]: Possibly a very amateur misunderstanding, but if it were possible for an object to appear out of nowhere and remained and rest, then its position would not have moved at all. Since the equation for calculation of GPE relies on change in position rather than the position itself, if this was zero then surely the gravitational potential energy would be zero, too? However, if it were on Earth and you dropped it, it would fall. Wouldn't that suggest that the object does in fact have GPE? Or perhaps is this just an impossible theoretical situation?
Thanks in advance for any help!

REPLY [1 votes]: Yes.
Gravitational potential does not require a change in position, it requires a hypothetical change in position. Wikipedia defines gravitational potential as the energy "that would be needed to move an object to that location from a fixed reference location" (emphasis mine). Two masses always have gravitational potential with respect to each other. It's possible to calculate the potential energy of a mass with respect to some point even in total absence of any information of how it got there, whether it was positioned through normal methods, or was magically put into place by some physics-defying method. A mass in a gravitational field will fall when released - to suggest that a mass spontaneously appearing in a gravitational field might not have potential energy implies that the mass can defy gravity. Assuming the only thing "special" about the mass is its spontaneous appearance, it will be affected by gravity just like every other bit of matter.
Furthermore, gravitational energy is always computed with respect to some particular point. It's not mathematically possible for a mass to have zero potential energy with respect to all points, as that implies there is zero potential energy difference between any of those points, which is clearly not the case. A mass cannot have 0 PE with respect to the earth and 0 PE with respect to the sun, as that implies there is 0 PE between the earth and the sun, which is obviously untrue. Note that we can calculate the earth's PE with respect to the sun regardless of whether the earth was positioned by hard physics, or placed by a divine being. Potential energy describes a state, not the process that led to it. Once the object magically appears, the state is indistinguishable from one where the object was placed normally. The object is there, so it has potential energy - how it got there is irrelevant.
Of course, a mass "spontaneously appearing" may be physics-defying in itself, but if we take that at face value and assume all other physics operate as usual, the mass must have potential energy.


TITLE: Adding fractions with exponents
QUESTION [0 upvotes]: $$3^5 + {1\over3^5}=?$$
My first instinct was to rewrite the second term as $3^{-5}$. Since the base is $3$, rewrite as $3^{5+-5}$. It simplifies to $3^0= 1$. Apparently this is incorrect. Can anyone please explain why?

REPLY [1 votes]: This is incorrect because the exponent rule that you were thinking of is: $$a^b\cdot a^c = a^{b + c}$$ So if you had $3^{5}\cdot 3^{-5}$ then you could use that rule:
$$3^{5}\cdot 3^{-5} = 3^{0}$$.

REPLY [1 votes]: As has already been pointed out $a^b+a^c\neq a^{b+c}$. That means in your case the best you can do is try and get a common denominator for both numbers and add them. Try and see if you know how.

REPLY [0 votes]: There is not much you can do in terms of simplifying these exponents. You could rewrite $3^{5}$ as $\frac{3^{10}}{3^{5}}$ then simplify with common denominators to
$$\frac{3^{10}+1}{3^{5}}$$


TITLE: Induction prove, how to come to $n\cdot(n+1)$
QUESTION [1 upvotes]: I am trying to solve an induction problem. Here are the steps for the example.
Prove this equation $$
1\cdot2 + 2\cdot3 + 3\cdot 4 + 4\cdot 5+\dots + \cdots +(n-1)\cdot n  ={n\cdot(n-1)\cdot(n+1)\over 3 }
$$ for $n=2,3,4,5$ and prove that the equation is right for all natural numbers $n\ge 2$ with induction.
${}{}$
induction beginning:
$$
\sum_{i=2}^n = {n\cdot(n-1)\cdot(n+1)\over 3}
$$
-> this is clear to me!
induction hypothesis:
$$
\sum_{i=2}^n = {n\cdot(n+1)\cdot(n+2)\over 3}
$$
->here you just put n+1, also clear to me
prove:
$$
\sum_{i=2}^n = {(n-1)\cdot n\cdot(n+1)\over 3} + n\cdot(n+1) = {n\cdot(n+1)\cdot(n+2)\over 3}
$$
-> not clear
Then you put up the fuction to prove it, however I do not understand, why you add $n\cdot(n+1)$ and how to come to $n\cdot(n+1)$?
Thx for your answer!!!

REPLY [3 votes]: For the left side of your equations, you should show what is being added, as $$\sum_{i=2}^n(i-1)i$$
For your induction beginning, you just check for $n=2$, getting $1 \cdot 2=\frac {2 \cdot 1 \cdot 3}3=2$
Then the induction hypothesis is $$\sum_{i=2}^n (i-1)i= {n*(n-1)*(n+1)\over 3}$$
(note the difference from what you wrote)
Now you want to prove it for $n+1$ which would be $$\sum_{i=2}^{n+1} (i-1)i= {(n+1)*n*(n+2)\over 3}$$
So we have $$\sum_{i=2}^{n+1} (i-1)i=\sum_{i=2}^n (i-1)i+n(n+1)={n*(n-1)*(n+1)\over 3}+n(n+1)$$
The reason you add $n(n+1)$ is that is the new term when you raise the upper limit of the sum to $n+1$.  Your induction hypothesis shows you how to sum up to $n$, which you need to make use of.  Now see if you can work the right side to get the right side of your goal.

REPLY [0 votes]: You want to prove $$\sum_{k=1}^{n} k(k+1) =\frac{n(n+1)(n+2)}{3}$$
induction beginning:
For $n=1$
$$
\sum_{k=1}^1 k(k+1)=2= {1\cdot(1+1)\cdot(1+2)\over 3}
$$
holds
induction hypothesis:
Assume $$
\sum_{k=1}^{n} k(k+1) =\frac{n(n+1)(n+2)}{3}
$$
prove:
$$\sum_{k=1}^{n+1} k(k+1) =\frac{(n+1)(n+2)(n+3)}{3}$$
We have
$$
\sum_{k=1}^{n+1} k(k+1) =\sum_{k=1}^{n} k(k+1) +(n+1)(n+2)=\frac{n(n+1)(n+2)}{3} +(n+1)(n+2)...(\text{easy algebra})...=
 \frac{(n+1)(n+2)(n+3)}{3}
$$


TITLE: When Does $\operatorname{Span}(A+B)=\operatorname{Span}(A)+\operatorname{Span}(B)$?
QUESTION [3 upvotes]: let there be $A,B \in V$.

$$A+B=\{a+b: a\in A,b \in B\}$$
Does $\operatorname{Span}(A+B)=\operatorname{Span}(A)+\operatorname{Span}(B)$?
I think it is right, even if there are vectors that are both in $A$ and $B$.

REPLY [3 votes]: Your notation is unclear. You probably mean $A$ and $B$ are subsets of $V$. If we denote by $A+B=\{a+b: a\in A, b\in B\}$, then a vector $v$ belongs to $\def\span{\operatorname{span}}\span(A+B)$ if it can be written as
$$
v=\alpha_1(a_1+b_1)+\alpha_2(a_2+b_2)+\dots+\alpha_n(a_n+b_n)
$$
for scalars $\alpha_1,\alpha_2,\dots,\alpha_n$ and $a_1,a_2,\dots,a_n\in A$ and $b_1,b_2,\dots,b_n\in B$. In this case
$$
v=(\alpha_1a_1+\alpha_2a_2+\dots+\alpha_na_n)+
(\alpha_1b_1+\alpha_2b_2+\dots+\alpha_nb_n)
$$
and this proves $\span(A+B)\subseteq\span(A)+\span(B)$.
If the vector space $V$ is finite dimensional, in order to have the reverse inclusion the condition is
\begin{align}
\dim\span(A+B)&=\dim(\span(A)+\span(B))\\
&=\dim\span(A)+\dim\span(B)-\dim(\span(A)\cap\span(B))
\end{align}
This is not true in general; an easy example is $A=\{u\}$, $B=\{-u\}$, where $u\ne0$. Then $A+B=\{0\}$, while $\span(A)=\span(B)=\langle u\rangle$ and, clearly,
$$
\{0\}=\span(A+B)\ne\span(A)+\span(B)=\langle u\rangle.
$$


TITLE: Cosine Summation Lemma
QUESTION [1 upvotes]: Just to give some background, I am attempting to solve part (b) of this Fourier matrix - multiplicity of eigenvalues? problem from Artin's Algebra textbook. If I can prove the following lemma, I should be able to finish my work.
Lemma: Let $m$ and $n$ be positive integers such that n does not divide m. Then $\sum_{l=0}^{n-1} cos(2πml/n) = 0$
I was able to prove this for even n. I was also able to prove it for the special case n=3, but  I am stumped on how to prove the general case. Please help me!

REPLY [1 votes]: Observe the angle-sum identities \begin{align*} \sin(lx + \frac 12 x) &= \sin lx \cos \frac x2 + \cos lx \sin \frac x2 \\  \sin(lx - \frac 12 x) &= \sin lx \cos \frac x2 - \cos lx \sin \frac x2 \end{align*}
lead to
$$ \cos lx \sin \frac x2 = \frac 12 \left[  \sin(lx + \frac 12 x) - \sin(lx - \frac 12 x) \right].$$
You can sum both sides from $l=0$ to $l=n-1$ and use the telescoping nature of the sum to obtain
$$
\sum_{l=0}^{n-1} \cos lx \sin \frac x2 = \frac 12 \sum_{l=0}^{n-1} \left[  \sin(lx + \frac 12 x) - \sin(lx - \frac 12 x) \right] = \frac 12 \left[ \sin (n-\frac 12) x - \sin( -\frac 12 x) \right].$$
The difference-of-sines formula gives you
$$ \sin (n-\frac 12)x - \sin(-\frac 12 x) =  2 \cos \left(\frac{n-1}{2}\right)  x \sin nx.$$
Thus
$$\sum_{l=0}^{n-1} \cos lx = \frac 1{ \sin \frac x2} \left[ \cos \left(\frac{n-1}{2}\right)  x \sin nx \right].$$
With $x = \dfrac{2\pi m}{n}$ you have $$\sin nx = \sin 2\pi m = 0$$ and $$\sin \frac x2 = \sin \frac{\pi m}{n} \not= 0$$ because $n$ does not divide $m$. Thus the sum is zero.


TITLE: Relative measurement error in addition / subtraction formula
QUESTION [0 upvotes]: if the Absolute Measurement Error = Wrong Value - Correct Value

positive/negative sign is neglected since a posisitve calculation
error is as bad as a negative calculation error.

and if the
Relative Measurement Error = Absolute Measurement Error ÷ Correct Value
and the value is determind by adding or subtracting two different values. For example:
first   volume= 5. correct number was: 4.5 = (4.5±.5)
secound volume= 6. correct number was: 4   = (4±2)
Final Volume = Second Volume - First Volume.
What is the relative measurement error of the final volume in the previous calculation?

REPLY [0 votes]: Uncorrelated errors add in quadrature, so if:
$$ V = V_1- V_2 $$
then
$$ \delta V^2 = \delta V_1^2 + \delta V_2^2$$
Thus:
$$ \delta V = \sqrt{0.5^2 + 2^2} = 2.06 $$
This is a special case of: If $z$ is some function of $x$ and $y$:
$$ z = f(x,y) $$
then
$$ \delta z^2 = \big(\frac{\partial f}{\partial x}\delta x\big)^2 + \big(\frac{\partial f}{\partial y}\delta y\big)^2 $$


TITLE: To what extent does a torsor determine a group
QUESTION [12 upvotes]: Let $k$ be a field, and suppose $G$ is a group-scheme over $k$ (I am happy to assume that $k=\mathbb{Q}$ and that $G$ is affine). A $G$-torsor over $k$ is a non-empty $k$-scheme $T$ equipped with an action $a:G\times T\to T$, such that $(a,\pi_2):G\times T\to T\times T$ is an isomorphism. There is an induced morphism
$$
\begin{align*}
T\times T\times T&\to T,\\
(t_1,t_2,t_3)&\mapsto a\bigg(\pi_2\big((a,\pi_2)^{-1}(t_1,t_2)\big),t_3\bigg)=``t_1t_2^{-1}t_3".
\end{align*}
$$
I would like to know to what extent the map $T\times T\times T\to T$ determines  $G$. I mean the following:

For which pairs of groups $G$ and $G'$ does there exist a $G$-torsor $T$, a $G'$-torsor $T'$, and an isomorphism $T\cong T'$ compatible with the maps $T\times T\times T\to T$ and $T'\times T'\times T'\to T'$?

REPLY [11 votes]: Under fairly general exactness assumptions on the ambient category, one can perform the construction of a group from $(T,m)$, if $T$ is an object equipped with a heap structure $m:T^3\to T$, i. e. an associative Maltsev operation. That is, $m$ must just satisfy (diagrammatic versions of) $m(x,x,y)=m(y,x,x)=y$ and $m(m(x,y,z),t,u)=m(x,y,m(z,t,u))$.
This is done exactly as one does construct a vector space from an affine space:
one gets a group as a quotient $T\times T\twoheadrightarrow G$ (which becomes the division map of the torsor $T$ over $G$, $(x,y)\mapsto x/y$ where $x/y$ is the unique element of $G$ with $(x/y)y=x$).
The quotient is by identifying $(x,y)$ with $(m(x,y,z),z)$ for all $x,y,z$. That is, one forms the coequalizer $T^3\rightrightarrows T^2\twoheadrightarrow G$ of appropriate morphisms.
The action of $G$ on $T$ is determined by $(x/y)\cdot z=m(x,y,z)$, and gives a $G$-torsor. It is an easy exercise to do this "without elements". Moreover any other group over which $T$ is a torsor and such that $m(x,y,z)$ is $(x/y)\cdot z$ for that structure too, is isomorphic to $G$, by an isomorphism compatible with the actions on $T$.
In fact switching sides one also has another group $G'$ such that $T$ is a $G$-$G'$-bitorsor.
This even works without global support assumption: defining support $S$ of $T$ as the coequalizer $T\times T\rightrightarrows T\twoheadrightarrow S$ of the projections $T\times T\rightrightarrows T$, one gets a group and a torsor structure in the slice category over $S$.
There are way too many references to choose from, spread across decades, so I think all this must be considered folklore.


TITLE: Help with equality of complex integrals
QUESTION [2 upvotes]: I need to prove this equality of integrals...but i dont know how to begin, so if anyone can give an idea...
Let f a continuous function on $\overline{D}=\{z : |z|\leq 1\}$. Then:
$$\overline{\int_{|z|=1} f(z) dz} = - \int_{|z|=1} \frac{\overline{f(z)}}{z^2} dz$$
And then ihave to calculate that
$$\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}dz$$
for every a.

REPLY [1 votes]: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
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 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\ol{\int_{\verts{z}\ =\ 1}\fermi\pars{z}\,\dd z}
     =-\int_{\verts{z}\ =\ 1}{\ol{f(z)}}\,{\dd z \over z^{2}}:\ {\large ?}}$

Note that
  \begin{align}&\ol{\int_{\verts{z}\ =\ 1}\fermi\pars{z}\,\dd z}
=\int_{\verts{z}\ =\ 1}\ol{\fermi\pars{z}}\,\dd\ol{z}
\end{align}

\begin{align}
&\mbox{However,}
\\[3mm]&1 = \verts{z}^{2} = z\,\ol{z}\quad\imp\quad
0 = \pars{\dd z}\ol{z} + z\,\dd\ol{z}\quad\imp\quad\dd\ol{z}
=-\,{\ol{z}\,\dd z \over z}=-\,{\verts{z}^{2}\,\dd z \over z^{2}}
=-\,{\dd z \over z^{2}}
\end{align}

$$
\ol{\int_{\verts{z}\ =\ 1}\fermi\pars{z}\,\dd z}
=-\int_{\verts{z}\ =\ 1}\ol{\fermi\pars{z}}\,{\dd z \over z^{2}}
$$


TITLE: Is it true that $\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S = \{ 0 \}$ for arbitrary linear maps $S, T$?
QUESTION [0 upvotes]: Suppose $U$ is a finite-dimensional vector space, that $S ∈ L(V, W )$, and that $T ∈ L(U, V )$. I'm trying to prove that $\operatorname{dim} \operatorname{null}(ST ) = \operatorname{dim} \operatorname{null}(T ) + \operatorname{dim} (\operatorname{range}(T ) ∩ \operatorname{null}(S))$.
First I realized that for a vector $x$ to be in $\operatorname{null}(ST)$, it can
(1) be in $\operatorname{null} T$, in which case $Sx=0$ (since $S$ is a linear map) and thus $x \in \operatorname{null}(ST)$
(2) get mapped to an element of $\operatorname{range}T$ that happens to be in the null space of $S$.
Hence $\operatorname{null}T$ and $\operatorname{range}(T ) ∩ \operatorname{null}(S)$ combine to make $\operatorname{null}(ST)$, i.e.  $$\operatorname{null}T + \operatorname{range}(T ) ∩ \operatorname{null}(S) = \operatorname{null}(ST).$$
Now, taking $\text{dim}$s, I get
$$\operatorname{dim}\operatorname{null}T + \operatorname{dim}(\operatorname{range}(T ) ∩ \operatorname{null}(S)) - \operatorname{dim}(\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S) = \operatorname{dim}\operatorname{null}(ST).$$
It seems like for the theorem to hold, we must have that $\text{dim}(\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S) = 0$, i.e. that $\text{null} \, T \cap \text{range} \, T \cap \text{null} \, S  = \{ 0 \}$ for arbitrary linear maps $S, T$. Could someone please point me in the right direction as to how go about conceptualizing and proving this?

REPLY [1 votes]: Say $u_1,\dots,u_n \in U$ is a basis of the kernel of $T$ and $v_1,\dots,v_m \in V$ is a basis of the space formed by intersecting the range of $T$ with the kernel of $S$. As the $v_i$s are in the range of $T$, they have preimages in $U$. Pick one preimage for each $v_i$;  you can group them with the earlier $u_i$s to get a collection of linearly independent (why?) vectors $u_1,\dots,u_{n+m}\in U$. This implies that the nullity of $ST$ is at least $n+m$.
Conversely let $u\in U$ be an arbitrary vector in the kernel of $ST$. As you noted, it is either in the kernel of $T$ or is mapped by $T$ into the kernel of $S$. Thus it is in the span of $u_1,\dots,u_{n+m}$ (does it matter that we chose specific preimages?), and so the nullity is at most $n+m$.