Explanation of Set Notation In a set, for example the following one

$$\{{(x_{1},x_{2}) \in \mathbb{R}^{2}}|x_{1}+x_{2}=1\}$$

does the notation $(x_{1},x_{2}) \in \mathbb{R}^{2}$ means that $x_{1}$ and $x_{2}$ are points in $\mathbb{R}^{2}$ space or it means that they are the coordinates of the points in that space?

For example if I have two points $p_{1}=(px_{1},py_{1})$ and $p_{2}=(px_{2},py_{2})$ in $\mathbb{R}^{2}$ does it mean that

$$px_{1}+py_{1}=1$$

or that

$$p_{1}+p_{2}=1$$

The set $$\Big\{ (x_1,x_2)\in\mathbb{R}^2\mid x_1+x_2=1\Big\}$$
 considers all points in $\mathbb{R}^2$ (that is, ordered pairs with real coordinates) such that they obey a property (in this case, the sum of their coordinates is $1$).

For example, point $\left(\dfrac12, \dfrac12\right)$ belongs to such set, because $\dfrac12 + \dfrac12 = 1$.

If you consider the set of points in $ \mathbb{R}^2 $ that do not obey this property, you can write
$$ \Big\{ (x_1,x_2)\in\mathbb{R}^2\mid x_1+x_2\neq1\Big\}.$$

Set builder notation, like what you are asking about here, is usually in the form of $\{x\in S | \phi(x)\}$, where the elements of the set are of the form $x$, they belong to the ambient set $S$, and they all satisfy the formula $\phi(x)$.

Here, our set reads $\{(x_1,x_2)\in\mathbb{R}^2 | x_1+x_2=1\}$. Our elements are ordered pairs, belonging to $\mathbb{R}^2$, with the property that their sum of the coordinates is $1$.


How to prove that these two functions do not intersect? I have this function $\;g(x)-x\;$ on the domain $\;0<x<\frac{1}{3}\;$, where
$g(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)+1}}{2 \sqrt{2} \pi  \cosh (\pi  x)}+\frac{18 \sqrt{2} x^2 \left(9 \pi  x^3+4 \tanh (\pi  x)\right)}{2 \sqrt{2} \pi }.$
I need to prove that $\;g(x)\;$ does not intersect $\;x\;$ over the mentioned domain. Calculating the first derivative does not work since it makes the problem more complicated. The plot of the functions is attached.
Any hint or suggestion is welcome.

Partial answer
Denote
$$g_1(x)=\frac{\left(9 x^2+1\right) \cosh ^{-1}\left(\frac{36 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)}{\left(9 x^2+1\right)^2}\right) \sqrt{81 x^4+54 x^2+\left(1-9 x^2\right)^2 \cosh (2 \pi  x)+1}}{2 \sqrt{2} \pi  \cosh (\pi  x)}$$ and
$$g_2(x) = \frac{18 \sqrt{2} x^2 \left(9 \pi  x^3+4 \tanh (\pi  x)\right)}{2 \sqrt{2} \pi }$$
$g_1, g_2$ are positive maps and we have $g = g_1+g_2$.
$g_2(x)/x$ is increasing on $(0, 1/3)$ as a product of positive increasing maps. Moreover $g_2(0.06) / 0.06 >1.0$. Hence we have the partial result that $g(x) - x$ is positive on $(0.06, 1/3)$. Remains to prove the result on $(0, 0.06]$.

$$ g(0)=0, g'(0)=1 $$
$$t(x)= x, t(0)=0,t'(0)=1$$
so there is a tangential contact for the two curves with two coinciding intersections as seen in your graph. It is confirmed that at the double root there is common slope$=1$


Reduced Cost in Network Simplex Algorithm On page 5 of the slide,

[T]he reduced cost of a non-basic arc $(i, j)$ is the sum of the costs of the arcs forming a cycle with $(i, j)$ in the current tree solution.

Why is that the case?

http://www.unc.edu/depts/stat-or/courses/provan/STOR724_web/lect14_simp.pdf
(bottom of page 13)

It's the sum of costs but directed consistent with (i,j), i.e. if the arc is going in the same direction as (i,j), it counts as positive, and is negative otherwise. It makes more sense that way since you want to add arcs that create negative cycles in order to decrease the overall cost.


Contour Integration Part I'm trying to evaluate the following integral, and I'm getting stuck on one part.  Here's the integral:

$$\int_{-\infty}^\infty \frac{\sin(x)}{x(x^2+1)} dx$$

Basically, I'm converting this to the complex plane and performing a contour integration over the top half of the plane (semi-circle).  Further, I'm looping around the singularity at z=0.  Now, I'm fine with all of the integrals except for the integral involving the loop around the singularity.  For this guy, I end up with the following.  I have no idea how to evaluate this in the limit as r goes to 0.  Any idea  how to proceed?

$$\int_{\pi}^0 \frac{e^{ire^{i\theta}}}{r^2e^{i2\theta}+1} d\theta$$

Write
$$f(z)=\frac{e^{iz}}{z(z^2+1)}$$
and let $C$ be the clockwise semi-circular contour of radius $r$ about $0$.  Now $f$ has a simple pole at $z=0$ with residue $1$, so
$$f(z)-\frac{1}{z}$$
has a removable singularity at $z=0$.  Therefore, for a suitable constant $c$, the function
$$g(z)=\cases{f(z)-1/z&if $z\ne0$\cr c&if $z=0$\cr}$$
is continuous.  Now
$$\left|\int_C g(z)\,dz\right|\le (\pi r)\max_{|z|\le r}|g(z)|\to (\pi)(0)|c|=0$$
as $r\to0$, and
$$\int_C\frac{1}{z}\,dz=\int_\pi^0 \frac{ie^{i\theta}}{e^{i\theta}}\,d\theta=-\pi i\ .$$
Therefore
$$\int_C f(z)\,dz=\int_C \Bigl(g(z)+\frac{1}{z}\Bigr) dz\to -\pi i$$
as $r\to0$.

$\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}
\color{#00f}{\large%
\int_{-\infty}^{\infty}{\sin\pars{x} \over x\pars{x^{2} + 1}}\,\dd x}&=
\Im\pp\int_{-\infty}^{\infty}{\expo{\ic x} \over x\pars{x^{2} + 1}}\,\dd x
=
\Im\int_{-\infty}^{\infty}{\expo{\ic x} \over x^{2} + 1}\bracks{%
{1 \over x + \ic 0^{+}} + \ic\pi\,\delta\pars{x}}\,\dd x
\\[3mm]&=
\Im\int_{-\infty}^{\infty}{\expo{\ic x} \over \pars{x + \ic 0^{+}}\pars{x^{2} + 1}}
\,\dd x + \pi
=\Im\bracks{2\pi\ic\,{\expo{\ic\pars{\ic}} \over \pars{\ic + \ic 0^{+}}
\pars{\ic + \ic}}} + \pi
\\[3mm]&=-\pi\expo{-1} + \pi
=\color{#00f}{\large{\pars{-1 + \expo{}}\pi \over \expo{}}} \approx 1.9859
\end{align}

  $\ds{\Large\tt\mbox{ADDENDA}}$ ( Contour Integration ):
  \begin{align}
\color{#00f}{\large%
\pp\int_{-\infty}^{\infty}{\sin\pars{x} \over x\pars{x^{2} + 1}}\,\dd x}
&=
\lim_{\epsilon \to 0^{+}}\Im\bracks{%
\int_{-\infty}^{-\epsilon}{\expo{\ic x} \over x\pars{x^{2} + 1}}\,\dd x
+
\int_{\epsilon}^{\infty}{\expo{\ic x} \over x\pars{x^{2} + 1}}\,\dd x}
\\[3mm]&=\lim_{\epsilon \to 0^{+}}\Im\bracks{%
2\pi\ic\,{\expo{\ic\pars{\ic}} \over \ic\pars{\ic + \ic}}\
-\
\overbrace{\int_{\pi}^{0}
{\exp\pars{\ic\epsilon\expo{\ic\theta}}
\over
\epsilon\expo{\ic\theta}\pars{\epsilon^{2}\expo{2\ic\theta} + 1}}
\,\pars{\epsilon\expo{\ic\theta}\ic\,\dd\theta}}
^{\ds{\to -\ic\pi\ \mbox{when}\ \epsilon \to 0^{+}}}}
\\[3mm]&=-\pi\expo{-1} + \pi
=\color{#00f}{\large{\pars{-1 + \expo{}}\pi \over \expo{}}} \approx 1.9859
\end{align}


I'm confused on how the chain rule is applied in a problem I'm working in book So the problem it shows a equation that I know that they are applying the chain rule, but I just don't know exactly they applied it:

$$\frac{dv}{dx} = \frac{dv}{dt} * \frac{dt}{dx} = \frac{dv}{dt} * \frac{1}{v}$$

What are the intermarried steps involved here to get the last part of this expression?

Could someone help me understand the steps of the applied chain in this particular problem? thank you in advance

You are interested in the change in velocity as a function of position.

Velocity is typically written as a function of time, and so is position. That is, we write $v(t)$ and $x(t)$. To get the first inequality, they simply applied the chain rule to those functions.

To get the second equality, they used the inverse function theorem combined with the chain rule. You can see the derivation of that result here.

It’s worth noting that there’s one more simplification that can be made. $\frac{dv}{dt}=a(t)$, so we have

$$\frac{dv}{dx}=a(t)+\frac{1}{v(t)}=\frac{v(t)a(t)+1}{v(t)}$$

$dv/dx = dv/dA dA/dx$ for any A, but here you apply it with $t$.

Also, $dt/dx = 1/(dx/dt) = 1/v$.


Computing with Lebesgue integrals This problem comes from Royden's Real Analysis, 4th ed., pg 84, #19:

  For a number $\alpha$, define $f(x)=x^\alpha$ for $0<x\le 1$ and
  $f(0)=0$. Compute $\int_0^1 f$.

MY WORK:

  I know that if $\alpha\ge 0$, then $f(x)=x^\alpha$ is a bounded
  function on $[0,1]$ and is Riemann integrable. Thus it is Lebesgue
  integrable and the two integrals are equal. So when $\alpha\ge 0$ we
  have
  $$
\int_0^1
 x^\alpha=\frac{x^{\alpha+1}}{\alpha+1}\Bigg\vert_0^1=\frac{1}{\alpha+1}.
$$ For the case when $-1<\alpha<0$, I want to define an increasing
  sequence of nonnegative measurable functions on $[0,1]$. So let
  $$
f_n=\begin{cases}x^\alpha&x\in[1/n,1]\\0&x\in[0,1/n)\end{cases} $$ and
  I want to consider the sequence of integrals $\{\int_0^1 f_n\}$. For
  each $n$ we have $f_n(x)$ is a bounded function on $[1/n,1]$ and is
  Riemann integrable. Thus for each $n$ it is Lebesgue integrable and
  the two integrals are equal. So
  $$
 \int_0^1f_n=\int_0^{\frac{1}{n}}f_n+\int_{\frac{1}{n}}^1f_n=\int_0^{\frac{1}{n}}0+\int_{\frac{1}{n}}^1x^\alpha=\frac{x^{\alpha+1}}{\alpha+1}\Bigg\vert_{\frac{1}{n}}^1=\frac{1}{\alpha+1}-\frac{\left(\frac{1}{n}\right)^{\alpha+1}}{\alpha+1}\le\frac{1}{\alpha+1}.
$$ Thus the sequence $\{\int_0^1 f_n\}$ is bounded by
  $\frac{1}{\alpha+1}$ for all $n$. Then by Beppo Levi's Lemma, $f_n$
  converges pointwise to the measurable function
  $$
 f(x)=\begin{cases}x^\alpha&x\in(0,1]\\0&x=0\end{cases} $$ and $$
 \int_0^1 f=\lim_{n\to\infty}\int_0^1
 f_n=\lim_{n\to\infty}\frac{1}{\alpha+1}-\frac{\left(\frac{1}{n}\right)^{\alpha+1}}{\alpha+1}=\frac{1}{\alpha+1}.
 $$

QUESTION:

  I am not sure what to do for the case when $\alpha\le -1$. Any ideas?
  I am certain that it must diverge to $\infty$, but I am not sure how
  to use the tools at my disposal to verify this.

For $x \in [0,1]$ and $\alpha \leqslant -1 $ we have $x^{\alpha} \geqslant x^{-1}$.

Hence,

$$\int_{[0,1]}x^{\alpha}\geqslant \int_{[0,1]}x^{-1}= \infty.$$

To show that the Lebesgue integral of $x^{-1}$ is infinite, consider the sequence of simple functions

$$\phi_n = \sum_{j=1}^nj \chi_{((j+1)^{-1},j^{-1})}.$$

Note that $0 \leqslant \phi_n(x) \leqslant x^{-1}$ and

$$\int_{(0,1]}\phi_n = \sum_{j=1}^n j\left(\frac1{j}- \frac1{j+1}\right)=\sum_{j=1}^n \frac1{j+1}\to_{n \to \infty} \infty$$

Recall in Calculus:

For $\alpha\leq -1$

$\int_0^1x^{-5} = lim_{n-->0}$$ \int_0^1x^{-5}=$$ lim_{n-->0}$ $-\frac{1}{4x^4}|_0^1=$ $-\frac{1}{4}lim_{n-->0}$ $(1-\frac{1}{n^4})=\infty$

I honestly think that this should be enough for $\alpha\leq -1$. This is Riemann integrable, right?


Intuitive explanation of the tower property of conditional expectation I understand how to define conditional expectation and how to prove that it exists.

Further, I think I understand what conditional expectation means intuitively. I can also prove the tower property, that is if $X$ and $Y$ are random variables (or $Y$ a $\sigma$-field) then we have that

$$\mathbb E[X] = \mathbb{E}[\mathbb E [X | Y]].$$

My question is: What is the intuitive meaning of this? It seems quite puzzling to me.

(I could find similar questions but not this one.)

First, recall that in $E[X|Y]$ we are taking the expectation with respect to $X$, and so it can be written as $E[X|Y]=E_X[X|Y]=g(Y)$ . Because it's a funciton of $Y$, it's a random variable, and hence we can take its expectation (with respect to $Y$ now). So the double expectation should be read as $E_Y[E_X[X|Y]]$.

About the intuitive meaning, there are several approaches. I like to think of the expectation as a kind of predictor/guess (indeed, it's the predictor that minimizes the mean squared error).

Suppose for example that $X, Y$ are two (positively) correlated variables, say the weigth and height of persons from a given population. The expectation of the weight $E(X)$ would be my best guess of the weight of a unknown person: I'd bet for this value, if not given more data (my uninformed bet is constant). Instead, if I know the height, I'd bet for $E(X | Y)$ : that means that for different persons I'd bet a diferent value, and my informed bet would not be constant: sometimes I'd bet more that the "uninformed bet" $E(X)$ (for tall persons) , sometime less. The natural question arises, can I say something about my informed bet in average? Well, the tower property answers: In average, you'll bet the same.

The end result is the same no matter how you slice it.

Think of the inner term of the RHS : E[E[X|Y]] as integrating over a specific direction, then the outer expectation integrates over the remaining directions.  This obviously skips important details but that's how I intuitively rationalize it.


Compact sets and Kuratowski limit I have been struggling with the following claim:

Let $A_n$ be a sequence of compact sets and $A$ a compact set. $A=\lim\sup_n A_n=\lim\inf_n A_n$ iff $d_H(A_n,A)\to 0$ where $d_H(.,.)$ is the Hausdorff metric.

$\lim\inf$  and $\lim\sup$ are defined by $\lim\inf_nA_n=\left\{y\in Y:\forall \varepsilon>0,\exists N, n\geq N\quad \implies B_{\varepsilon}(y)\cap A_n\neq\emptyset\quad \right\}$ and, $\lim\sup_n A_n=\left\{y\in Y: \forall\varepsilon>0,\forall N,\exists n\geq N \text{ so that } B_\epsilon(y)\cap A_n\neq\emptyset \right\}$.

Also $d_H (X,Y)=\max\left(\inf\left\{\epsilon>0:Y\subset\bigcup_{x\in X}B_\epsilon(x) \right\}, \inf\left\{\epsilon>0:X\subset\bigcup_{y\in Y}B_\epsilon(y)   \right\}  \right)$.

$\Rightarrow$: Let $\varepsilon>0$ be given. For each $a\in A$ $\exists N(a,\varepsilon)$ so that $n\geq N$ implies $B_\varepsilon(a)\cap A_n\neq\emptyset$. By compactness of $A$ we can find finite $a_1,a_2\dots a_k$ and associated $N_1,N_2\dots N_k$ so that $A\subset\cup_{i=1}^k B_\varepsilon(a_i)$. Let $N=\max_{1\leq i\leq k}N_i$. I want to show $n\geq N$ implies $A_n\subset A^\varepsilon$ but couldn't manage.

$\Leftarrow$: I want to show $A\subset\lim\inf A_n\subset\lim\sup A_n\subset A$. Let $\varepsilon>0$ be given. There exists $N$ such that $d_H(A_n,A)<\varepsilon$ for $n$ large enough. Pick $a\in A$, so $a\in A_n^{\varepsilon}$  which implies $B_{\varepsilon}(a)\cap A_n\neq\emptyset$ for all such large $n$'s. Thus $a\in\lim\inf A_n$. To finish this side, I need to show $\lim\sup A_n\subset A$  yet again couldn't find the solution. Thanks for any help!

For the direction

$$A = \limsup_n A_n = \liminf_n A_n \implies d_H(A_n,A) \to 0,$$

you need some additional hypothesis on the ambient space, namely that it is compact.

Without that hypothesis, the implication does not hold. An explicit counterexample using compact subsets of $\mathbb{R}$ is $A = [0,1]$, and $A_n = [0,1] \cup \{n\}$ for $n\in \mathbb{N}$. Generally, if the ambient (metric) space is not compact, it contains a sequence $(x_n)_{n\in \mathbb{N}}$ without accumulation point, and then $A = \{x_0\}$ and $A_n = \{x_0,x_n\}$ is a counterexample.

So let's suppose that the ambient space is compact.

What you have so far proves that for large enough $n$, you have $A \subset A_n^{2\varepsilon}$. For the converse inclusion, that $A_n \subset A^\varepsilon$ for all large enough $n$, you use the compactness of the ambient space. Suppose $A_n \setminus A^\varepsilon \neq\varnothing$ for infinitely many $n$. Then

$$F_n = \overline{\bigcup_{k=n}^\infty A_k\setminus A^\varepsilon}$$

is a decreasing sequence of nonempty compact sets, hence $F = \bigcap\limits_{n\in\mathbb{N}} F_n \neq \varnothing$. Now, what can be said about the points in $F$?

For the other direction,

$$d_H(A_n,A) \to 0 \implies A = \limsup_n A_n = \liminf_n A_n,$$

you have correctly shown that $A \subset \liminf_n A_n$. Now let $x\in \limsup_n A_n$. Fix $\varepsilon > 0$. Choose $N$ such that $d_H(A_n,A) < \varepsilon/2$ for $n \geqslant N$. Since $x \in \limsup_n A_n$, there is an $n_\varepsilon > N$ such that $A_{n_\varepsilon} \cap B_{\varepsilon/2}(x) \neq\varnothing$. But, for $y \in A_{n_\varepsilon}$, we also have $B_{\varepsilon/2}(y) \cap A \neq \varnothing$, and hence $B_\varepsilon(x) \cap A \neq \varnothing$. That means $x \in A^\varepsilon$. Since $\varepsilon$ was arbitrary, it follows that

$$x \in \bigcap_{\varepsilon > 0} A^\varepsilon = A.$$

Let's think about this.  Some directions don't work.  In space $A=[0,1]$, which is compact, let $A_n = \{ k/2^n : k=0,1,\dots,2^n\}$.  Then $d_H(A_n,A) \to 0$.  But $\limsup A_n = \liminf A_n = \bigcup A_n$ is not compact at all.


Find the Probability that duration of one run is longer than another One day a run begins at 5:31 p.m. and ends at 5:46 p.m. The following day another run begins at 5:31 p.m. and ends at 5:47 p.m. We have a watch to measure the runtime but it shows only hours and minutes (not seconds). What is the probability that the run the first day lasted longer then the second day?

I keep getting that the run on Day1 can last from 14:01 to 15:59 minutes and Day2 can last from 15:01 to 16:59 minutes. Then:
$$
P(Day2<Day1) = [indpt]=  \int\int f_X(x)*f_Y(y) dxdy =
$$
and assuming that the duration of the runs have a uniform distribution f = 1/(b-a) = 1/(59/30), I get:
$$
=\int\int (30/59)*(30/59) dxdy = (30/59)^2*0.95*0.95≈0.23
$$
when I integrate from 15+(1/60) to 15+(58/60), i.e. the period of time where Day2 can be smaller than Day1. This is incorrect since the answer should be 1/24 ≈ 0.0416, can anyone please show me where I'm wrong/ how to solve it?

We have four times involved so define an r.v. for each. The range for each of the times is $1$ minute and it is simpler to use minutes as our units. We define the random variables as:

\begin{align}
X &= \text{Day 1 Start Time - 5:31pm (in minutes)} \\
Y &= \text{Day 2 Start Time - 5:31pm (in minutes)} \\
Z &= \text{Day 1 End Time - 5:46pm (in minutes)} \\
W &= \text{Day 2 End Time - 5:46pm (in minutes)} \\
\end{align}

So the pdfs are $f_X(x)=f_Y(y)=f_Z(z)=f_W(w)=1$ and the ranges are $X,Y,Z\in(0,1)$ and $W\in(1,2)$.

We want the probability that $Y-X\gt W-Z$. So we integrate as follows:

\begin{align}
P(Y-X\gt W-Z) &= \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=1-(y-x)}^{1} \int_{w=1}^{z+(y-x)} 1\cdot 1\cdot 1\cdot 1\;dw\;dz\;dy\;dx \\
&= \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=1-(y-x)}^{1} (z+y-x-1) \;dz\;dy\;dx \\
&= \int_{x=0}^{1} \int_{y=x}^{1} \left[z^2/2+yz-xz-z\right]_{z=1-y+x}^{1}  \;dy\;dx \\
&= \int_{x=0}^{1} \int_{y=x}^{1} \left(y^2/2-xy+x^2/2\right)  \;dy\;dx \\
&= \int_{x=0}^{1} \left[y^3/6-xy^2/2+x^2y/2\right]_{y=x}^{1}  \;dx \\
&= \int_{x=0}^{1} \left(1/6-x/2+x^2/2-x^3/6\right) \;dx \\
&= \left[x/6-x^2/4+x^3/6-x^4/24\right]_{x=0}^{1} \\
&= 1/24.
\end{align}

It seems to me the answer should be about $\frac{1}{8}$.

You're correct that the valid potential lengths are 14:01 - 15:59 and 15:01 - 16:59 for day 1 and day 2, respectively.  Just to make the numbers nicer let's recenter to 0, so that the times are 0:00 - 1:58 for day 1 and 1:00 - 2:58 for day 2.  Note the possible time in seconds for day 1's length is 118 seconds and ranges from 0 to 118 in seconds, while day 2 ranges from 60 to 178 seconds, also with total length 118 seconds.

Now let $X \sim Unif(0,118)$ and $Y \sim Unif(60,178)$ be random variables corresponding to the lengths of day 1's and day 2's runs, respectively.  Thus the random variables live on the square $[0,118] \times [60,178]$ and we want to know $P(X > Y)$.  This is simply the area of the triangle in the bottom right of this plot divided by the total area, which you can see is about $\frac{1}{8}$.

Specifically, the joint pdf is
$$
f_{X,Y}(x,y) = \frac{1}{118^2}
$$
and the probability is
\begin{align*}
P(X > Y) & = \int_{60}^{118} \int_y^{118}\frac{1}{118^2} \, dx \, dy \approx 0.1208.
\end{align*}


Is $S^0 = \{x \in \mathbb{R} \ |\ \|x\| = 1 \}$ path connected in the trivial topology? Is $S^0 = \{x \in \mathbb{R} \   |\  \|x\| = 1 \}$ path connected in the trivial topology?

I'll add some definitions here for brevity

  Definition (Path): Given points $x, y$ of a space $X$, a path in $X$ from $x$ to $y$ is a continuous map $f: [a, b] \to X$ of some closed interval in the real line into $X$ such that $f(a) = x$ and $f(b) = y$.

  Definition (Path Connectedness): A space $X$ is said to be path connected if every pair of points can be joined by a path in $X$.

My Attempted Proof

We have clearly that $S^0 = \{-1, 1\}$. I assert that $S^0$ is path connected in the trivial topology $\mathcal{T} = \{ \emptyset, \{-1, 1\}\}$. Pick $[a, b] \subset \mathbb{R}$ such that $a \in \mathbb{Q}$ and $b \not\in \mathbb{Q}$.

If we define $f : [a, b] \subset \mathbb{R} \to \{-1, 1\}$ such that $$f(x) = \begin{cases}-1 \ \ \ \text{if} \ \ \ x \in \mathbb{Q}\\
\ \ \ 1 \ \ \ \text{if} \ \ \ x \not\in \mathbb{Q} \end{cases}$$
for $x \in [a, b]$, then $f$ is clearly continuous as $f^{-1}\left[\{-1, 1\}\right] = [a, b]$ which is open in the subsace topology. Furthermore $f$ is a path since $f(a) = -1$ and $f(b) = 1$, hence $(S^0, \mathcal{T})$ is path connected $\ \square$

Is my proof correct? This seemed like a slightly counter-intuitive example that I constructed, or perhaps it is trivial, I'm not so sure.

Your proof is correct, but your example of $f$ is more complicated than is necessary.  Suppose you have any function $f:[a,b]\to S^0$.  To check that $f$ is continuous, you just need to know that $f^{-1}(\emptyset)$ and $f^{-1}(\{-1,1\})$ are both open.  But $f^{-1}(\emptyset)=\emptyset$ and $f^{-1}(\{-1,1\})=[a,b]$, no matter what $f$ is (why?).  Since $\emptyset$ and $[a,b]$ are both open in $[a,b]$, this means $f$ is automatically continuous.

So actually every map $f:[a,b]\to S^0$ is continuous (using the trivial topology on $S^0$).  Can you find a simpler example of such an $f$ that will give a path from $-1$ to $1$?

(Also, strictly speaking, to prove $S^0$ is path-connected, you need to prove not just that there is a path from $-1$ to $1$, but also that there is a path from $1$ to $-1$, from $1$ to $1$, and from $-1$ to $-1$.  Can you explain how to get a path from $1$ to $-1$ given a path from $-1$ to $1$?  And can you explain why it is trivial that there is a path from any point to itself?)

Hi I made an edit but please ignore it now :). I think that the topology you have defined is not a topology because it does not contain the whole interval $[-1,1]$. However if you consider it to be a topology on $\{-1,1\}$ then your proof is correct.


Simple Field extensions. Suppose $[L:K]=p$ $p$=prime .Then there exists $u$ such that $L=K(u)$

Proof.If  $[L:K]=p$ $p$=prime then for $M$ an  arbitrary field extension must hold that  $$[L:K]=[L:M][M:K]$$ if $[L:K]=prime$ then either $[L:M]=1$ or $[M:K]$=1 if the first one is  true then $L=M$ for arbitrary field extension so it is true for $L=K(u)$ .

Now if the second one is true $M=K$ and i cant find argument that can prove that exists $u$ such that $L=K(u)$ only that for any extension i get K=M therefore K(u)=K.

THat is my question on how to solve the second part

Take a element u in L which is not in K consider the field K(u) which contains K properly and using multiplicative properties of degree we can show that degree of K(u) over K is 1 or p since is not degree 1 hence it must be p and hence it is equal to L itself because it contained in L also has same degree.


Show polynomial is irreducible Show that $x^4 + 4x^3 + 6x^2 + 2x + 1$ is irreducible over the field of rational numbers Q.
Now I've tried substituting x-1 for x to show that it is irreducible, but I'm having a hard time. This is the correct way to go about it, I know, but I don't know how to execute the problem.

Just take the polynomial modulo $2$. By the RRT, $x^4+1$ is irreducible so we are done.


Is this a good expression of the claim using logical notation? Four Square Theorem: Every positive integer can be written as a sum of four integer squares. Expressed in logical notation :
$$\forall n>0 =a_{0}^{2} + a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$$

In natural language: for all integers $n$, if $n > 0$, then there are integers $a_0, a_1, a_2, a_3$ such that $n = a_0^2 + a_1^2 + a_2^2 + a_3^2$. Symbolically:
$$\forall n \in {\mathbb Z}: n > 0 \to \exists a_0, a_1, a_2, a_3 \in {\mathbb Z}: n = a_0^2 + a_1^2 + a_2^2 + a_3^2.$$
If the context makes it clear that your talking about integers, you could also say
$$\forall n: n > 0 \to \exists a_0, a_1, a_2, a_3: n = a_0^2 + a_1^2 + a_2^2 + a_3^2.$$
While I'm at this, since $0$ can be written as $0^2 + 0^2 + 0^2 + 0^2$ and since negative numbers are definitely not the sum of four squares, I'd quantify over the natural numbers and leave out the $n > 0$ condition. That makes it a different, but equivalent statement.
$$\forall n \in {\mathbb N} \exists a_0, a_1, a_2, a_3 \in {\mathbb N}: n = a_0^2 + a_1^2 + a_2^2 + a_3^2.$$


Precalculus/Trigonometric Functions of Sine, Cosine, and Tangent with given parameters? for my precalculus class I was given an assignment for extra credit however it is some material that I have yet to cover or learn as far as sine, cosine, and tangent go. Below is the prompt that I was given:

If sin α = 3/13 where 0 < α < π/2 and cos β = -2/9 where π < β < 3π/2, determine the following:

sin(α - β)
cos(β - α)
tan(2β)
cos(4β)
cot(α + β)

I am not quite sure of the first step to take on these. I wasn't able to find much online either due to my search terms or what not. I hate to upload such a blank question however I am at a loss for how to approach these questions...

They require you to use the identities: $$\sin(\alpha\pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$$ $$\cos (\alpha\pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$$
From these one can infer the double-angle formulae:
$$\sin 2\alpha = 2\sin \alpha\cos \alpha$$ $$\cos 2\alpha = \cos^2 \alpha -\sin^2 \alpha = 2\cos^2 \alpha -1 =1-2\sin^2 \alpha$$ $$\tan 2\alpha = \frac {2\tan \alpha}{1-\tan^2\alpha}$$
To find the value of $\sin\beta$ or $\cos\alpha$, draw a triangle with the lengths of two sides given by the fractions $\frac 3 {13}$ or $\frac {-2} 9$. Use Pythagoras' Theorem to find the length of the third side, and infer the value of $\sin\beta$ or $\cos\alpha$.

You could use the CAST diagram to work out whether $\sin\beta$ and $\cos\alpha$ are positive or negative, by looking at where the angles $\alpha$ and $\beta$ lie in a circle.

If you want calculate this, you can use this site: http://www.milefoot.com/math/trig/22anglesumidentities.htm.

With those formulas your work is just analyze the signals of the functions in these intervals, can you do it or you don't know how analyze the signals ?


In a non-commutative monoid, is the left inverse of an element also the right inverse? Assume there is an element $a\in M$, where $M$ is a non-commutative monoid. If there exists $a b\in M$ such that $a * b = n$, where $n$ is the neutral, does it follow that $b * a = n$?

I want to say no, because the axioms which define a group shouldn't be redundant ($a * b = b * a = n$ is the 4th axiom). But the following logic seems to prove that the inverse is automatically associative:

$b * n = b$

$b * (a * b) = b$

$(b*a) * b = b$

Thus for the last line to be true, $b*a$ has to evaluate to $n$.

The answer is no. The bicyclic monoid is a counter-example. However, in a finite monoid, the answer is yes.

In your question, you called $M$ a non-commutative monoid. It should follow just from that, that the operations included in $M$ are not commutative. Thus, $a*b$ is not necessarily equal to $b*a$.

Normally, when someone twists my arm for an example of an associative but non-commutative binary operation, I answer "matrix multiplication."


Are Gödel's incompleteness theorems really about primitive recursive functions? Any formulation of Gödel's incompleteness theorems seems to involve arithmetic. Why is arithmetic so fundamental? After thinking about the issue a little bit, I came to the conclusion that the theorems are about more general objects called primitive recursion functions: it just so happens that the arithmetical operations are the most familiar (even to a first-grader) examples of these objects. In other words, arithmetic is used to state the theorems in a way most people would understand, but really the most technical statement must involve primitive recursive functions. Question: am I right in this conclusion?

If you are looking specifically at the word "arithmetic", it may help know that, although "arithmetic" in general mathematics today refers to the four basic operations, the meaning of "arithmetic" in formal logic settings is the theory of the natural numbers. There is first-order arithmetic (e.g. Peano arithmetic), second-order arithmetic, Presberger arithmetic, etc.

Similarly, "analysis" in logical settings refers to the theory of the real numbers, which is not the way that term is used in general mathematical settings.

So primitive recursive functions are not something separate from "arithmetic", in modern logic. In fact there is an entire theory called "primitive recursive arithmetic" that focuses on them. Primitive recursive functions can also be developed in stronger theories such as Peano arithmetic.

1) Gödel's theorems necessarily involve arithmetic, due to the method of the proof. The basic idea is to get a statement to refer to itself, and this is done by having it refer to its Gödel number, an encoding of the statement into a natural number. To refer to the natural number, the system must be about arithmetic.

2) One commonly overlooked detail about Gödel's theorems is that they deal with computable axiom systems: ones where a computer could decide whether or not a given statement is an axiom of the system. This computability means decidability by a recursive function. Primitive recursive is not required, just recursive. To quote Wolfram MathWorld which quotes Gödel 1931, " "To every $\omega$-consistent recursive class $\kappa$ of formulas, there correspond recursive class-signs $r$ such that neither ($v$ Gen $r$) nor Neg($v$ Gen $r$) belongs to Flg($\kappa$), where $v$ is the free variable of $r$".

Therefore I conclude that your conclusion is close to being right but is still not quite right.


Solving $(1.44)^t=t^{1.44}$ I've been trying to solve the equation  $(1.44)^t=t^{1.44}$, but other than the obvious solution ($t=1.44$) I haven't had much luck manipulating this into something useful. By taking the log of both sides I'm able to get $\dfrac{\ln t}{t}=\dfrac{\ln 1.44}{1.44}$, but then I'm left with essentially the same problem--my variables are in two different "places" and I can't figure out how to combine them. I can also use exponents to rewrite this as $t^{\frac{1}{t}}=e^{\frac{\ln 1.44}{1.44}}$ or $t^{t^{-1}}=e^{\frac{\ln 1.44}{1.44}}$.

I also tried rewriting the equation as $(1.44)^t-t^{1.44}=0$ and factoring out $1.44-t$, but it quickly turned into a complete mess. Any thoughts on how to approach this? I know two other solutions exist by looking at the graphs of the two functions.

One 'obvious' solution is of course $t=1.44$ for any other solutions you need to use a numerical method, because as you realized too, you cannot solve the equation for $t$ analitically.

Here is a plot. Showing the two curves (there is no real value for $t^1.44$ when $t<0$, so that curve is only plotted for positive $t$.

You can also take logs. Here is the plot for that.

Unfortunately, I cannot think of any way of getting the exact value of the larger solution in terms of elementary functions (or even, at the moment, "special" functions).

[Added later] Just to be clear, Mike's solution (at 0839Z 12 Sep 14) is not an analytic solution, although there may be one if someone could just think of it.


Black Scholes $ ( \;)^+$ Notation Question Came across an exercise involving the Black Scholes formula which uses:

$E[(S_{T}-K)^{+}|F_{t}]$

What is the $\;(\;)^{+}\;$ in the expression? Is this some kind of operator?

Thanks.

The expression $(x)^+$ is defined as $(\bullet)^+ : x \mapsto \max(0,x)$ for $x\in\mathbb{R}$.

It simply tells you that you take the positive payoff, that is the difference between current price and the strike price, otherwise if you get a negative, the result is zero. When you are dealing options, it shows you that when $S_T<K$, at time $T$ the option is worthless.


Negation of a statement about convergence How one can negate this statement:

A real sequence converges to a limit $x$ if and only if every subsequence also converges to the limit $x$.

I see no better negation than

"It is false that a real sequence converges to a limit $x$ if and only if every subsequence also converges to the limit $x$."

A logically equivalent one is

"A real sequence converges to a limit $x$ and some subsequence does not converge to the limit $x$, or, a real sequence does not converges to a limit $x$ and every subsequence converges to the limit $x$."

Both are equivalent to false.

We can negate as follows

$$\lnot(A\iff B) \equiv A\iff \lnot B$$

that is

a real sequence converge to a limit x if and only if exists a subsequence which doesn't converge to the limit x


line is the perpendicular bisector of a segment obtained by reflecting twice through two rays Given two rays, L and M, with common origin O, and a point Q inside the acute angle formed by the rays, reflect Q across L to obtain Q' and then reflect Q' across M to obtain Q''.  Similarly, reflect Q across M to obtain P' and then reflect P' across L to obtain P''.
Show that the line through O and Q is the perpendicular bisector of the segment joining P'' and Q''.
How do you do this?
It is easy to show that Q and all reflected points lie on the circle with center O that passes through Q.  I thought of using reflection across the line through O and Q and then use a symmetry argument but couldn't see just how to implement this idea.
I would really appreciate any help. Thanks

Thanks Emmad. As you said, the result follows. The fact you mention combined with the following show that the triangles that need to be shown are congruent in fact are:
the angle Q''Q'Q equals the angle P''P'Q so the arcs they intercept are the same.
Thank you for your answer


countable intersection of closed convex bounded subsets reflexive banach space is non empty. If $X$ is a reflexive Banach space and $(C_n), n \in \mathbb{N}$ is a sequence of closed convex bounded sets with $C_{n+1}$ contained in $C_n$ for all $n \in \mathbb{N}$. How does one show that the countable intersection of $C_n$ for $n \in \mathbb{N}$ is not the empty set?

Hint: Do you know the Eberlein-Shmulyan theorem?

the proof can be seen in Introduction to Banach Space Theory by Robert E. Megginson.


If $0\lt y \le 1$, prove that there exists a unique positive real number $x$ such that $x^2=y$ I'm stumped. I don't want an entire solution, just a hint.

If $0\lt y \le 1$, prove that there exists a unique positive real number $x$ such that $x^2=y$

The section in the book I'm on is the least upper bound property

Entire answer. Don't read it all......Let $A=\{z>0 :z^2\leq y\}.$ We have $A\ne \phi$ because $y/2\in A$ because $$0<(y/2)^2=y (y/4)\leq y(1/4)<y.$$   Observe  that $$(1)......(0<z\land z^2>y)\implies A\subset (0,z)\implies $$ $$\implies z \text { is an upper bound for } A.$$  $$\text {(2)...... Now let } x=\sup A.$$ (From Latin supremum, in math it's a synonym for $lub.$) We know $x$ exists because $A\ne  \phi$ and because $1$ is an upper bound for $A$. We show that $x^2$ can't be more than $y$ and can't be less than $y$. First, suppose $y<x^2.$ There exists positive integer $n$ for which $$(x^2-y>1/n)  \land (x>1/n).$$ Let $x'=x-1/(2n)$.  Since $0<x\leq 1$ we have  $$x'^2-y=(x^2-y)-(x/n)+1/(4n^2)\geq (x^2-y)-(1/n)+1/(4n^2)>$$ $$>1/n-1/n+1/4n^2>0.$$ But $x'>0$ so by (1),  $x'$ is an upper bound for $A$ with $x'<x=\sup A$, which is absurd. Therefore we cannot have $y<x^2.$....... Second, suppose $y>x^2$. Then there exists a positive integer $n'$ for which $$y-x^2>1/n'.$$ Let $x''=x+1/(4n')$ .Since $x\leq 1$, we compute $$(x'')^2=x^2+x/(2n')+1/(16n'^2)\leq x^2+1/(2n')+1/(16n'^2)<x^2+1/n'<y $$ which implies $x''\in A.$ But $A$ cannot have a member $x''$ which is greater than $\sup A=x.$ So we cannot  have $y>x^2$..... Therefore $x^2=y$..... The positive number $y$ cannot have two positive square roots $x,x_1$ because if $ [0=y-y=x^2-x_1^2=(x+x_1)(x-x_1)]\land [x+x_1\ne 0]$ then $x=x_1.$

Assume that there are two possible solutions, say $x_1 > 0$ and $x_2 > 0$ such that
$$x_1^2 = y = x_2^2$$
Can you now proceed to show that if both $x_1$ and $x_2$ are positive, then $x_1 = x_2$?


Constraints on variables to form orthonormal columns I have a series of equations \begin{align} u_{11} &= \frac{zc_1}{x} \frac{\sin(\delta + \omega)}{\sin \alpha} \\
u_{21} &= \frac{zc_2}{x} \frac{\sin \omega}{\sin \alpha} \\
u_{31} &= \frac{tc_1}{x} \frac{\sin \delta }{\sin \alpha} \\
u_{41} &= 0 \end{align}

and \begin{align} u_{13} &= \frac{zc_1}{y} \Bigg( \cos(\delta+\omega)+\frac{\sin(\delta + \omega)}{\tan \alpha} \Bigg)\\
u_{23} &= \frac{zc_2}{y} \Bigg( \cos(\omega)+\frac{\sin( \omega)}{\tan \alpha} \Bigg) \\
u_{33} &= \frac{tc_1}{y} \Bigg( \cos(\delta)+\frac{\sin(\delta)}{\tan \alpha} \Bigg) \\
u_{43} &= \frac{tc_2}{y} \end{align}

Where I know that $$0 <x,y<1$$ $$0 \leq z,t,c_1,c_2 \leq 1$$ $$0<\alpha<\pi/2$$ $$0 \leq \delta, \omega \leq \pi/2$$ and $$x^2+y^2=z^2+t^2=c_1^2+c_2^2=1$$ and all parameters are real. My goal is to find out the constraints needed to make the $(u_{11},u_{21},u_{31},u_{41})$ and $(u_{13},u_{23},u_{33},u_{43})$ orthonormal vectors. That is, given $(x,y,z,t,\alpha,\omega)$, can one find $c_1,c_2,\delta$ that will make these columns orthonormal? We therefore impose that $$u_{11}^2+u_{21}^2+u_{31}^2=1$$
$$u_{13}^2+u_{23}^2+u_{33}^2+u_{43}^2=1$$ and
$$u_{11}u_{13}+u_{21}u_{23}+u_{31}u_{33}=0$$ however expanding these expressions do not allow me to solve for $c_1,c_2$ or $\delta$. Is there an efficient way to go about this?

With some additional border conditions, this problem could be directly transformed into the modal solution of a physical problem. Which is the original problem bringing these expressions? That could help to clarify a better solution.

As is, none of the given expression can be simplified in a simple way enabling to give  "closed" inequalities on the given variables.

By applying the constraints, it is straightforward to find:
$$
\mathscr{C}_{11}:
(zc_1)^2(s\delta\omega)^2+(zc_2)^2(s\omega)^2+(tc_1)^2(s\delta)^2
=(xs\alpha)^2\\
\mathscr{C}_{33}:
(zc_1)^2(s\delta\omega\alpha)^2+(zc_2)^2(s\omega\alpha)^2+(tc_1)^2(s\delta\alpha)^2+(tc_2)^2(s\alpha)^2
=(ys\alpha)^2\\
\mathscr{C}_{13}:
(zc_1)^2s\delta\omega s\delta\omega\alpha+(zc_2)^2s\omega s\omega\alpha+(tc_1)^2s\delta s\delta\alpha
=0
$$
where $s\phi_1\cdots\phi_n$ and $c\phi_1\cdots\phi_n$ represent the $\sin$ and $\cos$ functions applied over the sum of the $\phi_i$ angles denoted right in front of it, not to be confused with $c_1$ and $c_2$, variables part of the problem.

Reordering between the $c_1$, $c_2$ terms:
$$
\mathscr{C}_{11}:
c_1^2 (z^2(s\delta\omega)^2+t^2(s\delta)^2)
+c_2^2 z^2(s\omega)^2
= (xs\alpha)^2\\
\mathscr{C}_{33}:
c_1^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)
+c_2^2 (z^2(s\omega\alpha)^2+t^2(s\alpha)^2)
= (ys\alpha)^2\\
\mathscr{C}_{13}:
c_1^2(z^2s\delta\omega s\delta\omega\alpha+t^2s\delta s\delta\alpha)
+c_2^2(z^2s\omega s\omega\alpha)
=0
$$

From Cramer's on $\mathscr{C}_{11}$ and $\mathscr{C}_{33}$:
$$
\Delta=
(z^2(s\delta\omega)^2+t^2(s\delta)^2)  (z^2(s\omega\alpha)^2+t^2(s\alpha)^2)-
z^2(s\omega)^2  (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)\\
\Delta_1=
(xs\alpha)^2  (z^2(s\omega\alpha)^2+t^2(s\alpha)^2)-
(ys\alpha)^2  (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)\\
\Delta_3=
(xs\alpha)^2  (z^2(s\delta\omega)^2+t^2(s\delta)^2)  -
(ys\alpha)^2  z^2(s\omega)^2
$$

The $c_1$ and $c_2$ values are the square roots from:
$$
c_1^2={\Delta_1 \over \Delta}=
{
(xs\alpha)^2  (z^2(s\omega\alpha)^2+t^2(s\alpha)^2)-
(ys\alpha)^2  (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)
\over
(z^2(s\delta\omega)^2+t^2(s\delta)^2)(z^2(s\omega\alpha)^2+t^2(s\alpha)^2)-
z^2(s\omega)^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)}\\
c_2^2={\Delta_3 \over \Delta}=
{
(xs\alpha)^2 ( z^2(s\delta\omega)^2+t^2(s\delta)^2)  -
(ys\alpha)^2  z^2(s\omega)^2
\over
(z^2(s\delta\omega)^2+t^2(s\delta)^2)(z^2(s\omega\alpha)^2+t^2(s\alpha)^2)-
z^2(s\omega)^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)}
$$

Note that the term $s\delta\omega\alpha$ must be negative in order the signs be preserved (!). This is the only term which can be negative after a $\sin$ operator involving angles in the $(0,pi/2)$ range.

And from $\mathscr{C}_{13}$ we have a expression able to be numerically solved for $\delta$:
$$
\mathscr{C}_{13}:
\Delta_1^2
(z^2 s\delta\omega s\delta\omega\alpha+t^2 s\delta s\delta\alpha)
=
-\Delta_3^2
(z^2 s\omega s\omega\alpha)
$$
Hence:
$$
(
x^2  (z^2 s\omega\alpha^2
+t^2 s\alpha^2)-
y^2  (z^2 s\delta\omega\alpha^2
+t^2 s\delta\alpha^2)
)^2
(z^2 s\delta\omega s\delta\omega\alpha
+t^2 s\delta s\delta\alpha)
=
-(
x^2  (z^2 s\delta\omega^2
+t^2 s\delta^2)  -
y^2  z^2 s\omega^2
)^2
(z^2 s\omega s\omega\alpha)
$$

Which, spite of the closed solution for $c_1(\delta)$ and $c_2(\delta)$, cannot be solved in a closed for the last parameter $\delta$.


$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?

Simplify before applying the rule...
$$
\lim_{x\to 0} \frac{\tan x- \sin x}{x^3} = \lim_{x\to 0}\frac{\sin x - \sin x \cos x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0}\frac{1-\cos x}{x^2} = \lim_{x\to 0}\frac{1-\cos x}{x^2}
$$
and then apply the rule
$$
 \lim_{x\to 0}\frac{1-\cos x}{x^2}= \lim_{x\to 0}\frac{\sin x}{2x} = \frac 12.
$$


$p$ is prime and $n>1$ integer such that $p|{n \choose k} , \forall 1\le k \le n-1$ , then $n$ is a power of $p$? Let $p$ be a prime and $n>1$ be an integer such that $p|{n \choose k} , \forall 1\le k \le n-1$ , then is it true that $n$ is a power of $p$ ?

From the given condition, I am only able to derive that $p|n$ . (I know that the converse of the claim I make , is true)

Please help.

HINT:

Indeed, see Lucas theorem

ADDED:

Lucas' theorem states: If $n= \sum a_i p^i$ and $k=\sum b_i p^i$ are base $p$ expansions then
$$\binom{n}{k} \equiv \prod \binom{a_i}{b_i} \pmod p$$

Now if $n$ is not a $p$ power there exists $1\le k <n$ so that $b_i \le a_i$ for all $i$. Note that $a_i \le p-1$. We get $\binom{n}{k} \not \equiv 0 \pmod p$.


Proving there are no integer solutions for $3x^2=9+y^3$ Prove there are no $x,y\in\mathbb{Z}$ such that $3x^2=9+y^3$.

Initial proof

Let us assume there are $x,y\in\mathbb{Z}$ that satisfy the equation, which can be rewritten as $$3(x^2-3)=y^3.$$ So, $$3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3.$$ As $3 \mid -3$, it follows that $$3 \mid x^2 \Rightarrow 3^2 \mid x^2.$$ Say $x^2=3^2 \cdot a^2$ for some $a \in \mathbb{Z}$. Then $$\begin{align*}x^2-3 &= 3^2 \cdot a^2 - 3 \\&= 3 \cdot  (3 \cdot a^2 - 1)\end{align*}$$ As $3^2 \mid x^2-3$, it follows that $3 \mid 3 \cdot a^2 - 1$. It is obvious that $3 \mid 3 \cdot a^2$, so it follows that $3 \mid -1$, which is false. Therefore, the assumption that there are $x,y\in\mathbb{Z}$ was wrong.

Alternative proof

$$
\begin{align*}
3x^2&=9+y^3\\
3(x^2-3)&=y^3
\end{align*}
$$

So, $3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3 \Rightarrow 3^2 \mid x^2 \Rightarrow 3^2 \not\mid x^2 - 3$ and we are done.

Both proofs are essentially the same, except for the fact that the second proof is much shorter. I think the first proof is much more clear, but it takes a bit longer to write down.

Would the second proof suffice in a math competition, or is it too short?
This problem was part of the Dutch finals of 1978. I figured it out in under a minute, which is quite unusual for me, so now I wonder: is my proof correct?

Your second proof starts without an assumption and this is not a matter of length. The word "Assume" before the first formula would suffice. This oversight would probably be forgiven if your work ended with a contradiction and the consequences which it does not.

Between $3^2\mid x^2-3$ and $3^2 \mid x^2$, you should add at least $3\mid x$ and ideally $3\mid x^2-3 \Rightarrow 3\mid x^2 \Rightarrow 3\mid x$.

You are highly likely to lose some points on that step, although it is not certain that you would.

The difference between you first and second version is NOT the shortness and lack of words, but the fact that you left out that step.

The second issue is that you do not say in your second proof that there are no integer solutions. You are certain to lose points on that. So, for example, you could circle the two contradictory assertions, make arrows to the next line and write "Contradiction. There are no solutions."

It is not an issue of length, but whether every step is addressed in some way.

Assume that the expression has integer solutions, or atleast one for that matter:

Work in modulus $3$, if:

$$3(x^2-3)=y^3$$

$y^3$ must be a multiple of $3$. Then we can set $y=3k$ where $k$ is an integer. Now we have:

$$27k^3=3(x^2-3)$$

Dividing both sides by $3$ gives us $9k^3=x^2-3$
Taking $-3$ to the other side shows us that $x^2$ is a mutiple of $3$. Thus $x$ is a multiple of $3$. Then we can set $x=3t$ where $t$ is an integer. This gives us:

$$9k^3+3=9t^2$$
Dividing both sides by $3$ again leaves us with this:
$$3k^3+1=3t^2$$
Which is impossible because $3t^2$ is a multiple of $3$ and $3k^3+1$ is not.

This is a contradiction.
Thus the proof is complete.


What is the *middle* digit of $3^{100000}$? The decimal representation of $3^{100000}$ has $47713$ digits. What is the $23857^{th}$ digit - i.e. the one in the $10^{23856}$'s place?

There are lots of questions on this site asking for the last digit of various large exponents, which are well handled by modular arithmetic. The last digit of $3^{100000}$ is $1$ by such methods. There are some fewer number of questions asking for the first digit of a number, which can be handled by considering the fractional part of the base ten logarithm of the number. The first digit of the number is $1$.

Either method can be modified to give the first few or last few digits, but they don't generalize well beyond that, and there's no obvious way to combine the methods.

Is there a method to find arbitrary digits of small integers raised to large ones? Is there any reason to suspect that computing a digit in the middle is, in terms of computational complexity, a more difficult problem than computing digits on the edge?

In contrast with other programming languages from its time, Pascal (the programming language)
also supports a set type, implemented as a bit pattern. The bit patterns associated with
the Pascal set type are $256$ bits wide, but this limitation is not essential and can been replaced
with other (larger) values eventually.
See Wikipedia for a reference.
A rather detailed description of the set type implementation can be found
as well .
So we have the following practice: a bit pattern in a computer is a set type
We also know that

a bit pattern in a computer is a natural number type
Indeed, everybody knows that a natural number can be represented as a binary i.e. a bit pattern.
The word "type" has been employed here in order to avoid confusion with other
(i.e the standard mathematical) "set" and "number" definitions.More precisely: the
hereditarily finite sets
are in one-to-one correspondence with the natural numbers. And the latter fact is independent of
computers.Examples.
$$
\begin{array}{l}
0 = 000 = \{\} \\
1 = 001 = \{0\} = \{\{\}\} \\
2 = 010 = \{1\} = \{\{\{\}\}\} \\
3 = 011 = \{0\; 1\} = \{\{\}\{\{\}\}\} \\
4 = 100 = \{2\} = \{\{\{\{\}\}\}\} \\
5 = 101 = \{0\; 2\} = \{\{\}\{\{\{\}\}\}\} \\
6 = 110 = \{1\; 2\} = \{\{\{\}\}\{\{\{\}\}\}\} \\
7 = 111 = \{0\; 1\; 2\} = \{\{\}\{\{\}\}\{\{\{\}\}\}\} \\
\cdots
\end{array}
$$
The above is related to the following reference, by Alexander Abian and Samuel LaMacchia:

ON THE CONSISTENCY AND INDEPENDENCE OF SOME SET-THEORETICAL AXIOMS

If the curly brackets $\{\}$ are replaced by square brackets $\left[\,\right]$ then another
important fact is observed:
a set type is is a natural number type is a sorted natural numbers array type
If now we devise an equivalent of the elementary number operations with sorted arrays,
then we have virtual unlimited precision at our disposal. That this approach indeed works, shall be demonstrated at hand of the OP's question.
Here is a link to the complete (Delphi Pascal) program that does the job:

Meelo.dpr

And here is a link to the number $3^{100000}$ itself, which is too large to fit into
MSE's margins:

drie5.txt

The screen output of the program is the number of digits, the first, the middle and the last digit:

47713
1 2 1

So, indeed, as Lucian says in a comment: it's "obvious" that the digit in the middle is a $\large\, 2$ .

Note.  A power like $\,3^{100000}\,$ sounds quite impressive, but with a smart
algorithm, the number of operations is only $\,\ln_2(100000)\approx17$ . For
real numbers $\,x\,$ and a natural $\,n\,$ it goes as follows:

function power(x : double; n : integer) : double;
var
  m : integer;
  p, y : double;
begin
  m := n; y := x; p := 1;
  while m > 0 do begin
    if (m and 1) > 0 then p := p * y;
    m := m shr 1; { m := m / 2 }
    y := y * y;
  end;
  power := p;
end;

Wikipedia reference: Efficient computation with integer exponents .

BONUS. In view of the above, the following answer is interesting:

The purpose of the $\sf ZFC$ Axiom of Infinity

Reference is made to a
paper by Kaye and Wong,
where on page 499 we read:

  It was observed in 1937 by Ackermann [1] that $\mathbb{N}$ with the membership relation
  defined by $n \in m$ iff the $n$th digit in the binary representation of $m$ is $1$
  satisfies ZF$-$inf. This interpretation, formalized in ZF with $\omega$ in place of $\mathbb{N}$
  yields a bijection between $\omega$ and the collection $V_\omega$ of hereditarily finite sets.

Using a computer program (that just calculates the whole number and extract the required digit), I found that the answer is 7.

This is the program (which is in Haskell):

digits :: Integer -> [Int]
digits = map (read . (:[])) . show

main = print ((digits (3^100000)) !! 23857)


Proof of Fermat's Little Theorem using Primitive Roots I just learned about primitive roots today, and then I thought of this proof of Fermat's Little Theorem. Seeing that most proofs of this theorem aren't simple, I think I'm either completely wrong in my application of primitive roots (must have missed something fundamental, having just learned about them), or primitive roots are extremely powerful. Which one is it?

Proof: We want to prove that if $\gcd(a,p)=1$, with $p$ a prime, then $a^{p-1}\equiv 1\pmod p$. Since $p$ is prime there exists a primitive root $\operatorname{mod}\, p$, say $j$. It is well known that we can write every least residue $\operatorname{mod}\, p$ as a power of $j$, so e.g. we can write $a\equiv j^k$. Thus, it suffices to prove that $a^{p-1}\equiv j^{k(p-1)}\equiv 1\pmod p$. But this is obvious because $j^{k(p-1)}\equiv (j^{p-1})^k\equiv (1)^k\equiv 1\pmod p$ since $\operatorname{ord}_p(j)=p-1$ by definition.

QED

Thanks for your help guys!

You have given a correct proof of Fermat's Little Theorem.

The reason you have not seen this proof before is twofold:

Other proofs of Fermat's Little Theorem naturally generalize to proofs of Euler's Theorem. Euler's Theorem cannot be proven using primitive roots, since there are not primitive roots in general. Further, from the perspective of abstract algebra, both Euler's Theorem and Fermat's Little Theorem are manifestations of an extremely basic theorem of group theory: Lagrange's Theorem.
The existence of a primitive root is a very non-trivial fact. (But it is independent of Fermat's Little Theorem, I believe.) Looking again from the perspective of abstract algebra, the existence of primitive roots for certain groups $U_n$ amount to completely understanding the structure of these groups and showing they are cyclic. This is not so easy.

A proof from scratch is simple.

There are obviously $p-1$ possible non-zero remainders if you divide a number relatively prime to $p$ by $p$, namely $1,2,\dots,p-1$. Now consider the remainders of $a,2a,3a,\dots,(p-1)a$. They must all be different, because if any two were the same then $p$ would divide their difference. That is impossible since $\gcd(a,p)=1$.

So they must be some arrangement of $1+k_1p,2+k_2p,\dots,p-1+k_{p-1}p$ for some integers $k_i$. Multiplying them all together we get $1\times2\times3\dots\times(p-1)=a^{p-1}\times1\times2\dots\times(p-1)\bmod p$. We can divide by $(p-1)!$ since it is not a multiple of $p$, to get $a^{p-1}=1\bmod p$.


Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? I am reading Tom Apostol's Analysis and come across this theorem.

Should $a \leq b$ if $a\leq b+\epsilon$ for all $\epsilon >0$?

I don't doubt the proof in the book but I don't understand the intuition or geometric explanation behind this. Could somebody shed some light on this equation? I just started studying analysis on my own.$\ \ $

Draw a number line. Mark the point $b$. Where can you mark $a$? Every number greater than $b$ may be written as $b+\varepsilon$ for some $\varepsilon >0$. Then $a\leqslant b+\varepsilon$ says every number greater than $b$ is also greater than $a$. Thus, you erase all what comes after $b$. The only choices left are the numbers to the left or $b$ itself.

Assume for contradiction $a > b$. However we can pick $\epsilon = (a-b)/2 > 0$ so that $a > b + \epsilon$, a contradiction of our original assumption $a \le b + \epsilon$. Imagine the real line:

$$
....\underbrace{b ......... b+\epsilon}_{\text{dist of } (a-b)/2} ......... a ....
$$

The same trick works for $a - \epsilon \le b + \epsilon$ using $\epsilon = (a-b)/3$.


Validity of $\exists x \forall y: x \le y$ in set of integers and set of natural numbers If I say $\exists x\in \mathbb{N} $ such that $\forall y \in \mathbb{N},$ $x \leq y$, this is a true statement. I am assuming that the set of natural numbers is defined for $0\leq x < \infty $.

But if I say the same thing for the set of all integers:
$\exists x\in \mathbb{Z} $ such that $\forall y \in \mathbb{Z},$ $x \leq y$ is a false statement, and I need help with my wording here: It is false because there does not exist an integer $x$ such that there are no integers less than $x$.

Am I correct in my logic here?

The statement is saying that there is a smallest element. This is true for natural numbers, false for integers. [...] [Your] claims are correct.

-- Ted (in a comment).

Well it seems that your logic is false but you have to say why do you think there is no such number as x. Cause it exists .


Is this Integration From First Principles notation standard outside of A Level Maths? I teach A Level Maths in the UK. We are required to do some 'introduction' to integral from first principles as part of the specification (link, page 25 is the interesting part).
In a previous exam question (Paper 2, June 2018), this was essentially the question:

Suppose we have the curve $y= \sqrt{x}$

The point $P(x,y)$ lies on the curve. Consider a rectangle with height $y$ and width $\delta x$.

Calculate $\displaystyle \lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \,\delta x$

The answer involves us recognising $$ \lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x = \int_4^9 \sqrt{x} \, dx$$ and evaluating the integral.
Is this notation standard?
To me, it doesn't make sense. How can you have $x=4$ to $9$ as the limits on the sum, for example? A sum only works over integral values.
In addition, one could easily give meaning to the limit as $\displaystyle \lim_{\delta x \rightarrow 0}\sum_{x = 4}^9 \sqrt{x} \, \delta x = \lim_{\delta x \rightarrow 0}( \sqrt{4}(\delta x) + \sqrt{5} (\delta x) + \cdots + \sqrt{9} (\delta x))$ which should be $0$ as $\delta x \rightarrow 0$.
The reason I am so confused is that this question and notation appears on a Pearson A Level Maths Exam paper. It is a regulated qualification. There are very qualified people out there that have deemed this to make sense and be used to assess the understanding of thousands of A Level students in the UK.

In my experience as a UK mathematician, this isn't standard notation. The aim is to convey $x$'s step size is $\delta x$ rather than $1$. I'd recommend writing the limit as $\lim_{\delta x\to0}\sum_{k=0}^{5/\delta x}\sqrt{4+k\delta x}\delta x$ - no, wait, make it $\lim_{N\to\infty}\sum_{k=0}^N\sqrt{4+5k/N}(5/N)$. This limit is the stated integral; in fact, if we set the maximum $k$ to $N-1$ instead (which won't change the limit), it's a Riemann sum.

Just change the equals sign to an arrow. Altogether, it should sufficiently demonstrate the intention to change a discrete sum to a more accurate continuous sum. The equals sign is indeed improper, so a simple arrow addresses this without exceeding A-Level needs. It is also reasonable ink usage, for a generally trivial incongruity.


Lebesgue point of density on $[0,1]$ and Dynkin's theorem The problem defines a density point $x\in[0,1]$ for a Borel set $A\subset [0,1]$ if $$ \lim_{\varepsilon \rightarrow 0^+} \frac{\mu([x-\varepsilon,x+\varepsilon]\cap A)}{2\varepsilon}=1.$$Denote all the density point of $A$ to be $A^*$. The problem asks
to show $$A^*\text{ is Borel}, \mu(A\,\Delta\, A^*)=0, \forall \text{Borel } A\subset[0,1]$$
where $\mu$ is the Lebesgue measure and $\Delta$ means symmetric difference.

The point here is that the professor asks to use Dynkin's $\pi-\lambda$ theorem to prove and he note that interval $I$ would have $\mu(I\,\Delta\, I^*)=0$.

I manage to show $A^*$ is Borel but do not know how to show the second part.  I let $\pi$ system be all the intervals in $[0,1]$   and the $\lambda$ system to be $S=\{A\text{ is Borel}:\mu(A\,\Delta\, A^*)=0\}$. I got trouble in checking that if $A\in S$ then $A^c\in S$ and if $A_1,\ldots,A_n,\ldots\in S$, $\bigcup A_n\in S$.

Indeed, I am not sure that whether Dynkin's theorem could be applied here or make the problem easier. Or we still need to go through the proof using Vitali covering lemma.

Let $C$ be the the collection of intervals. We know that for any $I \in C$ the Theorem holds.

Use that the $\sigma$-algebra generated by $C$ (the Borel sets) coinsides with the Dynkin class generated by $C$.


A merchant claims a loss of 4% on coffee, but uses weight equal to 840g instead of 1000g. What profit/loss the merchant actually makes? A merchant claims a loss of 4% on coffee, but uses weight equal to 840g instead of 1000g. What profit/loss the merchant actually makes?

Merchant is giving goods worth Rs 840g for Rs 1000. So he is making a profit of $1000-840/840=400/21$.
Now given he makes a loss. Applying successive percentage
$400/21-4-1600/21=1248/21$
That is loss. But my answer is not right

Suppose the cost price of coffee is 1 Rupee per gram.

When he sells 1000 grams, he claims to make a loss of 4%. So, he claims that he is selling 1000 Rs goods at 960 Rs.

He is actually selling 840 grams. So, his actual cost price is 840 Rs.

Gain = 960 - 840 = Rs. 120

Profit % = (120/840)*100 = 14.28%

You can check out problem 2 here, it also deals with cheating merchants / faulty weights.


Variance as a sum of conditional variances So, for a random variable $X$ and an event $A$, the following expression of the mean value of $X$ as a sum of conditional expectations is valid, or at least, i could prove it for discrete random variables:

$$ E(X) = E(X|_A) P(A) + E(X|_{A^c})(1-P(A)) $$

My question is, is there a similar expression for variances? I remember my teacher using such a thing once, something like:

$$ Var(X) = Var(X|_A) P(A) + Var(X|_{A^c})(1-P(A)) + \ ... $$

but i forgot it and i can't find it anywhere. Even the one above, i haven't found it anywhere i looked.

If you mean total probability theorem, using the law of total variance, we have:
$$\mathrm{Var(X)} = \mathrm{E}[\mathrm{Var}(X|A)] + \mathrm{Var}(\mathrm{E}[X|A]) .$$ It is somehow well-known equation, therefore, I don't include the proof. Hope that helps!


Finding the general solution of $u u_{xy} - u_x u_y = 0,$ In the book of Berg, at page 4, at the end of the introduction, as an exercise, it is asked to find the general solution of

$$u u_{xy} - u_x u_y = 0,$$

however, considering the fact that the book haven't showed any method or anything at all, how can we find the general solution of this PDE ?

Any solution of the form $u(x,y) = f(x)g(y)$ is valid, since

$$ uu_{xy} - u_xu_y = fg\cdot \frac{df}{dx}\frac{dg}{dy} - \frac{df}{dx}g\cdot f\frac{dg}{dy} = 0 $$

is always true


Every number $2^N > 4$ can be expressed by the sum of 2 primes? For example the first cases are:

$2^3= 8 = 3+5$

$2^4= 16 = 3+13$

and so on ...

Yes, every such number can be written as a sum of two primes. No, nobody knows how to prove it.

32=2^5.It is of the form 4k.we know that primes of the form 4a+1,4b-1. Now we write  32=4a+1+4b-1.then a+b=8. By solving a+b=8, we get the pairs (a,b)=(3,5),(7,1) for which 4a+1,4b-1 are primes. In this way we can find 2 primes for every power  2. For ex. 22 is of the form 4k+2. Now we write 22= 4a+1+4b+1where 4a+1 and 4b+1 are primes now we get a+b=5 For (a,b)=(1,4),we get 22=5+17.now we write 22=4a-1+4b-1 From this we get a+b=6For (a,b)=(1,5),(3,3) we can write 22=3+19=11+11.If N =4n=p+q where p, q are primes of the form 4a+1 and 4b-1 If N=4n+2=4a+1+4b+1=4p-1+4q-1 where 4a+1,4b+1,4p-1,4q-1 are primes.  For given any even integer N, we can  find 2 primes as the sum is N.This is an elementary approach to write  for any given even integer as a sum of 2 odd primes.


Is the sum and difference of two irrationals always irrational? If $x$ and $y$ are irrational, is $x + y$ irrational? Is $x - y$ irrational?

The short answer to your question is that is not necessarily true. For instance, $\sqrt{2}, \sqrt{2}-1,1-\sqrt{2}$ are all irrational but $$\sqrt{2} + (1-\sqrt{2}) = 1 \in \mathbb{Q}$$ and $$\sqrt{2} - (\sqrt{2}-1) = 1 \in \mathbb{Q}$$

However, it is worth noting that if $x$ and $y$ are irrational, then either $x+y$ or $x-y$ is irrational i.e. $x+y$ and $x-y$ cannot be both rationals. The proof for this is given below.

Proof

If both $x+y$ and $x-y$ are rational, then we have that $x+y = \dfrac{p_1}{q_1}$ and $x-y = \dfrac{p_2}{q_2}$, where $p_1,p_2 \in \mathbb{Z}$ and $q_1,q_2 \in \mathbb{Z} \backslash \{0\}$.

Hence, $x = \dfrac{\dfrac{p_1}{q_1} + \dfrac{p_2}{q_2}}{2} = \dfrac{p_1q_2 + p_2q_1}{2q_1q_2}$ and $y = \dfrac{\dfrac{p_1}{q_1} - \dfrac{p_2}{q_2}}{2} = \dfrac{p_1q_2 - p_2q_1}{2q_1q_2}$.

Now $p_1q_2 + p_2q_1, p_1q_2 - p_2q_1 \in \mathbb{Z}$, whereas $2q_1q_2 \in \mathbb{Z} \backslash \{0\}$.

This contradicts the fact that $x$ and $y$ are irrationals. Hence, if $x$ and $y$ are irrational then either $x+y$ is irrational or $x-y$ is irrational.

Below are some statements worth knowing.

1 Sum of two rationals is always a rational.

Proof: Let the two rationals be $\dfrac{p_1}{q_1}$ and $\dfrac{p_2}{q_2}$, where $p_1,p_2 \in \mathbb{Z}$ and $q_1,q_2 \in \mathbb{Z} \backslash{0}$. Then $$\dfrac{p_1}{q_1} + \dfrac{p_2}{q_2} = \dfrac{p_1q_2 + p_2q_1}{q_1q_2}$$
where $p_1q_2 + p_2q_1 \in \mathbb{Z}$ and $q_1 q_2 \in \mathbb{Z} \backslash \{0\}$. Hence, the sum is again a rational.

2 Sum of a rational and an irrational is always irrational.

Proof: Let the rational number be of the form $\dfrac{p}{q}$, where $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$ while the irrational number be $r$. If $r + \dfrac{p}{q}$ is a rational, then we have that $r + \dfrac{p}{q} = \dfrac{a}{b}$ for some $a \in \mathbb{Z}$ and $b \in \mathbb{Z} \backslash \{0\}$. This means that $r = \dfrac{a}{b} - \dfrac{p}{q} = \dfrac{aq-bp}{bq}$ where $aq-bp \in \mathbb{Z}$ and $bq \in \mathbb{Z} \backslash \{0\}$. This contradicts the fact that $r$ is irrational. Hence, our assumption that $r + \dfrac{p}{q}$ is a rational is false. Hence, $r + \dfrac{p}{q}$ is a irrational.

3 Sum of two irrationals can be rational or irrational.

Example for sum of two irrationals being irrational

$\sqrt{2}$ is irrational. $\sqrt{2} + \sqrt{2} = 2 \sqrt{2}$ which is again irrational.

Example for sum of two irrationals being rational

$\sqrt{2}$ and $1-\sqrt{2}$ are irrational. (Note that $1-\sqrt{2}$ is irrational from the second statement.) But, $\sqrt{2} + (1-\sqrt{2}) = 1$ which is rational.

Proof of sum of two irrational nos. is rational:

Let one irrational no. be 0.989889888 and other irrational no. be 0.121221222. Then 0.989889888+0.121221222=1.1111111 bar.

Proof of sum of two irrational no. is irrational:

Let one irrational no. be 1.4142135.... and other ben1.7350208....
If we add both of them we get 3.14626436..... which is irrational

Hence we conclude that sum of two irrational nos. can be rational or irrational


The real numbers $x,y$ and $z$ are such that $x-7y+8z=4$ and $8x+4y-z=7$. What is the maximum value $x^2-y^2+z^2?$ The real numbers $x,y$ and $z$ are such that
$x-7y+8z=4$ and $8x+4y-z=7$. What is the maximum value of $x^2-y^2+z^2?$

From those equations I got:

$12z-5x=13y$

$12x+5z=13$

$12y+5=13z$

$12-5y=13x$

I know that $5,12,13$ is a pythag triplet but I don’t know what to do next. I think lagrange multipliers could be used but there should be a solution that doesn’t require calculus

Hints, suggestions and solutions would be appreciated.

Taken from the 2014 KIMC https://chiuchang.org/imc/wp-content/uploads/sites/2/2018/01/2014-IWYMIC-Individual.x17381.pdf

Solving the system
$$x-7y+8z=4$$
$$8x+4y-z=-7$$
we get
$$y=\frac{12}{5}-\frac{13}{5}x$$
$$z=\frac{13}{5}-\frac{12}{5}x$$
and we get
$$x^2-y^2+z^2=1$$
Very NICE!

Below please find an image of the surface $x^2-y^2+z^2=1$ together with the (thick red) line of intersection of the two given planes.


Evolution by death and immigration of Poisson distributed population This is quite an interesting problem, but I'm not sure how to go about doing it. I know that by using some basic Poisson properties I can figure it out but I'm failing to see how. It goes like this:

  A population comprises of $X_n$ individuals at time $n=1,2,3...$ Suppose that $X_0$ has Poisson ($\mu$) distribution. Between time $n$ and time $n+1$ each of the $X_n$ individuals dies with probability $p$, independently of the others. The population at time $n+1$ is formed from the survivors together with a random number of immigrants who arrive independently according to a Poisson ($\mu$) distribution. What is the distribution of $X_n$? I feel that I'm close but can't quite get it. It sounds like I could set up a series and then maybe prove it by induction.

Let's consider time $n$. We have $X_{n+1} = X_n - D_n + I_n$, where $\{I_n\}$ are iids with $\operatorname{Poi}(\mu)$ distribution, and $D_n$ is the number of deaths. For each individual, the probability of dying is $p$. The total mortality is the sum of $X_n$ independent $\operatorname{Bernoulli}(p)$ random variables, thus $D_n \sim \operatorname{Binom}(X_n, p)$. Now consider the probability generating function of $X_n$:
$$
   \mathbb{E}\left(z^{X_{n+1}}\right) = \mathbb{E}\left(z^{X_{n} - D_n}\right) \underbrace{ \mathbb{E}\left(z^{I_n}\right)}_{\exp(\mu (z-1))} = \exp(\mu (z-1)) \mathbb{E}\left(z^{X_n} \, \mathbb{E}\left(z^{-D_n}|X_n\right)\right) = \\ \exp(\mu (z-1)) \mathbb{E}\left((p+(1-p) z)^{X_n}\right)
$$
That is
$$
    \mathcal{P}_{n+1}(z) = \mathrm{e}^{\mu(z-1)} \mathcal{P}_{n}\left(p+(1-p)z\right) \qquad \mathcal{P}_0(z) = \mathrm{e}^{\mu(z-1)}
$$
Solving:
$$
     \mathcal{P}_{n}(z) = \exp\left( \mu (z-1) \frac{1-(1-p)^n}{p}\right)
$$
Thus $X_n$ is Poisson distributed with mean
$$
  \mathbb{E}(X_n) = \mu \frac{1-(1-p)^n}{p}
$$

$X_0$ is $\mbox{Poisson}(\mu)$.

Mean number of $X_1$ is $\mu (1 - p)$ + $\mu$.  My claim is that it is also Poisson.

If you write out a few more terms, you'll soon see that $X_n$ is $\mbox{Poisson}\left(\mu\frac{1 -  (1-p)^{N+1}}{p}\right)$


Prove that $\int _e^\infty \frac{\ln(x)}{x^p} dx$ is divergent for $p \le1$. Prove that $\int _e^\infty \frac{\ln(x)}{x^p} dx$ is divergent for $p \le1$.

So my textbook divides the problem into first case $p=1$ and integrates and cases $p<1$ in which it uses integration by parts. However, isn't my method also valid (and much faster)? It is:

$$\int _e^\infty \frac{1}{x^p} dx \le\int _e^\infty \frac{\ln(x)}{x^p} dx$$

The integral on the left diverges by the so called $p$-series test for all $p \le 1$ and so does the integral on the right by the comparison test.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\int x^{\alpha}\,\dd x & = {x^{\alpha + 1} \over \alpha + 1}
\\[3mm]
\int x^{\alpha}\ln\pars{x}\,\dd x & =
{x^{\alpha + 1}\ln\pars{x} \over \alpha + 1} -
{x^{\alpha + 1} \over \pars{\alpha + 1}^{2}}
\\[3mm]
\int {\ln\pars{x} \over x^{p}}\,\dd x & =
{x^{1 - p}\ln\pars{x} \over 1 - p} -
{x^{1 - p} \over \pars{1 - p}^{2}}
\end{align}

In order to avoid the divergence when $x \to \infty$, it's clear that $p$ must be $> 1$.


How does one phrase the construction of the tensor product categorically? Given $U,V \in Vect$ we can build the tensor product as $U \otimes V := F(U \times V) / F(N)$ where $F:Set \rightarrow Vect$ is the free functor and where

$N:={(au,v) - a(u,v), (u,av) - a(u,v), (u+u',w) - (u,w) - (u',w), (u,v+v') - (u,v) - (u,v')  : a \in K; u, u' \in U; v, v' \in V }$

Then we have  the projection $p:U \times V \rightarrow U \otimes V$. The universalism of the projection, says that it is initial amongst all morphisms whose kernel includes $F(N)$. This means that these morphisms will also be bilinear. Thus we see that the tensor morphism is initial amongst all morphisms that are bilinear.

How does one phrase this in the terminology of universal morphisms? The Wikipedia page gives an example of the tensor algebra where they start with a forgetful functor $U:K-Alg \rightarrow K-Vect$.

I don't see how something similar can work here since the tensor product is an element of $Vect$. Yet it seems to me that the notion of a universal morphism here is appropriate.

The solution relies upon the fact that the tensor product is the coproduct in the category of commutative algebras. A proof of this is here in an answer in Math.SE by Omar Antolin-Carmenera.

Then we need only note that the categorical product has a formulation as a universal morphism, in fact, a terminal morphism as pointed out in the second example on the Wikipedia article on universal morphisms (note that there is a mistake in the detail of the proof, though the idea is the right one).

Finally, we just need take the dual of this construction to formulate the coproduct as an initial morphism.


Determine the whether the sequence $a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$ is convergent or divergent. Consider the recursively defined sequence $a_n = 1$

$$a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$$

Is the sequence convergent?

This is my attempt:

First, we prove that the sequence is positive, and monotonically increasing, using induction.

$\textbf{Base Case:}$ $a_1 = 1$ and $a_2 = 2$. $a_1 < a_2$ and $a_1,a_2 > 0$.

$\textbf{Inductive Hypothesis:}$ $a_n < a_{n+1}$ where $a_n, a_{n+1}$.

$\textbf{Inductive Step:}$ We prove that $a_n < a_{n+1} \Rightarrow a_{n+1} < a_{n+2}$.

By our induction hypothesis:

$$a_n < a_{n+1}$$

$$-a_n > -a_{n+1}$$

$$-\frac{1}{a_n} < -\frac{1}{a_{n+1}}$$ (True by our IH since $a_n, a_{n+1} > 0$ and thus, $-a_n, -a_{n+1}$ share the same sign).

$$3-\frac{1}{a_n} < 3-\frac{1}{a_{n+1}}$$

$$a_{n+1} < a_{n+2}$$.

Note that $a_{n+1} > 0$ by our IH, so $a_{n+2} > 0$.

Now we prove that the sequence is bounded.

Observe that $a_n$ is monotonically increasing, which means that $ - \frac{1}{a_n}$ is monotonically increasing as well and it is upped bounded by $0$. Thus, $3-\frac{1}{a_n}$ is upper bounded by $3$.

We have a sequence that is monotone and bounded. Hence, by the Monotone Convergence Theorem, This sequence converges.

I was wondering if this method is correct.

A shorter approach, look at the function $f(x)=3-\frac{1}{x}$ because this function "generates" the sequence, i.e. $f(a_n)=a_{n+1}$. This function is ascending, because $f'(x)=\frac{1}{x^2}$ and $\frac{1}{3}<a_1<a_2 \Rightarrow 0< \color{red}{f(a_1)\leq f(a_2)} \Rightarrow \frac{1}{3}<a_1<\color{red}{a_2 \leq a_3}$ and by induction $\frac{1}{3}< a_{n} \leq a_{n+1}$. So, the sequence is ascending and positive.

Given $a_n > \frac{1}{3} > 0, \forall n$, then $\frac{1}{a_n}>0 \Rightarrow a_{n+1}=3-\frac{1}{a_n} < 3 \Rightarrow 0<a_{n+1} < 3, \forall n$. So the sequence in bounded.

In your proof of $a_{n+1}>a_n$ you used that $a_n>0$, but it is not proven.

I like the following reasoning.
$$a_{n+1}-\frac{3-\sqrt5}{2}=3-\frac{3-\sqrt5}{2}-\frac{1}{a_n}=\frac{3+\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3-\sqrt5}{2}}-\frac{1}{a_n}>0$$ by induction because $a_1=1>\frac{3-\sqrt5}{2}.$
$$a_{n+1}-\frac{3+\sqrt5}{2}=3-\frac{3+\sqrt5}{2}-\frac{1}{a_n}=\frac{3-\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3+\sqrt5}{2}}-\frac{1}{a_n}<0$$ by induction because $a_1=1<\frac{3+\sqrt5}{2}.$

Thus, for all natural $n$ we got:
$$\frac{3-\sqrt5}{2}<a_n<\frac{3+\sqrt5}{2}.$$
In another hand, $$a_{n+1}-a_n=3-a_n-\frac{1}{a_n}=\frac{\left(a_n-\frac{3-\sqrt5}{2}\right)\left(\frac{3+\sqrt5}{2}-a_n\right)}{a_n}>0,$$
which says that $a$ converges.


infinite monoid H that is not a free monoid and contains a free monoid as a submonoid Let $H= \langle h_1, \ldots , h_n \rangle$ ($n>1$) be an infinite monoid that is not a free monoid. Does $H$ contain an isomorphic copy of a free monoid as a submonoid?
EDIT. It is a natural question. The corresponding question for groups is the Burnside problem.

Let $h$ be an element of $H$. If the submonoid generated by $h$ is not free, then there exist two integers $n \geqslant 0$ and $p > 0$ such that $h^n = h^{n + p}$. Thus $H$ is a torsion monoid. There exists a two-generated infinite monoid which is torsion, thus the answer to your question is negative. Consider for instance the quotient of the free monoid $\{a,b\}^*$ by the relations $x^3y = yx^3 = x^3$, for all words $x$ and $y$. This is a monoid with zero $M$ in which every $x^3 = 0$ is an identity. To prove it is infinite, consider the Prouhet-Thue-Morse infinite word. This infinite word is known to be cube-free, which means that it contains no factor of the form $uuu$, where $u$ is a nonempty word. It follows that each finite prefix of this infinite word corresponds to an element of $M$, and thus $M$ is infinite.
The corresponding problem for groups is the famous Burnside problem which also has a negative, but difficult, answer, first given by Golod and Shafarevich.

Hint: The only commutative free monoid is $(\Bbb N,+)$.


Gradient of $f(x,y)=\sqrt{x^2(1+y)+y^2+4}$ I have to solve the gradient of $f(x,y)=\sqrt{x^2(1+y)+y^2+4}$ as a part of a larger task. I know how to do this with partial derivatives but I was wondering if there are simpler ways to find the gradient since the nature of $f$ makes partial derivation a little messy?

You could use some cunning notation, for example:

$$\frac{d f}{d x}=\frac{x(1+y)}f$$

and etc.


Problems that become easier in a more general form When solving a problem, we often look at some special cases first, then try to work our way up to the general case.

It would be interesting to see some counterexamples to this mental process, i.e. problems that become easier when you formulate them in a more general (or ambitious) form.

Motivation/Example

Recently someone asked for the solution of $a,b,c$ such that $\frac{a}{b+c} = \frac{b}{a+c} = \frac{c}{a+b} (=t).$

Someone suggested writing this down as a system of linear equations in terms of $t$ and solving for $a,b,c$. It turns out that either (i) $a=b=c$ or (ii) $a+b+c=0$.

Solution (i) is obvious from looking at the problem, but (ii) was not apparent to me until I solved the system of equations.

Then I wondered how this would generalize to more variables, and wrote the problem as:
$$
\frac{x_i}{\sum x - x_i} = \frac{x_j}{\sum x - x_j} \quad \forall i,j\in1,2,\dots,n
$$

Looking at this formulation, both solutions became immediately evident without the need for linear algebra (for (ii), set $\sum x=0$ so that each denominator cancels out with its numerator).

Consider the following integral $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx$. All of our attempts at finding an anti-derivative fail because the antiderivative isn't expressable in terms of elementary functions.

Now consider the more general integral $f(y) = \displaystyle\int_{0}^{1}\dfrac{x^y-1}{\ln x}\,dx$.

We can differentiate with respect to $y$ and evaluate the resulting integral as follows:

$f'(y) = \displaystyle\int_{0}^{1}\dfrac{d}{dy}\left[\dfrac{x^y-1}{\ln x}\right]\,dx = \int_{0}^{1}x^y\,dx = \left[\dfrac{x^{y+1}}{y+1}\right]_{0}^{1} = \dfrac{1}{y+1}$.

Since $f'(y) = \dfrac{1}{y+1}$, we have $f(y) = \ln(y+1)+C$ for some constant $C$.

Trivially, $f(0) = \displaystyle\int_{0}^{1}\dfrac{x^0-1}{\ln x}\,dx = \int_{0}^{1}0\,dx = 0$. Hence $C = 0$, and thus, $f(y) = \ln(y+1)$.

Therefore, our original integral is $\displaystyle\int_{0}^{1}\dfrac{x^7-1}{\ln x}\,dx = f(7) = \ln 8$.

This technique of generalizing an integral by introducing a parameter and differentiating w.r.t. that parameter is known as Feynman Integration.

A nice example appeared on this web site today: Every prime number $p\ge 5$ has $24\mid p^2-1$ .

As posed, the problem sounds like it might be difficult. But it is very easy to show the more general result that every $n$ of the form $6k\pm 1$ has the required property.


In how many words the letter of word RAINBOW be arranged so that only 2 vowels always remain together? My Approach:

RAINBOW has 4 Consonants and 3 vowels.

Out of 3Vowels 2 vowel are selected  and  arranged in 3P2 ways

and the rest letters are arranged in 5! ways(1vowel and 2 consonants)

The Required arrangement is: 3P2*5!=720

But the Ans is 2880

Second Approach:

Out of 7 possible ways subract the one having either all vowel together or no vowel are together.

7!-(5!*3!+4!*5P3!)=2880  // i got the answer through this approach

What i have done wrong in ist approach and why the approach was wrong?

Let $V$ be a vowel and $C$ be a consonant.

In any word (for example: V V C C C V C) there are $4!$ possibilities to rotate $C$'s, and $3!$ possibilities to rotate $V$'s. This yields $3! \cdot 4! = 144$ possibilities to rotate $C$'s and $V$'s in a word.

When two $V$'s are in the first and second place, there are $4$ ways to put the third $V$ (same goes when two $V$'s are at the end). When two $V$'s are in the middle, there are $3$ ways to put the third $V$. You have 4 situations when two $V$'s are in the middle. So the number of permitted settings is $4 + 3 + 3 + 3 + 3 + 4 = 20$.

Now the total number of possiblities is $20 \cdot 144 = 2880$.


Real part of $\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}$ How can I compute the real part of
\begin{equation*}
\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}, \quad \text{where}\ \theta \in \mathbb{R}?
\end{equation*}

Maybe it's a silly question, but I'm feeling rusty with complex numbers...

For that very expression there's a trick to introduce $\alpha = \frac{\theta}{2}$ and noticing that

$$\frac{1- e^{i (n+1) \theta}}{1-e^{i \theta}} = \frac{e^{i (n+1)\alpha} (e^{i (n+1)\alpha} - e^{-i (n+1)\alpha})}{e^{i\alpha} (e^{i\alpha} - e^{-i\alpha})} = e^{in\alpha} \frac{\sin((n+1)\alpha)}{\sin \alpha}$$

I'll let you conclude...

hint: $$e^{i\theta} = \cos \theta + i\sin \theta$$, $$e^{(n+1)i\theta} = \cos(n+1)\theta+i\sin(n+1)\theta$$


How to prove that a matrix is invertible $\iff$ invertible at right $\iff$ invertible at left? Let $A$ a $n\times n$ invertible at left. In fact, I just want to prove that it's invertible at right (the rest is obvious). All what I can say is that there is a $B$ s.t. $BA=I.$
To prova $AB=I$, I have problem. I have that $$AB=AB^2A=BA^2B,$$
but I can't conclude that it's $I$.

Assuming $A\in \mathbb{R}^{n\times n}$, let $f_A : \mathbb{R}^n \to \mathbb{R}^n, x\mapsto Ax$ be the associated linear map. Since $BA = I$, we have $f_B\circ f_A = \operatorname{id}$, so $f_A$ is injective. But then $f_A$, as a linear map between finite-dimensional vector spaces of same dimension, is also surjective, hence bijective and so $A$ is invertible.

This is because $n = \operatorname{dim}(\ker f_A) + \operatorname{dim}(\operatorname{im} f_A)$. Since $f_A$ is injective, $\operatorname{dim}(\ker f_A) = 0$, so $\operatorname{dim}(\operatorname{im} f_A) = n$, hence $f_A$ is surjective.

If you can try $B = \text{adj}(A)/\det(A)$ and show by brute force computation that $BA = I = AB$.

You would only need $B^jA_i = \delta_{ij}$ and $A^iB_j = \delta_{ij}$. (Superscript means row vector and subscript meaning column vector index, and $\delta_{ij}$ means indicator of $i = j$).

You only need to think about the relationship between the definitions of $\text{adj}$ and $\det$. See https://en.wikipedia.org/wiki/Adjugate_matrix

If additionally you know how transpose relates to inverse, I think you only need $B^jA_i = \delta_{ij}$.


Derivation of the Hypergeometric Distribution The derivation of the hypergeometric refers to the following example: An urn contains N white balls, M black balls and we draw $n\le N+M$ balls without replacement. Let X be the number of white balls, we are looking for the probability that $X=x$ where $x$is an arbitrary fixed integer.

If we number the balls from $1$ to $N+M$, we may think about an outcome being a subset of $\{ 1,.....,N+M\}$, therefore, $\Omega$ is the family of all subsets with n points, the total number of outcomes being:$ \# \Omega= {N+M\choose{n}}$.

Now, we can consider that all possible outcomes are equally likely, so that P is uniform on $\Omega$. The quantity X is a random variable.

I get confused here:

The set $X^{-1}(\{x\})$ contains $N \choose x$$ M \choose {n-x}$ points. I really don't understand what this set is? Could someone give me an example of this?

To take an example, suppose there are $N=3$ white balls, $M=2$ black balls, you draw $n=3$ balls without replacement, and you are looking at the probability that the drawn balls have $x=2$ white and $n-x=1$ black.

Then if the five original balls are labelled $w_1,w_2,w_3,b_1,b_2$ then the possibilities for the three drawn balls of which two are white are:   $$X^{-1}(\{x\}) =$$ $$ \{\{w_1,w_2,b_1\}, \{w_1,w_3,b_1\}, \{w_2,w_3,b_1\}, \{w_1,w_2,b_2\}, \{$w_1,w_3,b_2\}, \{$w_2,w_3,b_2\} \}$$

with cardinality $6= {3 \choose 2}{2 \choose 1}$ as expected.

How many ways to pick exactly $x$ white balls?  You have to pick $x$ of the $N$ white balls (which can be done in ${N\choose x}$ ways and for each such way there are $M\choose n-x$ ways to pick the remaining $n-x$ balls from the $M$ black balls.


Eigenvalues with positive real parts imply instability Given an autonomous ode $\dot{x}=F(x)$ with $0$ being its equilibrium point, and all eigenvalues of $DF(0)$ have non-zero real parts.
I have learned that if the real parts of eigenvalues are all negative , then the system is asymptotically stable near $0$. Now I guess that if one of the eigenvalues has positive real part then the system is unstable near $0$, but I can't give a proof, can anyone help me?
Many thanks!

A proof of this result is given in Prof. Lebovitz's notes on Ordinary Differential Equations Chapter 8, and I paste his proof here.

Theorem 8.4.2 Consider the autonomous dynamical system
\begin{equation}
 \dot x = Ax + g(x),~~~~x\in\Omega
 \label{system}
\end{equation}
where $\Omega$ is an open region in $\mathbb R^n$ which contains the origin, $g\in C^1(\Omega)$ and $g(x) = o(|x|)$ as $x\rightarrow0$. If one of the eigenvalues of $A\in\mathbb R^{n\times n}$ has a positive real part, then the origin is an unstable equilibrium point.

The proof is accomplished by splitting the the eigenvalues of $A$ into positive and negative parts. Let's assume the eigenvalues of $\lambda_1,\cdots,\lambda_{n_1}$ of $A$ have positive real parts, while the remaining eigenvalues $\lambda_{n_1+1},\cdots,\lambda_n$ have non-positive real parts. Let
\begin{equation}
 \sigma = \min\{\Re(\lambda_1),\cdots,\Re(\lambda_{n_1})\}
 >0
\end{equation}
be the lower bound of the real parts of $\lambda_1,\cdots,\lambda_{n_1}$.
Now we apply the similarity transformation $B = P^{-1}AP$ to obtain the Jordan normal form of $A$, which is given in the block form
\begin{equation}
 B = \begin{bmatrix}
  B_1 & 0 \\
  0 & B_2
 \end{bmatrix} \in \mathbb C^{n\times n}
\end{equation}
and the matrices $B_1\in\mathbb C^{n_1\times n_1}$ and $B_2\in\mathbb C^{(n-n_1)\times(n-n_1)}$ satisfy

$B_1$ and $B_2$ possess the Jordan structure: eigenvalues appear on the diagonal, and a certain number of constants occur along a secondary diagonal;
Each eigenvalue of $B_1$ has a
positive real part whereas no eigenvalue of $B_2$ has a positive real part;
The constants along the secondary diagonal are
all equal to $\nu$, which is a fixed positive constant chosen to satisfy $\nu<\sigma/8$.

We make the change of variables $y = P^{-1}x\in D$, where $D:=\{P^{-1}x:x\in\Omega\}$ is a subset of $\mathbb C^n$. Note that the transformation matrix $P\in\mathbb C^{n\times n}$ may contain complex entries, hence $D$ can be not contained in $\mathbb R^n$. $h(y) = P^{-1} g(Py)$ is then well-defined as a $C^1$ mapping in $D$. We fix a constant $\epsilon_0 < \sigma/8$ and choose $\delta_0>0$ such that
\begin{equation}
 0<|y|<\delta_0\Longrightarrow |h(y)|<\epsilon_0|y|
\end{equation}
The original dynamical system now can be written as
\begin{align*}
 \dot y_1 & = B_1 y_1 + h_1(y) \\
 \dot y_2 & = B_2 y_2 + h_2(y)
\end{align*}
Define the norm of $y_1\in\mathbb C^{n_1}$ and $y_2\in\mathbb C^{n-n_1}$ by
\begin{equation}
 R_1 = |y_1| = \bigg(\sum_{i=1}^{n_1}
 |y^{(i)}|^2
 \bigg)^{\frac12},~~~~
 R_2 = |y_2| = \bigg(\sum_{i=n_1+1}^{n}
 |y^{(i)}|^2
 \bigg)^{\frac12}
\end{equation}
where $y^{(i)}$ denotes the $i$-th entry of $y\in\mathbb C^n$. Then we can verify
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}R_1^2(t) =
2\Re\bigg\{
\sum_{j,l=1}^{n_1}
B^{(j,l)} \overline{y^{(j)}} y^{(l)} +
\sum_{j=1}^{n_1} \overline{y^{(j)}} h^{(j)}(y)\bigg\}
\label{dR1}
\end{equation}
In the equation above, we have

The diagonal part
\begin{equation}
 2\Re\bigg\{
 \sum_{j=1}^{n_1} B^{(j,l)} |y^{(j)}|^2
 \bigg\} \geqslant 2\sigma \sum_{j=1}^{n_1} |y^{(j)}|^2 = 2\sigma R_1^2
\end{equation}
The off-diagonal (secondary diagonal) part
\begin{align*}
 \bigg|2\Re\bigg\{
  \sum_{j=1}^{n_1-1}
  B^{(j,j+1)} \overline{y^{(j)}}
  y^{(j+1)}
 \bigg\}\bigg| & \leqslant
 2\sum_{j=1}^{n_1-1}
 |B^{(j,j+1)}||y^{(j)}||y^{(j+1)}| \\
 & \leqslant 2\nu\sum_{j=1}^{n_1} |y^{(j)}|^2 = 2\nu R_1^2
\end{align*}
The nonlinear part
\begin{align*}
 \bigg|
 2\Re\bigg\{
 \sum_{j=1}^{n_1} \overline{y^{(j)}} h^{(j)}(y)
 \bigg\}
 \bigg| & \leqslant 2\sum_{j=1}^{n_1} |y^{(j)}| |h^{(j)}(y)| \\
 & \leqslant 2 |y_1| |h_1(y)| \\
 & \leqslant 2\epsilon_0 R_1 (R_1+R_2)
\end{align*}

Hence we conclude
\begin{equation}
 \frac{\mathrm{d}}{\mathrm{d} t}R_1(t) \geqslant
 (\sigma - \nu -\epsilon_0) R_1 - \epsilon_0 R_2
\end{equation}
In a similar way, we can prove that
\begin{equation}
 \frac{\mathrm{d}}{\mathrm{d} t}R_2(t) \leqslant (\nu+\epsilon_0) R_2 + \epsilon R_1
\end{equation}
Putting these together gives
\begin{equation}
 \frac{\mathrm{d}}{\mathrm{d} t}(R_1-R_2) \geqslant
 (\sigma - 2\nu - 2\epsilon_0) R_1 - 2\epsilon_0 R_2
\end{equation}
Recall the choices $\nu,\epsilon_0 < \sigma/8$, we have
\begin{equation}
 \frac{\mathrm{d}}{\mathrm{d} t}(R_1 - R_2) \geqslant \frac{\sigma}2(R_1-R_2)
\end{equation}
Let's choose the initial data such that $R_1>R_2$ at $t=0$, then the result above yields
\begin{equation}
 \lim_{t\rightarrow\infty} (R_1(t)-R_2(t)) = +\infty
\end{equation}
A contradiction to the stability.

You are comparing two systems of differential equations, the linear system
$$
\dot v=Av
$$
and the non-linear system
$$
\dot x=F(x)
$$
with $F(0)=0$ a stationary point at the origin and $DF(0)=F'(0)=A$ as the Jacobian. If $F$ is sufficiently smooth, say twice differentiable at least, the vector fields look similar close to the origin.
Now if one eigenvalue of $A$ is real and positive, or a conjugate eigenpair has a positive real part, then there are solutions of the linear system that connect the stationary point with infinity moving outwards. The question asks if the non-linear system is similar enough that solutions close-by inherit enough of this divergence to prevent this point from being stable.
The Hartman-Grobman theorem answers this in the most comprehensive way (lesser tools may be able to also answer this with more effort) by the claim that there is some diffeomorphism $\Phi$ mapping a neighborhood of the origin in the linear system to a neighborhood of the stationary point of the non-linear system (here also the origin) with $Φ(x)=x+O(|x|^2)$ so that $$x(t)=Φ(e^{At}v_0)$$ is the solution of $$\dot x=F(x)~~\text{ with }~~x_0=Φ(v_0)\iff v_0=Φ^{-1}(x_0)$$ for $x_0$ close to the fixed point $0$ and $t$ small enough.
Now if $A$ has an eigenvalue $\lambda=\alpha+i\beta$ with positive real part $α$, then there will be an eigenvector $u+iw$ leading to real solutions $e^{αt}(\cos(βt)u-\sin(βt)w)$ for the linear equation and thus
$$
x(t)=Φ(v(t))=Φ\bigl(e^{αt}(\cos(βt)u-\sin(βt)w)\bigr)
$$
(for a real positive eigenvalue just set $β=0$).
Now if you fix some point $x(t_*)$ on this solution that is still inside the range of the diffeomorphism, then the points $x_0=x(t_0)$, $t_0<t_*$ can be selected arbitrarily close to the origin (select $β(t_*-t_0)\in 2\pi\Bbb Z$ to get points on the same ray to the origin). Thus you get initial points arbitrarily close to the origin that have solutions that pass through the same point $x(t_*)$ (relatively) far away from the origin. This is then a contradiction to both definitions of stability and asymptotic stability.


Are there any shortcuts to tell if a square matrix is orthogonal? So, if one is asked if a given matrix $A$ is symmetric, one could compute $A^T$ and check if $A^T=A$, however you can also simply check the symmetric entries accross the diagonal and see if they are equal (i.e whether $A_{ij}=A_{ji}$).

If one wishes to check if a square matrix is orthogonal, one could compute all the dot products of it's rows and columns, which, assuming an $n\times n$ matrix, requires $2\times\binom{n}{2}$ dot products (and this is without taking into account that as $n$ increases, so does the number of arguments in the sum of each dot product).

So, is there a shortcut that one may use to tell that a given matrix is orthogonal without having to compute all the dot products?

Computing the dot product is the same as computing the matrix product $A^TA$ and checking it is equa to the identity matrix $I$, that, with the naive algorithm, costs $O(n^3)$ FLOPS.

With fast matrix multiplications, though, you can get down to $O(n^{2.3728639})$  FLOPS, so you can gain something substantial in time.


Has triangle an angle? I read axiomatic geometry and found the following definitions:

Points $A$ and $B$ and all those points that lie between those points is a line segment.

If $AB$ and $AC$ are two rays that does not belong to the same line, then the union of those rays forms an angle.

If three points are not collinear, they form a triangle. Line segments $AB$, $BC$ and $CA$ are the sides of a triangle.

Now, has a triangle an angle? I mean if triangle has line segments but not rays then I guess triangle has no angles unless we extend the line segments to the rays. Am I correct?

Yes, the angles in a triangle are defined from extending the segments into rays.

A triangle has... three angles! That's one definition of the triangle.

However, there are degenerate triangles, when the three points are colinear, but all different one from the others. Then one angle is $\pi$ while the other tow are $0$.


What is $\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$? What is $\displaystyle\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ ?
Find an asymptotic expansion of $\displaystyle \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ as $x\to 0$

One the one hand, $\sqrt{x}\to 0$, but $\displaystyle\frac{\ln n}{1+n^2 x}\sim \ln n$ which suggests that $\sum_{n=2}^\infty \frac{\ln n}{1+n^2 x}$ "diverges" to $\infty$.
I haven't been able to tell which term dominates here, let alone the asymptotic expansion.
Here's the graph of $x\to \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ for $x\in (0,0.5)$
This suggests the limit is $\infty$ and the estimate should be $1/x^\alpha$ for some positive $\alpha$.

If we look at the corresponding integral, we get

$$\int_2^\infty \frac{\sqrt{x}\ln t}{1 + t^2x}\,dt = \int_{2\sqrt{x}}^\infty \frac{\ln u - \ln \sqrt{x}}{1+u^2}\,du.$$

Both functions,

$$\frac{\ln u}{1+u^2} \quad\text{and}\quad  \frac{1}{1+u^2}$$

are Lebesgue integrable over $(0,\infty)$, so for the integral, we have the asymptotic development

\begin{align}
I(x) &= \int_0^\infty \frac{\ln u}{1+u^2}\,du - \frac{\pi}{4}\ln x - \int_0^{2\sqrt{x}}\frac{\ln u}{1+u^2}\,du + \frac{1}{2}\ln x\int_0^{2\sqrt{x}} \frac{1}{1+u^2}\,du\\
&= C - \frac{\pi}{4}\ln x + \frac{1}{2}\arctan (2\sqrt{x})\ln x - 2\sqrt{x}\ln (2\sqrt{x}) + 2\sqrt{x} + O(x)\\
&= C - \frac{\pi}{4}\ln x + 2(1-\ln 2)\sqrt{x} + O(x).
\end{align}

Now let's look at the difference between the series and the integral.

$$\frac{\sqrt{x}\ln n}{1+n^2x} - \int_{n}^{n+1} \frac{\sqrt{x}\ln t}{1+t^2x}\,dt = \sqrt{x}\int_n^{n+1} \frac{(t^2-n^2)x\ln n}{(1+n^2x)(1+t^2x)} - \frac{\ln \frac{t}{n}}{1+t^2x}\,dt.$$

We always have $0 \leqslant \ln \frac{t}{n} \leqslant \ln \frac{3}{2}$ on $[n,n+1]$, so we can estimate

$$0 < \int_2^\infty \frac{\sqrt{x}\ln \frac{t}{\lfloor t\rfloor}}{1+t^2x}\,dt < \ln \frac{3}{2} \int_0^\infty \frac{du}{1+u^2}.$$

For the other term, noting that $n \geqslant \frac{2}{3}t$ and $t^2 - n^2 < 2t+1 < 3t$, we can estimate

\begin{align}
0 &< \int_2^\infty \frac{x^{3/2}(t^2-\lfloor t\rfloor^2)\ln \lfloor t\rfloor}{(1+\lfloor t\rfloor^2x)(1+t^2x)}\,dt\\
&< \int_2^\infty \frac{x^{3/2}3t\ln t}{\bigl(1+\frac{4}{9}t^2x\bigr)(1+t^2x)}\,dt\\
&= 3\sqrt{x}\int_{2\sqrt{x}}^\infty \frac{u\ln \frac{u}{\sqrt{x}}}{\bigl(1+\frac{4}{9}u^2\bigr)(1+u^2)}\,du.
\end{align}

The latter tends to $0$ when $x\to 0$, due to the $\sqrt{x}$ factor, so the difference between the series and the integral remains bounded.

Overall

$$\sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2x} = -\frac{\pi}{4}\ln x + O(1).$$

I know we already have an answer, but I would argue this way:

for all $x$ and $n\ge 2$ we have

$$
\frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}\ln{n}}{n^2(1+x)}
$$

Then

$$
\sum_{n \ge 2} \frac{\sqrt{x}\ln{n}}{1+n^2x} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{\ln{n}}{n^2} > \frac{\sqrt{x}}{1+x} \sum_{n \ge 2} \frac{1}{n^2} = \frac{\sqrt{x}}{1+x}c
$$

Once the series $\sum_{n \ge 2} \frac{1}{n^2}$ is convergent.
This shows you that goes to infinity faster than $\frac{1}{\sqrt{x}}$, when $x \rightarrow 0 $.


How do I find the powerset of $A\cap B$? $A = \{0,1\}$

$B = \{1,2\}$

My Working :

$P(A\cap B)= P(\{\varnothing, \{1\}\}) = \{\varnothing,\{1\},\{\varnothing,\{1\}\}\}$

But the correct answer is $P(A\cap B)= \{\varnothing , \{1\}\}$.

The intersection of $A$ and $B$ is simply $\lbrace 1 \rbrace$, which has only $\lbrace \rbrace$ and $\lbrace 1 \rbrace$ as subsets, and so the power set is $\lbrace\lbrace \rbrace, \lbrace 1\rbrace \rbrace$. The empty set is a subset of $A$ and $B$, but it is not a member of either set. Therefore, the empty set is not a member of $A \cap B$. Similarly, $\lbrace 1\rbrace$ is a subset of, not a member of, $A$ and $B$, so it is not a member of the intersection.

$P(A\cap B)=P(A)\cap P(B)$ as any subset of $A\cap B$ has to be a subset of both $A$ and $B$, and the converse is trivial.