Round Table Adjacent Seats Problem If there are 10 seats in a round table and 7 people are already sitting in some random pattern(edit: uniform distribution, thank you for the correction) what is the probability that the next two people entering the room will get to sit in adjacent seats?

I am very challenged in trying to start to think about the problem. I would sincerely appreciate any thoughts and I will definitely indicate my progress thereafter.

We assign seats at random, by putting place mats labelled $1,2,3,\dots, A, B, C$ at random in front of the $10$ chairs. The place mats labelled $A,B,C$ indicate where any person apart from $1$ to $7$ may sit.  A pair of lovers now enters the room, and they wish to sit together. We find the probability that they can do so without asking anybody to move.

The only thing that matters is the placement of $A, B, C$. And since the table is round, all that matters is the placement of $B$ and $C$ relative to $A$. So we have $9$ empty spaces, and have to choose $2$ of them. There are $\binom{9}{2}$ choices, all equally likely.

We count the bad choices. For a choice to be bad, it must involve choosing $2$ non-adjacent seats from the $7$ not next to $A$. There are $\binom{7}{2}$ ways to choose $2$ seats, of which $6$ give an adjacent pair. So there are $15$ bad choices, out of the $\binom{9}{2}$ choices. Thus the probability there will be an adjacent pair is $\frac{21}{36}$.

Consider 10 chairs arranged in a circle.
The number of ways of leaving three seats empty is C(10,7) = 240.

Of these, the number of ways in which atleast 2 of the empty seats are adjacent is 10.8 = 80.
(Think of this situation as having 10 fixed spaces. There are 10 ways to choose a pair of adjacent spaces, considering the first and last also as adjacent-one for each space being the left space of the chosen pair. There are 8 ways to choose one of the remaining 8 spaces)

Hence, the required probability is 80/240 = 1/3.


Generate Random Coordinates in a Randomly Shaped Convex Polytope I am trying to generate random trial coordinates to insert points in a given convex three-dimensional domain. Such domain is defined by its surfaces, of which vertices are known. I am using a pseudo-random generator for the coordinates, how can I check if the point is inside the randomly shaped polytope?
This is straightforward for box domains, but I am struggling with randomly shaped ones.
Thanks.

Have a look at the MATLAB code I wrote that addresses this problem. https://au.mathworks.com/matlabcentral/fileexchange/34208-uniform-distribution-over-a-convex-polytope


Grandi's series contradiction This is the Grandi's series: $1-1+1-1+1-1+\dots$

The series can be equal to $0$

$$(1-1)+(1-1)+(1-1)+\dots=0+0+0+\dots=0,$$

or to $1$

$$1-(1-1)-(1-1)-(1-1)-\dots=1-0-0-0\dots=1,$$

or to $1/2$

$$S=1-1+1-1+1-\dots,\quad\quad    S=1-(1-1+1-1+1-1+1-...)$$
$$\Rightarrow   S=1-S\Rightarrow     2S=1\Rightarrow    S=1/2$$

Isn't this a contradiction? The integers are closed under addition and subtraction, but we get a fraction. Why?

The issue here is that addition of a finite number of terms is a nice, well-defined operation; addition of an infinite number of terms need not be.

In particular, rearranging terms in an infinite sum can actually change the sum. The series you have presented does not converge - and so discussing it as having a value is incorrect to begin with.  However, there are series which converge, and which STILL cannot be rearranged without changing the value: this is true for series that are called conditionally convergent - that is, the series converges, but the sum of absolute values of the terms does not.

Interestingly, for conditionally convergent series, it turns out that you can rearrange the terms to give any limit that you like.

At the end of the day, the big thing to remember is this:  addition of an infinite number of terms cannot be assumed to be associative or commutative, unless you know that the sum of the absolute values of the terms converges - a property known as absolute convergence.

$$S(5) = 1-1+1-1+1 = 1$$
then   $1-S(5) = 1-1+1-1 = S(4)$.  So  $1-S(n) = S(n-1)$.  If $S(5)=S(4)$ (which is not), then $S(5)=1/2$, wrong, and similar to $S=1/2$. Grandi assumes  $S(\infty)=S(\infty-1)=S$, so that $S=1/2$.

The question is: Is it really $S(\infty)=S(\infty-1)$?. Of course $\infty-1= \infty$, but this is not the case, now $\infty$ is not a value, is the order of the variable which still will take values of $1$ or $-1$, even their order is $\infty$.

So $S(\infty)$ does not equal $S(\infty-1)$. That's what I believe.


Why does $\cos(x)(\cos(x) - 1) = 0$ have solutions $x=\pi/2$ and $x=3\pi/2$ (as well as $x=0$)? The original equation is
$$2 - \sin^2(x) = 2 \cos^2 (x/2) \tag{1}$$
I used the Half-angle formula for $\cos$ as well as the Pythagorean identity for sine and cosine to reduce it and ended up with
$$\cos(x)(\cos(x) - 1) = 0 \tag{2}$$
The book has the answers as $x=\pi/2, 3\pi/2, 0$. I understand the $0$, but I thought the other answer would be $\pi$ since on the unit circle $\pi$ has a cosine of $1$?

If it's not mentioned that $x \in [0,2\pi]$, then the equation has a lot of other solutions as well, given by the trigonometric equations
$$\cos x = 0 \text{ and } \cos x = 1$$
for the first equation, the general solution is $x = n\pi + \frac{\pi}{2}$ and for the second equation, the general solution is $x = 2n\pi$. $\pi$ is not a solution as $\cos \pi = -1$. Infinite solutions exist because $\cos(x)$ is a periodic function. Have a look at the graph of $\cos x$ to clear things up:


The free monad over an endofunctor on $\mathcal{Set}$ Let $F$ be an endofunctor on the category $\mathbf{Set}$. How can I construct the free monad over $F$? Can this construction be generalized to other categories than $\mathbf{Set}$?

Let's investigate the specific case of free monads over $\newcommand\catname\mathbf\newcommand\Set{\catname{Set}}\Set\newcommand\Mnd{\catname{Mnd}}\newcommand\into\hookrightarrow\newcommand\id{\mathrm{id}}$, and the relationship to the Haskell free monad.

Free Monads in General:

First the general case of what a free monad is. (There are more general notions, but this corresponds to the Haskell notion.) Note that I am not writing the forgetful functor from monads in $C$, $\Mnd_C$, to the endofunctor category, $[C,C]$, explicitly.

Let $F: C\to C$ be an endofunctor. A monad $(T,\mu_T,\iota_T)$ is a free monad for $F$ if for monads $(S,\mu_S,\iota_S)$ we have a natural isomorphism
$$[C,C](F,S) \simeq \Mnd_C(T,S).$$
The functor on the right is corepresentable, so a natural transformation
$$\Mnd_C(T,S)\to [C,C](F,S)$$
is determined by an element $\alpha$ of $[C,C](F,T)$ by Yoneda's lemma. The free monad is usually considered to be the pair $(T,\alpha)$.

The Haskell Free Monad:

Let $\newcommand{\Hask}{\catname{Hask}}\Hask$ be the "category" of Haskell types (not quite a category, but it's close enough for our discussion).

Then the definition of the Free monad type in Haskell looks like this

data Free f a
  = Pure a
  | Free (f (Free f a))

instance Functor f => Functor (Free f) where
  fmap g (Pure x) = Pure (g x)
  fmap g (Free y) = Free (fmap (fmap g) y)

instance Functor f => Monad (Free f) where
  return = Pure
  (Pure x) >>= g = g x
  (Free y) >>= g = Free (fmap (>>= g) y)

For the non-Haskellers, this translates to
if $F:\Hask\to\Hask$ is a functor, the free monad functor $T$ is defined by the equation $T-=-\amalg FT-$.

Then the natural inclusion $A\into TA$ is the identity for the monad, and the multiplication $\mu: T^2A\to TA$ is defined by observing $T^2A=TA\amalg FT^2A$ and defining $\mu$ to satisfy $\mu = \id \amalg F\mu$.

You'll notice we had to solve several equations here, which aren't problems in $\Hask$ because types and functions can be defined recursively, so these equations define valid types and functions.

We should also show that this definition of Free f gives a free functor for f. The natural transformation from f to Free f is just Free . fmap return, but I'll leave the verification out for now.

Translating to Set:

Now let's translate this to $\Set$. It seems like a reasonable guess that solving the equations for $T$ and $\mu$ are where we might run into problems. I'll have to return to this later though, so I'll post this partial answer.

Now we are working with $F : \Set\to \Set$.
For each ordinal $\alpha$ define functors $F_\alpha$, with
$F_0=\newcommand\Id{\mathrm{Id}}\Id$,
$F_{\alpha+1}=-\sqcup FF_\alpha -$,
and $F_\beta = \newcommand\colim{\operatorname{colim}} \colim_{\alpha \in \beta} F_\alpha$ when $\beta$ is a limit ordinal.
When $\alpha \le \beta$, there are natural maps $F_\alpha\to F_\beta$, which
are constructed at the same time as $F_\beta$, so that the colimit makes sense for limit ordinals. The only map that needs to be constructed explicitly is the map $\iota_\alpha: F_\alpha\to F_{\alpha+1}$.
$\iota_0 : F_0\to F_1$ is clear, this is just the inclusion of $X$ in $X\sqcup FX$ for all $X$.
If $\alpha=\gamma+1$, then $F_\alpha - = -\sqcup FF_\gamma -$.
Then by induction $F\iota_\gamma : FF_\gamma \to FF_\alpha$, so
$$\iota_\alpha = \id \sqcup F\iota_\gamma$$ works.
On the other hand, if $\alpha$ is a limit ordinal, for $\gamma \in \alpha$,
define $j_\gamma : F_\gamma \to F_{\alpha+1}$ by the following formulas.
$j_0$ is the inclusion of $F_0=\Id$ into $F_{\alpha+1}$,
$$j_{\delta+1} = \id \sqcup F (F_\delta\to F_\alpha),$$
and for limit ordinals $\delta$,
$$j_{\delta} =  F_\delta \newcommand\toby\xrightarrow\toby{\iota_{\delta}} F_{\delta+1}\toby{j_{\delta+1}} F_{\alpha+1}.$$
It's not hard to show that these $j$s commute with the $\iota$s, and thus induce the desired map $\iota_{\alpha}: F_\alpha\to F_{\alpha+1}$.

This completes our construction of the functors $F_\alpha$.
Note that since we have $F\to F_1$, and $\Id\to F_0$, for all $\alpha \ge 1$, we have maps $F\to F_\alpha$ and $\Id \to F_\alpha$. We can think of these as being approximations to a solution to our equation.

We also need an approximation of our multiplication.
Let $i_{\alpha,\beta} : F_\alpha\to F_\beta$ be the maps constructed above.
Define $\mu_{\alpha,\beta}: F_\alpha F_\beta \to F_{\alpha+\beta}$ recursively. (Side note, ordinal addition is noncommutative, but I don't remember which order is correct. I think I may want $\beta+\alpha$ here actually, but this can be deduced from the definition of $\mu$ that I give.)
We have
$$\mu_{0,\beta} = \id,$$
$$\mu_{\alpha+1,\beta} = i_{\beta,\alpha+\beta+1} \sqcup F\mu_{\alpha,\beta},$$
and in the limit ordinal case,
$$\mu_{\alpha,\beta} = \colim_{\gamma\in\alpha} i_{\gamma+\beta,\alpha+\beta}\circ \mu_{\gamma,\beta}.$$

Then if the sequence $F_\alpha$ stabilizes eventually, we see that we can solve the equations defining the free monad in $\Set$.

There's quite a bit more to add here, but once again, I'll have to leave it at a partial answer for now.


Much less than, what does that mean? What exactly does $\ll$ mean?

I am familiar that this symbol means much less than.

...but what exactly does "much less than" mean? (Or the corollary, $\gg$)

On Wikipedia, the example they use is that $1\ll 9999999999$
But my thought on that is that $10^{10^{10^{11}}}\ll 10^{10^{10^{11}}}+9999999999$, based on the same logic. But I am confused because the numbers are comparitively.

Logically, I am really kinda spent dealing with this issue of mine. Thanks for any input.

The entire point is that it's NOT very clear what "much less than" is. What is "much less than"? In some contexts $1\ll 2$ and in others $ 1\not \ll 2$ but $1 \ll 100000$ and in others, even $1\not\ll 100000$. What precisely makes something "much less than"?

In most contexts, $\ll$ is used in approximation. For example: for $0<x\ll 1$, $(1+x)^n\approx 1+nx$.

So how small does $x$ have to be? Well small enough for the approximation to be "good enough"! What is good enough? It seems like I'm just pushing off the question, and that's indeed what I'm doing!

The entire point isn't whether something is "small enough" (big enough), or if the approximation is "good enough", the point is control. Can you control the error in a manageable way.

For the example $0<x \ll 1, (1+x)^n \approx 1+nx$. You can make $(1+x)^n$ as close as you'd like to $1+nx$ by making $x$ small enough! I think this is a more rigorous meaning.

$\ll$ is imprecise in the sense that you don't know how "small" something is. But it is precise in the sense that it implies that there is some control. There is some way of making the error in the approximation, argument, etc. as small as you'd like, provided that you make $x$ small enough.

Par example: Choose a priori positive number $N>1$ such that $0<x\ll y$. If $\frac{y-x}{x}>N$ for then $y>x\cdot(1+N)$. With repect to N, your 2nd example does not fit this requirement: $10^{10^{10^{10^{11}}}}\ll10^{10^{10^{10^{11}}}}+9999999999$. I had in mind $\frac{N-O}{O}\times 100\%$, the relative change has to overcome a number greater than one. If my profits this month is USD10,000 and next month it is USD20,000 then it is much greater if you have choose a priori p.e. $N=\frac{3}{2}$. One can play with the apriori choosen N to figure out a bit more.


Homotopy between a closed geodesic and a closed curve. Let $M^{n}$ a Riemannian manifold orientable with positive curvature and even dimension. Let $\gamma$ a closed geodesic. Prove that $\gamma$ is homotopic to a closed curve whose lenght is strictly less than $\gamma$. I would any tips for solve this problem, because , i honestly i do know which tool use.

Thanks.

Consider a parallel translation $P : T_pM\rightarrow T_pM$ along a
closed geodesic $\gamma$ where $\gamma (0)=\gamma (l)=p$

Since $M$ is orientable, then $P$ is orientation preserving isometry
on $T_pM$ so that it has a fixed point $v$. When
$\gamma(t,s)=\exp_{\gamma (t)}\ sV(t)$ where $V$ is a parallel field
s.t. $V(0)=v$, then ${\rm length}\ \gamma(t,s)<{\rm length}\
\gamma(t,0)$ for $s>0$, since $M$ has a positive curvature.

Either I didn't understand you, or the statement is false. There exist non- simply connected positively curved manifolds of even dimension (e.g. product of two lens spaces). So choose any non-trivial class in fundamental group and consider the shortest (smooth) loop in this class by minimizing the length functional. It will be geodesic.


Does this sequence of random variables satisfy the Law of Large Numbers? Notation: $S_n = X_1 + \ldots + X_n$

Definition $1$ : The sequence of random variables $X_1, X_2, \ldots$ satisfies the Weak Law of Large Numbers if $$\frac{S_n - E[S_n]}{n} \overset{p}{\to} 0$$

Definition $2$: The sequence of random variables $X_{1}, X_{2}, \ldots$ satisfies the Strong Law of Large Numbers if $$\frac{S_n - E[S_n]}{n} \overset{\text{a.s.}}{\to} 0$$

Exercise: Let $(X_{n})$ be a sequence of independent random variables with $P(X_n = n^2) = \frac{1}{n^3}$, $P(X_n = 0) = 1 - \frac{1}{n^3}$. Does this sequence satisfy the Weak Law of Large Numbers? Does it satisfy the Strong Law of Large Numbers?

I've tried to find the distribution for $S_n$, but this approach didn't work. Any suggestions/hints on how to do this exercise are really appreciated.

Note that $E(S_n)=\sum_{j=1}^nEX_j=\sum_{j=1}^n\frac 1j$ behaves as $\log n$, hence $\frac{E(S_n)}n$ converges to $0$.

Let $E_k:=\{X_k=k^2\}$. Since $\sum_j\frac 1{j^3}$ is convergent, by Borel-Cantelli's lemma, $P(\limsup_j E_j)=0$. This means that there is $\Omega'\subset \Omega$ of probability $1$ such that for each $\omega\in\Omega'$, there is $N(\omega)$ for which $X_k(\omega)\neq k^2$, hence $X_k(\omega)=0$ whenver $k\geqslant N(\omega)$. This means that $\frac{S_n(\omega)}n=\frac
1n\sum_{j=1}^{N(\omega)}X_j(\omega)\to 0$.

Note that $E[S_n]$ is the harmonic series.


For which complex numbers does a given series converge During studying for an exam, I came across the following exercise:

For which $z \in \mathbb{C}$ does the series
$$\sum\limits_{n=1}^{\infty} \frac{n^n z^n}{n!}$$
converge?

By using the ratio test (or alternatively: calculating the radius of convergence), I can see that the series converges for $|z|<\frac{1}{e}$ and diverges for $|z|>\frac{1}{e}$. However, I am not able to find out what happens if $|z|=\frac{1}{e}$. Does anyone have an idea?

Claim: if $|z|=\frac1e$ but $z\ne\frac1e$, then the series converges.

Write $z=e^{i\theta-1}$. We can show the claim using Dirichlet's test, with $a_n = n^n/(e^nn!)$ and $b_n = e^{in\theta}$, if we can prove that $\{n^n/(e^nn!)\}$ is a decreasing sequence whose limit equals $0$. (The geometric series formula for $\sum_{n=1}^N e^{in\theta}$ shows that the boundedness condition is satisfied.) Equivalently, one needs to show that
$$
\bigg\{\log \frac{e^nn!}{n^n} \bigg\} = \bigg\{ \sum_{k=1}^n \log k - (n \log n - n) \bigg\}
$$
is increasing and tends to $\infty$. This can be done by looking at the difference sequence
\begin{multline*}
\bigg( \sum_{k=1}^n \log k - (n \log n - n) \bigg) - \bigg( \sum_{k=1}^{n-1} \log k - ((n-1) \log (n-1) - (n-1)) \bigg) \\
= \log n - \int_{n-1}^n \log x\,dx.
\end{multline*}
That difference is the area of the almost triangle-shaped region above $y=\log x$ and $y=\log n$ between $n-1\le x\le n$; it is certainly greater than the area of the rectangle with upper-left corner $(n-1,\log n)$ and lower-right corner $(n-\frac12,\log(n-\frac12))$, which one can show (using the mean value theorem) is greater than a constant times $\frac1n$. This lower bound for the difference sequence forces the original sequence to increase to infinity by comparison with the harmonic series.

The original series diverges when $z=\frac1e$, but that's difficult to motivate without knowing Stirling's approximation. Nevertheless you can show, using a similar method, that
$$
\frac{n^n}{e^nn!}\cdot n
$$
increases to infinity, which is enough to show that $\sum \frac{n^n}{e^nn!}$ diverges.

Apply ratio test:
$$\left| {\frac{z \cdot n}{{n+1}}}\right| $$
Converges when $|z|<1$, diverges when $|z|\le1$


2 How do I solve this equation involving a logarithm? 2 How do I solve this equation involving a logarithm? 3

Use the properties of the log function:

$\log_a {\frac{b}{c}} = \log_a b -\log_a c$

and $\log_a x = \log_a b \times \log_b x$.


Find the left hand limit of $f(x)$ at $x = 4$ if $f(x) =\frac{|x-4|}{x-4}$ if $x \neq 4$ and $f(x)=0$ if $x=4$ I just began limits for an introductory knowledge of calculus for Physics and I encountered this question in my Mathematics textbook.
So, what I know about limits is that if
$$\lim_{x \rightarrow a^-}f(x) = m$$
It means that as $x \rightarrow a$ from the left side, $f(x) \rightarrow m$ and if it was $x \rightarrow a^+$ instead of $x \rightarrow a^-$, then that would mean that as $x \rightarrow a$ fropm the right side, $f(x) \rightarrow m$.

I also know that if
$$\lim_{x \rightarrow a^-}f(x) = \lim_{x \rightarrow a^+}f(x)$$
It means that
$$\lim_{x \rightarrow a}f(x) \text{ exists and else it does not exist}$$
One last thing that I know about limits is that the statement $x \rightarrow a^-$ can also be written as $x = a-h$, where $h > 0$ and $h \rightarrow 0$ and in the case of $x \rightarrow a^+$, $x = a + h$, $h>0$ and $h \rightarrow 0$
Now, on to the question, it is given that
$$f(x) = \begin{cases} \dfrac{|x-4|}{x-4}&,x \neq 4\\ 0&,x=4 \end{cases}$$
So, as $x \rightarrow 4$d, shouldn't the value of $f(x)$ approach 0, since $f(4)$ is defined and equal to $0$?
Here's how it's done in my book :
$$\lim_{x \rightarrow 4^-}f(x) = \lim_{h \rightarrow 0} f(4-h)$$
$$ = \lim_{h \rightarrow 0} \dfrac{|4-h-4|}{4-h-4} = \lim_{h \rightarrow 0} \dfrac{|-h|}{h}$$
Since $h > 0$, so $-h < 0$ and $|-h| = -(-h) = h$ which means that
$$\lim_{h \rightarrow 0} \dfrac{|-h|}{h} = \lim_{h \rightarrow 0} \dfrac{h}{-h} = \lim_{h \rightarrow 0} -1 = -1$$
I think that I have some misconception either about limits or about functions

Please let me know
Thanks!

When you compute a limit at a certain point, the value of the function at that point does not matter (otherwise, there would be no point computing the limit).
Now,
$$\lim_{x\to4^-}\frac{|x-4|}{x-4}$$ only involves values of $x$ that are less than $4$. But
$$x<4\implies|x-4|=-(x-4)$$ so that
$$x<4\implies\frac{|x-4|}{x-4}=-1.$$
The limit of a constant is that constant.
By a similar reasoning, $\lim_{x\to4^+}\dfrac{|x-4|}{x-4}=1$, and the ordinary limit does not exist.

Intuitively, the limit is the value that you would "naturally" assign to $f$ at $a$ if $f(a)$ was unknown, and it need not be equal to the actual $f(a)$. (In case of a discrepancy, we say that there is a discontinuity.) When we are not able to assign a "natural" value, the limit is said not to exist.

You can start by defining explicitly the function f :
$f(x)  = -1 \quad \forall x > 4 \quad and f(x)  = 1 \quad \forall x < 4$, by the way we can immediatly confirm there is no limit of f on left hand of x = 4.


prove that $n$ is prime or $n=4$ when $n|4((n−1)!+1)$ Let $n \gt 1$ be an integer such that :
$n|4((n-1)!+1)$
Prove that
$n=4$ or $n$ is prime

Since $4((n-1)!+1)$ has an explicit factor of $4$, plainly $2,4$ divide it, noting that $2$ is a prime. If $n\ne 2,4$ then $n|4((n-1)!+1)$ is true when $n|((n-1)!+1)$. Another way of saying that is $((n-1)!+1) \equiv 0 \bmod n$
If $n$ is prime, Wilson's theorem tells us that $(n-1)! \equiv -1 \bmod n$, which would imply the desired result $((n-1)!+1) \equiv 0 \bmod n$
It remains to be shown that if $n$ is not prime, the proposition fails. If $n>4$ is not prime, it is factorable: $n=ab$ where $2\le a,b \le \frac{n}{2}<n-1$. Since both $a,b$ are factors that appear in the factorial product $(n-1)!$, it is the case that $n\mid(n-1)!$ hence $n\not \mid((n-1)!+1)$ and we are done.


$ \inf_{ \text {open sets covering E} } ( \sum \text{length}(I_n) ) \leq \inf_{ \text {closed sets covering E} } ( \sum \text{length}(J_n) ) $ Let $E$ be a subset of $\mathbb R $, $J_n$ be a closed cover of $E$, where $\forall n, J_n = [a_n,b_n]$ and we build the open cover $I_n$ like this : $ \forall n, I_n = ]a_n - \frac {\epsilon }{2^{n+1}} , b_n + \frac {\epsilon }{2^{n+1}} [ $.

  I'm struggling to prove that the outer measure on open sets is
  inferior to the one on closed sets.  So here I'd like to show that :

  $$ \inf_{ \text {open sets covering E} } ( \sum \text{length}(I_n) ) \leq
 \inf_{ \text {closed sets covering E} } ( \sum \text{length}(J_n) ) $$

Why am I struggling?

For now I've proved this:

$$ \sum \text{length}(I_n) =  \sum \text{length}(J_n) + \epsilon$$
here I'm stuck.

Please help me to complete this proof. Moreover, I ask for not giving me a proof using "it works for all closed sets so we take the infinimum on the RHS". I'm asking for something really formal.

Let $\{J_n\}$ be a covering of $E$ by closed intervals. Let $\epsilon >0$. Expand $J_n$ to an open interval $I_n$ such that length of $I_n$ equals length of $J_n +\frac {\epsilon }{2^{n}}$. Then $(I_n)$ is an open covering of $E$ so LHS $\leq \sum \text {length of } I_n \leq \sum \text {length of } J_n +\epsilon$. Taking infimum over all $(J_n)$ we get LHS $\leq RHS +\epsilon$. Since $\epsilon$ is arbitarry we get LHS $\leq $ RHS.

I wrote this answer. One can also find a similar answer to this question in the Measure Theory and Integration by G. de Barra.

Let $\epsilon > 0$.
Let $(J_n)$ covering of $E$ with closed sets, such that :

$$ \sum length(Jn) - \epsilon \leq \inf_{ \text {closed sets covering E} } ( \sum \text{length}( \tilde{I}_n) )$$
which exists by definition of the infimum.
We then fix $(I_n)$ s.t. :
$$ J_n \subset I_n $$
$$length (I_n) = (1+ \epsilon) length( J_n) $$

so $$ \sum length( J_n) = \frac 1 {1+ \epsilon} \sum length(I_n) \leq \inf_{ \text {closed sets covering E} } ( \sum \text{length}( \tilde{I}_n) ) + \epsilon =: M$$

however, $$E \subset \bigcup I_n  \ \ \text{      open sets } \implies
\inf_{ \text {open sets covering E} } ( \sum \text{length}( \tilde{I}_n) )
\leq \sum length(I_n) \leq  (1 + \epsilon ) M $$


parabola passes through $4$ points Finding maximum number of parabola which passes through the point $A(1,2)\;,\; B(2,1)\;\;,(3,4)\;\;,(4,3)$

what i try

from these $4$ point one can imagine that parabola symmetrical about $y=x$ line

so axis of parabola along the line $y=x$ and directrix along the line $x+y+c=0$, where $c$ is any constant

did not understand how do i solve further

help me to solve it please

You rightly noticed that the four points are symmetrical around $y=x$.
So change the variables accordingly
$$
\left\{ \matrix{
  \xi  = y - x \hfill \cr
  \eta  = y + x \hfill \cr}  \right.\quad  \Leftrightarrow \quad \left\{ \matrix{
  x = {{\eta  - \xi } \over 2} \hfill \cr
  y = {{\eta  + \xi } \over 2} \hfill \cr}  \right.
$$

The four points become
$$
\left( {1,3} \right),\left( { - 1,3} \right),\left( {1,7} \right),\left( { - 1,7} \right)
$$

The parabola's axis is the $\eta$ axis, so its formula is
$$
\eta  - a = k\xi ^{\,2}
$$

In any case it is clear that, since the points form a rectangle, there cannot be a single parabola passing through all of them.

Hint:

A parabola is of the form $y = ax^2 + bx + c$

We are told that it passes through the points $(1,2),(2,1)$ and $(3,4)$.  (I suggest ignoring $(4,3)$ for now and revisit it later)

That means that plugging in those $x$ and $y$ values, the fact that it passes through $(\color{blue}{1},\color{red}{2})$ implies that $\color{red}{2} = a\cdot \color{blue}{1}^2 + b\cdot \color{blue}{1} + c$.  Similarly we can find the equations that are implied by the parabola passing through the other points.

We have then this system of equations:

$$\begin{cases}2 = a + b + c\\1 = 4a + 2b + c\\ 4 = 9a + 3b + c\\\end{cases}$$

Now... this is a system of three linear equations and three unknowns which can by solved by standard means.

(Now, consider including $(4,3)$ as well.  Does this affect the answer?)


Finding set of limit points as $ x \rightarrow 0 $ If the function $ f $ is defined as
$$
f(x)
=
\begin{cases}
|x|^{\phi - 1} \sin(1/x), & x \neq 0 \\
0, & x = 0
\end{cases}
$$
where $ \phi > 0 $, how does one find the set of limit points of $ f(x) $ as $ x \rightarrow 0 $, for each $ \phi > 0 $?

So, $\lim_{x \to 0+} x^{\phi - 1}\sin(\frac{1}{x}) = \lim_{x \to 0+} x^{\phi-2} x\sin(\frac{1}{x})$. As $x\sin(\frac{1}{x}) \to 1$ as $x \to 0$ that part has no importance, limit exists where limit $x^{\phi - 2}$ exists and has same value as that limit, and the latter limit is zero when $\phi > 2$, 1 when $\phi = 2$ and doesn't exist when $\phi < 2$


Simultaneous equations $x+2y-z=2 \\
x-y+z =5 \\
3x+3y-z=\mu$

The question is for what value of the parameter $\mu$ has the system (a) no solutions, (b) one solution, and (c) infinitely many solutions.

Row reduced echelon form is

$A =\begin{pmatrix} 1& 0& 1/3& 0\\ 0 &1 &-2/3 &0\\  0& 0& 0& 1\end{pmatrix}$

But I don't know how to use that.

EDIT: I want to know how to read the condition on $\mu$ from the matrix.

So we start here for no solution we just need to have just a parallel line so we just need to see for constant for 1 solution the determinant  of equations should be 0 . Which proves their concurrency and for infinitely many solutions $x_1/x_2=y_1/y_2=z_1/z_2=c_1/c_2$. Thus you get the parameters for the unknown.

if you add twice the first equation and the second equation, you have $3x + 3y -z = 9$ and comparing with the third equation $3x + 3y - z = \mu$ have infinitely many solutions if $\mu = 9$ and has no solutions otherwise.


General factorization of $x^n-y^n$ (n is real) I am looking for a general expansion of $x^{n}-y^{n}$ with $x,y>0$ and $n$ being real. I came across the following formula (Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$)

$$x^n-y^n = (x-y)(x^{(n-1)}+x^{(n-2)}y+...+y^n)$$

Does it hold true for $n \in \mathbb{R}$?

If you write $x = y + t$, then you can use the binomial series on $x^n = (y+t)^n$, so

$$\eqalign{ x^n - y^n &= \sum_{k=1}^\infty {n \choose k} y^{n-k} t^k\cr
&=\sum_{k=1}^\infty {n \choose k} y^{n-k} (x-y)^k\cr
&= (x-y) \sum_{k=1}^\infty {n \choose k} y^{n-k} (x-y)^{k-1}}  $$
The series converges for $|x-y| < |y|$.

You could factor out $x - y$, giving you:

$$(x - y)(x^{n-1} + x^{n-2}y + ... xy^{n - 2} + y^{n - 1}) = \\(x - y)\sum_{i=0}^{n-1}x^{i}~y^{n - 1 - i}$$

Edit: The question was changed to include this factorization after this answer was posted.


How can I show that $a^n|b^n \Rightarrow a|b$ How can I show the following
  $$a^n|b^n \Rightarrow a|b$$ with $a,b$ integers.

$$a^n|b^n \Rightarrow b^n=m \cdot a^n \Rightarrow b^n=(m\cdot a^{n-1}) \cdot a\qquad(1)$$
How can I continue? Do I maybe have to suppose the opposite and arrive at contradiction?
$$\text{So } a \nmid b \Rightarrow b=q\cdot a+r$$
Replacing this at the relation $(1)$ could I conlude to something to get a contradiction?
Or is there an other way to prove this??

Hint $ $ Clear if $\,a=0.$ Else $\,(\color{#c00}{b/a})^n \! = k\in\Bbb Z\,\Rightarrow\, b/a\in \Bbb Z,\,$ because the $ $ Rational Root Test $ $ implies that $ $ any $ $ rational $ $ root $ $ of $\ \ \color{#c00}x^n - k\ $ is integral.

Assuming, for contradiction, that $a \nmid b$, there must be some prime $p$ to some power $m$ such that $p^m | a$, but $p^m \nmid b$. Then $p^{nm} | a^n$, but $p^{nm}\nmid b$ by simple counting of prime factors.


Symbol for close to but below Let’s say x=0.97 and I want to symbolically emphasize that X is below but close to 1. Is there a symbol that I can use which denotes that situation? Thanks.

$\lesssim$ (MathJax/LaTeX: \lesssim) is sometimes seen meaning that.

In the case of limits I have seen $lim_{x\to1^-}$, meaning that $x$ is approaching $1$ from below.


Question on cofactors If the row sums of a symmetric matrix of size 4 by 4 are all $0$, then why are all the cofactors of the matrix equal?

Thanks in advance for any helpful answers.

There is no reason to consider a $4\times 4$ matrix in particular. When $n=1$, there is only one cofactor so nothing to prove. The result is easy to figure out for $2\times 2$ matrices, and that's maybe a good starting point.

  Then one realizes that more generally, as observed by user1551, it suffices that $A$ be an $n\times n$ matrix over an arbitrary field such that the row sums and the column sums be null (the latter being equivalent to the former for symmetric matrices).

In short: we use the adjugate formula $A\cdot\mbox{adj}(A)=(\det A)I_n$ to prove that the rows and the columns of $C=\mbox{adj}(A)^T$, the matrix of cofactors, are all in the span of  $(1,\ldots,1)$, which is the nullspace of $A$ and $A^T$ in the only nontrivial case.

Proof: the row condition means that the sum of the columns of $A$ is null. Whence the columns the columns are linearly dependent and $\mbox{rank} A\leq n-1$. In particular, $\det A=0$.
Let $C$ be the matrix of the cofactors of $A$. Then the transpose $C^T$ is the adjugate of $A$ and, as is well-know $AC^T=(\det A)I_n$. Therefore
$$
AC^T=0\quad\iff \quad \sum_{k=1}^na_{ik}c_{jk}=0\quad \forall 1\leq i,j\leq n.
$$

If $\mbox{rank} A\leq n-2$, then all the cofactors are zero and the result follows. So we assume that $\mbox{rank}\, A=n-1$. That is, the nullspace of $A$ has dimension $1$. But we already know that $v=(1,\ldots,1)$ is in $\ker A$ by assumption. So $\ker A$ is the one-dimensional span of $v$. Note that $\ker A^T$ also has dimension $1$ and is spanned by $v$ as-well by assumption.

Now looking at the above, we see that it means that every row $R_j$ of $C$ is in the nullspace of $A$, whence $R_j=r_jv=(r_j,\ldots,r_j)$.

Since the adjugate matrix of $A^T$ is the transpose of the adjugate matrix of $A$, we get $A^TC=0$ (or simply by taking the tranpose of $C^TA=0$). From that, we deduce that the columns $C_j$ of $C$ are of the form $C_j=c_jv=(c_j,\ldots,c_j)$.

Starting from $c=c_{1j}$ for every $j$ (first row), we see that $c_{ij}=c$ ($j$th column) for every $i$. That is, all the coefficients of the cofactor matrix are equal as desired. QED.

Here is a simpler proof. In general, for any $n\times n$ singular matrix $A$, $\operatorname{adj}(A)$ is either zero (when the rank of $A$ is $<n-1$) or in the form of $uv^T$ (when the rank of $A$ is $n-1$), where $u,v$ are basis vectors in respectively the right and left null spaces of $A$.

Now, in your case, $A$ is a singular matrix because its row sums and column sums are zero. If $\operatorname{rank}(A)<n-1$, its adjugate matrix is zero and we are done. If $\operatorname{rank}(A)=n-1$, its left or right null spaces are one-dimensional, and hence both null spaces are spanned by the all-one vector $e$. It follows that $\operatorname{adj}(A)$ is a multiple of $ee^T$ and all cofactors of $A$ are identical.


How can I prove that Let we have the following ultrametric space $(z,|.|_2)$ such that if $x=r.2^n$ then $|x|_2=2^{-n}$ how can I prove that the topology produced by this metric isn't discrete topology ?

Let's try being more precise. You are defining an ultrametric $d$ on the set $\mathbb{Z}$ of integers by setting
$$
|x|_2=\begin{cases}
0 & \text{if $x=0$}\\[3px]
2^{-n} & \text{if $x\ne0$ and $x=2^nr$, with $r$ odd}
\end{cases}
$$
and $d(x,y)=|x-y|_2$.

Now consider the sequence $(2^n)_{n\in\mathbb{N}}$. Is it convergent?

Show that there is some $y\in\Bbb Z$ such that $\{y\}$ is not open. For a hint, hover over the spoiler box below:

  Note that $|x-0|_2=|x|_2$ for any $x\in\Bbb Z.$


Taking the Derivative of Both Sides of an Equation If we have an equation like

y = x^2

This implies that

y’ = 2x

If we have an equation like

x = 4x^2

and we take the derivative of both sides we get

1 = 8x

With the solution x = 1/8, which is not the solution to the original equation. This is instead the value where the slopes of both sides of the equation are equal.

So when (and why) is an equation resulting from the derivation of both sides of some original equation implied by the original equation.

Your problem comes from thinking that you take derivatives of equations. But you don't. You take derivatives of functions.

For example, you can think of the equation
$$
x^2 -1 = (x-1)(x+1)
$$
as telling you two different ways to write the same function of $x$. When you differentiate the two expressions using the rules for derivatives you will get the same answer, in two different forms.

The equation in your question,
$$
x = 4x^2 ,
$$
does not say that two functions are equal. It asks for the value of $x$ at which those two functions happen to have the same value. That's just a number. It makes no sense to take the derivative.

The solutions of $f(x)=g(x)$ are the points of intersection of $f$ and $g$; the solutions of $f'(x)=g'(x)$ are the points where $f$ and $g$ have parallel tangents.

In our case  $f(x)=x$ and $g(x)=4x^2$.  The meet for $x\in\{0,1/4\}$.  Now $f'(x)=g'(x)\iff x=1/8$, for which $x$ the functions are parallel (but not meet).


Is the group of invertible elements in $\mathbb{Z}/8\mathbb{Z}$ isomorphic to $\mathbb{Z}/4\mathbb{Z}$ or to the symmetry group of the rectangle? Problem: Is the group $g(8) = \{[1], [3], [5], [7]\}$ isomorphic to $\mathbb{Z}_4$ or to the symmetry group of the rectangle?

Attempt: I know that $g(8)$ is isomorphic to $\mathbb{Z}_4$ because I compare the multiplication tables, and they look the same.

I know that $g(8)$ is not cyclic, and I know that the symmetry group of the rectangle is not cyclic since every non-identity element has order 2. Does that mean they are not isomorphic?

Can anyone help me verify the last part.
Thank you.

There are only two groups of order four, up to isomorphism: $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$. You can distinguish them because $\mathbb Z_4$ has two elements of order four, one element of order two and the identity. On the other hand, $\mathbb Z_2 \times \mathbb Z_2$ has three elements of order two and the identity.

We need to look at groups of $4$ elements: the set of invertible  elements of $\mathbb Z_8$ is a multiplicative group of order four, and there are only two such groups of order four, up to isomorphism: $\mathbb Z_4,$ or the Klein-4 group.
So your group must be isomorphic to $\mathbb Z_4$ or else to the Klein-4 group.
Since your group is not cyclic, it cannot be isomorphic to $\mathbb Z_4$, which is cyclic. (Any group that is isomorphic to a cyclic group is necessarily also cyclic).
That leaves only the Klein-4 group, which is NOT cyclic: I'd suggest you compare the group tables of your group and the Klein-4 group to convince yourself that they are isomorphic. The Klein-4 group is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$.


Why is $x=2 \implies (x-2)(x-3)=0$ false? Let $P(x)$ be the equation $x=2$
and $Q(x)$ be the equation  $(x-2)(x-3)=0$

By definition of implication I see that $P(x)$ implies $Q(x)$...

As I see it, any premise that is false can give any consequence.
With $x=2$ both sides of the arrow are true.. so implication is true.

It is clear that $Q(x)$ implies $P(x)$ and it is also clear that $P(x)$ and $Q(x)$ are not equivalent. Since $Q(x)$ implies $P(x)$ and $P(x)$ and $Q(x)$ are not equivalent it follows that $P(x)$ cannot imply $P(x)$.

Where am I wrong?

EDIT: I knew that $P(x)\Leftrightarrow Q(x)$ was not valid. And I thought wrongly that $P(x)\Leftarrow Q(x)$ is valid which should, if it was valid, imply that $P(x)\nRightarrow Q(x)$. But testing $P(x)\Rightarrow Q(x)$ by the truth table showed $P(x)\Rightarrow Q(x)$ is valid which confused me. I was simply wrongly thinking $P(x)\Leftarrow Q(x)$ which is the correct answer. I have also posted a soultion below, which typo is corrected.

$$(x-2)(x-3)=0 \iff x = 2 \ \, \text{or} \, \ x = 3$$

If $x=2$, then the hypothesis of the RHS is satisfied and hence we have the LHS.

I don't see why that is supposed to be false.

For the case $P(x)\Rightarrow Q(x)$:

If we choose $x=2$ then we got true on both sides of the arrow. If we choose $x\neq 2$ then P(x) is false and Q(x) can be false (if $x\neq 3$) or true (if $x=3$).
The implication $P(x)\Rightarrow Q(x)$ is therefore valid.

Following explanation will explain why I was wrong. This is for the case $P(x)\Leftarrow Q(x)$:

If we choose $x=2$ then we got true on both sides of the arrow.
If we choose $x=3$ then we got Q(x) is true and P(x) is false. Since this scenario is possible, it means the implication $P(x)\Leftarrow Q(x)$ is not valid, i.e. we have $P(x)\nLeftarrow Q(x)$.

A bit more intuitive and more self explaining similar example is when we consider $x=3 \Rightarrow x^2=9$ which is fully valid. What if we change the direction of the arrow in the implication? Changing direction of the arrow, gives us $x=3 \nLeftarrow x^2=9$ since $x=-3$ is true for $x^2=9$ but not for $x=3$.


generalized way of finding minimum value of a function? $f(x)=\frac{x^{2}-1}{x^{2}+1}$ for every real number for $x$, the minimum value of $f$ is what?
How  can I find the minimum value of this function.I only know trial and error method, but it's not a generalized way.

Please tell me a generic way to solve this type of problem

This case is so simple, it can be solved without even using calculus.
Write $f(x)$ as
$$f(x)= 1- \frac{2}{x²+1} $$
The minimum value of this function clearly occurs when the fraction on the right is greatest, which clearly occurs when the value of the denominator is least, which happens when $x=0$ Hence, the minimum value of the function is $f(0)=-1$.

Others have covered how to solve the problem if $f$ is differentiable.
But what if it's not?

Employ a computer and find a local minimum using the hill climbing algorithm. Or use a variant of it like "Random-restart hill climbing" for an approximation of the global minimum.


Proving the Mean Value Theorem Theorem

  The Mean Value Theorem: Let there be a function $f$ which is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $f'(c)=\dfrac{f(b)-f(a)}{b-a}$

I'd like to prove the Mean Value Theorem, assuming Rolle's Theorem has already been proved.

Proof

Let $f$ be the function whose graph is the chord between the endpoints $(a,f(a))$, and $(b,f(b))$.

Then let $g(x)=\dfrac{f(b)-f(a)}{b-a}(x-a)+f(a)$, and let $h(x)=f(x)-g(x)$, so $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Note that $h(a)=h(b)=0$, and so the hypotheses of Rolle's Theorem are met, and thus there exists a $c\in(a,b)\ni h'(c)=0$.

$$0=h'(c)=f'(c)-g'(c)=f'(c)-\dfrac{f(b)-f(a)}{b-a}$$ $$\implies f'(c)=\dfrac{f(b)-f(a)}{b-a}\tag*{$\blacksquare$}$$

Is the above proof valid (and are there any suggestions for improvement?)

Another way to prove the MVT:

Let us define a function $\phi$:[a,b]$\rightarrow IR$ by
$$\begin{matrix}
f(x)&g(x)&h(x) \\
f(a)&g(a)&h(a)\\
f(b)&g(b)&h(b)
\end{matrix}$$
Observe that $\phi (a)$=$\phi (b)$=0. Hence by Rolle's theorem $\exists$ c $\in$ (a,b) such that $\phi^{'} (c)$=0.
$$\begin{matrix}
f^{'}(x)&g^{'}(x)&h^{'}(x) \\
f(a)&g(a)&h(a)\\
f(b)&g(b)&h(b)
\end{matrix}$$
is equal to 0.Take h(x) to be a constant function,g(x)=x and expand the determinant to get the desired result.


The hat-matching problem, probability of at least $k$ matches the setup: $N$ people arrive at a party all of whom are wearing hats. We collect all the hats and then redistribute them. What is the probability that at least $k$ of the party members receive their own hats back?
There are a total of $\binom{N}{k}$ number of ways to make a subset of $k$ people out of these $N$ people. let's label each of these $\binom{N}{k}$ subsets with a unique number between $1$ to $\binom{N}{k}$, arbitrarily. let $\space T_i \space$ be the event that all people in the subset labeled $i$ have received their own hats back.
Then, the probability that at least $k$ people receive their own hats back is $$P(\bigcup\limits_{i=1}^{\binom{N}{k}} T_{i}) = \sum_{i=1}^{\binom{N}{k}} P(T_{i}) - \sum_{i_1 < i_2}P(T_{i_1}T_{i_2}) + \dots + (-1)^{n+1}\sum_{i_1 < i_2<\dots<i_n}P(T_{i_1}T_{i_2}\dots T_{i_n}) + \dots + (-1)^{\binom{N}{k}+1}P(T_{1}T_{2}\dots T_{\binom{N}{k}})$$
question1: is this correct? is there any way to take this calculation forward?
question2: is there a simpler way to find the probability that at least $k$ people receive their own hats back
(other than $\sum_{i=k}^N\text{probability that at exactly k people receive their own hats back}$, where the probability that exactly k people receive their own hats back  is $\frac{\binom{N}{k} D_{N - k}}{N!}$, where $D_n = n! \cdot \sum_{i=0}^{n}\frac{(-1)^i}{i!}$)

I think you can get a pretty straightforward answer by modifying @Brian M. Scott 's answer in your other question(the one linked to this post) for exactly $k$ people.

Suppose that we’ve put aside a set $S$ of $k$ men who are to get their own hats back; there are $(N-k)$! permutations that return their own hats to all $k$ of the men in $S$ (and possibly others as well). Let $T$ be the set consisting of the remaining $N-k$ men, and for each $i\in T$ let $A_i$ be the set of permutations also giving $i$ his own hat back. To get the number of permutations that return their own hats only to the $k$ men in $S$, we need to subtract $∣\cup_{i\in T} A_i∣$, and that’s exactly what we’ve done.

Since we don't care for only but for at least there's no need to subtract. So, there $(N-k)!$ permutations for at least $k$ in a given set of size $k$, $N\choose k$ ways to pick such a set and we need to divide by the total number of permutations, that is $N!$ to find the propability. Putting all together,
$$ p = \frac{(N-k)! {{N}\choose k}}{N!}  = \frac{1}{k!}$$
Edit: Apparently, that is wrong since for $k=1$ the probability can't always be equal to 1. I think the correct answer is  $ p = \frac{(N-k)!} {N!}$ because we don't care about defining the set since we don't subtract but this is not a solid proof.


Exercise 1.1.18 in Hatcher's Algebraic Topology Background

I am currently trying to solve exercise 1.1.18 in Hatcher's Algebraic Topology. The part of the exercise I am interested in is the following:

  Using the technique in the proof of Proposition 1.14, show that if a space $X$ is obtained from a path-connected subspace $A$ by attaching an n-cell $e^n$ with $n ≥ 2$, then the inclusion $A \hookrightarrow X$ induces a surjection on $\pi_1$ .

I know that $i:A \hookrightarrow X$ induces a homomorphism $i_*:\pi_1(A)\rightarrow \pi_1(X)$, so I only need to show this is a surjection. I think I understand the idea of the proof, which is to show that every loop $f\in \pi_1(X)$ is homotopic to a loop which is contained entirely in $A$. Hatcher's suggestion is to follow the proof $\pi_1(S^n)=0$ for $n\geq 2$, meaning that we should be able to push the sections of $f$ which are in the attached $n$-cell $e^n$ out. This is causing me a bit of trouble.

Attempt

Since $X$ is defined to be the result of attaching an $n$-cell to $A$ via some attaching map $\varphi:\partial D^n\rightarrow X$, it has the form $X=A \amalg e^n/\sim$, where $x\sim \varphi(x)$ for all $x \in \partial D^n$. Note first that since $A$ and $e^n$ are path connected, the adjunction space $X=A\cup_\varphi e^n$ is path connected. As such, our choice of base point does not affect the structure of $\pi_1(X)$, so let $x_0 \in A$ be the base point of $\pi_1(X)$ we are working over. Let $f \in \pi_1(X,x_0)$. Let $E=\text{Int}(e^n)$ and consider $f^{-1}(E)$. This is an open subset of $(0,1)$, so it is the union of a possibly infinite collection of subsets of $(0,1)$ of the form $(a_i,b_i)$. Let $f_i$ denote the restriction of $f$ to $(a_i,b_i)$. Note that $f_i$ lies in $e^n$ and, in particular, $f(a_i)$ and $f(b_i)$ lie on the boundary of $e^n$, so they are elements of $A$. For $n\geq 2$ we can homotopy $f_i$ to the path $g_i$ from $f(a_i)$ to $f(b_i)$ that goes along the boundary of $e^n$, which is homeomorphic to $S^{n-1}$, so it is path connected for $n\geq 2$. Since $e^n$ is homeomorphic to $D^n$, where $n\geq 2$, it is simply connected so $f_i$ and $g_i$ are homotopic. Repeating this process for all $f_i$, we obtain a loop $g$ homotopic to $f$ such that $g(I)\subseteq A$.

What really bothers me about this is how I could homotopy form a homotopy from $f$ to $g$ consisting of possibly infinitely many individual homotopies from $f_i$ to $g_i$. I believe I need there to only be finitely many $f_i's$, but I don't see how to show it.

Note: This is not homework.

Proof with Yoyo's suggestion

Since $X$ is defined to be the result of attaching an $n$-cell to $A$ via some attaching map $\varphi:\partial D^n\rightarrow X$, it has the form $X=A \amalg e^n/\sim$, where $x\sim \varphi(x)$ for all $x \in \partial D^n$. Note first that since $A$ and $e^n$ are path connected, the adjunction space $X=A\cup_\varphi e^n$ is also path connected. As such, our choice of base point does not affect the structure of $\pi_1(X)$, so let $x_0 \in A$ be the base point of $\pi_1(X)$ we are working over. Let $f \in \pi_1(X,x_0)$. Let $E=\text{Int}(e^n)$ and consider $f^{-1}(E)$. This is an open subset of $(0,1)$, so it is the union of a possibly infinite collection of subsets of $(0,1)$ of the form $(a_i,b_i)$. Let $x \in E$ and let $U$ be an open ball around $x$ in $e^n$. Like before, $f^{-1}(U)$ is an open subspace of $(0,1)$, so it is of the form $(c_i,d_i)$ for possibly an infinite number of $i$'s.  The preimage $f^{-1}(x)$ is a closed subspace of $I$, hence it is compact. The intervals $(c_i,d_i)$ form an open cover of $f^{-1}(x)$, hence a finite collection of these intervals cover $f^{-1}(x)$. Let $f_i$ denote the restriction of $f$ to $(c_i,d_i)$. Note that $f_i$ lies in the closure of $U$ and, in particular, $f(a_i)$ and $f(b_i)$ lie on the boundary of $U$. For $n\geq 2$ we can homotopy $f_i$ to the path $g_i$ from $f(a_i)$ to $f(b_i)$ in the closure of $U$ which is disjoint from $x$, since the closure of $U$ is homeomorphic to $D^{n}$, a closed and convex subset of $\mathbb{R}^n$. This gives a homotopy from $f$ to the path $g$, which is disjoint from $x$. Since $g$ is not surjective, it is homotopic to a path $h$ contained entirely in $A$.

Since $\pi_1(S^n) =0 $ for $n>1$ so all elements of $\pi_1(X)$  can be realized as an element os $\pi_1(A)$.


Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?

Theorem.  $\forall x\geq8, x$ can be represented by $5a + 3b$ where  $a,b \in \mathbb{N}$.

Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\
x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\
x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$

Inductive step:

$n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\
b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\
c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\
\\
S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$

Basis stays true, because $8,9,10 \in S$

Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.

If $x \in a_n$ then $x+1 \in b_x$,

If $x \in b_x$ then $x+1 \in c_x$,

If $x \in c_x$ then $x+1 \in a_x$.

I can't prove that but it's obvious. What do you think about this?

Proof by induction.
For the base case $n=8$ we have $8=5+3$.
Suppose that the statement holds for $k$ where $k\gt 8$. We show that it holds for $k+1$.

There are two cases.

1) $k$ has a $5$ as a summand in its representation.

2) $k$ has no $5$ as a summand in its representation.

For case 1, we delete "that $5$" in the sum representation  of $k$ and replace it by two "$3$"s ! This proves the statement for $k+1$.

For case 2, since $k\gt 8$, then $k$ has at least three "$3$"s in its sum representation. We remove these three $3$'s and replace them by two fives! We obtain a sum representation for $k+1$. This completes the proof.

Lets call $T$ every multiple of three: $T$ satisfies the condition.  Numbers are $T, T-1,$ or $T-2$.

for $n  > 3\cdot 2+1$:

$$n=T \implies n=3k $$
$$n=T-1 \implies  n= 3\cdot 2 - 1  + 3(k-2) =  5 + 3(k-2)$$
$$n=T-2 \implies  n= 3\cdot 4 - 2  + 3(k-4) = 10 + 3(k-2)$$


Why do we use "if" in the definitions instead of "if and only if"? I often write my notes as logical statements and constantly wonder why people use only the "if" direction in the definitions instead of the "if and only if". Consider:

"A homomorphism $\phi$ is said to be an isomorphism if it is an injection  and a surjection." would translate to:

$homomorphism(\phi) \wedge injection(\phi) \wedge surjection(\phi) \implies isomorphism(\phi)$

By adding the following statement consistent with the previous one:

$homomorphism(\phi) \wedge \neg injection(\phi) \wedge surjection(\phi) \implies isomorphism(\phi)$

we could deduce

$homomorphism(\phi) \wedge surjection(\phi) \implies isomorphism(\phi)$

which is obviously not what we intended to express in the definition, however the intention does not imply the deed.

From the strict logical perspective do you agree that all definitions of this form are "flawed" owing to the lack of the attention to details? Or is there some reason which I have not managed to observe that explains why "if" is used in lieu of "if and only if"?
Would you accept if I write in papers definitions in the iff version? I.e.
"A homomorphism $\phi$ is an isomorphism if and only if it is an injection and a surjection."

That is a good question. It used to bother me too until I saw it in some undergraduate textbook say "following the standard convention of writing if instead of if and only if in definitions ... ", do not remember the exact wording but that is basically what the book said. Since that time this never bothered me, but whenever I teach classes I always make my definitions as if and only if statements.

"If an only if" means "necessary and sufficient", which is a double implication.

"If" alone means"only if" or "sufficient", which is a single implication.


Functions involving real values If $f$ is a real valued differentiable function satisfying

$|f(x)-f(y)|<(x-y)^2$ for all real $x$ and $y$

and $f(0)=0$

then $f(1)$ equals:

Hint: First you don't need the assumption differentiability. You can prove that $f' =0$ (use definition of differentiability). This shows $f$ is constant, and then $f(1) = f(0)=0$

We can simpify it as f(x)-f(y)/x-y= x-y
If we take that x approaches to y it implies that derivative of f is zero
Since f'(x)= f(x)-f(y)/x-y
Now, if f'(x) is zero then we can say f is a constant function since derivative of a constant function is zero.
It implies f(0)=f(1)=0


Second-Price Sealed-Bid Auction Consider 2 individuals who are interested in one indivisible object. Each player $i$ has a valuation $v_i > 0$ for the object. Assume $v_1 \geq v_2$. In this scenario, each player submits a bid $b_i \geq 0$. The individual with the highest bid wins the auction and obtains the object at a price equal to the second highest bid, in case of a tie, the object goes to player 1. The payoff functions for each player are given by

$$u_1(b_1, b_2) =  \left\{
  \begin{array}{l l}
    v_1 - b_2 & \quad \text{if $b_1 \geq b_2$}\\
    0 & \quad \text{if $b_1 < b_2$}
  \end{array} \right. $$ and
$$u_2(b_1, b_2) =  \left\{
  \begin{array}{l l}
    v_2 - b_1 & \quad \text{if $b_2 > b_1$}\\
    0 & \quad \text{if $b_2 \leq b_1$}
  \end{array} \right. $$

Am I correct if I say that in this case, the best reply functions are given by:

$$\beta_1(b_2) =  \left\{
  \begin{array}{l l}
    \{b_1 | b_1 \geq b_2\} & \quad \text{if $b_2 < v_1$}\\
   \{b_1 | b_1 < b_2\}  & \quad \text{if $b_2 > v_1$} \\
[0, \infty]  & \quad \text{if $b_2 = v_1$}
  \end{array} \right. $$

and

$$\beta_2(b_1) =  \left\{
  \begin{array}{l l}
    \{b_2 | b_2 > b_1\} & \quad \text{if $b_1 < v_2$}\\
   \{b_2 | b_2 \leq b_1\}  & \quad \text{if $b_1 > v_2$} \\
[0, \infty]  & \quad \text{if $b_1 = v_2$}
  \end{array} \right. $$

These look right to me.  Why did you doubt them?

The second price auction has a dominant strategy equilibrium in which each player bids their true valuation. Given the other player bids $b_{-i}$, consider the 3 possible cases for player $i$:

$v_i > b_{-i}$. Player $i$ should bid any $b_i \in (b_{-i}, \infty)$ and get payoff $v_i - b_{-i}$. Otherwise he gets $0$.
$v_i = b_{-i}$. Player $i$ is indifferent between any bids.
$v_i < b_{-i}$. Player $i$ should bid any $b_i \in (-\infty, b_{-i})$ and get payoff $0$. Otherwise he gets negative payoff.


How prove this $S_{n}\neq 0$ let $n$ is postive integer numbers,and let
$$S_{n}=\sin{1}-\sin{4}+\cdots+(-1)^{n-1}\sin{(3n-2)}$$
show that
$$S_{n}\neq 0, \forall n\in N^{+}$$

My try: maybe this problem use
$$2\sin{x}\sin{y}=\cos{(x-y)}-\cos{(x+y)}$$

\begin{align*}2\sin{\dfrac{3}{2}}S_{n}&=2\sin{1}\sin{\dfrac{3}{2}}-2\sin{4}\sin{\dfrac{3}{2}}+\cdots+2(-1)^{n-1}\sin{(3n-2)}\sin{\dfrac{3}{2}}\\
&=\cos{\dfrac{1}{2}}-\cos{\dfrac{5}{2}}-\cos{\dfrac{5}{2}}+\cos{\dfrac{11}{2}}+\cdots
\end{align*}

I think  this is nice problem, Thank you

From Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$, we know $\sin\theta=\Im~ e^{i\theta}$. Then
$$S_n=\sum_{k=1}^n(-1)^{k-1}\sin(3k-2)=\Im~\sum_{k=1}^n(-1)^{k-1}e^{i(3k-2)}$$
The right hand side is actually the sum of geometric series with common ratio $-e^{3i}$. We have
$$S_n=\Im\frac{e^i(1-(-e^{3i})^n)}{1-(-e^{3i})}=\Im\frac{e^i-(-1)^ne^{i(3n+1)}}{1+e^{3i}}\neq0$$
The last step is a direct consequence of rationalize denominator of complex number.

go with product to sum

then calculate sum of $n$th of sigma

at then end you have $6n-7=+-1+4k\pi$
so

$6k-8=4k\pi$ divide by 4 -->$k$ is not an integer

$6k-6=4k\pi$ divide by 4 -->$k$ is not an integer

so $s(n)$ is not zero


Prove that the determinant represents a straight line. Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=1$, then show that the equation

represents a straight line.

Expansion of determinant is tedious. I can see that matrix is symmetric but not able to use this fact. Could someone help me with this?

Notice that the inner matrix can be written as

$$T = \mathrm{a}\mathrm{x}^T + \mathrm{x}\mathrm{a}^T - \mathrm{a}^T\mathrm{x},$$

where $\mathrm{a} = (a, b, c)$ and $\mathrm{x} = (x, y, 1)$ are column vectors. This gives

$$T\mathrm{a} = \mathrm{x}, \qquad T\mathrm{x} = \|\mathrm{x}\|^2 \mathrm{a}, \qquad T\mathrm{v} = -\langle \mathrm{a}, \mathrm{x} \rangle \mathrm{v} \tag{*}$$

whenever $\mathrm{v}$ is perpendicular to both $\mathrm{a}$ and $\mathrm{x}$. So if $\mathrm{a}$ and $\mathrm{x}$ are linearly independent, $\text{(*)}$ completely determines the representation of $T$ w.r.t. the basis $\{\mathrm{a}, \mathrm{x}, \mathrm{a}\times\mathrm{x}\}$, from which we compute as

$$\det T = \|\mathrm{x}\|^2 \langle a, x\rangle.$$

By appealing to the continuity or by a similar idea as before, we can check that this relation continues to hold when they are dependent. Now the conclusion can be easily drawn from this formula.

\begin{eqnarray*}
 Det\left[
\begin{array}{ccc}
ax-by-c&bx+ay&cx+a \\
bx+ay&-ax+by-c&cy+b\\
cx+a&cy+b&-ax-by+c \\
\end{array}
\right]=(x^2+y^2+1)(a^2+b^2+c^2)(ax+by+c).
\end{eqnarray*}
The first two factors are positive so $\color{red}{ax+by+c=0}$.


Number of 5-step self avoiding walks I am trying to compute the number of 5-step self avoiding walks on dimension d. I have computed the number of 1-step SAW which is 2d. For 2-steps it is $2d(2d-1)$. For 3-steps it is $2d(2d-1)^2$, which are easy to compute.

For computing the number of 4-step SAWs I looked to d=2. There are $4*3*3*3 = 108$ possible walks but in 8 cases the walk returns to the origin so there are 100 possible SAWs. I guess the number would be $2d(2d-1)^3-2^{d-1}$

Anyone know how to compute it for 5-step self avoiding walks?

As the number of times a walk of length 4 intersects with itself without going directly backwards is the amount of times it changes it's direction 3 times in the same way (only changing the direction to the left for example) this amount is 2d(2d-2) therefore the number would be 2d(2d-1)^3 - 2d(2d-2).
For walks of length you can use the same logic to find the amount of walks which intersect at the fourth or fifth step.


Pontryagin class of self-dual forms on a 4-manifold Let $X$ be an oriented Riemannian 4-manifold. The bundle of 2-forms $\wedge^2 X$ can be decomposed into the bundle of self-dual and anti-self-dual forms, $\wedge^2_+ X \oplus \wedge^2_- X$, using the Hodge star. I would like to show that $$p_1(\wedge^2_+ X)=p_1(X)+2e(X)$$

This is part of Exercise 10.1.3(a) in Stipsicz and Gompf's '4-Manifolds and Kirby Calculus'. The exercise also asks the reader to show that $w_2(\wedge^2_+ X)=w_2(X)$, which I could do, using the splitting principle (see the answer to Second Stiefel-Whitney class of self-dual two forms of four manifolds). For reference, if $TX=E_1 \oplus E_2 \oplus E_3 \oplus E_4$, then $\wedge^2_+ X=(E_1 \otimes E_2) \oplus (E_1 \otimes E_3) \oplus (E_1 \otimes E_4)$.

I tried to use the same method for this part. Now the splitting principle does not hold for real bundles over $\mathbb{Z}$ coefficients, so I decompose $TX \otimes \mathbb{C}$ instead. Let $TX \otimes \mathbb{C}=E_1 \oplus E_2 \oplus E_3 \oplus E_4$, then $(\wedge^2_+ X) \otimes \mathbb{C}=\wedge^2_+ (T^*X \otimes \mathbb{C})=(E^*_1 \otimes E^*_2) \oplus (E^*_1 \otimes E^*_3) \oplus (E^*_1 \otimes E^*_4)$. Denote $a_i=c_1(E_i)$, we have$$p_1(X)=-c_2(TX \otimes \mathbb{C})=-\sum_{i<j} a_ia_j$$ $$p_1(\wedge^2_+ X)=-(a_1+a_2)(a_1+a_3)-(a_1+a_2)(a_1+a_4)-(a_1+a_2)(a_1+a_4)$$
Taking the difference, I must show that $-2e(X)=3a_1^2+a_1a_2+a_1a_3+a_1a_4$. In fact, since $X$ is orientable, $\wedge^4 TX \otimes \mathbb{C}=E_1 \otimes E_2 \otimes E_3 \otimes E_4$ is trivial, so $a_1+a_2+a_3+a_4=c_1(E_1 \otimes E_2 \otimes E_3 \otimes E_4)=0$. So what I really need to show is $$e(X)=-a_1^2=-c_1(E_1)^2$$
The problem is, I have no idea why this true. In fact, I am inclined to believe that I have made a mistake somewhere, due to the asymmetry of this formula.

Any help with the original question, following my approach or not, is welcomed!

EDIT: The result is proved using Chern-Weil theory on P.195 of Walschap's "Metric Structures in Differential Geometry". But I'm still curious for an algebraic topology proof.

I thought about this problem for a while and there could be an explanation like this:

First of all, we need an interpretation of the first Pontryagin class as an obstruction class. For a rank $n$ bundle $E^n$ over $X$, we choose $n$ generic sections $\sigma_1,\cdots,\sigma_n$. Consider the points $x$ where they span a subspace of $E_x$ of rank $n-2$. Such points $x$ form a co-dimension 4 cycle of $X$. Its Poincare dual is $P_1(E)$. For details, see the link https://mathoverflow.net/questions/117036/what-is-geometrically-the-pontryagin-class.

In our case, $n=3$ or $4$ and $\dim X=4$. So the set of points $x$ is finite and discrete.

To prove $P_1(\Lambda^+X)=P_1(TX)+2e(X)$, we choose a generic section $v$ of $TX$. Then away from the zero locus of $v$, we have a bundle isomorphism:

$
\phi:\Lambda^+X|_{X-Z(v)}\to v^{\perp} \subset TX|_{X-Z(v)}
$

This means whenever we have three generic sections $\sigma_1,\sigma_2,\sigma_3$ of $\Lambda^+X$, we can apply $\phi$ and get sections of TX over $X-Z(v)$. Then add $v$, we get four sections.

Suppose $Z(v)=\{x_1,\cdots,x_m\}$. For each of them, we remove a four-ball around it. Outside these balls, $\{\sigma_1,\sigma_2,\sigma_3\}$ and $\{\phi(\sigma_i),v\}$ define the same obstruction cycles for $P_1$. Therefore, in order to compute $P_1(\Lambda^+X)-P_1(TX)$, it suffices to work with these four balls and we can deal with them separately.

In the end, we reduce to the case when $X=D^4$ and we prescribe a framing of $\Lambda^+X$ over $\partial X=S^3$. We take the section $v$ to be $v(x)=x$ (or $(x_1,x_2,x_3,-x_4)$, if the origin is a negative zero) for $x\in D^4\subset \mathbb{R}^4$ and so it vanishes at the origin. I guess it is not hard to verify the relation in this case, if we take care of the sign.

Do you think this is valid?


Integrate $e^{ax}\sin(bx)?$ Is there a general formula for finding the primitive of $$e^{ax}\sin(bx)?$$

I've done this manually with $a=9$ and $b=4$ using Euler's formulas. But it takes a bit of time. Is there a pattern here?

There is a pattern. Differentiating a function of the form $e^{ax}\sin (bx)$ yields a linear combination of a function of the same form, and a function $e^{ax}\cos (bx)$. The analogous property holds for functions $e^{ax}\cos (bx)$. So the primitive of $e^{ax}\sin (bx)$ will be a linear combination of $e^{ax}\sin (bx)$ and $e^{ax}\cos (bx)$ (plus a constant).

It remains to find the coefficients.

$$\begin{align}
\frac{d}{dx}\left(e^{ax}(m\sin (bx) + n\cos (bx)\right) &= e^{ax}\left(a\bigl(m\sin(bx) + n\cos(bx)\bigr) + \bigl(bm\cos(bx) - bn\sin(bx)\bigr)\right)\\
&= e^{ax}\left((am - bn)\sin (bx) + (an+bm)\cos(bx)\right)
\end{align}$$

Now solve the linear system

$$\begin{align}
am - bn &= 1\\
an + bm &= 0.
\end{align}$$


If $G$ is a finite group, $H$ is a subgroup of $G$, and there is an element of $G/H$ of order $n$, then there is an element of $G$ of order $n$ $G$ is a finite group, $H$ is a normal subgroup. There is an element in $G/H$ of order $n$. Prove that there is an element in $G$ of order $n$.

Note: I have proved that there exists an element in $G$, say $b$, such that $b^n=e$. I'm stuck in proving that $n$ is the smallest element, i.e. $n$ is the order of $b$.

Please provide a new proof if possible!

One can prove the following proposition (see below):

  If $G$ is a finite group and $H$ is a normal subgroup then $\forall g\in G$, $o(gH)\mid o(g)$, where $o(g)$ denotes the order of $g$.

Suppose $o(g)=l$, $o(gH)=n.$ By using above proposition

$$n\mid l\Rightarrow \exists r\ , \space l=nr.$$

Now let $b:=g^r$ and suppose $o(b)=k$. We have

$$b=g^r\Rightarrow b^n=g^{nr}=g^l=e$$

$$\Rightarrow k\mid n \qquad(*)$$

On the other hand

$$b^k=e\Rightarrow g^{kr}=e\Rightarrow l\mid rk\Rightarrow rn\mid rk$$

$$\Rightarrow n\mid k \qquad(**)$$

By considering $(*)$ and $(**)$, $k=n$ is concluded.

Proof of the proposition:

Here's a slightly more general statement that might actually isolate the fact better.

  Let $f:G\rightarrow G^{\prime}$ be a homomorphism. Then the order of $f(g)$ divides the order of $g$.

In your case, you are just talking about the projection $f:G\rightarrow G/H$.

Suppose $o(g)=n$ and $o(f(g))=m$, so $g^n=e$ by cosidering that $f$ is homomorphism one can conclude:

$$f(g^n)=f(e)=e\Rightarrow (f(g))^n=e \Rightarrow m\mid n.$$

(Reminder: $g^k=e$ implies that the order of $g$ divides $k$.)

o(g)/o(gH).Hence o(g)/n. i.e. o(g)=nt, for some positive integer t.
Now, o(g^t)=o(g)/GCD[t,o(g)]=nt/GCD[t,nt]=nt/t=n.
Hence our required element in G is g^r.


How to calculate relative error when true value is zero? How do I calculate relative error when the true value is zero?

Say I have $x_{true} = 0$ and $x_{test}$. If I define relative error as:

$\text{relative error} = \frac{x_{true}-x_{test}}{x_{true}}$

Then the relative error is always undefined. If instead I use the definition:

$\text{relative error} = \frac{x_{true}-x_{test}}{x_{test}}$

Then the relative error is always 100%. Both methods seem useless. Is there another alternative?

If this is based on any kind of real-world situation, then there should be multiple $x_{test}$ measurements, i.e. a distribution.  Then it will have a standard deviation, or at least quantiles, and you can define the distance from the mean of the $x_{test}$ to $x_{true}$ in terms of these.  E.g., $(\mu_{test} - x_{true}) / \sigma_{test}$ will give you a sort of 'relativized error'.  You can also apply standard statistical tests for significance, e.g. the t-test.

How about $$
\frac{|x_\text{true}-x_\text{test}|}{\mu_\text{true}}
$$
This would give a percentage relative error of the predicted data in respect to the mean of the real data.


When are angles negative? In earlier grades we learnt that in for instance an equilateral triangle all angles are 60 degrees. Now in high school I am taught that measuring angles anti-clockwise and clockwise make a difference.

So does that mean in an equilateral triangle one angle is 60 which is measures anti-clockwise and one is -60 degrees which is clockwise and.. I don't know, I am pretty confused.

How does it add to 180 then?

I was taught about this negative angle concept in trigonometry in unit circles which I understood very well but why is it if not applicable to any geometric figure?

It looks like I am lacking some common sense.

The following line segment on the number line has length four.

    ---------
0 1 2 3 4 5 6 7

The following arrow drawn on the number line represents a displacement of $-4$.

    <--------
0 1 2 3 4 5 6 7

The following arrow drawn on the number line represents a displacement of $4$.

    -------->
0 1 2 3 4 5 6 7

You can do the same thing with angles; sometimes you only care about the magnitude, but other times you pick an orientation: you decide which leg of the angle you start at and which leg you end at. When we orient angles, we typically pick counterclockwise to be the positive direction, and clockwise to be the negative direction.

Even more interesting is an idea of angular position versus angular displacement. If we consider an angle whose vertex is the origin and one leg (the starting leg) is the positive $x$ axis, then putting the other leg at $270^\circ$ would be the negative $y$ axis, but $-90^\circ$ would also be the negative $y$ axis.

The idea that $270^\circ$ and $-90^\circ$ represent the same angle is the idea of angular position.

However, there is a difference between actually going counterclockwise $270^\circ$ around the plane from the positive $x$ axis to the negative $y$ axis, and going $90^\circ$ clockwise (i.e. $-90^\circ$ counterclockwise) from the positive $x$ axis to the negative $y$ axis. When this difference matters, we call it "angular displacement". We could even consider going completely around the origin to return to the $x$ axis then continuing another three quarters the way around: this would be a displacement of $630^\circ$!

As an addendum, if we orient the two legs of the angle so that they point in opposite directions in relation to the vertex, then we usually consider the leg pointing away from the vertex to be the starting leg, and the leg pointing towards the vertex to be the ending leg.

This way, if we orient the three sides of a triangle so that the sides all point in a counter-clockwise direction around the triangle, then all of the angles are oriented counter-clockwise too.

Angle is a measurement and can't be negative, and it is used to define a certain point on the trigonometric circle. In the trigonometric circle, counterclockwise rotation is denoted the positive sign and counterclockwise is denoted the negative sign. for example, moving 60 degrees in the positive direction is completely identical to moving 300 degrees in the negative direction, as they both define one point.
Note that the choice of + or - is completely arbitrary and makes no difference; what really matters is the absolute value of angular displacement.


Is there a way to axiomatize the category of sets and relations? The system of axioms known as ETCS axiomatizes the category of sets and functions. Does anyone know of a way to axiomatize the category (and/or allegory) of sets and relations?

From Categories, Allegories by Freyd, Scedrov:

  2.414. If $\mathbf{C}$ is a topos, then $\mathscr{Rel}(\mathbf{C})$ is a power allegory. Conversely, if $\mathbf{A}$ is a unitary tabular power allegory, then $\mathscr{Map}(\mathbf{A})$ is a topos.

So a reasonable approach would be to start with "$\mathbf{A}$ unitary tabular power allegory", then translate well-pointed, NNO, and AC into the allegory language.

Here's one way presented in at least one publication:

Objects are pairs $(X,\rho)$ with $X$ a set and $\rho \subset X \times X $ is a relation on $X$.

For a pair of objects $(X,\rho)$ and $(Y,\sigma)$, arrows are maps $f: X \to Y$ that preserve the relation, ie, $x_1 \rho x_2 \implies f(x_1) \sigma f(x_2)$ for $x_1, x_2 \in X$. (Are arrows set functions or relations?)

(Further, there is an induced relation between parallel pairs of arrows but I'll have to recall the details)

This is for example exactly how Schreider and Sossinsky defines the category of tolerance spaces, ie, sets with a reflexive, symmetric relation on them.

Note the relations here are not general:

(1) the relations are binary

(2) are "endo" since they're defined on the set $X$, as opposed to relating distinct sets ($\xi \subset X \times Y$)

(3) are 2-valued.

I don't know if considering arrows that reflect relation rather than preserve it is still the same category (dualized).

Also you wrote "axiomatize" so I think this answer suffices. If you has asked for basic properties such as what limits exist, that's another story...


Help proving inequality 3. $\delta$ is the standard Euclidean valuation on $\mathbb{Z}[i]$. For each of the following pairs $a,b \in \mathbb{Z}[i]$, find $q,r \in \mathbb{Z}[i]$ such that $a = qb + r$, where either $r = 0$ or $\delta(r) \leq \frac12\delta(b)$. Verify that $r$ satisfies this condition.

$a = 5$, $b= (2+3i)$

$\dfrac{5}{2+3i} = \dfrac{10}{13} - \dfrac{15}{13i}$

Take $q = 1-i$. Then $5 = (2+3i)(1-i)-1$, so $r = -i$.

How does one prove $\delta(r) < \frac12\delta(b)$? Any help appreciated.

Thank you.

Also, the solution in my book is $\delta(r) = 1 < \frac{11}{2} = \frac12\delta(b)$.

I don't know to get this.

I think that the standard Euclidean valuation on $\mathbb{Z}[i]$ is just defined as $\delta(x+yi) = x^2 + y^2$. If this is true, then

$\delta(r) = \delta(-i) = (-1)^2 = 1$

and

$\delta(b) = \delta(2 + 3i) = 2^2 + 3^2 = 13$, so $\frac12\delta(b) = \frac{13}{2}$.

So I'm not sure why the book says $\frac12\delta(b) = \frac{11}{2}$.


What is the Direction of a Zero (Null) Vector? To be more precise, I am interested in knowing if the intuition that a Euclidean zero vector does not have a particular direction is actually correct, and if there is a rigorous formulation that would back it up.

Wikipedia's entries on the zero vector seems to agree with that intuition, but as always, one shouldn't blindly trust Wikipedia: in one place, it is stated that a zero vector "is orthogonal to all other vectors with the same number of components," while in another, it is stated that "two vectors can be considered orthogonal if and only if their dot product is zero, and they have non-null length." Correct me if I am wrong, but these two statements contradict each other.

This question popped up in my head when I heard someone arguing that since a Euclidean vector is defined as a geometric entity that has both a magnitude and direction, and since a zero vector is a vector with length 0, then it is only fitting for a zero vector to have a "direction 0." I am personally inclined to say that a zero vector does not have a particular direction, but as I said, I would like to know if there is a rigorous formulation that would lead to this conclusion.

And to put the question in a less "discussion-inducing" form, is there an agreement on the direction of a Euclidean zero vector?

The zero vector has no particular direction; this is consistent with the fact that it is orthogonal to every other vector.  (It doesn't really make sense to say it has "direction 0", since direction is not a magnitude; "direction 0" makes no more sense than "direction 1"
or "direction 5.873".)

Alternatively, you could say that it points in every direction, but with zero magnitude,
since if you take any vector and multiply it by zero, you get the zero vector.  "Every
direction" is the same as "no particular direction"; it's just a different way of phrasing
things.

This is one flaw with the traditional description of a vector as being a pair consisting of
a magnitude and a direction:  For the zero vector, the magnitude is zero, but the direction
is arbitrary.

I think it is possible for a zero magnitude vector to have direction.  Consider a particle moving in a circle at constant speed. Velocity is variable because direction changes.  magnitude of acceleration is zero because speed is constant. Acceleration (directed towards center) is variable because direction changes every instant. Clearly acceleration is present with magnitude zero.


estimate the limit of $(\cos(4x))^\frac{1}{x^2}$ I was looking for a way to estimate the limit of $\left(\cos\left(4x\right)\right)^\frac{1}{x^2}$ as x goes to zero using lohospital laws, which I didn't consider to be possible given there is no 0/0 here...how can one do this using lhospital?

hint: consider $$\lim_{x \to 0}(\cos(4x))^{\frac{1}{x^2}}=\exp\left(\lim_{x \to 0}\frac{\ln(\cos(4x))}{x^2}\right)$$

It is well-known that, if $ \lim f(x) = \exp \left( \lim \ln f(x) \right) $, by continuity. Hence, $$\lim_{x \to 0}(\cos(4x))^{\frac{1}{x^2}}=\exp\left(\lim_{x \to 0}\frac{\ln(\cos(4x))}{x^2}\right)$$Now, using l'Hopital's rule gives the inner limit to be $$ \begin {align*} \lim_{x \to 0} \frac {\frac{-4 \sin (4x)}{x\cos(4x)}}{2x} &= -2 \cdot \lim_{x \to 0} \frac {\sin (4x)}{x \cos (4x)} \\&= -2 \cdot \lim_{x \to 0} \frac {\tan(4x)}{x} \\&= -2 \cdot \lim_{x \to 0} \frac {4\sec^2(4x)}{1} \\&= -8 \cdot \sec^2 (0) = -8. \end {align*} $$Hence, the answer is $e^{-8}$. $\Box$


Evaluating $\lim\limits_{n \to \infty }\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}} $ Help me please with the limit:

$$\lim_{n \to \infty }\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}}$$

Thanks!

$$\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}}  \leq \sqrt[n]{\dfrac1{n!}}$$ Now our good old Stirling's formula will do the job.

With AM-GM the sequence is smaller than $\frac{H_n}{n}$ with Hn is équivalent ln (n) , then the limite is zéro.


Differentiation of Function Defined as Integral of Multiple Variables I am stuck on a question regarding finding the derivative $f'(0)$ of a function $f(x)$ defined as $f(x) = \int_{-\infty}^{e^{2x}} e^{\frac{-1}{2}(y-sin(x))^2}\, dy $ for $x \epsilon {\Bbb R}$.

I am vaguely aware that Leibniz' Integral Rule is the starting point for this question, but I don't know how to deal with the $-\infty$ in the lower bound of the integral. (https://en.wikipedia.org/wiki/Leibniz_integral_rule)

How do I compute this integral? Many thanks for any help given.

We have that $$f(x) = \int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\mathrm d y.$$ Thus, $$f'(x) = \left(e^{2x}\right)'\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}[-\left(y-\sin x\right)](-\cos x)\mathrm d y=2e^{2x}\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\left(y-\sin x\right)(\cos x)\mathrm d y.$$

Now set $x=0$ to obtain $f'(0).$


Proving $y=\lfloor x\rfloor$ doesn't have a primitive function Prove that $y=\lfloor x\rfloor$ doesn't have a primitive function.

I have the proof from a book here but I don't understand it:

  Suppose that there is $F(x)$ such that $F'(x)=\lfloor x\rfloor$. We
  have that $y=\lfloor x\rfloor$ has a jump discontinuity (or type 1)
  but $F'(x)$ has only an essential discontinuity (or type 2), a
  contradiction.

I don't understand, why does $F'(x)$ has an essential discontinuity?

The proof is silently appealing to Darboux’s theorem:

  If $I$ is an open interval, and $f:I\to\Bbb R$ is a differentiable function, then $f'$ has the intermediate value property. That is, if $a,b\in I$, $a<b$, and $y$ is between $f'(a)$ and $f'(b)$, then there is an $x\in[a,b]$ such that $f'(x)=y$.

This means that a derivative cannot have a jump discontinuity and therefore cannot be the floor function. In particular, suppose that $F'(x)=\lfloor x\rfloor$, and take $I$ to be the interval $(-1,2)$. Then $0,1\in I$, and

$$F'(0)=\lfloor 0\rfloor=0<\frac12<1=\lfloor 1\rfloor=F'(1)\;,$$

but there is no $x\in[0,1]$ such that $F'(x)=\frac12$, contradicting Darboux’s theorem.

A fairly 'wild' answer would be for $x>0$:
$$F(x)=\sum_{n=1}^{\infty}  \delta(x-n)$$


bilinear functional on complex vector space Given a complex vector space S, if g: S $\times$ S $\rightarrow$ $\mathbb{C}$ is a symmetric bilinear functional, can we say there always exist v $\in$ S and w $\in$ S such that g(v,w) = 0?

Yes.  Just take either $v$ or $w$, or even both, to be the zero vector.


Sketch the graph of $F_4^n(x)$ on the unit interval, where $F_4(x) = 4x(1-x)$. Conclude that $F_4$ has at least $2^n$ periodic points of period $n$. Sketching the graph I found that $F_4$ has exactly $2^{n-1}$ point wich prime period $n$. My $F_4^n$ graph look like a $|sin(x)|$ graph with $2^n$ intersections on $Id$, but I removed the $2^{n-1}$ to remove points that have period less then n. What is wrong?

Given the following functions:
$\varphi(x)=-4x+2$,
$\varphi^{-1}(x)=\frac{x-2}{-4}$
and $g(x)=x^2-2$
we have:
$$F_4(x)=(\varphi^{-1} \circ g\circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ g^{\circ n} \circ \varphi)(x) \tag{1}$$

Further $g(x)$ can be written as
$$g(x)=(\gamma^{-1} \circ T_2\circ \gamma)(x)$$
where $T_2(x)=2x^2-1$ (a Chebyshev polynomials), $\gamma(x)=\frac{x}{2}$ and $\gamma^{-1}(x)=2x$. Then
$$F_4(x)=(\varphi^{-1} \circ \gamma^{-1} \circ T_2\circ \gamma \circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma^{-1} \circ T_2^{\circ n}\circ \gamma \circ \varphi)(x) \tag{2}$$

Chebyshev polynomials satisfy
$$T_n(T_m(x))=T_{nm}(x) \Rightarrow T_2^{\circ n} (x)=T_{2^n}(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma^{-1} \circ T_{2^n}\circ \gamma \circ \varphi)(x) \tag{3}$$

Periodic points of period $n$ are fixed points of $\circ n$ composition.
Fixed points of $T_{2^n}(x)$ are fixed for $F_4^{\circ n}(x)$ "through" $\gamma(x)$ and $\varphi(x)$, both linear functions, i.e.

Proposition 1. $T_{2^n}(x)=x \iff F_4^{\circ n}(y)=y$
where $y=(\varphi^{-1} \circ \gamma^{-1} )(x)$

We should also observe that if $y \in [0,1]$ then
$$x=(\gamma \circ \varphi) (y) = -2y+1 \in [-1,1] \tag{4}$$

For Chebyshev polynomials we have $T_m(\cos{\alpha})=\cos{(m\alpha)}$. From $(4)$ we can assume that $\exists \alpha$ s.t. $x=(\gamma \circ \varphi) (y)=\cos{\alpha}$. As a result, for proposition 1 we will be looking at
$$T_{2^n}(\cos{\alpha})=\cos{(2^n \alpha)}=\cos{\alpha}$$
which means for $k\in\mathbb{Z}$
$$2^n \alpha = \alpha +2k\pi \Rightarrow \alpha =\frac{2k\pi}{2^n-1}$$
or
$$2^n \alpha = -\alpha +2k\pi \Rightarrow \alpha =\frac{2k\pi}{2^n+1}$$
and we found 2 sequences
$$x_{1,k} = T_{2^n}(x_{1,k})=T_2^{\circ n} (x_{1,k})
\text{, where } x_{1,k}=\cos{\left(\frac{2k\pi}{2^n-1}\right)}$$
$$x_{2,k} = T_{2^n}(x_{2,k})=T_2^{\circ n} (x_{2,k})
\text{, where } x_{2,k}=\cos{\left(\frac{2k\pi}{2^n+1}\right)}$$

Applying proposition 1
$$y=(\varphi^{-1} \circ \gamma^{-1} )(x)=\frac{1-x}{2}$$
and considering that $\frac{1-\cos{\alpha}}{2}=\sin^2{\frac{\alpha}{2}}$, we have
$$y_{1,k} =\sin^2{\left(\frac{k\pi}{2^n-1}\right)}$$
$$y_{2,k} =\sin^2{\left(\frac{k\pi}{2^n+1}\right)}$$
Some of these values will repeat, for example
$$y_{1,2^n-k}=y_{1,k-1} \text{, so we can limit } k=\overline{1,2^{n-1}}$$
$$y_{2,2^n-k}=y_{2,k+1} \text{, so we can limit } k=\overline{0,2^{n-1}-1}$$
which makes $2^n$ altogether.

Incomplete and with typos ... but contains a few hints and references.

Given the following functions:
$$\varphi(x)=-4x+2,\varphi^{-1}(x)=\frac{x-2}{-4}, g(x)=x^2-2$$
we have:
$$F_4(x)=(\varphi^{-1} \circ g\circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ g^{\circ n} \circ \varphi)(x)$$
Further to this, function $\gamma(x)=x+\frac{1}{x}$ has the property that
$$(g \circ \gamma)(x)=x^2+\frac{1}{x^2}=\gamma(x^2)\Rightarrow (\gamma^{-1} \circ g \circ \gamma)(x)=x^2\Rightarrow \\g(x)=(\gamma \circ t \circ \gamma^{-1})(x), t(x)=x^2$$
As a result
$$F_4(x)=(\varphi^{-1} \circ \gamma \circ t \circ \gamma^{-1} \circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma \circ t^{\circ n} \circ \gamma^{-1} \circ \varphi)(x) \tag{1}$$
Effectively, by introducing $\gamma$ we have to exclude $x=0$ perioding point, which is not of period $n$ though.
Periodic points of $t(x)=x^{2}$ are periodic for $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$. Complex roots of $x^{2^n-1}=1$ are periodic points for $t(x)$ (and  $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$) and, except $x=1$, all the other roots have period $n$ (there is an entire chapter in  this book, starting with page 6, dedicated to this function). To see all these working, let's take $x=\sin^2{\frac{\alpha}{2}} \in (0,1)$ then (I will omit the details as those are easy to check)
$$\varphi(x)=2\cos{\alpha}$$
$$\gamma^{-1}\left(2\cos{\alpha}\right)=e^{i\alpha}$$
$$t^{\circ n}\left(e^{i\alpha}\right)=e^{i2^{n}\alpha}$$
$$\gamma\left(e^{i2^{n}\alpha}\right)=2\cos{\left(2^{n}\alpha\right)}$$
$$\varphi^{-1}\left(2\cos{\left(2^{n}\alpha\right)}\right)=\sin^2{\left(2^{n-1}\alpha\right)}$$
or
$$F_4^{\circ n}\left(\sin^2{\frac{\alpha}{2}}\right)=\sin^2{\left(2^{n-1}\alpha\right)} \tag{2}$$
Complex roots of $z^{2^n-1}=1$ are of the form
$$z_k=\exp{\left(\frac{2i\pi k}{2^n-1}\right)},k=\overline{0..2^n-1}$$
thus taking $$\alpha=\frac{2\pi k}{2^n-1}$$
we have
$$F_4^{\circ n}\left(\sin^2{\frac{\pi k}{2^n-1}}\right)=\sin^2{\left(2^{n-1}\frac{2\pi k}{2^n-1}\right)}=\sin^2{\left(\frac{(2^{n}-1+1)\pi k}{2^n-1}\right)}=\\
\sin^2{\left(k\pi + \frac{\pi k}{2^n-1}\right)}=\sin^2{\frac{\pi k}{2^n-1}},k=\overline{0..2^n-1} \tag{3}$$
Note: we could simply apply (from $(2)$) $\sin^2{\frac{\alpha}{2}}=\sin^2{\left(2^{n-1}\alpha\right)} \Rightarrow \frac{\alpha}{2}+k\pi=2^{n-1}\alpha$ and arrive to the same conclusion, but without conjugacy functions reaching this point would require having a good intuition.
However
$$F_4^{\circ n}\left(\sin^2{\frac{\pi (2^n-k)}{2^n-1}}\right)=\sin^2{\left(\frac{\pi (2^n-k)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (2^n-1-k+1)}{2^n-1}\right)}=\\
\sin^2{\left(\pi - \frac{\pi (k-1)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}=F_4^{\circ n}\left(\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}\right) \tag{4}$$
this last statement reduces the number of periodic points with period $n$ to $2^{n-1}$.
Note: $(4)$ can also simply be obtained from $\frac{2\pi k}{2^n-1}=\frac{2\pi m}{2^n-1} + p\pi$, where $0\leq k,m \leq 2^n-1$ and concluding that $p=1$ and $k-m=2^n-1$.


Showing that $|A\cap B|/|A \cup B| + |B\cap C|/|B \cup C| - |A\cap C|/|A \cup C| \leq 1$ for finite sets $A,B,C$. If $A$, $B$ and $C$ are finite sets, prove that
  $$
 \frac{|A\cap B|}{|A \cup B|}
 + \frac{|B\cap C|}{|B \cup C|}
 - \frac{|A\cap C|}{|A \cup C|}
 \leq 1.
$$

It seem's simple, but I tried it for a long time and cannot get it out. Maybe I can use some optimization methods to calculate it, but that's not what I want...

I'm referring to the Venn diagram above. The indicated variables denote the numbers of elements in the particular component, and are integers $\geq0$.

We have to prove that
$$\eqalign{&{z+y_3\over x_1+x_2+y_1+y_2+y_3+z}+{z+y_1\over x_2+x_3+y_1+y_2+y_3+z}\cr
&\qquad\leq1+{z+y_2\over x_1+x_3+y_1+y_2+y_3+z}\ .\cr}\tag{1}$$
During the proof we shall use several times the fact that the function
$$t\mapsto{a+t\over b+t}\qquad(t>0)$$
is monotonically increasing when $0\leq a\leq b$.

As $x_2$ is not appearing on the RHS of $(1)$ we may assume $x_2=0$. Furthermore, the  LHS of $(1)$  decreases when $y_2$ increases, whereas  the mentioned principle  shows that the RHS of $(1)$ is an increasing function of $y_2$. It follows that we may assume $y_2=0$ as well.

Put $y_1+y_3+z=:s$. Then we have to prove
$${s-y_1\over s+x_1}+{s-y_3\over s+x_3}\leq 1+{s-y_1-y_3\over s+x_1+x_3}\ .\tag{2}$$
Here the RHS can be rewitten as
$${s-y_1+x_3\over s+x_1+x_3}+{s-y_3+x_1\over s+x_3+x_1}\ .$$
Now by the aforementioned principle we have
$${s-y_1\over s+x_1}\leq{s-y_1+x_3\over s+x_1+x_3}\ ,$$
and similarly for the second parts. This proves $(2)$, which is the "worst case" of $(1)$.


Why do vectors often have 1 dimenision as 1 In textbooks and school, why is it that vectors are often shown to be n x 1 or 1 x n matrices? I have rarely seen other dimensioned matrices representing matrices.

\begin{bmatrix}
    1 & 2 & 3 & 4 & 5\\
    3 & 4 & 5 & 6 & 7\\
  \end{bmatrix} is usually a matrix but

\begin{bmatrix}
    1 \\
    3 \\
    4 \\
  \end{bmatrix}
is usually a vector. Why?

Simplicity of notation. Using a 2x3 matrix allows you to represent 6 quantities with the understanding that there is some sort of "grouping" within those 6 quantities (2 sets of 3 quantities that are in some way "different"). When we work with vectors we are not usually concerned with such an internal grouping. Note, however, that there is nothing stopping you from viewing a 1xn set of scalars as a matrix or as a vector. You must simply make clear the type of mathematical object you are working with. After all, you could represent a 2x2 matrix as a column of 4 quantities, right? but that wouldn't be a very convenient or simple notation.

Yes, usually column vectors ($1\times n$ matrices) are used for vectors. But vectors are also very commonly regarded as such matrices, e.g. the matrix product $A\cdot v$ is everywhere used.

For another example, we can also build a matrix (of range $1$) by the matrix product $v^Tv$.

So, in many cases it is extremely useful to apply matrix operations for (column and/or row) vectors simply regarding them as matrices.


Is there a set $A$ such that $|\mathbb Z|<|A|<|\mathbb R|$ is undecidable? CH guarantees that the statement $|\mathbb Z|<|A|<|\mathbb R|$ is false for all $A$, but since $\sf CH$ is undecidable it might still be possible that there exist a set $A$ for which the statement is undecidable without the CH or it's negation.

My intuition says that there should be such a set because in $\sf ZFC+\neg\sf CH$ there would exist a set $A$ such that $|\mathbb Z|<|A|<|\mathbb R|$, but in $\sf ZFC$ only the set $A$ could not be an counterexample of $\sf CH$ and it could not be countable or be as large as $\mathbb R$ because it then wouldn't fulfil it's property in $\sf ZFC+\neg\sf CH$.

If it's true that such a set exists, to what extent can such a set be described (in a concrete way)?

: there exist a set A for which the statement is undecidable

This is a notoriously ambiguous claim because ZFC generally does not prove statements with concrete sets sets as parameters. One has to be careful with things like this as otherwise there could happen an absolute loss of sense. One of possible precise formulations of your para-question is this.

Is there a formula $F(x)$ of the language of ZFC such that

1) ZFC proves there is a unique $x$ st $F(x)$, and

2) ZFC proves that if CH fails then the $x$ of item 1 is a set of cardinality $\aleph_1$.

If this is what you ask then it answers in the positive by $F(x)$ being "$x=$ all countable ordinals" by obvious reasons. If you are looking for something else then pls specify it in unambiguous terms.

I am not entirely sure what your question is but the CH is one of my favorite topics in mathematics due to the breadth of mathematical logic surrounding it, so all I can offer is some resources for more information.

The Wiki page for the CH is incredibly well written and contains a lot of information that may help you, particularly this..

"Solomon Feferman (2011) has made a complex philosophical argument that CH is not a definite mathematical problem. He conjectures that CH is not definite according to this notion, and proposes that CH should therefore be considered not to have a truth value. Peter Koellner (2011b) wrote a critical commentary on Feferman's article.

The Wiki page also has many links to important relevant information such as the axiom of constructibility.


How to compute the limit,$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos1/x}$ How to compute the limit,
$$\lim_{x\rightarrow 0}\frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}$$

Rewrite $$\lim_{x\to 0}\frac{3x^2-3x\sin x}{x^2+x\cos\frac{1}{x}}=\lim_{x\to 0}\frac{3x^2-3x\sin x}{x(x+\cos\frac{1}{x})},$$ we will deduce the nature of $x+\cos\frac1x$ near $0$.
There are monotone decreasing sequence $a_n>0$ with $\lim_{n\to\infty}a_n=0$ such that $\cos\frac 1{a_n}=(-1)^n$.
Let $a_n\ll 1$, we have $a_{n+1}+\cos \frac 1{a_{n+1}}$ have a different sign from $a_{n}+\cos \frac 1{a_{n}}$. Since $(x+\cos\frac{1}{x})$ is continuous, there is a zero point between $a_{n+1}$ and $a_n$.
Hence there are infinite undefined points near $0$. Therefore, the limit does not exist.
Notice that this answer is only available with definition $\lim_{x\to x_0}f(x)=A\Leftrightarrow\forall \varepsilon>0\exists\delta>0\ s.t.\ 0<|x-x_0|<\delta\implies|f(x)-A|<\varepsilon$

For any $\delta > 0$ there exists an $0<x<\delta$ such that $x - \cos \frac {1}{x} = 0$

As $x-\cos \frac 1x$ is continuous, the are points close to these zeros such that $ x-\cos \frac 1x $ is arbitrarily close to zero, and $\frac {3x^2 - 3x\sin x}{x(x-\cos \frac 1x)}$ is arbitrarily large.


Is a function a special kind of relation? Is a function a "special kind of relation", or, does it "describe a specific relation"?

My text on discrete mathematics explains:

  A relation is a subset of a Cartesian product and a function is a
  special kind of relation.

But it would make more sense to me if a function described a relation as a subset of the Cartesian product.

My thoughts being:
Given a function, f(x) = y, we can compute a set of (x,y) coordinates within the Cartesian plain. And this set of coordinates would be the relation that is the subset of the Cartesian product.

Am I confused? Could anyone help explain how a function IS a relation?

A function is a specific kind of relation.

The best way to understand this is with the help of an example.
So, let us take a couple of sets - set $A$ = {1, 2, 3} and set $B$ = {$a$, $b$, $c$}. Thus the set $A \times B$ will have 9 elements.

We can choose $2^9 = 512$ different subsets of $A \times B$. Each of this subset is a relation between $A$ and $B$. So, $\phi$ is a relation, {$(1, a), (1, c), (2, b)$} is also a relation. $A$ is called the domain and $B$ is called the co-domain.

A function is also a subset of $A \times B$ (hence a relation), but it has constraints. For every  element in set $A$, there should be exactly one element in set $B$. More concretely, for every element $x$ in set A, there is exactly one $(x, y)$ in $f$ for some $y \epsilon B$

For example, {$(1, a), (2, b), (3, b)$} is a function, but {$(1, a), (2, b)$} is not because there is no entry of the form (3, *). Also, {$(1, a), (1, b), (3, b), (2, b)$} is also not a function because 1 has two values it is mapping to.

A function (from set $A$ to set $B$) is a relaction of special kind in which  $(x,y_1)=(x,y_2) \Rightarrow y_1=y_2$ for every $x \in A, y_1,y_2 \in B$.


Hard time figuring out word problem Computers are beginning to lose power. The number of computers losing power doubles every minute. At 10 minutes half of the computers have lost power. After how many minutes will all computers be without power?

The only thing I could come up with is

$$y- (2^9)x = .5y$$

Or a summation?

$$y-x2^i \text{ from 1 to 9?}$$

I have no clue.

Is it even solvable?

The wording is slightly ambiguous, but the second sentence is probably intended to mean that the total number of computers that have lost power doubles every minute. If so, you don’t need to set up any sort of function: if half of the computers have lost power after ten minutes, and one minute later twice as many will have lost power, then in $11$ minutes all of the computers will have lost power. If you want to set up a formula, let $n_1$ be the number of computers losing power in the first minute. Then after $t$ minutes $n_12^{t-1}$ computers will have lost power. You’re told that $$n_12^{10-1}=\frac12n\;,$$ where $n$ is the total number of computers; multiplying by $2$ yields $n_12^{10}=n$, and $n_12^{10}$ is the number of computers that have lost power after $10+1=11$ minutes.

The wording could also be read as meaning that if $n_1$ computers lose power in the first minute, then $2n_1$ lose power in the second minute, $4n_1$ in the third minute, and so on. In that case the number that have lost power after $t$ minutes is

$$n_1+2n_1+2^2n_1+\ldots+2^{t-1}n_1=n_1\sum_{k=0}^{t-1}2^k=n_1(2^t-1)\;,$$

and we have $$n_1(2^{10}-1)=\frac12n\;.$$

Then $n=2046n_1$, so after $t$ minutes the number of computers that have lost power is

$$n_1(2^t-1)=\frac1{2046}n(2^t-1)\;.$$

This is equal to $n$, the total number of computers, when $t$ satisfies the equation $2^t-1=2046$, or $2^t=2047$. Since $2^{11}=2048$, this occurs just a hair before $11$ minutes have passed; numerically, $t\approx10.999295387$ minutes.

The answer is $t=10\sqrt{2}$.

The initial assumption is that $\frac{dy}{dt}$ is negative and doubles every minute, hence, $\frac{dy}{dt}=-2t(L)$ , where L is a positive integer that represents the initial number of computers that lost power.

Integrating gives $y=-L(t^2)+c$.

Solving with the initial condition [ $y(10)=.5y(0)$ ] gives:

$y=L(200-(t^2))$, which leads to the only reasonable solution: that $y=0$ when $t=10\sqrt{2}$.


elements of nested sets I was thinking about elements of the power set, and I know that for a power set $P=\{\{\},\{a\},\{b\},\{a,b\}\}$ that $\{\},\{a\},\{b\},\{a,b\}$ are all elements of $P$. However, if you have another set containing sets, such as $A=\{1, 2, \{3, 4\}\}$: is 3 an element of $A$?


Can Extraneous Roots be Introduced by Elimination? Suppose you have two equtaions:
$$2xy + y^2 = 0$$
$$x^2 + 2xy + 1 = 0$$
Subtracting the second from the first yields $y^2 - x^2 - 1 = 0$. Isolating y, we discover that $y = \pm\sqrt{x^2 + 1}$. However, by inspection we can wee that the $2xy$ term in both equations must be negative, which means that a single value of x cannot have both a positive and negative corresponding y-value. (i.e. if x is positive, then y must be negative). It seems that the equation $y = \pm\sqrt{x^2 + 1}$ contains an "extraneous root", but I'm struggling to wrap my mind around how that could be. After all, I solved for y w.r.t. x without squaring both sides. Could anyone help me understand what is going on?

The problem is that your reasoning isn't reversible. Your two equations together imply $y^2-x^2-1=0$, but the converse is not true: that equation does not imply your original system.

Compare your question to the following argument. The system $$x+y=1$$ $$2x+y=1$$
yields, by subtraction, that $x=0$. But this second equation allows $y$ to be arbitrary! So we get a whole lot of "extraneous solutions": $(0,y)$, for any $y$. The problem, of course, is that the equation $x=0$ does not, in turn, imply the original system. I take it you do not find this situation puzzling.

You generally lose information when you replace a system of equations with a linear combination of them. The combination will include the solutions of your original equation, but it likely will include non-solutions, as well.

The form of the reasoning is: "any solution of the original equations will be a solution of this equation, too." Yes, but then you've only found a superset containing the solution set. To characterize the solution set exactly, you must worry about the converse.

You have taken a square root. Whenever you take a square root you can introduce an extraneous root. Even for the simple system $x^2=y^2$, $x=y$ if you took a square root of the first equation you'd get an extraneous root. You must always consider all the equations as they all contain information about the solution. If you had gone about solving it via elimination or substitution (rather than merging the two equations together randomly like you did) then you would have not encountered this.

If you solve your equation fully (and use complex numbers) you get two pairs of roots:
$$\left(\frac{1}{\sqrt3},-\frac{2}{\sqrt3}\right), \left(-\frac{1}{\sqrt3},\frac{2}{\sqrt3}\right),(i,0),(-i,0)$$
The second pair of roots does fit your reduced equation (with a small rearrangement), namely $x=\pm\sqrt{y^2-1}$.


Which of the following are complete residue systems modulo $11$? Which of the following are complete residue systems modulo 11?

$(a)\quad  0,1,2,4,8,16,32,64,128,256,512$

$(b)\quad  1,3,5,7,9,11,13,15,17,19,21$

$(c)\quad  2,4,6,8,10,12,14,16,17,20,22$

$(d)\quad  -5,-4,-3,-2,-1,0,1,2,3,4,5$

I have the answer which is that they are all complete residue systems modulo 11. However, I do not fully understand the definition of residue system or how to determine what sets make up a complete residue system?

To be a complete residue system modulo $n$, you need a set of $n$ integers, no two of which are congruent modulo $n$.

The correct answer is (b). 1, 3, 5, 7, and 9 are in their own remainder classes mod 11 while 11 is in cl(0), 13 in cl(2), 15 in cl(4), 17 in cl(6) 19 in cl(8), and 21 in cl(10). We have members of every remainder class  listed hence a complete residue system. The key is to know that all we need for a complete residue  system are members of every remainder class. Thomas defines a complete residue class as a set $S=\{r,s,\dotsc\}$ such that for every integer $k$ there exists a member of $S$, call it $m$, such that $k$ is congruent to $m$ mod $n$ or in the problem at hand mod 11. The complete residue system partitions the set of integers. It's important to realize that complete residue systems are not unique. All that's needed is that the set (the complete residue class) include any member of every remainder class.