Planes of a dodecahedron divide the space $\mathbb R^3$ In how many disjunct parts divide the planes of a regular dodecaheadron the space $\mathbb R^3$?
I encountered this task at university where we discussed it a lot, but nobody found a solution.

Does anybody know the solution? May it work with graph theory?

Here's a graphical count, based on intersecting the configuration with spheres of different sizes -- though in order to make things visible actually I'm normalizing the size of the spheres, such that the images here actually shows a sphere intersected by the planes of a dodecahedron of increasing size.

The 12 planes of the dodecahedron come in 6 parallel pairs, and in the figures, the regions of each sphere are color-coded according to (and labeled with) the number of pairs where the region is between the two parallel planes.

The first image shows a sphere far from the dodecahedron. Each of the 6 pairs of planes become a narrow band around the sphere, and the bands create the graph of an icosidodecahedron. Each face, edge, and vertex of the icosidodecahedron corresponds to an unbounded region of space -- there are $12$ of type 0, $20$ of type 0', $60$ of type 1, and $30$ of type 2.

As we zoom in towards the dodecahedron (or the dodecahedron become larger), the edges of each band move away from each other. The first change that happens is that the the 0' regions disappear and are replaced by $20$ cells of type 3. These new cells are the triangular spires of the great stellated dodecahedron.

Nothing new happens in the first of these pictures, except that the bands keep growing wider.

In the second picture the bands have just met in groups of five, and the last of the 0 and 1 regions disappear. Instead we get cells that are between 4 and 5 pairs of parallel planes.

The new yellow regions of type 4 merge, and we can see there are $30$ such cells, just as there were thirty 2s before, corresponding to the edges of either an icosahedron or a dodecahedron. In fact each of these cells is an irregular tetrahedron that shares one edge with the central dodecahedron, and the opposite edge with a great dodecahedron (whose edge figure looks like an icosahedron).

Removing the yellow tedrahedra from the great dodecahedron leaves a small stellated dodecahedron, whose pentagonal spires are the $12$ magenta 5 cells that came into being together with the 4s.

Finally a single red 6 cell appears: the original dodecahedron in the middle.

So how many regions were that? Summing everything we get
$$ \underbrace{(12+20)+60+30}_{\text{unbounded}}+\underbrace{20+30+12+1}_{\text{bounded}} = 185 $$

I found 432 in the book
"Les nombres remarquables" by Francois Le Lionnais (Herman, 1983)
(which is a list of interesting numbers), but there is no reference or proof. Among these 432 parts 192 are bounded.


Find: $~Cov(X,Y)~$ and $ρ_{X,Y}$ Flip a coin thrice ($3$ times), and let  $X$  and  $Y$  denote the number of heads in the first two flips, and in the last two flips, respectively. For example, if the coins turn up $h,h,t$ then  $X=2$  and  $Y=1$ , while if they turn up $t,t,h$ then  $X=0$  and  $Y=1$ . Find: $Cov(X,Y)$  and $ρ_{X,Y}$ .

I Have tried this approach but my answer comes wrong(ans $~Cov(X,Y)=1.522 ,~~X,~Y =0.5~$) please some correct me.

Let Z be the count of heads in the middle flip. On noting, among other things, that X,Y are conditionally independent given $~Z~$, you may now use the Law of Total Covariance $$\mathsf {Cov}\big(X,Y\big)=\mathsf E\big(\mathsf {Cov}(X,Y\mid Z)\big)+\mathsf{Cov}\big(\mathsf E(X\mid Z),\mathsf E(Y\mid Z)\big)$$

$$\rho_{\small X,Y}=\dfrac{\mathsf{Cov}(X,Y)}{\surd\mathsf {Var}(X)\cdot\surd\mathsf{Var}(Y)}$$

Using the law of total covariance as suggested in the question, the first term is zero because the conditional covariance of $X$ and $Y$ given $Z$ is zero.  For the second term, we note $$\operatorname{E}[X \mid Z] = Z + p, \quad \operatorname{E}[Y \mid Z] = Z + p,$$ where $p$ is the probability of observing heads for a single toss.  Here, we are assuming tosses of the same coin are independent and identically distributed.  Then $$\operatorname{Cov}[Z + p, Z + p] = \operatorname{Var}[Z + p] = \operatorname{Var}[Z] = p(1-p).$$  It follows that $$\operatorname{Cov}[X, Y] = p(1-p).$$  I leave the rest as an exercise.

You make things quite strangely. Moreover, no one could have guess that $\rho_{X,Y}$ denotes the correlatin, so be precise next time.

Hint on how to solve your problem

Find the joint distribution $f_{X,Y}(k,\ell):=\mathbb P\{X=k,Y=\ell\}$, $k,\ell\in \{0,1,2\}$.
Then, $$\text{Cov}(X,Y)=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y],$$
where $$\mathbb E[XY]=\sum_{k,\ell\in \{0,1,2\}}k\ell\mathbb P\{X=k,Y=\ell\}.$$
I let you do for the correlation.


If $a+b+c=1$ then $\sum\limits_{cyc}\frac{a}{\sqrt[3]{a+b}}\leq\frac{31}{27}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ and $a+b+c=1$.
Prove that:
$$\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\leq\frac{31}{27}$$

The equality occurs for $(a,b,c)=\left(\frac{19}{27},\frac{8}{27},0\right)$.
This inequality is similar to the following inequality, which was proposed by Walther Janous.

For all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\neq0$ prove that:
$$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq\frac{5}{4}\sqrt{x+y+z}$$

My proof:
By Cauchy-Schwarz $$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or
$$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$
$$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or
$$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$
$$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.
If we want to use a similar way for the starting inequality, we need to use Holder, which gives very big numbers.
Maybe there is another reasoning?
Thank you!

With computer, here is an ugly solution:

WLOG, assume that $c = \min(a, b, c).$
Since $x^{2/3}, x \ge 0$ is concave, we have
\begin{align}
\frac{b}{\sqrt[3]{b+c}} + \frac{c}{\sqrt[3]{c+a}}
&= \frac{b}{b+c}(b+c)^{2/3} + \frac{c}{c+a}(c+a)^{2/3}\\
&\le \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^{2/3}.
\end{align}
On the other hand, we have
$$\Big(\frac{1+5a+5b}{5+a+b}\Big)^3 - (a+b)^2 = \frac{(a^2+2ab+b^2+18a+18b+1)(1-a-b)^3}{(5+a+b)^3} \ge 0$$
which results in
$(a+b)^{2/3} \le \frac{1+5a+5b}{5+a+b}$.

Thus, it suffices to prove that
$$\frac{a}{a+b}\frac{1+5a+5b}{5+a+b} +  \Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^{2/3} \le \frac{31}{27}.$$
It suffices to prove that
$$\Big(\frac{31}{27} - \frac{a}{a+b}\frac{1+5a+5b}{5+a+b}\Big)^3\ge
\Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)^3\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^2. $$
After homogenization, it suffices to prove that
$$(a+b+c)^2\Big(\frac{31}{27} - \frac{a}{a+b}\frac{(a+b+c)+5a+5b}{5(a+b+c)+a+b}\Big)^3\ge
\Big(\frac{b}{b+c} + \frac{c}{c+a}\Big)^3\left(\frac{\frac{b}{b+c}}{\frac{b}{b+c} + \frac{c}{c+a}}(b+c) + \frac{\frac{c}{c+a}}{\frac{b}{b+c} + \frac{c}{c+a}}(c+a)\right)^2.$$
If suffice to prove that $f(a,b,c) \ge 0$ for $a, b, c \ge 0; \ c = \min(a,b,c)$ where $f(a,b,c)$ is a polynomial (a long expression).

We apply the Buffalo Way. There are three possible cases:

1) If $c = 0$, we have $f(a,b, 0) = 216a(a+28b)(8a-19b)^2(a+b)^5 \ge 0$.

2) If $c > 0$ and $c \le a \le b$, let $a = c + s, \ b = c+s+t; \ s,t \ge 0$.
$f(c+s, c+s+t, c)$ is a polynomial in $c, s, t$ with non-negative coefficients.

3) If $c>0$ and $c \le b\le a$, let $b = c+s, \ a = c+s + t; \ s,t \ge 0$.
We have
\begin{align}
&f(c+s+t, c+s, c) = g(c,s,t) + 216(s+t)(29s+t)(11s-8t)^2(2s+t)^5\\
&\quad +  108(146777s^4+13368s^3t-97257s^2t^2+42286st^3+7200t^4)(2s+t)^4c
\end{align}
where $g(c,s,t)$ is a polynomial with non-negative coefficients. Note that
$$146777s^4+13368s^3t-97257s^2t^2+42286st^3+7200t^4 \ge 2\sqrt{146777 \cdot 7200} s^2 t^2 + 2\sqrt{13368\cdot 42286}s^2t^2 - 97257s^2t^2 \ge 0.$$

The desired result follows. We are done.

I have an half proof :
We have :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$

We need a lemma :

Lemma
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then  we have :
$$\sum_{cyc}0.5(a+b)^{\frac{2}{3}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$

Proof : It's Jensen's inequality apply to the concave function $x^{\frac{2}{3}}$.
Now we prove :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then  we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \frac{a+b}{2(a+b)^{\frac{1}{3}}}+\frac{b+c}{2(b+c)^{\frac{1}{3}}}+\frac{a+c}{2(a+c)^{\frac{1}{3}}}$$

Proof : The inequality is equivalent to :
$$\frac{b-a}{2(a+b)^{\frac{1}{3}}}+\frac{c-b}{2(c+b)^{\frac{1}{3}}}+\frac{a-c}{2(a+c)^{\frac{1}{3}}}\leq 0$$
Wich is true with SOS method or rearrangement inequality .
Finally we have :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then  we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}} <\frac{31}{27}$$

Conclusion :
So the real case to threat is when $b\geq a\geq c $.


A sequence whose consecutive elements get closer faster then a geometric progression is a Cauchy sequence Let $K \in (0,1)$, and $d(x_{n+1}, x_n) \leq K^n l$ where $l = d(x_0, x_1)$. That is, let $\{x_n\}$ be a sequence whose consecutive elements don't get apart faster than a geometric progression that converges (that's why $K \in (0,1)$.
How can I prove that such sequences are Cauchy sequences? I know it intuitively, as the elements keep getting closer, which is the definition of a Cauchy sequence, but I'm having a hard time to prove it formally...

To show that it is indeed Cauchy, we have for $n>m$:
$$\begin{align}d(x_m, x_n) &\le d(x_m, x_{m+1}) + d(x_{m+1}, x_{m+2}) + \cdots + d(x_{n-1},x_n)
\\~\\&\le K^ml+K^{m+1}l+\cdots + K^{n-1}l
\\~\\&< \sum_{k=m}^\infty K^kl
\\~\\&= \frac l{1-K}K^m
\end{align}$$
for sufficiently large $m$, $d(x_m, x_n)$ can be as small as we want.

Hint: A geometric series $\sum_{n\ge0}k^n$ converges for $k\in(0,1)$, hence is Cauchy.


$T$ is bounded iff $S\circ T$ is bounded. Let $X , Y, Z$ are Banach spaces. $T:X\to Y$ be a linear map and $S:Y\to Z$ be a one-one linear map which is bounded. Show that $T$ is bounded iff $S\circ T$ is bounded.

One side is trivial that I know. But how to prove that if $S\circ T$ is bounded then $T$ is bounded too.

I think I got this one.

Since $X, Y, Z$ are Banach spaces, so we can use the Closed Graph theorem. We want to show $T$ is bounded. We prove $Gr(T)$ is closed. So take $(x_{n}, T(x_{n})) \to (x,y)$. That means $x_{n} \to x$ and $T(x_{n})\to y$. We need to prove $T(x)=y$.

Now we have since $S\circ T$ is bounded $S\circ T(x_{n}) \to S\circ T(x)$. Again since $S$ is bounded we have $S(T(x_{n})\to S(y)$. So by uniqueness of limit we have $S(T(x))=S(y)$. Since $S$ is one-one $T(x)=y$.

Hope this is fine!

The result is true if $S$ is bijective, otherwise it is false, $S$ is invertible and bounded implies that $S$ is open by the open mapping theorem. We deduce that $S^{-1}$ is continuous, hence bounded. Write $U=S\circ T$, and suppose that $U$ is bounded, we have $T=S^{-1}\circ U$ is bounded as product of two bounded maps.


How to show $\frac{d}{dt}(\det(I+At)=\text{Tr}(A)$? How to show $\frac{d}{dt}\det(I+At)\mid_{t=0}=\DeclareMathOperator{\Tr}{Tr}\Tr(A)$? I can't find a easy way to prove that.

You just need to look at the coefficient before $t$ in the polynomial obtained by expanding the determinant. Now try to identify where the terms $at$ come from. You should note that all the off-diagonal elements contain $t$, so there should not be too many such terms.

You can start with $2\times 2$ or $3\times 3$ to get the idea if you still stuck.

Hint: Note that the characteristic polynomial of $A$ is

$$\det(A-tI)=t^n-\mbox{Tr}(A)t^{n-1}+O(t^{n-2}) \; .$$


Proving $ \int_1^{t}\frac{1}{x}\exp(-4(\sqrt{t+1}- \sqrt{x+1}))dx \leq \frac{c}{\sqrt{t}} $? Is there a constant $c$ such that
$$\int_1^{t}\frac{1}{x}\exp\left[-4\Big(\sqrt{t+1}- \sqrt{x+1}\Big)\right]\mathrm{d} x
\le \frac{c}{\sqrt{t}}$$
for all $t \geq 1$?
My Approach: In simulation, this holds for $c = 2$. I have tried to prove this by breaking the integral into two parts $\int_1^{\lambda t}$ and $\int_{\lambda t}^{t}$ for some $\lambda < 1$. For the first part the exponential term dominates and $1/x \leq 1$. For the second part $1/x \leq 1/\sqrt{\lambda t}$ and the integral can be calculated. Any ideas?

HINT.-The answer is YES.
We have equivalently $$\frac{\sqrt t}{e^{4\sqrt{1+t}}}\int_1^t\frac{dx}{xe^{\sqrt{x+1}}}\lt c$$ and
$$\int_1^t\frac{dx}{xe^{\sqrt{x+1}}}\lt\int_1^{\infty}\frac{1}{x^2}dx=1$$
and since  $\dfrac{\sqrt t}{e^{4\sqrt{1+t}}}\lt1$ we can finish.
(one has actually $\dfrac{\sqrt t}{e^{4\sqrt{1+t}}}$ is very small and even its integral between $1$ and $\infty$ is approximately equal to $0.003968$).
$$*****************************************************************************$$
I have made a slip, noticed by River Li, and, by regulation, I have not been able to delete my answer as I wanted because it has already been accepted.
I have no willingness now to work with this problem in detail but I can give the following HINT: we can reduce the problem to a single function $F(x)=f(x)g(x)$ where
$$f(x)=\frac{\sqrt x}{e^\sqrt {x+1}}\\g(x)=\int_1^x\frac{e^\sqrt{t+1}}{t}dt$$
Now every numerical value of $F(a)$ for $a\ge 1$ is very small (despite the divergence of $g(x)$ and due to the fast convergence to $0$ of $f(x)$ and it can be ensured that
$F (x)\le 0.005$ (but this value of the constant $ c $ is not the best)

$$I=e^{-4\sqrt{t+1}}\int_1^t\frac{e^{4\sqrt{x+1}}}{x}dx$$
the integrand is:
$$f(x)=\frac{e^{4\sqrt{x+1}}}{x}$$
and we know that:
$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$
so for a positive $x$ (which we have here) we can say that:
$$e^x<\sum_{n=0}^N\frac{x^n}{n!},N<\infty$$
so for our:
$$e^{4\sqrt{x+1}}<1+4\sqrt{x+1}$$
so we can say that within our domain:
$$f(x)<\frac{1+4\sqrt{x+1}}{x}$$
since this function is decreasing for an increasing $x$ i.e. $x\in[1,t],f'(x)<0$
we know that the minimum is at $f(t)$ and the maxima is at $f(1)$, using these to approximate the integral.

Alternatively, looking at the function:
$$x\in[1,t],g(x)=\sqrt{t+1}-\sqrt{x+1}$$
we notice that $g(x)\ge0\therefore \exp(-4g(x))\le1$


infinity in the extended reals I learned about the extended reals a month ago in my measure theory class, and while the syntax for handling $+\infty$ or $-\infty$ is well defined, the mathematical properties of $+\infty$ in the extended real numbers hasn't been distinguished from the $+\infty$ that is derived from the naturals and used for sequences among other things.
To make things precise I shall denote $\infty_{\mathbb{N}}$ to denote the infinity used for sequences and $\infty_{\mathbb{\bar{R}}}$ for the infinity used in the extended reals.
Now, I shall state what is common to both of them:

They are both non-computable.
The notion of a definite length or magnitude can't be made precise for either of them.
We can define a syntax of arithmetical rules that's similar for both of them. Ex:

$\forall x,y \in \mathbb{R}, \infty_{\mathbb{\bar{R}}} +x = y+ \infty_{\mathbb{\bar{R}}}$
$\forall x,y \in \mathbb{N}, \infty_{\mathbb{N}} +x = y+ \infty_{\mathbb{N}}$
However, $\infty_{\mathbb{N}}$ for sequences like $(x_n)_{n=1}^N$ where $x_N = \sum_{n=1}^{N} 2^{-n}$ can be defined within the context of a while loop that never ends. This can be made clear by the following pseudo-code:
while $N \in \mathbb{N}$:
$x_N = \sum_{n=1}^{N} 2^{-n}$
$N:=N+1$
This loop, which never terminates, is what computer scientists call an infinite loop. Now, what happens if $u_n = n \pi$ and $y_n = 2n$.
At every increment of the loop, $y_n$ is an integer whereas $u_n$ is irrational. Yet, an analyst would say that $\lim_{n\to\infty} u_n = \lim_{n\to\infty} y_n$. However, given that $\infty_{\mathbb{N}}$ originates from the positive integers does $\infty_{\mathbb{N}}$ simply refer to the set of naturals whose magnitudes we haven't bother to make precise?
Likewise, an analyst would say that:
$\lim_{n\to\infty} \frac{n+1}{n+2}=1$
but it would be more precise to say that $\lim_{n\to\infty} \frac{n+1}{n+2}=1 +\epsilon$ where $|\epsilon|$ is the smallest positive rational that's greater than zero. I think the same could be said about all other limits using $\infty_{\mathbb{N}}$. The epsilon almost vanishes but it doesn't. Unless  $\infty_{\mathbb{N}}$ is a mathematical object that doesn't originate from the positive integers.
But if that's the case, why have mathematicians defined arithmetical operations for $\infty_{\mathbb{N}}$?
Now, as I have tried to reflect on $\infty_{\mathbb{\bar{R}}}$ it appears that there are problems with $\infty_{\mathbb{N}}$. These problems aren't unrelated but in my opinion must be resolved first. In fact, I don't think they are separate questions.
But, if it's possible to resolve the questions I've asked, I think it's natural to ask a further question. How is $\infty_{\mathbb{\bar{R}}}$ semantically different from $\infty_{\mathbb{N}}$? By this, I mean, what are its precise mathematical properties that distinguish it from $\infty_{\mathbb{N}}$?
Right now, if I go through the list of arithmetical operations that are defined for $\infty_{\mathbb{\bar{R}}}$, it appears that these aren't sufficient to show that $\infty_{\mathbb{N}} \neq \infty_{\mathbb{\bar{R}}}$.
Note 1: 5xum points out that my notion of a smallest positive rational that's greater than zero is clearer within the context of non-standard analysis, and not within standard analysis. As a result of this discussion I shall try to learn non-standard analysis.
Note 2: This is not clear in the two answers so far but Stefan Perko shows in the comments below that $\infty_{\mathbb{N}} = \infty_{\mathbb{\bar{R}}}$ for the following reason:
Given that $\infty :=\infty_{\mathbb{\bar{R}}}$ is the greatest element of $\mathbb{\bar{R}}$. Assume $\infty_{\mathbb{N}} \in \mathbb{\bar{R}}$ such that $\infty_{\mathbb{N}} \geq n$ for all naturals $n$. Then $\infty_{\mathbb{N}} \geq x$ for all reals $x$ by the Archimedian property. So it must be the same as $\infty$ (if it was a real there would be a greater real).

The thing is that $\infty_{\mathbb N}$ isn't really an element of any set. When talking about $\infty$ in sequences, saying $$\lim_{n\to\infty} a_n=L$$ simply means that for any $\epsilon>0$, there exists some $N\in\mathbb N$ such that $|a_n-L|<\epsilon$ if $n>N$.

So there is no $\infty_{\mathbb N}$ to speak of at all. The $\infty$ there is simply a symbol that describes a behavior, while in the extended reals, $\infty$ is an actual element of the set.

$\infty_{\mathbb{N}}$ and $\infty_{\overline{\mathbb{R}}}$ are not different notions - they're both just symbols standing in for "bigger than anything". Either can be used as an element of a set (for example, in the extended reals) but neither can be treated as "really" a number. For example, in the extended reals, no one ever tries to define $\infty - 1$, or (even worse) $\infty - \infty$. So think of the extended reals as the reals together with a couple of random objects - say, a horseshoe and a chipmunk - and a list of rules for what to do if you run into one of them in your calculations.

Another thing: "Origin" isn't really a thing to most mathematicians. The fact that your $\infty_{\mathbb{N}}$ is derived from the naturals and $\infty_{\overline{\mathbb{R}}}$ from the reals doesn't mean they're different. For example, $\pi$ can be derived from a circle or from the sum $\sum_{i=0}^{\infty}\frac{4(-1)^i}{2i+1}$; that doesn't mean we have two different versions of $\pi$, just two different ways of talking about the same thing.

And finally: $\infty_{\mathbb{N}}$ isn't just a stand-in for a natural we haven't specified yet - it's a very precise notion. $\lim_{n \to \infty}f(n) = L$ means "as $n$ grows without bound, $f(n)$ gets arbitrarily close to $L$". That "without bound" is $\infty_{\mathbb{N}}$ - it's not saying the magnitude we're interested in hasn't been specified, it's saying that there is no fixed magnitude.


Average length of the longest segment This post is related to a previous SE post  If a 1 meter rope ….    concerning average length of a smallest segment.

A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.
My answer is 11/18.  Here is how I do it:

Here we have two independent random variables $X,Y$, both uniform on $[0,1]$. Let
$A=\min (X,Y), B=\max (X,Y)$ and $C=\max (A, 1-B, B-A)$. First we want to find the probability
density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then
$$ F_C(a) = P(C\le a)=P(A\le a, 1-B\le a, B-A\le a).$$  By rewriting this probability as area in the unit square, I get
$$F_C(a)=\left\{\begin{array}{ll} (3a-1)^2 & \frac{1}{3}\le a\le \frac{1}{2}\\ 1-3(1-a)^2 & \frac{1}{2}\le a\le 1\end{array}\right.$$ from which it follows that
$$f_C(a)=\left\{\begin{array}{ll} 6(3a-1) & \frac{1}{3}\le a\le \frac{1}{2}\\ 6(1-a) & \frac{1}{2}\le a\le 1\end{array}\right.$$ Therefore
the expected value of $C$ is
$$\int_{1/3} ^{1/2}6a(3a-1) da+\int_{1/2} ^{1}6a(1-a) da= \frac{11}{18}.$$

My questions are:

(A) Is there a "clever" way to figure out this number 11/18?

(B) What is the answer if the rope is divided into $n>3$ segments?

The answer to (B) is actually given in both Yuval Filmus' and my answers to the question about the average length of the shortest segment.  It's $$\frac{1}{n} H_n,$$ where $H_n = \sum_{k=1}^n \frac{1}{k},$ i.e., the $n$th harmonic number.

"Clever" is of course subjective, but here's an argument for (A) in the $n$-piece case.  At least there's only one (single-variable) integration in it. :)
If $X_1, X_2, \ldots, X_{n-1}$ denote the positions on the rope where the cuts are made, let $V_i = X_i - X_{i-1}$, where $X_0 = 0$ and $X_n = 1$.  So the $V_i$'s are the lengths of the pieces of rope.

The key idea is that the probability that any particular $k$ of the $V_i$'s simultaneously have lengths longer than $c_1, c_2, \ldots, c_k$, respectively (where $\sum_{i=1}^k c_i \leq 1$), is $$(1-c_1-c_2-\ldots-c_k)^{n-1}.$$  This is proved formally in David and Nagaraja's Order Statistics, p. 135.  Intuitively, the idea is that in order to have pieces of size at least $c_1, c_2, \ldots, c_k$, all $n-1$ of the cuts have to occur in intervals of the rope of total length $1 - c_1 - c_2 - \ldots - c_k$.  For example, $P(V_1 > c_1)$ is the probability that all $n-1$ cuts occur in the interval $(c_1, 1]$, which, since the cuts are randomly distributed in $[0,1]$, is $(1-c_1)^{n-1}$.

If $V_{(n)}$ denotes the largest piece of rope, then
$$P(V_{(n)} > x) = P(V_1 > x \text{ or } V_2 > x \text{ or } \cdots \text{ or } V_n > x).$$  This calls for the principle of inclusion/exclusion.  Thus we have, using the "key idea" above,
$$P(V_{(n)} > x) = n(1-x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \cdots + (-1)^{k-1} \binom{n}{k} (1 - kx)^{n-1} + \cdots,$$
where the sum continues until $kx > 1$.

Therefore,
$$E[V_{(n)}] = \int_0^{\infty} P(V_{(n)} > x) dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \int_0^{1/k} (1 - kx)^{n-1} dx = \sum_{k=1}^n \binom{n}{k} (-1)^{k-1} \frac{1}{nk} $$
$$= \frac{1}{n} \sum_{k=1}^n \frac{\binom{n}{k}}{k} (-1)^{k-1} = \frac{H_n}{n},$$
where the last step applies a known binomial sum identity.

A similar, but a bit more complex problem, that highlights how important it is to understand the exact wording of a formulation:

cut a rope (or break a rod) at a random point. Set aside a piece. Cut (or break) the remaining piece at another random point.

Question is the same (expected length of the longest segment).

That's how I initially understood the problem..
Got stuck in algebra and closed-form anti-derivatives..

Same principle, heavier calculations.
I got an answer $2\ln2 - \ln3 + 3/8 \approx 0.663$, that matches what a MatLab script gives for $1000$ tests.


Proving a group of order $77$ has a subgroup of order $7$ without Sylow theorem. The question is

  Show if $G$ has order 77 then $G$ has a subgroup of order 7. Without using Sylow Theorems.

Attempt sketch:

Let $x \in G$. By Lagrange's theorem the order of $x$ is either $1, 7, 11$. Suppose $x \neq e$ Then $x$ has order $7$ or $11$. Now suppose $|x| = 7$. Then $x$ is the generator of a group of order $7$ and we are done. Suppose now that there does not exist an element of order $7$ in G. Then $G$ cannot have order $77$ since $G$ would be cyclic of order $11$ Hence $G$ has an element of order $7$ and thus a subgroup of order $7$.

Please excuse the horrible way this is written.

All elements of $G$ must have orders $1$, $7$, $11$, $77$ by Lagrange's Theorem. If $G$ has an element of order $7$, then we're done. If it has an element $a$ of order $77$, then it also has an element of order $7$, namely $a^{11}$.

Therefore we may assume that all elements besides $e$ have order $11$. In that case, $G$ is the union of several subgroups of order $11$, whose pairwise intersections are all $\{e\}$. Thus the cardinality of $G - \{e\}$ must be a multiple of $10$. But this number is actually $76$, a contradiction.

Let $a(\neq e)\in G$ then order of an element divides order of the group ,so possible values of $o(a)=1,7,11,77$.

Now it is not possible that $G$ has all elements of order $11$ because if so then let $o(a)=11,o(b)=11$;take $H=\langle a \rangle  ,K=\langle b \rangle  $ as $o(H)=o(K)=11 $ and $H\cap K=\{e\}$ then $o(HK)=\dfrac{o(H)o(K)}{o(H\cap K)}=121$.Now $HK\subseteq G$ so do you see a contradiction?

Thus $G$ must have atleast one element of order $7$.

If $o(a)=7\implies $ we have a cyclic group generated by $\langle a \rangle  $ of order $7$.

If $o(a)=77$.Then $G$ is cyclic and so $G$ has a unique subgroup corresponding to its each divisor.

DONE.


Diophantine equation of second degree $x^2+y^2+z^2=2t^2$ How to solve this Diophantine equation of second degree?
Solution, references, anything. I will be very grateful.

$$x^2+y^2+z^2=2t^2$$

Thank you.

You can start with $x,y,z$ any Pythagorean triple, then take $t=z$.  There are more.

For every $t$, $x^2+y^2+z^2=2t^2$, defines a 3-Dimensional sphere with radius $\sqrt{2}t$. Thus, every point on the surface of the sphere is a solution for the diophantine equation. If you are interested in characterizing all the solutions, this should help
\begin{align}
r&=\sqrt{2}t \\
x&=r\sin{(\theta)}\cos(\phi) \\
y&=r\sin{(\theta)}\sin(\phi) \\
z&=r\cos{(\theta)} \\
\end{align}
where $0\leq \theta \leq \pi$ and $0 \leq \phi \leq 2\pi$. Thus, substituting any value for $t$ , $\theta$ and $\phi$ in the above mentioned equations will give you a solution.


Expectation Poisson Distribution A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and $10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. What is the expected amount paid to the company under this policy during a one-year period?

I know how to calculate the expectation and what the series is. I'm having problems with the summations. I know it should involve:

$$\sum_{k=2}^{+\infty} \frac{(1.5)^k}{k!}$$

Let $X$ be the number of snowstorms occurring in the given year and let $Y$ be the amount paid to the company.  Call one unit of money $\$ 10{,}000$.

Then $Y$ takes the value $0$ when $X=0$ or $X=1$, the value  $1$ when $X=2$, the value $2$ when $X=3$, etc..

The expected payment is
$$\eqalign{
\Bbb E(Y)
&=\sum_{k=2}^\infty (k-1)P[X=k]\cr
&=\sum_{k=2}^\infty (k-1) e^{-1.5}{(1.5)^k\over k!}\cr
&=\sum_{k=1}^\infty (k-1) e^{-1.5}{(1.5)^k\over k!}\cr
&=
\sum_{k=1}^\infty k e^{-1.5}{(1.5)^k\over k!}
-\sum_{k=1}^\infty  e^{-1.5}{(1.5)^k\over k!}\cr
&=\underbrace{ \sum_{k=0}^\infty k e^{-1.5}{(1.5)^k\over k!}}_{\text{mean of } X} -
\biggl(-e^{-1.5}+\underbrace{\sum_{k=0}^\infty  e^{-1.5}{(1.5)^k\over k!}}_{=1}\biggr)\cr
&=1.5+e^{-1.5}- 1\cr
&=0.5+e^{-1.5}\cr
&\approx .7231\,\text{units}.
}$$

If they got $\$10,000$ every time, it would be $\$10,000\cdot(1.5)$.  From that subtract $\$10,000$ times the probability that there's exactly one such storm, which is $1.5e^{-1.5}$.  So you get
$$
\$15,000 - \$10,000\cdot1.5e^{-1.5}.
$$


Taylor Polynomial of degree 5 centered around x=x I was wondering if this was possible. For a taylor series polynomial shouldn't it be centered around $x_0$ instead of $x$?

The question was asking for a Taylor polynomial of degree five in the form of $f(x+h,y)$ in $x+h$ centered around $x$, keeping $y$ as a constant, so I wrote something like:

$$f(x+h)\approx\sum_{n=0}^{5}\frac{f^{(n)}(x)}{n!}(x+h)^n$$

But I think it should be something like this:

$$f(x+h)\approx\sum_{n=0}^{5}\frac{f^{(n)}(h)}{n!}(x+h)^n$$

Is this correct or am I missing something?

The idea is to treat $x+h$ as a variable.  Let's let $z=x+h$. Now

$$f(z,y)\approx\sum_{n=0}^5\frac{f^{(n)}(x,y)}{n!}(z-x)^n=\sum_{n=0}^5\frac{f^{(n)}(x,y)}{n!}h^n.$$

The last expression comes from substituting $x+h$ for $z$.


Prove $(A \cup B)' = A' \cap B'$ I would like some assistance in verifying this proof?

(I understand the last conjecture about "symmetry" is probably shaky, but I just want to know if the first part is right since going backwards will more than likely be the same as is usually the case in such proofs).

We seek to prove $(A \cup B)' = A' \cap B'$.

Suppose we have the universal set $U$.

We know $x \in (A \cup B)'$.

$\implies x \in U - (A \cup B)$.

Thus $x \in U \land x \notin (A \cup B)$.

$x \notin A \lor x \notin B \implies x \notin (A \cap B)$.

So $x \in U \land (x \notin (A \cap B))$.

$\implies x \in U \land (x \notin A \land x \notin B) $.

$\implies (x \in U \land x \notin A) \land (x \in U \land x \notin B)$

$\implies x \in (U-A) \land x \in (U-B)$

$\implies x \in A' \land x \in B'$

$\implies x \in A' \cap B'$

By the symmetry of the situation we see that $x \in A' \cap B' \implies x \in (A \cup B)' $.

Therefore the sets are subsets of one another, therefore they are equal.

Your proof is good.

Instead of saying "by the symmetry of the situation", which is somewhat ambiguous, you could replace the implication arrows to go both ways, i.e. $\Leftrightarrow$. That is the standard way to say what you mean. Just make sure that the proof is then readable backwards as well.

However, if you are unsure, it is easier to avoid mistakes if you write the forward case $\Rightarrow$ and backward case $\Leftarrow$ separately.

We can prove using the contradiction method.

Let us assume there is an $x$ such that $x\in(A\cup B)'$ and $x\notin (A' \cap B')$.
As $x \in (A\cup B)'$, thus $x\notin (A \cup B)$. This would mean that, $x\notin A$ and $x\notin B$. Thus, $x\in A'$ and $x\in B'$. Which implies that $x\in (A'\cap B')$. But that is a contradiction.

Thus there exists NO element $x$ such that $x\in(A\cup B)'$ and $x\notin (A' \cap B')$. Thus both sets are equal.


Product of Polynomials is Zero How would I prove that if two polynomials (over reals) have a product of zero, then one of those polynomials must be zero?

A different proof.  All nonzero polynomials have a high order term $a_nx^n$ with $a_n\ne0$, $n\ge 0$ integer. (High order means that all terms in the polynomial have an exponent $\le n$; for example in $a+bx+cx^2$ the high order term is $cx^2$.) Suppose both of your polynomials have high order terms, $a_nx^n$ and $b_mx^m$ respectively, with $a_n\ne 0$ and $b_m\ne 0$. Then the term $a_nb_mx^{n+m}$ appears in the product, and what is more it is the only term where $x$ appears with exponent $n+m$.  However, by our hypothesis $a_nb_m$ has to be zero.  The product of two nonzero reals is nonzero.  This is the desired contradiction.

The proof is easy using any computer language, versus using a mathematical language.  Note also, this proof resolves as something being true, not something which must be further observed to be true, as would be the case with a proof resembling    0 = p(a) * p(b) ...

proof =  ( (0==p(a)*p(b)) == (0==p(a)||0==p(b)) ) ? true : false;

This method examines the product as well as the terms.


Maximum product of distances to the vertices of a square Suppose a unit square $ABCD$. Where is point $P$, inside or on the square, such that $Q=PA\cdot PB\cdot PC\cdot PD$ is maximized?

I solved this numerically with a computer, and curiously and counterintuitively (after all, there's symmetry), the answer is not the middle of the square but rather the midpoint of any of the four edges (for the former $Q=\frac{1}{4}$ while for the latter $Q=\frac{5}{16}$).

Two questions arise: firstly, how does one solve this problem using calculus (without a computer), and secondly, is there a nice explanation for why the maximum does not occur at the center of the square?

Starting with the second question, I would intuitively put the point in the middle if I were to minimize the distance to all four vertices. Perhaps your intuition betrayed you?

Nonetheless, onto the first question:

Assume that the square's vertices are the points $(0,0), (1,0), (1,1), (0,1)$. Therefore a point $P = (x, y)$ inside the square has $0 \leq x \leq 1; 0 \leq y \leq 1$. Also, the distance of the point to the vertices can be found with the pythagorean theorem.

Distance to $(0, 0)$ is $\sqrt{x^2 + y^2}$;
Distance to $(1, 1)$ is $\sqrt{(1-x)^2 + (1-y)^2}$;
Distance to $(0, 1)$ is $\sqrt{x^2 + (1-y)^2}$;
Distance to $(1, 0)$ is $\sqrt{(1-x)^2 + y^2}$;

Thus you want to maximize the function

$$D(x, y) = \sqrt{x^2 + y^2}\cdot\sqrt{(1-x)^2 + (1-y)^2}\cdot\sqrt{x^2 + (1-y)^2}\cdot\sqrt{(1-x)^2 + y^2}$$

We can use derivatives to find all the candidate points for the maximum and then find it; If we derive $D$ and find the zeroes of its gradient, we see that its only zero is at $(\frac12, \frac12)$. We then check if it is a maximum, and we see it is not. Now, the square is limited and compact and our $D$ function is continuous, thus because of the Weierstrauss Theorem, $D$ must have a maximum on the square. If we could not find it with our test, it is because the function must be growing as we step away from the center; therefore the maximum occurs at the border of the square, in one edge.

Because of the symmetry of our function, one can understand that the values at the edges are the same, so the $4$ edges will have a point that maximizes the distance function. One can then use intuition to understand that the farther away from the two vertices of that edge, the better. If "intuition" is not good enough for you, one can then again use calculus. Let me know if you think it is needed.


Composition of two axis-angle rotations Please note that I am not referring to Euler angles of the form (α,β,γ). I am referring to the axis-angle representation, in which a unit vector indicates the direction axis of a rotation and a scalar the magnitude of the rotation.

Let $(\hat{n_1},\theta_1)$ refer to the first rotation and $(\hat{n_2},\theta_2)$ refer to the second rotation. What is the value of the first rotation followed by the second rotation, in axis-angle representation?

I understand that the composition of two rotations represented by quaternions $q_1$ and $q_2$ is equal to their product $q_2q_1$. Is there a way to find the composition of axis-angle rotations (without having to convert them to quaternions, multiply them, and convert them back to axis-angle) in a similar manner? Is there a simplified formula for this operation?

The formula is given in this excerpt from a journal paper.  It was discovered by the French mathematician Olinde Rodrigues in 1840, which was before the invention of vectors or even quaternions (which were invented before vectors).

The composition of $\alpha\hat{l}$ and $\beta\hat{m}$ (where the second rotation is applied and then the first is applied) is given by $\gamma\hat{n}$, where $\cos\frac{\gamma}{2} = \cos\frac{\alpha}{2}\cos\frac{\beta}{2} - \sin\frac{\alpha}{2}\sin\frac{\beta}{2}\hat{l}\cdot\hat{m}$ and $\sin\frac{\gamma}{2}\hat{n}=\sin\frac{\alpha}{2}\cos\frac{\beta}{2}\hat{l}+\cos\frac{\alpha}{2}\sin\frac{\beta}{2}\hat{m}+\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\hat{l}\times\hat{m}$.

As a sanity check, it's easy to see that when $\hat{l}=\hat{m}$, then it's easy to see that $\gamma=\alpha+\beta$ and $\hat{n}= \hat{l}=\hat{m}$.

In any case, these formulas are proven in detail in this chapter of Simon Altman's book "Rotations, Quaternions, and Double Groups", but it basically boils down to this spherical triangle:

See also this related result proven by William Rowan Hamilton after he invented quaternions.

Look at the following link: Axis–angle representation


Complexification of manifolds Let $M$ be a smooth manifold and $TM\otimes \mathbb C$ be its complexified tangent bundle. Is there always a complex manifold $N$ s.t  $TM\otimes \mathbb C$ is its tangent bundle? If yes, is $M$ embedded as a smooth submanifold in $N$?

If the real dimension of $M$ is odd, so is the real dimension of $TM\otimes \mathbb{C}$.


Is this procedure on Harmonic Series correct? $\require{cancel}$I need to evaluate the following limit:
  $$\lim_{n\to\infty}\sum_{np}^{nq}\frac1k \quad \text{where $\quad p\geq1,q\geq1$}$$

The way I tackled this problem is as follows:

Recall the relation:

$$\sum_{k=1}^n\frac1k = \ln(n)+\gamma+o(1) \quad \text{as $n\to\infty$}$$

Therefore, assuming $q\geq p$, the initial limit can be expressed as:

$$\begin{align*}
\lim_{n\to\infty}\sum_{np}^{nq}\frac1k &= \lim_{n\to\infty}\left(\sum_{k=1}^{nq}\frac1k-\sum_{k=1}^{np-1}\frac1k\right)\\
&=\lim_{n\to\infty}\left(\ln(nq)+\cancel{\gamma}-\ln(np-1)\cancel{-\gamma}+o(1)\right)\\
&= \lim_{n\to\infty}\ln\left(\frac{nq}{np-1}+o(1)\right)=\ln\left(\lim_{n\to\infty}\left(\frac{\cancel{n}q}{\cancel{n}\left(p-\frac1n\right)}+o(1)\right)\right)=\boxed{\ln\left(\frac{q}{p}\right)}
\end{align*}$$

  Are the result and procedure correct? Also, do I need to prove the initial identity (for a Homework Assignment)?

The approach in the OP is fine.

  I thought it might be instructive to present another way forward that doesn't rely on the asymptotic expansion of the Harmonic number, but rather relies on only straightforward analysis and Riemann sums.  To the end, we proceed.

We simply write

$$\begin{align}
\sum_{k=np}^{nq}\frac{1}{k}&=\sum_{k=0}^{n(q-p)}\frac{1}{k+np}\\\\
&=\frac{1}{np}\sum_{k=0}^{n(q-p)}\frac{1}{1+\frac k{np}}\tag 1
\end{align}$$

We recognize that the right-hand side of $(1)$ is the Riemann sum for

$$\begin{align}
\frac1p\int_0^{q-p}\frac{1}{1+x/p}\,dx&=\log\left(1+\frac{q-p}{p}\right)\\\\
&=\log(q/p)
\end{align}$$

Hence, we see that the coveted limit is

  $$\lim_{n\to \infty}\sum_{k=np}^{nq}\frac{1}{k}=\log(q/p)$$

as expected!

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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  $\ds{\lim_{n \to \infty}\sum_{np}^{nq}{1 \over k}:\ {\large ?}
\quad\mbox{where}\quad p \geq 1\,,\ q \geq 1}$.

\begin{align}
\lim_{n \to \infty}\sum_{np}^{nq}{1 \over k} & =
\lim_{n \to \infty}\pars{\sum_{k = 1}^{nq}{1 \over k} -
\sum_{k = 1}^{np - 1}{1 \over k}} =
\lim_{n \to \infty}\pars{H_{nq} - H_{np - 1}}\qquad
\pars{~H_{m}:\ Harmonic\ Number~}
\\[5mm] & =
\lim_{n \to \infty}\bracks{\Psi\pars{nq + 1} - \Psi\pars{np}}\qquad
\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] &
\lim_{n \to \infty}\bracks{\ln\pars{nq + 1} - \ln\pars{np}} =
\bbx{\ds{\ln\pars{q \over p}}}
\end{align}


Let $m,n \in\mathbb Z$. If $m \le n \le m$ then $m = n$. I an learning proofs with $\mathbb{N}$.
Here are my axioms:

a)If $m,n \in\mathbb N$ then $m + n \in\mathbb N$

b)If $m,n \in\mathbb N$ then $mn \in\mathbb N$

c) $0 \notin\ \mathbb N$

d) For every $m \in\mathbb Z$, we have $m \in\mathbb N$ or $m = 0$ or $-m \in\mathbb N$

Definition: $m > n  \Leftrightarrow m - n \in\mathbb N$.

By the way, I would greatly appreciate if someone could please explain to me why this is. My strategy is to use a contradiction.

Proposition 1: (that I have proven)

For $m \in\mathbb Z$, one and only one of the following is true: $m \in\mathbb N$, $-m \in\mathbb N$, $m = 0$.

I have proven $1 \in\mathbb N$ and: Let $m,n,p \in\mathbb Z$. If $m < n$ and $n < p$ then $m < p$ by deriving $p - m = (p - n) + (n - m) \in\mathbb N$.

I have also proven: For each $n \in\mathbb N$ there exists $m \in\mathbb N$ such that $m > n$

Proposition 2:

Let $m,n \in\mathbb Z$. If $m \le n \le m$ then $m = n$.

I get confused because of the equality within the $\leq$. How could I approach this? By separating $<$ and $=$ ?

Any hints would be greatly appreciated.

First proposition

If $m \in \mathbb N$ and $-m \in \mathbb N$ then $ 0 \in \mathbb N$, absurd.
If $m \in \mathbb N$ and $ m = 0$, absurd.
Similar to $-m$.

Therefore it folows the affirmation.

Second proposition

If $m \leq n \leq m$ then $m - n \in \mathbb Z$ and by the first proposition only one of those $3$ holds, then if $m - n \in \mathbb N$ then $m>n$ by definition, which is a contradiction. Similar if $n-m \in \mathbb N$.

Thus the only possibility is $m - n = 0$.

Suppose $m \ne n$. That means: $m < n < m$. Then by definition, $m−n∈N$ and $n-m∈N$. By axiom (a): $0=m-n+n-m∈N$. Came to contradiction by axiom (c), so $m$ is indeed equal to $n$.


Solve $y '' + 3y ' + 2y = \frac{1}{e^{x} + 1} $ I am trying to solve this differential equation.
$$y'' + 3y' + 2y = \frac{1}{e^{x} + 1}$$

I know that I have to solve $$x^2+3x+2=0$$

when the solutions of this equation are $(x_1,x_2) = (-2,-1)$

so $$y_0(x) = c_1 e^{-2x}+c_2e^{-x}$$

Then I am looking for a solution of $y(x) = K \frac{1}{e^x +1}$ and I calculate $y'$ and $y''$.

The problem is that I end up to nowhere. Can someone help me?

let homogeneous solutions of this problem y1 and y2. and y1=exp(x) and y2=exp(2x)
and let the nonhomogeneous solution of this problem yp.
and let yp u1*exp(x) +u2*exp(2x)
then, wronskian of exp(x) and exp(2x) = exp(3x)

then, differentiation of u1 = -exp(-3x)exp(2x)(1/(exp(x)+1))
and differentiation of u2 = exp(-3x)exp(x)(1/(exp(x)+1))

u can integrate differentiation of u1 and u2 and u can get u1, u2

finally, the solution is c1*y1+c2*y2+yp


Need help understanding a paragraph from a book "If in the obvious equalities $(k+1)^3−k^3=3k^2+3k+1$, for the different values $k=1,2,…,n−1$, we add the left and the right sides separately, we obtain the equation $n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$, where $σ_n=1^2+2^2+…+(n−1)^2$."

I'm stuck trying to understand what the author has done in the paragraph below. Is it possible to explain it using high school-level math? For instance, what is meant with "for the different values $k=1,2,…,n−1$, we add the left and the right sides separately"? I don't understand what he does after that either.

This equality

$$(k+1)^3−k^3=3k^2+3k+1,$$

is true for every $k$, in particular it is true when $k=1$,$k=2$,etc... Writing all these equations, we get

$$\begin{array}{rcl}
(1+1)^3−1^3&=&3\cdot 1^2+3\cdot 1+1\\
(2+1)^3−2^3&=&3\cdot 2^2+3\cdot 2+1\\
&\vdots &\\
((n-2)+1)^3−(n-2)^3&=&3\cdot (n-2)^2+3\cdot (n-2)+1\\
((n-1)+1)^3−(n-1)^3&=&3\cdot (n-1)^2+3\cdot (n-1)+1\\
\end{array}$$
Then you just sum all these equations (right hand sides with right hand sides, left hand sides with left hand sides) to find

$$\color{red}{2^3}-1^3+\color{blue}{3^3}-\color{red}{2^3}+4^3-\color{blue}{3^3}+\ldots +\color{green}{(n-1)^3}−(n-2)^3+n^3−\color{green}{(n-1)^3} = n^3-1$$

for the left hand side (note that this is called a telescopic sum). Now for the other side of the equation, note that
$$3\cdot 1^2 + 3 \cdot 2^2+ \ldots +3\cdot (n-2)^2+3\cdot (n-1)^2 = 3 \sigma_n$$
where $σ_n=1^2+2^2+…+(n−1)^2.$ Furthermore
$$3\cdot 1 + 3 \cdot 2+ \ldots +3\cdot (n-2)+3\cdot (n-1)= 3(1+2+\ldots +(n-1)) = 3\frac{n(n-1)}{2}. $$
(see the note for a proof of this fact) and
$$\underbrace{1+ 1+\ldots +1 + 1}_{(n-1) \text{ times}} = n-1.$$ So you finally get
$$n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$$
NOTE: For any $k \in \mathbb{N},$ we have
$$\begin{array}{rcl}2(1+2+\ldots+(k-1)+k) &=& (1+2+\ldots+k)+(k+(k-1)+\ldots+2+1)\\ &=& (1+k)+(2+(k-1))+\ldots +((k-1)+2)+(k+1)\\ &=& k(k+1)\end{array}$$

He performed a summation over the given value of k on each side. You may want to look into series to get a clearer idea.


What is the cardinality of the set of all non-measurable sets in $\Bbb R^n$? The cardinality of the set of all measurable sets in $\Bbb{R}^n$ can be shown to be the same as the power set of $\Bbb{R}$ by looking into Cantor set.

  Denote $M=$$\{$$Ω⊆\Bbb{R}^n:Ω$ is measurable$\}$, then $card(M)≤card(2^\Bbb{R})$. The Cantor set is measurable, has measure zero and has the same cardinality of $\Bbb{R} ⇒$ every subset of Cantor set is measurable $⇒ card(M)≥card(2^\Bbb{R} )⇒card(M)=card(2^\Bbb{R} )$.

I am wondering what is the cardinality of the set of all non-measurable sets in $\Bbb{R}^n$? I know a theorem guarantees every non-empty open sets in $\Bbb{R}$ has a non-measurable subsets, is this useful to show the cardinality of all non-measurable sets in $\Bbb{R}^n$? Thank you!

Given that $(0,1)$ has a non-measurable subset, adding points in $(1,2)$ can't make it measurable.  There are $2^{\mathfrak c}$ subsets of $(1,2)$, so take your non-measurable subset of $(0,1)$ union each subset of $(1,2)$.

EDIT: I am trying to do a Schroeder Bernstein:
Let N be the collection of non-measurable subsets.
1)There are at most $2^{\mathbb c}$ non-measurable subsets, since there are at most $2^{\mathbb c}$ subsets. This is the point I was trying to make in 1), though not in the clearest way.
2) There are at least $2^{\mathbb c}$ non-measurable subsets, by a translation argument.

1)Well, the total number of  subsets of $\mathbb R^n$ is $|2^{|\mathbb R|}$| and the cardinality of all open sets is $|\mathbb R|$ . Now you have an expression $|A|=|B \cup C|$, where A represents all sets and $B$ represents all measurable sets and $B \cap C = ${}. and you know two infinite cardinalities. Now, choose your set theory and do your cardinal arithmetic. EDIT: this was not a clear way of making

*So $|N|\leq 2^{\mathbb c} $

2)Let's work in $\mathbb R^1$. I claim there are $2^{\mathbb R}$ non-measurable subsets.
  Let $T$ be non-measurable. Then, for every non-zero Real a, the set $T+a$:={$t+a: t \in T$} is also non-measurable; if it was measurable, then $T=T+a-a$ would be the translation of a measurable set and so measurable. Since there are $\mathbb R$ choices for $a$, this means there are at least $2^{\mathbb R}$ non-measurable subsets of the Reals.

** So $|N| \geq 2^{\mathbb c} $

From *, ** , and Cantor-Shroeder-Bernstein, there are $2^{\mathbb c}$ non-measurable subsets.

EDIT: My apologies, this is unifinished, I don't have time at the moment to
finish it. IfIa m unable to finish it /correct it in a few days, I will just
delete it.


Related Rates - Differentiation An airplane is flying at an altitude of 8 miles and passes over a radar station. When the airplane is 12 miles from the base of the station, the radar detects that its horizontal distance is changing at a rate of 320 mph. Find how fast the airplane is flying at this point in time.

The question asks to find the speed of the airplane. I know $\displaystyle speed = \frac{distance}{time}$. And, in this case, if I were to draw a right triangle with the radar station and airplane, I get the distance as $\sqrt{80}$ miles. How do I then find time? Do I even have to find time?

If it helps, the answer is 160 mph

This question is dumb

Answer is 238 mph

CALCULUS IS LIFE

this question is bad because no life because no calculus

dx/dt = 320
dy/dt = 0
solve for change in hypotenuse

x^2 + y^2 = C^2
differentiate with respect to time


Closed form of the integral $$I=\int_{-1}^1 \frac{\sin\left(\frac{\sinh x}{x}\right)\cdot\log\left(\frac{1+x}{1+x^2}\right)}{x} \space\text{d}x$$

According to Wolfram Alpha, the integral comes out to $$I=2.1607...$$

I don't have a clue on how to solve this because of how the $x$ value is deeply nested here. An answer I am looking for does the following:

a) Showing each step. All of your steps also much use only the real plane. I.E. Real Analysis

b) You don't have to state theorems in your work, but if what you did is not obvious, just explain.

Thanks for any help.

EDIT: this integral has a specific application. That application is that I am just beginning to work on nested integral arguments. The further I get in my study, I consistently give myself challenge problems. Strangely, after a while of trying to crack this, I got nowhere! I just wanted to see how to do this and where my mistake lay. (On original trial of the problem, I tried to substitute $x^2$ with $-u^2$ and ignored the fact that $x \not\to -u$)

I will give some of my own estimates.. Note that

$$
\begin{align*}
\int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log\left(\frac{1+x}{1+x^2}\right) \mathrm{d}x
& = \int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \Bigl[ \log\left( 1 + x\right) -  \log\left(1 + x^2\right) \Big] \mathrm{d}x \\
& = \int_{-1}^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log(1+x) \mathrm{d}x
\end{align*}
$$
Due to the symmetry of the last integral. For example split it at $x=0$ and use the substitution $x \mapsto -x$ on the integral over $[-1,0]$. Again we can map the integral onto $[0,1]$. Introducing the function
$$
f(x) = \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log(1 + x)
$$
The integral can then be written as
$$
\begin{align*}
\int_{-1}^{1} f(x) \,\mathrm{d}x
& = \int_{0}^{1} f(x) + f(-x)\,\mathrm{d}x \\
& = \int_0^1 \frac{1}{x} \sin\left(\frac{\sinh x}{x}\right) \log\left(\frac{1+x}{1-x}\right)\,\mathrm{d}x
\end{align*}
$$
The reason for these algebraic manipulations is that the taylor expansion of the last expression converges much faster. It only contains $\sin x$ and $\log(1 \pm x)$ which are both very well known. We have

$$
\log \frac{1+x}{1-x} = 2\sum_{k=0}^\infty \frac{x^{1+2k}}{1+2k}\\
\frac{\sinh x}{x} = \sum_{j=0}^\infty \frac{x^{2j}}{(1+2j)!}
$$
Combining these and finding the taylor expansion (allowed since $|x|\leq 1$ we get
$$
f(x) \sim 2 \sin(1) + \frac{x^2}{3} \bigl(2 \sin(1) + \cos(1)\bigr)+ \frac{x^4}{180} \bigl(67 \sin(1) + 23 \cos(1)\bigr)
$$
Integrating from $0$ to $1$ we get
$$
\int_{0}^{1} f(x) \,\mathrm{d}x = \frac{689}{300} \sin(1)+\frac{41}{300}\cos(1) \approx 2.006419677
$$
Which is - if I dare so myself - a darn good estimate. Seeing from the taylor expansion that the answer will always be on the form $a \sin 1 + b \cos 1$, one can improve these constants somewhat. For example
$$
\frac{268}{125} \sin 1 + \frac{33}{50} \cos 1
$$
Which is good for 9 decimals give or take.


If $f(x/n)\to0$ when $n\to\infty$, for every $x$ and $f$ is continuous, then $f(x)\to0$ when $x\to0$, or not? Let $f:(0,\infty)\to \Bbb R$ be continuous and such that for each $x>0$ the sequence $\{f(\frac{x}{n})\}\to 0$.

  Does it imply that $\lim_{x\to 0^+} f(x)=0$?

My try:Fix $x>0$.Since $f$ is continuous then $\lim _{n\to \infty}f(\frac{x}{n})=f(\lim_{n\to \infty} \frac{x}{n})=f(0)\implies f(0)=0$

Since $f$ is continuous and $f(0)=0\implies \lim _{x\to 0^+} f(x)=0$

Is the solution correct?In the book it is  given to use the Baire Category Theorem.

Expanding on Moya's comment, here is a big hint.

Fix $\epsilon>0$. Let
$$
K_n = \{x>0:|f({x}/m)|\le\epsilon \text{ for all } m\ge n\}
$$
The idea is that for all $x$, the sequence $f(x/n)$ gets close 0 eventually, so we define $K_n$ to be the set of $x$ for which is this sequence is close to zero specifically at time $n$.

Next, prove that

The sets $K_n$ are closed.
The union of $K_n$ is $\mathbb R^+$.

Since (one form of) the Baire Category Theorem states that $\mathbb R^+$ cannot be written as a countable union of nowhere dense closed sets, you can conclude that one of the sets $K_n$ is somewhere dense, which for closed sets means it contains an interval $[x_1,x_2]$, for some $x_1<x_2$.

This is huge progress. Now, we have series of solid blocks of points $$[x_1/n,x_2/n],[x_1/(n+1),x_2/(n+1)],[x_1/(n+2),x_2/(n+2)]\dots,$$ where for any $x$ in these intervals, $|f(x)|<\epsilon$. Use this fact to find a $\delta>0$ such that $|f(x)|<\epsilon$ whenever $|x|<\delta$.

I'll give you an outline:

Start with the definition of a limit.

$$\forall \epsilon \text{ } \exists \delta \text{ s.t. } |f(x)|\leq \epsilon \text{ whenever } x\in (0,\delta).$$

And the definition of $\{f(\frac{x}{n})\}\to0$

$$\forall \epsilon \text{ } \exists N \text{ s.t. } |f(\frac{x}{n})|\leq \epsilon \text{ whenever } n>N.$$

And continuity (*edit, accidentally wrote the definition for uniform continuity)

$$\forall \epsilon,x \text{ } \exists \delta \text{ s.t. } |f(x)-f(c)|\leq \epsilon \text{ whenever } |x-c|<\delta.$$

First force $|f(x)|\leq \frac{\epsilon}{2}$, then force $|f(x)-f(c)|\leq \frac{\epsilon}{2}$, that gives that $f(c)<\epsilon$.  You'll have to be a bit more rigorous with the deltas, let me know if I should elaborate.


$$x^4+2x^3-x-1=\left(x^2+x-\frac{1}{2}\right)^2-\frac{5}{4}$$
and the rest is smooth.

We can get it by the following way.
$$x^4+2x^3-x-1=(x^2+x+k)^2-((2k+1)x^2+(2k+1)x+k^2+1)$$
for all value of $k$ and for $k=-\frac{1}{2}$ we'll get  a difference of squares.

hint

Let $f (x)=x^4+2x^3-x-1$.
$f (0)=-1,f (1)=1,f (-2)=1$
there is at least one root in $(0,1) $ and

one in $ (-2,0)$.

$$f''(x)=12x^2+12x=12x (1+x)\geq 0$$

there are only two real roots.


Need clarification on Independent Events Probability question In the World Series of baseball, two teams A and B play a sequence of games against each other, and the first team that wins a total of four games becomes the winner
of the World Series. If the probability that team A will win any particular game against team B is 1/3, what is the probability that team A will win the World Series?

I can solve this easily my way, by summing the probabilities of winning on the 4th, 5th, 6th, and 7th game. My textbook offers a different solution and I'm trying to understand why it works. They say that

$\sum \matrix{7 \\ i=4} \pmatrix{7 \\ i} \left(\displaystyle\frac{1}{3}\right)^i \left(\displaystyle\frac{2}{3}\right)^{7-i} $

Would anyone kindly explain why that is a valid approach?

This problem can be formulated as the equivalent one:

Team A wins the majority of 7 games

Thus, you compute the probability of winning 4, 5, 6 or 7 games out of 7. This yields the desired result.

Your approach, as well as the textbook's approach, are equally valid. Here's how.

I'll just reframe the question to save some space and to avoid complexity but still maintaining the nature of the problem.

Suppose instead of 4, the winner is declared as soon as any team wins 3 games.

Let $p=\displaystyle\frac{1}{3}$ and $q=\displaystyle\frac{2}{3}$ such that $p+q=1$

Let W denote that team A wins a game and L denote that team A loses a game.
Following are the scenarios where team A wins (It's not very hard to figure out that playing 5 games are enough in any tournament to find a winning team).

Possible tournaments where A wins exactly 3 games ($\pmatrix{5 \\ 3}$ = 10 tournaments)

W,W,W,L,L
W,W,L,W,L
W,W,L,L,W
W,L,W,W,L
W,L,W,L,W
W,L,L,W,W
L,W,W,W,L
L,W,W,L,W
L,W,L,W,W
L,L,W,W,W

Possible tournaments where A wins exactly 4 games ($\pmatrix{5 \\ 4}$ = 5 tournaments)

W,W,W,W,L
W,W,W,L,W
W,W,L,W,W
W,L,W,W,W
L,W,W,W,W

Possible tournaments where A wins exactly 5 games ($\pmatrix{5 \\ 5}$ = 1 tournament)

W,W,W,W,W

The list mentioned above is the textbook's approach.

P(team A winning) = $\sum \matrix{5 \\ i=3} \pmatrix{5 \\ i} p^i * q^{5-i}$

Notice the bold W? Well, that is your approach. The bold W means that you are not considering the outputs after W (or, the teams don't play once team A wins 3 out of 5 games).

To see how it is equivalent, we'll rearrange the 3 lists into another set of lists (preserving the order number).

 Team A winning on the 3rd game
1. W,W,W,L,L

 11. W,W,W,W,L

12. W,W,W,L,W

16. W,W,W,W,W

 Team A winning on the 4th game
2. W,W,L,W,L

13. W,W,L,W,W

4. W,L,W,W,L

14. W,L,W,W,W

7. L,W,W,W,L

15. L,W,W,W,W

 Team A winning on the 5th game
3. W,W,L,L,W

5. W,L,W,L,W

6. W,L,L,W,W

8. L,W,W,L,W

9. L,W,L,W,W

10. L,L,W,W,W

Let's calculate the probability when team A is winning on the 3rd game (This event occurs when either of 1, 11, 12 or 16 occurs).
P(team A winning on 3rd game) = $p^{3}*q*q+p^{3}*p*q+p^{3}*q*p+p^{3}*p*p$

P(team A winning on 3rd game) = $p^{3}*[q^{2}+2pq+p^{2}]$

P(team A winning on 3rd game) = $p^{3}$

The intuition here is that once team A has won on the 3rd game, the result of the remaining two games does not change the outcome of team A winning the tournament. Imagine telling the players to continue playing the games even after the match has been decided, just for fun.

Similarly, team A winning on 4th game happens when either of 2, 13, 4, 14, 7 or 15 occurs

P(team A winning on 4th game) = $p^{2}*q*p*[p+q] + p^{2}*q*p*[p+q] + p*q*p^{2}*[p+q]$

P(team A winning on 4th game) = $3*p^{3}*q$

Once team A has won on the 4th game, the result of the remaining game does not change the outcome of team A winning the tournament. Imagine telling the players to continue playing the game even after the match has been decided, just for fun.

On the same line, team A winning on 5th game happens when either of 3, 5, 6, 8, 9 or 10 occurs
P(team A winning on 5th game) = $6*p^{3}*q^{2}$

This gives: P(team A winning) = $p^{3} + 3 * p^{3}*q + 6*p^{3}*q^{2}$; which is equivalent to the formula derived using your approach, i.e.,

P(team A winning) = $\sum \matrix{4 \\ i=2} \pmatrix{i \\ 2} p^{3} * q^{i-2}$

Takeaway here is that there is no harm in teams playing all the 5 games (or even more number of games) once the winner is declared in the first 3 or 4 or 5 games because, for example, P(Team A will win the tournament by playing n more games | Team A has already won first 3 games) = 1

For more clarification refer to section 4 on page no. 9 here.


If a function is discontinuous on $\mathbb Q$, is it necessarily discontinuous on $\mathbb R \setminus \mathbb Q$? Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is discontinuous on $\mathbb Q$. Is $f$ necessarily discontinuous on $\mathbb R \setminus \mathbb Q$?

Let $\{x\}$ represent the fractional part function. Consider the function $$f(x) = \sum\limits_{n=1}^{\infty} \frac{\{nx\}}{n^2} \quad (x \text{ real}).$$ Then $f$ is continuous at every irrational point even though it is discontinuous at every rational point.

Here is simple example of a function continuous at all irrational numbers but discontinuous at every rational number.

For $x\in\Bbb Q$ write $x=\frac pq$ as reduced fraction and define $f(x)=\frac1q$; for  $x\notin\Bbb Q$ define $f(x)=0$.

Okay, so it seems like that is called Thomae's function (which I had not heard of before), and by a ridiculous number of other names for such a straightforward function. I'll use this as an excuse to add the function that I was thinking about before this simpler example came in the way, a function that does have some interesting behaviour at the irrational numbers.

In fact that other function is initially defined as a function on the positive irrational numbers only. Each such number$~x$ has a unique continued fraction expansion, giving a sequence $(a_n)_{n\in\Bbb N}$ of positive integers (except that $a_0$ is allowed to be zero too). One defines $f(x)=y$ where $y$ has continued fraction expansion $(a_{2n})_{n\in\Bbb N}$, obtained by dropping the numbers at odd index. This defines a continuous surjective function from $\Bbb R_{>0}\setminus\Bbb Q$ to itself; continuity is a consequence of the facts (1) that for every nonempty open interval there is a finite initial sequence such that all continued fractions extending it lie in the interval, and (2) that for every $N$ and every irrational number$~x$ there is a neighbourhood of $x$ in which all continued fractions expansions coincide with that of $x$ for the first $N$ terms.
This function$~f$ is in fact one coordinate of a continuous bijective function $\Bbb R_{>0}\setminus\Bbb Q\to(\Bbb R_{>0}\setminus\Bbb Q)^2$ with a similarly defined other coordinate (using the odd-index terms of the continued fraction). It is not hard to see that this function cannot be extended continuously at any rational number; therefore one will get a function of the desired type if one extends the definition in any manner that does not destroy the continuity at the irrational numbers. The simplest way to do that is to use essentially the same continued-fraction procedure to define $f$ at rational values; only the expansion now is finite (so $f$ will take a rational value) and has two possibilities of which one must choose one. How the details are settled does not matter.


Prove $ab \leq (\frac{a+b}{2})^2$ $\forall a,b \in{R^+}$ As the title says I am trying to prove that $ab \leq (\frac{a+b}{2})^2$ $\forall a,b \in{R^+}$.

I tried using induction but the induction step never ends up simplifying the way I want it to. I may be looking at this question in the wrong way or I am just not seeing the "trick" in the induction step. Any help is appreciated! Thanks.

edit: The 2 is inside the parenthesis

Induction is not the right method, since $a,b \in \mathbb R$.

$ab \leq \frac{(a+b)^2}{2} \iff 2ab \le (a+b)^2 \iff 2ab \le a^2+2ab+b^2 \iff a^2+b^2 \ge 0$.

As @dxiv suggests, Result follows from AM-GM inequality, Which says,
If $\{a_i\}$ be a sequence with $a_i \in \mathbb{R}^+ $,

Then, $\frac{1}{n}\sum_{\forall i} a_i \geq \prod_{\forall i} a_i^\frac{1}{n}$ Where. $n$ is the length of the sequence.

Let, $\{a,b\}$ be a sequence, with $a, b \geq 0$; and $n = 2$

Then,

$\frac{a + b}{2} \geq \sqrt{ab} \implies \frac{a + b}{2}^2 \geq {ab}$


Finding the inverse of the function $f(k, x) = k^{x}x.$ Recently, I have been looking at the function $f(x) = e^{x}x,$ where its inverse is the Lambert W function.

I was intrigued by the fact that it is rather hard to calculate its solution, in comparison with other functions.

This intrigue encouraged me to develop my own function that would trace the inverse functions of functions of the form $f(x) = k^{x}x.$

$$g(k, 0, x) = k,$$

$$g(k, n, x) = \dfrac{1}{2} \log_{k}\left(\dfrac{k^{g(k, n - 1, x)} x}{g(k, n - 1, x)}\right),$$ and

$$\lim_{n \to \infty}{g(k, n, x)} = f^{-1}(k, x),$$ where

$$f(k, x) = k^{x}x.$$

Please, feel free to try it for yourself.

My question is, has this function ever been found before, and, if not, is it helpful in any way?

I will include a 'proof' of this, if you would like.

The inverse of $f(k,x)$ can be found in terms of the Lambert W. \begin{align*} y&=k^xx \\
y &= e^{x \ln k}x \\
y\ln k &= e^{x \ln k}x\ln k \\
W(y\ln k) &= x\ln k \\
x &= \frac{W(y\ln k)}{\ln k} \end{align*}

You have that

$$f(k,x) = \frac{1}{\ln(k)} e^{x\ln(k)} (x\ln(k)) = y$$

Hence, the inverse of your function is

$$\frac{1}{\ln(k)} W(y\ln(k))$$

Indeed,

$$f\left(k,\frac{1}{\ln(k)} W(y\ln(k)) \right) = \frac{1}{\ln(k)}\exp \left( \ln(k)\frac{1}{\ln(k)} W(y\ln(k)) \right)\ln(k)\frac{1}{\ln(k)} W(y\ln(k)) $$

$$= \frac{1}{\ln(k)}\exp \left( W(y\ln(k)) \right) W(y\ln(k))$$

$$= \frac{1}{\ln(k)}y\ln(k) = y$$

So your function can easily be expressed with Lambert function (and vice-versa)

$$W(y) = \ln(k) f^{-1}(\frac{x}{\ln(k)}, k )$$


A Game Theory Puzzle I encountered this game theory puzzle that I have no idea how to solve.

    In , two players take turn to fill in $1,\dots, 9$ in the circles (each number can only be used once). The player who can make one of the edges to have sum equals $20$ wins the game.

I haven't taken any game theory course, but I have read about Zermelo's theorem. As I believe, without strong reasons, that the game should have a winning strategy, I have tried to consider a case where the game ends in a draw and try to deduce backwards that at some point at least one of the players have made some suboptimal moves. And that if they choose to play optimally, it is possible to guarantee a win. However, since there are too many combinations, I have no idea how to show this generally.

Play 5 in a corner. When your opponent plays $a$, play $10-a$ in the symmetrical spot on the same edge.

in this game the person, who puts the first value in circle can always win, but he has to make pair of 10 on each side on which the second user is putting a value.

always remember sum of 20 is never going to be calculated on the side having 5 in circle, and if the 5 is at edge then only sum of 20 can be achieved from its opposite side


Deterministic finite automaton algorithm I am having a little trouble understanding this question.

For a DFA M = (Q, Σ, δ, q0, F), we say that a state q ∈ Q is reachable if there
exists some string w ∈ Σ∗ such that q = δ∗(q0, w).

Give an algorithm that, given as input a DFA expressed as a five-tuple M =
(Q, Σ, δ, q0, F), returns the set of all of M’s reachable states.

Can anyone help me understand and answer this question?

Someone hands you a DFA $M=\langle Q,\Sigma,\delta,q_0,F\rangle$, i.e., a set of states, an input alphabet, a transition function, an initial state, and a set of acceptor states. You’re to come up with an algorithm — a systematic procedure — that takes this information as input and produces as output a list of the reachable states of $M$.

A first easy observation is that you can ignore $F$ altogether: you don’t care which states are accepting, just which ones can actually be reached from $q_0$ with some input. You will need to use the other four items, however. A second easy observation is that $q_0$ is reachable, since $\delta^*(q_0,\epsilon)=q_0$, where $\epsilon$ is the empty word.

If $\Sigma=\{\sigma_1,\ldots,\sigma_n\}$, the natural thing to do is to start by running $M$ with all $n$ one-symbol inputs; that will tell you which states are reachable in one transition. Then you could systematically test inputs of length $2$:

$$\sigma_1\sigma_1,\sigma_1\sigma_2,\ldots,\sigma_1\sigma_n,\sigma_2\sigma_1,\sigma_2\sigma_2,\ldots,\sigma_2\sigma_n,\ldots,\sigma_n\sigma_n\;.$$

And you could go on to longer and longer inputs. The only real question is how to know when to stop: at what point can you be sure that you’ve found all of the reachable states?

HINT: If there are $m$ states, how long a word must you input in order to guarantee that the DFA will visit some state twice while reading the word?

I hope that this algorithm will be useful for you:

algorithm: determine the set if reachable states of an DFA;
   intput: an DFA M = (Q, Σ, δ, q0, F);
   output: an list of reachable state;
   begin
      create a list result and insert q0;
      create a stack s and push q0;
      while s is not empty
         do
            q = pop(s);
            if q != q0 then
               insert q in result;
            for each symbol a of Σ
               do
                  q' = δ(q, a);
                  if q' does not belongs to result then
                     push(s,q');
               done
            done
      return result;
   end.


Is the following proposition true or false? The next proposition is in a Japanese algebra text written by a Japanese mathematician who is famous in algebraic geometry.

But the proof is obviously not correct.

Then, is the next proposition true or false?

  Let $k \subset k_1 \subset k_2$ be fields. If an element $\alpha\in k_2$ is algebraic over $k$, it is algebraic over $k_1$. If $\alpha$ has degree $n$ over $k$, its degree over $k_1$ divides $n$.

The first part is true and the second part is false.

  If an element $\alpha\in k_2$ is algebraic over $k$, it is algebraic over $k_1$.

This is true because a polynomial over $k$ is a polynomial over $k_1$.

  If $\alpha$ has degree $n$ over $k$, its degree over $k_1$ divides $n$.

This is not true. Let $k = \mathbb{Q}$, $k_1 = \mathbb{Q}(\sqrt[3]{2})$,$k_2 = \mathbb{Q}(\sqrt[3]{2}, \omega)$, where $\omega = e^{2\pi i / 3}$ as usual, and let $\alpha = \omega \sqrt[3]{2}$.
Then:

The minimal polynomial of $\alpha$ over $k = \mathbb{Q}$ is $x^3 - 2$, which has degree $3$.
The minimal polynomial of $\alpha$ over $k_1 = \mathbb{Q}(\sqrt[3]{2})$ is $(x/\sqrt[3]{2})^2 + (x/\sqrt[3]{2}) + 1$ (because $\omega^2 + \omega + 1 = 0$), which has degree $2$.

And you can see that $2$ does not divide $3$.

The proposition is true.  For $\alpha$ to be algebraic over $k$, there must be a polynomial with coefficients in $k$ with $\alpha$ as a root.  Since each coefficient is also in $k_1$, the first assertion follows.  For $\alpha$ to have degree $n$ over $k$, then the dimension of $k_2$ over $k$, denoted [$k_2$ : $k$], is equal to $n$; then there is a theorem which asserts [$k_2$ : $k$] = [$k_2$ : $k_1$] [$k_1$ : $k$], and the second assertion follows.


Confusion about differential forms Consider the 2-form
$$\omega=\sum_{i=1}^ndx^i\wedge dy^i \,$$
on $\mathbb{R}^{2n}$, with coordinates $(x^1,\ldots,x^n,y^1,\ldots,y^n)$.

I want to be able to compute $\omega(\alpha,\beta)$, for $\alpha,\beta\in T_p\mathbb{R}^{2n}$.

I don't understand how to calculate, for example,
$$\omega\left(\left(\frac{\partial}{\partial x_i}\right)\Big\vert_p,\left(\frac{\partial}{\partial y_j}\right)\Big\vert_p\right)\, (*)$$
I know for $1$-forms that $dx^i\frac{\partial}{\partial x_j}=\delta_{ij}$.
I think that $(*)$ is also probably going to be $\delta_{ij}$, but I don't know how to show it.

The key thing here is the definition of $\wedge$: depending on convention,
$$ dx \wedge dy = dx \otimes dy - dy \otimes dx, $$
or it's half that. Going with this definition, we then have
$$ (dx^i \wedge dy^i)\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial y^k} \right) = (dx^i \otimes dy^i)\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial y^k} \right) - (dy^i \otimes dx^i)\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial y^k} \right) \\
= dx^i\left( \frac{\partial}{\partial x^j} \right) dy^j\left( \frac{\partial}{\partial y^k} \right) - dy^i\left( \frac{\partial}{\partial x^j} \right) dx^j\left( \frac{\partial}{\partial y^k} \right). $$
The latter term is zero since the $x$ and $y$ are different coordinates. The duality implies that the first term is $\delta^i_j \delta^i_k $, and summing this over $j$ gives $ \delta_{jk}$. One can do exactly the same thing for $\left( \frac{\partial}{\partial y^j} , \frac{\partial}{\partial x^k} \right)$, $\left( \frac{\partial}{\partial x^j} , \frac{\partial}{\partial x^k} \right)$ and $\left( \frac{\partial}{\partial y^j} , \frac{\partial}{\partial y^k} \right)$ (finding the former is $-\delta_{ik}$ and the latter two zero), and then use linearity on $\sum_{i} \alpha^i \frac{\partial}{\partial x^i} + \alpha'^i \frac{\partial}{\partial y^i} $ to arrive at $\omega(\alpha,\beta) = \sum_i (\alpha^i \beta'^i-\beta^i\alpha'^i) $.


How can I check for the accuracy of numerical result to optimization problem? How can I check for the accuracy of numerical result to optimization problem?

Or when is this possible?

Intuitively it could be possible at least to some extent, when one knows how to find analytic solutions or analytic approximations.

However, what if one just starts with some numerical solution, but the problem is e.g. non-linear. Then is there anyway to know, whether one can trust the optimization routine to have found the correct answer and not some other?

If you have an objective function in low dimensions (e.g., one or two scalar variables) then the simplest way of checking your optimising value is to plot the function and look at the optimising value.  You should be able to see if the optimising value you have calculated is actually at a minimising/maximising point.  If your objective function is in higher dimensions then this gets trickier, but you can always hold some argument values constant (at the optimising point) and check the others with plots.


What is the meaning of "$X \mapsto AX$ is a surjective mapping from $\mathbb{R}^n$ to $\mathbb{R}^n$"? If square matrix $A$ has determinant $1$, then $X \mapsto AX$ is a surjective mapping from $\mathbb{R}^n$ to $\mathbb{R}^n$.

What does $X \mapsto AX$ mean??

Is it equivalent to say $T(x)=Ax$?

Because then $A$ would be invertible, and for a given $Y \in \mathbb{R}^n$, we have: $X = A^{-1}Y$, then $T(X) = T(A^{-1}Y) = AA^{-1}Y = I_nY = Y$, showing $T$ is surjective.


If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of these triples as $8^2 + 15^2 = 64 + 225= 289 = 17^2$. Other examples of this triple are $(3, 4, 5)$ and $(5, 12, 13)$.

  Using Proof by Contradiction, show that: If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple.

We must suppose that $(a+1,b+1,c+1)$ is in fact a Pythagorean triple, and that $(a,b,c)$ is, too. Then we have $$(a+1)^2+(b+1)^2=(c+1)^2\tag{1}$$ and $$a^2+b^2=c^2.\tag{2}$$

Expand $(1)$--using for example that $(a+1)^2=a^2+2a+1$--and then use $(2)$ to eliminate all the squared terms from the resulting equation. You should be able to conclude (after gathering the $a,b,c$ terms on one side and other terms on the other) that $1$ is an even number.

Suppose both statement are true. Then

$(a+1)^2+(b+1)^2=(c+1)^2$

and as well

$a^2+b^2=c^2$

From 1.

$a^2+2a+1+b^2+2b+1=c^2+2c+1$

$\implies 2a+2b=2c-1$

But this does not satisfy for Pythagorean triplet 3,4,5 or for any other like 5,12,13.

Hence our supposition is wrong.


Can area of rectangle be greater than the square of its diagonal? Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?

The answer given to me is area of 486 m2. This is the explanation given to me

  Is it possible to have a rectangle of diagonal 18 m and area greater than the area of a square of side 18 m ?

The area of the square built on the diagonal must be at least twice the area of the rectangle:

$\hskip 4 cm$

A wall has a thickness.

Let's say the rectangle of the wall has width a and length b on the outside, so 2a + 2b = 72 or a + b = 36.

Assume the wall has a uniform thickness t. Then we can calculate the length of the diagonal as $(a - 2t)^2 + (b - 2t)^2 = d^2$. Given that d = 18, this lets us calculate t based on d, and the area of the wall is (2a + 2b - 4t) * t or (72 - 4t) * t.

The solution is $t = 9 ± (9a - a^2/4 - 40.5)^{1/2}$. For a solution to exist, we need $9a - a^2/4 - 40.5 ≥ 0$ or $162^{1/2} ≥ |a - 18|$ or $18 - 162^{1/2} ≤ a ≤18 + 162^{1/2}$, so a is roughly between 5 and 31.

We can exclude the solution $t = 9 + (9a - a^2/4 - 40.5)^{1/2}$ - the walls cannot be more than 9 thick. Therefore $t = 9 - (9a - a^2/4 - 40.5)^{1/2}$. We can show that the calculated thickness is always ≥ 0.

This t can be used to calculate the area of the wall as (72 - 4t) * t.

Alternatively, if the wall has zero thickness, then the diagonal is not 18. Everything follows from a false statement, so the wall can have any area.


How is it true that zero is neither a positive number nor a negative number? At first, the number zero looked like it was positive to me because positive numbers can be written with or without a plus sign to the left of them, but it's false.  I was surprised when I heard that zero is neither positive nor negative, but it's still a number and it's still even.  At least I know it's in between the positive and the negative numbers, so that must be why.  Also, because zero is neither negative nor positive, it's also known as neutral.  Am I on the right track?  Also, is there such thing as ±0, since it's neutral?  That's why I put the plus/minus sign there.  Is this how zero is neither negative nor positive?  I can see happy faces in your answers!

Would it make sense to adjust the definition of positive or negative so that one of them includes $0$? The following pretty theorems

The product of a positve and a negative number is negative
The product of two negative numbers is positive
The product of two positive numbers is positive

would require ugly adjustments, e.g.

The product of a positive number and a negative number is negative, except when the positive number happens to be zero, in which case the result is zero, hence positive

The emergence of ugly theorems is often a hint that the definitions are bad (as in: not very useful - recall that definitions cannot be "wrong")

This is just a matter of definition.

If you define positive numbers as all the real numbers $x$ such that $x=|x|$, then $0$ is a positive number. If you define positive numbers as all the real numbers $x$ such that $x=|x|$ except $0$, then $0$ is not a positive number.

Added: Maybe I have to admit, sometimes it is not only a matter of definition, but also a matter personal preference.


What is the expected probability of a basket to be frequent on average? I was reading on frequent pattern mining algorithm and came up with the following question. Let a company have $10^4$ different products and there are $10^9$ transactions each containing exactly 10 different products. If for each transaction a product is uniformly chosen, then what is the probability that a given fixed transaction of size 10 is chosen for $10^3$ times among $10^9$ transactions?

This is self study and the problem is stated in Slide 9 in here

P(10)  9990! *  10! / 1000!

= 10! / 10000 x 9999 x 9998 x 9997 x 9996 x 9995 x 9994 x 9993 x 9992 x 9991   = 3.645 * 10 ^ (-34)

with 10^9 transactions, you have a poisson distribution with parameter 10 ^(-25) so the chances of 1 particular transaction is extremely negligible.  The chance of getting 1000 occurrences of a transaction is

P(1000) = $\lambda ^ {1000} / 1000! e^{-\lambda} $

log(P(1000)) = 1000 log(3.645 * 10 ^ (-34)) - (1000 log 1000 - 1000) = - 76994 - 5907 = -82901

that's natural log, as log 10 it is -36003

so my answer is $P(1000) = 10 ^ {-36003}$


Exponential Generating function Use an exponential generating function to determine how many ways there are to make an r-arrangement of pennies, nickels, dimes and quarters with at least one penny and an odd number of quarters.  coins of the same denomination are identical.

You need exponential generating functions if you want to consider the order of the coins. You have the following generating functions:

Pennies: At least one, $\frac{z}{1!} + \frac{z^2}{2!} + \ldots = e^z - 1$
Dimes: Any number, $1 + \frac{z}{1!} + \frac{z^2}{2!} + \ldots = e^z$
Quarters: An odd number, $\frac{z}{1!} + \frac{z^3}{3!} + \ldots
                              = \frac{e^z - e^{-z}}{2}$

Thus the exponential generating function for $n$ coins is:
$$
(e^z - 1) \cdot e^z \cdot \frac{e^z - e^{-z}}{2}
  = \frac{1}{2} \left(
                  e^{3 z} - e^{2 z} - e^z + 1
                \right)
$$
We want:
\begin{align}
n! [z^n] \frac{1}{2} \left(
                       e^{3 z} - e^{2 z} - e^z + 1
                     \right)
  &= \frac{n!}{2} \left(
                    \frac{3^n}{n!} - \frac{2^n}{n!} - \frac{1}{n!} + [n = 0]
                  \right) \\
  &= \frac{3^n - 2^n - [n \ne 0]}{2}
\end{align}
(Here $[n = 0]$ is Iverson's bracket, i.e., it is 1 if $n = 0$ and 0 otherwise).

I am going to use regular generating functions instead of exponential ones because I think that is simpler - I'm not sure why the question asks for exponential specifically. I will make series where the exponent $n$ in $x^n$ is the number of cents produced and the coefficient of $x^n$ is the number of possible ways to make those $n$ cents out of the coin(s) in question.

With at least one penny there is one way to make 1 cent, 1 way to make 2 cents etc... so the GF is $$(1)x+(1)x^2+\cdots=\frac{x}{1-x}$$ With an odd number of quarters there is one way to make 25 cents, one way to make 75 cents, etc... so the GF is $$(1)x^{25}+(1)x^{75}+(1)x^{125}+\cdots=\frac{x^{25}}{1-x^{50}}$$ With nickels and dimes the GF's are respectively $$1+x^5+x^{10}+\cdots=\frac{1}{1-x^5}$$ and $$1+x^{10}+x^{20}+\cdots=\frac{1}{1-x^{10}}$$ We multiply (multiplication gives the total number of possible ways to make some number of cents with all the possible combinations) these functions to get $$\frac{x^{26}}{(1-x)(1-x^5)(1-x^{10})(1-x^{50})}$$ So to find the number of ways to make $n$ cents you would take the coefficient of the $x^n$ term in that series: $$\text{#ways to make $n$ pennies}=[x^n]\left(\frac{x^{26}}{(1-x)(1-x^5)(1-x^{10})(1-x^{50})}\right)$$

You should probably use a CAS to write that as a power series, or else use partial fractions to decompose it into pieces.


Refuting the Anti-Cantor Cranks I occasionally have the opportunity to argue with anti-Cantor cranks, people who for some reason or the other attack the validity of Cantor's diagonalization proof of the uncountability of the real numbers, arguably one of the most beautiful ideas in mathematics. They usually make the same sorts of arguments, so years ago I wrote up this FAQ to deal with them. Unfortunately, it's still hard to get anywhere with these people; the discussion frequently turns into something of this form:

ME: Suppose there is an ordered list containing all the real numbers. Then we can read off the diagonal entries and construct a real number that differs in the Nth decimal place from the Nth real number on the list. This real number obviously cannot be in the list. So the list doesn't contain all the real numbers.

THEM: Of course your proposed number is not on the list; it's not a well-defined real number.

ME: What do you mean? I gave you the exact procedure for constructing it. You take the Nth real number on the list, and you make it differ from that number in the Nth decimal place.

THEM: But if we really have a list of all the real numbers, then your proposed number has to be somewhere in the list, right?

ME: Yes, of course, so let's say it's in the 57th place. Then it would have to differ from itself in the 57th place, which is impossible!

THEM: Exactly, it's impossible! Your definition requires that it differs in some place from itself, which is impossible, so your definition is bad.

ME: But you're only saying that it's impossible on the basis of the assumption that there's a complete list of real numbers, and the whole point is to disprove that assumption.

THEM: But we're doing this proof under that assumption, so how can we make a definition that runs contrary to that assumption?

ME: But that definition is a good one regardless of whether there are countably or uncountably many reals. It is a complete, algorithmic, unambiguous specification of the real number. What else could you want?

THEM: I want the definition to be both unambiguous and non-contradictory, and your definition is contradictory!

ME: Forget about the purported complete lists of real numbers for a moment. Don't you agree that for any list of real numbers, complete or incomplete, it's possible to construct a real number that differs in the Nth place from the Nth number on the list?

THEM: No, it's only possible to construct such a real number if that real number isn't on the list, otherwise it's a contradictory definition.

ME: Don't you see that the contradiction is not the fault of my perfectly good definition, but rather the fault of your assumption that there are countably many real numbers?

THEM: No, I don't.

ME: But what if we took our putative complete list of real numbers, and fed it in line by line into a computer with an algorithm that spits out, digit by digit, a real number that differs in the Nth digit from the Nth number on the list? Would such a computer program work?

THEM: No it wouldn't, the computer program would hit the place on the list where the number being constructed would reside, and then it would get crash, because it can't choose a digit for the number that differs in the nth place from itself.

ME: Argh!

So how do I stop going in circles and convince them that they're wrong?

Any help would be greatly appreciated.

Thank You in Advance.

The formulation "If $f:\mathbf{N} \to \mathbf{R}$ is an arbitrary mapping, then $f$ is not surjective" clearly fixes the original list of real numbers and sets aside the potentially combatitive issue of whether or not the list is all of $\mathbf{R}$.

Significantly, the argument is no longer by contradiction, but by direct implication: The diagonal procedure constructs a real number not in the image of $f$. Perhaps this may help circumvent the sense of double-talk presumably conveyed in first positing the existence of an enumeration of $\mathbf{R}$, then arguing that some infinite decimal is not on said list.

First off, there are formally verified proofs of the theorem. In fact, every major theorem proof system or proof explorer has proved it, and it is number 22 in  the classic "100 theorems" benchmark: http://www.cs.ru.nl/~freek/100/

The problem is one may not accept the axioms and rules of inference.

However, that  is not the only problem. One may not even accept the statement as it is stated. I will show an alternative statement, which may be what the cranks have in mind.

Consider a Gödel numbering or any similar scheme from the naturals to the formulas of set theory. Clearly, this numbering lists all the possible formulas of set theory.

Since real numbers are merely formulas of set theory, the encoding will define  every real number. This last statement is bound to be controversial, but nonetheless it's a valid definition of what it means to be a real number and your cranky friends could be utilizing it. (In fact, if I am understanding Asaf correctly, interestingly enough, this is the usual definition: Undefinable Real Numbers )

No idea why this got a down-vote and a delete request. The point was not brought forward in the previous answers and is based on Hamkin's answer: https://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numb#answer-44129

Simply substitute definable real number in place of real number and the diagonalization proof no longer works. Note that each real number may be definable.


A doubt about an improper integral After a class about improper integrals (in high school) I was wondering about the integral $$\displaystyle\int\limits_0^1\sin{\frac{1}{x}}$$

but I found out that I'm unable to calculate the corresponding indefinite integral, is there a way to compute this integral by hand?

p.s.
after the edits this question is different from the original one, I figured out that my problem was different from the one I was originally asking

Since the function $x\mapsto\sin\frac1x$ is continuous on the interval $(0,1]$ and
$$\left|\sin\frac1x\right|\le1$$
then the integral is convergent.

Set $y=x^{-1}$. Then

\begin{equation*}
\int_{0}^{1}dx\sin \frac{1}{x}=\int_{1}^{\infty }dy\frac{1}{y^{2}}\sin y
\end{equation*}
which is well-behaved.


What is the difference between maximal flow and maximum flow? I have tried a lot on internet, but I am unable to get a good answer on the difference between maximal and maximum flow in case of network flow. Anybody has an idea? with example would be really helpful.

A maximum flow is a flow that attains the highest flow value possible for the given network.
A maximal flow is a flow whose value cannot be increased without decreasing the flow along some arc.

All maximum flows are maximal flows. Not all maximal flows are maximum flows.

Figure 3.9 in the Bang-Jensen and Gutin textbook, first edition illustrates an example of a flow that is maximal but not maximum.

The difference is very much analogous to the difference between a global optimum and a local optimum.

Both terms refer to the same thing. In the Wikipedia article on the maximum flow problem both terms are used equivalently. There you can also find some references to scientific publications that use both terms.


Can we use Peano's axioms to prove that integer = prime + integer? Every integer greater than 2 can be expressed as sum of some prime number greater than 2 and some nonegative integer....$n=p+m$. Since 3=3+0; 4=3+1; 5=3+2 or 5=5+0...etc it is obvious that statement is true.My question is: Can we use Peano's axioms to prove this statement (especially sixth axiom which states "For every natural number $n$, $S(n)$ is a natural number.")?

Note: My usual definition of $\mathbb{N}$ includes $0$, as for example in Halmos's Naive Set Theory. This does not matter here. The proposed result asks for an expression of natural numbers greater than $2$ (which means it doesn't matter if you consider $0$ a natural number or not) as a sum of a prime greater than 2 and a nonnegative integer; the proof below shows that every nonnegative integer (in particular, every "natural number if you don't include $0$") is the sum of a prime greater than two and a "natural-number-including-zero" (that is, a "nonnegative integer").

Let $S=\{n\in\mathbb{N}\mid n\leq 2\text{ or there exists a prime }p\gt 2\text{ and a natural }m\text{ such that }n=p+m\}$.

Note that $0$, $1$, and $2$ lies in $S$.

Assume that $k\in S$; we want to prove that $s(k)$, the successor of $k$, lies in $S$.

If $k\lt 2$, then $s(k)\leq 2$, so $s(k)\in S$.
If $k=2$, then $s(k) = k+1 = 2+1 = 3+0$ is the sum of a prime, $3$, and a natural number, $0$, so $s(k)\in S$.
If $k=p+m$ for some prime $p$ and some natural number $m$, then
$$s(k) = s(p+m) = p+s(m),$$
and by Peano's postulates, since $m\in\mathbb{N}$ then $s(m)\in\mathbb{N}$. So $s(k)$ is the sum of a prime and a natural number, hence $s(k)\in S$.

By Peano's Fifth Postulate (Induction), $S=\mathbb{N}$. This proves that every natural number greater than $2$ is the sum of a prime and natural number. $\Box$

If you don't consider the natural numbers to include zero, it is very simple to translate the proof above: the definition of $S$ just has to replace the word "natural" with "nonnegative integer":
$$S = \{n\in\mathbb{N}\mid n\leq 2\text{ or there exists a prime }p\gt 2\text{ and a nonnegative integer }m\text{ such that }n=p+m\}.$$

Then the first line of the proof can omit the observation that $0\in S$, and where it says "natural number" afterwards, just replace with "nonnegative integer", and the observation that if $m$ is a nonnegative integer, then it is either a natural-greater-than-1, so $s(m)$ is a natural number (under either definition), or else $m=0$ in which case "$s(m)$" denotes $1$, which is a natural number (under either definition).

Yes we can use the Peano's axiom to prove that integer = prime + integer. Think of $0 + a = a$.


Points in plane with every pair having at least two equidistant points? I was given this question in person by a fellow trainee at the downtime of an IMO training session, which made me think this problem is Olympiad related. I am interested in the solution as much as the origin!

$\textbf{Problem:}$Does there exist a non-empty finite set of points $P$ in $\mathbb{R}^2$ such that for each pair of points $x$, $y$ in $P$ there exists at least two distinct points $a$ and $b$ in $P$ such that both $a$ and $b$ are equidistant from $x$ and $y$.

To be honest, in over a decade and getting a degree in Mathematics, I have not made much headway into this problem. I tried to do some counting based on the number of pairs and pigeonhole to get the conclusion that there must be a point that is the circumcenter of a triangle made up of points in $P$. Assuming $P$ is a minimal solution (in terms of size of $P$) does not seem to help that much.
EDIT: originally the question left open whether it was in the plane or space, but the space case is trivial.
EDIT EDIT: added a note to say the set should be non-empty. Note that we do not need at least two points, because when we say "pairs of points" we don't say "distinct pair of points"!

A picture is worth a thousand words:

... but I will write a few words, nevertheless. This configuration has many symmetries and interesting properties, some of which I will describe here; however, I recommend trying to explore the configuration and figure them out on your own before reading further.
The eight points are arranged in two squares of different side lengths, with the same centre, and at a 45 degree angle from each other, the internal square coloured red and the external coloured blue. There are also eight equilateral triangles with vertices from the eight points: each side of the internal red square is completed to an outward-pointing equilateral red triangle, and each side of the external blue square is completed to an inward-pointing blue triangle.
All twelve red segments have equal lengths, as do all twelve blue segments. These equalities suffice to verify that any pair of points has a pair of two equidistant points. Up to symmetries, there are only six cases to check: two where both vertices are from the
blue square (adjacent or opposite); two where both vertices are from the red square (adjacent or opposite); and two where each vertex is from a different square (and the distance between the two vertices is either red or blue). The final two cases are the nicest and most interesting, and the latter of which is shown in the diagram, where the green perpendicular bisector of the two green points passes through a different pair of points, obtained from the first pair by a 90 degree rotation.

I do not know the origin of the problem, but I have encountered it before—on the xkcd forums (or "fora"), of all places, at least 10, maybe 15 years ago. I had recently recalled this problem and had presented it to some of our IMO students. My attempt to look up the original formulation and my original answer led me here. Sadly, the xkcd fora seem to have been offline since August 2019, and apparently they were not archived in full (or at least, I am unable to find such an archive), and thus some small pages in the history of this question might now be forever lost...

Let $P$ be the finite set of points above mentioned.
We have, by definition, that for every pair of points $x$ and $y$, there is another different pair $a$ and $b$ that are located in the line that passes through the midpoint of $x$ and $y$ and is perpendicular to the line segment $\overline{\rm xy}$ connecting such points.
Let $L$ be the set of all lines that can be constructed by passing through each pair of points in $P$ (this is a much bigger set, but still finite). Now consider the pair of lines $l_1,l_2\in L$ which present the smallest nonzero angle between them (there can be more than one, but it's enough to consider one pair). Then there is a pair of points $x_1\in l_1$ and $x_2\in l_2$ which doesn't satisfy the condition, i.e., there is no other pair $a,b$ equidistant from $x_1$ and $x_2$, which leads to a contradiction.
Note that if such pair $a,b$ existed, the condition "$l_1$ and $l_2$ having the smallest nonzero angle between them" would not hold, as the angle between $l_1$ or $l_2$ and the line perpendicular to the segment $\overline{\rm x_1x_2}$ (the line passing through $a$ and $b$) would be smaller.
Therefore, $P$ and $L$ must be both infinite sets.


Show that $(y-2x+3)^3=(y-x+1)^2$ is a solution of the differential equation $(2x-4y)dx+(x+y-3)dy=0$ I want to show that
$$(y-2x+3)^3=(y-x+1)^2$$
is a solution for:
$$(2x-4y)dx+(x+y-3)dy=0$$
what I did so far is:
$$\frac{dy}{dx}=\frac{4y-2x}{x+y-3}$$
any suggestions?

$(y-2x+3)^3=(y-x+1)^2$

differentiating this we have

$3(y-2x+3)^2(dy-2dx)=2(y-x+1)(dy-dx)$

using the original equation and substituting we have

$3(y-x+1)(dy-2dx)=2(y-2x+3)(dy-dx)$

$(y+x-3)dy+(-4y+2x)dx=0$

$(2x-4y)dx+(y+x-3)dy=0$

$\fbox{}$

Now it's time to abuse notation.

"Cross multiply" to get

$$(x+y-3)\,dy = (4y-2x)\,dx$$

Then re-arrange to get

$$-(4y-2x)\,dx + (x+y-3)\,dy = 0\implies (2x-4y)\,dx + (x+y-3)\,dy = 0$$


$\gcd(m,n)$ divides $\binom{m}{n} ?$ One of my students asked me this question.

Can anyone please explain the necessary and sufficient conditions such that

$\gcd(m,n)$ divides $\binom{m}{n} ?$

Is it true that $\gcd(m,n) | \binom{m}{n} \iff \gcd(m,n)=1$ or $m =kn$ ?

I'm sure of it in backward direction. I need help about the forward case?

For primes $p$, Kummer's theorem tells you the $p$-adic order of $m \choose n$.  What you want is that this is at least the $p$-adic order of $g = \gcd(m,n)$ for each prime dividing $g$.  In particular, note that the $p$-adic order of ${m \choose n}$ is the same as that of ${mp \choose np}$.

EDIT: Thus one way to find examples where $1 < \gcd(m,n) \mid {m \choose n}$ is to start with $m_0, n_0$ coprime, find a prime power $p^k$ dividing ${m_0 \choose n_0}$, and take $m = p^k m_0$, $n = p^k n_0$.

$\gcd(m,n) | \binom{m}{n} \iff gcd(m,n)=1$ or if $gcd(m,n)\neq1$ then the prime factorization of $n$ and $m$ must complain with the following:

$n = p^{\alpha_{1}}q^{\alpha_{2}}\dots w^{\alpha_{n}}$

$m=p^{\alpha_{1}+i}q^{\alpha_{2}+j}\dots w^{\alpha_{n}+k}x$ with  $i,j,k,x\geq 1$

Let's explain it:

$\binom{m}{n} = \frac{m!}{n!(m-n)!} = \frac{m(m-1)\dots(m-n+1)(m-n)!}{n!(m-n)!} = \frac{m(m-1)\dots(m-n+1)!}{n!}$

If $gcd(m,n) = n$ and $m=p^{\alpha_{1}+i}q^{\alpha_{2}+j}\dots w^{\alpha_{n}+k}x$ then

$\frac{p^{\alpha_{1}+i-1}q^{\alpha_{2}+j-1}\dots w^{\alpha_{n}+k-1}x(m-1)\dots(m-n+1)!}{(n-1)!}$

then you want to divide this binomial by $gcd(m,n)=n$ so:

$\frac{p^{\alpha_{1}+i-1}q^{\alpha_{2}+j-1}\dots w^{\alpha_{n}+k-1}x(m-1)\dots(m-n+1)!}{p^{\alpha_{1}}q^{\alpha_{2}}\dots w^{\alpha_{n}}(n-1)!} \Rightarrow \frac{p^{\alpha_{1}+i-2}q^{\alpha_{2}+j-2}\dots w^{\alpha_{n}+k-2}x(m-1)\dots(m-n+1)!}{(n-1)!}$

so it's satisfied when $i,j,k \geq 1$