Prove $\|x\|+\|y\|+\|z\|+\|x+y+z\| \ge \|x+y\|+\|y+z\|+\|z+x\|$ Assuming $x,y,z$ are complex numbers, or vectors, prove
$$\|x\|+\|y\|+\|z\|+\|x+y+z\| \ge \|x+y\|+\|y+z\|+\|z+x\|$$

I've tried replacing $x+y$, $y+z$, $z+x$ with $a, b, c$ respectively to see if it is any easier to prove it and this is the new form:
$$2(\|a\|+\|b\|+\|c\|)\le\|a+b+c\|+\|a+b-c\|+\|b-a+c\|+\|b-a-c\|$$

My hunch is that showing that $d(r,s)=\|r\|+\|s\|-\|r+s\|$ is a distance might help using $d(r,s)\le d(r,t)+d(t,s)$.

I found these links that seem to have the solution:
https://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers

Inequality for absolute values

After squaring of the both sides we need to probe that:
$$\sum_{cyc}\left(\|x\|\|x+y+z\|+\|y\|\|z\|\right)\geq\sum_{cyc}\|x+y\|\|x+z\|,$$ which is true by the triangle inequality for complex numbers:
$$\|x\|\|x+y+z\|+\|y\|\|z\|=\|x^2+xy+xz\|+\|yz\|\geq$$
$$\geq\|x^2+xy+xz+yz\|=\|x+y\|\|x+z\|.$$
It seems that for vectors it's not true.


Understanding the behaviour of order statistics of samples of uniform distribution Let $P$ denote any continuous distribution with density $p$ on $[0,1]$ and $Q$ the uniform distribution on $[0,1]$ whose density is $1$.

Let $X_1,\ldots,X_n$ be $n$ i.i.d samples drawn from the distribution $Q$ and $X_{(1)},\ldots,X_{(n)}$ be their order statistics and let $Y$ be a random variable distributed acccording to $P$. Define weights as:
$$
w_i=\frac{p(X_{(i)})}{q(X_{(i)})}= p(X_{(i)}), 1 \leq i \leq n
$$
and
$$
\tilde{w}_i = \mathbb{P}(Y \in [X_{(i)},X_{(i+1)}))= F_{P}(X_{(i+1)})-F_P(X_{(i)}).
$$

I am trying to understand if there exists any notion of similarity or convergence between the random variables $w_i$ and $\tilde{w}_i$ as $n \to \infty$.

Any help towards this would be appreciated.

The reason why I encountered this is because of the observation that if I use a Gaussian Kernel Density estimator with weights $w_i$ and $\tilde{w}_i$ respectively, both the weights are able to fit the pdf $p(x)$ perfectly across a range of distributions $P$ and $Q$. I want to know if this is a manifestation of some underlying fact about the above defined weights. I am attaching the plots for completeness:

In short, the two weights sequences you define are asymptotically equivalent in probability. I put a sketch of proof, see below.

But I'd recommend to have a closer look at Chapter 7 of [David H.A., Nagaraja H.N. Order Statistics (2003)]. I believe the topics discussed there are pretty close to the current one.

Proposition. $\forall\rho>0\;\;\mathbb{P}(|w_i-\tilde{w}_i|>\rho)\to 0, \;n \to \infty$.

Proof.

a) A basic fact of order statistics is that $X_{(i)}$ converges in probability to $\frac{(i-1)/n\;+\; i/n}{2}$ as $n\to\infty$.

b) If we denote $\delta_{i+1}=X_{(i+1)}-X_{(i)}$, then it's possible to get $\delta_{i+1}=\textit{o}(1/n)$ and

$$ F_{P}(X_{(i+1)}) = F_P(X_{(i)}) + p(X_{(i)})\cdot\delta_{i+1} + \textit{o}(1/n^2)
$$
using the Taylor approximation; the notation here is $p(\bullet)\equiv F^{\prime}_P(\bullet)$.

c) Finally,

$\mathbb{P}(|w_i-\tilde{w}_i|>\rho) = \mathbb{P}(|p(X_{(i)}) - \left(F_{P}(X_{(i+1)}) - F_P(X_{(i)})\right) |>\rho) = \mathbb{P}(|p(X_{(i)})\cdot\left(1-\delta_{i+1} \right) + \textit{o}(1/n^2)|>\rho).
$

(We just used b) and then a).) As we see, the left-hand side of the inequality is of order $\textit{o}(1/n)$ (in probability, see a)). Thus, we can choose appropriate (big enough) $n^\ast$ given any positive $\rho$, which means that the probability of the event $\left\{\ldots >\rho\right\}$ tends to zero. This concludes the proof.


What's the difference between a probability mass function and a discrete probability distribution? The Wikipedia page on "probability mass function" says "The probability mass function is often the primary means of defining a discrete probability distribution" (emphasis added).  I was under the impression that those two terms were completely synonymous.  What's the difference between them?

The difference is subtle. I think it helps to realize that a (probability) distribution is a more general concept. A distribution is something intrinsic to a random variable and determines/describes the behaviour of it. In contrast, a probability mass function (PMF) is one way to concretely describe the distribution quantitatively. Things to note:

In general, a distribution does not necessarily have a PMF or a probabilty density function (PDF). However, a discrete distribution always has a PMF, and a PMF always describes a discrete distribution, and therefore the two things are basically the same.
You should think of a PMF as just one way to describe/define a dicrete distribution. We could also use the cummulative density function for example, or any other exotic description we can think of.
Maybe you can very loosely think like this: The distribution is what carries out the random experiments, and it might look at the PMF to decide what to do, or it might look at something else.

The probability mass function is a function with whom you can calculate probability of a random variable.

Discrete probabilty distribution is about a way a random variable can behave in terms of the values it can assume.


Let $f,g$ be continuous functions. If we can always can get x,y s.t. $f(x)<g(x)$ and $g(y)<f(y)$, prove that $f(a)=g(a)$. The whole question: Let $f,g:X\rightarrow \mathbb{R}$ be continuous at $a$. Supposing that for every neighborhood $V$ of $a$ exists points $x,y$ such that $f(x)<g(x)$ and $g(y)<f(y)$, prove that $f(a)=g(a)$.

I'm stuck. I've tried to fix some $\epsilon$ and by continuity get some neighborhood of $a$ and so get the points $x,y$ given by hyphotesis. But i couldn't work with them succesfully. I also tried the contrapositive, but i couldn't go much further.

Any hints on how to solve this problem?

Consider the continuous function $h(x) = f(x) - g(x)$ over $X$.  We note that $\lim_{x \to a}h(x) = h(a)$, call this limit $L$.  Every neighborhood of $a$ contains an $x,y$ such that $h(x) < 0 < h(y)$.  However, by the definition of a limit: for any $\epsilon > 0$, we can select a neighborhood $V$ such that the $x,y \in V$ must also satisfy $|h(x) - L|< \epsilon$ and $|h(y) - L| < \epsilon$.

Thus, for every $\epsilon > 0$, we have $-\epsilon < L < \epsilon$.

Let $ h(x) = f(x) - g(x)$. Consider intervals $(a -1/n ,a + 1/n)$ around a .
Thus by intermediate value theorem on the intervals we get a sequence of $ b_n$ such that $h(b_n) = 0$. Now the intersection of these intervals is just one point a and $ b_n \to a$, thus $h(a) = 0 \Rightarrow f(a) = g(a)$


Base (topology) with closed intervals I am curious why it's a problem to define a base using closed sets?

For example, my book uses the definition under "Constructing Topologies from Bases" as specified at http://en.wikibooks.org/wiki/Topology/Bases, as opposed to the "definition" listed on this page.  I don't see why closed intervals are a problem for example, the point ${1} \in [0,1], [1,2]$ so in particular $ {1}  \in [0,1]\cap[1,2]=[1,1]=\{1\}$
I realize that topologies consist of "open sets" but why can't closed sets be a base for (larger) open sets for a topology.... or more generaly, why can't topologies be constructed using closed sets.

$\newcommand{\ms}{\mathscr}$As is pointed out in the post linked from amWhy’s comment, one can construct a topology using closed sets. Recall that $\ms{T}\subseteq\wp(X)$ is a topology on $X$ iff

$\varnothing,X\in\ms{T}$;
$\bigcup\ms{U}\in\ms{T}$ whenever $\ms{U}\subseteq\ms{T}$; and
$U\cap V\in\ms{T}$ whenever $U,V\in\ms T$.

Suppose that $\ms T$ is a topology on $X$, and let $\ms C=\{X\setminus U:U\in\ms T\}$, the set of closed sets in $\langle X,\ms T\rangle$. Then it’s immediate from the De Morgan laws that $\ms C$ satisfies the following conditions:

$\varnothing,X\in\ms C$;
$\bigcap\ms F\in\ms C$ whenever $\ms F\subseteq\ms C$; and
$H\cup K\in\ms C$ whenever $H,K\in\ms C$.

It’s also clear that a family $\ms C\subseteq\wp(X)$ is the family of closed sets of some topology on $X$ iff $\ms C$ satisfies these conditions.

Next, recall that a family $\ms B\subseteq\wp(X)$ is a base for a topology on $X$ iff it satisfies the following conditions:

$\bigcup\ms B=X$, and
if $B_0,B_1\in\ms B$ and $x\in B_0\cap B_1$, then there is a $B_2\in\ms B$ such that $x\in B_2\subseteq B_0\cap B_1$.

In this case $\left\{\bigcup\ms U:\ms U\subseteq\ms B\right\}$ is a topology on $X$, and we say that $\ms B$ is a base for $\ms T$.

By looking at the complements of members of a base for a topology on $X$, we can see how the notion of a base for the closed sets ought to be defined. A family $\ms X\subseteq\wp(X)$ is a base for the closed sets of a topology on $X$ iff

$\bigcap\ms D=\varnothing$, and
if $D_0,D_1\in\ms D$ and $x\notin D_0\cup D_1$, then there is a $D_2\in\ms D$ such that $x\notin D_2\supseteq D_0\cup D_1$.

In this case $\left\{\bigcap\ms H:\ms H\subseteq\ms D\right\}$ is the family of closed sets of a topology on $X$, specifically, of the topology for which $\{X\setminus D:D\in\ms D\}$ is a base.

Now we can ask whether the closed intervals in $\Bbb R$ are a base for the closed sets of some topology on $\Bbb R$. They certainly cover $\Bbb R$. However, it’s not necessarily true that if $I_0$ and $I_1$ are closed intervals not containing $x$, then there is a closed interval $I_2$ such that $x\notin I_2\supseteq I_0\cup I_1$. For instance, let $I_0=[0,1]$, $I_1=[3,4]$, and $x=2$: any closed interval that contains $[0,1]\cup[3,4]$ necessarily also contains $2$. On the other hand, the collection of all subsets of $\Bbb R$ of the form $(\leftarrow,a]\cup[b,\to)$ with $a<b$ is a base for the closed sets of the usual topology on $\Bbb R$.

One can of course also ask whether the closed intervals of $\Bbb R$ are a base for the open sets of some topology on $\Bbb R$. In fact they are, but that topology isn’t the usual one: it’s an easy exercise to show that it’s the discrete topology. (HINT: If $a<b<c$, then $[a,b]\cap[b,c]=\{b\}$.)

The simple answer is that if you wanted closed intervals to form a basis of some topology, then you would, for example, have to allow ${2}$ (which is the intersection of $[1,2]$ and $[2,3]$) to be a closed interval itself. But look at the question in your link. It says:
Show that the collection $\mathcal{C}=\{[a,b]:a,b\in\mathbb{R},a<b\}$ of all closed intervals in $\mathbb{R}$ is '''not''' a base for a topology on $\mathbb{R}$.
It categorically says that [2,2] is not a valid basis element. So, I guess that should clear it out.
However, as a fun fact, if you were to allow singletons, your topology is just the discrete topology. Cheers!


Is $4\underbrace{999 . . . 9}_{224 ({\rm times})}$ prime? Is $4\underbrace{999 . . . 9}_{224 ({\rm times})}$ prime?

I wanted to find smallest prime its sum of digits is $2020$. I started with small primes; the smallest three digits prime its sum of digits is 22 is $499$; four digits is $4999$ with sum of digits 31, five digit is $49999$ with sum of 40.For the sum $2020$ we have:

$2020=224\times 9+4$ and desired number can be of the form  $4\underbrace{999 . . . 9}_{224 ({\rm times})}$ . So this number has at least 225 digits. If it is not prime we have to search for numbers with number of digits more than 225 which of course have digits less than 9 and first digit probably less than 4.   I could not check it with my computer. I have these questions:

1- is  $4\underbrace{999 . . . 9}_{224 ({\rm times})}$  primes?

2- are numbers of the form $499 . . . 99$  always primes? If so what is theoretical reason? If not   what is conditions  for it to be prime?

Update: the closed form of these numbers is $N=5\times 10^n-1=5(10^n-1)+ 4$, $n ≥ 2$ if n is even we have:

$10^{2k}-1=(10^k-1)(10^k+1)$

Since $[10^n-1, 5, 4]=1$ N can be a prime, but brute force gives a counter result. If n is odd N can be composite.

A computer search finds $4259\mid 5\times10^{224}-1$. I know of no elegant proof of this, just brute force.

is prime
the form $N=5 \times 10^n-1$ is not prime if $n \neq1,7,13,19,25,...,n-1 \neq 6b$


Drawing a region in the complex plane Let $a,b \in \mathbb{C}$, $b \neq 0 $. and let

$$ G_0 = \{ z \in \mathbb{C} : Im ( \frac{ z -a }{b} ) = 0 \} $$

Question is to draw this set. My attempt: Let $z = (x,y) , a = ( \alpha, \beta) , b = ( \alpha', \beta') $.  After doing computation I find that the $y$ coordinate of the complex number $w = \frac{ z -a }{b} $ is

$$ \frac{-(x - \alpha) \beta' + (y- \beta) \alpha'}{(\alpha')^2 + (\beta')^2} = Im(w)$$

In particular, $Im(w) = 0 $ implies

$$ (x-\alpha)\beta' = (y-\beta) \alpha' $$

iff

$$ y = \beta' x + ( \beta - \frac{ \alpha \beta'}{\alpha'} )$$

So, we have a line with slope $\beta'$ and $yintercept$ $\beta - \frac{ \alpha \beta'}{\alpha'} $. This is a correct drawing? thanks

Hint. A complex number has imaginary part equal to zero f and only if it is a real number.


Need fractional differential equations resource. I'm studying linear fractional differential equations. Can some one give me a simple resources about the fractional differential equations. Is there any explanation for :
$$
\frac{\mathit{\lambda}}{\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}}\mathop{\int}\limits_{t0}\limits^{t}{{\mathrm{(}}{t}\mathrm{{-}}{s}{\mathrm{)}}^{{q}\mathrm{{-}}{1}}}{x}_{0}\mathrm{(}s\mathrm{)}ds
$$
How did become like this
$$
\frac{{x}_{0}\hspace{0.33em}\mathit{\lambda}}{\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}}\mathop{\int}\limits_{0}\limits^{1}{{\mathrm{(}}{t}\mathrm{{-}}{t}_{0}}{\mathrm{)}}^{{2}{q}\mathrm{{-}}{1}}{\mathrm{(}}{1}\mathrm{{-}}\mathit{\sigma}{\mathrm{)}}^{{q}\mathrm{{-}}{1}}{\mathit{\sigma}}^{{q}\mathrm{{-}}{1}}{d}\mathit{\sigma}
$$
Where:
$$
{x}_{0}{\mathrm{(}}{t}{\mathrm{)}}\mathrm{{=}}\frac{{x}_{0}{\mathrm{(}}{t}\mathrm{{-}}{t}_{0}{\mathrm{)}}^{{q}\mathrm{{-}}{1}}}{\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}}
$$

You can solve it there by clicking the dsolve button in mathHandbook.com

reference on http://drhuang.com/science/mathematics/fractional_calculus/fractional_differential_equation.htm


Determining convergence or divergence of series $ \sum_{n=1}^{\infty} \frac{(-1)^n}{ n+\sin (n)} $ I am wondering the convergence or divergence following series

$$
\sum_{n=1}^{\infty}
\frac{(-1)^n}{ n+\sin (n)}
\\
$$

My 1st attempt is 'alternating series test'

$$
$$

But, $$\frac{1}{n+\sin(n)}$$

isn't monotone decreasing. SO I failed...

$$
$$

Please give me some advice.

Thanks in advance.

\begin{align*}
\frac{(-1)^n}{n+\sin n}&=\frac{(-1)^n}{n}+\Bigl(\frac{(-1)^n}{n+\sin n}-\frac{(-1)^n}{n}\Bigr)\\
&=\frac{(-1)^n}{n}-\frac{(-1)^n\sin n}{(n+\sin n)n}.
\end{align*}

$\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n}$ converges conditionally.
$\displaystyle\sum_{n=1}^\infty\frac{(-1)^n\sin n}{(n+\sin n)n}$ converges absolutely.

Easier: see that $-1 \leq \sin n \leq 1 $, hence the bounds on the summand are $\frac{(-1)^n}{n-1}$ and $\frac{(-1)^n}{n+1}$ which converge by alternating test (Leibniz test) and hence your original series converges.


Determining if a random variable is lognormal I'm struggling to understand a part of the solution to this question: If $X$ and $Y$ are lognormal random variables, is their product $XY$ lognormally distributed?

The solution I have suggests looking at $ln(XY) = ln(X) + ln(Y)$. However, it says that $XY$ being lognormal depends on if $ln(X)$ and $ln(Y)$ are joint Normal. That is, $ln(X) + ln(Y)$ may not be normally distributed even if they are individually normally distributed implying $XY$ will not be lognormal.

But I don't understand how $ln(X) + ln(Y)$ could be possibly non-Normal if each of them are Normal, as the sum of two normals is a normal.

The sum of two joint normal distributed random variables is again normal.

It is important that the random variables are actually joint normal, as the assertion can fail otherwise. The classical example is the following:

Let $X$ and $Z$ be independent random variables, where $X$ is standard Gaussian and $Z$ is Rademacher distributed, i.e. $P(Z = 1) = P(Z = -1) = \frac{1}{2}$.

Then the random variables $X$ and $Y = ZX$ are both normal [and even uncorrelated!], but their sum isn't normal since $P(X + Y = 0) = P(X(Z + 1) = 0) = P(Z=-1)=\frac{1}{2}$.

As a counter-example to your statement that the "sum of two normals is a normal":

Suppose $W \sim N(0,1)$, $S=1$ when $|W| \lt 1$, and $S=-1$ when $|W| \ge 1$. Then $V=SW $ is also normally distributed as $N(0,1)$ but $|W+V| \lt 2$ so in this case the sum $W+V$ is not normally distributed.


How is this property called for mod? We have a name for the property of integers to be $0$ or $1$ $\mathrm{mod}\ 2$ - parity.

Is there any similar name for the remainder for any other base? Like a generalization of parity? Could I use parity in a broader sense, just to name the remainder $\mathrm{mod}\ n$?

Actually there is a standard name: residue.

There are $5$ residues modulo $5$, namely $0,1,2,3,4$.
Every prime greater than $3$ falls into only $2$ residue-classes modulo $6$.

Either AlessioDV's good answer, or you could say: "of the form $nq+r$". as a representation of $\equiv r \pmod n$ for example a number $N$ with $$N\equiv 1\pmod 3$$ has property $N=3q+1$.


How can I calculate $E(N_n)$ and $V(N_n)$ with $N_n = \min(X_1, X_2,\cdots, X_n)$? Consider ($X_1, X_2,\cdots, X_n$) a random sample of a population of $X$ as a Weibull distribution of parameters $(0, \delta, 2), \delta\in\mathbb R^+$, (in short , $X \sim W (0, \delta, 2)$ where $X$ has distribution function $F(x) = 1 - e^{-\left(\frac x\delta\right)^2}$.
Consider $$\overline{X}_n = \frac1n\sum\limits_{i = 1}^n X_i\text{ and }N_n = \min(X_1, X_2, \cdots, X_n)$$

I already calculate $E(X)$ and $V(X)$ but now I'm trying to calculate $E(N_n)$ and $V(N_n)$. How can I calculate $E(N)$ and $V(N)$? I have no idea...

The solution is

  $$E(N_n) = \frac{\sqrt\pi}2\frac\delta{\sqrt n}\text{ and }V(N_n) = \left(1 - \frac\pi4\right)\left(\frac\delta{\sqrt n}\right)^2 = \left(1 - \frac\pi4\right)\frac{\delta^2}n.$$

enter image description here

For any $t>0$, we have
$$
\{\min(X_1,\ldots,X_n)>t\} = \bigcap_{i=1}^n\{X_i>t\},
$$
and so
$$
\mathbb P(N_n > t) = \mathbb P\left(\bigcap_{i=1}^n\{X_i>t\}\right).
$$
Since the $X_i$ are independent,
$$\mathbb P\left(\bigcap_{i=1}^n\{X_i>t\}\right) = \prod_{i=1}^n \mathbb P(X_i>t).
$$
Since the $X_i$ all have $W(0,\delta,2)$ distribution,
$$
\prod_{i=1}^n \mathbb P(X_i>t) = \mathbb P(X_1>t)^n = (1-F(t))^n = e^{-\left(\frac x\delta\right)^{2n}}.
$$
It follows that
$$
\mathbb E[N_n] = \int_0^\infty (1-F_{N_n}(t))\ \mathsf dt = \int_0^\infty e^{-\left(\frac x\delta\right)^{2n}}\ \mathsf dt = \delta\cdot \Gamma \left(1+\frac{1}{2 n}\right),
$$
and
$$
\mathbb E[N_n^2] = \int_0^\infty 2t(1-F_{N_n}(t))\ \mathsf dt = \int_0^\infty 2te^{-\left(\frac x\delta\right)^{2n}}\ \mathsf dt = \delta^2 \Gamma \left(1+\frac{1}{n}\right).
$$
Therefore
\begin{align}
\operatorname{Var}(N_n) &= \mathbb E[N_n^2] - \mathbb E[N_n]^2\\
&= \delta^2 \Gamma \left(1+\frac{1}{n}\right) - \left(\delta\cdot \Gamma \left(1+\frac{1}{2 n}\right)\right)^2\\
&= \delta ^2 \left(\Gamma \left(1+\frac{1}{n}\right)-\Gamma \left(1+\frac{1}{2 n}\right)^2\right)
\end{align}


Proving the set of the strictly increasing sequences of natural numbers is not enumerable. How would one proceed to prove this statement?

  The set of the strictly increasing sequences of natural numbers is not enumerable.

I've been trying to solve this for quite a while, however I don't even know where to start.

As other answers note, there are lots of fancy ways to prove this. But we can always go back to the basics. A straightforward diagonalization proof-by-contradiction suffices. Suppose there is such an enumeration. Maybe this is it:

1 --> 1, 2, 3, 5, ...
2 --> 4, 5, 7, 100, ...
3 --> 1, 2, 3, 8, ...
4 --> 2, 4, 5, 6, ...

Now take the first number of sequence one, and add one to it.  That's our first number: 2.

Now take the second number of sequence two - 5 - and the number from the previous step - 2. Take the larger and add one: 6.

Now take the third number of sequence three - 3 - and the number from the previous step - 6. Take the larger and add one: 7.

Now take the fourth number of sequence four - 6 - and the number from the previous step - 7. Take the larger and add one: 8.

Keep doing that and construct the sequence of monotone increasing naturals:

2, 6, 7, 8, ...

By assumption, this sequence is in our enumeration, but where can it be? It cannot be at spot n for any n because by its construction the nth element of this sequence is larger than the element at spot n of the nth sequence.

That's a contradiction, and therefore there cannot be any such enumeration.

Let $\{{s_i}_j\}$ be a countable list of strictly increasing sequences; define $\{c_i\}$ via $c_i = \max (c_{i-1}, {s_i}_i)+1$ and .... presto, Cantor!

But it  could be simpler (depending on one's idea of simple) to reduce to things we already know are uncountable.

For simplicity, we can consider the sequence of differences between terms and not have to worry about the terms being increasing. i.e $a_{i+1} > a_i$ so $b_{i+1} = a_{i+1} - a_i > 0$ so $b_{i+1} \in \mathbb N$ and if $b_0 = a_0$. We have a one-to-one correspondence between $\{$ all increasing sequence of natural numbers $\}$ and $\{$ all sequences of natural numbers $\}$.

$\{$ all sequences of natural numbers $\} \supset \{$ all sequences of 1.... 10$\} \cong \{$ all sequences of 0....9 $\} \cong \{$all real numbers between 0 and 1$\}$ which is uncountable.


How can I explain topology to my grandmother? I was recently look at a post on tex.stackexchange about explaining $\LaTeX$ to the OP's grandmother. I was wondering, could the same thing be done for topology? Except in this case the "grandmother" is me. I have not fully understood the gist of topology and its capabilities. To my understanding, topology is the study of spaces but how does that translate into equations and variables? Anything would be helpful.

Thanks

Topology, aka "rubber-sheet geometry", is when a teacup is identical to a donut but there is no way a teacup could ever be like this.

Topologists worry a lot about odd rings and bottles, some of them are quite concerned by knots while others try to comb hairy balls. All in all, these are rather strange characters...

Firstly, a topology is nothing but a subset of the power set of some set. This special set is required to satisfy some conditions(conditions that we intuit from some "concrete" objects, mainly Euclidean spaces) by definition. So this way, a topology, as a "compact"(do not confuse with term "compact space", I used its daily meaning) set, carries lots of information about the whole big set(which is your space). You can ask what kind of information we can get out of it? That is the crutial point indeed! Since a topology is a set of sets, it tells you in which way your points are connected/interrelated. For instance, due to its topology, some part of your space may perform wild behaviour while some other parts are kind of tame etc. Moreover, you can read from the topology where "too many" points are accumulated and where the less are at and so on...

A few words on "doughnot = coffee mug" issue which is always given as a cliche example to a beginner of topology... You consider, as a topologist, a doughnot and a coffee mug as the same; the same in the sense that you care only in which way the points of both shapes are related. Here, for example, you forget about "distance" matter(the distance between any two points) as we have stepped up in a generalization(abstraction) level.

I think the key point is, to mention again, a topology is a core which one can reach the information of how the points are "related".

(Not an organized entry but I would be appreciated if I could contribute a bit.)


To prove that the limit of a bi-variate function is nonexistent at a point It has been asked to evaluate $$\lim_{(x,y)\to(0,0)} \frac{x^3 + y^3}{x-y}$$ if it exists at all and otherwise to disprove it.

After much thoughts , I came up with an idea of substituting $y$ with $x-mx^3$ which devolved the limit to $(2/m)$ and thus served my purpose of disproving the existence of a limit.

But, thinking of such half-weird substitutions take some reasonable amount of time; the luxury of which is seldom available at examinations. So, what are  other better methods to disprove the existence of this limit?

Any general algorithm (??) on disproving the existence of bi-variate limits (without indulging into trick-substitutions) will be also appreciated.

The standard and more effective method to show (by hand calculation) that a limit doesn't exist is to find at least two different paths with different limits as you have done.

Of course we can apply the epsilon-delta method using the same two paths but that way shouldn't be so different form the method you have already used to prove that limit doesn't exist.

Note that with some practice and a good strategy, figure out such "half-weird substitutions" don't take much time and it is indeed a correct an effective method to proceed. For the proper strategy to follow, refer also to the related

What is $\lim_{(x,y)\to(0,0)}\frac{ (x^2y^2}{(x^3-y^3)}$?


There exists $c \in [0,1]$ for which $\int_{0}^{1}\sin(x^3) = \int_{0}^{c}\sin(x^2)$ T/F: There exists $c \in [0,1]$ for which $\int_{0}^{1}\sin(x^3) = \int_{0}^{c}\sin(x^2)$

I know the answer it true, and I already saw the proof. What I don't get is this:

$\sin$ is monotonically increasing in $[0, 1]$, so from the monotonocity of integrals:

$$\sin(x^3) \leq \sin(x^2) \quad \forall x \in [0, 1]$$
$$\int_{0}^{1}\sin(x^3) \leq \int_{0}^{1}\sin(x^2) \quad \forall x \in [0, 1]$$
$$L = \int_{0}^{1}\sin(x^3) \leq \int_{0}^{c}\sin(x^2) + \int_{c}^{1}\sin(x^2) = R \quad \forall x \in [0, 1]$$

So if there exists such $c$, it means that:

case 1: $L < R \implies \int_{c}^{1}\sin(x^2) < 0 \quad$  which is false.
case 2: $L = R \implies \int_{c}^{1}\sin(x^2) = 0 \implies c = 1 \implies \int_{0}^{1}\sin(x^3) = \int_{0}^{1}\sin(x^2) \quad$ which is again, false

So what's going on?

Let $L=\int_0^1\sin(x^3)\,dx$ and consider
$$
f(t)=\int_0^t \sin(x^2)\,dx
$$
Then $f(0)\le L\le f(1)$ and by the intermediate value theorem $L=f(c)$ for some $c\in[0,1]$.

About your doubts. It doesn't make sense to say
$$
\int_0^1\sin(x^3)\,dx\le\int_0^1\sin(x^2)\,dx
\qquad \text{for all $x\in[0,1]$}
$$
as the integrals don't depend on $x$.

Then, indeed,
$$
L\le\int_0^c\sin(x^2)\,dx+\int_c^1\sin(x^2)\,dx
$$
just says that
$$
\int_c^1\sin(x^2)\,dx\ge0
$$

Hint : put $f(t)=\int_0^t \sin(x^2)dx$ to conclude


To prove $f(x)=\lim_{n\to \infty}{x^{2n} \over x^{2n}+1 }$ is discontinuous at $x=\pm1$ I came across this question to show that the given $f(x)$ is continuous at all points on $R$ except $x=\pm1$

$$f(x)=\lim_{n\to \infty}{x^{2n} \over x^{2n}+1 }$$

I know that to be continuous at $x=1$,
$$\lim_{x\to 1}f(x)=f(1)$$
I found $f(1)$ to be ${1 \over 2}$,but I am stuck as to how to calculate $\lim_{x\to 1}f(x)$ as there are two variables involved but its a question on a single variable calculus book.

The simplest way is likely to compute the pointwise limit at every $x$.

When $\lvert x\rvert<1$, $x^{2n}\to0$ as $n\to\infty$, so that
$$
f(x):=\lim_{n\to\infty}\frac{x^{2n}}{x^{2n}+1}=\frac{0}{0+1}=0.
$$

When $\lvert x\rvert >1$, $x^{2n}\to+\infty$ as $n\to\infty$; so, we can write
$$
f(x):=\lim_{n\to\infty}\frac{x^{2n}}{x^{2n}+1}=\lim_{n\to\infty}\frac{1}{1+\frac{1}{x^{2n}}}=\frac{1}{1+0}=1.
$$

Finally, when $x=\pm1$,
$$
f(x):=\lim_{n\to\infty}\frac{1}{1+1}=\frac{1}{2}.
$$

Now, the fact that this function is discontinuous at $x=\pm1$ is clear.

When $x\not=\pm1$ then the limit equals $1$.  Therefore the limit $f(x)$ as $x$ approaches $\pm1$ equals one.   Otherwise at $x=\pm1$, $f(x)=1/2$.  Such a function is not continuous at $x=\pm1$.


Calculate the value of the sum $1+3+5+\cdots+(2n+1)$ I have been thinking about this for a long time, may I know which step of my thinking is wrong as I do not seems to get the correct answer. If I am not going towards the right direction, may I get some help thanks!

My attempt:

Let $S = 1+3+5+\dotsb+(2n+1)\label{a}\tag{1}.$

Then I rearrange S from the last to first terms:

$S = (2n+1)+(2n-1)+(2n-3)+\dotsb+1\label{b}\tag{2}.$

Adding the two series $(1)+(2)$:

$$2S = (2n+2)+(2n+2)+(2n+2)+\dotsb+(2n+2),$$

I have $n$ copies of $(2n+2)$.

Therefore:

$2S = n(2n+2)$

$S = n(n+1)$.

Build a square in the following manner. On your first step, place $1$ block. On your second step put $3$ blocks around that, and on your third step put $5$ blocks around what you have, and so on. I think it should be pretty easy to see that the sum of the blocks is the area of the square for the step you are on.

In other words, the sum of the first $n$ odd integers is

$$
(2(1) - 1) + (2(2) - 1) + \dotsb + (2(n) - 1) = n^2.
$$

Image source: google "sum of first $n$ odd numbers."

A tip: evaluate the first few partial sums.

$$1, 4, 9, 16, 25, 36, 49\cdots$$

Nothing familiar ?


Quadratic Congruence (with Chinese Remainder Thm) How do we solve quadratic congruences such as:

  $x^2 \equiv11 \pmod{39}$

I know I must use the chinese remainder theorem with $p = 13, 3$ but I've only done linear examples and am unsure about how to do quadratic ones.

$$x^2 \equiv 11 \pmod{39} \implies x^2 \equiv 2 \pmod 3 \implies \text{No solution}$$

EDIT
In general, when you want to solve for $$x^2 \equiv a \pmod n \,\,\, (\spadesuit)$$ and if $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, the idea is to first solve for $$x^2 \equiv a \pmod {p_l^{a_l}} \,\,\, (\clubsuit)$$ You have a solution for your original problem $(\spadesuit)$ iff you have a solution for each $l \in \{1,2,\ldots,k\}$ in $(\clubsuit)$. Once you find solution for each $l$ in $(\clubsuit)$, put them together using Chinese Remainder theorem.
For instance, if you have $x^2 \equiv 23 \pmod {77}$, then we need to look at $x^2 \equiv 23 \pmod 7$ and $x^2 \equiv 23 \pmod{11}$ i.e. $x^2 \equiv 2 \pmod 7$ and $x^2 \equiv 1 \pmod{11}$.
$$x^2 \equiv 2 \pmod7 \implies x \equiv \pm 3 \pmod 7$$ Similarly, $$x^2 \equiv 1 \pmod{11} \implies x \equiv \pm 1 \pmod{11}$$
Hence,
$$x \equiv 3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 45 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 67 \pmod{77}$$
$$x \equiv 3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 10 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 32 \pmod{77}$$
Hence,
$$x \equiv \pm 10, \pm 32 \pmod{77}$$

$x^2\equiv11(\pmod{39}$

$39$ is a composite number $3\cdot13$

so now problem is :

$x^2\equiv11\pmod3$ and $x^2\equiv11\pmod{13}$.

if $x^2\equiv a\pmod{p}$ is a problem then it has solution if $a^{((p-1)/2)} \pmod {p}=1$

and have no solution if ^((p-1)/2) mod(p)=p-1
1)x^2=11mod(3) (p-1)/2=1
 11^1 mod(3)=2=p-1 so no solution
2)x^2=11 mod(13) (p-1)/2=6
 11^6 mod(13)=12=p-1 so no solution


stuck in combinatorics exercice , need of help Bob has :

12 Hats
12 shirts
5 Pants
4 Jackets
3 belts Wearing 1 pair of pants and 1 shirt is OBLIGATORY.
4 out of the 12 Shirts are warm so no jacket has to be with them

5 out of the 12 Shirts Can be worn with a Jacket (not necessarily)

The remaining 3 shirts out of the 12 are light to they MUST be worn with a jacket on (Obligatory)

3 pants out of 5 must be worn with belts on

Finally Wearing a hat is OPTIONAL

in how many Ways can bob get dressed ??

MY ANSWER : Upper body: (4 + 5*5 + 3*4)*13 = 533. Here the 5*5 represents the 5 shirts that can be either (a) with one of the 4 jackets or (b) with no jacket. The final 13 is the 12 hats and no hat.

Lower body: other 2 pants CAN be worn with belts. Hence, 2 + 3*3 = 11.

Finally, 533*11 = 5863.

Lower body: $3\cdot 3 + 2\cdot 4 = 17$ since 3 pants must be worn with one of 3 belts, and the other 2 pants have the extra choice of no belt.


What is a universal property? Sorry, but I do not understand the formal definition of "universal property" as given at Wikipedia.

To make the following summary more readable I do equate "universal" with "initial" and omit the tedious details concerning duality.

  Suppose that $U: D \to C$ is a functor from a category $D$ to a category $C$,
  and let $X$ be an object of $C$.

  A universal morphism from $X$ to $U$ [...] consists of a pair $(A,\varphi)$
  where $A$ is an object of $D$ and $\varphi: X \to U(A)$ is a morphism in $C$, such
  that the following universal property is satisfied:

  Whenever $Y$ is an object of $D$ and $f: X \to U(Y)$ is a morphism in $C$, then
  there exists a unique morphism $g: A \to Y$ such that the following
  diagram commutes:

  $\hspace{5cm}$

What kind of definition is this? Instead of "such that the following universal property is satisfied" one can equivalently say "such that the following property is satisfied". So how can this be a definition of "universal property"?

Unfortunately, not even Awodey in his Category Theory gives a concise definition of "universal property".

  Where do I find a really concise definition of "universal property"?

EDIT: I wonder why the attitude "you only have to understand the concrete examples, and the abstract notion will pop out by itself" seems to be accepted in this context. This reminds me of Augustine of Hippo:

  What, then, is time a universal property? If no one ask of me, I know; if I wish to
  explain to him who asks, I know not.

I agree with you that this is not about “concrete examples.” More about language. I apologize if my story is elementary, but there is really nothing complex.

Maybe you do not realize that “$A$ has universal property” is the same as “$A$ is a universal object” is the same as “an object $A$ is universal.” These are different names of the same term. So the definition of a universal object also defines universal property.

Consider the definition of an initial object. “…an initial object is an object… such that…” “Initial” is a property of objects, thus this definition defines a property. Properties are named not only by adjectives (e.g. “transitive”, “injective”), but also by nouns (e.g. “equivalence”, “injection”; “a function $f$ is an injection” is the same as “a function $f$ is injective”). In contrast, the definition of average, i.e. $(x, y)\mapsto \frac{x+y}{2}$, defines not a property.

Consider the shorter definition in Wikipedia which you did not cite:

  An initial morphism from $X$ to $U$ is an initial object in the category
  $X \downarrow U$.

This definition defines a property because it uses the definition of an initial object. The longer definition in Wikipedia which you cited is the shorter definition with the terms “initial object” and “comma category” unfolded.

“The universal property of the quotient group” is not a definition, it is a theorem which says that the quotient group $G/N$ is an initial object in a category defined as:

object: $(X, f)$ where $X$ is a group and $f:G\to X$ and $N\subseteq ker(f)$;
morphism of type $(X_0, f_0)\to (X_1, f_1)$: $g:X_0\to X_1$ such that $g\circ f_0 = f_1$.

I have essentially seconded lhf's answer, but he/she did not construct the category. I just can not find explicit construction of this category in textbooks.

Wikipedia's definition of the universal property does not include the universal property of the quotient group as a particular case. The problem is that in Wikipedia's definition $f$ is a morphism, but in the case of groups $f$ is a homomorphism such that $N\subseteq ker(f)$. IMHO Wikipedia's definition is not general enough.

P. S. I prefer “initial” and “terminal” over “universal”. A universal object is an initial object or a terminal object depending on context. Therefore, any text involving “universal” forces a reader to guess a precise meaning.

Given a diagram or graph a universal property says for the pair (U f) U is a object and f is a unique arrow, a U-cone is terminal or intial in f, if for any other cone over the U-cone the latter must factor through the former in particular through the unique arrow f.


Set of logical statements with contradiction. Let $\Sigma_1, \Sigma_2$ be two finite sets of statements in propositional calculus s.t  $\Sigma_1\cup\Sigma_2$ has a contradiction. Prove that there is a statement A s.t $$\Sigma_1\vdash{A}\  and\  \Sigma_2\vdash{\neg{A}}$$

is it still true with infinite  $\Sigma_1,\Sigma_2$?

My attempt:

in case one of the sets is $\phi$ or one of the sets has a contradiction its obvious...
So, we left with the case where both aren't empty and both without contradiction.

I didn't manage to prove this... Any kind of help will be appreciated Thanks!

Let $A_1,\ldots,A_n\in\Sigma_1$ and $B_1,\ldots,B_m\in\Sigma_2$ the sentences that generate the contradiction.

Let $A=A_1\wedge\ldots\wedge A_n$. It is clear that $A_1\wedge\ldots\wedge A_n\vdash A$. Since propositional calculus is sound, $A\wedge(B_1\wedge\ldots\wedge B_m)$ must be false in $\Sigma_2$, or, equivalently, $(B_1\wedge\ldots\wedge B_m)\to\neg A$ must be true. Since propositional calculus is complete, $(B_1\wedge\ldots\wedge B_m)\vdash\neg A$.

For the infinite sets case, I have been thinking, but no result so far.

There are a couple of problems with your puzzle.

the idea that $\Sigma_1, \Sigma_2$ be two finite sets of statements is in itself allready a contradiction, if they are closed onder modus ponens and so they are infinite sets , there are just an infinite number of theorems that belong to both sets.
If contradiction of a set means that  $ (\Sigma_1\cup\Sigma_2\vdash {A} $  and $ \Sigma_1\cup\Sigma_2\vdash{\neg{A}}$ you cannot tell which if $ A$ is from $\Sigma_1 $  or from $ \Sigma_2$ so how can you proof that $\Sigma_1\vdash {A} $?


Alternative definition of Gamma function? The Gamma function is defined in terms of an integral as

  The notation $Γ(t)$ is due to Legendre. If the real part of the complex number $t$ is positive $(Re(t) > 0)$, then the integral
  $$
    \Gamma(t) = \int_0^\infty x^t e^{-x}\,\frac{{\rm d}x}{x}
$$
  converges absolutely, and is known as the Euler integral of the second kind (the Euler integral of the first kind defines the Beta function).

Can it be equivalently defined in terms of a recursive relation as
$$ \Gamma(t+1)=t \Gamma(t)$$ $$\Gamma(1)=1$$
with some non-redundant conditions?

Thanks!

Baby Rudin gives the following definition:

$\forall x\in(0,+\infty):\ \Gamma(x+1)=x\Gamma(x);$

$\forall n\in{\Bbb N}:\ \Gamma(n+1)=n!;$

$\log\Gamma$ is convex in $(0,+\infty)$.

There are an infinitude of functions interpolating $n!$, a few interesting alternatives are discussed here (Bernoulli's, Hadamard's, and Luschny's).


Mathematical Fallacy Analysis I am interested in mathematical fallacy and found some cases about it. I am one of education major college student, and of course I am afraid that students in school will encounter it, especially the lack of understanding ones. Here is the sample.
We already know that $(-1)^3 = -1$. Yet, I will show you that it is not a true fact.

  $(-1)^3 = (-1)^{\frac{6}{2}} = ((-1)^6)^{\frac{1}{2}} = 1^{\frac{1}{2}} = 1$

So, the conclusion is $-1 = 1$.
Most of students are easily trapped by such a imaginary number cases and absolute value properties, and that mathematical fallacies I shall look at are in that areas.
What I want to know is that: are there any method, study theory, approachment, or anything, which can be used by teacher to make students have capabilities to analyze mathematical fallacy in solution steps and it is more better if they can think critically, implied that they just don't memorize certain math subject's properties.

The example given (or others like it) is exactly what students should be encouraged to keep in mind when they work with fractional exponents, because it demonstrates how easy it is to go astray if you don't pay attention to the precise statement of mathematical laws.

The fallacy, of course, lies in the naive expectation that the "associative" law $z^{ab}=(z^a)^b$ holds for all numbers $z$, not just the positive reals.  (Note, I'm assuming here that $a$ and $b$ are real numbers; the situation is even more delicate if $a$ and $b$ are complex.) It could be worth having students concoct their own examples of the fallacy, so that they can take "ownership" of it. For students familiar with function notation, it might also be worth presenting it in the form of an abstract question:  Is $E(E(x,y),z))$ necessarily equal to $E(x,M(y,z))$? (I trust it's obvious what operations $E$ and $M$ refer to.)


solution using synthetic geometry I managed to solve this problem only using complex numbers but I'd like to solve it using synthetic geometry and I can't.
Can someone help me to solve this problem using  synthetic geometry?

Let $ABC$  an acute triangle with $AB > AC$ . Let $O$ its circumcenter and let $D$ the midpoint of $BC$. The circle of diameter $AD$ intersects again  $AB$ and $AC$ in $E$ and in $F$, respectively. Let $M$ the midpoint of $EF$.
Prove that $MD$ is parallel to $AO$.

This is my solution. But, as I wrote above, I'd like to solve it using synthetic geometry and I can't.

Setting the origin of the plane in O and the points A, B and C on the circumference of unit radius, we have  that

$ a \bar{a} = 1 \text{;} \ b \bar{b} = 1 \text{;} \ c \bar{c} = 1 $.

Because $ D $ is the midpoint of $ BC $, we can write

1) $ d = \dfrac{b + c}{2} $

and because $ AD $ is a diamtere of the new circle then said $ Q $ his midpoint we have

2) $ q = \dfrac{a + d}{2} = \dfrac{2a + b + c}{4}$

Said $ M_{1} $ the projection of $ Q $ on $ AB $, then

$ m_{1} = \frac{1}{2} \left[ \left( \dfrac{\bar{q} - \bar{a}}{\bar{b} - \bar{a}}\right) (b - a) + a + q \right] $

but

$\dfrac{1}{\bar{b} - \bar{a}} = \dfrac{1}{\dfrac{1}{b} - \dfrac{1}{a}} = \dfrac{ab}{a – b} $

and so

$ m_{1} = \frac{1}{2} \left[ \bar{q} ab (-1) - \dfrac{ab}{a} (- 1)  + a + q \right] \Rightarrow $

$m_{1} = \frac{1}{2} \left( a + b + q - ab\bar{q} \right) $

In the same way, said $ M_{2} $ the projection of $ Q $ on $ AC $ we have

$ m_{2} = \frac{1}{2} \left[  \dfrac{\bar{q} - \bar{a}}{\bar{c} - \bar{a}} (c - a) + a + q \right] \Rightarrow $

$ m_{2} = \frac{1}{2} \left( a + c + q - ac\bar{q} \right) $

Because $ Q $ is the center of the new circle passing through $ A \text{,} D \text{,} E \text{,} F $ we have that $ M_{1}Q $ is axes of $ AE $ and that $ M_{2}Q $ in axes of $ AF $.

So we have that $ M_{1} $ is midpoint of $ AE $ and that $ M_{2} $ is midpoint of $ AF $.
So we can write

$ m_{1} = \dfrac{a + e}{2} \ \Rightarrow \ e = 2m_{1} – a $

and also

$ m_{2} = \dfrac{a + f}{2} \ \Rightarrow \ f = 2m_{2} - a$

The point $ M $ is defined as midpoint of $ EF $  so we have

$ m = \dfrac{e + f}{2} = \dfrac{2m_{1} + 2m_{2} - 2a}{2} = m_{1} + m_{2} - a$

therefore replacing $ m_{1} $ and  $ m_{2} $ we have

$m = \frac{1}{2} a + \frac{1}{2} c + \frac{1}{2} q + \frac{1}{2} a + \frac{1}{2} b + \frac{1}{2} q - \frac{ab\bar{q}}{2} - \frac{ac\bar{q}}{2} - a \ \Rightarrow $

$m = \dfrac{b + c}{2} + q - a\bar{q} \dfrac{b + c}{2}$

We know, furthemore, that $ \bar{q} $ is:

3) $\bar{q} = \dfrac{2\bar{a} + \bar{b} + \bar{c}}{4} = \frac{1}{4} \left( \frac{2}{a} + \frac{1}{b} + \frac{1}{c} \right) = \dfrac{2bc + ab + ac}{4abc}$

Now we write the equation of the parallel line through $D$ parallel to $AO$:

let $z$ be a generic point of this line it is possible to write

4) $ \dfrac{z - d}{a - 0} = \dfrac{\bar{z} - \bar{d}}{\bar{a} – 0} $

Replacing $d$ with the expression of 1) and taking into account that $ \bar {a} = \frac{1}{a} $, we get

5) $z - \dfrac{b + c}{2} = a^{2} \left( \bar{z} - \dfrac{\bar{b} + \bar{c}}{2} \right) = a^{2}\bar{z} - \dfrac{a^{2} (\bar{b} + \bar{c})}{2}$

If the $ M $ point belongs to this line, $ m $ must satisfy equation 5). Substituting the value of $ m $ we have

$\dfrac{b + c}{2} + q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right)  - \dfrac{b + c}{2} = a^{2}\left( \dfrac{\bar{b} + \bar{c}}{2} + \bar{q} - \bar{a}q \cdot \dfrac{\bar{b} + \bar{c}}{2} \right) - \dfrac{a^{2} (\bar{b} + \bar{c})}{2}   \Rightarrow $

$ q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) = a^{2}\bar{q} - \dfrac{aq( b + c)}{2bc} \ \Rightarrow $

$ q \left( 1 + \dfrac{ab + ac}{2bc} \right) = \bar{q}a \left( a + \dfrac{b + c}{2} \right)$

Substituting the values of  $ q $ in 2) and the value of $ \bar{q} $ in 3) we have

$ \dfrac{(2a + b + c)(2bc + ab + ac)}{8bc} = \dfrac{a (2bc + ab + ac) (2a + b + c)}{a \cdot 8bc} $

which is, obviously, an identity, so $ MD $ is parallel to $ AO $, as we wanted to prove.

Hints:

Construct squares $EDPR$ and $FDQS$ outwardly on the sides $ED$ and $FD$ of $\triangle DEF$. Let the line $PQ$ intersect lines $AB, AC$ at $U, V$ respectively .

First show that $MD \perp \operatorname{line} PQ = \operatorname{line}\ UV$. Therefore if $l$ is the line through $A$ that is parallel to $MD$ then $l \perp \operatorname{line} UV$.

Using $\triangle DEF \cong \triangle ABC$ (use ratio $AB/AC$ and $\angle A$) and $\operatorname{line} DP \parallel \operatorname{line} AB$ show that $\angle AUV = \angle ACB$. Since $\angle OAB = \frac\pi 2 - \angle ACB$, conclude that $\operatorname{line} AO \perp UV$.

Thus lines $AO$ and $l$ through $A$ are both perpendicular to the line $UV$, which implies that $\operatorname{line} AO = l \parallel \operatorname{line} MD$.

I was not able to solve it using synthetic geometry either. Is there anyone who can solve it and post a solution using synthetic geometry methods?


How can I solve $4x + 51y = 9$ using congruences? I'm given: $4x+51y=9$.

I am given a hint that when we use $4x=9 \pmod{51}$ we get $x = 15 + 15t$, and also if we use the congruence $51y=9 \pmod 4$ we get $y=3+4s$. They say it's handy to then find the relation between $s$ and $t$.

I have no idea how they got those suggestions and I need to know how to do that too.

I'm really stuck :(


Choosing the variance of a random normal variable to fulfill some criteria Suppose you create a random normal variable $X$ with mean zero and variance $\sigma^2$. You wish to choose $\sigma^2$ such that 80% of the time (or $a$% of the time, to make it more general), $X$ is between -1 and 1 (or between $-b$ and $b$ to make it more general). How to calculate $\sigma^2$?

First, note that

$$\mathbb{P}(-1 \leq X \leq 1) = \mathbb{P}(X \leq 1) - \mathbb{P}(X \leq -1). $$

Furthermore, we have that $\mathbb{P}(X \leq -1) = \mathbb{P}(X \geq 1) = 1 - \mathbb{P}(X \leq 1)$ by symmetry of the density of the normal cdf. Hence, we obtain
$$\mathbb{P}(-1 \leq X \leq 1) = 2 \mathbb{P}(X \leq 1) - 1 = 0.8 \Leftrightarrow  \mathbb{P}(X \leq 1) = 0.9$$
Standardizing yields
$$\mathbb{P}(X \leq 1) = \mathbb{P}\left(Z \leq \frac{1}{\sigma}\right) = 0.9 ,$$
where $Z \sim N(0,1)$. Looking in the table for the $z$-scores of the normal distribution, we find that
$$\frac{1}{\sigma} = 1.282 \Rightarrow \sigma \approx 0.78$$

Hint:

Find $\sigma^2$ such that:
$$\mathbb{P}(-1 \leq X \leq 1) = 0.8,$$
knowing that $X \sim N(0,\sigma^2)$.


What are the hyperbolic rotation matrices in 3 and 4 dimensions? So the hyperbola-preserving transformation in 2 dimensional space is given by the matrix
\begin{pmatrix}
\cosh(\phi) & \sinh(\phi) \\
\sinh(\phi) & \cosh(\phi) \end{pmatrix}

I'm wondering what such a matrix would be in 3 dimensional space (so that it preserves 2 dimensional hyperboloids) and 4 dimensional space (so that it preserves 3 dimensional hyperboloids). Sources or derivations would be appreciated. Thank you!

In a way, your transformation matrix is a variation of a common 2d rotation matrix

$$\begin{pmatrix}\cos\phi&-\sin\phi\\\sin\phi&\cos\phi\end{pmatrix}\;.$$

Where the above preserves the unit circle $x^2+y^2=1$, yours preserves the hyperbola $x^2-y^2=1$. The unit circle here corresponds to the unit sphere in 3d. There are many ways to describe 3d rotations, but one very common one is to describe them as a product of rotations around the coordinate axes. You can do the same for your hyperboloid as well.

For example, the one-sheeted hyperboloid $x^2+y^2-z^2=1$ has rotational symmetry around the $z$ axis. So you'd have these three “rotation” matrices:

$$
\begin{pmatrix}
\cos\alpha & -\sin\alpha & 0 \\
\sin\alpha & \cos\alpha & 0 \\
0 & 0 & 1
\end{pmatrix}
\qquad
\begin{pmatrix}
\cosh\beta & 0 & \sinh\beta \\
0 & 1 & 0 \\
\sinh\beta & 0 & \cosh\beta
\end{pmatrix}
\qquad
\begin{pmatrix}
1 & 0 & 0 \\
0 & \cosh\gamma & \sinh\gamma \\
0 & \sinh\gamma & \cosh\gamma
\end{pmatrix}
$$

Each of them preserves the hyperboloid, so a product of them will preserve it as well. The two-sheeted hyperboloid $z^2-y^2-x^2=1$ is preserved by the above matrices, too. If you want $x^2-y^2-z^2=1$ instead, you have to change coordinates, so that the rotation around $x$ becomes a regular rotation while the other two use hyperbolic functions.

I don't know whether it would make sense to translate any of the other rotation formalisms (like axis-angle or quaternions) into something preserving a hyperboloid. Probably things would become too complicated to make this useful.

In four dimensions, you can try the same approach. You have $\binom42=6$ coordinate planes, and for each of them you can give a possible rotation matrix. If the signs of the coordinates in the equation of your hyperboloid are the same, you use a regular rotation, otherwise you use hyperbolic functions. So for example if you have $x^2+y^2+z^2-w^2=1$ then any matrix modifying the $w$ coordinate would use hyperbolic functions, while those that just involve two of $x,y,z$ are regular rotations using circular trigonometric functions.

You can always transform any formula for spheric rotation into a formula for hyperbolic rotation by using the mathematical identities:
$$\cos x=\cosh ix$$
$$i\sin x=\sinh ix$$
$$\cosh x=\cos ix$$
$$\sinh x=-i\sin ix$$


A 10 digit positive number and ordered triplets A 10 digit positive number is said to be a “LearnHub”number if its digits are all distinct and it is a multiple of 11111. How many “LearnHub” numbers are there?

Find the number of ordered triplets (a, b, c) of positive integers for which LCM (a, b) =1000, LCM (b, c) = 2000 and LCM (c, a) = 2000.

How do i do this ? I have worked a lot on them.

Note that $1000=2^35^3$ and $2000=2^45^3$. What does this tell you about the prime factors of $a$, $b$, and $c$? Do you know what the LCM is of, say, $p^rq^s$ and $p^tq^u$, if $p$ and $q$ are primes?

Can you see why any number with all 10 digits must be a multiple of 9? Can you see why any number that's a multiple of 9 and a multiple of 11111 must be a multiple of 99999? Now, $99999=100000-1$, so if you mutiply it by the 5-digit number abcde, you get abcde00000-abcde. From this you should be able to work out the possibilities for abcde to make the result contain all 10 digits.

EDIT: Of course, there's always the computer programmer's solution, where you just ask your computer to look at all the 10-digit multiples of 11111 and count up how many have all 10 digits.

$1000 = 2^3 * 5^3$

$2000 = 2^4 * 5^3$

Since LCM of $a$ and $b$ is $1000$, they at the most contain $2$ three times and $5$ three times.

Also the LCM of terms containing $C$ is $2000$, the term $c$ should contain $2$ four times, as $a$ and $b$ can contain $2$ only three times.

so $c$ should either be

$2^4 = 16$

$2^4 * 5^1 = 80$

$2^4 * 5^2 = 400$

or $2^4 * 5^3 = 2000$.

Either $a$ or $b$ should have $2^3$. Hence fixing one as $2^3$ other can take values from $2^0$ to $2^3$.
Hence total combination= $2*4 - 1$ (the one combination is $2^3,2^3$ which we took twice)$= 7$

2 of the three should have power of $5$ as $5^3$, whereas the third can take 4 values in terms of powers of $5$ $(5^0$ to $5^3)$. Hence total no of such arrangement = $(3!/2! *4) -2$(this is for $5^3,5^3,5^3$ which we took thrice) = $12 -2 =10$

Hence ordered triplets will be $10*7 = 70$


Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$ Find a derivable function $f$ for which $f(x) - f(x-1) =\alpha ( f(x-1) - f(x-2) )$.

My initial conditions would be:
$$\begin{align*}
f(0) &= 0\\
f(1) &= \beta
\end{align*}$$
and $\alpha < 1$
and my domain $[0,+\infty[$

Basically, if looping on integers, every increment will be $\alpha$ times the previous increment, but I want a derivable function.

for example, for $\beta = 0.5$ and $\alpha = 1$:

$$\begin{align*}
f(2) &= 1.5\\
f(3) &= 1.75\\
f(4) &= 1.875
\end{align*}$$
I want to be able to evaluate $f(5.7)$ for instance.

If $\alpha=0$ then any periodic function of period $1$ is a solution. If $\alpha\ne0$ write $\alpha=e^\lambda$ for some $\lambda\in{\mathbb C}$. Then any function $f$ of the form
$$f(x):=g(x)+e^{\lambda x} h(x)$$
with $g$ and $h$ periodic of period $1$ is a solution.

If you take a possible solution $f$ to your problem (e.g. Dids) and multiply it by a differentiable function $g$ with $g(x)=g(x+1)$ and $g(0)=1$, e.g. $g(x)=cos(x\cdot 2 \pi)$ you get a new solution $fg$ which also solves your problem. Hence there is no reasonable way to evaluate $f(5.7)$ with only this information (but there is a reasonable way to evaluate $f(n)$ where $n$ is an integer).


Negating the statement $\exists x \in \Bbb R$ so that $x$ is not an integer, $x > 2016$, and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$ There exists a real number $x$ so that $x$ is not an integer, $x > 2016$, and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$.

I would like clarification on how to negate this. My idea of negation is for all real numbers $x$, so that $x$ is not an integer, $x>2016$ and  $\lfloor x^2\rfloor = \lfloor x\rfloor ^2$.

I'm tempted to say for all $x$ so that $x$ is an integer, $x> 2016$, but $\lfloor x^2 \rfloor \neq \lfloor x\rfloor ^2$.

Sometimes, representing the situation symbolically can help. Let's define

$p(x)$ means "x is not an integer"
$q(x)$ means "x is greater than 2016"
$r(x)$ means "$\lfloor x^2\rfloor = \lfloor x\rfloor^2$"

Now your original statement is

$$\exists x \in \mathbb{R} : p(x) \wedge q(x) \wedge r(x).$$

We can get a simplified version of its negation using the rules of boolean algebra as follows:

$$\begin{align*}
\neg \left[ \exists x \in \mathbb{R} : p(x) \wedge q(x) \wedge r(x) \right]&\iff \forall x \in \mathbb{R} : \neg \left[p(x) \wedge q(x) \wedge r(x)\right] \\
&\iff \forall x \in \mathbb{R} : \neg p(x) \vee \neg q(x) \vee \neg r(x)\\
\end{align*}$$

To translate back into more natural language:

  Every real number $x$ has at least one of the following three properties, possibly more: it is an integer, it is less than or equal to 2016, and/or $\lfloor x^2\rfloor \neq \lfloor x \rfloor ^2$.

I read this question somewhat differently than the other two answerers. The phrasing "so that" in the original statement is sloppy at best. Usually, you would have something like "$\exists x\in\mathbb{R}$, where $x$ is not an integer, ..." In fact, in a reasonably decent text, I think you would more likely come across something like "$\exists x\in\mathbb{R}\setminus\mathbb{Z}$." All this to say, I do not think that "$x$ is an integer" should be treated as its own statement or claim. A better phrasing of your original statement, I believe, would be something along the lines of the following:

  There exists a noninteger real $x$ for which $x>2016$ and $\lfloor x^2 \rfloor = \lfloor x \rfloor^2$.

The above statement becomes easy to express in a more symbolically formal way:
$$
(\exists x\in\mathbb{R}\setminus\mathbb{Z})(x>2016\land\lfloor x^2 \rfloor = \lfloor x \rfloor^2).
$$
The symbolic negation then becomes easy:
$$
(\forall x\in\mathbb{R}\setminus\mathbb{Z})(x\leq2016\lor\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2).
$$
Finally, the linguistic equivalent, no doubt more meaningful, becomes the following:

  For all noninteger real $x$, either $x\leq2016$ or $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.


Finding the unknown variable What is the value of $x$ in $x^{x}=25$? How can this be solved in the easiest way of all? I just couldn't deduce any idea regarding where to start.

\begin{align}
x^x &= 25 \\
x^2 &= 25
\end{align}
possibilities, $x=5$ or $x=-5$ because $5^2 = 25$ and $-5^2 =25$
description:
\begin{align}
-5\times-5
&=(-1)\times5\times(-1)\times5 \\
&=(-1)\times(-1)\times5\times5 \\
&=1\times5\times5 \\
&=25
\end{align}


Parametric Equation of a Circle in 3D Space? So, my dilemma here is... I have an axis. This axis is given to me in the format of the slope of the axis in the x,y and z axes.

I need to come up with a parametric equation of a circle. This circle needs to have an axis of rotation at the given axis with a variable radius.

I've worked on this problem for days, and still haven't come up with a solution.

I'm using this circle to map the path of a satellite, programmed in C.

And help would be greatly appreciated.

Thanks!

Let $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$ be two unit vectors perpendicular to the direction of the axis and each other, and let $(c_1,c_2,c_3)$ be any point on the axis. (If ${\bf v} = (v_1,v_2,v_3)$ is a unit vector in the direction of the axis, you can choose ${\bf a} = (a_1,a_2,a_3)$ by solving ${\bf a} \cdot {\bf v} = 0$, scaling ${\bf a}$ to make $\|{\bf a}\| = 1$, then letting ${\bf b} = {\bf a} \times {\bf v}$.)

Then for any $r$ and $\theta$, the point $(c_1,c_2,c_3) + r\cos(\theta)(a_1,a_2,a_3) + r\sin(\theta)(b_1,b_2,b_3)$ will be at distance $r$ from $(c_1,c_2,c_3)$, and as $\theta$ goes from $0$ to $2\pi$, the points of distance $r$ from $(c_1,c_2,c_3)$ on the plane containing $(c_1,c_2,c_3)$ perpendicular to the axis will be traced out.

So the parameterization of the circle of radius $r$ around the axis, centered at $(c_1,c_2,c_3)$, is given by
$$x(\theta) = c_1 + r\cos(\theta)a_1 + r\sin(\theta)b_1$$
$$y(\theta) = c_2 + r\cos(\theta)a_2 + r\sin(\theta)b_2$$
$$z(\theta) = c_3 + r\cos(\theta)a_3 + r\sin(\theta)b_3$$

If your circle has a unit normal vector < cos(a), cos(b), cos(c)> then depending on c, you have:

< cos(t-arcsin(cos(b)/sin(c)))/sqrt(sin(t)^2 +cos(t)^2*sec(c)^2), sin(t-arcsin(cos(b)/sin(c)))/sqrt(sin(t)^2 +cos(t)^2*sec(c)^2), cos(t)*sin(c)/sqrt(cos(t)^2+sin(t)^2*cos(c)^2>  0

< sin(t)*sin(a), -sin(t)*cos(a), cos(t)>    cos(c)=0

and of course, < cos(t), sin(t), 0>   sin(c)=0

The first two will have t=0 be the maximum point


Finding a Pythagorean triple $a^2 + b^2 = c^2$ with $a+b+c=40$ Let's say you're asked to find a Pythagorean triple $a^2 + b^2 = c^2$ such that $a + b + c = 40$. The catch is that the question is asked at a job interview, and you weren't expecting questions about Pythagorean triples.

It is trivial to look up the answer. It is also trivial to write a computer program that would find the answer. There is also plenty of material written about the properties of Pythagorean triples and methods for generating them. However, none of this would be of any help during a job interview.

How would you solve this in an interview situation?

Assuming you do have a pen and paper, you could substitute $c = 40 - a - b$ into the first equation to get

$$a^2 + b^2 = (40 - a - b)^2 = a^2 + b^2 + 1600 - 80(a + b) + 2ab.$$

Rewriting this equation, you get

$$a + b - 20 = \frac{ab}{40}.$$

From this it follows that $ab$ has to be a multiple of $40$, i.e., one of them is a multiple of $5$. That narrows it down to only a few options...

If that's still too much brute-force, you could also note that $a + b > 20$ from the above equation, and $a + b < 27$, since $c$ has to be the largest of the three. This leaves only the three pairs

$$\{(5,16),(10,16),(15,8)\}.$$

Looking at the earlier equation, you see the third pair is the right one.

This is a rect triangle with a length of 40.
for example triangle (3,4,5) with a length of 12, you can get the solution: (120/12,160/12,200/12).
the point is that you know some patterns in advanced. (5,12,13) etc...


How many $3$-digit numbers are divisible by $4$? I want to find out how many numbers can be formed taking the digits from zero to nine without repetition such that they are divisible by four (of 3 digits).

My Work: I find out the number pairs for the ones and ten's places which are divisible by four (like $04, 40, 08, 80, 12, 16,$ etc). These are $22$ in number. For any pair, for the hundred place, I have $8$ digits left (including $0$). So, my rough calculation becomes $8\times 22 = 176$. Then, I find out the number pairs without zero ($16$ in number)  .

Thus, my final answer becomes $176 - 16\times 1 = 160$. Am I correct ?

How many three-digit positive integers with distinct digits are divisible by $4$?

If this was the intended question, then you are indeed correct.  Your explanation could be improved if you made clear that you were subtract the $16$ strings with leading digit zero.

For a number to be divisible by $4$, the number formed by its last two digits (appending leading zeros as necessary) must be divisible by $4$.

\begin{array}{c c c c c}
\color{red}{00} & \color{blue}{04} & \color{blue}{08} & 12 & 16\\
\color{blue}{20} & 24 & 28 & 32 & 36\\
\color{blue}{40} & \color{red}{44} & 48 & 52 & 56\\
\color{blue}{60} & 64 & 68 & 72 & 76\\
\color{blue}{80} & 84 & \color{red}{88} & 92 & 96
\end{array}

Of the $25$ such numbers, three of them have a repeated digit ($\color{red}{00}, \color{red}{44}, \color{red}{88}$).  Of the remaining $22$ choices, exactly six include the digit $0$ ($\color{blue}{04}, \color{blue}{08}, \color{blue}{20}, \color{blue}{40}, \color{blue}{60}, \color{blue}{80}$).

Since the leading digit of a three-digit positive integer cannot be $0$, in the absence of the restriction that the digits cannot be repeated, we would have nine choices for the leading digit.  Given that restriction, we have $9 - 1 = 8$ choices for the leading digit when $0$ is one of the final two digits and $9 - 2 = 7$ choices for the leading digit when $0$ is not one of the final two digits.  Since there are six admissible choices for the final two digits that include the digit $0$ and $22 - 6 = 16$ admissible choices for the final two digits that do not include the digit $0$, the number of three-digit numbers that are divisible by $4$ in which no digit is repeated is
$$8 \cdot 6 + 7 \cdot 16 = 48 + 112 = 160$$

HINT:
There are $\lfloor2.5-\epsilon\rfloor=2$ single digit multiples of four. (excluding $0$)

There are $\lfloor25-\epsilon\rfloor-2=22$ two digit multiples of four.

There are $\lfloor 250-\epsilon\rfloor-22-2=225$ three digit multiples of four.

You now have to subtract any cases where there are repeated digits.


How is the material conditional treated in Natural Deduction? I'm confused by the definition of the material conditional.

In my implementation of propositional-logic I have the following definition of the material conditional:

$$\frac{P\to{Q}}{\neg(P\land\neg{Q})}\quad\small\text{[MaterialConditionalElimination]}$$

$$\frac{\neg(P\land\neg{Q})}{P\to{Q}}\quad\small\text{[MaterialConditionalIntroduction]}$$

I should that this is a definition because there is (effectively) a bi-directional inference rule. That is to say, whenever you encounter $\neg(P\land\neg{Q})$ you can replace it by $P\to{Q}$ and vice-versa.

However, according to Wikipedia, only the first of these rules is found in minimal logic. But I cannot see how you can derive the second from the first with the addition of the principle of explosion. Also, it is the second that serves, if one rule only can be taken, as a definition, because that is the rule that introduces the new $\to$ connective.

Update: It seems that the material conditional does not find its way into propositional logic when defined in this natural deductive style. I have therefore taken it out of the aforementioned implementation.

Further update: Well, it appears that it does, but only at the classical level. At this level, however, it is equivalent to logical consequence $\Rightarrow$ and therefore I am opting to leave it out still.

The implication sign does not denote material implication in intuitionistic logic. The Wikipedia page does not say what you think it does. It says that $\lnot P \lor Q$ (not $\lnot (P \land \lnot Q)$) entails $P \to Q$ in intuitionistic logic, but not in minimal logic (which is true but the reverse entailment is not provable in intuitionistic logic). In intuitionistic logic (and hence also in minimal logic) $\lnot P \lor Q$ is not equivalent to $\lnot(P \land \lnot Q)$ and neither of those is equivalent to $P \to Q$.

To see that $\lnot P \lor Q$ is strictly stronger than $P \to Q$ in intuitionistic logic, take $P \equiv Q$ for $P$ a variable, then $P \to P$ is provable, but $\lnot P \lor P$ expresses the intuitionistically unacceptable law of the excluded middle. To see that $P \to Q$ is strictly stronger than $\lnot(P \land \lnot Q)$, take $Q$ to be a variable and take $P = \lnot\lnot Q$, then $\lnot(\lnot\lnot Q \land \lnot Q)$ is provable, but $\lnot\lnot Q \to Q$ is the intuitionistically unacceptable principle of double-negation elimination.

Let me try to answer my own question, although I am far from sure that I have it right. Perhaps this is just another way of asking it.

The first rule, the introduction rule...

$$
\frac{\neg(P\land\neg{Q})}{P\to{Q}}
$$

...it looks like should be taken as a given in minimal logic. This is debatable, however, if you consider $\to$ to be a primitive of the logic, and therefore not requiring a rule for its definition at all.

What Wikipedia says is that, additionally, in minimal logic $P\to Q$ logically entails $\neg(P\land\neg{Q})$. I believe that one way of stating this is in the form of the second inference rule, the elimination rule:

$$
\frac{P\to{Q}}{\neg(P\land\neg{Q})}
$$

Therefore the statements $P\to Q$ and $\neg(P\land\neg{Q})$ can be seen as being entirely equivalent, since we effectively have a bi-directional inference rule.

If drawing attention to this equivalence seems churlish, consider that introduction and elimination rules are not generally symmetric in this manner. Consider, for example, the introduction rule for logical consequence:

$$
\frac{[P]\;...\;Q}{P\Rightarrow{Q}}
$$

Turning this rule on its head is non-sensical, although many (including me) might carelessly view the antecendant and consequent as being equivalent in day to day usage. However, of course the corresponding rule is not the introduction rule just turned on its head, it is Modus Ponens:

$$
\frac{P\Rightarrow{Q}\;\;P}{Q}
$$

So the rules concerned with the material conditional seem to stand out to me against all the other standard (whatever quite that means) rules in propositional logic in that they seem to define nothing more than a syntactic equivalence.

Put another way, it seems to me that $P\to Q$ can be viewed as nothing more than a convenient shorthand for $\neg(P\land\neg Q)$ and it is this usage for which I am seeking clarification.


Finding parametric equations for curve of intersection I am having difficulty finding parametric equations of the curve of intersection of $z = x^2 - y^2$ and cylinder $x^2 + y^2 = 1$

I am aware that the first equation represents a hyperboloid, and the second equation represents a cylinder. I think for the second one I can just do $x = cos(t)$ and $y = sin(t)$. Not sure about the other one.

We have the two surfaces

$$
x^2-y^2 = z\\
x^2+y^2 = 1
$$

Solving for $x^2, y^2$ we have

$$
x^2 = \frac 12(1+z)\\
y^2 = \frac 12(1-z)
$$

so a parametric representation can be read as

$$
(x(s),y(s),z(s)) = \left(\pm\sqrt{\frac 12(1+s)},\pm\sqrt{\frac 12(1-s)},s\right)
$$

for $-1\le s \le 1$

Attached a plot with the four leafs

For the first one use that $$\cosh^2(t)-\sinh^2(t)=1$$
And since$$x^2=z+y^2$$ we get

$$z+2y^2=1$$ so....
$$z=1-2y^2$$


Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy $ f\left(m^{2}+m n\right)=f(m)^{2}+f(m) f(n) $ Question -

Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the equation
$$
f\left(m^{2}+m n\right)=f(m)^{2}+f(m) f(n)
$$
for all natural numbers $m, n$

by putting $m=1$ and $f(1)=k$ we get
$f(n+1)=k^2 + kf(n)$

then hint says use $3^2 + 3.1 = 2^2 +2.4$ to get polynomial relation for k..
i am not getting how to use this hint ...i think i am missing some very easy tricks to get at this which i have not lernt yet ...

any help will be appreciated

thankyou

Putting $n=1$ in the condition for $f$ gives
$$f(m^2+m)=f(m)^2+kf(m)$$
Now set $m=3$. By the hint, we have
$$f(3^2+3)=f(2^2+2\cdot 4)=f(2)^2+f(2)f(4)$$
which gives us the condition
$$f(3)^2+kf(3)=f(2)^2+f(2)f(4)$$
You should be able to find $f(2)$,$f(3)$ and $f(4)$ in terms of $k$ by using your condition for $f(n+1)$.

Hope this helps.

The function which you are searching for has the following properties.
$$f(x+y) = f(x)+f(y)$$
and
$$f(x*y) = f(x)*f(y)$$
The only function which fits into these constraints is identity function, i.e,
$$f(x)=x$$


Does $\lim_{n \to \infty} |a_n|=0$ and $\limsup_{n \to \infty} |b_n|<\infty$ imply $\lim_{n\to\infty} |a_nb_n|=0$? I would like to know if the following statement is true:

Let $(a_n)\subset \mathbb{C}$ and $(b_n)\subset \mathbb{C}.$
If $$\lim_{n \rightarrow \infty} |a_n|=0$$ and $$\limsup_{n \rightarrow \infty} |b_n|\leq \alpha<\infty,$$ then
$$\lim_{n \rightarrow \infty}|a_n b_n|=0.$$

Thank you very much.

Masik

Yes, $\limsup |b_n|\le \alpha$ means that $|b_n|<\alpha+1$ eventually. This means that $0\le |a_nb_n|\le |a_n|(\alpha+1)$ eventually and the RHS $\to0$ as $n\to\infty$.

first of all we know that
$lim_{n \to \infty} | b_n | <= M  \ for M \in N$.
there for if we assume that
so $ 0 <= lim_{n \to \infty} |a_n| \ lim_{n \to \infty}| b_n | <= lim_{n \to \infty} |a_n| \ M = 0 $.


Solve for x $ \sin 30° \sin x \sin 10° = \sin 20° \sin ({80°-x}) \sin 40° $ $ \sin 30° \sin x \sin 10° = \sin 20° \sin ({80°-x}) \sin 40° $

I tried transformation formulas , $ 2\sin a \sin b $ one. I know the value of sin 30° but what about others?

Original problem

In triangle ABC, P is an interior point such that $ \angle PAB = 10°. \angle PBA = 20° PAC = 40°  \angle PCA = 30° $ then what kind of triangle it is ?  I solved it till I got stuck here.

Assume AB = 1.

Apply sine law to PAB to get $PB = 2 \sin 10^0$.

Apply sine law to PAC to get $PB = 2 \sin 20^0$ and $PC = 4 \sin 20^0 \sin 40^0$.

Apply cosine law to PBC to get $BC$ in terms of those angles. Note that $\cos 100^0 = - \sin 10^0$.

$BC^2 = 4 \sin^2 10 + 16 \sin^220^0\sin^2 40^0 + 16 \sin 20^0 \sin 40^0 \sin^210^0$

Suggestion:- Convert all angles in $BC^2$to $\sin 10^0$ (or $\sin 20^0$) by compound angle formula. Hope that the result is 1.

Like Solve $\frac{\sin(xº)\sin(80º)}{\sin(170º - xº) \sin(70º)} = \frac {\sin(60º)}{\sin (100º)}$

$$\dfrac{\sin(80-x)}{\sin x}=\dfrac{\sin30\sin10}{\sin20\sin40}$$

$$\implies\sin80\cot x-\cos80=\dfrac{\sin30\sin10}{\sin20\sin40}=\dfrac{4\sin10\sin80}{2\sin(3\cdot20)}$$

using https://brainly.in/question/3475250

$$\implies\cos10\cot x=\sin10\left(1+4\cos10\tan30\right)$$

$$\implies\sqrt3\cot x=\sqrt3\tan10+4\sin10$$

Like Simplifying $\tan100^{\circ}+4\sin100^{\circ}$

set $A=360 n-3x$ in $2\sec A\sin x+\tan x=-\tan A$

$$2\sec3x\sin x+\tan x=\tan3x$$

Here $x=10\implies2\cdot\dfrac2{\sqrt3}\sin10+\tan10=\dfrac1{\sqrt3}\iff4\sin10+\sqrt3\tan10=1$

Can you take it from here?


sufficient condition for derivative to be continuous? let $f:\mathbb{R} \rightarrow\mathbb{R}$ be a continuous function in $x_0\in\mathbb{R}$.

In addition, $f'(x)$ is defined for every $x\in(x_0-\delta,x_0+\delta),x_0\notin(x_0-\delta,x_0+\delta)$

In addition, i know that $\underset{x_0}{lim}    f'(x)=L$

is it true that $f'(x)$ is continuous  at every  $x\in(x_0-\delta,x_0+\delta),x_0\notin(x_0-\delta,x_0+\delta)$

Assuming I understand your notation, no.

Let
$$
g(x) = \begin{cases} x^2 \sin \frac{1}{x} & x \ne 0 \\ 0 & x = 0\end{cases}
$$

$g$ is everywhere differentiable, but the derivative is not continuous.

Now let
$$
f(x) = g(x - \frac{\delta}{2}) + g(x + \frac{\delta}{2})
$$
It's everywhere differentiable, but the deriv is not continuous at the points $\pm \delta/2$.

$f'(x)$ must not be continuous at every $x \in (x_0 - \delta, x_0 + \delta)$.

Example: May be $rect_\delta(x-x_0)$ The rectangle function with Center at Point $x_0$ and broadness $\delta$. Then $f'(x)=rect_\delta(x-x_0)$ is discontinuous at $x_0= \delta/2$ and $x_0=- \delta/2$. The function $f(x)$, however is continuous.


Greatest common divisor of real analytic functions Consider two real-valued real analytic functions $f$ and $g$. I want to prove that there exists a greatest common divisor $d$, which is a real analytic function. By greatest common divisor, I mean the following:

Common divisor: There exist real analytic functions $q_1, q_2$ such that $f = dq_1, g = dq_2$, and
Greateast: If there is any other function $d_1$ that satisfies 1. above, then there exists a real analytic function $q_3$ such that $d = d_1q_3$.

I am guessing that a proof could be derived from the Taylor series expansion, but I am not sure how to proceed.

The relevant data are the roots (with multiplicity).

A GCD of two analytic functions $f$ and $g$ will be an analytic function $h$ such that for each $a \in \mathbb{R}$ one has $\operatorname{ord}_a (h) = \min \{\operatorname{ord}_a (f), \operatorname{ord}_a (g) \}$ where $\operatorname{ord}_a$ denotes the multiplicity of the root at $a$ (setting it $0$ if there is no root).

Put differently the order is the index where the Taylor series expansion around $a$ actually starts.

Algebraically the ring of real analytic functions is a Bézout domain, see for example Ring of analytic functions,  meaning every finitely generated ideal is principal and in particular any two elements admit a GCD (see GCD domain)

What do you mean by "greatest"? I.e., what would $\max\{\sin x, \cos x\}$ be?

Check out the development leading up to GCD among integers: It starts with the division algorithm, i.e., for all $a, b \in \mathbb{Z}$ there are $q, r \in \mathbb{Z}$ with $0 \le r \le \lvert b \rvert$ such that $a = b q + r$. Compare to the same situation for polynomials, where now you have $a(x) = b(x) q(x) + r(x)$, where the degree of $r$ is less than the degree of $b$. From this fundamental relationship you develop the idea of greatest common divisor (be it in the sense of greatest absolute value for integers or largest degree for polynomials). This assumes you have an integral domain with a valuation function (absolute value, degree) and a division algorithm. Real valued functions with pointwise operations (sum, product) aren't an integral domain (there are zero divisors, functions that aren't zero that multiply to the zero function).


Regular Expression for Simple Language I'm having trouble writing a regular expression given the following $\{a, b, c\}$ which produces the set of strings of length 3. I don't really understand how to restrict the length of the string. Obviously you could have 3 $a$'s, or 3 $b$'s or 3 $c$'s, but you could have $aab, aac, aba, \dots $

Is it something like $a^* \cup b^* \cup c^*$?

The regular expression you have given gives the language
$$\{\varepsilon, a, aa, aaa, \dots\} \cup \{\varepsilon, b, bb, bbb, \dots\} \cup \{\varepsilon, c, cc, ccc, \dots\}$$
$$= \{\varepsilon, a, aa, aaa, \dots, b, bb, bbb, \dots, c, cc, ccc, \dots\},$$
which is clearly not what you are after.

Using only union ($\cup$), concatenation ($\circ$), and the Kleene star ($^*$), you could do something like
$$(a \cup b \cup c) \circ (a \cup b \cup c) \circ (a \cup b \cup c).$$

I think it can be simply written as $((a+b+c)^3)^*$ since $(a+b+c)^3$ will produce all strings of length $3$.


Using $\gcd(a,b)$ to find gcd of other values $\gcd(a^2,b)$ and $\gcd(a^3,b)$ If $\gcd(a,b) = p$, a prime, what are the possible values of $\gcd(a^2,b)$ and $\gcd(a^3,b)$?

Through examples I've been able to find the answer, but I don't know how to come up with a proof.

Update: I think gcd of either is $p^2$, but again this is just by working examples.

I'm not familiar with much number theory. I've just started the second section of my textbook, so all I know is just divisibility and primes.

You do not have enough information to answer unambiguously. The most we can say is that $\mathrm{gcd}(a^2,b)$ is either $p$ or $p^2$, and $\mathrm{gcd}(a^3,b)$ is one of $p$, $p^2$, and $p^3$.

For suppose $\mathrm{gcd}(a,b)=p$. Then there are integers $x,y$ such that $ax+by=p$, so (squaring this expression) $a^2x^2+b(2axy+by^2)=p^2$, showing that a linear combination of $a^2,b$ with integer coefficients is $p^2$, which implies that $\mathrm{gcd}(a^2,b)$ divides $p^2$. Since $\mathrm{gcd}(a,b)$ divides any combination of $a,b$ and therefore of $a^2,b$, then $p$ divides $\mathrm{gcd}(a^2,b)$.

Now: $\gcd(p,p^2)=p$ and $\gcd(p^2,p^2)=p^2$, so certainly $p^2$ is possible. On the other hand, $\gcd(p,p)=p$ and $\gcd(p^2,p)=p$, so $p$ is possible as well.

A similar analysis, with similar examples, gives the result for $\mathrm{gcd}(a^3,b)$.

It is easy to prove that is $\gcd(a, b) = m$ then $\gcd(a / m, b / m) = 1$, $\gcd(k a, k b) = k \gcd(a, b)$, and if $\gcd(a, b) = \gcd(a, c) = 1$ then $\gcd(a, b c) = 1$. Splice those together.


Separating 18 people into 5 teams A teacher wants to divide her class of 18 students into 5 teams to work on projects,
with two teams of 3 students each and three teams of 4 students each.

a) In how many ways can she do this, if the teams are not numbered?

b) What is the probability that two of the students, Mia and Max, will be on the same team?

[This is not a homework problem.]

Assume first that the teams are labelled, with the two teams of three having labels A and B, and the three teams of four having labels C, D, and E.

There are $\binom{18}{3}$ ways to choose the people who will be on Team A. For each of these ways, there are $\binom{15}{3}$ ways to choose Team B. For every way to choose Teams A and B,  there are  $\binom{12}{4}$ ways to choose Team C, and  then $\binom{8}{4}$ ways to choose Team D, and finally, if we like, $\binom{4}{4}$ ways to choose Team E.

However, when we remove the labels, the number of choices for the $2$  three-person teams gets divided by $2!$, and the number of choices for the $3$ four-person teams gets divided by $3!$, for a total of
$$\frac{\binom{18}{3}\binom{15}{3}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{2!3!}.$$

For the Mia and Max problem, we could count. But we prefer to work directly with probabilities.

Imagine that the people first get divided into $2$ groups, of $6$ and $12$, to make up the two kinds of team.

The probability that Mia is selected for the group of $6$ is $\frac{6}{18}$. Given that this has happened, the probability Max is chosen for the same team is $\frac{2}{17}$. Thus the probability Mia and Max are in the same team of three is
$$\frac{6}{18}\cdot\frac{2}{17}.$$
Similarly, the probability that Mia and Max are in the same group of $4$ is
$$\frac{12}{18}\cdot\frac{3}{17}.$$
Add.

Remark: Doing problem (b) by counting and dividing is not difficult, just a little more messy-looking. Note that the probabilities are the same for the labelled teams case as for the unlabelled case. It is easier not to make a mistake by counting the number of ways that Mia and Max can be on the same labelled team, and dividing by the number of labelled teams.

Starting with the "binomial(18, 3)+binomial(15, 3)+binomial(12, 4)+binomial(8, 4)+binomial(4, 4)" solution I got 1837 possible combinations, but if you start with dividing into the 4 teammember teams first I get "binomial(18, 4)+binomial(14, 4)+binomial(10, 4)+binomial(6, 3)+binomial(3, 3) = 4292 possible combinations", no idea if it's right or not just something to double check.


Is the number of Lebesgue covering dimensions a topological property? Is it correct to characterize the number of dimensions that a topological space has as a 'topological property'? By 'dimensions,' I have in mind Lebesgue covering dimensions.
If the answer is yes, then why? If the answer is no, then what kind of a property is the number of dimensions?

Isn’t it obvious from the definition that the covering dimension is a topological invariant? it’s defined purely in terms of open sets and set-theory concepts.

There are many different definitions of dimension in different contexts. Some are topological, most are not.  Look up: topological dimension.


If $a > b$, is $a^2 > b^2$? Given $a > b$, where $a,b ∈ ℝ$, is it always true that $a^2 > b^2$?

Given that $ a > b $, it is not always true that $ a^{2} > b^{2} $. One counterexample would be that one of $ a $ or $ b $ is negative, say $ a = 1 $, and $ b = -1 $. Then $$ a^2 = 1^{2} = 1 $$ and $$ b^{2} = (-1)^{2} = 1 $$ making $ a^2 = b^2 $, a contradiction of our assumption that $ a^{2} > b^{2} $.


Estimation, bias, and mean square error Let $X$ be a continuous random variable with pdf
$f(x) =\frac{1}{2}(1+ \theta x)$, for $-1 < x < 1$, and $-1 < \theta < 1$

(a) Show that $E(X) = \frac{\theta}{3}$.

(b) Given a random sample of size $n$ from a population with pdf $f(X)$, consider the estimator $\hat{\theta} = 3\bar{X}$. Find the variance, bias, and mean squared error of $\hat{\theta}$.

For (a) I simply integrate $x f(x)$

For (b) I'm not sure it is related to part (a) or the pdf at all.
I'm not quiet sure how to proceed.

I know that $\operatorname{Var}(\hat{\theta})=E((\hat{\theta}-E[\theta])^2)$

$\operatorname{Bias}(\hat{\theta})=E(\hat{\theta}-\theta))$, and

$\operatorname{MSE}=\sqrt{\operatorname{Var}(\hat{\theta})}$

Do I just need to manipulate the definitions?

$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$
\begin{align}
\E(\bar X) & = \E\left( \frac {X_1+\cdots+X_n} n \right) \\[8pt]
& = \frac 1 n \E(X_1+\cdots+X_n) = \frac 1 n (\E(X_1)+\cdots+\E(X_n)) \\[8pt]
& = \frac 1 n \Big( n\E(X_1) \Big) = \E(X_1). \\[25pt]
\var( \bar X ) & = \var \left( \frac {X_1+\cdots+X_n} n \right) \\[8pt]
& = \frac 1 {n^2} \var(X_1+\cdots+X_n) = \frac 1 {n^2} (\var(X_1)+\cdots+\var(X_n)) \\[8pt]
& = \frac 1 {n^2} \Big( n \var(X_1) \Big) = \frac 1 n \var(X_1).
\end{align}

Based on the above, you should be able to find the bias and the mean squared error.


The elementary coordinate geometry of polynomials? Of rational expressions? Of radicals? With a few colleagues, we're trying to design an (intermediate) algebra course (US terminology) where we stress the interplay between algebra and geometry. The algebraic topics we would like to cover are (1) linear equation in two variables, (2) quadratic equations in two variables, (3) polynomials in one variable, (4) rational functions in one variable (though we're not sure we want to introduce functions), (5) radicals.

For (1) and (2) there are obvious geometric counterparts: lines and conic sections.

Question: Are there natural geometric counterparts for (3), (4) and (5)?  Are there elementary geometric constructions that naturally lead to these algebraic objects?

Side question: Are there (affordable) textbooks or lecture notes out there which have this kind of approach?

This seems rather ambitious for most U.S. college intermediate algebra courses, which typically are not-for-credit remedial courses that lie below the level of college algebra and precalculus courses. Nonetheless, here are some things I've used in precalculus courses that might of use.

To see what the graph of something like $yx^2 + 2y^3 = 3x + 3y$ looks like using a (standard implicit-incapable) graphing calculator, solve for $x$ in terms of $y$ using the quadratic formula (or $y$ in terms of $x$ when possible, but I'm giving an example where we don't have the option of solving for $y$) and then graph both solutions simultaneously as if $x$ and $y$ were switched. That is, if you're using one of the TI graphing calculators, then enter for y1= and y2= the following:

y1  =  (3+(9-4x(2x^3-3x))^(1/2))/(2x)
y2  =  (3-(9-4x(2x^3-3x))^(1/2))/(2x)

(Note for would-be editors: Please don't LaTeX the expressions above, as what I've given is what the calculator input should be.)

To account for the fact that you're graphing the inverse relation, rotate what you see 90 degrees counterclockwise and then reflect the rotated result across the vertical axis. Equivalently, you can reflect what you see about the line $y=x$, but I suspect what I first suggested is easier for students to carry out.

Here's a more elaborate example. Suppose we want to know what the graph of $y^4 - 4xy^3 + 2y^2 + 4xy + 1 = 0$ looks like. (Yes, I know about the Newton polygon method, but let's not go there.) Although this can be solved for $y$ in terms of $x$, it is rather difficult to do so and the result is somewhat difficult to interpret graphically by hand. You'll get the 4 different expressions $y = x \pm \sqrt{x^2 - 1} \pm \sqrt{x^2 - x} \pm \sqrt{x^2 + x}$, where the 4 sign permutations are $(+,+,+),$ $(+,-,-)$, $(-,+,-)$, and $(-,-,+)$. On the other hand, it is easy to solve for $x$ in terms of $y$ and the result is $x = \frac{\left(y^2+1\right)^2}{4y\left(y^2-1\right)}$, which can be graphed by hand using standard methods for graphing rational functions.

For graphs of polynomial functions, especially when given as (or easily put into) factored as linear and real-irreducible quadratics with real coefficients (and probably best to mostly avoid using real-irreducible quadratics, at least at the beginning), you can discuss how their graphs locally look at each $x$-intercept and how their graphs roughly look globally by using "order of contact with the $x$-axis" notions (which you don't have to define precisely, of course) and end behavior. For example, since $y = (x+2)^3 (2x+1)^2 x (3-x)$ has the form $y = (x)^3(2x)^2(x)(-x) + \;$ lower order terms, or $y = -4x^7 + \;$ lower order terms, a zoomed out view of the graph will look like the graph of $y = -4x^7$, so the graph "enters at the upper left of quadrant 2" and "exits at the lower right of quadrant 4". Also, the graph passes through the $x$-axis at $x=-2$ "in a cubic fashion" so that locally at this zero the graph looks like a version (translated and reflected, the latter because in going from left to right the graph passes from positive $y$-values to negative $y$-values) of the graph of $y = x^3$. The same kind of analysis leads to what the graph locally looks like at the other $x$-intercepts, which I'll assume you know what I'm talking about by now since this is (or at least it used to be) a fairly standard topic in precalculus courses.

In the case of rational functions, one topic that could be investigated is linear fractional transformations (as they're called in complex analysis), or quotients of linear (= affine) functions, by looking at them through the lens of precalculus transformations of the graph of $y = \frac{1}{x}$. Although you definitely don't want to consider the general case, here's the general case version, where I'm assuming $c \neq 0$. (The first equality comes from a 1-step long division calculation.)

$\frac{ax+b}{cx+d} = \frac{a}{c} + \frac{b - \frac{ad}{c}}{cx+d}$

$= \frac{a}{c} + \frac{\frac{b}{c} - \frac{ad}{c^2}}{x+\frac{d}{c}}$

$= \frac{a}{c} + \frac{1}{c^2}(bc-ad)\left[ \frac{1}{1 - \left(-\frac{d}{c}\right)}\right]$

Note that this shows the graph of $y = \frac{ax+b}{cx+d}$ can be obtained from the graph of $y = \frac{1}{x}$ by a horizontal translation to the right by $-\frac{d}{c}$ units, followed by a vertical stretch by a factor of $\frac{1}{c^2}(bc-ad)$, followed by a vertical translation up by $\frac{a}{c}$ units. For a possibly useful reference, see Edward C. Wallace, "Investigations involving involutions", Mathematics Teacher 81 (1988), pp. 578-579. [See also these related letters in the Reader Reflections column: Thomas Edwards (MT 83, p. 496), Larry Hoehm (MT 83, p. 496), Andrew Berry (MT 95, p. 406), and Sidney H. Kung (MT 97, pp. 227 & 242).] While I'm on the topic, here's an "intermediate algebra appropriate" situation where a linear fractional transformation arises. For which number or numbers $x$, if any, can we find a number whose sum with $x$ equals its product with $x$? If we denote the desired number by $y$, then the condition becomes $x + y = xy$, or $y = \frac{x}{x-1}$.

For 5, for square roots, it seems almost too obvious to use the hypotenuse of right triangles. For higher roots, diagonals on cubes of higher dimensions?

For 3, a quadratic in one variable is also a conic section. For higher degrees...so this is for high school right? ...yeah this one isn't obvious.


Prove that if $\chi(G-u-v)=\chi(G)-2$ for every vertices $u, v$ ($u \ne v$) then G is complete graph. I'm trying to prove this by contradiction:
if $G$ isn't complete graph, there must exist vertices $u,v \in V(G)$ for which edge $uv \notin E(G)$. Then $u$ and $v$ must have the same color in proper coloring $C$ and now I'd like to prove that by removing $u, v$ we have $\chi(G-u-v) = \chi(G) - 1$ (that would be the contradiction), but I'm not sure whether it's true.

Consider the contrapositive of the statement instead:

If $G$ is not complete then $\exists u, v \neq u \text{ s.t. } \chi(G - u - v) \neq \chi(G) - 2 \quad \quad (1)$.

To prove (1), note that since $G$ is not complete, $\exists u, v \neq u$ with the same color in a proper minimal coloring of the graph; for such a pair of vertices we have that $\chi(G - u - v) \geq \chi(G) - 1$ and therefore $\chi(G - u - v) \neq \chi(G)-2$.

The accepted answer assumes a great deal about the existence of 2 vertices of the same color in the proper coloring of an incomplete graph $G$. It is better to assume less. Consider this proof by contradiction instead:

Assume $G$ is not a complete graph with $\chi(G)=n$ and that the statement holds. Then, $\exists$ vertices $u$, $v\neq w$ in $G$ s.t. $u$ is not adjacent to $v$. Remove these vertices and attain a graph $G-u-v$. By assumption, $\chi(G-u-v) = \chi(G) - 2 = n - 2$. Now, add back 2 vertices $u$, $v$ that are not adjacent to one another. This graph $G$ can clearly be colored in at most $(n-2) + 1 = n - 1$ colors since the maximum degree of any vertex is $\delta(G) = n - 1$, but this is a contradiction since $\chi(G) \leq n - 1 \implies n \leq n - 1$, a clear falsehood. Thus, $G$ is a complete graph if $\chi(G-u-v) = \chi(G) - 2$. $\Box$


how many permutations of {1,2,...,9} How many permutations of {1,2,…,9} are there such that 1 does not immediately precede 2, 2 does not immediately precede 3, and so forth up to 8 not immediately preceding 9? One obvious example of such a permutation might be 987654321, but there are many others, such as 132465879 or 351724698.

For $\{1,2,3,\dots,n\}$, the numbers are tabulated here, along with lots of information, formulas, references, links, and whatnot. For $n=9$, it says 148329.

Let us do it step by step

First take total permutation = 9!(With out any constraint and no repetition)

Ok then as we know 12 in a sequence is not allowed so calculate total number of permutation when there is 12 = 7!

There are total 8 pairs so Ans should be 9!-8*(7!)=322560


Multiplicative Inverse Element in $\mathbb{Q}[\sqrt[3]{2}]$ So elements of this ring look like
$$a+b\sqrt[3]{2}+c\sqrt[3]{4}$$
If I want to find the multiplicative inverse element for the above general element, then I'm trying to find $x,y,z\in\mathbb{Q}$ such that
$$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(x+y\sqrt[3]{2}+z\sqrt[3]{4})=1$$
I can see that expanding gives me the system
$$ax+2cy+2bz=1$$
$$bx+ay+2cz=0$$
$$cx+by+az=0$$
I don't want to solve this using matrices because I know it will turn ugly.  Is there a more elegant way to approach the inverse calculation to avoid the ugly calculation?  The only thing I thought of was setting the bottom two equations equal to each other
$$bx+ay+2cz=cx+by+az$$
Which seems to indicate that
$$a=b, b=c, a=2c$$ but this would make me think $a=b=c=0$ and thus a multiplicative inverse does not exist.

First let me remark that there is a general abstract argument which shows that if $L/K$ is a field extension and $a \in L$ is algebraic over $K$, then the $K$-algebra $K[a]$ is a field. Namely, $K[a]$ is an integral domain  which is also a finite-dimensional vector space over $K$. This implies for $0 \neq x \in K[a]$ that the linear map $K[a] \to K[a]$, $y \mapsto x \cdot y$ is surjective, since it is injective, which means that $x$ is invertible. To get a constructive proof, we just have to sit down and make the linear algebra argument here explicit.

The underlying vector space of $\mathbb{Q}[\sqrt[3]{2}]$ has basis $1,\sqrt[3]{2},\sqrt[3]{4}$. For some fixed non-zero element $x=a + b \sqrt[3]{2} + c \sqrt[3]{4}$, let's write down the linear map $y \mapsto x \cdot y$ in terms of this basis:

$~~~\,1 \mapsto a + b \sqrt[3]{2} + c \sqrt[3]{4}$

$\sqrt[3]{2} \mapsto 2c + a \sqrt[3]{2} + b \sqrt[3]{4} $

$\sqrt[3]{4} \mapsto 2b+2c\sqrt[3]{2} + a \sqrt[3]{4} $

The corresponding matrix is:

$$\begin{pmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{pmatrix}$$

From linear algebra we know how to invert matrices, for example via Cramer's rule. In this case, we get:

$$\frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot \begin{pmatrix} a^2 - 2bc &  2b^2 - 2ac &  4c^2 - 2ab \\  2c^2 - ab &  a^2 - 2bc &  2b^2 - 2ac \\ b^2 - ac &  2c^2 - ab &  a^2 - 2bc \end{pmatrix}$$

The determinant $a^3 - 6abc + 2b^3 + 4c^3$ has been computed via the Rule of Sarrus and  the cofactors have been computed by the usual formula for $2 \times 2$-determinants.

This matrix represents the linear map $y \mapsto x^{-1} \cdot y$ with respect to our basis. Thus, to get $x^{-1}$, we just have to evaluate at $1$, and we get:

$$x^{-1} = \frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot ((a^2 - 2bc) + (2c^2 - ab) \sqrt[3]{2} + (b^2 - ac) \sqrt[3]{4})$$

Of course, this method works quite generally. For example, for $x=a + b \sqrt[3]{p} + c \sqrt[3]{p}^2$ we have:

$$x^{-1} = \frac{1}{a^3 - 3pabc + pb^3 + p^2 c^3} \cdot ((a^2 - pbc) + (pc^2 - ab) \sqrt[3]{p} + (b^2 - ac) \sqrt[3]{p}^2)$$

Try these:
Rationalizing the denominator 3
Deradicalization of denominators


Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.

OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$

After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$

How do I proceed from here to end up with something squared $- 1$?

You might note that for any $N$ at all, $$(N-1)(N+1) = N^2-1$$ and so is one less than a perfect square. I presume you have seen this before.

Then you could take $N = M+1$ in the formula above, and get $$M\cdot(M+2) = (M+1)^2 - 1$$ is one less than a perfect square.

And here you have $M = n^2+3n$, don't you?

It seems like the way to attack this that doesn't require guessing is this:  Start with $N (N-1) (N-2) (N-3) = N^4 + 6 N^3 + 11 N^2 + 6 N = M^2 - 1$  Since the product is "near" $N^4$, $M$ has to be "near" $N^2$.  And it's very likely that $M$ is a polynomial in $N$.  So set $M = N^2 + aN + b$.  Then $M^2-1 = N^4 + 2aN^3 + (2b+a^2)N^2 + 2abN + (b^2-1)$.  Setting that into the first equation and equating powers of $N$ gives $2a = 6, 2b+a^2 = 11, 2ab = 6, b^2-1 = 0$, which has the unique solution $a = 3, b = 1$.  So $M = N^2 + 3N + 1$ and $N (N-1) (N-2) (N-3) = (N^2 + 3N + 1)^2 -1$.


What's the behavior of $\displaystyle\sum_{n=1}^\infty (z+\sqrt{5}+2i)^{n!}$ outside its radius of convergence? I want to check the behavior of $$\displaystyle\sum_{n=1}^\infty (z+\sqrt{5}+2i)^{n!}$$ outside its radius of convergence.

I've tried to use the ratio test as follows: $$\left|\frac{(z-\sqrt{5}+2i)^{(n+1)!}}{(z-\sqrt{5}+2i)^{n!}}\right|=|(z-\sqrt{5}+2i)^{nn!}|$$
This will converge to zero, if $|z-\sqrt{5}+2i|<1$.

However, what's about the cases $|z-\sqrt{5}+2i|=1$. How can I show convergence or divergence in this case?

EDIT: I've updated my question to make things more clear. Why I was down-voted for that?

Evidently, this is a power series around $-\sqrt 5 -2i$, so you really should make a change of variable and look at the power series $$\sum z^{n!}.$$

The ratio test for series will indeed give you the correct radius of convergence 1, as you have written in your question.

However, the ratio test never helps you for boundary behaviour.

On the boundary, you usually use one of the following approaches:

terms do not even converge to zero, therefore divergent
terms are positive for $z=R$ and converge (because of known series, integral test, etc), therefore absolute convergence on boundary
terms converge for $z\not=R$ but on boundary due to Leibniz test/Abel criterion/Raabe test etc.

If we wish to known the radius of convergence of $$\sum_{n=0}^\infty z^{n!}=z+z+z^2+z^6+z^{24}+\cdots, $$ then we calculate, $$\frac{1}{R} = \limsup \sqrt[n]{|\text{$n^{th}$ coefficient}|} = \limsup \sqrt[n]{\{0,2,1,0,0,0,1,0,0,0,\ldots\}}=1. $$ Rudin Principles of Mathematical Analysis - page 69 - Theorem 3.39


Value of $x^2 \sin (\frac1x)$ at $x=0$ What is the value of $y=x^2 \sin(1/x)$ at $x=0$?

I see that $x^2 =0$ but $\sin (1/x)$ is undefined.

More generally: if a function made up of a product of functions,  like $y= a(x)b(x)c(x)d(x)\dots$, then at a specific value of $x$ if one of the functions is $0$ and all the other functions are undefined, does it still mean that $y=0$?

The function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x = 0$ because "$1/0$" does not exist. Note also that this function is continuous on $\mathbb{R}^{\ast}$.

However, this is not all we can say. Since the function $x \, \longmapsto \, \sin \Big( \frac{1}{x} \Big)$ is bounded, one can prove that :

$$ \lim \limits_{x \to 0} x^{2} \sin \Big( \frac{1}{x} \Big) = 0.$$

As a consequence, even though the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x=0$, it has a finite limit as $x$ goes to $0$. So, this function can be extended to a new function, say $\overline{f}$, defined on $\mathbb{R}$ the following way :

$$ \overline{f}(x) =
\begin{cases} x^{2} \sin \Big( \frac{1}{x} \Big) & \text{if } x \neq 0 \\[2mm]
0 & \text{if } x = 0
\end{cases}
$$

This new function $\overline{f}$ is defined at $x=0$ and continuous on $\mathbb{R}$. In a very unformal, non-mathematical way, we could say that the value of the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ at $x=0$ is $0$. But this is not really the truth !

Pretty sure if one of the functions is undefined $y$ must also be undefined.

Pretty sure this is a question about limits though where $y\rightarrow 0$ when $x\rightarrow 0$.