Do values attached to integers have implicit parentheses? Given $5x/30x^2$ I was wondering which is the correct equivalent form.

According to BEDMAS this expression is equivalent to
$5*\cfrac{x}{30}*x^2$
but, intuitively, I believe that it could also look like:
$\cfrac{5x}{30x^2}$

I asked this question on MathOverflow (which was "Off-topic" and closed) and was told it was ambiguous. I was wondering what the convention was or if such a convention exists. According to Wikipedia the order of operations can be different based on the mnemonic used.

I am going to play Devil's Advocate on this one, because I've given it a little bit of thought after a tangentially related question on MathOverflow about bad notation. There is a good lexical argument to be made for the "intuitive" reading of $5x/30x^2$ as $\frac{5x}{30x^2}$.

Disclaimer. The following argument is not intended to communicate why expressions such as $a/bc$ are unambiguous. By definition, if there is no strong socially agreed upon convention, it is therefore ambiguous. However, I will argue for the hypothesis that almost anyone raised from an early age to read a European language, who quickly reads such an expression without reflecting too much, will probably understand $a/bc = \frac{a}{bc}$. Therefore, there exists a reasonable opportunity to establish such a convention.

One of the responders to the question of bad notation complained about the usage of juxtaposition for multiplication: he thought it introduced too much ambiguitiy. (For instance, does $f(x+y)$) refer to the product of a scalar $f$ with $x + y$, or the value of a function $f$ evaluated at $x + y$, or something else?) In the comments to that answer, I speculated that the reason for this convention is that it reduces the problem of parsing a mathematical expression such as $ax^2 + bx + c$; to the previously-solved problem (for Europeans, anyway) of parsing a written sentence in a European language, which breaks sentences into words. Juxtaposed variables form nice little cohesive "words". This hypothesis can even be extended to account for exponents and subscripts — more notation in which we don't use operator 'symbols' — to play the role of intra-word apostrophes or other diacritical marks.

Under this hypothesis, I can make a strong case for prefering the reading $a/bc$ as $\frac{a}{bc}$. Unlike multiplication or exponentiation, we introduce an actual symbol — a prominent punctuation mark, if you will, as intrusive as the $+$ symbol — into the expression. This divides the expression into two "words", $a$ and $bc$, upon which we perform the operation of division; thus leading to the reading $a/(bc)$. This is especially likely in the case where $a$ is effectively a place-holder, as in $1/2x$, where for "fluent speakers of mathematics" the $1$ is only there as a placeholder to allow the concept of multiplicative inverse to be expressed. Add to this the fact that it makes no sense at all to write $a/bc$ if you wish to convey the meaning $ac/b$, and one can almost safely say that anyone who writes $a/bc$ means $\frac{a}{bc}$.

The above is meant as a descriptivist (as opposed to prescriptivist) argument for how people read and write mathematics. If this is going to happen anyway, and if (as appears to be the case) it doesn't entail any heavy modification of our existing conventions, why not accept it in order to allow more flexibility in our notational repertoire? We can cut down on parentheses, and get rid of this phony ambiguity between something which can be better expressed anyway and an expression which at present we can only confidently convey with parentheses or with extra vertical space.

Now, of course, there is no solid existing agreement about whether to read $a/bc$ as $\frac{a}{bc}$ or as $\left(\frac ab\right)c$, except to accept that latter because of BEDMAS or what-have-you. So if your job is to evaluate your student's ability to communicate, then you should conclude that they aren't doing such a good job. But if you are trying to evaluate what they mean — and if you believe that they learned to read a European language from a tender age — then chances that when they write $5x/30x^2$, they mean $\frac{5x}{30x^2}$.

I think that the ambiguity originates, not from what is written, but from whom does the writing.

One who is completing her math assignment should write $\frac{a}{b}$ on her paper.
One who is shingling your roof should write $1-1/2"\times 25 \div 7$ on a piece of plywood.


Generalized Euclid's lemma for factorial rings. Let $A$ be a factorial ring. Is the following result true in this generality: if $a|bc$ and $a$, $b$ are coprime then $a|c$? It would be great if this were true, but I am not at all true. I think the classical proof should go through

As we are working in a factorial ring (what I learned to call a unique factorization domain), we have that every element can be written as a product of irreducibles and if two such products are equal then there is a bijection between terms with each an associate of the other.

Furthermore, for definition of coprime, we mean that if $a$ is coprime to $b$ that no divisor of $a$ is also a divisor of $b$ and vice versa (except for a unit).  I.e. in the factorizations, none of $a$'s factors are associate to any of $b$'s factors.

Let $a = a_1a_2\dots a_n$, $b = b_1b_2\dots b_o$, $c=c_1c_2\dots c_p$

$a|bc \Rightarrow ka = bc\Rightarrow k_1k_2\dots k_ma_1a_2\dots a_n=b_1b_2\dots b_oc_1c_2\dots c_p$

Since $a$ is coprime to $b$, that means that none of $a_1,a_2,\dots,a_n$ are associates to $b_1,b_2,\dots,b_o$, however each of $a_1,a_2,\dots,a_n$ must be associate to something on the righthandside of the equation, implying that each of $a_1,a_2,\dots,a_n$ are associates to some collection of $c_1,c_2,\dots,c_p$ (not necessarily all).

As such, by a permutation of indices and by grouping of units, we may assume without loss of generality that $a_1 = uc_1, a_2 = c_2, a_3 = c_3,\dots, a_n = c_n$ with $n\leq p$ and $u$ a unit.

Thus, $a(u\prod_{i=n+1}^p c_i) = c$ and thus $a|c$.

If by "factorial ring" you mean a unique factorisation domain, then the proof is easy. Just note that if $p \mid a$ and $p$ is prime, then $p \nmid b$ by hypothesis, so $p \mid c$.

This is because by definition a prime element is a $p$ such that if $p \mid bc$ then either $p \mid b$ or $p \mid c$. Also note that in a UFD every irreducible element, i.e. every element which is not the product of two non-invertible elements, is prime.


Odds of winning contest I enter a contest every week with $850$ others that I have a random chance of being selected as the winner.  There is only one winner.  If I play for $850$ weeks, should I have a $100\%$ chance of being a winner?

The probability of beeing selected at least once is One minus the probability of beeing never selected (converse probability)

$P(X\geq 1)=1-P(X=0)$,

where $X$ is the random variable of beeing selected in 850 weeks. It is binomial distributed:

$X\sim Bin(850,p)$

$P(X=0)=\binom{850}{0}\cdot p^0\cdot (1-p)^{850}=(1-p)^{850}$

p is the probability of beeing selected in one week.

$1-P(X=0)$ is $1$ if $P(X=0)=0$.

We can assume that $p<1$. It is possible that $(1-p)^{850}$ becomes $0$ ?

The winning chances are independent of each other. So, for every match you have chance of 1 for 850 members. Even if you play for infinite times, there is no guarantee for 100% winning. So for 850 matches you have probability of winning as

  1/(850)^ 850

The above case is probability for winning in all the 850 matches.

The probability for winning atleast once is binomial. It is

  1 - ( losing all the times ).

Losing all the times is  (no. of players - 1) / no. of players raised to power of no. of matches played


Finding the number of Spanning Trees of a Graph $G$ This is my first question on the Mathematics StackExchange site, so please do tell me if I have not adhered to any conventions in this post.

Ok, so if I have a graph $G$, with say $6$ vertices and $7$ Edges, how would I determine how many possible spanning tree's are there?

I know this is probably a very simple question, but I am very new to Graph Theory.

Thanks.

One of my favorite ways of counting spanning trees is the contraction-deletion theorem. For any graph $G$, the number of spanning trees $\tau(G)$ of $G$ is equal to $\tau(G-e) + \tau(G / e)$, where $e$ is any edge of $G$, and where $G-e$ is the deletion of $e$ from $G$, and $G/e$ is the contraction of $e$ in $G$. This gives you a recursive way to compute the number of spanning trees of a graph.

Another rule, is that if you have a graph with a cut-vertex (a vertex which removal would separate the graph), then the number of spanning trees can be computed on each side of the cut-vertex, and then multiplied together.

There are some more examples of rules, for complete graphs, complete bipartite graphs and others in Section 5.2 here http://www.student.dtu.dk/~dawi/01227/01227-GraphTheory.pdf. It also contains an appendix (D) of small graphs and their number of spanning trees, which is useful if you use the contraction-deletion theorem.


Do moments define distributions? Do moments define distributions?

Suppose I have two random variables $X$ and $Y$. If I know $E\left[X^k\right] = E\left[Y^k\right]$ for every $k \in \mathbb N$, can I say that $X$ and $Y$ have the same distribution?

This question is known as (indeterminate) moment problem and has been first considered by Stieltjes and Hamburger. In general, the answer to your question is: No, distributions are not uniquely determined by their moments.

The standard counterexample is the following (see e.g. Rick Durrett, Probability: Theory and Examples): The lognormal distribution

$$p(x) := \frac{1}{x\sqrt{2\pi}} \exp \left(- \frac{(\log x)^2}{2} \right)$$

and the "perturbed" lognormal distribution

$$q(x) := p(x) (1+ \sin(2\pi \log(x))$$

have the same moments.

Much more interesting is the question under which additional assumptions the moments are determining. @StefanHansen already mentioned the existence of exponential moments, but obviously that's a strong condition. Some years ago Christian Berg showed that so-called Hankel matrices are strongly related to this problem; in fact one can show that the moment problem is determinante if and only if the smallest eigenvalue of the Hankel matrix converges to $0$. For a more detailed discussion see e.g. this introduction or Christian Berg's paper.

A moment is an integral of a weighted probability distribution, which is a number. So the mapping is "many-to-one" which is not invertible. So an integral cannot serve as the definition for an integrand.


Is there a formula for the roots of a Quintic Equation? I can get my head around this so someone explain it please.

$(1)$ From Galois theory it is known there is no formula to solve a general quintic equation.

But it is known a general quintic can be solved for the 5 roots exactly. Back in 1858 Hermite and Kronecker independently showed the quintic can be exactly solved for (using elliptic modular function). Also I think they're maybe other solution for the quintic which means a formula for each of the 5 roots.

So why is the claim in Galois theory that there is no formula to solve it?
I know I am missing something here because the above $(1)$ is an established result.

So what is the value in saying, using Galois theory we do not have a formula for the 5 roots? Since for practical purposes we can actually find the 5 roots each time using say for example the formula based on elliptic modular functions.

Galois-theory only says that there is no general formula to solve a quintic equation in terms of radicals. That is, there is no formula only using the arithmetic operations "sum, multiplication etc. and taking the $n$-th root".

For instance for the polynomial $x^5 - 4x + 2$ it is known that it has a root that is not expressible in the above mentioned operations (as its Galois-group is $S_5$). (Edit: Another example is $x^5 + x + 1$ which also has Galois-group $S_5$. If you Wolframalpha this polynomial you see nicely how four of its roots can be expressed by radicals, but the fifth can't.)

The solution you mean is the solution using Bring radicals - wikipedia-article here: https://en.wikipedia.org/wiki/Bring_radical -, which is not a contradiction, as it is not expressed in form of radicals (in the sense of $n$-th roots of something).

I'd like to share that the principal quintic $x^5+ax^2+bx+c=0$ is algebraically solvable by the "Fórmula Luderiana para Equação Quíntica" (Luderian Formula for Quintic Equation) just check it out at slideshare for details.


Proof that $\overline{A}=A\cup \partial A$ Let $(X,d)$ be a metric space and $A\subset X$. Prove that $\overline{A}=A\cup \partial A$ (that is, the closure of $A$ is the union of $A$ with its boundary).

Let $x\in \overline{A}$, then $x\in A$ or $x\not\in A$. Suppose $x\not\in A$, then $\forall\varepsilon >0, B_\varepsilon(x)\cap X\setminus A\ne \emptyset$. Since $\overline{A}$ is closed, there is a sequence $(x_k)_k\subset A$ converging to $x$. Since $(x_k)_k\subset A$, for every $x_k$ there exists $\varepsilon_{x_k}$ such that $B_{\varepsilon_{x_k}}(x_k)\subset A$. Thus $\forall \varepsilon >0$, $B_\varepsilon(x)\cap A\ne \emptyset$. Hence, $x\in \partial A$. Thus $\overline{A}\subset A\cup \partial A$.

For the other direction, suppose $x\in A\cup \partial A$, then $x\in A$ or $x\in \partial A$. If $x\in A$ then $x\in\overline{A}$. If $x\in \partial A$ then $\forall\varepsilon > 0$, $B_\varepsilon(x)\cap A\ne \emptyset \ne B_\varepsilon(x)\cap X\setminus A$. Since $B_\varepsilon(x)\cap A\ne \emptyset$, there are infinitely many points in $B_\varepsilon(x)$ of some sequence contained in $A$. Let $\varepsilon'<\varepsilon$, then there are infinitely many points in $B_\varepsilon'(x)$ of some sequence contained in $A$. Proceeding by induction and taking $\varepsilon$ smaller and smaller, we see that for every $\varepsilon > 0$ there are infinitely many terms in $B_\varepsilon(x)$ of some sequence contained in $A$. Inductively, this means that there is a sequence contained in $A$ which converges to $x$. Thus, $x\in\overline{A}$.

Please let me know if you find my proof correct. Maybe it's a bit over-complicated, but it would be interesting to know if it's OK.

Note that $\partial A = \overline{A} \setminus A^\circ = \overline{A} \cap (A^\circ)^c$. Then $$A \cup \partial A = A \cup \left(\overline{A} \cap (A^\circ)^c\right) = (A \cup\overline{A}) \cap (A \cup (A^\circ)^c).$$ Since $A \subseteq \overline{A}$, $A \cup \overline{A} = \overline{A}$. Further, $\overline{A} \subseteq A \cup (A^\circ)^c$ so the result follows.

Write A= U {B(x,r(x))| x is in A} . cl A= U{ cl B(x,r)| x is in A} . Any boundary point is arbitrarily close to some x in A so it is within cl B(x,r). Thus every boundary point is in the closure.


Minimum value of quadratic equation I know that $ax^2+bx+c$ if $a < 0$ it has maximum value and we find $m = -b/2a$
but if $a>0$ it should have minimum value how can find it?
Thanks in advance.

It's the same $m = \frac{-b}{2a}$. One way you can notice this is as follows: Suppose the 2nd degree polynomial $ax^2+bx+c$ has $a>0$. Then multiply through by $-1$ and you'll have a 2nd degree polynomial with $a<0$. As you pointed out, the second polynomial will have minimum at $x = \frac{-b}{2a}$. This must mean the same value of $x$ gives a maximum in the first polynomial.

It is very simple.

If $a<0$ implies that $ax^2+bx+c$ has a maximum, then $-ax^2-bx-c$ (such that $-a>0$) has a minimum, and these two functions are mirrors of each other around the horizontal axis.


Asymptotic standard normal distribution I need to solve the following exercise.

  Assume that $X_\lambda$ is Poisson distributed with mean $\lambda$ . Show that $Y(\lambda) = \frac{X_\lambda - \lambda}{\sqrt{\lambda}}$ is asymptotic standard normal distributed for $\lambda \to \infty$.

Any help would be really appreciated.

Not quite sure if the one on Wiki is right or if I understand it, but anyway I'm basing my answer on CLT in Larsen and Marx (an elementary probability book):

Anyway, this answer might be weird or very wrong or stupid but...

Consider iid $X_i$ ~ $Poi(\lambda/n)$.

Then $X := \sum_{i=1}^{n} X_i$ ~ $Poi(\lambda)$

Then

$$(\frac{1}{n} \sum_{i=1}^{n} X_i - \mu) / (\sigma/\sqrt{n})$$

$$= (\frac{1}{n} \sum_{i=1}^{n} X_i - \lambda/n) / (\sqrt{\lambda}/n)$$

$$= (\sum_{i=1}^{n} X_i - \lambda) / (\sqrt{\lambda})$$

$$= Y(\lambda)$$

Now I'm thinking we have to come up w/ some kind of relationship between n and $\lambda$ s.t. as $n \to \infty$, $\lambda \to \infty$

Guess 1:

$$= (\sum_{i=1}^{(\frac{\lambda}{E[X_i]})} X_i - \lambda) / (\sqrt{\lambda})$$

I was thinking as $\lambda \to \infty \iff n \to \infty$ as $\lambda = n E[X_i]$ but then again $E[X_i]$ is a function of both n and $\lambda$

Guess 2:

$P(X_i = k) \ge 0 \to (\frac{\lambda}{n})^k \ge 0$

Not sure how to proceed. I guess $\exists a > 0$ s.t. $\lambda^k = a n^k$ ?


How to solve the inequality $x^2>10$ using square roots? Solve the inequality: $$x^2>10$$

How am I supposed to do this? It doesn't make sense when I take into account that if $x^2=10$ then $x=+\sqrt{10}$ and $x=-\sqrt{10}$

But how am I supposed to apply this to an inequality, I would get $x>\sqrt{10}$ and $x>-\sqrt{10}$

But for some reason this just doesn't make sense to me.  Can someone explain it to me mathematically, instead of just having to memorize these kinds of things?

Sketch the graph of $x^2$ (it's a parabola opening upwards with vertex in $(0,0)$) and sketch the line $y=10$.

They intersect in $x=-\sqrt{10}$ and $x=\sqrt{10}$, and the sketch immediately gives the solution to the inequality:

$$x<-\sqrt{10} \vee x>\sqrt{10}$$

One way to think about this is as a graph.  What happens if you plot $y= x^2$?  You get a parabola.  Now, for which values of $x$ is $y > 10$?  The answer is $x>\sqrt{10}$ and $x<\sqrt{10}$.

You can see a graph like this here:  http://www.wolframalpha.com/input/?i=x%5E2+%3D+10


How to find $\phi$ for triple integral spherical coordinates Having trouble calculating bounds and how to do it generally for other terms.

I've been asked to find the volume of the solid bounded by the sphere

$x^2 + y^2 + z^2 = 4$

and the cones

$z = \sqrt{x^2 + y^2}$

$z = -\sqrt{3x^2 + 3y^2}$

I assume that $0 <= \rho <= 2$ and $0 <= \theta <= 2\pi$, but I don't know how to find $\phi$ for either cones.

Using cylindrical coordinates, we have the equations $$\begin{align}
z^2&=4-r^2,\\
z&=r\\
z&=-\sqrt 3 r
\end{align}.$$

The first equation gives $4-r^2\ge 0,$ or $0\le r\le 2.$ Then the last two give $-\sqrt 3 r\le z\le r.$ Finally, we have $0\le \phi\le 2π,$ since all the equations are independent of $\phi.$

So you want to calculate the $$\int_0^{2π}\int_0^2\int_{-\sqrt 3 r}^r{r\mathrm dz \mathrm dr\mathrm d\phi.}$$


orthogonal vectors of vector $(a,b)$ which is of same length as that of vector $(a,b)$ We have a vector $(a',b')$ which is orthogonal to vector $(a,b)$ and has same length as that of vector $(a,b)$ and lies counterclockwise of vector $(a,b)$.If vector $(a,b)$ is represented by $1i+ 1j$ .Then what should be vector $(a',b')$? How do we solve for vector $(a',b')$?

Might not be what you looking for, but here's an alternative approach exploiting the ease of rotating in the complex plane.

Identify $(a,b)$ with the complex number $z=a+bi$. The vector orthogonal to $(a,b)$ in counterclockwise direction, is the vector you get by rotating over 90° in the same direction. In the complex plane, this rotation corresponds to the simple multiplication with $i$, so the corresponding complex number is:
$$z' = iz = i(a+bi) = bi^2+ia = -b+ai$$
and this corresponds to the vector $(a',b') = (-b,a)$.

M and M' is orthogonal since the angle between two vectors is 90 degrees. Just rotate your point 90 degrees in clockwise direction


Combinatorics Question of finding number of ways for choosing correct answer In a multiple choice test there are six questions, each have four alternative answers of which only one answer is correct. If a candidate answers all the questions by choosing one answer for each question then number of ways to get exactly four correct answer is

There are $\binom{6}{4} = 15$ ways to choose $4$ question to answer right, $1^4$ ways to choose right answer for each of them and $3^2$ ways to answer two remaining questions. So there are $\binom{6}{4}\cdot 1^4\cdot 3^2 = 135$ ways to answer exactly $4$ questions right.


How to prove that this function is integrable on $[0,1]$ Here I tried to find two step functions, one of them is less than $f$ on $[0,1]$ whereas one of them is greater than $f$ on the same closed interval, to prove this function is Riemann-integrable on this interval, however, I could not find such two step functions to help me show this function is integrable. May I get some hint about the solution to this problem?

Hint:

$$\int_0^1 (-1)^{[1/x]} \, dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^{[1/x]} dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^k dx \\ = \sum_{k=1}^{\infty} (-1)^k \left[\frac1{k}-\frac1{k+1} \right]=\sum_{k=1}^{\infty}  \frac{(-1)^k}{k(k+1)}$$

Alternatively consider a partition $P = (0, 1/n, \ldots, 1/2, 1)$.

The difference between upper and lower Riemann sums is $1/n$ since

$$\sup_{(1/(k+1),1/k]} f(x) - \inf_{(1/(k+1),1/k]} f(x) = (-1)^k - (-1)^k = 0,$$

and

$$\sup_{[0,1/n]} f(x) - \inf_{[0,1/n]} f(x) = 1.$$

By choosing $n$ sufficiently large, this difference can be made smaller than any $\epsilon > 0$ and $f$ is Riemann integrable.

First try to sketch the graph.

Denote the upper and the lower Riemann sums for a partition $P$ of a function $f$ by $U(P,f)$ and $L(P,f)$ respectively.

We say that $f$ is Riemann integrable if for any $\varepsilon >0$ there is a partition $P$ such that $$\big| U(P,f)-L(P,f)\big|< \varepsilon$$

Observe that the upper and the lower sums are always equal for a constant function.

Pick $\varepsilon >0$. There is a natural number $N$ such that $\frac 2 {N} < \varepsilon$

Consider the partition $0 < \frac 1 N < \frac 1 {N-1} < \frac 1 {N -2} < \cdots < 1$

Now since $f$ is constant on $( \frac 1 2 , 1]$ and $( \frac 1 3, \frac 1 2 ]$ and $\dots$ and $( \frac 1 N,\frac 1 {N-1}]$, upper and lower sums will be equal in these intervals.

Also, $f$ is bounded above by $1$ and below by $-1$ gives that

$$\big| U(P,f) - L(P,f) \big| \leq \bigg| \big(1 \cdot \frac 1 N + \text{ some other terms} \big) - \big( -1 \cdot \frac 1 N + \text{ some other terms} \big) \bigg| = \frac 2 N < \varepsilon$$

Many steps are missing in the proof above.


What is $57^{46}$ divided by 17? $$\frac{57^{46}}{17}$$

What is best and quickest way to evaluate an approximate answer to this by hand?

It really depends what you mean by "approximate answer". Since you tagged arithmetic...

Notice 57 = 3 . 19, you can say that $57^{46} / 17 = 3 \cdot 57^{45} \cdot \frac{19}{17}$. So if you know what $57^{45}$ is, you can approximate to within 2% if you multiply that by 3.3.

Now, compute by hand $57^5 = 601692057$, which is less than 0.3% off from $6\cdot 10^8$. Observe that $57^{45} = (57^5)^9$. Using the binomial theorem you see that you can approximate that by $(6\cdot 10^8)^9 = 10 077 696 \cdot 10^{72}$ to within 3%. Rounding off the lower digits won't matter much in the error, so you have $57^{45} \sim 10^{79}$ to 3%.

So putting it all together you have that to within 5%,

$$ 57^{46} \sim 3.3 \cdot 10^{79} $$


Writing SQL code to recognize when three input lengths form a triangle I tried to do a SQL code that recognize automatically when we have a triangle from three different length inputs $A$, $B$, and $C$ for an HackerRank contest but it seems I'm wrong in my attempt. I ouput the following results :

10 10 10 Equilateral
11 11 11 Equilateral
30 32 30 Isoceles
40 40 40 Equilateral
20 20 21 Isoceles
21 21 21 Equilateral
20 22 21 Scalene
20 20 40 Isoceles
20 22 21 Scalene
30 32 41 Scalene
50 22 51 Scalene
20 12 61 Not A Triangle
20 22 50 Not A Triangle
50 52 51 Scalene
80 80 80 Equilateral

One was guilty : $(20,20,40)$ but may I say a triangle is an object with three sides where the sum of two is at least equal or strictly equal to the last side it doesn't change anything : I'm wrong.

Here is the code I tried :

SELECT A,B,C,
    CASE WHEN A=B AND A=C THEN 'Equilateral'
         WHEN (A+B>=C AND A+C>=B AND B+C>=A) AND
              (A=B OR A=C OR B=C) THEN 'Isoceles'
         WHEN (A+B>=C AND A+C>=B AND B+C>=A) THEN 'Scalene'
         ELSE 'Not A Triangle'
    END
FROM TRIANGLES

But even with strict inequalities the following output is wrong :

10 10 10 Equilateral
11 11 11 Equilateral
30 32 30 Isoceles
40 40 40 Equilateral
20 20 21 Isoceles
21 21 21 Equilateral
20 22 21 Scalene
20 20 40 Not A Triangle
20 22 21 Scalene
30 32 41 Scalene
50 22 51 Scalene
20 12 61 Not A Triangle
20 22 50 Not A Triangle
50 52 51 Scalene
80 80 80 Equilateral

Update June, 2nd

One answer that works is the following :

SELECT
    CASE WHEN A + B > C AND A+C>B AND B+C>A THEN
        CASE WHEN A = B AND B = C THEN 'Equilateral'
             WHEN A = B OR B = C OR A = C THEN 'Isosceles'
             WHEN A != B OR B != C OR A != C THEN 'Scalene'
        END
     ELSE 'Not A Triangle'
END FROM TRIANGLES;

With the following output:
As you can see, it's the same. I don't understand :

Equilateral
Equilateral
Isosceles
Equilateral
Isosceles
Equilateral
Scalene
Not A Triangle
Scalene
Scalene
Scalene
Not A Triangle
Not A Triangle
Scalene
Equilateral

As @uniquesolution says in the comments, the better approach is to ensure trianglehood first, and classify triangles second.  So:

If any two sides fail to exceed the length of the third, it is not a triangle.  Else...
If all three sides are equal, the triangle is equilateral.  Else...
If two sides are equal, the triangle is isosceles.  Else...
The triangle is scalene.

TRY THIS:
SELECT CASE
    WHEN A = B and B = C and C = A THEN 'Equilateral'
    WHEN (A=B or A=C or B = C) and (A+B > C AND A+C>=B AND B+C>=A)THEN 'Isosceles'
    WHEN (A<>B and B<>C and C<>A) and (A+B > C AND A+C>=B AND B+C>=A) THEN 'Scalene'
    ELSE 'Not A Triangle'
END AS Dummy
FROM TRIANGLES;

Which returns
Equilateral
Equilateral
Isosceles
Equilateral
Isosceles
Equilateral
Scalene
Not A Triangle
Scalene
Scalene
Scalene
Not A Triangle
Not A Triangle
Scalene
Equilateral


How to prove that $\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1$? How can one prove this identity?

  $$\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1$$

There is a formula for $\zeta$ values at even integers, but it involves Bernoulli numbers; simply plugging it in does not appear to be an efficient approach.

$$
\begin{align}
\sum_{n=1}^\infty\frac{\zeta(2n)}{2^{2n-1}}
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac2{k^{2n}2^{2n}}\tag{1}\\
&=2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{(4k^2)^n}\tag{2}\\
&=2\sum_{k=1}^\infty\frac1{4k^2-1}\tag{3}\\
&=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\tag{4}\\[6pt]
&=1\tag{5}
\end{align}
$$
Explanation:
$(1)$: expand $\zeta(2n)=\sum\limits_{k=1}^\infty\frac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum

$\newcommand{\angles}[1]{\left\langle{#1}\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace{#1}\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack{#1}\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left({#1}\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert{#1}\right\vert}$

\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\zeta\pars{2n} \over 2^{2n - 1}}} =
\sum_{n = 2}^{\infty}{\zeta\pars{n} \over 2^{n - 1}} - \sum_{n = 1}^{\infty}{\zeta\pars{2n + 1} \over 2^{2n}}
\\[3mm] = &\
-\sum_{n = 2}^{\infty}\pars{-1}^{n}\zeta\pars{n}\pars{-\,\half}^{n - 1} -
\sum_{n = 1}^{\infty}\bracks{\zeta\pars{2n + 1} - 1}\pars{\half}^{2n}\ -\
\underbrace{\sum_{n = 1}^{\infty}\pars{\half}^{2n}}_{\ds{1 \over 3}}
\\ = &\
-\bracks{\Psi\pars{1 + z} + \gamma}_{\ z\ =\ -1/2}
\\[3mm] & - \bracks{%
{1 \over 2z} - \half\,\pi\cot\pars{\pi z} - {1 \over 1 - z^{2}} + 1 - \gamma - \Psi\pars{1 + z}}_{\ z\ =\ 1/2} - {1 \over 3}
\\[8mm] = &\
-\Psi\pars{\half} - {2 \over 3}\ +\
\underbrace{\Psi\pars{3 \over 2}}_{\ds{\Psi\pars{1/2} + 1/\pars{1/2}}} -
{1 \over 3} = \color{#f00}{1}
\end{align}

  $\Psi$ and $\gamma$ are the Digamma function and the Euler-Mascheroni constant, respectively. We used the Digamma Recurrence Formula $\ds{\Psi\pars{z + 1} = \Psi\pars{z} + 1/z}$ and the identities $\mathbf{6.3.14}$ and $\mathbf{6.3.15}$ in this link.


Concave Convex $e^{-x}-e^{-2x}$ My main goal is to see here if my discusison on concave/convex is good enough.
Problem
Find the extreme points (max and/or min) for $$f(x)=e^{-x}-e^{-2x}.$$   Also discuss concavity/convexity.
Attempt
Extreme points
Solve $f'(x) = 0   -e^{-x} + 2e^{-2x}  = 0$
Multiply by $-e^{2x}$ and get $e^x - 2  = 0 \quad \Rightarrow \quad e^x  = 2 \quad \Rightarrow \quad x  = \ln(2).$
Then: $f(\ln(2))  = e^{-\ln 2} - e^{-2\ln(2)}  = e^{\ln(2^{-1})} - e^{\ln(2^{-2})} = e^{\ln(\frac{1}{2})} - e^{\ln(\frac{1}{4})}  = \frac{1}{2} - \frac{1}{4}  = \frac{1}{4}.$
There is an extreme point at $\left(\ln(2), \frac{1}{4}\right).$
Second derivative test: $f''(x)  = e^{-x} - 4e^{-2x}.$
$f''(\ln(2))  = e^{-\ln(2)} - 4e^{-2\ln 2}  = e^{\ln(2^{-1})} - 4e^{\ln(2^{-2})}  = e^{\ln(\frac{1}{2})} - 4e^{\ln(\frac{1}{4})}.$
$=  \frac{1}{2} - 4\left(\frac{1}{4}\right)  = \frac{1}{2} - 1  = -\frac{1}{2}.$
In $x_0=\ln(2)$ we have $f'(x_0)=0$ and $f''(x_0)<0.$
Thus, $\left(\ln(2), \frac{1}{4}\right)$ is a maximum point.
Concave/Convex
$f''(\ln(4)) = 0$. And $f''$ is negative to the left of $\ln(4)$, and positive to the right, so $f$ is concave for $x<\ln(4)$ and conve for $f>\ln(4)$.

My main goal is to see here if my discusison on concave/convex is good enough.

Yes, your discussion is very good and very clear.

!

$f′′(\ln(4))=0$

$f′′$ is negative to the left of $\ln(4)$
 and positive to the right


categorical interpretation of quantification Many constructions in intuitionistic and classical logic have relatively simple counterparts in category theory.  For instance, conjunctions, disjunctions, and conditionals have analogues in products, coproducts, and exponential objects.

More esoteric concepts still permit reasonably simple categorical interpretations.  For instance, the modal operator $\Box$ (necessity) can be described as a monoidal endofunctor (see, for instance, Paiva & Ritter's Basic Constructive Modality (PDF)).

What are the similar constructions that are suitable interpretations for universal and existential quantification in first-order logic?

There are two concepts in category theory that correspond to $\land$ and $\lor$:

(Binary) categorical products and coproducts, as you mention.
Meets and joins as operations on the subobjects of a given object (satisfying the Beck-Chevalley condition).

Similarly, there are two ways to express logical quantifiers as categorical notions. In the following I assume a category $\mathcal{C}$ which has whatever structure is necessary to make sense of what I am saying (I can be more precise if needed).

For every arrow $f : A \to B$ there is a pullback functor $f^{*} : \mathcal{C}/B \to \mathcal{C}/A$ which takes an arrow $g : D \to B$ to its pullback $f^{*}(g) : f^{*}(D) \to A$ along $f$. Here $\mathcal{C}/X$ is the slice category over $X$. There is another functor $\Sigma_f : \mathcal{C}/A \to \mathcal{C}/B$ defined on an object $h : E \to A$ as $\Sigma_f(h) = f \circ h$. Then $\Sigma_f$ is left adjoint to $f^{*}$. Sometimes $\mathcal{C}$ is such that $f^{*}$ also has a right adjoint $\Pi_f$ of $f^{*}$. When such structure exists, $\mathcal{C}$ is said to be locally cartesian closed. The existential quantifier corresponds to $\Sigma_f$ and the universal one to $\Pi_f$. More precisely, universal quantification $\forall x \in X . \phi(x, y)$ corresponds to $\Pi_{\pi_1}(\phi)$ where $\pi_1 : X \times Y \to Y$ is the first projection and $\phi : Z \to X \times Y$.
There is a subobject functor $\mathsf{Sub} : \mathcal{C}^{\mathrm{op}} \to \mathsf{Poset}$ which takes an object $A$ to the poset of its subobjects and an arrow $f : A \to B$ to the monotone map $f^{*} : \mathsf{Sub}(B) \to \mathsf{Sub}(A)$ which takes a subobject represented by a mono $m : X \to B$ to the subobject represented by $f^{*}(m) : f^{*}(X) \to A$, the pullback of $m$ along $f$. When the posets $\mathsf{Sub}(A)$ are (distributive) lattices, we can interpret $\land$ and $\lor$ as meets and joins. To interpret the quantifiers we again ask for the left and right adjoints of $f^{*}$. The left adjoints exist when $\mathcal{C}$ has stable image factorizations, and these give the existential quantifiers. The right adjoints give the universal quantifiers. There is a little detail called the Beck-Chevalley condition which requires that the adjoints are stable under pullbacks. (In logic the condition corresponds to the fact that substitution commutes with quantifiers.)

We can also express the inference rules for quantifiers in a form that makes it obvious that they are adjoints. Consider the following rule:
$$\frac{x : X, y : Y \mid \phi \vdash \psi}{y : Y \mid \phi \vdash \forall x : X . \psi}$$
This says: "if $\phi$ entails $\psi$, where $x$ and $y$ are free variables of type $X$ and $Y$ respectively, then $\phi$ entails $\forall x : X . \psi$ with free variable $y : Y$." (Note that $x$ cannot occur free in $\phi$ because $\phi$ appears on the bottom line in a context where only the free variable $y$ is allowed.) This is the introduction rule for $\forall$. But we can turn it around to get the elimination rule:
$$\frac{y : Y \mid \phi \vdash \forall x : X . \psi}{x : X, y : Y \mid \phi \vdash \psi}$$
This is not the elimination rule as usually stated, which is
$$\frac{y : Y \mid \phi \vdash \forall x : X . \psi \qquad a : X}{y : Y \mid \phi \vdash \psi[a/x]},$$
but we can obtain this one from ours by using the rule of substitution: from $\forall x : X . \psi$ first derive $\psi$, and then substiute $a$ for $x$.

Now to be completely precise, we should consider $\phi$ above the line to be different from the one below the line because one appears in the context with variables $x : X$ and $y : Y$, and the other in the context with just $y : Y$. So let us write $[x]^{*}(\phi)$ for the formula $\phi$ considered in the context with the variable $x$ added to it (this is known as weakening in logic, although here we are adding stuff to the typing context, whereas in logic there is also weakening that adds hypotheses). With this notation our rules read as
$$\frac{x : X, y : Y \mid [x]^{*}(\phi) \vdash \psi}{y : Y \mid \phi \vdash \forall x : X . \psi}$$
and
$$\frac{y : Y \mid \phi \vdash \forall x : X . \psi}{x : X, y : Y \mid [x]^{*}(\phi) \vdash \psi}.$$
But this has the form of an adjunction between $[x]^{*}$ and $\forall x : X$! We also see the correspondence between logic and categories clearly now: weakening $[x]^{*}$ corresponds to pullback along the projection $\pi_1 : X \times Y \to X$, and $\forall x : X$ corresponds the functor $\Pi_{\pi_1}$.

I leave it as an exercise for you to figure out how the existential is a left adjoint to weakening. So the moral of the story is that universal and existential quantifiers are rigth and left adjoints of weakening.

References:

Steve Awodey: Category theory, Oxford University Press. This has a nice and easy explanation of what I wrote above.
Bart Jacobs: Categorical Logic and Type Theory. This is nuclear weapons, it contains everything and more about the subject, including generalizations that unify the above two views.

It's a good exercise to see the natural bijections worked out in some detail, but as I noted in your other reply (Categorical semantics explained – what is an interpretation?), padawan, there is an additional minor detail that you left out, which should be made clear.
While the natural bijection for the $∧$ instantiates to bijections between the hom sets $(C,A∧B)$ and $(C,A)×(C,B)$ for $A,B,C ∈ |××^+|$, for consistency the natural bijection for $∀$ can't instantiate to bijections between $(C,∀A)$ and $^T(C,A)$ for $A ∈ |^T| = ||^T$ and $C ∈ |^+| = ||$: the $C$ on the right has to be lifted to $||^T$ as a constant function. This is implemented by Curry's $K$ combinator, represented here as a functor $_T:  → ^T$ that embeds the objects and morphisms of $$ as constant functions in $^T$ (i.e. $_T C: t∈T ↦ C$ for $C∈||$ and $_T f: t∈T ↦ f$ for $f ∈ (A,B)$ and $A,B ∈ ||$.
Therefore, the correct statement of the natural bijection ... which produces the details you spelled out ... is actually $(C,∀A) ⇔ ^T(_TC,A)$, for $C ∈ ||$ and $A ∈ |^T|$. This forms an adjunction pair $(∀_T,_T)$ for each set $T$ that is being considered: be they the finite sets $T = ∅$ or $T = 2 = \{0,1\}$ or the infinite sets $T$ of terms in a language or families of such sets if the language is multi-sorted.
This is a natural launching point for considering possible generalizations, where the set $T$, or family of such sets, is replaced by other kinds of objects, perhaps even by categories.


cardinality of Galois group in $\mathbb{Q}$($\zeta_n$) Let $\mathbb{Q}$($\zeta_n$) be some cyclotomic field, where $\zeta_n$ is a n-th root of unity.

I already managed to show that $\mathbb{Q}$($\zeta_n$) is an Galois extension, but now i struggle to show that
$[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$, where $\varphi(n)$ is Euler's totient function.

For any $\sigma \in Gal(\mathbb Q(\zeta_n)/\mathbb Q)),$ $\sigma(\zeta_n)$ must be a primitive $n$-th root of unity, because $\sigma$ is a $\mathbb Q$-automorphism. Since $\zeta_n$ generates $\mathbb Q(\zeta_n)/\mathbb Q$, it follows that this group has as many elements as the set of primitive $n$-th roots of unity, that is $\varphi(n)$, where $\varphi$ denotes Euler's totient function.


What's wrong with this equal probability solution for Monty Hall Problem? I'm confused about why we should change door in the Monty Hall Problem, when thinking from a different perspective  gives me equal probability.

Think about this first: if we have two doors, and one car behind one of them, then we have a 50/50 chance of choosing the right door.

Back to Monty Hall: after we pick a door, one door is opened and shows a goat, and the other door remains closed. Let's call the door we picked A and the  other closed door B. Now since 1 door has already been opened, our knowledge has changed such that the car can only be behind A or B. Therefore, the problem is equivalent to: given two closed doors (A and B) and one car, which door should be chosen (we know it's a 50/50 thing)?
Then, not switching door = choosing A, and switching door = choosing B. Therefore, it seems that switching should be equally likely, instead of more likely.

Another way to think: no matter which door we choose from the three, we know BEFOREHAND that we can definitely open a door with a goat in the remaining two. Therefore, showing an open door with a goat reveals nothing new about which door has the car.

What's wrong with this thinking process? (Note that I know the argument why switching gives advantage, and I know experiments have been done to prove that. My question is why the above thinking, which seems legit, is actually wrong.) Thanks.

Your error is a common one: You have confused the event that the host reveals door $C$ with the event that there is a goat behind door $C$. This is distinction enough to see: if the car is, in fact, behind door $A$, then the host will only reveal door $C$ with a probability of $50\%$, whereas there is definitely a goat behind door $C$.

In particular, let $C_{\text{goat}}$ be the event that there's a goat behind door $C$, and $C_{\text{revealed}}$ be the event that door $C$ is revealed, and $A_{\text{car}}$ be the event that there is a car behind door $A$. Then, one can see that $A_{\text{car}}$ and $C_{\text{revealed}}$ are independent of each other, since the host reveals door $B$ or $C$ randomly with $50\%$ probability if there's a car behind door $A$, and if there's not a car behind door $A$, the host is forced to pick the door which hasn't got a car behind it - which is $B$ or $C$ with probability $50\%$. Thus, one concludes that
$$P(A_{\text{car}}|C_{\text{revealed}})=P(A_{\text{car}})=\frac{1}3.$$
The calculation you've done (annotated with a $\neq$ sign where things go wrong) is:
$$P(A_{\text{car}}|C_{\text{revealed}})\neq P(A_{\text{car}}|C_{\text{goat}})=\frac{1}2$$

Assume A event that selected door have prize
$P(A)$ = $\frac{1}{3}$
And B event that selected door don't have prize
$P(B)$ = $\frac{2}{3}$
Assume $S$ to be event that he switch doors and get right prize door
And $S'$ to be event that he didn't switch doors
So $P(S)$ = $P(A)$*P($\frac{S}{A}$) + $P(B)$*P($\frac{S}{B}$)
= $\frac{1}{3}$$*$$0$ + $\frac{2}{3}$$*$$1$
= $\frac{2}{3}$
$P(S')$ = $\frac{1}{3}$
Clearly we can see that the probability of winning the prize after switching is more, so switching is advisable.


Derivative of the function $f(x) = |x+2|$ How can I get the equation of the derivative $f(x) = |x+2|$ ?

I have already graphed the original function $|x+2|$ and the derivative function, but I'm not sure how to find the derivative, the equation and if it is differentiable at any points. Question says that the answer can be piece wise. Steps and help would be greatly appreciated.

Use the fact that $|u|$ is $u$ if $u>0$ and $-u$ if $u<0$.  So when $x>-2$ we have $f'(x)=(x+2)'=1$ and when $x<-2$ we have $f'(x)=(-x-2)'=-1$.  $f'(-2)$ is undefined, which is easily seen using the limit definition.

if you need a formula for the derivative of $|x+2|$ it is $$\dfrac{x+2}{|x+2|}, x \neq -2.$$

the reason for this is $x + 2$ is $x + 2$ for $x > -2$ so the derivative is $1.$
similarly, $|x+2| = -x - 2$ for $x < -2.$ therefore the derivative is $-1$ for $x < -2.$ that is the piecewise definition.


What makes elementary functions elementary? Is there a mathematical reason (or possibly a historical one) that the "elementary" functions are what they are?  As I'm learning calculus, I seem to focus most of my attention on trigonometric, logarithmic, exponential, and $n$th roots, and solving problems that have solutions which are elementary functions.  I've been curious why these functions are called elementary, as opposed to some other functions that turn up rather naturally in mathematics.  What is the reason that these functions take up most of our attention, and is there a reason that some additional functions are not included amongst the elementary functions?  In other words, what property or properties do these functions possess that separates them from non-elementary functions (if there is one)?

As Sivaram Ambikasaran mentioned the description on Wikipedia is fine.

I believe the class of elementary functions, $E$, is commonly thought of as a construction of the form

All polynomials are in $E$
The exponential and the logarithm function is in $E$
The sine and cosine functions are in $E$.
$E$ is closed under addition, subtraction, multiplication, division and composition (finitely many operations of these).
$E$ is the smallest set with the properties 1-4.

This applies to both real or complex valued functions.

Edit 1:

Some examples of functions that are elementary

$f(x)=1-x^2$
$s(x) = \sqrt{x}$ (see addendum below)
$g(x)=\arctan x$ (see addendum below)
$U(x)=\sin\frac{1}{\log(1+x^2)}$
$A(x)=|x| = \sqrt{x^2}$
$_2F_1(1,1,2,x) = \log(1-x)$ (a Gauss Hypergeometric notation that ends up in a elementary function)

Some examples of functions that are not elementary

The Sine integral $\operatorname{si}(x)=\int_0^x\frac{\sin t}{t}dt$
The Error function $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt$
The Cantor function
The characteristic function of an interval.

Edit 2:

The nomenclature elementary function is of course used since the functions that are elementary can be deduced using finitely many applications of elementary operations on basic "high school functions".

Edit 3:

Convinced by comments - removed closed under inversion in step 5. Added log in step 2.

Addendum (2013-02-28):

To see that the square root function $s(x)=\sqrt{x}$ is elementary using the above definition just note that
$$ x\mapsto \frac{1}{2}\log x = \log x^{1/2}=\log\sqrt{x}$$
and hence $$s(x)=\sqrt{x}=\exp(\log\sqrt{x})$$ is elementary too.
Perhaps $\arctan$ is a bit harder to deduce from the other steps in the construction of $E$, first note that
\begin{eqnarray}
\tan x=\frac{\sin x}{\cos x}= \frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}=
-i\frac{e^{2ix}-1}{e^{2ix}+1}=\\
-i\frac{e^{2ix}+1 -2}{e^{2ix}+1} = -i\frac{e^{2ix}+1 }{e^{2ix}+1}-i\frac{-2}{e^{2ix}+1}=\\-i +i\frac{2}{e^{2ix}+1}
\end{eqnarray}

Hence if we solve for, $e^{2ix}$ in the above identity we get
\begin{eqnarray}
e^{2ix}=\frac{2}{1-i\tan x}-1= \frac{2-(1-i\tan x)}{1-i\tan x}= \frac{1+i\tan x}{1-i\tan x}
\end{eqnarray}
and taking the complex logarithm (more precisely the principal branch of the logarithm) we get $$x=\frac{1}{2i}\log\frac{1+i\tan x}{1-i\tan x}$$
That is $$t\mapsto \arctan t =\frac{1}{2i}\log\frac{1+it}{1-it}$$
belongs to $E$.

The term elementary function is rather historical than mathematical. Elementary functions are those which were first studied in the 17th and 18th centuries, built up with the basic arithmetic operations:

polynomials
exponentials and logarithms
trigonometric functions


Image of Boundary of Half Closed plane under Diffeomorphism Let $S: A\to B$ be diffeomorphism, where A and B are an open subset of closed half-space in $R^n= \{x\in \mathbb R^n|x_1\geq 0\}$.$\partial H=\{x\in R^n|x_1=0\}$. Then image of $\partial H \cap A $ is $\partial H \cap B $ under S.

It is clear that $\partial H \cap A $ does not contain any interior point. I thought to use the inverse function theorem.

Suppose on contrary it maps to some interior point. Then there exist nhbd of that point which is homeomorphic to the inverse image of the map under restriction. which is a contradiction to the fact that domain does not contain any open nbhd.

Is this argument is enough?

Any help will be appreciated.

which is a contradiction to the fact that domain does not contain any open nbhd.

What do you mean by that? Clearly the domain, being $\partial H$ contains lots of open neighbourhoods. No, this is not enough.

One of the possible solutions follows from the homotopy theory. So assume that there is a homeomorphism $S:A\to B$ that maps an interior point to a boundary point. In other words we have a homeomorphism $f:D\to C$ where $D$ is an open Euclidean ball in $\mathbb{R}^n$ and $C$ is some open subset of $\mathbb{R}^n_{\geq 0}$ that has a nonempty intersection with boundary $\partial H$. Note that $D$ is contractible and therefore so is $C$. But $D\backslash\{p\}$ is not contractible for any point $p\in D$ (it is a homotopy sphere). However $C\backslash\{q\}$ remains contractible when $q$ belongs to the boundary. This can be derived from the fact that $C'\backslash\{q\}$ is a star domain where $q$ is boundary point and $C'$ and open ball around $q$ in $\mathbb{R}^n_{\geq 0}$.


Probability and independence I am trying to solve a particular probability question.

I have a fair 10-sides die, whose sides are labelled 1 through 10. I am trying to find the probability of rolling a multiple of 5 or an odd number.

I find the probability as:

P(multiple of 5) OR  P(odd number)=P(multiple of 5) + P(odd number)-[P(multiple of 5) AND P(odd number)]=2/10+5/10-[(2/10)(5/10)]=6/10 (which is the correct answer)

Notice that I used the assumption that rolling a multiple of 5 is independent of rolling an odd number, since I essentially multiplied their probabilities to get the answer. However, rolling a multiple of 5 INVOLVES rolling an odd number in one case, namely rolling a 5.

So did I get the correct answer by fluke? Can the multiplication rule also be used to describe non-independent events sometimes?

No, it's not a fluke, in the sense that $P(A\cap B)=P(A)P(B)$ is the definition of independence of $A$ and $B$, so if it holds, as it does here, $A$ and $B$ are necessarily independent. In another sense, it was a fluke, in that it might not have been obvious that the events multiple of 5 and odd are independent: this depends on the fact that there are exactly as many odd as non-odd multiples of 5 between 1 and 10.

So you got lucky, this time: your two events really are independent. What would happen if you'd been asked about a 9-sided die? Or a 10-sided die, but with the events multiple of 3 and even?

In general, arithmetic progressions like $(na+c)_{n\in\Bbb{N}}$ and $(nb+d)_{n\in\Bbb{n}},$ e.g. odds when $a=c=1$ or multiples of 5 when $b=5,c=0,$ are "independent" in $\Bbb{N}$ in the sense of natural density whenever $a$ and $b$ are relatively prime. To bring this fact down to the finite realm of dice with $k$ sides, we need $k$ to be a common multiple of $a$ and $b$. In fact 10 is a common multiple of 2 and 5, and that indicates a bigger reason why you lucked out with this computation.

Actually when we through a fair die each outcome is identically independently distributed, so  there is no need to assume multiple of 5 is independent of rolling an odd number. Two events are said to be dependent if they effect occurrence or non-occurrence of other event. In your case both events are Independent.


Finding $\sqrt{a+ib}$ How do to find the square root of a complex number?

$z=a+ib$

$z=\sqrt{3+i4}=a+ib$

remove the square by squaring both side

$a^2+2iab+b^2=3+i4$

$a^2+b^2=3$

$2ab=4$

So I just solve the simultaneous equation. But my friend is telling me this way I am doing is wrong. Can anyone give a hand, thank in advance

Much easier: If you want to find $\sqrt{a+ib}$ start by writing $$a+ib=re^{it}$$for some $r>0$ and $t\in\mathbb R$.


Vector space over an infinite field which is a finite union of subspaces Let $V=\bigcup_{i=1}^n W_i$ where $W_i$'s are subspaces of a vector space $V$ over an infinite field $F$. Show that $V=W_r$ for some $1 \leq r \leq n$.

I know the result "Let $W_1 \cup W_2$ is a subspace of a vector space $V$ iff $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$."

Now can I extend this to some $n$ subspaces.

I have some answers here
& here.

So before someone put it as a duplicate I want to mention that I want a proof of this problem using basic facts which we use in proving the mentioned result.

The general result  is that if $\lvert K\rvert\ge n$, $V$ cannot be the union of $n$ proper subspaces (Avoidance lemma for vector spaces).

We'll prove that if $V=\displaystyle\bigcup_{i=1}^n W_i$ and the $W_i$s are proper subspaces of $V$, then $\;\lvert K\rvert\le n-1$.

We can suppose no subspace is contained in the union of the others.

Pick $u\in W_1\smallsetminus\displaystyle\bigcup_{i\neq1} W_i$, and $v\notin W_1$. The set $v+Ku$ is disjoint from $W_1$, and it intersects each $W_i\enspace(i>1)$ in at most $1$ point (otherwise $u$ would belong to $W_i$). As this set is in bijection with $K$, there results that $K$ has at most $n-1$ elements.

$V$  is  a  vector  space  over  an  infinite  field.
If $\ \ $   $\cup_{i=1}^{n} W_{i}=V$  is  a vector space  itself  then  from  what  you  already  know we  can  write $$W_{1}\subseteq W_{2}\subseteq W_{3}........\subseteq W_{n}$$  possibly  with  a  re-ordering  of  the  subspaces  $W_{i}$'s .

Then  what  do  we  see  here?  $W_{n}$  containing  the  other  subspaces  of  $V$. So  intuitively  we  could  suspect  that  $V=W_{n}$  might  be  the  thing .

Let's  prove  it. $$W_{n}\subseteq V$$  is known.  For  the  reverse  inclusion ; if possible  let  $$V\neq W_{n}$$ let $w$  be  a  vector such  that $$w\in V \ \  but\ \    w\notin W_{n}$$  . Now  $$\cup_{i=1}^{n} W_{i}=V$$  implies  $w$  is  in  $W_{i}$  for  some  $i=1,2...,n$. But  $$W_{i}\subseteq W_{n}$$ for  all  $i=1,2,....n$ .  Hence $$w\in W_{n}$$ . A  contradiction.  Thus  $$V\subseteq W_{n}$$  must  be  true. Hence  proved  $$V=W_{n}$$


Show bounded and convex function on $\mathbb R$ is constant How can we show that a bounded and convex function on $\mathbb R$ is constant? Derivatives are of no use since the function does not have to differentiable. I saw an answer here I think a while ago but did not understand it at all.

Since derivatives are useless, we would have to use the definition and somehow show that the function lies between two values which are equal to each other. But I am unable to progress any further.

Suppose that $f$ is a convex function that is not constant. Then there must be $a$ and $b$ with $f(a)<f(b)$. Without loss of generality, assume that $a<b$.

Then for any $c>b$ we must have, due to convexity,
$$ f(c) \ge f(a) + (c-a)\frac{f(b)-f(a)}{b-a} $$
which grows without bounds as $c\to \infty$. So $f$ cannot be bounded.

I don't think that any of this is true: consider the function f(x) = \exp(-x) on [0,\inf). This function is strictly convex and bounded but not constant.


negative number divided by positive number, what would be remainder? my question is

  If $-27$ is divided by $5$, what would be the remainder?

There are two conventions, depending on whether you allow the remainder to be negative.

Either you don't

$$-27 = -6 \times 5 + 3$$

or you do

$$-27 = -5 \times 5 + (-2)$$

Note that whatever convention you choose, the two possibilities for the remainder will always differ by $5$.

suppose you have a loan of 27 i.e -27.what will the loan amount which you have to pay after dividing the loan as much as possible.
after dividing the loan among 5 people each with -5 you will be still left with -2.so -2 must be the remainder.


How can I to solve $ \cos x = 2x$? I would like to get an approx. solution to the equation: $ \cos x = 2x$, I don't need an exact solution just some approx. And I need a solution using elementary mathematics (without derivatives etc).

Take a pocket calculator, start with $0$ and repeatedly type [cos], [$\div$], [2], [=]. This will more or less quickly converge to a value $x$ such that $\frac{\cos x}2=x$, just what you want.

if $\cos x=2x$ which means
$$x=\frac {1}{2.22131587} \,\mathrm{rad},$$

$$x=28.65957881\,\mathrm{Grad},$$

$$x=25.79362093\,\mathrm{Degree}$$

point:
$$\cos x=1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+\cdots$$

$0 < x < +\frac {1}{2.5}\,\mathrm{rad}$ because $\cos x=2x$

This is also a mathematical method, i mean Estimate the amount through math without using a computer or calculator!.

$$\cos x \simeq 1-\frac {x^2}{2!}=2x,$$

$$x^2+4x-2=0,$$

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

where the $a=1$ and $b=4$ and $c=-2$ then $x\simeq 0.45$


Random digraphs and expected number of neighbors What is the expected number of neighbors of a vertex in a random digraph $D (n,p)$?

The degree of a vertex in a digraph is defined to be the minimum of in-degree and out-degree of that vertex.

If $X$ is the random variable that counts neighbors of a vertex $v$, then $x=\min \{x^+,x^-\}$, where $x^+$ and $x^-$ are random variables that count out-neighbors and in-neighbors of $v$.

Since both $x^+$ and $x^-$ are binomial random variables, it is easy to find CDF, but I am still not sure how to use that to find the expected number of neighbors.

Let x be the number of neighbors of a vertex v in a random digraph D (n,p) . Then E (X)= (n-1)[1-(1-p)^2].  I think this works correctly. I just figured this out.


Group with exactly two subgroups of index 2 I am looking for a group $G$ such that $G$ has exactly two subgroups of index 2.

I have searched by GAP but I couldn't find it.

Such a group does not exist. Suppose the group $G$ has subgroups $N \ne M$ of index two. Clearly they are both normal, and $M N = G$. Then
$$
\lvert G : M \cap N \rvert
=
\lvert G : M \rvert
\cdot
\lvert M : M \cap N \rvert
=
\lvert G : M \rvert
\cdot
\lvert M N : N \rvert
=
\lvert G : M \rvert
\cdot
\lvert G : N \rvert
=
4.
$$
So $G / M \cap N$ is a group of order $4$, which is isomorphic to the Klein four-group $V$, as $a^{2} \in M \cap N$ for each $a \in G$.

Since $V$ has three subgroups of index $2$, $G$ has at least three such subgroups, by the correspondence theorem.

$G=S_3\times S_3$ would work. The subgroups of index two being $\mathbb{Z}_3\times S_3$ and $S_3\times\mathbb{Z}_3$.

Edit: Removed last sentence due to an error pointed out by Tobias Kildetoft.

Edit: $G$ is not an example of such a group as Andreas Caranti points out.


Invertibility of Toeplitz operator in $\ell_1$ Suppose we have a Toeplitz operator
$$
T(a) = \begin{bmatrix} a_{0} & a_{-1} & a_{-2} & \ldots & \ldots &a_{-n+1} &\dots \\\\
a_{1} & a_0 & a_{-1} & \ddots & & \vdots \\\\\ a_{2} & a_{1} & \ddots & \ddots & \ddots& \vdots \\\\
\vdots & \ddots & \ddots & \ddots & a_{-1} & a_{-2}\\\\
\vdots & & \ddots & a_{1} & a_{0}& a_{-1} \\\\ a_{n-1} & \ldots & \ldots & a_{2} & a_{1} & a_{0} \\\\
\vdots
\end{bmatrix}
$$
where $a=(a_k)$, $k\in \mathbb{Z}$.

In the book by Hagen, Roch and Silbermann you can find the conditions when this operator is invertible in $\ell_2$, theorem 1.31, in page 48.

My question is whether there similar conditions for existence of invertibility in space $\ell_1$?

The condition in the book is the following. Denote $\mathbb{T}$ the unit circle in $\mathbb{C}$. Let $a\in L^{\infty}(\mathbb{T})$ and let

$$a_k=\frac{1}{2\pi}\int_0^{2\pi}a(e^{i\theta})e^{-ik\theta}d\theta$$

Theorem. 1.31 (Coburn's theorem) Let $a\in C(\mathbb{T})$. Operator $T(a)$ is invertible in $\ell_2$ if and only if $0\notin a(\mathbb{T})$ and if $\mathrm{wind}\text{ }a(\mathbb{T})=0$.

Here $\mathrm{wind}\text{ }a(\mathbb{T})$ is a winding number of the curve $a(\mathbb{T})$, provided with the orientation inherited by the usual counter-clockwise orientation of the unit circle, around the origin. (The definition taken from the book)

the theory is pretty much complete in all the lp spaces. relevant sources include:

Analysis of Toeplitz Operators By Albrecht Böttcher, Bernd Silbermann (Ch 6.)

an algebraic formula for the inverse: gohberg-semencul

books by gohberg-feldman, gohberg-kaashoek, gohberg-goldberg.

A general framework is the theory of Fredholm operators on Banach
spaces. For the toeplitz operator, the winding number is just
the "fredholm index". A fredholm operator, when invertible, has zero index.
but the converse statement does not hold.

if the function is lp and only piecewise continuous on T, one of the sources
(i think, gohberg-feldman) provides a way to fill the gaps of a(T) by circles.
now, coburn's condition generalizes as follows: this modified curve should
avoid the origin.


When to do u-substitution and when to integrate by parts I'm in my first semester of calculus, so the problems I'm facing are about as hard as those on KhanAcademy calculus playlist. I'm currently doing integration, a somewhat difficult part of the course. Doing derivatives is mechanics; finding the integral is an art.

The two main techniques showed are

u-substitution
integration by parts

My question is: are there any rules of thumb (preferably with a logical reason behind it) of when to use which?

Always do a $u$-sub if you can; if you cannot, consider integration by parts.

A $u$-sub can be done whenever you have something containing a function (we'll call this $g$), and that something is multiplied by the derivative of $g$.  That is, if you have $\int f(g(x))g'(x)dx$, use a u-sub.

Integration by parts is whenever you have two functions multiplied together--one that you can integrate, one that you can differentiate.

My strategy is to try to "play it out" in my mind and try to see which one will work better.  The best way to get better at these sorts of integrals is to practice large sets of each type.  Then, you start to think "Oh--this looks like a u-sub!" or, "maybe by-parts is better for this."  Practice is really the best way to get better at recognizing each type.

One thing you should remember is choose that function as u which when differentiated yield a less difficult function or may decrease its power or some such thing.  Also if we have cos or sin then differentiation of function will bring back the original \int uv in some form. What I feel while teaching is the "Tabular method" which is following Bernoulli's formula of \int uv.


Lagrange mean value theorem for two variables - visualization and intuition behind it The two-variable version of the Lagrange mean-value theorem says that given a function $f(x,y)$,
$$f(\vec{p_o} + \vec h)-f(\vec {p_o})=df(\vec {p_{\theta}})$$
Where $\vec p_{\theta}=\vec p_o + \theta \vec h $
with $\theta \in (0,1).$

I don't understand this theorem, neither do I see the intuition behind it.

Is there a simple way to visualize it? If not, could you come up with a practical example where this theorem could be used?

For a function $f(x,y):\mathbb{R^2}\to \mathbb{R}$ the MVT

$$f(\vec{p_o} + \vec h)-f(\vec {p_o})=df(\vec {p_{\theta}})$$

has an interpretation very similar to MVT for functions of one variable.

Indeed we have that

$$df(\vec {p_{\theta}})=|\vec h|\frac{\partial f(\vec p_{\theta})}{\partial \vec v}$$

with $\vec v = \frac{\vec h}{|\vec h|}$ and thus

$$df(\vec {p_{\theta}})=|\vec h|\frac{\partial f(\vec p_{\theta})}{\partial \vec v}=|\vec h|\langle\nabla f(\vec p_{\theta}),\frac{\vec h}{|\vec h|}\rangle =\langle\nabla f(\vec p_{\theta}),\vec h\rangle$$

From the first expression the MVT can be expressed as

$$\frac{f(\vec{p_o} + \vec h)-f(\vec {p_o})}{|\vec h|}=\frac{\partial f(\vec p_{\theta})}{\partial \vec v}$$


Number of possible products If I have set (2,3,4,5,6,7,8,9), how many possible products are there of any three numbers from the set. Repeats are allowed e.g. 2*2*2, 2*3*3, etc.
I don't know how to solve this other than brute force. Thx a lot.

The number of products consisting of $3$ elements with repetition from the $8$ element set
  \begin{align*}
\{2,3,4,5,6,7,8,9\}=\{2,2^2,2^3,2\cdot 3, 3,3^2,5,7\}
\end{align*}
  is
  \begin{align*}
\binom{8+3-1}{3}=\binom{10}{3}=\color{blue}{120}
\end{align*}

Here the strategy is to count  the products which  occur more than once manually and then subtract this number from $120$.

We can see at a glance  that  products having two elements from $\{5,7\}$ cannot have two different representations. Therefore candidates have either a factor $5$ or $7$ or none of them.

We need some kind of systematic way to find all possible products with multiple representations. In order to do so we arrange all candidates  according  to  two  criterias:

The product contains either a  factor $5$ or  $7$ or none  of them
We sort candidates by increasing  powers of  $2$

  Products containing  $5$ or $7$:

  We only need to list two factors, since the third one is either $5$ or $7$. We obtain
  \begin{align*}
\text{factor }2\ :&\qquad 2\cdot 3^2=(2\cdot 3)\cdot 3&(90,126)\\
\text{factor }2^2:&\qquad  2^2\cdot3=2\cdot(2\cdot 3)&(60,84)\\
&\qquad 2^2\cdot  3^2=(2\cdot  3)(2\cdot 3)&(180,252)\\
\text{factor }2^3:&\qquad  2^3\cdot 3=2^2\cdot (2\cdot 3)&(120, 168)\\
\text{factor }2^4:&\qquad 2\cdot 2^3=2^2\cdot 2^2&(80,112)\\
\end{align*}

  We see there are $5$ possibilities  to  write   a product  with two different representations. Since the third factor is either $5$ or  $7$ we have a total of $\color{blue}{10}$  possibilities. The bracketed values on the right of each line give the number with factor  $5$ resp. $7$.

  Products containing neither $5$ nor $7$:

  \begin{align*}
\text{factor }2\ :&\qquad   2\cdot  3\cdot  3^2=(2\cdot 3)\cdot 3\cdot 3=54\\
&\qquad 2\cdot 3^2\cdot 3^2=(2\cdot 3)\cdot   3\cdot 3^2=162\\
&\qquad \rightarrow 2\text{  multiples}\\
\text{factor  }2^2:&\qquad 2\cdot 2\cdot 3^2=2^2\cdot 3\cdot 3=2\cdot (2\cdot 3)\cdot 3=36\\
&\qquad 2^2\cdot 3\cdot 3^2=2\cdot (2\cdot 3)\cdot 3^2=(2\cdot 3)(2\cdot 3)\cdot 3=108\\
&\qquad 2^2\cdot 3^2\cdot 3^2=(2\cdot 3)(2\cdot 3)\cdot 3^2=324\\
&\qquad \rightarrow 5\text{  multiples}\\
\text{factor  }2^3:&\qquad 2\cdot 2^2\cdot 3=2\cdot 2\cdot (2\cdot 3)=24\\
&\qquad 2\cdot 2^2 \cdot  3^2=2^2\cdot(2\cdot3)\cdot      3=2^3\cdot 3\cdot  3=2\cdot(2\cdot3)(2\cdot 3)=72\\
&\qquad 2^3\cdot 3\cdot   3^2=2^2\cdot (2\cdot  3)\cdot 3^2=(2\cdot  3)(2\cdot 3)(2\cdot 3)=216\\
&\qquad \rightarrow 6\text{  multiples}\\
\text{factor }2^4:&\qquad 2\cdot2^3\cdot    3=2^2\cdot2^2\cdot    3=2\cdot 2^2\cdot (2\cdot 3)=48\\
&\qquad 2\cdot 2^3\cdot  3^2=2^2\cdot2^2\cdot   3^2=2^3\cdot(2\cdot3)\cdot3=2^2\cdot   (2\cdot      3)(2\cdot  3)=144\\
&\qquad \rightarrow 5\text{  multiples}\\
\text{factor  }2^5:&\qquad   2\cdot 2\cdot2^3=2\cdot2^2\cdot2^2=32\\
&\qquad   2^2\cdot   2^3\cdot 3=2\cdot  2^3\cdot (2\cdot  3)=2^2\cdot   2^2\cdot   (2\cdot  3)=96\\
&\qquad       2^2\cdot  2^3\cdot 3^2=2^3\cdot   (2\cdot    3)(2\cdot   3)=288\\
&\qquad \rightarrow 4\text{  multiples}\\
\text{factor }2^6:&\qquad     2\cdot  2^2\cdot 2^3=2^2\cdot    2^2\cdot 2^2=64\\
&\qquad 2^3\cdot 2^3\cdot 3=2^2\cdot  2^3\cdot (2\cdot 3)=192\\
&\qquad \rightarrow 2\text{  multiples}\\
\text{factor }2^7:&\qquad 2\cdot 2^3\cdot 2^3=2^2\cdot 2^2\cdot 2^3=128\\
&\qquad \rightarrow 1\text{  multiple}\\
\end{align*}
  giving a total of $2+5+6+5+4+2+1=\color{blue}{25}$ multiple product representations.

  We conclude the number of different products is $$120-10-25=\color{blue}{85}.$$

$8^3$ (3 slots, 8 possibilities).

EDIT: This doesn't account for product degeneracy.


Expected value of product of two random variables If we have a process defined as:

$Y_t =  ε_t + Y_{t-1}$

How do you determine the expected value:

$E[Y_tY_{t-1}]$

Since $Y_t = Y_0 + \sum_{i=1}^t \epsilon_i$, if $Y_0 = 0$ we get $Y_t = \sum_{i=1}^t \epsilon_i$. Furthermore you said $\epsilon_i \sim N(0,1)$, and since under the assumption of independence of the $\epsilon_i$ the sum of (standard) normal random variables is again a normal random variable (with the means and variances added) we get $Y_t = \sum_{i=1}^t \epsilon_i \sim N(0,t)$. Writing out the expected value a bit, we get

$$\mathbb{E}[Y_t Y_{t-1}] = \mathbb{E}[(Y_{t-1} + \epsilon_t)Y_{t-1}] = \mathbb{E}[Y_{t-1}^2] + \mathbb{E}[\epsilon_t Y_{t-1}].$$

From $Y_{t-1} \sim N(0,t-1)$ it follows that $\mathrm{Var}(Y_{t-1}) = \mathbb{E}[Y_{t-1}^2] - \mathbb{E}[Y_{t-1}]^2 = \mathbb{E}[Y_{t-1}^2] = t - 1$. Finally, since $\epsilon_t$ and $Y_{t-1}$ are independent and symmetric around $0$, it follows that

$$\mathbb{E}[Y_t Y_{t-1}] = \mathbb{E}[Y_{t-1}^2] + \mathbb{E}[\epsilon_t Y_{t-1}] = (t - 1) + 0 = t - 1.$$

Without the assumption of independence of the $\epsilon_i$, however, this does not work.


Evaluate $\int \frac{1}{\sqrt{1-\sin^4{x}}}dx$ $$f(x) = \int \frac{1}{\sqrt{1-\sin^4{x}}}dx$$

I tried this by breaking the denominator as $\sqrt{(\cos^2x)(1+\sin^2x)}$ and then trying to make the integral in forms of $\sec x$ and $\tan x$. But I couldn't succeed.

Can somebody please help me out?

Taking $t = \tan \frac{x}{2}$ and having $\sin x = \frac{2t}{1+t^2}$ we come to irrational function.


Cardinality of the set of bijective Lebesgue-measurable functions on the unit intervall Let $I= [0,1]$ be the unit intervall. It is well known that the cardinality of $$B = \{ f: I \to I \mid f \text{ bijective} \}$$ and $$L = \{ f: I \to I \mid f \text{ Lebesgue-measurable} \}$$ is $2^{|\mathbb{R}|}$.

Question: What is the cardinality of $B \cap L$ ?

The cardinality of the set all bijective, Lebesgue-measurable functions $f: I \to I$ is $2^{|\mathbb{R}|}$.

Proof: Wlog we can take $I = [-1, 1]$.

The idea is to associate each measurable subset $M \subseteq (0,1]$ a measurable bijection $f_M$. The construction of $f_M$ follows the accepted answer of Cardinality of real bijective functions/injective functions from $\mathbb{R}$ to $\mathbb{R}$.

Let $\chi_M: [0,1] \to I$ be the characteristic function of $M$

\begin{equation}
    \chi_M(x)=
    \begin{cases}
      1, & \text{if}\ x \in M \\
      -1, & \text{if}\ x \not\in M
    \end{cases}
  \end{equation}
and let $$f_M: I \to I, \, x \mapsto \chi_M(|x|)\cdot x.$$
$f_M$ satisfies $f_M \circ f_M = \text{id}_I$ and is thus a bijection. Moreover, $f_M$ can be represented by
$$f_M = (\chi_M \circ \text{abs}) \cdot \text{id}_M$$
as a composition and product of measurable functions (note: the characteristic function of a measurable set is measurable) and is therefore measurable.

The cardinality of measurable subsets of $(0,1]$ is $2^{|\mathbb{R}|}$. In fact, the Cantor set $C$ has cardinality $|\mathbb{R}|$ and has measure $0$. Hence all of the $2^{|\mathbb{R}|}$ subsets of $C$ are also measurable (with measure $0$). Since the assignment $M \to f_M$ is injective, we have found $2^{|\mathbb{R}|}$ bijective, measurable functions $f_M$ and since the cardinality of $I^I$ is $2^{|\mathbb{R}|}$, there are  $2^{|\mathbb{R}|}$ bijective, measurable functions $I \to I$. q.e.d.

Remark: It's easy to see that $f_M$ is non-measurable, if $M$ is non-measurable. Since there are $2^{|\mathbb{R}|}$ non-measurable subsets of $I$ (in ZFC), the same proof shows:

  The cardinality of the set of all bijective, non-Lebesgue-measurable functions $f: I \to I$ is $2^{|\mathbb{R}|}$.

Let $C$ be the Cantor set, or any other Lebesgue-null set of cardinality $\mathfrak{c}$  (I'll write $\mathfrak{c}$ instead of $|\mathbb{R}|$).  Let $g : C \to C$ be any bijection; there are $2^\mathfrak{c}$ of them.  For each $g$, consider the function $f_g : I \to I$ defined by
$$f_g(x) = \begin{cases} g(x), & x \in C \\ x, & x \notin C.\end{cases}$$
Then each $f_g$ is clearly a bijection, and $f_g = f_h$ iff $g=h$.  Moreover $f_g$ is Lebesgue measurable because it is a.e. equal to the measurable function $f(x) = x$.  So there are at least $2^\mathfrak{c}$ measurable bijections, and you already know there are at most that many.


prove that $N$ is divisible by $1,2,\ldots,k$ which $k+1$ is the lowest prime number after $N$ Suppose $n$ is a natural number ($n\ge 5$) and $k+1$ is the lowest prime number that is greater than $n$
prove that $A_i \mid n!$ which $A_i$ are these numbers:
$1,2,\ldots,k$

The statement is obvious for $A_i\leq n$. By Bertrand's postulate, $n<A_i\leq k<2n$.

Suppose that $A_i$ doesn't divide $n!$. Let's assume also that $A_i$ is the minimum with this property. Then there exists a prime $p$ and some $\alpha\geq 1$ such that $p^\alpha|A_i$ and $p^\alpha$ doesn't divide $n!$. In fact, $A_i=p^\alpha$.
Indeed, if $A_i$ has other prime factor, say $A_i=mp^\alpha$, then $(m-1)p^\alpha$ divides $n!$ since $A_i$ is the minimum that doesn't. But this is a contradiction since $p^\alpha$ doesn't divide $n!$.

But $p^{\alpha-1}=p^\alpha/p\leq A_i/2<n$, so $n!$ is a multiple of $p^\beta$, where
$$\beta=\sum_{r=1}^{\alpha-1} r$$
that is
$$\beta=\frac{\alpha(\alpha-1)}2$$

But it must be $\alpha-1\leq\beta<\alpha$, that is, $\beta=\alpha-1$ and $\alpha=2$. (Note that $\alpha>1$ because $A_i$ is composite).

Again by the minimality of $A_i$, if $p\neq 2$, $2p$ divides $n!$ and $p^{\beta+1}=A_i$ divides $n!$ and we arrive at a contradiction. If $p=2$ then $A_i=4$, which contradicts $n\geq 5$.

Hint:
$Ai\mid n!, 1\le A_i \le n$ is obvious. Since the next prime is $k+1$, it is obvious that all numbers from $n+1$ to $k$ are composite and have factors less than n, i.e. $A_i = A_jA_k, 1\le j, k \lt i, \forall (n+1)\le A_i\le k$ Also, $n! = 1\cdot2\cdot3\cdots A_j\cdots A_kc\cdots n$, which clearly shows what is required. For numbers that are powers of other numbers, we can use Bertrand's Postualte.


Fair and unfair coins There are 4 fair coins and 1 unfair coin that has only heads. We choose a coin and flip it three times. The result is HHH. What is the probability that the fourth flip is H?

First, use Bayes' rule to determine the probability of having selected the unfair coin. Call $UC$ the event "we selected the unfair coin" and $FC $  "we selected a fair coin".

$$P(HHH) = P(HHH|UC)P_0(UC) + P(HHH|FC)P_0(FC) = 1\times\frac{1}{5}+\frac{1}{8}\times\frac{4}{5} = \frac{3}{10}$$
$$P(UC|HHH) = \frac{P(HHH|UC)  P_0(UC)}{P(HHH)} = \frac{1}{5}\times\frac{10}{3} = \frac{2}{3}$$
$$P(FC|HHH) = 1-P(UC|HHH)=\frac{1}{3}$$

So we have:

$$P(H|HHH) = P(H|HHH,UC)P(UC) + P(H|HHH,FC)P(FC) = \frac{2}{3}\times1 +\frac{1}{3}\times\frac{1}{2} = \frac{5}{6} $$

$ p(H|HHH) = p(H|unfair) + 4 \cdot p(H|fair)= p(unfair) + p(fair) \cdot p(H) = \dfrac{1}{5} + \dfrac{4}{5} \dfrac{1}{2} = \dfrac{6}{10}$


The proof of $e^x \leq x + e^{x^2}$ How can we prove the inequality $e^x \le x + e^{x^2}$ for $x\in\mathbb{R}$?

Hmmm.... I will give it a try. Consider the function $f(x) = e^{x²} + x - e^x$, $x \in \mathbb{R} $. Find the global minimum of this function. It turns out to be zero. Does that prove it ?

Note: The second derivative is $e^{x^2} (4 x^2+2)-e^x$ which is strictly positive.  $e^{x^2}$ becomes less than $x$ between $0$ and $1$ but that difficulty is overcome by multiplying by 2 (*)

(*)That difficulty is not overcome just by multiplying by 2, I learned it the hard way. You still have to prove it !

I did not answer the question, but I will keep the post here. Reminding me of my once failed efforts!


What kind of graph is this? A pie chart? Besides a gross use of Comic Sans, what kind of graph is this? I've seen it used for population statistics before, but I don't know what it's called.

My first instinct was to just say "bullet chart" due to process of elimination. After researching what a bullet chart is, which includes thresholds and targets, I'm now leaning toward some kind of unorthodox pie chart since it's dealing solely of percentage makeup of the whole.

I'm still not convinced that this is simply called a "pie chart," but it's all I got.

This is a Partition Chart which is similar in data structure to a Pie Chart.

Here's how you make it in excel -> http://peltiertech.com/ineffective-chart-partition-chart/

The squares of each colour correspond to a certain value. Aggregate these squares together and you've got a pie chart! The total number of squares is also a constant, like with Pie charts total slices together always make 360 degrees.

IMHO it is a bar graph as seen from bird perspective


How to determine if a vector field is the curl of another vector field? I want to see if the Stokes theorem can be applied to a given flux integral of the vector field $F$.  To do so, I need to determine if the vector field $F$ is the curl of some other vector field.  How would I determine this?

I found an algorithm that seems to work out, but perhaps someone can advise me where the divergence free comes into-play to power this algorithm.

Let $\textbf F$ be a Euclidean vector-field such that $\nabla \cdot \textbf{F}=0$. If there exists a vector-field $\textbf A$ such that $F=\nabla \times A,$ then we have the following set of equations. Additionally, I have found that they have the corresponding solutions, where $\oplus $ is the exclusive or symbol:
$$\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=F_x: A_z=\int F_x dy \oplus A_y=-\int F_x dz $$

$$-\left(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z}\right)=F_y: A_z=-\int F_ydx \oplus A_x=\int F_y dz$$

$$\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=F_z: A_y=\int F_zdx \oplus A_x=-\int F_zdy$$

These exclusive or $\oplus $ solutions yield two choices for $\textbf A$

You can determine whether a vector field can be written as the curl of another vector field (in $\mathbb{R}^3$) by looking at it's divergence.  Assume a vector field $F$ can be written as the curl of another vector field, call it $G$.  Then $F = \text{curl}~G$.  Take the divergence of $F$, and say $\text{div}~F \not= 0$.  Then, as implied by Clairaut's Theorem, $\text{div}~F = \text{div}(\text{curl}~G) = 0$, which contradicts $\text{div}~F \not= 0$.  So therefore $F$ is not the curl of another vector field if $\text{div}~F \not= 0$.  So if $\text{div}~F = 0$, then the field is the curl of another field.  It also means the vector field is incompressible (solenoidal)!

S/O to Cameron Williams for making me realize the connection to divergence there.


Work and time, when work is split into parts I'm stuck on a particular type of work and time problems.

For example,

1) A,B,C can complete a work separately in 24,36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?

A simpler version of the same type of problem is as follows:

2) A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?

My attempt at problem 2:

A's 1 day work=1/14 and B's 1 day work= 1/21

Assume that it takes 'd' days to complete the entire work when both A and B are working together. Then,

(1/14 + 1/21)*d= 1

-> d=42/5 days.

But it is stated that 3 days before the completion of the work, A left. Therefore, work done by both in (d-3) days is:

(1/14 + 1/21)*(42/5 - 3)= 9/14

Remaining work= 1- 9/14 = 5/14 which is to be done by B alone. Hence the time taken by B to do (5/14) of the work  is:

(5/14)*21 = 7.5 days.

Total time taken to complete the work = (d-3) + 7.5
                                      = 12.9 days.

However, this answer does not concur with the one that is provided.

My Understanding of problem 1:

Problem 1 is an extended version of problem 2. But since i think i'm doing problem 2 wrong, following the same method on problem 1 will also result in a wrong answer.

Where did i go wrong?

You asked where you went wrong in solving this problem:

  A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is?

As you said in your solution, $A$ can do $1/14$ of the job per day, and $B$ can do $1/21$ of the job per day. On each day that they work together, then, they do $$\frac1{14}+\frac1{21}=\frac5{42}$$ of the job. Up to here you were doing fine; it’s at this point that you went astray. You know that for the last three days of the job $B$ will be working alone. In those $3$ days he’ll do $$3\cdot\frac1{21}=\frac17$$ of the job. That means that the two of them working together must have done $\frac67$ of the job before $A$ left. This would have taken them

$$\frac{6/7}{5/42}=\frac67\cdot\frac{42}5=\frac{36}5\text{ days}\;.$$

Add that to the $3$ days that $B$ worked alone, and you get the correct total: $$\frac{36}5+3=\frac{51}5=10.2\text{ days}\;.$$

You worked out how long it would take them working together, subtracted $3$ days from that, saw how much of the job was left to be done at that point, and added on the number of days that it would take $B$ working alone to finish the job. But as your own figures show, $B$ actually needs $7.5$ days to finish the job at that point, not $3$, so he ends up working alone for $7.5$ days. This means that $A$ actually left $7.5$ days before the end of the job, not $3$ days before. You have to figure out how long it takes them to reach the point at which $B$ can finish in $3$ days.

Added:

  1) A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed?

Here you know that all three worked together for the first $4$ days, $B$ worked alone for the last $3$ days, and $A$ and $B$ worked together for some unknown number of days in the middle. Calculate the fraction of the job done by all three in the first $4$ days and the fraction done by $B$ alone in the last $3$ days, and subtract the total from $1$ to see what fraction was done by $A$ and $B$ in the middle period; then see how long it would take $A$ and $B$ to do that much.

A better approach to the problem.

Take the LCM of 14 and 21 which will give you the total amount of work.
LCM (14,21) = 42.

A completes in 14 days.So he does 42/14 in 1 day.Similarly B does 42/21 in 1 day.
A=3,B=2.Since A left 3 days prior to the completion of the work, so his 3 days work is absent.Lets add his 3 days pending work to the total amount of work.

In 1 day he does 3 units of work, so in 3 days he will do 9 units of work.

Total work = 42+9 = 51.

Both will now take 3+2 to complete the total work.

A+B = 51/5 = 10.2.


Prove $\lim _{ n\rightarrow \infty }{ { x }_{ n }^{ k } } ={ \left( \lim _{ n\rightarrow \infty }{ { x }_{ n } } \right) }^{ k }$ I'm trying to prove that the limit of the sequence $x_n^k$ is the same as the limit of $x_n$ all raised to the $k$th power.  Prove $$\lim _{ n\rightarrow \infty  }{ { x }_{ n }^{ k } } ={ \left( \lim _{ n\rightarrow \infty  }{ { x }_{ n } }  \right)  }^{ k }$$

Assume $\displaystyle\lim_{n\to\infty}x_n=L$ and $\displaystyle\lim_{n\to\infty}y_n=M$. Then, $(x_n)$ is bounded and thus there is a constant $c>0$ such that
$$|x_n|\leq c,\quad\forall\ n\in\mathbb N.$$

Furthermore, given $\varepsilon>0$ there is $n_0\in\mathbb  N$ such that
$$n>n_0\quad\Rightarrow\quad |x_n-L|<\frac{\varepsilon}{2|M|},\quad|y_n-M|<\frac{\varepsilon}{2c} .$$

As a consequence,
$$n>n_0\quad\Rightarrow\quad |x_ny_n-LM|\leq |x_n||y_n-M|+|M||x_n-L|<c\frac{\varepsilon}{2c}+|M|\frac{\varepsilon}{2|M|}=\varepsilon.$$

This proves that
$$\lim_{n\to\infty} (x_n\cdot y_n)=LM=\left(\lim_{n\to\infty} x_n\right)\left(\lim_{n\to\infty} y_n\right)$$
Therefore,
$$\lim_{n\to\infty}x_n^k=\lim_{n\to\infty}(x_n\cdot x_n\cdots x_n)=\left(\lim_{n\to\infty} x_n\right)\left(\lim_{n\to\infty} x_n\right)\cdots \left(\lim_{n\to\infty} x_n\right)=\left(\lim_{n\to\infty} x_n\right)^k$$

For $\epsilon > 0$, there is a $M_1 \in \mathbb{N}$ that satisfies:

$$|x_n - L| < \epsilon, \forall n > M_1$$

So $x_n$ converges to $L$. We need a similar result for $x_n^k$ and $L^k$. So, for any $\epsilon' > 0$, we want to show existence of a $M_2 \in \mathbb{N}$ that satisfies:

$$|x_n^k - L^k| < \epsilon', \forall n > M_2$$

But:

$$|x_n^k - L^k| = |(x_n - L)(x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1})|$$

$$= |x_n - L| |x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|$$

Take $\epsilon = \frac{\epsilon'}{|x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|}$. By hypothesis, there existis $M_1 \in \mathbb{N}$ that satisfies:

$$|x_n - L| < \epsilon, \forall n > M_1$$

Therefore:

$$|x_n^k - L^k| = |x_n - L| |x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|\\ < \frac{\epsilon'}{|x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|} \cdot |x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|$$

$$|x_n^k - L^k| < \epsilon'$$

for all $n > M_1$. Taking $M_2 = M_1$ gives the result.


Einstein summation confusion I was trying to evaluate the following expression:

$$\frac{1}{\sqrt{u_iu_i}}.$$

I know the author used this as short notation for

$$\frac{1}{\sqrt{u_1^2+u_2^2+u_3^2}}$$

and I always thought of it as this expression. Now, I can rewrite

$$u_i^{-1/2}u_i^{-1/2}.$$

Using the summation convention (https://en.wikipedia.org/wiki/Einstein_notation)

$\sum_{i=1}^{3}c_ix^i=c_ix^i$

this would result in:

$u_i^{-1/2}u_i^{-1/2}=\sum_{i=1}^{n}u_i^{-1/2}u_i^{-1/2}=u_1^{-1/2}u_1^{-1/2}+u_2^{-1/2}u_2^{-1/2}+u_3^{-1/2}u_3^{-1/2}.$

Which is clearly not what was intended by the author.

  My question: Which intrepretation is right and which is wrong? I would be glad > if someone could give a trustable source.

Think of it like this(*)

$$
\frac{1}{\sqrt{u_i u_i}} = \frac{1}{\sqrt{\sum u_i u_i}} = \frac{1}{\sqrt{u1^2 + u_2^2 + u_3^2}}
$$

(*)Important: There is however an abuse of notation here: $x_ix_i$ does not expand to $x_1^2 + x_2^2 + x_3^2$ using Einstein's convention, because it violates one rule: one upper index sums with one and only one lower index. In this expression there're only lower indices! You can fix this by introducing a Kronecker delta

$$
\frac{1}{\sqrt{u_i\delta^{ij}u_j}} = \frac{1}{\sqrt{u1^2 + u_2^2 + u_3^2}}
$$

$u^iu_i$ is shortened for $\sum_i u_iu_i$ ; which is $u_1^2+u_2^2+u_3^2$ when we implicitly have $i\in\{1,2,3\}$

Thus $(u^iu_i)^{-1/2} = (u_1^2+u_2^2+u_3^2)^{-1/2}$ because we resolve the Einstein notation before applying the exponent to the series.

This is quite distinct from $u^iv_i$ in which the exponent is the bound variable of the series.

PS: Also note that when using Einstein notation $(u^iv_i)^2 \neq (u^i)^2(v_i)^2$ as the square of a series is not equal to a series of squares.

$$(u^iv_i)^2=  (u_1v_1+u_2v_2+u_3v_3)^2~\neq~ (u_1+u_2+u_3)^2(u_1+u_2+u_3)^2=(u^i)^2(v_i)^2$$

Similarly $(u^iu_i)^{-1/2}\neq (u^i)^{-1/2}(u_i)^{-1/2}$.   Exponents do not distribute over series.


Proving that the sum $\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$ Proving that the sum

  $$\sum_{n=1}^{\infty }\frac{\sin^m(n\pi/2)}{n^2}=\frac{\pi^2}{8}$$
  When $m$=integer even number

I know that the $\frac{\pi^2}{8}$ comes from $\sum_{n=0}^{\infty }\frac{1}{(2n+1)^2}$ and I know how to prove it but I don't know how to prove the above sum  especially with the power $m$

Hint: Determine the expression $\sin^m{(n\pi/2)}$. From this point of view you can do some machinary work for solving the series.

After some calculations you get the series
$$\sum _{k=1}^{\infty } \frac{1-(-1)^k}{2k^2}$$

Finally you have to compute the series.
Maybe some Fourier series are useful or you are able to split the series in smart partial series.


Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:

$$
x^3-3x+1=0
$$

Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?

$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$

We have
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} = \frac{a^2c+b^2a+c^2b}{abc}
$$
Since from Vieta's relations we know
$$
a+b+c =0,\quad ab+bc+ca =-3,\quad abc =-1,
$$
our goal is to calculate
$$
s = a^2c+b^2a+c^2b.
$$
Let's introduce
$$
p = ac^2+ba^2+cb^2.
$$
Than we have
$$
0 = (ab+bc+ac)(a+b+c) = p+s+3abc
$$
and $p+s = 3$.

Now let's multiply
$$
s\cdot p = a^3b^3 + a^3c^3 +b^3c^3 + 3(abc)^2+ abc(a^3+b^3+c^3)
$$
Since $a,b,c$ are the roots of polynomial the last equation can be rewritten as
$$
sp = (3a-1)(3b-1)+(3a-1)(3c-1)+(3b-1)(3c-1) + 3(abc)^2
+ abc(3(a+b+c)-3)=
$$
$$
=9(ab+ac+bc)-6(a+b+c)+3 + 3(abc)^2 + abc(3(a+b+c)-3) =-27+3+3+3=-18.
$$

So, $s+p=3$ and $sp=-18$. From here one can deduce that $s= 6$. (see @mathlove answer)

Required,
$= a/b + b/c + c/a$

By Cross multiplication,

$= (a^2bc+b^2ac+c^2ab)/(abc)$

$= abc  (a+b+c) /(abc)$

$ = (a+b+c)   ..........1$

Well known

Property Relation:-

{If α1, α2,α3 ...  αn are the roots of  the equation

$f(x)= a_0x_n  +a_1x_{n-1}  +a_2x_{n-2} +...+a_{n-1}x + a_n =0$, then

$f(x)= a_0 (x-α_1)(x-α_2)(x-α_3)... (x-α_n)$

Equating both the RHS terms we get,

$a_0x_n  +a_1x_{n-1}  +a_2x_{n-2} +...+a_{n-1}x + a_n = a_0(x-α_1)(x-α_2)(x-α_3)... (x-α_n)$

Comparing coefficients of $x_{n-1}$   on both sides, we get

  $S1 = α_1 + α_2+α_3 +... + α_n = ∑α_i  = -a_1/ a_0$

or, S1= - coeff. of $x_n-1$/coeff. of $x_n$

Comparing coefficients of xn-2   on both sides, we get

  $S2 = α_1 α_2+ α_1α_3 +...  = ∑α_i α_j  = (-1)^2a_2/ a_0$
                                         $i≠ j$

or, S2= (-1)2 coeff. of $x_{n-2}$/coeff. of $x_n$

Comparing coefficients of xn-3   on both sides, we get

  $S3 = α_1 α_2α_3+ α_2α_3α_4 +...  = ∑α_i α_j α_k  = (-1)^3a_3/ a_0$

If $a$ $b$ $c$ are the roots of
     $x^3-3x+1=0$

then

$abc = 1$

$a+b+c = 0$

$ab + bc + cd = -1$

Substituting in eqn 1, we get
Ans: 0


$L \cong K(\sqrt{x}) \cong K[X]/(X^2-x)$ for $x \in K$ I am trying to prove that if $L$ is a quadratic extension of a field $K$ with characteristic differing from 2, then $L \cong K(\sqrt{x}) \cong K[X]/(X^2-x)$ for $x \in K$.

I tried making a isomorphism from $K(\sqrt{x})$ to $K[X]/(X^2-x)$ given that $X^2-x=0$, so $f(X)^2-f(x)=0$, by using the abc-formula I got 1 and 0. These both go to 0 in $f(X)^2-f(x)=0$, but then the function would not be bijective.

Could anyone help me with a method that should work in order to prove these isomorphisms?

Let $L$ be a quadratic extension of $K$. Then $K(\alpha)=L$ for all $\alpha\in L\ \backslash \ K$ (because $\alpha$ is an algebraic element with degree $2$). We have to prove that exists $x\in K$ such that $L = K(\alpha) = K(\sqrt{x})$.

Take the minimum polinomial $\mu_\alpha(X) = X^2 + bX + c$ of $\alpha$. Since $char K\neq 2$ we have:
$$
\alpha = \dfrac{-b+\sqrt{b^2-4c}}{2}
$$
Then $K(\alpha)=K(\dfrac{-b+\sqrt{b^2-4c}}{2}) = K(\sqrt{\Delta})$. Take $\Delta=x$ and you are done.

The last isomorphism is given by the evaluation map:
\begin{gather}
K[X]\longrightarrow K[\sqrt{x}]\\
X \longmapsto \sqrt{x}
\end{gather}
That is surjective and the kernel is $(X^2 - x)$ that is maximal because $x$ is not a square in $K$.

You map $\sqrt{x}\longmapsto X$, and there is no other element of $K(\sqrt{x})$ which maps to $X$, so your mapping is bijective.


how to evaluate $\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}\text{d}x$ I was solving a physics problem and eventually the problem boiled down to solving the following integral:

$$\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}\text{d}x$$

I have already tried substitutions like $\sin x=t^2$ , $\sin x=t$ and have tried using the properties of definite integrals given on http://www.sosmath.com/calculus/integ/integ02/integ02.html but I could not solve this integral. Please help!

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$\ds{{\cal I} \equiv \int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}}:\ {\large ?}}$

With $x = \arcsin\pars{z^{2}}$ we'll get:
$$
{\cal I} = 2\int_{0}^{1}{\dd z \over \root{1 - z^{4}}}\,.\quad\mbox{With}\ z = t^{1/4}\,,\quad
{\cal I} = \half\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t
$$

Then
$$
{\cal I}=\half\,{\rm B}\pars{{1 \over 4},\half}=\half\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}}
={\root{\pi} \over 2}\,\Gamma^{2}\pars{1/4}\,{\sin\pars{\pi/4} \over \pi}
$$
$$\color{#00f}{\large%
\int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}}
=
{1 \over 4}\,\root{2 \over \pi}\,\Gamma^{\,2}\pars{1 \over 4}}
\approx 2.62206
$$

So, which answer is prefereable?
$$
\sqrt{2}\;\mathrm{K}\left(\frac{1}{\sqrt{2}}\right) =
\frac{1}{2}\;\mathrm{B}\left(\frac{1}{4},\frac{1}{2}\right) =
\frac{\pi^{3/4}}{\sqrt{2}\,\Gamma(3/4)^2}
$$


Proving Hölder's Inequality Let $f,g,\alpha:[a,b]\rightarrow \mathbb{R}$ with $\alpha$ increasing and $f,g \in \mathscr{R}(\alpha)$, and $p,q>0$ with $\frac{1}{p}+\frac{1}{q}=1$. Prove that $$\left|\int_a^b f(x)g(x)d\alpha\right|\leq \left(\int_a^b \left|f(x)\right|^p d\alpha \right)^{1/p} \left(\int_a^b \left|g(x)\right|^q d\alpha \right)^{1/q}$$

I am using Young's inequality, which states that for $a,b>0$, $uv\leq \frac{1}{p}u^{p}+\frac{1}{q}v^{q}$. This gets me as far as showing that $$\left|\int_a^b f(x)g(x)d\alpha\right|\leq \int\left( \frac {1}{p}|f(x)|^p +\frac{1}{q}|g(x)|^q\right)d\alpha$$

But here I'm stuck. I'm vaguely thinking that I could use the fact that $\frac {1}{p}|f(x)|^p +\frac{1}{q}|g(x)|^q$ is a convex combination and so if I do some Jensen's inequality type thing, but I can't figure out a way to make it work out.

Suppose $\displaystyle\int_a^b \left|f(x)\right|^p d\alpha\neq 0$ and  $\displaystyle\int_a^b \left|g(x)\right|^q d\alpha\neq 0$. Otherwise, if $\displaystyle\int_a^b \left|f(x)\right|^p d\alpha=0$, then $f\equiv 0$ a.e. and the Holder's inequality is trivial in this case.

Now applying Young's inequality with $u=\displaystyle\frac{|f(x)|}{(\int_a^b \left|f(x)\right|^p d\alpha)^{\frac{1}{p}}}$ and $v=\displaystyle\frac{|g(x)|}{(\int_a^b \left|g(x)\right|^q d\alpha)^{\frac{1}{q}}}$, we have
$$\frac{|f(x)|}{(\int_a^b \left|f(x)\right|^p d\alpha)^{\frac{1}{p}}}\cdot\frac{|g(x)|}{(\int_a^b \left|g(x)\right|^q d\alpha)^{\frac{1}{q}}}\leq\frac{1}{p}\frac{|f(x)|^p}{\int_a^b \left|f(x)\right|^p d\alpha}+\frac{1}{q}\frac{|g(x)|^q}{\int_a^b \left|g(x)\right|^q d\alpha}.$$
Integrating it from $a$ to $b$ with respect to $d\alpha$, we obtain
$$\frac{\int_a^b|f(x)||g(x)|d\alpha}{(\int_a^b \left|f(x)\right|^p d\alpha)^{\frac{1}{p}}(\int_a^b \left|g(x)\right|^q d\alpha)^{\frac{1}{q}}}\leq\frac{1}{p}+\frac{1}{q}=1$$
which implies that
$$ \tag{1}\int_a^b|f(x)||g(x)|d\alpha\leq\left(\int_a^b \left|f(x)\right|^p d\alpha \right)^{1/p} \left(\int_a^b \left|g(x)\right|^q d\alpha \right)^{1/q}.$$
Now the inequality which we want to prove follows from $(1)$ and the inequality
$$\left|\int_a^b f(x)g(x)d\alpha\right|\leq\int_a^b|f(x)||g(x)|d\alpha.$$

In the vast majority of books dealing with Real Analysis, Hölder's inequality is proven by the use of Young's inequality for the case in which $p , q > 1$, and they usually have as an exercise the question whether this inequality is valid for $p =1$ which means that $q = \infty$.
Well, if $f \in L_{1}$ and $g \in L_{\infty}$ then,
\begin{equation}
\Vert f \Vert_{1} = \int_{a}^{b} \vert f(x) \vert dx < \infty \text{  and  } \Vert g \Vert_{\infty} = \sup_{x \in [a,b]} \lbrace g(x): x \in [a,b] \rbrace < \infty.
\end{equation}
So, using properties of the absolute value, we have
\begin{eqnarray}
\left\vert \int_{a}^{b} f(x)g(x) dx \right\vert &\leq& \int_{a}^{b} \vert f(x)g(x) \vert dx \\
&=& \int_{a}^{b} \vert f(x)\vert \vert g(x) \vert dx \\
&\leq&  \int_{a}^{b} \vert f(x)\vert \sup_{x \in [a,b]} \lbrace \vert g(x) \vert : x \in [a,b] \rbrace dx \\
&=& \sup_{x \in [a,b]} \lbrace \vert g(x) \vert : x \in [a,b] \rbrace \int_{a}^{b} \vert f(x)\vert dx \\
&=& \Vert g \Vert_{\infty} \Vert f \Vert_{1}.
\end{eqnarray}
This is the answer to the mentionated case.


Meaning of negative mutual information From wiki, I know negative multiple mutual information exists. But I can't understand its meaning. Does someone help me have a better understand? For example, if I have three variables which have negative mutual information, if they have a lower mutual information, what's it mean?

Mutual information (MI) measures how much two variables are inter-dependent.
So, higher the MI, more similar the variables.
Two variables could be, for example, the intensity values of two greyscale images.
But many algorithms use the matching cost, i.e. how much two variables are different. Hence, the minus sign.