why do independent variables have zero covariance? I can see it how happens from the expression using $E[XY]=E[X]E[Y]$, but I can't understand it intuitively.
$$cov(x,y)=E[(x-E[x])(y-E[y])]$$ for independent random variables x and y.

How do you guys understand the logic behind it being $0$?
Figures explaining would be just awesome.

Because $X$ and $Y$ are independent, the value of one does not affect the value of the other. So consider the expectation over each of the two variables separately, ie. first consider:

$E_X[(X - E(X))(Y - E(Y))]$

Because they're independent, nothing that happens inside the expectation really affects what happens to $Y$, which is as good as if it were a constant (just like if you take the partial derivative of something with respect to $x$, you treat $y$ as a constant). So we pull the $Y$ bit out, giving us $E_X[X - E(X)] \times (Y - E(Y))$, but the bit on the left is "the expected difference between $X$ and its expected value" which is of course $0$. You can do the same thing with the expectation over $Y$, and you can also apply the law of iterated expectations to show that the whole thing must thus be zero.

By the definition of Covariance, generally:

$$\mathsf {Cov}(X,Y)~{=~\mathsf E((X-\mathsf E(X))(Y-\mathsf E(Y))
\\ = \mathsf E(XY-X\mathsf E(Y)-\mathsf E(X)Y+\mathsf E(X)\mathsf E(Y))
\\ = \mathsf E(XY)-\mathsf E(X)\mathsf E(Y)
}$$

We have in the case of independence

$$\begin{align}\mathsf E(XY)~&=~\mathsf E(~\mathsf E(XY\mid X)~) &\text{Law of Total Expectation}\\[1ex]&=~\mathsf E(~X~\mathsf E(Y\mid X)~) \\[1ex] &=~ \mathsf E(X~\mathsf E(Y)) & \text{independence}\\[1ex] &=~ \mathsf E(X)~\mathsf E(Y) \end{align}$$

Then when independent the covariance is zero.

Alternatively we could say

$$\begin{align}\mathsf E\Big(\big(X-\mathsf E(X)\big)\big(Y-\mathsf E(Y)\big)\Big) ~&=~\mathsf E\Big(\mathsf E\Big(\big(X-\mathsf E(X)\big)\big(Y-\mathsf E(Y)\big)~\Big\vert~ X\Big)\Big) & \text{Law of Total Expectation}
\\[1ex] &=~\mathsf E\Big(\big(X-\mathsf E(X)\big)~\mathsf E\big(Y-\mathsf E(Y)\mid X\big)\Big)
\\[1ex] &=~\mathsf E\Big(\big(X-\mathsf E(X)\big)~\mathsf E\big(Y-\mathsf E(Y)\big)\Big) &\text{by independence}
\\[1ex] &=~\mathsf E\big(X-\mathsf E(X)\big)~\mathsf E\big(Y-\mathsf E(Y)\big)
\\[1ex] &=~\big(\mathsf E(X)-\mathsf E(X)\big)~\big(\mathsf E(Y)-\mathsf E(Y)\big)
\\[1ex] &=~0 \end{align}$$

In general, when $X,Y$ are independent and $g,h$ are monovariate functions. $$\mathsf E(g(X)\cdot h(Y))=\mathsf E(g(X))\cdot\mathsf E(h(Y))$$


14th card is ace clarificarion. There are 2 questions.

I need to find probablity that 14th card is ace and that first ace occurs on 14th card.

For the first part, I though it like this.

Choose any of 1 ace in ways = ${4 \choose 1}$

And choose 13 cards from remaining 51 cards= ${51 \choose 13}$

Choose 14 cards from 52 = ${52 \choose 14}$

Probability = $\frac{{4 \choose 1}{51 \choose 13}} {{52 \choose 14}}$

For Second Part

I removed all four aces, cards remaining  = 48.
Choose 13 from these = ${48 \choose 13}$
And any one ace = ${4 \choose 1}$
Probability = $\frac{{4 \choose 1} {48 \choose 13}} {{52 \choose 14}} $

Both answers do not match.

First part:
$$\frac{\binom 41 \binom {51}{13} 13!}{\binom {52}{14}14!}$$

Second part:
$$\frac{\binom 41 \binom {48}{13} 13!}{\binom {52}{14}14!} $$

NOTE:

Order matters.($13!$ and $14!$)

Your answer consider these two(and many more) cases the same:
$$[2,3,\color{green}{4,5},6,7,8,9,10,J,Q,K,2,A]\leftrightarrow[2,3,\color{red}{5,4},6,7,8,9,10,J,Q,K,2,A]$$
because you only select those $13$ cards, not arrange them before the $14$th ace card

For (1), your denominator is wrong, because of the $14$ cards chosen order is important (for at least one of the cards).  (1) would be easier if you imagined picking the $14$th card first (or if you prefer, laying out the first $13$ cards face down, and then turning up the $14$th card.

For (2) your denominator is again wrong.  You might try doing this one with order being important throughout the problem.  (I.e., choose a non-ace for slot 1, choose a non-ace from the remaining cards for slot 2, etc.)


Describe the nonzero integer solutions to the equation $a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 =0$ Can someone describe all the integer solutions to the above equation such that $abcdefg\neq 0$ ?

I do not think there is a known way of describing all of them, but here you'll find some further reading: http://mathworld.wolfram.com/CubicNumber.html


Exponents and mod arithmetic Is it true that if $a\equiv b \bmod n$, the $c^a \equiv c^b \bmod n$? I'm not quite sure how to prove this myself.

Counterexample:

$$3\equiv0 \mod 3$$

But

$$2^3 \not \equiv 2^0 \mod 3$$

A counterexample:

Consider when $n = 15$, and $2 \equiv 17 \bmod 15$. Then $2^2 \equiv 4 \bmod 15$ compared to $2^{17} \equiv 2 \bmod 15$ (and this last one is easy because $2^8 \equiv 1 \bmod 15$ by Euler's theorem).


If I start off with 1 bacteria, and it triples every hour, how many do I have after 24 hours? I'm in 9th grade, and I'm having some difficulties completing a math assignment in my Algebra 1 class. My question is about exponential growth.

  If I start off with 1 bacteria, and it triples every hour, how many will I have after 24 hours? If you answer this, please walk me through the steps. It would be greatly appreciated!

One more question. What would the formula for this situation be? ($y=a(1+r)^x$)

Let $a$ be the initial amount of bacteria you have. Then each hour you are going to multiply by $3$. So it is going to look like $a(3)$ for $1$ hour, $a(3)(3)$ for $2$ hours, $a(3)(3)(3)$ for $3$ hours and so on. In order to simplify this you can say you have $a(3)^x$ where $x$ is the amount of hours that pass. The formula for this will be $y=a(3)^x$.

The formula you gave is the exponential growth function where $a$ is the initial amount, $x$ is the amount of hours (or whatever time interval you are using) passes, and $r$ is the growth rate that is given in a percentage/decimal.

In this case, $r$ would be $200$% or $2.00$. Then you would have $y=a(1+2.00)^x=a(3)^x$.

Hope this helps!

Assume the amount of bacteria is a function of the form $y(x)=A e^{rx}$, with $r,A$ unknown, and $x$ being time in hours. Since we start off with 1 bacteria, this means $y(0)=Ae^0=A=1.$ The statement about it tripling every hour means the following. Fix a time $x$. Then $y(x+1)=3y(x)$. In other words, with $A=1$, we have $e^{r(x+1)}=3e^{rx}.$ Use the fact that $e^{r(x+1)}=e^{rx}e^r.$ Then the $e^{rx}$ cancel on both sides and we get $e^r=3$.

Hence, $r=\ln(3)$. Therefore, the original function is $y(x)=e^{\ln(3)x}$, (which actually reduces to $(e^{\ln(3)})^x=3^x.$) Can you finish the rest?


Find a differentiable function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f_{xy}\neq f_{yx}$. Find a differentiable function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that the second-order partial derivatives $f_{xy}(0, 0)$ and $f_{yx}(0, 0)$ exist but are not equal.

So I need a function of $x$ and $y$ where $f_{xy}(0, 0)\ne f_{yx}(0, 0)$.

So looking at $f(x,y)=x^2y^2$:

$f_x(x,y)=2xy^2$

$f_{xy}(x,y)=4xy$

$f_y(x,y)=2x^2y$

$f_{yx}(x,y)=4xy$

This function obviously fails, but I'm trying to find an example and I'm having trouble. I obviously need an some terms added, but the added terms will be killed off by the second partial derivative... So any hints?

As explained in the comments, because of the Schwarz's theorem, the function cannot have continuous second partial derivatives at $(0,0)$.

Here there is an example.

And here (page 20) there is a reference for the following interesting facts.

There exists a function $f$, the mixed second derivatives of which exist at every point but such that $f_{xy}\neq f_{yx}$ on a set of positive measure.
There exists a function $f$, the mixed second derivatives of which exist almost everywhere and such that $f_{xy}\neq f_{yx}$ almost everywhere.


Solving equation with infinite exponent tower How to solve this equation for $x$ where $a>0$? The exponent tower goes on forever:

$$a=x^{x^{x^{.^{.^{.}}}}}$$

My Calculus book gives the following reasoning:

$$ln(a)=x^{x^{x^{.^{.^{.}}}}}ln(x)=a\,ln(x)$$

To conclude that: $$x=a^\frac{1}{a}$$

Why is this correct?

Equation:
$$a = {x^{x^{x^{.^{.^{.}}}}}}$$

Take $ln$ of both sides, then use the power rule in right and plug $"a"$ in ${x^{x^{x^{.^{.^{.}}}}}}$:

$$ln(a) = ln({x^{x^{x^{.^{.^{.}}}}}})$$
$$ln(a) = {x^{x^{x^{.^{.^{.}}}}}} \cdot ln(x)$$
$$ln(a) = a \cdot ln(x)$$

Divide both sides by "$a"$, then rearrange left side for using power rule afterwards:
$$\frac{ln(a)}{a} = ln(x)$$
$$\frac{1}{a}ln(a) = ln(x)$$

Use power rule in left, then take $exp$ of both sides and lastly, rewrite left side as a root:
$$ln(a^\frac{1}{a}) = ln(x)$$
$$a^\frac{1}{a} = x$$
$$\sqrt [a]{a} = x$$

Your book is right. Don't consider that $x^n$ is infinite. It is finite here in this case, and equal to $a$ as given in the question. Your book has a very good explanatory logic.


What is the relationship between Calculus and Vector Calculus? Im a highschool student, and i read a book about calculus.
And Im curious about vector calculus.
Vector Calculus = vector + calculus  ??
is it right? if so, is calculus a higher concept of vector calculus?

Not really.
Etymologically, calculus refers to pebbles of chalk (latin calculi) that were used in counting and simple computations (like with an abacus). Today, calculus could best be taken to mean "way of computation". This way, we have

vector calculus
differential calculus
propositional calculus

and more. But when we use "calculus" alone, we tend to use it as a short for  "infinitesimal calculus".


In how many ways 20 identical balls can be put into 5 distinct bins such that each bin contains at least 2 balls? Firstly, I have to find the total no of ways in which 20 identical balls can be put into 5 distinct bins.
Then, I have to subtract the no of cases when no of balls in at least one bin is less than 2.
But, I'm not getting how to find that no.
If possible, plz suggest me various ways to approach and solve this kind of problems.
Plz help me with the complete solution to this problem.

As the balls the identical, I think we can just start the process with each box having two balls in each box. The identical nature of the balls tells us it can be done in only 1 way. So we are left with the task of filling 10 identical balls in 5 boxes.

The answer 5^20 is not right for identical balls, think of a less severe situation with 4 balls and 2 boxes, the problem is essentially asking in how many ways can you put the 4 balls in two different containers.
Let's say the balls are b1,b2,b3 and b4 and say one 1 way of filling it is,

b1,b2 goes to container A
b3,b4 goes to container B

another way of filling it proposes

b1,b3 goes to container A
b2,b4 goes to container B

But in our case b1=b2=b3=b4 thereby both the cases are equivalent.
I'm fact, there are 4C2 equivalent ways of filling the container 2 balls at a time.
Same goes to the case of the original problem of filling 5 container with 20 identical balls, so there will be less number of possibilities.

Firstly give two balls to each box then use stick method to assign the remaining balls.
The answer is 11C4+2(11C3)+11C2+11C1=726
Ask me if you don't get this!


Why do we care if a set forms a basis? I am taking an advanced linear algebra course and am once again confused by a lot of the concepts. I understand that the definition of a basis is a set of vectors that spans the vector space and is linearly independent. Can anyone provide any intuition on why this matters?

I can visualize (kind of) what it means for a set to span a vector space, and I understand (from regression analysis) why it is important that vectors are linearly independent, but I don't really understand why it matters if they are both, ie why it matters if they form a basis.

For reference I am an undergraduate interested in statistics and data analysis. I have taken courses in mathematical statistics, regression analysis and am currently enrolled in time series analysis. I am somewhat familiar with PCA. Any intuition that can be provided through any of those lenses would be much appreciated.

We care because mathematicians are inherently lazy, and having basis usually enables you to check things for a very small number of cases while being able to generalize to the whole space...

Take linear transformations as an example. If you know how a linear transformation behaves for the basis vectors, you automatically know how it behaves for the whole space.

There is an important theorem:

All Basis have the same size.

This allows one to define the dimension of a vector space as this size.

The dimension is the fundamental invariant of a vector space.


Strategy for 7 up 7 down game This is a dice game where the host throws the dice and the players bet on 3 numbers - numbers below 7, 7 and numbers above 7. On numbers below and above 7 the players get double their stake and on 7 the players get 5 times their stake. For example if i bet 100 bucks on "below" and 7 comes, I will lose, but if 6 comes, then i win 200 bucks. Similarly if someone has betted 100 bucks on 7 and if 7 comes, he wins 500 bucks.

What is the best way to bet in this game?

As they say, "the only winning move is not to play." I assume there are two dice. In that case, the expected value of all three bets is the same, and it's negative: on average, you lose $\frac{1}{6}$ of what you bet.

Some details: when rolling $2$ dice the probabilities of getting a sum of $2, 3, 4, ... 12$ are $\frac{1}{36}, \frac{2}{36}, \frac{3}{36}$, and so on, increasing until $7$ and then decreasing. This is a nice exercise if you've never done it; you just need to count the number of ways a pair of numbers between $1$ and $6$ can add up to a given number. It follows that the probability of getting a sum of $7$ is $\frac{1}{6}$, and by symmetry the probability of getting above $7$ or below $7$ are each $\frac{5}{12}$.

It follows that if you bet $7$ then on average you multiply your stake by $\frac{5}{6}$, whereas if you bet either below or above then on average you also multiply your stake by $\frac{5}{6}$. So all three options are equally bad, in the long run.

Some meta-lessons to learn from this computation:

Generally the expected value of a game like this is negative, or the host would never bother to offer it. For example, the expected value of every game at a casino and the lottery has to be negative, or else the casino (resp. the institution running the lottery) would not make money. This is as much a lesson about incentives as it is about probability.
If one of the options was better than the others, at some point people would be aware of this and play only that option.

Dont worry about all these mathematics, make it simple, gambling is all about sheer luck. Doubling the loss bets isn't a good idea, I have lost many times doing that, even after 15 deals the side won't change, this can happen anytime, this is very risky attempt for someone playing with limited amount, and this idea will be successful for the one who have the money backed for the betting. Who knows how the cards are shuffled, then there is no point of permutations, combinations and probabilities. Please dont misguide people with this kind if advices. For me gambling is sheer luck to add some more money to our hard earned money  what I say is accept both loss and profit. And do not gamble with commitments, if u do that chances of losing is more than 100%


Proving a limit equals zero I have to prove, without L'hopital rule, the following limit:
$$\lim_{x \to \infty}\sqrt{x} \sin \frac{1}{x} =0$$

I tried doing a variable change, setting $t=\frac{1}{x}$ and reaching the following: $$\lim_{t\to 0} \sqrt{\frac{1}{t}} \sin t $$

But I can't prove neither. Tried the second version with the squeeze theorem, but I can't prove the limit.

Thanks!

See that
$|\sin x|\leq x$ while $x\to0$ and here $\dfrac1x\to0$ so
$$-\dfrac1x\leq\sin\dfrac1x\leq\dfrac1x$$
multiply RH and LH by $\sqrt{x}$ gives us
$$-\sqrt{x}\dfrac1x\leq\sqrt{x}\sin\dfrac1x\leq\sqrt{x}\dfrac1x$$
or
$$-\dfrac{1}{\sqrt{x}}\leq\sqrt{x}\sin\dfrac1x\leq\dfrac{1}{\sqrt{x}}$$
finally the sandwich theorem concludes that
$$\lim_{x\to\infty}\sqrt{x}\sin\dfrac1x=0$$


Equivalent way to write existential quantifier? Let $L(x,y)$ be the statement, "$x$ loves $y$."

If the universe U is the set of all people,

can the statement, "There is exactly one person whom everybody loves."

or, equivalently (according to my textbook):     $∃x[∀y(L(y,x))  ∧  ∀z(∀y(L(y,z) → z=x)]$

be written as (my answer):

[($∃x∀yL(y,x)) ∧ (∀z∃y¬L(y,z))]     \mbox∣~  x≠z, z∈U$     ?

The original statement is the answer from my textbook, which I think seems pretty weird.  That's why I'm trying to rewrite it.
So, if my answer is not a 'better' way to write "There is exactly one person whom everybody loves.", what is the most logical way to write this statement?

The statement from the book
$$\exists x [ \forall y L(y,x) \land \forall z ( \forall y L(y,z) \rightarrow z = x ) ]$$

is a good way to express this. To see this you might write the property of a person to be loved by everyone as
$$P(x) := \forall y L(y,x)$$

Then the statement becomes
$$\exists x [ P(x) \land \forall z ( P(z) \rightarrow z = x ) ]$$

which means that there is a person $x$ with property $P$, and for any person $y$ with property $P$, that person $y$ must already be $x$.

This might further be abbreviated
$$\exists! x P(x)$$

$$\color{red}(\exists \color{red}x~\forall y~ Lyx\color{red}) \land (\forall z~\exists y \lnot Lyz) \text{ s.t. } \color{red}x \ne z$$

Look at the red parts.  The first $x$ is inside the parenthesis (quantified), the second is outside (unquantified).  The second $x$, grammatically, refers to a different variable than the first due to how you wrote it.

(Btw ignore the complaints about s.t., it is well understood and fine to use if done grammatically correctly).

Here are some ways to transform the formula:

$$\exists x~\bigg( (\forall y ~ Lyx)  \land (\forall z~\forall y ~Lyz \to z = x)\bigg)$$

First it would appear you'd like to use the contrapositive:

$$\exists x~\bigg( (\forall y ~ Lyx)  \land (\forall z~\forall y ~ z \ne x \to \lnot Lyz )\bigg)$$

Since $x \ne z$ doesn't reference $y$, the $\forall y$ can be moved (this is a formula from the prenex technique):

$$\exists x~\bigg( (\forall y ~ Lyx)  \land (\forall z~z \ne x \to \forall y ~ \lnot Lyz )\bigg)$$

One thing you can also notice is that you have a formula of the form $\forall a Pa \land \forall b Qb$, which can be written $\forall c Pc \land Qc$ :

$$\exists x~\bigg( (\forall w ~ Lwx)  \land (\forall w~w \ne x \to \forall y ~ \lnot Lyw )\bigg)$$

$$\exists x~\bigg( \forall w ~ (Lwx  \land (w \ne x \to \forall y ~ \lnot Lyw ))\bigg)$$

There's other manipulations you can do, but more big picture, there are a lot of ways to write "the predicate $P$ is inhabited exactly once".  One way attributed to Dijkstra is as $\exists x (\forall y~P(y) \text{ iff } x = y)$, which in your case becomes:

$$\exists x (\forall y~(\forall z ~ Lzy) \text{ iff } y = x)$$


Notation conventions for subsets of widely used sets I want to express "$x$ is rational and not equal to $0$"
$x\neq 0 \in \mathbb{Q}$ seems awkward to me since $x\neq 0$ is a logical statement, not a rational number. Options I can think of include:
$$x\in \mathbb{Q} \setminus 0$$
$$x\in \mathbb{Q} \land x \neq 0$$
$$x \in \mathbb{Q} | x\neq 0$$
Is there a standard convention? I'm trying to develop a sensible, standard notation to use amongst my classes.

One alternative notation you can use is
$$x\in \mathbb Q^\times.$$
Here $\mathbb Q^\times$ denotes all elements of $\mathbb Q$ which are invertible in $\mathbb Q$. Since $\mathbb Q$ is a field, this is all non-zero elements of $\mathbb Q$. However be aware that this is not always equal to the set without zero; for example $\mathbb Z^\times = \{-1,1\}$ because for any other $n\in\mathbb Z\setminus\{0\}$, $\frac{1}{n}\notin\mathbb Z$.
Yet another not uncommon notation writes the inequality without the leading $x$ as index, that is,
$$x\in\mathbb Q_{\ne 0}.$$
This notation has the advantage that it works also for a whole bunch of other sets, like $\mathbb R_{<0}$ for the negative real numbers, $\mathbb Q_{\ge 0}$ for the nonnegative rational numbers or $\mathbb Z_{>42}$ for the integers larger than $42$.
However if only a single number is excluded, I personally would prefer the notation
$$x\in\mathbb Q\setminus\{0\}$$
(note the braces around $0$, which are missing in your post). At least in cases where I explicitly would use formulas; otherwise I'd just write:

$x$ is a non-zero rational number.

Any set excluding $0$ is generally notated with a superscript asterisk.  This notation comes from multiplicative groups.
$$\Bbb Q\setminus\{0\} = \Bbb Q^*$$


Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$ $a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?

I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the cubic equation non-negative.
The discriminant of $x^3 + A x^2 + B x + C=0$ is
$A^2 B^2 - 4 B^3 - 4 A^3 C + 18 A B C - 27 C^2$
Is there an easier way?

Yes, we need to find $ab+ac+bc=7$ before.
Now, $a$, $b$ and $c$ are roots of the equation:
$$(x-a)(x-b)(x-c)=0$$ or
$$abc=x^3-5x^2+7x.$$
Now, $$(x^3-5x^2+7x)'=(x-1)(3x-7),$$ which gives that a maximal value of $abc$ for which the equation $$abc=x^3-5x^2+7x$$ has three real roots holds for $x=1$, which gives:
$$\max_{a+b+c=7,a^2+b^2+c^2=11}{abc}=3.$$
The equality occurs for example, for $(a,b,c)=(1,1,3).$

As you did not exactly clarify what you mean by abc (their product?) I can only give you halve an answer. I assume you do not want a integer solution rather than a continous one and also real solutions.
With some calculations you will find:
$$b_{1/2}=\frac{1}{2}(+/- \sqrt{-3a^2+10a-3}-a+5) $$
and
$$c_{1/2}=\frac{1}{2}(+/- \sqrt{-3a^2+10a-3}-a+5) $$
This means you can define $f_{1/1}(x\in \mathbb{R}):=a*b_1*c_1$, $f_{1/2}(x\in \mathbb{R}):=a*b_1*c_2$ and so on (all combinations) and from them just calculate the maximum by using standard analysis methods.


How to prove that () Δ () ≠ ( Δ ) for any 2 sets. So the question is how can I prove that for any 2 Sets, A and B, the symmetric difference between the power set of A and the power set of B IS NOT equal to the powr set of A symmetric difference B.
() Δ () ≠ ( Δ )
I have been stuck on this question for a while and I can't find a way to prove it.

The idea is that the empty set ($\varnothing$) lies in the RHS, but not in LHS. And thus the assertion is proved.
In details, let $A$ and $B$ be a set. Since any power set contains the empty set, we have that $\varnothing \in P(A),P(B),P(A\Delta B)$. By the definition of symmetric difference, we have $\varnothing \notin P(A)\Delta P(B)$. Hence, $P(A)\Delta P(B)\neq P(A\Delta B)$, as desired. QED

The symmetric difference of two sets is the set of elements that are in both sets but not in their intersection and is given by:
$$ A Δ B = A \cup B \backslash A \cap B $$
The powerset of a set is the set of all elements of $S$ including the empty set and the set itself and is represented by $P(S)$.
The cardinality of $P(S)$ is given by $$|P(S)| = 2^{|S|}$$
We need to prove that $$() \space Δ \space () ≠ ( \space Δ \space )$$
$$P(A \cup B) \backslash P(A \cap B) ≠  P((A \cup B) \backslash (A \cap B)) $$
Let $X = A \cup B$ and $ Y = A \cap B$. We need to show that:
$$P(X) \space \backslash \space P(Y) ≠ P(X \backslash Y)$$
We can prove that two sets are not equal by showing that they have a different number of elements. Let $|X| =x, |Y|=y$.
$$|X \backslash Y| = x - y$$
$$|P(X) \space \backslash \space P(Y)| = 2^x - 2^y$$
Similarly, on the right hand side,
$$|P(X \backslash Y)| = 2^{x-y}$$
We can prove that $$2^x - 2^y ≠ 2^{x-y}$$ by using a counter example. Suppose $x=1, y=2$
$$2^1 - 2^2 = -3 ≠ 2^{1-2}=1/2$$


How many different answer keys are possible? A professor writes $40$ discrete mathematics true/false questions. Of the statements in these questions, $17$ are true. If the questions can be positioned in any order, how many different answer keys are possible?

No. of questions=40. so, 40! ways of arrangements.
No. of ways of arranging answers= 2^40( two choices true and false)
total no. of keys possible= 40!*2^40


Linear independence of $\sin(x)$ and $\cos(x)$ In the vector space of $f:\mathbb R \to \mathbb R$, how do I prove that functions $\sin(x)$ and $\cos(x)$ are linearly independent. By def., two elements of a vector space are linearly independent if $0 = a\cos(x) + b\sin(x)$ implies that $a=b=0$, but how can I formalize that? Giving $x$ different values? Thanks in advance.

Hint: If $a\cos(x)+b\sin(x)=0$ for all $x\in\mathbb{R}$ then it
is especially true for $x=0,\frac{\pi}{2}$

If Cos(x) and sin(x) are linearly independent, then this implies.

acos(x)+bsin(x)=0 if and only if (for all values of x) a=b=0.

But!! For x=pi/4 rad; cos(pi/4)=sin(pi/4)= sqrt(2)/2 and for a=1 and b=-1
we get: (1)*sqrt(2)/2 +(-1)*sqrt(2)/2 which is sqrt(2)/2-sqrt(2)/2=0.
Therefore: cos(x) and sin(x) cannot be linearly independent.


proof that $b^{\frac{m}{n}}=b^{\frac{p}{q}}$ if $\frac{m}{n}=\frac{p}{q}$ So I just solved this problem, however I'm not sure if my proof works, can someone tell me if it's correct or not, also I suspect there is a much simpler way to prove it so don't hesitate to post your own proof.

The problem is: if $m,n,p,q$ are integers, $n\gt 0,q\gt 0,$ and $r=\frac{m}{n}=\frac{p}{q},$ prove that $$(b^m)^\frac{1}{n}=(b^p)^\frac{1}{q}.$$

First, I had to show that $b^\frac{1}{an}=(b^\frac{1}{n})^\frac{1}{a}$, which can be done by simply noting that $((b^\frac{1}{n})^\frac{1}{a})^{an}=(((b^\frac{1}{n})^\frac{1}{a})^a)^n=(b^\frac{1}{n})^n)=b,$ and since there exists a unique positive nth root for every integer we must have $b^\frac{1}{an}=(b^\frac{1}{n})^\frac{1}{a}$.

Now since $r=\frac{m}{n}=\frac{p}{q}$, without loss of generality we can write $m=ap, $ and $n=aq$ for some integer $a$ thus $(b^m)^\frac{1}{n}=(b^{ap})^\frac{1}{aq}=(((b^p)^a)^\frac{1}{a})^\frac{1}{q}=(b^p)^\frac{1}{q}.$

My main problem with this proof is that I feel like I have cheated a bit with adding the factor $a$ since it's a bit more number theoretic, and I just started doing analysis. Also I think there should be a much easier way to do it.

From the comments For a positive real $b$ and a positive integer $n$, $b^{1/n}$ is defined to be the unique positive real such that $(b^{1/n})^n = b$.

Try $\dfrac{m}{n}=\dfrac{p}{q}=r$.

Then $b^{\frac{m}{n}}=b^r=b^{\frac{p}{q}}$.

As I said in the comments, the factor $a$ doesn't necessarily exist: consider the example of $3/3 = 2/2.$

However, if you let $d = \gcd(m,n)$ such that $m = dm'$ and $n = dn'$, then $m/n = m'/n'$ and you can apply your argument to show that $(b^m)^{1/n} = (b^{m'})^{1/n'}$. Similarly, if you let $e = \gcd(p,q)$, $p = ep'$, $q = eq'$, then $(b^p)^{1/q} = (b^{p'})^{1/q'}$.

But of course $m/n = p/q \implies m'/n' = p'/q'$, and besides $\gcd(p',q') = \gcd(m',n') = 1$. But it's a standard fact every (positive) fraction has a unique representation with coprime (positive) numerator and denominator (use Euclid's lemma four times and the fact $x \mid y \text{ and } y \mid x \implies x = y$ for positive $x$, $y$), so $m' = p'$ and $n' = q'$ and the claim is proven.


Showing $a \le b$ if $a \le b+\varepsilon$, for all $\varepsilon \gt 0$ So I think this is the last problem I have and I'm not thinking I'm doing it properly.

  Let $a,b$ be real numbers and suppose for all $\varepsilon \gt 0, a \le b+\varepsilon$.  Show that $a \le b$.

Assume $b\lt a$.  Then $b+\varepsilon \lt a + \varepsilon$ and
$$a \le b+\varepsilon \lt a+\varepsilon$$
If I can show that there is a contradiction I can prove it, but I'm having trouble with the contradiction.

You could prove the contrapositive of this question:

  Suppose that $b<a$. Then there exists $\varepsilon>0$ so that $b+\varepsilon<a$.

Assume that $a>b$. Then we have $a>b+\epsilon$ for $\epsilon=\frac {b-1} \epsilon$


Find a basis of $M_2(F)$ so that every member of the basis is idempotent Let $V=M_{2\times 2}(F)$ (the space of 2x2 matrices with coefficients in a field $F$). Find a basis $\{A_1,A_2,A_3,A_4\}$ of $V$ so that $A_j^2=A_j$ for all $j$.

My attempt. Let $A_j$ be $$\begin{pmatrix} a_1 & a_2\\ a_3 & a_4\\ \end{pmatrix}. $$

We want to have $A_j^2=A_j$, so $$\begin{pmatrix} a_1^2+a_2a_3 & a_1a_2+a_2a_4\\ a_1a_3+a_3a_4 & a_3a_2+a_4^2\\ \end{pmatrix} = \begin{pmatrix} a_1 & a_2\\ a_3 & a_4\\ \end{pmatrix}. $$

If we let $a_1=1$ and $a_2=a_3=a_4=0$ then the matrix $$\begin{pmatrix} 1 & 0\\ 0 & 0\\ \end{pmatrix} $$ meets the property. Similarly if we let $a_4=1$ and $a_1=a_2=a_3=0$ the matrix $$\begin{pmatrix} 0 & 0\\ 0 & 1\\ \end{pmatrix} $$ also meets the property, but I´m having trouble finding the other two matrices, can you help me please? I would really appreciate it :)

Hint.  If $a_2=0$ and $a_3=0$ then the matrix is
$$\pmatrix{a_1&0\cr0&a_4\cr}\ ,$$
which is no good because it is a linear combination of the two matrices you have already.  So you must have $a_2$ or $a_3$ or both non-zero.  To keep things as simple as possible, try starting with $a_3=0$ and $a_2\ne0$. . .

Good luck!

An idempotent basis for V is:
$\{A_1, A_2, A_3, A_4\} =\{[1, 0; 0, 0], [1, 0; 1, 0], [0, 1; 0, 1], [0, 0; 0, 1]\}$

To see this, first of all notice that A1^2 = A1, A2^2 = A2, A3^2 = A3, and A4^2 = A4.
Next, to show that $\mathrm{ span}\{A_1, A_2, A_3, A_4\} = V$, notice that A2- A1 = [0, 0; 1, 0] and A3 - A4 = [0, 1; 0, 0]. Thus we have that each matrix v = [a1,a2;a3,a4] in V can be written as a linear combination of A1, A2, A3, A4 as follows. v = a1(A1)+a2(A3-A4)+a3(A2-A1)+a4(A4).
Moreover, 0 = [0 0; 0 0] = a1(A1)+a2(A2)+a3(A3)+a4(A4) = [a1+a2, a3, a2, a3+a4] => a1 = a2 = a3 = a4 = 0. Therefore, A1, A2, A3, and A4 are also linearly independent.
So we can conclude that A1, A2, A3, and A4 are a basis for V.


Differentiable function such that $\lim_\limits{x\to \infty}f(x)=\infty$ Let $f:(1, \infty) \to \mathbb{R}$ be a differentiable function such that $$f'(x)=\frac{x^2-(f(x))^2}{x^2(1+(f(x))^2)}$$
  for all $x>1$. Prove that $\lim_\limits{x \to \infty}f(x)=\infty$

From the given relation, I got that $f$ is infinitely differentiable. I tried to get an inequality between $x$ and $f(x)$, and so I tried to prove that the derivative doesn't change signs, so that $f$ is monotonic, but I don't think this works. Also, I noticed that if $f(x)=x$ for a given $x$, then $f'(x)=0$. I also don't see why the domain is given as $(1,\infty)$. Perhaps this is supposed to hint to something.

Claim 1: $f$ is unbounded above.

Proof:
By contradiction. Suppose $f(x)\le M$ for all $x\in(1,\infty)$. Then $$
(1+f(x)^2)f'(x)=1-\frac{f(x)^2}{x^2}\ge\frac12,\quad x\ge M\sqrt2.
$$
Integrating
$$
f(x)+\frac13\,f(x)^3\ge \frac{x}{2}+C,\quad x\ge M\sqrt2.
$$
The function $h(u)=u+u^3/3$ is strictly increasing and has an inverse $h^{-1}$ and $f(x)\ge h^{-1}(x/2+C)$ for large $x$. Since $\lim_{u\to\infty}h^{-1}(u)=+\infty$, we reach a contradiction with the fact that $f$ was assumed bounded.

Claim 2: $f$ has at most one critical point, which is a strict minimum.

Proof: $\xi\in(1,\infty)$ is a critical point of $f$ if and only if $f(\xi)=\xi$. Derivating the equation satisfied by $f$ we get
$$
f''(\xi)=\frac{(2\xi-2f(\xi)f'(\xi))\xi^2(1+f(\xi)^2)}{\xi^4(1+f(\xi)^2)^2}=\frac{2}{\xi(1+f(\xi)^2)}>0.
$$
If $f$ had two critical points, both would be minimums, and there would be a maximum between them, an impossibility.

Finally, from Claims 2 it follows that $f$ is eventually increasing, and since it is unbounded by Claim 1, we must have $\lim_{x\to\infty}f(x)=+\infty$.

If $f(x)=0$ then $f'(x)=1$ and if $f'(x)=0$, then $f''(x)>0$. So $f$ has at most one local minimum, and either $f'$ is always negative (then so is $f$, and $f$ tends to $-\infty$ faster than $-x$) or $f'$ changes sign once from negative to positive or $f'$ is always positive. So $f$ is increasing on $(a,\infty)$ for some $a\ge 1$. Either $f$ tends to $\infty$ or $f$ is bounded. However, if $f$ is bounded, $f'(x)\to 1$ as $x\to\infty$, a contradiction.


Intuition behind $i^{i}$. My query is about the $i^{i}$ , where $i$ is defined to be the imaginary unit, and $i \in C$.

I know the proof of this value, we just have to substitute $i$ as $\large{e^{i\left(\frac{\pi}{2}+2n\pi\right)}}$. Where $n$ is any integer because all these values corresponds to "$i$" right?

Which gives us, $i^{i}=\large{e^{-\frac{\pi}{2}+2n\pi}}$ Where $n=0$, is the principal value. $i.e.,$ $e^{-\frac{\pi}{2}}$.

I mean this result always blow my mind, to be honestly speaking. Not only that $i^i$ despite being a complex number raised to a complex number. It also attains a real value! And that too not just one, but many!

My question would be that first, is this result strictly valid in accordance to mathematics? Are there any loopholes in defining it as such? Here's what I think it could be of importance http://www.cut-the-knot.org/do_you_know/complex.shtml as some quotes in this page indicates.

Secondly, is there any intuition behind why could it be so, even if it could, the idea that it can take infinitely many values on $\mathcal R$ just by changing "$n$" is still hard kind of.

I know that I added a "contradictory" comment as you will see, but a "rigorous" intuition will be,
$i^i=(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})^i$
Now, by De Moivre's Theorem,
$i^i=\cos i\frac{\pi}{2}+i\sin i\frac{\pi}{2}$
By thee Euler's Formula,
$i^i=e^{i^2\frac{\pi}{2}}$
$i^i=e^{-\frac{\pi}{2}}$
$i^i=\frac{1}{e^\frac{\pi}{2}}$
Which, according to my calculator is, $0.207879576...$.
Which is confirmed by google.
P.S.:I didn't prove it for $2n\pi$, just a case of it.
As far as you want the intuition, think the reverse, isn't it normal for the complex root of a real number to be complex? For example, $2^{\frac1i}$ is complex. So, the question you are asking is just the reverse of it.
I hope it makes sense.


If $X$ is a $n$ by $k$ real matrix, and we know that $X^{T}X$ is invertible, is $X$ invertible as well? If $X$ is a $n$ by $k$ real matrix, and we know that $X^{T}X$ is invertible, we know that $rank(X^{T}X) = n$ by $n$. Can we say that $X$ is invertible as well and hence has rank $n$? thanks!

$$\bf X=\begin{bmatrix}1\\0\end{bmatrix}\,:\mathbb R\to\mathbb R^2\\{\bf X}^T\bf X=I:\mathbb R\to\mathbb R $$
But X is not surjective so has no right inverse.

No, if I understand the question right, it is not possible to say that $\mathbf{X} $ is invertible and also $\mathbf{X}^T\mathbf{X}$ is only invertible when $\mathbf{X}$ has full column rank. In that case, you can find the Moore Penrose pseudo inverse of $\mathbf{X}$.


$f \circ g =\operatorname{ id}$ and $g \circ f \neq \operatorname{id}$? Are there two functions $f$ and $g$ s.t.
$$f \circ g = \operatorname{id}$$
but
$$g \circ f \neq \operatorname{id}?$$
Could someone give an example or a proof that this is impossible?

This must be trivial, but I can't figure it out. :) Thanks!

Hint: Consider the two functions on the integers:
$$
f(x)=\left\lfloor\frac x2\right\rfloor
$$
and
$$
g(x)=2x
$$

It is possible to define the functions via matrices, in which case the product of the matrices is not commutative - I don't know if that is quite what you are looking for. You could also define the functions by $f(x) = x^2$, $g(x) = \sqrt{x}$, where $f:\mathbb{R} \to [0 , \infty)$ and $g: [0 , \infty) \to [0 , \infty)$. Here, $f(g(x)) = x $ but $g(f(x)) = |x|$. There are of course other examples of similar type.


Finding the basis for a subspace of P3 How would I find a basis for the set $V \subset P_{3} \subset C[-1,1]$ defined as all $f \in P_{3}$ with $f(-1)=0$ and $\int_{-1}^{1}f(x)dx = 0$? Should I be looking for a specific polynomial that satisfies both of the conditions of $V$?

Yes, you should. We know that every $f\in P_3$ is of the form $a+bx+cx^2+dx^3$; then $f(-1)=0$ implies that $a-b+c-d=0$.
Then $\int_{-1}^1f(x)dx=ax+\frac b2x^2+\frac c3 x^3+\frac d4x^4\Bigg|_{-1}^1=2a+\frac{2c}3=0$
We can simplify these down by finding the null space of the corresponding matrix $\begin{bmatrix}1&-1&1&-1\\2&0&\frac23&0\end{bmatrix}\sim\begin{bmatrix}1&0&\frac13&0\\0&1&-\frac23&1\end{bmatrix}$, and so the null space is spanned by $\begin{bmatrix}1\\0\\-3\\-2\end{bmatrix}$ and $\begin{bmatrix}0\\1\\0\\-1\end{bmatrix}$.
Therefore, a basis for $V$ is the polynomials $1-3x^2-2x^3$ and $x-x^3$.


Metamath set variable a set but not a set? Huh? In Metamath, "every set variable is a set" is a theorem.  What to make of a statement like "not exists x ( x not equal x )"?  This can be proved starting with "for all x ( x = x )", applying double negation, then quantifier negation.  In "not exists x ( x not equal x )", the set variable x is a set just because it is a set variable, but also it is not a set because of what the sentence says (and the sentence is a theorem).  Which view is right?  What should I make of this?

One difference I notice between the theorems named
vex
and nex
is that $x$ is a free variable in one but is a bound variable in the other.
Specifically, vex states that
$$ \vdash x \in V. $$
In this statement, $x$ is a free variable.
So I think a more careful statement of the theorem in mathematical English
might be, "Every free set variable is a set."
Once you put a quantifier in front of a variable,
for example in $ \vdash \lnot \exists x \varphi, $
the variable becomes bound and the rules of the quantifier
govern what is possible.
In particular, when we use $\lnot \exists,$ saying that something
does not exist, many properties (or lack of properties) that
would cause trouble if the thing existed can no longer cause trouble.
So, for example, $x$ must be a set if there is any such thing as $x,$
but if there is no such thing, we can just as easily call it a set
and not a set at the same time.
I'm reminded of an old joke:

A customer in a butcher shop asks the price of roast beef, and on hearing the answer, complains:
"Why is the price so high? The price is a dollar lower at the shop across the street."
"Why don't you go buy your roast beef there, in that case?"
"I can't, they're all out of roast beef."
"Well, if I were out of roast beef, I could lower my price by a dollar."

Perhaps someone with deeper insight into logical notation
can explain the meaning of the variables in the
vex and nex theorems further.

You have actually recognized the problem introduced by Russell's paradox that motivates some set theorists to study set theory in terms of a modal structure obtained by its many possible models.  The problem is the distinction between what a quantifier is "supposed to mean" and how Russell's paradox interferes with the representation of classical quantifiers in the square of opposition.

If you examine the formal sentences associated with the square of opposition in the link,

https://plato.stanford.edu/entries/square/#ModRevSqu

you can see the problem with "Every set is not an element of itself" and interpreting the universal quantifier in terms of sets.  Free variables carry implicit universal quantification on the basis of what quantifiers are "supposed to mean".

It is best to remember that metamath is intended to execute logical transformations.  So, you may view the system as either assuming that set theory has a model or that it is treating set theory in a purely formal matter.

If you would like to consider the problem of negative existentials further,
look at the link,

https://plato.stanford.edu/entries/existence/#FreRusExiNotProInd


f(a) = inverse of a is an isomorphism iff a group G is Abelian $G$ is a group and $f:G \rightarrow G$ is a function defined as $f(a)=a^{-1}$ where $a^{-1}$ is the inverse of $a$ under the group operation. Prove that $f$ is an isomorphism if and only if $G$ is abelian.

I understand that I have to prove $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. How might I do that?

Reference: Fraleigh p. 49 Question 4.40 in A First Course in Abstract Algebra

First note that it is a bijection of $G$ onto $G$ no matter what. So this boils down to $G$ Abelian if and onbly if $f$ is a homorphism.

If $G$ is Abelian, I think you can show that $f$ is a homomorphism.

Now if $f$ is a homomorphism
$$
ab=(a^{-1})^{-1}(b^{-1})^{-1}=(b^{-1}a^{-1})^{-1}=(f(b)f(a))^{-1}=(f(ba))^{-1}=((ba)^{-1})^{-1}=ba.
$$

$f\colon G → G$ defined by $f(x)=x^-1$
$f$ isomorphism $⇒$ f is a 1-1 homomorphism
$∀ x,y∈G, xy = ((xy)^{-1})^{-1} = ((y^{-1})(x^{-1}))^{-1}
 = (f(y)(x))^{-1} = f(yx)^{-1}$  (∵$f$ is a homomorphism)
$= ((yx)^{-1})^{-1}$ (by mapping defining)
$= yx (∵ (a^{-1})^{-1} = a)
∴ xy = yx ⇒ G$ is abelian


Laplace's equation on a square domain with a central point reservoire Could someone please tell me the solution to this problem. I have $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$ on the square domain $-L<x<L, -L<y<L$ with homogeneous Dirichlet boundary conditions: $$u(L,y)=u(-L,y)=u(x,L)=u(x,-L)=0$$ and further condition $u(0,0)=1$.

I have tried separation of variables but quickly realised that it would only give me the trivial $u=0$.

This is not for some kind of homework, I am solving the heat equation numerically and would just like to know what the analytical steady state solution is so I may compare.
Thanks in advance!

There's only one solution to the problem
$$\begin{cases}
\Delta u =0 & \Omega, \\
u=0 & \partial \Omega
\end{cases}
$$
when $\Omega$ is a domain with Lipschitz continuous boundary, such as the square. This solution is the identically vanishing one, as you can see in various ways; one of them is observing that the energy integral
$$\int_{\Omega} \lvert \nabla u\rvert^2\, dx$$
must vanish.

The world of piecewise functions can have innovative possibilities.

You can introduce some jumping points in the normal continuous functions so that the derivatives of the modified functions do not change.

In fact the solution of this PDE can be $u(x,y)=\begin{cases}1&\text{when}~x=y=0\\0&\text{otherwise}\end{cases}$ .


Correct way to solve a limit. I have the following limit,

$$\lim_{n\rightarrow \infty}\left \{ \frac{\left ( n+1 \right )\left ( n+2 \right )...3n}{n^{2n}} \right \}^{\frac{1}{n}}$$

My procedure of solving (which is wrong).

Step 1: I break up the expression in the following manner,

$$\lim_{n\rightarrow \infty }\left  \{ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right )...\left ( \frac{3n}{n} \right ) \right  \}^{\frac{1}{n}}$$

Step 2: I apply the limits,

$$\left \{ \left ( 1+\frac{1}{\infty } \right ) \left ( 1+\frac{2}{\infty } \right )...\left ( 2+\frac{1}{\infty } \right )...3\right \}^{\frac{1}{\infty }}$$

which makes it, $$\left \{ 1^{\infty } *2^{\infty }*3\right \}^{0}$$

I am stuck here and could not proceed. Can somebody please help me out?

A simpler approach is to use the following theorem on sequences:

  Theorem: If $a_{n}$ is a sequence of positive terms such that $\lim\limits_{n \to \infty}\dfrac{a_{n + 1}}{a_{n}} = L$ then $\lim\limits_{n \to \infty}\sqrt[n]{a_{n}} = L$.

Let $$a_{n} = \frac{(n + 1)(n + 2)\cdots(n + 2n)}{n^{2n}}$$ and the sequence in question is $b_{n} = \sqrt[n]{a_{n}}$. We have
\begin{align}
\frac{a_{n + 1}}{a_{n}} &= \frac{(n + 2)(n + 3)\cdots(3n + 3)}{(n + 1)^{2n + 2}}\cdot\frac{n^{2n}}{(n + 1)(n + 2)\cdots(n + 2n)}\notag\\
&= \frac{3(3n + 1)(3n + 2)}{(n + 1)^{2}}\left(\frac{n}{n + 1}\right)^{2n}\notag\\
&= 27\cdot\dfrac{\left(1 + \dfrac{1}{3n}\right)\left(1 + \dfrac{2}{3n}\right)}{\left(1 + \dfrac{1}{n}\right)^{2}}\cdot\left\{\left(1 + \dfrac{1}{n}\right)^{n}\right\}^{-2}\notag\\
&\to \frac{27}{e^{2}}\text{ as }n \to \infty\notag
\end{align}
and therefore the sequence in question $b_{n} = a_{n}^{1/n} \to 27/e^{2}$ as $n \to \infty$.


Do infinity and zero really exist? From the first day that I entered college, I was wondering about the relationship between some basic mathematical abstract concepts and nature. I'm going to explain them here and you may find them a little bit opinion-based, but it's not. So please guide me in case you have good scientific-based answers/examples.

Infinity:
Does infinity exist in reality? We all know it is an abstract concept but what I'm curious to know is if there are any physical phenomena out there that can show, stimulate or somehow help us understand the concept of infinity in reality/nature/the physical world.

Zero:
What about Zeno's paradox? In nature (our physical world) there is a "smallest distance". It's about $1.6 \times 10^{-35}$ meters. Another example would be when someone says "there are three apples on the desk, if you take all of them there are $0$ apples on the desk". Obviously, it's an abstract concept, but what I would like to know is that if there is any observable physical event that can anyhow show us the concept of nothingness or absolute zero. E.g., we don't have absolute zero temperature in thermodynamics, absolute zero distance between 2 points in mechanics, or absolute zero gravity in a given space, etc.

Axiom of empty set: This axioms states:

  "There is a set such that no element is a member of it."

One can imagine an empty set in nature as an absolutely empty box. But one can see it the other way around: the element "nothing" is there (quoting from Prof. Lawrence Krauss's book "nothing is something"). So, the same way one can say the element "nothing" is a member of a set. Obviously, it's an abstract concept and this may look like a play on words, but it's also an interesting paradox.

$\mathbb R$?! (the set of real numbers): There are millions of mathematical theorems which are based on $\mathbb R$! I wonder if there are any non-countable physical phenomena in the real world/nature?

Again, I know these mathematical concepts are just abstractions and help us to solve real-world problems! However, I'm more interested in the relation between these abstract concepts and our physical world/nature.

Claim: At least one of $0$, $\infty$ exists.

Proof: If $\infty$ exists, we are done.  If not, the number of instances of $\infty$ is $0$.

I am actually somewhat serious.  This is an adaptation of a certain philosopher's (unfortunately I cannot remember who) ontological proof of the existence of the empty set, which is itself an extraordinarily perceptive and witty riff on St. Anselm's ontological argument for the existence of God.  Strangely, while theists and atheists alike tend to find St. Anselm's argument intriguing but not convincing (I seem to recall that Anselm's paper begins by apologizing -- to God -- for making the argument), when you invert it to prove the existence of nothingness, it seems quite convincing!

The empty set and 0 are easy. The empty set is a set which has no proper subsets and its cardinality is 0. The important concept here is that if you have 0 apples on your desk I can not take an apple from your desk, and I shall be able to see that.

If you had 5 apples on your desk there would be a 1 to 1 mapping between the set of digits on my hand and the set of apples on your desk, with 0 elements left in either set. These sets would both have cardinality 5. You can define <, =, and > by considering which, if either, set has elements left over after mapping.  I can now take an apple from your desk and when you map the proper subset I leave against your hand you will see you are missing an apple (or you have grown an extra digit).

Infinity is trickier. The set of even integers is a proper subset of the set of integers and there is a 1 to 1 mapping between the set of integers and the set of even integers, both sets have the same cardinality. This property defines infinite, and in this case both are countable.

There is a 1 to 1 mapping between the set of integers and:
  the set of rational numbers;
  the set of irrational numbers which are the root of an integer polynomial, so these are also countable.

The set of transcendental numbers is much larger than any of these sets and is called uncountable (i.e. there is not a 1 to 1 mapping between the set of integers and the set of transcendental numbers, there are transcendental numbers left over when the set of integers is exhausted). The set of real numbers is made up of the sets of integers, rational numbers, irrational numbers which are the root of an integer polynomial , and transcendental numbers and so it is uncountable.

The next known larger infinity is the set of all single value functions (see http://www.math.utah.edu/~pa/math/sets/FgtR.html).


Simplify and rewrite in term of i The problem click here

I believe the answer is (b) I tried it in my calculator and it's by far the closest answer from the rest of the other options but it could also be (a) and that's where I need your help guys.

Could it be non above?, Because it sure seems like it in the calculator.


Looking for a rational sine function for an acute triangle I have a triangle $(5,6,7)$, that is $a=5$, $b=6$, $c=7$ and I want to find the coordinates of the vertices on the  $xy$ plane in a manner that I can implement in a program or a spreadsheet. I let $(x_{ac},y_{ac})=(0,0)$. I let $(x_{bc},y_{bc}=(7,0$). Then, from the law of cosines, to get $(x_{ab},y_{ab})$, I let

$$x_{ab}=a*\cos(\alpha)=a*\frac{a^2+c^2-b^2}{2ab}=5*\frac{25+49-36}{2*5*7}=\frac{38}{14}=2.714285714285714$$
and this appears to be the correct value according to answers to my earlier question.

I can use this sine identity to get the value for $y_{ab}$:
$$y_{ab}=a\sin(\alpha)=a\sqrt{1-\cos^2(\alpha)}=a\sqrt{1-\left(\frac{a^2+c^2-b^2}{2ab}\right)^2}$$
 but I would like to find a $rational$ solution (without the square root) like the cosine function I used for $x_{ab}$ (if one exists).
Can the law(s) of sines (or cosines) be manipulated to produce a sine value using only side lengths without the radical?

This is possible only for acute triangles of a special type.

For example, a triangle with sides $a=10,\ b=17,\ c=21$ can be constructed as follows.

Let $n = 2$ and $c\in Ox.$ Then:
$$a = 2(2^{2n-2}+1) = 10,\quad a_x = 2(2^{2n-2}-1) = 6,\quad a_y= 2^{n+1}=8,$$
$$b = 2^{2n}+1 = 17,\quad b_x = 2^{2n}-1 = 15,\quad b_y = a_y=8,\quad c = a_x+b_x = 21,$$
wherein
$$\sin(\alpha)=\dfrac {a_y}{a} = \dfrac45,\quad \cos(\alpha)= \dfrac{a_x}{a} = \dfrac35,$$
$$\quad \sin(\beta) = \dfrac{b_y}{b} = \dfrac8{17},\quad \cos(\beta)=\dfrac{b_x}{b} = \dfrac{15}{17}.$$

The reason is that for any $n\geq 2,\quad k=0\dots n-1$ there are Pythagorean triples
with the sides $2^k(2^{2n-2k}\pm1)$ and with the common branch $2^{n+1}.$

More interesting examples can be constructed using known Pythagorean triples.
Some of them are given in the table below
$$\begin{vmatrix}
a & a_x & a_y=b_y & b & b_x & c= a_x+b_x\\
13 & 5  & 12 & 15 & 9 & 14\\
13 & 5  & 12 & 20 & 16 & 21\\
26 & 10 & 24 & 25 & 7 & 17\\
25 & 7 & 24 & 30 & 18 & 25\\
5 & \dfrac75 & \dfrac{24}5 & 6 & \dfrac{18}5 & 5\\
\end{vmatrix}$$

$$\color{brown}{\textbf{About triangle calculation}}.$$

Let $a=5,\ b=6,\ c=7,\ t=a_x,$ then
$$a^2-a_x^2 = b^2-b_x^2,$$
$$a^2-t^2 = b^2-(c-t)^2,$$
$$a_x=t=\dfrac{a^2-b^2+c^2}{2c} = \dfrac{19}7,\quad b_x= c-t = \dfrac{30}7,$$
$$a_y=b_y=\sqrt{a^2-a_x^2}=\sqrt{b^2-b_x^2}=\dfrac{12}7\sqrt6,$$
$$A=(0,0),\ B=(a_x, a_y) = \left(\dfrac{19}7,\dfrac{12}7\sqrt6\right),\ C=(0,c)=(0,7).$$

$$\color{brown}{\textbf{About triangular pyramid calculation (earlier question)}}.$$

Let
$$l_A=AD=9,\ l_B=BD=8,\ l_C=CD=10,\ x=D_x,\ y=D_y,\ h=D_z.$$
Then
$$l_A^2-(x^2+y^2) = l_B^2 - ((x-a_x)^2+(y-a_y)^2) = l_C^2 - ((c-x)^2+y^2)=h^2,$$
$$\begin{cases}
2a_xx+2a_yy = l_A^2 -l_B^2 + a^2\\[4pt]
2cx=l_A^2-l_C^2+c^2\\[4pt]
h=\sqrt{l_A^2-(x^2+y^2)}
\end{cases}\rightarrow
\begin{cases}
x=\dfrac{l_A^2-l_C^2+c^2}{2c} = \dfrac{15}7\\[4pt]
y = \dfrac{l_A^2 -l_B^2 + a^2 -2a_xx}{2a_y} = \dfrac{81-64+25-570/49}{\dfrac{24}7\sqrt6} = \dfrac{31}{21}\sqrt6\\[4pt]
h=\sqrt{l_A^2-x^2-\dfrac{\left(l_A^2 -l_B^2 + a^2 -2a_xx\right)^2}{4(a^2-a_x^2)}}  = \color{brown}{\sqrt{\dfrac{190}{3}}}
\end{cases}$$
(see also Wolfram Alpha).

And trigonometry is not needed.

I think I would look for the intersection of the cricles

$x^2 + y^2 = 25\\
(x-7)^2 + y^2 = 36\\
x^2 -14x + 49 + y^2 = 36\\
x^2 - 14x + y^2 = -13$

Subtract the last line from the first line.

$14x = 38\\
x = \frac {38}{14}$

And substitute into either equation

$\frac {38^2}{14^2} + y^2 = 25\\
y^2 = \frac {4900 - 1444}{196}\\
y = \frac {\sqrt {3456}}{14}\\
y = \frac {12\sqrt {6}}{7}$


Power Rule for Indefinite Integrals To prove  $\int x^p \, dx = \frac{x^{p+1}}{p+1} + C$, my calculus textbook writes:

$$F '(x) = \frac{d}{dx} \left(\frac{x^{p+1}}{p+1} +C\right) = \frac{d}{dx} \left(\frac{x^{p+1}}{p+1}\right)+\frac{d}{dx}(C)=\frac{(p+1)x^p}{p+1}+0=x^p.$$

I am confused on how they take the derivative of $\dfrac{x^p+1}{p+1}$ without using the quotient rule. Can someone please explain to me why it is that they apply the power rule to the numerator but seem to completely ignore the denominator?

$p$ is a constant, so $\frac{1}{p+1}$ is a constant, so $\frac{x^{p+1}}{p+1}=\frac{1}{p+1}x^{p+1}$.  Just because there's a quotient doesn't mean we need to think of it as a quotient.

By definition,
$$
{d \over dx}x^n = \lim_{h \rightarrow 0} {(x+h)^n - x^n \over h}.
$$
Now, by Newton's binomial theorem,
$$
(x + h)^n = x^n + n h x^{n-1} + (\mbox{2nd and higher powers of $h$}).
$$
Subtracting $x^n$ and dividing by $h$, we get
$$
{(x+h)^n - x^n \over h} = n x^{n-1} + (\mbox{1st and higher powers of $h$})
$$
Now, letting $h \rightarrow 0$ on the right-hand side leaves just $nx^{n-1}$.


Least and most significant bit calculation using bitwise operations I am working on a project and I need to calculate the least significant bit (LSB) and most significant bit (MSB) of integers.

Suppose $x$ is an $n$-bit unsigned integer ($n=16, 32$ or $64$). We know that $y=x \ \& \ ($~$x+1)$ clears all the bits of $x$ except for the LSB. This is lightning fast, just three operations. Is there something similar for the MSB? What is the fastest way to compute it?

Here is a way that works in $\log(|n|)$ where |n| is the number of bits needed to represent $n$.
Let's say we have a 32-bit integers.

MST(int x)
{
    x|=(x>>1);
    x|=(x>>2);
    x|=(x>>4);
    x|=(x>>8);
    x|=(x>>16);
    x++;
    x>>=1;
    return x;
}

The reason why this works is that the first 5 lines set all bits right to the mst to 1. By adding one to the number we flip them all (including mst) to zero and put a one the left of them all. we shift this one to the right (and hence it's now in the position of mst) and return the number.

I just came across this hack from an old book about chess programming:

$$
y=XOR(x,x-1)=00...001111...11,
$$

where the leftmost $1$ in $y$ is the leftmost $1$ of $x$, ie the MSB of $x$. Then we can add $1$ and shift right by $1$. I am not an expert but I think it's faster than what we 've seen here.


Two cards are drawn without replacement from an ordinary deck, find the probability.. Two cards are drawn without replacement from an ordinary deck, find the probability that the second is a red card, given the first is a red card.

P (2nd Red Card / 1st Red Card) = 13/52 * 12 * 51 = 1/17 - is this correct?

Your answer is not correct, no.  (For starters: note that there are 26 red cards in a standard 52-card deck.)

Let $R_1$ be the event that the first card is red, and $R_2$ the event that the second card is red.

If the first card is red, then when you go to draw your second card there are a total of $26-1=25$ red cards in the deck, and a total of $52-1=51$ cards overall. So, we should have
$$
P(R_2\mid R_1)=\frac{25}{51}.
$$
If you don't understand this intuition, we could also go about this using the conditional probability formula:
$$
P(R_2\mid R_1)=\frac{P(R_1\text{ and }R_2)}{P(R_1)}.
$$
Now
$$
P(R_1)=\frac{26}{52}=\frac{1}{2},
$$
since half the cards are red, and
$$
P(R_1\text{ and }R_2)=\frac{26}{52}\cdot\frac{25}{51}.
$$
So, we find
$$
P(R_2\mid R_1)=\frac{\frac{26}{52}\cdot\frac{25}{51}}{\frac{26}{52}}=\frac{25}{51},
$$
as claimed above.

Can the alternate form of conditional probability be used here i.e.

$$P(R_2 | R_1) = \frac{n(R_1 \bigcap R_2)}{n(R_1)}$$

We've promoted this method since the product rule hasn't been taught yet.

thanks


exponential integral on n-dimensional unit sphere I'd like to compute the expectation of $\exp(\sum x_i)$ under the uniform distribution on the sphere with radius $r> 0$. To this end, I would like to evaluate the following integral:
$$\int_{\large{x_1, \ldots, x_n,\ \sum x_i^2\  =\  r}}\exp\left(\sum_{i=1}^n x_i\right)\,\mathrm{d}x_1 \ldots \mathrm{d}x_n $$
That is, the exponential of the sum of the (Cartesian) coordinates on the sphere.
Maple cannot do it. It is easy to find results about integrating polynomials on the sphere, and I found something about integrating just $\exp(x_1)$ on the $n$-dimensional sphere, but nothing on the exponential of the full sum.  Any help would be greatly appreciated!

If you know how to integrate $\exp(x_1)$ you know this integral, as follows. Write $e_1$ for the standard basis vector $(1, 0, 0, \dots)$ of $\mathbb{R}^n$, where we think of $x = (x_1, \dots x_n) \in \mathbb{R}^n$ as a vector, and write $v = (1, 1, 1, \dots)$ for the all-ones vector. Let me also replace your $r$ with an $r^2$ so that $r$ is the length $\| x \|$, which will make things nicer. You want to compute
$$I(v, r) = \int_{|x| = r} \exp(\langle v, x \rangle) \, dV_x$$
(where $dV_x$ is shorthand I just made up for the volume element $dx_1 \dots dx_n$) and you say you know how to compute
$$I(e_1, r) = \int_{|x| = r} \exp(\langle e_1, x \rangle) \, dV_x.$$
But by spherical symmetry it's clear that $I(v, r)$, as a function of $v$, depends only on the length of $v$! So we can in fact reduce these integrals to each other. Write $v = su$ where $\| u \| = \frac{v}{\| v \|}$ is a unit vector and $s = \| v \|$ is the length of $v$ (which in this case is $\sqrt{n}$). Then
$$\langle v, x \rangle = \langle su, x \rangle = \langle u, sx \rangle$$
and so making the change of coordinates $y = sx$ (so $y_i = sx_i$) we have that
$$I(v, r) = \int_{|y| = sr} \exp(\langle u, y \rangle) \frac{dV_y}{s^n} = \frac{1}{s^n} I(u, sr) = \frac{1}{\sqrt{n}^n} I(e_1, \sqrt{n} r).$$


How to find $\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x}\,dx \,$? How to find
$$\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x}\,dx \,\,?$$

The integrand $ \frac{\cot x}{\cot x + \csc x} $ is not defined at $x =0$. But the function is bounded on $(0 , \frac{\pi}{2}]$.
$$\lim _{x \to 0} \frac{\cot x}{\cot x + \csc x} = 0$$  So this is not an improper integral.
My Attempt :
$$\int_0 ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x} = \lim_{t \to 0} \int_t ^ \frac{\pi}{2} \frac{\cot x}{\cot x + \csc x} = \lim_{t \to 0}  \left[(\frac{\pi}{2} - 1)+ (\tan{\frac{t}{2} - t})\right] = \left(\frac{\pi}{2} - 1\right)$$.
I know how to find the anti-derivative of the integrand. I first found out the anti-derivative of the integrand in $[t , \frac{\pi}{2}]$ , where $0 < t < \frac{\pi}{2}$. Let's say it is $\phi(t)$. Then I find $\lim_{t \to 0} \phi(t)$.
I am not sure if this is a right way. Can anyone please check it?

HINT:  Take , $$\cot x- \csc x=t$$
then, $$\cot x+ \csc x=\frac{1}{t}$$
and $$\cot x=\frac{1}{2}\left[t+\frac{1}{t}\right]$$
I think you can proceed from here.


How to show that an open map $f $ implies the surjectivity of $f'$ in a dense set Let $f$ be a $C^1$ map from $U\to \mathbb{R}^m$, where $U$ is an open set in $\mathbb{R}^n$, $n\geq m$. Then we know that if $f'$ is surjective everywhere, then $f$ is open.

My question is whether the converse is valid,  if the answer is no, can we prove the following weaker conclusion:

If $f$ is open, $f'$ is surjective in a dense set of $U$

I expect that the answer is negative when $n>m$; that is, $C^1$ regularity is not enough. An important paper on the subject is A singular map of a cube onto a square by Robert Kaufman, J. Differential Geom. 14 (1979), no. 4, 593–594 (1981). In this two-page note Kaufman constructs a $C^1$ map from $[0,1]^3$ onto $[0,1]^2$ whose derivative has rank at most $1$ at every point. Therefore, the set of points where the derivative is surjective is empty. A more elaborate construction  by S.M. Bates in On the image size of singular maps I takes Hölder continuity of the derivative into account.

Although in these examples the image of $f$ contains an open subset of $\mathbb R^m$, I do not know if they are, or can be modified into, open maps. A starting point would be to go through the papers that cite Kaufman's example, some of which are very recent.

If f is defined on na open subset of Rm with values on Rn and is of class C1, then f open implies that f'(x) is surjective on na open and dense subset of U.


Legendre symbol $(-3/p)$ where $p = 1 \mod 3$ Suppose $p = 1 \bmod 3$, prove the following statements:

prove that $x^2 + x + 1 = 0 \mod p$ has a solution
Prove that $\left(\frac{-3}{p}\right) = 1$ if $p = 1\mod 3$
Determine the discriminant of $x^2 + x + 1$
Prove using 2,3 that $\left(\frac{-3}{p}\right) = -1$ if $p = -1\mod 3$

This is what I've tried by each question:

prove $x^2 + x = -1 \mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{\frac{p-1}{2}} = -1 \mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.
Note the following:
$(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 \mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p.
-3
?

You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p \equiv 1 \mod 3$ then there is an $x \ne 1$ for which  $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.
Agree with your own answer except for the typo:
$$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(\frac{-3}{p})=1$
Indeed $D=1^2-4\cdot 1\cdot 1 = -3$. Effectively 2. says that $\sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=\frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.
If $p \equiv -1 \mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $\sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(\frac{-3}{p})=-1$


what is $d(y/x)$ in $ d\log(y/x)=\frac{d(y/x)}{(y/x)}$? For example what is meant by $d(y/x)$ in right hand numerator in the following?

$$ d\log(y/x)=\frac{d(y/x)}{(y/x)}$$

Is the same as $\frac{dy}{dx}$? If not, what does it mean?

No, it is not the same

$$
{\rm d}\left(\frac{y}{x}\right) = \frac{{\rm d}y}{x} - \frac{y}{x^2}{\rm d}x \not= \frac{{\rm d}y}{{\rm d}x}
$$

For any variable z, d(z) is a differential. Differentials are explained well in many textbooks and internet articles.


Estimating a set of probabilities for sequences of independent events. Assume a line of obstacles, if dropping a marble at obstacle position $n$ it should stop with (as close to a given probability $p(n)$ as possible) somewhere before obstacle $n+m(n)$. Assuming we have $m(n)$ and $p(n)$ sampled at least some values of $n$. How can we estimate the independent probabilities of stopping (or passing) of each individual obstacle?

Let us assume the event of passing a given obstacle is independent of having passed any previous or passing any subsequent obstacles, but that they are always encountered in the same order: $$1\to 2\to 3 \to \cdots \to N$$

Found a way that works. Following the ideas in this question, we approximate the integrals with riemann sums (although we already have a discrete case with explicit product/sums). All $m(n)$ are 2 and the probabilities $p(n)$ given by the vector $\bf d$:

$${\bf v} = 2^\left[\min_{\bf v}\left\{\|{\bf Tv} - \log_2({\bf d})\|_2^2\right\}\right]$$

$${\bf T} = \left[\begin{array}{cccccccccccc}
1&1&0&0&0&0&0&0&0&0&0&0\\
0&0&1&1&0&0&0&0&0&0&0&0\\
0&0&0&0&1&1&0&0&0&0&0&0\\
0&0&0&0&0&0&1&1&0&0&0&0\\
0&0&0&0&0&0&0&0&1&1&0&0\\
0&0&0&0&0&0&0&0&0&0&1&1
\end{array}\right], \log_2({\bf d}) = \frac{1}{2}\left[\begin{array}{c}-1\\-2\\-3\\-4\\-5\\-6\end{array}\right]$$

So $$\cases{{v_1}^1\cdot {v_2}^1 = 2^{-1/2}\\{v_3}^1\cdot {v_4}^1 = 2^{-1}\\\vdots}$$

With a small isotropic Tikhonov regularizer with $\lambda = 10^{-6}$ gives us the solution:

$$\left[\begin{array}{c}
0.840896488111804\\
0.840896488111804\\
0.707106903718765\\
0.707106903718765\\
0.594603712056721\\
0.594603712056721\\
0.500000173286739\\
0.500000173286739\\
0.420448389772112\\
0.420448389772112\\
0.353553574391631\\
0.353553574391631
\end{array}\right]$$

I we plug back in, we see several well known facts, for example $\sqrt{1/2} = 0.7071069\cdots, \sqrt{1/4} = 0.5$.

So we can build linear least squares problems encoding exponents instead of coefficients as numbers. As the system is underdetermined we happen to get the solution where pairs are equal, but bear in mind many other constraints can be put. Simultaneous overlapping constraints could probably be useful in some application..?


How to show that a function is strictly decreasing. The function $f \colon \mathbb{R} \to \mathbb{R}$ is defined by $f(x)=\frac{x^2+2x}{x^2-1}$.
How would you show that $f(x)$ is a strictly decreasing function.

Actually your function $f(x) = \frac{x^2+2x}{x^2-1}$ is NOT strictly decreasing in its whole domain $\mathbb{R} \smallsetminus \{-1,1\}$. For instance, $f(\frac{1}{2}) = \frac{5}{4}(-\frac{4}{3}) = -\frac{5}{3} <  \frac{8}{3} = f(2)$ but $\frac{1}{2} < 2$.

It is true that $f(x)$ is strictly decreasing in $(-\infty, -1)$, and in $(-1,1)$, and in $(1, +\infty)$. You can prove that because the derivative $f'(x) = -2\frac{x^2+x+1}{(x^2-1)^2} < 0$ for all $x \in \mathbb{R} \smallsetminus \{-1,1\}$ and $f(x)$ is continuous in $(-\infty, -1)$, and in $(-1,1)$, and in $(1, +\infty)$. But this does not imply that $f(x)$ is strictly decreasing in $\mathbb{R} \smallsetminus \{-1, 1\}$ since $x = -1$ and $x = 1$ are infinite discontinuities. More precisely, $\lim_{x\to \pm 1^-} f(x) = -\infty$ and  $\lim_{x\to \pm 1^+} f(x) = +\infty$.


An argument for showing no simple group of order 504 in $S_7$ In this example, the first argument is that suppose the simple subgroup $H$ has order $504=2^3 3^2 7$, then $H\subset A_7 $, is there a more general theorem for this result? Because the example didn't make any argument here which made it looks trivial.

For me, I have to make the following argument: if $H$ is not in $A_7$ then $HA_7 =S_7$ and the third (diamond) Isomorphism Theorem would imply that $|H : H\cap A_7|=2$ and thus $H$ is not simple, a contradiction.

And is there a more general statement about this result?  Given an
 arbitrary group and its two simple subgroups of index say $2$ and $10$.
 Can we say anything about these two subgroups like containment?

Any subgroup of $S_{n}$ either contains all even permutation or half odd half even permutation. Consider $H \cap A_{7}$. Now If in H the latter case happens, then the even permutations of $H$ form a normal subgroup of $H$ , a contradiction to the simplicity of $H$. Hence the result.

A more general result can be likethis

If G is a simple group, and if there exists a subgroup of index $m$ in $G$ say $H$, that is , $[G:H]=m$, then $G$ can be embedded inside $A_{m}$.

You already showed that $G$ must be a subgroup of $A_7$. Since $|A_7|=2520=5\cdot504$, you want to consider the action of $A_7$ on the $5$ cosets of $G$ in $A_7$: Clearly, $A_7$ acts non-trivially, so $A_7/K$ must be isomorphic to a non-trivial subgroup of $A_5$, where $K$ is the kernel of the operation. But $A_7$ is simple, so $K$ must be $1$ which is an obvious contradiction.


Spectrum of an element of a non unital C*-algebra I know  that spectrum of an element $x$ of a  unital C*-algebra  is nonempty. I like to find an example of a non unital C*-algebra that has an element with empty spectrum, if it exists.

Motivation

I saw a theorem about "spectrum of an element x of a unital C*-algebra is nonempty" so I wanted to know if the unital assumption is essential.

Spectrum

For Banach algebras the spectrum is always nonempty.

(This is due to the Neumann series.)

For algebras the spectrum may be empty.

(Consider the algebra of polynomials.)

Extensions

For Banach algebras the spectrum depends on extensions.

(Consider the Banach algebra of the bilateral shift.)

For C*-algebras the spectrum is independent of extensions.

(This is due to the spectral permanence.)

Reference

An investigation was done in: Extensions: Spectrum


Show that the curve $x^2+y^2-3=0$ has no rational points Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.

Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.

Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.

Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.

In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.

Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped  into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.

Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.

Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.

We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.

Suppose to the contrary that there is a rational solution of the equation. Then there exist integers $a$, $b$, and $q$, with $q\ne 0$, such that $a^2+b^2=3q^2$, and $a$, $b$, and $q$ have no common factor greater than $1$.

Note that $a$ and $b$ must both be divisible by $3$. For if an integer $m$ is not divisible by $3$, then $m^2$ has remainder $1$ on division by $3$. So if one or both of $a$ and $b$ is not divisible by $3$, then $a^2+b^2$ has remainder $1$ or $2$ on division by $3$, and therefore cannot be of the shape $3q^2$.

Thus both $a$ and $b$ are divisible by $3$. It follows that $q$ is divisible by $3$, contradicting our assumption that $a$, $b$, and $q$ have no common divisor greater than $1$.

Here is a more geometric flavored proof:
$x^2+y^2-3=0 \iff x^2+y^2 = \sqrt{3}^2$ is a circle with radius $\sqrt{3}$ centered at the origin. Think of the points along the circle as polar coordinates $(r, \theta)$, i.e.
\begin{equation}(\sqrt{3}, \theta) \text{ where } 0 \leq \theta \leq 2\pi \end{equation}
The formulas for converting a polar coordinate into a cartesian coordinate is just right triangle trigonometry:
\begin{align*}
  x &= r\cos\theta \\
  y &= r\sin\theta
\end{align*}
Since $r=\sqrt{3}$, we have that
\begin{align*}
  x &= \sqrt{3}\cos\theta \\
  y &= \sqrt{3}\sin\theta
\end{align*}
Then you can say that $\sqrt{3}$ multiplied by any number between $-1$ and $1$ is irrational.


How does this probability transformation question simplify? Suppose that X is unif(-1,1) and Y= X^2. Then $f_y(y) = f_x(-\sqrt{y}) *|-1/(2\sqrt{y}) | + f_x(\sqrt{y})*|1/(2\sqrt{y}) | $

this makes sense to me, but then they simplify it to $1/(2\sqrt{y})$

which to me doesn't make sense, wouldn't it be $1/(\sqrt{y})$ since both $f_x(+/- \sqrt{y})$ become 1, they uniform so the pdf is 1/(b-a), which is in both cases 1?

The uniform is over the range -1 to 1, so the $f_x(x) = \dfrac{1}{2}$, not 1.


Why can I solve an impossible equation using linear algebra? I am currently learning matlab and linear algebra side by side and I stumbled upon this example from mathworks

A = [1 2 0; 0 4 3];
b = [8; 18];
x = A\b

x = 3×1

     0
4.0000
0.6667

which in my mind translates to

$$
A = \left[
\begin{matrix}
1 & 2 & 0 \\
0 & 4 & 3
\end{matrix}\right]
B = \left[
\begin{matrix}
8 \\ 18
\end{matrix}\right]
x = \left[
\begin{matrix}
a \\ b \\ c
\end{matrix}\right]
$$

$$
Ax = \left[
\begin{matrix}
1 & 2 & 0 \\
0 & 4 & 3
\end{matrix}\right] \times \left[
\begin{matrix}
a \\ b \\ c
\end{matrix}\right]
= \left[
\begin{matrix}
a + 2b \\ 4b + 3c
\end{matrix}\right]
$$

which boils down to
$$
\left[
\begin{matrix}
a + 2b \\ 4b + 3c
\end{matrix}\right] = \left[
\begin{matrix}
8\\ 18
\end{matrix}\right] \Rightarrow \begin{matrix}a + 2b = 8 \\4b + 3c = 18\end{matrix}
$$

which is an equation with 3 unknown (a, b and c) with two equations, which is impossible! Yes there is a solution
$$
x = \left[
\begin{matrix}
0 \\ 4 \\ 2/3
\end{matrix}\right]
$$

How can I solve an impossible equation (three unknown and two equations) using linear algebra?

If there are fewer equations than unknowns, usually there are many solutions. It is not impossible, but indeterminate.

An extreme example is this:  one unknown, but no equation !

The set of solutions $S = \{ x \vert Ax=b \}$ will be one of three cases:

No solution, $S = \emptyset$, $b \not\in \{ Ax \mid x\in V\}$, where $A: V \to W$.
One solution, $S = \{ y \}$
Infinite many solutions

The "impossible case" is 1., but your system is of case 3.


How do you maximize a polynomial over an integer domain? I am working on maximizing the polynomial $f:\mathbb{R}^N \rightarrow \mathbb{R}$ $$f(v):=\prod_{i=1}^{N}( v_i+\alpha_i^2)$$ over integer $n$-partitions of $P$, $n\leq N$: $$\left\{v \left|\, v_i \in \{0,1,2,\dots, N\} \ ; \sum_i v_i = P \right. \right\}$$

This can be done by brute-force in application, but that seems very awful for even moderate $P, N$.

An implicit form for the continuous case is done in Lemma II of this paper: http://ee.ucd.ie/~mark/papers/ISSC_Water_Pouring.pdf

If you find the continuous solution and round it, the error will be less than I think $(0.5+A)^N$, $A$ being the largest of all the $\alpha^2$s corresponding to a nonzero weight. (Haven't looked very closely yet but this is what I got after some algebra). This is a very rough bound. Is it possible to get closer in general?

Can anyone suggest any reading   on the topic?

If $N$ and $P$ are not too large, and @Macavity is correct about the solution minimizing the largest deviation of $v_i+\alpha_i^2$, the problem appears to be rather easily solved using mixed-integer linear programming. However, I do not always get the correct solution.

I've developed a test below in the MATLAB Toolbox YALMIP (disclaimer, developed by me) using Gurobi to solve the MILPs.

To check if the solution is correct (for very small $N$), I use a much more expensive approach and create a binary expansion of each integer, i.e, $v_1 = b_{01} + 2b_{11}+4b_{21}...$. With that, we can exploit the fact that a product of expressions involving binaries, can be linearized by introducing a large amount of new variables and constraints (that is why I limit this to $N=4$). Hence, a MILP is obtained again. This linearization is available in YALMIP in the command binmodel. The code runs for a while and then terminates with a counter example, typically for $P=2$

ok = 1
while ok
    % Random data for small instance
    N = 4;
    P = ceil(rand(1)*N^2);
    alpha = ceil(rand(1,N)*sqrt(N));

    % Create a binary expansion
    d = 1+floor(log2(N));
    b = binvar(d,N,'full');
    v = (2.^(0:1:d-1))*b;

    % Solve by by binary linearization. Expensive!
    Model = [sum(v)==P, v <= N];
    [Objective,cut] = binmodel(prod((alpha.^2+v)));
    solvesdp([cut,Model],-Objective);
    trueobj = double(Objective);
    v1 = double(v);

    % Solve by integer model and simple objective
    v = intvar(1,N);
    sdpvar smallest largest
    Model = [sum(v)==P,0 <= v <= N, smallest <= alpha.^2+v <= largest]
    solvesdp(Model, largest-smallest)
    objalternative = prod(alpha.^2+double(v));
    v2 = double(v)

    % Check for failure
    ok = abs(trueobj-objalternative)<=1e-8
    if ~ok
        trueobj
        v1
        v1+alpha.^2
        objalternative
        v2
        v2+alpha.^2
        break
    end
end

Hopefully I've missed something simple, as the MILP model with deviation of $v_i+\alpha_i^2$ is really fast and terminates in 0s for $N=200$. The known correct method with the binary expansion is intractable for $N>5$ or something like that. Obviously, a problem is that the distance measure leads to non-unique solutions here

Here is one of the problematic cases

P =

     2

alpha^2 =

     1     1     1     4

trueobj =

    16

v1 =

     0     1     1     0

v1+alpha^2 =

     1     2     2     4

objalternative =

    12

v2 =

     2     0     0     0

v2+alpha^2 =

     3     1     1     4

With the lagrange multipliers, I get :

$\forall i \ \ v_i=\frac{(P+\sum\alpha_i^2)}{N}- \alpha_i^2$.

What do you think ?


What is the probability of obtaining a hand with TWO PAIRS in a standard 5 card game of poker? What is the intuition behind, 'What is the probability of obtaining a hand with TWO PAIRS in a standard 5 card game of poker?'

I know the solution is,

$$\frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\cdot\binom{11}{1}\binom{4}{1}}{\binom{52}{5}} = 0.047539$$

I understand the reasoning behind $\binom{11}{1}\binom{4}{1}$. That is, the number of ways a single card (After the two pairs have been selected) can be selected is $11\cdot4$.

However, the $\binom{13}{2}\binom{4}{2}\binom{4}{2}$ part confuses me a little. Why doesn't $\frac{\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}\cdot\binom{11}{1}\binom{4}{1}}{\binom{52}{5}}$ work? I've read that it counts solutions twice, but I don't understand how? How is that $\binom{13}{2}$ - 78 cards can be selected amongst $\binom{4}{2}\binom{4}{2}$ - 36 suits?

First pick the ranks of the first pair and second pair. There are 13 choices, then choose which two you want of those of each rank. This gives $\binom{13}{2}\binom{4}{2}\binom{4}{2}$ in total.

$\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}$ doesn't work because choosing one rank and then choose another separately implies that the pairs aren't the same type in the sense that the order is not interchangeable. For example, it is saying that $AAKKQ$ is not the same as   $KKAAQ$. Also, you are choosing two per rank, not one, thus it should be $\binom{4}{2}$.

The key to correctly applying the combinatorics is to acknowledge how many kinds of MULTIPLICITY - four-suit ("quad"), three-suit ("triple" or "set"), two-suit ("pair" ), or one-suit ("singleton") - exist in a hand AND how many times EACH one occurs.

For example, in a five-card full-house hand, TWO multiplicities exist, once each, a triple and a pair. Thus TWO ranks are selected from the thirteen available using $\binom{13}{2} = 78$. Then, each selected rank (say, K and Q) can be used in EITHER multiplicity to make DISTINCT full-house hands - KKK full of QQ, or vice versa - which is SECOND choice, $\binom{2}{1}=2$.

But, in a six-card full house of the form AAABBB of ranks A and B, both of which are triples (and, of course, the higher of which is strategically played as the rank "full of" the lower) the second choice would not exist, so the "$\binom{2}{1}=2$" multiplication would not be performed.

Likewise in a six-card three-pair scenario $\binom{13}{3} = 286$ would indeed select the three ranks of each pair, but because they are ALL pairs (the same multiplicity) no second choice under "$\binom{3}{1}=3$" or "$\binom{3}{2}=3$" could possibly attach.

But a seven-card THREE-pair hand with ONE singleton would require $\binom{13}{4}\binom{4}{3} = 715*4 = 2860$ ways to fully assign ALL ranks to ALL possible variants.

One final example: ten-card four-of-a-kind of the form AAAABBBCCD: it contains four multiplicities - a quad, a triple, a pair, and a singleton, once each. So $\binom{13}{4}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1} = 17160$ ways to conceive of distributing all selectable ranks through all multiplicities.

If it had been a quad of the form AAAABBCCDD, the result would be only $\binom{13}{4}\binom{4}{1}\binom{3}{3} = 2860$ distinct possible rank-multiplicity assignments. This explains why our (intuitive) experience is that triples and quads with singletons, rather than with multiple pairs, are the most likely.


Absolute value in Linear programming problem I have this problem, where I have to reformulate it as a linear programming problem.

Minimize   $ 2x_1 +3 \vert {x_2-10}\vert $

subject to $\vert {x_1+2}\vert + \vert {x_2}\vert ≤5  $

The subject is new to me. Can anyone help?

Alternatively, the constraint $\vert {x_1+2}\vert + \vert {x_2}\vert ≤5$ corresponds to:
$$\pm(x_1+2)\pm x_2 \le 5 \iff
\begin{cases}\ \ \ \ \ x_1+2+x_2\le 5\\
-(x_1+2)+x_2\le 5\\
\ \ \ \ \ x_1+2-x_2\le 5\\
-(x_1+2)-x_2\le 5\\
\end{cases}$$
The feasible region is:

$\hspace{3cm}$

Since $x_2<10$, then the original problem can be reformulated as:
$$\text{Minimize} \ \ 2x_1+3(10-x_2) \ \ \text{subject to:}\\
 \begin{cases}x_1+x_2\le 3\\
x_1-x_2\ge -7\\
x_1-x_2\le 3\\
x_1+x_2\ge -7\\
\end{cases}$$
The answer is: $f(-2,5)=11$. To be verified: Original and Reformulated.


I having trouble using partial fractions. For the integral below I have to use partial fractions, however I am at a lost on how to do so.
$$\int\frac{dt}{t^2-t-20}$$

The farthest I have gotten to is factoring the denominator to $(t+5)(t-4)$. Thanks in advance.

Note that the signs you wrote are wrong.

$$\frac{1}{t^2-t-20}=\frac{1}{(t\color{red}{-}5)(t\color{red}{+}4)}=\frac 19\left(\frac{1}{t-5}-\frac{1}{t+4}\right)$$


Applied Trigonometric Differentiation Problem I'm stuck with the above problem (see image). I understand, i think, that i need to find the maximum height of the triangle using line AB as its base. To do that i need to find a function for the height, differentiate it, and then solve for when the differential is equal to 0 (max distance from AB).

Apparently the answer is 6 - sqrt(3) feet. I can't figure out how to get that.

I've included another image of my initial approach, but am getting stuck because this seems like a multivariable problem where the height is dependent on both the angle,theta, and x. i'm not studying multivariable calculus so i must be missing something. Really appreciate the help!

It seems the answer to be purely geometric. The problem is symmetric. With this and using your notations, 2x+2y=L and h^2=y^2-x^2, where L=8 ft. Physics has nothing to do with it.

If pulley C is in a position level with pulley B then the weight will hang 4ft directly below B.

If the pulleys are frictionless then the system will stabilise as an isosceles triangle with 'base' AB and sides AC and BC each of 3ft length ... which gives the vertical distance as the x in a 1,x,3 right-angled triangle, which is sqrt(8), or 2sqrt(2).


Basic arithmetic - trick question. I have the following question:

A baker filled a measuring cup with $3/4$ cup of water. He poured $1/2$ of the water into the batter, and then spilled $1/8$ of the water on the floor.
How much water will the baker need to add to what is left in the cup to have 50% more than what he started with?

Now, these are the possible answers given by the question:

$1/8$ cup.
$3/8$ cup.
$1/4$ cup.
$1/2$ cup.
$7/8$ cup.

Here's what I did:
We start with $$3/4 \rightarrow 6/8,$$ after pouring the water into the batter $$6/8 - 1/2 \implies 6/8 - 4/8 = 2/8$$ and after spilling the water $$2/8 - 1/8 = 1/8.$$
Now, to find how much we need to add, we add $50\%$ of $3/4$ to the original $3/4$, that is $$(1/2)(3/4) + 3/4 = 3/8 + 6/8 = 9/8.$$
So the total amount of water we need to add is $$9/8 - 1/8 = 8/8.$$ But as you can see this is not one of the possible options.
I don't know if I'm doing something wrong or if the answers are incorrect, I suspect that I'm reading something wrong and there's something I'm just not seeing. I ask for your help with this. Thanks in advance.

Briefly: this question, as presented, is poorly worded.

Less Briefly: you subtracted $\frac12$, but you should have multiplied by $1-\frac12$ (that is, by $\frac12$).

In Detail: the baker poured half "of the water". In ordinary English, we would mean that to mean 'half of the water he had in the measuring cup', not '$\frac12$ a cup of water'. If you proceed from here you get the answer $\frac78$, which is presumably what was intended.

However, the question then goes on to say that he spilled $\frac18$ "of the water" on the floor. Following the logic from before, we would expect this to mean 'an eighth of the water he had in the measuring cup', although this phrase is now itself ambiguous (since it could he 'an eighth of the original amount' or 'an eighth after pouring half'). Neither interpretation gets you an answer on the list, unfortunately.

I did it like this simply. ( agree wording can be a pain for some)

1.) started with 3/4 and he puts half into batter leaving him with 1/2 of 3/4 = 3/8

2.) he spilled 1/8 on the floor so left with  3/8-1/8 = 2/8 { keep it in this form}

3.) 50% more than he had (3/4) = i do itthis way since we know half of 3/4= 3/8 from above add it to 3/4  = so 3/4 + 3/8 = 9/8

4.) so to get to 9/8 from what he has left 2/8 we need to add back 7/8


Explicit description of $\Bbb Q_p \cap \bar{\Bbb Q}$ Note that we can embed $\Bbb Q_p$ into $\Bbb C$, as it is discussed here. But as far as I understand, this embedding sends the power
series to transcendental elements, so we can't certainly embed $\Bbb Q_p$ into $\bar{\Bbb Q}$ (I love typing barBbbQ). And when we complete $\Bbb Q$ with the $p$-adic norm, then we get, for example, $(p-1)$-th roots of unity (to see this, just note that $x^{p-1} - 1$ factors in $\Bbb Z / p$, so by Hensel's lemma it factors in $\Bbb Q_p$).

So the question is: do we get anything else? What?

Tautologically, you want to describe the number fields $K \subset \overline{\mathbf{Q}}$ which embed in $\mathbf{Q}_p$. It's certainly not going to be possible to describe such fields completely. For example,
consider the polynomial:

$$f(x) = x(x-1)^{n-1} + p g(x),$$

where $g(x)$ is an arbitrary polynomial. For a random choice of $g(x)$, the polynomial will be irreducible and (generically, by Hilbert irreducibility) have splitting field $S_n$. On the other hand, Hensel's Lemma will ensure that $K = \mathbf{Q}[x]/f(x)$ embeds into $\mathbf{Q}_p$.

For a Galois extension $K/\mathbf{Q}$, the condition is equivalent to asking that $K$ is unramified at $p$, and that the Frobenius element (conjugacy class) $\mathrm{Frob}_p \in G = \mathrm{Gal}(K/\mathbf{Q})$ is the identity.

For example, if $K/\mathbf{Q}$ is abelian, then the ramification condition implies that the conductor $N$ is prime to $p$, that is, $K \subset \mathbf{Q}(\zeta_N)$. The largest subfield of $\mathbf{Q}(\zeta_N)$ which embeds into $\mathbf{Q}_p$ corresponds, by Galois theory, to the largest quotient on which $\mathrm{Frob}_p$ is trivial. Yet

$$\mathrm{Frob}_p = [p] \in (\mathbf{Z}/N \mathbf{Z})^{\times} = G
= \mathrm{Gal}(\mathbf{Q}(\zeta_N)/\mathbf{Q}).$$

The corresponding extension has Galois group $(\mathbf{Z}/N \mathbf{Z})^{\times}/[p].$

For example:

$\mathbf{Q}(\zeta_N) \hookrightarrow \mathbf{Q}_p$ if and only if $[p]$ is trivial in $G$, or equivalently, if $p \equiv 1 \mod N$.
There will be infinitely many abelian fields of any degree inside $\mathbf{Q}_p$ --- to generate extensions of degree $M$, for example, choose $N$ to be divisible by at least two distinct primes of the form $1 \mod M$. Then $G$ surjects onto $(\mathbf{Z}/M \mathbf{Z})^2$, and hence $G/[p]$ surjects onto $\mathbf{Z}/M \mathbf{Z}.$
A version of the inverse Galois problem predicts that there exists (arbitary many) fields $K$ with Galois group any finite group $G$ in which any (fixed) prime $p$ splits completely.

The set $\mathbb Q_p\cap \overline{\mathbb Q}$ is simply the set of $p$-adic numbers which satisfy a polynomial with coefficients in $\mathbb Q$. In fact you can see the two sets living inside the same one, namely $\overline{\mathbb Q_p}$.


Using the basic laws of exponent I have some problems with this question. Please help me. Thanks

  Simplify given expression$$
a^2 (abc)^{-2} a^3 b^7
$$

What are exponents of $a$, $b$, and $c$?

I get $3,5,-2$ as exponents of $a,b,c$ respectively. Is it correct?

Assuming that you'd like to simplify the problem then
$$ a^2 (abc)^{-2} a^3 b^7= \frac{a^2  a^3 b^7}{(abc)^2}= \frac{a^{2} a^3b^7}{a^2b^2c^2}= \frac{ a^3b^7}{b^2c^2} = \frac{ a^3b^5}{c^2}=a^3b^5c^{-2} $$
Here's a MathJax Tutorial for future reference.


Total variation distance of probaiblity measures Let $E$ be a set and $\mathcal E\subseteq 2^E$ with $\emptyset\in\mathcal E$. If $\mu:\mathcal E\to\mathbb R$ with $\mu(\emptyset)=0$, then $$\operatorname{Var}_\mu(B):=\sup\left\{\sum_{i=1}^n\left|\mu(B_i)\right|:n\in\mathbb N\text{ and }B_1,\ldots,B_n\in\mathcal B\text{ are disjoint with }\biguplus_{i=1}^nB_i\subseteq B\right\}$$ for $B\subseteq E$. If $E\in\mathcal E$, let $$\left|\mu\right|:=\operatorname{Var}_\mu(E).$$

  Assume $(E,\mathcal E)$ is a measurable space and $\mu,\nu$ are probability measures on $(E,\mathcal E)$. Is it possible to show that $$\left|\mu-\nu\right|=\sup_{B\in\mathcal E}\left|\mu(B)-\nu(B)\right\|\tag1$$ or is there a counterexample?

I was able to show the claim assuming that $\mu$ and $\nu$ are both absolutely continuous with respect to a common reference measure. Is it possible to show $(1)$ in general?

You are off by a factor of 2: $\| \mu-\nu\| =2\sup_{B\in\mathcal E}|\mu(B)-\nu(B)|$. For this it's important that $\mu $ and $\nu$ are probabilities, so that $\mu(E)-\nu(E)=0$.

Later addition: Let $\lambda=\mu+\nu$ and let $f$ and $g$ be the Radon-Nikodym densities of $\mu$ and $\nu$ with respect to $\lambda$. It's not hard to check that $\|\mu-\nu\|=\int_E|f-g|\,d\lambda$.

First, because $\mu(E)=\nu(E)=1$,  if $B\in\mathcal E$ then,
$$
2|\mu(B)-\nu(B)|=|\mu(B)-\nu(B)|+|\mu(B^c)-\nu(B^c)|\le\operatorname{Var}_{\mu-\nu}(E)=\|\mu-\nu\|.
$$
This shows that $\|\mu-\nu\|\ge 2\sup_{B\in\mathcal E}|\mu(B)-\nu(B)|$.

For the reverse inequality,
$$
\eqalign{
\|\mu-\nu\|&=\int_E|f-g|\,d\lambda\cr
&=\int_{\{f>g\}}(f-g)\,d\lambda+\int_{\{f<g\}}(g-f)\,d\lambda\cr
&=(\mu-\nu)(\{f>g\})+(\nu-\mu)(\{f<g\})\cr
&=|(\mu-\nu)(\{f>g\})|+|(\mu-\nu)(\{f<g\})|\cr
&\le 2\sup_{B\in\mathcal E}|\mu(B)-\nu(B)|.
}
$$

Your claim that the equality holds when the measures are absolutely continuous w.r.t. some measure is wrong. You can check that it fails even in simple cases like $\mu (A)=m(A\cap (0,1))$, $\nu (A)= m(A\cap (-1,0))$ where $m$ is Lebesgue measure. [Any two probability measures are absolutely continuous w.r.t some probability measure, e.g. $\frac {\mu +\nu} 2$].

If $\mu =\delta_a$ and $\nu =\delta_b$ with $a \neq b$ then $|\mu-\nu|=2$ and $|\mu(B)-\nu(B)| \leq 1$ for all $B$.

[$|\mu-\nu| \geq |(\mu-\nu)(\{a\})|+|(\mu-\nu)(\{b\})|=2$].

More generally if $P$ and $Q$ are any two probability measures with $P\perp Q$ the $|P-Q|=2$ and $|P(B)-Q(B)| \leq 1$ for all $B$.
What is true in general is RHS $\leq$ LHS $\leq 2$RHS.


Integer polynomial on integer values: dense mod $\pi$? I was trying to answer the question at How to show that $\lim_{n \to \infty} \sin(n^2)$ does not exist?

I was able to show that $\sin(n^2)$ does not tend to a limit, but am still unable to show that $n^2$ itself is dense mod $\pi$. Can this be deduced in a straightforward way from the famous theorem that the integers are dense mod $\pi$?

Also, are all integer polynomials dense modulus an irrational number? I struggle to find the relevant literature online.

Yes, I think so. Answering loosely: say we want an integer $n$ so that $n^2$ is "close to" a certain value $r$ mod $\pi$. Let $s = \sqrt{r}$. Because the integers are dense mod $\pi$, there is an integer $k$ so that $k$ is "close to" $s$. Let $n = k$. $n^2 \mathrm{mod} \pi = k^2 \mathrm{mod} \pi = (k \mathrm{mod} \pi)^2$. So $n^2$ is "close to" $s^2 = r$ mod $\pi$.


Is it correct to call the probability mass function by the name “discrete probability density function”? Commonly, the probability density function (PDF) is used when dealing with continuous random variables, while the probability mass function (PMF) is used for discrete random variables.
This is the reason they are called “density function” and “mass function” respectively.
However, my professor would talk about a “continuous PDF” and a “discrete PDF”, instead of a PDF and a PMF.
It seems that my professor is not the only one to use the term “discrete probability density function”. It is also used in these UBC lecture notes.
Is it correct to call the probability mass function by the name “discrete probability density function”

Let $X$ be a discrete random variable with probability mass function $p_X : \mathcal{X} \to [0,1]$, where $\mathcal{X}$ is a discrete set (possibly countably infinite). Random variable $X$ can be thought of as a continuous random variable with the following probability density function

$$f_X (x) = \sum_{x_k \in \mathcal{X}} p_X  (x_k) \, \delta (x - x_k)$$

where $\delta$ is the Dirac delta. One could call $f_X$ a discrete probability density function, as its support is a discrete set. However, do note that $f_X$ is a generalized function, not a proper function.


$A_1,A_2$ fulfill property, but their sum $A_1+A_2$ does not Let $V$ be a real, reflexive, separable Banach space.
Are there operators $A_1,A_2: V \to V^*$ that fulfill the property
\begin{cases}
    u_n \rightharpoonup u \\
    A_iu_n \rightharpoonup b \\
    \limsup_{n \to \infty} \langle A_iu_n,u_n\rangle\leq\langle b,u\rangle
\end{cases}implying $A_iu=b$
for a sequence $(u_n)_{n \in \mathbb N}$ in $V$, $b \in V^*$, but their sum $A_1+A_2$ doesn't?

My work:

$(A_1+A_2)u_n=A_1u_n+A_2u_n \rightharpoonup2b \in V^*$ and
$\limsup\, \langle (A_1+A_2)u_n,u_n \rangle=\limsup \langle A_1u_n,u_n \rangle+\limsup \langle A_2u_n,u_n \rangle \leq 2 \langle b,u\rangle$
So I think I need to find $A_1,A_2$ fulfilling the property but $(A_1+A_2)u \neq 2b?$

Here is a counterexample. Let $V = \mathbb{R}$ and
$$
A_1(x)=
\begin{cases}
\frac1x & \text{if } x \ne 0 \\
42 & \text{if } x = 0
\end{cases}
$$
and
$$
A_2(x)=
\begin{cases}
-\frac1x & \text{if } x \ne 0 \\
23 & \text{if } x = 0
\end{cases}
$$
Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.

No, this is not possible.

Using the linearity of $A$ in these properties and an inequality involving $\limsup$,
it can be shown that $A_1+A_2$ fulfills the same properties.