A challenge by R. P. Feynman: give counter-intuitive theorems that can be translated into everyday language The following is a quote from Surely you're joking, Mr. Feynman.  The question is: are there any interesting theorems that you think would be a good example to tell Richard Feynman, as an answer to his challenge?  Theorems should be totally counter-intuitive, and be easily translatable to everyday language.  (Apparently the Banach-Tarski paradox was not a good example.)

  Then I got an idea.  I challenged
  them: "I bet there isn't a single
  theorem that you can tell me - what
  the assumptions are and what the
  theorem is in terms I can understand -
  where I can't tell you right away
  whether it's true or false."

  It often went like this: They would
  explain to me, "You've got an orange,
  OK? Now you cut the orange into a
  finite number of pieces, put it back
  together, and it's as big as the sun.
  True or false?"

  "No holes."

  "Impossible!

  "Ha! Everybody gather around! It's
  So-and-so's theorem of immeasurable
  measure!"

  Just when they think they've got
  me, I remind them, "But you said an
  orange! You can't cut the orange peel
  any thinner than the atoms."

  "But we have the condition of
  continuity: We can keep on cutting!"

  "No, you said an orange, so I
  assumed that you meant a real orange."

  So I always won. If I guessed it
  right, great. If I guessed it wrong,
  there was always something I could
  find in their simplification that they
  left out.

Every simple closed curve that you can draw by hand will pass through the corners of some square.  The question was asked by Toeplitz in 1911, and has only been partially answered in 1989 by Stromquist.  As of now, the answer is only known to be positive, for the curves that can be drawn by hand. (i.e. the curves that are piecewise the graph of a continuous function)

I find the result beyond my intuition.

For details, see http://www.webpages.uidaho.edu/~markn/squares/  (the figure is also borrowed from this site)


Maxima and Minima of Sin(x)/x I am trying to calculate the maximum and minimum points (between $-3\pi$ and $3\pi$) of $$f(x)=\frac{\sin(x)}{x}$$
I have found the derivative of the function and let it equal to zero.
$$f'(x)=\frac{x\cos(x) - \sin(x)}{x^2}$$
$$f'(x)=0$$
$$\frac{x\cos(x) - \sin(x)}{x^2}=0$$
$$x\cos(x) - \sin(x)=0$$
$$x\cos(x)=\sin(x)$$
$$x=\tan(x)$$

I am unaware as to how to find $x$. I assume that once I find $x$, I can use a sign diagram or second derivative test to determine the minimum and maximum values. Any help would be highly appreciated.

Equations which contains polynomial and trigonometric functions do not show explicit solutions and numerical methods should be used to find the roots.

The simplest root finding method is Newton; starting from a reasonable guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}$$ So, in your case, $$F(x)=x \cos(x)-\sin(x)$$ I think it is better to let it under this form because of the discontinuities of $\tan(x)$.

You can notice that if $x=a$ is a root, $x=-a$ will be another root. So, let us just focus on $0\leq x \leq 3\pi$. If you plot the function, you notice that, beside the trivial $x=0$, there are two roots located close to $5$ and $8$. These would be the guesses.

Using $F'(x)=-x \sin (x)$, the iterative scheme then write $$x_{n+1}=x_n-\frac{1}{x_n}+\cot (x_n)$$ Let us start with $x_0=5$; the method then produces the following iterates : $4.50419$, $4.49343$, $4.49341$ which is the solution for six significant figures.

I let you doing the work for the other solution.

You don't need to solve the equation for $x$ to find the extreme point. You know that extremes happen when $x = \tan(x)$ and, therefore, taking the inverse $\tan$ on both sides, when $\mathrm{atan}(x) = x$. I will write $x_e$ to note the points where $f(x)$ has an extreme. Then

$\displaystyle f(x_e) = \frac{\sin(x_e)}{x_e} = \frac{\sin[\mathrm{atan}(x_e)]}{x_e}$.

But

$\displaystyle \sin[\mathrm{atan}(x_e)] = \frac{x_e}{\sqrt{1 + x_e^2}}$,

so, for each $x_e$ that is the point of an extreme, the value of the function at a point is

$\displaystyle f(x_e) = \frac{1}{\sqrt{1 + x_e^2}}$.

An alternative would be

$\displaystyle f(x_e) = \frac{\sin(x_e)}{x_e} = \frac{\sin(x_e)}{\tan(x_e)} = \cos(x_e)$, which is actually the same result, since, using again that $x_e = \mathrm{atan}(x_e)$, we have

$\displaystyle \cos[\mathrm{atan}(x_e)] = \frac{1}{\sqrt{1 + x_e^2}}$.

In any case, you see that
$\displaystyle f(x_e) = \frac{1}{\sqrt{1 + x_e^2}}$ is maximum at $x_e = 0$, as it decreases everywhere else. At this point, $f(0) = 1$.


How to prove a function is not onto? Let $f : Z\to Z$ be the function defined by $f(x) = 3x + 1$. Prove that $f $ is not onto, using a proof by contradiction. (Choose an integer $n$, and then prove ($\forall m \in Z$)($f(m) ≠ n$) by contradiction.)

so far I have:
\begin{gather*}
    1.\quad \text{Let}\quad y  = -1     \qquad   \text{assumption}\\
    2. \quad \text{Let}\quad f(x) = -1       \qquad \text{hypothesis}\\
    3. \quad 3x+1 = -1        \qquad \text{Definition of $f (2)$}\\
    4. \quad 3x  = -2            \qquad \text{Algebra}\\
    5. \quad x = -2/3          \qquad \text{Algebra}
\end{gather*}

What else is needed? I don't know where to go from here?

Your work shows that the only real number $x$ for which $f(x)=−1$ is $x=−2/3$. In particular, there is no integer $n$ such that $f(n)=−1$. Since the domain of $f$ is $\mathbb Z$, the set of integers, you can conclude that $f$ is not surjective.

Let $y=3x+1$ be any element in $\mathbb Z$, then $x=\frac{y-1}{3}$. So,
$$f(x) = f\left(\frac{y-1}{3}\right) = 3\frac{y-1}{3} + 1 = y$$
Thus, by Definition of onto function we conclude that given function is onto.


Sensitivity and Specficity (Conditional Probability problem) A medical diagnostic test with sensitivity (true positive rate) of
.95 and specificity (true negative rate) of .90. The disease in question is rare
and occurs in the population with the probability 1/10,000 . A patient is
diagnosed with the disease.
Given the diagnosis, what is the probability that the patient actually has the
disease?

I understand how to determine the True Positive Rate and the True Negative Rate with their specific equations, however i am unsure as to how find the final probability.

Thank you

Let B be positive diagnosis.

Let A be person is sick.

$$P(A|B)= \frac{P(A \cap B)}{P(B)} = \frac{0.0001\cdot0.95}{0.0001\cdot0.95+ 0.9999\cdot0.1}\approx 0.000949$$


Is an undefined equality vacuously true? To quote Wikipedia,

  [..] equation is an equality containing one or more variables. Solving the equation consists of determining which values of the variables make the equality true.

Consider a univariate equation $f(x) = 0$ with $f:D \mapsto \mathbb{R}$ and say we want to solve the equation over $x \in A$. As I understand it, for every $x \in A$ we have a condition $f(x) = 0$ and we have to verify whether it is true. But if $\exists x \in A: x \not\in D $, then there is no condition! Does that mean that the equality is vacuously true? I can't see why not.

But in school, if we solve something like $\frac{1}{x} = 0$, the answer is there is no solution, and, specifically, $x = 0$ is not a solution. This contradicts the above reasoning.

Regarding equations, we have to separate identities, like : $(x^2-1)=(x-1)(x+1)$ from equations like : $x^2+1=0$.
In the first case, we are stating :

$\forall x \ [(x^2-1)=(x-1)(x+1)]$,

i.e. we are saying that in e.g. the domain $\mathbb N$ of natural numbers, the formula is true.
This is not so for $x^2+1=0$; the formula $\forall x \ (x^2-2=0)$ is obviously false in $\mathbb N$.
What we are "searching for" when we ask to solve the above equation is the truth-value of the formula :

$\exists x \ (x^2-2=0)$.

This formula is false in $\mathbb N$, while it is true in $\mathbb R$.
Thus, we can consider : $\exists x \in \mathbb N \ (x^2-2=0)$.
But this is not a case of vacuously true statement :

A statement is "vacuously true" if it resembles the statement $P \to Q$, where $P$ is known to be false.
Statements that can be reduced to this basic form include the following universally quantified statements:

$\forall x \in A:Q(x)$, where the set $A$ is empty.

The equation we are considering : $\exists x \in \mathbb R \ (x^2-2=0)$ has not the said form, because it is not universally quantified.

When $x\not\in D$ the formula $(\forall x)[x\in D \Rightarrow f(x)=0]$ is vacuously true.


Are there any number $n$ such that $a_n = 0 \mod (2n + 1) $ where $a_0 = 1, a_1 = 4, a_{n + 2}=3 a_{n + 1} - a_{n}$? Define the sequence $a_n$ by the following.

$$a_0 = 1, a_1 = 4,$$
$$a_{n + 2}=3 a_{n + 1} - a_{n}$$

$a_n ≠ 0 \mod (2n + 1)$ for $1 \le n \le 10^5 $.
Are there any number $n$ such that $a_n = 0 \mod (2n + 1)$?

No, there are no others. You have the original Lucas numbers of odd index. The device of using $2n+1$ means they are asking about the Lucas numbers divisible by their index, see https://oeis.org/A016089  It seems this is proved in the article Noam mentions, (in any case it is in the Examples) with easier to read in Somer. The two books with articles by Somer were edited by Gerald E. Bergum, Andreas N. Philippou, and A. F. Horadam.

There are plenty of even indexes that divide the Lucas number, but the only odd one is $1$ itself.

Note that $L_{n+2} = L_{n+1} + L_n$ gives $L_{n+1} = L_{n+2} - L_n,$
$$ L_{n+3} = L_{n+2} + L_{n+1} = L_{n+2} +  L_{n+2} - L_n = 2 L_{n+2} - L_n , $$
$$ L_{n+4} = L_{n+3} + L_{n+2} = 2 L_{n+2} - L_n +L_{n+2} = 3 L_{n+2} - L_n.  $$

HINT: Solving the difference equation (use Z Transform):
$$
a_{n+2} = 3 a_{n+1} - a_n \quad\Rightarrow\quad
a_n = \left(\dfrac{3-\sqrt{5}}{2} \right)^n c_1 + \left(\dfrac{3+\sqrt{5}}{2} \right)^n c_2
$$
By conditions $a_0 = 1$ and $a_1 = 4$:
$$
\begin{align}
c_1 + c_2 &= 1 \\
\left( \dfrac{3}{2} - \dfrac{\sqrt{5}}{2} \right) c_1 + \left( \dfrac{3}{2} + \dfrac{\sqrt{5}}{2} \right) c_2&= 4
\end{align}
$$
$$
\therefore \quad c_1 = \dfrac{1-\sqrt{5}}{2} \quad c_2 = \dfrac{1+\sqrt{5}}{2}
$$
Thus:
$$
\begin{align}
a_n &= \left(\dfrac{3-\sqrt{5}}{2} \right)^n \left(\dfrac{1-\sqrt{5}}{2}\right) + \left(\dfrac{3+\sqrt{5}}{2} \right)^n \left(\dfrac{1+\sqrt{5}}{2}\right) \\
\end{align}
$$
Perhaps this can help you a little bit, since putting you in understanding the structure of the numbers you want to test.


$\frac{1}{\sqrt{-1}}=\sqrt{-1}$? I have trouble to comprehend what my mistake is in the following calculation:

If we set $\sqrt{-1}$ to be the new number with the property that $(\sqrt{-1})^2 = -1$ then I can write $$\frac{1}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{-1}.$$

But we also have (and I know this is the correct result) $$ \frac{1}{\sqrt{-1}}\cdot\frac{\sqrt{-1}}{\sqrt{-1}}=\frac{\sqrt{-1}}{-1}=-\sqrt{-1}$$

What am I missing? Thanks.

The incorrect step is $$\frac{1}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}$$
This kind of relations that you learned very soon in your life are so anchored in your thinking that you don't even think that they need to be re-checked whenever the settings are different. Complex numbers behave differently to positive real numbers in that respect.

The problem is due to the use of the same symbol $\sqrt{\cdot}$ with different meanings.

Let's separate the meanings by distinguishing:

$\sqrt[\text{real}]{\cdot}$ is an operator defined on non-negative real numbers, that associates to $x$ the unique non-negative real number $y$ such that $y \cdot y = x$;
$\sqrt[\text{new}]{-1}$ is a symbol used to indicate a new number, that satisfies the property $\sqrt[\text{new}]{-1}\cdot \sqrt[\text{new}]{-1} = -1$.

The first chain of equations you wrote translates to:

$$\displaystyle\frac{1}{\sqrt[\text{new}]{-1}}=\frac{\sqrt[\text{real}]{1}}{\sqrt[\text{new}]{-1}} \overset{!}{=} \frac{\sqrt[\text{real}]{1}}{\sqrt[\text{real}]{-1}} \overset{!}{=} \sqrt[\text{real}]{\frac{1}{-1}} \overset{!}{=} \sqrt[\text{real}]{-1},$$

but the steps marked with $\overset{!}{=}$ are not valid since $\sqrt[\text{real}]{-1}$ is not defined.


How to prove $|a^{2} + b^{2} + c^{2} - 2\left( ab+bc+ac\right)| \le \frac{1}{27}$ Let $a,b,c$ are $3$ edge of a triangle and $\sqrt{a} +  \sqrt{b} +  \sqrt{c} = 1.$

How to prove $|a^{2} + b^{2} + c^{2} - 2\left( ab+bc+ac\right)|  \le \frac{1}{27}$? Can this be proved with simple way?

Let $a=\frac{y+z}{18}$, $b=\frac{x+z}{18}$ and $c=\frac{x+y}{18}$.

Hence, $x$, $y$ and $z$ are positives, $\sum\limits_{cyc}\sqrt{x+y}=3\sqrt2$ and we need to prove that $xy+xz+yz\leq3$,
which we can prove by the Contradiction method.

Let $x=ku$, $y=kv$ and $z=kw$, where $k>0$ and $uv+uw+vw=3$.

Hence, $k>1$ and $3\sqrt2=\sum\limits_{cyc}\sqrt{x+y}=\sqrt{k}\sum\limits_{cyc}\sqrt{u+v}>\sum\limits_{cyc}\sqrt{u+v}$,

which is contradiction because we'll prove now that $\sum\limits_{cyc}\sqrt{u+v}\geq3\sqrt2$.

Indeed, we need to prove that

$2(u+v+w)+2\sum\limits_{cyc}\sqrt{(u+v)(u+w)}\geq6\sqrt{3(uv+uw+vw)}$ or

$6(u+v+w)-6\sqrt{3(uv+uw+vw)}\geq\sum\limits_{cyc}\left(\sqrt{u+w}-\sqrt{v+w}\right)^2$ or

$\sum\limits_{cyc}\frac{3(u-v)^2}{u+v+w+\sqrt{3(uv+uw+vw)}}\geq\sum\limits_{cyc}\frac{(u-v)^2}{\left(\sqrt{u+w}+\sqrt{v+w}\right)^2}$, which is obvious because

$3\left(u+v+2w+2\sqrt{w^2+uv+uw+vw}\right)\geq u+v+w+\sqrt{3(uv+uw+vw)}$.

Done

Yes , this can be proved using Lagrange's multiplier..
Let
$$f(a,b,c)=a^2+b^2+c^2-2(ab+bc+ca)+k(\sqrt{a}+\sqrt{b}+\sqrt{c}-1)$$
Solve the following four equations:
$$\frac{df}{da}=0$$
$$\frac{df}{db}=0$$
$$\frac{df}{dc}=0$$
$$ \sqrt{a}+\sqrt{b}+\sqrt{c}=1$$

You will get a set of values for $(a,b,c)$
Put these values of (a,b,c) in $f$

You get the minimum and maximum values


HINT: from your last line, consider the two different cases, according with $n$ odd and even.

I tried what @Joe suggested on the problem statement itself. I'm getting the right answer for even $n$ but the answer for odd $n$ is equal upto a sign.


Closed subset of a manifold is a submanifold if there exists a smooth retract. My question is regarding a specific exercise that I'm not really sure how to approach. Suppose I have a smooth manifold $M$ and let $A$ be a closed subset. Now suppose that there exists a smooth function $f:M \to A$ such that $f|_A = id$ (smooth retract).

I want to show that $A$ is a submanifold, i.e, it is Hausdorff, second countable and locally euclidean. Hausdorff and second countable are both hereditary properties, so only the latter is left.

My attempt was to prove that if $(U, \varphi)$ is a chart around $x \in A \subset M$ for $M$, then $\left(f(U), \varphi|_{f(U)}\right)$ is the desired chart. However, I'm not sure if $f(U)$ is open. I probably should use that $A$ is closed and $f$ is a retract, but I couldn't connect both facts.

There is no need for a solution, just a hint is enough.

PS: John Lee suggested that I should ask a question instead of leaving the exercise for the community, so here it is. I rewrote my previous post here. :)

Assume $M$ connected. Note that $A=Im(f)$ is therefore connected. Let $a\in A$, $f'(a)$ is a linear projection on $T_aM$. As the rank of a projection is a contious function on the set of projection (it is the trace), along $A$ the rank of $f'$ is constant $r$. Now, let us check that it is constant in a neigbourhood of $A$ . As $f\circ f= f$, $f'(f(p))\circ f'(p)= f'(p)$, and $f'(f(p))$ induces a isomorphism on $Im ( f'(p))$, so that the rank of $f'(p)$ is less than or equal to $r$. But the rank can only increase in the neigbourhood of $A$, so it is constant. If you know the "constant rank theorem" you are done. If not let us argue as follows.

Now choose a point $a\in A$. In order to study the image of $f$ near this point, let us choose  coordinates $x_1,...,x_n$ near $a$ such that $f'(a)$ is the projection onto $x_{r+1}==x_n=0$, and let us consider the new system of coordiantes $y_i=x_i\circ f$ for $i\leq r$, $y_i=x_i$  if $i> r$. There are functions $F_j, j>r$ such that $f(y_1,..y_n)=(y_1,,..,y_r, F_{r+1},...F_{n})$ Writing that the rank of $f'$ is $r$ we see that ${\partial F_j \over \partial y_k }=0$ if $k>r$. Or $f(y_1,...y_n)= (y_1,..,y_r, F_1(y_1,..) ..)$. So near $a$ the image of $F$ is just the graph of the function $F=(F_{r+1},...F_n$), a submanifold.

This not true. Consider two segments $C_1, C_2\subset R^2$ which are not parallel such that $C_1\cap C_2$ is a point. Take $A=C_1\bigcup C_2$ it is not a smooth manifold you have a singular point at the intersection. Consider $U_i$ a tubular neigborhood of $C_i, i=1,2$ diffeomorphic to $C_i\times [0,1]$. $X=U_1\cap U_2$ retracts to $A$ and is a manifold.


How does one refute this ultrafinitist argument? From Wikipedia:

  Edward Nelson criticized the classical conception of natural numbers because of the circularity of its definition. In classical mathematics the natural numbers are defined as $0$ and numbers obtained by the iterative applications of the successor function to $0$. But the concept of natural number is already assumed for the iteration. In other words, to obtain a number like $2\uparrow\uparrow\uparrow 6$ one needs to perform the successor function iteratively, in fact exactly $2\uparrow\uparrow\uparrow 6$ times to $0$.

Like most people, I am not an ultrafinitist. However, I do not quite know how to refute this argument. What are some criticisms of it?

I am agnostic about most questions of the philosophy of mathematics, but in this case, I think Nelson's argument is incoherent: a primitive shepherd who doesn't have any abstract concept of number can count sheep being herded into a fold by making marks on a stick and count the sheep out again by crossing the marks off (and this is how our human notions of cardinal numbers probably developed). There is no circularity involved.

To talk meaningfully about notations like $2 \uparrow\uparrow\uparrow 6$, you must have admitted a notion of definition by recursion that (as an ultrafinitist) Nelson can't accept. It seems to be me to be incoherent for an ultrafinitist to say anything more than "I can't accept that a primitive shepherd could ever have enough sticks".

Nelson's critique is groundless. Peano's axioms are just a list of the essential properties of the set of natural numbers, essential in the sense that all other known properties can, in theory, be derived from these few properties. There is nothing "circular" about them. So successful have they been in this regard that these essential properties have come to define the set of natural numbers $\mathbb{N}$.
The natural numbers and many of their properties have been known for millennia across diverse cultures. Can any reasonable person deny that:

(1) $0$ is a natural number.
(2) Every natural number has a unique successor.
(3) Different natural numbers cannot have the same successor.
(4) No natural number has a successor that is $0$.

Now, some might quibble about the 5th axiom, the principle of mathematical induction:

(5) For all subsets $P \subset\mathbb{N}$, if (a) $0\in P$, and (b) for all $x\in P$, we also have the successor of $x$ being in $P$, then $P=\mathbb{N}$.

With only a bit more reflection, this, too, passes for a reasonable property of the set of natural numbers. There is nothing weird or "unnatural" about it. In fact, induction can be shown to hold on some subset of any non-empty set $X$.
Proof sketch: Let $X$ be an arbitrary, non-empty set. There exists at least one function $f: X \to X$, namely the identity function. Given any function $f:X\to X$ and $x_0\in X$, we can construct a subset $N\subset X$ such that $N=\{x_0, f(x_0) f(f(x_0)), \cdots \}$ and show that induction holds on $N$, namely that:

For all subsets $P \subset N$, if (a) $0\in P$, and (b) for all $x\in P$, we also have the successor of $x$ being in $P$, then $P=N$.

See my formal proof at http://www.dcproof.com/InductionMinRequirementsV2.htm


Why is it that Complex Numbers are algebraically closed? I find it curious that Complex Numbers give enough flexibility to be algebraically closed, where the reals, rational numbers do not.  For the reals it is easy to see that they cannot be used to solve equations like $x^2 + 1 =0$.  Geometrically, one can look at the number line as see that any $x$ squared yields a positive number which when added to one cannot get you back to zero.  In the complex case, however, we are working with the plane.  In this case exponents stretch and rotate any given $x$.  It is easy to therefore see in the particular circumstance that if $x=i$ that $x^2$ rotates it to $-1$ which when added to one yields the desired result (i.e. $0$).  So because the Complex Numbers are algebraically closed, I conclude that any polynomial equation with complex coefficients my be solved by choosing one or more $x$'s in the plane and rotating them and stretching them such that they will combine using the given coefficients to produce the RHS.

Question: Why is it that we do not need a larger space than the plane to solve Complex polynomial equations?

I have tried to find a sufficient answer through Google, but was not able.  I also searched M.SE and could not find a sufficient answer.  I am not a mathematician, so I am looking for an intuitive answer if possible.

Of course there are many proofs, and perhaps some others will post the most attractive proofs,  but I think you are looking for an intuitive explanation that would somehow make the result seem less surprising.

One such explanation, I think, is the simple observation that reals already go a long way towards being algebraically closed---they are a real closed field---since every odd-degree polynomial over $\mathbb{R}$ has a root in $\mathbb{R}$. This follows immediately from the intermediate value theorem, since in the large scale every odd degree polynomial moves from $-\infty$ to $\infty$ or conversely and hence must cross the axis.

One definition of the complex numbers is that they are the algebraic closure of the reals.

In other words, start with the reals and write down some degree-$d$ polynomial that has fewer than $d$ roots (incl multiplicities).  Call one of these roots $x$.  Now extend the reals with $x$.  After you continue this (infinite) process, you have $\mathbb{C}$.

If you're willing to buy this as the definition of $\mathbb{C}$, then it's trivial to see that it's algebraically closed.


$a_n = n + na_{n-1}$. This is $a_n = n + n(n-1) + n(n-1)(n-2) + \ldots + n!$. In turn, $\frac {a_n}{n!} = \frac{1}{(n-1)!} + \frac{1}{(n-2)!} + \frac{1}{(n-3)!} + \ldots + 1$, which approaches the Taylor series for $e$.


Algebra; Functions & Graphs Given:

Why is -2 non-inclusive?

The function is strictly increasing on any interval $J\subset(-\infty,-2]$, so the suggested answer is true since $(-\infty,-2)\subset(-\infty,-2]$.

The derivative of the function (the slope) is 0 at x = -2.

The definition of an increasing function requires that the slope be strictly positive. :)


Prove the module of a holomorphic function cannot be $\frac{K}{\cosh x}$ Prove that there does not exist any holomorphic function $f(z)$, such that $|f(z)| =\frac{K}{\cosh x}$ where $K>0$ is a constant, and $x=Re(z)$.

My idea is to use the maximum modulus principle. $|f(z)| =\frac{K}{\cosh x} = \frac{2K}{e^x+e^{-x}} \leq K$. An entire function with bounded modulus must be a constant. This leads to a contradiction.

My question is, in my attempt, I assume $f$ to be holomorphic on $\mathbb{C}$. Is it possible to prove the same result when $f$ is holomorphic on $\Omega$ where $\Omega$ is a domain? And is it possible not to use maximum modulus principle?

Thank you for any help!

For a solution which does not need the maximum modulus principle and works on more general domains, see Complex function does not have a modulus equal to a function of its real part.


About deformations of Lie algebra structures and Nijenhuis bracket In a previous post Lie bracket as a tensor I asked about the formalism of deformed Lie brackets, how a tensor could satisfy the Leibniz rule. This is question is solved and now I want a proof of the next result:

Let $M$ be a finite-dimensional manifold and let $[\cdot,\cdot]$ be the usual Lie bracket of the $\mathbb R$-algebra of vector fields over $M$. Consider an endomorphism $N\in \Gamma^{\infty}(TM\otimes T^*M)$. Then, the bracket

$$
[X,Y]_N = [N(X),Y] + [X,N(Y)] - N([X,Y])
$$

satisfies the Jacobi identity (and then defines a Lie bracket) if and only if $N$ is a Nijenhuis tensor, i.e.

$$
\frac{1}{2}[N,N]_{FN}(X,Y) = [N(X),N(Y)]-N([N(X),Y]) - N([X,N(Y)]) + N^2([X,Y]) = 0
$$

for all $X,Y$ in $\Gamma^{\infty}(TM)$.

One proof can be found in Deformations on Lie algebra structures, of A. Nijenhuis and R. W. Richardson, (1967), publised on Journal of Mathematics and Mechanics, T. 17, No. 1, pp. 89-105.

However, I have been looking for at Internet and i have not found anything. And also, the library of my university neither have it.

EDIT___________________________________________

After big efforts and many computations finally I founded this equality:

$$
\oint_{X,Y,Z} [X,[Y,Z]_N]_N = -\frac{1}{2} \oint_{X,Y,Z} \Bigg(T_N(X,[Y,Z]) + [X, T_N(Y,Z)] \Bigg) ,
$$

where I am naming

$$
T_N(X,Y) = \frac{1}{2} [N,N]_{FN}(X,Y)
$$

and

$$
\oint_{X,Y,Z} f(X,Y,Z) = f(X,Y,Z) + f(Y,Z,X) + f(Z,X,Y)
$$

the cyclic sum that appears in the Jacobi identity.

So one direction is clear. But, I am not sure

$$
\oint_{X,Y,Z} \Bigg(T_N(X,[Y,Z]) + [X, T_N(Y,Z)] \Bigg) = 0
$$

implies

$$
T_N(X,Y)=0
$$

for all vector fields $X$ and $Y$.

Can anybody helps me?

Thanks.

Well... as is usual, I will answer my own question. The statement is more general than my question. In fact, $(TM,[\cdot,\cdot]_N,N)$ is a Lie algebroid if and only if $N$ is a Nijenhuis operatoor. The proof is as follows.

First, check $N$ is a homomorphism if and only if $T_N=0$:

$$
N([X,Y]_N)= N([N(X),Y]) + N([X,N(Y)]) - N^2([X,Y]) = [N(X),N(Y)] \, , \quad \forall X,Y\in{\frak X}(M) \, .
$$
This expresion is equivaent to say $T_N=0$.

After, consider the derivation $D$ defined on $T^*M$ by means of this new bracket and the anchor map $N$:

$$
D \alpha \, (X_1,\dots, X_{p+1}) = \sum_{i=1}^{p+1} (-1)^{i+1} N(X_i)(\alpha(X_1,\dots,\hat{X}_i,\dots,X_{p+1}) + \sum_{i<j} (-1)^{i+j} \alpha([X_i,X_j]_N,X_1,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_{p+1}) \, .
$$

As a derivation of degree 1, it can be write as (Frölicher-Nijenhuis decomposition theorem)

$$
D = \mathcal{L}_K + i_L
$$

where $K$ is $TM$-valued 1-form and $L$ is a $TM$-valued 2-form. Studing the behavior of $D$ over functions and 1-forms one can see that $L=0$ and $K=N$, so

$$
D=\mathcal{L}_N \, .
$$

Finally we have to study the operator $\mathcal{L}^2_N$. We know

$$
\mathcal{L}^2_N = 0 \Longleftrightarrow T_N = 0 \, ,
$$

since

$$
T_N=\frac{1}{2} [N,N]_{FN} = \frac{1}{2} [\mathcal{L}_N,\mathcal{L}_N](f) = \mathcal{L}_N^2 (f) \, .
$$

A short computation shows

$$
D^2\alpha (X,Y,Z) = \mathcal{L}^2_N \alpha(X,Y,Z) = \oint_{X,Y,Z} N([X,Y])(\alpha(Z) ) - \oint_{X,Y,Z} [N(X),N(Y)](\alpha(Z)) + \oint_{X,Y,Z} \alpha([[X,Y]_N,Z]_N) \, ,
$$

thus

$$
\oint_{X,Y,Z} \alpha([[X,Y]_N,Z]_N)  = 0 \Longleftrightarrow N([X,Y]_N)= N([N(X),Y]) + N([X,N(Y)]) - N^2([X,Y]) = [N(X),N(Y)] \Longleftrightarrow T_N = 0
$$


$xy-zw$ is an irrreducible element in $\mathbb C[x,y,z,w]$ Prove that $xy-zw$ is an irreducible element in the polynomial ring $\mathbb C[x,y,z,w]$.

My attempt was:

Consider the homomorphism from $\mathbb C[x,y,z,w]$ to $\mathbb C$ induced by the map $x,y,z,w$ onto $1,2,1,2$ respectively.

Since $\mathbb C$ is an integral domain the ideal generated by $xy-zw$ is prime and hence irreducible as $\mathbb C[x,y,z,w]$ is an integral domain.

Is this correct?

Thanks.

See this question for a general proof that the generic determinant is an irreducible polynomial. I will adapt the proof to this special case.

I will suppose $C$ is an integral domain (maybe you meant $\Bbb C$; that is a particular integral domain). Suppose $xy-zw=ab$ with $a,b\in C[x,y,z,w]$. Since the degree of $xy-zw$ in $x$ is $1$, one of $a,b$ must be of degree $1$ in $x$ and the other of degree $0$ (it does not contain $x$); by symmetry we may suppose $a$ is of degree $1$, say $a=px+q$ with $p,q\in C[y,z,w]$ and $p\neq0$. Now $b$ cannot contain either $z$ or $w$, lest the leading term in that variable produce a term in the product $pxb$ divisible by $xz$ respectively by $xw$ and which cannot be canceled by any other terms in the product $ab$, contradicting the fact that $ab=xy-zw$ contains no such terms. So $a$ is of degree $1$ in $z$ and $w$ as well as in $x$. Repeating the argument with $z$ in the place of $x$, we see that $b$ cannot contain any $y$ either (as $xy-zw$ has no terms divisible by $yz$), so it is a constant in $C$. But that constant divides the coefficients of all terms in $ab$, which means it is invertible, so the decomposition $ab$ is trivial. As a consequence $xy-zw$ is an irreducible polynomial in $C[y,z,w]$.

I used the fact that $C$ is an integral domain in the argument that the leading term of $b$ produces a (nonzero) term of $pxb$. I haven't tried to see if one can construct a ring with zero divisors over which $xy-zw$ factors non-trivially. But in any case question about factorisation usually suppose integral domains.

Consider $\phi: \mathbb{C}[x,y,z,w] \longrightarrow \mathbb{C}[t_1,t_2,t_3]$ where $\phi(x)=t_1$, $\phi(y)=t_1t_2t_3$, $\phi(z)=t_1t_3$, $\phi(w)=t_1t_2$. Now show that $\phi$ is surjective onto $\mathbb{C}[t_1,t_1t_2,t_1t_3, t_1t_2t_3]$ with the ideal $\langle xy-zw\rangle$ as the kernel.


Particle swarm optimization I don't know where to start. Like, I don't know how to plug the info into the algorithm. Show two iterations of particle swarm optimization (neighborhood approach) method.

  Mathematically show two iterations of a PSO method. For the problem the optimum is at $(0,0)$, the objective function is the distance from the optimum, and the three particles are initialised at $(4,3)$, $(-5,6)$, and $(-3, -7)$. (It is understood that everyone will have different random numbers, this is ok)

Alright, so the idea behind particle swarm optimization is this:

We have a problem, but its hard to solve. So lets take numerous guesses at a solution and call them "particles". Lets call our particles $v_{i}$ where $i$ is an index variable for the particles. Luckily, we have a function which tells us how good our current guesses are (in this case the distance function). So lets try and improve our guesses.

We know which guess is currently best, so we should want our particles to explore that area thus we should move our particles closer to the best known solution. Lets call this variable $g$.

However, we also want to explore other areas, and thus we will also move particles based on there own "memory" which is their best known position. Lets call this variable $p_{i}$, where $i$ is referring to the $i$th particles best known position.

Also usually we want to have a random component which allows our particles to explore completely new areas, this is usually called velocity. Lets call this variable $v$.

In addition to this, we will randomize how much we consider each of these components, so individual particles will behave independently (and thus explore the solution space better). Lets call these vectors $r_{1},r_{2}$ (note these are in the range $(0,1)^{d}$ where $d$ is the dimension of the space we are working in)$.

So our when we improve our guesses (running an iteration of PSO), we are going to replace our particles with

$v_{i} = r_{1}g + r_{2}p_{i} +v$

So lets try for your example. Let's give names to our particles,  $v_{1} = (4,3)$, $v_{2} = (-5,6)$, $v_{3} =(-3,7)$

From your initialized particles, $(4,3)$ is the closest to $(0,0)$ and thus it is our global best solution, so $g = (4,3)$.

Each particle has only seen one place so currently $p_{i} =v_{i}$

Now in practice you would want to take random variables to get your weighting, for my purposes I will have all the random variables $=.5$ (in all components, and $v =0$.

For notation I will say $v_{i,new}$ as the updated particle and $v_{i,old}$ for the non-updated particle.

we have

$v_{1,new} = [.5,0.5]^{t}(4,3) + [0.5,0.5]^{t}(4,3) +0 = (4,3)$
$v_{2,new} =  [.5,0.5]^{t}(4,3) + [0.5,0.5]^{t}(-5,6) +0 = (-1/2,9/2)$
$v_{3,new} =  [.5,0.5]^{t}(4,3) + [0.5,0.5]^{t}(-3,7) +0 = (-1/2,5)$

Now we check, did any our particles improve? Well because of how I picked the random variables, $v_{1}$ didn't change.

However both the other particles got closer to (0,0) (you can verify this with the distance function). Therefore we update
$p_{2} = (-1/2,9/2)$ and $p_{3} = (-1/2,5)$

Also note that v_{2,new} is closer to (0,0) than our previous best guess $(4,3)$ therefore we update $g$ to $g= v_{2,new}$.

Now we can just keep continuing this procedure and that is particle swarm optimization. Hope that helps.


Visual proof that $\lim_{x\to \infty} \frac{x^r}{a^x} = 0$ I am looking for a visual proof (in this sense: https://mathoverflow.net/questions/8846/proofs-without-words) for the fact that

$$
    \lim_{x\to \infty} \frac{x^r}{a^x}  = 0
$$

($r > 0,a>1$)

Or at least for a simple (high school level) argument that makes this relation plausible (but it doesn't have to be a rigerous mathematical proof - I know to proof it rigerously, but that's not what I am looking for).

Every time $x$ increases by $1$, the bottom gets multiplied by $a$ and the top gets multiplied by $$\left(\frac{x+1}{x}\right)^r = \left(1+\frac{1}{x}\right)^r \approx 1+\frac{r}{x}$$

So the bottom keeps growing by the same factor while the factor that the top grows by gets smaller and smaller, and will eventually get smaller than $a$. In the long run (i.e. as $x \to\infty$) the bottom will win out, driving the whole expression to $0$. (This does make an appeal to a binomial theorem approximation.)

It's not intuitive, and probably doesn't have the "Aha!" appeal of a visual proof, but I think it would help high schoolers start to think about comparing different rates of growth and help them build their intuition.

And if I were presenting this to high schoolers I would definitely start off showing how it worked in a specific simple example, like $\frac{x^2}{2^x}$. Write ten terms of $x^2$, note that they're increasing, but then write ten terms of $2^x$ and note how the crush the terms on the top. Then make the generalization, and maybe note that if $a$ is close to $1$ and $r$ is large it will take more terms for the bottom to win, but it will eventually, and we are going to $\infty$.

I guess that we can do the following:
$$ 0 = \lim_{x \rightarrow \infty} \frac{x^r}{a^x} =  \lim_{x \rightarrow \infty} \frac{\ln x^r}{\ln a^x} = \lim_{x \rightarrow \infty} \frac{r \ln x}{x \ln a} = \frac r {\ln a} \lim_{x \rightarrow \infty} \frac{\ln x}{x} = \lim_{x \rightarrow \infty} \frac{\ln x}{x} $$

Using the last limit we can make visual proof.


Why are absolute value signs ignored in the indefinite integral $\int x \sec^{-1}(x) dx$? I have seen that the integral $\int x \sec^{-1}(x) dx$ is evaluated in the following way using partial integration:

$$\int x \sec^{-1}(x) dx =\dfrac{x^2}{2}\sec^{-1}(x)-{\displaystyle\int}\dfrac{x}{2\sqrt{x^2-1}}\,\mathrm{d}x$$

That is, $\frac{d}{dx} \sec^{-1}(x)=\frac{1}{x\sqrt{x^2-1}}$ is used.

However, it is actually known that $\frac{d}{dx} \sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}$.

Why can the absolute value signs be omitted?

EDIT:
I have also seen this integral but it doesn't solve my problem:

$$\int{{\frac{{du}}{{u\sqrt{{{{u}^{2}}-1}}}}}}\,=\text{arcsec}\,\left| u \right|+C$$

As you have suspected, the absolute value cannot be omitted. So the proposed answer only works for $x > 1$.

For a more precise computation, we may recognize the derivative of $\operatorname{arcsec}$ as

$$ \frac{d}{dz}\operatorname{arcsec}(z) = \frac{1}{z^2\sqrt{1-\frac{1}{z^2}}}. $$

This formula holds on $\{z\in\mathbb{C} : |z| > 1 \}$. Then we can compute the integral as

\begin{align}
\int z \operatorname{arcsec}(z) \, \mathrm{d}z
&= \frac{z^2}{2}\operatorname{arcsec}(z) - \int \frac{1}{2\sqrt{1-\frac{1}{z^2}}} \, \mathrm{d}z \\
&= \frac{z^2}{2}\operatorname{arcsec}(z) - \frac{z}{2}\sqrt{1 - \frac{1}{z^2}} + \mathsf{C}.
\end{align}

Again, this holds on all of $\{z \in \mathbb{C} : |z| > 1\}$, and in particular, this reduces to

$$ \int x \operatorname{arcsec}(x) \, \mathrm{d}x = \frac{x^2}{2}\operatorname{arcsec}(x) - \frac{\operatorname{sign}(x)}{2}\sqrt{x^2 - 1} + \mathsf{C} $$

for real $x$. Obviously this is different from what we get when we compute the first line in OP's question.


Chinese remainder theorem for divisibility problem Find the smallest $n$ so that:

$n$ divided by $3$ gives remainder $2$, $n$ divided by $11$ gives remainder $8$, $n$ divided by $27$ gives remainder $5$, $n$ divided by $47$ gives remainder $40$.

I'm not sure if it is $2813$, but that is the number I used to generate this problem. Anyways, I'm trying to figure out how to use the Chinese remainder to solve this problem step by step. I've seen "tricks" where they use the LCM and other things, but I need to know the way with the Chinese remainder theorem for my exam.

3 is   not needed, it is taken care of by 27.  Begins with

$$ 27 \cdot (-2) + 11 \cdot 5 = 1 $$
$$ 297 \cdot 22 + 47 \cdot (-139) = 1 $$

$$ 8 \cdot 27 \cdot (-2) + 5 \cdot 11 \cdot 5 = -432 + 275 = -157 $$
is $8 \pmod {11}$ and $5 \pmod {27}.$

$$ 40 \cdot 297 \cdot 22 + (-157) \cdot 47 \cdot (-139) = 1287041 $$
is $40 \pmod {47}$ and $-157 \pmod {297}$

Now, as you knew,
$$  1287041 \equiv 2813 \pmod {13959},  $$
so the smallest answer is your 2813.

Note
$$ 27 \cdot 11 \cdot 47 = 13959 $$
$$  1287041 = 92 \cdot 13959 + 2813   $$

LATER

Looking at the other answers, I can see how there is an advantage to choosing positive representatives in the middle stages. In this case
$$ -157 + 297 = 140 \equiv -157 \pmod {297}  $$
The next step changes to
$$ 40 \cdot 297 \cdot 22 + 140 \cdot 47 \cdot (-139) = 261360-914620 = -653260 $$
Well, not much better


Function in H(curl) $\cap$ H(div), but not in H1 it is well known, that for a non-convex domain $\Omega$ the space  $H^1(\Omega, \mathbb{R}²)$ is a proper subset of $H(curl) \cap H(div)$.
Here, $H(curl) = \{v \in L²(\Omega)², \nabla\times v = \partial_1v_2-\partial_2v_1 \in L²(\Omega)\} $ and $H(div)=\{v \in L²(\Omega)², \text{div } v = \partial_1v_1+\partial_2v_2  \in L²(\Omega)\}$.

I am looking for an example, where $ v \in H(curl) \cap H(div)$, but not in $H^1(\Omega, \mathbb{R}²)$. Furthermore, I want to have the constraint div v=0. Lets consider $\Omega=(-0.5,0.5)²\backslash [0,0.5]²$.

In Paper, there might be an example given by equation (5.2)
$v=\nabla \times (r^\frac{2}{3} \text{ cos}(\frac{2}{3} \theta-\frac{\pi}{3}))$. Is this one an example, I am looking for?

Thank you very much for your help.

The example you give from the paper is a little bit obscure to me. I propose another solution/perspective, which is quite general (but does not provide an explicit expression, at least in principle) for what concerns counter-examples in the context of $H(\text{curl}),H(\text{div})$. The main idea is to go back to the Laplace equation, which is very easy to deal with, and everything is known about it.

Consider the (weak) solution $u \in H^1(\Omega)$ of the boundary value problem

$$
\begin{cases}
\Delta u = 0 \qquad \text{in  } \Omega \\
u = g \qquad \text{on } \ \partial \Omega,
\end{cases}
$$
where $g \in H^{1/2}(\partial \Omega)$. It is well-known that $u \notin H^2(\Omega)$ if the domain is not enough regular (non-convex, with non-smooth boundary...).
Define now

$$ \mathbf{w} := \nabla u; $$
then $\mathbf{w} \in \mathbf{L}^2(\Omega) $ since $u \in H^1(\Omega)$, $\operatorname{curl} \mathbf{w} = \mathbf{0}$ since $\mathbf{w}$ is a gradient and $\operatorname{div} \mathbf{w} = \operatorname{div}(\nabla u)= \Delta u = 0$ by construction.

On the other hand, it can happen that $\mathbf{w} \notin (H^1(\Omega))^3$, because $u$ is not necessarily in $H^2(\Omega)$: see remark above.

The equation (5.2) from above is indeed the example, I was looking for. See for example the book of Peter Monk on Finite Element Methods for Maxwell's equation on page 76 for the explanation.


Find max/min of the following function Find the minimum e max distance (in $R^2$)
) between the point $Q =
(
3/
2
, −
3/
2
)$
and the set
$$B = \{(x, y) ∈ R^2
: yx = 1, x ≥ 0, y ≥ 0\}$$

In other words I have to find max /min points of the function

$${(x-3/2)^2 + (y+3/2)^2}$$

The set $B$ is clearly neither bounded nor convex. If I use Lagrange I can only find local min/max, but how do I show they are global? In a previous question :Does the following function admit a maximum? you suggested me to maximize/minimize the $x$ component and the $y$ component independently, but it is not clear to me if I can do it in this exercise as well. It seens to me that choosing $x=3/2$, which clearly minimizes the first component of the sum, would restrict the choice of $y$ as $yx = 1 $.

Is there any way to show that the point found using lagrange is a global min? Possibly without using the bordered hessian method?

(Note that you have a typo in your function, if the point is $(3/2,-3/2)$ then the second minus in your function should be a plus)

Using Lagrange multipliers is overkill here. In any case, Lagrange multipliers will give you critical points, which then you can check to see if they are the actual max/min.

Here, $x,y>0$ (otherwise $xy\ne1$), and you have that $y=1/x$. So your function becomes
$$
\left(x-\tfrac32\right)^2+\left(\tfrac1x+\tfrac32\right)^2,\ \ \ \ x>0.
$$
Now you can do this using one-variable calculus.

Lets say y->0 and x->$\infty$ then value of above expression -> $\infty$ (so yeah no maximum).

$(x-\frac{3}{2})^2 +(y+\frac{3}{2})^2$(You have made a mistake here or given the point wrong) = $x^2 +y^2 +\frac{9}{2} + 3(y-x) = z^2 +3z +\frac{13}{2}$.

(as $x=\frac{1}{y}$, $(y-\frac{1}{y})$=z, z can be any real number , the quadratic equation have its minimum at $\frac{-3}{2}$ which is the required answer)

So minimum value is $\sqrt{\frac{17}{4}}$


Intertwined differential equation Given $$\frac{dx(t)}{dt}=x(t)+y(t),\quad \frac{dy(t)}{dt}=x(t)-y(t),\quad \frac{dx(t=0)}{dt}=0$$. What is  $ x(\infty)/y(\infty)?$
I differentated first equation, do proper substitutions and obtained solutions as $$x(t)=A\cosh (t\sqrt{2})\quad y(t)=A(\sqrt{2}\sinh (t\sqrt{2})-\cosh (t\sqrt{2})$$. My evaluation of   $ x(\infty)/y(\infty)=1+\sqrt{2}$. My question is (1). Is the answer correct?
(2).Is there a simpler method of evaluating $ x(\infty)/y(\infty)$?

In the same spirit as @Aryadeva, you also have
$$x'+y'=2(x+y) \implies x+y=Ce^{2t}$$ that is to say
$$x=\frac{1}{2} \left(C e^{2t}+K\right)\qquad \text{and} \qquad y=\frac{1}{2} \left(C e^{2t}-K\right)$$ The condition for $x$ gives $C=0$; so $x=K$ and $y=-K$.

Your answer is totally correct, but I'm surprised nobody has mentioned this method yet and I think it is interesting.
One can represent this system using a matrix. Let $\underline{r}(t)=[x(t),y(t)]^{\mathrm{T}}$. Then,
$$\dot{\underline{r}}=\begin{bmatrix}
1 & 1\\
1 & -1
\end{bmatrix}\underline{r}$$
First let's define the matrix exponential: For a square matrix $\mathbf{X}$, define
$$\exp(\mathbf{X})=e^\mathbf{X}=\sum_{k=0}^\infty\frac{1}{k!}\mathbf{X}^k$$
It indeed follows the usual notion of the derivative, that is for a scalar $t$,
$$\frac{\mathrm{d}}{\mathrm{d}t}\exp(\mathbf{X}t)=\mathbf{X}\exp(\mathbf{X}t)$$
So, let
$$\mathbf{M}=\begin{bmatrix}
1 & 1\\
1 & -1
\end{bmatrix}$$
Then we can restate our system as
$$\dot{\underline{r}}=\mathbf{M}\underline{r}$$
We make the guess solution $\underline{r}(t)=\underline{c}\exp(\mathbf{M}t)$, with $\underline{c}$ being some constant vector. Then, we can see that
$$\dot{\underline{r}}=\mathbf{M}\underline{c}\exp(\mathbf{M}t)=\mathbf{M}\underline{r}$$
As required. Furthermore, we may notice that $\underline{r}(0)=\underline{c}\exp(0\mathbf{M})=\underline{c}$. Defining $\underline{r}_0=\underline{r}(0)$, we have
$$\underline{r}(t)=\underline{r}_0\exp(\mathbf{M}t).$$
It can be shown that this can equivalently be written as
$$\underline{r}(t)=c_1 e^{\lambda_1 t}\underline{u}_1+c_2 e^{\lambda_2 t}\underline{u}_2$$
With $\lambda_1,\lambda_2$ being the eigenvalues of the matrix $\mathbf{M}$ and $\underline{u}_1,\underline{u}_2$ being the respective eigenvectors. And of course $c_1,c_2$ are some constants to be determined by the initial conditions. One can compute the eigenvalues/eigenvectors of $\mathbf{M}$ to be:
$$\lambda_1=-\sqrt{2} ~;~ \lambda_2=\sqrt{2}$$
$$\underline{u}_1=\begin{bmatrix}
1-\sqrt{2}\\
1
\end{bmatrix} ~;~ \underline{u}_2=\begin{bmatrix}
1+\sqrt{2}\\
1
\end{bmatrix}$$
Which leads to
$$x(t)=d_1 e^{-\sqrt{2}t}+d_2 e^{\sqrt{2}t}$$
$$y(t)=c_1 e^{-\sqrt{2}t}+c_2 e^{\sqrt{2}t}$$
With of course $d_1=(1-\sqrt{2})c_1$ and $d_2=(1+\sqrt{2})c_2$. What happens when we impose the condition $\dot{x}(0)=0?$ First we compute:
$$\dot{x}(t)=-\sqrt{2}~d_1 e^{-\sqrt{2}t}+\sqrt{2}~d_2 e^{\sqrt{2}t}$$
So,
$$\dot{x}(0)=\sqrt{2}(d_2-d_1)$$
Thus we require $d_2=d_1$ or $(1-\sqrt{2})c_1=(1+\sqrt{2})c_2$. Let $d_1=d_2=C/2$ thus
$$x(t)=C\cosh(\sqrt{2}t)$$
$$y(t)=C\left(\sqrt{2}\sinh(\sqrt{2}t)-\cosh(\sqrt{2}t)\right)$$
So
$$\lim_{t\to\infty}\frac{y(t)}{x(t)}=\lim_{t\to\infty}\frac{\sqrt{2}\sinh(\sqrt{2}t)}{\cosh(\sqrt{2}t)}-1=\sqrt{2}-1$$
$$\left(\lim_{t\to\infty}\frac{y(t)}{x(t)}\right)^{-1}=\lim_{t\to\infty}\frac{x(t)}{y(t)}=\frac{1}{\sqrt{2}-1}=1+\sqrt{2}.$$


Find stopping times $S$ and $T$ of Brownian motion $B$ such that $S\le T$, $E[S]$ finite, and $E[B(S)^2]>E[B(T)^2]$ This is exercise 2.12 of Peter Morters and Yuval Peres' book Brownian Motion:

  Find two stopping times $S\le T$ with $E[S]<\infty$ such that $E[B(S)^2]>E[B(T)^2]$.

I considered about deterministic stopping times, but it does not work. And by Wald's Lemma, it seems we need to find a stopping time $T$ with $E[T]=\infty$. While Wald's second lemma says $E[B(S)^2]=E[S]$. May I get some hint about it?

Your answer is correct. In general if $S=\inf\{t:|B(t)|=a\}$ and $B(0)=x$, then $\mathbb{E}[S]=a^2-x^2$. You can find the expectation of your $S$ by shifting but maybe just make $S=\inf\{t:B(t)\in\{-1,1\}$ instead.


Command like DedekindEta for Sage In Mathematica we have the command DedekindEta for computing values of the Dedekind eta function.

Is there an equivalent command for Sage or Maple?

I believe that this reference manual section may be what you're looking for, and in particular (as a modular form, I guess) the function you seek seems to be here, roughly.


What calculation to use to work out parts per 100 at a particular speed. At work we have product mixing into a tank.
I need to write (and understand) a program that will mix these products to the correct ratio in the tank.

For example; it may be a simple at 98.6% product A and 1.4% product B.
Product A flows into the tank at 2000L/h while product B only flows at 0-50L/h.

How do I calculate the speed in which B should flow?

I would like to understand the calculation as there could be different ratios including more products.

EDIT*
After reading Hennings answer and believing it to be correct (and it way we'll still be) I may need some clarification.

Although it makes sense, what confuses me is that it may not take the full hour for product A to reach its set point, it may only take 10 or 15 minutes. Maybe it could even take 1.5hrs.

Does the same logic apply or does it then need to be multiplied/divided by time to yield the correct answer.?

EDIT*
Before wasting time reading further do know this:
Although flow may not stay the same, I will not be using a set amount in my equation, it will take the REAL value and calculate as it goes. In other words it would have no fixed numbers, it would look like this:

B_flow = (A_flow/A_level) * B_level

Ok more info:
So this is the process as I know it.

Product A is the main ingredient, it will always have a bulk percentage of 90+%. This will flow at a variable rate, somewhere around the mark of 2000L/h.

Product B - F are smaller products that will flow at a rate of 0-50L/h - this is the variable that I need calculated.

There is many possible end-products that may be produced, but here is a very basic example:

Product A flows in at 2000L/h filling 98% of the tank over the course of 15 minutes (this is also an unknown variable, depending on flow rate and the overall wanted percentage of the product).

I need to calculate what rate any other product would need to flow at in order to have its respective amount in the tank at the end of the same time bracket (ie. 15 minutes).

So if I wanted 98% product A and 2% product B. How would B's flow rate be calculated?

Please let me know if there is more information required.

In one hour, $98.6$ liters of A flow into the tank $\frac{2000}{98.6}$ times. This is the same number of times $1.4$ liters of B should flow in. So the desired flow rate of B is
$$ \frac{2000}{98.6}\cdot 1.4 = 28.4 $$
liters per hour.

(This assumes the desired percentages are volume ratios, of course. If they are mass ratios, and A and B have different densities, you need more data).

I am assuming that the products are fluids and that they are getting into the tank through tubes.

We define flux of a material is the amount of material getting into the tank per unit time. If $\rho$ is the density of the material, and $v(r)$ is its speed through the pipe, then the flux is
\begin{equation}
\phi = \int_0^R \rho v(r) 2\pi r dr
\end{equation}
If you now have two materials and their flux in the tank should be such that $\phi_1 = \alpha \phi_2$ then,
\begin{equation}
\int_0^{R_1} \rho_1 v_1(r) 2\pi r dr = \alpha\int_0^{R_2} \rho_2 v_2(r) 2\pi r dr
\end{equation}
Since $\alpha$ is fixed, by your requirements, $\rho_1$ and $\rho_2$ are material parameters, you can change either $R_1$, $R_2$ or $v_1$, $v_2$ or both to match the two sides of the equation. For a tube of circular cross section, we can assume $v$ to be of the form,
\begin{equation}
v(r) = \frac{G}{4\mu}(R^2 - r^2),
\end{equation}
where $G$ is the pressure gradient driving the flow and $\mu$ is the dynamic viscosity of the material.


How to prove determinant for second order tensors? I was asked in an exam to prove the following expression,
$$ \det(A) = \dfrac16 \epsilon_{ijk}\epsilon_{lmn}A_{il}A_{jm}A_{kn}$$
I know that it can be written in the following form,
$$ \det(A) = \epsilon_{ijk}A_{i1}A_{j2}A_{k3}$$
I don't know how to proceed beyond it.

Will someone please help me, Thank you.

Just start with right hand side and expand it. There are 6 dummy indices (i,j,k,l,m,n). So if you expand them as per summation - you will get to the answer.


What graph with 7 vertices that doesn't contain $K_3$ as subgraph has the maximal number of edges? is there graph with vertices 7 that not contain $K_3$ as subgraph and have biggest edge?

is this question trying to ask to remove all triangle in K7 graph?

for this problem , is this possible to solve like this

$\frac{7.6}{2}-\frac{3.2}{2}=18 edges$
but ofourse it contain triangle(?)

here is the possible graph that i can think of..

edit: i think once again that, biggest one is bipartite graph, such as $K_{3,4}$ with 12 edges and 7 vertices .(?)

Indeed, $K_{3,4}$ is correct.
To prove this, let's solve for $n$ vertices.
First note that any bipartite graph is triangle-free (as it's an odd length cycle).
I'll show that the graph with the most edges is bipartite:

Let $G = (V,E)$ be a triangle-free graph such that $|V|=7$ and let $v\in V$ such that $\deg_G(v)=\varDelta_G$, the vertex with the highest degree in $G$.
Create the following bipartite graph $H=(L,R,E{'})$:
$L=N_G(v)$, the set of neighbors of $v$, $R=G/N_G(v)$, $E^{'}=\{(v_i,u_i)|v_i\in L,u_i\in R\}$, i.e., the complete bipartite graph between the neighbors of $v$ to the rest of the vertices.
Now, let $u\in V$  be a vertex in $G$. If $u\in N_G(v)$ then $\deg_H(v)=|N_G(v)|=\deg_G(v)\ge \deg_G(u)$.
If $u\notin N_G(v)$ then $\deg_G(v)\le |V/N_G(v)|$ as $u$ can't be connected to another vertex in $N_G(v)$ for it would create a triangle. But, $\deg_H(v)=|V/N_G(v)|$ so $\deg_H(u)\ge \deg_G(u)$.

We got that for each vertex $u\in V$ it holds that $\deg_H(u)\ge \deg_G(u)$, so $$E(H)=\frac{\sum_{u\in V} {\deg_H(u)}}{2}\ge \frac{\sum_{u\in V} {\deg_G(u)}}{2} = E(G)$$

As we got that every triangle-free graph has a bipartite graph with at least as many edges, it is sufficient to take the bipartite graph with $n$ vertices with the highest number of edges.

So the biggest graph is obviously of the form $K_{k,l}$ where $k+l=n$. Let $\alpha$ be the unique number such that $k=\frac{n}{2} +\alpha, l=\frac{n}{2}-\alpha$ and we got that the number of edges is $\frac{n^2}{2}-\alpha^2$ so to maximize it we must choose $\alpha = 0$ when $n$ is even and $\alpha = \frac{1}{2}$ when $n$ is odd, and the number of edegs is $\lfloor \frac{n^2}{4} \rfloor$.

If we let $n=7$, we get that the maximal graph is indeed $K_{\frac{7}{2} + \frac{1}{2}, \frac{7}{2}-\frac{1}{2}}=K_{4,3}$ with $12$ edges.

This graph $K_{\lceil n/2 \rceil, \lfloor n/2 \rfloor}$ has a name: the Turán graph. You can read more about it here.


Surface Area of a Hypercube I am interested in computing the surface area of an $n$-dimensional hypercube and am interested in a reference or an answer which defines the notion of surface area for higher dimensional polytopes as I am trying to compute the surface area of an infinite family of duoprisms and knowing the surface area of an $n$-dimensional hypercube would be very useful to my understanding.

That is, what is the surface area of an $n$-dimensional hypercube with side length $s$, and how can you think about surface area of higher dimensional polytopes in general?

EDIT: In regarding as to whether I am referring to "surface area" or "surface volume", I am interested in understanding any $k$-dimensional version of surface hyper(area/volume) for an $n$-dimensional polytope. I think it makes sense that the $n$-dimensional version of this quantity is the volume, the $(n-1)$-dimensional version would be surface "area", and there are $(n-2),...,1$-dimensional versions of this idea. Is there any way to understand this with differential forms perhaps?

Expand the polynomial $(x+2)^n$.  For example, $(x+2)^3=1x^3+6x^2+12x+8$.  Read off the coefficients: A cube has 1 cube, 6 faces, 12 edges and 8 corners.

The answer is $$2 \frac{d}{dx}(V_{n}(L))$$
or, equivalently:
$$2nx^{n-1}$$
where $V_{n+1}(L)$ is the Volume of an n-dimensional sphere.

As to why, think about slightly increasing the size of a cube for example. How much new volume has been created? it's going to approach 3x$^2$ times the tiny change in x, because each face of the cube will be expanded, and all the other bits will converge to zero.

It's hard to explain, but here's a good video to watch to get a feel for what I'm talking about.

https://www.youtube.com/watch?v=S0_qX4VJhMQ

However, this only covers half of what the surface area of these shapes is, because it's only increasing with respect to half of the total sides. So, multiplying by two will give your final answer.

This similar tactic can be used to find the surface areas of other shapes too!
for instance, if you imagine slightly increasing the radius of a sphere, and make them both concentric, then subtract the overlapping area, what you're left with is a sort of shell, which as the slight nudge becomes smaller, becomes close to the true surface area, times the size of that small nudge.


Using the quotient rule to find $f'$ where $f(x) = \dfrac{6x}{5 \ln(x)}$ Below is a problem and the process by which the first derivative is found:

I used the quotient rule but did not get the same result. Can someone explain this to me?

Here is what I got after using the quotient rule:

Your answer is the same as your professor's. $\frac{30\ln{x} - 30}{(5\ln{x})^2} = \frac{30\ln{x} - 30}{25(\ln{x})^2} = \frac{6\ln{x} - 6}{5(\ln{x})^2}$. Splitting the numerator and simplifying slightly, we get $\frac{6}{5(\ln{x})} - \frac{6}{5(\ln{x})^2} = \frac{6}{5}\left((\ln{x})^{-1} - (\ln{x})^{-2}\right)$.

Just because answers look different doesn't mean they are. If your answer looks different from someone else's, you should compare your steps against theirs, not assume that one answer is right and the other is wrong.


Show that $\prod _{_{r=1}}^{\infty }\left[1+\frac{x}{r}\right]e^{-x/r}$ converges for all finite x (except for the zeros of 1+x/r) Background: This is Arfken et all Mathematical methods 12.5.8

There is a hint to write the nth factor as $1+a_n$

An infinite product $\prod^\infty(1+a_n)$ diverges or converges with $\sum (a_n)$. Therefore the first term will diverge by comparison with the harmonic, and the second term will converge but I don't know how to proceed with the combined product.
Also, what is meant by the zeros of 1+x/r?

For $r>|x|$ take logarithms
$$\ln\left[\left(1+\frac xr\right)e^{-x/r}\right]
=-\frac xr+\sum_{n=1}^\infty(-1)^n\frac{x^n}{nr^n}=-\frac{x^2}{2r^2}
+O((x/r)^3).$$
This means that the sum of $\ln\left[\left(1+\frac xr\right)e^{-x/r}\right]$ over all these $r$ is bounded by a multiple of the convergent
series $\sum1/r^2$, so it converges.

Stealing the credit off of Lord Shark's answer and putting it in a form that I understand: Instead of taking the ln just expand the exponential into it's series and multiply the product out into a sum:

$$\prod _{_{n=1}}^{\infty }\left[1+\frac{x}{n}\right]e^{-x/n}=(1+\color{red}{\sum_n \frac{x}{n}}+\sum \frac{x^2}{n_1 n_2}\dots)(1-\color{red}{\sum\frac{x}{n}}+\sum\frac{x^2}{2!n^2}\dots)$$
The red terms in the product cross multiply with 1 and sum to zero, so we get a sum of convergent sums: $$(1+\sum\frac{x^2}{2!n^2}+\sum \frac{x^2}{n_1 n_2}\dots)$$ with each of these sums and higher orders converging with comparison to $\sum\frac{1}{n^2}$ so the final answer that they converge was correct.


Positive Integer Cover Who knows whether or not union of the following 5 equations for $x,y=0,1,2,3,4,...$ covers natural numbers fully after a certain integer?
$A=10xy+x+y$ (non zero $x$ and $y$)
$B=10xy+7x+3y+2$
$C=10xy+9x+9y+8$
$D=10xy+x+3y$ (non zero $y$)
$E=10xy+7x+9y+6$
(Tips: we know the set $A\cup B\cup C$ will never cover $\Bbb N$ fully, whether one by one correspondence of the elements of the set $D\cup E$ will help it to cover $\Bbb N$ fully after a certain integer?)


Is it mathematically possible to solve this problem? $$\vec{y}= \log{\frac{\vec{x}}{\sqrt[3]{x_1 \cdot x_2 \cdot x_3}}}$$
Here, $x$ and $y$ are always vectors of length 8 and all elements of $x$ are greater than $0$. $x_1$, $x_2$, and $x_3$ are the first three elements of $x$; is it mathematically possible to determine these elements given you only have $y$?  I should note that $x$ is known for some of my dataset, which could be useful to find a solution for the other portion of the data where $x$ is unknown.

We have
$$x_i=\sqrt[3]{x_1x_2x_3}\,e^{y_i}.$$
Multiplying the equations for $i=1,2,3$, you get the compatibility condition
$$1=e^{y_1+y_2+y_3}.$$ Without it, no solution exists.
Then using  ratios,
$$x_2=x_1e^{y_2-y_1}$$
$$x_3=x_1e^{y_3-y_1}$$ and you are free to choose $x_1$.

I think what you are asking is that for vectors:
$$y=\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix},x=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$$
then is there a way to calculate:
$$y=\begin{pmatrix}\log\frac{x_1}{(\prod x)^{1/3}}\\\log\frac{x_2}{(\prod x)^{1/3}}\\\log\frac{x_3}{(\prod x)^{1/3}}\end{pmatrix}$$
notice that:
$$y_1+y_2+y_3=\log(x_1x_2x_3)-3\log(\prod x)^{1/3}=\log(\prod x)-\log(\prod x)=0$$
and since we know $|y|=8$ we have the system:
$$y_1+y_2+y_3=0$$
$$y_1^2+y_2^2+y_3^2=64$$


If graph G has 2n+1 vertices where every vertex has degree at least n, G is connected So I need to show that each pair of vertices is joined by a path, or show that there is a vertex joined to every other vertex by a path.

But.. how?

Let $u,v$ be any two vertices. If they aren't connected directly, then $u$ has at least $n$ neighbours and $v$ has at least $n$ neighbours. If all mentioned vertices were distinct, we would get $2n+2$ vertices. Can you conclude from here?

Let $e$ be the number of edges in $G$.Use Handshaking theorem $\sum_1^{2n+1}deg(v)=2e$

$\rightarrow[n(2n+1)/2]\le{e}$..............................................$(1)$

Use the fact that a graph with $n$ vertices that contains more than $(n-1)(n-2)/2$ edges is connected.Replacing $n$ by $2n+1$ in this, we see that $(1)$ holds true, hence $G$ is connected.


Absolute value of complex number in Numerical Recipes In numerical recipes in C, absolute value of complex number $a+ib$ is implemented as $b*\sqrt{1+\left(\frac{a}{b}\right)^2}$ if $|b|$ is greater than $|a|$ and $a*\sqrt{1+\left(\frac{b}{a}\right)^2}$ if $|a|$ is greater than $|b|$. Does the numerical error in the absolute value calculation reduce when doing this. If so, how?

One point in favor of these formulas is that if you just do $\sqrt{a^2+b^2}$ there's a risk of overflow or underflow when you square the $a$ and $b$. The NR formulas will avoid this right up (or down) to the limit of the floating-point representation.

That is a very strange way of defining the 'absolute value of a complex number' (should be called the modulus of the complex number).

Really, you don't need those conditions. Even if you used the 'wrong' formula, you will still get the exact same answers. In fact, the given formulas introduce conditions about $a$ and $b$ being non-zero, which is most certainly unnecessary (since the modulus can still be computed even if either $a$ or $b$ are zero).

Also, I don't understand what you mean by the 'numerical error' of the modulus. There is no numerical error, the modulus is not an approximation. It gives an exact value.


Locus of points whose distance between two 3D lines is at a given ratio I'm aware that a very similar question has been asked before (Question 888147) but this one is its 3D version, which doesn't seem to easily generalize.

I'm given two lines $\ell_1$ and $\ell_2$ that intersect at the origin, and I'd like to know what the locus of the points such that $$\text{d}(X,\ell_1)=k\cdot\text{d}(X,\ell_2)$$ looks like.

I know that if a point $X$ has this property, multiplying it by a scalar yields another point with it, which leads me to suspect the shape I'm looking for is some kind of double cone (or two of them, as analogy with the 2D case makes me suspect), but I'm clueless on how to prove this.

Since I don't even have a formula for distances between points and lines in 3D in terms of the coordinates, I really don't know where to start.

A formula for the distance from a point to a line shows that the condition $\operatorname d((x, y, z), l_2) = k \operatorname d((x, y, z), l_1)$ gives a quadric surface. The quadric is symmetric wrt the plane containing $l_1$ and $l_2$. If we already know that the cross-section by this plane is a pair of lines and also know that the quadric is not a pair of planes for $k \neq 1$, it follows that the only possibility is a cone.

If we choose a coordinate system such that $l_1$ goes through the origin in the $x$ direction and $l_2$ goes through the origin in the direction $(\cos \phi, \sin \phi, 0)$ and rotate the lines by the angle
$$-\arctan(k^2 - \cos 2 \phi, -\sin 2\phi)/2$$
around the $z$ axis from the positive $x$ direction to the positive $y$ direction to eliminate the $x y$ term, the equation of the cone becomes
$$\frac {1 - k^2 + \rho} 2 x^2 + \frac {1 - k^2 - \rho} 2 y^2 +
 (1 - k^2) z^2 = 0, \\
\rho = \sqrt {1 + k^4 - 2 k ^2 \cos 2 \phi}.$$

For any couple of distinct lines, it is always possible to find some coordinates transformation such that the 2 lines, in the new coordinate system, lies over 2 of the 3 axis, let's say over x and y axis.
Now, the locus of points at a distance $d$ from a line is a cyclinder, so we have the intersection of 2 cylinders.
The intersection looks like a kind of "bent" circle.
Varying the diameter of one cylinder and keeping the ratio with the other cylinder constant, we got the projection from the origin of this bent circle.
So, yes, in the end we have something like a double cone.
Going back to original coordinate system, we retain a cone like shape.


How many ways to order vectors in $\{0,1\}^n$? How many different rankings can be produced for the vectors in $\{0,1\}^n$ that also respect the usual $\geqq$ ordering of vectors (defined below)?

I want to produce a complete ordering where, for $x\neq y$, if $x_i \geq y_i$ for all $i=1,\dots,n$ then $x$ is greater than $y$ according to that new vector ordering.

This leaves some freedom in the ordering where I could have $(1,0,0) > (0,1,1)$ as a possibility. $(1,0,0) < (0,1,1)$ is also allowed.

Example: For $n=2$, there are two rankings.

$(1,1) \geq (0,1) \geq (1,0) \geq (0,0)$ and $(1,1) \geq' (1,0) \geq' (0,1) \geq' (0,0)$.

You have to first rank the entries of the vectors. As in the first ranking of your example the second entry has 'higher importance' for the order (-> higher priority for the actual ranking of the whole set) than the first entry.

In your second ranking it is the first entry which dominates for the purpose of ranking.

How many possibilities are there to rank the $n$ entries there are?

The answer to this (and your) question is $n!=n(n-1)\cdots$


Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$ Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$

I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{ \frac{h}{1}}
= \frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}
$$
In the numerator maybe I can continue with a difference of squares, but I'm a little confused.


Understanding Lagrange's Theorem (Group Theory) I am beginning with Abstract Algebra and I'm trying to understand Lagrange's Theorem. The theorem reads

  For any finite group $G$, the order of every subgroup $H$ of $G$ should divide the order of $G$.

It seems simple and I've used it to solve some exercices but I believe I'm missing the essence of it. Is there an example that can help me understand it better, or visualize it? Is there a geometric interpretation?

The left-multiplication map $x\mapsto ax$ is bijective on $G$; its injectivity follows from the cancellative property of the group's operation, $ah=ag\iff h=g$, and overall bijectivity is a consequence of the fact that it has an inverse map, $x\mapsto a^{-1}x$. I like to view a subgroup $H\le G$ as a "puck" and the overgroup $G$ as a "air-hockey table" on which $H$ resides, and to move $H$ around we apply left multiplication by various elements. If you left-multiply by an element $a\in H$, you have not moved the puck at all since $a\in H\iff H=aH$.

Every element $g\in G$ is in some coset, or left translate, of $H$ - in particular, $g=ge\in gH$ since we know that $e\in H$. Thus, the collection of all translates (possible places for the puck to be positioned in) of $H$ "cover" the entire air-hockey table. It remains, then, to investigate the nature of the overlaps between positions, i.e. the intersections of distinct cosets. Here is the proof that cosets that overlap nontrivially must in fact be identical, put into visual form:

$\hskip 0.6in$

This means the cosets of $H$ partition the group $G$. As left multiplication is bijective, every coset is the same size, so each "looks" the same from the viewpoint of cardinality. Continuing with the idea of an air hockey table, this tells us the puck positions tile it, so we have something like:

$\hskip 1.2in$

The most fundamentally basic meaning imputed to multiplication of natural numbers is the following: if Alice has $n$ bags each containing $m$ apples, then she has $n\times m$ apples total. Similarly, our group $G$ is covered by some number $[G:H]$ of disjoint cosets, each containing $|H|$ elements, so $|G|=[G:H]\times|H|$. Note that this is even true on the level of arbitrary infinite cardinals. Thus, $|H|$ is a divisor of the order $|G|$: Lagrange's theorem.

The converse is not globally true: not every divisor $d$ of $n=|G|$ corresponds to a subgroup $H\le G$ of size $|H|=d$. Sylow theory however yields a local version of a converse: for every prime power $q=p^r$ that is a divisor $q\mid n$, there is a $p$-subgroup $H$ of size $|H|=q$.

For example If order of G is 12 then we can find only subgroups of order 1,2,3,4,6 and 12 which are divisiors of 12

It means that we can not find a subgroup of  other order except above orders


a simple question about topological space X is a topological space, $ A\subseteq B\subseteq X $, if $A$ is a nowhere dense subset of $B$ , then $A$ is a nowhere dense subset of the whole space $X$?

Is this right? I thind it's right, but I can't prove it. Maybe we need to add some contions: e.g. $B$ is a closed subset of $X$, and/or $A$ is (also) a closed subset of $X$.

If $A$ is nowhere dense in $B$, then $\operatorname{cl}_BA$ has empty interior in $B$, and therefore $B\setminus\operatorname{cl}_BA$ is a dense open set in $B$. There is then a $U$ open in $X$ such that $U\cap B=B\setminus\operatorname{cl}_BA$. Let

$$V=U\cup(X\setminus\operatorname{cl}_XB)\;;$$

clearly $V$ is open in $X$, and you should have no trouble showing that $V$ is dense in $X$. Finally, $V\cap A=\varnothing$, so $A$ must be nowhere dense in $X$.

If $\overline{B}$ has empty interior, then since $\overline{A} \subseteq \overline{B}$, $\overline{A}$ must have empty interior also (otherwise contradicting the nowhere-denseness of $B$).


is complex number under absolute value a group? I have just started going over abstract algebra.

One of the question is

$*$ is defined on $\mathbb C$ such that $a*b=|ab|$

I tried to check three axioms : 1) Associativity 2) identity 3) inverse

I found out with long computation that Associativity works.

For identity, $a*e=e*a=a$ , where $a*e=|ae|=a$

I am quite confused how to find out the value of e, since this is a complex number.

Can anyone show me why this is a group or not a group?

It’s not a group as it fails to have an identity.

Hint. Look at $a = -1$ and the range of the defined operation “$*$”.

It is not group. Problem is with a=-s s$\in R^+$.


Prove that $\forall n \in N $ , $ \sum_{i=1}^n i = \frac{n(n+1)}{2}$, counting in two ways the number of shaded squares in the diagram Prove that $\forall n \in  N $  $$ \sum_{i=1}^n i = \frac{n(n+1)}{2}$$ counting in two ways  the number of shaded squares in the diagram.

I have been thinking about this, but I can't understand how prove it counting the shaded squares. Can i assume that $ 1+2+3+...+n $ is the number of shaded squares?
I can't see it clearly.

The left-hand side sums over the numbers of shaded squares per row. The right-hand side is the number of shaded squares in an $n\times(n+1)$ rectangle; exactly half are shaded, by an order-$2$ rotational symmetry.

Here's an idea for you, which seems to explain the famous legend about how Gauss came up with a solution at his school:
Call the sum of all shaded squares $\;S\;$ , and we now count them in two ways:
(*) Number of shaded squares by row, from to to bottom:
$$\;S= 1 +2 +3+\ldots+n\;\;(*)$$
(**) And now let's count the number of shaded squares per column from left to right:
$$S=n+(n-1)+(n-2)+\ldots+2+1 \;\;(**)$$
Now put both sums above one over the other and sum them up column after column:
$$\begin{cases}S=1&+&2&+&3&+\ldots+&(n-1)&+&n\\{}\\
S=n&+&(n-1)&+&(n-2)&+\ldots+&2&+&1\\{}\\
--&--&--&--&--&--&--&--&--&--&--\\{}\\
2S=(n+1)&+&(n+1)&+&(n+1)&\ldots&(n+1)&+&(n+1)\end{cases}$$
You got, in fact, $\;2S=n(n+1)\;$ ...and now divide by two and we're done.


directional derivatives in (0,0) I've this function : $f(x,y)= \dfrac{(1+x^2)x^2y^4}{x^4+2x^2y^4+y^8}$ for $(x,y)\ne (0,0)$ and $0$ for $(x,y)=(0,0)$

It's admits directional derivatives at the origin?

A directional derivative needs a direction. Maybe you travel vertically straight through $(0,0)$, along the line $x=0$. Then your function is constantly $0$, so of course that particular directional derivative exists.

Otherwise, travel in the direction of the line $y=kx$. Then away from $(0,0)$ your function is $$f(x)=\frac{k^4(1+x^2)x^6}{x^4+2k^4x^6+k^8x^8}=\frac{k^4(1+x^2)x^2}{1+2k^4x^2+k^8x^4}$$ which is either identically $0$ if $k=0$, or otherwise is quadratic in nature as $x\to0$. In both cases, the derivative is $0$ at $x=0$.

So all directional derivatives for this function exist at $(0,0)$, and they are all equal to $0$.

Note though what happens if you approach $(0,0)$ along the nonlinear path $x=y^2$...

A function admits directional derivative at a point if its gradient $\nabla{f}$ exists at that point. The gradient of your function is given by,
$$\nabla{f}=\left(\begin{array}{cc} -\frac{2\, x\, y^4\, \left(2\, x^6 + 3\, x^4 - 2\, y^4 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} & \frac{4\, x^2\, y^3\, \left(x^6 + x^4 + x^2 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} \end{array}\right)$$
which doesn't exist at $(x,y)=(0,0)$, i.e. the origin, since you have $\frac{0}{0}$ which is undefined.


Solution to exercise 1.3H of Vakil's 'Foundations of Algebraic Geometry' I have been trying to solve exercise 1.3H of Ravi Vakil's AG notes for quite a while now, and I just cannot seem to do it. The question asks me to prove (from the concrete definition, presumably without much heavy categorical machinery) that the tensor functor is right exact.

This question has been asked before (see Proving that the tensor product is right exact) but the answers given only contain hints to help with finding the solution, and unfortunately I cannot seem to get the answer even with these hints, no matter how hard I try.

I feel as if I am not really grasping what equality means for tensor products, and I really feel I need to see a (mostly completed) proof to rectify this.

Edit: Rings are assumed commutative.

Let $A\to B\to 0$ be an exact sequence of modules; we must show $A\otimes M\to B\otimes M\to 0$ is exact. An element of $B\otimes M$ is of the form $\sum b_i\otimes m_i$, where the representation is unique up to bilinear rewriting. But this is the image of $\sum a_i\otimes m_i$, where each $a_i$ maps to $b_i$.

That's pretty much it. To make sure this wasn't too facile, let's check whether such a proof would incorrectly prove left exactness, supposing now $0\to A \stackrel{f}{\to} B$ were exact; if $\sum f(a_i)\otimes m_i=0$, there's no reason we should have each $f(a_i)=0$. So there's no problem.

Here's a less hands on proof that still isn't categorically very sophisticated: in the situation of the first paragraph we want to prove $A\otimes M\to B\otimes M$ is an epimorphism, so that any two maps $g,h:B\otimes M\to C$ which are equal on $A\otimes M$ are equal. But maps out of the tensor product are equal if and only if the represent the same bilinear map out of $B\times M$; since $A\times M\to B\times M$ is a surjection of sets, the claim follows.


Prove that the $\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$ Can anyone share a way of proving the following limit? I'm looking for an idea that can be used at the Calc I level, using the epsilon delta formal definition.

$$\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$$

Thanks.

Using the binomial theorem, you can show the following extension to the triangle inequality
$$
\left|\sqrt[n]{x_n}-\sqrt[n]{a}\right|\le\sqrt[n]{|x_n-a|}
$$
which allows a simple proof of the limit.

Proof: Assume first $a≤b$. Then expand $(\sqrt[n]a+\sqrt[n]{b-a})^n$ by the binomial theorem and only keep the two pure powers terms $a+...+(b-a)$ to get the inequality $(…)^n≥b$, or $\sqrt[n]{b-a}\ge \sqrt[n]b-\sqrt[n]a$. Then change the roles of $a$ and $b$ to obtain the other case of the absolute values.

The function $x \mapsto x^{1/n}$ is continuous everywhere.


Inner product in Sobolev spaces Many books state the following:

Let $u, v \in W^{1,2}(\Omega)$ (a Sobolev space), the scalar product is:

$$u \cdot v = \int_{\Omega} uv \; dx + \int_{\Omega} \nabla u \nabla v \; dx $$

with $x = (x_1,x_2)$

Is that a definition or it is derived from any relationship? If is a relationship how to find it?

Is that also an inner product?

An answer I've heard follows:

One way (there are others) to analyze solutions in modern PDE analysis is to:

Specify a space of functions with certain regularity (e.g. all functions on a bounded domain that have continuous derivatives up to second order)
Imagine a PDE as an operator on this function space.

We then use tools from functional analysis to say something about whether a solution exists (maybe play some game about examining how sequences in the domain get mapped to the range under the operator defined in 2).

We need to be careful about the spaces and operators we define if we want to play this game though. Consider our first guess for how we might want to play:

Consider continuous functions on a bounded interval under the sup norm
Consider the derivative operator

If we consider the sequence $f_n(x)=\sin(nx)$, The derivative operator will give the following bound:
$$f'_n(x) = n \cos(nx) \\ ||f'_n|| \leq n ||f_n||$$
Which is to say, the derivative operator is not a continuous operator (I skipped a step, but stated the punchline). Continuous operators over spaces are things we need in order to play our game though -- think about what they do: if we have a sequence in the domain, a continuous operator allows us to say something about how the limiting object in the domain passes to the range.

So we'd like to manufacture complete spaces of functions over which we can define continuous maps. Sobolev spaces $W^{k,p}$ are one technology we have for this. What we do: control (via $L^p$ norm) how badly the first $k$ derivatives of a function can behave. These norms are created so that we can create continuous operators over these spaces.

That $W^{1,2}$ is Hilbert is coincidental, in general these things are Banach spaces.


Functor that does not preserve monic and epic In Jacobson’s book BAII, he gave two exercises to show that there exists functor that does not preserve monic or epic.

Ex-1.Let M and N be monoids as categories with a single object. Show that in this identification,a functor is a homomorphism of M into N.

Ex-2.Use Ex-1 to construct a functor $F$ and a monic(epic) $f$ such that $F(f)$ is not monic(epic)

I have solved Ex-1. However, I don’t know how to construct concrete example to solve Ex-2.
I do hope someone can give me some hints. Thank you very much!

Hint: Let $M$ be a free monoid.

 A simple example is with $M=(\Bbb N,+),\ \,N=(\{0,1\},\max)$ and $F(m)=\min(1,m)$.
 $1$ is cancellable in $M$ but not in $N$.

Hint: Find a monoid, and a homomorphism which takes an element with a left cancellative property to an element that doesn't.


Let P and Q represent formulas. Would stuff like $(P \wedge \neg P) \models Q$ make sense? I confess, I am in a state of total confusion right now. And I am still struggling to grasp the underlying distinction between the normal material implication, $\Rightarrow$, and the notion of semantic consequence, $\models$.

I've seen stuff like $(P \wedge \neg P) \models Q$ appear in a book I'm currently reading (A First Course in Logic, by Shawn Hedman), but at the end of the day, it had me wondering what the difference is between material implication and semantic consequence. I mean, really.

We all know that the sentence $(P \wedge \neg P) \Rightarrow Q$ is a tautology. And in a material sense (only thinking in terms of truth values and not in terms of how the two formulas P and Q are connected semantically), this is totally acceptable to me. But to write stuff like $(P \wedge \neg P) \models Q$  blows my entire understanding of the difference between $\Rightarrow$ and $\models$ right out the window. I thought the double turnstile symbol is only to be used in a sense that has more to do with the underlying meaning/interpretation behind the sentences? Like, if we say $P \models Q$ (where P and Q are sentences), then wouldn't it mean that we can see a clear logical connection which allows us to accept that Q follows from P?

If translated into words, the sentence "formula Q is a semantic consequence of $P \wedge \neg P$" is incomprehensible to me. It doesn't have that logical connection that I thought must accompany every use of the double turnstile symbol (for instance, I can easily accept $(P \wedge Q) \models P$, because the logical connection is there, after all if its stated that P and Q are both true, then logically, P must be true). If no logical connection is required for the use of $\models$, then how is it different from the normal material implication?

It feels like I'm missing something crucial, like a bigger picture, or a wider generalization of what the double turnstile symbol represents.

"We all know that the sentence (P∧¬P)⇒Q is a tautology."

I certainly don't know this and gasp when I see statements like this made.  Anyone who claims to is also, simply put, wrong.  Why things work that way helps to illuminate things.

To say that something is a tautology is to imply that something is an object-level sequence of symbols.  If you check your definition of a well-formed formula, or whatever equivalent term gets used, (P∧¬P)⇒Q is not well-formed.  Thus, (P∧¬P)⇒Q is not a tautology, because a tautology is by definition a well-formed formula.  ((P∧¬P)⇒Q) is a tautology.

On the other hand, (P∧¬P)⊨Q is a meta-language construct.  It is not well-formed in the object-language, and since the arity of the predicate |= seems to vary, there might not exist a corresponding well-formed formula (though maybe not also).  Also, |= is a predicate, while ((P∧¬P)⇒Q) doesn't have any predicates.

The difference might also get illuminated by looking at other uses of |=.  For instance, I think you agree that, "{p, (p⇒q)} |= q" makes sense.  Let's suppose that |= is no different from ⇒.  Then, "{p, (p⇒q)} |= q" is no different from "{p, (p⇒q)} ⇒ q".  There's at least three problems,

"{p, (p⇒q)} ⇒ q" is not well-formed, since all well-formed formal implications begin with a '(' and end with a ')'.
q is a proposition, but what we have on the left side of the arrow is a set.  But, sets of propositions are not propositions.  Nor are propositions sets of propositions.  So, again, "{p, (p⇒q)} ⇒ q" is not well-formed, and it's not even apparent how to make into a well-formed formula, or even necessarily the case that it can get made into a well-formed formula, since it would even two different types of entities needing to fall under the same roof... so to speak.  That sounds like a possible category error.

Pece has also commented that:

"... ⊨ is a relation between finite sequences of wffs on the left and wff on the right, while ⟹ is a binary connective in the language that is used to construct formulas."


Vector spaces involving square matrices Friends! Do you know some interesting vector spaces involving square matrices? I'm trying to do my first research work in coding theory, and I plan it to be related to square matrices. Thank you. :)

Let $A\in\mathbb{R}^{n\times n}$ be a nonsymmetric and nonsingular matrix with zero/nonzero element structure that has nonzero elements on the main diagonal, i.e., $\alpha_{i,j}\neq 0$, the first superdiagonal, i.e., $\alpha_{i,i+1}\neq 0$, the first subdiagonal, i.e., $\alpha_{i-1,i}\neq 0$, the fourth superdiagonal, i.e., $\alpha_{i,i+4}\neq 0$, and the fourth subdiagonal, i.e., $\alpha_{i-4,i}\neq 0$. For $n = 15$ this has the form
$$
 A =
\begin{pmatrix}
* & *  & 0  & 0  & * & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
* & *  & *  & 0  & 0 & * & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & *  & *  & *  & 0 & 0 & * & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0  & *  & *  & * & 0 & 0 & * & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
* & 0  & 0  & *  & * & * & 0 & 0 & * & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & *  & 0  & 0  & * & * & * & 0 & 0 & * & 0 & 0 & 0 & 0 & 0 \\
0 & 0  & *  &  0  & 0 & * & * & * & 0 & 0 & * & 0 & 0 & 0 & 0 \\
0 & 0  & 0  & *  & 0 & 0 & * & * & * & 0 & 0 & * & 0 & 0 & 0 \\
0 & 0  & 0  & 0  & * & 0 & 0 & * & * & * & 0 & 0 & * & 0 & 0 \\
0 & 0  & 0  & 0  & 0 & * & 0 & 0 & * & * & * & 0 & 0 & * & 0 \\
0 & 0  & 0  & 0  & 0 & 0 & * & 0 & 0 & * & * & * & 0 & 0 & * \\
0 & 0  & 0  & 0  & 0 & 0 & 0 & * & 0 & 0 & * & * & * & 0 & 0 \\
0 & 0  & 0  & 0  & 0 & 0 & 0 & 0 & * & 0 & 0 & * & * & * & 0 \\
0 & 0  & 0  & 0  & 0 & 0 & 0 & 0 & 0 & * & 0 & 0 & * & * & * \\
0 & 0  & 0  & 0  & 0 & 0 & 0 & 0 & 0 & 0 & * & 0 & 0 & * & * \\
\end{pmatrix}
$$
Your task is to implement a factorization routine that given $A$ in a suitably efficient data structure returns the factors $L$ and $U$ where $L$ is unit lower triangular and $U$ is upper triangular. In order to guarantee that pivoting is not required you may restrict $A$ to be a diagonally dominant matrix. You must also write routines that solve systems $Lv = f$ and $Uv = f$ for arbitary vectors $v$ and $f$ and $L$ and $U$ matrices that result from your factorization routine. Your code must be as efficient as possible in terms of the number of comptuations and the storage required.
Your solution must include:

A discussion of the structure of the factors $L$ and $U$.
A description of your factorization algorithm and triangular system solving algorithms.
A derivation of the computational complexity of your algorithms.
A description and justification for an efficient data structure to hold A and to return $L$ and $U$. Be specific about your data structure’s characteristics and how you access elements in the data structure that correspond to specific elements in $A$, $L$ and $U$.
A discussion of the design of the experiments to verify the code works correctly. Be specific about the metrics used to assess the results generated by the code and the way you generate examples.
A discussion of the nonzero patterns of $L$ and $U$ if partial pivoting (in a column) was introduced to the algorithm and how it would affect your data structure. You do not have to implement pivoting.


Is it possible for this puzzle to be solved without the use of computational software? This puzzle was found from the Hot Network Questions on the right. A repost of a question in a way (from Mathematica SE), but I was wondering if the following puzzle could be done:

A quick translation:

Can anyone solve this? It's so hard!
Fill in the numbers $1–9$ in the boxes, the numbers cannot repeat.

It has already been solved in the given link, the answers being (hidden for those who wish to try and solve it):

 $17\times4=68+25=93$

The answer to the Mathematica question stated there's a way to solve it without brute-forcing, but I didn't understand it too well. How would one—if possible—solve this without using too complex softwares?

You can reduce the effort to a level that can be handled without a computer.

Let's write the calculation as

\begin{array}{cc}
&A&B\\
\times&&C\\\hline
&D&E\\
+&F&G\\\hline
&H&I
\end{array}

There are strong restrictions on $A$ and $C$, since their product must be single-digit. We can't have $C=1$, since otherwise $DE=AB$. So either $A=1$, or $(A,C)$ must be one of $(2,3)$, $(2,4)$, $(3,2)$ or $(4,2)$. We can exclude $(2,4)$ and $(4,2)$: This implies $D=8$, $F=1$ and $H=9$, but then adding $E$ and $G$ mustn't result in a carry, which is impossible, since $I$ can't be $8$ or $9$ and $E+G$ must be at least $3+5$.

So unless $A=1$, $(A,C)$ must be either $(2,3)$ or $(3,2)$. Also $B\notin\{1,5\}$. That leaves $10$ possible triples for $(A,B,C)$, which each determine values for $D$ and $E$, and it shouldn't be too much effort to go through those $10$ cases, find the $4$ remaining options for $F$, $G$, $H$ and $I$ and show that they don't work out.

That leaves us with $A=1$. In that case the options for $(C,B)$ can be enumerated as $(2,7)$, $(2,8)$, $(2,9)$, $(3,4)$, $(3,6)$, $(3,8)$, $(3,9)$, $(4,3)$, $(4,7)$, $(4,8)$, $(4,9)$, $(6,3)$, $(6,7)$, $(6,9)$, $(7,2)$, $(7,4)$, $(7,6)$, $(7,8)$, $(7,9)$, $(8,2)$, $(8,3)$, $(8,4)$, $(8,7)$, $(8,9)$. That's $24$ more cases, for a total of $34$ cases to be checked.

$1 7* 4$

$=6 8
+2 5$

$9 3$

is the right answer!

  $$17*4 = 68 + 25 = 93$$

There may be more combinations too!


How to find correlation between two random variables? What is the way to find a correlation between two variables, where $X_1,X_2,X_3$ are independent and random variable is a linear equation of these variables. Also, how to find an value of $a_1$, so that two variables are uncorrelated? For example if variables are given as $X_1+2X_2$ and $3X_1+aX_2$. Thanks for any advice!

We have $X_1,X_2$ and $X_3$ independent.

Let $U=a_1X_1+a_2X_2+a_3X_3$ and $V=b_1X_1+b_2X_2+b_3X_3$

$$Corr(U,V)=\frac{cov(U,V)}{\sqrt{var(U)var(V)}}=\frac{\sum_{k=1}^{3}{\sum_{j=1}^{3}{a_kb_jcov(X_k,X_j)}}}{\sqrt{cov(U,U)cov(V,V)}}$$

if $ k \ne j$, $cov(X_k,X_j)=0$

$$Corr(U,V)=\frac{a_1 b_1var(X_1)+a_2 b_2var(X_2)+a_3b_3var(X_3)}{\sqrt{cov(U,U)cov(V,V)}}$$

$$cov(U,U)=a_1^2var(X_1)+a_2^2var(X_2)+a_3^2var(X_3)$$
$$cov(V,V)=b_1^2var(X_1)+b_2^2var(X_2)+b_3^2var(X_3)$$

Therefore,

$$Corr(U,V)=\frac{a_1 b_1var(X_1)+a_2 b_2var(X_2)+a_3b_3var(X_3)}{\sqrt{\left(a_1^2var(X_1)+a_2^2var(X_2)+a_3^2var(X_3)\right)\left(b_1^2var(X_1)+b_2^2var(X_2)+b_3^2var(X_3)\right)}}$$

Now, if you want the correlation between $X_1+2X_2$ and $3X_1+aX_2$ to be zero, in other words, their covariance should be nil:

$$0=cov(X_1+2X_2,3X_1+aX_2)=3cov(X_1,X_1)+2acov(X_2,X_2)=3var(X_1)+2avar(X_2)$$

because $X_1$ and $X_2$ are independent, we have $cov(X_1,X_2)=0$

You choose
$$a=-\frac{3var(X_1)}{2 var(X_2)}$$


Exercise at the Beginning of Part II in Fulton's Book on Young Tableaux In Fulton's Book Young Tableaux, there's an Exercise at the beginning of part II for which I cannot find a solution (there doesn't seem to be one for this exercise in my copy of the book). It reads:

  Exercise. Show that, if $e_1,\ldots,e_m$ is a basis for the ($\mathbb{C}$-vectorspace) $E$, then the images of the vectors $(e_i\wedge e_j)\otimes e_k$, for all $i<j$ and $i\le k$, form a basis of \[ E^{(2,1)} := \left.{\textstyle\bigwedge^2E}\otimes E\middle/\left((u\wedge v)\otimes w - (w\wedge v)\otimes u - (u\wedge w)\otimes v\:\middle|\:u,v,w\in E\right)\right.. \]

First, I do not see why, for $i<j<k$, the equality \[(e_i\wedge e_j)\otimes e_k = (e_k\wedge e_j)\otimes e_i + (e_i\wedge e_k)\otimes e_j\] already holds in $\bigwedge^2 E\otimes E$. But even that aside, I do not see how to solve the exercise. Any help would be welcome.

It is enough to show two things:

1) The listed element generate the space.

We know that $(e_i \wedge e_j) \otimes e_k$ with no restraints on the indices generate the space, and we can clearly restrict to $i < j$ by the antisymmetric behaviour of the wedge product. Now, if $i > k$, we use the identity:

$$((u∧v)\otimes w=(w∧v)⊗u+(u∧w)⊗v$$

for $u=e_i$, $v=e_j$ and $w=e_k$.

This gives
$$((e_i∧e_j)\otimes e_k=(e_k∧e_j)⊗e_i+(e_i∧e_k)⊗e_j= (e_k∧e_j)⊗e_i-(e_k∧e_i)⊗e_j,$$

which shows that the vector is actually a linear combination of two of our given vectors. Therefore, they are indeed a generating set.

2) The listed elements are independent.

Note that the relations in the ideal are multilinear in $u$,$v$,$w$, so it is enough to use the relations for the basis vectors to generate the full ideal.

Now, suppose that you have a linear combination of the given vectors that gives zero in the quotient space. This means that the linear combination lies in the ideal which means that the linear combination is a linear combination of relations in the ideal.

You can now look at the smallest index of an $e_i$ that occurs in this expression and check that its coefficient is 0.

You should know that the $e_i \wedge e_j (i<j)$ are a basis of $\bigwedge^2E$, so that after tensoring you have a basis
$$
e_i \wedge e_j \otimes e_k \quad(1\leq i< j \leq n,1\leq k \leq n)
$$
for $\bigwedge^2E \otimes E$.

All you have to show is that the elements where $k<i$ are in the subspace $V$ generated by
$$
(u\wedge v) \otimes w - (w\wedge v)\otimes u - (u\wedge w)\otimes v\
$$
but look at

$$
(e_i \wedge e_j ) \otimes e_k \equiv (e_k \wedge e_j)\otimes e_i + (e_i\wedge e_k)\otimes e_j \quad (\text{mod }  V)
$$

and other permutations of the indices. Assume $k<i<j$. Can you show that the left side is $\equiv 0$ mod $V$?

EDIT: this is not an answer, the necessary computations are more difficult than I expected!!


When does $E(Y\mid X_1,X_2)=E(Y\mid X_2)$ imply $Y\perp X_1\mid X_2$? $Y\perp X_1\mid X_2$, by which I mean that $Y$ is independent of $X_1$ conditional on $X_2$, implies $E(Y\mid X_1,X_2)=E(Y\mid X_2)$. However, the reversed implication does not hold in general since conditional expectations do not in general fully characterize conditional distributions. Nevertheless, under certain assumptions, the reversed implication is true. For example, suppose $Y$ is a dummy variable. Then the distribution of $Y\perp X_1\mid X_2$ is characterized by $E(Y\mid X_1,X_2)=P(Y=1\mid X_1,X_2)$, which gives the reversed implication. But I believe this is not the only example. So, when exactly does $E(Y\mid X_1,X_2)=E(Y\mid X_2)$ imply $Y\perp X_1\mid X_2$? For example, if $E(Y\mid X_1,X_2)=E(Y\mid X_2)$ imply $Y\perp X_1\mid X_2$, does this imply that some property about the random variables is necessarily satisfied?
I found this exercise in a book on causal inference, and I was not sure what kind of answer the author was after.
One idea is that whether the implication is true or not is related to if moments other than the first moment of $Y$ given $(X_1,X_2)$ do not depend on $X_1$.

The question is essentially equivalent (with an added conditional) to the following one:
The implicancy $E[X | Y ] = E[X] \implies X \perp Y$ (where $\perp$ denotes independence) is in general false. Under which additional assumptions it's true?
There is no general answer to this. Practically speaking, there are two important cases:

$(X,Y)$ are jointly gaussian (not the same as "$X,Y$ are both gaussian"!)
$(X,Y)$ are both Bernoulli (takes values in $\{0,1\} \times \{0,1\} $)

This, because in both cases above no correlation implies independence (and no correlation is weaker than $E[X | Y ] = E[X]$).
Assuming the first case is the most typical/important, going back to the original question, we can assert:
If $(Y,X_1)$ conditioned on $X_2$ are jointly gaussian, then $E(Y\mid X_1,X_2)=E(Y\mid X_2) \implies Y\perp X_1\mid X_2$
Or slightly simpler and more in particular:
If $(Y,X_1,X_2)$ are jointly gaussian, then $E(Y\mid X_1,X_2)=E(Y\mid X_2) \implies Y\perp X_1\mid X_2$