Sequence of M-Lipschitz functions Let $\alpha$ , $\beta \in \mathbb R$ with $\alpha < \beta$, $M \geq 0$ and $(f_n)$ a sequence of M-Lipschitz functions of [$\alpha , \beta$] in $\mathbb R$. If $(f_n)$ converges simply (or "pointwise" in English I think) to a function $f$ on [$\alpha , \beta$], show that $f$ is also M-lipschitz.
We have that $(f_n)$ converges simply so $\forall x \in [\alpha , \beta], \forall \epsilon > 0, \exists n_0$ such that $\forall n > n_0, \lvert f_n(x) - f(x) \rvert \leq \epsilon$
And we also have that the $f_n$ are M-Lipschitz so $\forall x,y \in [\alpha , \beta], \lvert f_n(x) - f_n(y) \rvert \leq M\lvert x-y \rvert$
We want to show that $\forall x,y \in [\alpha , \beta], \lvert f(x) - f(y) \rvert \leq M\lvert x-y \rvert$
I tried to write $\lvert f(x) - f(y) \rvert = \lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert$
Then we have $\lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert \leq \epsilon + M\lvert x-y \rvert + \epsilon$ but it is not really the form we wanted and I think I made some mistakes with the quantifiers. Can someone help me ? (The correction solves this exercise just by saying : when going to the limit the inequality stays true. I tried to find something more precise and clear).

If $f_n$ is $M$-Lipschitz (with $M$ that does not depend on $n$) then for every $x,y$ in your domain you have
$$
|f_n(x)-f_n(y)|\leqslant M|x-y|\quad \forall n\in\mathbb{N}.
$$
But the absolute value is a continuous function, as well as the sum of real numbers. Therefore, if $f_n$ converges pointwise to $f$ you can pass to the limit for $n\to\infty$ in the previous relation to get
$$
|f(x)-f(y)|\leqslant M|x-y|.
$$

If $x \leq M|x-y| + \epsilon$ for every $\epsilon$, then $x \leq M|x-y|$.
Proof:
Suppose $x > M|x-y|$.
Then $x> M|x-y| - \epsilon$ for every $\epsilon > 0$. Hence $M|x-y| - \epsilon  < x\leq M|x-y| + \epsilon \implies$
$-\epsilon <x-M|x-y| \leq \epsilon \implies x-M|x-y|=0$ (which is proven using a basic contradiction).
Hence $x = M|x-y| \implies x > x$, a contradiction.
Your problem uses $2\epsilon$ which is a special case of this.


Probability of at least one king in a 13-card hand? What is the probability of drawing 13 cards and having at least one king?

Here is what I came up with:

$$1 - \frac{\begin{pmatrix}
48\\
13
\end{pmatrix}}{\begin{pmatrix}
52\\
13
\end{pmatrix}}$$

Is this correct?
Thanks.

Your answer is indeed correct.

There are $\binom{52}{13}$ hands with $13$ cards that can be selected from a standard deck.

The number with no kings is $\binom{48}{13}$ since we must select $13$ of the $52 - 4 = 48$ cards that are not kings.

Thus, the probability that no kings are selected is
$$\Pr(\text{no kings}) = \frac{\dbinom{48}{13}}{\dbinom{52}{13}}$$
Since the probability that at least one king is selected is found by subtracting the probability of no kings from $1$, we obtain your answer
$$\Pr(\text{at least one king}) = 1 - \frac{\dbinom{48}{13}}{\dbinom{52}{13}}$$

$\begin{pmatrix}52 \\ 13\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 52 card deck.  $\begin{pmatrix}52 \\ 48\end{pmatrix}$ is the number of 13 card hands that can be dealt from a 48 card deck- with the kings missing.  So your $\frac{\begin{pmatrix}52 \\ 13\end{pmatrix}}{\begin{pmatrix}52 \\ 48\end{pmatrix}}$ is the probability that a 13 card hand randomly dealt from a 52 card deck has no kings and, yes, $1- \frac{\begin{pmatrix}52 \\ 13\end{pmatrix}}{\begin{pmatrix}52 \\ 48\end{pmatrix}}$ is the probability that such a hand does contain at least one king.


Is there any constant $c$ satisfying the inequality Is there any positive constant $c\in \mathbb{R}$ such that for any nonzero $a,b\in \mathbb{R}^n$,

$$\frac{1}{2}\|a-b\|^2-\frac{1}{\pi}\sqrt{\|a\|^2\|b\|^2-\langle a,b\rangle^2}+\langle a,b\rangle\frac{\cos^{-1}(\rho)}{\pi}\geq c\|a-b\|^2$$

where $\rho=\frac{\langle a,b\rangle}{\|a\|\,\|b\|}$.

I checked extreme cases where $a,b$ are orthogonal or in the same or opposite directions. It seems that there is a positive $c$, satisfying the above inequality.

We have two vectors $a$, $b$ enclosing an angle $\phi\in[0,\pi]$, and are told to prove that
$${1\over2}|a-b|^2-{1\over\pi}|a|\,|b|\sin\phi+|a|\,|b|\cos\phi\,{\phi\over\pi}\geq c|a-b|^2$$
for a universal $c>0$. This claim can be written as
$$|a|\,|b|{\sin\phi-\phi\cos\phi\over\pi}\leq\left({1\over2}-c\right)|a-b|^2\ .$$
Now one can verify (by drawing the graphs) that
$$0\leq{\sin\phi-\phi\cos\phi\over\pi}\leq{1\over2}(1-\cos\phi)\qquad(0\leq\phi\leq\pi)\ .$$
Therefore it is sufficient to prove
$${1\over2}|a|\,|b|(1-\cos\phi)\leq\left({1\over2}-c\right)|a-b|^2\ .\tag{1}$$
Note that by the cosine theorem
$$|a-b|^2=|a|^2+|b|^2-2|a|\,|b|\cos\phi=\bigl(|a|-|b|\bigr)^2+2|a|\,|b|(1-\cos\phi)\ .$$
This proves
$${1\over2}|a|\,|b|(1-\cos\phi)\leq{1\over4}|a-b|^2\ ,$$
so that $(1)$ is true with $c={1\over4}$.

Your inequality is equivalent to  find $c\ge 0$ such that  for all $\theta $, $$
\|a\|\|b\|\cos(\theta)\theta\geq \|a\|\|b\|\sin(\theta)+{\pi}(c-0.5)||a-b||^2$$
($\langle a,b\rangle=\|a\|\|b\|\cos(\theta)$)
 so
the inequality should hold with $c=0$

$\|a-b\|^2=\|a\|^2+\|b\|^2-2\|a\|\|b\|\cos(\theta)$
replacing we would get $$\|a\|\|b\|(\sin(\theta)-\theta\cos(\theta)+\pi\cos(\theta))\le \frac{\pi}{2}(\|a\|^2+\|b\|^2)$$ so by AGM inequality for $\|a\|=\|b\|$ we require $\sin(\theta)-\theta\cos(\theta)+\pi\cos(\theta)-\pi\le 0$ which isn't true for all $\theta$.


Using trapezoidal rule to calculate an improper integral. I have numerically calculated the integral $$\int_{-1}^{1}\frac{e^{-x^2}}{\sqrt{1-x^2}}dx$$ using Gauss-Legendre and Gauss-Chebyshev quadrature. Now, I am asked to calculate the integral using the trapezoidal rule and compare the different methods. I had previously used the trapezoidal method for other integrals, however, with this one, I can't evaluate the function at the limits of the integral, since the denominator is $0$.

How can I calculate this integral using the trapezoidal rule? Is it even possible?

Hint: Calculate

$$\int_{-\varepsilon}^{\varepsilon}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}dx$$

using the trapezoidal rule. The expression should be a function of $\varepsilon$. Then calculate $\varepsilon \to 1$ for that expression.


Differential equation degree doubt $$\frac{dy}{dx} = \sin^{-1} (y)$$ The above equation is a form of $\frac{dy}{dx} = f(y)$, so degree should be $1$.  But if I write it as $$y = \sin\left(\frac{dy}{dx}\right)$$ then degree is not defined as it is not a polynomial in $\frac{dy}{dx}$. Please explain?

The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:

$$\frac{dy}{dx}=\sqrt[3]{x}\tag1$$
$$\left(\frac{dy}{dx}\right)^3=x\tag2$$

The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $\mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.

The problem is when you try to find degree of i.e. $$y=e^{y'}
\quad\text{or}\quad
y=\sin\left(\frac{dy}{dx}\right)\tag{a,b}$$
There exist a formula that allow you define a degree of non-polynomials, namely $$\deg\;f(x)=\lim_{x\to\infty}\frac{\log|f(x)|}{\log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $\infty$.

To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example,
$\frac{dy}{dx} = y, \frac{dy}{dx} = y^2 + 1$ are of the first order.
$ \frac{d^2 y}{dx^2} = y, \frac{d^2 y}{dx^2} = (\frac{dy}{dx})^3 + 1$ are of second order.

Your differential equation $\frac{dy}{dx} = \sin^{-1}y$ is first order even if it can be written in the equivalent form $y = \sin(\frac{dy}{dx}).$


Find a subset that forms a basis for the span of a set Let $p_1 = 1 - 2t - t^2$, $p_2 = t + t^2 + t^3$, $p_3 = 1 - t + t ^3$ and $p_4 = 3 + 4t + t^2 + 4t^3$.

Let $S$ be the set of these four functions. Find a subset of $S$ that is a basis for the span of $S$.

So I've started out by taking these functions, making them vectors and putting them into a matrix:

$$
\begin{bmatrix}
0 & -1 & -2 & 1\\
1 & 1 & 1 & 0\\
1 & 0 & -1 & 1\\
4 & 1 & 4 & 3
\end{bmatrix}
$$

Then I reduced it:

$$ \sim
\begin{bmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}
$$

I guess my question is, is this the correct way to do this? Where do I go from here? What does a zero row mean for a basis?

A better way is to put them into as matrix as columns instead of as rows. The elementary row operations preserve linear dependence relations among the columns, so when you get to reduced form, if, say, columns 1, 2, and 4 are a basis for the column space, then $p_1,p_2,p_4$ are a basis for the original space.

Yes, its the right way, assuming that you reduced your matrix without any mistakes. Now, the zero row mean, that this row are linear dependable on the first three rows, hence your basis for $S$ is $\text{Span}\{p_1,p_2,p_3\}$.


Differential algebra and differential-algebraic equations Could you give me some information about differential algebra? What is it about?

Differential-algebraic equations (DAEs) are polynomials with complex coefficients and the unknown variables are $z, x, x'$.

Is this correct?

What is the difference between them and the ODEs?

Two possible solutions of DAEs $C^{\infty}$ functions and formal power series, right?

$$$$

EDIT:

About differential algebra I found also this link.

Does this mean that Differential algebra is about differential equations over a field or a ring?

These are two distinct and not very related concepts (though not completely unrelated).

Differential algebra is the study of differential rings and fields and related structures. Let me briefly mention some things about differential fields to give you some idea about what differential algebra is about.

A differential field is a field $\mathbb F$ together with a function $\partial : \mathbb F \to \mathbb F$, called a derivation, that satisfies the product rule:
$$\partial(xy) = \partial(x)y + x\partial(y)$$
An element $x \in \mathbb F$ is called a constant of the differential field if
$$\partial(x) = 0.$$
The set of all constants form a subfield of the differential field.

An example of a differential ring is $\mathbb R(t)$, the field of rational functions in $t$ over $\mathbb R$, with the derivation $\frac{d}{dt}$, differentiation with respect to $t$. The constants of $\mathbb R(t)$ is $\mathbb R$.

Elements of differential algebra are used in e.g. differential Galois theory and symbolic integration.

A differential algebraic system of equations is a system of equations where some equations are algebraic equations and some are differential equations. The equations need not be polynomial. I say system of equations, because if it is not a system of equations, i.e. there is only one equation, it will either be purely algebraic or differential.

An example of a DAE system is the equations describing the motions of a planar pendulum, having position $(x,y)$, velocity $(u,v)$, all functions of time $t$, with length $L$:
$$\begin{align}
\dot x &= u \\
\dot y &= v \\
\dot u &= \lambda x \\
\dot v &= \lambda y - g \\
L^2 &= x^2 + y^2
\end{align}$$
as you can see, the last equation is algebraic and not differential.

I have explained the difference between an ODE and a DAE here: What is the difference between an implicit ordinary differential equation and a differential algebraic equation?

As an aside, your description of DAEs ("polynomials with complex coefficients and the unknown variables are $z,x,x'$") got me thinking of holonomic functions, and while they are not exactly what you described, they do come close. A holonomic function $y(t)$ is a function that satisfies
$$a_r(t) y^{(r)}(t) + a_{r-1}(t) y^{(r-1)}(t) + \dots a_1(t)y'(t) + a_0(t)y(t) = 0$$
where each $a_i(t)$ is a polynomial in $t$.

as a PhD of DAE I want you to read these 3 books in this order:

Numerical Solution of Initial-Value Problems in Differential-Algebraic Equations (Classics in Applied Mathematics) 2nd Edition
by K. E. Brenan (Author), S. L. Campbell (Author), L. R. Petzold (Author)

Singular Systems of Differential Equations Vol. I and II,
by S. L. Campbell (Author)


Proof that $2^{10}+5^{12}$ is a composite number A hint or full answer will help a lot, because I have no idea what to do.

Hint: This can be expressed as the difference of two squares in a reasonably easy way.


Proof question of integers. Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2$.  Show that $\frac{a^2+b^2}{ab+1}$ is the square of an integer.

Numberphile did a detailed 2 part video on the question:
Part 1: https://www.youtube.com/watch?v=Y30VF3cSIYQ
Part 2: https://www.youtube.com/watch?v=L0Vj_7Y2-xY


Bounded sequence has no convergent subsequence How can you prove that in the metric space $(\mathbb{R},d)$ where $d(x,y)=|\arctan{x}-\arctan{y}|$ the sequence $(x_n)=n$ is bounded but it has no convergent subsequence ?

Edit 1. Can I say that $(x_n)$ is bounded because $d(n,0)\leq \frac{\pi}{2}$ ?

Edit 2. to see that $x_n$ has no convergent subsequence, I suppose by contradiction that there exists a subsequence $x_{\varphi(n)}$ which converges to $\ell$ it means that $d(x_{\varphi(n)},\ell)\rightarrow 0$ it means that $|\arctan(x_{\varphi(n)})-\arctan(\ell)|\rightarrow 0,$ so $\frac{\pi}{2}-\arctan{\ell} =0$ but $\frac{\pi}{2}-\arctan{\ell}\neq 0$ contradiction.

Thank you.

$1)$ $x_n$ is bounded as the maximum distance between $x$ and $y$ is $|1-(-1)|=2$ as $-1 \leq |\arctan x|\leq 1$.

$2)$ Suppose that $x_n$ has a convergent subsequence $x_{n_k}$. Suppose further that $\displaystyle \lim_{k \rightarrow \infty}x_{n_k}=x_*$ say.

Then $d(x_{n_k},x_*)=|\arctan n_k-\arctan x_*|> \arctan  (x_*+1)-\arctan x_*$ for $n_k > x_*+1$.

Contradiction.

Boundedness is simple, by your own comment.

To show it has no convergent subsequences:

Assume it has a convergent subsequence, $(x_{n_k})$, with limit $K$. Then $n_k\ge n=x_n$. Now take $\epsilon=\frac{\frac{\pi}{2}-\arctan{K}}{2}$. Then take $N$ s.t. $\arctan{N}>\arctan{K}+\epsilon$ (possible again by your own comment out the limit of $\arctan$). Then we easily see that for every $n_k\ge N$, $d(x_{n_k},K)>\epsilon$ (because $\arctan$ is increasing). This is a contradiction, so $(x_n)$ cannot have converging subsequences.


Let $A = \{1\}, B = \{3\}$ Suppose $f = \{(1,3),(5,6)\}$ Is $f: A \rightarrow B$? Let $A = \{1\}, B = \{3\}$

  Suppose $f = \{(1,3),(5,6)\}$ Is $f$ a function from $A$ to $B$?

The definition of the function, say $F$, that the book gives:

$$\tag 1 \forall a \in A \exists! b \in B ((a,b) \in F)$$

Then $F: A \rightarrow B$

Coming back to our example: although we have $(5,6) \in f$ and $5 \notin A$ and $6 \notin B$, but, if I get it correctly, main condition of the function is not violated, hence I believe that yes, $f$ is a function from $A$ to $B$.

I've tried to answer my own question, but I need your verification:  is $f$ a function from $A$ to $B$?

Usually a function $f$ is from $A$ to $B$, we have $$f\subseteq A\times B,$$
so I would say that $f$ is not $A\to B$.

If that's your only definition, then technically it meets it, but it's not standard.

The domain and range of $f$ are not $A$ and $B$ respectively, but instead are $\{1,5\}$ and $\{3,6\}$ respectively.


Confusion with the eccentricity of ellipse Confusion with the eccentricity of ellipse. On wikipedia I got the following in the directrix section of ellipse.

  Each focus F of the ellipse is associated with a line parallel to the minor axis called a directrix. Refer to the illustration on the right, in which the ellipse is centered at the origin. The distance from any point P on the ellipse to the focus F is a constant fraction of that point's perpendicular distance to the directrix, resulting in the equality e = PF/PD. The ratio of these two distances is the eccentricity of the ellipse. This property (which can be proved using the Dandelin spheres) can be taken as another definition of the ellipse.
  Besides the well-known ratio e = f/a, where f is the distance from the center to the focus and a is the distance from the center to the farthest vertices (most sharply curved points of the ellipse), it is also true that e = a/d, where d is the distance from the center to the directrix.

It is given that $e=\frac fa=\frac ad$

In my book it was only given that $e=f/a$ (in my book there is nothing given about directrix of an ellipse).

My question

Knowing that $e=f/a$ how can I get $e=a/d$ and $e=PF/PD$?

If I understood your question correctly, you're essentially asking how one can find the equation for the directrix if one only has the equation for an ellipse with a given eccentricity.

You start with the equation below
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\qquad(1)
$$
where $a$ and $b$ are positive real-valued constants.

If you then define two points on the $x$ axis, $F$ and $F'$, by their coordinates,
$$
F \equiv (\epsilon a, 0)
\qquad(2)
$$
$$
F' \equiv (-\epsilon a, 0)
\qquad(3)
$$
for some as-yet undetermined real value $\epsilon \ge 0$, and then require that the sum of the distances from $F$ to $P$ and from $P$ to $F'$ be equal to $2a$ - for any arbitrary point $P$ on the ellipse - you'll find out that that is only possible if you choose $\epsilon$ to satisfy
$$
\epsilon = \sqrt{1 - \frac{b^2}{a^2}}
\qquad(4)
$$

So far, the above has nothing to do with the directrix. It's just a way to construct the foci and find the eccentricity, starting from some equation and then requiring that the curve described by that equation must have the fundamental property of an ellipse (namely, that the sum of the distances from any point on the ellipse to the two foci is a constant).

Now, let's see how we can prove the directrix property. Imagine that there is a vertical line at $x=d$ for some as-yet-undetermined real value $d \ge 0$ and let's see if we can find a value of $d$ such that the directrix property is satisfied. The focus for positive $x$ is $F$ (see its coordinates above), and an arbitrary point $D$ in the would-be directrix has coordinates $D = (d, y)$. The directrix property mandates that
$$
\epsilon = \frac{\overline{PF}}{\overline{PD}}
\qquad(5)
$$

Now, let's look at the square of the distances involved, since that gets rid of the square roots:
$$
\overline{PF}^{\,2} = (x - \epsilon a)^2 + (y - 0)^2 = (x - \epsilon a)^2 + y^2
$$

But $P$ lies on the ellipse so $(x,y)$ satisfies $(1)$. Therefore, eliminating $x^2$ in favour of $y^2$, but leaving the $x$ term alone, we have
$$
\overline{PF}^{\,2} =
a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax
$$

How about $\overline{PD}^{\,2}$? Note that $D$ and $P$ have the same $y$ coordinate so
$$
\overline{PD}^{\,2} = (x - d)^2 + (0)^2 = (x - d\,)^2
$$

Here's the crux now. Can we find a value of $d$ such that $(5)$ is true?
Imposing
$$
\frac{\overline{PF}^{\,2}}{\overline{PD}^{\,2}} =
\frac{a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax}{(x - d\,)^2} = \epsilon^2
$$
we get
$$
a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax = \epsilon^2\,(x - d\,)^2
$$

Note that the complicated expression above is something like this:
$$
(\mbox{terms independent of $x$ and $y$}) + (\mbox{terms involving $y^2$}) - 2\epsilon ax + 2\epsilon^2 xd = 0
$$
because the $x^2$ terms can be removed using $(1)$. Since the above has to be true for all $x$, the only hope for $(5)$ to be possible is if we choose $d$ such that
$$
2\epsilon ax - 2\epsilon^2 xd = 0
\qquad\Rightarrow\qquad
d = \frac{a}{\epsilon}
$$

With this choice, it's then not hard to show that the terms not written above also vanish so, indeed,
$$
\epsilon = \frac{\overline{PF}}{\overline{PD}}
$$
is true, provided we make that choice for $d$.

  So, what's the conclusion? The conclusion is that an ellipse has two equivalent properties, namely, (A) the sum of the distances from the foci to any arbitrary point on the ellipse is a constant, and (B) the ratio of (the distance from an arbitrary point on the ellipse to one of the foci) to (the distance from that point to a fixed vertical line parallel to the semi-minor axis) is equal to the eccentricity of the ellipse.

  And we also proved that $\epsilon = a/d$.

  (Technically, I only proved that (A) above implies (B). We'd also have to prove that (B) implies (A) for the equivalence I referred to be proven. That's an exercise I leave for the reader...)

Hint: If the eccentricity $e$ & the major axis $2a$ of an ellipse are known then we have the following

Distance of each focus from the center of ellipse
$$=\text{(semi-major axis)}\times \text{(eccentricity of ellipse)}=\color{red}{ae}$$
Distance of each directrix from the center of ellipse
$$=\frac{\text{semi-major axis} }{\text{eccentricity of ellipse}}=\color{red}{\frac ae}$$


Rationals as a direct sum of two proper subgroups Prove that the group $(\mathbb{Q},+)$ is not isomorphic to the direct sum of two proper subgroups.

My approach: Suppose that $\mathbb{Q}=A\oplus B$ where $A,B$ - proper subgroups of $\mathbb{Q}$. By the definition it means that each $q\in \mathbb{Q}$ has unique representation, i.e. if $q=a_1+b_1$ and $q=a_2+b_2$ then $a_1=a_2$, $b_1=b_2$. Since $1\in \mathbb{Q}$ and $0\in A$ then $1=0+1$ where $1\in B$. Since $2=1+1\in B$ hence $1=(-1)+2$ and so $-1\in A$. We get two distinct representations, contradiction.

Remark: Please do not duplicate this question. I have sen some topics with this question but i did not saw the solution similar to above.

What's wrong with this solution?

You assert $1 \in B$, but there's no reason to think this is true.

Here's one way to do it.

We claim $(\mathbb{Q}, +)$ can not be the direct product of two non-trivial subgroups.

Suppose we two subgroups $H, K$ of $\mathbb{Q}$ such that $Q \cong H \times K$. Then, there would be a homomorphism $h : \mathbb{Q} \rightarrow K$ with kernel $H$. But since any non-zero homomorphism from $(\mathbb{Q}, +)$ to $(\mathbb{Q}, +)$ is always bijective, we conclude $H = \{0\}$, which is a contradiction.


How to find the point on a parabola where x and y are equal? On a parabola how could i find the point at which the y and x points are equal and meet on a point of the graph, algebraically?

Substitute $x$ for $y$ in the parabola equation and solve for $x$.

Replace x with y or y with x and solve for the unknown variable. Then, the value you get will be valid for x=y.


How do I find this limit? $\lim_{n\to\infty}\frac{(n+1)^2}{4^n}$ $$\lim_{n\to\infty}\frac{(n+1)^2}{4^n}$$ The answer says zero is the limit, how this is I do not understand, I've tried H'opitals rule , didn;t work out for me, anyone have any suggestions on how to solve this limit ?

By l'Hospital's rule, you have
$$
\lim_{n\to\infty}\frac{(n+1)^2}{4^n}=\lim_{n\to\infty}\frac{2(n+1)}{4^n\cdot\log 4}=\lim_{n\to\infty}\frac{2}{4^n(\log 4)^2}=\frac{2}{(\log 4)^2}\lim_{n\to\infty}\frac{1}{4^n}=0.
$$

$4^n$ increases much faster than $(n+1)^2$ and when $n$ tends to infinity, the denominator will be very large compared to the numerator. Hence the limit is $0$.


How to tell if it's Anti-Symmetric or Not Anti-Symmetric? For each of the following, state whether the relation is reflexive, symmetric,
antisymmetric, transitive or an equivalence:

(i) R = { (a, a), (b, c), (c, b), (d, d) }

My answer:

It is not Reflexive.

It Symmetric.

[Anti-symmetric or not Anti-symmetric?]

Is not Transitive.

My lecturer gave me this formula where it shows,

Anti-Symmetric: (a, b) & (b, a), a=b

Not Anti-Symmetric: (a, b), (b, a), a≠b

I still do not know how to tell between these two.

Geometrically the idea behind symmetry is that there is an axis (at least one) which you could fold the object you're looking at along and everything on one side of the axis would match everything on the other side of the axis (think of a square with a line drawn from one corner to the opposite corner: if you fold along that line then the edges of the square on one side match up to the edges on the other side).

Anti-symmetry is the idea that there is no such matching at all -- there is nothing on one side of any axis that matches up to something on the other.

In terms of relations then, anti-symmetry means that if $aRb$, i.e. $a$ relates to $b$ in some way, then $bRa$ cannot be true unless $a=b$.  Because if $aRb$ and $bRa$ then we have a matching, and anti-symmetry says there are no matchings.

It's a bit easier to see with a concrete example: let our relation $R$ be $\subseteq$.  Given two sets $A$ and $B$, we see that if $A\subseteq B$ and $B \subseteq A$ then $A=B$.  So being a subset of another set is anti-symmetric -- two sets can't be subsets of each other (which would be symmetry) unless they're the same set.

The issue with telling if a relation is antisymmetric or not is that you're usually not looking at the relation as a subset of a set's product with itself. You're usually given a slightly more abstract formulation of the relation.

A relation $R$ would be anti-symmetric if, whenever $(a,b)$ and $(b,a)$ are in $R$, we have $a=b$.

In your proposed relation, the $(a,a),(d,d)$ pairings satisfy this trivially, but what about $(b,c)$ and $(c,b)$? It depends. Can we assume $b \ne c$? If so, it's not antisymmetric. Is there no reason to assume we cannot have $b=c$? If $b=c$, then it is symmetric.

However, if $b=c$, then we'd effectively have a repeat ordered pair in the set since $(b,c)=(c,b)$, which we usually don't like unless we're working with multi-sets or whatever. Plus why even state the problem like this because this is clearly ambiguous as heck.

Anecdotally, this would also mean the relation is reflexive and transitive if $b=c$.

So, to conclude a slightly long diatribe, my gut instinct is that $a,b,c,d$ are meant to be distinct, and thus $R$ is not anti-symmetric outright since $x = y$ (where $x,y \in \{a,b,c,d\}$) is never satisfied. But it's ambiguous and depends on the assumptions you're allowed to make.

A nice example to make a point out of what I meant by anti-symmetric being a more typical concern in more abstract formulations of relations: we say $a \mid b$ if and only if $a$ divides $b$. You can easily show from definition this is antisymmetric as a relation on the integers.

Suppose $a \mid b$ and $b \mid a$. Then $b = na$ and $a = mb$ for some integers $m,n$. But then that means $b=nmb$ or $a=mna$ by substitution of one into the other, so it has to hold that $m=n=1$. But then $b=a$.


Does $n!$ divide $ n^n$? Today while I was reading on how to shuffle an array I came across a statement that claims we shall not swap an array entry with the whole array range when shuffling the array otherwise we end up with a biased result. Click here for details.

They demonstrated that by saying that $n!$ does not divide $n^n$ because $(n-1)$ divide $n!$ but at the same time $(n-1)$ does not share any prime factor with $n$, if I understood it right.

However, I did not grasp the "proof" given, so, can you help me figure out how to demonstrate it in a more rigorous approach?

In the comments it is detailed how to find a prime $p$ such that $p \; \mid \; n-1$ but $p \; \nmid \; n^n$.

Suppose it were true now that $(n-1)! \; \mid \; n^n$.
Then we would have
$$
p \mid n -1 \mid (n-1)! \mid n^n
$$
so $p \mid n^n$, which is impossible.

$\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With Induction, you will use:
$$
{\pars{n + 1}^{n + 1} \over \pars{n + 1}!} = {\pars{n + 1}^{n} \over n!}
={n^{n} \over n!}\,{\pars{n + 1}^{n} \over n^{n}}
={n^{n} \over n!}\
\underbrace{\pars{1 + {1 \over n}}^{n}}
_{\ds{\not\in\ {\mathbb N}_{+}}}
$$
$\imp\quad n!\ \not\vert\ n^{n}$


How to solve the determinant without expanding? Here is the the determinant equation:
$$
  \begin{vmatrix}
   1 & bc & bc(b+c)\\
   1 & ca & ca(c+a)\\
   1 & ab & ab(a+b)\\
  \end{vmatrix} = abc\begin{vmatrix}
                      a & 1 & b+c\\
                      b & 1 & c+a\\
                      c & 1 & a+b\\
                     \end{vmatrix}=0
$$
I am new and beginner in matrices and determinants.So please explain the steps of  the answers.

It's a famous rule of determinants that

  If we multiply a row (column) of a matrix A by a number, the determinant of A will be multiplied by the same number.

Suppose we are comfortably in $\mathbb{R}$ and suppose also that $a,b,c \ne 0$, then multiply and divide the first row for $a$, the second for $b$ and the third for $c$
$$
  \frac{abc}{abc}\begin{vmatrix}
   1 & bc & bc(b+c)\\
   1 & ca & ca(c+a)\\
   1 & ab & ab(a+b)\\
  \end{vmatrix} = \frac{1}{abc}\begin{vmatrix}
                      a & abc & abc(b+c)\\
                      b & abc & abc(c+a)\\
                      c & abc & abc(a+b)\\
                     \end{vmatrix}
$$

Now you can conclude beacuse taking out the $(abc)^2$ factor that multiplies the  second and third column you get

$$
   abc\begin{vmatrix}
                      a & 1& (b+c)\\
                      b & 1 & (c+a)\\
                      c & 1 & (a+b)\\
                     \end{vmatrix}
$$

EDIT

At this point, to conclude that the determinant is $0$ you can simply use the Sarrus' rule or, without expanding, you can subtract the first row to the second and third row to get

$$
abc
\begin{vmatrix}
  a & 1 & b+c\\
  b-a & 0 & a-b\\
  c-a & 0 & a-c\\
\end{vmatrix} = abc[(a-b)(c-a) - (a-c)(b-a)] = 0
$$

Notice that in the header of the proof I supposed that $a,b,c \ne 0$, but now we can conclude even if one between $a,b,c = 0$, because the property is satisfied as well.

We have
$$\begin{vmatrix}
                      a & 1 & b+c\\
                      b & 1 & c+a\\
                      c & 1 & a+b\\
                     \end{vmatrix} = \begin{vmatrix}
                      a & 1 & a+b+c\\
                      b & 1 & b+c+a\\
                      c & 1 & c+a+b\\
                     \end{vmatrix} = 0$$
Here the first columm was added to the third. Thus it remains to verify the first equation in your question.

Case 1: $abc = 0$.

In that case at least one of $a, b ,c$ is $0$ and $$
 \begin{vmatrix}
   1 & bc & bc(b+c)\\
   1 & ca & ca(c+a)\\
   1 & ab & ab(a+b)\\
  \end{vmatrix}$$
has at least two rows of the form $(1 \quad 0 \quad 0)$ so that the first equation follows.

Case 2: $abc \ne 0$.

Then
$$
 abc \begin{vmatrix}
   1 & bc & bc(b+c)\\
   1 & ca & ca(c+a)\\
   1 & ab & ab(a+b)\\
  \end{vmatrix} = \begin{vmatrix}
                      a & abc & abc(b+c)\\
                      b & abc & abc(c+a)\\
                      c & abc & abc(a+b)\\
                     \end{vmatrix} = (abc)^2\begin{vmatrix}
                      a & 1 & b+c\\
                      b & 1 & c+a\\
                      c & 1 & a+b\\
                     \end{vmatrix}
$$
In the first step the factor $abc$ was shifted into the determinant by multiplying the first row with $a$, the second with $b$ and the third with $c$. In the second step the factor (abc) was extracted from columns two and three.

Dividing by $abc \ne 0$ yields the desired result.


Removing finite points from closed set. I have the following homework problem:

"Let $S$ be a closed set and let
$$
S_1 = \{x_1, x_2,..., x_n\} \subset S
$$
Is $S \smallsetminus S_1$ also a closed set? If true, prove the statement. If false, give a counterexample".

I think I have a counterexample to the statement, but I'm not completely sure of its validity:

Let $S =\{p\} $ a set composed of a single point. Since $S$ contains only one point, $S$ is closed. Let $S_1 =S$. Since any set is a subset of itself, then it holds true that $S_1 \subset S$. We then have:
$$
S \smallsetminus S_1 = S \smallsetminus S = \emptyset
$$
Since $ \emptyset$ is an open and closed set, in particular, it is open. Hence the statement is false.

My question is if it's valid to just assume that the empty set is in particular open and "ignoring" that it's also closed. Does this counterexample hold?

No, your example doesn't work because the empty set is closed. Being open does not prevent a set from being closed, as they are not mutually exclusive conditions. However you could just remove $0$ from the interval $[0,1]$ to obtain a counterexample.

This works if you want to show it's not true in general. More generally, if there exists a one point set in the space that is not open, then the whole space minus that point is a counterexample. Thus the discrete topology is actually the only situation where the statement holds.

Recall that in order for a set to be closed, it must contain all of it's accumulation points.

Well, a point $x$ of a set $A$ is said to be an accumulation point if for any neighborhood of $x$, the neighborhood contains at least one other point in the set $A$.

We know that any finite set is clearly not infinite. So, you can generate an extremely small neighborhood about any point in a finite set and get infinitely many points not in the original set.

Your question doesn't specify if this set is infinite or finite. Assuming finite $S$, then yes, any $S\setminus S_1$ would contain no accumulation points and be closed.

If $S$ is infinite, closed, and you remove any part it, you're removing accumulation points and thus ensuring the set is no longer closed. However, this is only true unless the infinite set has no accumulation points to begin with.

Another problem arises with the topology being used here. It's not specified, but in a general case this is the structure.


Detection of self intersection point of curve What numerical procedure is be adopted to detect self-intersecting  parametrized points $ [x(t), y(t) ] $ in  $ \mathbb R^2 $ ?

Observation : @ roots ( t= 2, t=-1 ) parabola has double value with respect to cubic. How to build an algorithm ?

I think the way to find out  is that: $\exists t_1 \neq t_2: (x(t_1),y(t_1)) = (x(t_2),y(t_2))$. For if $r(t) = (x(t),y(t))$ is a continuously differentiable function on $\mathbb{R}$, then if $r'(t) = 0$ for some $t$ then you have a self intersection point.

Example: $r(t) = (t^2-t, t^3-t), t \in \mathbb{R}$.

We have $r(0) = (0,0) = r(1)$.


Countable product of metric spaces is metrizable [General Metric] I know that if we have a countable collection of metric spaces $\{(X_n,\rho_n)\}_{n=1}^{\infty}$ then $X=\Pi^{\infty}_{n=1}X_n$ is a metric space with metric $\rho((x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}})=\sum_{n=1}^{\infty} \dfrac{\rho_n(x_n,y_n)}{2^n [1+\rho_n(x_n,y_n)]}$.

Also I have proved that $X=\Pi^{\infty}_{n=1}X_n$ is a metric space with metric $\rho_\beta((x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}})=\sum_{n=1}^{\infty} \dfrac{\rho_n(x_n,y_n)}{\beta^n [1+\rho_n(x_n,y_n)]}$ where $\beta \in \mathbb{N}$ and $\beta \geq 2$.

But I need to have it from a book or any bibliographic reference, exists some bibliographic reference to this problem with the metric $\rho_\beta$?

In the book "Le cours de la APM III, Elements de topologie" by André Revuz and Germaine Revuz (Association des professeurs de Mathématiques de l'Enseignement Public PARIS-1966), you find the statement of exercise 20 (page 38) as follows:

Exercise 20.-Produit dénombrable d'espaces métriques.

1º Si $d$ est une distance definie sur un ensemble $E$, il en est de même de $\delta$ définie par $$\delta(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ et $\delta$ définie la même topologie que $d$.

2º Soit $E_n$, avec $n\in\mathbb N$,une famille dénombrable d'espaces métriques et $E=\prod\{E_n; n\in\mathbb N\}$ leur produit cartésien. A la distance $d_n$ de $E_n$, on associe la distance $\delta_n$ définie en 1º;en on définit sur $E\times E$ (avec $x=\{x_n\}$ et $y=\{y_n\}$ l'application $\delta$ dans $\mathbb R^+$ donnée par
$$\delta(x,y)=\sum_{n=1}^{\infty}\frac{1}{2^n}\delta_(x_n,y_n)$$  Montrer que $\delta$ est une distance sur $E$, et que la topologie qu'elle définit est la topologie produit.

This exercise 20 is completely developed on page 172. If you want, I will print the detailed solution in two and a half pages of the book. Where there are also good references is in the analysis books of Jean Dieudonné and those of Laurent Schwartz.

The proof is the same for both cases. All we need is that $\sum_n \frac{1}{\beta^n}$ converges. See my proof here; convergence of such a series is used at one place in the proof.


QR factorization: $A=QR$ and $R=Q^TA$ gives $Q Q^T = I$ I am learning linear algebra and I am on QR factorization: $A = QR$. A solution to $R$ is $R = Q^TA$, where $Q$ and $A$ have size $m * n$ and $R$ is $n * n$, which means that $Q$ is not necessarily square.
My question:
Since $A = QR$, if we substitute the R in $A = QR$ with $Q^TA$ then we get $A = QQ^TA$, which means $QQ^T = I$. However, $QQ^T$ shouldn't be equal to $I$ unless $Q$ is square. So there must be something wrong with my deduction above. What went wrong?
Update: I emailed professor Strang at MIT and found the answer

I have chosen to delete my original answer and write a new one. I do this since neither my first answer nor the answer the OP has provided address the differences between the full and reduced $QR$ factorizations, which have come into play here and probably caused some misunderstandings. I first the briefly describe the difference, before I address what the question is really about, namely the product $QQ^\top$
The reduced $QR$ decomposition.
A reduced $QR$ decomposition is a composition of $A$ into a product $A= QR$, where $Q$ is an $m \times n$ matrix with orthonormal columns and $R$ an $n \times n$ upper triangular matrix. Note that $A$ needs to be either square ($m=n$) or tall ($m > n$) since the columns of $Q$ need to be linearly independent. The columns of $Q$ form an orthonormal basis of the columns space of $A$.
The full $QR$ decomposition. The full $QR$ decomposition is obtained by extending the orthonormal basis for the column space of $A$ to an orthonormal basis of the entire space. In this case $R$ will be an upper triangular $m \times n$.
Now, in both cases $Q^\top Q = I$ since the columns of $Q$ are orthonormal. However in the first case, the opposite product is not the identity. As Strang points out in his lecture notes, $QQ^\top$ will be the projection onto the column space of $Q$ (or equivalently $A$), which when $Q$ does not span the entire space, will not be $I$. In the case of the full decomposition however, $Q$ is square and spans the whole space, so we have $Q^\top Q = Q Q^\top = I$. Please note that there is a typo in Strang's notes, as the projection matrix should be
$$P = Q (Q^\top Q)^{-1} Q^\top.$$
The OP is right that you get that $A = QQ^\top A$ in any case. However, you can not "cancel out" matrices in the same way as with numbers. You need to multiply both sides by the inverse, which does not exist if $Q$ is not square. The reason why $QQ^\top A = A$ in any case, is that, as the OP has pointed out in his answer, $QQ^\top$ projects into the column space of $A$, and the columns of $A$ will therefore be eigenvectors of $QQ^\top$.

will be identity but may not be $n\times n$ could be $m\times m$ as well. $Q$ is $m\times n$, $R$ is $n\times n$, $A$ is $m\times n$ and $Q^T$ is $n\times m$. so $Q^TQ$ and $QQ^T$ is well defined and makes complete sense.


Prove that if $f$ is Riemann integrable on $[a,b]$ then so is $f^2$ Prove that if $f$ is Riemann integrable on $[a,b]$ then so is $f^2$.

I have already proved that a function is Riemann integrable if and only if it is bounded and continuous a.e. If $f$ is bounded, then so is $f^2$. If $f$ is continuous a.e. then so is $f^2$ because it is a composition of a continuous function and a function that is continuous a.e.

But what if I was asked to prove the proposition directly (without refering to the above theorem)? Is that easy or technical?

Lemma: If function $f:[a,b]\rightarrow \mathbb{R}$ is Riemann integrable then it is bounded on $[a,b]$.

Proof:

It is clear that $f$ is integrable if and only if for every $\epsilon>0$ there is $\delta>0$ such that $|S_1-S_2|<\epsilon$ whenever $S_1$ and $S_2$ are Riemann sums corresponding to partitions of $[a,b]$ of diameter less than $\delta$.

Choose $\epsilon>0$ and aparition of $[a,b]$ such that for arbitrary $x_i^{'}, x_i^{''} \in [x_{i-1},x_i]$, $i=1,\ldots, N$ we have

$$\Bigg|\sum_{i=1}^n (f(x_i^{'})-f(x_i^{''}))(x_{i}-x_{i-1}) \Bigg |< \epsilon$$

If we apply this inequality to the special case where, for some fixed index $j=1,\ldots, N$, we have $x_i^{'}= x_i^{''}$ if $i\ne j$ and $x_j^{''}= x_j$, we get

$$|(f(x_j^{'})-f(x_j))(x_{j}-x_{j-1})|<\epsilon$$
implying
$$|f(x_j^{'})|<\frac{\epsilon}{x_{j}-x_{j-1}}+|f(x_j)|$$

This last inequality holds for all $x_j^{'}\in [x_{j-1},x_j]$, thus $f$ is bounded on $[x_{j-1},x_j]$. Therefore $f$ is bounded on all $[a,b]$.

Since $f$ is Riemann integrable on $[a,b]$ there is an $M$ with $|f(x)|\leq M$ for all $x\in[a,b]$. The function $g(x):=f^2(x)$ then satisfies
$$|g(x)-g(y)|=|f(x)+f(y)|\>|f(x)-f(y)|\leq 2M \>|f(x)-f(y)|$$
for arbitrary $x$, $y\in[a,b]$. This implies that  any test you can think of to establish the integrability of $f$ will also be passed by $g$.

Consider the counter-example:  $f$ on $[0,1]$ defined by $f(x)=1/\sqrt{x}$ for $x\not=0$ and $f(0)=0$.  In this case $f$ is Riemann integrable but $f^2$ is not.


Filling in fractions I am trying to solve the problem in the below photo

My try:
Let $S_N$ be the set $\{N,2N,3N\}$. These numbers will be the deominators of the fractions. $S_{11},S_{12},\cdots, S_{15}$ would not work as deominators as they use the same digit(s) twice or more. $S_{16}$ will not work either as the digits $5$ and $7$ are used as numerators. Both these digits need to be in the fraction with the smallest deominator or the fractions do not have the same deominators and the sum cannot be $1$. As for the others, when including the constraint that all of the deominaters can be simplifyed to $N$, none of the sets $S_{17},S_{18},S_{19}$ could make a sum larger than $1$. The sets $S_{20},S_{21},\cdots, S_{33}$ reuse a digit/digits, and $S_{34}$ and over contain three digit numbers.

By then, I have no more clues as to which numbers I should have as deominators.

Update: It turns out that up to the obvious permutations this problem has a unique solution whether you interpret the adjacent digits in the denominator as multiplication or concatenation. Who knew math could be so much fun.

Multiplication

I'm not sure yet of a better way to go about this, but for fractions to add up to an integer, you need the denominators to be "compatible" in some sense. E.g., you can't add some number of fifths with some number of sevenths and get an integer unless you started with two integers. With that in mind, your denominators should have most, if not all of the small primes ($2$ and $3$ for this problem).

  $$\frac1{3\cdot6}+\frac5{8\cdot9}+\frac7{2\cdot4}$$

Concatenation

The guiding principle is the same here. You need compatible denominators, so you want them to all have a roughly even distribution of the small primes in your problem. This is hampered partially by the difficulties in finding the prime factors of the sum of two numbers ($10a+b$ for digits $a$ and $b$), but the numbers in question are small and there aren't many of them, so just throwing everything even and most everything with a factor of $3$ into the denominator and hoping for the best reduces the number of things you have to check substantially.

  $$\frac5{34}+\frac7{68}+\frac9{12}$$


Topology -- Continuity and the induced topology. Here is my question.

Let $X=Y=\Bbb R$, with the usual topology. Let $A=[0,1]$ and topologize $A$ with the induced topology from $X$.

Does there exist a continuous function from the topological space $A$ onto $Y$?

Why or why not?

I'd argue No, since we'd be going from a closed to open set. But I also thought that any subset of a topology is an open subset.

Any help would be appreciated.

No: $[0,1]$ is compact, $\mathbb R$ is not compact, and the image of a compact set under a continuous function is always compact.

The extreme value theorem, from real analysis, states that any continuous function from a compact set to the reals is bounded (and has a maximum). A bounded function cannot be surjective onto $\Bbb R$.


Given a quadrilateral, prove it is a square. Consider this game.

  You have a perfect ruler and a perfect protractor. You are also able
  to observe perfectly if two lines are parallel.

  Is it possible to prove that a given quadrilateral is a square, using
  exactly 3 steps?

For example, a proof using 4 steps would be this.

  Observe that opposite sides are parallel (two steps).
  The angle between the diagonals is 90 degrees.
  The angle between any two adjacent sides is 90 degrees.

My gut feeling is that it is not possible to do this in 3 steps and that the minimum would be 4. But how would one prove this?

I don't believe this is possible.

A quadrilateral has 4 degrees of freedom, to similarity: place the polygon so it has two of its vertices at $(0,0)$ and $(1,0)$.  Then the other two points are free to be (almost) anywhere.  In order to find one of these points we must make two measurements, and the other is completely independent, so there is no measurement that can be used to constrain both.

Three steps :

1- the diagonals meet at their midpoint (or is it two steps one for each diagonal) : this ensures we have a parallelogram

2- the diagonals are of equal length

3- the diagonal meet at $90$ degrees.


Given two points on a unit sphere, how to express their angular difference in spherical coordinates? Given two points on a unit sphere in spherical coordinates representation ($\theta$ is longitude, $\phi$ is latitude in $[0,\pi]$), I want to express their angular difference in spherical coordinates.

In detail: given two points $p_1,p_2$ on the meridian, the first is under the second. Making the first the new north pole but leaving the meridian at $\theta =0°$ should give me a spherical coordinate with $\phi = |\phi_1 - \phi_2| $ which seems obvious, but the new longitude $\theta$ needs to become $180°$ although $\theta_1 = \theta_2 = 0$ (meridian).

If the answer is to convert both coordinates to Euclidian (X,Y,Z), then subtract and then reconvert to spherical coordinates I would not be surprised. But I would be very glad, if you could state why such a difference as in $ \mathbb{R}^{3}$ is not possible in spherical coordinates. What property do they miss?

Edit: these images should clarify my intentions.

The left image shows an angular difference. But I understand that a coordinate-wise difference is not sufficient. To show an example, please look at the right image.

$A$ becomes the new north pole. Thereby $B's$ polar coordinate changes to $\theta(B-A)$. But additionally the azimuthal angle of $B'$ becomes $\phi(B)=$ 180°. Imagine another extreme case where $\phi(B)$ started with 90°. It would end up at the same 90° after making $A$ the new north pole.

Given two points on a sphere $(x_1,y_1,z_1)=(\sin \phi_1 \cos \theta_1,\sin \phi_1 \sin \theta_1,\cos \phi_1 )$ and $(x_2,y_2,z_2)=(\sin \phi_2 \cos \theta_2,\sin \phi_2 \sin \theta_2,\cos \phi_2 )$. Take the dot product of these to obtain the cosine of the angle $\alpha$ between them relative to the center of the sphere. We have
\begin{eqnarray*}
\cos \alpha = \cos \phi_1 \cos \phi_2 +\sin \phi_1 \sin \phi_2 ( \cos \theta_1 \cos \theta_2 +\sin \theta_1  \sin \theta_2) \\
\cos \alpha = \cos \phi_1 \cos \phi_2 +\sin \phi_1 \sin \phi_2 \cos (\theta_1 - \theta_2 ).
\end{eqnarray*}
\begin{eqnarray*}
 \alpha = \color{blue}{\cos^{-1} \left( \cos \phi_1 \cos \phi_2 +\sin \phi_1 \sin \phi_2 \cos (\theta_1 - \theta_2 )\right)}.
\end{eqnarray*}

The original question says "θ is longitude, ϕ is latitude in [0,π]", but in the associated diagram, θ is shown as colatitude, and ϕ as longitude.
Which is correct for the following discussion?


Find the volume formula of a simplex proving that $\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_n=\frac 1{n!}$ Definition
If $x_0,...,x_n$ are $(n+1)$ affinely indipendent point of $\Bbb R^n$ (which means that the vectors $(x_1-x_0),...,(x_n-x_0)$ are linearly independent) then simplex determined by them is the set
$$
S:=\Biggl\{x\in\Bbb R^n: x=\alpha_1v_1+...+\alpha_nv_n, \sum_{i=1}^n\alpha_i\le1\,\,\,\text{and}\,\,\,\alpha_i\ge0\,\,\,\text{for all}\,i\Biggl\}
$$
where $v_i:=(x_i-x_0)$ for each $i>0$.

So with the previous definition I try to show that the volume of a symplex $S$ is given by the formula
$$
v(S)=\Big|\frac{1}{n!}\det\big[(x_1-x_0),...,(x_n-x_0)\big]\Big|
$$
for each $n\in\Bbb N$.
So if $x_0,...,x_n$ are $(n+1)$ point affinely independet we define the transformation $h:\Bbb R^n\rightarrow\Bbb R^n$ through the condition
$$
h(x):=A\cdot x+x_0
$$
where $A$ is the matrix whose $j$-th column is the vectors $(x_j-x_0)$ for each $j=1,...,n$. So now we observe that the transformation $h$ carries the simplex
$$
E:=\{x\in\Bbb R^n:x_1+...+x_n\le 1\,\,\,\text{and}\,\,\, x_i\ge 0\,\,\,\text{for all}\,i\}
$$
onto the simplex $S$ generated by the points $x_0,...,x_n$. So if we prove that
$$
v(E):=\frac 1{n!}
$$
for all $n\in\Bbb N$ then by the change variable theorem (it is easy to verify that $h$ is a diffeomorphism)
$$
v(S)=\int_S 1=\int_E|\det A|=\frac 1{n!}|\det A|
$$
for each $n\in\Bbb N$.
So let's start to prove by induction that
$$
v(E)=\frac 1{n!}
$$
for each $n\in\Bbb N$.
So if $n=1$ then $E=[0,1]$ and so clearly the formula trivially holds. So we suppose that the formula holds for $(n-1)$ and we prove that it holds for $n$.
So we have to prove that
$$
\int_E1=\frac 1{n!}
$$
and to do this we will use Fubini's formula. So I have to prove that
$$
\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n=\frac 1{n!}
$$
but unfornuately I don't be able to prove it.
For sake of completeness I point out that it seems that here there is a similar solution to that I gave (see the answer of the professor Blatter) but I don't fully understand it. In particular the solution I linked says that if we define
$$
E_\xi:=\{x\in\Bbb R^n:x_1+...+x_{n-1}\le1,\,\,\,\text{and}\,\,\,x_1,...,x_{n-1}\ge 0\,\,\,\text{and}\,\,\,x_n=\xi\}
$$
for any $\xi\in[0,1]$ then $E=\bigcup_{\xi\in[0,1]}E_\xi$ and so if we observe that the projection of $E_\xi$ is a $(n-1)$ dimensional simplex then $\int_E 1=\int_0^1(1-x_n)^{n-1}v(E_\xi)\,dx_n=\frac 1 n(1-x_n)^nv(E_\xi)=\frac 1{n!}$ that complete the proof but I don't understand how to prove effectively last equality.
So I ask to prove the last equality and then I ask to prove that $h[E]=S$ too. So could someone help me, please?

With a slight generalization, it becomes easy to prove by induction that for each $a$, we have: $$\int_0^a \int_0^{a-x_n} \cdots \int_0^{a-(x_n+\cdots+x_2)} dx_1 \cdots dx_n = \frac{a^n}{n!}.$$
The base case $n = 1$ is trivial.  For the inductive case, we have using the inductive hypothesis that the desired integral is equal to $\int_0^a \frac{(a-x_n)^{n-1}}{(n-1)!} dx_n = \left. -\frac{(a-x_n)^n}{n!}\right|_{x_n=0}^{x_n=a} = \frac{a^n}{n!}$.
(And in fact, via the transformation that you outlined in the question, the general case given above can be proved from the special case $a=1$.  This would give another way to proceed in a proof by induction: use the transformation along with the inductive hypothesis to conclude that the function being integrated with respect to $x_n$ is $\frac{(1-x_n)^{n-1}}{(n-1)!}$.)

First of all I summarise to follow what I above defined to point out some things.
Definition
If $x_0,...,x_n$ are $(n+1)$ affinely independent point of $\Bbb R^n$ then the simplex determinated by them is the set
$$
S:=\Biggl\{x\in\Bbb R^n: x=x_0+\alpha_1v_1+...+\alpha_nv_n,\,\sum_{i=1}^n\alpha_i\le1\,\,\,\text{and}\,\,\,\alpha_i\ge 0\,\,\,\text{for all}\,i\Biggl\}
$$
where $v_i:=(x_i-x_0)$ for each $i>0$.
So with this definition it is clear that a simplex can be always individuated by one point $x_0$, that we call origin, and by vectors of a base $v_1,...,v_n$, that we call directions. Finally we call the points $(x_0+v_1),...,(x_0+v_n)$ vertex.
So let's start solve the question.
Theorem
If $S$ is a simplex generated by the affinely independent points $x_0,...,x_n$ of $\Bbb R^n$ then
$$
v(S)=\Big|\frac{1}{n!}\det\big[v_1,...,v_n\big]\Big|
$$
where $v_i:=(x_i-x_0)$ for each $i=1,...,n$.
Proof. So if $x_0,...,x_n$ are $(n+1)$ affinely independent points then we define the tranformation $h:\Bbb R^n\rightarrow\Bbb R^n$ through the condition
$$
h(x):=A\cdot x+x_0
$$
where $A$ is the matrix whose $j$-th column is for each $j=1,...,n$ the vector $v_j$ as above defined. First of all we observe that the transformation $h$ is a diffeomorphism because the vectors $v_1,...,v_n$ are linearly independet so that the matrix $A$ is not singualr. Then we observe that the trasformation $h$ map the simplex $E$ with origin the intersection $O$ of the axes and with directions the vector of canonical base $\mathcal E:=\{e_1,...,e_n\}$ onto the simplex $S$ because $h$ map the vertex of $E$ in the vertex of $S$. Now we suppose that
$$
v(E)=\frac 1{n!}
$$
so that we observe that in this particular case from the change of variables theorem it would follow that
$$
v(S)=\int_S 1=\int_E |\det A|=|\det A|\int_E 1=|\det A|v(E)=\frac 1{n!}|\det A|
$$
and so the theorem would hold. So let's prove the theorem using induction. So clearly if $n=1$ then $S=[a,b]$ for some $a,b\in\Bbb R$ and so the formula trivially holds. So we suppose that the formula holds for $(n-1)$ and we prove that it holds of $n$. So we know that it is sufficient to prove that
$$
\int_E 1=\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n=\frac 1{n!}
$$
and exactly we can do this knowing that the theorem holds for any siplex of $\Bbb R^{n-1}$ by the inductive hypotesis. So fixing $x_n$ let be $\xi:=(1-x_n)$ and we define
$$
T:=\big\{x\in\Bbb R^{n-1}:x_1+...+x_{n-1}\le\xi\,\,\,\text{and}\,\,\,x_i\ge 0\,\,\,\text{for all}\,\,\,i=1,...,(n-1)\big\}
$$
that is $T$ is the simplex of $\Bbb R^{n-1}$ with origin the intersection $O$ of axes and with directions  those vectors that are proportional to the vectors of canonical base and of length $\xi$. So by the inductive hypotesis we know that
$$
v(T)=\Biggl|\frac 1{(n-1)!}\det[\xi e_1,...,\xi e_n]\Biggl|=\frac 1{(n-1)!}(1-x_n)^{n-1}
$$
and moreover by definition of volume
$$
v(T)=\int_T 1=\int_0^{1-x_n}...\int_0^{1-(x_2+...+x_{n-1}+x_n)}1dx_1...dx_{n-1}
$$
so that we conclude that
$$
\int_0^1\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}dx_n=
\\
\int_0^1\Biggl[\int_0^{1-x_n}...\int_0^{1-(x_n+...+x_2)}1\,\,\,dx_1...dx_{n-1}\Biggl]dx_n=
\\
\int_0^1\frac 1{(n-1)!}(1-x_n)^{n-1}dx_n=-\frac 1{n!}(1-x_n)^n\Biggl|_0^1=\frac 1{n!}
$$
and so the theorem holds.


What is the difference between linear and affine function I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated

A linear function fixes the origin, whereas an affine function need not do so. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else.

Linear functions between vector spaces preserve the vector space structure (so in particular they must fix the origin). While affine functions don't preserve the origin, they do preserve some of the other geometry of the space, such as the collection of straight lines.

If you choose bases for vector spaces $V$ and $W$ of dimensions $m$ and $n$ respectively, and consider functions $f\colon V\to W$, then $f$ is linear if $f(v)=Av$ for some $n\times m$ matrix $A$ and $f$ is affine if $f(v)=Av+b$ for some matrix $A$ and vector $b$, where coordinate representations are used with respect to the bases chosen.

Answer for any confused French reader.
In France/French, the distinction between linear and affine appears to be different from other countries.
An affine function in the French sense is a any function that may be written
$f(x) = a + bx$
where $a$ and $b$ are independent of $x$ (not necessarily Real). For example, $f(n) = a + bn$, where $n \in N$ is an affine function over the Natural number set $N$.
A linear function in the French sense is an affine function that passes through the origin, that is $a=0$ and $f(x)=bx$ for some number $b$ independent of $x$.
Reference: wiki/Fonction_affine


Cubic diophantine equation with a prime $x^3 + y^3 + z^3 - 3xyz = p$. Question: Find all triple positive integers $(x, y, z)$ so that
$$x^3 + y^3 + z^3 - 3xyz = p,$$
where $p$ is a prime number greater than $3$.

I have tried the following: The equation factors as

$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p.$$

Since $x + y + z> 1$, we must have $x + y + z = p$ and
$$x^2 + y^2 + z^2-xy-yz - zx = 1.$$
The last equation is equivalent to
$$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2.$$
Without loss of generality you can assume that $x ≥ y ≥ z$, we have $xy ≥ 1$ and $xz ≥ 2$, implying
$$(xy)^2 + (yz)^2 + (zx)^2 ≥ 6> 2.$$

Who can help me and correct me, thank you.

Result: If $p>3$ is a prime number and  $x$, $y$ and $z$ are positive integers such that
$$x^3 + y^3 + z^3 - 3xyz = p,$$
then if $p\equiv1\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\tfrac{p-1}{3},\tfrac{p-1}{3},\tfrac{p+2}{3}\right),$$
and if $p\equiv2\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\tfrac{p+1}{3},\tfrac{p+1}{3},\tfrac{p-2}{3}\right).$$

Proof: As you already note, the equation can be expressed as
$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p,$$
which immediately shows that, because $x$, $y$ and $z$ must be positive,
$$x+y+z=p\qquad\text{ and }\qquad x^2 + y^2 + z^2-xy-yz-zx=1.\tag{1}$$
The latter can be rewritten as
$$(x-y)^2+(x-z)^2+(y-z)^2=2,$$
which shows that two of the three numbers are the same, and the third differs from them by only $1$. That is to say, without loss of generality we have
$$x=y=z\pm1.$$
Plugging this back into the first equation found at $(1)$ shows that
$$p=x+y+z=3x\pm1,$$
and so we find that $x=\tfrac{p\mp1}{3}$. As $x$ must be an integer we see that only one of the two choices of the $\pm$-sign is possible, depending on whether $p\equiv1\pmod{3}$ or $p\equiv2\pmod{3}$.

Conversely, a routine check shows that if $p\equiv\pm1\pmod{3}$ then the triplet of positive integers
$$(x,y,z)=\left(\tfrac{p\mp1}{3},\tfrac{p\mp1}{3},\tfrac{p\pm2}{3}\right),$$ and its three distinct permutations satisfy the equation
$$x^3 + y^3 + z^3 - 3xyz = p.$$

Strangely enough, the solution is finite.

for the equation:

$$X^3+Y^3+Z^3-3XYZ=q=ab$$

If it is possible to decompose the coefficient as follows:  $4b=k^2+3t^2$

Then the solutions are of the form:

$$X=\frac{1}{6}(2a-3t\pm{k})$$

$$Y=\frac{1}{6}(2a+3t\pm{k})$$

$$Z=\frac{1}{3}(a\mp{k})$$

Thought the solution is determined by the equation Pell, but when calculating the sign was a mistake. There's no difference, but the amount should be. Therefore, the number of solutions of course.

I may be wrong, though. We still need to check other options.


If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$? Prove, disprove, or give a counterexample:

  If $\mathcal{P}(A)=\mathcal{P}(B)$, then $A=B$.

Assume $\mathcal{P}(A)=\mathcal{P}(B)$. Since we know $A \subseteq A$, we know $A \in \mathcal{P}(A)$.

Since $\mathcal{P}(A)=\mathcal{P}(B)$, we know that $A \in \mathcal{P}(B)$.

Therefore, $A \subseteq B$ and $A=B$.

Is this proof okay?

Edit: I should note this isn't probability, $\mathcal{P}$ is the power set.

You can prove: $$\bigcup \mathcal{P}(A)=A$$$$\bigcup\mathcal{P}(B)=B$$ and by hypothesis you have $\mathcal{P}(A)=\mathcal{P}(B)$ therefore $$B=\bigcup\mathcal{P}(B)=\bigcup\mathcal{P}(A)=A$$

If we know $A$ we can determine unequivocally $\mathcal{P}(A)$ and if we know $\mathcal{P}(A)$ we can determine unequivocally A, as element of  $\mathcal{P}(A)$ which includes any subset , so if $A=B$ then $\mathcal{P}(A)=\mathcal{P}(B)$ and if $A\neq B$ then $\mathcal{P}(A)\neq\mathcal{P}(B)$ what completes the proof of your theorem.

Another way of proving:
$$(\mathcal{P}(A)=\mathcal{P}(B))\implies [(A\in\mathcal{P}(B)) \wedge (B\in\mathcal{P}(A))]\implies [(A \subseteq B) \wedge (B \subseteq A)]\\
[(A \subseteq B) \wedge (B \subseteq A)] \implies [(A\cup B=B) \wedge (B\cup A=A)]\implies (A=B)$$

PS. -1 as rating of my answer is more against reputation of math.stackexchange.com , than mine.


Any natural number is greater than or equal to its product of digits. I was randomly thinking and I stumbled upon this question in my mind. I was thinking of numbers, and I observed that any number is greater than or equal to the product of digits in many cases. I was thinking, is this true for all natural numbers or for only a few cases? Please help.

Let $k$ be the number of digits, and let $a$ be the first digit (assume $a>0$).  Then the product of the digits is bounded above by $a\times 9^{k-1}$.  The value of the number itself is bounded below by the total value of that first digit: $a\times 10^{k-1}$.

Let k be the number of digits, and ak-1 ak-2 .... a0 are the digits of the number.

for example, Let the number is 265 then k=3 , a2 = 2, a1=6, a0 = 5.

Then product of digit of the number is            ak-1 * ak-2 * ... * a0

The actual value of number is                         ak-1 * 10k-1 + ak-2 * 10k-2 +....+a0 * 100

Compare product of digit of the number and 1st part of actual value of the number

          ak-1 * ak-2 * ... * a0

          ak-1 * 10k-1

In product of digit of the number after first digit (ak-1) there are product of k-1 digit such that 0<=digit<10.

In actual value of number after first digit (ak-1) there are product of k-1  10.

10k-1 always grater than product of k-1 number in {0-9}.

Therefore,

If k>1 the number  is always grater than product of digit of the number.

And if k=1 we can't find the product of digit of the number because product is binary operator.

Any natural number is greater than or equal to its product of digits.


Should I start with complex variables before fourier series? My math professor has said that it would be useful to start with complex analysis before learning fourier series in the signals and systems course(I'm an undergrad EE). Do you agree with that and why ?!

(I'm planning to learn complex variables from Brown/Churchil)

Well, Fourier series are more conveniently expressed in complex form as series

$$\sum_{n=-\infty}^\infty c_ne^{inx}\;,\;\;n\in\Bbb N\;,\;\;c_n\in\Bbb C$$

I think complex analysis is way too much for the above, as things can be put in real form in a  more or less easy way, but this may heavily depend on how deep that course of your intents to dig in these matters.

Anyway, a hefty real analysis background, particularly in infinite series, both numerical and of functions, could be enough, together with some basic of complex numbers.

There are two forms of fourier series namely trigonometrical and complex form
As you are an EE undergrad you would be dealing mostly with complex form of fourier series which further leads to development of Fourier transform .

the basic relation that is used is that of euler ie

e^ix = cosx + isinx

and no major part of complex theory is required


Slice of pizza with no crust The following question came up at a conference and a solution took a while to find.

Puzzle. Find a way of cutting a pizza into finitely many congruent pieces such that at least one piece of pizza has no crust on it.

We can make this more concrete,

Let $D$ be the unit disc in the plane $\mathbb{R}^2$. Find a finite set of subsets of $D$, $\mathcal{A}=\{A_i\subset D\}_{i=0}^n$, such that

for each $i$, $A_i$ is simply connected and equal to the closure of its interior
for each $i, j$ with $i\neq j$, $\operatorname{int}(A_i)\cap \operatorname{int}(A_j)=\emptyset$
$\bigcup\mathcal{A}=D$
for each $i,j$, $A_i=t(A_j)$ where $t$ is a (possibly orientation reversing) rigid transformation of the plane
for some $i$, $\lambda(A_i\cap\partial D)=0$ where $\lambda$ is the Lebesgue measure on the boundary circle.

Note that we require only that $\lambda(A_i\cap\partial D)=0$ and not that $A_i\cap\partial D=\emptyset$. I know of a solution but am interested in what kinds of solutions other people can find, and so I welcome the attempt.

Here is another with 12 pieces, but all pieces have the same orientation:

$\hspace{32mm}$

Using this idea, the pizza can be divided into $6n$ equal pieces with the same orientation for any $n$. However, to have some pieces with no crust, we need $n\gt1$. Above is $n=2$, here is $n=3$:

$\hspace{32mm}$

To cut a pizza like this, a blade shaped like, and as long as one sixth of, the circumference of the pizza would be most useful, since all of the cuts are this size and shape.

Here is Mathematica code that will generate these sliced pizzas for any $n$:

Pizza[n_] :=
 Module[{g, arcs = {Thickness[1.3/400], Circle[{0, 0}, 1]}},
  For[i = 0, i < 6, For[j = 0, j < n, AppendTo[arcs,
     Rotate[Rotate[Circle[{-1, 0}, 1, {0, Pi/3}],
       j Pi/3/n, {-1/2, Sqrt[3]/2}], i Pi/3, {0, 0}]]; ++j]; ++i];
  Show[Graphics[arcs], ImageSize -> 400,
   PlotRange -> 1.01 {{-1, 1}, {-1, 1}}]]

Motivation

I thought of the construction of a regular hexagon: you draw a circle with a compass, and then mark arcs on the circle whose chords are the radius of the circle. Due to the properties of equilateral triangles, each arc is exactly $1/6$ of the circumference of the circle, and the chords of those arcs form a regular hexagon. At each vertex of the hexagon, the compass will span to the next vertex (by construction) and to the center of the circle (again, by construction).

Connecting each vertex to the center with arcs centered at the previous vertex, we get the circle tiled by $6$ curvy triangles with congruent sides; two convex sides and one concave side. The centers of the convex sides are the opposite vertices of the curvy triangle. Since the chords of the curved sides have a length $1$ radius, we can trace out the interior convex sides with a congruent arc rotating about the opposite vertex.

$\hspace{32mm}$

Since we can sweep out these $6$ triangles with these congruent arcs, we can split up the curvy triangles into any number of congruent pieces with these arcs.

If this violates the parameters in a clear way, consider this a teaching opportunity.  Would this count:


Calculating a Cardgame I was send here from stackoverflow because they thought maybe you can help me.

Here my original post: https://stackoverflow.com/questions/26799476/a-faster-way-then-doing-14-for-loops

What I want:

Maybe someone knows the Smartphone game "Elemental Kingdoms" If not i will try to explain.

You got a whole bunch of cards and runes. And you choose 10 cards and 4 runes. This is then your deck.

The enemy got a Demon (a very strong card which you have to damage as hard as possible).

Know you draw every time its your turn a card. A Card got Cooldown ( time until it can be played which reduces every round) if it has has cooldown 0 it can be played. Its automaticly placed on the first free place on your field.

Cards can attack cards in front of them. although everycard got abilitys which are giving them aditional attack or resistance or stuns....

Its over if the demon is dead (doesnt matter for the simulation) or when the player got no cards left or is health is 0. Health is lost if no card is in front of the demon that could take the damage or if the demon got a ability through that health is lost.

Know i wanted to create a Simulator which creates every deck, simulates the fight serveral times and at the end gives out the Deck which dealt the most damage to the demon.

The Problem with that is: There are more than 200 cards which can be used (and the same card could be 10 times in the deck) so there are to many possible decks.

So i wondered if someone knows if it is possible to do what i want mathematically or should i just leave it?

I appreciate everything that helps me :)

It sounds like you have far too many choices to list all the possible decks.  In cases like that one usually resorts to generating random decks as a sample of all possible decks.  It sounds like you are drawing your deck with replacement.  If there are $200$ cards, you throw $10$ random numbers from $1$ through $200$ and those are the cards in your deck.  Pick your four runes the same way.  Run your simulation with this deck.  Do this lots of times, keeping track of the best results.

Now you can just accept the best deck you found.  It will probably not be the best possible, but it shouldn't be bad.  Another approach is to print out the best few dozen decks found with their scores and look at them.  You may find some regularities there.  For example, there might be some particular card that is in every one, because it is particularly strong.  You can then go back and redo the simulation with that card always picked first.  You could also collect statistics over the cards.  For each card, you might compute the average score of all the decks it is in.  Some strong or weak ones may stand out.  If you run a lot of trials, you could even compute average scores for decks with a given pair of cards.  Since there are tens thousands of pairs, you would need millions of simulations, but if they are fast that is OK.

Basically you are using your simulator as a research tool.  You collect data, study it, make a hypothesis about what makes a strong deck, and test that hypothesis.  Have fun. It is a good way to learn programming.

what i think the writer is trying to say is:

is their a way to enter all of your exact cards into a simulator and have the simulator tell you the best result for your entire inventory

example :

list of 200 exact cards
vs

demon

simulator tells you the best 10 cards you have to obtain a maximum possible score


$\alpha \cap \beta = \varnothing$: An Analysis on Set Disjointness Question: Prove: $A\cap B$ and $A\setminus B$ are disjoint and that $A=(A\cap B)\cup (A\setminus B)$.

In the above question I would like to know how to come to the desired conclusion. What is the idea here?

For the first:

Suppose $x \in A\cap B$ and $x \in A\backslash B$. The first statement implies $x\in B$, while the second implies $x\not\in B$ which is a contradiction.

For the second:

Suppose $x\in (A\cap B)\cup (A\backslash B)$.Then $x\in A \cap B$ or $x\in A\backslash B$. Both cases imply, $x\in A$ and so $(A\cap B)\cup (A\backslash B)\subseteq A$.

Now suppose $x\in A$. If $x\in B$, then $x\in A\cap B$ and if $x \not\in B$ then, $x\in A\backslash B$. Both cases imply $x \in (A\cap B)\cup (A\backslash B)$ and so $A\subseteq (A\cap B)\cup (A\backslash B)$.

First: They are asking to prove $(A \cap B) \cap (A \setminus B) = 0$. Knowing that $A \setminus B$ is equal to simply $A, x \epsilon A$ and $x$ is NOT an element of $B$, we can continue:

The intersection of A and B includes elements both in A and B. $A \setminus B$ has elements in A not in B. Thus, the intersection is null.


Correlation and Moment generating function Let $M=E[e^{t\cdot X}]$ be the moment-generating function of the random vector $X$ in $\mathbb{R}^n$. Then is it true that
$$
E[(X -EX)^s] = \left. \frac{\partial \ln{M}}{\partial t^s} \right|_{t=0}
$$
where $s \in \mathbb{N}^n$ and
$$
X^s = X_1^{s_1}\cdots X_n^{s_n},  \qquad \frac{\partial}{\partial t^s} =\frac{\partial}{\partial t_1^{s_1}} \cdots \frac{\partial}{\partial t_n^{s_n}}
$$

I only checked this for $|s| = s_1+\cdots+s_n \le 3$, but it seems to be true.

EDIT: If $|s|=1$, then the formula doesn't hold. However, it seems to hold for at least $|s|=2,3$

EDIT 2: Realized that the formula doesn't hold for $|s|=4$. It seems that the $|s|=2,3$ cases were indeed a fluke

EDIT 3: I realized that what I'm essentially asking is the cumulant expansion. The "fluke" is actually documented on wikipedia.

Let $n=1$. Then $\frac d {dt} \ln M(t)=\frac {M'(t)} {M(t)}$. At $t=0$ this has the value $EX$ nor $E(X-EX)$.


For $\mathbb{R}$, does every open cover have a "countable" subcover? I am reading intro theory on Heine-Borel where the equivalent statement to "a set $E$ contained in $\mathbb{R}$ is compact" (closed and bounded) is that "every open cover for $E$ has a finite subcover". Here, they give an example of how an open set $A$ cannot have a finite subcover:

  Consider $A$ to be the open interval $(0,1)$. Then for each point $x$ in $(0,1)$, consider the infinite collection of sets $O_x = \{(\frac{x}{2},1) : x \in (0,1)\}$ whose union serves as an open cover for $(0,1)$. But it is impossible to find a finite subcover for $O_x$ because for any minimum value of $x$ in $(0,1)$ we can find $0 \le y \le \frac{x}{2}$ where $y$ is not contained in the finite union of sets.

I assume this weakness lies in the fact that due to the density of natural numbers in $\mathbb{R}$, we can always find such a $y$ that is smaller than $\frac{x}{2}$ and thus a finite union of sets is insufficient.

But what if we had a countable union of sets—the same cardinality as that of the natural numbers. Then, on $\mathbb{R}$, must every open cover have a "countable" subcover?

The answer to your question is yes, in a strong way.

The key observation is that the topology on $\mathbb{R}$ is countably generated - there is a countable set $\{U_i: i\in\mathbb{N}\}$ of open sets such that any open set $V$ is a union of $U_i$s. (Take the $U_i$s to be the open intervals with rational endpoints.) Call these the basic opens.

Now suppose $X\subseteq\mathbb{R}$ and $\mathcal{C}=\{C_j: j\in J\}$ is an open cover of $X$. Let $$K=\{i: U_i\subseteq C_j\mbox{ for some $j\in J$}\}$$ be the set of (indices of) basic opens contained in elements of $C_j$. $K$ is countable, so we can turn this around: for each $i\in K$, let $D_i$ be some element of $\mathcal{C}$ with $U_i\subseteq D_i$.

There are countably many $D_i$s, and it's easy to see that $\bigcup D_i=\bigcup \mathcal{C}$; so they form a countable subcover of $X$. This shows that any subset of $\mathbb{R}$ has the countable subcover property! In topological terminology, the usual topology on $\mathbb{R}$ is hereditarily Lindelof, and the proof above used the fact that $\mathbb{R}$ is second countable (second countability implies hereditary Lindelofness). Second countability is a very strong property, but most commonly-encountered topological spaces have it.

We say a topological space $X$ is Lindelof if every open cover of $X$ has a countable sub-cover.

For a topological space $X$ you have

$X$ is second countable implies $X$ is Lindelof.
$X$ is second countable implies $X$ is Separable.

And if $X$ is metrizable then the above three properties are equivalent.


How to go about computing an ideal in a polynomial ring I am stuck on how to characterize the elements in $(5,x+7)$ in the polynomial ring $\mathbb{Z}[x]$. I seriously can't even figure out if $x$ is in this ideal. I know that elements in the ideal look like $5h(x)+(x+7)g(x)$ for $h(x),g(x)\in \mathbb{Z}[x]$, and then I tried to expand this out, but I got some huge mess that I didn't know how to work with. In particular I am trying to figure out whether this ideal is maximal or prime or neither, and I thought the best way to do that was to find some elements that are not in the ideal so I could maybe find some zero divisors in $\mathbb{Z}[x]/(5,x+7)$.

I think your best bet is to read @Servaes answer on this post:
Is the ideal $(2,X+1)\subset\Bbb{Z}[X]$ prime, maximal or neither?
Then convince yourself that this quotient ring is a field (that you've probably seen before) and conclude that your ideal is maximal.


Series, computing the limit $\lim\limits_{n\to\infty}\frac{1+2+\cdots+n}{n^3}$ How to compute the following limit? The series is given by

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

Thanks for your help...


How often should I compound my Algorand (or other proof of stake crypto assets)? Note to editors/mods: Please help me tag this appropriately! I'm not really a math wiz... is it multivariable calculus?
See this note from the Algorand FAQ about how collecting rewards on staked assets works with compounding.

The participation rewards are calculated automatically, however the compounding effect is not automatic. This is because the rewards are calculated from the last recorded balance on the blockchain, the easiest way to force rewards compounding is to send a zero Algo payment transaction to the target address on a frequent, recurring basis. This transaction will trigger the commit of all accrued rewards and record them to the on-chain balance of the account.

As of this question, the network fee of a 0.00 ALGO transaction to myself is .001 ALGO, and the APR for rewards is ~5.827% (converted from ~6% APY). Let's say I have some amount of ALGO principal, P, and I decide to claim my rewards every d days.
It stands to reason then that every time I claim my rewards, the resulting interest, I is:
I = P * ( d * .05827 /365)-.001
Every d days, P goes to P+I.
How do we find, for a given value P, which value d is the optimal compound interval such that we maximize the rate that P increases?
My suspicion is that as P increases sufficiently high, it will make sense to compound much more frequently. It's been 11 years since I took Calculus 1 and 2, and this seems like an interesting problem. It seems like there's some recursive nature to this that makes it challenging.

Instead of writing d for the number of days between collecting interest, let's write n for the number of times a year you do it.  Also write r for the annualized interest rate and f for the fee per transaction.  At subsequent collections the balance is

$P$,
$P(1 + r / n) - f$,
$(P(1 + r / n) - f)(1 + r / n) - f
= P(1 + r / n)^2 - f[(1 + r/ n) + 1]$,
$P(1 + r / n)^3 - f[(1 + r/ n)^2 + (1+r/n) + 1]$,
and so on.

At the end of the year, then, we have
$$P_{new} = P\left(1 + \frac{r}{n}\right)^n - f\left[\left(1 + \frac{r}{n}\right)^{n-1} + \ldots + \left(1+\frac{r}{n}\right) + 1\right]$$
and of course we want to maximize this quantity.
Summing the geometric series on the right, we can write this as
$$P_{new} = P\left(1 + \frac{r}{n}\right)^n - \frac{1 - (1 + r/ n)^n }{1 - (1+r/n)} f$$
which we can simplify to
$$P_{new} = P\left(1 + \frac{r}{n}\right)^n - \left[\left(1 + \frac{r}{n}\right)^n - 1\right]\frac{f}{r} n$$
or, equivalently, to
$$P_{new} = \left(P - \frac{f}{r} n\right)\left(1 + \frac{r}{n}\right)^n + \frac{f}{r} n$$
It's tedious but not difficult to write out an expression for the derivative with respect to n, but unfortunately the roots of the derivative are not given by an elementary function of $P$, $r$, and $f$.

However, let's take a look at what would happen for some of the values you've given.  You say that $f$ is 0.001 and $r$ is 5.827%.  Let's say you have the equivalent of \$1 million, which is apparently around 2.95 million of these.
Calculating by brute force, the optimal number of compounding periods per year is 33, which corresponds to something like once every eleven days.  This will result in a balance of ALGO 3,126,343 or roughly \$1,061,875 assuming a constant exchange rate.
On the other hand, let's say that instead of 33 times a year you do it six times a year.  This will result in a final balance of ALGO 3,126,124, or roughly \$1,061,631, which is \$244 less than you made compounding 33 times per year.
In fact, let's say you just do it three times a year.  Then you'll end up with a final balance of ALGO 3,125,256.92 or roughly $1,061,337.25, about \$538 less than you would have made with 33 compounding periods.  That represents less than a 1% difference in your total profit, or about 5 basis points of your initial investment.  In other words, these are amounts that would be swamped by small fluctuations in exchange rates.
Of course nobody said that these need to be equally spaced throughout the year like we've been assuming.  But that's even more work for what I hope is clear is a fairly paltry reward.

By the way, they could do away with all this nonsense if they simply accounted for compounding when they added in the interest.  All they'd have to do is give you an interest payment of $P(e^{r \Delta t} - 1)$ rather than an interest payment of $P r \Delta t$.  I guess this way they collect more 0.001 ALGO fees, though...

Thinking in terms of the number of days between claims makes sense when the principal is much larger than the daily interest accrued.
In such a scenario, you want to claim at the point that you would make more from the interest in the days between claims on the claimed coin than you you lose in fees by claiming it.
For a low fee coin like Algorand, that works out to claiming quite often. With a principal of 1000, it's "best" to claim roughly once a week. With a principal over 400,000 it's optimal to claim every day. But as others mentioned, the difference here is small, claiming daily vs claiming yearly works out to the equivalent of an extra 0.1-0.2% interest. It's probably not worth it if you have only \$1,000 staked as claiming weekly vs yearly will net you an extra \$1-2.
Another thing to consider is that certain coins and staking methods do actually automatically compound the interest but the coins are not added to your wallet (as that would require an extra transaction that would incur a fee). For instance, I believe cardano does this as part of their protocol. The only reason to claim the staking rewards there is if you intend to remove all your coins from the staking pool.


Does the n'th derivative of a smooth transition have to have more than n+1 roots for some n? I'm wondering if the space
$$S=\left\{f\in C^{\infty}([0,1])\ \ :\ \ f(0)=0\bigwedge f(1)=1\bigwedge\forall n\in\mathbb{N},\ f^{(n)}(0)=f^{(n)}(1)=0\right\}$$
contains an element $f_0$ such that $f_0^{(n)}$ has at most $n+1$ roots for all $n\in\mathbb{N}_0$.

Intuitively I'm somewhat inclined to think it's false. Do anyone know if this holds, or how I should go on about finding out?

If $f_0^{(n)}$ has $n+1$ roots, then by the mean value theorem $f_0^{(n+1)}$ has $n$ distinct roots in the interior of the support in addition to the boundary points being roots. Hence by induction $f_0^{(n)}$ must have $n+1$ roots for all $n\in\mathbb{N}$.

As to whether the lower bound is non-strict, I think maybe the way to go is to assume that $(a_i)_{i\in\mathbb{N}}$ is a sequence with $a_i \in S$ and $a_i^{(i)}$ having at most $i+1$ roots. Because if those properties imply that not all derivatives of $a_i$ remain bounded as $i\to\infty$, then no such sequence can have an element of $S$ as limit, and so the lower bound must be strict.

I made some numerically experiments with linear programs which indicated the existence of such a sequence where also $\sup_x a_i^{(1)}(x)<2$. So several derivatives must be considered simultaneously.

I would guess that the answer to your question is yes and my candidate for a function would be as follows. Let us work on the interval $[-1,1]$ and consider the bump function
$$ \psi(x) = \exp\left(\frac {-1}{1-x^2} \right)\,.$$
Then all derivatives at the endpoints vanish,
$$ \psi^{(n)}(-1) = \psi^{(n)}(1) = 0\,.$$
Now define
$$ f(x) = a \int_{-1}^x \psi(y) \,dy\,,$$
and choose $a \in \mathbb R$ such that $f(1) = 1$.

To prove your required properties one would need explicit expressions for the derivatives of $\psi$ which involve polynomials of high degrees. I believe that intuitively $\psi$ and its derivatives have no more roots than necessary, but I have not checked the details.


Phragmen-Lindelöf theorem, question from Conway, chapter VI Page 141, Question 3:

  Let $G=\{z:|\operatorname{Im} z| < \pi/2\}$  and suppose $f:G\rightarrow C$ and $\limsup|f(z)| \leq M$ on $w$ in the boundary of $G$. Also, suppose $A < \infty$ and $a < 1$ can be found such that $|f(z)| < \exp[A \exp(a|\operatorname{Re} z|)]$ for all $z$ in $G$. Show that $|f(z)|\leqslant M$ for all $z$ in $G$.

Thanks in advance.

Consider $f\circ\log (z)$ and apply Corol. 4.2 of Chapter 6 in Conway's book.

Corollary 4.2   Let $a\geq \frac{1}{2}$ and put $G=\{z: |arg z|<\frac{\pi}{2a}\}$. Suppose that $f$ is analytic on $G$ and there is a constant $M$ s.t. $\limsup_{z\to w}|f(z)|\leq M$ for all $w$ in $\partial G$. If there are positive constants $P$ and $b<a$ s.t. $|f(z)|\leq P \exp(|z|^b)$ for all z with $|z|$ sufficiently large, then $|f(z)|\leq M$ for all $z$ in $G$.


Prove that $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor$ Prove that $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor \,\,\forall x,y \ge 0$
My attempt so far:
Suppose that either $x$ or $y$ is greater than $0$.
Now $\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor 4x\rfloor + \lfloor x\rfloor +\lfloor 4y\rfloor + \lfloor y\rfloor \ge \lfloor 4x\rfloor + \lfloor 4y\rfloor +1\ge \lfloor 4x+4y\rfloor \ge \lfloor 3x+y\rfloor + \lfloor x+3y\rfloor$
For $x,y \in [0,1)$!

Proof that
$\lfloor 5x\rfloor + \lfloor 5y\rfloor \ge \lfloor {3x+y}\rfloor + \lfloor {x+3y}\rfloor$
is FALSE.
Let $x=^-1$ and $y=0$ then the inequality above becomes $^-5 + 0 \ge ^-3 + ^-1$; which is FALSE.


Using an 8-way splitter, extract 1/12-th of something I have a pile of dirt. And I have an 8 way splitter.

I can split into 8 and recombine splits as many times as I wish.

  What is the easiest run of splitting/combining I can do to pull out exactly 1/12th of the dirt?

(Not entirely serious, but too long for a comment.)

There is no solution in the strict sense, as noted in the comments already, since any legal split or recombine leaves piles which are fractions having denominators that are powers of $2\,$. Therefore, it is not possible to isolate a pile of size $1/12$ in finitely many steps.

However, assuming one has enough time and patience to work through infinitely many steps, consider that division by $4$ is available as a split-by-$8$ followed by a recombine-$2\,$. Then, first split the whole pile by $4\,$, and set $1/4$ aside. Further split one of the remaining $1/4$ piles by $4$ and add it to the $1/4$ pile previously set aside. Repeat infinitely many times (or until the atoms of dirt no longer split ;-)), then the pile which collected all the intermediate splits will have a size of:

$$
\frac{1}{4}+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^3 + \cdots \;=\; \frac{1}{4} \cdot \frac{1}{1-\cfrac{1}{4}} \;=\; \frac{1}{3}
$$

Finally, divide that pile once more by $4$ to get $\;\displaystyle \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}\,$.

Not sure if this is the quickest, but was the most obvious to me.

Pound the splitter down:  you now have 8 piles of 1/8.
Combine three piles, then pound the splitter down. (this gives you 24s, seems like we're getting somewhere!)  Repeat this with another 3 piles.  You now have 16 piles of 1/24 and 2 piles of 1/8.
Combine the two piles of 1/8, with 3 piles of 1/24.  Pound the splitter down on that. You now have 24 piles of 1/24.
Combine two piles of 1/24 and you have your 1/12th.
Eat the dirt, for extra credit.  ;-)


Integration: Branch cuts Can someone show me how to calculate this integral using branch cuts ?

$$\int_0^{\infty}\Big(\frac{x}{1-x}\Big)^{\frac{1}{3}}\frac{1}{1+x^2}dx$$

Let
$$f(z)=\frac{|z|^{1/3}e^{i\varphi_1/3}|z-1|^{-1/3}e^{-i\varphi_2/3}}{1+z^2}$$
where $\varphi_1=\arg{z}$, $\varphi_2=\arg(z-1)$, $0\le\varphi_1,\varphi_2\le2\pi$. Integrating $f(z)$ over a dumbbell contour, we get
\begin{align}
\int_\gamma f(z)\ dz
&=\int^1_0\frac{x^{1/3}e^{i(0)/3}(1-x)^{-1/3}e^{-i(\pi)/3}}{1+x^2}dx+\int^0_1\frac{x^{1/3}e^{i(2\pi)/3}(1-x)^{-1/3}e^{-i(\pi)/3}}{1+x^2}dx\\
&=-i\sqrt{3}\int^1_0\frac{x^{1/3}(1-x)^{-1/3}}{1+x^2}dx\\
&=2\pi i\left(\sum_\pm\operatorname*{Res}_{z=\pm i}f(z)-\color{grey}{\operatorname*{Res}_{z=0}\frac{(1-z)^{-1/3}}{1+z^2}}\right)\\
&=2\pi i\left(\frac{e^{\pi i/6}\times2^{-1/6}\times e^{-\pi i/4}-e^{\pi i/2}\times2^{-1/6}\times e^{-5\pi i/12}}{2i}\right)\\
&=-i\frac{\pi}{\sqrt[3]{4}}\left(\sqrt{3}-1\right)
\end{align}
which implies
$$\int^1_0\frac{x^{1/3}(1-x)^{-1/3}}{1+x^2}=\frac{\pi}{\sqrt[3]{4}}\left(1-\frac{1}{\sqrt{3}}\right)$$
For the remaining integral, one can apply the substitution $x\mapsto 1-x^3$ to get
\begin{align}
\int^1_0\frac{(x-1)^{-1/3}}{1+x^2}
&=-\int^1_0\frac{3x}{(x^3-1)^2+1}dx\\
&=-\sum_{\omega\in S}\frac{1}{2(\omega^4-\omega)}\int^1_0\frac{x-\sigma+it}{(x-\sigma)^2+t^2}dx\\
&=-\sum_{\omega\in S}\frac{1}{2(\omega^4-\omega)}\left[\frac{1}{2}\ln\left((x-\sigma)^2+t^2\right)+i\arctan\left(\frac{x-\sigma}{t}\right)\right]^1_0\\
&=-\sum_{\omega\in S}\frac{\ln\left(\frac{(1-\sigma)^2+t^2}{\sigma^2+t^2}\right)+2i\left(\arctan\left(\frac{1-\sigma}{t}\right)+\arctan\left(\frac{\sigma}{t}\right)\right)}{4(\omega^4-\omega)}
\end{align}
where $\omega=\sigma+it$, $\sigma\in\mathbb{R}$ and $$S=\left\{\sqrt[6]{2}e^{\pi i/12},\sqrt[6]{2}e^{7\pi i/12},\sqrt[6]{2}e^{3\pi i/4},\sqrt[6]{2}e^{5\pi i/4},\sqrt[6]{2}e^{17\pi i/12},\sqrt[6]{2}e^{23\pi i/12}\right\}$$
Writing out the $6$ terms manually then using Mathematica to simplify, we find that the sum (which is equivalent to a hypergeometric function as pointed out by Lucian) evaluates to
\begin{align}
\frac{3}{2}{}_3\mathrm{F}_2\left(\left.\begin{matrix}1,1,\frac{1}{2}\\\frac{4}{3},\frac{5}{6}\end{matrix}\right|-1\right)
=&\ \frac{\pi\sqrt[3]{2}}{12}(3+\sqrt{3})+\frac{\ln{2}}{3\sqrt[3]{4}}+\frac{\ln(\sqrt[3]{2}-1)}{4\sqrt[3]{4}}-\frac{\ln(2+\sqrt[3]{4}+\sqrt[3]{16})}{2\sqrt[3]{4}}\\
&+\frac{\sqrt{3}}{2\sqrt[3]{4}}\operatorname{artanh}\left(\sqrt{\frac{3}{5-\sqrt[3]{4}}}\right)+\frac{\sqrt{3}-1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}+\sqrt[3]{16}-1}{\sqrt{3}+1}\right)\\
&-\frac{\sqrt{3}+1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}-\sqrt[3]{16}+1}{\sqrt{3}-1}\right)-\frac{\arctan(1+\sqrt[3]{2})}{\sqrt[3]{4}}
\end{align}
Therefore,
\begin{align}
\int^\infty_0\frac{x^{1/3}(1-x)^{-1/3}}{1+x^2}dx
=&\ -\frac{\pi\sqrt[3]{2}}{12}(3+\sqrt{3})-\frac{\ln{2}}{3\sqrt[3]{4}}-\frac{\ln(\sqrt[3]{2}-1)}{4\sqrt[3]{4}}+\frac{\ln(2+\sqrt[3]{4}+\sqrt[3]{16})}{2\sqrt[3]{4}}\\
&-\frac{\sqrt{3}}{2\sqrt[3]{4}}\operatorname{artanh}\left(\sqrt{\frac{3}{5-\sqrt[3]{4}}}\right)-\frac{\sqrt{3}-1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}+\sqrt[3]{16}-1}{\sqrt{3}+1}\right)\\
&+\frac{\sqrt{3}+1}{2\sqrt[3]{4}}\arctan\left(\frac{\sqrt{3}-\sqrt[3]{16}+1}{\sqrt{3}-1}\right)+\frac{\arctan(1+\sqrt[3]{2})}{\sqrt[3]{4}}\\
&+\frac{\pi}{\sqrt[3]{4}}\left(1-\frac{1}{\sqrt{3}}\right)
\end{align}


Order between one to one functions and their inverses Let $f,g :R \to R $ one to one functions such that
$f(x)< g(x), \forall x \in R $

Is it true that $f^{-1}(x)>g^{-1}(x), \forall x \in R$??

I'd say yes, thinking at their graphs: if the graph of f is below  the graph of g, when we take the symmetry with respect to the line $y=x$ the graph of $f^{-1}(x)$ will be above the graph of $g^{-1}$. But i am interested in a rigorously formulated proof . If it is true...

No, if the graph of $g(x)$ is above the graph of $f(x)$, then the graph of $g^{-1}(x)$ is to the right of the graph of $f^{-1}(x)$. If these functions are increasing, then being to the right of another function is the same as being below it. However, if they are decreasing, then being to the right of another function is the same as being above it.

For a concrete counterexample, take $f(x)=-x$ and $g(x)=-x+1$. Then $f^{-1}(x)=-x$ and $g^{-1}(x)=-x+1$, so $g^{-1}(x)>f^{-1}(x)$ for all $x$.

Right!
I'd say like this:
Let $f^{-1}(x)=a, g^{-1}(x)=b$
So we have $x=f(a)=g(b)$

Then from the hypothesis $g(b)>f(b)$ so $f(a)>f(b)$

And if we assume that f is strictly increasing it follows that $a>b$
So it's sufficient to add the condition that $f$ is increasing,
 $g$ doesn't need to be.


Alternatives to Jech for forcing via Boolean algebras As the title suggests, apart from Jech's Set Theory, are there any other noteworthy books or notes which cover forcing through Boolean algebras? All other texts I've seen use the poset approach (Kunen, Weaver, etc).

I'm going through the chapter on Jech's book, and having somewhere else to look for when I get stuck would probably be useful.

Try Rosser's book, Independence Proofs.

Also, once you understand that these methods involve semiregular topologies, look at the section on the relatively prime integer topology in Counterexamples in Topology by Steen and Seebach as well as the embedding of minimal Hausdorff topologies into semiregular topologies in Extensions and Absolutes of Hausdorff Spaces by Porter and Woods.  With regard to the latter, you will find that there are three kinds of extensions to H-closed spaces which need not be the same.  This will seem somewhat disconnected from set theory.  However, in so far as set theory presumably encompasses the "philosophy of the infinite", this three-fold multiplicity seems related to the fact that the free orthomodular lattice on countably many generators can be embedded into the free orthomodular lattice on three generators.

I had been led to these resources by my own curiosity and Jech's presentation of Boolean-valued forcing.  If formulating a complete Boolean algebra from a given separative partial order involves topologies formed from regular open sets, then one should be asking questions of what role semiregular topologies play in foundations.  One cannot do this without learning about instances of semiregular topologies elsewhere in mathematics.


Is this equation homogenous or inhomogeous? I have the following differential equation in my perturbation theory notes

$y'' + 2y' = -2y$

$y(0) = 0$

It says in the following section that this equation is inhomogeneous. But I thought equations are inhomogenous if they are of the form

$ay'' + by' + cy = F(x)$ with $F(x) \ne 0$

So why is the above equation described as inhomogeneous? Is it a mistake..or is it something to do with a boundary condition being specified..or is it something else?

your first equation $y''+2y'+2y=0$ is homogeneous and has the solution
$y \left( x \right) ={\it \_C1}\,{{\rm e}^{-x}}\sin \left( x \right) +{
\it \_C2}\,{{\rm e}^{-x}}\cos \left( x \right)
$


Coin Flips and Hypothesis Tests Here's a problem I thought of that I don't know how to approach:

You have a fair coin that you keep on flipping. After every flip, you perform a hypothesis test based on all coin flips thus far, with significance level $\alpha$, where your null hypothesis is that the coin is fair and your alternative hypothesis is that the coin is not fair. In terms of $\alpha$, what is the expected number of flips before the first time that you reject the null hypothesis?

Edit based on comment below: For what values of $\alpha$ is the answer to the question above finite? For those values for which it is infinite, what is the probability that the null hypothesis will ever be rejected, in terms of $\alpha$?

Edit 2: My post was edited to say "You believe that you have a fair coin." The coin is in fact fair, and you know that. You do the hypothesis tests anyway. Otherwise the problem is unapproachable because you don't know the probability that any particular toss will come up a certain way.

EDIT: This answer was unclear for OP at first, so I tried to make it clearer through a new approach. Apparently it arose another legitimate doubt, so I tried now to put both answers together and clarify them even more. (Still I might be wrong, but I'll try to express myself better)

What you look for, is the expected number of tosses before we do a Type I error (rejecting $H_0$ when it was true). The probability of that is precisely $\alpha$ (that's another way to define it).

So $P(Type\ I\ error)=\alpha$.

Let $X_n$ be the event of rejecting $n^{th}$ test.

Now, $E[X_1]=\alpha$ stands for the expected number of games (a game is starting to test in the way we do a new coin) where $H_0$ was rejected on the first throw. $E[X_1+X_2]=E[X_1]+E[X_2]$ is the expected number of games where $H_0$ is rejected either on the first or the second throw. Note that with most $\alpha$ this will be lower than $1$, so the expectation for a single game is not to reject $H_0$ yet.

When do we expect to have rejected $H_0$? Precisely when the number of expected games in which we reject $H_0$ is $1$. Therefore, we look for $n$ such as
$$
E[X_1+X_2+...+X_n]=1\\
E[X_1+X_2+...+X_n]=E[nX_1]=nE[X_1]=n\alpha=1\\ n=\frac{1}{\alpha}
$$

The other answer goes like this: Let the variable $T$ count the number of tests before rejecting one. We look for $E[T]$.

Also, using previous notation, $P(X_n)=\alpha(1-\alpha)^{n-1}$ (I'm aware this implies independence between the events $X_n$ and $X_{n-1}$ but since I'm looking for the expected value, for the linearity of the Expected Value, it shouldn't be a problem, though I'm aware I'm not being polite with notation).

$$E[T] = \sum_{n=1}^{\infty}nP(X_n) =
\sum_{n=1}^{\infty}n\alpha(1-\alpha)^{n-1}=
\alpha\sum_{n=1}^{\infty}n(1-\alpha)^{n-1}=\\
\alpha\sum_{n=0}^{\infty}(n+1)(1-\alpha)^{n}=
\alpha(\sum_{n=0}^{\infty}n(1-\alpha)^{n}+\sum_{n=0}^{\infty}(1-\alpha)^{n}) =
\alpha(\frac{1-\alpha}{\alpha^2}+\frac{1}{\alpha}) \\
E[T]=\frac{1}{\alpha}
$$