Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$ Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$

a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$

$a_n =  \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$

$b_n =  \frac{1}{2\pi}\int_0^r r\sin(nr)\,dr = \frac{(-1)^n + 1}{2n}$

$f(x) = \frac{\pi}{8} - \sum_n [\frac{((-1)^n - 1)\cos(nx)}{2\pi n^2} + \frac{((-1)^n + 1)\sin(nx)}{2n}]$

The prof asked us to use this Fourier series to prove that $\pi^2/8 = 1+1/3^2+1/5^2+1/7^2+\cdots$. How do I do this?

This would be one way of going about it but using a different fourier series:
$$f(x)\ = 1+\sum_{n=1}^{\infty}\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}\cos\left(\frac{\pi nx}{2}\right)$$
Notice how $$\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}$$ is 0 when n is even, so we can just  use odd numbers instead with $(2n+1)$.
So we can make a new formula with only odds and we can do this by letting $n =(2n+1)$, and because we don't want the even results we can resolve the numerator to be equal to $2$.
$$f(x)\ =\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\cos\left(n\pi x\ +\frac{\pi x}{2}\right)$$
Then setting $x = 0$ gives:
$$0=\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$
After rearranging:
$$\frac{\pi^{2}}{8}=\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$


Prove that $A$ is an open set and $B$ is a closed set. Suppose that $E \subseteq \mathbb R^m$. Let $$A=\{x \in \mathbb R^m: \rho (x, E) < r\}, B=\{x \in \mathbb R^m: \rho (x, E) \le r\}.$$

Prove that $A$ is an open set and $B$ is a closed set.

Thanks ahead:)

The arbitrary union of open sets is open.  We certainly know that an open ball $B_r(x)$ (of radius $r$ centred at $x$) is an open set.  Behold:

$$A=\bigcup_{x\in E} B_r(x)$$

$B^c=\{x : \rho(x,E) > r \}$. So if $x\in B^c$, let $d=\rho(x,E)$ then $d>r$ so pick $\varepsilon >0$ such that $d -\varepsilon > r$ then by reverse triangle inequality $B_{\varepsilon} (x) \subset B^c$ thus the set is closed (it's complement is open). To see the last claim: Choose $x' \in \partial B_{\varepsilon}(x) $ then by the reverse triangle inequality we have $\rho(x',E) \ge \rho(x,E) -\rho(x,x')=d-\varepsilon > r$ (which proves the claim)

Gvien $B_E(p,r)$ (with $r > 0$ y $p \in E$) y $q \in B_E(p,r)$, we choose $r′ =r − \rho(p,q)$, thus we have that $B_E(q,r′) \subseteq B_E(p,r)$, which is what you wanted to proof.


Finding the boundaries on $\rho$ in spherical coordinates For spherical coordinates, how would we find the boundary on $\rho$ for the volume enclosed by the surfaces $z = \sqrt{x^2 + y^2}$ and $x^2 + y^2 + (z-a)^2 = a^2$? I'm not sure how to algebraic show this, as I can't really do it visually either. Visually, I know it should depend on the value of $\phi$, but that's about it.

I was able to find the boundary on $\rho$ visually, when it was constant bounds as in this question I had submitted earlier: Spherical coordinates and finding the boundaries for $\rho$.

I disagree with the answer posted, and believe the correct answer is indeed $2a\cos\phi$. Here is why. Let us view this region in the $x$-$z$ plane, and since it is symmetric about the $z$ axis, this is all we need to look at. Taking some line in some direction specified by $(\theta,\phi)$, how far does $\rho$ go? Below I have plotted what this looks like with $a=2$:

$\rho$ goes as far as the surface $x^2+y^2+(z-a)^2=a^2$, so we solve for this.
$$\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+(\rho\cos\phi-a)^2=a^2$$
$$\implies\rho^2\sin^2\phi+\rho^2\cos^2\phi-2a\rho\cos\phi=0$$
$$\implies\rho=2a\cos\phi$$

Also, we can check that this is correct. Take the angle $\pm\pi/4$. This is along one of the red lines in my plot. These lines go from $(0,0)$ to $(\pm a,a)$, and so have length $\sqrt{(\pm a)^2+a^2}=\sqrt2a$. Meanwhile using our formula, $2a\cos(\pi/4)=2a\times\frac1{\sqrt2}=\sqrt2a$ as well.

We can also check what happens when we take the angle $\phi=0$. In the image, this of course goes from $(0,0)$ to $(0,2a)$, which has length $2a$. This agrees with $2a\cos(0)=2a$. So our answer seems to be right.

This surface is just a cone bounded from 0, at the origin, to $2a$, in the positive z direction.  In spherical coordinates, the equation of an right circular cone is $\phi = \frac{\pi}{ 4}$.  The minimum value for $\rho$ can be calculated by dropping a perpendicular line from the intersection point of the cone and the sphere.  For this right triangle, $\rho^2 = a^2 + a^2$, so $\rho = \sqrt{2}a$ here.

For the volume integral, the angle $\phi$ will go from $- \frac{\pi}{ 4} $ to $\frac{\pi}{ 4}$, and in doing so will hit $\phi = 0$.  At this point, $\rho$ is simply a vertical line up the $z$ axis through the sphere.  So, it is evident that at this point $\rho$ is at its maximum value:  $2a$.


Computation of conditional probabilities of composite events Let $P(A) = 0.5, P(B|A) = 0.2$ and $P(A \cup B) = 0.8$

Find

a) $P(A \cap B)$
b) $P(A | B^c)$
c) $P(A \cap B^c | A \cup B)$

So for

(a) I know that P( B|A ) = $P(A \cap B)$ / P(A). So $P(A \cap B)$ =  .5 * .2 = .1.

I'm confused as to what (b) is asking for. The probability of A given the complement of B, so I think the complement of B is 0.7 but I'm still confused as to how I'm supposed to approach this problem. Same with (c)

Hint$$p(B/A)=\frac{p(A\cap B)}{p(A)}\implies p(A\cap B)=\cdots\\
p(A\cup B)=p(A)+p(B)-p(A\cap B)\implies p(B)=\cdots\\
p(A/B^c)=\frac{p(A\cap B^c)}{p(B^c)}=\frac{p(A)-p(A\cap B)}{1-p(B)}=\cdots$$

You should use a diagram tree and then fill each node with the values you have.


A square pool is surrounded by a concrete deck A square pool of area $144 \, \text{m}^2$ is surrounded by a concrete deck of area $25 \, \text{m}^2$. What is the perimeter of the outside of the deck?

So, the side of square pool $=\sqrt{144\ M^2}=12\ M$

Assuming the deck to be also square,

if the side the deck is $a\ M,$ the area of the deck with the pool $=a^2\ M^2$

So, the area of the deck $=(a^2-144)\ M^2$

Can you take it home from here?

the area of the pool is 144m^2. The area of the concrete area surrounding the pool is 25m^2. The total area of the concrete block and the pool is 144m^2+25m^2=169m^2.

Now to find the length of each side of a square that includes the pool and concrete.

sqrt(169m^2)= 13 m

Perimeter of the square is 13m
Formula for perimeter of square is P=4a

Formula for square

P=4a
P=4(13m)
P= 52m

Verify

12m + 2x= 52
x=20


Partial derivative of $\sin(xz)$: Is this making sense and is it a proper use of notation? The exercise here is to calculate $\dfrac{\partial}{\partial x} \sin(xz)$

The usual way is just considering $z$ as a constant and deriving, so it would be $z\cos(xz)$.

However, I want to solve it this way:
Define $a=xz$, then $\sin(a) = \sin(xz)$

$\dfrac{\partial}{\partial x} \sin(a) = \dfrac{\partial}{\partial a} \sin(a) \dfrac{\partial}{\partial x} xz = z\cos(a) = z\cos(xz)$.

Is this second way correct? My friend said it is not making sense and not a proper use of notation. But I do not understant why...

(Just to not leave this unanswered)

As comments suggested, this is a fair use of chain rule and there is nothing wrong. The other given answers are talking about problems that doesn't exist.

Yes.. but you are trying to replicate the chain rule in derivatives with the problem being it's not applied correctly. You can not use /∂a for Sin(a) and /∂x for xz.

You'll still get the same answer if you correct your technique slightly -

Here you go -

The Chain Rule says:
derivative of f(g(x)) = f'(g(x))g'(x)

where

f(g) = sin(g)

Considering a=xz
Taking Sine on both sides
sin(a) = sin(xz)

g(x) = xz

∂/∂x sin(a)=∂/∂x (Sin (xz))

-- Applying Chain Rule in Derivatives

0 = Cos(xz).∂/∂x(xz)

=zCos(xz)


Existence $a,b \in \mathbb{R}$ such that $\det(aA+bB) <0 $? Problem: Let $A,B \in \mathcal{M}_n(\mathbb{R})$, and $AB=BA, \det(AB) \not=0$

Existence $a,b \in \mathbb{R}$ such that  $\det(aA+bB) <0 $?

Attempt: if $\det A <0$ or $\det B>0$, we chose $a=1,b=0$ or $b=1,a=0$.

If $\det A>0$ and $\det B>0$ then $\det A^{-1}>0$ and $\det B^{-1}>0$

*) if $n$ is odd number then $\det(aA+bB) <0 \Leftrightarrow \det A^{-1}\det(aA+bB)=\det(a+bA^{-1}B)<0$, we chose $b=1$ and $\det(a+X)$ with ($X=A^{-1}B \in \mathcal{M}(\mathbb{R})$) is a polynomial has deg is an odd. Hence, there exists $a$ such that $\det (a+X)<0$

**) if $n$ is even, I can not go further :(

If $n=2k$ pick $A=B=I_n$. Then
$$\det(aA+bB)=(a+b)^{2k} \geq 0$$

Thank you very much, so that it's not true if $n=2k$.

If this problem has changed by

Problem: Let $A,B \in \mathcal{M}_n(\mathbb{R})$, and $AB=BA, det(AB) \not=0$

Existence $a,b \in \mathbb{R}, a^2+b^2 \not= 0$ such that  $det(aA+bB) =0 $?


Applying Gronwall inequality for $\frac{d}{dt}\|u(t)\|_{L^{q}}^q \leq f(t) + C\|u(t)\|_{L^{q}}^{q-2}$ Suppose we have a function $u(x,t)$, where $x \in \mathbb R^n$ and $t>0$ denotes time. And suppose $q>2$, then how to apply Gronwall to the inequality below?
$$\frac{d}{dt}\|u(t)\|_{L^{q}}^q \leq f(t) + C\|u(t)\|_{L^{q}}^{q-2}.$$
Here $f(t)$ is any function and $C$ is a constant, both of which do not matter here since my question is how to use Gronwall when we have a power.

I just found the answer myself but the idea was inspired by Alex Francisco who gave his answer above. So we just use Young's inequality, i.e. we consider $\|u(·, t)\|_q^{q-2}=1\times\|u(·, t)\|_q^{q-2} \leq C + C_1\|u(·, t)\|_q^{q}$, where $C$ and $C_1$ are constants.

$\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}$Note that for any $a, b, x > 0$, by AM-GM,\begin{align*}
&\peq ax^2 + \frac{b}{x^{q - 2}} = (q - 2) · \frac{ax^2}{q - 2} + 2 · \frac{b}{2x^{q - 2}}\\
&\geqslant q \left( \frac{ax^2}{q - 2} \right)^{\frac{q - 2}{q}} \left( \frac{b}{2x^{q - 2}} \right)^{\frac{2}{q}} = q \left( \frac{a}{q - 2} \right)^{\frac{q - 2}{q}} \left( \frac{b}{2} \right)^{\frac{2}{q}}. \tag{1}
\end{align*}
By setting $\left( \dfrac{q - 2}{a} \right)^{\frac{q - 2}{q}} = q \left( \dfrac{b}{2} \right)^{\frac{2}{q}} = c^{-\frac{2(q - 2)}{q}}$, where $c > 0$ is a constant, (1) becomes$$
(q - 2) c^2 x^2 + \frac{2}{q^{\frac{q}{2}} c^{q - 2} x^{q - 2}} \geqslant 1.
$$
Thus for any $c > 0$,$$
x^{q - 2} \leqslant (q - 2) c^2 x^q + \frac{2}{q^{\frac{q}{2}} c^{q - 2}}. \quad \forall x \geqslant 0 \tag{2}
$$

Because $\dfrac{\d}{\d t} (\|u(·, t)\|_q^q) \leqslant f(t) + C \|u(·, t)\|_q^{q - 2}$, then for any $c > 0$,$$
\frac{\d}{\d t} (\|u(·, t)\|_q^q) \leqslant f(t) + C \|u(·, t)\|_q^{q - 2} \leqslant f(t) + \frac{2C}{q^{\frac{q}{2}} c^{q - 2}} + (q - 2) c^2 C \|u(·, t)\|_q^q.
$$
Now Gronwall's inequality can be applied.


simple systems of equations problem Choose h and k such that the system has 1) no solution, 2) a unique solution, and 3) many solutions. Give separate answers for each part.

x-3y=1,
2x+hy=k

For 1) and 3), isn't that impossible? And for 2) can't h and k be any real number?

You have that $x-3y=1$, $2x+hy=k$. When you multiply the first equation with $2$, you get $2x-6y=2$. Subtrating the two equations, you get, $2x+hy-(2x-6y)=k-2\Rightarrow (h+6)y=k-2$. If $h=-6$ and $k\neq 2$, you get $0y\neq 0$, so a contradiction. Therefore, for $h=-6, k\neq 2$, there are no solutions. If you take $h=-6$ and $k=2$, you get $0y=0$, so $y$ can have any value. If $h\neq -6$, it follows that $y=\frac{k-2}{h+6}$ and $x=1+3\frac{k-2}{h+6}$, therefore you get a unique solution.

You could solve this, using determinants and matrix for systems, if you want a general algorithm.

You wish to analyze the system
\begin{array}{rcrcr}
x & - & 3\,y & = & 1 \\
2\,x & + & h\,y & = & k
\end{array}
To do so, note that $h\neq-6$ implies
$$
\DeclareMathOperator{rref}{rref}\rref
\begin{bmatrix}
1&-3&1\\
2&h&k
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 3\,\frac{k-2}{h+6}+1 \\
0 & 1 & \frac{k-2}{h+6}
\end{bmatrix}
$$
This means that the system has a unique solution when $h\neq-6$. Do you see why?

Can you find the reduced row-echelon form of the system when $h=-6$?


Does the least value of $\cos A + \cos B + \cos C$ exist, where A, B, C are angles of a triangle? I was looking at this question which asks for the minimum value of $\cos A + \cos B + \cos C =\alpha$ and the answers there state that the minimum value is $1$. This value exists for a degenerate triangle.
But in a similar question which asks for the minimum area ($A$), for a given semi-perimeter ($s$), the possible values of $A$ are
$$\frac {s^2}{3√3} \tag 1$$
in the case when $a=b=c$,
and
$$\frac {s^2}{4} \tag 2$$
in the case when $s=s-a=s-b=s-c$.
But we reject $(2)$ here as then the triangle wouldn't be an "appropriate" triangle (as my teacher said).
Even though in both $\alpha$'s as well as $(2)$'s case the triangles become degenerate but one of them is acceptable and the other is not. Why so?
Also, if possible please point out the corrections to any conceptual mistake that I may have.
Thank you!

We'll begin with a proof, possibly not the shortest possible one, of an identity @HarishChandraRajpoot quoted:$$\begin{align}\cos A+\cos B-\cos(A+B)&=1+\sin A\sin B-(1-\cos A)(1-\cos B)\\&=1+4\sin\tfrac{A}{2}\sin\tfrac{B}{2}\cos\tfrac{A+B}{2}\\&=1+4\sin\tfrac{A}{2}\sin\tfrac{B}{2}\sin\tfrac{C}{2}.\end{align}$$This can be made arbitrarily close to $1$, but must exceed it for a nondegenerate triangle, so $1$ is an infimum but not a minimum obtainable in nondegenerate triangles. If $A,\,B$ are both small, to quadratic order$$\cos A+\cos B-\cos(A+B)\approx1-A^2/2+1-B^2/2-1+(A+B)^2/2=1+AB.$$On the other hand, the supremum is an achievable maximum. Since$$\frac{d^2}{dx^2}\ln\sin x=\cot^\prime x=-\csc^2 x<0,$$by Jensen's inequality$$\ln\sin\tfrac{A}{2}+\cdots\ge3\ln\sin\frac{\pi}{6}=-3\ln 2\implies 1+4\sin\tfrac{A}{2}\cdots\ge1+4e^{-3\ln2}=\tfrac32,$$with equality iff the triangle is equilateral. (Note that if we'd tried using Jensen's inequality directly on a sum of cosines, the second derivative $-\cos x$ would have had a different sign for obtuse angles, which is why we had to work in terms of half-angles.)

Does not exist.
For $a=b=1$ and $c\rightarrow2^-$ we see that
$$\cos\alpha+\cos\beta+\cos\gamma\rightarrow1.$$
We'll prove that $$\cos\alpha+\cos\beta+\cos\gamma>1.$$
Indeed, we need to prove that:
$$\sum_{cyc}\frac{b^2+c^2-a^2}{2bc}>1$$ or
$$\sum_{cyc}(a^2b+a^2c-a^3)>2abc$$ or
$$(a+b-c)(a+c-b)(b+c-a)>0,$$ which is obvious.
The value $1$ does not occur, which says that the minimal value does not exist.
Now, we see that $\sum\limits_{cyc}\cos\alpha\rightarrow1^+$ for any $a+b\rightarrow c^+$.
Your case with $p-a=p-b=p-c\rightarrow p$ is impossible because $a$, $b$ and $c$ are positives.
For $a=b=c$ we'll get a maximal value of the sum.


Gold bar inspection question There is a vault containing (N) 10,000 gold bars.

A number of different bars (I) can be chosen at random and inspected for purity.  Assume all inspections come back indicating the bar inspected is pure.

How many inspections are needed to be able to say there is a 90% chance that 99.9% of the gold bars are pure?

Let $M$ represent the number of impure golden bars in the vault. The relevant probability statement is $Pr[No\ impure\ bars\ after\ I\ inspections|M\leq10]\geq0.9$. Is there some way to simplify the left side of the equation in terms of the given variables? Then you should be able to solve for $I$.


Finding side of triangle? this is probably a basic question but I haven't taken a geometry class since high school

My professor wants us to create a program that tells the user to enter three points (x1,y1),(x2,y2), and (x3,y3) to ultimately find the area of a triangle

she wants us to use this formula; s=(side1+side2+side3) then area^2 = ( s(s-side1)(s-side2)(s-side3) )

My question is how do I get the sides??

Another way to approach to the solution. Is use the following:

  Let $A=(x_1,y_1)$, $B=(x_2,y_2)$ and $C=(x_3,y_3)$ points in a plane. The area of the triangle $ABC$ is $$[ABC]=\frac12 \left| \begin{matrix} 1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{matrix}\right|$$

Then you only have to find a determinant, and if you do not want negative area, apply absolute value to the final answer.

Use the co-ordinates given to you by the  user. Assuming they are in order or appearance, x =x2 - x1 gives you the horizontal length of the side, and y = y2-y1 gives you the vertical distance, from there you take sqrt ( x^2 + y^2) and volia, you have your side length. Repeat for the other sides and then use the formula your professer gave


Proving the convergnce of a sequence So, I have to prove that the sequence defined as
$a_{n+1}=\frac{6(1+a_n)}{7+a_n}$ converges and then find the limit.

I have few questions; Do i have to assume that $a_n \geq 0$ or $a_n \leq0$. Because if i assume that $a_n \geq 0$, I can prove that

$\left|a_{n+1}-2\right| \leq \frac{4}{7}\left|a_{n}-2\right|$, and by induction, it would be easy to show that $a_n \to 2$.

Any help would be appreciated. Is there a way to prove convergence without knowing the signs of $a_n$?

If you want an easier way to find what value the limit converges to, consider that by the definition of convergence, as $n\to\infty$, $a_{n+1} = a_n$.

Thus, let's let $a_{n+1} = a_n = L$.

Substituting this in, we get:

$$a_{n+1}=\frac{6(1+a_n)}{7+a_n}$$
$$L = \frac{6(1+L)}{7+L}$$
$$L = \frac{6 + 6L}{7+L}$$
$$7L + L^2 = 6 + 6L$$
$$L ^2 + L - 6 = 0$$
$$(L+3)(L-2) = 0$$

Therefore, $L = -3, 2$

Since all of the terms in this sequence are positive, $L = -3$ s an extraneous solution.

Therefore, the sequence converges to $2$.

As for proving that the sequence converges, assume that $a_n \ge 0$ and use induction.

As Varun showed, the fixed points obeying $L=6(1+L)/(7+L)$ are $+2$ and $-3$. For $a_0\gt -3$, one may see that one is getting close to $L=+2$. For $a_0\gt -3$, the subsequent terms are getting more negative, but they inevitably drop below $a_n\lt -7$ at some point, and for $a_0\lt -7$, $a_1$ may already be calculated to be positive i.e. $a_1\gt - 3$, because the graph of $y=6(1+x)/(7+x)$ is a hyperbola, and one starts to converge towards the limit $L=2$, anyway.

As Arpit wrote, Contraction Mapping Theorem guarantees the existence of a limit. The fixed points $L=+2$ and $L=-3$ are the only possible limits. And the sequence diverges away from $L=-3$ for almost all $a_0$. The only exception occurs when $a_0=-3$ when all $a_n=-3$ and $L=-3$ is also the limit. For all other values of $a_0$, the limit has to be $L=+2$.


Metric Space non-negative A common proof of the non-negativity of a distance in a metric space is the following (I found one like this on Wikipedia):

$d(x,y)+d(y,x) ≥ d(x,x)$

$d(x,y)+d(x,y) ≥ 0$

$2d(x,y) ≥ 0$

$d(x,y) ≥ 0$

However, the very first step seems to assume non-negativity. If we don't assume non-negativity then it is possible that

$d(x,y)+d(y,x) < 0$

It seems wrong that we assume the conclusion in the first step of the proof...

Thanks in advance.

The first step is triangle in equality. [Put $z=x$ in $d(z,x) \leq d(z,y)+d(y,x)$ ]. If you assume that all properties of a metric except non-negativity are satisfied then  this proof does show that $d(x,y) \geq 0$ for all $x,y$.

The second step uses the fact that $d(x,x)=0$. The third one uses symmetry: $d(x,y)=d(y,x)$.

The distance in a space $X$ is, by definition, a function $d:X \mathrm{x} X\to\mathbb{R}^+$. So the non-negativity of $d$ is given in the definition itself.


Probability of drawing a pair from a poker hand, unordered with replacement? I am wondering what is the probability with which you can draw a pair in a 5-card hand from a standard 52-card deck, if order does not matter in the context of cards in the hand, and if the cards can be replaced with every draw?

To find the probability of a one-pair hand, it will be easier if we take the order of drawing into account. It is true that all we care about is what hand we end up with. But taking order into account will get us the right answer without the need to look at special cases.

There are $52^5$ equally likely sequences of $5$ cards. We count the number of $5$-card sequences that satisfy the "one pair" condition. Counting "hands" will not work unless we are very careful, for not all hands are equally likely.

The two locations in the draw where the  pair will be  can be chosen in $\binom{5}{2}$ ways. Once we have chosen the two locations, the kind we have two of can be chosen in $13$ ways. Once we have done that, the leftmost chosen "pair" location can be filled in $4$ ways, and for each such way we can fill the other pair location in $4$ ways, for a total of $4^2$.

Now it is time to deal with the remaining $3$ positions. There are $48$ ways to fill the leftmost empty position. For each of these, we can fill the next position in $44$ ways, and then the last position can be filled in $40$ ways.

The number of one-pair sequences is therefore
$$\binom{5}{2}(13)(4^2)(48)(44)(40).$$
Divide by $52^5$ for the probability.

Remark: We interpreted "the cards can be replaced at every draw" to mean that the drawn card is replaced in the deck before the next draw. So the model is that we draw a card, record its value, replace and shuffle, doing this a total of $5$ times. We want the probability that among the $5$ recorded cards, there are exactly two cards (possibly identical) of one kind, and $3$ useless cards all of different kinds.

"There are 525 equally likely sequences of 5 cards." This is already erroneous. We have 52 possibilities for the 1st card, but this leaves only 51 for the next choice, 50 for the next choice, and so on. So there are 52*51*50*49*48 possible 5-card draws.

Other errors are also present.


Constrained optimization problem? Mats in a yoga studio Suppose you had a yoga studio measuring $21 \times 13$ ft, and you want to fill it with yoga mats measuring $6 \times 2$ ft, under the constraint that yoga mat edges are at least $1$ ft apart from other yoga mat edges, and that all yoga mat edges are at least $0.5$ ft away from the nearest wall.
I did this in a silly geometric way, by rearranging rectangles like a puzzle and seeing how many I could fit in the studio. My best shot was $13$, and that is the correct answer apparently. The (horribly not-to-scale) sketch below shows the overall organization: The long edge of $6$ yoga mats should line both long sides of the yoga studio ($3$ mats on each side), and in the middle you can just fit $7$ mats (long side by long side).
I would like to approach this problem from a constrained optimization perspective. First, this would confirm that the optimal number of mats that would fit is $13$ (i.e., I wouldn't have to take the book's word for it), and second, I want to be able to approach geometric optimization problems like this routinely in a principled way, rather than resort to heuristics.

This is NOT an approach via constrained optimization, but rather a way to confirm that the optimal value is indeed $13$. Hence I am not actually answering your question in full, but this is too long to fit in a comment.
Since the yoga mats must be $0.5$ ft from the walls and $1$ ft from each other, we can "expand" the yoga mats so they have an extra $0.5$ ft on each side.
Now these expanded yoga mats, with dimensions $3$ ft $\times\ 7$ ft and an area of $21$ ft$^2$, can stick to the walls and each other.
How many of these yoga mats can we fit? The maximum is precisely $\dfrac {21\times 13}{3\times 7} = 13.$
EDIT: I'm a little bit worried about the extra $0.5\times 0.5$ corners generated from this method. However in this case, they only contribute $1$ ft$^2$ for each mat, so the maximum remains $\left\lfloor \dfrac {21\times 13}{20}\right\rfloor = \lfloor 13.65\rfloor =13$.

Each mat requires a 0.5 ft border on each side.  The required space for each mat is  21 sq ft.  The yoga studio is $21\times 13$ sq ft.
If you have found a way to fit 13, it must be optimal, as you have no wasted space.


How can QR-decomposition be used for low-rank estimation? In a coursebook I'm reading, some algorithms used to calculate the QR-decomposition of a matrix are introduced, namely Gram-Schmidt orthogonalization and the use of Givens rotations with or without column pivoting.

A question that has been posed on previous exams asks how the QR-decomposition can be used to do low-rank approximation of a matrix. This is unclear to me. I understand how this can be done using singular value decomposition, but cannot seem to grasp how the same can be achieved using the QR-decomposition.

Thanks in advance!

You want a "rank revealing" QR decomposition, see https://math.berkeley.edu/~mgu/MA273/Strong_RRQR.pdf or https://people.csail.mit.edu/yujia/assets/pdf/rrqr_slides.pdf or https://www.irisa.fr/sage/wg-statlin/WORKSHOPS/LEMASSOL05/SLIDES/QR/Guyomarch.pdf (or even how rank-revealing QR factorization determine the rank of the matrix)


How does one find the Zariski closure of a set? I've started to learn algebraic geometry this week (so I do not have much knowledge in the subjet) and, after reading the definition of the Zariski closure $V(I(S))$ of a set $S$, I've tried to do the following exercises without much success:

  Find the Zariski closure of the following sets:

  1) $\{(n^2,n^3): n \in \mathbb{N} \} \subset \mathbb{A}^2(\mathbb{Q})$

  2) $\{(x,y): x^2+y^2 < 1 \} \subset \mathbb{A}^2(\mathbb{R})$

Any help with these? In general, how does one usually attacks this kind of problem?

Thank you!

Since the first part has been dealt with in the comments:

Assume $f(X,Y)=\sum_{i,j\ge 0}a_{i,j}X^iY^j\in\mathbb R[X,Y]$ is a polynomial with $f(x,y)=0$ whenever $x^2+y^2<1$.
Let $\alpha\in\mathbb R$ and consider the polynomial $g(T)=f(T,\alpha T)\in\mathbb R[T]$. Then $g(t)=0$ whenever $t^2(1+\alpha^2)<1$. As there are infinitely many such $t$, $g$ must be the zero polynomial. We conclude that each of its coefficients
$$\sum_{i+j=k}a_{i,j}\alpha^j $$
is zero - no matter what $\alpha\in\mathbb R$ we pick. Hence each of the polynomials $\sum_{i+j=k}a_{i,j}X^j$ has infinitely many roots, hence is the zero polynomials, hence all $a_{i,j}$ are zero. We conclude $f=0$.

It So question is “find Zariski closure for infinite set of 1-dimension objects (points) in two-dimension space”, and the answer is f=0. Isn’t that obvious? To some extent. We need projection to 1st dimension (x,  and to find “closure” there, it will be f(x, y) = g(y) space. Because non-trivial polynomial of (x) can have only finite number of roots. Then second projection f(x,y) = h(x). And intersection of these spaces is f(x,y) = 0.


Finding the closest distance between a point and a surface Assuming that I am given a matrix (A) and a vector (b) which represents a surface, say

\begin{bmatrix}2&0&1\\0&1&0\\1&0&2\end{bmatrix} and \begin{bmatrix}4\\0\\2\end{bmatrix}

and the equation representing the surface as:
$r^TAr+b^Tr=1$

If I am asked to find the closest distance between point "a" and the surface, represented by a position vector of

\begin{bmatrix}-1\\0\\0\end{bmatrix}

What am I supposed to do?

Below is what I've attempted:

Using this as an example, I tried to rewrite the equation representing the surface into quadratic form, in this case:

$x^T Ax=3$

where $r=x+a$ (note: $a$ is the point where I am trying to find it's distance from the surface).

I'm thinking about rotating the axis, however I am stuck here (my teacher told me to try not consider a rotation matrix).

My first idea for what to do: First, rewrite your equation $$r^TAr + b^Tr = 1 $$ into $$(r-c)^TA(r-c) = d$$
(You can find a suitable $c$ by multiplying out $(r-c)^TA(r-c)$, rewriting $r^TAc=c^TAr$ and solving $b^T = -2c^TA$. Then $d$ must be $1+c^TAc$ -- though I haven't triple-checked my reasoning for sign errors; caveat lector).

$A$ is real symmetric and therefore orthogonally diagonalizable; it is easy to see that its eigenvalues are $1,1,3$ and therefore your surface is an oblate spheroid centered on $c$.

Now the natural thing to do would be to transform $a-c$ by the orthogonal diagonalizing matrix, so you end up with $a-c$ in a coordinate system where $A=\operatorname{diag}(1,1,3)$. Then by symmetry the shortest line betwen $a$ and the spheroid would be in the plane that contains the transformed point and the $z$-axis, and we've reduced the problem to finding the shortest distance between a point and an ellipse. That seems to have no nice closed-form solution, but at least it is now easy to parameterize the ellipse and find a minimum numerically by setting the derivative of the squared distance to $0$.

However, this works because the diagonalizing matrix for $A$ can be chosen to be orthogonal -- that is, a rotation matrix -- which sounds like what your teacher is asking you not to do.

It is possible that you're supposed to use Lagrange multipliers instead.

Closest distance between a plane and a line goes like this:
If plane is ax+by+cz=d where a,b,c,d are constants and the point is (h,k,l)
Then shortest distance is
(ha+bk+cl-d)/root(a^2+b^2+c^2)


Comparing a series expansion to polynomial regression So I don't have a great background in mathematics but I have a quick and hopefully simple question for you guys.

I'm a graduate student and I'm doing some polynomial regression on some thermodynamic data to create a model and I'm kind of struggling a bit with understanding some of the differences between a series expansion and a regression. Maybe it's a subtlety or maybe I'm just ignorant.

But anyway. Let's say I take a Maclaurin series expansion to third order:

$$f(x) = f(0) + f^\prime(0) x + \frac{1}{2!}f^{\prime\prime}(0) x^2 + \frac{1}{3!}f^{\prime\prime\prime}(0)x^3 + O(x^4)$$

But this is just a polynomial:

$$f(x) = a + bx + cx^2 + dx^3$$

where the coefficients are the derivatives of $x$ and their associated coefficients.

In the series expansion, the coefficients are defined to be the $n^{th}$ derivatives computed at $0$.  But when I compare this to the polynomial expansion, can I really make the identity that each of the coefficients is proportional to the derivative computed at $0$? Or are the values of the coefficients more globally defined since they have been obtained through regression?

The reason I ask is because it has important implications in the language I will use to describe the values of the coefficients in the polynomial fit.

Any help or insight would be greatly appreciated.

Edit.

I have a model where I can compute the internal energy of a system $U$ as a function of the number of particles $N$ and then some extensive property of the system $\xi$, so $U \equiv U(N,\xi)$.

I am ultimately interested in calculating the chemical potential $\mu$, which is the change in the internal energy as particles are added to they system, this is expressed as taking a derivative with respect to $N$:

$$ \mu(N,\xi) = \left(\frac{\partial U}{\partial N}\right)_{\xi}$$

So since I am able to calculate $U(N,\xi)$, I can compute $\mu(N,\xi)$ by fitting a 2D surface (polynomial) to the data and then computing its derivative with respect to $N$.

$$ U(N,\xi) = U(0,0) + U_{ ξ } \xi + U_N N + \frac{1}{2!}U_{ξξ} ξ ^2 + \frac{1}{2!}U_{NN} N^2 + U_{N ξ }N ξ  + \dots$$

This can be rewritten as
$$ U(N,\xi) = U(0,0) + \left(\frac{\partial U(0,0)}{\partial \xi}\right)_{N} \xi + \mu(0,0) N + \frac{1}{2}\left(\frac{\partial^2 U(0,0)}{\partial ξ^2}\right)_N ξ ^2 + \frac{1}{2}\left(\frac{\partial \mu(0,0)}{\partial N}\right)_{\xi} N^2 + \left(\frac{\partial \mu(0,0)}{\partial \xi}\right) N ξ  + \dots$$

So the derivatives in this model have physical meaning, since they represent thermodynamic variables and derivatives of thermodynamic variables.

So in this case,

$$ \mu(N, ξ ) = \mu(0,0) + \left(\frac{\partial \mu(0,0)}{\partial N}\right)_{\xi} N + \left(\frac{\partial \mu(0,0)}{\partial \xi}\right) ξ + \dots $$

The main question above was if I perform a polynomial regression on my $U(N, ξ )$ data, can I reasonably identify each of the polynomial's coefficients with the $\partial_i U$ values which are supposed be computed at $N =  ξ  = 0$ in the Taylor expansion, or will these coefficients correspond to something else since they are obtained by regression.

I suppose the fitted coefficients should truly correspond to the $N =  ξ  = 0$ case.

I'm not sure if I understand your question properly but...

Take for example series of $\exp$:

$$e^x=a+b\,x+c\,x^2+d\,x^3+...$$

Plug $x=0$ to find $a$:

$$a=e^0=1$$

Differentiate both sides:

$$\frac{d}{dx}e^x=\frac{d}{dx}(a+b\,x+c\,x^2+d\,x^3+...)$$
$$e^x=b+2c\,x+3d\,x^2+...$$

And plug $x=0$ to find $b$:

$$b=e^0=1$$

Differentiate again and plug $x=0$:

$$e^x=2c+6d\,x+...$$
$$2c=e^0=1\implies c=\frac12$$

Repeat to find other coefficients.

Seems more than $~3~$ years has happened but maybe this may be useful for someone.

I have tested what you asked on a trial function: $~~~f(x)=x*e^x~$.

I fitted the function using a polinomial function of eight order, and then I obtained the Taylor series expansion around $~x=0~$.

If you plot both function for $~x~$ on $~[0:1]~$ you can see that the functions are not the same.


Does $HK \cong H \times K$ imply that $H$ and $K$ satisfy the Internal Direct Product criteria? Let $H$ and $K$ be subgroups of a group $G$, and suppose that $HK \cong H \times K$.

Does it then follow that that $HK$ is the internal direct product of $H$ and $K$? In other words, does it follow that both $H$ and $K$ are normal in $HK$, and that $H \cap K = \{e\}$?

Not necessarily - it depends on what the isomorphism is.

Say $G=\mathbb{Z}_2^{\omega}$ and take $H=\mathbb{Z}_2\times \mathbb{Z}_2 \times 0 \times 0 \times 0 \dots\;\; (\simeq \mathbb{Z}_2 \times \mathbb{Z}_2),$ $K=0 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \dots \;\;(\simeq \mathbb{Z}_2^{\omega})$.

Then $HK=G=\mathbb{Z}_2^{\omega}\simeq (\mathbb{Z}_2\times \mathbb{Z}_2) \times \mathbb{Z}_2^{\omega} \simeq H \times K$, but $H \cap K = 0 \times \mathbb{Z}_2 \times 0 \times 0 \dots \;\; \neq 0$.

However, assuming both $K,H$ are normal in $G$, you can say the following:

  Assume that the map $H \times K \rightarrow HK, \;\; (h,k) \mapsto hk$
  is an isomorphism. Then $H \cap K = \{e\}$.

Yes that's right since the intersection is the identity e. And  h and k are normal subgroup it will follow that h×k =hk


Combinatorial Argument for a Binomial idenity The following problem appeared in Putnam Competition, 1965:

Prove that
$$ \sum_{r=0}^{\lfloor \frac{n-1}{2}\rfloor} \left(\frac{n-2r}{n} \binom{n}{r}\right)^2 = \frac{1}{n} \binom{2n-2}{n-1} $$
where $\lfloor x \rfloor$ is the integer part of $x$. This is not difficult to prove using Algebra. Since the right hand side is the $(n-1)$th Catalan number, is there a Combinatorial argument that would establish the result?

The well-known result that
$$ \sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}n $$
has an easy bijective combinatorial proof based on counting the
$2n$-digit binary words with $n$ 1's and $n$ 0's
by splitting it into two $n$-digit words and counting
the number of 1's and 0's in each.

Similarly, the Catalan number result
$$ \sum_{k=0}^{\lfloor n/2\rfloor}
\left(\binom{n}{k}\!-\!\binom{n}{k\!-\!1}
\right)^2 \!=\!
\sum_{k=0}^{\lfloor n/2\rfloor} T(n,k)T(n,k)
= C_n $$
has an easy bijective combinatorial proof. The numbers being squared are the
triangular sequence OEIS A008315. Note the A008315 comment

  T(n,k) is the number of n-digit binary words (length n sequences on {0,1}) containing k 1's such that no initial segment of the sequence has more 1's than 0's.

Note that one combinatorial interpretation of $C_n$ is that it is the number
of $2n$-digit binary words such that the number of 1's and 0's are both
equal to $n$ and such that no initial segment of the word has more 1's than 0's.

Now split each $2n$-digit binary word into two $n$-digit words as before.
The first $n$ digits has the required relation between the 1's and 0's.
Now reverse the order of the last $n$ digits and change all the 0's to 1's
and vice versa. This also has the required relation between the 1's and 0's.

As a comment by Mike Earnest indicates, it is possible to split each
$2n$-digit word into unequal parts. Thus, by similar
reasoning as above, the generalized identity
$$ \sum_{k=0}^{\lfloor r/2\rfloor}
 T(r,k)T(2n-r,k+n-r) = C_n $$
where $\,0\le r\le n\,$ is proved.

You can simplify the expression to
$$
\sum_{r=0}^{b}\binom{n}{r}^2  - \sum_{r=1}^{b}\binom{n}{r}\binom{n-1}{r-1}
$$
which are unlikely to exist in closed form, because most binomial sums don't. You can get an approximation.

For example, you can divide the first sum by $\binom{n}{b}^2$ to get
$$
\frac{S_b}{\binom{n}{b}^2} = 1 + \frac{\binom{n}{b-1}^2}{\binom{n}{b}^2} + \ldots + \frac{1}{\binom{n}{b}^2}
$$
Each $k^{th}$ term simplifies to
$$
\Big(\frac{b}{n-b+k}\Big)^2
$$
for which you can find bounds since you know $\frac{n-1}{2}\leq b = \lfloor\frac{n-1}{2}\rfloor \leq\frac{n}{2}$, and hence the asymptotic bounds on the sum.

EDIT: for $n geq 2$ I compared every term to its upper-bound (rewrite $L=e^{\log L}) e^{-\frac{4k}n}$ and the corresponding integral is equal to
$$
\frac{n}{4}\Big(e^{-\frac{4}{n}} - e^{-2} \Big)
$$
Now multiply by $\binom{n}{b}^2$ to get the upper bound on the first sum:
$$
\sum_{r=1}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{r}^2 \leq \binom{n}{\lfloor \frac{n-1}{2} \rfloor}^2 \frac{n}{4}\Big(e^{-\frac{4}{n}} - e^{-2} \Big)
$$


Functional equation: $ f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17$ $f: \mathbb R \to \mathbb R, f(x^2+x+3)+2f(x^2-3x+5)= 6x^2 -10 x + 17
\forall x \in \mathbb R $, then find the function $f(x)$

I have:

$f(15/4)= 16/3$ (both quadratics intersect at $x=1/2$)

$f(3)= 3$ (by substituting $x=0$ and $x=1$ and then solving the simultaneous equations obtained)

$f(5)= 7$

I am not getting anything fruitful from these.

Could anyone provide me a hint on how to proceed?

Edit:

Someone had commented (now it's deleted) that we can assume $f(x) = ax +b$, how can we do that? I got the right answer using that.  Is it always okay to assume that way? When is it a reliable assumption?


Special values of $j$-invariant Let $j(\tau)$ be Klein's absolute invariant defined for $\tau \in \mathbb{H}$ by
$$j(\tau) = q^{-1} + 744 + 196884q + 21493760q^2 + 864299970q^3 + \cdots$$
with $q := e^{2\pi i\tau}$.
Are there any known special values of $j(\tau)$ for which $\textrm{Re}(\tau)$ is irrational? Wikipedia has a list of special values but in all of those cases $\textrm{Re}(\tau)$ is rational.

Since it wasn't specified that $\tau$ need to be an algebraic number, then yes, there are infinitely many special values of $j(\tau)$ where $\Re(\tau)$ is irrational. However, $\tau$ has a hypergeometric closed-form.

It is known that the equation,
$$j(\tau) = n$$
can be solved for $\tau$ in terms of the hypergeometric function. Using Method 4, we find,

$$\tau = \frac{_2F_1\big(\tfrac16,\tfrac56,1,1-\alpha\big)}{_2F_1\big(\tfrac16,\tfrac56,1,\alpha\big)}\sqrt{-1}$$

where

$$\alpha=\frac{1+\sqrt{1-\frac{1728}n}}{2}$$

For example, negating the year $n=-2017$, then,

$$\tau \approx -0.273239 + 0.6868913\,i$$

such that,

$$j(\tau)=-2017$$

The real values of the j-invariant are of interest in theoretical physics (for instance, Witten’s 3d gravity j-invariant is associated to black hole entropy). Now observe that the real number $\phi$, meaning the golden ratio $\phi$, is a special root of the (degree 10) polynomial that one obtains when writing $j$ as $A+iB/ C +iD$ and looking for real parts (using the usual ratio involving the modular discriminant). This is one of few integer values of $27 j$ (watch out for the normalisation factors), like the other special values on the ribbon graph (dessins of Grothendieck).

Why integer values are special is a much more difficult question!


Is every embedded submanifold globally a level set? It's a well-known theorem (Corollary 8.10 in Lee Smooth) that given a smooth map of manifolds  $\phi:M\rightarrow N$ and a regular value $p\in N$ of $\phi$, the level set $\phi^{-1}(p)\subset M$ is a closed embedded submanifold. Is the converse true? That is, given an embedded submanifold $S\subset M$, is there necessarily a manifold $N$, smooth map $\phi:M\rightarrow N$, and regular value $p\in N$ of $\phi$ such that $S=\phi^{-1}(p)$?

Prop. 8.12 in Lee Smooth shows that this is true locally; specifically,

  Let $S$ be a subset of a smooth $n$-manifold $M$. Then $S$ is an embedded $k$-submanifold of $M$ if and only if every point $p\in S$ has a neighborhood $U\subset M$ such that $U\cap S$ is a level set of a submersion $\phi:U\rightarrow\mathbb{R}^{n-k}$.

(and any level set of a submersion is of course the level set of a regular value). I feel like this is the kind of question where, if there is a counterexample, it probably is very simple, but I wasn't able to come up with one.

A closed submanifold $S\subset M$ of codimension $k$ is the inverse image of a regular value of a smooth map $f:M\rightarrow S^k$ if and only if it has trivial normal bundle.

One implication is explained in evgeniamerkulova's answer (and still holds with $S^k$ replaced by any other manifold of dimension $k$).

For the other one, pick a tubular neighbourhood $U$ of $S$ in $M$. Then $U$ is diffeomorphic to $S\times\mathbb{R}^k$, because $S$ has trivial normal bundle. You can now define a map $U\rightarrow \mathbb{R}^k$ by $(x,v)\mapsto v$ and extend it to $M$ by mapping the complement of $U$ to infinity. You can then approximate the resulting map by a smooth map that has $0$ as a regular value and whose zero set is $S$. (This is exercise 4.6.5 in Hirsch's Differential Topology).

If $M$ is compact, you can find a tubular neighborhood $U$ of $M$, which is diffeomorphic to the normal bundle $\mathcal N$ of $M$ in $N$, by some diffeo $h:\mathcal N\to U$. Now pick a metric on $\mathcal N$, and let $q:\mathcal N\to\mathbb R$ be the square of the corresponding norm. If $\phi:\mathcal N\to\mathbb R$ is a smooth function such that $\phi(v)\sim\tfrac1{q(v)}$ outside of the unit balls in each fiber, with the approximation better and better as $q(v)$ grows, and constantly equal to $1$ inside the balls of radius $\tfrac12$ in each fiber, then the product $\phi\cdot q$ composed with $h$ gives you a map $U\to\mathbb R$, which extends to a smooth map $N\to\mathbb R$ (with value $1$ outside of $U$), whose zero-set is precisely $N$.

If $M$ is not compact, you can have problems: for example, consider the map $t\in\mathbb R\mapsto ((1+e^t)\cos t,(1+e^t)\sin t)\in\mathbb R^2$. Any function vanishing in its image also vanishes on the unit circle.


Finding the asymptotes of the trajectory of a point parameterized by $(x,y)=\left(\frac{t}{1-t^2},\frac{t-2t^3}{1-t^2}\right)$ for $t\geq 0$ Let $(x,y)$ be Cartesian coordinates in the plane and suppose a moving point has coordinates $$x=\dfrac{t}{1-t^2}, \quad y=\dfrac{t-2t^3}{1-t^2}$$ at time $t\ (t\geq 0)$. Describe the trajectory of the point and find asymptotes.

Solution: I sketched the trajectory of the point and it was not so difficult. Firstly we need to sketch $x(t)$ and $y(t)$ then we need to "combine" them in order to get the graph of $(x,y)$.
But I have some issues with finding asymptotes.
Definition: The line $c_0+c_1x$ is called an asymptote of the graph of the function $y=f(x)$ as $x\to - \infty$ (or $x\to +\infty$) if $f(x)-(c_0+c_1x)=o(1)$ as $x\to -\infty$ (or $x\to+\infty$).
Proposition: The line $y=kx+b$ is an oblique asymptote for the function $f(x)$ as $x\to +\infty$ if and only if $\lim \limits_{x\to +\infty}\dfrac{f(x)}{x}=k,\ k\in \mathbb{R}$ and $\lim \limits_{x\to +\infty} (f(x)-kx)=b, \ b\in \mathbb{R}$.

Since $\lim \limits_{t\to 1}\dfrac{y(t)}{x(t)}=-1$ and $\lim
 \limits_{t\to 1} ((y(t)+x(t))=2$, the line $y=-x+2$ is an asymptote
for both ends of the trajectory, corresponding to $t$ approaching $1$.
It is also clear that the line $x=0$ is a vertical asymptote for the
portion of the trajectory corresponding to $t\to+\infty$.

This is an excerpt from Zorich's book.
Our parametric curve $(x(t),y(t))$ defines a function $f:\mathbb{R}\to \mathbb{R}$. Zorich is doing limit computation in terms of $t$. Can anyone explain in a rigorous way why it implies that $\lim \limits_{x\to +\infty}\dfrac{f(x)}{x}=-1$ and $\lim \limits_{x\to +\infty} (f(x)+x)=2$.
Would be very grateful for help!

The trajectory is defined for $t \ge 0$ and $t \neq 1$. Let $I= [0,1) \cup(1, \infty)$.
$x$ is a strictly increasing continuous bijection $[0,1) \to [0,\infty)$ and a strictly increasing continuous bijection
$(1,\infty) \to (-\infty,0)$. Hence $(x(t), y(t))$, $t \in I$ defines a function $f:\mathbb{R} \to \mathbb{R}$.
It is straightforward to check that the maps $x:[0,1) \to [0,\infty)$ and $x:(1,\infty) \to (-\infty,0)$ are actually homeomorphisms.
It is straightforward to check that $\lim_{s \to \infty} \phi(s) = L$ iff $\lim_{t \uparrow 1} \phi(x(t)) = L$.
Then we have $\lim_{s \to \infty} {f(s) \over s} = \lim_{t \uparrow 1} {f(x(t)) \over x(t)} = \lim_{t \uparrow 1} {y(t) \over x(t)} = \lim_{t \uparrow 1} (1-2t^2) =-1$, and similarly,
$\lim_{s \to \infty} {f(s) + s} = \lim_{t \uparrow 1} {f(x(t)) + x(t)} = \lim_{t \uparrow 1} {y(t)+ x(t)} = \lim_{t \uparrow 1} 2t =2$.
Note that it is also true that $\lim_{t \to 1} {y(t) \over x(t)} = \lim_{t \to1} (1-2t^2) =-1$ and $\lim_{t \to 1} {y(t)+ x(t)} = \lim_{t \to 1} 2t =2$, and using the
$x:(1,\infty) \to (-\infty,0)$ portion of the curve, we can show in a similar manner that
$\lim_{s \to -\infty} {f(s) \over s} = -1$ and
$\lim_{s \to -\infty} {f(s) + s} = 2$.
Elaboration
Take the case where $L$ is finite for example. Also, note that I am dealing with $x:[0,1) \to [0,\infty)$ here. Note that $x$ is strictly increasing on this domain:
Suppose for all $\epsilon>0$ there is some $S$ such that if $s > S$ then $|\phi(s)-L| < \epsilon$. Now let $t_0= x^{-1}(S), \delta = 1-t_0$ and note that if $0< |1-t| < \delta$ (and implicitly we have $t \in [0,1)$) we have $|\phi(x(t))-L| < \epsilon$.
The other direction is similar, suppose we have some $\delta>0$ such that
if $0< |1-t| < \delta$ (and implicitly $t \in [0,1)$) then $|\phi(x(t))-L| < \epsilon$. We can assume that $\delta<1$. Let $S=x(1-\delta)$, then if $s > S$ there is some $t \in (1-\delta,1)$ such that $x(t)=s$ and so
$|\phi(s)-L| = |\phi(x(t))-L| < \epsilon$.
The case for infinite $L$ is similar.

hint
With cartesian equation $ y=f(x) $, we look for oblic  asymptotes Only when
$$\lim_{|x|\to\infty}f(x)=\pm\infty$$
we look for
$$\lim_{x\to\infty}\frac{f(x)}{x}=a$$
and
$$\lim_{x\to\infty}(f(x)-ax)=b$$
This cartesian representation can be seen as a  special parametric one if we put
$$x(t)=t \;\text{ and } \;y(t)=f(t).$$
(In  parametric representation with the variable $ t $, You need $ t_0 $, such that both $ x(t) $ and $ y(t) $ go to infinity when $ t $ goes to $ t_0$.
then
$$\lim_{x\to \infty} \frac{f(x)}{x}$$
becomes
$$\lim_{t\to t_0}\frac{y(t)}{x(t)}$$
and
$$\lim_{x\to \infty}(f(x)-ax)$$
is replaced by
$$\lim_{t\to t_0}(y(t)-ax(t))$$
In your case $ t_0\in\{-1,1\}$ and knowing that $ t\ge 0$, we have only the case $t_0=1$.


Are linear combinations of vectors that form a basis still a basis? V is a subspace of $R^n$ and $B={u,v,w}$ is a basis for V and

$x = 2u-3v+w$

$y = u+v$

$z = -v-w$

Explain why ${x,y,z}$ are a basis for V

Can I say that:
Any linear combinations of vectors that form a basis are still a basis?

you can see that $$\pmatrix{x\\y\\z} = \pmatrix{2&-3&1\\1&1&0\\0&-1&-1}
\pmatrix{u\\v\\w}$$ and you need to show that the coefficient matrix in invertible. you can do that either by row reducing it see that you have three pivots or you could find the determinant not zero.

No. Consider the linear combination of the canonical basis (0,1)+(1,0) and (1,0)+(0,1). They are linearly dependent and hence do not form a basis for R^2


Power Towers, and Notation for Iterated Exponentiation So far, we use the symbol
$$\sum$$
to denote sums, and
$$\prod$$
to denote products. But is there any such notation for exponentiation?

Has any research been done about exponentiation of this type, where the numbers in the "power tower" form some sequence?
$$a_1^{{a_2}^{{...}^{a_n}}}$$
And it need not be a finite sequence, I might add...

Does anybody know of any examples that we know how to evaluate? Does anybody know how to determine convergence or divergence in these "power towers"? And can anybody suggest or point me towards any existing way to denote something like this?

Let me give an example. Suppose we have the power tower defined by the sequence
$$a_n=\frac{3}{2^n}$$
Does anybody know how to evaluate the infinite power tower defined by this sequence (it seems to converge) ? How about a partial one?

We already know how to evaluate some infinite power towers where $a_n$ is a constant, like
$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}=2$$

Does anybody know of any resources about this to point me towards, or any original insights or ideas about this concept?

Thanks!

Creutz/Sternheimer ([5] and [6] below) use $\Xi$ for the exponentiation analog of $\Sigma$ (summation) and $\Pi$ (product). They were probably unaware of [4], but a similar usage appears in [4] for the corresponding transfinite operations on ordinals --- $\Xi_{0}$ is for $\Sigma$ (transfinite sums), and $\Xi_{1}$ is for $\Pi$ (transfinite products), and $\Xi_{2}$ is for transfinite "bottom-up evaluated" exponential towers of ordinals, and $\Xi_{3}$ is for transfinite "bottom-up evaluated" bottom-up-tetration towers of ordinals, etc. (When defining these higher operations for ordinals, the top-down evaluation method does not work very well -- see this question and my comments under it.)

The papers below (except for [4]) are the most relevant papers I know of for what you are mainly asking about, namely convergence issues for limits of power towers of the form $a_1^{{a_2^{{...}^{{a_n}^{...}}}}}.$

[1] David Francis Barrow (1888-1970), Infinite exponentials, American Mathematical Monthly 43 #3 (March 1936), 150-160.

[2] Wolfgang Joseph Thron (1918-2001), Convergence of infinite exponentials with complex elements, Proceedings of the American Mathematical Society 8 #6 (December 1957), 1040-1043.

[3] Donald Lewis Shell (1924-2015), On the convergence of infinite exponentials, Proceedings of the American Mathematical Society 13 #5 (October 1962), 678-681.

[4] Arthur Leonard Rubin (1956-$_{---}$) and Jean Estelle Rubin (1926-2002), Accumulation functions on the ordinals, Fundamenta Mathematicae 70 #2 (1971), 205-220.

[5] Michael John Creutz (1944-$_{---}$) and Rudolph Max Sternheimer (1926-2000), On the convergence of iterated exponentiation---I, Fibonacci Quarterly 18 #4 (1980), 341-347.

[6] Michael John Creutz (1944-$_{---}$) and Rudolph Max Sternheimer (1926-2000),  On the convergence of iterated exponentiation---II, Fibonacci Quarterly 19 #4 (1981), 326-335.

[7] Irvine Noel Baker (1932-2001) and Philip Jonathan Rippon (??-$_{---}$), Iterating exponential functions with cyclic exponents, Mathematical Proceedings of the Cambridge Philosophical Society 105 #2 (March 1989), 357-375.

here's my research (somewhat as it goes), if you have a finite power tower like:

$a^{{(ab)}^{(ab^2)}}$

then we have $(ab)^{(ab^2)}$ as a power by the power rules used for integers at least this equals:

$$a^{(ab^2)}b^{(ab^2)}=(a^a)^{(b^2)}(b^{(b^2)})^a=({(a^a)^b})^b({(b^b)^b})^a$$ this is our exponent if we then raise a to it we get:
$$\Large a^{(a^{(ab^2)}b^{(ab^2)})}=a^{((a^a)^{(b^2)}(b^{(b^2)})^a)}=a^{(({(a^a)^b})^b({(b^b)^b})^a)}= (a^{({(a^a)^b})^b }\Large )^{({(b^b)^b})^a}= ???$$

it's a work in progress I guess for me. this example was for a geometric sequence.


Proving f Riemann integrable implies |f| Riemann integrable by contradiction. I know that there exists a theorem telling us that if a real valued function $f$ defined over an interval $[a,b]$ is Riemann integrable, then |$f$| is too (moreover I believe the converse to be false). I've tried to look at the proofs, but they seem a bit beyond my ability at the moment. My first thought was to suppose there existed a Riemann integrable function, let us say $g$, such that |$g$| was not Riemann integrable, and show that this leads to some sort of contradiction.
I thought about this for a while and have not been able to think of such a function/proof. Would anybody know of such an example?

Let $a,b\in\mathbb{R}$ with $a<b$. Suppose $f:[a,b]\to\mathbb{R}$ is bounded and Riemann integrable. It suffices to show that for any interval $I\subseteq[a,b]$, we have that $$\sup_{x\in I}|f|-\inf_{x\in I}|f|\leq\sup_{x\in I}f-\inf_{x\in I}f $$
where $\sup_{x\in I}g(x)$ is defined as $\sup\{g(x)\in\mathbb{R}:x\in I\}$ if $g$ is a function that is bounded on $I$.
Note that for $x\in I$, we have $f(x)\leq |f(x)|$.
Break this into three cases:
Case 1:
$\inf_{x\in I}f(x)\geq 0$. In this case, we have that $f(x)\geq 0$ for all $x\in I$. This means $|f(x)|=f(x)$ so $$ \sup_{x\in I}|f|=\sup_{x\in I}f\text{ and }\inf_{x\in I}|f|=\inf_{x\in I}f$$ which implies that $$\sup_{x\in I}|f|-\inf_{x\in I}|f|=\sup_{x\in I}f-\inf_{x\in I}f $$
Case 2:
$\sup_{x\in I}f(x)\leq 0$. In this case, we have that $f(x)\leq 0$ for all $x\in I$, thus $|f(x)|=-f(x)$. Then we get that $$\sup_{x\in I}|f|=-\inf_{x\in I}f \text{ and }\inf_{x\in I}|f|=-\sup_{x\in I}f$$ so in particular, we have
$$\sup_{x\in I}|f|-\inf_{x\in I}|f|=-\inf_{x\in I}f-\big(-\sup_{x\in I}f\big)=\sup_{x\in I}f-\inf_{x\in I}f$$
If you want Case 2 to follow relatively quickly, it would suffice to show the following:
Let $A\subseteq\mathbb{R}$ be non-empty and bounded. Define $-A=\{-a\in\mathbb{R}:a\in A\}$. Then $\sup(-A)=-\inf(A)$ and $\inf(-A)=-\sup(A)$.
Case 3:
$\inf_{x\in I}f\leq 0\leq\sup_{x\in I}f$. I will leave this case to you, as it requires some sub-cases based upon the magnitudes of $\inf_{x\in I}f$ and $\sup_{x\in I}f$.
With this it should be easier to finish the proof.

For instance $f(x)=\begin{cases}\dfrac x{1-x}\,,x\in[0,1)\\\dfrac{x}{1+x}\,,x\in(-1,0]\\0\,,x=\pm1\end{cases}$, would provide a counter-example.  We have $\int_{-1}^1f\rm dx=0$, but $\int_{-1}^1|f|\rm dx=\infty$.


How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other? I have a homework problem in my textbook that has stumped me so far. There is a similar one to it that has not been assigned and has an answer in the back of the textbook. It reads:

How many ways are there for $8$ men and $5$ women to stand in a line so that no two women stand next to each other?

The answer is $609638400$, but no matter what I try I cannot reach that number. I have tried doing $2(8!5!/3!)*(8!/5!)$ since each woman must be paired with a man in order to prevent two women getting near each other. But of course, it's the wrong answer.

What am I doing wrong here?

For this type of problem first consider the position of the men, then the position of the women.In general first consider the the group with more people.

How many possible ways are there to arrange eight men in a row? It will be $ ^8P_8 = 8! = 40320$

Now as no two women stand next to each other,we can imagine the situation as

                   * M * M * M * M * M * M * M * M *

Hence, we need to find how many ways we can arrange $5$ women in the $9$ possible (as shown above) places,this is actually $ ^9P_5 = 9*8*7*6*5 = 15120$

Now applying the fundamental law of counting (precisely product rule), total number of possible arrangements satisfying both constraints is: $15120 * 40320 = 609638400$ which is your required/desired answer.

M * M * M * M * M * M * M * M * .

there are 9 "*" .
8 "M"  .
now select 5 * from 9 * then == c(9,5)


How to calculate the number of palindromes of a given number of characters? A palindrome is a set of characters which equal the same thing forwards and backwards, for example; abccba.
For a set of a given amount of  characters (we can use 9 as an example) how would you calculate the amount of palindromes of characters mathematically?
This doesn't mean the words that can be formed but rather characters which are the same at start characters and end characters.
What formula can be used to calculate this for simple lengths like 9 with lowercase letters?

It depends (slightly) on whether you want an odd number of digits in your string, or an even number of digits in your string.

Let's say that the "alphabet" contains $N$ different "characters".  (These could be the digits 1-9, or the letters A-Z, or any other set of distinguishable symbols.)  You want to know how many palindromes of a given length there are.

If the string length is to be even:  Say the string length is $2k$.  You can choose each of the first $k$ characters to be anything you want; once you've chosen those, the rest of the string is forced on you by the required symmetry.  So there are $k$ free choices, each drawn from a set of $N$ characters; that means there are $N^k$ possibilities.

If the string length is to be odd:  Say the string length is $2k+1$.  You can choose each of the first $k+1$ characters to be anything you want; once you've chosen those, the rest of the string is forced on you by the required symmetry.  So there are $k+1$ free choices, each drawn from a set of $N$ characters; that means there are $N^{k+1}$ possibilities.

So, for example, if your alphabet consists of the 26 lowercase letters a-z, and you want a string with 9 characters, then $N=26$ and the string has length $2k+1$ with $k=4$; therefore the number of possible palindromes is $26^5=11,881,376$.

include

include

include

using namespace std;
int main()
{
    string s,s1,s2;
    cin>>s;
    s1 = s.substr(0,s.length()/2+1);
    int x = (s.length()%2 ==0 ) ? s.length()/2 : s.length()/2 + 1;
    s2 = s.substr(s.length()/2,x);
    string s3 = s2;
    for(int i=0;i

for(int i=0;i<s1.length() ; i++)
{
    if(s1[i]!=s3[i])
    {
        s3 = s3.substr(0,i)+ s1[i] +s3.substr(i,s3.length()-i);

        if(s1[i+1]!= s3[i+1])
       { cout<<"NA";
        return 0;
       }
    }

}
s2=s3;
 for(int i=0;i<s3.length()+1;i++)
{
    s2[i]=s3[s3.length()-i-1];
}
cout<<s1<<s2.substr(1,s2.length());
return 0;

}


Integrating $ 3 \int \frac{\cos^5 x}{\sin x} dx$? I need to solve the integral $$3 \int  \frac{\cos^5 x}{\sin x} dx$$

I'm not sure how to proceed from here. I think Integration by Parts may be useful, but I'm not entirely sure what to make $u$ and $dv$.

A start: Note that $\cos^4x=(1-\sin^2 x)^2$. Let $u=\sin x$.

$$
\int \frac{\cos^5x}{\sin x}\,dx
=
\int\frac{\cos^4x \cdot \cos x}{\sin x}\,dx
$$
$$
=
\int \frac{(1-\sin^2 x)^2  \cos x}{\sin x}\, dx
$$
Then put $t=\sin x$ then $dt=\cos x\, dx$

Then the integral is equal to
$$\int\frac{(1-t^2)^2}{t}\, dt$$


Summing $ \sum _{k=1}^{n} k\cos(k\theta) $ and $ \sum _{k=1}^{n} k\sin(k\theta) $ I'm trying to find

$$\sum _{k=1}^{n} k\cos(k\theta)\qquad\text{and}\qquad\sum _{k=1}^{n} k\sin(k\theta)$$

I tried working with complex numbers, defining $z=\cos(\theta)+ i \sin(\theta)$ and using De Movire's, but so far nothing has come up.

HINT:

$$\sum_{1\le k\le n}\left(\cos k\theta+i\sin k\theta\right)=\sum_{1\le k\le n}e^{i k\theta}=e^{i\theta}\cdot\frac{e^{in\theta}-1}
{e^{i\theta}-1}$$

Differentiate  wrt $\theta$ and equate the real & the imaginary parts

$$\text{On differentiation, the left hand becomes }\sum_{1\le k\le n}k\left(-\sin k\theta+i\cos k\theta\right)$$

Alternatively, if $S=\sum_{1\le k\le n}k\cdot a^k=a+2\cdot a^2+\cdots+(n-1)\cdot a^{n-1}+n\cdot a^n$

$a\cdot S=a^2+2\cdot a^3+\cdots+(n-1)\cdot a^n+n\cdot a^{n+1}$

$$\text{On subtraction,}(a-1)S=n\cdot a^{n+1}-(a+a^2+\cdots+a^{n-1}+a^n)=n\cdot a^{n+1}-\frac{a(a^n-1)}{a-1}$$

$$\implies S=\sum_{1\le k\le n}k\cdot a^k=\frac{n\cdot a^{n+1}}{a-1}-\frac{a(a^n-1)}{(a-1)^2}$$

Putting $a=e^{i\theta},$

$$S=\sum_{1\le k\le n}k\cdot e^{ik\theta}=\frac{n\cdot e^{i(n+1)\theta}}{e^{i\theta}-1}-\frac{e^{i\theta}(e^{in\theta}-1)}{(e^{i\theta}-1)^2}$$ (The derivative method should bring us here,too)

$$S=\frac{ne^{i(n+1)\theta}}{e^{i\frac\theta2}(e^{i\frac\theta2}-e^{-i\frac\theta2})}-\frac{e^{i\theta}(e^{in\theta}-1)}{\{e^{i\frac\theta2}(e^{i\frac\theta2}-e^{-i\frac\theta2})\}^2}$$

As $e^{ix}-e^{-ix}=2i\sin x,$  $$S=\frac{ne^{i\frac{(2n+1)\theta}2}}{(e^{i\frac\theta2}-e^{-i\frac\theta2})}-\frac{(e^{in\theta}-1)}{(e^{i\frac\theta2}-e^{-i\frac\theta2})^2}$$

$$S=\frac{n\left(\cos\frac{(2n+1)\theta}2+i\sin\frac{(2n+1)\theta}2\right)}{2i\sin\frac\theta2}-\frac{(\cos n\theta+i\sin n\theta-1)}{(2i\sin\frac\theta2)^2}$$

$$=\frac{\cos n\theta-1}{4\sin^2\frac\theta2}+\frac{n\sin \frac{(2n+1)\theta}2}{2\sin\frac\theta2}+i\left(\frac{\sin n\theta}{4\sin^2\frac\theta2}-\frac{n(\cos\frac{(2n+1)\theta}2}{2\sin\frac\theta2}\right)$$

Again, $$S=\sum_{1\le k\le n}k\cdot e^{ik\theta}=\sum_{1\le k\le n}k(\cos k\theta+i\sin k\theta)=\sum_{1\le k\le n}k\cdot\cos k\theta+i\sum_{1\le k\le n}k\cdot\sin k\theta$$

Now, equate the real & the imaginary parts

All identities are coming from  Wikipedia page for trigonometric identities. The main identity that we start with is:
$$
\sum_{k=1}^n \cos(kx)= \frac{ \sin((n+\frac{1}{2}) x)}{2\sin(x/2)} -1
$$
Get a derivative with respect to $x$, and we have
\begin{eqnarray*}
\sum_{k=1}^n k \sin(kx)&=&  \frac{  \cos(x/2) \sin((n+\frac{1}{2})x)- (2n+1) \cos((n+\frac{1}{2})x) \sin(x/2)}{4\sin^2(x/2)}\\
&=&  \frac{-2n \cos((n+\frac{1}{2})x) \sin(x/2) + \left(  \cos(x/2) \sin((n+\frac{1}{2})x)- \cos((n+\frac{1}{2})x) \sin(x/2)\right)}{4\sin^2(x/2)}\\
&=&  \frac{-2n \cos((n+\frac{1}{2})x) \sin(x/2) +   \sin(nx)}{4\sin^2(x/2)}\\
&=& \frac{ -n\sin( (n+1)x) + n\sin(nx) +    \sin(nx)}{4\sin^2(x/2)}\\
&=& \frac{ -n\sin( (n+1)x) + (n + 1)   \sin(nx)}{4\sin^2(x/2)}.
\end{eqnarray*}


Statistics-Math-Probability An urn is filled with 10 green balls, 4 red balls, and 2 orange balls.  Three balls are selected without replacement.  Calculate the following probabilities:

(a) P(at least one ball is red)
I think the answer is: (4/16)(3/15)(2/14) = .007 or 1/140

(b) P(exactly one ball is orange)
I think the answer is: (2/16)(14/15)(13/14) + (14/16)(2/15)(13/14) + (14/16)(13/15)(2/14) = 13/40 or .325

(c) P(at most one ball is green)

(d) P(neither ball is red)

For part (a), I would calculate the probability that $no$ red ball is selected, and subtract the result from one.

For part (b), you are correct, but notice that your three terms are equal. Why is this? Try to find the pattern and generalize it.

For part (c), you can take the sum of $P$(exactly one ball is green), whose calculation mirrors that of part (b), plus $P$(no ball is green), whose calculation mirrors that of part (a).

For part (d), you just need to find P(no ball is red), which you should be able to recycle from part (a).

Answer:

Total number of ways you can select 3 balls = ${16\choose3}$

Part a) $$=1-\frac{{12\choose3}}{{16\choose3}} = 0.6071$$

Part b)$$=\frac{{2\choose1}{14\choose2}}{{16\choose3}}=0.325$$

Part c) $$=\frac{{10\choose0}{6\choose3}+{10\choose1}{6\choose2}}{{16\choose3}}=0.3035$$

Part d) $$=\frac{{12\choose3}{4\choose0}}{{16\choose3}}=0.39285$$


The proof of the abc conjecture I recently heard that there was a workshop on Inter-universal Teichmuller theory in the Clay institute from 7-11 December 2015.This field of mathematics offers a potential proof of the abc conjecture.If anyone knows, what is the progress of the confirmation of Mochizuki's papers?

I attended a talk by Brian Conrad at my University recently regarding the $abc-$Conjecture. He informed us that the only people who have the faintest idea about Mochizuki's proof are Mochizuki himself and some of his closest understudies who claim its validity. Apparently they are completely unable to explain the work to outsiders. You may have already known that.

Here is a link to Conrad's own words: http://mathbabe.org/2015/12/15/notes-on-the-oxford-iut-workshop-by-brian-conrad/.

Most mathematicians when they read a new paper they make a diagonal reading first to see what it is about and get a glimpse at the kind of methods and ideas used, and then concentrate on the parts which they feel they need more clarification. This is what happens when experts in one field read a paper in their field of expertise.I once had a professor who told that just by reading the first few pages of an article in his area, he could determine the relevance of what was being discussed and whether the paper was likely to have errors or not.

However, as Mochizuki points out on page 9 of his report http://www.kurims.kyoto-u.ac.jp/~motizuki/IUTeich%20Verification%20Report%202014-12.pdf , there seems to be no one right now that has sufficient expertise in absolute Anabelian Geometry to be able to look at IUT papers with the kind of insight and ease an Arithmetic Geometer looks at other papers in Arithmetic Geometry. For researchers to read their papers, they would have to become complete students again: learn the paper line by line, definition by definition, lemma by lemma like a student does when he's taking a course in a new subject.

Most researchers, are simply not willing to do that and arrogantly categorize IUT as too abstract - thge truth is that they simply don't have enough background!!

There have been 3 mathematicians - Hoshi, Yamashita and Saidi who have gone through all of the papers of IUT a total of 12 times and they all confirm that the theory is correct.


About rings having finite number of non-invertible elements Let $(A, +, \cdot)$ a ring and $D$ the set of non-invertible elements
  of $A$. We know $a^2=0, \forall a \in D$.

  Prove:

  $axa=0, \forall x \in A, \forall a \in D$
  If $D$ is finite having at least two elements, then there is $c \in D, c \ne 0$ such that $ac=ca, \forall a \in D$

I was able to prove 1) as follows:

First, let's notice $ax \in D, \forall x \in A$ therefore $ax + a \in D$.

Then $0 = (ax + a)^2= (ax)^2 + aax + axa + a^2=axa$

I can't prove the second one. Any help in appreciated.

UPDATE

To prove $ax \in D$ suppose $ax=y$ where $y$ is invertible. Then, $aax=ay$ therefore $ay=0$ so $a=0$ and $y=0$ contradiction

UPDATE 2

Example of such ring: $\mathbb{Z}_4$

Define $[x,y]=xy-yx$ in $R$.

Claim I: For all $a\in D$ and $x\in R$ we have $a[a,x]=[a,x]a=0$. In particular $[a,[a,x]]=0$. Proof: Immediate consequence of part (a).

Claim II: If $a,b\in D$, then $[a,b]=ab-ba\in D$.

Proof by contradiction: Suppose $x$ is a two-sided inverse to $[a,b]$. Then $$1=[a,b]x\Longrightarrow a=a[a,b]x=0$$
This means $0=[a,b]$ is invertible which is absurd. Also $[a,b]^2=0$ is again an immediate consequence of claim I. Therefore $[a,b]\in D$.

Suppose $D$ has $n\geq 3$ elements. Take $a,b\in D$ such that $[a,b]\neq 0$ (if we can't do this there is nothing to prove). Then $[a,[a,b]]=[b,[a,b]]=0$ meaning $[a,b]$ commutes with $a,b$. Choose $c\in D$ such that $[c,[a,b]]\neq 0$ (again if no such $c$ exists we are done), then $[c,[c,[a,b]]]=0$. Also, by Jacobi identity (this bracket is obviously a Lie bracket)
$$0=[a,[c,[a,b]]]+ [c,[[a,b],a]]+[[a,b],[c,a]]=
[a,[c,[a,b]]]
$$
again because of claim I. This shows that $[c,[a,b]]$ commutes with all $a,b,c$. Continue in this fashion until you run out of elements in $D$ (or you have gotten lucky somewhere in the process).

Just want to know how are you saying that $ax\in D$ ,$a\in D$  gives you $ax+a\in D$. If you can do so then i guess you can proceed in the following manner to prove B.

Take a non-zero $c$ and set $s=ac-ca $ for any $a$ in $D$. so $as=a^2c-aca=0-0=0$(by the first you have $axa=0$. Now by hypothesis you have $1$ in $D$. so $1+a\in D$ so $1+a+a=1+a+a+a^2=(1+a)^2 \in D$ so $1+a+a=0$.$s+as+as=0$. But you have $as=0$ so $s=0$. Hence $s=ac-ca=0$ .Thus proven


Two functions cannot have the same derivative at not interior points. What discuss below is a reference form the text "Analysis on Manifolds" by James Munkres.

First of all we observe the the interior of $U$ is not empty when $U$ is too: for details see here.
So by the first definition to we can suppose that the derivative of $\alpha$ exist only in $\text{int}(U)$ but strangerly Munkres state that $D\beta(x)=D\alpha(x)$ for any $x\in U$. So how explain this statement?
Could it be a typo? So could someone help me, please?

I'm not sure I properly understand what your problem is, but Munkres nowhere explicitly writes down the equation $\ D\alpha(x)=D\beta(x)\ $, or implies that $\  D\alpha(x)\ $ exists at a boundary point of $\ U\ $ in any sense other than what he is here showing can be taken to be its definition there. The point of his Lemma $23.2$ is that if $\ \beta_1\ $ and $\ \beta_2\ $ are any $\ C^r\ $ extensions of $\ \alpha\ $ to an open subset of $\ \mathbf{R}^k\ $, then $\ D\beta_1(x)= D\beta_2(x)\ $ for all $\ x\in U\ $.  Thus, even though the normal definition of $\ D\alpha(x)\ $ cannot be used to define it on the intersection of $\ U\ $ with the boundary of $\ \mathbf{H}^k\ $ , you can nevertheless take its definition to be $\ D\beta(x)\ $ there, where $\ \beta\ $ is any extension of $\ \alpha\ $ to an open subset of $\ \mathbf{R}^k\ $.

This, "First of all we observe the the interior of U is not empty when U is too".  That is not true,  the singleton set U= {x} is non-empty but its interior is empty.


Evaluate the integral $\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$ $\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$

Okay so I converted this into polar form because I was told to do so
I got the integral of $(7r\cos\theta + 3r\sin\theta)rdrd\theta$ where $0\le \theta \le \pi/4$ and $0\le r \le \sqrt{18}$

I think I'm making a mistake solving this integral. I keep getting $72$ which is incorrect.

Work:
Taking the first integral I get $7r^3\cos\theta/3 + r^3\sin\theta$ from $0$ to $\sqrt{18}$
then I get $7(\sqrt{18})^3 \cos\theta/ 3 + \sqrt {18}^3\sin\theta$
Then I took the second integral and got $7(\sqrt{18})^3 \sin\theta/ 3 - (\sqrt {18})^3\cos{\theta}$
I plugged in the values of pi/4 and o and got

$126 - 54 + (\sqrt{18}^3) = 148.367532368147$

Sorry, while typing the work I found my mistake. Thanks everyone for trying to help me out :)

Let $I$ denote the integral value. By calculating the integral by the polar form, we have:
\begin{equation}
I=\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}7r^2\cos(\theta)+3r^2\sin(\theta)drd\theta\\
=\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}7r^2\cos(\theta)drd\theta+\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}3r^2\sin(\theta)drd\theta\\
=\int_0^{\frac{\pi}{4}}\cos(\theta)d\theta\int_0^\sqrt{18}7r^2dr+\int_0^{\frac{\pi}{4}}\sin(\theta)d\theta\int_0^\sqrt{18}3r^2dr\\
=\frac{\sqrt{2}}{2}\times126\sqrt{2}+(1-\frac{\sqrt{2}}{2})\times54\sqrt{2}\\
=72+54\sqrt{2}
\end{equation}

I'm not sure about polar form, but you can evaluate the integral as such:

$\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$

$\int\limits_{0}^{3}(\frac{7}{2}x^2 + 3y\cdot x)|_{\sqrt{18-y^2}}^y$ $dy$

$\int\limits_{0}^{3}(\frac{7}{2}y^2 + 3y^2))-(\frac{7}{2}(18-y^2) + 3y\cdot \sqrt{18-y^2})$ $dy$

$\int\limits_{0}^{3} 10\cdot y^2-3y\cdot \sqrt{18-y^2}-63$ $dy$

It should be relatively easy to solve from there using more basic, one-variable methods.


How many ways are there to give 21 different books to 3 people, A, B, C, so that A's + B's = 2C's books? Only thing I found is that A and B will take 14 books, while C has to get 7. Any thoughts?
Thanks in advance.
Edit:
So the question is not answered, we are looking for ways to give the books to the three people with the above conditions, though I might have come up with a solution, which I'd be glad to discuss about.
Solution (?):
First of all, we give 7 books to C, but how many ways are there? C(n,r) = C(21, 7). Then the remaining 14 books have to be given to A and B, but B is actually not having any impact, you can give A the 14 with $2^n$ -1 ways, n = 14. B will either way take the remaining books A didn't take.
Multiplying those 2 numbers we take the required ways.
Any thoughts?

If A gets k books out of a selection of 21, then B must get the rest of the 14 books (14-k) from the rest of the selection (21-k). Determining the books for A and B also determines the books for C.
$\displaystyle\sum_{k=0}^{14}\binom{21}{k}\binom{21-k}{14-k} = 1,905,131,520$

Hint:
Obviously you can take a book from A and give it to B (or the other way round) while still satisfying the conditions.
Write the two conditions as $$A+B+C=21$$ $$A+B=2C$$
You have two equations and three unknowns, so there is not a single unique solution, but in this example you can find $C$ and $A+B$ and it is not difficult to do so.  You are already most of the way there.


Prove that the unit ball in $X$ is not compact Let $X$ bet the set of all sequences $\{a_n\}_{n=1}^\infty$ in
  $\mathbb R$ with $\lim_{n \to \infty} a_n = 0$. For any $\{a_n\},
 \{b_n\} \subset X$, we define a metric

  $$d(\{a_n\}, \{b_n\}) = \sup\{|a_n - b_n|, n = 1,2,\dots \}$$

  Prove that the unit ball $\{\{a_n\}\in X: d(\{a_n\}, 0) = \sup_{n \in
 \mathbb N} |a_n| \le 1\}$ in $X$ is not compact

I think this unit ball is both bounded and closed in metric space, so it is compact. But the question asks me to prove that it is not compact. Please help.

One of the more intuitive definitions/characterizations of compactness in a metric space $(X,d)$ is the following:

A set $K \subseteq X$ is compact if and only if every sequence $\{a_k\} \subseteq K$ has a convergent subsequence.

So in order to show that the unit ball in the space of all sequences is not compact, you have to find a sequence (of sequences) in the unit ball that does not contain a convergent subsequence.

Consider 0= (0,0,0,……………) € l2
Consider the closed ball, B [0,1].
Clearly B [0,1] is a closed and bounded subset of l2 . (Any closed ball in a metric space is a closed set and diam B[0,1]


Theorem for some structure that it cannot be axiomatized in any language? Godel’s theorem states that there cannot be an enumerable axiomatization of arithmetic $(\mathbb N,0,1,+)$ in first order logic. By implication, there cannot be any axiomatization in any language that is complete.

But this doesn’t mean that arithmetic cannot be axiomatized at all. E.g. we have second order Peano axioms that characterizes arithmetic up to isomorphism. It just means that we cannot derive all propositions from this axiomatization (which is possible because second-order logic is not complete).

My question is: is there a different theorem, similar in spirit to Godel’s, for some mathematical structure that can be constructed explicitly, that there cannot be an axiomatization that characterizes it up to isomorphism, in any language? (I.e. not first order, or second order, or higher order, or anything else that I don’t know about).

Intuitively, I would guess there might be, because: There is only a countable amount of possible enumerable axiomatizations in all languages combined, since there is only a countable amount of Turing machines. On the other hand there is an uncountable amount of structures. So there must be some structure that cannot be axiomatized. My question is whether there is an explicit constructable structure where we know that it cannot be.

A crucial point in your question is the following broad claim:

There is only a countable amount of possible enumerable axiomatizations in all languages combined, since there is only a countable amount of Turing machines.

There's an implicit restriction here to logics which are "computably describable." Of course this is necessary to hope for a positive answer, but it's not immediately clear what this should mean - especially since you explicitly want to include things like second-order logic whose semantics are absurdly complicated.
One possible idea is the following: roughly, say that a definable logic consists of a computable set of things we call sentences and a semantics given by a formula in the language of set theory. This encapsulates all the higher-order logics, computable infinitary logic, and so forth. I don't know of a single candidate for a natural logic which has countably many sentences but is not a definable logic in the above sense, so this seems reasonable.
It turns out, however, that it is consistent that every structure can be pinned down up to isomorphism by some sentence in some definable logic. This is a consequence of the existence of pointwise-definable models; the obstacle to the "obvious" proof of contradiction here is that definable logics can be arbitrarily complicated to define in terms of the Levy hierarchy so we can't definably diagonalize against them (per Tarski).
In fact, even handling second-order logic alone is quite difficult. While a quick counting argument shows that there must be a countable structure which is not pinned down up to isomorphism by any computable set of second-order sentences, there is no known candidate, and in fact it is consistent with $\mathsf{ZFC}$ that every countable structure is pinned down up to isomorphism by its full second-order theory (see the discussion here). And any attempt to develop a "Godel-flavored" analysis of second-order logic immediately runs into the problem that the entailment relation for second-order logic is not second-order definable in arithmetic (in contrast with the situation for first-order logic, which follows from the completeness theorem).

That said, there is a version of Godel's first incompleteness theorem which applies to any (set-sized) logic.
First, let me prove my statement above about the complexity of the second-order entailment relation. The syntax for $\mathsf{SOL}$ is computable, and there is a second-order sentence $\theta$ which pins down $\mathbb{N}$ up to isomorphism. If the second-order entailment relation were second-order definable over $\mathbb{N}$, the full second-order theory of $\mathbb{N}$ would be second-order definable in $\mathbb{N}$. This would give us a second-order-definable-over-$\mathbb{N}$ listing of the second-order-definable-over-$\mathbb{N}$ sets; diagonalizing against that listing would give us a contradiction. (This last sentence is really just Tarski's argument, but I think this phrasing makes it easier to think about.)
This generalizes in a precise, if messy, way to any context where we have a logic $\mathcal{L}$ and a structure $\mathbb{A}$ sufficient for developing the syntax of $\mathcal{L}$ (this generalizes $\mathbb{N}$). Then either $\mathcal{L}$-entailment is too complicated to define in $\mathbb{A}$, or no good approximation to the $\mathcal{L}$-theory of $\mathbb{A}$ (this generalizes "no consistent completion of $\mathsf{Q}$") is "nicely definable" in $\mathbb{A}$.
Incidentally, this idea leads to a proof of the usual first incompleteness theorem which is "purely model-theoretic" in a precise sense, see here (although I don't know its exact provenance). And while the added generality seems useless for the most part, I did use the more general version above to prove a small result about infinitary logics in a relatively short way.

If I understand the question correctly, you can always construct a halting problem.  You can make a second-order system first order by importing the second-order concepts as first-order axioms, and then you just have the halting problem with your additional axioms.
For instance, let's say that you have a set of operations A.  You can construct a halting problem based on A.  Now, let's say that you have another set of operations B, which includes all of A plus a halting oracle for A.  This set of operations will then itself be subject to a halting problem.
I think Turing covers this in "Systems of Logic Based on Ordinals".


Only prime number for $p^2+2$ to be prime as well Prove that the only prime number $p$ for $p^2+2$ to be prime as well is $p=3$

If $p\neq 3$ then it has form $3k\pm 1$ so $p^2+2=(3k\pm 1)^2+2=9k^2\pm6k+3=3(3k^2\pm 2k+1)$ so wont be prime

hint: If $q = p^2 + 2$ is a prime number, then $p = 3k+1$ or $p = 3k+2$


Where should concatenation take place in the order of operations? Concatenation being the joining of two integers by their numerals, or, more plainly, just smooshing the numbers together (denoted $a \ || \ b$ .) Examples: $5 \ || \ 2 = 52$, and $20 \ || \ 17 = 2017$.

My question is where do you think concatenation should go in the order of operations? And why?

Or, how would you evaluate $5 \times 1 \ || \ 2 + 4$ ? And why?

$5 \times 1 \ || \ 2 + 4 = (5 \times 1) \ || \ 2 + 4 = 5 \ || \ 2 + 4= 52 + 4 = 56$,
$5 \times 1 \ || \ 2 + 4 = 5 \times (1 \ || \ 2) + 4 = 5 \times 12 + 4= 60 + 4 = 64$.

Edit: I intend on using concatenation in expressions with exponentiation, multiplication/division, and addition/subtraction. No parenthesis, that would be too obvious, since concatenation would probably have to come after parenthesis in the order of operations.

Consider that whenever you write a numeral with more than one digit, that's concatenation, denoted by the empty string, and it always has higher precedence than arithmetic operators.  $12+12$ is $24$, not $132$, and $12 \times 12$ is $144$, not $122$.  So one argument is that if you're going to give concatenation an alternate symbol, the simplest choice is to keep the precedence the same.

(This is a bit more complicated when multiplication is also denoted by the empty string, in which case for example $3(1+2)$ means $9$, not $33$, and $10x$ means ten times $x$, not $10^{2+\lceil \log_{10}{x} \rceil}+x$.  The ambiguity is resolved by making the empty string denote concatenation when both operands are digit strings and multiplication otherwise.  But that's not a universal convention, think about those puzzles where you're supposed to solve for the unknown digits, like $10x + y0x = x10$.)

On the other hand it may be more aesthetic to associate precedence with the appearance of the glyph.  The double vertical bar I think creates a sense of wide spacing, so it looks like it should be a low precedence operator.

perl has a separate operator for concatenation (.) and it is given a low precedence.  Personally I prefer $\cdot$ (\$\cdot\$) to the double vertical bar, and I would prefer it to have a high precedence.  But there isn't really one right way to do it.  I think it's possible to walk a narrow path and avoid either overusing parentheses or tediously explaining your usage: just use it enough times in a context where there is only one sensible interpretation of the precedence, and an attentive reader can figure it out.

It is quite rare to use concatenation in the same expression with multiplication or addition.  I would just shoot anybody who wrote $5 \times 1||2+4$ and if they survived offer them a lifetime supply of parentheses.  My real answer is that there is no convention because concatenation is a recreational operation while addition and multiplication are real mathematical operations.   Anything you do with concatenation is dependent on the base you are operating in, and real math doesn't do that.

I enjoy puzzles that involve the digits of numbers, but in a sense that is not mathematics.


Terminating condition of Simplex Method - Stronger termination conditon My textbook states "If there are no negative values in the top row of the Simplex tableau, then we have reached optimality"

That seems intuitive enough. However, I am wondering if the following, stronger statement is also true?

"We have reached optimality if and only if there are no negative values in the top row of the Simplex tableau".

Your stronger statement is true if the solution is non-degenerate. (Provided that the objective function is maximize.)


Why does the Fibonacci Series start with 0, 1? The Fibonacci Series is based on the principle that the succeeding number is the sum of the previous two numbers. Then how is it logical to start with a 0? Shouldn't it start with 1 directly?

One key number-theoretical reason for starting the sequence $(0,1)$ instead of $(1,1)$ is that it makes the divisibility property of the Fibonacci sequence more straightforward to state; i.e., that $F_k$ divides $F_{nk}$ for any $k,n$.  If you start with $F_0=1$ instead of $F_0=0$ then this breaks down (for instance, in that numbering $F_2=2$ but $F_4=5$) and a lot of results have to be presented with indices shifted.  This has to do, roughly with the representation of $F_n$ as $\frac{1}{\sqrt{5}}\left(\phi^n-\varphi^n\right)$ (with $\varphi=\frac{1}{\phi} = \phi-1$); the fact that the exponents 'match up' with the index leads to straightforward arguments for the various divisibility properties.

$0$ should not be considered as a Fibonacci number. Accordingly to Zeckendorf's theorem, every positive integer can be uniquely written as the sum of one or more distinct non-consecutive Fibonacci numbers. Taking $0$ as a Fibonacci number is therefore like taking $1$ as a prime number!

By the way, the divisibility property still holds taking $F_1=F_2=1$, so I don't agree with the argument given in the first answer.


Show that if $a,b \in \Bbb R$ then $\max\{a,b\} = \frac12(a+b+|a-b|)$ and $\min\{a,b\} = \frac12(a+b-|a-b|)$

how would you go about solving this?

I started with suppose $a \leq b$

Also, show min{a,b,c} = min{min{a,b},c}.

How would I go about showing that?

You can use the fact that $$\min\{x,y\}=-\max\{-x,-y\}$$ to get the result about mins from the one about maxes.


Reparametrize a function as a polynomial Assume I have a function continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $\lim_{x\rightarrow\pm \infty} f(x)=\pm \infty$.

I am looking for some continuous and strictly monotonous increasing function $g:\mathbb{R}\rightarrow \mathbb{R}$ with $g(0)=0$, such that $f(g(r))=p(r)$ is a polynomial. That is $g$ can be used to reparametrize $f$ into a polynomial $p$.

Example: If I take $f(x)=|x|$, then I can choose $g(r)=x\cdot|x|$, such that $f(g(r))=|x\cdot|x||=x^2=p(r)$.

I don't have a proof, but I think such a $g$ should always exist, given the conditions on $f$. If not, please let me know, which conditions I am missing.

Is there a general method for finding such a $g$? My main problem is, that I don't know what the resulting polynomial $p$ should look like, except for the degree of the polynomial.

Example: Let $f(x)=|x-a|+b$. I know that $p$ must have the form $p(r)=\beta_2 r^2+\beta_1 r + |a| + b$ from the form of $f$. But I can neither find the coefficients $\beta$ nor the function $g$ that allows for this transformation. I tried the approach $|g(r)-a|+b=\beta_2r^2+\beta_1x+|a|+b$, but I cannot solve this because I don't know $\beta$ and $|\cdot|$ is not invertible (so I can't just solve for $g$).

You cannot do this if $f$ has  zeros at  $0$ and distinct points $a_n$ tending to $0$ and non-zero at other points. Since $f(x)=p(g^{-1}(x))$ in some interval around $0$ we see that $f$ cannot have at zeros accumulating at $0$. [ $g$ continuous and strictly increasing so it is has an inverse on some interval $(-r,r)$].

The answer is essentially

$$g(r)=f^{-1}(p(r))$$ where $p$ is a polynomial that you can choose relatively freely.

More specifically, you can decompose $f$ in monotonic branches, that are each invertible, and choose a polynomial that follows the same decomposition, with extrema every time the growth direction changes.

E.g. the function $|x|$ is decomposed in a decreasing branch $-x$ in $(-\infty,0)$, and an increasing one $x$ in $(0,\infty)$. Then the polynomial $(r+2)^2$ has a similar behavior (decreasing to zero until $r=-2$ then increasing).

So we map the first branch to $(x+2)^2$ by the inverse transform of $-x$, and the second branch by $x$ (it turns out that these functions are their own inverses):

$$g(r)=\begin{cases}r<2\to-(r+2)^2,\\r>2\to (r+2)^2.\end{cases}$$

Notice that the ordinates of the local extrema must match those of the given function, but the abscissas a free, provided the order is kept.

Let us play this game with

$$\begin{cases}x<-1&\to x+2,\\-1<x<1&\to-x,\\x>1&\to x-2\end{cases}.$$

We can map it to $\dfrac{r^3-3r}2$, that also has the local extrema $(-1,1)$ and $(1,-1)$.

Now

$$g(r)=\begin{cases}r<-1&\to\dfrac{r^3-3r}2-2,\\-1<r<1&\to-\dfrac{r^3-3r}2,\\r>-1&\to\dfrac{r^3-3r}2+2.\end{cases}$$

The function $g$:


Well-orderings of $\mathbb R$ without Choice The question is about well-ordering $\mathbb R$ in ZF. Without the Axiom of Choice (AC) there exists a set that is not well-ordered. This could occur two ways: a) there are models of ZF in which $\mathbb R$ is well ordered and some other set isn't; or b) there are models in which $\mathbb R$ itself can not be well-ordered. Both possibilities are consistent with ZF.

Do I have this right?

If so then I am confused about the following argument that $\mathbb R$ must be well-ordered even in the absence of choice.

Let $\mathfrak c$ be the cardinality of $\mathbb R$. It's easy to show that $\mathfrak c = 2^{\aleph_0}$.

Now (this is the part that's got me confused) in set theory cardinals are defined as certain ordinals. Therefore $\mathfrak c$ is some ordinal; which immediately induces a well-order on $\mathbb R$.

What is wrong with this argument? I can imagine a couple of scenarios but without much confidence.

$\mathfrak c$ is a cardinal but it is possible that it's not well-ordered. But if it's a cardinal that's not an ordinal, then what is it? This scenario seems unlikely.
Or perhaps $\mathfrak c$ is not always well-defined. When people say, "Let $\mathfrak c$ be the cardinality of $\mathbb R$," they typically neglect to prove that such a thing exists. Perhaps there are models of ZF in which $\mathbb R$ has no well-defined cardinality at all. This seems like the only sensible way out, in which case there are a lot of misleading expositions out there.

I found several Stackexchange questions about properties of well-orderings of $\mathbb R$ under AC; but nothing about this particular question. According to this article the reals might not be well-ordered in Solovay's model in which all sets of reals are Lebesgue-measurable; but this requires an inaccessible cardinal. What happens if we take the inaccessible away?

In general what is the correct way to understand well-ordering the reals in ZF?

Tl;dr: the cardinality of a set need not be a cardinal. Yes, this is terrible terminology.

The former possibility is what happens if $\mathbb{R}$ is not well-orderable. $\mathfrak{c}$ is always well-defined; it's just that, in the absence of choice, it need not be an ordinal.

Note that this means that we need to be careful with our terminology when working in just $ZF$: we need to distinguish between cardinals in the sense of initial ordinals (equivalently, cardinalities of well-orderable sets) and cardinalities in general.

For that matter, we need to be a little careful how we define a possibly-non-well-orderable cardinality! We use the notion of rank. The rank of a set is defined in terms of the cumulative hierarchy. We define $V_\theta$, for every ordinal $\theta$, inductively as follows:

$V_0=\emptyset$,
$V_\lambda=\bigcup_{\beta<\lambda} V_\beta$ for $\lambda$ limit, and
$V_{\alpha+1}=\mathcal{P}(V_\alpha)$.

Then by the Axiom of Foundation, we can show that for all $A$ there is some $\theta$ such that $A\in V_\theta$. The least such $\theta$ is the rank of $A$. Moreover, each $V_\theta$ is a set.

  If $X$ is a non-well-orderable set, then we let $\vert X\vert$ be the set of all sets $Y$ such that $(i)$ $Y$ is in bijection with $X$, and $(ii)$ the rank of $Y$ is as small as possible (no set of strictly smaller rank is in bijection with $X$).

That is, we take a Fregean approach - a cardinality is a set of all sets of a given size - but use the notion of rank to make the result a set, instead of a proper class (this is called Scott's trick).

Note that, however, this clashes slightly with the definition of the cardinality of a well-orderable set; but this doesn't really cause any problems.

By the way, I think Solovay's model is a bit of a red herring, here. The most straightforward way to break the well-orderability of $\mathbb{R}$ is Cohen's original model; Jech's big book has a good exposition of it. This does require forcing, though, so is a serious chunk of work to get through.

According to the Wikipedia article Cardinal number, in ZFC, the cardinality of any set is defined to be the smallest ordinal number that is equivalent to that set, that is, there is a bijection from that ordinal number to that set. That's probably because in ZFC, it can be proven that for all sets, there is a smallest ordinal number that is equivalent to that set. The problem with your proof that the set of all real numbers can be well ordered is that you need the axiom of choice to prove that the cardinality of all sets can be defined in that way and therefore that $\mathfrak c$ is an ordinal number according to that definition.


Examples of useful, insightful, and interesting hand-waving It seems to me that some hand-waving (by which I mean some arguments that aim at giving some form of intuition on the problem even at expenses of complete rigour [and not mnemonics for high-schoolers or totally bogus oversimplistic smoke curtains]) may be really useful at times to get some insight on a problem.

For example, in Levi's Mathematical Mechanic there are many intelligent examples of problems where some physical intuition (while not perfectly rigorous) may help yield some result and even converted into a formal argument.

So, I would like to collect a "big list" of "useful" (and possibly somewhat sophisticated), insightful and interesting heuristics and hand-waving arguments (which may also include some reference to physical principles e.g. see this)

Answers by Christian Blatter often contain A little bit of Physics.
This one is a wonderful example:

How find this inequality $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$

A parabola is the trajectory $\,\vec{r}\,$ of a particle with initial position
$\vec{s}$ , initial velocity $\,\vec{v}\,$ and constant acceleration $\,\vec{a}$ .
This leads to the representation
$\,\vec{r}(t) = \vec{s} + \vec{v}\, t + \vec{a}\, t^2$ , as has been employed in:

Proving that for each two parabolas, there exists a transformation taking one to the other

What is wrong with this method for a rotated and shifted parabola?

The following answer is inspired by the physical - what is "slimness" - and the physics / mechanics of solid bodies
- Moments of inertia .
With respect to the latter, any - "slim" or "fat" - 2-D body can be thought as an ellipse (of inertia). Then there is a wonderful relationship between the physical, the physics, and the mathematics of
Steiner ellipses :

How fat is a triangle?

Couldn't really distinguish between physics intuition and a mathematical argument at this place:

Proof of the extrinsic to intrinsic rotation transform

I use the following in my calculus class to find limits of various quotients when the variable approach zero. They are:

$\begin{eqnarray}(BIG + small)^r &=& BIG^r + r\ BIG^{r-1} small + \cdots\\
\sin(small) &=& small + \cdots\\
\cos(small) &=& 1 - {1 \over 2}small^2 + \cdots\\
\tan(small) &=& small + \cdots\\
e^{small} &=& 1 + small + \cdots\\
\ln(1+small) = &=& small + \cdots
\end{eqnarray}$


A Manifold Contained in Another QUESTION: Let $M$ be a $k$-manifold-without-boundary in $\mathbb R^n$ and $N$ be another manifold-without-boundary in $\mathbb R^n$ such that $M\subseteq N$.

Assume that there exists a point $\mathbf p\in M$ such that for each open set $U$ in $\mathbb R^n$ which contains $\mathbf p$, there is a point $\mathbf q\in U$ such that $\mathbf q\in N\setminus M$.

Then can $N$ possibly be a $k$-manifold?

Intuitively it seems obvious that the dimension of $N$ should be greater than $k$ but I am unable to make it into a proof.

Can somebody help?

Thanks.

I think the answer is no; if $M$ is a stand-alone manifold and $ M \subset N$ , then if $p$ is a point in $M \subset N$ , then $p$ is in a chart $(U, \phi )$ for $M$ , and $U \subset M$ (inclusion is strict unless the manifold has just 1 chart.), and all open sets containing $p$ are intersections of charts, each of which is contained in $M$.

EDIT: Let $p$ be in the manifold. Then there is a coordinate patch $(U_p, \phi_p)$containing $p$. Now, $U_p$ is open in the manifold, so that there is an open ( in the manifold topology) set $V_p$ containing $p$, and ( by openness of $V_p$ ) $V_p$ is strictly contained in $U_p$ . So, in a manifold, the condition you described is not possible.


Why isn't there an $x$ such that $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$? If $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$, then $x$ is equal to

$$\frac{-\pi}{2}\le \sin^{-1}x \le \frac{\pi}{2}$$
$$\frac{-\pi}{10}\le \frac{\sin^{-1}x}{5} \le \frac{\pi}{10}$$

This show that no value of such $x$ exist.
I want to know why this is happening.

Now in this approach,
$$\frac{\sin^{-1}x}{5}=\frac{\pi}{4}$$
$$\sin^{-1}x=\frac{5\pi}{4}$$
Why this is happening?

I think this is just happening because of principal value  of $\sin^{-1}x$.

Indeed, the range for $\arcsin(x)$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$, so no real $x$ satisfies $\arcsin(x)=\frac{5}{4}\pi$.

$\sin^{-1} x = \frac{5 \pi}{4}$ gives $x=\sin \frac{5 \pi}{4} = \sin (\pi + \frac{\pi}{4})$, then $x = -\sin \frac{\pi}{4} = -\frac{1}{\sqrt 2}$. Your error is restricting $\sin^{-1} x$ to the values $-\frac{\pi}{2} \le \sin^{-1} x \le \frac{\pi}{2}$ (the principal value is within this range).


Number Theory : Primes not in Twin Primes I was working through some basic number theory questions , when I came across :

  Show that there are infinitely many primes that are not one of the primes in a pair of twin primes

How can I go about solving it ? I have absolutely no idea ...

All primes $p$ and $q$ of the form $p = 23+30n$, $q = 37 + 30n$ where $n$ is natural, are unpaired.

To proof this, think of all of the factors of 30 and why 25 and 35 will not be prime. Know also, all prime pairs greater than 3, like 5,7 use the forms 6n-1, 6n+1 to make a pair.

With the exception of 7, numbers ≡ {7, 23} mod 30, which sequence as 7 {+16+14} {repeat ... ∞} cannot possibly be twin primes given their closest possible proximity to other prime numbers is +4 and -4, respectively. A modulo 30 factorization wheel, which is populated by n ≡ {1, 7, 11, 13, 17, 19, 23, 29} (mod 30) (aka, natural numbers not divisible by 2, 3, or 5), parses all twin prime candidates > (5,7) into 3 distribution channels, and makes this clear:

n ≡ {11,13} mod 30 (at 132° & 156°)
n ≡ {17,19} mod 30 (at 204° & 228°)
n ≡ {29, 1} mod 30 (at 348° & 12°).

Ref: primesdemystified.com/twinprimes