Where is the fault in my proof? I had some spare time, so I was just doing random equations, then accidentally came up with a proof that showed that i was -1. I know this is wrong, but I can't find where I went wrong. Could someone point out where a mathematical error was made?

$$(-1)^{2.5}=-1\\
(-1)^{5/2}=-1\\
(\sqrt{-1})^5=-1\\
i^5=-1\\
i=-1$$

Your mistake is that you have "$(-1)^{5/2} = -1$". It actually holds that $(-1)^{5/2} = i$ since you get by euler identity that
$$(-1)^{5/2} = {e^{i\pi}}^{5/2} = e^{5/2 i\pi} = i.$$
Furthermore you shouldn't write $\sqrt{-1} = i$ because the root isn't defined for negative values and you can get all sorts of wrong proofs by using the rules for square roots in combination with this notation.


Shortest path on a sphere I'm quite a newbie in differential geometry. Calculus is not my cup of tea ; but I find geometrical proofs really beautiful. So I'm looking for a simple - by simple I mean with almost no calculus - proof that the shortest path between two points on a sphere is the arc of the great circle on which they lie. Any hint ?

Edit: Or at least a reference ?

Here's a geometric observation that can hardly be called a "proof", but may be appealing nonetheless.

If $p$ and $q$ are distinct points of the sphere $S^{n}$, if $C:[0, 1] \to S^{n}$ is a "shortest path" joining $p$ to $q$, and if $F:S^{n} \to S^{n}$ is a distance-preserving map fixing $p$ and $q$, then $F \circ C$ is also a shortest path (because the length of $F \circ C$ is equal to the length of $C$).

Assume $q \neq -p$. If you believe there exists a unique shortest path from $p$ to $q$, it's not difficult to see that the "short" great circle arc is the only candidate: Every point not on the great circle through $p$ and $q$ is moved by some isometry of the sphere that fixes $p$ and $q$.

If you're thinking specifically of $S^{2}$, reflection $F$ in the plane containing $p$, $q$, and the center of the sphere is an isometry, and $f(x) = x$ if and only if $x$ lies on the great circle through $p$ and $q$.

(A similar argument "justifies" that the shortest path between distinct points of the Euclidean plane is the line segment joining them.)

An image worth a thousand words....

a movie a thousand images, and an equation a thousand movies..... :)

Stare at the figure and try to draw a path shorter than that between
$A$ and $B$, or $B$ and $C$, or any of the fourth vertices with any of the
other three.  As you said, you want a geometrical argument. It should not take time to convince you that the paths drawn on this figure (and the paths you can draw in your mind for the diagonals $A-C$ and $B-D$) are all connected to an arc that is centered at the origin. Those are parts of big circles.


Does $f(dx)$ have any meaning? Simple question, does a differential $dx$ have any meaning composed in a function $f$, such as $\sqrt{dx}$, where $f(x)\neq x$?

Sort of. There is a dual concept to differentials, that of a "tangent vector", which is not unreasonable to think of as a kind of infinitesimal.

While $\mathrm{d}x$ is supposed to denote a differential, many unfortunately use the notation when they wish to speak of an infinitesimal. :(

Anyways, if $f$ is differentiable at a point $a$ and $\epsilon$ is an infinitesimal, then we have

$$ f(a+\epsilon) = f(a) + f'(a) \epsilon $$

Note this is a literal equality and not merely an approximation, as this kind variety of infinitesimal satisfies $\epsilon^2 = 0$.

Since $dx=\Delta x$, we may view $f(dx)$ as $f(\Delta x)$.


solve $\sin(3x - 4) = \cos(7x)$ I am attempting to solve the equation $$\sin(3x + 4) = \cos(7x),$$ with all numbers in degrees.

My process is as such:

$\cos(7x) = \sin(90 - 7x)$

$\sin(3x + 4) = \sin(90 - 7x)$

$3x + 4 = 90 - 7x + 360n$; (where n is an integer and $360n$ is added due to cycling)

$x = \frac{94}{10} + 36n$

However, when I graph, I see that there is another answer that I have not solved for, which is $\frac{133}{2} + 90n$. How can I achieve this answer?

Draw a graph of $\sin x$.

You will quickly see that $\sin a = \sin b$ does not imply $a \equiv b \pmod{360}$ - just see $\sin 60 = \sin 120$.

The rest is left to the reader.

$$\sin(3x+4)=\sin3x\cos4+\cos3x\sin4$$ and \begin{align}\cos(7x)+\cos90&=2\cos\frac{(90+7x)}{2}.\cos\frac{(90-7x)}{2}\\
&=\cos^2(45)-\sin^2(7x/2)\\
&=\frac{1}{4}-\sin^2(7x/2)\end{align} now you have two sin equations can you the graphs from here if you think cos and sin are bit difficult.


How do you find the second moment of the beta distribution? I'm required to show $ E(Y^2) = \dfrac{\alpha(\alpha + 1)}{(\alpha + \beta + 1)(\alpha + \beta)} $ for the beta distribution using the definition of expectation.

Now so far I have $ \int\limits_0^1 {y^2  \dfrac{\Gamma\left( \alpha + \beta \right)}{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)} y^{\alpha-1}(1-y)^{\beta-1} dy} $ and I simplified it so that I pulled the gamma constants out front of the integral and combined $ y^2y^{\alpha-1} $ to be $ y^{\alpha+1} $. I'm not too sure where to continue from here... can anyone help me out?

Try looking at the kernel of the integral - in other words, ignore all the constant factors, focus on the bits involving the variable you are integrating with respect to.

Do you recognise it?

It's a good idea whenever you see the PDF of a distribution to pay attention to what its kernel is too. So forget about its normalization factor. Now, if a PDF can be written as the product of a normalizing factor $N$ and a kernel $k(x)$, then because I know:

$$\int_{-\infty}^{\infty}f_X(x)dx=N\int_{-\infty}^{\infty}k(x)dx=1$$

I also know that:

$$\int_{-\infty}^{\infty}k(x)dx=\frac{1}{N}$$

So, learn to recognize your PDF kernels! If you see an integral with the same kernel, but a different constant factor, then you can easily evaluate it:

$$\int_{-\infty}^{\infty}A\cdot k(x)dx=\frac{A}{N}$$

Extra hint: the beta distribution has support [0, 1] so we only need our integrals to have those limits. If $X \sim Beta (\alpha,\, \beta)$ then $f_X(x)=\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha -1}(1-x)^{\beta -1}$ so the kernel is just $x^{\alpha -1}(1-x)^{\beta -1}$.

You have $ \int\limits_0^1 {y^2  \dfrac{\Gamma\left( \alpha + \beta \right)}{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)} y^{\alpha-1}(1-y)^{\beta-1} dy} $ which has kernel $y^{\alpha+1}(1-y)^{\beta-1} $. Which PDF is this the kernel of, and with what parameters?

you have to use the recurrent formula
First gross moment
µ= B(a+1,b)/B(a,b)= a/(a+b)

2nd gross moment
M20= B(a+2,b)/B(a,b) = B(a+2,b)/(B(a+1,b){ B(a+1,b)/B(a,b)= {(a+1)/(a+1+b)}{a)/(a+b)
If you need the variance

V= M20 -µ^2

This is a presnt of Agustín F. Correa from Argentina


The derivative of $e^x$ using the definition of derivative as a limit and the definition $e^x = \lim_{n\to\infty}(1+x/n)^n$, without L'Hôpital's rule Let's define
$$
e^x := \lim_{n\to\infty}\left(1+\frac{x} {n}\right)^n, \forall x\in\Bbb R
$$

and

$$
\frac{d} {dx} f(x) := \lim_{\Delta x\to0} \frac{f(x+\Delta x) - f(x)} {\Delta x}
$$

Prove that

$$
\frac{d} {dx} e^x = e^x
$$

using the definition of $e^x$ and derivation above, without using L'Hôpital's rule or the "logarithm trick" and/or the "inverse function derivative trick".

$$
\left( \frac{d} {dx} f^{-1}(x)= \frac{1} {\left(\frac{d}{d(f^{-1}(x))}f(x)\right)(f^{-1}(x))}\right)
$$

Or equivalently prove that the following two definiton of $e$ are identical
$$
1)\space\space\space\space e =\lim_{n\to\infty}(1+\frac{1} {n})^n
$$

$$
2) \space\space\space\space e\in\Bbb R,\space\space(\frac{d} {dx} e^x)(x=0) = 1
$$

What I've got is
$$
\frac{d}{dx}e^x=e^x \lim_{\Delta x\to0}\frac{e^{\Delta x} - 1} {\Delta x} =
e^x \lim_{\Delta x\to0}\lim_{n\to\infty}\frac{\left(1+\frac{\Delta x}{n}\right)^{n}-1}{\Delta x} =
e^x \lim_{\Delta x\to0}\frac{e^{0+\Delta x}-e^0}{\Delta x}
$$

If i assume that $n\in\Bbb N$ I could use binomial theorem but I didn't got much out of it. Wolframalpha just uses L'Hospital rule to solve it, but I am looking for an elementary solution. What I'm interested in is basically is the equivalence of the two definition of $e$ mentioned above. And I'd like to get a direct proof rather than an indirect(I mean which involves logarithms or the derivatives of invers functions).

I look forward getting your aswers.

One needs to establish the following two properties:

1) $\displaystyle \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$

2) $\displaystyle e^{x + y} = e^{x}\cdot e^{y}$

and it turns out that both of these can be derived (albeit with some minor difficulty) using the definition $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$

We start with 1) first and that too with limitation $x \to 0+$. We have

$\displaystyle \begin{aligned}\lim_{x \to 0+}\frac{e^{x} - 1}{x} &= \lim_{x \to 0+}\dfrac{{\displaystyle \lim_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} - 1}}{x}\\
&= \lim_{x \to 0+}\lim_{n \to \infty}\frac{1}{x}\left\{\left(1 + \dfrac{x}{n}\right)^{n} - 1\right\}\\
&= \lim_{x \to 0+}\lim_{n \to \infty}\frac{1}{x}\left\{\left(1 + x + \dfrac{(1 - 1/n)}{2!}x^{2} + \dfrac{(1 - 1/n)(1 - 2/n)}{3!}x^{3} + \cdots\right) - 1\right\}\\
&= \lim_{x \to 0+}\lim_{n \to \infty}\left(1 + \dfrac{(1 - 1/n)}{2!}x + \dfrac{(1 - 1/n)(1 - 2/n)}{3!}x^{2} + \cdots\right)\\
&= \lim_{x \to 0+}\lim_{n \to \infty}(1 + \phi(x, n))\end{aligned}$

where $\phi(x, n)$ is a finite sum defined by $$\phi(x, n) = \frac{(1 - 1/n)}{2!}x + \cdots + \frac{(1 - 1/n)(1 - 2/n)\cdots(1 - (n - 1)/n)}{n!}x^{n - 1}$$ For fixed positive $x$ the function $\phi(x, n)$ is a increasing sequence bounded by the convergent series $$F(x) = \frac{x}{2!} + \frac{x^{2}}{3!} + \cdots$$ Hence the limit $\lim_{n \to \infty}\phi(x, n)$ exists and let say it is equal to $\phi(x)$. Then $0 \leq \phi(x) \leq F(x)$. Now let $x < 2$ and then we can see that $$F(x) \leq \frac{x}{2} + \frac{x^{2}}{2^{2}} + \frac{x^{3}}{2^{3}} + \cdots = \frac{x}{2 - x}$$ Hence $\lim_{x \to 0+}F(x) = 0$ and therefore $\lim_{x \to 0+}\phi(x) = 0$.

We now have

$\displaystyle \begin{aligned}\lim_{x \to 0+}\frac{e^{x} - 1}{x} &= \lim_{x \to 0+}\lim_{n \to \infty}1 + \phi(x, n)\\
&= \lim_{x \to 0+}1 + \phi(x)\\
&= 1 + 0 = 1\end{aligned}$

From this it follows that $\lim_{x \to 0+}e^{x} = 1$. To handle the case for $x \to 0-$ we need to use another trick. We show that for $x > 0$ we have $$\lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^{-n} = e^{x}$$ Clearly we have

$\displaystyle \begin{aligned}\left(1 - \frac{x}{n}\right)^{-n} - \left(1 + \frac{x}{n}\right)^{n} &= \left(1 + \frac{x}{n}\right)^{n}\left\{\left(1 - \frac{x^{2}}{n^{2}}\right)^{-n} - 1\right\}\\
&< e^{x}\left\{\left(1 - \frac{x^{2}}{n}\right)^{-1} - 1\right\} = \frac{x^{2}e^{x}}{n - x^{2}}\end{aligned}$

and this last expression tends to $0$ as $n \to \infty$ and hence $$\lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^{-n} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = e^{x}$$ Taking reciprocals we see that $$\lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^{n} = \frac{1}{e^{x}}$$ or in other words $e^{-x} = 1/e^{x}$ for $x > 0$ and by duality it holds for $x < 0$ also. Thus we can see that if $x \to 0-$ then we can write $x = -y$ so that $y \to 0+$ and then

$\displaystyle \begin{aligned}\lim_{x \to 0-}\frac{e^{x} - 1}{x} &= \lim_{y \to 0+}\frac{e^{-y} - 1}{-y}\\
&= \lim_{y \to 0+}\frac{e^{y} - 1}{y}\frac{1}{e^{y}} = 1\cdot 1 = 1\end{aligned}$

Thus we have established two properties of $e^{x}$ namely $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\, e^{-x} = \frac{1}{e^{x}}$$ The second property allows us to consider only positive arguments of the exponential function. Thus to establish the fundamental property $e^{x + y} = e^{x} \cdot e^{y}$ we need to consider $x, y > 0$ (for $x = y = 0$ it is obviously true). We can see that

$\displaystyle \begin{aligned} f(x, y, n) &= \left(1 + \frac{x}{n}\right)^{n}\left(1 + \frac{y}{n}\right)^{n} - \left(1 + \frac{x + y}{n}\right)^{n}\\
&= \left(1 + \frac{x + y}{n} + \frac{xy}{n^{2}}\right)^{n} - \left(1 + \frac{x + y}{n}\right)^{n}\\
&= \left(1 + \frac{x + y}{n}\right)^{n}\left\{\left(1 + \frac{xy}{n(n + x + y)}\right)^{n} - 1\right\}\\
&< e^{x + y}\left\{\left(1 + \frac{xy}{n^{2}}\right)^{n} - 1\right\}\\
&= e^{x + y}\left\{\frac{xy}{n} + \frac{(1 - 1/n)}{2!}\left(\frac{xy}{n}\right)^{2} + \cdots\right\}\\
&< e^{x + y}\left\{\frac{xy}{n} + \left(\frac{xy}{n}\right)^{2} + \cdots\right\}\\
&= e^{x + y}\frac{xy}{n - xy}\end{aligned}$

This shows that for fixed $x, y > 0$ the function $f(x, y, n) \to 0$ as $n \to \infty$. And therefore we have established $e^{x}e^{y} - e^{x + y} = 0$. Now we can easily show that

$\displaystyle \begin{aligned}\frac{d}{dx}e^{x} &= \lim_{h \to 0}\frac{e^{x + h} - e^{x}}{h}\\
&= \lim_{h \to 0}e^{x}\cdot\frac{e^{h} - 1}{h} = e^{x} \cdot 1 = e^{x}\end{aligned}$

If I am understanding the question correctly, I believe this should do it.


Proof By Induction - Divisibility by $7$. I am attempting:

My solution is: But I am not sure where I am going wrong. The answer I get is not divisible by 7.

HINT:

You have to prove the truth of $p(k+1)$ using $p(k)$, so you have to take out something from $p(k$) and then apply it to $p(k+1)$ to establish its truth.

As you have assumed that $p(k)$ is true. So, $4^{k+1}+5^{2k-1}$ must be divisible by 7 say it is $7m$ where $m$ is an integer. So you get $4^{k+1}+5^{2k-1}=7m$. Some flipping will give you $5^{2k-1}=7m-4^{k+1}$. Now How does $p(k+1)$ looks like??

It will look like $4^{k+2}+5^{2k+1}$. If we prove that $4^{k+2}+5^{2k+1}$ is divisible by $7$ then we are done. Try using  $5^{2k-1}=7m-4^{k+1}$ to proceed further.

Before considering your proof, let's gather some insight from a simpler proof using congruences. Below the inductive step follows very simply by using $\,\rm{\color{#C00}{CPR}} = $ Congruence Product Rule to multiply the first two congruences

$$\begin{align}
{\rm mod}\,\ 7\!:\qquad\  4\,\ &\equiv\,\  5^{\large 2}\\[0.3em]
4^{\large K+1}&\equiv -5^{\large 2K-1}\ \ \ {\rm i.e.}\ \ P(K)\\
\overset{\rm{\color{#C00}{CPR}}}\Longrightarrow\ \ \ 4^{\large K+2}&\equiv -5^{\large 2K+1}\ \ \ {\rm i.e.}\ \ P(K\!+\!1)
\end{align}$$

The common inductive proofs using divisibility in other answers effectively do the same thing, i.e. they repeat the proof of the Congruence Product Rule in this special case, but expressed in divisibility vs.  congruence language (e.g. see here).  But the product rule is much less arithmetically intuitive when expressed as unstructured divisibilities, which greatly complicates the discovery of the inductive step. I explain this at length in other answers, e.g. see here.

If congruences  are unfamiliar then you can instead use the rule in divisibility form as below. This will allow you to structure the induction in the above intuitive arithmetical Product Rule form.

$$\begin{align} {\rm mod}\,\ m\!:\, A\equiv a,\, B\equiv b&\ \ \,\Longrightarrow\,\ \ AB\equiv ab\qquad\text{Congruence Product Rule}\\[3pt]
m\mid A-a,\ B-b&\,\Rightarrow\, m\mid AB-ab\qquad\text{Divisibility Product Rule}\\[4pt]
{\bf Proof}\quad (A-a)B+a(B&-b)\, = AB-ab\end{align}$$

To finish the proof that you started we can proceed as follows

$$\begin{align} f(k\!+\!1) - f(k) &=\, 3 \cdot 4^{\large k+1}+ \color{#0a0}{24}\cdot  5^{2k-1}\\
&=\, 3 \cdot 4^{\large k+1}+ \color{#0a0}3 \cdot  5^{2k-1} + \color{#0a0}{21}\cdot  5^{2k-1}\\
&=\, 3\, f(k) + 7n\\
\Rightarrow\qquad f(k\!+\!1)\, &=\, 4\, f(k) + 7n\\[0,3em]
\Rightarrow\ \ 7\mid f(k\!+\!1)\,\ &{\rm if}\,\  7\mid f(k), \ \ {\rm i.e.}\ \ P(k\!+\!1)\ \ {\rm if}\ \ P(k)
\end{align}$$

Note that the above says that $\ f(k\!+\!1)\equiv 4\,f(k)\ \pmod{7}\,$

so an easy induction shows that $\ f(k)\equiv 4^{\large k-1}\, f(1)\pmod 7,\ $ so $\ 7\mid f(k)\iff 7\mid f(1)$

Now how using the Product Rule as above makes it much clearer that incrementing the index amounts simply to multiplication by $\,4,\,$ when viewed modulo $\,7.\,$ Once that innate arithmetical structure hs been revealed, the proof is easy.


Is $\{(a, b, c) \in \mathbb{C}^3 : a^3 = b^3\}$ a subspace of $\mathbb{C}^3$ I have two questions to solve:

Is $\{(a, b, c) \in \mathbb{R}^3 : a^3 = b^3\}$ a subspace of $\mathbb{R}^3$?
Is $\{(a, b, c) \in \mathbb{C}^3 : a^3 = b^3\}$ a subspace of $\mathbb{C}^3$?

For the first one, I proved it is. Then for the second part, I found almost no difference. So I am not sure if I am on the right track. I mean, I believe there must be some differences. But I can't figure it out.

Try with $(e^{i\frac{π}{3}} , -1,0)$ and $(1,1,0)$.
It's sum won't satisfy the property $a^{3}=b^{3}$


I can multiple matricies in any order? I am trying to complete some fairly simple matrix algebra for a homework task.

Is it possible to multiple matrices in any order?

Is the method for the below correct?

Thanks

In general, for matrices $A$ and $B$,
$$AB \neq BA,$$
so we can't change the order of the matrices in an expression. However,
$$(AB)C=A(BC)=ABC,$$
so we can perform the multiplication operations in any order.

This is because we can remove all the brackets from the expression for a sequence of multiplications by repeatedly applying the rule. For example:

$$(A((BC)(DE)))F$$
$$=A((BC)(DE))F$$
$$=A(BC)(DE)F$$
$$=ABC(DE)F$$
$$=ABCDEF$$

However the terms were originally grouped, we end up with $ABCDEF$.

The algebraic explanation is that the set of $n\times n$-matrices $M_n(R)$ over a commutative ring $R$ (entries) forms a ring with unity (unit matrix). This includes that the multiplicative monoid is associative, i.e.,
$A(BC) = (AB)C$ for any matrices $A,B,C\in M_n(R)$.


What are the issues in modern set theory? This is spurred by the comments to my answer here.  I'm unfamiliar with set theory beyond Cohen's proof of the independence of the continuum hypothesis from ZFC.  In particular, I haven't witnessed any real interaction between set-theoretic issues and the more conventional math I've studied, the sort of place where you realize "in order to really understand this problem in homotopy theory, I need to read about large cardinals."  I've even gotten the feeling from several professional mathematicians I've talked to that set theory is no longer relevant, and that if someone were to find some set-theoretic flaw in their axioms (a non-standard model or somesuch), they would just ignore it and try again with different axioms.

I also don't personally care for the abstraction of set theory, but this is a bad reason to judge anything, especially at this early stage in my life, and I feel like I'd be more interested if I knew of some ways it interacted with the rest of the mathematical world.  So:

What do set theorists today care about?
How does set theory interact with the rest of mathematics?
(more subjective but) Would mathematicians working outside of set theory benefit from thinking about large cardinals, non-standard models, or their ilk?
Could you recommend any books or papers that might convince a non-set theorist that the subject as it's currently practiced is worth studying?

Thanks a lot!

Set theory today is a vibrant, active research area,
characterized by intense fundamental work both on set
theory's own questions, arising from a deep historical
wellspring of ideas, and also on the interaction of those
ideas with other mathematical subjects. It is fascinating and I would encourage anyone to learn more about it.

Since the field is simply too vast to summarize easily,
allow me merely to describe a few of the major topics that
are actively studied in set theory today.

Large
cardinals.
These are the strong axioms of infinity, first studied by
Cantor, which often generalize properties true of $\omega$
to a larger context, while providing a robust hierarchy of
axioms increasing in consistency strength. Large cardinal
axioms often express combinatorial aspects of infinity,
which have powerful consequences, even low down. To give
one deep example, if there are sufficiently many Woodin
cardinals, then all projective sets of reals are Lebesgue
measurable, a shocking but very welcome situation. You may
recognize some of the various large cardinal
concepts---inaccessible, Mahlo, weakly compact,
indescribable, totally indescribable, unfoldable, Ramsey,
measurable, tall, strong, strongly compact, supercompact,
almost huge, huge and so on---and new large cardinal
concepts are often introduced for a particular purpose.
(For example, in recent work Thomas Johnstone and I proved
that a certain forcing axiom was exactly equiconsistent
with what we called the uplifting cardinals.) I encourage
you to follow the Wikipedia link for more information.

Forcing.
The subject of set theory came to maturity with the
development of forcing, an extremely flexible technique for
constructing new models of set theory from existing models.
If one has a model of set theory $M$, one can construct a
forcing extension $M[G]$ by adding a new ideal element $G$,
which will be an $M$-generic filter for a forcing notion
$\mathbb{P}$ in $M$, akin to a field extension in the sense
that every object in $M[G]$ is constructible algebraically
from $G$ and objects in $M$. The interaction of a model of
set theory with its forcing extensions provides an
extremely rich, intensely studied mathematical context.

Independence Phenomenon. The initial uses of forcing
were focused on proving diverse independence results, which
show that a statement of set theory is neither provable nor
refutable from the basic ZFC axioms. For example, the
Continuum Hypothesis is famously independent of ZFC, but we
now have thousands of examples. Although it is now the norm
for statements of infinite combinatorics to be independent,
the phenomenon is particularly interesting when it is shown
that a statement from outside set theory is independent,
and there are many prominent examples.

Forcing Axioms. The first forcing axioms were often
viewed as unifying combinatorial assertions that could be
proved consistent by forcing and then applied by
researchers with less knowledge of forcing. Thus, they
tended to unify much of the power of forcing in a way that
was easily employed outside the field. For example, one
sees applications of Martin's
Axiom
undertaken by topologists or algebraists. Within set
theory, however, these axioms are a focal point, viewed as
expressing particularly robust collections of consequences,
and there is intense work on various axioms and finding
their large cardinal strength.

Inner model
theory.
This is a huge on-going effort to construct and understand
the canonical fine-structural inner models that may exist
for large cardinals, the analogues of Gödel's
constructible universe $L$, but which may accommodate large
cardinals. Understanding these inner models amounts in a
sense to the ability to take the large cardinal concept
completely apart and then fit it together again. These
models have often provided a powerful tool for showing that
other mathematical statements have large cardinal strength.

Cardinal characteristics of the
continuum.
This subject is concerned with the diverse cardinal
characteristics of the continuum, such as the size of the
smallest non-Lebesgue measurable set, the additivity of the
null ideal or the cofinality of the order $\omega^\omega$
under eventual domination, and many others. These cardinals
are all equal to the continuum under CH, but separate into
a rich hierarchy of distinct notions when CH fails.

Descriptive set
theory.
This is the study of various complexity hierarchies at the
level of the reals and sets of reals.

Borel equivalence relation
theory.
Arising from descriptive set theory, this subject is an
exciting comparatively recent development in set theory,
which provides a precise way to understand what otherwise
might be a merely informal understanding of the comparative
difficulty of classification problems in mathematics. The
idea is that many classification problems arising in
algebra, analysis or topology turn out naturally to
correspond to equivalence relations on a standard Borel
space. These relations fit into a natural hierarchy under
the notion of Borel reducibility, and this notion provides
us with a way to say that one classification problem in
mathematics is at least as hard as or strictly harder than
another. Researchers in this area are deeply knowledgable
both about set theory and also about the subject area in
which their equivalence relations arise.

Philosophy of set theory. Lastly, let me also
mention the emerging subject known as the philosophy of set
theory, which is concerned with some of the philosophical
issues arising in set theoretic research, particularly in
the context of large cardinals, such as: How can we decide
when or whether to adopt new mathematical axioms? What does
it mean to say that a mathematical statement is true? In
what sense is there an intended model of the axioms of set
theory? Much of the discussion in this area weaves together
profoundly philosophical concerns with extremely technical
mathematics concerning deep features of forcing, large
cardinals and inner model theory.

Remark. I see in your answer to the linked question
you mentioned that you may not have been exposed to much
set theory at Harvard, and I find this a pity. I would
encourage you to look beyond any limiting perspectives you
may have encountered, and you will discover the rich,
fascinating subject of set theory. The standard
introductory level graduate texts would be Jech's book Set
Theory and Kanamori's book The Higher Infinite, on large
cardinals, and both of these are outstanding.

I apologize for this too-long answer...

i find this talk of some area of math being disconnected from another area, and that this being a bad thing truly bizarre. not many areas of mathematics pop up in other areas in truly significant ways. mathematical concepts do, of course groups show up almost everywhere, the concept of continuity and etc. but so does the concept of union, subset, quotient, projection, or the concept that a property restricted to a domain defines an object, or the concept of infinity, Borel sets, reducibility between two mathematical structures, complexity of a mathematical classification and etc. these are concepts that come from mathematical logic that are everywhere. everyone uses them, and some use without giving a proper credit to the role logicians have played in development of these concepts.

i know many algebraist in my department who never seem to use Birkhoff's Ergodic Theorem, in fact many are not even aware of it, does this mean ergodic theory is useless?

does every person in analysis use sophisticated cancelation theories from group theory, most likely not, they often use the concept of group and work with that concept but i would be very surprised if small cancelation theory is everywhere in analysis. set theory is no different than this. our basic concepts and operations are everywhere, like unions, intersections, projections, infinity and etc are everywhere, and we have sophisticated theories that sometimes show up in interesting places (like recently in C^* algebras and even in theoretic financial math (when building certain kinds of limits)...), over time they will show up in more and more places, set theory is still very young.

it is difficult to understand what makes one feel like the need to express himself in this way towards any subject.


find a topology where the sequence $\left(\frac1n\right)$ converges to $1$ can someone help me to find a topology where the sequence $\left(\frac1n\right)$ converges to an unique value which is $1$?

I was thinking of the Trivial topology $(X,T)$ but it's not the wanted topology because the sequence converges to every point of $X$.

Paint a gigantic "$0$" symbol on $1$, and paint a gigantic "$1$" symbol on $0$. Now use the standard topology, except use the symbols you painted on $0$ and $1$ instead of the usual meanings of $0$ and $1$.

To put this another way, let $f : \mathbb{R} \to \mathbb{R}$ be the function defined by $f(0)=1$, $f(1)=0$, and $f(x)=x$ if $x \ne 0,1$. Define $U \subset \mathbb{R}$ to be open in the topology $T$ if and only if $f(U)$ is open in the standard topology.

In the topological space (*) induced by the metric $$d(x,y) = \inf_{n\in \mathbb{Z}} |x-y-n|, \qquad (x,y) \in (0,1]^2$$
then $$\lim_{n \to \infty} d(1,\frac1n) =0$$

(*) Called the circle $\mathbb{S}_1$ or $\mathbb{R}/\mathbb{Z}$


A geometry problem with a cube (solid geometry) I ask you to solve this problem, because in the book and I have different answers
Point $M$ is the midpoint of the edge $AB$ of the cube $ABCDA_1B_1C_1D_1$. Find the distance between the straight lines $A_1M$ and $B_1C$, given that the edge of the cube is equal to $a$.
Please help me. It's Translated from Russian and it's the full context of the problem.
The answer in the book is $(\sqrt{6}/3)a$.
But my answer is $2a\sqrt3$.

Let $H$ be the point on $A_1M$ such that $A_1H/HM=2$, and $K$ be the point on $B_1C$ such that $CK/KB_1=2$. Then it is not difficult to show that segment $HK$ is perpendicular to both $A_1M$ and $B_1C$ and is thus the distance between the lines.
To compute the length of $HK$ it may be useful to note that $HK={2\over3}GF$, where $G$ and $F$ are the midpoints of $A_1A$ and $B_1C_1$ respectively.

HINTS

Take cube edges as parallel to x-,y-, z- axes.
Use vectors to find position vectors of red lines.
Normalize for length. Minimum distance lies along their cross product to find distance between skew vectors.


Paradox in Theory of probability of transition to polar coordinates Let there are two independent random variables $X$,$Y$ with normal distribution. Vector $(X, Y)$ can be considered as a random point on the plane. Let $R$ and $\phi$ polar coordinates of this point.
assuming that $X =Y $ we get that the distribution $R^2 = 2X^2$ coincides with the distribution of the square of a random quantity multiplied by 2.
At the same time, provided that $\phi = \pi/4$ or $\phi = 5\pi/4$ distribution of random variables $R^2 = X^2 + Y^2$ is the same as the distribution of the sum of squares of two independent standard normal values. Therefore, we got different distributions when $X=Y$ and when $\phi = \pi/4$ or $\phi = 5\pi/4$ and it's a paradox. What's the catch?

The catch is that as soon as you assumed that $X=Y$, they are not independent anymore.


Dirichlet Function Pointwise Convergence We say that a function $f$ is Baire Class $1$ if there is a sequence of functions $f_i \to f$ pointwise where each $f_i$ is continuous.

The set of discontinuities of a Baire Class $1$ function $f$ must be a first-category (meagre) set of points.

The Dirichlet function $f(x) = \begin{cases} 1 & x \in \mathbb Q \\ 0 & x \notin \mathbb Q \end{cases}$ is not Baire class $1$ since it is discontinuous everywhere.

But can't we write the function as $f_i(x) = \lim_{i\to\infty} \cos(i!\pi x)^{2i}$, each of which is a continuous function that converges pointwise to the Dirichlet function?  What is wrong with this?

Let $0\le n_1<n_2<...<n_k<\dots$ and define $$x=2\sum_{j=1}^\infty \frac{1}{n_j!}$$

Basically, we will pick $\{n_i\}$ to grow "fast enough" so that $\limsup_i \left(\cos n_i!\pi x\right)^{2n_i}>0$

Now $n_i!x = 2K + 2\sum_{j>i} \frac{n_i!}{n_j!}$ for some integer $K$.  And:$$0\leq \sum_{j>i} \frac{n_i!}{n_j!} < \frac{2}{n_i^{n_j-n_i}}(1+\frac{1}{n_j}+\frac{1}{n_j^2}+..) = \frac{2}{n_i^{n_j-n_i}}\frac{n_j}{n_j-1}$$

So $n_i!x\pi = 2K\pi + z$ where $0\leq z < \frac{2\pi}{n_i^{n_j-n_i}}\frac{n_j}{n_j-1}$.

It's gonna be a little messy, but if you choose a reasonable sequence of $n_i$, then you'll be able to show that $$\limsup_{i\to\infty}\left(\cos n_i!\pi x\right)^{2n_i}>0 $$

In particular then, $\left(\cos j!\pi x\right)^{2j}$ cannot converge to zero.

The set of discontinuities of a Baire Class 1 function f must be a first-category (meagre) set of points. Hence Dirichlet is not Baire 1
$f(x)=\lim_{i\to\infty} \lim_{j\to\infty} (\cos(i!πx)^{2j})$. This proves Dirichlet function is Baire 2 function.


Why is the complex domain of cosine naturally a sphere? Near the end of this MAA piece about elliptic curves, the author explains why the complex domain of the cosine function is a sphere: since it's periodic, its domain can be taken as a cylinder, wrapping up the real axis. And because cosine of $\theta\pm i\infty$ is $\infty$, the two ends of the cylinder can be identified with a single point $\infty$. Ok, great, but this sounds to me like a pinched torus. Can I have a clearer explanation why this is a sphere?

Let me start by saying you are right in a sense, and I think the article is at best being unclear, but the larger point that the equation of ellipse cuts out a sphere is also right.  There are a few things going on, hence a long answer below.

What one is trying to describe is the topology of a set defined by a quadratic equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Of course it is crucially important to explain what values $x$ and $y$ are allowed to take. In the context to the article, that means whether one allows only real values  or complex ones as well (this is the question of choosing the field of coefficients), and whether one allows infinite values and in what sense (this is a choice of compactification). Both of these choices affect the topology of the solution set. If only finite real values are allowed the result is the familiar ellipse in the plane (which has topology of a circle); now one has various options about how to treat adding infinite values of $x$ and $y$. However the (real) ellipse does not go to infinity in any way (it stays in finite part of the plane; or, in other words, it is already compact so needs no compactification; algebraically, no real vector with large modulus can solve  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$). Thus for this equation over real numbers the second choice is irrelevant. However (spoiler!) it is not irrelevant over complex numbers.

To  better see what's going on let's stick to the real variables for now, but consider the equation $uw=1$ instead. The solutions to this in $\mathbb{R}^2$ define a hyperbola and this has topology of 2 lines. Now if we compactify $\mathbb{R}^2$ to the sphere - treating all points at infinity as one - we get one extra point, and the topology of solution set is a wedge of two circles. If we treat $u$-infinities and $v$-infinities as different (compactify each $\mathbb{R}$ to $S^1$, so $\mathbb{R}^2$ is compactified to $S^1\times S^1$), then we get 2 extra points ("$(\infty, 0)$" and "$(0, \infty)$") and a topology of a single circle. We can also compactify $\mathbb{R}^2$ to $\mathbb{R}P^2$ (and still get a circle), or to $\mathbb{D}^2$ (and get two closed segments) or any other number of things, some more natural than others. To finish with the discussion over $\mathbb{R}$, note that if we consider $(u-v)(u+v)=1$ instead of $uv=1$, the two curves in $\mathbb{R}^2$ are isomorphic, but in the compactification $S^1\times S^1$ we get now a wedge of two circles instead of a single circle. This is because the coordinate change of $\mathbb{R}^2$ given by $u_n=u-v$, $v_n=u+v$ taking curve $(u-v)(u+v)=1$ to $u_n v_n=1$ does not extend to these compactifications. This never happens for $\mathbb{R}P^2$, which is one big reason why this is usually the prefered choice of compactification.

Now over complex numbers, the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ has a solution set in $\mathbb{C}^2$ which is a cylinder (aka tangent bundle of $S^1$). Then we can compactify by adding one point for "all infinity" and get a "pinched torus" with both infinities of the cylinder filled in by that one extra point; or by adding a point at infinity to each coordinate separately - this is what is done in the article - and still get a "pinched torus", since $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ forces $x$ to have infinite modulus as soon as $y$ does and vice versa, so the only added point is $(\infty, \infty)$. However, if we compactify to $\mathbb{C}P^2$ we get two extra points ($[a:ib:0]$ and $[a:-ib:0]$), and topology of a sphere. This is in a sense the "right" compactification of the image, sitting inside the "right compactification" $\mathbb{C}P^2$.

Finally, we should note that this is not so much about the domain of $\cos$ but more about the image of $z \to (a\cos z, b\sin z)$ in $\mathbb{C}^2$ and its compactifications. Whether we can extend $\cos z$ or $\sin z$ or $(\cos z, \sin z)$ to a map of some compactification of $\mathbb{C}$ to somewhere depends on where (which extension/compactification of $\mathbb{C}$ or $\mathbb{C}^2$) we are mapping to, and does not really have so much intrinsic meaning.

Trigonometric functions are circular functions.  Complex cosine is:
cos z = (e^(iz) + e^(-iz))/2    where    z = x + iy   where  x and y are real numbers and z is a complex number.  Suppose we restrict z so modulus z = 1 = modulus (x + iy).  Hence, the image of z is the unit circle in the complex plane.  What if we were to look at the map  X = (x, y)?  Let X = (0, 0)  correspond to an origin.  Then, apparently, the image cos(z) = cos(0) = 1. The unit sphere has radius = 1.  Consider any rotation of the plane we choose to call the complex plane passing through any point we have chosen to call the origin, namely X = (0, 0) with the property modulus z = 1. Again, this is the unit circle on that particular complex plane.   When modulus z = r for any non-negative r, we have an entire complex plane which may be mapped onto the unit sphere.  So much for the prelims.  Perhaps making more sense  write z = (e^x)(e^(iy) which has modulus r = e^x.  Then realize cos(z) has modulus 1 and is entire as (x,y) range over all reals.  Recalling the pre-image of cos(z) was z = x + iy, the domain of cos(z) may be mapped onto the unit sphere by starting at z, drawing a secant to the North Pole of the unit sphere centered at (0, 0, 1) and calling the point P on the unit sphere where the secant first pierces the unit sphere starting from z in the complex plane and repeating this for every z in the complex plane.  The result is the points P trace out the unit sphere, so that by reverse stereographic projection, we have the unit sphere of all such points P as our domain for cos(z).  For finite r = e^x, the North Pole of the unit sphere mapping gets omitted.  Now let x approach infinity, so r = e^x approaches infinity.  Identify the North Pole of the unit sphere with the point at infinity, and I believe the stereographic mapping for z = (e^x)(e^iy) becomes onto for all real x and y.  I hope this helps.


Why is it that when proving trig identities, one must work both sides independently? Suppose that you have to prove the trig identity:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}=\sin\theta$$

I have always been told that I should manipulate the left and right sides of the equation separately, until I have transformed them each into something identical. So I would do:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}$$
$$=\frac{\sin\theta(1 - \sin^2\theta)}{\cos^2\theta}$$
$$=\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$
$$=\sin\theta$$

And then, since the left side equals the right side, I have proved the identity. My problem is: why can't I manipulate the entire equation? In this situation it probably won't make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated.

I understand, of course, that I can't simply assume the identity is true. If I assume a false statement, and then derive from it a true statement, I still haven't proved the original statement. However, why can't I do this:

$$\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}\not=\sin\theta$$
$$\sin\theta - \sin^3\theta\not=(\sin\theta)(\cos^2\theta)$$
$$\sin\theta(1 - \sin^2\theta)\not=(\sin\theta)(\cos^2\theta)$$
$$(\sin\theta)(\cos^2\theta)\not=(\sin\theta)(\cos^2\theta)$$

Since the last statement is obviously false, is this not a proof by contradiction that the first statement is false, and thus the identity is true?

Or, why can't I take the identity equation, manipulate it, arrive at $(\sin\theta)(\cos^2\theta)=(\sin\theta)(\cos^2\theta)$, and then work backwards to arrive at the trig identity. Now, I start with a statement which is obviously true, and derive another statement (the identity) which must also be true - isn't that correct?

Another argument that I have heard for keeping the two sides isolated is that manipulating an equation allows you to do things that are not always valid in every case. But the same is true when manipulating just one side of the equation. In my first proof, the step

$$\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}$$
$$=\sin\theta$$

is not valid when theta is $\pi/2$, for example, because then it constitutes division by zero.

Why can't I manipulate the entire equation?

You can. The analytical method for proving an identity consists of starting with the
identity you want to prove, in the present case
$$
\begin{equation}
\frac{\sin \theta -\sin ^{3}\theta }{\cos ^{2}\theta }=\sin \theta,\qquad \cos \theta \neq 0
\tag{1}
\end{equation}
$$
and establish a sequence of identities so that each one is a consequence of
the next one. For the identity $(1)$ to be true is enough that the following
holds
$$
\begin{equation}
\sin \theta -\sin ^{3}\theta =\sin \theta \cos ^{2}\theta   \tag{2}
\end{equation}
$$
or this equivalent one
$$
\begin{equation}
\sin \theta \left( 1-\sin ^{2}\theta \right) =\sin \theta \cos ^{2}\theta
\tag{3}
\end{equation}
$$
or finally this last one
$$
\begin{equation}
\sin \theta \cos ^{2}\theta =\sin \theta \cos ^{2}\theta   \tag{4}
\end{equation}
$$

Since $(4)$ is true so is $(1)$.

The book indicated below illustrates this method with the following identity
$$
\frac{1+\sin a}{\cos a}=\frac{\cos a}{1-\sin a}\qquad a\neq (2k+1)\frac{\pi
}{2}
$$

It is enough that the following holds
$$
(1+\sin a)(1-\sin a)=\cos a\cos a
$$

or
$$
1-\sin ^{2}a=\cos ^{2}a,
$$

which is true if
$$
1=\cos ^{2}a+\sin ^{2}a
$$

is true. Since this was proven to be true, all the previous indentities
hold, and so does the first identity.

Reference: J. Calado, Compêndio de Trigonometria, Empresa Literária
Fluminense, Lisbon, pp. 90-91, 1967.

I think it can be done.  We are proving that LHS =RHS,  as by assuming so,  you arrive at a Universal truth.

Similarly,  by assuming it's not true,  we arrive at a contradiction ,  which is called,  proof by contradiction.


Can every uncountable subset of $\mathbb R$ be split up this way? For me this question is like a fish that anytime when I (seem to) catch it, manages to slip out of my hands again.

  If $U$ is an uncountable subset of $\mathbb R$ then can it be shown that some $x\in\mathbb R$ exists such that $U\cap(-\infty,x)$ and $U\cap(x,\infty)$ are both uncountable?

Thanks in advance.

Let $a$ be the supremum of all $x$ such that $U\cap (-\infty,x)$ is countable ($a=-\infty$ if there are none). Let $b$ be the infimum of all $x$ such that $U\cap (x,\infty)$ is countable, and $\infty$ if there are none. Now $a=b$ would imply that $U$ is countable so we must have
$a<b$ and any $a<x<b$ will satisfy your demands.

I could be off here, but the following comes to mind:

If $U$ has no neither first nor last item, for instance if it's set $\mathbb{Z}$, then any splitting the subset $U$ to $U_1\in\langle-\infty,x\rangle\cap U$ and $U_2\in\langle x,\infty\rangle$, both $U_1$ and $U_2$ are uncountable.

However, if the set has a first or a last item, for example set $\mathbb{N}$, then $U_1$ xor $U_2$ would be countable.

It would be impossible to prove that both subsets are uncountable without further specifying that $U$ doesn't have first nor last item. And if it so, any $x\in\langle-\infty,\infty\rangle$ would satisfy the criteria because $U\in\langle-\infty,\infty\rangle$ and $x\in U$


Prove that there exists a rational number raised to an irrational number that is an irrational number Prove: There exists $a \in \mathbb{Q}$ and $b \in \mathbb{R}\smallsetminus \mathbb{Q}$ such that $a^b \in \mathbb{R} \smallsetminus \mathbb{Q}$.

I've tried using $\log_23$, $\sqrt 2$, and $\frac{1}{\sqrt 2}$ for the irrational number, but couldn't find a way to prove $a^b$ was irrational.

Is there a way to prove this without using Gelfond–Schneider theorem?

Well, either $2^{\sqrt{2}}$ is irrational and we are done, or $(2^\sqrt{2})^{\sqrt{2}/4}=\sqrt{2}$ is an irrational, which is a rational to an irrational power.

Let $p$ be a rational in the form $\frac{a}{b}$ and $q$ be an irrational number, now,
$$p^q=1+\frac{\left(q\ln p\right)}{1!}+\frac{\left(q\ln p\right)^2}{2!}...$$
For all $p \gt 1$, Each term in series is irrational thus $p^q$ must be irrational.


The square of a standard Normal random variable I am having a bit of trouble with this:

Let $U=Z^2$ where Z is a standard Normal random variable with pdf:

$$f_z(z) = \frac{1}{\sqrt{2\pi}} e^{\frac{-z^2}{2}}$$

I want to use the inversion method but have thus far only learned to use this when functions are strictly increasing or decreasing. Since a standard normal distribution function is strictly increasing and then strictly decreasing I thought perhaps I could find some way to use this method.

I have the final answer as $f_u(u)= \frac{1}{\sqrt{2\pi}\sqrt{u}}e^{\frac{-u}{2}}$ for $u>0$

But I am not very comfortable with the process of getting to that answer. I used a bit of a walk through and made some assumptions about what was happening.

Could anyone help me understand how I should use the above information to reach this answer?

Here is a way to get the answer using more basic principles.

$$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty.$$  This much you already know.  Then $$\begin{align*} F_U(u) &= \Pr[U \le u] \\ &= \Pr[Z^2 \le u] \\ &= \Pr[-\sqrt{u} \le Z \le \sqrt{u}] \\ &= \Pr[Z \le \sqrt{u}] - \Pr[Z \le -\sqrt{u}] \\ &= F_Z(\sqrt{u}) - F_Z(-\sqrt{u}), \end{align*}$$  where $F_Z(z)$ is the cumulative distribution function for $Z$.  Now differentiate with respect to $u$, taking care to use the chain rule:  $$f_U(u) = \frac{f_Z(\sqrt{u})}{2\sqrt{u}} - \frac{f_Z(-\sqrt{u})}{-2\sqrt{u}} = \frac{f_Z(\sqrt{u}) + f_Z(-\sqrt{u})}{2\sqrt{u}}.$$  Now substitute back into the density for $Z$: $$f_U(u) = \frac{1}{\sqrt{2\pi}}\frac{2e^{-u/2}}{2\sqrt{u}} = \frac{e^{-u/2}}{\sqrt{2\pi u}}, \quad u > 0. $$  The distribution of $U$ is known as the chi-square distribution with $1$ degree of freedom.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}
\frac{\expo{-z^{2}/2}}{\root{2\pi}}\delta\pars{u - z^{2}}\,\dd z}
=2\Theta\pars{u}\,\frac{\expo{-u/2}}{\root{2\pi}}\int_{0}^{\infty}
\frac{\delta\pars{z - \root{u}}}{2\root{u}}\,\dd z
\\[5mm]&=\color{#66f}{\large\Theta\pars{u}\,\frac{u^{-1/2}\ \expo{-u/2}}{\root{2\pi}}}
\end{align}


Étale cohomology of projective space I have some very basic question about étale cohomology.
Namely I would like to compute the étale cohomology of of the projective space over the algebraic closure of $\mathbb F_q$ along with its Frobenius operation:

$$H^i(\mathbb P^n_{\mathbb F},\mathbb Z /l)$$

I would expect that it vanishes for $i>2n$ or odd $i$ and is $\mathbb Z/l$ with Frobenius operation by multiplication by $q^{i/2}$ otherwise.
Using the Gysin sequence I can check, that the cohomology groups look as expected, however I don't know how to compute the operation of the Frobenius.

So my questions are:

How does one compute the Frobenius operation on cohomology in this example?

What are general techniques to compute Frobenius action on $l$-adic or étale cohomology?

You could count points and use the Lefschetz fixed point formula.


Theorem of liouville Consider two entire functions with no zeroes and having a ratio equal to unity at infinity. Use Liouville’s Theorem to show that they are in fact the same function.

My attempt
Consider $h(z) = f(z)/g(z)$.
First of all, $h$ is entire, since $f$ and $g$ are entire, and $g(z)$ is nonzero for all $z$ in $\mathbb{C}$.

The fact that $\lim_{z→\infty} h(z) = \lim_{z→\infty} f(z)/g(z) = 1$ suggests that $h(z)$ is bounded as well.

Why: Since $\lim_{z→\infty} f(z)/g(z) = 1$, there exists $N > 0$ such that $|f(z)/g(z) - 1| < 1$ for all $|z| > N$.
(Note that I am taking $\epsilon = 1$ for concreteness.)

Then, $|f(z)/g(z) - 1| \geq ||f(z)/g(z)| - 1|$
$\implies ||f(z)/g(z)| - 1| < 1$
$\implies |f(z)/g(z)| - 1 < 1$ or $1 - |f(z)/g(z)| < 1$
$\implies 0 < |f(z)/g(z)| < 2$.

That is, $|f(z)/g(z)|$ is bounded above by $2$ for $|z| > N$.
Moreover, we know that $|f(z)/g(z)|$ has a maximum $M$ on $|z| \leq N$ by maximum modulus principle (or simply from $|z| \leq N$ being compact). Hence, $|f(z)/g(z)|$ is bounded above by $\max \{2, M\}$ for all z in $\mathbb{C}$.

Hence, $h(z)$ is constant by Liouville's Theorem, i.e. $h(z) = f(z)/g(z) = c$ for some constant $c$.

Since this is true for all $z$ in $\mathbb{C}$, taking the limit as $z\to \infty$ yields $c = 1$.
Hence $f(z) = g(z)$ for all $z$ in $\mathbb{C}$.

Is correct my work?

Since both are entire functions without zeros, $h(z) = \frac{f(z)}{g(z)}$ is an entire function. $\lim_{z \rightarrow \infty} \frac{f(z)}{g(z)} \rightarrow 1$ where $\infty$ can be interpreted as the point at infinity, implies that $h$ is a bounded. A bounded entire function is constant $c$. $h(z) = c$ implies that $f(z) = c g(z)$. However $\lim_{z \rightarrow \infty} h(z) = 1$ implies that $c = 1$.


Proving $\phi$ is continuous In this video the lecturer gave an example, that is:

$\phi : M\to N$, where $M$ is equipped with arbitrary topology $\mathcal O_M$, and $N$ is equipped with the chaotic topology $\{\varnothing, N\}$. Then any $\phi$ is continuous.

It's clear that the preimage of $\varnothing$ must be $\varnothing$ and thus a open set in $M$, but why the preimage of $N$ is an open set in $M$?

The preimage $\phi^{-1}(N) = \{x \in M : \phi(x) \in N\}$ is equal to $M$ and $M$ is open by definition of a topology.

Any map into a space with the trivial topology is continuous (as is any map from a discrete space), which can be easily seen, for instance, by applying the definition that the preimage of any open set is open.


Can epsilon be a matrix? Question

In the following expression can $\epsilon$ be a matrix?

$$ (H + \epsilon H_1) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) = (E |m\rangle + \epsilon E|m_1\rangle + \epsilon^2 E_2 |m_2\rangle + \dots) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) $$

Background

So in quantum mechanics we generally have a solution $|m\rangle$ to a Hamiltonian:

$$ H | m\rangle = E |m\rangle $$

Now using perturbation theory:

$$ (H + \epsilon H_1) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) = (E |m\rangle + \epsilon E|m_1\rangle + \epsilon^2 E_2 |m_2\rangle + \dots) ( |m\rangle +\epsilon|m_1\rangle + \epsilon^2 |m_2\rangle + \dots) $$

I was curious and substituted $\epsilon$ as a matrix:

$$ \epsilon =
\left( \begin{array}{cc}
0 & 0 \\
1 & 0 \end{array} \right) $$

where $\epsilon$ now, is the nilpotent matrix, we get:

$$ \left( \begin{array}{cc}
H | m \rangle & 0 \\
H_1 |m_1 \rangle + H | m\rangle & H |m_1 \rangle \end{array} \right) = \left( \begin{array}{cc}
E | m \rangle & 0 \\
E_1 |m_1 \rangle + E | m\rangle & E |m_1 \rangle \end{array} \right)$$

Which is what we'd expect if we compared powers of $\epsilon$'s. All this made me wonder if $\epsilon$ could be a matrix? Say something like $| m_k\rangle \langle m_k |$ ? Say we chose $\epsilon \to \hat I \epsilon$

then there exists a radius of convergence. What is the radius of convergence in a general case of any matrix?

I would say there's nothing preventing you from using a matrix as perturbation of a matrix equation as long as you let the limit of the norm converge to $0$.

Yes and No. When I say yes I mean it is possible in several ways. But it does not make sense.

As usual for a physicist you did not specify what the space is in which your states (kets) live and thus not what the $H$ and $H_1$ are. But of course they are meant to be operators which you can consider to be generalizations of matrices to infinite dimensions. So when you assume $\epsilon$ to be a matrix you could as well absorb this into $H_1$.

Further you put $\epsilon$ to be a constant matrix. Then you can just leave the $\epsilon$ away. The epsilon is there to control the perturbation $H_1$. If you set $\epsilon$ to be a constant you directly solve the problem.


Differential equation with no unique solution Can someone help me find solutions for the following differential equation: $x'=-t\,\text{sign}(x)\sqrt{|x|}$ with $x(\tau)=\xi$.

With $x$ a function of $t$ and $\tau$ and $\xi$ constants.

It is possible to compute solutions where $x(t)$ is not changing its sign. Since you wanted to find 'solutions' and not 'all solutions' I will not investigate the case where $x(t)$ changes its sign. (It seems strenuous.)

Now let's take the IVP
$$\begin{cases} x'&=-t\,\text{sign}(x)\sqrt{|x|} \\ x(\tau)&=\xi \end{cases}$$

Assuming $x(t)>0$ for all $t$:

We get $x'=-t \sqrt{x}$ which has the solution (by separation of variables) $$x(t)=\frac{1}{16}(t^2-A)^2>0$$
with $x'(t)=\frac{1}{4}(t^2-A)t$ where $A>0$ is a constant that has the properties:

$x'(t)<0$ i.e. $\begin{cases} \text{for $t>0$: }~t^2-A<0 \Rightarrow t^2<A \Rightarrow |t|<\sqrt{A} \Rightarrow 0<t<\sqrt{A} \\ \text{for $t<0$: }~t^2-A>0 \Rightarrow t^2>A \Rightarrow |t|>\sqrt{A} \Rightarrow t<-\sqrt{A} \end{cases}$
$x(\tau)=\xi$ i.e. $\begin{cases} \text{for $\tau>0$: }~\tau^2-A=- 4 \sqrt{\xi}<0 \Rightarrow A=\tau^2 + 4\sqrt{\xi} \\ \text{for $\tau<0$: }~\tau^2-A= 4 \sqrt{\xi}>0 \Rightarrow A=\tau^2 - 4\sqrt{\xi} \\ \text{for $\tau=0$: }~ 0-A=-4\sqrt{\xi}<0 \Rightarrow A=4\sqrt{\xi} \end{cases}$

The first condition is problematic for arbitrarily large $t$. Since $A$ is specified by the second condition we get by the first condition that $x(t)=\frac{1}{16}(t^2-A)^2$ is only a solution for $$t \in \begin{cases} \left(-\infty,-\sqrt{\tau^2+4 \sqrt{\xi}} \right) \cup \left(0,\sqrt{\tau^2+4 \sqrt{\xi}} \right) &\text{ if } \tau>0\\
\left(-\infty,-\sqrt{\tau^2-4 \sqrt{\xi}} \right) \cup \left(0,\sqrt{\tau^2-4 \sqrt{\xi}} \right) &\text{ if } \tau<0 \wedge  \tau^2>4 \sqrt{\xi} \\ \left(-\infty,-\sqrt{4 \sqrt{\xi}} \right) \cup \left(0,\sqrt{4 \sqrt{\xi}} \right) &\text{ if } \tau=0 \end{cases}$$

Assuming $x(t)<0$ for all $t$:

We get $x'=t\sqrt{-x}$ i.e. $y'=-t\sqrt{y}$ for $y=-x$. So we are back in our first case just with a minus and the solution has to stay negative.

If $t\mapsto x(t)>0$ is a solution defined in some interval $J$ then $$\left(2\sqrt{x(t)}+{1\over2}t^2\right)'={x'(t)\over\sqrt{x(t)}}+t=0\ ,$$
hence $$2\sqrt{x(t)}+{1\over2}t^2={1\over2}c^2$$
for some $c>0$. This allows to conclude that
$$\sqrt{x(t)}={1\over4}(c^2-t^2)\qquad(-c<t<c)\ ,$$
or
$$x(t)={1\over16}(c^2-t^2)^2\qquad(-c<t<c)\ .\tag{1}$$
It is easy to check that these functions, together with their negatives, indeed solve the given ODE, and at the same time their graphs fill the complete $(t,x)$-plane, minus the line $x=0$, in the desired way.

The function $$x(t):\equiv0\tag{2}$$ is a solution as well, but a special one: The Lipschitz-assumption of the general existence and uniqueness theorem is not fulfilled in the points $(\tau,0)$ with $\tau\ne0$. Therefore it does not come as a surprise that IVPs with such a point as initial point have three solution germs starting there: the solution $(2)$ and two arcs of type $(1)$ going off towards $t=0$. In direction $|t|\to\infty$ there is just the trivial behavior $(2)$. See the following figure:


$f$ strictly increasing does not imply $f'>0$ We know that a function $f: [a,b] \to \mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f'>0 \mbox{ on} (a,b)$ , f is strictly increasing on $[a,b]$. Is there any counterexample that shows the converse fails?

I have been trying to come up with simple examples but they all involve functions that are discontinuous or has derivative $f'=0$ which does not agree with the hypothesis hmmm

Consider $f(x)=x^3$ on $[-1,1]$. It is strictly increasing, but has zero derivative at $0$.

Definition of Strictly Increasing:
if $f(x)$ is strictly increasing then,

$\forall x\forall y : y>x, f(y)> f(x) $

Definition of Monotonically increasing:
if $f(x)$ is monotonically increasing then,
$\forall x\forall y : y\geq x, f(y)\geq f(x) $

Definition of Derivative:
$f'(x) = \lim_{∆x\to 0} \frac{f(x+∆x)-f(x)}{∆x} $

Postulate: $f(x)$ is a Strictly Increasing function
$\iff$ $f'(x) >0$

Proof: WLOG, let $ ∆x\overset{\underset{\mathrm{def}}{}}{=} y-x$

Thus, $y = x + ∆x $

$f'(x) = \lim_{(y-x) \to 0} \frac{f(y)-f(x)}{y-x} $

$\because y > x, ∆x >0 $ and $ y-x > 0 $

also, $\because f(y) > f(x), f(y) - f(x) > 0 $

Thus $ \lim_{∆x\to 0} \frac{f(x+∆x)-f(x)}{∆x} > 0 $

$ \therefore f'(x) > 0$ QED

Using the same logic, $f(x)$  is monotonic $\iff f'(x) \geq 0$


How many $6$ digit numbers can you make with the numbers $\{1, 2, 3, 4, 5\}$ so that the digit $2$ appears at least 3 times? I can't seem to understand.

I thought about it this way. We take a first example:

$222aaa$ for each blank space we have $5$ positions, so in this case the answer would be $125$ ($5 \times 5 \times 5$) numbers.

Now the position of the twos can change so we calculate the numbers for that taking this example:

$222aaa$ and the possibilities for this are the number of ways you can arrange $6$ digits ($6!$) and divide by repetitions so divided by $2 \cdot (3!)$ so answer $= 20$

So final answer should be $125 \cdot 20 = 2500$. But this answer is wrong and I don't understand why.

Let ${S_n}$ be the number of ways you can have the  digit 2 appears ONLY 3, 4 ,5 , and 6 times where $ n = 3 , 4 , 5, 6$.  We can then add $S_3 + S_4 + S_5 + S_6$ to get our answer.  To find $S_n$, note that there are ${6}\choose {n}$ ways of arranging the 2's, and for each of those there are $6-n$ places for the other digits; however, we will not include 2 in these other digits or else we will end up overcounting.  So we $4^{(6-n)}$ different numbers we can make.

So $S_3 =$ ${6}\choose{3} $$4^3$

$S_4 =$ ${6}\choose {4}$$4^2$

$S_5 =$ ${6}\choose{5} $$4 = 24$

$S_6 =$ ${6}\choose{6}$ $=1$

Now just add these numbers up.

Since you have only $5$ numbers, you can try it making cases

  Case $1.$ When there are $3~2's$

$3$ spaces left

  Case $1.a)$  When all the three numbers are different

Total numbers= $4\times 3\times 2=24$

Total permutation=$\frac{6!}{3!}=120$

Total cases=$24\times 120=2880$

  Case $1.b)$ When two of three numbers are same

Total numbers=$12$

Total permutation=${6!}{3!2!}=60$

Total cases=$12\times 60=720$

  Case 2. When there are $4 ~2's$

$2$ places left

  Case $2.a)$ When the two numbers are different

Total numbers=$4\times 3=12$

Total permutation=$\frac{6!}{4!}$

Total cases=$12\times 30=360$

  Case $2.b) When the two numbers are same

Total numbers=$4$

Total permutation=$\frac{6!}{4!2!}=15$

Total cases=$4\times 15=60$

  Case $3.$ When there are five $2's$

Total cases=$4\times 6=24$

  Case $4.$ When there are  six $2's$

Total cases=$1$

Adding all will give $2880+720+360+60+24+1=4045$


Derivative Notation: Let $f(x)=x$. Is $f′(x^2)=1$ or is $f′(x^2)=2x$? Let $f(x)=x$.

Is $f'(x^2)=1$ or is $f'(x^2)=2x$?

In other words, using this notation, are we evaluating the derivative at a $x^2$?

If you consider the function $f$ of a variable $x$, then $f'$ is likewise a function. When you write $f'(a)$, you evaluate that function at $a$. If $a$ happens to be $x^2$, then you evaluate at $x^2$. It has nothing to do with differentiating with respect to $a$ or $x^2$ or anything else. So, in your example, $f'(a)=1$ no matter what you put in for $a$ when $f(x)=x$.

If you consider $g(x)=x^2$ and the composition $(f\circ g)(x) = f(x^2)$, then
$(f\circ g)'(x) = f'(g(x))g'(x) = f'(x^2)\cdot 2x$.

As $f(x)=x$, then

$$\left(f(x^2)\right)'=(x^2)'.$$

The $'$ denotes differentiation on $x$.


Calculate limit of $\sqrt[n]{2^n - n}$ Calculate limit of $\sqrt[n]{2^n - n}$.
I know that lim $\sqrt[n]{2^n - n} \le 2$, but don't know where to go from here.


$ef$ theorem for inertial degree and ramification index $L/K$ is a finite algebraic extension of fields. $O_{P}$ and $O_{\mathcal{P}}$ are DVRs with $P$ and $\mathcal{P}$ maximal ideals respectively, their quotient fields are $K$ and $L$ respectively. Let $e$ be the ramification index of $\mathcal{P} / P$ and $f$ be the inertial degree of $\mathcal{P}/P$.
I am working on showing $ef \leq n=[L: K]$ (without separable condition). Not quite understand several details in the proof.
The proof in the book is: Let $\Pi$ be a generator of $\mathcal{P}$ and choose $\omega_{1},\ldots,\omega_{m}$ such that their modulo $\mathcal{P}$ are linearly independent over $O_{P}/P$. Then show $\omega_{i}\Pi^{j}$ are linearly independent over $K$.
Take a relation
$$\sum_{j=0}^{e-1}\sum_{i=1}^{m}a_{ij}\omega_{i}\Pi^{j}=0.$$
If $a_{ij}$ are not all zero, we can assume the are all in $O_{P}$ and at least one of them is not in $P$ (I understand they could be put in $O_{P}$ since K is the quotient field, but why at least one not in $P$?).
Then consider the elements
$$A_{j}=\sum_{i=1}^{m}a_{ij}\omega_{i}.$$
If some $a_{ij}\not \in P$, then $A_{j}$ is a unit in $O_{\mathcal{P}}$. Otherwise, $A_{j}$ is divisible by $\pi$, the generator of $P$ and so $\text{ord}_{\mathcal{P}}(A_{j}) \geq e$.
Thus, $\text{ord}_{\mathcal{P}}(\sum_{j=0}^{e-1}A_{j}\Pi^{j}) =j_{0}$ for some $j_{0} < e$ (how to get this conclusion? Just becasuce you have some $A_{j}$ with $0$ order and then multiply with $\Pi^{j}$?). Finally this is a contradiction since $\sum_{j=0}^{e-1}A_{j}\Pi^{j}=0$ (why is it a contradiction? Isn't $0$ has $\mathcal{P}$ order less than e?).
Thanks for any kind help you offer!

First question: you have a relation
$$\sum\limits_{i,j} a_{ij} \omega_i \Pi^j =0$$
for some $a_{ij} \in K$, not all zero, and you want to justify the claim that you can multiply this equation by some $c \in K$ so that all of the $ca_{ij}$ are in $\mathcal O_P$ with at least one of these not in $P$.  The only way this would not be possible is if all the nonzero $a_{ij}$ had the same valuation.  That would mean you could choose a $c \in K$ such that all the nonzero terms $ca_{ij}$ were units in $\mathcal O_P$.
Taking the equation $$\sum\limits_{i=1}^m \sum\limits_{j=0}^{e-1} ca_{ij} \omega_i \Pi^j = 0 \tag{1}$$
modulo $\mathcal P$ yields the equation $\sum\limits_{i=1}^{m} ca_{i0} \omega_i \equiv 0 \pmod{\mathcal P}$.  The linear independence of the cosets $\omega_i + \mathcal P$ over the field $\mathcal O_P/P$ implies that all the $ca_{i0}$ are in $P$.  Since the nonzero terms $ca_{i0}$ are units in $\mathcal O_P$, this means that all the $ca_{i0}, 1 \leq i \leq m$ must be $0$.
Equation (1) is then
$$\sum\limits_{i=1}^m \sum\limits_{j=1}^{e-1} ca_{ij} \omega_i \Pi^j = 0.$$
Divide this equation by $\Pi$ and reduce modulo $P$ again.  The same argument as in the previous paragraph implies that all the terms $ca_{i1}, 1 \leq i \leq m$ are all zero.  Iterating this argument shows that all the terms $ca_{ij}$ have to be zero.  This contradicts the assumption that not all the $a_{ij}$ were zero.
Second question:
For each $A_j, 0 \leq j \leq e-1$, we have either $\operatorname{ord}_{\mathcal P}(A_j) =0$ or $\operatorname{ord}_{\mathcal P}(A_j) \geq e$ as they explain.  Since there is at least one $a_{ij}$ which is not in $P$, this means that there is at least one $j$ for which $\operatorname{ord}_{\mathcal P}(A_j) = 0$.
The usual rule
$$\operatorname{ord}_{\mathcal P}(x+y) \geq \operatorname{Min}\{ \operatorname{ord}_{\mathcal P}(x), \operatorname{ord}_{\mathcal P}(y)\}$$
is written for nonzero $x, y \in L$, but it continues to make sense if we make the convention that $\operatorname{ord}_{\mathcal P}(0) = \infty$.  Remember this is an exact equality if $\operatorname{ord}_{\mathcal P}(x) \neq \operatorname{ord}_{\mathcal P}(y)$.  Let $S$ be the set of $0 \leq j \leq e-1$ such that $\operatorname{ord}_{\mathcal P}(A_j) = 0$ (we know it is nonempty), and let $T$ be the set of $0 \leq j \leq e-1$ such that $\operatorname{ord}_{\mathcal P}(A_j) \geq e$.  If we set
$$X_S = \sum\limits_{j \in S} A_j \Pi^j, X_T = \sum\limits_{j \in T} A_j \Pi^j$$
then we have $$\sum\limits_{j=0}^{e-1} A_j \Pi^j = X_S + X_T = 0.$$
Since all the $\operatorname{ord}_{\mathcal P}(A_j)$ for $j \in S$ are $0$, we see that $\operatorname{ord}_{\mathcal P}(A_j \Pi^j) = j$ for all $j \in S$.  Thus all the terms $A_j \Pi^j$ for $j \in S$ have distinct valuations, and so
$$\operatorname{ord}_{\mathcal P}(X_S)= j_0$$
where $j_0$ is the smallest number in $S \subseteq \{0,1, ... , e-1\}$.
Since all the terms $A_j \Pi^j : j \in T$ have order $\geq e$, the element $X_T$ has order $\geq e$.  This implies
$$\operatorname{ord}_{\mathcal P}(X_S + X_T) = j_0 < e.$$
In particular, $\sum\limits_{j=0}^{e-1} A_j \Pi^j$ does not have infinite valuation, so it cannot be zero.

$L,K$ are discretely valued field, the valuation $v$ is surjective $L\to \Bbb{Z}\cup \infty$ and $K\to e\Bbb{Z}\cup \infty$.
$\Pi$ is chosen such that $v(\Pi)=1$.
$\omega_i$ are taken in $O_\mathcal{P}$ to be a $O_P/P$ basis of $O_\mathcal{P}/\mathcal{P}$, ie. $m=f$.
The $O_P/P$ independence of the $\omega_i$ gives that
$$v(\sum\limits_{i=1}^m a_{ij} \omega_i) = \inf_i v(a_{ij})$$
Thus $v(\sum\limits_{i=1}^m a_{ij} \omega_i \Pi^j) \in j+ e\Bbb{Z}\cup \infty$.
For $v(b)\ne v(c)$, $v(b+c)=\inf( v(b),v(c))$. Thus $$v(\sum_{j=0}^{e-1} \Pi^j\sum\limits_{i=1}^m a_{ij} \omega_i )=\inf_j v(\Pi^j\sum\limits_{i=1}^m a_{ij} \omega_i)=\inf_j (j+\inf_i v(a_{ij}))$$
That it is $\infty$ means that each $v(a_{ij})=\infty$ ie. each $a_{ij}=0$.


Why $U\le A \times B$ does not imply $U=\left(A\cap U\right) \times \left( B \cap U \right)$? Let $A,B$ be groups.Can you explain why $U\le A \times B$ does not imply $U=\left(A\cap U\right) \times \left( B \cap U \right)$
this is an exercise in the book of the theory of finite groups an introuduction written by H.Kurzweil. the meaning of each symbol may as follows.
$A\times B$ is the direct product of $A$,$B$. $U$ is the subgroup of $A\times B$ , thus $U= \lbrace\left(a,b\right)|a \in A,b\in B \rbrace$, $A \cap U=\lbrace a_1|\left(a_1,b_1\right) \in U,a_1\in A  \rbrace$,in the same way we could know $\left( B \cap U \right)$

How about a simple example such as $A=B\ne1$ and $U=\{(x,x)\mid x\in A\}$?

Think about it intuitively.

Take $A,B = \mathbb{R}$. The LHS $U$ is "some set of points in the plane" whereas the RHS is "all possible x-coordinates from points in $U$ with all possible y-coordinates from points in $U$".

Clearly $U\subseteq$ RHS but the RHS can be bigger.

For a specific example let $U = \{(0,1),(1,0)\}$. Then the RHS is $\{(0,0),(0,1),(1,0),(1,1)\}$.

The point is that the object on the RHS is combining things componentwise that might not have existed when we chose "special" points to be in $U$.


Quadratic Diophantine equation - Find all integer solutions I have a Quadratic  Diophantine equation.

I am sorry I didn't get how to show correctly. So I used a
picture

How can I find all integer solutions? Should I do something with Z(14)?

The integer equation $11 x^2 - 14 y^2 = 1$ is impossible. However, I do not see that a proof is available for a beginner. There is no simple proof with congruences. There is one using, in effect, continued fractions.

The principal genus of this discriminant is two forms,
$$ x^2 - 154 y^2 $$
$$ 2 x^2 - 77 y^2,  $$
and your form $11x^2 - 14 y^2$ is $SL_2 \mathbb Z$ equivalent to $2x^2 - 77 y^2$ and simply does not integrally represent $1.$

The other genus is their negatives,
 $$ -x^2 + 154 y^2 $$
$$ -2 x^2 + 77 y^2,  $$

Here are a list of the class group, Gauss-Lagrange reduced forms, and the Lagrange cycle of your form

    1.             1          24         -10   cycle length            10
    2.            -1          24          10   cycle length            10
    3.             2          24          -5   cycle length             8
    4.            -2          24           5   cycle length             8

  form class number is   4

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 11 0 -14

  0  form             11           0         -14  delta      0
  1  form            -14           0          11  delta      1
  2  form             11          22          -3

          -1          -1
           0          -1

To Return
          -1           1
           0          -1

0  form   11 22 -3   delta  -7     ambiguous
1  form   -3 20 18   delta  1
2  form   18 16 -5   delta  -4
3  form   -5 24 2   delta  12
4  form   2 24 -5   delta  -4     ambiguous
5  form   -5 16 18   delta  1
6  form   18 20 -3   delta  -7
7  form   -3 22 11   delta  2
8  form   11 22 -3

  form   11 x^2  + 22 x y  -3 y^2

minimum was   2rep   x = 5   y = 39 disc 616 dSqrt 24  M_Ratio  4.760331
Automorph, written on right of Gram matrix:
2419  5148
18876  40171
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

If you are asking for $(x, y)$ pairs that solve the equation: $$11x^2-14y^2=1$$ you can use the on solver at here to show that there are no solutions in integers.


Intermediate value theorem for $\sin x.$ How is intermediate value theorem valid for $\sin x$ in  $[0,\pi]$?

It has max value $1$ in the interval $[0,\pi]$ which doesn't lie between values given by $\sin0$ and $\sin\pi$.

Let $f\colon[a,b]\longrightarrow\mathbb R$ be a continuous function and let $y\in\mathbb R$. The intermediate value theorem  says that if $f(a)\geqslant y\geqslant f(b)$ or if $f(a)\leqslant y\leqslant f(b)$, then there is a $c\in[a,b]$ such that $f(c)=y$. But it says nothing if $y$ lies outside the interval bounded by $f(a)$ and $f(b)$. So, there is no contradiction here.

In general, any time you have a theorem, "If ____, then ____," it really means, "If ____, then ____ and maybe some other stuff not mentioned here also happens."
Because when you're working in just about any branch of mathematics,
no matter how thoroughly you describe the implications of any mathematical fact there is always some other possible case you could have said something about but didn't.

The only things you can rule out are the things the theorem explicitly says cannot happen.
Unless a theorem says "and there are no other values outside this interval,"
don't assume all the values are inside the interval.

In your example, the value $\sin(\pi/2)=1$ is part of the "other stuff"
that "also happens."


Finding Angle using Geometry In an equilateral triangle $ ABC $ the point $ D $ and $ E $ are on sides $ AC $ and $AB$  respectively, such that $ BD $ and $ CE $ intersect at $P$ , and the area of the quadrilateral $ ADPE $ is equal to area of  $ \Delta BPC $ find $ \angle BPE $.
This question when I first tried looked easy and I was also able to guess the answer but when I tried to proof, was not able to work it out. I want some help. Thank you. No image was provided in question. I am attaching my drawing.

The hint.

Prove that $$\left(S_{\Delta BPC}\right)^2=S_{\Delta APB}S_{\Delta APC}.$$
I got $$\measuredangle BPE=60^{\circ}.$$

Let $S_{\Delta BPC}=a$, $S_{\Delta PAC}=b$ and $S_{\Delta PAB}=c$.

Thus, $$\frac{S_{\Delta PEB}}{c}=\frac{BE}{AB}=\frac{BE}{BE+EA}=\frac{1}{1+\frac{EA}{BE}}=\frac{1}{1+\frac{b}{a}}=\frac{a}{a+b},$$
which gives $$S_{\Delta PEB}=\frac{ac}{a+b}.$$
Similarly, $$S_{\Delta PDC}=\frac{ab}{a+c}.$$
Thus, $$S_{AEPD}=b+c-\frac{ab}{a+c}-\frac{ac}{a+b}=\frac{bc}{a+c}+\frac{bc}{a+b}.$$
Id est, $$\frac{bc}{a+c}+\frac{bc}{a+b}=a$$ or
$$a^2=bc$$ or $$\frac{a}{b}=\frac{c}{a}$$ or
$$\frac{BE}{AE}=\frac{AD}{CD},$$ which gives $$DC=AE,$$
$$\Delta AEC\cong\Delta CDB,$$ which gives $$\measuredangle BPE=\measuredangle DBC+\measuredangle ECB=\measuredangle DBC+60^{\circ}-\measuredangle ACE=60^{\circ}.$$

IF there exists a set of points P for which the two areas are equal, then one such point P lies on the vertical altitude (axis of symmetry), thus we only need to find that point.  Dividing each of the two areas in half with the vertical altitude, the problem can be stated as Area(BPS) = Area (APE), where S is the foot of the altitude (the midpoint of BC).  The midpoint (centroid?) of the triangle (with altitudes) divides the area of the whole triangle into six similar triangles, including the two above.  So the midpoint of the triangle satisfies the condition, thus the angle is 60 degrees.


building transformation matrix from spherical to cartesian coordinate system How to arrive at the following from given $ x  = r\sin \theta \cos \phi, y  = r\sin \theta \sin \phi, z=r\cos\theta $

$$ \begin{bmatrix}
A_x\\
A_y\\
A_z
\end{bmatrix}
= \begin{bmatrix}
\sin \theta \cos \phi & \cos \theta \cos \phi  & -\sin\phi\\
\sin \theta \sin \phi & \cos \theta \sin \phi  & \cos\phi\\
\cos\theta & -\sin\theta  & 0
\end{bmatrix}
\begin{bmatrix}
A_r\\
A_\theta\\
A_\phi
\end{bmatrix}$$

Also how show that
$$ \begin{bmatrix}
\hat i\\
\hat j\\
\hat k
\end{bmatrix}
= \begin{bmatrix}
\sin \theta \cos \phi & \cos \theta \cos \phi  & -\sin\phi\\
\sin \theta \sin \phi & \cos \theta \sin \phi  & \cos\phi\\
\cos\theta & -\sin\theta  & 0
\end{bmatrix}
\begin{bmatrix}
\hat e_r\\
\hat e_\theta\\
\hat e_\phi
\end{bmatrix}$$
How to change $(a,b,c)$ into spherical polar coordinates and $ (r ,\theta, \phi)$ into cartesian coordinates using this matrix?
Thank you!!

I can partially answer this. I believe your first matrix is not the correct general transformation matrix for cartesian to spherical coordinates because you are missing factors of $\rho$ (the radial coordinate), as well as some other incorrect pieces. So it is not clear what you are trying to show.

If you are trying to derive the general transformation matrix from spherical to cartesian, it is:
$$\begin{bmatrix}
A_x\\
A_y\\
A_z
\end{bmatrix}
= \begin{bmatrix}
\sin \theta \cos \phi & \rho \cos \theta \cos \phi  & -\rho \sin \theta \sin\phi\\
\sin \theta \sin \phi & \rho \cos \theta \sin \phi  & \rho \sin\theta \cos\phi\\
\cos\theta & -\rho \sin\theta  & 0
\end{bmatrix}
\begin{bmatrix}
A_r\\
A_\theta\\
A_\phi
\end{bmatrix}
$$
This matrix is formed from the derivative matrix in the following way:
$
\sum_{(i,j)} \frac {\partial x^i} {\partial u^j}
$
where
$$x^1  = \rho\sin \theta \cos \phi, x^2  = \rho\sin \theta \sin \phi, x^3=\rho\cos\theta$$
and
$u^1 = \rho, u^2 = \theta, u^3 = \phi
$
(Note: the superscripts are indices, not exponents).
If you write out all the derivatives, you get this matrix:

$$
\begin{bmatrix}
A_x\\
A_y\\
A_z
\end{bmatrix}
= \begin{bmatrix}
\frac {\partial (\rho \sin \theta \cos \phi)} {\partial \rho} &
\frac {\partial (\rho \sin \theta \cos \phi)} {\partial \theta} &
\frac {\partial (\rho \sin \theta \cos \phi)} {\partial \phi}\\
\frac {\partial (\rho \sin \theta \sin \phi)} {\partial \rho} &
\frac {\partial (\rho \sin \theta \sin \phi)} {\partial \theta} &
\frac {\partial (\rho \sin \theta \sin \phi)} {\partial \phi}\\
\frac {\partial (\rho \cos \theta)} {\partial \rho} &
\frac {\partial (\rho \cos \theta)} {\partial \theta} &
\frac {\partial (\rho \cos \theta)} {\partial \phi}
\end{bmatrix}
\begin{bmatrix}
A_r\\
A_\theta\\
A_\phi
\end{bmatrix}
$$
...which evaluates to the correct general transformation matrix I listed above.

get the elements of the jacobian matrix, then divide them to their respective scale factors


A question on a compact space Show:

  If the closure of every discrete subset of a space is compact then the whole space is compact.

Thanks advance:)

I realize that this is homework, but I don’t see any way to give a useful hint.

Let $X$ be a Hausdorff space in which the closure of every discrete subset is compact. Clearly $X$ contains no infinite closed subset and is therefore countably compact. Suppose that $X$ is not compact. Then there are a regular uncountable cardinal $\kappa$ and an open cover $\mathscr{U}=\{U_\xi:\xi<\kappa\}$ such that $U_\xi\subsetneqq U_\eta$ whenever $\xi<\eta<\kappa$, and $\mathscr{U}$ has no finite subcover. For each $\xi<\kappa$ let $x_\xi\in U_{\xi+1}\setminus U_\xi$, and let $Y=\{x_\xi:\xi<\kappa\}$; clearly $Y$ is right-separated (i.e., initial segments are relatively open).

Let $\mathscr D=\{D\subseteq Y:D\text{ is discrete}\}$. For $D_0,D_1\in\mathscr{D}$ define $D_0\preceq D_1$ iff $D_0\subseteq D_1$, and $\xi<\eta$ whenever $x_\xi\in D_0$ and $x_\eta\in D_1\setminus D_0$. (In other words, $D_0\preceq D_1$ iff $D_0$ is an initial segment of $D_1$ with respect to the well-order on $Y$ induced by the ordinal subscripts.) Clearly $\langle\mathscr{D},\preceq\rangle$ is a partial order. Suppose that $\mathscr{C}$ is a chain in $\langle\mathscr{D},\preceq\rangle$. Let $D=\bigcup\mathscr{C}$, and suppose that $x_\xi\in D$. Fix $C\in\mathscr{C}$ with $x_\xi\in C$; $C$ is discrete, so there is an open nbhd $V$ of $x_\xi$ such that $V\cap C=\{x_\xi\}$. Let $x_\eta$ be any other element of $D$; if $\eta<\xi$, then $x_\eta\in C$, so $x_\eta\notin V$, and if $\eta>\xi$, then $x_\eta\notin U_\xi$. Thus, $V\cap U_\xi$ is an open nbhd of $x_\xi$ that contains no other element of $D$. Since $x_\xi$ was an arbitrary element of $D$, $D$ is discrete, i.e., $D\in\mathscr{D}$. Clearly $C\preceq D$ for all $C\in\mathscr{C}$, so $D$ is an upper bound for $\mathscr{C}$ in $\mathscr{D}$, and by Zorn’s lemma $\mathscr{D}$ has a maximal element $M$.

Let $K=\operatorname{cl}M$; $M$ is discrete, so $K$ is compact. Moreover, the maximality of $M$ implies that $M$ is dense in $Y$ and hence that $K\supseteq Y$. $\mathscr{U}$ is an open cover of $K$, so it has a finite subcover, and since $\mathscr{U}$ is an increasing nest of open sets, there is some $\eta<\kappa$ such that $Y\subseteq K\subseteq U_\eta$, which is absurd, since for example $x_{\eta+1}\in Y\setminus U_\eta$. This contradiction shows that $X$ must be compact.

(The Zorn’s lemma argument can of course be replaced by a straightforward transfinite recursion to construct $M$.)

Continuing what Bryce said above .it implies every infinite set is not discrete.so given a
infinite set there is a limit point x such that every neighbourhood of it contains all the points of the sequence.So x is a complete accumualation point.Hence every infinite subset of X has a complete accumulation point.Hence X is compact


Non-trivial continuous limits of repeated iteration This is not an exercise in any book, I've just thought for a while about this idea and couldn't figure it out, so I thought people at Math.SE might help me.

Define a function transform as follows. For a function $f$ and each x in its domain, define a T-transform $T(f)(x)$ to be the limit of the sequence formed by repeated iteration with $f$ of x $( x, f(x), f(f(x)), f(f(f(x))), ...)$, and if the sequence doesn't converge, define $T(f)(x)$ to be zero. Are the only possible continuous T-transforms of functions from reals to reals constant functions and the identity function? In other words, is there a non-trivial continuous T-transform of some function? I suspect there is none, but I don't know how to prove it.

There are functions $f$ with continuous, nontrivial $T(f)$. Let
$$
f(x) = \begin{cases}x/2 & x\le 0\\x & 0\le x\le 1\\ (x+1)/2 & x\ge 1\end{cases}
$$
Then
$$
T(f)(x)= \begin{cases}0& x\le 0\\x & 0\le x\le 1\\ 1 & x\ge 1\end{cases}
$$
Edit: Here is another nice example. Let $g(x)$ be a triangle wave, defined by setting $g(4k+1)=1$ and $g(4k+3)=-1$ for all integers $k$, and interpolating linearly between. Then $g$ satisfies $g(g(x))=g(x)$, implying that $T(g)(x)=g(x)$, so $T(g)$ is the same triangle wave.

Take $f:x\mapsto|x|$ and we will have

$$T(f)(x)=|x|$$

Likewise for $f:x\mapsto-|x|$.


Has error correction been "solved"? I recently came across Dan Piponi's blog post An End to Coding Theory and it left me very confused. The relevant portion is:

  But in the sixties Robert Gallager looked at generating random sparse syndrome matrices and found that the resulting codes, called Low Density Parity Check (LDPC) codes, were good in the sense that they allowed messages to be transmitted at rates near the optimum rate found by Shannon - the so-called Shannon limit. Unfortunately the computers of the day weren't up to the task of finding the most likely element of M from a given element of C. But now they are. We now have near-optimal error correcting codes and the design of these codes is ridiculously simply. There was no need to use exotic mathematics, random matrices are as good as almost anything else. The past forty years of coding theory has been, more or less, a useless excursion. Any further research in the area can only yield tiny improvements.

In summary, he states that the rate of LDPC codes is very near channel capacity—so near that further improvements would not be worth the while.

So, my question is: What does modern research in error-correcting codes entail? I noticed that Dan did not mention what channel the rate of LDPC codes approach the capacity of, so maybe there exist channels that LDPC codes don't work well on? What other directions does modern research in the field explore?

This is largely true in the sense that long LDPC codes are within a fraction of a dB of the Shannon limit in AWGN channel. Their design (by a process called density evolution - don't look at me, I don't know the math) also seems to be relatively simple. It depends on stochastic math as opposed to algebra/combinatorics of finite fields/rings.
My limited experience only allows me to make the following remarks:

The block length of an LDPC code needs to be relatively large. This
is one of the reasons they were not chosen to LTE standard
(European/Asian 4G cellular standard). My understanding about the
`best' code in a radio channel is (as a function of the block
length): less than 100 bits => use an algebraic block code, 40-400
bits => use a convolutional code, 300-20000 bits => use a turbo
code, 5000 bits+ => use an LDPC code (the intervals overlap as I am
not willing to bet the family fortune on any one of them). IIRC this is how it is done in LTE (release 9), except that in that cellular standard the longest data block has 6144 bits (so handled with a turbo code).
The design of an LDPC code needs to take into account that it is usually not practical to base it on a random Tanner graph with the prescribed weight distribution. In particular, when it is supposed to be used on a handheld device. The routing problem of the belief propagation algorithm becomes prohibitive otherwise. So the LDPC codes specified for radio communication standards a few years back (DVB-S/T2 in Europe, Mediaflo in the US) were designed around a particular decoding circuitry, or rather the Tanner graphs had to come from a certain kind of a family. I have never been truly up to speed with these aspects, so some progress may have been made.
When higher order modulation is used there are further problems and considerations that need to be taken into account. It is not entirely obvious that we should stick to LDPC codes based on bits alone in that case, because the channel will create dependencies among the LLRs of individual bits. But the BP algorithm becomes very complicated, if we want to feed something other than LLRs of individual bits to it.
Getting rid of the error floors is still a challenge. Alas, I'm not up to speed with that either. Possible workarounds involve using catenated code with a light weight algebraic outer code handling the residual errors.

Caveat: I have only studied this cursorily (enough to implement an LDPC decoder and simulate/analyze some of the codes). These are recollections and impressions rather than hard facts. You can get the minutes of the relevant meetings of workgroup in charge of this within LTE from the web pages of 3GPP (=third generation partnership project).
Other types of codes will remain to be useful for specific purposes. There is no substitute to RS (or AG, if you need longer blocks), when you only have hard byte errors. LDPC (and Turbo) codes are designed with soft bit errors (=reliability estimates of received bits) in mind. This is a valid assumption in all wireless communication. Also network coding appears to be a hot topic currently. There the code operates at the level of data packets that may or may not arrive at all. Also look up Raptor code, if this type of problems are of interest to you.
Algebraic coding theory is not dead yet, but it does concentrate on problems other than designing ECCs for bulk data transmission. An algebraic machinery yielding LDPC code designs guaranteed to have very low error floors (= high minimum Hamming distance) may be out there, but I don't know what it would be.

The problem has been solved since 1993 with the advent of turbo codes. Tribute to Mr Berrou and Glavieux who discovered those iterative decoders and the concepts of exchanging extrinsic information. Then McKay (unfortunately he just passed away) rediscovered the LDPC codes (Gallager codes).
Today the research is focused mainly in which system has the lowest complexity (very important for the future networks).
Recently also polar codes have been discovered bu Anrikan but I have doubts how the will beat turbo or ldpc....


Calculate limits without L'Hospital's Rule Questions:

(A) Calculate

$$\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{2x - \pi}$$

(B) Calculate

$$\lim_{x ~\to~0^{+}} x^{3}e^{\frac{1}{x}}$$

without using L'Hospital's Rule.

Attempted solution:

(A)

Using the definition of $\cot x$ gives:

$$\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{2x - \pi} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{\cos x}{\sin x}}{2x - \pi} = \lim_{x \to \frac{\pi}{2}} \frac{1}{\sin x} \cdot \frac{\cos x}{2x - \pi}$$

Since the first term will turn out to be 1, I carry on without it and making the substitution $y = x - \frac{\pi}{2}$ as well as the fact that $\cos x = \cos (-x)$:

$$= \lim_{y \to 0} \frac{\cos (y+\frac{\pi}{2})}{2y} = \lim_{y \to 0} \frac{\cos (-y-\frac{\pi}{2})}{2y}$$

Using the subtraction formula for cosine gives:

$$ = \lim_{y \to 0} \frac{\cos(-y) \cos \frac{\pi}{2} + \sin(-y) \sin{\frac{\pi}{2}}}{2y} = \lim_{y \to 0} \frac{\sin(-y)}{2y} =$$
$$ \lim_{y \to 0} \frac{- \sin(y)}{2y} = -\frac{1}{2}\lim_{y \to 0} \frac{\sin(y)}{y} = -\frac{1}{2} \cdot 1 = -\frac{1}{2}$$

Does this look reasonable?

(B)

I know that the limit does not exist (it "becomes" $\infty$). I figure that $e^{x}$ is continuous function with respect to x and so:

$$\lim_{x ~\to~0^{+}} x^{3}e^{\frac{1}{x}} = \lim_{x ~\to~0^{+}} x^{3} \cdot \lim_{x ~\to~0^{+}} e^{\frac{1}{x}} = 0 \cdot e^{\lim_{x ~\to~0^{+}} \frac{1}{x}} $$

I am not sure where to go on from here. Will it be an argument based on $\frac{1}{x}$ growing faster towards infinity than $x^3$ shrinks towards $0$? Or is there a smart algebraic trick that can be used?

In many cases, the limit appears to give a "zero times infinity" expression, but after figuring out the secret steps you can change it to something that you can calculate the limit from. How do you know when to stop trying and declare that the limit does not exist?

Your first calculation looks good as far as I can tell.

You cannot write $$\lim_{x\to0^+}x^3e^{1/x}=\lim_{x\to0^+}x^3\cdot\lim_{x\to0^+}e^{1/x}$$ because the limit of a product should only be said to equal the product of the limits when all three limits exist. In general, one or both limits on the right might not exist and yet the limit on the left could still exist. It might help to let $z=1/x$, and note that $$\lim_{x\to0^+}x^3e^{1/x}=\lim_{z\to\infty}\frac{e^z}{z^3}$$ Then can you show that for large $z$, $e^z>cz^4$? One method is to

show that $e^x>1$ for large enough $x$
show that for some $x_0$, $e^{x_0}>x_0$, and then since we have already established that $e^x$ has a larger derivative than $x$, it follows that for large enough $x$, $e^x>x$
repeat to show $e^x>\frac{1}{2}x^2$
repeat to show $e^x>\frac{1}{6}x^3$
repeat to show $e^x>\frac{1}{24}x^4$

So  $$\lim_{x\to0^+}x^3e^{1/x}=\lim_{z\to\infty}\frac{e^z}{z^3}>\lim_{z\to\infty}\frac{cz^4}{z^3}=\infty$$

A To prove $\displaystyle\lim_{x \to \infty} \frac{\sin x}{x} = 1$ just look at the wedge of a circle and notice the the arc and the height are about the same.  With an appropriate change of angles you can get

I think in your cases it might be useful that $\displaystyle \lim_{x \to 0} \frac{\tan x }{x} =1$