Expectation Random Variables Say $X$ to be uniformly distributed from $[0,1]$. Say $k_1$ and $k_2$ to be two non negative constants (that is, they take values from $[0,+inft]$.

I want to compute the expectation of the following max involving the two constants:

$$E[\max(X+k_1,k_2)]$$

(that is, the expectation of the max between the sum "random variable plus a constant k1" and "a constant k2").

As someone suggested:

let $Y=X+k_1$ be the uniform variable on $[k_1,k_1+1]$. Then your variable is $\max \{ Y,k_2 \}$. Cases:

If $k_2 < k_1$, then $\max \{ Y,k_2 \} = Y$, so you have $E(Y)=\frac{2k_1+1}{2}$.
If $k_2 > k_1+1$, then $\max \{ Y,k_2 \}=k_2$, so you have just $k_2$.
If $k_1 \leq k_2 \leq k_1+1$, then $\mathbb{P}(Y \leq k_2) = k_2-k_1$.

how to handle the last case?

Let $Y=X+k_1$ be the uniform variable on $[k_1,k_1+1]$. Then your variable is $\max \{ Y,k_2 \}$. Cases:

If $k_2 < k_1$, then $\max \{ Y,k_2 \} = Y$, so you have $E(Y)=\frac{2k_1+1}{2}$.
If $k_2 > k_1+1$, then $\max \{ Y,k_2 \}=k_2$, so you have just $k_2$.
If $k_1 \leq k_2 \leq k_1+1$, then $\mathbb{P}(Y \leq k_2) = k_2-k_1$.

Can you handle the last case?

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

  Note that $\ds{\max\pars{a,b} = {a + b + \verts{a - b} \over 2}}$. Then,

\begin{align}&\color{#66f}{\large{\mathbb E}\bracks{\max\pars{X + k_{1},k_{2}}}}
={\mathbb E}\bracks{X + k_{1} + k_{2} + \verts{X + k_{1} - k_{2}} \over 2}
\\[5mm]&={1 \over 4} + {k_{1} + k_{2} \over 2}
+\half\,\color{#c00000}{{\mathbb E}\pars{\verts{X + k_{1} - k_{2}}}}
\end{align}

  \begin{align}&\color{#c00000}{{\mathbb E}\bracks{\verts{X + k_{1} - k_{2}}}}
=\int_{0}^{1}\verts{X + k_{1} - k_{2}}\,\dd X
\\[5mm]&=\verts{1 + k_{1} - k_{2}}
-\int_{0}^{1}X\sgn\pars{X + k_{1} - k_{2}}\,\dd X
\\[5mm]&=\verts{1 + k_{1} - k_{2}} - \half\,\sgn\pars{1 + k_{1} - k_{2}}
+ \int_{0}^{1}X^{2}\,\delta\pars{X + k_{1} - k_{2}}\,\dd X
\\[5mm]&=\verts{1 + k_{1} - k_{2}} - \half\,\sgn\pars{1 + k_{1} - k_{2}}
+ \pars{k_{2} - k_{1}}^{2}
\int_{0}^{1}\delta\pars{X - \bracks{k_{2} - k_{1}}}\,\dd X
\\[5mm]&=\color{#c00000}{%
\verts{1 + k_{1} - k_{2}} - \half\,\sgn\pars{1 + k_{1} - k_{2}}
+ \pars{k_{2} - k_{1}}^{2}\Theta\pars{k_{2} - k_{1}}\Theta\pars{1 - k_{2} + k_{1}}}
\end{align}

\begin{align}
&\color{#66f}{\large{\mathbb E}\bracks{\max\pars{X + k_{1},k_{2}}}}
\\[5mm]&=\color{#66f}{\large{1 \over 4} + {k_{1} + k_{2} \over 2}
+\half\,\verts{1 + k_{1} - k_{2}} - {1 \over 4}\,\sgn\pars{1 + k_{1} - k_{2}}}
\\[5mm]&+ \color{#66f}{\large\half\,\pars{k_{2} - k_{1}}^{2}
\Theta\pars{k_{2} - k_{1}}\Theta\pars{1 - k_{2} + k_{1}}}
\end{align}

  $\ds{\delta\pars{x}}$ is the
  Dirac Delta Function and $\ds{\Theta\pars{x}}$ is the
  Heaviside Step Function .


Why isn't the perfect closure separable? Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, a graduate course p.301, says:

  On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

The intuition I can provide is that in an algebraic extension $F\subset K$ of characteristic $p$  , the field $K$ will be a separable extension of its purely inseparable extension $K_p$ over $F$ whenever $trdeg_{\mathbb F_p} F\leq 1$. This is why my counterexamples on MathOverflow  here and here have $F=\mathbb F_p(u,v)$ , a purely transcendental extension of $\mathbb F_p$ of degree two.

I can't prove my intuition but it implies that  Zev's example won't work (he didn't say it did, so he made no mistake), and here is a proof that indeed it doesn't.

The key point is that is that if $\mathbb F_q$ is a finite field of characteristic $p$, the Frobenius map $\mathbb F_q \to \mathbb F_q: x\mapsto x^p$ is surjective and this implies that for $f(T)\in \mathbb F_q [T]$ we have
$ (f(T))^{1/p}\in \mathbb F_q [T^{1/p}]$,  because of  the freshman's dream
$(a+bT+\ldots)^{1/p}=a^{1/p}+b^{1/p}T^{1/p}+\ldots $ !
This implies that

$$(\mathbb F_q(T))^{p^{-\infty}}=\mathbb F_q(T,T^{p^{-1}},T^{p^{-2}},\ldots)$$

And now the rest is easy. If $F= \mathbb F_p(T)$ and
$K=(\mathbb F_{p^2}(T))^{p^{-\infty}}=\mathbb F_{p^2}(T,T^{p^{-1}},T^{p^{-2}},\ldots \quad )$ we have
$K_p=\mathbb F_p(T,T^{p^{-1}},T^{p^{-2}},\ldots \quad )$  and $K$ is a separable extension of the perfect closure $K_p$ of $F$ in $K$ . Indeed $K$ is just the simple separable extension $K=K_p(g)$ of $K_p$, where $g \in \mathbb F_{p^2}$ is any element such that $\mathbb F_{p^2}=\mathbb F_{p}(g)$. [ The element $g$ is of course separable over the (perfect !) field $\mathbb F_p$, so it is a fortiori separable over $K_p$].

Edit  What I called my intuition above has now been proved by ulrich on MathOverflow here. He proves  that if $trdeg_{\mathbb F_p} F\leq 1$, any algebraic extension $F\subset K$ has the property that $K$ is separable over its
purely inseparable extension $K_p$ over $F$.

Hint: let $g(X)$ be irreducible and separable; let $f(X)=g(X^p)$; adjoin one root of f to your field; study the extension.


Limit involving arctan I'm working on a problem dealing with convergence in distribution of a sequence of random variables (in particular, $1/n$ times the maximum of $n$ Cauchy random variables) and have ended up with needing to take the limit of $$\left(\frac{\text{tan}^{-1}(nt)}{\pi}+\frac{1}{2} \right)^n $$ as $n\rightarrow \infty$.

I tried rewriting it, for example, as
$$\frac{1}{2^n}\left(\frac{n\cdot2\text{tan}^{-1}(nt)/\pi}{n}+1 \right)^n $$
hoping to get something that looks like $e$, but haven't really been able to come up with anything.

The Taylor series expansion at infinity for $\tan^{-1}(nt)$ looks to be

$\frac{\pi}{2} - \frac{1}{nt} + O(\frac{1}{n^3})$.

for $t \geq 1.$
So plugging in to your equation should give you $(1 - \frac{1}{n\pi t})^n$, and taking the limit gives you $e^{-\frac{1}{\pi t}}$.

If you take the log of this function, this limit diverges for every t, which implies this limit diverges too.


How to find the particular solution of the differential equation. Original equation $$x^8y'+7y=e^\frac{1}{x^7}\quad y(1)=e$$ This is the original equation with the initial condition. How would i separate to take the integral? This is what I have done so far but I am not sure if it right and if it is I do not know how to continue.
$$y'+\frac{7y}{x^8}=\frac{e^\frac{1}{x^7}}{x^8}$$

Your steps are correct so far:
$$ \frac{dy}{dx} + 7x^{-8}y = e^{x^{-7}}x^{-8}$$

You can solve this using the method of integrating factors:

Your integrating factor (IF) is:

$$ \text{IF} = e^{\int 7x^{-8}}dx = e^{-x^{-7}}$$

You then multiply through by IF:

$$ e^{-x^{-7}}\frac{dy}{dx} + 7x^{-8}ye^{-x^{-7}} = x^{-8} $$

$$  \frac{d}{dx}\left(ye^{-x^{-7}}\right) = x^{-8}$$

Integrating both sides wrt $x$:

$$ ye^{-x^{-7}} = \int x^{-8}dx = -\frac{1}{7}x^{-7} + c$$

where $c$ is a constant.

$$ y = -\frac{1}{7}x^{-7}e^{x^{-7}} + ce^{x^{-7}}$$

You can then use $y(1) = e$ to find an expression for $c$ and hence a solution to the differential equation.


Why Hilbert changes the property of a set in his Infinite hotel? I'm not a mathematician, so my question may look a bit lame to most of you.

In the Infinite Hotel paradox we are dealing with infinite set of pairs (room/guest). A main property of a pair is 50/50 ratio.
A set of pairs inherits the properties of its members, right?

So, an infinite set of pairs also have 50/50 ratio.

Now, Hilbert brakes the infinite set of pairs in to two sets in order to insert one guest in the newly created infinite set of guests.
But even divided in to two sets, the 50/50 ratio of the broken infinite set of pairs, cannot magically disappear, right? Because even after we take the guests out of the rooms we'll still have 50% rooms and 50% guest.

If my assumption is correct, we will brake the 50/50 ratio by adding one quest, and the rooms will become less than 50%, which will leave no room for the new guest.

Am I missing something here, and if yes, what?

Foreword:

Reiterating what is said in comments and hopefully elaborating on a few key points:

The specific mathematical concepts behind the underlying problem are those of cardinality of sets and bijective functions.  These formalized concepts mirror how we think in the finite case and allow us to think beyond that as well in the infinite cases.

The purpose of the "Hilbert's Hotel" example is to show explicitly via way of visualization how our intuition from finite sets does not directly translate to infinite sets.

What is a bijection and what is meant by cardinality?

A bijective function is a function from a domain to a codomain where the following properties are satisfied:

It is a function, i.e. every element in the domain maps to exactly one output in the codomain (e.g. each person can be in only one room at a time)
It is a surjective function, i.e. every element in the codomain is mapped to by at least one element in the domain (e.g. every room has at least one person in it)
It is an injective function, i.e. every element in the codomain is mapped to by at most one element in the domain (e.g. no room may have more than one person in it)

We say that if there exists a bijective function between two sets, that they "have the same cardinality" (i.e. they have the "same number of elements").  If the sets are called $A$ and $B$, we write that they have the same cardinality as $|A|=|B|$.  (Other popular options exist, such as $\#A=\#B$ or $A\sim B$, etc... I will stick with $|A|=|B|$ through the rest of this post).

For example, if we have $50$ people and $50$ rooms, then we can come up with a bijective function pairing the people to the rooms so that exactly one person goes in each room.

What are some results for the finite case?

A few properties that you should be well familiar with for finite sets:

If $|A|=|B|$ and $A'\supsetneq A$, then $|A'|\gneq |B|$.

For example, if we have $50$ people and $50$ rooms, then we have a bijective function pairing people to rooms, but as soon as we add a $51$'st person, then we cannot come up with such a pairing.  $51$ people and $50$ rooms, we will by the pigeonhole principle be forced to have some room with more than one person in it.

Similarly, if we had $50$ people and $51$ rooms, then we would have at least one empty room.

Indeed, these observations lead us to the following result:

If $A$ and $B$ are both finite sets and there exists an injective function from $A$ to $B$ which is not surjective then $|A|<|B|$ and no bijection exists between $A$ and $B$.  Similarly if $A$ and $B$ are both finite sets and there exists a surjective function from $A$ to $B$ which is not injective then $|A|>|B|$ and no bijection exists.

The very point of Hilbert's Hotel is to show that the above result does not hold for infinite sets.

What really is the hotel story meant to portray?

Letting $P=\{p_1,p_2,p_3,\dots\}$ and $R=\{r_1,r_2,r_3,\dots\}$ be (countably) infinite sets and letting $f:P\to Q$ given by $f(p_i)=r_i$ be a bijective function between them...

(here, $P$ represents our set of hotel guests and $R$ our set of rooms, our bijection just says that the $i$'th person is in the $i$'th room)

...and let $P'=P\cup \{p'\}$ where $p'\notin P$...

(if we add another new person to the hotel)

...then the function $g$ between $P'$ and $R$ given by $g(x)=\begin{cases}r_1&\text{if}~x=p'\\r_{i+1}&\text{if}~x=p_i\end{cases}$ is a bijection, thereby proving that $P'$ and $R$ are of the same cardinality despite $P'\supsetneq P$.

(by shifting each person down a room and giving the new person the first room, we successfully pair the rooms and people again leaving no person without a room and leaving no room without a person such that each room has exactly one person)

Note that if we tried to do the same thing in a finite setting, then we would have had the "last person" try to change rooms to the next room, but there would not be a "next room" for him to go to.  Since we are working with infinite sets however, such a "last person" does not exist and so this does not present a problem.

Could we do it without making people change rooms first?

Given $P=\{p_1,p_2,\dots\}$ and $R=\{r_1,r_2,\dots\}$ be our infinite sets, letting $P'=P\cup \{p'\}$ where $p'\notin P$, and letting $f~:~P\to R$ given by $f(p_i)=r_i$ be our bijective function between the two, you are asking if there exists a bijective function $g$ from $P'$ to $R$ such that $g(p_i)=f(p_i)=r_i$ for all $p_i\in P$.

The answer is no, we can not.  Suppose that $g(p')\in R$.  Then $g(p')=r_i$ for some $i$.  But, since $f$ is a bijection, $f^{-1}$ exists, and so $g(p')=r_i=g(f^{-1}(r_i))=g(p_i)$ and so $r_i$ is mapped to by two elements, $p'$ and $p_i$.

That is to say, if we do not move the people around in the rooms, and tell our new person to find some room to enter, it would necessarily already have someone in it.

Again, this would have in the finite case shown to us that no bijection exists between the sets, but in the infinite case as shown earlier this does not matter, a bijection still exists.

How about that 50/50 ratio thing?

When babies first learn to count, they point to each object they count and say the next number in sequence.  So long as they don't accidentally point to the same thing more than once and don't skip or mess up any numbers in their sequence, they are successful at saying how many of an object there are.

Extending this method of counting and taking out the middle man of using numbers, we could also check to see if there are the same number of one type of object as another by pointing to a pair, one of objectA and one of objectB, before taking them away, checking to make sure that we pair each of them up successfully.

That is in essence coming up with a bijection between the set of objects in the first set with the set of objects in the second set.

Further, in the finite case, we notice that if we have $|A|$ elements in set $A$ and we have $|B|$ elements in set $B$ and $A$ and $B$ are disjoint sets, then $\frac{|A|}{|A\cup B|}=\frac{1}{2}=50\%$.

We also notice in the finite case that if $|A|\neq |B|$ then $\frac{|A|}{|A\cup B|}\neq \frac{1}{2}$.

As we move to the infinite case, things can get much weirder.  If $A$ is infinite, then $|A|$ is infinity, but arithmetic with infinity is not well defined.  We would have had $\frac{|A|}{|A\cup B|}=\frac{\infty}{\infty}$ which could be literally anything.  You might try to say that it would be better written as $\frac{\infty}{2\cdot \infty}=\frac{1}{2}$ but that is not entirely correct.

You might try to say for instance that "Only half of all positive integers are even" and so there are "more" positive integers than positive even integers.  In terms of that $\{1,2,3,4,\dots\}\supsetneq \{2,4,6,\dots\}$ you are correct, the positive integers strictly contains the positive even integers as a subset.  In terms of cardinality however, they are considered to be the same size.  That is because the function $f:\Bbb N\to 2\Bbb N$ given by $f(n)=2n$ is a bijection between those sets.

You might also say that there are "more positive integers which are not multiples of 3 than positive integers which are multiples of 3."  In terms of relative density when looking at the numbers written in the usual order, $1,2,\color{blue}{3},4,5,\color{blue}{6},7,8,\color{blue}{9},\dots$ you might be correct.  But, notice that I could have written the numbers in a different order: $1,\color{blue}{3},2,\color{blue}{6},4,\color{blue}{9},5,\color{blue}{12},7,\color{blue}{15},\dots$ where I alternate between writing the next multiple of $3$ and the next non-multiple of $3$.

Having a convenient way to pair things is nice, but if the pairing you are looking at doesn't have the desired property it does not automatically preclude a different pairing from existing in the infinite case even though it does in the finite case.

Hi @JMoravitz and all who took part in solving this problem.
I'm writing an answer to my question, because it gives me more word space, and because I don't see a need for further discussion.
First I'd like to thank you all for the patience and for taking from your precious time to answer.

Thank you!

I did really hope that there is something which I don't see in the problem and I asked the question with the sincere hope to get an answer which defeats all my arguments; I like to learn.
If the moderator does not close the topic you can comment and I'll respond, but there is no really much left to be said.

Mathematician have the highest IQ, but in many cases the intelligence lead us to negligence which results in mistakes.
I'm not mathematician and probably not as intelligent as you guys, but I work as a hobby on logical problems, language deception (logical problems due to improper use of language and mind stereotypes) and human's psychology.
I came to the conclusion that most of the logical fallacies and the paradoxes for that matter come from improper use of properties or missing properties.

After our discussion I can say with greater certainty that Hilbert's paradox is a real solvable paradox and it comes from improper use of properties and deception based on mind stereotypes.
I understand the point with the bijection which JMoravitz is making  but before we come to the pairing which is the easy part, we have to solve a much difficult problem - the 50/50 percent ratio in set of pairs. Because it is always on our way.

Three points to solve the paradox

1. If we vacate a room, we cannot use it for new guest unless we brake the infinity. How so? Because ∞+1=∞ rooms, and we already have infinite number of guests on the other side of the set (50/50 ratio is still present). Hence the conclusion that although it seams that we have an empty room we don't really have one, because the number on both sides is always equal - infinite.

2. Not taking in account the above, we actually make the number of the hotel guests finite (if there is empty room it is because the guests are finite) which automatically makes the number of the occupied rooms finite + 1 empty room. Now the number of the rooms is with 1 greater than the number of the guests and we can accommodate one guest. But in this case we don't work with infinity, do we? If we don't agree with this then point 1 applies.

3. We can ignore point 1 and use the obvious argument against point 2 (which would be wrong) that after we move the guests we end up with infinite occupied rooms and one empty room in the hotel (!)
Well, here is the fallacy - the hotel is defined as infinite countable number of occupied rooms. Let's lose the "hotel" word which is deceiving us and let's use "infinite set of pairs". If we have one element which is not paired, does it belong to the set? Of course not.
So, the "empty" room does not belong to the hotel because it is not a member of the set of pairs. And this is what I call the "language deception". The "hotel" word is bringing deception.

Let me explain, why the room does not belong to the hotel.

What happens if we move the guests with one room down, instead of moving them up? The guest from room №1 leaves the room, and the room is taken by the guest from room №2, the guest from №3 goes in room №2, the guest from №4 goes in №3 and so on (it is the opposite of vacating a room)
Now one guest left the hotel. Is this person part of the hotel guests? Of course not. So is the case with the empty room - it does not belong to the hotel.

By taking one guest out of the hotel we can create another hotel if we find a room for the guest. And we already know how to create another room for the new hotel, don't we ;)

Every paradox has a solution and if the solution is wrong, there is solution for the wrong solution.
By following the "property rule" we can solve any paradox. You guys can try me on that ;)

Once again, thank you!
It was a great pleasure for me.


Example of a function that is not twice differentiable Give an example of a function f that is defined in a neighborhood of a s.t. $\lim_{h\to 0}(f(a+h)+f(a-h)-2f(a))/h^2$ exists, but is not twice differentiable.

Note: this follows a problem where I prove that the limit above $= f''(a)$ if $f$ is twice differentiable at $a$.

Take $f(x)=x^2\sin(x^{-1})$ on $\mathbb{R}-\{0\}$ and set $f(0)=0$. You can prove this is differentiable at zero, but not twice differentiable there. That said, $f(-h)=-f(h)$ and $f(0)=0$ so that $f(h)+f(-h)-2f(0)=0$.

Consider the function $f(x) = \sum_{i=0}^{n}|x-i|$. This function is continuous everywhere but not differentiable at exactly n points. Consider the function
$G(x) = \int_{0}^{x} f(t) dt$. This function is differentiable, since $f(t)$ is continuous due to FTC II. Now $G'(x) = f(x)$, which is not differentiable at $n$ points.


Hopf gradings on complex commutative group rings Let $G$ be a finite abelian group.
The complex group ring $\mathbb{C}G$ admits a structure of Hopf algebra when the multiplication is the usual multiplication in a group ring and the co-multiplication is defined by $\Delta (g)=g\otimes g$ and
the anti-pod is $S(g)=g^{-1}$ for any $g\in G$.

Now a Hopf-grading on a Hopf algebra is a grading of the algebra which "respects" the multiplication and the co-multiplication.
However, it seems that this is not enough and there is some extra compatibility condition which I do not understand.

I tried reading about it in Montgomery book "Hopf algebras and their actions on rings", but I still can't fully understand it.

So for now, I want to leave the general definition and to concentrate only on the Hopf algebra $\mathbb{C}G$ as above.

I hope someone can tell me what are all the Hopf-gradings on $\mathbb{C}G$, and if this is too much work, I would like to know just what conditions should I check and perhaps a small example (for example $\mathbb{C}C_4)$.

Thanks in advance for any help.

There is two notion of gradation - one is based on the filtration on vector space, but the second is related to group action.

If $H$ is a Hopf algebra over field $k$, then it is $\mathbb{N}$-filtered $k$-Hopf algebra iff there exist filtration $\{H_i\}_{i\in \mathbb{N}}$ on $H$ such that

$H_iH_j\subset H_{i+j}$
$\Delta H_n \subset \sum\limits_{k=0}^{n} H_{k}\otimes H_{n-k}$
$S(H_i)\subset H_i$

If $H\cong \bigoplus\limits_{i\in\mathbb{N}}H_{(i)}$ for some linear supspaces $H_{(i)}$ of $H$, and the above conditions hold if we replace $H_i$ by $H_{(i)}$ then we have $\mathbb{N}$-graded $k$-Hopf algebra. It turns out that if $H$ is connected ($H_{(0)}\cong k$) then the condition for $S$ is unnecessary and $S$ can be uniquely determined from other components of bialgebra.

The above definition can be generalize for directed sets instead of $\mathbb{N}$. Such sets should be also monoid.

The second notion of gradation is a gradation by group (Montgomery Hopf algebras and their actions on rings, 10.5.1). Suppose we have finite abelian group $G$. Then it is know that $\hat G \cong G$, so there exist bicharacter $\lambda$ on $G$. Hopf algebra $H$ (in some braided monoidal category) is called $G$-graded $\lambda$-Hopf algebra (or $G$-graded Hopf algebra with respect to given bicharacter $\lambda$) if it has gradation (as a vector space) $H\cong \bigoplus\limits_{g\in G} H_g$ and

$H$ is a $G$-graded algebra $\left(\right.H_gH_h\subset H_{g\cdot h}$, where $\cdot$ is a multiplication in $G\left.\right)$
$H$ is a $G$-graded coalgebra $\left( \Delta(H_g)\subset \sum\limits_{p\cdot q =g}H_p\otimes H_q \right)$
$\Delta$ and $\varepsilon$ are algebra maps with $m_{H\otimes H}$ given by
$$m_{H\otimes H}\left(a_{g_1}\otimes b_{h_{1}},a_{g_2}\otimes b_{h_{2}}\right)=\lambda(h_1,g_2)a_{g_1}a_{g_{2}}\otimes b_{h_{1}}b_{h_{2}},$$
where $A\cong \bigoplus\limits_{g\in G} a_{g}$ and $B\cong \bigoplus\limits_{h\in G}b_h$.
$S\circ m=m\circ \Psi \circ (S\otimes S)$, where $\Psi$ is a braiding in our braided monoidal category.

Often in this context is it assumed that $G$-graded Hopf algebra is connected, so we also need $\varepsilon(H_g)=0$ unless $g=e$ ($e$-neutral element in $G$).

As an example let us take $kG$ as a Hopf algebra. In the natural way it forms $kG$-comodule algebra ($H$-comodule algebra $A$ is a $k$-algebra which is simultaneously $H$-comodule with a coaction $\delta$ which is an algebra homomorphism). Indeed, take $\delta=\Delta$.

One can easily check that any $kG$-comodule $A$ is a $G$-graded $k$-module. We have coaction $\tilde{\delta}(a)=\sum\limits_{g\in G}a_g\otimes g$ on $A$, where $a_g\in A_g$ and $A\cong \bigoplus\limits_{g\in G}A_g$. In this case we have $1_A\in A_e$ and for any $a\in A_g,\ b\in A_h$ we have an equality
$$\tilde{\delta}(ab)=ab\otimes gh \in A_{gh}$$
Hence $A$ is a $G$-graded $k$-algebra. Converse statement is also true.

If you are interested in the second notion you can find a lot of informations in this article and in section 3 here. Some aspects of gradation on $\mathbb{C}G$ you can find also in this article.


Stats 101 - Exam Question Fail, or just tricky to answer? This appeared on a friends Stats 101 exam:

  You have a random sample of 2000
  respondents who's mean age is 47.3. A
  friend of yours, (who is knowledgeable
  about such things) claims that
  republican voters are, on average,
  older and demonstrate a mean age of
  50.   Using your data, demonstrate if he is correct, or incorrect to a 90%
  confidence level

The natural reaction for most of us here is that there isn't enough data in the question to even guess at the accuracy of the statement. When asked, the professor said "It isn't relevant to the problem".

Are we being trolled or is there a way, mathematically speaking, to state that there's insufficient information for a conclusion... or /gasp/ is there actually an answer ?

edit

I just now saw that there is a stackexchange specifically for Stats :( If someone has the ability, I don't mind this being moved to a more appropriate forum.

So you want the confidence interval to be of the following form:

$$50 \pm 1.65 \cdot \frac{s}{\sqrt{2000}}$$ where $\frac{s}{\sqrt{2000}}$ is the standard error of the mean. We need to find $s$. Given the data, we can't find $s$ exactly. But we can probably use a minimum variance unbiased estimator. One such estimator for $s$ is the following: $$K_n = \sqrt{\frac{n-1}{2}} \frac{\Gamma(\frac{n-1}{2})}{\Gamma(\frac{n}{2})}$$ where $\Gamma(n)$ is the gamma function. Also, we don't know the number of republicans in the sample. So this interval just gives a tighter estimate (i.e. less conservative).

I'm sure the average age of Republican voters, together with  a confidence level, can be found somewhere on the internet. Find these figures and see whether your friend is right. For all we have been told, the 2000 respondents may have been drawn from the subscribers to a fishermen's magazine.


Functions (set theory) I'm having troubles proving that if $f,g \subset A \times B$ are functions, then $f\cap g$ is also a function.

Thanks for your help!

Kind regards,

Phi.

Definition: let $h, E, F$ a sets, we define: $$h \text{ is relation if }\,\big[\forall x \in h:\exists y,\exists z: x=(y,z)\big] \\\operatorname{dom}(h):=\{x|\exists y:(x,y)\in h\} \\ \operatorname{cod}(h):=\{x|\exists y:(y,x)\in h\} \\ h \text{ is function if }\, \big[h\text{ is relation} \wedge  \forall x,\forall y,\forall z: (x,y) \in h \wedge (x,z)\in h \to y=z\big] \\ h: E\to F \text{ for }\big[ h\text{ is function} \wedge  \operatorname{dom}(h)= E \wedge \operatorname{cod}(h)\subseteq F\big]$$

  Thereom: let be $f,g$ functions and $f \subseteq (A \times B)\supseteq g$: $$ (f \cap g) \text{ is function}$$

  Proof:

  $(f\cap g)$ is relaction because $f \subseteq (A \times B)\supseteq g$
  let be $x,y,z$: $$ \begin{align}(x,y) \in (f \cap g) \wedge (x,z)\in (f\cap g) \leftrightarrow&  \\ \big((x,y) \in f \wedge (x,y) \in g\big) \wedge \big((x,z) \in f \wedge (x,z) \in g\big) \leftrightarrow& \\ \big((x,y) \in f \wedge (x,z) \in f\big) \wedge \big( (x,y) \in g \wedge (x,z) \in g\big)  \to& \\\  (y=z) \wedge (y=z) \text{ because } f \text{ and } g \text{ are by hyp. functions} \to& \\y=z \,\text{ (Q.E.D)} \end{align}$$
  naturally $(f \cap g) \subseteq (A\times B)$, in fact $$ \begin{align} x \in (f\cap g) \leftrightarrow x  \in f\wedge x \in g  \to& \\  x \in (A\times B) \wedge x \in (A \times B) \,(\text{because } f \subseteq (A \times B)\supseteq g) \to&\\ x \in (A \times B) \,\text{ (Q.E.D)}\end{align}$$

$f$ and $g$ are functions which mean that for each $a\in A$ there is a unique $b\in B$ such that $f(a)= b$. And similarly for $g$. First show that $f\cap g$ is a subset of $A\times B$ then show that it has the uniqueness property that $f$ and $g$ have. The trick is to show that for all $x$ in the domain of $f\cap g$ it is either in the domain of $f$ and the domain of $g$. Thus it will inherit the uniqueness property from $f$ and $g$.


Inverse of continuous function on unbounded domain Consider the following property: "Let $A$ be a closed bounded set of $\mathbb{R}$. Suppose that $f:A \rightarrow \mathbb{R}$ is a continuous injective function. Then $f^{-1}:f(A) \rightarrow A$ is also continuous."

I am looking for an example of a continuous injective function $f:A \rightarrow \mathbb{R}$ on a closed but unbounded set $A$ of $\mathbb{R}$ for which $f^{-1}$ is not continuous, in order to get an idea of how important the conditions for this property are. I can't think of a good example though. Any thoughts?

I don't see anyway you will get such examples. Coz it's beyond the theorem. Note that if $A$ is compact and $f$ is continuous then $f(A)$ will be also compact.
Moreover is it possible if $[a,b]$ is closed and not bounded? Whereas  $a-\epsilon$ and $b+\epsilon$ are it's lower and upper bounds. Remember Every continuous function in $[a,b]$ is bounded.


V.I. Arnold says Russian students can't solve this problem, but American students can -- why? In a book of word problems by V.I Arnold, the following appears:

The hypotenuse of a right-angled  triangle  (in  a  standard  American  examination) is 10 inches, the altitude dropped onto it is 6 inches. Find the area of the triangle.

American school students had been coping successfully with this problem for over a decade. But then Russian school students arrived from Moscow, and none of them was able to solve it as had their American peers (giving 30 square inches as the answer). Why?

Here's the book. I assume the answer is some joke at the expense of the Americans, but I don't get it. Possibly a joke about inches? Anyone?

There is no such right triangle. The maximum possible altitude is half the hypotenuse (inscribe the triangle into a circle to see this), which here is $5$ inches. You would only get $30$ square inches if you tried to compute the area without checking whether the triangle actually exists.

This is from a paper full of trick questions, so likely, is criticizing the American students. The crux of the problem is whether the discrepancy is noticed, and how it is handled.

In middle school, kids are taught a simplified formula,
Triangle Area $\Delta = \dfrac{1}{2}\times b\times h$, or Area equals Base times Height.

In Geometry, the full version is taught,

$$\Delta = \dfrac{1}{2}\times b\times a$$
$$\implies \Delta = \dfrac{1}{2}\times \text{base (any side of triangle)} \times \text{altitude (line perpendicular to base and going to opposite vertex)}$$

The definitions of base and altitude are critical and repeatedly taught. They have been standard since the time of Euclid, but teaching the simplified version causes confusion.

Using this formula, the $30 \,\mathrm{in^2}$ area would be true for a triangle with hypotenuse of $10\,\mathrm{in^2}$ and corresponding altitude of $6\,\mathrm{in^2}$.

Kids are taught that the sides of a right triangle with hypotenuse length $10 \,\mathrm{in}$ and a side of $6 \,\mathrm{in}$ is a Pythagorian Triple, $6, 8, 10$. For this $6:8:10$ triangle, $6$ and $8$ are perpendicular and thus altitudes of each other, the Area $\Delta = \dfrac{1}{2}\times 6\times 8 = 24 \,\mathrm{in^2}$. The altitude for the hypotenuse can be found by $\Delta = 24 = \dfrac{1}{2}\times 10\times \text{altitude}$, and $\text{altitude} = 4.8 \,\mathrm{in}$.

Thus, using simple tools taught to the students, the "altitude $6$" triangle cannot be a right triangle, since the right triangle with side $6$ and Hypotenuse $10$ has a side of $6$, not the altitude to the hypotenuse. Expecting complicated proofs from secondary students is unreasonable, but applying the Pythagorean Theorem and area formulae are standard.

The typical student probably sees the $6$ and $10$ as parts of a Pythagorean Triple and "knows" it is a right triangle, then finds the obvious area without further thought. Thus, answering the question as 30 could imply sloppy work or lack of knowledge or understanding.

Alternately, it could be that the American students saw the discrepancy, and made a judgement call on how to answer a test question, assuming a typographical error.

PS. Yuan's inscribing the triangle in a circle to determine the maximum possible altitude to the hypotenuse is simple, brilliant, and uses concepts taught in Geometry!


Is this method of indefinite integration correct? $\int{ dx\over12+5\tan(x)}$ I am integrating:

$$ \int{ dx\over12+5\tan(x)} $$

I proposed $x = \arctan(u)$, replacing $dx$ by $du \over 1+u^2$, so  the integral becomes:

$$ \int{du \over (1+u^2)(12+5u)} $$

Which can be integrated using partial fractions and then I eventually get that the anti-derivative is:

$$ {5 \over 169} \ln |12+5u| - {5 \over 338}\ln|1+u^2| + {12 \over 169}\arctan (u) + C $$

And finally going back to the original variable substituting $u = \tan(x)$

So my question is if this method is correct?

Hint:

Multiply numerator &   denominator by  $\cos x$

Write $$\cos x=A(12\cos x+5\sin x)+B\cdot\dfrac{d(12\cos x+5\sin x)}{dx}$$


Deriving covariance of sample mean and sample variance $\newcommand{\cov}{\operatorname{cov}}$I want to find the covariance between the sample mean and the sample variance, I think I am along the right track but am not sure.

Suppose we have i.i.d. random variables $X_1, X_2, \ldots, X_n$.

Define

$$\overline{X} = \frac{1}{n} \sum_{i=1}^n X_i \text{ and } S^2 = \frac{1}{2n(n-1)} \sum_{i=1}^n \sum_{j=1}^n (X_i - X_j)^2 $$

I want to find their covariance $\cov(\overline{X}, S^2)$. Here is what I have done:

\begin{align}
& \cov(\overline{X}, S^2) = \cov\left(\frac{1}{n}\sum_{i=1}^n X_i, \frac{1}{2n(n-1)} \sum_{i=1}^n\sum_{j=1}^n (X_i - X_j)^2 \right) \\
= {} & \frac{1}{2n^2(n-1)} \sum_{i=1}^n \cov(X_i,\sum_{k=1}^{n}\sum_{j=1}^{n} (X_k - X_j)^2) \\
= {} & \frac{1}{2n^2(n-1)} \sum_{i=1}^n \cov(X_i,2\sum_{k=1}^{n} (X_i - X_k)^2)
\end{align}

(because $\cov(X_i, (X_a - X_b)^2)$ is zero if $i$ isn't $a$ or $b$)

$$= \frac{1}{n^2(n-1)} \sum_{i=1}^{n} \sum_{k=1}^n \cov(X_i, (X_i - X_k)^2).$$

Now, I compute $\cov(X_i, (X_i - X_k)^2)$:

for $ i = k, \cov(X_i, (X_i - X_k)^2) = 0$

for $ i \neq k,$

\begin{align}
& \cov(X_i, (X_i - X_k)^2) = \cov(X_i, X_i^2 - 2X_iX_k + X_k^2) = \cov(X_i, X_i^2) - 2\cov(X_i, X_iX_k) \\[8pt]
= {} & E(X^3) - E(X)E(X^2) -2E(X)E(X^2) + 2(E(X))^3 \\[8pt]
= {} & E(X^3) -3E(X)E(X^2) + 2(E(X))^3.
\end{align}

When I plug this in, I do not get the answer that I want which is $E(X-E(X))^3$, can anyone please tell me where I went wrong?

UPDATE: I plugged it in again and it is actually correct! Sorry for being so careless. Thanks to anyone who has read the question. Should I delete this post?

I hope this will help you to find a mistake.

https://www.tandfonline.com/doi/pdf/10.1198/000313007X188379


Central limit theorem on packs of variables I'm trying to solve the following exercise:

  Let $\mu$ be a probability distribution on $\mathbb{R}$ having second moment $\sigma^2<\infty$ such that if $X$ and $Y$ are independent with law $\mu$ then the law of $(X+Y)/\sqrt{2}$ is also $\mu$. Show that $\mu =\mathcal{N}(0,1)$
  Hint: apply the central limits theorem to packs of $2^n$ variables

My attempt:

So let $Z_n=(X_1,Y_1)+\cdots+(X_n,Y_n)$, then $\mathbb{E}(Z_n)=n\mathbb{E}(Z_n)$ then for $n\to \infty$
$$T_n=\frac{Z_n-n\mathbb{E}(Z_n)}{\sqrt{\sigma^2n}}\xrightarrow{\mathcal{D}}\mathcal{N}(0,1)$$
converges in distribution to the normal distribution.

Now I don't see the connection how to proof $\mu=\mathcal{N}(0,1)$. I also do not understand what "packs" of $2^n$ variables are. Is it $Z_n=(X_n,Y_n)+\dots$?

I think it must be proved that $\mu=\mathcal N(0,\sigma^2)$ but for convenience I will also preassume that $\sigma=1$

If $\phi$ denotes the characteristic function then:$$\phi(t)=\phi\left(\frac{t}{\sqrt2}\right)^2$$

Note that this can be repeated to arrive at $\phi(t)=\phi(\frac{t}2)^4$ and can be repeated again.

Actually with this it can be shown that $X$ and $2^{-\frac12n}(X_1+\cdots+X_{2^n})$ have equal distribution if the $X_i$ are independent and all mentioned random variables have distribution $\mu$.

It is not really necessary to use the characteristic function to come to this conclusion. You could just observe that $([X_1+Y_1]/\sqrt2+[X_2+Y_2]/\sqrt2)/\sqrt2$ again has $\mu$ as distribution, and so on - if the $X_i$ and $Y_i$ are independent and have $\mu$ as distribution.

Also we have $0$ as expectation, since $\nu=(\nu+\nu)/\sqrt2$ implies $\nu=0$.

Applying the CLS on the $X_i$ you will find that the constant $\mu$ must convergence in distribution to $\mathcal N(0,1)$.

This can only be the case if $\mu=\mathcal N(0,1)$.

Hint: without relying on the CLT, you can use characteristic functions.

Let $f$ be the characteristic function of the law $\mu$.

You can easily show that  $f$ satisfies this functional equation which can be solved to find that $f$ is the characteristic function of $\mathcal{N}(0,1)$.


How to solve $\int (2x^2 + 1) e^{x^2} \, dx$ using elementary integration and integration by parts? $$\int (2x^2+1)e^{x^2} \, dx$$

It's part of my homework, and I have tried a few things but it seems to lead to more difficult integrals. I'd appreciate a hint more than an answer but all help is valued.

Start by expanding the integrand.

$$\begin{align}
I&=\int \left(2x^2e^{x^2}+e^{x^2}\right)\,dx\\
&=\int 2x^2e^{x^2}\,dx+\int {e^{x^2}}\,dx\tag{a}\\
&=x e^{x^2}-\int 2x^2 e^{x^2}\,dx+\int 2x^2 e^{x^2}\,dx\\
&=x e^{x^2}+C
\end{align}$$

$(\text{a})$: Use integration by parts on the second integral with $u=e^{x^2}$ and $dv=dx$.

Here is a hint; you can not analytically evaluate $\int \exp(x^2) dx$.
Read more about that on wiki.
So your final answer should contain terms with $\int \exp(x^2) dx$.


Show $f$ is a surjective isometry. Give $f:(\mathbb{R}^2,d^{1}) \rightarrow (\mathbb{R}^2,d^{\infty})$ let $f$ be defined $$f(x_1,x_2)=(x_1+x_2,x_1-x_2)$$ My ideal is to try to show that $f$ preserves the origin and that $f$ is continuous and injective, thereby implying $f$ is surjective. I wonder if it is possible to show that $$d^{1}(x,y) = d^{\infty}(f(x),f(y))$$ to show that $f$ is an isometry, but I do not know how to manipulate the terms, for example what does $d^{\infty}(f(x_1,x_2),f(y_1,y_2))$ even look like? At any rate, here is my work thus far.

I can show that $f$ preserves the origin.$ f(0,0)=
  \left[ {\begin{array}{cc}
   1 & 1 \\
   1 & -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}0\\0\end{array} } \right]=(0,0).$

Moreover, $f$ is invertible so that,$ f^{-1}(0,0)= \frac{1}{2}
  \left[ {\begin{array}{cc}
   1 & 1 \\
   1 & -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}0\\0\end{array} } \right]=(0,0). $
For any $a,b \in R^{2}$ we have
$ f(a,b)=
  \left[ {\begin{array}{cc}
   1 & 1 \\
   1 & -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}a\\b\end{array} } \right]=(a+b,a-b).$

So that, $ f^{-1}(a+b,a-b)= \frac{1}{2}
  \left[ {\begin{array}{cc}
   1 & 1 \\
   1 & -1 \\
  \end{array} } \right] \left[ {\begin{array}{cc}a+b\\a-b\end{array} } \right]= \left(   \frac{(a+b)-(a-b)}{2},\frac{(a+b)+(b-a)}{2} \right)= \left( \frac{2a-b+b}{2}, \frac{a+2b-a}{2}\right) = (a,b)$

I am pretty certain that this shows that $f$ is a surjective isometry.I believe I have shown that $f$ and $f^{-1}$ are injective with the origin preserved, also both spaces have the same dimension and the kernel of $f$ contains only the origin, therfore $f$ is a surjective isometry. But, something feels off.

So I have been able to find a solution.

I must show that $$d^{1}(x,y) = d^{\infty}(f(x),f(y)).$$ I have that $$f(x) = f(x_1,x_2)= (x_1+x_2,x_1-x_2)$$ On the L.H.S. of the first equation we have $$d^{1}(x,y)= |(x_1-y_1)|+|(x_2-y_2)|$$ which is the Taxi Cab metric on $\mathbb{R}^2.$ On the R.H.S. we have $$ d^{\infty}(f(x),f(y)) =  d^{\infty}(f(x_1,x_1),f(y_1,y_2))= d^{\infty}((x_1+x_2,x_1-x_2),(y_1+y_2,y_1-y_2))$$ So that we get $$max:\{|(x_1+x_2)-(y_1+y_2)|,|(x_1-x_2)-(y_1-y_2)|\}$$
and this is $$max:\{|(x_1-y_1)+(x_2-y_2)|,|(x_1-y_1)-(x_2-y_2)|\}$$ and it is clear that  $$d^{1}(x,y) = d^{\infty}(f(x),f(y)).$$ The rest is shown above by showing that $f$ is invertible.


Does inverse image of idempotent element contain an idempotent element? Suppose $f\colon \oplus A_i\to\oplus B_j$ is a ring homomorphism, where $A_i,B_j$ are local rings. Both sides have finitely many summands. Suppose an idempotent $x$ (i.e. $x=x^2$) is in the image of $f$. Can we always choose an idempotent in $f^{-1}(x)$?

In fact, we want to lift the idempotents from $(\oplus_{i=1}^nA_i)/I$, where $I$ is an ideal of $\oplus_{i=1}^nA_i$, to $\oplus_{i=1}^nA_i$. Since $I=\oplus_{i=1}^nI_i$ with $I_i\subset A_i$ ideals, the question reduces to the following

  Let $A$ be a local ring. Then the idempotents lift modulo any ideal $J\subset A$.

If $a+J$ is idempotent, then since $A/J$ is also local we get $a\in J$ or $1-a\in J$.

Since $B_i $ are local, so the set of idempotents of $B_i $ has only 0 and 1. Thus $x=(x_1,x_2,...,x_n,...) $ where $x_i $are 0 or 1 and almost all of them are 0. So $x$ has an idempotents preimage.


Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$ Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .

$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfrac{\sqrt{3}}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
&=\dfrac{1.73}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
\end{align}$

$\begin{align}\sin 18^{\circ}&=\sin (30-12)^{\circ}\\~\\
&=\sin (30^{\circ})\cos (12^{\circ})-\cos (30^{\circ})\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{\sqrt3}{2}\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{1.73}{2}\sin (12^{\circ})\\~\\
\end{align}$

$\begin{align}\tan 23^{\circ}&=\dfrac{\sin (30-7)^{\circ}}{\cos (30-7)^{\circ}}\\~\\
&=\dfrac{\sin (30)^{\circ}\cos 7^{\circ}-\cos (30)^{\circ}\sin 7^{\circ}}{\cos (30)^{\circ}\cos 7^{\circ}+\sin (30)^{\circ}\sin 7^{\circ}}\\~\\
\end{align}$

is their any simple way,do i have to rote all values of of $\sin,\cos $ from $1,2,3\cdots15$

I have studied maths upto $12$th grade.

This may make a nice challenge for you. Use a regular pentagon to find the $\sin 18^\circ$.


Why is it that linear fractional transformations cannot take the upper half-plane to the first quadrant? Problem: Is there a linear fractional transformation that takes the upper half-plane to the first quadrant?  If yes, write it down.  If no, give proof.

Attempt:

We have that a linear fractional transformation $f$ is a transformation of form

$$
f(z) = {az + b \over cz + d} \text{ s.t. } ad - bc \ne 0
$$

and where $a,b,c,d \in \mathbb{C}$.
Note by another theorem, $f$ must send circles and lines to circles or lines.
I think (but am not certain) that linear fractional transformations must take boundaries to boundaries.
Hence if the supposition were true, we would have that the $x$-axis of $\mathbb{C}$ is sent to the boundary of the first quandrant (i.e., the union of the positive reals, zero, and the pure imaginary numbers from $0$ upwards).  This violates (2).
Then the argument is complete.

Is this correct?

No. Linear fractional transformations naturally act on the Riemann sphere ℂP1 by biholomorphisms. (In fact, all biholomorphisms of ℂP1 are linear fractional transformations, but we don’t need it.) A biholomorphism, i.e. a holomorphic map with holomorphic inverse, is also a diffeomorphism: any Riemannian surface is also a smooth 2-dimensional real manifold.

A half-plane on ℂ is a half of Riemann sphere bounded by a great circle. Its boundary is smooth. The boundary of a quadrant isn’t smooth.

This was a conceptual proof that used “higher mathematics”, but I’m sure one can also present an elementary proof.


How to solve this least square problem effectively? I want to solve the least square problem,
$\min\|Ax-b\|_2$, but the condition number of $A'*A$ is very large, How can I solve this problem effectively?

You should try regularization. Several common types are Tikhonov regularization and $\ell^1$ regularization. See this book for a machine learning perspective.

Many common regularizer are of the form $f(x, \lambda)$ where $f(x,\lambda)$ is a non-negative function with tuning parameter $\lambda\geq 0$ to which you solve $\arg\min_x ||Ax-b||_2^2 + f(x,\lambda)$. If $ f=0$, you get the original (unregularized) problem. If $f(x,\lambda) = \lambda ||x||_2$ you get Tikhonov regularization (which can be efficiently implemented via say, the SVD, by applying a certain function to the singular values) (also known as ridge regression). If you have $f(x,\lambda) = \lambda ||x||_1$, this will make solutions of $x$ to be sparser (you may see this under the name LASSO in machine learning-- this is $\ell^1$ regularization).

The Tikhonov regularization problem has a nice form: $x = (A^T A + \lambda I)^{-1} A^T b$, or equivalently, if $A=USV^T$ is the SVD of $A$ with singular values $\{\sigma_i\}$, estimate $x$ by  $V D U^T b$ where $D$ has diagonal $\{ \sigma_i/(\sigma_i^2 + \lambda^2)\}$.

The LASSO problem typically doesn't have a nice solution (the orthogonal case in the original paper is the only one I know of)but can easily done with common convex programming packages , e.g, cvx, or with soft thresholding (see the original paper for details, or the papers on ISTA/FISTA (FISTA=Fast Iterative Shrinkage Thresholding Algorithm)).

Hint: Check Moore–Penrose pseudoinverse.


Conditional expectation of a bivariate normal distribution Can you help me?

Find the conditional expectation $\mathbb{E}[X|Y]$ if $(X,Y)$ possesses a bivariate normal distribution.

Is $\mathbb{E}[X|Y=y]=\mu_X+\sigma_X\rho(\frac{\displaystyle y-\mu_Y}{\displaystyle \sigma_Y})$ the solution?

My question: Is the same $\mathbb{E}[X|Y=y]$ and $\mathbb{E}[X|Y]$?

$$ \mathbb{E}[X|Y]=h(Y) $$
where
$$h(y)=\mathbb{E}[X|Y=y]$$
So yes, it's somewhat the same, but not quite.

For future reference, here's derivation of this formula.
We'll suppose that $\sigma_X, \sigma_Y\neq 0$.We have that $$(X, Y) \sim N\left((\mu_X,\mu_Y), \begin{bmatrix}
     \sigma_X^2 & \rho\sigma_X\sigma_Y \\
     \rho\sigma_X\sigma_Y & \sigma_Y^2
\end{bmatrix}\right) $$
Then
$$\mathbb{E}[X|Y=y] = \frac{\int_{-\infty}^{\infty}x\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_X)^2}{\sigma_X^2}+\frac{(y-\mu_Y)^2}{\sigma_Y^2}-\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_Y^2}\right]\right)\text{d}x}{\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_X)^2}{\sigma_X^2}+\frac{(y-\mu_Y)^2}{\sigma_Y^2}-\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_Y^2}\right]\right)\text{d}x}\\ =
\frac{\int_{-\infty}^{\infty}x\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_X)^2}{\sigma_X^2}-\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}+\frac{\rho^2(y-\mu_Y)^2}{\sigma_Y^2}\right]\right)\text{d}x}{\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_X)^2}{\sigma_X^2}-\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}+\frac{\rho^2(y-\mu_Y)^2}{\sigma_Y^2}\right]\right)\text{d}x}\\=
 \frac{\int_{-\infty}^{\infty}x\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{x-\mu_X}{\sigma_X}-\rho\frac{y-\mu_Y}{\sigma_Y}\right]^2\right)\text{d}x}{\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2(1-\rho^2)}\left[\frac{x-\mu_X}{\sigma_X}-\rho\frac{y-\mu_Y}{\sigma_Y}\right]^2\right)\text{d}x}\\=
 \frac{\int_{-\infty}^{\infty}\left(x+\mu_X+\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y}\right)\exp\left(-\frac{1}{2(1-\rho^2)}\frac{x^2}{\sigma_X^2}\right)\text{d}x}{\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2(1-\rho^2)}\frac{x^2}{\sigma_X^2}\right)\text{d}x} =
\mu_X+\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y}$$
Because
$$\int_{-\infty}^\infty x\exp\left(-\frac{1}{2(1-\rho^2)}\frac{x^2}{\sigma_X^2}\right)\text{d}x = 0 $$
This gives the final result
$$\mathbb{E}[X|Y] = \mu_1+\rho\frac{\sigma_X}{\sigma_Y}(Y-\mu_2) $$

Sort ... of

$\mathbb{E}[X|Y=y]=\mu_X+\sigma_X\rho(\frac{\displaystyle y-\mu_Y}{\displaystyle \sigma_Y})$

but

$\mathbb{E}[X|Y]=\mu_X+\sigma_X\rho(\frac{\displaystyle \color{red}{Y}-\mu_Y}{\displaystyle \sigma_Y})$

No.

$E[X|Y=y]$ is a number since $(Y=y)$ is an event. If $\mathbb P(Y=y) > 0$, it is defined as

$$E[X|Y=y] = \frac{1}{\mathbb P(Y=y)}\int_{Y=y} X d \mathbb P$$

O/w, if Y is a continuous random variable,

$$E[X|Y=y] = \int_{\mathbb R} x f_{X|Y}(x|y) dx$$

where

$$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$

assuming $f_{X|Y}(x|y), f_{X,Y}(x,y)$ and $f_Y(y)$ exist.

For a continuous random variable that is not absolutely continuous (pdf doesn't exist), I have no idea.

$\mathbb{E}[X|Y]$ aka $\mathbb{E}[X|\sigma(Y)]$ is a random variable that depends on the value of $Y$ . It is defined as the random variable $Z$ s.t.

$Z$ is $Y$-measurable or $\sigma(Z) := \sigma(\mathbb{E}[X|Y]) \subseteq \sigma(Y)$
$\forall G \in \sigma(Y)$,

$$\int_G Z d \mathbb P := \int_G \mathbb{E}[X|Y] d \mathbb P = \int_G X d \mathbb P$$

Say $Y$ takes values in $\mathbb N$.

Then $\sigma(Y) = \sigma((Y=1),(Y=2),(Y=3), ...)$ and

$$\mathbb{E}[X|Y] = \mathbb{E}[X|Y=1]1_{Y=1} + \mathbb{E}[X|Y=2]1_{Y=2} + \cdots = \sum_{i=1}^{\infty} \mathbb{E}[X|Y=i]1_{Y=i}$$

2 implies that:

$\forall y \in \mathbb{N}$,

$$\int_{Y=y} Z d \mathbb P := \int_{Y=y} \mathbb{E}[X|Y] d \mathbb P = \int_{Y=y} X d \mathbb P$$

Hopefully, it is clear that in the above case and in general, if $Y=y$, then $E[X|Y=y] = \mathbb{E}[X|Y]$.


Is $(\sqrt{-x})^2$ equal to $x$ or $-x$? I have read How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$? and Is $\sqrt{x^2}=|x|$ or $=x$? Isn't $(x^2)^\frac12=x?$, but I can't understand what the value of $(\sqrt{-x})^2$ is when using complex numbers. My understanding is that it can be either

$x$, since

$$(\sqrt{-x})^2 = \sqrt{-x} \cdot \sqrt{-x} =  \sqrt{(-x)\cdot(-x)} = \sqrt{x^2} = x$$
or $-x$, since

$$(\sqrt{-x})^2 = (\sqrt{x} \cdot i)^2 = \sqrt{x}^2 \cdot i^2 = x \cdot (-1) = -x$$

Is this reasoning wrong? If so, why is it wrong and what is the correct solution?

This is because the property $\sqrt{xy} = \sqrt{x}\sqrt{y}$ that we're using does not hold for $x,y \in \Bbb C$, only for $x,y \in \Bbb R$. The complex square root is not a function per se, it is a multi-valued function.

We use the polar representation $x = re^{i\theta}$. This gives:

$-x = (-1)\cdot x = e^{i\pi} \cdot re^{i\theta} = r e^{i(\pi + \theta)}$

$\sqrt{-x} = r^{1/2} e^{1/2\cdot i (\pi + \theta)} = r^{1/2} e^{i(\pi/2 + \theta/2)}$

$(\sqrt{-x})^2 = r e^{i(\pi + \theta)} = -x$.


bijective $C^{1}$-image Let $F: \mathbb{R^{3}}\to \mathbb{ R}$ a $C^{1}$-image and assume that $(dF) (x, y, z)\neq 0$ as soon as $F (x, y, z) = 0$.
Define $O= \{(x, y, z) ∈ \mathbb{R^{3}}| F (x, y, z) = 0\}$.
Prove that every point of $O$ has an environment that looks like a piece of the xy plane in $\mathbb{R^{3}}$.
I think this means that I have to prove that for every point $(a, b, c) \in O$ there open parts $U, V \subset \mathbb{R^{3}}$ exist and a bijective $C^{1}$- image $\phi: U → V$ with $C^{1}$-inverse such that $(0, 0, 0) \in U, (a, b, c) \in V$ and $O \cap V = \phi (U \cap (\mathbb{R^{2}} \times {0})$
My attempt:
Let's take $(a,b,c) \in O$ random. Let's assume that (1x3)- matric $(dF)(x,y,z)$ is of rang $1$.
Because of the implicit function theorem there exist

an open part $V\subset \mathbb{R^{3}}$ so $(a,b,c)\in V$
an open part $U \subset \mathbb{R^{2}}$ with $(0,0) \in U$
a $C^{1}$-image $\phi:U\to \mathbb{R}$ with $\phi((0,0))=(a,b,c)$

and $ V \cap O=\phi(U)$ and the image of $(d\phi)(y)$ equals de kern of $(dF)(\phi(y))$ for all $y\in U$.
Because the condition that $(dF)(x,y,z) \neq 0$, $U$ needs to be a part of $\mathbb{R^{3}}$. This means that $(0,0,0) \in U$ and $O \cap V=\phi(U\cap (\mathbb{R^{2} \times {0} ))}$
I'm not that sure that the conclusions I made about the condition $(dF)(x,y,z) \neq 0$ are correct. Or maybe I started the question wrong. Can someone check my answer and see what I do wrong or right.

I have written an answer here, but am unsure if it is correct. Hence I am waiting there for verification.

Here is an alternative, more straightforward solution: By the regular level set theorem, $O=F^{-1}(0)$ is an embedded submanifold of codimension 1. Hence it has a smooth structure which makes it into a manifold of dimension 2.


Let $Y = \max(X, 0)$, then median of $Y$ is _____. Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = \max(X, 0)$ where $\max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.

My try:
Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.
Can you please explain in other way?

You have for $X$ a continuous random variable with median $0$

$P(X \le 0)=\frac12$
$P(X \gt 0)=\frac12$

so, since $X \le 0 \implies Y=0$ and $X\gt 0 \implies Y \gt 0$,

$P(Y =0)=\frac12$
$P(Y \gt 0)=\frac12$

so, moving towards the definition of a median,

$P(Y \le 0) \ge \frac12$
$P(Y \ge 0) \ge \frac12$

so the median of $Y$ is $0$

As mean =0 and variance  σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
Mean=Median=Mode.

As given that mean is 0.

Hence Median and mode will be 0


Non-uniqueness of solutions of the heat equation For the heat equation $(\partial_t-\partial_x^2)f(t,x)=0$ defined on $[0,T)\times(-\infty,\infty)$, to obtain uniqueness of the initial value problem, usually it is required to limit the growth of the potential solution at infinity, i.e. $|f(t,x)|<\exp(c\cdot x^2)$.
My question is, if we do not impose any such conditions, is uniqueness no longer valid? In particular, is there a well known example of a function $f(t,x)$ that satisfies the heat equation on $[0,T)\times(-\infty,\infty)$, $f(0,x)=0$, but $f$ is not identically zero?

What if we relaxed the conditions a little bit, and only required that $f$ satisfies the heat equation in $(0,T)\times(-\infty,\infty)$ and is continuous on $[0,T)\times(-\infty,\infty)$, is there an example in this case?

The type of solutions you are looking for were constructed by Tychonov. You can find them in chapter 7 of Fritz John's classical book Partial Differential Equations.

Uniqueness follows also from one sided inequalities like in Widder's theorem: if $f(x,t)$ is a non-negative solution of the heat equation with $f(x,0)=0$, then $f\equiv0$. You can find a proof again in chapter 7 of Fritz John's book.

Of course use separation of variables:

Let $f(t,x)=T(t)X(x)$ ,

Then $T'(t)X(x)-T(t)X''(x)=0$

$T'(t)X(x)=T(t)X''(x)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore f(t,x)=C_1x+C_2+\int_sC_3(s)e^{-ts^2}\sin xs~ds+\int_sC_4(s)e^{-ts^2}\cos xs~ds$ or $f(t,x)=C_1x+C_2+\sum_sC_3(s)e^{-ts^2}\sin xs+\sum_sC_4(s)e^{-ts^2}\cos xs$

This is already the general solution on $[0,\infty)\times(-\infty,\infty)$ , not only on $[0,T)\times(-\infty,\infty)$ .

If further $f(0,x)=0$ exist, it is the best to choose $f(t,x)=C_1x+C_2+\int_0^\infty C_3(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds$

Then substitute $f(0,x)=0$ :

$C_1x+C_2+\int_0^\infty C_3(s)\sin xs~ds+\int_0^\infty C_4(s)\cos xs~ds=0$

$C_1x+C_2+\mathcal{F}_{s,s\to x}\{C_3(s)\}+\mathcal{F}_{c,s\to x}\{C_4(s)\}=0$

$\mathcal{F}_{s,s\to x}\{C_3(s)\}=-C_1x-C_2-\mathcal{F}_{c,s\to x}\{C_4(s)\}$

$C_3(s)=C_1\delta'(s)-\dfrac{2C_2}{\pi s}-\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}$

$\therefore f(t,x)=C_1x+C_2+C_1\int_0^\infty\delta'(s)e^{-ts^2}\sin xs~ds-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_1x+C_2-C_1\Biggl.\dfrac{d}{ds}(e^{-ts^2}\sin xs)\Biggr|_{s=0}-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_1x+C_2-C_1x-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds=C_2-\dfrac{2C_2}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_4(s)\}\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s)e^{-ts^2}\cos xs~ds$


Over a commutative unital ring, does $\det A = 0$ imply that $A\mathbf{x} = 0$ has a non-zero solution? This is a standard theorem in linear algebra over fields that is still true when working over an integral domain.

  For a commutative unital integral domain $R$, let $A$ be an $n\times n$ matrix with entries in $R$. The system of linear equations $A\mathbf{x}=0$ has a non-zero solution if and only if $\det(A) = 0$.

The forward direction of this statement is not true though if we are working over a ring $R$ that is not an integral domain, a counter example being
$$
R = \boldsymbol{Z}_6
\qquad
A = \begin{pmatrix} 3 & 0  \\ 0 & 3 \end{pmatrix}
\qquad
\mathbf{x} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}
\,.
$$

But what about the other direction? Does $\det(A) = 0$ imply that $A\mathbf{x} = 0$ has a non-zero solution over an arbitrary commutative unital ring? I've played around for a bit, and have yet to find a counterexample.

Let $A$ be square of size
$n \times n$, and $\det A=0$. Then $A\,\textrm{adj}(A)=0$ where $\textrm{adj}(A)$ is the adjugate of $A$.
As long as $\textrm{adj}(A)\neq 0$ then taking some column of $\textrm{adj}(A)$ gives you $x$ with
$Ax=0$.

But what if adj$(A)=0$? In this case all minors of $A$ vanish. Find
a largest square submatrix of $A$ with nonzero determinant. We can assume
that it's size $r$ by $r$ and fills the top left corner of $A$.
Let $B$
be the top-left $r+1$ by $r+1$ submatrix of $A$. Then $\det B=0$ but
adj$(B)\ne0$. Let $y$ be the $(r+1)$-th column of adj$(B)$ and $z$ be the column
vector of height $n$ got by appending zeroes below $y$. Then $z\ne0$
(due to the non-vanishing of the top left $r$ by $r$
determinant) and I claim that $Az=0$.

The top $r+1$ entries of $Az$ are certainly zero. This is the identity
$B\,$adj$(B)=0$. If we look at another entry, say the $s$-th then
it is zero, essentially by replacing the bottom row of $B$ by the first $r+1$-th
entries of the $s$-th row of $A$, and noting that the new matrix has zero
determinant, as it is obtained from a submatrix of $A$ of size $r+1$
by $r+1$ by elementary row operations. As all submatrices of $A$ of this
size have vanishing determinant, this completes the proof that $Az=0$.

Consider $\pmatrix{2&3\cr 2&3}$ in $\mathbb{Z}/6$ its determinant is $0$, and $2x+3y=0$ implies that $2x=3y$. But the multiples of $2$ are $\{0,2,4\}$ and the multiples of $3$ are $\{0,3\}$.


If there are entire $G_k$s such that $f=\exp\circ\exp\circ\cdots \circ\exp\circ G_k$ ($k$ times), must $f$ be constant? I am a French guest and I hope that my English isn't too bad...

So here is my issue:
I consider an entire function $f$ which satisfies the following property for each complex number $z\in \mathbb{C}$:

$\forall ~ k \in \mathbb{N}^*$, there exists an entire function $G_k$ that satisfies
$$f(z)=\exp_k(G_k(z))$$
where $\exp_k$ denotes $\exp \circ \exp \circ ...\circ \exp$, $k$-times.

In other words, I can take (as many times as I wish) the $\log$ of my function $f$, and it will always give an entire function that doesn't vanish on $\mathbb{C}$.

Is $f$ a constant? (surely different from $0$...)

Thanks everyone!

(PS: This forum is so cool!)

I must say that at first I was quite skeptical  that such function could exist, but it turned out that one can prove its existence.  I have written down the proof but I will give here only sketch it (you should not have any problems computing it for yourself, nice exercise)

Let us define $b_1=2\pi \rm i m_1$ and $\displaystyle b_n=2\pi \rm i m_n+ \log b_{n-1}$ and
$$f_n(z)=\exp_{n}\left[b_n+ \frac{z}{b_1b_2\ldots b_{n-1}}\right],$$
where $\exp_{n}[z]=\exp\circ\cdots\circ\exp(z)$ $n$-times. One must find a sequence of  integers $m_k>0$ such that  the sequence  $f_n(z)$ will converge on every compact set. With some more effort one can prove that limit function $g(z)=1+z+ \mathcal{O}(z^2)$ have the desired property.

I think that the concept of 'order of an entire function' can help you to get an answer.
(See definition in W. Rudin : Exercise 2, Ch. 15 of ; or alternatively here : Wiki - Entire function )

Namely, your entire function $f$ is clearly of infinite order (well, at least for k>1), while constant functions -like polynomials- are entire functions of order zero ...

NB: of course I'm not considering the trivial case where the $G_k$ are constant (and non-zero), in which case $f$ would be constant as well ...


What is a constructive proof of  $\lnot\lnot(P\vee\lnot P)$? Glivenko's theorem says that $\lnot\lnot P$ is a theorem of intuitionistic logic whenever $P$ is a theorem of classical logic. Is it closely related to the so-called Gödel–Gentzen negative translation which embeds classical logic into intuitionistic logic.

Since $P\vee\lnot P$ is a well-known theorem of classical logic, I expect that by Glivenko's theorem, $\lnot\lnot(P\vee\lnot P)$ is provable in intuitionistic logic.  But I cannot find a proof! I must be overlooking something obvious.

If $\lnot\lnot(P\vee\lnot P)$ is indeed provable in IL, I would like to see a natural deduction or sequent calculus proof of it, or especially a construction of a typed $\lambda$-calculus term with the type $((P\vee(P\to \bot))\to\bot)\to\bot$.

If it is not provable in IL, what have I misunderstood?

Addendum: I cross-checked the proposition with the usual model, letting $P$ be a subset of $\mathbb R$, and interpreting $\vee$ as set union and $\lnot x$ as the interior of the complement of $x$, and the proposition $\lnot\lnot(P\vee\lnot P)$ did seem to come out as all of $\mathbb R$ for any choice of $P$, so if I have made a mistake here too I would like to know what subset of $\mathbb R$ is the counterexample.

Per the BHK interpetation, a proof of $(P \lor (P \to \bot)) \to \bot$ is pair of procedures, one for each disjunct in the hypothesis.

The first, $s_1$, is a procedure which takes a proof of $P$ as input and produces a proof of $\bot$. In other words, the first is a proof of $P \to \bot$.
The second, $s_2$, takes any proof of $P \to \bot$ as input and produces a proof of $\bot$. In other words, it is a proof of $(P \to \bot) \to \bot$.

By composing these to form $s_2(s_1)$, we obtain a procedure which takes no input and produces a proof of $\bot$.

The procedure just described (composition) takes any proof of $(P \lor (P \to \bot)) \to \bot$ and produces a proof of $\bot$. This means that $((P \lor (P \to \bot))\to \bot) \to \bot$ is provable.

This description shows how to write a $\lambda$ expression of the appropriate type. All that it does is to break the input into its two components and run the second one with the first as input. In particular, if we view $(P \lor \lnot P) \to \bot$ as a type $p(n,s)$ such that $\lambda s.p(1,s)$ is of type $P \to \bot$ and $\lambda\,s.p(2,s)$ is of type $\lnot P \to \bot$ then $$\lambda\, p(n,s) . p(2,\lambda\, s.p(1,s))$$ is of type $((P \lor \lnot P) \to \bot) \to \bot$

How about:
Define g:((A+(A->0)->0)->0  by g(f) = finr(finl)

The type of functions A->0 is the negation of A and  A+(A->0) is the
type of A or not(A). The construction of g  is the proof we want.
inl:A -> A+(A->0) is the left inclusion. inr is the right inclusion.
I hope it is clear now.  The notation comes (to me) from the HoTT book.


Six People Permutation in Linear Arrangement With Exceptions I can't seem to wrap my head around this. My professor hasn't done anything like this in class, and all I've found on the internet on this are circular arrangements.

We have Arnie, Betty, Carl, Debbie, Ernie, and Felicia

Ernie and Felicia refuse to be seated by each other AND Arnie and Betty refuse to be seated beside each other. How many line-ups are possible?

The idea would be to use the principle of inclusion and exclusion. Count all permutations ($6!=720$), then count ($=X$) all the permutations where E and F sit next to each other, count ($=Y$) all the permutations where A and B sit next to each other, finally count ($=Z$) all the permutations where A and B sit next to each other and E and F also sit next to each other.
The number you are looking for is then
$$720 - X - Y + Z$$
To calculate $X$, there are 5 ways to choose the neighbouring seats for E and F $((1,2), (2,3),\ldots,(5,6))$, 2 ways to seat E and F there, and finally $4!=24$ to seat the remaining four people in the remaining four seats. So we get
$$ X=5 \times 2 \times 24 = 240$$.
The calculation for $Y$ is exactly the same, as it is exactly the same problem, so we get $Y=240$.
To calulate $Z$, we need to determine which pairs of neighbouring seats are compatible for the (E/F) and (A/B) pairs. That's because if (E/F) sit in the seats $(1,2)$, then (A/B) can't sit in $(2,3)$.
So we get (while listing the earlier pair first) the following list of compatible pairs:

$(1,2)$: $((3,4), (4,5), (5,6))$
$(2,3)$: $((4,5), (5,6))$
$(3,4)$: $((5,6))$

This means there are $3+2+1=6$ pairs of 'compatible' neighbouring seats to take for (E/F) and (A/B).
We can assign the 2 pairs of seats to the 2 pairs of people in 2 ways. Once (E/F) has their pair of seats, they can sit there in 2 ways. The same goes for (A/B). The remaining 2 people (C and D) can also take the remaining 2 seats in 2 ways.
So we get
$$ Z= 6 \times 2 \times 2 \times 2 \times 2 = 96$$ (corrected due to comment)
So the number you are looking for, the number of permutations where neither E and F nor A and B sit next to each other is
$$ 720 - 240 - 240 + 96 = 336$$

Okay this is wrong. Leaving it up for the comments.

There are 2 times (6 choose 2) = 30 ways to pick where Ernie and Felicia sit.

Then 2 times (4 choose 2) =12 ways to choose where Arnie and Betty sit.

Then just 2 ways to seat the remaining Carl, Debbie

Multiply these together to get 720


Length of a curve without function? I need to find the length of this curve:

I don't have a function but I do have 3 sets of coordinates:

$(0, 51)$,
$(337, 674)$,  and
$(1022, 1022)$

The $(337, 674)$ set refers to the tangent and it makes the curve be equidistant.

I know nothing beyond basic math, however I'm working on this tablet pressure sensitivity thing and there's no other way out. I have to find the length of the curve or figure out the function which creates that curve.

Thank you fellas!

If you know (or can make good estimates for) other points on the curve, you can make decent approximations using triangles.

E.g. if you divide it up into 1, 2, or 4 triangles (you can use however many you want; more triangles is more accurate), it might look like this:

!!

To find the length of the curve, just add the lengths of the hypotenuses of the triangles. For each triangle, just take the $x$ length and $y$ length and plug them into $L = \sqrt{x^2 + y^2}$.

Use a Riemann sum for the arc length integral, making estimates of the slope at test points.

Sorry if you are unfamiliar with that vocabulary, but you say "I can't believe Math hasn't come up with a way to solve my issue", and in fact it has, and it uses this vocabulary.


Convert complex to polar cord Write the following numbers in the polar form $^{}, −<≤$:

$r$ = ?, $\theta$ =?

$$-\dfrac{\sqrt{7}(1+i)}{\sqrt{2}+ i} $$

When i used wolfram I got $r = 2.16025$ and $\theta = -170.264^{°}$. However, when I input this in WebAssign I got the right answer for $r$ but the wrong answer for $\theta$. I then tried doing $tan(-170.264^{°}) =.171579$ (in degree) and $tan(-170.264) = -.710889017$ (in radians), and they were both wrong. I'm lost what should be $\theta$ ?

Considering all factors in turn, the modulus will be

$$\frac{\sqrt7\sqrt2}{\sqrt3}$$

and the argument

$$-\pi+\frac{\pi}4-\arctan\frac1{\sqrt2}.$$

($-\pi$ is chosen rather than $\pi$ so that the sum be in the allowed range.)

Firstly, you can rewrite $\frac{−\sqrt7(1+i)}{\sqrt2+i}$ as $-\frac{\sqrt7+\sqrt14}{3} + i(\frac{\sqrt7 - \sqrt14}{3})$. Then use that $re^{i\theta} = r(cos(\theta) + isin(\theta))$ and you get that $$\begin{equation*}
 \begin{cases}
    r\cos(\theta) = -\frac{\sqrt7+\sqrt14}{3}\\
   r\sin(\theta) = \frac{\sqrt7 - \sqrt14}{3}
 \end{cases}
\end{equation*}$$
$$or$$
$$\begin{equation*}
 \begin{cases}
    \tan(\theta) = -\frac{\sqrt7-\sqrt14}{\sqrt7+\sqrt14}\\
   r^2 = \frac{14}{3}
 \end{cases}
\end{equation*}$$


Probability of tossing five coins and getting at least one head here is my dilemma. I want to know the probability of getting at least one head in five coins being tossed one after the other. Could you help me get the logic of this as it involves both mutually exclusive and independent events? Also, if possible please explain the logic with permutations and combinations. It would be much helpful.

Lenin

Hint: look at the complementary event, $A=\{\text{getting 5 tails}\}$. If you know how to compute $\mathbb{P}(A)$, then the probability you are looking for is $1-\mathbb{P}(A)$.

As an alternative (more complex) way of proving it, you can consider the 5 disjoint events $B_1,\dots,B_5$, where $B_i$ is defined as the event of getting exactly $i$ heads out of 5 (independent and fair) coin tosses. Then, observing that there are exactly $\binom{5}{i}$ sequences of $5$ draws containing $i$ heads (why?), and that each of them has probability exactly $p^i p^{5-i} = \frac{1}{2^5}$ (the last equality, as we assume $p=1/2$ since the coin is not biased), you get that the probability you are looking for is
$$
\mathbb{P}\left(\bigcup_{i=1}^5 B_i\right) =\sum_{i=1}^5 \mathbb{P}\left( B_i\right) = \frac{1}{2^5} \sum_{i=1}^5 \binom{5}{i} = \frac{1}{2^5}\left(\sum_{i=1}^5 \binom{5}{i} - \binom{5}{0} \right) = \frac{1}{2^5}\left(2^5 - 1 \right)
$$
where the first equality relies on the fact that the $B_i$'s are disjoint.

In an intuitive way, P(E)=$\frac{Favorable \ Outcomes}{Total \ Outcomes}$

So the to count the favorable outcomes: There are two possible outcomes for each toss. And there are 5 total tosses. So for each outcome there will be another outcome, so we'll multiply $2*2*2*2*2=2^5$

Sample Space or $n(S)=2^5$

Now think of getting one or more heads. We know that sum all probabilities is $1$. So we think of the other side. Why don't we calculate the number of possibilities for no head? And that would just be $\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{32}$

Probability of getting no head=$\frac{1}{32}$

So $P(E)=1-\frac{1}{32}$ =
$\frac{31}{32}$

So this probability will be the required one.


Fair coin toss and Bayes You have a coin and your prior assumption is that its probability of heads $\theta$ is chosen from a uniform distribution on $[0, 1]$. You toss the coin 10 times and get 6 heads. What is the estimate of $\theta$?
I figured that it has to be $\frac{6}{10}$ but is there a theorem or rule that can upend my guess?

One way to solve this is by using Maximum Likelihood Estimation. From Wikipedia (emphasis mine):

From the vantage point of Bayesian inference, MLE is a special case of maximum a posteriori estimation (MAP) that assumes a uniform prior distribution of the parameters (in our case, $\theta$).

Let $x_i$ denote the outcome (head = 1 / tails = 0) in a coin flip and let $X_n = (x_1, x_2, \dots, x_n)$ be a sequence of $n$ flips. We have that $m = \sum_{i=1}^n x_i$ is the number of heads in $n$ flips.
By Bayes' Theorem, we have that:
$$
f(\theta|X_n) = \frac{f(X_n|\theta) \cdot f(\theta)}{f(X_n)}
$$
Assuming $\theta \sim \mathsf{Uniform}(0,1)$ we get:
$$
f(\theta|X_n) \propto f(X_n|\theta)
$$
This means that the most probable $\theta$ (i.e. MAP) that explains the sequence $X_n$ can be found by maximizing the likelihood $f(X_n|\theta)$ (i.e. MLE).
The likelihood of observing sequence $X_n$ given that the coin is parametrized by $\theta$ is:
$$
f(X_n|\theta)=\theta^m \cdot (1-\theta)^{n - m}
$$
Now, the maximum likelihood estimator of $\theta$ is:
$$
\widehat{\theta}_{\text{MLE}}(X_n) = \arg\max_\theta \big( f(X_n | \theta) \big) = \arg\max_\theta \big( \log f(X_n | \theta) \big)
$$
Note that maximizing $f(X_n|\theta)$ is the same as maximizing $\log f(X_n|\theta)$ since $\log$ is monotonously increasing, and in this setting it's much more convenient to work with $\log$.
Therefore:
$$
\log f(X_n|\theta) = \log \big(\theta^m \cdot (1-\theta)^{n - m} \big) = m \cdot \log \theta + \left(n-m\right) \cdot \log(1-\theta)
$$
Now we can maximize:
$$
\frac{\partial}{\partial \theta} \log f(X_n|\theta) =
\frac{m}{\theta} - \frac{n-m}{1-\theta} =
\frac{m\cdot(1-\theta)-\theta \cdot (n-m)}{\theta \cdot (1-\theta)} = 0
$$
Which finally gives:
$$
\boxed{\theta = \frac{m}{n}}
$$

I don't know of any pertinent probability theorem; my knowledge of probability is limited.  Also, I have no opinion on whether Peter's comment is accurate.  I would attack the problem as follows:
Suppose that $n$ is a fixed integer (e.g. 10).  Further suppose that $x \,\in \{0, 1, 2, \cdots, n\}$ and assume that the chance of a heads on any one toss is $x/n.$  Calculate the probability of getting exactly 6 heads out of 10 tosses as a function of $x.$ Denote this probability as $P_n(x).$
Then, Bayes theorem should kick in.  That is, let $F_n(x)$ denote the (relative) probability that the chance of a heads on any one toss is $x/n.$  Then $F_n(x) = \frac{P_n(x)}{P_n(0) + P_n(1) + \cdots + P_n(n)}.$  This will give you the relative value of $F_n(x).$
I presume that the interpretation of "estimated probability for chance of a heads" is interpreted as $\sum_{x=0}^n \,\frac{x}{n}F_n(x).$
The above is a key point, where I may well be mistaken.  If so, someone, please advise.
Assuming that everything (so far) in my answer is accurate, then the ultimate answer should simply be the limit as $n \to \infty$ of the estimated probability, as a function of $n.$
It wouldn't surprise me if the calculation (as $n \to \infty$) results in the summation of an infinite # of very small ranges being replaced by an integration formula.


why is this economics parabola sideways? switch x and y axis Why is this "parabola" sideways?

Normally, f(x)=y, would be that x is graphed horizontally, and you would write that "wage rate is a function of hours worked", which is how I would normally think about wages and hours.  However, this graph seems to read that the hours worked are a function of the rate.

Perhaps a graph of the wage itself against hours might be more useful.

Is there some point to graphing this "sideways"?

(Of course, it might not be a parabola, I'm just guessing.)

The wikipedia entry says it's "S-shaped", which I interpret to mean it goes up and down, maybe more like a sine-wave or something along those lines.  But why sideways?

This is not a mathematics question but rather an economics question. It is tradition in economics to have in the Y-axis the independent variable and in the X-axis the dependent variable.

Here wage is the independent variable (set by market forces) and hours worked is the dependent variable (workers choose how much to work at the prevailing wages). It may not be realistic but this what the graph is suppose to depict.

The reason it bends backwards is that as wages become too high, workers prefer to work less and take more leisure time.

Please note that $y=f(x)$ meaning $y$ is a dependent variable and $x$ is an independent variable and this convention is always used. In economics traditionally for simplicity's sake we take $p=f(qd)$ or $p=f(qs)$ which is not correct. In economics the correct form i.e. $qd,qs=f(p)$ should be used.

Dr. M. Ahsanuddin


Why are addition and multiplication commutative, but not exponentiation? We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.

As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation
where
$H_0(a,b) = b+1$   (successor op)
$H_1(a,b) = a+b$   (addition op)
$H_2(a,b) = ab $   (multiplication op)
$H_3(a,b) = a^b$   (exponentiation op)
$H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )
etc.

Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$

Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?

Thanks.

My personal opinion is that the exponential is not naturally regarded as the next step in the progression from addition to multiplication, so there's no reason to expect it to share properties with the other two.

Notice what happens if you demand that all of your quantities have units. Addition is an operation you do to two quantities with the same units: for example, you can add two distances to get another distance. Multiplication is an operation you do to two quantities with units $a$ and $b$ to get a quantity with units $ab$; for example, you can multiply two distances to get an area.

However, you cannot exponentiate two unital quantities. In computing $a^b$, the quantity $b$ needs to be unitless or else there's no sensible way to assign a value to the result. This tells us that $a$ and $b$ are being treated very differently, so there's no reason to expect that switching them is a physically meaningful operation.

As I've discussed on math.SE before, the exponential has many generalizations in mathematics, and in most of them the base and the exponent are very different objects: in fact, in most of them the base is just $e$.

For any two positive real numbers, their product to the base 2 can be described as the exponential to the base 2 of the sum of their logs to the base 2. We could define another function $f$ for all positive integers $x$ and $y$ by $f(x, y) = 2^{2^{\log_2(\log_2x) + \log_2(\log_2y)}}$. This operation is commutative and distributes over multiplication. However, it is a different function than exponentation. The exponential to the base 2 of a product is not the product of the exponentials to the base 2. The exponential to the base 2 of a sum is the product of the exponentials to the base 2. On the other hand, multiplication distributes over addition. This shows that exponentation is not the double exponentation to the base 2 of the sum of the double logs to the base 2. Thus, exponentation doesn't necessarily have to be commutative.


Finding $\int_0^1 \left\{\frac1x\right\}\left\{\frac1{1-x}\right\}\left\{1-\frac1x\right\}dx$ where $\{x\}$ is fractional part of $x$ I was able to evaluate this from $0$ to $\frac{1}{2}$ but the middle part becomes too hard to evaluate under $\frac{1}{2}$ to $1$. I only have a high school level of understanding of maths, please make it easier to understand.

$$\int_0^1 \left\{\frac1x\right\}\left\{\frac1{1-x}\right\}\left\{1-\frac1x\right\}\,\mathrm{d}x=?$$

where $\{x\}$ is fractional part of $x$.

I haven't slogged through all the calculations, but I can explain how to do it.  Actually, if you were able to compute the integral from $0$ to $\frac12$, you've got the basic idea.  you just have to modify it to compute the integral from $\frac12$ to $1$.

The basic idea of course is to split $[0,1]$ into intervals on which the integer parts of the various expressions are constant.

$$\begin{align}
&\left\lfloor\frac1x\right\rfloor=n \iff\\
&n\leq \frac1x<n+1\iff\\
&\frac1{n+1}<x\leq\frac1n
\end{align}$$
In particular, the integer part is $1$ for $\frac12<x\leq1$, and of course the integer part of $1-\frac1x$ is $-1$.

Similarly, $$\begin{align}
&\left\lfloor\frac1{1-x}\right\rfloor=n\iff\\
&n\leq\frac1{1-x}<n+1\iff\\
&\frac1{n+1}<1-x\leq\frac1n\iff\\
&-\frac1n\leq x-1<-\frac1{n+1}\iff\\
&\frac{n-1}n\leq x<\frac n{n+1}
\end{align}$$
In particular, the integer part is $1$ for $0\leq x<\frac12$.

Therefore, we need to split the interval $[0,\frac12]$ up into intervals of the form $\left(\frac1{n+1},\frac12\right]$, which I assume you did, and we need to split the interval $\left(\frac12,1\right]$ up into intervals of the form $\left[\frac {n-1}n, \frac n{n+1}\right)$.

Consider the integral

$$I_k = \int_{\frac{1}{k+1}}^\frac{1}{k}\{\frac{1}{x}\}\{1-\frac{1}{x}\}\{{\frac{1}{1-x}}\}dx$$

For this range of $x$, we have

$$\{\frac{1}{x}\} = \frac{1}{x} - k \\
\{1 - \frac{1}{x}\} = k+1 - \frac{1}{x} \\
\{\frac{1}{1-x}\} = \frac{1}{1-x} -1$$

Now, substituting, we have

$$I_k = \int_{\frac{1}{k+1}}^\frac{1}{k}\left(\frac{1}{x} - k\right)\left(k+1 - \frac{1}{x}\right)\left(\frac{1}{1-x} - 1\right)dx$$

You can use partial fractions to decompose this into $\ln$ terms to get $I_k$, and then sum $I_k$ to get $I$


If $g(x) = \sqrt{10 − x}$ What is the Domain for $(g\circ g)(x) = g(g(x))$? What is the domain for $10 − \sqrt{10 − x} ≥$ ?

The expression you need to compute the domain of is
$$
g(g(x)=\sqrt{10-g(x)}=\sqrt{10-\sqrt{10-x}}
$$
Thus you have two conditions to fulfill:
\begin{cases}
10-x\ge0\\[1ex]
10-\sqrt{10-x}\ge0
\end{cases}
The first condition is equivalent to $x\le 10$; the second one to
$$
\sqrt{10-x}\le 10
$$
that can be squared, because both sides are surely non negative:
$$
10-x\le 100
$$
So the answer is…

$\sqrt{10-x}\le10$ gives $x\ge-90$ but be carefull $x\le10$ must be hold.


Use lagrange multipliers to calculate the maximum and minimum $f(x,y,z)=x^2y^2z^2$ constrained by $x^2+y^2+z^2=1$

$\nabla f_x$ $=$ $2xy^{2}z^{2}$,
$\nabla f_y$ $=$ $2yx^{2}z^{2}$,
$\nabla f_z$ $=$ $2zx^{2}y^{2}$

$\nabla g_x$ $=$ $2x$,
$\nabla g_y$ $=$ $2y$,
$\nabla g_z$ $=$ $2z$

setting the sets equal to each other and multiplying each to get the equality where $\nabla f_{x..z} = 2x^2y^2z^2$ gives me(namely multiplying $x$ on $\nabla f_x = \lambda g_x$...and so on) :

$\lambda2x^2 = \lambda2y^2=\lambda2z^2 \to x=y=z$ after division and taking the roots. So:

$x=y=z=a \to 3a^2=1 \to a= \pm \frac{1}{\sqrt{3}}$

Thus a maximum at $f\Big(\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}}\Big) = \frac{1}{27}$.

However looking at wolrfram alpha, it also gives a minimum at $f\Big(-\frac{1}{2},-\frac{\sqrt{3}}{2},0\Big) = 0$. I would like to know how there are more solutions from $\lambda 2x^2=\lambda 2y^2=\lambda 2z^2$ than just x=y=z

Using the corrections from posters lets say I let $x=0$ now I have $y^2+z^2=1$ Then if I wanted one scenario just to show there is a minimum I could say $y^2=z^2$ and then say $a^2+a^2=1 \to 2a^2=1 \to a^2=\frac{1}{2} \to a=\pm\frac{1}{\sqrt{2}}$ therefore have a minimum at $f\Big(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\Big)$ Is this correct. I feel the only correct solution thought is to use set notation and proof to show there are infinitely many cases. But does my above asnwer for one of those cases hold?

You have three equations from the first order conditions:

$$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z $$

Suppose $x=0$. Then the first equation is satisfied, and the other two equations imply that $\lambda=0$ (since we cannot have $x=y=z=0$ as that does not satisfy the constraint). This gives us infinitely many solutions with $\lambda=x=0$, since we can choose any $y$ and $z$ such that $y^2+z^2=1$. (Geometrically, these solutions are just intersections of the constraint set - a sphere of radius $1$ centred at the origin - and the $x=0$ plane. The intersection is a circle of radius $1$ centred at the origin in the $x=0$ plane.) Mathematically, the set of such minimizers is:

$$\{(0,y,z)\in\mathbb{R}^3\ \ |\ \ y^2+z^2=1\}$$

Similar arguments hold for the cases $y=0$ and $z=0$ (everything is symmetric). These critical points are all minimizers of $f$ on the constraint set. The minimum is $0$. (Notice that $\lambda=0$ here because even if we did not have the equality constraint, these points would minimize $f$.)

You lost all these solutions because you inadvertently assumed all the variables were nonzero when you divided the equations by one another.

Now suppose that $x$, $y$ and $z$ are all nonzero. Then $\lambda\neq0$ (this is implied by any of the three equations). Now since all the variables are nonzero, we can divide the equations by $2x$, $2y$ and $2z$ respectively to get

$$y^2z^2=\lambda  \qquad x^2z^2=\lambda \qquad x^2y^2=\lambda.  $$

Then dividing each equation by each other equation we find that we require $$x^2=y^2=z^2.$$

Plugging into the constraint then implies that $$x^2=y^2=z^2=\frac{1}{3}$$

and $\lambda =1/9$ (from any of the three equations). This gives us $8$ solutions, which are maximizers of $f$ on the constraint set. The maximum is $1/27$.

You didn't consider the case $x = y = \lambda = 0, z = 1$  which satisfies the nabla equations, and gives the minimum value for $f$ of $0$.


How to prove the convergence of $u_{n+1}=\frac{\sin(u_n)}{n+1}$? I am first asked to prove that for all n, $u_n \in [0,1]$ for $u_0=1$. This is my attempt:

Let's first study the function defined as $f_x(n)=\frac{\sin(x)}{1+n}$ where $x$ is fixed and $x \in[0,\frac{2}{\pi}]$. We can immediately see that $0\leq f_x(n) \leq \frac{\pi}{2(n+1)}$.

We also know that $\sin$ is strictly decreasing when $x$ tends to $0$ on the interval [0,1]. Thus we can now say a few things:

$f_x(n)$ is decreasing
As $f_x(n) \leq \frac{\pi}{2(n+1)}$
$\forall n \geq 1$ $\frac{\pi}{2(n+1)}\leq 1 $
Thus $\sin(\frac{\pi}{2(n+1)})$ is decreasing
Thus $\sin(\frac{\pi}{2(n+1)}) \leq \sin(\frac{\pi}{2n}) $
Thus $u_{n+1} \leq u_n$
Finally as $u_0=1$ we have $u_n \leq 1$
As $\sin(x) \geq 0 \forall x \in [0,\frac{\pi}{2}]$ we have $un>0$, thus $u_n \in [0,1]$ for all n.

I think my reasoning can be pretty confusing, but I am incapable of finding something clearer. My thoughts are pretty disorganised for such cases.

Now I need to show that $\sum u_n$ converges. But here I don't know how to proceed on doing this.

Here is a straightforward approach, if you want to have an answer following your approach, then leave a comment.

We prove inductively that $0≤u_n≤\frac{1}{n!}$ for all $n\in\mathbb{N}_0$. This is true for $n=0$, so assume it to be true for an $n\in\mathbb{N}_0$ fix. Then using that whenever $0≤x$ then $0≤\sin(x)≤x$ we obtain
$$
0≤u_{n+1}=\frac{\sin(u_n)}{n+1}≤\frac{u_n}{n+1}\le\frac{1}{(n+1)!}.
$$
Thus the inequality holds for all $n$. By the squeeze theorem we obtain that $\lim_{n\to\infty}u_n=0$ and by comparison we obtain the convergence of $\sum_{n=0}^{\infty}u_n$.

Given that $u_0 = 1$, and
$$u_{n+1} = \frac{ \sin u_n }{n+1} \ \mbox{ for } \ n = 0, 1, 2, 3, \ldots,$$
we note that $u_0 \in [0,1]$, and if $u_n \in [0,1]$, then $$0 \leq u_n \leq 1 < \frac{\pi}{2},$$ which implies that
$$0 \leq \sin u_n < 1,$$
and hence $$0 \leq u_{n+1} < \frac{1}{n+1} \leq 1$$
for all $n = 0, 1, 2, 3, \ldots$. Here we have used the fact that the sine function is strictly increasing on $[0, \frac{\pi}{2}]$ and that $\sin 0 = 0$ and $\sin \frac{\pi}{2} = 1$.

Now, using the auxiliary function $f \colon \mathbb{R} \to \mathbb{R}$ defined by $f(t) = t - \sin t$ for all $t \in \mathbb{R}$, we note that
$$\frac{\mathrm{d} f(t)}{\mathrm{d} t} = 1- \cos t > 0$$ if $0 < t < 2 \pi$, showing that this function is strictly increasing on $[0, 1]$. Now $f(0) = 0$. So $f(t) \geq 0$ for all $t \in [0, 1]$. That is, $$t \geq \sin t \ \mbox{ for all } \ t \in [0, 1].$$

Thus, we obtain
$$u_{n+1} = \frac{\sin u_n}{n+1} \leq \frac{u_n}{n+1} \leq u_n$$
for all $n = 0, 1, 2, 3, \ldots$, thus showing that this sequence is monotonically decreasing.

Thus we have a monotonically decreasing sequence that is also bounded below. Hence this sequence is convergent.

Let $u$ denote the limit of this sequence. Then $u$ must satisfy
$$0 \leq u \leq 1 \ \ \mbox{ and } \ \ u = \lim_{n \to \infty} u_{n+1},$$
and therefore $u$ must be $0$.


Relationship between eigenvalues of matrix squared Let $A$ be symmetric positive definite. Suppose $\lambda$ is the largest eigenvalue of $A^T A $. Then we know $\lambda(A^T A ) = \lambda(A^2 )$

Does it follow that $\lambda(A^2) = [\lambda(A)]^2 $ ?

Hint: If $A=U\Lambda U^T$ where $\Lambda$ is the diagonal matrix of eigenvalues and $U$ is the orthonormal matrix of eigenvectors, then $A^2=U\Lambda^2 U^T$ (can you prove this?). Let $\lambda_{max}(.)$ denote the largest eigenvalue. Thus $\lambda_{max}(A)=\lambda_{max}(\Lambda)$. Also, $\lambda_{max}(A^2)=\lambda_{max}(\Lambda^2)$. Now can you think of any situation where $\lambda_{max}(\Lambda^2)\leq\lambda_{max}(\Lambda)^2$ for any diagonal matrix $\Lambda$.


How do Peano Axioms imply "nextness" with the successor? Going with this explanation of Peano's Axioms, I cannot understand how/where the successor function is definitively stated to be the very next number in the case of natural numbers. In this treatment, it says

  The successor of $x$ is sometimes denoted $Sx$ instead of $x′$. We
  then have $1=S0$, $2=S1=SS0$, and so on.

Again, I don't see from the axioms how very next is guaranteed. Neither further on when addition is defined "based on the axioms." It seems a successor function could map "nextness" in many ways, e.g. $Sx$ could be $x+56$, or whatever as long as $Sx$ was something "further up the natural number line." So yes, I know I'm missing something here. . . .

If you just restrict yourself to the first axioms listed on that page, then indeed you could say that the expressions $0,S(0), S(S(0)), ...$ denote $0,2,1,4,3,6,5,....$ respectively. That would still satisfy the axioms. But you would still have a structure that is isomorphic to the natural numbers.

Also, if you define:

$$\forall x \forall y (x < y \leftrightarrow (y=S(x) \lor \exists z (y = S(z) \land x < z))$$

(which I would think captures the notion of $<$),

then using only those very first axioms, you can prove that:

$$\forall x \neg \exists y (x < y \land y < S(x))$$

i.e. that there is no number between any number and its successor ... and thus that the successor indeed gives you the very next number.

Finally, if you use the addition and multiplication axioms stated later on on that website, then you can prove some results that are certainly very suggestive of $S(0)$ working in a way consistent with our concept of the number $1$.

For example, it follows from the Peano Axioms that $$\forall x ( x · S(0) = x)$$ .... which would make sense if $S(0)$ would take the role we normally reserve for $1$ rather than, say, $56$

Likewise, you can prove from the Peano Axioms that $$\forall x ( x · S(S(0))= x + x)$$ ... again, that would square with the way we think about numbers if $S(S(0))$ is seen as $2$, rather than $117$.

First, I am not alone in asking this question of "nextness." The following is from the Wikipedia treatment, Peano Axioms:

  However, considering the notion of natural numbers as being defined by
  these axioms, axioms 1, 6, 7, 8 do not imply that the successor
  function generates all the natural numbers different from 0. Put
  differently, they do not guarantee that every natural number other
  than zero must succeed some other natural number.

...which goes on to use the often included axiom of induction to rework the successor function into implying nextness as always "one more." But then the article describes addition as defined recursively by two phantom Peanos:

\begin{align}
    a + 0 & = a \\
 a + S(b) & = S(a + b)
\end{align}

If the second expression is considered an axiom, then yes, $S$ would have to be consistently a "one more" successor. That is, there cannot be an $S(b) + a$ that is different from an $S(c)$ where $c = a+b$. However, these addition axioms are not really original to the PAs.

We might fall back on insisting that a successor function cannot from one application to the next behave differently, i.e., (allow me to switch successor notation) $m \succ m'$ implying $m'-m \geq 1$, and then $n \succ n'$ yielding $(n'-n) \neq (m'-m)$ is a violation of the spirit of the successor function and its brother induction.

It is true that $m=n \iff m'=n'$ implies uniqueness of the successor function only in the sense of injection (as pointed out in a comment by Somos). We might say a successor function must be an injective function producing distinct results, e.g.,  $n'$ is distinct from the successors which produced $1,...,n$, but I'm still doubtful this implies "one more," i.e., $n'$ to $n''$ proceeds one "very next thing" in a strict order at a time.

However, we might be rescued by Kleene's (Introduction to Metamathematics; p.20) example where he leverages the PA that $x' \neq 0$, or, $0$ is not the successor to any natural number. We might set up a contradiction: $$0''''=0''$$ but this would be true only if $$0'''=0'$$ Applying this logic again, $$0'''=0'$$ only if $$0''=0$$ So yes, this is a contradiction; hence, we have butted up against the lower boundary of the successor concept, which, therefore, implies a successor being greater than since we can run any successor-based induction of $\mathbb{N}$ backwards to $0$. Ironically, this is a sort of reverse-induction to prove that the Peano successor is consistently "one more" demanding some sort of $\mathbb{N}$-like order.

All in all, I'm still stuck on this....


Convexity and Affineness In reading about convex optimization, the author states that all convex sets are affine.  Are affinity and convexity equivalent?  If I understand, both definitions incorporate the notion that a set is affine/convex iff for every two points in the set, the line connecting them is also in the set.  It seems as if they are as their respective definitions are almost identical.  I would appreciate a delineation of the differences between affine sets and convex sets.

A set $S$ is convex iff for every pair of points $x,y\in S$, the line segment $\overline{xy}$ joining $x$ to $y$ is a subset of $S$. $S$ is affine iff for every pair of points $x,y\in S$, the whole infinite line containing $x$ and $y$ is a subset of $A$. In the xy-plane, for instance, $S=\{(x,0):0\le x\le 1\}$ is a convex set but not an affine set: the smallest affine set containing $S$ is the whole $x$-axis.

Mathematically, $S$ is affine iff it contains every affine combination of its points, where an affine combination of the points $x_1,\dots,x_n\in S$ is any point of the form $$a_1x_1+a_2x_2+\cdots+a_nx_n$$ such that $$a_1+a_2+\cdots+a_n=1\;.$$

$S$ is convex iff it contains every convex combination of its points. Convex combinations are the special case of affine combinations in which all of the coefficients are non-negative. That is, a convex combination of the points $x_1,\dots,x_n\in S$ is any point of the form

$$a_1x_1+a_2x_2+\cdots+a_nx_n$$ such that $$a_1+a_2+\cdots+a_n=1$$ and $$a_1,a_2,\dots,a_n\ge 0\;.$$

You can think of a convex combination of points as a kind of weighted average of those points; an affine combination is then a weighted ‘average’ in which some of the weights are allowed to be negative. In particular, $$\{ax+by:a+b=1\text{ and }a,b\ge 0\}$$ is simply the set of points on the line segment from $x$ to $y$; the point $ax+yb$ is $b$ fraction of the way from $x$ to $y$. The set

$$\{ax+by:a+b=1\}\;,$$

on the other hand, includes the whole line through $x$ and $y$. The point $-x+2y$, for instance, does not lie between $x$ and $y$: instead, $y$ is between $x$ and $-x+2y$.

A set S in R^n is said to be convex if we consider any two distinct points from this set, then the line segment through these points must lie in the set we considered.On the other hand, By an affine set we mean a set A in R^n, in which if we consider any two distinct points from this set, then the line passing through these points must lie in the set we considered.
Clearly the difference is---- line through any two points and line segment any two points.


differentiation integral symbol I hopefully have a question which should be obvious to all but has been bothering me for sometime.

I have used the symbol $d$ for the differential entity of integration, for example, $$\int{x}{dx}=\frac{1}{2}x^2+c$$

Common to partial differential equations, I have typically used $d$ or $D$ to represent the total (or material or convective...) derivative as opposed to using $\partial$ to represent the partial derivative of a value.

My question is: why is $d$ still typically used when integrating partial differential equations, when in fact, I would argue that I am integrating with respect to a partial differential?

As an example, shouldn't $$\int_{x_{0}}^{x_{1}} \frac{\partial f(x,y)}{\partial x} dx$$
be written as $$\int_{x_{0}}^{x_{1}} \frac{\partial f(x,y)}{\partial x} \partial x$$
especially since I am only integrating over $x$?

Am I missing something, or is this just overlooked?

Thanks,
Jon

First, the derivative should always be written using an upright 'd' as in

$$ \int f(x) \mathrm{d}x $$

To address your question regarding the integration of a multivariable function, the integration can be seen as the integral with respect to each variable, just over a singleton bound.

$$ \int f(x,y)\mathrm{d}x = \int_{x_0}^{x_1}\int_{y_0}^{y_0}f(x,y) \mathrm{d}y \mathrm{d}x $$

So, are you doing a 'partial' integration? Not really. When you integrate only with respect to one variable, you're just integrating with respect to the second variable as well, but over an infinitesimal bound.