Logic and Sets Expressions The logical operations $\Rightarrow$ and $\Leftrightarrow$ correspond to relations on sets. What are these two relations? Is there anything in logic that could not be expressed in terms of set theory and vice versa?

To say that $P(x)\Rightarrow Q(x)$ is to say that if $x$ satisfies property $P$ then it satisfies property $Q$.

One can see this as $x\in P\Rightarrow x\in Q$, or in a compact notation: $P\subseteq Q$.

From this you can infer that $P\Leftrightarrow Q$ corresponds to $P\subseteq Q\land Q\subseteq P$, which essentially to say $P=Q$.

Similarly, if we think of $P(x)$ as $x\in P$, we have that $P(x)\land Q(x)$ is the same as to say $x\in P$ and $x\in Q$, or $x\in P\cap Q$; the $\lor$ corresponds the same way with $P\cup Q$; and lastly $\lnot P(x)$ is to say that $x\notin P$.

I will make a short story.... very long. I hope you will find it interesting/illuminating, but if you cannot really bear it, jump right to the last few lines for a summary.

I think one should have clear in mind the distinction between logic and ...anything else.
Logic comes first. Logic is the science of reasoning and as such it can be applied to anything that requires reasoning.
Set theory and mathematics (which is based on set theory) are just sample applications.
You (can/should) use logic in physics, law, politics, biology, criminal investigations, love matters, ect
We can say: mathematics is the language of nature, but logic is the language of mathematics therefore - not surprisingly - logic is the language of nature.  It may also be used as the language of imagination (as in model  theory, in a sense).
To create a logical theory of something, just decide what are your axioms and anything else will follow "logically".
The problem consists in creating the axioms for a specific application.
Unfortunately in the world education systems logic is taught after law, biology, mathematics,...if ever, under the assumption that people can instinctively reason clearly and correctly. This is often true...and often not.
Now, "set theory".
There are several set theories, some are even incompatible with each other. The first was even inconsistent (naive set theory), that is, it led to contradictory statements.
How do all these theories look like? well, nothing special in a sense: they are just a bunch of axioms with variables x,y,z... They differ from one another in the specific bunch of axioms. (Notice that I avoid saying "set" of axioms). You can see ZFC or NBG or Grothendieck-Tarski set theories formally described in Wikipedia. Nowhere it is written that the variables are called "sets". You just call them that way for lack of a better term. Actually in NBG they are supposedly called classes.
You look at these axioms and see ⇒ and ⇔ and affirm that they "correspond" to relations on sets. That is not really correct. All you can/should say is that the particular set theory that you are looking at uses the logical connectives  ⇒ and ⇔ in its axioms. But that is hardly surprising is it? It also uses "and" and "or" connectives for that matter. So what? Set theories are logical theories so they use logical connectives such as ⇒ and ⇔ and quantifiers. That's all.
Now we come to your question. Is there something in logic which is not a set? Of course! Some examples:
Take ZFC as your set theory and ask the theory if the collection of sets equal to themselves exists. More specifically: does an y exists such that all members x of y satisfy the relation x=x ? Well, following logic rules  from the axioms, the answer is no. So here we have an example of relation (x=x) which does not "correspond" to a "set".
Ok, we can change set theory to suit our needs. Take NBG theory. Here the y exists and it is modestly called "the universe" U. It is a class, a proper class. So NBG allows "bigger" collections of things than NFC and you may feel that NBG is all you need, but wait. Consider the collection made of just one element U. That is {U}. Simple collection apparently. Just one thing inside. Yet this collection does not exists in NBG! That’s because U is a proper class, which means it cannot be a member af another class. So the collection {U} cannot be a class. Ok, we can change set theory again to suit our needs. We use Grothendieck-Tarski (GT) theory. Now we can talk of many universes and collections of universes. The nice thing is that now everything is a "set" in some universe. it is a little bit like saying :"you can be anything you want in your imaginary parallel universe". Now it seems at last that GT has  a set for everything you might consider in logic, but wait...read the small print on the label: "Only works with pure sets". Pure sets are sets built on…the empty set. Oh dear!  After all this chat you discover that all “sets and classes and…” described in these grand theories are just built on …nothing really. The empty set. This miracle of imagination. Much ado about nothing.  The only constant symbol present in any set theory is the empty set. So , for example elementary particles do not exist even in GT (even if you define an elementary particle as a quantum field, you need to add the axioms of quantum field theory to “recover” the elementary particle).  GT alone is not enough to describe everyday things like electrons or people.  So you take a relation or a predicate in logic – for example Human(x) (meaning x is a human) and you think it describes a collection of all objects which are humans and ask yourself: is this a set in GT? The answer is no, because it is a collection of objects (people) which are not sets, so it cannot be a set. All you can say is that relations or predicate in logic describe “collections” (All x such that….) but that these collections in general do not follow the axioms of any “set theory”. We may further  say that GT and friends describe the axioms of collections of the membership relation which is thereby defined. These collections are called “sets” or “classes” and are built out of just one constant called “the empty set”.

To summarize:
Logic is the science of reasoning and is used everywhere reason is used. You create axioms, put them together in theories  and deduce some conclusions from them. You can create collections in logic (All x such that….). These collections should not in general be called anything else than “collections”. This is because there are certain theories called “ xyz set theory” which have given the word “set” a specific technical meaning, so they have “robbed” us of the term “set”.  These theories describe very special collections ultimately made of just one collection which has no members.


3-colored Icosahedron out of 90 Sonobe units I'm new to Modular-Origami. Currently I'm trying to build a Icosahedron out of 90 Sonobe units. I want to make it three-colored. But I'm having trouble building it so that there are no 2 colors next to each other. I know that it's possible for 6, 12 and 30 units.

But is it even possible to make a 3-colored Icosahedron out of 90 Sonobe units? And why if / if not? If it should be possible, is there any trick of building the Icosahedron? Is Algebra the solution or Graph theory?

Greetings

I think I can show you an example of 3-colored Dodecahedron. Sorry, not an Icosahedron. I've never seen Icosahedron of 90 Sonobe units. The white circles show the only axis of symmetry.


Proving equivalence of expectations I need to show that the expectation $\mathbb{E}_{X,Y}[(Y-h(X))^2]$ can be written in the form $\mathbb{E}_X[\dots] + \mathbb{E}_{X,Y}[\dots]$, but I am unsure how to proceed.

$X$ and $Y$ are jointly distributed r.v.'s, and $h(x)$ is an arbitrary function taking $X$ as input and returning a scalar. ("$\dots$" represents expressions that are not specified here but need to be determined).

Namely, I need to show:
$$
\mathbb{E}_{X,Y}[(Y-h(X))^2] = \mathbb{E}_X[(E[Y|X] - h(X))^2] + \mathbb{E}_{X,Y}[(Y-E[Y|X])^2]
$$

How might I proceed in such a proof dealing with these expectation expressions?

One trivial way is $E_X[0] + E_{X,Y}[(Y-h(X))^2]$. But presumably they want you to expand the square $(Y-h(X))^2$ and put any terms that do not involve $Y$ into the first expectation, and put the rest in the second expectation.

\begin{align}
&E[(Y-h(X))^2]\\
&= E[(Y-E[Y \mid X] + E[Y \mid X] - h(X))^2]\\
&= E[(Y-E[Y \mid X])^2] + E[(E[Y \mid X]-h(X))^2] + 2 E[(Y-E[Y \mid X])(E[Y \mid X]-h(X))].
\end{align}
It remains to show the third term is zero.
\begin{align}
&E[(Y-E[Y \mid X])(E[Y \mid X]-h(X))]\\
&= E\Big[E\big[(Y-E[Y \mid X])(E[Y \mid X]-h(X)) \mid X\big]\Big] & \text{tower rule}\\
&= E\Big[(E[Y \mid X]-h(X))E\big[(Y-E[Y \mid X]) \mid X\big]\Big] & \text{$E[Y \mid X]-h(X)$ are "constants" given $X$}\\
&= E\Big[(E[Y \mid X]-h(X))\underbrace{(E[Y \mid X] - E[Y \mid X])}_{=0}\Big] & \text{linearity, tower rule}\\
&=0.
\end{align}

Remark: This is a "Pythagorean theorem." Essentially, $E[Y \mid X]$ is the "orthogonal projection" of $Y$ onto the space of $X$-measurable functions. Also see the law of total variance.


Why is the square root of a negative number impossible? What is the algorithm to get the square root of a number?
Why does it not accept / allow for negative numbers?

I've always been told that the square root of minus one does not exist by my maths teacher, but I don't really get why?

JChau asked in a separate question if it's ever possible for the square root of a number to be negative, and another user moved for that to be closed as a duplicate of this one. It has since been deleted, but here is my answer to that other question, which is also pertinent here.

We say $x$ is a "square root" of $y$ if $x^2=y$. Thus, both $+7$ and $-7$ are square roots of $49$.

However for positive reals $x$, by definition the square root function applied to $x$ yields the positive square root. Often one will abbreviate "the square root function applied to $x$" or equivalently "the positive square root of $x$" as simply "the square root of $x$," if no confusion should arise. Therefore we have $\sqrt{49}=+7$, despite $-7$ also being a square root.

The square root function, like all bona fide functions, is single-valued rather than multi-valued, so if we were tasked with creating our own square root function from scratch we would have to make a choice between the two square roots of every positive number as the value the function takes; if we want to further impose continuity (and, subsequently, smoothness for $x>0$), we would end up having to set $\sqrt{x}$ to either always be the positive square root or always the negative square root. At this point it's an understandable choice to make it always the positive one.

The same kind of "having to make a choice" situation arises if one wants to define a square root function for complex numbers. We can no long impose the same kind of continuity conditions and get a straight answer - instead we have to form a sort of "barricade" in which the value of the square root jumps dramatically when we cross over this barricade. This is known as a branch cut.

The standard branch in $\Bbb C$ is where we consider the negative real axis as part of the quadrant above it but not part of the quadrant below it. In this setting, complex numbers when written in polar coordinates will have a phase (angle) in the interval $(-\pi,\pi]$ (note if you cross over the negative real axis, the phase will jump from one side of this interval to the other).

The standard branch normally comes up in the discussion of the logarithm, but it is connected to taking powers with complex numbers because $z^w:=\exp(w\ln z)$ for complex $w,z\in\Bbb C$ ($z\ne0$). The logarithm will be defined by $\ln (re^{i\theta})=(\ln r)+i\theta$, so the imaginary part of a logarithm will depend on which branch we have chosen. The default choice, usually unspoken, is the standard branch.

Using the standard branch, we have $\sqrt{re^{i\theta}}=\exp(\frac{1}{2}(\ln r+i\theta))=e^{\frac{1}{2}\ln r}e^{i\theta/2}=\sqrt{r}e^{i\theta/2}$. Thus the phase of $\sqrt{z}$ will be in $(-\pi/2,\pi/2]$ for all $z\in\Bbb Z\setminus0$. This precludes $\sqrt{z}$ from ever being a negative real number, or even to the left of the imaginary axis. However, other nonstandard choices of branch cuts can lead to $z^{1/2}$ taking values on the negative real axis.

Another word for $\ln$ and $\sqrt{}$ with the standard branch is the principal value.

The idea of branch cuts leads into more advanced complex analysis topics of monodromy (which pertains to "running around" a singularity, like crossing over the branch mentioned earlier) and also Riemann surfaces, which can be thought of as what we get when we refuse to cut the plane into branches and instead consider a function multi-valued and look at its graph (I am probably butchering that description though).

It is impossible to find the square root of negative one, or the square root of any negative number, because no number times itself can equal a negative number. For instance, if I try to find the square root of negative one, I start by attempting to multiply -1*-1, but that would give a solution of one, since a negative times a negative equals a positive. The next step is to multiply 1*-1 or -1*1, but while those equations do come up with an answer of one, 1 and -1 are not the same number; they just have the same absolute value, but if you try to multiply the absolute values, you will get positive one because positive*positive=positive. Same goes for -2,-3,-4,-5,-6, and every negative number. However, the square root of negative numbers is used in scientific and advanced mathematical calculations as represented by the letter i, which basically means "imaginary number" because it does not exist, however, how scientists could use i in calculations pertaining to the real world is puzzling, because i does not exist.


When should I classify a statement as "undefined" versus "vacuously true/false": implications for induction Although this question has sort of been asked before (Why aren't vacuous truths just undefined?), I think my question is emphasizing a somewhat different aspect. Rather than knowing "why" vacuous truths aren't just undefined...I want to know when one should treat a statement as undefined, instead of vacuously true (or vacuously false).

In Tao's Analysis I, there is the following passage:

  "To summarize so far, among all the objects studied in mathematics, some of the objects happen to be sets; and if $x$ is an object and $A$ is a set, then either $x \in A$ is true or $x \in A$ is false. (If $A$ is not a set, we leave the statement $x \in A$ undefined; for instance, we consider the statement $3 \in 4$ to be neither be true or false, but simply meaningless, since 4 is not
  a set.)"

So while this certainly makes sense...I am reluctant to claim that I comfortably know when to declare something as being vacuous versus undefined. If I were to rephrase this slightly, I could say: Prove to me that $3$ is not an element of $4$.

...Well, unfortunately, $4$ is not a set so I have no way of disproving your claim. However, this sounds vaguely similar to any sort of argument that invokes the "vacuously true" approach. (And for the situation of a base case of an induction proof, it is quite important to make the distinction between "undefined" and "vacuously true", I would think!) Could anyone offer an explanation that will clear this up for me?

Cheers~

As I understand, in classical logic, if the antecedent of any implication is false, then that implication is said to to be vacuously true. For any object $x$ and proposition $P$, for example, we have, the implication $x \in \emptyset \implies P$ being (vacuously) true. Likewise, $0=1 \implies P$.

A function $f: A \to B$ is said to be undefined for some object $x$ means only that $x \notin A$. I have never heard of a syntactically correct logical proposition being "undefined" or "meaningless."


Linear independency of vectors in Euclidean space I'm having trouble finishing my proof for the following problem:

  Let $V$ be a Euclidean vector space. Let $n\geq 1, v_1, ..., v_n, w \in V$ with the following properties:
  $$\forall 1\leq i,j\leq n, i \neq j: \langle v_i, w \rangle > 0 \text{ and } \langle v_i, v_j \rangle \leq 0.$$
  Show that $v_1, ... v_n$ are linearly independent.

Proof:
Suppose $\sum_{i=1}^n \lambda_iv_i = 0$ for $\lambda_i \in \mathbb R$.
We have $$0 = \langle \sum_{i=1}^n \lambda_iv_i, w\rangle = \sum_{i=1}^n \lambda_i \langle v_i, w \rangle$$
(I don't know if I get information out of this. I at least know that $\lambda_i$ can't be all positive since $\langle v_i, w \rangle > 0$ for all $i$)
But also, for all $1\leq j \leq n$: $$0 = \langle \sum_{i=1}^n \lambda_iv_i, v_j \rangle = \sum_{i=1}^n \lambda_i \langle v_i, v_j \rangle = \sum_{i=1\\i\neq j}^n \lambda_i \langle v_i, v_j \rangle$$
So we in fact have one summand less than in the first equation. Does setting them equal help? Assumably I am nearly done, but I don't see how to conclude. Any help appreciated!

First, it is important to interpret the given properties. While $\langle v_i,v_j\rangle\leq 0$ for $i\neq j$ refers to an angle of at least $90$ degree between each vectors $v_i$ and $v_j$ with $i\neq j$, you have $\langle v_i,w\rangle>0$ which yields an angle of less than $90$ degree between $v_i$ and $w$ for all $i$. If you try to imagine what the consequences are for $n=2$ or $n=3$ you get an idea for the proof.
The proof is seperated in three parts and maybe there is a shorter version...
The main idea is a proof by induction. If we consider $v_1,\ldots,v_{n+1}$ by induction hypothesis we get $v_1,\ldots,v_n$ linear independent. If $v_1,\ldots,v_{n+1}$ is not linear independent, we get $\lambda_1,\ldots,\lambda_n$ such that
$$
v_{n+1}=\sum_{i=1}^n\lambda_iv_i.
$$
From the fact $\langle v_{n+1},v_i\rangle\leq 0$ for $1\leq i\leq n$ we like to conclude $\lambda_i\leq 0$ for $1\leq i\leq n$, such that we get a contradition to $\langle w,v_i\rangle>0$ for $1\leq i\leq n+1$. But the problem is, that $v_1,\ldots,v_n$ are not orthogonal. Therefore we use the trick to orthogonalize the vectors and check, that the properties still holds. This is the context of claim 1 and proposition 1. Using this idea yields the proof of your statement at the end.
Claim 1: Let be $n\geq 2$ and $v_1,\ldots,v_{n-1}$ linear independent. If $u_1\ldots,u_{n-1}$ is the corresponding set of orthogonalized vectors, using the Gram-Schmidt process, then $\langle v_n,u_i\rangle\leq 0$ and $\langle w,u_i\rangle>0$ for $1\leq i\leq n-1$.

Proof of claim 1: Prove by induction.
Base case $n=2$: Since $u_1:=v_1$ the statement is obvious.
Inductive step: Let be $n+1>2$ and $v_1,\ldots,v_n$ linear independent and $u_1,\ldots,u_n$ the corresponding orthogonalized vectors with the Gram-Schmidt process. By renaming the vectors and the induction hypothesis
we get $\langle v_{n+1},u_i\rangle\leq 0$ and $\langle w,u_i\rangle>0$ for $1\leq i\leq n-1$. We consider
$$
\langle v_{n+1}, u_n\rangle=\left\langle v_{n+1},v_n-\sum_{i=1}^{n-1}\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}u_i\right\rangle
=\langle v_{n+1},v_n\rangle-\sum_{i=1}^{n-1}\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}\langle v_{n+1},u_i\rangle.
$$
By induction hypothesis we have $\langle v_n,u_i\rangle\leq 0$ for $1\leq i\leq n-1$ and therefore the RHS is nonpositive and we get $\langle v_{n+1},u_n\rangle \leq 0$. Further we have
$$
\langle w,u_n\rangle=\left\langle w,v_n-\sum_{i=1}^{n-1}\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}u_i\right\rangle
=\langle w,v_n\rangle+\sum_{i=1}^{n-1}\frac{\left(-\langle v_n,u_i\rangle\right)}{\langle u_i,u_i\rangle}\langle w,u_i\rangle.
$$
which yields $\langle w,u_n\rangle>0$ the same way.

Proposition 1: If $v_1,\ldots,v_{n-1}$ are linear independent and $u_1,\ldots,u_{n-1}$ is the corresponding set of orthogonalized vectors, using Gram-Schmidt process and there exists $\mu_1,\ldots,\mu_{n-1}\in\mathbb{R}$ such that
$$
v_n=\sum_{i=1}^{n-1}\mu_iu_i,
$$
then $\mu_i\leq 0$ for $1\leq i\leq n-1$.
Proof of proposition 1:
From $v_n=\sum_{i=1}^{n-1}\mu_iu_i$ we get $\mu_i=\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}$,
since $u_1,\ldots,u_{n-1}$ are orthogonal. Using claim 1 yields $\mu_i\leq 0$.

Proof of your statement: Prove by induction.
Base case
For $n=1$ there is nothing to proof, since $\langle v_1,w\rangle>0$ yields $v_1\neq 0$.
For $n=2$ from $0=\sum_{i=1}^2\lambda_i\langle v_i,w\rangle$ you get
$\lambda_1=0=\lambda_2$ and you are done or w.l.o.g. $\lambda_1<0<\lambda_2$. But $$0=\left\langle \sum_{i=1}^2 \lambda_iv_i, v_1 \right\rangle = \sum_{i=1}^2 \lambda_i \langle v_i, v_1 \rangle \Leftrightarrow -\lambda_1\langle v_1v_1\rangle= \lambda_2 \langle v_2, v_1 \rangle.
$$
The LHS is positive while the RHS is negative and we get a contradition. Therefore $\lambda_1=\lambda_2=0$ has to be true and $v_1,v_2$ are lineare independent.
Inductive step:
Now we consider $n+1>2$ and $v_1,\ldots,v_{n+1}\in V$. By induction hypothesis we get, that $v_1,\ldots,v_n$ are linear independent. Either $v_1,\ldots,v_{n+1}$ are lineare independent and we are done or we consider $u_1,\ldots, u_n$ the corresponding orthogonalized vectors by Gram-Schmidt process and there exists $\mu_1,\ldots,\mu_n$ such that
$$
v_{n+1}=\sum_{i=1}^n\mu_iu_i.
$$
We consider
$$
0=\left\langle w , v_{n+1}-\sum_{i=1}^n\mu_i u_i\right\rangle=\langle w,v_{n+1}\rangle+\sum_{i=1}^n(-\mu_i)\langle w, u_i\rangle.
$$
By proposition 1 and claim 1 we have $(-\mu_i)\langle w,u_i\rangle\geq 0$ for $1\leq i\leq n$ which yields a contradition since $\langle w,v_{n+1}\rangle>0$. Therefore we get, that $v_1,\ldots,v_{n+1}$ are lineare independent.

We know that inner product can't be negative, and if is 0, the vectors are orthogonal. Orthogonal vectors are, by definition, linearly independent.
As for the w information, i think it was given to tell you that for any v vector, it is not the 0 vector, which is orthogonal to any vector (if it was, the inner product and v and w was 0).


Finding the indefinite integral $\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$ This is a homework question, I tried many subtitutions but nothing worked for me...

$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$

Any clue will help.

Thanks.

Hint::

$$\int\frac{\sqrt{x+8}}{\sqrt{x-3}-\sqrt{x+3}}dx$$
$$=\int\frac{\sqrt{x+8}(\sqrt{x-3}+\sqrt{x+3})}{(\sqrt{x-3}-\sqrt{x+3})(\sqrt{x-3}+\sqrt{x+3})}dx$$
$$=\int\frac{\sqrt{x+8}\sqrt{x-3}}{6}dx+\int\frac{\sqrt{x+8}\sqrt{x+3}}{6}dx$$
$$=\int\frac{\sqrt{x^2+5x-24}}{6}dx+\int\frac{\sqrt{x^2+5x+24}}{6}dx$$
$$=\frac{I_1+I_2}{6}$$
$$I_1 = \int\sqrt{x^2+5x-24}$$
$$I_1= \int\sqrt{\left(x+\frac{5}{2}\right)^2-\frac{121}{4}}dx$$
Substitute $u = \left(x+\frac{5}{2}\right)dx$
$$I_1= \int\sqrt{u^2-\frac{121}{4}}du$$
Substitute $\frac{11\sec\theta}{2} = u;du = \frac{11\sec\theta\tan \theta}{2}d\theta$
$$I_1= \frac{121}{4}\int\tan ^2\theta  \sec \theta d\theta$$
$$I_1= \frac{121}{4}\left[\int\sec^3 \theta d\theta -  \int \sec \theta d\theta\right]$$
Using reduction formula
$$\int{sec^m \theta} = \frac{\sin \theta \sec ^{m-1} \theta}{m-1} + \frac{m-2}{m-1}\int\sec^{m-2}\theta d\theta$$

Substitute and Solve

Hint: If you multiply an divide by $\sqrt{x-3}+\sqrt{x+3}$, the result will be this, getting rid of the fraction and obtainin just two integrals of a single square root. That may be easier to solve:
$$\int\frac{\sqrt{x^2+5x-24}+\sqrt{x^2+11x+24}}{-6}=\int\frac{\sqrt{(x-3)(x+8)}+\sqrt{(x+3)(x+8)}}{-6}=$$
$$=\frac{-1}{6}\int\sqrt{(x-3)(x+8)}-\frac{1}{6}\int\sqrt{(x+3)(x+8)}$$


If $\left|z+\frac{2}{z}\right |=2$, then prove that max value of $|z|$ is $\sqrt 3 + 1$ (Corrected version) Then the expression will be
$$|z|+\frac{2}{|z|}\ge 2$$

Solving the quadratic inequality

$$|z|\ge1+i$$ and $$|z|\le1-i$$

I don’t know how to solve further. Please help

Thanks!

this is the corrected version of an identical question asked by me. I am sorry for the inconvenience caused

$|z^{2}+2|=2|z|$ gives $|z|^{2}-2 \leq 2|z|$ or $(|z|-1)^{2} \leq 3$. Hence $|z| \leq 1+\sqrt 3$. I will let you find a value of $z$ for which equality holds.

You have to choose $z$ to be $(1+\sqrt 3)e^{i\theta}$ for  some $\theta$.

Let $z=re^{i\theta}$. Then, $\left|z+\frac{2}{z}\right |=\left|r+\frac{2}{r}e^{-i2\theta}\right |=2$ leads to

$$\left(r+\frac2r \cos 2\theta \right)^2+\frac{4}{r^2}\sin^2 2\theta= 4$$

or,

$$r^2+\frac{4}{r^2}=8\sin^2\theta\le 8$$

Rearrange the inequality to get,

$$(r^2 -4 )^2\le 12$$

which leads to

$$r\le \sqrt{4+2\sqrt3}=1+\sqrt3$$


The values of constants in the Equation. $$ \frac{\int_0^{4\pi} e^t(\sin^6at+\cos^4at)\,dt}{\int_0^\pi e^t(\sin^6at+\cos^4at)\,dt}= L, $$
the question asks the value of $a$ and $L$.
My friend solved it by differentiating, but i didn't understood a single thing. May be it is a very simple one, but i don't know how to solve it.

Notice that the numerator can be written as
$$\int_{0}^{4\pi}e^t(\sin^{6}at+\cos^{4}at)dt=\int_{0}^{\pi}e^t(\sin^{6}at+\cos^{4}at)dt+\int_{\pi}^{2\pi}e^t(\sin^{6}at+\cos^{4}at)dt+\int_{2\pi}^{3\pi}e^t(\sin^{6}at+\cos^{4}at)dt+\int_{3\pi}^{4\pi}e^t(\sin^{6}at+\cos^{4}at)dt$$

Now substitute $t=k+\pi$ in the second integral on the R.H.S, to get $$\int_{\pi}^{2\pi}e^t(\sin^{6}at+\cos^{4}at)dt=\int_{0}^{\pi}e^{k+\pi}(\sin^{6}ak+\cos^{4}ak)dk$$
because $\sin^6ak=\sin^6(ak+\pi)$ and same for $\cos^4ak$
$$=e^\pi \times \int_{0}^{\pi}e^k(\sin^{6}ak+\cos^{4}ak)dk$$
Similarly for the other two integrals substitute $t=k+2\pi$ and $t=k+3\pi$ to get
$$\frac{\int_{0}^{4\pi}e^t(\sin^{6}at+\cos^{4}at)dt}{\int_{0}^{\pi}e^t(\sin^{6}at+\cos^{4}at)dt}=1+e^\pi+e^{2\pi}+e^{3\pi}$$
$$=\color{red} {\frac{e^{4\pi}-1}{e^\pi-1}}$$

Why the particular substitutions?

  These are done in order to make the limits of the integrals present in the  numerators and denominators the same while also keeping in mind the periodicity of the trigonometric functions.

We have$$\frac{\int_0^{4\pi} e^t(\sin^6at+\cos^4at)\,dt}{\int_0^\pi e^t(\sin^6at+\cos^4at)\,dt}= F(a)$$ It follows the equation $$F(a)=L$$
The calculation made by Nikunj makes us see that $F$ is independent of $a$.

Thus the answer is $a$  arbitrary and $L={\frac{e^{4\pi}-1}{e^\pi-1}}\approx 12.9512$


Secret santa probability 26 people put their name in a hat.
Each person picks a name out of the hat to buy a gift for.
If a person picks out themselves they put the name back into the hat

What is the probability of the last person picking themselves?

I only ask this because it just happened!!

I do not have an exact answer, but a Monte Carlo approach shows that the probability that the last person picks his own name is approximately 0.03351 (95% confidence interval 0.03316 to 0.03386).

It would be a mistake to simply count the possible sequences of draws and divide the number of cases in which the last person picks his own name by the total number of sequences, as in Arthur's proposed solution.  The reason is that the sequences are not all equally likely.  For example, in the case of three people, the first person may draw 2 or 3, each with probability 1/2.  If he draws a 2, the second person may draw a 1 or 3, each with probability 1/2.  But if the first person draws a 3, the second person must draw a 1.  So the sequences (2,1,3) and (2,3,1) each have probability 1/4, while the sequence (3,1,2) has probability 1/2.

Here is a Python function which simulates the process. The persons are numbered from 0 to npersons-1 for programming convenience. (The version of Python used was 2.7.1).

import random

def draw_gifts(npersons, ntrials):
    "Simulate drawing gifts with npersons, ntrials times; return number of failures"
    nfail = 0
    for t in range(ntrials):
        hat = [i for i in range(npersons)]
        for i in range(npersons-1):
            # person i draws from the hat
            # hat[0]..hat[i-1] contain numbers already drawn
            # hat[i]..hat[npersons-1] contain eligible numbers
            j = random.randint(i, npersons-1)
            # redraw if necessary
            while hat[j] == i:
                j = random.randint(i, npersons-1)
            temp = hat[i]
            hat[i] = hat[j]
            hat[j] = temp
        if hat[npersons-1] == npersons-1:
            nfail += 1
    return nfail

If the random number generator seed is first set to 1234 via "random.seed(1234)" and the function above is invoked via "draw_gifts(26, 1000000)" (simulating 26 persons and 10^6 trials), the returned number of failures, i.e. cases in which the last person draws his own number, is 33508.  A standard statistical formula for a confidence interval on proportions, based on a Normal approximation, yields a 95% confidence interval of 0.03316 to 0.03386 for the probability of failure. See http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval.

After the first $25$ people have picked names, there are two possibilities:

$\bullet$ We have a derangement of the first $25$ names, in this case the last person picks his own name.
$\bullet$ The last person picks the name of another person, and then we get a derangement of all names.

The number of derangements of $n$ objects is equal to $\left[ \frac{n!}{e}\right]$ (see link), so the probability of the last person picking his own name is equal to $$\frac{\left[ \frac{25!}{e}\right]}{\left[ \frac{25!}{e}\right] + \left[ \frac{26!}{e}\right]} = \frac{5706255282633466762357224}{154068892631103602583645049} \approx \frac{1}{27} - 2.4 \cdot 10^{-28}$$


Continuous time Markov processes on general state spaces Looking around I have found lots of material on continuous time Markov processes on finite or countable state spaces, say $\{0,1,\ldots,J\}$ for some $J\in\mathbb{N}$ or just $\mathbb{N}$. Similarly I have earlier worked with (discrete time) Markov chains on general state spaces, following the modern classic by Meyn & Tweedie.

My question concerns monographs on continuous time Markov processes on general state spaces, say some subset of $\mathbb{R}^k$, $k\in\mathbb{N}$. Are there any good references - preferably but not necessarily suited for an ambitious master student - on this topic?

Another classic in probability theory is William Feller's An Introduction to Probability Theory and its Applications, volume I and II, which has a decent coverage of the topic, perhaps it is worth looking up (more specifically chapter 10, vol. II), even though it is a bit old.

I just finished a course on stochastic calculus, so I still lack experience on continuous time markov processes, but I think I can give you some references I have looked at:

-Chapter 10 of Introduction to Stochastic Integration by Hui-Hsiung Kuo explains the Markov property (and gives examples in terms of the Brownian motion), and diffusion processes. If you have a background in stochastic calculus (this book or Oksendal's are good introductory books) I would start here.

-A more detailed exposition (but too advanced for me at the moment of writing) is the two volume book of Rogers and Williams, Diffussions, Markov Processes and Martingales.


Circle rotating within a circle (roulette) This was something in a course of mine I'm a bit too thick to see. If one takes a circle of radius $3$ and a circle of radius $1$, and rolls the smaller circle smoothly inside the larger one until the point $P$ which starts out in contact with the larger circle is in the same place, how can one prove that the smaller circle goes through exactly two rotations?

The curve generated by the path of the vector $\mathbf{P}$ (the vector from the center of the larger circle to $P$) is called a roulette - more specifically, a deltoid.

If you trust the picture, then you can see that the smaller circle does go through two rotations: letting $\mathbf{C}$ be the center of the smaller circle, the little vector pointing from $\mathbf{C}$ to $\mathbf{P}$ rotates around to pointing due right exactly twice, so the circle rotates twice also. But I don't know how to prove this.

This question came up because we wanted to find the parametric equation of $\mathbf{P}$. It'll be the sum of the vectors $\mathbf{C}$ and $\mathbf{P- C}$; the first one has equation $(2\cos(t), 2\sin(t))$, and if the smaller circle rotates around (clockwise) exactly two times, it'll have the equation $(\cos(2t), -\sin(2t))$. This would give the parametrization of the curve immediately. But how to prove that?

(More generally, it seems like a circle of radius $1$ rotating in a circle of radius $n$ will go around $n-1$ times.)

From the following figure it can immediately been read off that when the point  of contact $Q$ moves counterclockwise by the angle $\phi$ then the inner circle (as a rigid structure) rotates by the amount $2\phi$ clockwise. Therefore after a full turn of $Q$ the point $P$ has made $2$ turns.

The smaller circle is actually making 3 rotations in the larger Circle: when the Red Dot on the smaller Circle points to the right for the first time, it's actually half way through the second rotation.

The Red Dot pointing  due right is not actually significant of a full rotation. A full rotation of the smaller Circle is actually signified by the Red Dot touching the Outer Circle again.

It's more easily seen that there's three rotations of the smaller Circle if you start the smaller Circle in the bottom of the larger Circle, marking the two where they touch together and rotating the larger circle around until the two points come back together. You'll see that the smaller Circle actually next three rotations,


Find the surface area of the part of the cone $z=\sqrt{x^2+y^2}$ that lies below the plane $x+2z=3$ It seems as though I'll have to use some sort of integral (surface integral?) using the distance between the plane and the bottom of the cone as limits of integration. Unfortunately I'm not familiar with surface integrals.

Parametrize the surface, for example:
$$
x=x, \quad y=y, \quad z=\sqrt{x^2+y^2}; \quad (x,y)\in D
$$
where $D$ is the projection of  the intersection of $x+2z=3$ and $z=\sqrt{x^2+y^2}$ in the $xy$ plane. Can you write $D$ explicitly?
From this parametrization, it follows that the surface area equals
$$
A=\iint_D ||r_x \times r_y||\; dA
$$
where $r_x=(1,0,\frac{\partial \sqrt{x^2+y^2}}{ \partial x})$ and $r_y=(0,1,\frac{\partial \sqrt{x^2+y^2}}{ \partial y})$.

This is the general methodology. I am leaving the technical details for you to do ;)

The projection of the intersection of the cone and plane onto the $xy$ plane is an ellipse $$\frac{(x+1)^2}{4}+\frac{y^2}{3}=1.$$(Obtained by solving equations $z=\sqrt{x^2+y^2}$ and $x+2z=3$.) Using change of variables $u=\frac{x+1}{2},\,v=\frac{y}{\sqrt{3}}$ with the Jacobian $\frac{\partial(x,y)}{\partial(x,y)}=2\sqrt{3}$. Then we have $$\int\int_{\frac{(x+1)^2}{4}+\frac{y^2}{3}\leq 1}dydx=2\sqrt{3}\int\int_{u^2+v^2\leq 1}dvdu=2\sqrt{3}\pi$$.


Is interior points of a subset $E$ of a metric space $X$ is always a limit point of $E$? According to Rudin a point $p$ is said to be a limit point of a subset $E$ of a metric space $X$ if every neighborhood $N$ of $p$ contains a point $q \neq 0$ such that $q\in E$.

A point  a point $p$ is said to be an interior point of a subset $E$ of a metric space $X$ if there exist a neighborhood $N$ of $p$  such that $N \subset E$.

Examples given shows that there may be limit points which are not interior points : $(a,b)$. There may be sets which does not contain any of its limit points and no point of it is an interior point.

Is  interior points of a subset $E$ of a metric space $X$  is always a limit point of $E$ ? If $p$ is an interior point of $E$ then there exist a a neighborhood $N$ of radius $r$ such that $N \subset E$. Then any neighborhood of $p$ having radius less than $r$ will be contained in $N$ and hence contained in $E$. Any neighborhood of $p$ having radius greater than $r$ contains $N$ and hence intersect $E$ at some points other than $p$. Hence $p$ is a limit point of $E$. Is my argument is correct ?

Generally - In topology- this proof will not work as it does not have radius concept. But I want a counter example from metric space if I am wrong. Particularly a subset $E$ of $\mathbb{R}$ which has an interior point which is not a limit point of $E$.

If you are asking about $\mathbb{R}$ with the (standard) Euclidean metric, then the answer is yes:

Suppose that $X\subseteq\mathbb{R}$ and that $x\in X$ is an interior point.  Then, by the definition of an interior point, there is some open set $U$ so that $x\in U$ and $U\subseteq X$.  Since the topology on $\mathbb{R}$ is generated by open intervals, we know that there is an open interval $I$ so that $x\in I$, $I\subseteq U\subseteq X$.  Observe that for $n$ sufficiently large, $x+\frac{1}{n}$ is a sequence of points distinct from $x$ and in $I$ (and hence in $X$) that converge to $x$.  Therefore, $x$ is a limit point.

The reason that the argument above worked is that there are infinitely many points in any open interval.  If, instead, we use a different metric, the answer to your question is no:

Consider the discrete metric on $\mathbb{R}$, i.e.,
$$
d(x,y)=\begin{cases}0&x=y\\
1&x\not=y
\end{cases}.
$$
Suppose that $X\subseteq\mathbb{R}$ and that $x\in X$.  We observe that $x$ is an interior point because the open ball $B\left(x,\frac{1}{2}\right)=\{x\}\subseteq X$.  On the other hand, $x$ is not a limit point because the open ball $B\left(x,\frac{1}{2}\right)$ is an open set containing $x$ and no other points of $X$.

We know that every point in an open set is interior point right? Then, if a interior point is always a limit point, doesn't this imply that an open set is always closed? As open set consists of interior points and if they are limit points which are inside the set.


Given P(A|B) and P(A|C), how to get or strategically approach P(A|(B & C))? Given P(A|B) and P(A|C), how to get or strategically approach P(A|(B & C))?

Is there a way to approach this if it is not known whether B and C are dependent?
If not, how to get P(A|(B & C)) assuming B and C are independent, or, dependent?

Suppose two $0/1$ coins $X_1,X_2$ are thrown. Let $B$ be the event $X_1 = 0$, $C$ be the event $X_2 = 0$, $A^1$ be the event $X_1 + X_2 = 0$, and $A^0$ be the event $X_1 + X_2 = 1$. Then $$P(A^0|B) = P(A^1|B) = P(A^0|C) = P(A^1|C) = 1/2$$ whereas $P(A^0|B \land C) = 0$ and $P(A^1|B\land C) = 1$.

If A and B are independent from C, then P(C|A) = P(C|B). If they are dependent, then just follow Bayes' theorem for a three event senerio.


Partial functions - where can I learn more about this (heuristic, informal) system of conventions? Is there a name for the following (heuristic, informal) system of conventions for dealing with partial functions and undefined expressions? I'd like to know whether it has any undesirable quirks that I haven't picked up on.

There is an designated entity $U$ (intuitively, $U$ represents "undefinedness").
Functions can be partial, and every function $f$ corresponds to a set $\mathrm{dod}(f)$ called its domain of definition. Whenever $f$ is evaluated at some $x \notin \mathrm{dod}(f)$, it returns $U$. For example, if we're working in the real numbers, the expression $0^{-1}$ would return $U$. Furthermore, if $U$ is an input to a function, then $U$ is output. For example, in the real numbers, an expression like $U+4$ would equal $U$.
All relations are total and the logic is two-valued; and if a relation is given $U$ as an input, then $\mathbf{false}$ is output. For example, in the real numbers, an expression like $3 < U$ would be false. This has the slightly odd result that $U = U$ is false (less surprisingly, $U \neq U$).
The entity $U$ is taken as being outside the domain over which quantifiers range. For example, from $\forall x(x = x)$ we cannot deduce $U = U$.

So just to reiterate, I'm looking for

more information / a reference.

In particular, I'd like to know:

Does this system of conventions have a name?
Does it have any weird quirks that make it a bad system of conventions?

This topic has been the subject of much debate by people who worry about formal specification languages. There is a paper by Stoddart, Dunne and Galloway, Undefined Expressions and Logic in Z and B, that compares various approaches and gives some references that may be useful (See http://dx.doi.org/10.1023/A:1008797018928 - or search for it via google if you don't have access to SpringerLink).

"Free logic" as mentioned in a comment by Peter Smith may be of interest (see http://plato.stanford.edu/entries/logic-free/), but I don't think you are interested in having syntax for saying things like "such-and-such an expression is undefined".

Your proposal isn't very clear about this, but if you intend $U$ to be a semantic object that is not part of the syntax, then what you describe is very like the approach originally taken in Z (see Stoddart et al.). The main quirk is that it can be counter-intuitive: e.g., in Z, $1/0 = 1/0$ and $1/0 \not= 1/0$ are both false (because $\not=$ is defined in terms of set membership: $x \not= y$ means $(x, y) \in \{a, b : X \mathrel{|} \lnot a = b\}$).

Two claims are made here, among others:

(1) There is an undefined symbol U.
(2) If you evaluate a function outside its domain of definition, it returns U.

How are we to understand the second, given the first?
(a) It can't be that the function, for some argument $a$ "outside the domain of definition", returns the symbol 'U' - for that object would be a perfectly good value, so $a$ is in the domain of definition after all (i.e. is an object which, given to the function $f$ as argument, yields a value).
(b) It can't be either that the the function, for some argument  $a$ outside the domain of definition, returns what 'U' denotes, since by hypothesis there is no such thing since 'U' is undefined.
So we seem to be heading for straight nonsense based on use/mention confusion.
(c) But perhaps, more charitably, the idea in (2) is that 'U' functions like the word 'zilch'. So when a function is applied to an object outside the domain of definition it returns zilch. But note 'zilch' isn't undefined (any more than 'nothing' is undefined) -- it's in the dictionary! And if 'U' works like 'zilch' it won't be undefined. So we still won't have (1).


Position as function of time given velocity with respect to position. I have been able to solve this problem in a very clumsy way that I don't think is using the right steps.  In particular, I'm not sure where the assignments of the arbitrary constants are rightly applied.

Question:
Find the position $s$ as a function of $t$ from the given velocity $v = ds/dt$. Evaluate the constant of integration so as to have $s=s_0$ when $t=0$.

$$v = \sqrt{2gs},\ \ \   (g = \text{constant}).$$

My attempt and issues with the problem has involved:

$$v=\frac{ds}{dt}=\sqrt{2gs}$$

$$s^{-1/2} ds = (2g)^{1/2} dt$$

$$d(2*s^{1/2}) = d(t*(\sqrt{2g}) + C$$

Now, I can convert that to:

$$s=(1/2)gt^2 + C$$

And when $s=s_0$ at $t=0$,

$$s=(1/2)gt^2 + s_0$$

Substituting the right side of the original equation $V=\sqrt{2gs}$ doesn't seem to work. On the other hand, I can see that

$$v=\frac{ds}{dt}= \sqrt{2gs} = gt$$

so I am thinking there may be an approach using that equality to determine $C$ as well.

UPDATE:

This is the stated answer, but I’m not sure of the steps for getting the $ t\sqrt{2gs_0}$ term into the equation:
$$s = s_0 + t\sqrt{2gs_0} + \frac12gt^2$$

We have

$$v=\frac{ds}{dt}=\sqrt{2gs} \implies \frac{ds}{\sqrt s}=\sqrt{2g}\, dt \implies 2\sqrt s=\sqrt{2g}\, t+c$$

Edit to help the OP step by step:

$$\implies 4s=2gt^2+2c\sqrt{2g}\, t+c^2 \implies  s=\frac12gt^2+\frac12c\sqrt{2g}\, t+\frac14c^2$$

and from $s(0)=s_0$ we obtain $\frac14c^2=s_0 \implies \frac12 c=\sqrt{s_0}$ and therefore

$$s=\frac12gt^2+\sqrt{2gs_0}\, t+s_0$$


Why do disks on planes grow more quickly with radius than disks on spheres? In the book, Mr. Tompkins in Wonderland, there is written something like this:

  On a sphere the area within a given radius grows more slowly with the radius than on a plane.

Could you explain this to me? I think that formulas shows something totally different:

The area of a sphere is $4 \pi r^2$, but the area of a disk on the plane is $\pi r^2$.

UPDATE:
Later in this book there is:

  If, for example, you are on the north pole, the circle with the radius equal to a half meridian is the equator, and the area included is the northern hemisphere. Increase the radius twice and you will get in all the earth's surface; the area will increase only twice instead of four times if it were on a plane.

Could someone explain me this?

The claim is not that the (surface) area, $4 \pi r^2$, of a sphere of radius $r$ grows more slowly than the area, $\pi r^2$, of a disk of radius $r$ on the plane. Rather, the book is asserting that the area of a disk on a sphere grows more slowly with radius than does a disk on the plane.

If we pick a point $P$ on a sphere of radius $R$, we can ask which points on $R$ are a fixed distance $r$ from $P$ along the surface of the sphere---(for $0 < r < \pi R$) by symmetry these points comprise a circle. The area the book means is the region inside this circle (i.e., in the region bounded by the circle and containing $P$).

Now (for $0 \leq r \leq \pi R$), the central angle between $P$ and a point a distance $r$ from $P$ along the sphere is just $\frac{r}{R}$. Thus, if we declare $P$ to be the north pole of the sphere and use the usual colatitude-longitude coordinates on the sphere, we find that the area $A_{\text{sphere}}(r)$ of the region $D$ of points with a distance $r$ from $P$ is
$$A_{\text{sphere}}(r) = \iint_D dA = \int_0^{2 \pi} \int_0^{\frac{r}{R}} \sin \phi \,d\phi \,d\theta = 2 \pi R^2 \left(1 - \cos \frac{r}{R}\right).$$

Now, the rate of growth (w.r.t. radius $r$) of the area $A_{\text{plane}}(r)$ of a disk on the plane is
$$\frac{d}{dr} A_{\text{plane}} (r) = 2 \pi r,$$ and the rate of growth of the region $D$ with respect to the radius $r$ measured along the surface of the sphere is
$$\frac{d}{dr} A_{\text{sphere}} (r) = 2 \pi R \sin \frac{r}{R}.$$
Since $0 < \sin u < u$, for $u > 0$, we have for $0 < r < \pi R$ that
$$\frac{d}{dr} A_{\text{sphere}} (r) = 2 \pi R \sin \frac{r}{R} < 2 \pi R \left(\frac{r}{R}\right) = 2 \pi r = \frac{d}{dr} A_{\text{plane}} (r)$$
as claimed.

Note that no point is further than $\pi R$ from $P$ (and only the point $-P$ antipodal to $P$ is exactly that distance away). So, the claim is trivially true for $r > \pi R$: By that point, the sphere is already covered but the disk in the plane keeps growing.

Remark We can extract a little more information by inspecting the Taylor series of the ratio $$\rho(r) := \frac{A_{\text{sphere}}(r)}{A_{\text{plane}}(r)}$$ around $r = 0$. Expanding to third order gives
$$\rho(r) = 1 - \frac{1}{12 R^2} r^2 + O(r^4).$$ Now, we can read off the Gaussian curvature of the sphere at the point $P$ by inspecting the quadratic term; importantly, we can do this for any surface, so comparing the rates of growth of disks on general surfaces with those on the plane is important insofar as it yields a reasonably concrete interpretation of curvature.

If one interprets the area of a sphere as its surface area, then the formula would be $4\pi r^2$.
The area of a circle is $\pi r^2$.
Therefore the surface area of a sphere grows 4 units when the area of a circle grows 1 unit, which is not slower. I'm sure I'm misinterpreting this, though.


Why is the exterior set of $\mathbb R\setminus \mathbb Q$ a null set? Given the set  $\mathbb R\setminus \mathbb Q.$

The interior set is the collection of all the interior points, where the interior point of a set $S$ from $\mathbb R,$ is a point $x  \in  S,$ such that there exists an $ \varepsilon >0 $ to make an open set U that looks like $(x - \varepsilon, x + \varepsilon)$ such that $x \in U$ and $U \subset S$.

My explanation for Interior set being a null set (Please review)

So, for any arbitrary irrational point in the given set, If I form an open interval around that point of size $|x|<\varepsilon$, but that interval has no other point than the irrational point itself, so therefore a neighbourhood does not exist for the irrational point.

Why would the Exterior set be a null set?

(Exterior Set - Collection of all the exterior points of set S)

(Exterior Point - A number $a \in\mathbb R$ is said to be an exterior point of a set $S$ from $\mathbb R$ if there exist a neighbourhood of a which is contained in $S^c$)

Your solution for the interior set is iffy - you start in the right way by examining each $x\in\mathbb R\setminus\mathbb Q$, but then you do something strange by writing $|x|<\varepsilon$ - since you are not allowed to control what $x$ is in an argument like this and then say that some interval consists only of $x$, which is false.

Rather, what you should ask is the following:

  I've been given some $x\in\mathbb R\setminus\mathbb Q$. Is there any $\varepsilon>0$ such that the interval $(x-\varepsilon,x+\varepsilon)$ is a subset of $\mathbb R\setminus \mathbb Q$.

In simpler language, you are asking the following

  Is there any $\varepsilon>0$ such that every element of $(x-\varepsilon,x+\varepsilon)$ is irrational?

The answer to this is "No" because every open interval contains a rational number. So, to write a proof that the interior is empty, you would start as follows:

  We will show that the interior of $\mathbb R\setminus \mathbb Q$ is empty. To see this, fix $x\in\mathbb R\setminus \mathbb Q$. We claim that for any $\varepsilon>0$, the interval $(x-\varepsilon,x+\varepsilon)$ contains a rational number. ...

And then you would argue why this is true.

Note that the exterior of a set is just the interior of the complement - and to show that the interior of $\mathbb Q$ is empty, it suffices to show that every open interval contains an irrational number, which will follow very similar reasoning to the interior being empty.

It follows from the density of the rationals that the exterior of R/Q must be empty.


Prove that if $a,b\in \mathbb{R}$, $a>0$, then $\lim \frac{n^b}{(1+a)^n}=0$ as $n\to\infty$ As the title says, i want to show that $$\lim_{n\to \infty} \frac{n^b}{(1+a)^n}=0$$ where $a,b\in \mathbb{R}$ with $a>0$.

I tried to bound the sequence $x_n =  \frac{n^b}{(1+a)^n}$ and use the sandwish theorem, but have no results. My problem is the exponent in the denominator, maybe it could be bounded with Bernoulli inequality ($(1+a)^n\geq 1+na$) but then the problem is de numerator $n^b$.

I'm grateful for any hints for doing this problem, or some steps to clarify.

HINT:

Bernoulli's inequality won't suffice when $b\ge1$, but the binomial theorem will suffice.

From the binomial theorem, with $n>b$, we see that

$$(1+a)^n\ge \binom{n}{\lfloor b\rfloor +1}a^{\lfloor b \rfloor +1}$$

where $\lfloor \cdot \rfloor$ is the floor function.

Can you proceed?

By the ratio test, the sequence converges if
$$\frac{(n+1)^b}{(1+a)n^b}<1.$$

This is achieved when

$$n>\frac1{\sqrt[b]{1+a}-1},$$ a finite number.

E.g., with $\dfrac{n^3}{2^n}$, for $n>3$,

$$\frac{(n+1)^3}{2n^3}<\frac{125}{2\cdot64}<1.$$


Spectrum of adjacency matrix of complete graph Fooling around in matlab, I did an eigenvalue decomposition of the adjacency matrix of $K_5$.

 A = [0     1     1     1     1;
      1     0     1     1     1;
      1     1     0     1     1;
      1     1     1     0     1;
      1     1     1     1     0];

The eigenvalues are $4, -1, -1, -1, -1$. It now seems obvious that the largest eigenvalue for the adjacency matrix for $K_n$ would be $n-1$. What I don't understand is why $-1$ is an eigenvalue of multiplicity $n-1$. Is there any special meaning to this?

It is possible to give a lower bound on the multiplicities of the eigenvalues of the adjacency or Laplacian matrix of a graph $G$ using representation theory.

Namely, the vector space of functions $G \to \mathbb{C}$ is a representation of the symmetry group, and both the adjacency and Laplacian matrix respect this group action. It follows that each irreducible subrepresentation lies in an eigenspace of the adjacency and Laplacian matrices, so their dimensions give a lower bound on the multiplicities.

In this particular case, $K_n$ has symmetry group the full symmetric group $S_n$, and the corresponding representation decomposes into the trivial representation (this corresponds to the eigenvalue $n-1$) and an $n-1$-dimensional irreducible representation, so there can only be one other eigenvalue and it must have multiplicity $n-1$. Moreover, the sum of the eigenvalues is $0$ because the trace of the adjacency matrix is zero, so the remaining eigenvalue must be $-1$.

This is all just a very fancy way of saying that

  Any permutation of an eigenvector of $K_n$ is also an eigenvector with the same eigenvalue. If the eigenvector doesn't have all of its entries equal, then its permutations span a vector space of dimension $n-1$, so the remaining eigenvalue has multiplicity $n-1$.

Which is in turn just a very fancy way of saying that

  Any vector $v$ all of whose entries sum to zero is an eigenvector of the adjacency matrix $A_n$ of $K_n$; moreover, it is easy to see that $(A_n + I)v = 0$ for any such $v$, hence $A_n v = -v$.

That is a straightforward explanation but it misses the bigger lesson, which is that

  Highly symmetric graphs have eigenvalues of high multiplicity.

Another argument proceeds using the observation that, on the one hand, $\text{tr}(A^k)$ counts the number of walks of length $k$ which begin and end at the same vertex and, on the other hand,
$$\text{tr}(A^k) = \sum_{i=1}^n \lambda_i^k$$

where $\lambda_i$ are the eigenvalues. This sequence uniquely determines the eigenvalues (exercise), so to prove the desired result it suffices to prove that
$$\text{tr}(A^k) = (n-1)^k + (n-1)(-1)^k.$$

Now, the total number of walks of length $k-1$ is just $n(n-1)^{k-1}$, where the first $n$ counts the number of possible starting vertices and each factor of $n-1$ counts the possible next vertices. If such a walk ends at the original vertex (so is itself a closed walk of length $k-1$), it cannot be extended to a closed walk of length $k$; otherwise, it can be uniquely extended as such. Thus it follows that
$$n(n-1)^{k-1} = \text{tr}(A^{k-1}) + \text{tr}(A^k)$$

and then the desired result follows by induction from the initial condition $\text{tr}(A^0) = n$.

consider matrix $J_n$ has $n$ order of all ones. $A(K_n) = J_n - I_n$
the eigenvalues of $J_n$ are $n$ and $0$, the eigenvalues of $I_n$ is $1$.
apply the spectral mapping theory.


Spectrum restrictions in the signature consisting of just a single binary operation In the signature {*}, where * is an operator of arity 2, is there a theory whose spectrum is the set of prime powers?

Finite models of field theory have the cardinality of prime powers. Of course, they don't quite fit your signature.

However, the answer to this question demonstrates methods to capture both addition and multiplication in a single binary operation. For each constant and function defined, you will have to 'reverse skolemize' into a FOL sentence, 'and' them all together with unique variable names, and then 'and' all the axioms that use these functions into one sentence (this ensures the 'reverse skolemized' function is the same for each axiom).

Actually performing all this would be a nasty bit of work, but I believe this information is sufficient to demonstrate the answer to your question is an affirmative.

Add to group theory an axiom asserting that every element's order is a power of some prime $p$.

Theorem 1.1 in this paper:
http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cauchyapp.pdf


Vector expression for intersection of a reflected ray from a cylinder I am trying to find an expression for a reflected ray from a cylindrical surface, as shown in the sketch below.

$E$ is the starting point of the ray, $\hat d_{1}$ is the unit vector in the ray direction. $I$ is the intersection point on the cylinder, $\hat d_{2}$ unit vector in the reflected ray's direction, and $P$ is the intersection on the $Z=0$ plane. The cylinder is centered at the origin with radius $r$. Hence $I_{x}^{2} + I_{y}^{2} = r^{2}$.

I derived the following expression for the reflected ray direction in terms of the incoming ray:

$$\hat{d_{2}} = \hat{d_{1}} - 2(\hat{d_{1}}.\hat{n})\hat{n}$$

where $\hat n = (\frac{I_x}{r}, \frac{I_y}{r}, 0)$.

Also, I have:

$$P = I + \hat{d_{2}}t$$
$$P_{z} = 0$$

So, if $E$ and $I$ are known, I can solve for $P$. Thus I have $P$ in terms of $E$ and $I$.

My question is: If $E$ and $P$ are known instead, can I get an expression for $I$ in terms of $E$ and $P$? (I have not been able to derive it.)

Theory

Let $\mathbf{I}$ be a unit ray of light (we only care about direction) and let $\mathbf{\hat{u}}$ be a vector normal to a surface.

We can decompose our ray into a parallel and perpendicular component as follows:

$$\mathbf{I} = \text{I}_\shortparallel\mathbf{u}_\shortparallel + \text{I}_\perp \mathbf{u}_\perp \tag{1}$$

As the light is reflected by the surface, the component parallel to the surface is unchanged while the component normal to it changes sign. We then write an expression for the reflected ray, $\tilde{\mathbf{I}}$

$$\tilde{\mathbf{I}} = \text{I}_\shortparallel\mathbf{u}_\shortparallel - \text{I}_\perp \mathbf{u}_\perp \tag{2}$$

It is usually easy to obtain an expression for $\text{I}_\perp\mathbf{u}_\perp$ because $\mathbf{u}_\perp = \frac{\nabla f}{|\nabla f|}$ and $\text{I}_\perp = \mathbf{I} \cdot \mathbf{u}_\perp$.

We therefore want to get rid of the $\text{I}_\shortparallel\mathbf{u}_\shortparallel$ in $(2)$ by solving for it in $(1)$ and replacing. Doing that computation:

$$\text{I}_\shortparallel\mathbf{u}_\shortparallel = \mathbf{I}-\text{I}_\perp\mathbf{u}_\perp$$

so we arrive at

$$ \tilde{\mathbf{I}} = \mathbf{I} - 2\text{I}_\perp\mathbf{u}_\perp$$

where $\text{I}_\perp = \mathbf{I} \cdot \mathbf{u}_\perp$ and $\mathbf{u}_\perp = \frac{\nabla f}{|\nabla f|}$

leading to a final expression:

$$ \tilde{\mathbf{I}} = \mathbf{I} - 2\left(\mathbf{I} \cdot \frac{\nabla f}{|\nabla f|}\right)\frac{\nabla f}{|\nabla f|}$$

This is the general equation and will work for any surface normal at any point. You can use it without thinking at all.

Now in your case we are dealing with a cylinder. Cylinders can be nicely described with cylindrical coordinates but I will use cartesian coordinates: $x^2 + y^2 = r^2$. Then we know that the normal vector at any point on the cylinder is $\nabla f(x,y) = (2x, 2y, 0)$ where $f(x,y) = x^2 + y^2 = r^2$ (and is a level set of the function $z^2 = x^2 + y^2$).

Let $\mathbf{E}$ be a point such that ray $\mathbf{I}$ emanates from $\mathbf{E}$. $\mathbf{I}$ intersects $f$ at the point defined by $|\mathbf{p}_\text{intersection}| = |\mathbf{E} + \alpha \mathbf{I}| = r$. For multiple intersections we choose the point $\mathbf{p}$ closest to $\mathbf{E}$. We solve for this point of intersection with the cylinder as follows:

$$\begin{align}
p_x &= \alpha(E_x + I_x)\\
p_y &= \alpha(E_y + I_y)\\
p_z &= \alpha(E_z + I_z)
\end{align}
$$

so $$\alpha = \frac{r^2}{\sum (E_i + I_i)^2}$$

and we then compute the point of intersection $\mathbf{p} = \mathbf{E} + \alpha\mathbf{I}$.

We now know where the ray hits the cylinder and can compute the reflected ray $\tilde{\mathbf{I}}$. Now we must find where the reflected ray intersects an arbitrary plane. Consider some plane given by a normal $\mathbf{\hat{n}}$ and a point $\mathbf{r}_0$ such that if $\mathbf{r}$ lies on the plane $(\mathbf{r} - \mathbf{r}_0)\cdot \mathbf{\hat{n}} = 0$.
If the intersection point of $\tilde{\mathbf{I}}$ is not normal to $\mathbf{\hat{n}}$ then it will intersect our plane exactly at the point $\mathbf{w}$ defined implicitly by

$$(\mathbf{w} - \mathbf{r}_0)\cdot\mathbf{\hat{n}} = 0$$

Again we solve for the parameter, although this time it is a bit easier. With $\mathbf{w}$ given by $\mathbf{w} = \mathbf{p} + \beta \mathbf{\tilde{I}}$,

$$\begin{align}
(\mathbf{w} - \mathbf{r}_0)\cdot\mathbf{\hat{n}} &= 0 \\
(\mathbf{p} + \beta \mathbf{\tilde{I}} - \mathbf{r}_0)\cdot\mathbf{\hat{n}} &= 0 \\
\mathbf{p}\cdot\mathbf{\hat{n}} + \beta \mathbf{\tilde{I}}\cdot\mathbf{\hat{n}} - \mathbf{r}_0\cdot\mathbf{\hat{n}}  &=0\\
\end{align}
$$

so

$$\beta =  \frac{\mathbf{r}_0\cdot\mathbf{\hat{n}} - \mathbf{p}\cdot\mathbf{\hat{n}}}{ \mathbf{\tilde{I}}\cdot\mathbf{\hat{n}} }$$

with which we can compute $\mathbf{w}$, the point where the reflected ray intersects a plane.

Answer

Assume we are only given $\mathbf{E}$, the origin point of the ray, and $\mathbf{w}$, the point where the ray intersects some plane. Is the mapping from $\mathbf{E}$ to $\mathbf{w}$ invertible such that $\mathbf{I}$ is recoverable?

This is equivalent to asking whether each $\mathbf{w}$ is uniquely generated by the $\mathbf{I}$s. The answer to this question is no.

I can pick any $\mathbf{w}$ on a plane, any starting point $\mathbf{E}$, and any point on the cylinder, and there exists some $\mathbf{I}$ which emanates from $\mathbf{E}$ whose reflection on the cylinder intersects the plane at $\mathbf{w}$.

Find the midpoint of E and P, and project it on the cylinder. This gives I directly.
$$
I_{z} = \frac{1}{2}(E_{z}+P_{z}) \\
I_{x}= \frac{rx}{\sqrt{x^{2}+y^{2}}}\\
I_{y} = \frac{ry}{\sqrt{x^{2}+y^{2}}}
$$
where $x=\frac{E_{x}+P_{x}}{2}$ and $x=\frac{E_{y}+P_{y}}{2}$.


Hint:

Use Tringnometric transformation such that

x=sinu, y=secu and z = cotu

For any value of u you can find x,y,z >0 such that xyz = 1 by definition.

For a value of u = pi/6, x=.5, y = 1.154 and z = 1.732

Further, I computed LHS, for u = pi/4 and its value is .3365

Just to double check, I computed LHS for u = pi/3 and its value is .34552

In both sample cases, LHS is >= (1/3).

Thanks

Satish


Expectation (Boundaries on the Integral) In a model for hospital room charges X and hospital surgical charges Y for a particular type of hospital admission, the region of probability (after scaling units) is 0 ≤ y ≤ 2x+1 ≤ 3

The joint density function of X and Y is f(x, y) = 0.3(x + y)

Find the expected excess of surgical charges over room charges for an admission.

Am I solving a double integral here of 0.3(x+y) dxdy? Cause Im having trouble finding the boundaries on the integral?

You have the joint support of: $0 ≤ y ≤ 2x+1 ≤ 3$

That is $(0\leq y \leq 3) \times (\frac{y-1}{2}\leq x\leq \frac{3-1}{2})$, which gives you

$$\int_0^3 \int_{(y-1)/2}^{1} (...)\; 0.3 (x+y)\operatorname d x\operatorname d y$$

Alternatively we could use :

$$\int_{-1/2}^1 \int_{0}^{2x+1} (...)\; 0.3 (x+y)\operatorname d y\operatorname d x$$


A $\{0,1\}$-matrix with positive spectrum must have all eigenvalues equal to $1$ Here's a cute problem that was frequently given by the late Herbert Wilf during his talks.

Problem: Let $A$ be an $n \times n$ matrix with entries from $\{0,1\}$ having all positive eigenvalues. Prove that all of the eigenvalues of $A$ are $1$.

Proof:

   Use the AM-GM inequality to relate the trace and determinant.

Is there any other proof?

If one wants to use the AM-GM inequality, you could proceed as follows:
Since $A$ has all $1$'s or $0$'s on the diagonal, it follows that $tr(A)\leq n$.
Now calculating the determinant by expanding along any row/column, one can easily see that the determinant is an integer, since it is a sum of products of matrix entries (up to sign). Since all eigenvalues are positive, this integer must be positive. AM-GM inequality implies
$$det(A)^{\frac{1}{n}}=\left(\prod_{i}\lambda_{i}\right)^{\frac{1}{n}}\leq \frac{1}{n}\sum_{i=1}^{n}\lambda_{i}\leq 1.$$
Since $det(A)\neq 0$, and $m^{\frac{1}{n}}>1$ for $m>1$, the above inequality forces $det(A)=1$.
We therefore have equality which happens precisely when $\lambda_{i}=\lambda_{j}$ for all $i,j$. Combining this with the above equality gives the result.

Since $\det(A) \neq 0$, $A$ is nonsingular and all diagonals of $A$ must equal one.  Next, we know the relationship between the determinant and trace is
\begin{align*}
\det(A) &= \exp(tr(\ln(A))) \\
&= \exp(0) \\
&= \exp(\sum_i \ln(\lambda_i)) \\
\end{align*}.  Step 2 holds because all diagonals of $A$ are one, so their log is zero, and step 3 because natural logs are analytic. This suggests $$\sum_i \ln(\lambda_i) = 0$$, or, $$\prod_i \lambda_i = 1$$.

Then, because all diagonals of $A$ are one, $tr(A) = \sum_i \lambda_i = n$ must be true.  Hence, the AM-GM inequality states $$ \left[ \prod_{i} \lambda_i  \right]^{\frac{1}{n}} \leq \frac{1}{n} \sum_{i} \lambda_{i}$$, with equality holding only when $\lambda_i = \lambda_j$  $\forall i,j$. Thus, $$ 1^{\frac{1}{n}} = \frac{1}{n} \cdot n$$, and all eigenvalues of $A$ must be equal.  Since $\prod_i \lambda_i = 1$, all eigenvalues must be one.


Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$. Let $K$ be a Sylow subgroup of a finite group $G$. Prove that if $x \in N(K)$ and the order of $x$ is a power of $p$, then $ x \in K$.

This is how I tried...

Since $K$ is normal in $N(K)$, the factor group $$\frac{N(K)}{K}=\{K,xK,....\}
$$ is justified.

Then I took $$(xK)^{P^t}=x^{P^t}K=K$$.

This is where I am stuck.

You can also do this in a more elementary way not using the quotient group.

As already noted $K$ is normal in $N(K)$. This implies that $K$ is the only $p$-Sylow-subgroup of $N(K)$ (since $N(K)$ certainly has no $p$-subgroups of larger order and all $p$-Sylow-subgroups of $N(K)$ are conjugate under $N(K)$). Hence the group generated by $x$ (a $p$-subgroup of $N(K)$) lies in $K$ (since it must lie in SOME $p$-Sylow-subgroup of $N(K)$ and there is only one).

Clearly, $K$ is normal in $N(K)$ so we can talk about the factor group $\frac{N(K)}{K}$. Since $G$ is finite and K is a Sylow p-subgroup of $G$ so $\frac{|N(K)|}{|K|}$ doesn't have any power of $p$ and it is given that $|x|$ is a power of $p$ so $\gcd(|x|,\frac{|N(K)|}{|K|}) = 1$ $\implies \exists \ m,n\in \mathbb{Z}$ such that $|x|m + n\frac{|N(K)|}{|K|} = 1$. and since  $x\in N(K)$ we know that $x^{\frac{|N(K)|}{|K|}} \in K$. Now $x = x^{|x|m + n\frac{|N(K)|}{|K|}} = (x^{|x|})^m (x^{\frac{|N(K)|}{|K|}})^n = (x^{\frac{|N(K)|}{|K|}})^n \in K$ as $x^{\frac{|N(K)|}{|K|}} \in K$


Prove: If x≡y (mod m), then (m,x) = (m,y) The question is to prove if x≡y (mod m), then (m,x) = (m,y). I think that I should start by showing that m|x-y and by the definition of division x-y=mq for some integer q. If I let d=(m,x) then I know that d is a linear combination of m and x such that d=mr+xs for some integers r,s. I can substitute to get d=mr+(mq+y)s=mr+mqs+ys. So now I have d=m(r+qs)+ys. Does that show, by BeZout's identity that d=(m,y)? Or am I wrong?

Hint $\ $ If $\ d\mid m\ $ and $\ x-y = km\ $ then $\ d\mid x\iff d\mid y.\,$ Therefore $\,m,x\,$ and $\,m,y\,$ have the same set $\,S\,$ of common divisors $\,d,\,$ hence the same greatest common divisor (= max $S).$

We know m|x-y => mk=x-y. Also
(m,x)=d => m=dk1, x=dk2. If we put these in the first equation (mk=x-y) we have: dk1k=dk2-y => y=d(k2-k1k) => d|y. So d divides y. Therefore, d|(y,m) => (x,m)|(y,m). (1)

In the same way we can show that (y,m)|(x,m). (2)

Considering (1) and (2) we can say (x,m)=(y,m).


Pullback and Pushforward Isomorphism of Sheaves Suppose we have two schemes $X, Y$ and a map $f\colon X\to Y$. Then we know that $\operatorname{Hom}_X(f^*\mathcal{G}, \mathcal{F})\simeq \operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F})$, where $\mathcal{F}$ is an $\mathcal{O}_X$-module and $\mathcal{G}$ an $\mathcal{O}_Y$-module (and the Homs are in the category of $\mathcal{O}_X$-modules etc). This gives a natural map $f^* f_* \mathcal{F}\to \mathcal{F}$, just by setting $\mathcal{G}=f_* \mathcal{F}$ and looking at where the identity map goes.

Are there any well-known conditions on the map or sheaves that give this is an isomorphism? For instance, I was looking through a book and saw that the map is surjective if $\mathcal{F}$ is a very ample invertible sheaf (and maybe some more hypothesis on the map and $X$ and $Y$ were assumed as well).

This question may have general interest to students: What can be said about the canonical map
$\rho: \pi^*\pi_*\mathcal{E} \rightarrow \mathcal{E}?$
where $\pi: X \rightarrow S$ any map of schemes and $\mathcal{E}$ a quasi coherent $\mathcal{O}_X$-module?
Example. Let $S:=Spec(k)$ with $k$ a field and let $X$ be a scheme of finite type over $S$.
If $\mathcal{E}$ is a quasi coherent $\mathcal{O}_X$-module and $\pi: X \rightarrow S$ is the canonical map, it follows $\pi^*\pi_*\mathcal{E}\rightarrow \mathcal{E}$
is the following map:
M1. $\rho: \mathcal{O}_X\otimes \operatorname{H}^0(X, \mathcal{E}) \rightarrow \mathcal{E}.$
defined by $\rho(U)(e \otimes s_U ):= es_U \in \mathcal{E}(U)$ on the open set $U$. Here $e\in \mathcal{O}_X(U)$ and $s_U$ is the restriction of the global section $s$ to the open set $U$. Since $
\operatorname{H}^0(X,\mathcal{E})$ is a $k$-vector space it follows $\pi^*\pi_*\mathcal{E}$ is a free $\mathcal{O}_X$-module of rank $dim_k(\operatorname{H}^0(X,\mathcal{E}))$.
Example.If $\mathcal{E}$ is a rank $r$ locally trivial $\mathcal{O}_X$-module
we may associate a "geometric vector bundle" $\pi: \mathbb{V}(\mathcal{E}^*)\rightarrow X$ to $\mathcal{E}$, and the fiber $\pi^{-1}(p)\cong \mathbb{A}^r_{\kappa(p)}$ is affine $r$-space over $\kappa(p)$ - the residue field of $p$. The fibration $\pi$ is by Hartshorne,  Ex. II.5.18 a Zariski locally trivial fibration with affine spaces as fibers.
The map $\rho$ "evaluates" a global section $s$ in the fiber $\mathcal{E}(p)$ at a point $p\in X$. We get for every $p$ an element $\rho(p)(s) \in \mathcal{E}(p)$. Here $\mathbb{A}^r_{\kappa(p)} \cong Spec(Sym_{\kappa(p)}(\mathcal{E}(p)^*))$
Hence the map $\rho$ is surjective iff $\mathcal{E}$ is generated by global sections. I believe the following claim is correct:
Lemma. Let $k=A$ be any commutative ring with $\Gamma(X,\mathcal{E})$ a free $A$-module. If the map $\rho$ is an isomorphism it follows $\mathcal{E}$ is a free (or trivial) $\mathcal{O}_X$-module.
Proof. Let $S:=Spec(k)$ and let $\pi: X \rightarrow S$ be the structure morphism. There is by definition a map
M1. $\pi^{\#}: \mathcal{O}_S \rightarrow \pi_*\mathcal{O}_X$
and this gives a canonical map
$\rho: k:=\mathcal{O}_S(S) \rightarrow \Gamma(X,\mathcal{O}_X)$.
Hence $\Gamma(X,\mathcal{O}_X)$ is a $k$-algebra and left $k$-module.
It follows $\rho$ gives the abelian group $\Gamma(X, \mathcal{E})$ the structure of a $k$-module with a basis $\{e_i\}_{i\in I}$. Hence $\pi_*\mathcal{E}$ is a trivial $\mathcal{O}_S$-module on the basis $\{e_i\}_{i\in I}$. The pull back of a locally trivial $\mathcal{O}_S$-module is a locally trivial $\mathcal{O}_X$-module, and it follows there is an isomorphism
M2. $\pi^*\pi_*\mathcal{E} \cong \pi^*(\oplus_{i \in I}\mathcal{O}_S e_i) \cong \oplus_{i\in I}\mathcal{O}_X e_i$
since pull back commutes with direct sums and there is an isomorphism
$\pi^*(\mathcal{O}_S)\cong \mathcal{O}_X\otimes_{\pi^{-1}(\mathcal{O}_S)} \pi^{-1}(\mathcal{O}_S) \cong \mathcal{O}_X$
It follows $\pi^*\pi_*\mathcal{E}\cong \oplus_{i\in I} \mathcal{O}_X e_i$ is a free $\mathcal{O}_X$-module on the set $\{e_i\}_{i\in I}$. Hence if $\rho$ is an isomorphism it follows $\mathcal{E}$ is a free $\mathcal{O}_X$-module. QED
A comment on "free/trivial" modules. There seems to be confusion on what is meant by "free (or trivial) $\mathcal{O}_X$-module". I use the following definition:
Definition. Given any set $T:=\{e_i\}_{i\in I}$. We say a left $\mathcal{O}_X$-module $\mathcal{E}$ is "free (or trivial) on the set $T$" iff for any open set $U \subseteq X$ it follows
Free1. $\mathcal{E}(U) \cong \oplus_{i\in I}\mathcal{O}_X(U)e_i.$
The sum is the direct sum.
It has the following restriction maps: Given $V\subseteq U$ open subsets, it follows the restriction map
Free2. $\eta_{U,V}: \mathcal{E}(U) \rightarrow \mathcal{E}(V)$
is defined by $\eta_{U,V}(\sum_{i\in I} s_ie_i):=\oplus_{i\in I}(s_i)_Ve_i$.
Here $s_i\in \mathcal{O}_X(U)$ and $(s_i)_V\in \mathcal{O}_X(V)$ is the restriction of the section $s_i$ to the open set $V$. One checks $\mathcal{E}$ is a quasi coherent $\mathcal{O}_X$-module in general.
I believe with this definition of "free/trivial", the proof of the "Lemma/observation" above is correct. I agree - it is not difficult, but some students are not used to "sheaves" and $\mathcal{O}_X$-modules.
PS: An $\mathcal{O}_X$-module $\mathcal{E}$ is called "locally trivial of rank $r$" iff there is an open cover $\{U_i\}_{i\in I}$ with $\mathcal{E}_{U_i} \cong (\mathcal{O}_X)_{U_i}^r$ for all $i$. The module $\mathcal{E}$ is "trivial of rank $r$" iff we may choose $I:=\{1\}$ and $U_1:=X$. Hence when people use the notion "$\mathcal{E}$ is a trivial $\mathcal{O}_X$-module" this means $\mathcal{E}$ is a direct sum of copies of the structure sheaf $\mathcal{O}_X$.
Example: Let $k$ be a field and let $B:=k[x], E:=\oplus_{i\in I}k[x]e_i, S:=Spec(k)$ and $C:=Spec(B)$
Let $\pi: S \rightarrow S$ be the canonical map. It follows
$\pi^*\pi_*\mathcal{E} \cong B\otimes_k (\oplus_{i\in I}Be_i)\cong
\oplus_{i\in I} B\otimes_k B e_i \cong \oplus_{i\in I} k[x,y]e_i$
and the canonical map
$\phi : B\otimes_k E \rightarrow E$ defined by $b\otimes e:=be$ is not an isomorphism in this case since clearly $k[x,y] \neq k[x]$.
"If $\mathcal{F}=\mathcal{O}_X$, then the stereotypical example of when this is true is when $X=\mathbb{P}^n_Y$. More generally (but still with the $\mathcal{F}=\mathcal{O}_X$), the condition that "f is proper and all fibers are geometrically connected" may suffice, although I am not at all confident of that."
The fiber of the morphism $p: \mathbb{P}^n_Y \rightarrow Y$ over a point $s$ (with residue field $\kappa(s):=k$)
is the space $\mathbb{P}^n_{k}$ and projective space is "geometrically connected" in general. Let $K$ be the algebraic closure of $k$. There is an isomorphism $Spec(K) \times_{Spec(k)}\mathbb{P}^n_{k}\cong \mathbb{P}^n_{K}$, and $\mathbb{P}^n_{K}$ is connected for any field $k$.
If $\mathcal{E}$ is a finite rank locally trivial $\mathcal{O}_X$-module
we may consider $\pi:X:=\mathbb{P}(\mathcal{E}^*)\rightarrow Y$. If $Y:=Spec(k)$ with $k$ and $\mathcal{F}$ is a locally trivial $\mathcal{O}_X$-module that is not free it follows by the above Lemma that the map $\rho: \Gamma(X, \mathcal{F})\otimes \mathcal{O}_X \rightarrow \mathcal{E}$ is not an isomorphism.
The answer is: The canonical map $\rho$ is by the above argument(s) seldom an isomorphism.

Students might want to get more details since many books in algebraic geometry "leave the details to the reader".
Let $f:X\rightarrow S$ be a morphism of schemes with $V:=Spec(A)\subseteq S$ and $U:=f^{-1}(V):=Spec(B)$ and $\mathcal{E}$ a locally trivial $\mathcal{O}_X$-module of rank $r$ with $\mathcal{E}(U):=E$. Let
$\pi:Y:=\mathbb{V}(\mathcal{E}^*):=Spec(Sym_{\mathcal{O}_X}(\mathcal{E}^*))\rightarrow X$.
be the geometric vector bundle of $\mathcal{E}$ as defined in Hartshornes book Exercise II.5.18. (Note: Hartshorne does not dualize).
Let us understand the map $\rho$ in terms of this exercise.
It follows for any point $s\in U$ there is an isomorphism $\pi^{-1}(s)\cong \mathbb{A}^r_{\kappa(s)}$ where $\kappa(s)$ is the residue field of $s$. This follows from the following "functoriality argument":
$Sym_A(E^*)\otimes_A \kappa(s) \cong Sym_{\kappa(s)}(E^*\otimes_A \kappa(s)):=Sym_{\kappa(s)}(E^*(s))$.
Hence for any vector $v\in \mathcal{E}(s)$ it follows we get a canonical $\kappa(s)$-linear map
$u_v:\kappa(s)^*\rightarrow \mathcal{E}(s)$.
Taking duals we get a canonical map
$u_v: \mathcal{E}(s)^* \rightarrow \kappa(s).$
We get an element
$u_v\in Hom_{\kappa(s)}(\mathcal{E}(s)^*, \kappa(s))$ and by functoriality an element
$u_v\in Hom_{\kappa(s)-alg}(Sym_{\kappa(s)}(\mathcal{E}(s)^*), \kappa(s))$.
Hence the map $u_v$ gives a canonical $\kappa(s)$-rational point
$u_v \in \mathbb{V}(\mathcal{E}^*(s))(\kappa(s)) \cong \mathbb{A}^r_{\kappa(s)}(\kappa(s))$
since the map $u_v$ is a map of schemes
$u_v:Spec(\kappa(s)) \rightarrow Spec(Sym_{\kappa(s)}(\mathcal{E}^*(s))$
over $\kappa(s)$.
Lemma. The above construction gives a one-to-one correspondence between vectors $v\in \mathcal{E}(s)$ and $\kappa(s)$-rational points $u_v$ in  $\pi^{-1}(s)\cong \mathbb{V}(\mathcal{E}^*(s))$.
Proof: This follows from the construction above QED
Hence if we view a global section $t$ of $\mathcal{E}$ as a section of the projection morphism $\pi$, it follows the "value" of $t$ at the point $s$ is the element $u_{t(s)}\in \mathbb{V}(\mathcal{E}^*(s))(\kappa(s))$ from the Lemma corresponding to the value $t(s)\in \mathcal{E}(s)$.
The canonical map $\rho:f^*f_*\mathcal{E}\rightarrow \mathcal{E}$ gives a map on the fiber
$\rho(s): \Gamma(X, \mathcal{E})\otimes_{\mathcal{O}_{X,s}} \kappa(s) \rightarrow \mathcal{E}(s)$
with $v:=\rho(s)(t)$ for a global section $t$. The corresponding $\kappa(s)$-rational points $u_{v}\in \pi^{-1}(s)\cong \mathbb{A}^r_{\kappa(s)}$ is the evaluation of the global section $t$ at the point $s$.
Example. Let $C:=\mathbb{P}^1_k$ with $k$ a field and $\mathcal{L}_i:=\mathcal{O}(l_i)$ for $i=1,..,n$ and let $\mathcal{L}_d:=\mathcal{O}(d)$. Let $\mathcal{E}:=\oplus_i \mathcal{L}_i$ with $l_i\geq 1$ for all $i$. Consider the map $\rho$ in this case:
$\rho: \Gamma(C, \mathcal{E}\otimes \mathcal{L}_d)\otimes \mathcal{O}_C \rightarrow \mathcal{E}\otimes \mathcal{L}_d \rightarrow 0$. In this case the rank of $\mathcal{E}\otimes \mathcal{L}_d$ is $n$ for all choices of $d$.
There is an isomorphism
F1. $\Gamma(C, \mathcal{E}\otimes \mathcal{L}_d)\cong \oplus_n \Gamma(C, \mathcal{O}(l_i+d))$
hence the map $\rho$ is surjective for all $d\geq 0$, but it is not injective. The dimension in F1 can be arbitrary large.
Example. If $A$ is any commutative unital $k$-algebra with $k$ a field, and $I\subseteq A\otimes_k A$ the kernel of the multiplication map $m:A\otimes_k A\rightarrow A$ defined by $m(a\otimes b):= ab$, we may form the left and right $A$-module $J(l):=A\otimes_k A/I^{l+1}$. There are two maps $p,q:A\rightarrow J(l)$ defined by $p(a):=1\otimes a$ and $q(a):=a\otimes 1$. Let $E$ be a left $J(l)$-module. It follows $p^*p_*(E)\cong J(l)\otimes_A E$ and there is a short exact sequence
$0 \rightarrow J \rightarrow J(l)\otimes_A E \rightarrow^{\rho} E \rightarrow 0$
where $\rho:p^*p_*(E)\rightarrow E$ is the canonical map. Hence in this situation the map $\rho$ is always surjective but seldom injective. For affine schemes the map $\rho$ is always surjective.


$Set$, $Set^{op}$, categorical properties, and proving why these categories aren't isomorphic. This is going to be a bit of a long post, so fair warning.

So from what I could gather from some research, on the first link below, it goes that the category $Set$ is not isomorphic to its opposite, $Set^{op}$. My issues concern proving why.

One of the answers seems to make use of properties regarding the empty set. Namely, that, for all objects $X \in Set$, the arrow $X \rightarrow \emptyset$ exists if and only if $X = \emptyset$. This arrow would be isomorphic, clearly. There also exists an arrow $\emptyset \rightarrow X$ for all objects $X \in Set$.

So in $Set^{op}$, the dual idea reverses these arrows, i.e. for all $X$ we have an arrow $X \rightarrow \emptyset$. Apparently from what I could find at the second link below, this isn't an inherent violation ... for some technical reasons that I don't really understand. But I guess I'm basically mean to take the "reversing all arrows" part of the definition of opposite category very literally, and that it doesn't inherently break any rules by itself.

So my question is, then, how exactly do these become relevant to proving $Set$ is not isomorphic to its opposite? Because if we don't really have these violations, it seems trivial to define a functor $T$ such that

Object function: $X \mapsto T(X)$ as according to the below
Arrow function: $(f : A \rightarrow B) \mapsto (T(f) : T(A) \rightarrow T(B))$, where $T(A) = B$ and $T(B) = A$.

But obviously that's not right since it doesn't really invoke any assumptions about $Set$ and could be generalized to prove $\mathscr{C} \cong \mathscr{C}^{op}$ for any category $\mathscr{C}$.

Supposedly, according to a third post, this has something to do with the notion of "categorical properties." Namely, that - whatever these properties are, that term hasn't come up in my course at all - these are preserved by functors - which wasn't even mentioned when we covered functors.

So I'm trying to rationalize my way through this, but I can't be sure because ... well, I can't really figure out what a categorical property is, and how that applies in this case.

Does anyone have any nudges in the right direction for me?

Link 1 - Why is every category not isomorphic to its opposite?

Link 2 - What is the opposite category of $Set$?

Link 3 - Proof that $ \textbf{Set}\ $ is not isomorphic to its dual

Let me flesh out Andreas Blass's answer at one of the questions you linked to which you seem to be misunderstanding.

  Perhaps the easiest way to see that the category of sets isn't isomorphic to its dual is to observe that there is an object, namely the empty set, such that all morphisms into it are isomorphisms, but there is no object such that all morphisms out of it are isomorphisms.

What this is saying is that in $Set$ there exists an object $A$ (namely $\emptyset$) with the property that for every object $B$ and every morphism $f:B\to A$, $f$ is an isomorphism.  The claim is that this is a categorical property (preserved by isomorphisms of categories*): if a category $\mathcal{C}$ has such an object $A$ and $F:\mathcal{C}\to\mathcal{D}$ is an isomorphism of categories, then $\mathcal{D}$ also has such an object.

The proof is pretty trivial: just observe that $F(A)$ will have the same property in $\mathcal{D}$.  Indeed, suppose $B$ is an object of $\mathcal{D}$ and $f:B\to F(A)$ is a morphism.  Then $F^{-1}(f):F^{-1}(B)\to A$ is a morphism, so by the assumed property of $A$ it is an isomorphism.  Thus $f=F(F^{-1}(f))$ is also an isomorphism, as desired.

What's going on here is that we can freely "translate" between $\mathcal{C}$ and $\mathcal{D}$ by applying $F$ and its inverse, so any reasonable property of an object $A$ in $\mathcal{C}$ will also be true of the object $F(A)$ in $\mathcal{D}$.

OK, now let's get back to our story of $Set$.  In the category $Set$, there is an object $A$ with the property stated above (namely $A=\emptyset$).  If $Set$ and $Set^{op}$ were isomorphic, then there would exist some object $A'$ with the same property in $Set^{op}$.  Let's unravel what this means.  We have an object $A'$ of $Set^{op}$ (i.e., a set) such that for any object $B$ of $Set^{op}$ (i.e., a set) and any morphism $f:B\to A'$ (i.e., a function $f:A'\to B$), $f$ is an isomorphism (i.e., $f$ is a bijection).  Translating this back into ordinary statements about sets as indicated in parentheses, we have a set $A'$ such that for any set $B$, every function $f:A'\to B$ is a bijection.  However, no such set exists: for instance, if $A'$ is any set and $x$ is not an element of $A'$, then $B=A'\cup\{x\}$ is a set and the inclusion map $f:A'\to B$ is not a bijection.

So, no such object $A'$ of $Set^{op}$ exists, and so $Set^{op}$ is not isomorphic to $Set$.

*Actually, usually when people talk about "categorical properties", they really mean properties that are preserved by equivalences of categories, not just isomorphisms.  That is a more subtle story that I won't get into here, since your question is just about isomorphisms.

Part of this depends on how you define $\mathrm{Set}$ unfortunately.
A constructive $\mathrm{Set}$ has a dual interpretation as a programming language. Traditionally $\mathrm{Set}$ has been defined to be easy to use and give semantics to.
Pretty much most definitions of $\mathrm{Set}$ want the empty object $\emptyset$ to be a strict initial object. What is the programmatic interpretation of a map $X \rightarrow \emptyset$? If a map $1 \rightarrow X$ is a variable then what is its opposite? These maps are "labels" that handle control flow.
Under these sort of constructive interpretations a category with strict initial objects lacks exceptional control flow beyond the identity. For technical reasons combining control flow with exponential objects (or even a distributive category) cannot be given a constructive interpretation. Formal systems handling control flow usually use some kind of substructural logic.
There is nothing wrong with defining a category that has "labels". These categories have perfectly reasonable constructive interpretations. But these categories don't really resemble common definitions of $\mathrm{Set}$.
Reworking the foundations of math in some kind of substructural logic sounds very fun and interesting to me. But for technical reasons you can't combine many useful features of $\mathrm{Set}$ with control flow. And without "control flow" there must only be one map $X \rightarrow \emptyset$. So as fun and interesting $\mathrm{Set}^\mathrm{op}$ might be it can't be isomorphic with $\mathrm{Set}$.


Notation question about probability kernels I'm looking at a theorem in Kallenberg's Foundations of Modern Probability (1st ed). It's talking about probability kernels from $(T, \mathcal{T})$ to $(S, \mathcal{S})$, being functions $\nu : T \times \mathcal{S} \to [0,1]$ so that $\nu(t, \bullet)$ is a measure for all $t$ and $\nu(\bullet, B)$ is $\mathcal{T}$-measurable for all $B \in \mathcal{S}$.

He then states a theorem on p. 85 (first edition), which I am paraphrasing:

  Fix two measurable spaces $(S, \mathcal{S})$ and $(T, \mathcal{T})$, a $\sigma$-ﬁeld $\mathcal{F}
 \subset \mathcal{A}$, and a random element $\xi$ in $S$ such that
  $P(\xi \in \bullet \mid \mathcal{F}): \Omega \times \mathcal{S} \to
 \mathbb{R}$ is a kernel $\nu$. Further consider an
  $\mathcal{F}$-measurable random element $\eta$ in $T$ and a measurable
  function $f$ on $S \times T$ with $E|f(\xi, \eta)| < \infty$. Then

  $$E(f(\xi, \eta) \mid \mathcal{F}) = \int \nu(ds)f(s,\eta) \quad
 \text{a.s.}$$

What is he talking about with $\nu(ds)$? I thought $\nu$ was a function $\Omega \times \mathcal{S} \to [0,1]$...

The dependence on $\omega \in \Omega$ is hidden not just for $\nu$ but throughout that equation (and frequently beyond). A more explicit version would be

  $$ E[f(\xi,\eta) \mid \mathcal{F}](\omega) = \int_S \nu(\omega, ds) f(s, \eta(\omega)) \text{ for almost every $\omega \in \Omega$.}$$

Essentially in each “location” / “case” / “atom” (represented by $\omega$) that can be discerned by $\mathcal{F}$ in the probability space, we know that $\xi$ is distributed according to a restricted law $\nu(\omega, \mathord{\cdot})$, and we further know the value of $\eta$ is fixed because it is $\mathcal F$-measurable. So we can then just integrate $f(\cdot, \eta(\omega))$ against $\nu(\omega, \mathord{\cdot})$ over $S$ to find the conditional expectation.

(For search purposes, this is the "disintegration" theorem in Kallenberg.)

In the theorem, $\nu$ is defined to be the probability density kernel $P(\xi|\mathcal{F})$. This means that $\nu(ds)$  is the probability measure of $ds\in \Omega$. So, if you integrate over the entire domain space, $\int_{\Omega}{\nu(ds)} = 1$. So in the Theorem, it's the measure theoretic approach towards defining an expected value.


Sequence of continuous functions with bounded derivative Let $f_n$ be a sequence of continuous functions on $[0,1]$, and continuously differentiable on $(0,1)$. Assume $|f_n|\le 1$ and $f_n'\le 1$ $\forall x\in [0,1]$ and $n$. Then

$f_n$ is a convergent sequence in $C[0,1]$
$f_n$ has a convergent subsequence in $C[0,1]$

well, by Bolzano-Weirstrass theorem (every bounded sequence has a convergent subsequence) we can say $2$ is correct, I am not able to say true or false against $1$, please help.

Consider the case $f_n(x)=(-1)^n$ for all integer $n$ and all $x\in [0,1]$.
Take $f_n(x):=-x^n$. Then $|f_n(x)|\leq 1$ and $f'_n(x)=-nx^{n-1}\leq 0\leq 1$. We have that $f_n$ converges pointwise to the function which is $0$ in $[0,1)$ and $-1$ at $1$. A uniformly converging subsequence would converge to this map, which is not possible.

However, if we replace the condition "$\forall x\in[0,1], f'_n(x)\leq 1$" by "$\forall x\in[0,1], |f'_n(x)|\leq 1$", Arzelà-Ascoli theorem applies.


If a is less than 1 in absolute value, than n times a to the n'th power converges to zero Prove that if $|a|<1$ then
$$\lim_{n\to\infty}{n\cdot a^n}=0$$
Lopital's rule won't help here, and this problem appears in a book before lopital has been taught.

I noticed that:
$$\frac{a_{n+1}}{a_n}=a\frac{n+1}{n}\to a$$

So, can I conclude that it behaves like a geometric series for large n, so it converges to zero?

*Other solution methods are welcome.

You intuition is spot on. You have that $$\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=a<1$$

Now choose a number $\ell$ with $a<\ell<1$. By the definition of limit, there must exist $N>0$ such that whenever $n\geqslant N$ we have $$\frac{a_{n+1}}{a_n}<\ell $$

In particular when $k=1,2,3,\ldots$ $$a_{N+k}<\ell a_{N+k-1}$$

By repeated use of this we can write, for $k=1,2,\ldots,$ that  $$a_{N+k}<\ell ^k a_N$$

Note $N$ is fixed, so as $k\to\infty$ we get that $$\lim\limits_{k\to\infty}a_{N+k}\leqslant a_N \lim_{k\to\infty} \ell^k=0$$

Since we know $a_n\geqslant 0$; we conclude $\lim\limits_{k\to\infty} a_k=0$.

This is indeed the very idea behind D'Alambert's criterion

  Let $\langle a_n\rangle $ be a sequence of positive terms, and suppose that $$\lim_{k\to\infty}\frac{a_{k+1}}{a_k}=\ell <1$$
  Then $\lim\limits_{k\to\infty} a_k=0$. What's more, $\displaystyle\sum_{k\geqslant 0}a_k$ exists.

In fact, if you know about $\limsup$ and $\liminf$; we can give a finer criterion.

  If $$\ell =\limsup\limits_{n\to\infty}\frac{a_{n+1}}{a_n}<1$$ then the sequences is absolutely summable. If $$\ell'=\liminf\limits_{n\to\infty}\frac{a_{n+1}}{a_n}>1$$ then the sequence is not summable (the sum diverges).

Let $\frac{1}{b} = a$ where |b|>1 therefore the limit becomes $$\lim_{n\to\infty}{n\cdot \frac{1}{b^n}}=0$$ As n increases the absolute value of the denomenator increases and therefore the absolute value of the entire expression becomes smaller and therefore converges to zero.


Why dont we say if $f(U)$ is open for every open set $U$ in $A$, then $f$ is countinous? Definition of continuity in topology;

Let $f :A \to B$.
If $f^{-1}(U)$ is open in A for every open set U in B, we say f is countinous.

Why dont we say if $f(U)$ is open for every open set $U$ in $A$, then $f$ is countinous ?

Does the image of an open set under a continous mapping need to be open ?
 Thank you for your help.

No. This would make the constant functions ($f:X\to\mathbb R; \ f(x) = c$) discontinuous, since

$$f(U) = \{c\}$$
for $U\ne\emptyset$ open.

Consider the metric space $T=\left \{ 0,1 \right \}$ with the discrete metric $d(x,y)=1$ if $x \neq y$ and $d(x,y)=0$ if $x=y$.

It is easy to show that any subset of $T$ is open.

Take any metric space $X$ and any discontinuous function $f:X \rightarrow T$ to obtain a counterexample.


To find subbasis for a topology on X. Consider the discrete topology $\tau$ on $X:= \{ a,b,c, d,e \}$. Find subbasis for $\tau$ which does not contain any singleton sets.

The definition of subbasis is as follows:

  Definition: A subbasis $S$ for a topology on $X$ is a collection of subsets of $X$ whose union is $X$.

So let $S$ be equal to the  collection of $\{a,b\}$, $\{c,d\}$ and  $\{d,e\}$.

Clearly union of these three elements is $X$.

So should be $S$ - as defined - be taken as subbasis? Please check the answer I posted in comment.

Hint: You can write $\{a\}$ as $\{a,b\}\cap\{a,c\}$. Do the same with each of the elements of $X$.

The correct singleton sets should be $\{(a,b),(b,c),(c,d),(d,e),(a,e)\}$


Optimizing Projectile Arclength I was running through some old Putnam problems and came across one from the 1940 exam that asked the following:

  A stone is thrown from the ground with speed $v$ at an angle $θ$ to the horizontal. There is no friction and the ground is flat. Find the total distance it travels before hitting the ground. Show that the distance is greatest when $\sin θ \cdot\ln (\sec θ + \tan θ) = 1$.

My work:

We can describe the motion with the following vector:

$$\vec r(t)=\langle v\cos\theta \cdot t,v\sin\theta \cdot t-gt^2/2\rangle$$

We know the vector hits the ground when the $y$ component is equal to $0$. Solving for $t_0$, we get that the ball reaches the ground again at $t_0=\frac{2v\sin\theta}{g}$.

Let's set up an integral for the arc length $s$:

$$s=\int_0^{t_0}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$

Some quick differentiation and we can replace $\frac{dx}{dt}$ and $\frac{dy}{dt}$ with the following identities:

$$s=\int_0^{t_0}\sqrt{(-v\sin\theta)^2+(v\cos\theta-gt)^2}dt$$

Since we are trying to optimize $\theta$, if we take the derivative of the arc length, set it equal to $0$, and find the maximum, then we can solve for optimal $\theta$. We'll also replace $t_0$ with its identity in terms of $\theta$.

$$\frac{ds}{d\theta}=\frac{d}{d\theta}\int_0^{\frac{2v\sin\theta}{g}}\sqrt{v^2-2v\cos\theta\cdot gt+g^2t^2}dt$$

$$\frac{ds}{d\theta}=\frac{2v\cos\theta}{g}\sqrt{v^2-2v\cos\theta\cdot gt+g^2t^2}$$

Here I am unsure how to relate $t$ to $v$ and $\theta$. Do I use the identity for $t_0$, a kinematics equation, or simply treat $t$ as a constant? This, of course, assumes my work thus far has been valid.

Next I would either show that the identity holds after finding $\theta$ or use some relation involving $\theta$ and show that it's an identity to the listed equation.

Thanks for taking the time to read/respond.

WLOG, $v=1$ and $g=1$ (you can rescale time and space independently), the trajectory is

$$x=t\cos\theta,\\y=t\sin\theta-\frac{t^2}2,$$ and the total travel time is

$$2\sin\theta.$$

Then you want to maximize

$$L=\int_0^{2\sin\theta}\sqrt{(t-\sin\theta)^2+\cos^2\theta}\,dt=\int_{-\sin\theta}^{\sin\theta}\sqrt{t^2+\cos^2\theta}\,dt\\
=\sin^2\theta\int_{-1}^{1}\sqrt{u^2+\tan^2\theta}\,du\\
=\frac12\sin^2\theta\left.\left(u\sqrt{u^2+\tan^2\theta}+\tan^2\theta\log(u+\sqrt{u^2+\tan^2\theta})\right)\right|_{u=-1}^1\\
=\frac12\sin^2\theta\left(2\sec\theta+\tan^2\theta\log\frac{\sec\theta+1}{\sec\theta-1}\right).$$

The claim should follow by differentiation and simplification.

Instruction is as follows

1) Evaluate $s(\theta, v)$

2) Take the gradient of the expression and equate to zero $\nabla_{v, \theta}s(v, \theta)=0$

3) This gives two equations

4) Solve them

5) Find the Hessian
$$\begin{bmatrix}
\partial^{2}_{v, v}s(v, \theta) & \partial^{2}_{v, \theta}s(v, \theta) \\
\partial^{2}_{v, \theta}s(v, \theta) & \partial^{2}_{\theta, \theta}s(v, \theta)
\end{bmatrix}
$$
6) Evaluate the Hessain at the extremum's you've found in 4. Find the eigenvalues of the Hessian.

7) If all the eigenvalues of Hessian are positive then we have a minimum. If all the eigenvalues of Hessian are negavive then we have a minimum.


EdExcel GCSE question about Hannah and the sweets: show that $n^2-n=90$ This is my reconstruction of the EdExcel GCSE question that has caused such a Twitter storm in the UK in the last 24 hours, along with its solution.

Hannah has a bag containing $n$ sweets, 6 of which are orange. She eats two sweets at random from the bag.

The probability that the two sweets Hannah eats are both orange is $\frac{1}{3}$. Show that $n^2 - n = 90$.

Possible follow-up (I don't know if this was part of the original): how many sweets were there in total in the bag?

If both sweets that Hannah eats are orange, then of course the first one must be orange, and so must the second one. We can calculate the probability of that happening.

For the first sweet, there are $n$ in total, and 6 are orange, so the probability that the one she chooses is orange is $\frac{6}{n}$.

Once she's done this, there are now $n-1$ sweets left, 5 of which are orange. So the probability that the second one is also orange is $\frac{5}{n-1}$.

The overall probability (that both of these things happen) is then the product of these: $\frac{6}{n} \times \frac{5}{n-1}$.

Now, we are told in the question that the probability that they are both orange is $\frac{1}{3}$. So we have $$\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3}$$

If we multiply both sides by $n$, then by $n-1$, then by 3, we get $$6 \times 5 \times 3 = n(n-1)$$ or $$90 = n^2 - n$$ which is what we were asked to show.

As a follow-up, we can find the value of $n$, i.e., the total number of sweets in the bag. We need to solve $$n^2 - n - 90 = 0$$ which we can do by factorising: $$(n-a)(n-b) = n^2 - n - 90$$

We need two values whose sum is $1$ and whose product is $-90$. It is not hard to see that the values $a=10$ and $b=-9$ will do.

This means that there were either $10$ sweets or $-9$ sweets in the bag. Obviously there can't have been $-9$ sweets; so we conclude that there were $10$ sweets in the bag.

The chance of an orange sweet being chosen the first time is 6/n. Everyone got that right.
The chance of an orange sweet being chosen the second time is (6-6/n)/(n-1). Most people seem to have got that wrong.
If the chance of both sweets being orange is 1/3, then n X n=108, but that's impossible because n must be an integer. In fact the chance of both sweets being orange can never be 1/3. The question is in error either deliberately or by mistake.


Prove that when $x>0$, $(x+1/2)^3>3x^2+1/8$ without graphing Prove that when $x>0$,  $(x+1/2)^3>3x^2+1/8$ without graphing.

The only thing I have tried till now is graphing, from which I proved this, but how would you prove this mathematically?