subgroup of group of order $p^2$ Let $p\ge5 $ be a prime. Then

$\mathbb{F_p\times\mathbb{F_p}}$ has atleast five subgroups of order p.

Every subgroup of $\mathbb{F_p\times\mathbb{F_p}}$ is of the form ${H_1\times{H_2}}$ where ${H_1 , {H_2}}$ are subgroups of $\mathbb{F_p}$.

Every subgroup of $\mathbb{F_p\times\mathbb{F_p}}$ is an ideal of the ring $\mathbb{F_p\times\mathbb{F_p}}$

The ring $\mathbb{F_p\times\mathbb{F_p}}$  is a field.

1...true .Let p=5 then subgroup generated by (1,0) , (2,0) , (3,0) , (4,0) , (0,1),....are all subgroups of order 5.
2...not true consider subgroup generated by (x,x) which cannot be written as ${H_1\times{H_2}}$.
3...I am not able to conclude for this option
I know that ideals of ring corresponds to  normal subgroups of corresponding group . $\mathbb{F_p\times\mathbb{F_p}}$ is abelian group so all subgroups are normal so can i say that all subgroups are normal subgroups and hence ideal ?
4...not true because in Fp×Fp  (1,0)(0.1)=(0,0) hence  it is not integral domain   hence it is not field.

(1) is true.  Consider any element $x(k,n) = (nk\pmod p) \times k$ with $0\leq n<p$ and $k\neq 0\pmod p) $. It is easy to see that $x(k,n)$ gnerates a cyclic subgroup of $\Bbb{Z}_p \times \Bbb{Z}_p$.  It is also easy to show that the group generated by  $x(k,m)$ does not contain $x(k,n)$. Thus there are $p$ distinct subgroups of this form.

(2) is false. Consider the subgroup $\{ k \times k\}$ with $0\leq k<p$. This is not of the form $H_1 \times H_2$


Proof about fundamental class for M exists then M is compact. In this proof, I can't see why the claim"this image is compact" is true, and I don't understand why it implies $M$ is compact.

Image of a compact space under a continuous function is compact. You should know this right, I figure you took General topology?


Prove that every odd prime number can be written as a difference of two squares. Prove that every odd prime number can be written as a difference of two squares.
Prove also that this presentation is unique.
Is such presentation possible if p is just an odd natural number?
Can 2 be represented this way?

Answers

\3. Yes the presentation (i.e. odd numbers being written as differences of two squares) is possible for all odd natural number however the presentation may not be unique. For example, $57=11^2-8^2=29^2-28^2$.

\4. 2 can't be written as a difference of two squares because 4-1=3 and 1-1=0 and the difference of squares grows to integers larger that 3.

Can I get some help in proving questions 1 and 2?

$1$. Let $(x+y)(x-y) = p$

Since $p$ is prime, the smaller divisor has to be one, ie. $(x-y) = 1$, giving $2y+1 = p \implies y = \frac{p-1}{2}$ (you're guaranteed y is an integer because $p$ is an odd number).

So the only possible solution set is $x = \frac{p+1}{2}, y = \frac{p-1}{2}$

$2$. Uniqueness already established via reasoning above.

$3$. Possible, but it will be non-unique as $(x-y)$ can take on multiple values, e.g. $1$ or a single prime divisor of $p$ or a product of some (but not all) prime divisors of $p$.

$4$. No, because again $(x-y)$ = 1 is forced. But now you get $x = \frac{3}{2}$ which is non-integral. So no integer solution sets exist.

I found some thing very useful, the product of any two odd numbers can always be expressed as the difference of squares of two integers.

Let a,b belongs to Z+ , are two odd numbers,

Now a+b is always even, and [a+b]/2 certainly generates an integer and also [a-b]/2 generates another integer.

Now {[a+b]^2/4} -{[a-b]^2/4} = ab

Now consider the case of any odd prime number

Any prime number can be written as P X 1 since it cant be decomposed further.

So {[p+1]^2/4}-{[p-1]^2/4} = p

Any odd prime number can be expressed as the difference of squares of two integers. Of course its unique !!! as the numbers [P+1]/2 and [P-1]/2
are unique for a given prime number.
.


Problem in solving this book and pages question I was solving some old olympiad problems and I got that one. I m stuck at it.

"In a book with page numbers 1 to 100,some pages are torn off. The sum of the numbers on the remaining pages is 4949. How many pages are torn off??"

I tried to the summation formula(sum of all the pages is sum of all the natural numbers from 1 to 100,since the pages goes from 1 to 100)  and then try to eliminate some numbers (by hit and trial) to see whether the sum comes out to be 4949 or not but I wasn't so lucky. Hoping for help. Any suggestion is heartily welcome

We have that $1+\cdots+100=5050.$ So, if the remaining pages sum $4949$ the pages which are not in the book add up $101.$ Moreover, if $2a-1$ is not in the book then $2a$ is also not in the book. They add up $4a-1.$ It's not possible to torn out only a page because $4a-1=101$ has no integer solution. If we turn out two pages we have to solve $4a-1+4b-1=101$ which also doesn't have integer solution. So, assume $n$ pages are torn out. We have $$4a_1-1+\cdots+4a_n-1=101 (\iff 4(a_1+\cdots+a_n)=101+n).$$ Thus the number of pages can be $3,7,11, \dots$ because $101+n$ must be a multiple of $4.$ Now, note that if $n\ge 7$ then $$\dfrac{n(n+1)}{2}=1+\cdots+n\le a_1+\cdots+a_n=\dfrac{101+n}{4}< \dfrac{n(n+1)}{2}$$ gives a contradiction. So, $n=3.$

Edit

Let's prove of the last inequality for $n\ge 7:$
$$\dfrac{101+n}{4}< \dfrac{n(n+1)}{2}\iff 2n^2+2n>101+n\iff 2n^2+n>101.$$ Now, $$n\ge 7\implies 2n^2+n\ge 98+7=105>101,$$ and we are done.

Edit 2

In order to determine the pages we have to solve $$a_1+a_2+a_3=26.$$ Then, the pages are $2a_1-1,2a_1;$ $2a_2-1,2a_2;$ and $2a_3-1,2a_3.$

Now, solving $a_1+a_2+a_3=26$ is to get the partitions of $26$ into distinct parts (that is, $a_1,a_2$ and $a_3$ are different). According to Wolfram Alpha there are $165$ solutions. (See https://www.wolframalpha.com/input/?i=PartitionsQ(26).)

Start by summing up all the pages from $1$ to $100$ to see how much it would add up if no pages are torn. Then it should become clear.


Basic real analysis - Proving convergence of series I am not sure if I'm heading in the right direction for this proof. I started off with the fact that a series is convergent if it satisfies the Cauchy criterion so I have to show that the sequence an satisfies the Cauchy criterion. I begin to prove, "If the series bn converges then the series an converges. Since bn converges then bn satisfies the Cauchy criterion and there exists a bm such that the sequence bn converges to. From here I don't know where to go to show an satisfies the Cauchy criterion.

Cauchy criterion:

Suppose $\lim(a_n/b_n)=C\neq0.$

($\Rightarrow$)Assume $\sum a_n$ converge,

let there exists $K\in \Bbb N$ such that $C/2\le a_n/b_n \le 2C$, $\forall n\ge K$, i.e.
$$(C/2)b_n\le a_n\le 2Cb_n\text{ , } \forall n\ge K$$
$$(C/2)\sum_{n=K}^{\infty}b_n \le \sum_{n=K}^{\infty}a_n\le 2C\sum_{n=K}^{\infty}b_n$$
$\sum a_n=\sum_{n=1}^{K-1}a_n+\sum_{n=K}^{\infty}a_n$. So $\sum_{n=K}^{\infty}b_n$ converge, hence $\sum b_n$ converge.

Similar argument for another direction.


Show an equation has 2 solutions Show that $2+\ln(x)=x$ has two solutions at $x>0$

that is equivalent to $2+\ln(x)-x=0$ which mean that there are two values of $x$ where $f(x)=0$.

How should I approach this?

One approach would be to note that $f(x)=2+\mathrm{ln}(x)-x$ is continuous and note that:
$$f(\frac{1}{e^2})=2-2-\frac{1}{e^2}=-\frac{1}{e^2}<0$$
$$f(1)=2+0-1=1>0$$
$$f(e^2)=2+2-e^2<0$$
such that $f$ should have (at least) two zeros.

Note that this approach only shows existence of two solutions, not what they are.

Why bother messing around with analysis at all? Let's solve for $x$!

$ 2+\ln(x)=x$

$ -2 = -x+\ln(x) $

$e^{-2} = e^{-x+\ln(x)}$

$e^{-2} = e^{-x}e^{\ln(x)}$

$-e^{-2} = -xe^{-x} $

Notice here that on the right-hand side, we have a equation of the form $y=ae^a$, where $a<0$.

To solve for $a$ (and ultimately, x), we can use the Lambert W-function, which is defined as the inverse of the relation y. (I.e. $W(ae^a)=a$)

However for $a<0$ the Lambert W-Function is in fact bijective, and precisely bijective. The two solutions:

$W(-e^{-2})=-x$

and

$W_{-1}(-e^{-2})=-x$

So therefore:

$x=-W(-e^{-2})$

OR

$x=-W_{-1} (-e^{-2})$

Addendum:
Technically, if $-x ∈ (-1,0)$, then we must use the function $W_0(x)$ instead of the function $W_{-1}(x)$. But nonetheless there are two solutions because of the property of the Lambert W-function. It really isn't any different to solving an equation to the form $x^2=16$ and realizing that there must be two solutions, $x=-4, 4$.


Combination of linear functions that give the derivative operator Let $D$ be the derivative operator and $C^\infty$ the set of functions
  derivable infinitely many times.

  Here $f^n=f\circ f\circ\cdots\circ f\text{, }n\text{ times}$

It can be easily shown that there exists no linear function $f\in\mathcal{L}(C^\infty)$  such that $f^2=D$.

If I am not wrong, this can be easily extended to the non existence of a function $f\in\mathcal{L}(C^\infty)$ such that $\prod f=D$ as a finite product.

What can be said for an infinite product? How can one (dis)approve of the existence of a linear function $f\in\mathcal{L}(C^\infty)$ such that $f^n=D$ ?

It seems unlikely to me, but I have found no way to prove it.

Not claiming to understand the main question of the "infinite product", I could comment on the title and the linear non-existence claim. Fractional derivative is a linear operator, where one could write $D^{\frac{1}{2}}$ as a functional square root of the differentiation operator that when applied twice to a function will have the same effect as differentiation. (see further, e.g., the "Short Introduction to Fractional Calculus").


Is there positive such that the inequality is true? Let $A\in [0,1]$ and $B\in (0,1)$, then there is small $c>0$ such that $A-cB\geq 0$.
Is the statement true ?
My way if A is not zero that is clear, cause I can choose $c=\frac{A}{B}$. Moreover, $\frac{c}{2}$. But in case $A=0$ ?

From what I see, the relation $A-cB\geq 0$ is only true for $c>0$ when $A\in ( 0,1] ,\ B\in [ 0,1]$
Here it is
$ \begin{array}{l}
A-cB\geqslant 0,\\
\\
cB\leqslant A\Longrightarrow c\leqslant \frac{A}{B}\\
\\
0< c\leqslant \frac{A}{B} ,\ \frac{A}{B} \ cannot\ be\ zero\Longrightarrow A\neq 0,\ B\neq \infty ,-\infty \\
\Longrightarrow A\in ( 0,a] ,\ B\in [ b,c] ,\ where\ a,c,b\ are\ both\ finite\ numbers\ with \ a\geq0
\end{array}$

The only way that $0-cB \geq 0$ is that either $c $ or $B $ is equal to zero (because $cB\overset{!}{=}0$).
So no, this is not possible.


Bridgeland-Stability Condition: Why is the Harder-Narasimhan filtration unique? I'm trying to understand Bridgeland's notion of stability condition on a triangulated category as defined in Definition 1.1 of the paper Stability conditions on triangulated categories.

See also Definition 3.3 in the same paper.
The decompositions of a non-zero object as in Definition 3.3(c) are the Harder-Narasimhan filtrations.

  Question: Why is the Harder-Narasimhan filtration of any non-zero object unique (up to isomorphism of the semistable factors)?

Thank you for your help!

The short answer is that the filtrations are unique because they arise as factorization of an orthogonal, multiple factorization system: see here §4.1.

The even shorter answer is that you can prove it for triangulated categories, but it's unsatisfying.


Markov Chains as Autoregressive Processes Is there a simple way to approximate a Markov Chain as an Autoregressive Process, for instance, an AR(1) process?

I am aware that it is easy to approximate an AR(1) process with a Markov Chain, but I am not sure about the converse.

Thanks.

AR(1) processes are Markov chains only if the innovation/error term is independent of the past.


How do you solve for the limit of this series? I need to take the limit of this summation so that I kind find out whether it converges or diverges. The equation is:

$$\sum_{k=1}^\infty \frac{4}{k+4}$$

What I have tried so far is the following:

$$\lim_{k \rightarrow \infty} \frac{k+4}{(k+1)+4}$$

Which then gives us $\frac{4}{4} = 1$, however I know that this is incorrect. How do your properly take the limit of the summation?

If you're just trying to find out weather it converges or diverges, you don't need to actually find its limit.

Instead try a comparison test with the Harmonic Series $$\sum_{n=1}^{\infty}\frac{1}{n}$$

This definitely diverges as it's a p-series with p=1

For $k\geq 1$,
$$
\frac{4}{4+k}=\frac{1}{1+\frac{k}{4}}> \frac{1}{1+k}
$$

But, for $N\geq 1$
$$
\sum_{k=1}^{N}\frac{1}{1+k}=\frac{1}{2}+\frac{1}{3} + \ldots + \frac{1}{N+1}= \big(\sum_{k=1}^{N}\frac{1}{k}\big)+ \big(\frac{1}{N+1}-1\big)
$$

So, since the harmonic series diverges, given any $M>0$, you can find an index $N$ such that $\sum_{1}^{N} 1/k > M+1$, and since $1/(N+1)-1 > -1$, for that same index N, $\sum_{1}^{N}1/(1+k)>M$, and $\sum 1/(1+k)$ diverges.

Therefore, $\sum 4/(4+k)$ also diverges, by the comparison test.


Prove, by induction, that $5^n + 5 < 5^{n+1}$ for all $n\in\Bbb N$. Prove, by induction, that $5^n + 5 < 5^{n+1}$ for all $n\in\Bbb N$.

Attempt:

If $n = 1$, then $5^1 + 5 < 5^{2}$ => $10 < 25$ which is a true statement so the base case holds.

Assume $5^k + 5 < 5^{k+1}$ is true for some $k\in\Bbb N$.

How to prove it for $n = k+1$?


What is the maximal path of a tree? Could anyone explain obviously what the maximal path is ? Is it necessary for a tree that has two maximal paths that share no common vertex ?

I will assume that you meant a connected tree. Also, I am using maximal to mean "of maximal length," which may not be what you meant.

A path in a tree is a sequence of vertices, with no repeats, where successive vertices in the sequence are adjacent. If a path $P$ is at least as long as any other path, then that path is maximal. It is possible for there to be more than one maximal path. For example, in this tree,

     C
    /
A--B
    \
     D

both of the paths $A,B,C$ and $A,B,D$ are maximal, since they are of length $3$, and there are no paths of length $4$ or greater.

Note that these two paths share a common vertex. This is always true: any two maximal paths will share a common vertex.

Why? Suppose you had two disjoint paths, $P_1=v_1,v_2\dots,v_n$ and $P_2=w_1,w_2\dots, w_n$. Consider the pair $v_i,w_j$ for which the distance between $v_i$ and $w_j$ is minimal (here, distance means the length of the shortest path connecting them). Then there is a path $v_i,p_1,p_2,\dots,p_k,w_j$ which connects $v_i$ and $w_j$. None of the intermediate vertices $p_1,\dots,p_k$ can be in $P_1$ or the $P_2$, because of the minimality when we chose $v_i$ and $w_j$. This implies that $P_1$ and $P_2$ weren't maximal, since at least one of the following paths are longer:
$$
v_1,\dots,v_i,p_1,\dots,p_kw_j,\dots,w_n
$$
$$
w_1,\dots,w_j,p_k,\dots,p_1,v_i,\dots,v_n
$$
We have shown that disjoint paths are not maximal, which means that any two maximal paths intersect.

Maximal path in a graph G is a path P in G that is not contained in a longer path.

When a graph is finite, no path can extend forever, so maximal(non-extendible) paths exist.


Why does the Cauchy-Schwarz Inequality even have a name? When I came across the Cauchy-Schwarz inequality the other day, I found it really weird that this was its own thing, and it had lines upon lines of proof.

I've always thought the geometric definition of dot multiplication:
$$|{\bf a }||{\bf b }|\cos \theta$$ is equivalent to the other, algebraic definition: $$a_1\cdot b_1+a_2\cdot b_2+\cdots+a_n\cdot b_n$$
And since the inequality is directly implied by the geometric definition (the fact that $\cos(\theta)$ is $1$ only when $\bf a$ and $\bf b$ are collinear), then shouldn't the Cauchy-Schwarz inequality be the world's most obvious and almost-no-proof-needed thing?

Can someone correct me on where my thought process went wrong?

Side note: it's actually the Cauchy-Schwarz-Bunyakovsky inequality, and don't let anyone tell you otherwise.

The problem with using the geometric definition is that you have to define what an angle is. Sure, in three dimensional space, you have pretty clear ideas about what an angle is, but what do you take as $\theta$ in your equation when $i$ and $j$ are $10$ dimensional vectors? Or infinitely-dimensional vectors? What if $i$ and $j$ are polynomials?

The Cauchy-Schwarz inequality tells you that anytime you have a vector space and an inner product defined on it, you can be sure that for any two vectors $u,v$ in your space, it is true that $\left|\langle u,v\rangle\right| \leq \|u\|\|v\|$.

Not all vector spaces are simple $\mathbb R^n$ businesses, either. You have the vector space of all continuous functions on $[0,1]$, for example. You can define the inner product as

$$\langle f,g\rangle=\int_0^1 f(x)g(x)dx$$

and use Cauchy-Schwarz to prove that for any pair $f,g$, you have
$$\left|\int_{0}^1f(x)g(x)dx\right| \leq \sqrt{\int_0^1 f^2(x)dx\int_0^1g^2(x)dx}$$

which is not a trivial inequality.

Flip your argument around. The Cauchy-Schwarz inequality makes thinking about angles seem outmodish, vestigial, obsolete. You don't need protractors to do geometric analysis anymore. You just need algebra: just one quadratic inequality. So much of what your proof using $\cos \theta$ has been doing only requires a simple algebraic inequality, and neither $\cos$ nor $\theta$.

Why algebra is important is out of scope to this answer.


What is the symbolic form of "there does not exist a largest natural number " Other students in office hour said this is the correct form
$(\forall x)(\exists y)(y>x)$
{ for all x natural number, there exists y such that y is greater than x }

But
"there does not exist a largest natural number "
$\neg(\exists x)(x\text{ largest natural number})$

Am I even close ?

The formula $\forall x \exists y (y > x)$ reads "for any $x$ there is a $y$ which is larger than $x$" which says that any $x$ is not largest and therefore no $x$ is largest. You can also use the formula $\forall \equiv \neg\exists\neg$ to get

$$
\begin{align*}
\forall x \exists y (y > x) &\equiv \neg\exists x \neg \exists y (y > x) \\
&\equiv \neg \exists x \forall y \neg(y > x) \\
&\equiv \neg \exists x \forall y (y \le x)
\end{align*}
$$

which says that "there is no $x$ such that every $y$ is less than or equal to $x$". This better fits with the "there does not exist a largest natural number" phrasing.

Since you are interested in formalisms, I would suggest to rephrase the statement to be more formal. In Mathematical logic, one usually uses quantors (similarly, the negation operator) and parentheses in the following way:
$$
\forall x (\; \text{logical statement} \;)
$$
So, nesting this, your statement becomes
$$
\forall x \in \mathbb{N}\left(\,\exists y \in \mathbb{N}\left(\,y>x\right) \right)
$$
and the other statement, the one you were asking for, becomes
$$
\neg (\,\exists M \in \mathbb{N}
\,(\,\forall l \in \mathbb{N} \,(\,M \geq l))
$$
where I replaced your variable-symbol $x$ by $M$ and used $l$ as the variable-symbol for your natural number.


About preimage of a function. I'm trying to show that $f(f^{-1}(E))= E$ is not necessarily true. I want to use a non-injective function as an example, but what I want to know is if I can consider $E$ to be a set which is not a subset of the range of $f$. If so, I'd like to know why. Thank you!

Yes you can consider $E$ to be a set which is not a subset of the range of $f$. In fact $f(f^{-1}(E))\neq E$ if and only if $E$ is not a subset of the range. This is because $f(f^{-1}(E))=\text{range}(f)\cap E$ (can you prove it?).

This picture should avail. $a \in$ set X. $b, c \in$ set $Y.$ Your $E$ is my $S$ and $S \subsetneq Y$ here.

The pink bespeaks both $f^{-1}(S)$ and $f: X \to Y$ defined by $f(a) = b$. The green point is $c \in Y$.

Then $f(f^{-1}(\{c\}))=f(\varnothing)=\varnothing\neq\{c\}\quad\text{and}\quad f(f^{-1}(Y))=f(\{a\})=\{b\}\neq Y$.


"modulo" or "mod" is a binary operator.

In general, $a$ mod $b$ gives the remainder $r$ obtained on dividing $a$ by $b$, that is,

$a=b \times c + r$ implies  $a$ mod $b$ = $r$, where $a,c \in \mathbb{Z}$, $b\in \mathbb{Z^+}, r\in \mathbb{W}$ with $0≤r<b$.

When $a$ and $b$ leave the same remainder on dividing by $c$, we say that $a$ and $b$ are equivalent or congruent in the world or more appropriately the system of $c$, that is,

$a \equiv b \pmod c$ iff $a$ mod $c=b$ mod $c$, where $a, b\in\mathbb{Z}$ and $c\in\mathbb{Z^+}$.

Modulo Arithmetic is just basically the remainder.let's say that we do 7mod3, it is 4 because 7/4 has a remainder of 3. its really simple


Improper Riemann integral questions and determining whether they exist. Explain why
\begin{equation*}
\int_{0}^{1} \frac{dx}{x}, \quad \int_{1}^{\infty} \frac{dx}{x}
\end{equation*}
are improper Riemann integrals, and determine whether the limits they represent exist.
The improper Riemann integrals are the limits
\begin{equation*}
\begin{split}
\int_{0}^{1} \frac{dx}{x} &= \lim_{b\to 0^+} \int_{b}^{1} \frac{dx}{x} \\
&= \lim_{b\to 0^+} \left[\ln{(x)}\right]_{b}^{1} \\
&= \lim_{b\to 0^+} \left(-\ln{(b)}\right) \\
&= -\infty
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
\int_{1}^{\infty} \frac{dx}{x} &= \lim_{b\to \infty} \int_{1}^{b} \frac{dx}{x} \\
&= \lim_{b\to\infty} \left[\ln{(x)}\right]_{1}^{b} \\
&= \lim_{b\to\infty} \left(\ln{(b)}\right) \\
&= \infty.
\end{split}
\end{equation*}
So neither of these integrals exist.
My question is how do I explain why these are improper Riemann integrals? Thanks!

Despite the imtegrand becoming $\infty$ at say $(x=a)$. One way it becomes improper but convergent is when $0<p<1$ where $$I=\int_{a}^{b} \frac{dx}{(x-a)^p}<\infty.$$ For instance
$$\int_{0}^{1} \frac{dx}{\sqrt{x}}=2$$ is improper but convergent. The integral $$\int_{0}^{1} \frac{dx}{x^{1.01}} =\infty,$$ is improper and divergent.

Next, one more way and integral becomes improper but convergent when $q>1$ where the integral is $$J=\int_{1}^{\infty} \frac{dx}{x^q}.$$ For instance $$\int_{1}^{\infty} \frac{dx}{x^{1.01}}=100, \int_{0}^{\infty} \frac{dx}{\sqrt{x}}=\infty$$

Finally both integrals of yours are inproper but divergent the first one is so as $p=1$ and second one is so as $q=1$.

Notice that the well known integral $$\pi=\int{-1}^{1} \frac{dx}{\sqrt{1-x^2}}=\int_{-1}^{1} \frac{dx}{(1-x)^{1/2} (1+x)^{1/2}},$$ is improper but convergent because for both $x=\pm 1$, $p=1/2<1.$

The integral $$\int_{0}^{1} \ln x~ dx =x\ln x-x|_{0}^{1}=-1.$$ is improper but convergent, In this case one has to take the limit $x\rightarrow 0$ of the anti-derivative.

The improper integrals do not connect well to area under the curve. This may
also be seen as a defect in the theory of integration or these integrals are  called improper which may or may not be convergent.

There are many other ways and integral is improper but converget for instance
$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\pi/2,$$ despite the integrand existing only as limit when $x \rightarrow 0.$

The integral $$\int_{0}^{\infty} \frac {dx}{(1+x^4)^{1/4}} \sim \int^{\infty} \frac{dx}{(x^4)^{1/4}} =\infty$$ is divergen, we know this without actually solving this integral!

Most often studying the integrand near the point of singularity (discontinuity) or $\infty$ and making out the value of $p$ or $q$ helps in finding if it is
convergent.

In the first case the limit from the right of the function at 0 goes to infinity (it does not exist). In the second case the interval of integration is unbounded.


Prove that 17 divides 1111111111111111 (16 1's) and doesn't divide 11111111 I need to prove that $17$ divides $\underbrace{1111111111111111}_{\text{16 1's}}$ and doesn't divide $\underbrace{11111111}_{\text{8 1's}}$ by using congruence.

I know that $\underbrace{1111111111111111}_{\text{16 1's}}= \frac{10^{16}-1}{9}$ and that $10^{15}\equiv{1}\pmod {17}$. But how do I use these two facts to figure out if $17$ divides $1111111111111111$?


Possible values of nullity in 4x2 matrix Let $A$ be a 4 by 2 matrix. Explain why the rows of $A$ must be linearly dependent. What are the possible values of nullity(A)?

I understand the first part. I do not understand the second part.

The solutions say that the answer is : 1 or 2. But I said the answer was nullity=0,1 or 2.

Why 0?

Consider the matrix with row vectors [1,0] [0,1] [0,0] [0,0]. Rank(A)=2. Nullity(A) will therefore equal 0.

Am I right?

Yes, you're right.  It would seem that the book has some kind of typo.

Remember the definition of nullity - its the dimension of the null space. In other words, its the dimension of the solution set to [A|0], which always contains the zero solution. Therefore nullity(A) is not 0.


When is matrix multiplication commutative? I know that matrix multiplication in general is not commutative. So, in general:

$A, B \in \mathbb{R}^{n \times n}: A \cdot B \neq B \cdot A$

But for some matrices, this equations holds, e.g. A = Identity or A = Null-matrix $\forall B \in \mathbb{R}^{n \times n}$.

I think I remember that a group of special matrices (was it $O(n)$, the group of orthogonal matrices?) exist, for which matrix multiplication is commutative.

For which matrices $A, B \in \mathbb{R}^{n \times n}$ is $A\cdot B = B \cdot A$?

Two matrices that are simultaneously diagonalizable are always commutative.

Proof: Let $A$, $B$ be two such $n \times n$ matrices over a base field $\mathbb K$, $v_1, \ldots, v_n$ a basis of Eigenvectors for $A$. Since $A$ and $B$ are simultaneously diagonalizable, such a basis exists and is also a basis of Eigenvectors for $B$. Denote the corresponding Eigenvalues of $A$ by $\lambda_1,\ldots\lambda_n$ and those of $B$ by $\mu_1,\ldots,\mu_n$.

Then it is known that there is a matrix $T$ whose columns are $v_1,\ldots,v_n$ such that $T^{-1} A T =: D_A$ and $T^{-1} B T =: D_B$ are diagonal matrices. Since $D_A$ and $D_B$ trivially commute (explicit calculation shows this), we have $$AB = T D_A T^{-1} T D_B T^{-1} = T D_A D_B T^{-1} =T D_B D_A T^{-1}= T D_B T^{-1} T D_A T^{-1} = BA.$$

So there is no group of Matrix pairs that commute. It is
$$
AB = BA
$$
if and only if there is a polynomial
$$
p \in \mathbb{R}[x]
$$
such that
$$
p(A)=B.
$$
This can be proven using Jordan Normalform or by simple computing.


How to compute the recursive sequence with square? Consider a sequence defined by
$x_{n+1}^2=ax_n^2+bx_n+c$,where $a,b,c>0$ and $x_n >0$,
is there any general method to compute the exact form of $x_n$?
Thank you!

This doesn't define a sequence because $x_{n+1}$ could be
$+\sqrt{a x_n^2+b x_n+c}$
or
$-\sqrt{a x_n^2+b x_n+c}$.
Let's say you set $a=b=c=1$ and $x_0 = 1$, then the relation says
$x_1^2 = 1\cdot 1^2 + 1\cdot 1+ 1 = 3$.
This doesn't define $x_1$, it could be that $x_1 = \sqrt{3}$ or $x_1 = -\sqrt{3}$.
Do you have any information about the sign of $x_n$?

If you know $x_{n+1}$ then you can use the quadratic formula to find $x_n$.


Is this epsilon-delta proof that $\sin(x)$ is continuous circular? Prove that $\lim_{x\to a}\sin x=\sin a$, where $a$ is any real number.

Solution 13 here:

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html#SOLUTION13

claims to prove that $\sin(x)$ is continous, in other words: for every real $a$, $\lim_{x\to a}\sin(x)=\sin(a)$. The solution uses the Mean Value Theorem, which only works if the function in question is differentiable on some interval; in particular the function must be continous on some interval. So the solution is assuming that $\sin(x)$ is continous in order to prove that it is continous.

Is this solution bogus or am I missing something?

Note: I am not asking you to provide a correct proof. I already know of a correct proof (which does not use MVT).

You are right. That "proof" is totally wrong for the reason that you said. The same thing could be done with any function.

The answers you got are excessively literal.  They are at best technically correct on their own terms, and will gather some votes since it is fun to validate clever observation of errors in class material.  But it is misleading to describe the solution as circular.

By "prove" the author of the web page means "find an explicit function $D(x)$ so that setting $\delta = D(\epsilon)$ satisfies the epsilon-based definition of continuity or limit".   That is why the web page is titled
SOLUTIONS TO LIMITS OF FUNCTIONS USING THE PRECISE DEFINITION OF LIMIT.
The purpose of the page is to illustrate the technique of using the mean-value theorem to obtain epsilon-delta estimates.  It is not circular, in applying that technique, to assume a function is continuous and differentiable, in order to prove concrete estimates demonstrating continuity at some given point.

You can of course get a joke proof by reciting the epsilon-delta definition of continuity and saying "whatever $\delta$ makes the continuity assumption true!".  But that is not an estimate that can be computed on its own without the continuity assumptions.  If you ask for such computability as part of what is to be proved, then there is no circularity.

To very specifically answer the criticism,

  So the solution is assuming that $\sin (x)$ is continuous in order to prove that it is continuous.

No.  The solution is assuming that $f(x)=\sin x$ is continuous, differentiable, and with a known formula for its derivative; in order to prove that $|f'(x)|$ is bounded (which is the key to getting a computable bound on $\delta$ as a function of $\epsilon$, which is what the page means by "using the precise definition of limit").  There are examples of differentiable functions with unbounded derivatives on a finite closed interval. The existence of the examples means that assuming continuous etc is not enough to get the intended conclusion.

At most there is a criticism that "using the precise definition of limit" is defined only from the context of a class, book, or web page.  But it always implies computability (as in computability theory) and that can be taken as a very broad and sufficient definition for discussions like this one that parse things down to very fine details.


Coding the complex exp, sin and cos function I want to implement the complex exp, sin and cos function. The problem is the following:

For computing $\exp(z)$ I wanted to use $\exp(z) = \cos(z)+ i\cdot\sin(z)$. Then I first have to implement sin and cos.

For computing sin and cos I wanted to use the power series expansion

$\sin(z) = z - \dfrac{z^3}{3!}+\dfrac{z^5}{5!}-\dfrac{z^7}{7!}+\dots$

same for $\cos(z)$.

So I need to implement first how to compute $(x+iy)^n$. I thought I should use something like $z^n = (re^{i\varphi})^n$. The problem: I need the complex exp first.

Even if I use the power series expansion for exp I need the code for raising a complex number to a power. Where should I start?

I also thougt about $z^n = (re^{i\varphi})^n = r^n(\cos(\varphi n) +i\sin(\varphi n))$ (De Moivre's Formula) which allows me to compute the $n$-th power of $z$ without using exp. You should see the point here. I cant compute sin and cos without computing the $n$-th power of $z$.

If you have $z=a+bi$, you can find $r$ and $\theta$ for $z$ using
$$r=\sqrt{a^2+b^2} \quad \text{and} \quad \theta =\arctan(a,b)$$
where $\arctan(a,b)$ gives you an angle from $0$ to $2\pi$ by incorporating the sign of the value of $a$ and $b$. Then you can use the normal exponential since $r$ and $\theta$ are real numbers to obtain
$$z^n = r^n \cos(n\theta)+ir^n\sin(n\theta)$$

You do NOT want to use
the power series expansion
to compute sin and cos.

The first step is to do
a range reduction,
so that
the parameter is in
a restricted range
such as
$[0, \pi/4]$.

Then you should
investigate ways
of computing the
restricted sin and cos.

cordic is one thing
you might look up.

For serious computation,
do not roll your own
functions.


Center-commutator duality I'm reading this article by Keith Conrad, on subgroup series. I'm having trouble with a statement he does at page 6:

  Any subgroup of $G$ which contains $[G,G]$ is normal in $G$.

He says this as evidence that commutator and center play dual roles, since any subgroup of $G$ contained in $Z(G)$ is normal in $G$. Now, this I'm sure I understand, but I don't see how the quoted line holds.

What I have read is that $[G,G]$ is the least normal subgroup of $G$ such that the quotient is abelian, which seems related.

Also, while we're at it: are the center and the commutator "really" (as in, categorically) dual constructions? I'm quite a novice in category theory, so please excuse me if this question is trivial.

Rewritten to ease the "start-up" cost.

First, your first question:
The fact that any subgroup of $G$ that contains $[G,G]$ is normal in $G$ follows by the isomorphism theorems: if $H$ contains $N\triangleleft G$, then $H/N$ is a subgroup of $G/N$, and $H\triangleleft G$ if and only if $(H/N)\triangleleft (G/N)$. In the instant case, $N=[G,G]$ and $G/N$ is abelian, hence $H/[G,G]$ is necessarily normal in $G/[G,G]$ (since every subgroup of an abelian group is abelian), so $H$ is normal in $G$, as claimed.

Your second question is actually much more interesting (well, to me). The short answer is that the center and the commutator subgroup are not dual constructions in the categorical sense, but they are dual constructions in a different sense (which is related to a special kind of category that turns up a lot in General Algebra).
First, let me give "universal" definitions of both subgroups, so we can see how the two are directly connected to one another.
The commutator and the center.
As you note, the commutator subgroup of $G$ is the smallest normal subgroup $N$ of $G$ such that $G/N$ is abelian.
One way to characterize abelian groups is via identities: equations in the language of group theory that may or may not be satisfied by a group. For example, the identity "$xy=yx$" is not satisfied by every group (meaning, given a group $G$, it need not be true that for all $x,y\in G$ you have $xy=yx$, though it may be true  for some choices of $x,y\in G$). However, those groups that do satisfy this identity are precisely the abelian groups.
Any identity in group theory can be rewritten in the form $\mathbf{w}=1$, where $\mathbf{w}$ is a word in the language of group theory (a product of variables and their inverses); just move everything to the same side. For abelian groups, we have $(xy)(yx)^{-1}=1$, or $xyx^{-1}y^{-1}=1$. You may recognize the left hand side as none other than the commutator $[x,y]$ of $x$ and $y$.  Call it $\mathbf{w}(x,y)$; that is, $\mathbf{w}(x,y) = [x,y] = xyx^{-1}y^{-1}$.
So another way of defining the commutator subgroup of $G$ is to use the word $\mathbf{w}(x,y)$ that characterizes the class of all abelian groups (as "the groups for which every possible evaluation of that word equals the identity"): the commutator subgroup of $G$ is precisely the subgroup generated by all values that $\mathbf{w}(x,y)$ takes as $x$ and $y$ range over the elements of the group. Because we define the subgroup as "the subgroup generated by all values that the word $\mathbf{w}(x,y)$ takes", we say that this is a verbal subgroup.
Here is an interesting connection between the center and the commutator subgroup: you can think of the word $\mathbf{w}(x,y)$ as giving a function (not a homomorphism) $G\times G\to G$, that maps $(x,y)$ to $\mathbf{w}(x,y)$. Even though it is not a homomorphism, we may ask: does it factor through something? Is there a normal subgroup $N$ of $G$ such that the map $G\times G\to G$ factors through $(G/N)\times (G/N)$?
In other words: what is the set of all elements $z$ of $G$ such that, for all $x,y\in G$, $[xz,y] = [zx,y] = [x,zy] = [x,yz] = [x,y]$?
I'll spill the beans: it's the center. To see this, note that if $z$ has this property, then $[z,y] = [1,y] = 1$ for all $y\in G$, so necessarily $z\in Z(G)$. Conversely, if $z\in Z(G)$, then
$$[xz,y] = xzy(xz)^{-1}y^{-1} = xzyz^{-1}x^{-1}y^{-1} = xzz^{-1}yx^{-1}y^{-1}=xyx^{-1}y^{-1}=[x,y]$$
and similarly for $[zx,y]$, $[x,zy]$, and $[x,yz]$.
In this context, the center is refered to as the marginal subgroup of $G$ relative to the word $\mathbf{w}(x,y)$.  The term was coined by Phillip Hall; the idea is that these are elements that can be "pushed to the margins" when computing the values of $\mathbf{w}(x,y)$.
So here's a sense in which the commutator subgroup and the center are "dual": the commutator is the subgroup generated by all values of $\mathbf{w}(x,y)$, and the center is the subgroup of all elements that don't affect the values of $\mathbf{w}(x,y)$ when you introduce them as factors.
However, there are several ways in which the two constructions are not truly "dual" in a categorical sense:

The commutator subgroup is associated to the left adjoint of the forgetful functor from $\mathcal{A}b\mathcal{G}roup$ to $\mathcal{G}roup$. The left adjoint maps a group $G$ to $G/[G,G]$; that is, there is a natural bijection between the homomorphism from $G$ to an abelian group $A$, and the (abelian group) homomorphisms between $G/[G,G]$ and $A$. The group $G/[G,G]$ is left universal among all abelian groups into which $G$ maps, etc. All the things that are related to the fact that we have a left adjoint and that $\mathcal{A}b\mathcal{G}roup$ is a reflective subcategory of $\mathcal{G}roup$, as Qiaochu mentioned.

However, the center is not associated to a right adjoint.

The commutator subgroup is the equalizer of all maps from $G$ into abelian groups; the center, however, is not the coequalizer of all map from abelian groups into $G$.

While $[G,G]$ is always the smallest normal subgroup of $G$ such that the quotient is abelian, it is not true in general that $Z(G)$ is the largest normal subgroup of $G$ that is abelian.

The set of all marginal elements is already, directly, a subgroup; we don't need to talk about the "subgroup generated by all marginal elements of $G$". But the set of all values of $\mathbf{w}(x,y)$ is not necessarily a subgroup, we need to talk about the "subgroup they generate" to get the corresponding verbal subgroup.

Nonetheless, the construction of verbal and marginal subgroups does establish some "dual relation" between the two. In fact, that relation fits into a much larger picture of verbal and marginal subgroups (which also explains the "duality" between the upper and lower central series of a group, which is what Keith Conrad's notes point out).
Verbal and marginal subgroups in general
The commutator subgroup of $G$ is an example of a verbal subgroup of $G$, and the center is the corresponding marginal subgroup.
Varieties. The key to these notions is the idea of a variety of groups. A variety of groups is any collection of groups which is closed under subgroups, homomorphic images, and arbitrary direct products. That is, if $\mathfrak{V}$ is the class, then you require that if $H\lt G$ and $G\in\mathfrak{V}$, then $H\in\mathfrak{V}$. That if $\varphi\colon G\to K$ is an onto group homomorphism and $G\in\mathfrak{V}$, then $K\in\mathfrak{V}$; and that if $\{G_i\}_{i\in I}$ is an arbitrary family of groups with $G_i\in\mathfrak{V}$ for all $i\in I$, then $\prod_{i\in I}G_i\in\mathfrak{V}$.  (Note that since the empty product is the trivial group, a variety of groups is always nonempty.)
Examples of varieties are: the class of all groups; the class of all abelian groups; the class of all groups of exponent $n$ for a given $n$; the class of all nilpotent groups of class at most $c$ (for a given $c$); the class of all solvable groups of solvability length at most $s$ (for a given, fixed $s$).
Examples of collections that are not varieties are: divisible groups (subgroup of divisible need not be divisible); finite groups (arbitrary products of finite groups need not be finite); solvable groups (arbitrary products of solvable groups need not be solvable); torsionfree groups (homomorphic image of torsionfree groups need not be torsionfree).
Verbal subgroups. Given any variety of groups $\mathfrak{V}$, and a group $G$, there is a smallest normal subgroup of $G$, called $\mathfrak{V}(G)$, with the property that $G/\mathfrak{V}(G)\in\mathfrak{V}$. To see this, note that the family of all such normal subgroups is nonempty, since it contains $G$ itself; and given any family $\{N_i\}_{i\in I}$ of normal subgroups of $G$ such that $G/N_i\in\mathfrak{V}$ for all $i\in I$, then mapping $G$ to $\prod G/N_i \in\mathfrak{V}$ in the obvious way shows that $G/\cap N_i$ is a subgroup of a group in $\mathfrak{V}$, hence is itself in $\mathfrak{V}$. So the class of such subgroups has a smallest element (the intersection of all subgroups in the class). This subgroup is called the $\mathfrak{V}$-verbal subgroup of $G$.
(In the case where $\mathfrak{V}=\mathfrak{A}$, the variety of all abelian groups, the $\mathfrak{A}$-verbal subgroup is precisely the commutator subgroup).
The reason it's called a "verbal subgroup" is that to every variety there corresponds a set of "identities" that every group in $\mathfrak{V}$ satisfies, and such that every group that satisfies the identities is in $\mathfrak{V}$; these identities can be coded as words in the free group of countable rank (they are none other than the elements of $\mathfrak{V}(\mathbf{F}_{\infty})$, where $\mathbf{F}_{\infty}$ is the free group of countable rank), and the $\mathfrak{V}$-verbal subgroup corresponds to the subgroup of $G$ given by all the possible evaluations of these words in $G$ (equivalently, it is the subgroup of $G$ generated by all the images of $\mathfrak{V}(\mathbf{F}_{\infty})$ under all possible homomorphism $\mathbf{F}_{\infty}\to G$).
In general you don't need all the elements of $\mathfrak{V}(\mathbf{F}_{\infty})$, you only a set of elements that generates this subgroup as a fully invariant subgroup of $\mathbf{F}_{\infty}$ (see below for the definition of "fully invariant subgroup").
In the case of abelian groups, for instance, the identity that defines abelian groups is $xy=yx$ (every group in which every possible evaluation makes $xy=yx$ true is abelian, and in an abelian group every possible evaluation  makes $xy=yx$ true), which can be "coded" as the single word $xyx^{-1}y^{-1}$. This word completely defines the variety of abelian groups in the sense that a group is abelian if and only if every evaluation of $xyx^{-1}y^{-1}$ in the group is trivial. (Though abelian groups satisfy many other identities that are not true for arbitrary groups, they can be shown to be consequences of the fact that $[x,y]=1$ for all $x$ and $y$). The verbal subgroup corresponding to the class of all abelian groups is then precisely the subgroup generated by all the possible evaluations of this word, namely the commutator subgroup.
Verbal subgroups are very, very nice: they sit at the top of a hierarchy of subgroups, that is ordered in terms of how well they behave with respect to homomorphisms:

At the bottom we have the plain, "vanilla" subgroups.
Then we have the normal subgroup: a subgroup $H$ of $G$ is normal if and only if for every $\varphi\in\mathrm{Inn}(G)$, $\varphi(H)\subseteq H$. ($\mathrm{Inn}(G)$ are the inner automorphisms of $G$, the automorphisms of the form $x\mapsto g^{-1}xg$, for some fixed $g\in G$).
Then we have the characteristic subgroups: a subgroup $H$ of $G$ is characteristic if and only if for every $\varphi\in\mathrm{Aut}(G)$, we have $\varphi(H)\subseteq H$.
Then we have the fully invariant subgroups: a subgroup $H$ of $G$ is fully invariant if and only if for every $\varphi\in\mathrm{End}(G)$, we have $\varphi(H)\subseteq H$.
Verbal subgroups, however, have an even stronger property: for every group homomorphism $\varphi\colon G\to K$, between any two groupos, $\varphi(\mathfrak{V}(G))\subseteq\mathfrak{V}(K)$. That is, the image of a verbal subgroup under any homomorphism is contained in the corresponding verbal subgroup of the image. In particular, verbal subgroups are fully invariant (hence characteristic, hence normal).

(It so happens that a subgroup of a free group is fully invariant if and only if it is verbal, which is why when we are looking for a set of generators for $\mathfrak{V}(\mathbf{F}_{\infty})$ it is enough to find one that generates it as a fully invariant subgroup).
So verbal subgroups are absolutely the bee's knees of subgroup (at least, relative to homomorphisms). They are also related to the subcategory $\mathfrak{V}$ of $\mathcal{G}roup$ in exactly the same way as the commutator subgroup is related to $\mathcal{A}b\mathcal{G}roup$: the map $G\to G/\mathfrak{V}(G)$ is a functor from $\mathcal{G}roup$ to $\mathfrak{V}$, and it is the left adjoint of the forgetful functor from $\mathfrak{V}$ to $\mathcal{G}roup$.
Marginal subgroups are related to verbal subgroups. Given any element $\mathbf{w}\in\mathbf{F}_{\infty}$, $\mathbf{w}$ is a word in finitely many letters, say $\mathbf{w}=\mathbf{w}(\mathbf{x}_1,\ldots,\mathbf{x}_n)$. Given a group $G$, you can look at the subgroup generated by all the values of the word $\mathbf{w}$,
$$\mathbf{w}(G) = \Bigl\langle \mathbf{w}(g_1,\ldots,g_n)\Bigm| g_1,\ldots,g_n\in G\Bigr\rangle.$$
This is just the $\mathbf{w}$-verbal subgroup of $G$, of course. But you can consider the set of all elements of $G$ that "don't matter" relative to $\mathbf{w}$: that is,
$$\mathbf{w}^*(G) = \left\langle x\in G\left| \begin{array}{l}
\text{for all }g_1,\ldots,g_n\in G,\quad 1\leq i\leq n,\\
\mathbf{w}(g_1,\ldots,g_{i-1},xg_i,g_{i+1},\ldots,g_n) =\mathbf{w}(g_1,\ldots,g_n)\\\
\mathbf{w}(g_1,\ldots,g_{i-1},g_ix,g_{i+1},\ldots,g_n) = \mathbf{w}(g_1,\ldots,g_n).\end{array}\right.\right\rangle.$$
It's easy to check that $\mathbf{w}^*(G)$ is a subgroup of $G$, called the "marginal subgroup of $G$ relative to $\mathbf{w}$."  In fact, it is always characteristic.
(In the case of $\mathbf{w}(x,y) = [x,y]$, which yields the commutator subgroup as a verbal subgroup, has marginal subgroup equal to the center, as we saw above.)
Given a variety $\mathfrak{V}$, we can consider the marginal subgroup relative to $\mathfrak{V}$, $\mathfrak{V}^*(G)$, by taking the intersection of all $\mathbf{w}^*(G)$ for $\mathbf{w}\in\mathfrak{V}(\mathbf{F}_{\infty})$.
Marginal subgroups are not quite as nice as verbal subgroup: they are always characteristic, but seldom fully invariant, and they don't have nice functorial properties like the verbal subgroups do. But of course they are closely related to the verbal subgroups.
Other examples of marginal subgroups: if you consider the word $\mathbf{w}_c(x_1,\ldots,x_{c}) = [x_1,x_2,\ldots,x_c]$ (where $[a,b,c]=[[a,b],c]$), then the verbal subgroup $\mathbf{w}_c(G)$ is precisely the $c$th term of the lower central series of $G$; the corresponding marginal subgroup $\mathbf{w}_c^*(G)$ is the $c-1$st center of $G$, $Z_{c-1}(G)$, the $c-1$st term of the upper central series of $G$; and so, a group $G$ is nilpotent of class at most $c$ if and only if $\mathbf{w}_{c+1}(G) = \{1\}$, if and only if $\mathbf{w}_{c+1}^*(G) = G$; that is, if and only if $G_{c+1}=\{1\}$, if and only if $Z_c(G) = G$.  The word $\mathbf{w}_{c+1}$ defines to the variety of nilpotent groups of class at most $c$.  The fact that the upper central series occurs precisely as the marginal subgroups of the lower central series (which is defined as a series of verbal subgroups) is behind many of the nice "dualities" between the two.
As another example, if you take $\mathbf{w}(x) = x^n$, this word corresponds to the Burnside variety $\mathfrak{B}_n$ of all groups of exponent $n$ (groups in which every element is of exponent $n$; some call them "groups of exponent dividing $n$"), so named due to the Burnside and restricted Burnside problems; the Burnside problem for exponent $n$ is true if and only if $\mathbf{F}_k/\mathfrak{B}_n(\mathbf{F}_k)$ is finite for every $k$, where $\mathbf{F}_k$ is the free group of rank $k$. The marginal subgroup corresponding to $\mathbf{w}$ contains the set of all central element of exponent $n$, and those are the only central elements in it; that is, $\mathbf{w}^*(G)\cap Z(G) = \{x\in Z(G)\mid x^n = 1\}$. But in general it could be larger; for example, in a group of exponent $n$, $\mathbf{w}^*(G)=G$ even if $G$ is not abelian.
Some duality
So there is duality between verbal subgroups and the corresponding marginal subgroups, and it is exploited when dealing with things like the Schur multiplier, Baer invariants, and isologism, among other topics. You also see it at play when you look at nilpotent groups, and consider the upper and lower central series of $G$, which play kind of "dual roles" (one going up, the other going down; if $G$ is nilpotent of class exactly $c$, then $Z_{i+1}(G)\subseteq G_{c-i}$, where $G_{c-i}$ is the $(c-i)$th term of the lower central series of $G$, etc.)
For another example of a duality between the verbal and marginal subgroups, consider the following; given a variety $\mathfrak{V}$, the following are equivalent:

$G\in\mathfrak{V}$.
$\mathfrak{V}(G) = \{1\}$.
$\mathfrak{V}^*(G) = G$.

The first two equivalences are just from the definition of $\mathfrak{V}(G)$ as the smallest normal subgroup of $G$ for which $G/\mathfrak{V}(G)\in\mathfrak{V}$. To see the final equivalence, note that if $\mathbf{w}(x_1,\ldots,x_n)$ is an identity of $\mathfrak{V}$ (an identity that every group in $\mathfrak{V}$ must satisfy), then for all $g_1,\ldots,g_n\in\mathfrak{V}^*(G)$ we have
$$\mathbf{w}(g_1,\ldots,g_n) = \mathbf{w}(1g_1,\ldots,1g_n) = \mathbf{w}(1,\ldots,1) = 1.$$
Therefore, if $\mathfrak{V}^*(G)=G$, then $\mathfrak{V}(G)=\{1\}$. Converserly, if $\mathfrak{V}(G)=1$, then for all words $\mathbf{w}$ corresponding to $\mathfrak{V}$, all $x_1,\ldots,x_n\in G$, all $i$, $1\leq i\leq n$, and all $g\in G$ we have
\begin{align*}
1 &= \mathbf{w}(x_1,\ldots,x_{i-1},x_i,x_{i+1},\ldots,x_n) = \mathbf{w}(x_1,\ldots,x_{i-1},gx_{i},x_{i+1},\ldots,x_n)\\\
&= \mathbf{w}(x_1,\ldots,x_{i-1},x_ig,x_{i+1},\ldots,x_n),
\end{align*}
so $g\in\mathfrak{V}^*(G)$ for all $g\in G$.
References
A great place to learn about varieties of groups is Hanna Neumann's classic book, Varieties of Groups Ergebnisse der Mathematik und ihrer Grenzgebiete Band 37, Springer-Verlag, 1967. It is of course a bit dated (many of the problems posed were solved in the 70s, and there is work of Kleiman in the late 70s and early 80s that essentially shows that things can get pretty nasty when dealing with arbitrary varieties; I'm sad to say that the varietal point of view has all fallen by the way side in today's group theory).
Marginal subgroups were introduced in a very short note by Phillip Hall: Verbal and marginal subgroups, Phillip Hall, J. Reine Angew. Math. 182 (1940), pp. 156-157, MR 2, 125i.
If you want to see verbal and marginal subgroups play a major role and their interplay with homology, there is the tour de force by C.R. Leedham-Green and S. McKay, Baer invariants, isologisms, varietal laws, and homology, Acta Math. 136 (1976) no. 1-2, pp. 99-150, MR 0435250 (55 #8210). This one is very hard going, though, so since you are only beginning, you might want to set it aside for a while before trying to dig in.
From a slightly more general point of view, varieties of groups (and marginal subgroups) are the intersection of Group Theory and General Algebra (also called "Universal Algebra"); the idea of "identities" giving important subcategories of the collection of all your objects is central to General Algebra, and a lot of the universal and functorial properties of verbal subgroups are true in much more general settings.
You can get a pretty good introduction to the viewpoint of varieties of groups (if not to the specific properties they have by virtue of they being varieties of groups) with the oft-recommended (by me, anyway) General Algebra book by George Bergman. The files are in postscript, but you can get them converted to PDF with freely available software/websites like ps2pdf.

Here is a sort of "duality" (though very weak) between the center and commutator of a finite group, which occurred to me recently:

Let $\rm{Irr}(G)$ be the set of irreducible complex characters of the finite group $G$. Then the commutator subgroup is equal to $$\bigcap_{\chi \in \rm{Irr}(G),\ \chi(1)=1}\rm{ker}(\chi)$$

But if we look at something which in a certain sense is the complement of this, namely $$H = \bigcap_{\chi\in\rm{Irr}(G),\ \chi(1)\neq 1}\rm{ker}(\chi)$$ then this subgroup is contained in the center of $G$.

Proof (only of the last part): We know that the intersection of the kernels of all irreducible characters is trivial, so we get that $H\cap G' = \{ 1\}$. But it is a general fact that any normal subgroup that intersects the commutator subgroup trivially is central. To see this, let $N$ be such a subgroup, $x \in N$ and $g\in G$. Then the element $xgx^{-1}g^{-1}$ is in $N$ since $N$ is normal, and it is in $G'$ by definition, so it is $1$, and hence $N$ is central.

The above is of course not anywhere close to a proper duality, as we really look at a sort of complement instead, and we only get something contained in the center, rather than the entire center.

Remark: The above is a special case of a more general result, which states that if $m$ is the degree of some irreducible character of $G$ then $$\bigcap_{\chi\in\rm{Irr}(G),\ \chi(1)\neq m}\rm{ker}(\chi)$$ is abelian, and in fact, one can go further and leave out $2$ degrees and get something of derived length at most $2$, and even leave out $3$ and get something of derived length at most $3$ (though for this last result, one needs to assume that $G$ is solvable to begin with). These results are proved in the paper "Irreducible character degrees and normal subgroups" By Isaacs and Knutson.


What is the probability that you never lose this hypothetical dice game? You have a four sided die numbered 1-4 and are playing a game. On the first round, you roll the die once. If you get a 1 you lose the die. If you get a  2 you keep the die. If you get a 3 or 4 you get another, identical die. On the second round you roll each die that you have, and the same thing happens with each die. Once you finish rolling all your dice you move on to the third round and so on. If you have zero dice you lose the game. What is the probability that you never lose the game over infinite rounds?

Let $p$ be the probability of losing a die and all its "spawns" (the dice you got with it by rolling $3$ or $4$ and their spawns and so on). So we have

$$ p = \frac{1}{4} + \frac{1}{4} p + \frac{1}{2}p^2$$

This comes from the following reasoning: You can either lose the die immediately or keep the situation as is (you have the same $p$) or then you get a spawn and have to get rid of that too (hence the $p^2$).

Solving gives $p=\frac{1}{2}$ or $p=1$. Which one should we choose?

EDIT Other solutions have given resources for choosing the correct value, but for completion's sake I'll add some reasoning to my answer (also since I've learned something new myself and to reinforce it for my self :D).

I'll prove we should choose the smaller of the solutions (so in this case $p = \frac{1}{2}$).

Define

$X_n = $ "The number of dice we have after round $n$"

$A_n = \{X_n = 0\} =$ "We have lost all dice after round $n$"

$p_n = \mathbb{P}(A_n)$

We have $A_{n+1} \subset A_n$, so $p=\lim_{n\to\infty} p_n$.

Then define

$$f(x) = \frac{1}{4} + \frac{1}{4}x + \frac{1}{2}x^2.$$

Lemma 1

$$p_{n} = f(p_{n-1}).$$

Proof of Lemma 1: Notice that $f$ is the probability generating function (pgf) of $X_1$ and let's define $f_n$ to be the pgf of $X_n$. Then we have

$$f_n = f \circ f_{n-1}$$

This comes from the fact that if we have $k$ dice, they each "transform as" $f(x)$ and are independent so they produce a term $f(x)^k$ . So it goes like that: we keep composing f with itself. Now, lemma 1 follows easily, since $p_n = f_n(0)$. $$\tag*{$\blacksquare$}$$

Let $t$ be the smallest value on $[0,1]$ for which we have $t=f(t)$ (so $t=\frac{1}{2}$ in this case, but let's do this a bit more generally). Now, if we prove $p_n \leq t$ for all $n$ we are done, since $p$ is their limit. Let's do this inductively. The base case: $p_0=0 \leq t$. Inductive step: assume $p_n \leq t$. Since $f$ is increasing on $[0, 1]$,

$$p_{n+1} = f(p_n) \leq f(t) = t$$

and this finishes the proof.

Notice also, that $f'(1) = \mathbb{E}(X_1)$ is the slope of $f$ at $1$ and since $f'' \geq 0$, $f$ is convex so whether $\mathbb{E}(X_1)$ is bigger than $1$ or not decides whether the smallest solution will be $<1$ or $=1$. Because $f$ is a pgf, $f(1) = 1$ so $1$ will be the "default" solution if there arent' any smaller ones.

there's always the chance that with n dice you will roll all of them and get 1's, it's just a really small probability but it might happen.
Generally on each round you are expected to increase the number of dice you have by 25%.

EDIT: To answer the question more specifically, with n dice your chance of losing is $\frac1{4^n}$. With each round the expected increase of amount of dice you have is 25%, so over infinite rounds you will expect to have $n\to \infty$ which means $\frac1{4^n}\to0$


What is the proper geometrical name for a a rectangle with a semi-circle at each end? I'm talking about the shape made up of a rectangle with a semi-circle at each end.  Does it have a particular name? Does it begin with e?

Obround, apparently.  I don't know Wiktionary's source.  This definition of obround does not appear in OED, for example.  Googling indicates that this definition is commonly used for machine parts having this shape.

Another shape called " Bermuda Bottle" has meridian curvature proportional to $x$. Also ratio of shell curvatures is $2$ when rotated about minor axis, however minor axis measures only $\approx 0.6 $ times major axis.

EDIT1:

It is is same shape as filled parachutes.

The differential equation of Cornu Spiral by J.M. is not a mathematician is ( $s$ is arc ):

$$ \frac{d\phi}{ds} = s/a $$

and the Bermuda Bottle (not sketched) is

$$\frac{d\phi}{ds} = x/b $$

They are similar in appearance. Clothoid curve has a third order discontinuity at sharpest corners.


Little o(h) limit about h=0 I understand that generally if a function $f(h)$ is described as $o(h)$ that $f(h)$ has a smaller rate of growth than $h$ (like it would have to be $\sqrt{h}$). i.e. $\sqrt{h} = o(h)$, just like (for example) $4h=o(h^2)$. The notes I'm reading (CT3), however, states that:

A function $f(h)$ is described as $o(h)$ if:
$$\lim_{h \to 0}  \frac{f(h)}{h}  = 0 $$

but if I use, for example, $f(h)= \sqrt{h}$ which does have a slower growth rate then $h$ then the limit doesn't go to $0$. Is there a different meaning to little $0$ when its approaching $0$ compared to when it goes to infinity cause the only way that limit holds is if $f(h)$ goes to $0$ faster then $h$ does which I guess means $f(h)$ decreases faster then $h$ as $h \to 0$.
Anyway you can ignore my thoughts on the question but an explanation of the text I quoted would be greatly appreciated.

Yes, the little-o notation (and Landau symbols in general) behaves differently for $x\to 0$ and for $x\to\infty$.

When we're considering $x\to\infty$ (as you may be familiar with from analysis of algorithms), $\sqrt x$ grows slower than $x$ -- because for large $x$, the square root of $x$ is smaller than $x$ by a ratio that becomes ever more lopsided. Therefore in this context we say that $\sqrt x = o(x)$.

On the other hand, if we're considering $x\to 0$ (which is more common in analysis), then when $x$ is close to zero, $\sqrt x$ is larger than $x$ by a ratio that tends to infinity. Therefore in that context we say that $x = o(\sqrt x)$.

If would be less confusing to make it explicit which limit we're working with, and write something like
$$ \sqrt x = \mathop o_{x\to\infty}(x) \qquad\qquad\qquad x = \mathop o_{x\to 0}(\sqrt x) $$
However, in most practical uses of the notation, the limit is the same throughout the entire calculation, so repeating it for every $o$ would be tedious and distracting. So generally it is left implicit, though one should make it clear before one starts using asymptotic notation which limit one is considering.

Your starting point is wrong. $h^2=o(h)$ and not the reverse. $h=o(\sqrt{h})$ and not the reverse.The definition of the Landau notation in the vicinity of a point $x_0$ is $$f=o(g)\Leftrightarrow\lim_{x\to x_0}\frac{f(x)}{g(x)}=0$$


Set theory: how is $\left\lbrace 1, 1 \right\rbrace$ different from $\left\lbrace 1 \right\rbrace$? What set theory differentiate  sets: $\left\lbrace 1, 1 \right\rbrace$ from $\left\lbrace  1 \right\rbrace$; order does not matter?

Thank you in advance.

Addition/clarification: Thank you very much, everybody, for the providing idea of multi sets - I have found few papers and I hope it will be useful.

I am from microscopy/Imaging field without formal mathematical education, and I try to find the most basic mathematical object which can represent image. I was thinking that set would be the most basic/universal. It works well for just one measurement (or pixel), e.g. {1}; if we have two measurements (pixels), we can have two identical values {1, 1} from set of rational numbers. In classical set theory, it immediately collapses to {1}, so I lose my information. Practically it can happen if I measure number of photons from fixed point inside my object (cell) and I do not care about order of the measurements.

The idea of set is quite attractive as I can embed it in other sets, form universes, etc. On other side, I feel that image always requires some sort of order - except from the very basic measurements I have mentioned before. It can be ordered pair of set of rational numbers and structures, e.g. Cumulative hierarchy which represent pixels (or measurements) and order of the image. I am not sure how easy I can work with them.

Meanwhile I appreciate all your answers, comments and suggestions.

People often refer to these sorts of objects (containers where the order of the elements within them doesn't matter, just how many times each appears) as "multisets".

From any theory of ordinary sets you get a corresponding theory of multisets, and vice versa; an ordinary set can be thought of as just a multiset where nothing appears more than once, and conversely, a multiset can be thought of as just an ordinary set along with a function from that ordinary set to positive cardinalities (representing the number of repetitions of each element).

In elementary set theory, we do not distinguish $\{1,1\}$ from $\{1\}$, we simply characterize a set by the elements it contains. In a similar vein, we say that $\{1,2\} = \{2,1\}$. In general, two sets are equal (by definition) if and only if they contain the same elements, i.e., $A=B \iff \left[x \in A \iff x\in B\right]$. I'll leave it to you to verify that this condition is true for the above two examples.


Using addition theorem in conditional probability Two balls are drawn from an urn containing 2 white,3 red and 4 black balls one by one without replacement. What is the probability that at least one ball is red?

Your approach is valid, but there is a quicker way -- especially if we decide to increase the number of balls we draw.

Recall that $P(\text{at least one red ball}) = 1-P(\text{no red balls})$.

The probability that we don't draw a red ball the first time is $6/9$, since there are $6$ non-red balls, and $3$ red balls.

Now there are $8$ balls remaining: $5$ non-red balls and $3$ red balls. The probability that I don't draw a red ball the second time is $5/8$.

So $$P(\text{at least one red ball})=1-\left(\frac{6}{9}\right)\left(\frac{5}{8}\right).$$

Use the Hypergeometric distribution. If a random variable $X$ has Hypergeometric distribution: $X \stackrel{d}{=}Hg(n,R,N)$. Where $N$ is the size of the population, $n$ is the sample you take, and $R$ is the number of individuals from the population with the characteristic you are looking for. Then we have:

$$P(X=x)=\frac{\binom{R}{x}\binom{N-R}{n-x}}{\binom{N}{n}}.$$

You need to find $P(X\geq1)=1-P(X=0)$, $X \stackrel{d}{=}Hg(n=2,R=3,N=9).$


Sum of dice adding to less than $5$ I have seen some places saying that the sum of $2$ dice being thrown being less than $5$ can occur for the cases

$$(1,1),(2,2),(2,1),(3,1),$$

giving a probability of $\frac4{36}$ for the event.

However, I have also seen people counting the numbers twice, as there are two different ways to throw the dice, giving the cases

$$(1,1),(1,1),(2,2),(2,2),(2,1),(1,2),(3,1),(1,3),$$

and a probability of $\frac8{36}$.

Which way is correct?

Neither is correct.  The second approach is the correct one – you list all the ordered possibilities. Nevertheless, you have listed $(1,1)$ and $(2,2)$ twice.

Neither.

There is a single way to roll a sum of 2: roll $(1,1)$.

There are two ways to roll a sum of 3 and two ways to roll a sum of 4, as you pointed out, where the numbers are different. Notice that $(2,2)$ also gives a sum of 4.

So the total probability is $6/36$.

One nice way to see this visually is to make a table, one with the value of the first die along the columns and the value of the second die along the rows, and record their sum at each cell in the table. The table will have $6 \times 6 = 36$ cells and you'll see that the diagonal rolls $(1,1)$, $(2,2)$, etc. all only appear once, while all the others appear twice.


Using Mean Value Theorem to solve an inequality involving $\cos^{-1}(x)$ Question:
Using the Mean Value Theorem, show that for all $-\frac{1}{2}\lt a,b \lt \frac{1}{2}$ with $a\lt b$
$$\lvert \cos^{-1}(a)-\cos^{-1}(b)\rvert \lt \frac{2\sqrt{3}}{3}\lvert a-b\rvert$$

My attempt:
Let $f(x)= \cos^{-1}(x)$
$\cos^{-1}(x)\ \text{is continuous and differentiable on the interval so by the MVT,} $
$$\lvert \cos^{-1}(b)-\cos^{-1}(a) \rvert= \lvert b-a\rvert.\lvert f'(c)\rvert$$
$$\lvert f'(c) \rvert= \frac{1}{\sqrt{1-c^2}} \lt \frac{2\sqrt{3}}{3} $$
Therefore we have,
$$\lvert \cos^{-1}(a)-\cos^{-1}(b) \rvert= \lvert a-b\rvert.\lvert f'(c) \rvert \le 1.\lvert a-b\rvert \lt \lvert b-a\rvert \frac{2\sqrt{3}}{3}$$
Because $\lvert a-b\rvert= \lvert b -a\rvert$
Would this be correct?

No, this wouldn't be correct. The equality you wrote from the MVT is not right; if you take absolute values, then you must also have $|f'(c)|$ in the RHS; and just because $f'(c) < 1$ it doesn't mean its absolute value is also $<1$.


Horn–Schunck method. Explanation of iterative solution I am reading this paper (explanation of Horn-Shunck method for finding optical flow) and trying to understand it.
My stumbling block is obtainig solution of system of linear equations

I(x, y, t) - brightness of the image at point (x, y) at time t, alpha - just regularization coefficient, u and v - optical flow to find

Horn and Schunck proposed this iterative solution based on Gauss–Seidel method (page number nine in paper).
But my math skill is not enough to understand how to obtain this formula

Can anyone explain me this solution?

P.S. Sorry for my poor english

So uh... I'll try in a quick way, Sorry I don't have the time right now, I made a program using this method and that's what I got from doing it.
The last part is an interative algorithm, the next value depends on the last iteration value.

$h(u,v)^k =(u^(k+1),v^(k+1)) = $is the flow vector on an specific point $(x_o,y_o,t_o) $

$\bar{h}=(\bar{u}^k,\bar{v}^k) = $ is the local average flow vector, including all neighbors like this for example \begin{matrix}\\ \frac{1}{12} && \frac{1}{6} && \frac{1}{12} \\
\frac{1}{6} && 0 && \frac{1}{6} \\
\frac{1}{12} && \frac{1}{6} && \frac{1}{12} \\
 \end{matrix}

and $I_x = \frac{\partial I}{\partial x} = $ is the value of the derivative in the $I_x(x_o,y_o,t_o)$
 Hope this helps!!!

after that you need to understand how the condition for stopping the interative algorithm.

Greetings from Brasil!

Take a look at this. It explains the Gauss Seidel method
https://www.math.ust.hk/~mamu/courses/231/Slides/CH07_3B.pdf


Question about Monty Hall if you already knew a bad door beforehand So the setup of the scenario here is exactly the traditional Monty Hall problem. Except, before the game starts, you decide to cheat and open a door and it happens to be a goat. Before you can peek at the two other doors, the game begins, and you pick a door that's not the one you just peeked into (obviously, because you peeked a goat). Here it seems the probability to win a car is $\frac{1}{2}$, because you improved your original $\frac{1}{3}$ chance of winning a car by cheating.

Then the game host opens the same door you just peeked at before the game started (knowing it was a goat, like the traditional Monty Hall problem). Is it still a $\frac{2}{3}$ chance of winning the car if you switch, even though you already knew the informatiom beforehand? Or does it still remain $\frac{1}{2}$ because you already knew the information before, so you just screwed yourself out of the better $\frac{2}{3}$ probability by cheating?

If you originally picked a door with a goat, then the game host must pick the door you peeked at, because the other door contains the car. In this situation, it doesn't seem like anything changes because you already knew the door the game host opened beforehand by cheating.

If you originally picked a door with a car, the game host can open either other door, because they both have goats. If the game host opens the door you did not peek at and reveals a goat, then you know with $100\%$ probability that the door you originally picked contains the winning car. However, if the game host opens the door you peeked at, then it seems like the same situation in the paragraph I described before this one.

If you pick the car first, as you say, the host has 50% chance of confirming you have the car, and 50% of giving you no new information.

If you pick the unknown goat first, the host gives you no new information.

Ultimately the host has a $\frac{3}{4}$ chance of revealing the goat you already knew about.

If you're picking between those two randomly, then in $\frac{1}{4}$ of all cases, you pick the car, then the host reveals the second goat, so you win the car. In the other $\frac{3}{4}$ of cases, you're back to the $\frac{2}{3}$ - $\frac{1}{3}$ situation, since in $\frac{2}{3}$ of those $\frac{3}{4}$, you had the goat to begin with, and in the other $\frac{1}{3}$ you had the car.

$$P(\mathrm{car}|\mathrm{unknown\,goat\,revealed}) = 1$$

$$P(\mathrm{car}|\mathrm{known\,goat\,revealed}) = \frac{P(\mathrm{car}\cap\mathrm{known\,goat\,revealed})}{P(\mathrm{known\,goat\,revealed})} = \frac{1/4}{3/4} = \frac{1}{3}$$

You can expect to get the car $\frac{3}{4}$ of the time by switching if the goat you knew about is revealed and staying if not.

Of course, if you pick randomly between the two unpicked doors, you're a fool, because picking the door you know to be a goat forces Monty to show you the other goat and hence brand new car :)

No, the probability is now $1/2$ not $2/3$ because for it to be $2/3$ with the switching strategy you need a $2/3$ chance of originally chosing the wrong door, but by cheating you lowered that to $1/2$


Inequality in a Sobolev space In the Sobolev space $H^2(\mathbb{R}^2)$ is it true that there exists a constant $c>0$ such that
$$\|u\|_{L^2(\mathbb{R}^2)}\le c\|\nabla u\|_{L^2(\mathbb{R}^2)}\;,\quad\quad \text{for all}\quad u\in H^2(\mathbb{R}^2)?$$
Thank you in advance.

So this is called the Poincaré inequality. It is true on a bounded domain. When you are in $\mathbb{R}^d$ and not in a bounded domain, you need some additional weights (for example it works with Gaussian weights, or sufficiently decaying weights).
In the form you are asking, it is false. indeed, take $\langle x\rangle = \sqrt{1+|x|^2}$ and
$$
u(x) = \frac{1}{\langle x\rangle}
$$
Then $u∉ L^2$ (since $u^2 ∼ |x|^{-2}$ when $x\to\infty$), but
$$
∇u  = \frac{-x}{\langle x\rangle^3}
$$
therefore $∇u ∈ L^2$ since $|∇u|^2 = \frac{|x|^2}{\langle x\rangle^6} ∼ |x|^{-4}$ when $x\to\infty$.

No
$u=1|_{ [0,1]^2} $ indicatrice function of $ [0,1] ^2$
$u\in H^2(R^2)$
$\nabla u=0$, but $u $ no zero


Exponential Growth and Decay Question: A Bacteria Culture Contains 100 Cells and Grows at a Rate Proportional to its Size A bacteria culture contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.

a) Find and expression for the number of bacteria after $t$ hours

$P(t)=P(0)e^{kt}=100$

b) Find the number of bacteria after 3 hours.

$P(t)=P(3)e^{k3}=$ (confused as to how to set this up)

d) When will the population reach 10,000?

You know that "grows at a rate proportional to its size". Therefore you can write this in an equation:

$\frac{dP(t)}{dt}=kP(t)$ with some coefficient $k>0$. This simple differential equation has for solution (after separation of variables) $P(t)=Ce^{kt}$, where $C>0$ is a constant, that we determine from the initial condition: at $t=0$ you have 100 cells $\Rightarrow P(0)=100 \Leftrightarrow C=100$. You find $k$ with the second condition "After an hour the population has increased to 420", i.e $P(1)=420\Leftrightarrow 420=100 \times e^{k.1}\Rightarrow k=ln(4.2)$.

For b): after 3 hours you have $P(3)=100\times e^{k.3}=100 \times(e^{ln(4.2)})^3=100\times 4.2^3=7408.8$.

For c): We want to find $t$ such that $P(t)=10000\Leftrightarrow 100\times e^{kt}=10000\Leftrightarrow e^{kt}=100\Leftrightarrow 4.2^t=100$ or $t=log_{4.2}100=\frac{ln 100}{ln 4.2}\approx3.2$


3 Dices and the probabilty of me guessing 2-of them and 3-of them back to back. A weird thing just happened and I tried to solve the problem but I couldn't. My friend had 3 dices, I guessed 2 of them as 2, 4 and he rolled '2,4,5'. So I guessed it correct. Than I guessed 3 of them as '1, 3, 5' and he rolled that. We discussed what are the chances of me guessing the first one correct, and right after that guessing the 2nd one correct back to back, and we are stuck. Any help would be appreciated.

For your first guess you would have been right if your friend had rolled a $1$ with the $2$ and $4$.  Because you have three different dice, there are six ways this could have happened:
$$1,2,4\quad\hbox{or}\quad 1,4,2\quad\hbox{or}\quad 2,1,4\quad\hbox{or}\quad
  4,1,2\quad\hbox{or}\quad 2,4,1\quad\hbox{or}\quad 4,2,1\,.$$
The same thing happens if he had rolled a $3,5$ or $6$.  If the rolls were $2,2,4$ it's a bit different and there are only three possibilities:
$$2,2,4\quad\hbox{or}\quad 2,4,2\quad\hbox{or}\quad 4,2,2\,.$$
And the same for $2,4,4$.  So altogether you would have been right in $30$ cases and the chance of your first guess being right is $30/6^3$.  You can figure out the probability of the second guess being right in a similar way - it's actually easier than the first one - and then combine them to get your final answer.

Think about the number of combination that could occur and you still guess correctly.

So for your first guess the outcomes that result in a correct guess are below, where X is any number except 2 or 4:

Dice1   Dice2   Dice3
2       4       X
4       2       X
X       2       4
X       4       2
2       X       4
4       X       2
2       2       4
2       4       4
2       4       2
4       2       4
4       4       2
4       2       2

There are 6 combinations where you see a 2 and 4 and a different number, plus 6 combinations where we see only 4 and 2. The total number of combinations is therefore 6 * 4 = 24 (the six combinations above, times the 4 values that X can take) plus the 6 combinations of only 4 and 2. The total is 30 combinations where your guess is correct.

The total number of combinations for 3 dice is 1/6 * 1/6 * 1/6 = 1/216

So the probability of guessing 2 out of 3 dice correctly is 30/216

Hopefully you can see that for the case where you guess 3 out of 3 the probability is 6/216 (think about the above list but where X is a specific number and you cannot repeat any numbers).

So the total probability of your guesses both being correct is the probability of the first being correct multiplied by the probability of the second being correct

30/216  *  6/216 = 5/1296


Is my proof of the continuity of $\frac{1}{x}$ correct? I need to prove that $\frac{1}{x}$ is continuous, but my proof, as far as I did now, only applies to the interval $x\ge1$ wich is a progress. So, there's how I did:

$$\frac{|x-a|}{|x|}<|x-a|<\delta \tag{for $x>1$}$$

Then, by choosing $\delta=\epsilon|a|$ we have:

$$\frac{|x-a|}{|x|}<\epsilon|a|\implies\frac{|x-a|}{|x||a|}<\epsilon\implies\frac{|x-a|}{|xa|}<\epsilon \implies \frac{|a-x|}{|xa|}<\epsilon \implies \left|\frac{1}{x} - \frac{1}{a}\right|<\epsilon$$

This, implies that $\lim \limits_{x\to a}\frac{1}{x} = \frac{1}{a}$ for $x>1$. I did not think so much about the $x\le1$ case, but I would like if this proof is correct.

There are a few not very big things wrong with the proof, and a major problem: you do not explain the logic of the argument.

Let $a\ne 0$, and let $f(x)=\frac{1}{x}$. We want to show that $f$  is continuous at $x=a$.  In what follows, we will assume that $a\gt 0$. By putting in the appropriate absolute value signs, one can transform the proof into one that works for all $a\ne 0$.

We want to show that for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-a|\lt \delta$, then $|f(x)-f(a)\lt \epsilon$.

If following the argument poses difficulty, let $a=0.4$.  Now let $\epsilon \gt 0$ be given.

It is easy to see that for any $x\ne 0$,  we have $|f(x)-f(a)|=\frac{|x-a|}{a|x|}$.

We have control over $|x-a|$, since we have great freedom in selecting $\delta$. The number $a$ is fixed. We need to control the size of $x$, since if $x$ is close to $0$, the $x$ in the denominator may make $|f(x)-f(a)|$ big.

Now comes the first key step. We will make sure ultimately that $\delta\le \frac{a}{2}$. Why? Well, then if $|x-a|\le \delta$, then we will have $\frac{a}{2}\lt x\lt \frac{3a}{2}$. The important part is that we will have $x\gt \frac{a}{2}$, so $x$ will stay nicely away from  $0$.

Thus if $\delta\le \frac{a}{2}$, and $|x-a|\lt \delta$, we have
$$|f(x)-f(a)|\lt \frac{2}{a^2}|x-a|.$$
Now to make sure that $|f(x)-f(a)|\lt \epsilon$, it is enough to make sure, in addition to the $\delta \le \frac{a}{2}$, that $\delta\le \frac{a^2\epsilon}{2}$.

Finally, we choose our $\delta$: it is $\min\left(\frac{a}{2},\frac{a^2\epsilon}{2}\right)$. By our arguments above, if $|x-a|\lt \delta$, then $|f(x)-f(a)|\lt \epsilon$.

Remark: The restriction to $x\ge 1$ in the argument was an attempt to avoid small denominators, so you were aware of the small denominators problem. In your proof, we can make sure that $x\ge 1$ if $a\gt 1$ by picking $\delta$ small enough, for example $\delta\le \frac{a-1}{2}$. But you did not do that.

There is a problem with the assertion that if $\delta=\epsilon |a|$ then everything is OK. Imagine, to be  silly, that $a=2$ and $\epsilon=47$. (This is the silly part.) Then $x$ being within $\epsilon |a|$ of $a$ guarantees nothing,  $x$ could even be equal to $0$.

You may think the argument of the answer tedious, and overkill. True, and true. But I wanted to describe the issues in detail, since the basic pattern is fairly standard.

You are correct.And for cases where x is less then or equal to 1 function is not continuous,since it is not continuous at 0.

We can see that left-sided and right sided limit at 0 are not equal,thus it does not exist,thus function is not continuous for $x \le 1$


Identifying a quotient group (NBHM-$2014$) Let $\mathbb C^*$ denote the multiplicative group of non-zero complex numbers and let $P$ denote the subgroup of positive real numbers. Identify the quotient group.

My thought $$\frac{\mathbb C^*}{P}=\{P,-P,iP,-iP\}.$$ Is it right?

I think you can set this map $$f:\mathbb C^*\to U,~~z=a+ib\mapsto \frac{a}{|z|}+i\frac{b}{|z|}$$ wherein $U=\{z\in\mathbb C^*\mid|z|=1\}$. Show this map is surjective with $\ker f=\mathbb R^*_{>0}$.

I am not sure if this is in any way different than B.S's way. i am just omitting maps and trying to see just as a quotient.

We see  $x+iy\in \mathbb{C}$ as $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$

Now, As  $\sqrt{x^2+y^2}\in \mathbb{R}$ when we see $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$ modulo real numbers we would get  :

$$(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})\equiv (\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}}) \text { mod }\mathbb{R}$$

And then we have $|(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})|=\sqrt{(\dfrac{x}{\sqrt{x^2+y^2}})^2+(\dfrac{y}{\sqrt{x^2+y^2}})^2}=1$

So,for any $z\in \mathbb{C}$ we have $z=|z|(z')$ for $|z|\in \mathbb{R}$ and $|z'|=1$.

Thus $\mathbb{C}^*/P=\{z\in \mathbb{C} : |z|=1\}$


Box containing Red and black balls A box containing 4 red balls and 6 black balls . Three balls are selected randomly from the box one after the another, without replacement . The probality that the selected set has one red ball and two black balls is

I have a doubt in this problem .

There are 10 balls

Sample Space =  First i pick 1st ball X then 2nd ball X then 3rd ball

So Sample Space = 10 * 9 * 8

Ist trial I pick 1 red ball from 4 red balls = 4C1
2nd trial I pick 1 black ball from 6 black balls = 6C1
3rd trial I pick 1 black ball from 5 black balls = 5C1

a.) Probality = (4C1*6C1*5C1)/(10*9*8)

        OR

b.) I pick 1 red ball from 4 red balls =4C1
    I pick 2 black balls from 6 red balls = 6C2

Probality = (4C1*6C2)/(10*9*8)

Is it a or b ? Pleas elaborate ???

tl;dr

  A box containing 4 red balls and 6 black balls . Three balls are selected randomly from the box one after the another, without replacement . The probality that the selected set has one red ball and two black balls is

Do you think order matters? No. Why? Because if we select $RBB$ or $BRB$ or $BBR$ any of them occuring is favorable, and we are done, you might as well consider order but that would cancel eventually. So total possibilities in three turns:
$$\binom{10}3$$
Ways to do favorable event:
$$\binom62\binom41$$
Probability:
$$P=\frac{\binom62\binom41}{\binom{10}3}=\frac12=50\%$$

Now let me consider order:

Total ways:
$$10*9*8$$
Favourable ways:
$$RBB:\binom41\binom61\binom51\\
BRB:\binom61\binom41\binom51\\
BBR:\binom61\binom51\binom41$$
Or simply no. of ways of arranging $B,R,B$ times "you know":
$$\frac{3!}{2!}6*5*4=6*5*4*3$$
Probability is:
$$\frac{6*5*4*3}{10*9*8}=\frac12=50\%$$

a.) Probality = (4C1*6C1*5C1)/(10*9*8)

   OR

  b.) I pick 1 red ball from 4 red balls =4C1 I pick 2 black balls from 6 red balls = 6C2

  Probality = (4C1*6C2)/(10*9*8)

  Is it a or b ? Pleas elaborate ???

Neither, though (b) is closer.  It is important to observe that the favoured case must always be counted in the same manner as the universal case; and vice versa.

(b) is closer because the numerator is counting the ways to pick 1 of 4 red balls and 2 of 6 black balls.  Which is your favoured case.  Now for the denominator you must count the universal case in the same manner: ways to select any 3 of all 10 balls.

$$\mathbb P = \frac{{^4\mathsf C_1}\cdot{^6\mathsf C_2}}{^{10}\mathsf C_3}$$


Term for listing vertices of polygon in consecutive order If I have a polygon with vertices in some order $P_1, P_2, \ldots, P_n$, then the order may be such that $\overline{P_i P_{i+1}}$ is an edge of the polygon for each $i$, or they could be in some random order with no relation to the edges.

Is there an established term for the former case?

To be concrete, I want to write a sentence of the form "This method returns the vertices in _______ order," or something similar.

I could say something like "consecutive" or "sequential" but I think the meaning there is not quite intuitive.  In the case that the polygon is convex this would be equivalent to saying something like "in either clockwise or counterclockwise order," but I was hoping for something more general and less confusing.

I could also write a long explanation, but I'm hoping for something that's readable and unambiguous and fits in a single sentence.

Since you have the order dependent upon some index, why not "indicial order" or "index order" or "ordered by index?"


probability in the casino I have a question regarding probability.

Lets assume that roulette has just red and black, no green and an equal amount of both.

Lets also assume we start each game with a bet of 1 dollar.
We also have enough dollars to be able to double our bet up to 40 times.

Now an example game is we bet our 1 dollar on black.
If we lose we bet 2 dollars on black, the total expenditure is now 3 dollars.
If we lose again we bet 4 dollars on black, the total expenditure is now 7 dollars.

If we win we get 8 dollars back. 1 of which is a profit.

Remember we can lose 40 times in a row before we cannot double up any longer.

The odds of getting 2 reds in a row is 25% so getting 40 reds in a row would be very low.

My question is what is the probability of making a profit after one entire game ( a game is where you keep doubling until you can or you profit)

and after 1000 games.

After one game, you make a dollar unless you lose $40$ times in a row, which happens $2^{-40}$ of the time, about once in $10^{12}$ tries.  After $1000$ games, you have to win them all, so you will lose approximately $1000 \cdot 2^{-40}$ of the time, or about once in a billion tries.  This is not exact-it is actually smaller because some of the probability comes from two losses.  The chance of that is greater than zero, but very small.

The probability of you losing after an entire game is $\frac{1}{2}^{40}$, because you have to get 40 reds in a row to lose, and the probability of getting a red is $\frac{1}{2}$. The probability of making a profit is then $1-\frac{1}{2}^{40}$.

Starting with 1\$, the (maximum) amount of money you could lose after an entire game is $2^{40}-1$(lost 1\$ after round one, 3\$ after round two, 7\$ after round three...), if you win once, you earn 1\$. So playing this game you have a chance of $\frac{1}{2}^{40}$ to lose $2^{40}-1$\$, and a chance of $1-\frac{1}{2}^{40}$ to win 1\$, which means that if you repeated the full game forever (restocking to be able to play 40 rounds again), you would neither gain money nor lose money, because $\frac{1}{2}^{40}\times (2^{40}-1) = (1-\frac{1}{2}^{40})\times 1$, as worked out below:

Lose: $\frac{1}{2}^{40}\times (2^{40}-1) = \frac{1}{2}^{40}\times 2^{40} - \frac{1}{2}^{40} = 1 - \frac{1}{2}^{40}$

Win: $(1-\frac{1}{2}^{40})\times 1 = 1-\frac{1}{2}^{40}$

Lose = Win: $1 - \frac{1}{2}^{40} = 1 - \frac{1}{2}^{40}$


Can skew lines be perpendicular? In a math textbook for primary school, there's a picture like the one below. They explain which of the line segments are perpendicular and which are parallel to AD.

There's also a note stating that A1B1, D1C1, BB1 and CC1 are neither parallel nor perpendicular to AD.

I was wondering about that statement, because I learned that those four are perpendicular to AD. So I wrote to the book publisher asking about that. They replied that actually there are two conventions for this situation and they choose "not perpendicular" which, according to them, is more popular.

Is there really no consensus on that?

Edit:
I found another question here, related to a mathematical consensus: Is $0$ a natural number? and I'd accept an answer referring to some resources and clarifying whether the consensus exists or not.

According to Wikipedia, "skew lines are two lines that do not intersect and are not parallel."
Two sentences later it wrote "Two lines that both lie in the same plane must either cross each other or be parallel, so skew lines can exist only in three or more dimensions."
Therefore, skew lines (at least if you trust the Wikipedia definition) cannot be perpendicular.

I think, Since, A1B1 and AD are not on the same plane, we can't say they are parallel or perpendicular. But your title asks whether skew lines can be perpendicular or not? - I say, Yes, but if two lines are originally perpendicular and you skew them horizontally or vertically (or both) they are will not remain perpendicular.


Jargon for maximum/minumum absolute value in a set Given a group of numbers $-5,-3,1,2$, the maximum is 2, the minimum is -5.

What is the mathematical jargon for the maximum and minimum in absolute terms (i.e. -5 and 1 respectively)?

Basically, I want to say that:

'-5 is the [jargon to mean maximum in absolute terms] in the group of numbers'.

I know that 'absolute maximum' generally refers to a global maximum in a context of a function. Is the phrase 'maximum absolute' used or easily confused with another meaning?

I appreciate this is a very basic question which probably has a really obvious answer but the answer escapes me...

I would go with 'element with the greatest absolute value/magnitude', but as with any question like this, if there's any ambiguity about what you're saying, you're better off explaining it like you did here.


Is there a formula for the cosine of 1/5 of an angle? I'm trying to find a formula for $\cos \frac{x}{5} $ as follow.

By the elementary trigonometric identity, $\cos 5x = 16\cos^5 x - 20\cos^3 x + 5\cos x$

By putting $x=  \frac{x}{5}$ , and using the change of variable $ y = \cos \frac{x}{5} $, we get the quintic equation $ 16y^5 - 20y^3 + 5y - \cos x = 0$.

If we can solve this equation by radicals, then we can get a formula for $\cos \frac{x}{5} $

However, I can not see how we can resolve this equation. Is this a solvable quintic equation? If so, how can we solve it by radicals?

Depending on how you look at this situation, it is solvable by radicals. Let’s take your cosine, and as @mweiss does, let’s call it $C$. I should point out that the case $C=1$ is very special, since $16x^5-20x^3+5x-1=(x-1)(4x^2+2x-1)^2$, probably giving the Wolfram-generated roots that I (wrongly) disparaged in a comment.

In the general case, set $S$ to be the corresponding sine, $S=\sqrt{1-C^2}$, then $\zeta=C+iS$ is a point on the unit circle, and you’re talking about $\zeta^{1/5}$ and your answer is the real part of this. If you accept the fifth root of a real number as something understandable, I don’t think it can be justified to disparage the fifth root of a complex number.

To direct an abstract eye onto the problem, let’s take a general number $\kappa$, which will play the role of $C$, and look at the field $F_0=\Bbb Q(\kappa)$. Then we make a quadratic extension, adjoining $\sigma=\sqrt{1-\kappa^2}$, so that we have $F_1=F_0(\sigma)$, and then adjoin a fifth root of unity, $\omega_5$, where
$$
\omega_5=\frac{-1+\sqrt5}4+\frac i2\sqrt{\frac{5+\sqrt5}2}\,,
$$
and $F_2=F_1(\omega_5)$ is at worst quartic over $F_1$, with Galois group cyclic of order $4$, $2$, 0r $1$, depending on what $\kappa$ was.
Then adjoin $(\kappa+i\sigma)^{1/5}$ to get the final field $F_3$, and again this extension is Galois, with cyclic group of order $5$ or $1$. So, starting with $F_0=\Bbb Q(\kappa)$, we get a chain of radical extensions,
$$
F_0\subset F_1\subset F_2\subset F_3\,,
$$
each inclusion being a cyclic Galois extension. If you call $\xi=(\kappa+i\sigma)^{1/5}$, then its real part is $\frac12(\xi+1/\xi)$, the desired cosine.

p = 3, trisection
p >= 3
Python explanation


Given $a,\,b\in\mathbb{Z}$ relatively primes and $\,n \gt 0$, $n\in\mathbb{Z}$, calculate a) $\gcd(a+nb,a+(n+1)b)$; b) $\gcd(a+nb,a+(n+2)b)$. I was able to solve part a), proving by Mathematical Induction on $\,n\,$ that $\,\gcd(a+nb,a+(n+1)b) = 1$. (I can supply my solution if of interest).

However, I did'nt succeeded in tackling part b). Clearly, $\,n\,$ and $\,n+2\,$ always have the same parity, but noting this wasn't enough for me to find a solution. On the other side, I have performed a few numerical experiments that led me to suspect $\,\gcd(a+nb,a+(n+2)b)\,$ is equal to

1 if $\,a\,$ is odd and $\,b\,$ is even, with any $\,n > 0$;
1 is $\,a\,$ is even and $\,b\,$ is odd, with $\,n\,$ odd;
2 if $\,a\,$ is even and $\,b\,$ is odd, with $\,n\,$ even;
1 is $\,a\,$ is odd and $\,b\,$ is odd, with $\,n\,$ even;
2 is $\,a\,$ is odd and $\,b\,$ is odd, with $\,n\,$ odd;

I would appreciate any help in proving or disproving this conjecture.

$$\gcd(a+nb,a+(n+2)b)= \gcd(a+nb,2b).$$ Since $b$ is coprime to $a$ it is also coprime to $a+nb$ and therefore $$\gcd(a+nb,2b)= \gcd(a+nb,2).$$ The required gcd is therefore $1$ if $a+nb$ is odd and $2$ if $a+nb$ is even.

If $p$ is a prime which divides $a+nb$ and $a+(n+1)b$, it divides $a+(n+1)b-(a+nb)=b$, it divides $nb$ and $(a+nb)-nb=a$ contradiction since $gcd(a,b)=1$. We deduce that $gcd(a+nb, a+(n+1)b)=1$.

If $p$ divides $a+nb$ and $a+(n+2)b$, $p$ divides $a+(n+2)b-(a+nb)=2b$.

If $p$ is odd, $p$ divides $b$, $nb$ and $a+nb-nb=a$ contradiction. If $p=2$

$a+(n+2)b=a+nb+2b$ thus $a+nb$ is even, since $a$ and $b$ are odd, $nb$ is odd and $n$ is odd.

Thus if $n$ is odd $gcd(a+nb,a+(n+2)b)=2$ if it is even it is $1$.