In an abelian Group in which there are 2 subgroups of order m and n there is a subgroup of order mcm(m,n), how to prove? This is an Herstein's exercise

  If an abelian group has two subgroups of order m and n, prove that it
  also has a subgroup of order lcm(m,n).

I've solved the exercise right before this, which is an analogue statement for elements

  If an abelian group has two elements of order m and n, prove that it
  also has a element of order lcm(m,n).

I would like some hints (just give me half an hour) before a complete answer.

I'm trying to demonstrate that, given $H<G,K<G; m=\#H, n=\#K$ then $\#HK=r=$ lcm$(m,n)$

I know (/ I can prove) that $$\#HK={{\#H\#K}\over{\#(H\cap K)}}$$; so I would need to show, if my intuition is right this time, that $\#(H\cap K)=$ gcd$(m,n)$.

@Simon

Yes, I know that $\#H\cap K|\#H $ and $\#H\cap K|\#K $, but I don't know how to prove maximality. It sort of involves a converse of lagrange theorem, maybe. But Herstein implicitly wrote it was not needed.

@Qil

Ok, I believe I've proved what it's called Abelian groups fundamental theorem. I'll just sketch it here, so you can tell we if I'am using the concepts correctly.

Step 1 - Let $G$ be a group, and $H$ a normal subgroup. Then we can define an homomorphism to $G/H$ and consequently the isomorphism $$G\rightarrow H \times G/H.$$
Step 2 - Now, let G be an abelian group, all subgroups are normal. Let us consider $$\alpha_1=\max_G \{k:o(g)=k\} $$$$g_1\in G : o(g_1)=\alpha_1$$We know that $\alpha_1|n$ for Lagrange; we can now write, using step 1, the following isomorphism $$G\rightarrow \langle g_1\rangle \times G/\langle g_1\rangle$$
Step 3 - We repeat step two, substituting $G$ with $G/\langle g_1\rangle$ and find a non-increasing sequence $\alpha_1,\cdots,\alpha_k$ such that $n=\alpha_1\cdots\alpha_k$; each $\alpha_i$ corresponding to a cyclic group of order $\alpha_i.$

Step 4 - We know that every cyclic group of order $\beta$ has a subgroup of order $\beta_k$ iff $\beta_k | \beta$. We've obtained a decomposition of $G$ into cyclic groups of prime order.

Merging those bricks together leads us to any subgroup of order $d$ when $d|n$.

But would Herstein want us to prove this for that exercise?

If $G$ is an abelian group of order $d$, then for any positive divisor $k$ of $d$, there is a subgroup of $G$ of order $k$.
Apply (1) to $HK$: let $e=\# (H\cap K)$. Then $\# (HK)=mn/e$. As $e \mid \gcd(m,n)$ and $mn=\gcd(m,n)\mathrm{lcm}(m,n)$, we see that $\mathrm{lcm}(m,n) \mid \# (HK)$. So (1) implies that $HK$ has a subgroup of order $\mathrm{lcm}(m,n)$.


A square matrix $A$ is such that $\mathrm{rank}(A^{k}) = \mathrm{rank}(A^{k+1})$ A square matrix $A$ is such that $\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+1})$. Then $\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+i})$ for all $i$.

I think we prove this by induction on $i$.

Let $i = 2$. We show that rank $(A^{k})$ = rank $(A^{k+2})$.

Now, we know that rank $(A^{k+2})$ = rank $(A^{k+1}A)$ $\le$ min (rank$(A^{k+1})$, rank$(A)$) $\le$ rank $(A^{k+1})$ = rank $(A^{k})$.

However, I cannot proceed further.

Think geometrically. We have $A^{k+1}=A^kA$, so the image of $A^{k+1}$ is contained in the image of $A^k$. If the rank of the two are equal, that means the image is the same. Since $A^{k+1}=AA^k$, this means that $A$ is a bijection on said image.

so you know now that $rank (A^k)=rank (A^{k+1})=rank (A^{k+2})$.
you just can take $k'=k+1$  and do the same work for $A^{k'}$. You obtain $$rank (A^k)=rank (A^{k+1})=rank (A^{k'})=rank (A^{k'+1})=rank (A^{k'+2})= rank (A^{(k+1)+2}) = rank (A^{k+3}) $$ and so on


What is $\Bbb R^{\times}$? [unit group, ring to "times" power] I'm doing some sheets for my Abstract Algebra class and I can't seem to remember the group defined as $\mathbb{R}^{\times}$. It's obviously some variation of $\mathbb{R}$ but I'm away from college on reading week so can't ask my tutor. If someone could clear up the confusion I'd be grateful.

This is the usual notation for the unit group of the ring $R$, that is, the group of invertible elements using ring multiplication as the group operation. Note that this is usually not the same thing as $R \setminus \{0\}$, because most elements aren't invertible (if it is the same, your ring is a field, by definition). While you can talk about $R \setminus \{0\}$ under multiplication, it's usually just a monoid (or even a semigroup, if you permit rings without unity).

The notation is often used on the form $\Bbb R^*$ i.e. with a star and it means $\Bbb R\setminus \{0\}$ and we have $(\Bbb R^*,\times)$ is a multiplicative  group.


Is $\sum Y_j$ a measurable $\sigma(X_1,...,X_n)$ function? Let $X_n=\left(\sum^n_{i=1} Y_i\right)^2$, and I would like to know if $\sum^n_{i=1} Y_i$ is measurable in $\sigma(X_1,...,X_n)$.

If not, why? I tend to think that it's not. But I can't think of a justification for it...

Any help would be appreciated.

(Following copper.hat comment)

There are sets which are not Lebesgue measurable. Let $A$ be such a set.

Define a function $f(x)= \begin{cases}
      1, & x\in A \\
      -1, & x\in A^c
   \end{cases}
$. Clearly, f is not Lebesgue measurable. However, $f^2$ and $|f|$ are measurable.
In this way, we could have $(\sum Y_n)^2$ measurable, but not $\sum Y_n$

Let $(\Omega, \mathscr F, \mathbb P)$ be a probability space.

Consider $Y$, a mapping from $\Omega \to \mathbb R$ s.t. $X := Y^2 \in m\mathscr F$.

Let $\omega \in \Omega$. If we know $X(\omega)$, we don't necessarily know $Y(\omega)$. For example if $X(\omega) = 25$, then $Y(\omega)$ could be 5 or -5.

So $\{Y = 5\} \cup \{Y = -5\} = \{X = 25\} \in \mathscr F$.

It need not follow that either $\{Y = 5\}$ or $\{Y = -5\}  \in \mathscr F$.

Hence we cannot conclude that $Y \in m\mathscr F$.


Deriving line of best fit with least squares I know there are a lot of resources on this but I want to derive it.

So let's say we have $n$ known points of the form $x_k, y_k$ and we want a line $y = mx + b$ that best approximates the relationship. We approximate by minimizing the sum of the squared distances.

In other words we minimize

$s = \sum (y_k - (mx + b))^2 = \sum (y_k - mx - b)^2$

Set derivative to $0$:

$\frac{d}{dx} s = \sum -2m(y_k - mx - b) = 0$

Is this correct so far? I'm not quite sure where to go from here.

Remember that $x_k$ is your data for the independent variable. There are the same number of $x$'s as $y$'s. So you need an index. You are minimizing $ SS =\sum_k (y_k-(mx_k+b))^2 $

As another answer indicated your problem is you want to minimize in terms of $m$ and $b$. (You are estimating your coefficients. The data can't change.)

So you have $$\frac{\partial}{\partial m } SS = -2\sum_k x_k(y_k-(mx_k+b)) = 0 \\ \frac{\partial}{\partial b} SS = -2\sum_k (y_k-(mx_k+b)) = 0 $$
as the other answer indicated, you know the solution to these equations will be the unique minimum because $SS$ is an upward-opening paraboloid as a function of  $m$ and $k$.

Now solve these two equations for $b$ and $m.$  You may find it convenient to introduce the notation for the sample mean $ \bar x = \frac{1}{n}\sum_k x_k.$

I complained in the comments about an incorrect answer on here, but it did get something right. The treatment is simpler and more generalizable if you write instead of $y_k\approx mx_k+b$ the vector equation $$ y \approx X\beta $$ where $\beta = (b,m)^T$ are your coefficients and $X$ is an $n\times 2 $ matrix whose first column is all ones and whose second column is the vector of $x$'s.

(It is generalizable because you can easily add more regressors by adding additional columns to $X$ and corresponding coefficients to $\beta.$)

Then to get the least squares estimator for $\beta$ we minimize $SS = (y-X\beta)^2$ (where we're using somewhat abbreviated vector notation). The minimum is again given by setting the gradient to zero  $$0=\nabla_{\beta} (y -X\beta)^2 =-2 X^T(y-X\beta)$$ so we have $$ X^TX\beta=X^Ty \\\Rightarrow \beta = (X^TX)^{-1}X^T y$$
provided $X^TX$ is invertible.

(And all this invertibility requires is that $X$ is full column rank. In other words your regressors cannot be linearly dependent. So, for instance, in your case of a simple best fit line, invertibility fails only if the $x$'s are all the same... which would mean your points all lie on a vertical line.)

Note that $y = m x +b = \pmatrix{1 & x} \pmatrix{b \\ m}$

So make a matrix $A$ for all the $n$ points such as

$$ {\rm A} = \pmatrix{ 1 & x_1 \\ 1 & x_2 \\ \vdots \\ 1 & x_k \\ \vdots } $$

and a vector of the $y$ values

$$ {\bf y} = \pmatrix{ y_1 \\ y_2 \\ \vdots \\ y_k \\ \vdots} $$

Then you are trying to solve

$$ {\bf y} = {\rm A} {\bf c} $$

for ${\bf c}=\pmatrix{b \\ m}$.

The least squares method is

$$ {\rm A}^\top {\bf y} = ({\rm A}^\top {\rm A}) {\bf c} \Rightarrow$$

$$ \boxed{ {\bf c} =  ({\rm A}^\top {\rm A})^{-1} {\rm A}^\top {\bf y} } $$

Note that ${\rm A}^\star = ({\rm A}^\top {\rm A})^{-1} {\rm A}^\top$ is called the pesudo inverse of ${\rm A}$.

Expanded out this gives

$$ {\bf c} = \pmatrix{ 1 & {\bf x} \\ {\bf x}^\top & {\bf x}^\top {\bf x}}^{-1} \pmatrix{ {\bf y}  \\ {\bf x}^\top {\bf y} } $$

which you can expand further to get a direct expression. Here ${\bf x} = \pmatrix{x_1 & x_2 & \cdots & x_k & \cdots}^\top$.


Locally graded group with all proper subgroups abelian A group $G$ is said to be locally graded if every finitely generated nontrivial subgroup of $G$ contains a proper subgroup of finite index.

I have to prove that a locally graded group with all proper subgroups being Abelian is itself Abelian or finite. I can suppose that a group, $G$, is not finite and prove that it is Abelian, but I don't know how to proceed.

Thanks.

If $G\neq 1$, then $G$ being locally graded, it has a proper subgroup of finite index $H$. Hence $H$ is abelian; in particular $G$ is virtually abelian.

Assume that $G$ is not abelian and infinite. Then any two non-commuting elements of $G$ generate $G$; in particular $G$ is finitely generated.

Being finitely generated and virtually abelian, $G$ has a normal subgroup of finite index $N$ that is free abelian of finite rank. Pick a prime $p$ greater than $|G/N|$, and let $q>p$ be another larger prime. Then the finite group $H=G/N^{pq}$ also has all its proper subgroups abelian; it's an extension of $N/N^{pq}$ by $G/N$, which have coprime order; hence this extension is split, so $H=(N/N^{pq})\rtimes K$ with $K\simeq G/N$. Since $N/N^{pq}$ splits as the direct product of its Sylows $N/N^p$ and $N/N^q$, $H$ has the proper subgroup $N/N^q\rtimes K$. Hence $N/N^q\rtimes K\simeq G/N^q$ is abelian. Since this holds for every prime $q>p$ and $\bigcap_{q>p}N^q=1$, we deduce that $G$ is abelian, contradiction.

A group G is said to be locally graded if every finitely generated nontrivial subgroup of G contains a proper subgroup of finite index.


Can I solve x on this way? or is there a mistake I made or is the question wrong? $$\cos3x\sin2x=\cos3x$$
Can I just pull $\cos3x$ out of both sides and continue with $\sin2x=0$ which leaves us with $2x=0$?

We have
$$
\cos(3x) \sin(2x) = \cos(3x) \iff \\
\cos(3x) (\sin(2x) - 1) = 0 \iff \\
\cos(3x) = 0 \vee (\sin(2x) - 1) = 0 \iff \\
\cos(3x) = 0 \vee \sin(2x) = 1 \iff \\
3x = \frac{\pi}{2} + k \pi \vee 2x = \frac{\pi}{2} + m 2\pi \iff \\
x = \frac{\pi}{6} + k \frac{\pi} 3 \vee x = \frac{\pi}{4} + m \pi
$$
for arbitrary $k, m \in \mathbb{Z}$.

$\cos 3x \sin 2x=\cos 3x\implies \cos 3x=0$ or $\sin 2x=1$

The rest you can solve.$0/0$ is undefined.Keep that in mind.

If any bound on $x$ is given, then answer it appropriately.


Nonlinear ODE Abel equation of the first kind Edited: I have the following equation that represents a particular circuit. I am trying to get a solution that is practical for design purposes. I have went through some papers on Abel equations of the first kind, but I have not been able to get a reasonable solution. Thank you

$\frac{dy}{dx} + A_1y + A_3y^3 = F(x)$

Ok, let's say $F(x)$ is sinusoidal. It can be any waveform but let's say it is $F\sin(x)$ where F is just an amplitude.

$\frac{dy}{dx} + A_1y + A_3y^3 = F\sin(x)$

Hint (Before edit): I assume $A_1$, $A_3$ and $F$ to be constants. Then this differential equation is separable.

$$\dfrac{dy}{F-A_1y-A_3y^3}=dx$$

The solution is not very nice. But if you can narrow the range of $y$ you could simplify it.

EDIT: In the case of a general $F(x)$ it is very likely that there is no simple closed form solution to the ODE. If you are trying to solve a practical problem you could try to look how much $F(x)$ does change in the practical range of values for $x$. It might be, that $F(x)\approx \text{const.}$ in the practical range of values.


Steady states in a differential system Given the following model

$$\begin{cases} \frac{du}{d\tau} = \beta u \big(1- \frac{u}{v}\big)\\
\frac{dv}{d\tau} = v(1-v) - \alpha uv
\end{cases}$$

where $\alpha, \beta \geq 0$.

I'm trying to calculate and determine the stability of the steady states. The only one I can find though is $(u,v) = (0,1)$.

I've tried letting $\dot{u} = \dot{v} = 0$ but it doesn't make things easier.

This differential system is found in population dynamics.

You will find a general setting in the excellent document "Modeling Population Dynamics" (https://staff.fnwi.uva.nl/a.m.deroos/downloads/pdf_readers/syllabus.pdf): see equations (7.2) page 137.


Whats the size of the angle I have exams for primary teaching maths coming up. And the tasks aare always at high level, so I need some help from you. Please can you help me to solve this geometric task? w is the angle bisector.
Thank you so much, Sophia

If $w$ is a line bisecting an angle

We can find the angle in the upper-left hand corner of the diagram to be $51^\circ$ ($180-90-39=51$). Since we're assuming $w$ bisects that angle, the two angles between $w$ and the two relevant lines of the triangle are equal. This gives them each a value of $\frac{51}{2}=25.5$. Using the parallel postulate, we can then find $\varepsilon$ to be $180-(180-90-25.5)=115.5^\circ$.

If $w$ is an angle

If we assume $w$ is an angle (which it probably is not, due to its lack of angle notation), we can solve for $\varepsilon$ in terms of $w$.

We know that $A$ is $90^\circ$ ($A = 180-90$)

We can form a triangle with the known angle and angles $A$ and $B$. Since the angles of a triangle always sum to $180^\circ$, we can find $B$ as such: $$B = 180-39-A = 180-39-90 = 51.$$

We know that $C = w$ and with the knowledge that $B$ is $51^\circ$, we can find

$$D=180-51-w = 129-w$$

Then we can find

$$\varepsilon = 180-(129-w) = 51 + w$$

If you are ever informed of the value of $w$, just plug it into the last equation. For example, if you knew $w=10$, $\varepsilon=51+10=61$.

After long thinking I found the solution, its 115,5 degree. If anyone is interested in the way of solving, let me know.
Best and thanks,
Sophia


Modular equation with complex numbers: given $\displaystyle \Big\vert \frac{z-50}{z-2} \Big\vert = 5$ find $\vert z \vert $ I am asked the following question:

Suppose $z$ is a complex number. Find $\vert z \vert$.
$$\Big\vert \frac{z-50}{z-2} \Big\vert = 5$$

The book's solution:
Initially, I pretty much did the same as the book instructed, using all properties of modulus, substituting $\vert z \vert$ for $z \cdot z^*$ and so on (I'll just place the book's mark scheme):

Second (?) solution: I basically treated $z$ as an unknown:
$$\Big\vert \frac{z-50}{z-2} \Big\vert = 5$$
$$\frac{(z-50)^2}{(z-2)^2} = 25$$
$$z^2 -100z +2500 = 25z^2 -100z+100$$
$$z^2 = 100$$
$$\vert z \vert = 10$$
My question: I found that the answers are the same. But is this method valid in the first place? Is it ok for me to assume and treat z as an unknown variable?
Any help is highly appreciated

The first method looks okay.  The second method is not a correct approach.
$$\left|\frac{z-50}{z-2}\right|^2=\frac{z-50}{z-2} \cdot \frac{\overline{z}-50}{\overline{z}-2}$$

I disagree with the other answers, perhaps because I meta-cheat.
What the OP did, without realizing it, is assume that $z$ maybe expressed as $x + iy$.  The meta-cheating idea, is that if the problem is solvable, then the problem is still solvable if
$$y ~\text{is assumed to equal} ~0.$$
In effect, perhaps without realizing it, the OP treated the problem like a math puzzle rather than a math problem.  It was not a coincidence that the OP's method worked.
There seems to be confusion about the underlying mathematical basis for this specific meta-cheating.
I am asserting that the method is flawless when solving for |z| with respect to the complex variable z, as long as long as any satisfying values for the value to be solved for, (in this case |z|), each occur at least once in whatever
restricted domain that you are exploring.
The only reason that the meta-cheating involves restricting the domain to $\mathbb{R}$ is for ease of use. The principle will mathematically hold, regardless of what restricted domain you investigate, as long as |z| takes on all pertinent values there.
Naturally, the specific meta-cheating process for this problem involves identifying candidate values for z that may not work against the original equation.
Anytime you have a problem asking for you to solve for |z|,
then if there is only one value for |z| to be determined, the method can not fail.  This is because if no real solutions in the meta-cheated equation are uncovered, then you know immediately that in this specific problem, you must abandon the meta-cheating idea.
This answer (in effect) answers why the OP's (invalid) method works.  It is because he restricted the domain to be investigated to $\mathbb{R}$, and because all values to be solved for, |z|, each occurred at least once in the restricted domain.

A comment/question was asked:
For what values of $a,b,c$ will the meta-cheating approach lead to the wrong conclusion re solving for $|z|$ re
$$\frac{|z-a|}{|z-b|} = c.$$
Exploring, I squared both sides.  This may result in candidate values for $z$ that do not solve the original equation.
This resulted in
$$|z|^2(c^2 - 1) + 2\text{Re}(z)(a - bc^2) + (b^2c^2 - a^2) = 0.$$
In order for the meta-cheating idea to fail:
(1)
There must be at least two distinct values for |z|, such that each of them will have at least one underlying value of $z$ that satisfies the equation.
(2)
At least one of the two distinct values for |z| must never occur in the restricted domain of $z \in \mathbb{R}.$
Looking at the general problem, I actually have no opinion when the meta-cheating would fail, re the above two conditions both occurring.
Edit
However, given the latest downvote, I really can't help myself here:
In the above equation, which explored when the meta-cheating would or would not fail with respect to the equation
$$\frac{|z-a|}{|z-b|} = c$$
It is immediate that when $(a - bc^2) = 0$, which is the case in the OP's problem, you are mathematically guaranteed that there can only be one satisfying value for $|z|$, which is the value being asked for.
In such a situation, per discussion previously in this answer, the meta-cheating approach can not fail.
Come to think of it, when $(a - bc^2) = 0$, if the problem has any solutions at all, the locus of points $z$ that provide the appropriate value of $|z|$ will have to be a complete circle centered at the origin, which (further) guarantees that the locus of points will cross the $x$-axis.


By what rule or property this expression is equal to six? $$2^{-1} \equiv 6\mod{11}$$

Sorry for very strange question. I want to understand on which algorithm there is a computation of this expression. Similarly interested in why this expression is equal to two?

$$6^{-1} \equiv 2\mod11$$

It's because $2\cdot 6 \equiv 1 \mod 11$

It is part of the collection of integer pairs such that:

$$xy\equiv1\mod p$$

in this case with $p=11$.

For $p=11$, they are $1\cdot1,2\cdot6,3\cdot4,5\cdot9,7\cdot8$ and $10\cdot10$

Except for $1$ and $p-1$, each $x$ is paired to a unique $y$.


Properties of Nonnegative Orthant Let
$$\mathbb{R}_+^n = \{x = (x_1,...,x_n)| x_1 \geq 0, x_2 \geq 0,\cdots, x_n \geq 0 \}$$
be the nonnegative orthant of $\mathbb{R}^n$.

How can we demonstrate the properties of the monotonic cone?

1) Let $c \in \mathbb{R}^n$ be given. Show that $c^Tx \geq 0 \forall x \in \mathbb{R}_+^n$ if and only if $c \in \mathbb{R}_+^n$.

2) Let $x, y \in \mathbb{R}_+^n$. Show that $x^Ty=0$ if and only if $x_iy_i = 0$ for each $i=1,...n$

First, I was thinking if we show the monotone cone is self-dual then $c \geq 0$, and therefore $c \in \mathbb{R}_+^n$.

Another part of it, suppose if  $c \leq 0$, then $c$ not $\mathbb{R}_+^n$, and $c^Tx \leq 0$. But this is a contradiction.

Hints

1) One direction is obvious. For the other direction note that the canonical basis is nonnegative, that is $e^i \in \Bbb R_n^+$ for every $i$, where $(e^i)_j:=1$ if $i=j$ and $(e^i)_j:=0$ otherwise.

2) Again, one direction is obvious. For the other direction, suppose by contradiction that $x^\top y=0$ and $x_iy_i \neq 0$, then $x_iy_i>0$ as $x,y\in\Bbb R^n_+$ and so $x^\top y\geq x_iy_i>0$, a contradiction.


Prove: $\gcd(a,b) = \gcd(a, b + at)$. I know that $\gcd(a,b)$ divides $a$ and $b$, and must also then divide $(a)(t)$ ($t$ being some integer).  This makes sense to me, but how do I prove it?  It seems that the addition of $(a)(t)$ is a continuation of the linear combination of $\gcd(a,b) = av + bu$ for some $v$, $u$ being elements of $\mathbb{Z}$.

Any help?

HINT $\rm\ \ $ If $\rm\ c\ |\ a\ $ then $\rm\ c\ |\ b + a\ t\ \iff\ c\ |\ b\:.\ $ This implies that  $\rm\ \{a\:,\:b+a\ t\}\ $ and $\rm\ \{a\:,\: b\}\ $ have the same set of common divisors $\rm\:c\:,\:$ hence they have the same greatest common divisor.

Modly: $\:$ if $\rm\ a\equiv 0\ $ then $\rm\ b+a\ t\equiv 0\: \iff\: b\equiv 0\ \ \ (mod\ c)$

You're asking for a proof of the below.

let c = GCD(a,b); then c | a.t for integer t.

Let's refactor:

if c | a then c | a.t for integer t.

This is fundamental (it's very close to the definition of 'divisibility'), but to prove it:

if c | a then cx = a, for some integer x (this is the definition of divisibility)
if cx = a then cxt = at, for all integer t
c | cxt (again, the definition of divisibility)
if c | cxt, then c | at


What should I do if limits of integration are ones where Dirac Delta gets Infinite? I am working on an integral involving Dirac delta function. I am from Physics background. I know that if I integrate from a to b and in between there is a point where argument of Dirac delta function is zero then the answer to this integral is unity. What if at say 'a' the argument of Dirac Delta Function is zero. What should I do in that case? What if I have a Riemann Integrable Function along with Dirac delta function? Please help.

Consider a graph of the delta function (distribution) to obtain:

$$\int_a^b\delta(x-t)\,\mathrm dx
= \cases{0&$t<a,\text{ or }t>b$\\
         1&$a<t<b$\\
\color{red}{1/2}&$\color{red}{t=a},\text{ or }t=b$}$$

So that
$$\int_a^bf(x)\delta(x-a)\,\mathrm dx
= \tfrac12f(a)$$


Calculate absolute values with unknown constant I am to calculate all $x$ if $f(x) = g(x)$ and if

$$f(x)= |2x+2| + |3-2x|$$
$$g(x)= x + 14$$

How do I mix regular numbers with absolute values in such a sense? I thought I could calculate it like this:

$$|2x+2| + |3-2x| = 3x+2 + 3+2x = 3x+5$$

But then I realised that the end value of either of the absolute values is determined by $x$. It feels like a catch 22: how do I calculate this equation without knowing the value of $x$ until I've calculated it?

HINT: You can assume three cases:

$x \geq \dfrac 32$
$-1<x< \dfrac 32$
$x \leq -1$

1.
Find the roots of all |A| expressions(total of m):

x=-1, x=3/2

2.Partition the $\mathcal{R}$ (if you want to solve in Real numbers set) into m+1 sub domains

I    x <- 1
II   -1 =< x <= 3/2
III  x>3/2

3.Find the sign of |A| in each sub domain and solve for x
$I:$
$-(2x+2) + (3-2x) = x+14$
$-2x -2 + 3 -2x = x+14$
$-4x+1 = x+14$
$5x=-13$
$x=-13/5 \in I\checkmark$
$II:$
$(2x+2) + (3-2x) = x+14$
$+2x +2 + 3 -2x = x+14$
$5 = x+14$
$x=-9$
$x=-9 \notin II$
$III:$
$(2x+2)+ [-(3-2x)] = x+14$
$2x+2-3 +2x = x+14$
$4x-1 = x+14$
$3x=15$
$x=5 \in III\checkmark$

NOTE: in each sub domain check if the answer is in its corresponding sub domain


Prove a useful formula for computing expectations Suppose $X$ is a non-negative random variable, and $h$ is a non-decreasing function on $\mathbb{R}_+$ such that $h(0)=0$ and $h$ is absolutely continuous on each bounded interval. ($h(a) = \int_0^a h'(s) ds$ for all $0\leq a <\infty$) Then,
\begin{align}
\mathbb{E}[h(X)] = \int_0^\infty h'(s) P(X>s) dt.
\end{align}

I am thinking about partial integration, but it is not really obvious how to use this. Can somebody help me?

LINKED

Since $X\geqslant0$ a.s. and $h$ is increasing, we have
$$\mathbb E[h(X)] = \int_0^\infty (1-F_{h(X)}(x)\ \mathsf dx = \int_0^\infty (1-F_X(h^{-1}(y))\ \mathsf dy, $$
where $$h^{-1}(y)=\min\{x: h(x)=y\}. $$
Applying the change of variables $s=g^{-1}(x)$, we have
$$\mathbb E[h(X)]=\int_0^\infty(1-F_X(s))h'(s)\ \mathsf ds=\int_0^\infty h'(s)\mathbb P(X>s)\ \mathsf ds. $$

Alternatively, from @Did's hint we see that
$$
\mathbb E[h(X)]
= \mathbb E\left[\int_0^\infty h'(s)\mathsf 1_{\{X>s\}}\ \mathsf ds \right]=\int_0^\infty h'(s)\mathbb P(X>s)\ \mathsf ds.
$$


The limit of functions of two variables The limit of $f(x,y)=\frac{xy^2}{x^2+y^4}$ as $(x,y) \longrightarrow(0,0)$ is doesn't exist, because if we take two paths:

1) Along the path, $x=0$, $y\longrightarrow0$:
$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=0$$
2)  Along the path, $x=y^2$, $y\longrightarrow0$:
$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{y\longrightarrow0}\frac{y^2y^2}{(y^2)^2+y^4}=\frac{1}{2}$$

But if we use polar coordinate methods of evaluating limits of functions of two variable, we get the limit of the above function becomes zero, i.e.,
$x=r\cos(\theta),y=r\sin(\theta)$, we know that $x^2+y^2=r^2$ and this indicates that $r\longrightarrow0$ as $(x,y)\longrightarrow(0,0)$, therefore
$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\longrightarrow0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$
$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$
Please someone help me on this contradictory, why this is so happened?.

When you fixing $\theta$ and take limit as $ r \to 0$ you are taking limits along straight lines. It is not enough to take limits along these lines. When you take limit along the parabola $x=y^{2}$ we get a  different value for the  limit  so the limit of the function does not exist.

I asked this question a couple of days ago. I want to answer my question myself, I'm not fully satisfied with your answers. Let me get started by stating the following Theorem with out proof.

Theorem 1. If the limits of $f(x,y)$ and $g(x,y)$ exists as $(x,y)\rightarrow(0,0)$, then the limit of the product $f(x,y)g(x,y)$ exists as $(x,y)\rightarrow(0.0)$ and is equal to the product of individual limits.

Theorem 2. If the limits of $f(x,y)$ exists and the limits of $g(x,y)$ doesn't exist other than $\pm \infty$, then the limit of the product $f(x,y)g(x,y)$ doesn't exist.

Theorem 1 states that to be the limit of the product exists the limits of the individual functions must exist.

Now let me apply the Theorem 1 and 2 to see the above example has no contradiction with the two methods, using paths and using polar coordinate method. As we have seen, the limit doesn't exist since we got two different limits for the same function along two different paths. Now let's check why the limit is doesn't exist when we use polar coordinate.

Let $x=r\cos(\theta), y=r\sin(\theta)$, we know that $x^2+y^2=r^2$, this indicates that $r\rightarrow 0$ as $(x,y)\rightarrow(0,0)$. Therefore
$$\lim_{(x,y)\rightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\rightarrow 0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$
$$=\lim_{r\rightarrow 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\ne\bigg(\lim_{r\rightarrow 0}r\bigg)\bigg(\lim_{r\rightarrow 0}\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\bigg)$$
As you see the right hand equation, which is the product of two limits $\lim_{r\rightarrow 0}r$ which exists, but the limit $\lim_{r\rightarrow 0}\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}$ doesn't exist, that is why I put $\ne$ in between according to Theorem 1. Therefore in this case we can't apply Theorem 1 and by Theorem 2 the limit doesn't exist.


Prove that $\frac{1}{\phi}<\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx< \frac{24+\sqrt{2}}{41} $ I'm sure that's a coincidence, but the Laplace transform of $1/\Gamma(x)$ at $s=1$ turns out to be pretty close to the inverse of the Golden ratio:

$$F(1)=\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx=0.61985841414477344973$$

  Can we prove analytically that: $$\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx>\frac{1}{\phi}$$

The continued fraction of this number also starts very beautifully:

$$F(1)=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\dots}}}}}}}}}}=$$

$$=[0; 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 7, 2, 1, 1, 1, 2, 6, 1, 4, 7, 1, 3, 1, 1, 6, 1,\dots]$$

The question has no practical application, but everyone here loves Golden ratio questions, right?

  I'm sure there is no closed form for this integral, but if there is some useful transformation, I would like to see it as well.

If we replace every partial quotient in the CF after the first 5 2's by $2$, we obtain a great approximation to the integral:

$$F(1) \approx \frac{1}{41} \left(24+\sqrt{2}\right)$$

So, the more complicated question would be:

  Can we prove analytically that: $$\int_0^\infty \frac{e^{-x}}{\Gamma(x)} dx<\frac{1}{41} \left(24+\sqrt{2}\right)$$

The difference is about $4 \cdot 10^{-7}$.

“…if there is some useful transformation, I would like to see it as well.”  Not sure if the following is useful in regard to the problem you pose, however it extends results in the literature, though I’m sure it’s known, (see below) and may possibly help with a bound.
Define
$$F\left( s \right)=\int\limits_{0}^{\infty }{\frac{{{e}^{-sx}}}{\Gamma \left( x \right)}dx}$$
where s is assumed positive and real and then consider the following integral
$$\frac{1}{\Gamma \left( z \right)}=\frac{1}{2\pi i}\int\limits_{-\infty }^{\left( 0+ \right)}{{{e}^{t}}{{t}^{-z}}dt}$$
where the $\left( 0+ \right)$ notation depicts a contour looping around the negative real axis in a positive sense.  Swapping integration leads to
$$F\left( s \right)=\frac{1}{2\pi i}\int\limits_{-\infty }^{\left( 0+ \right)}{\frac{{{e}^{t}}}{s+\log \left( t \right)}dt}$$
Consider now
$$\int\limits_{{{C}_{1}}}^{{}}{\frac{{{e}^{t}}}{s+\log \left( t \right)}dt}$$
where the contour ${{C}_{1}}$ contains a loop integral about the negative real-axis, hooking around the origin connecting to a portion of  a large circle centred at a point $c>{{e}^{-s}}\in {{\mathbb{R}}^{+}}$ which in turn connects to a vertical line in the plane:
$$\int\limits_{{{C}_{1}}}^{{}}{\frac{{{e}^{t}}}{s+\log \left( t \right)}dt}=-\int\limits_{-R}^{\left( 0+ \right)}{\frac{{{e}^{t}}}{s+\log \left( t \right)}dt}+\int\limits_{{{\Gamma }_{R}}}^{{}}{\frac{{{e}^{c+{{\operatorname{Re}}^{it}}}}}{s+\log \left( c+{{\operatorname{Re}}^{it}} \right)}i{{\operatorname{Re}}^{it}}dt}+\int\limits_{\underset{\left( c>{{e}^{-s}} \right)}{\mathop{c-iR}}\,}^{c+iR}{\frac{{{e}^{t}}}{s+\log \left( t \right)}dt}$$
This is a Bromwich contour with a branch cut along the negative-real axis. By construction the contour integral picks up the residue which occurs at $t={{e}^{-s}}$.  Noting that the integral along the semi-circle vanishes as $R\to \infty $
$$\int\limits_{0}^{\infty }{{{e}^{-sx}}dx}\frac{1}{2\pi }\int\limits_{\underset{\left( c>{{e}^{-s}} \right)}{\mathop{-\infty }}\,}^{\infty }{{{e}^{c+it}}{{\left( c+it \right)}^{-x}}dt}=\underset{t={{e}^{-s}}}{\mathop{res}}\,\left( \frac{{{e}^{t}}}{s+\log \left( t \right)} \right)+\frac{1}{2\pi i}\int\limits_{-\infty }^{\left( 0+ \right)}{\frac{{{e}^{t}}}{s+\log \left( t \right)}dt}$$
Note that
$$\underset{t={{e}^{-s}}}{\mathop{res}}\,\left( \frac{{{e}^{t}}}{s+\log \left( t \right)} \right)={{e}^{\frac{1}{{{e}^{s}}}-s}}$$
Then consider $\int\limits_{{{C}_{2}}}^{{}}{{{e}^{c+iz}}{{\left( c+iz \right)}^{-w}}dz}$ where for now $0\le \operatorname{Re}\left( w \right)<1$.  There is a branch point at $z=ic$  and so let ${{C}_{2}}$ have a portion along the real-axis, a semi-circle in the UHP and a branch cut along the imaginary axis incorporating a small circle about the branch point.  As the radius of the semi-circle becomes large and the radius of the circle about the branch point becomes small, the integral over these contours vanishes leaving
$$\int\limits_{-\infty }^{\infty }{{{e}^{c+it}}{{\left( c+it \right)}^{-w}}dt}=2\sin \left( \pi w \right)\int\limits_{0}^{\infty }{{{e}^{-t}}{{t}^{-w}}dt}=2\sin \left( \pi w \right)\Gamma \left( 1-w \right)=\frac{2\pi }{\Gamma \left( w \right)}$$
for $0<\operatorname{Re}\left( w \right)<1$
The integral representation over the real-line is an entire function of w on$\mathbb{C}$and so the result can be extended, for example, to $\operatorname{Re}\left( w \right)\ge 0$.  Therefore
$$\int\limits_{0}^{\infty }{\frac{{{e}^{-sx}}}{\Gamma \left( x \right)}dx}={{e}^{\frac{1}{{{e}^{s}}}-s}}+\frac{1}{2\pi i}\int\limits_{-\infty }^{\left( 0+ \right)}{\frac{{{e}^{t}}}{s+\log \left( t \right)}dt}$$
Breaking up the loop integral, then
$$\int\limits_{0}^{\infty }{\frac{{{e}^{-sx}}}{\Gamma \left( x \right)}dx}={{e}^{\frac{1}{{{e}^{s}}}-s}}+\int\limits_{0}^{\infty }{\frac{{{e}^{-x}}}{{{\pi }^{2}}+{{\left( s+\log \left( x \right) \right)}^{2}}}dx}$$
whereupon making substitutions
$$\int\limits_{0}^{\infty }{\frac{{{e}^{-sx}}}{\Gamma \left( x \right)}dx}={{e}^{\frac{1}{{{e}^{s}}}-s}}+\int\limits_{-\infty }^{\infty }{\frac{{{e}^{x-s-{{e}^{x-s}}}}}{{{\pi }^{2}}+{{x}^{2}}}dx}={{e}^{\frac{1}{{{e}^{s}}}-s}}+\frac{1}{\pi }\int\limits_{-\pi /2}^{\pi /2}{{{e}^{\pi \tan \left( x \right)-s-{{e}^{\pi \tan \left( x \right)-s}}}}dx}$$
This generalises the result in Borwein et. al. “experimentation in mathematics” 2004 pg 287.
Note the following additional perhaps curious/worthless fact:
$${{e}^{\frac{1}{{{e}^{s}}}-s}}=\sum\limits_{n=0}^{\infty }{\frac{{{e}^{-sn}}}{\Gamma \left( n \right)}}$$
and so the whole thing smacks of Euler-Maclaurin.  Going down that path the following pops out
$$\int\limits_{0}^{\infty }{\frac{{{e}^{-st}}\psi \left( t \right)}{\Gamma \left( t \right)}dt}=-0.6198584141460518$$
which to 12 places has the same magnitude as your number $F\left( 1 \right)$.

Essentially, this can be proved by dividing the integral interval into $(0,1/100)$, $(1/100,10000)$ and $(10000,\infty)$ and bound the integrand $e^{-x}/ \Gamma(x)$ with simpler functions on each part.

For the lower bound we don't really need to get into the details. There are software that can automatically compute guaranteed numeric upper and lower bounds of integrals. This type of methods are called validated numerics. One such software is arblib. You can access it through SageMath.

The following code proves the lower bound.

low=RBF(CBF.integral(lambda x, _: (e^(-x))/gamma(x), 1/100, 5))
phi=RBF((1+sqrt(5))/2)
low > 1/phi

The output is True.

For the upper bound,
we use the following inequalities
$$
e^{-x}/\Gamma(x) \le x, \qquad x \in (0,1/100)
$$
and
$$
e^{-x}/\Gamma(x) \le e^{-x}, \qquad x \in (10000,\infty)
$$
So the following code proves the upper bound.

p1=RBF(CBF.integral(lambda x, _: x, 0, 1/100))
p2=RBF(CBF.integral(lambda x, _: (e^(-x))/gamma(x), 1/100,10000))
p3=RBF(integral(e^(-x), x,10000,oo))
up=RBF(1/41*(24+sqrt(2)))
up > p1+p2+p3

The output is again True.

If one want an completely analytic proof, for the lower bound one can use the inequality
$$
f(x):=e^{-x}/\Gamma(x) \ge f_{1,2,5}(x)
$$
where $f_{1,2,5}(x)$ is the Pade approximant of order $(1,2)$ at $x=5$. Since $f_{1,2,5}(x)$ is a rational function, one can do integrate it on $(1/100,5)$ exactly and get an analytic lower bound.


What is a primitive notion and an Axiom? I'm a beginner of learning about Classes, Sets, Types of Definitions(Intensional, Extensional and Ostensive) when I came across primitive notion and Axioms. Now I understand Axiom of Extensionality but the actual definition of an Axiom is just confusing, according to one source, it's any logical statement assumed to be true, and another says an axiom is the little bits of information that help us infer(because without these bits, we can't infer from nothing)

My second question is: What's a primitive Notion, and where is its relevance in sets and mathematical logic? Thanks for the help guys!

I think you may be hoping for more of an insight, but "primitive notion" simply means a syntactic element that appears in the Axioms with no definition in terms of simpler elements.  We have to start somewhere, right?  So in set theory the relation $\in$ for set membership is a primitive notion.

Similarly the Axioms are the statements with which we begin reasoning to prove Theorems in a theory.  The Axioms come without justification; they are assumed to be true for the purpose of the theory...  If the Axioms turn out to be inconsistent, then the theory they create is not so interesting.

Perhaps it will help to illustrate the primitive notions of set theory (membership $\in$ and identity $=$, though this last is often bundled with the logic of predicate calculus) if we discuss some notions that are not primitive because they are defined in terms of simpler notions.  We can think of these definitions as abbreviations for expressions made up, ultimately, of only primitive notions.

A simple example might be the subset relation $\subseteq$.  To say one term is a subset of another can be defined using only logical syntax and the primitive notion of membership:

$$ x \subseteq y \equiv_{def} \forall z (z \in x \implies z \in y) $$

The idea of this definition/abbreviation is that we can use the subset symbol anywhere that we would like to proceed as if replacing a statement about "subset" is to mean the more verbose statement, any member of the first term is also a member of the second term.

In this way one can actually build up an easily recognized framework for mathematics starting from the primitive notion of membership $\in$.

Just a comment -- but in the age of computer science, where we have exact and explicit machine-specification for absolutely everything, the notion that some concepts have to be left undefined (ie, "vague") seems completely misguided and unnecessary.  I personally think the concept of "primitive notion" as given by Alfred Tarski in the Wikipedia definition is a formula for bottomless confusion and wasted time.

When we set out to construct a given discipline, we distinguish, first of all, a certain small group of expressions of this discipline that seem to us to be immediately understandable; the expressions in this group we call PRIMITIVE TERMS or UNDEFINED TERMS, and we employ them without explaining their meanings. At the same time we adopt the principle: not to employ any of the other expressions of the discipline under consideration, unless its meaning has first been determined with the help of primitive terms and of such expressions of the discipline whose meanings have been explained previously. The sentence which determines the meaning of a term in this way is called a DEFINITION,...

https://en.wikipedia.org/wiki/Primitive_notion

Computer science is built on bits.  Algebra can be built on bits.  The real number line can be built on bits.  Any arithmetic operation can be built from bits.  Any logical operation can be built on bits.  Why do it any other way?

Thanks.


Minimum and Maximum number of triangulations of a polygon Triangulation of a simple polygon $P$ is a decomposition of $P$ into triangles by a maximal set of non-intersecting diagonals.  We also know that triangulation of a polygon is not neccessarily unique.
The question (taken from Computational Geometry in C by J. Rourke):

  Which polygons have the fewest number of distinct triangulations? Can
  polygons have unique triangulations? Which polygons have the largest
  number of distinct triangulations?

Note: I've already answered the second part by drawing a convex 4-gon with only 1 possible diagonal. The problem is the other two parts.

Convex polygons (ones with all internal angles less than $180^\circ$) have minimal polygons equal to the number of vertices minus 2.

Note: While these are all regular (same lengths and angles) they don't have to be. (It was just easier to draw.)

Any concave corners increase the number of polygons required. Many kinds of polygons have maximal triangles. Easy to draw examples would be stars (but lots of convolutions of these basic shapes are possible).

This is a very broad question and whole books could be written about this as there are so many cases to cover. I'd recommend reading up on computer graphic triangulation and computer graphic primitives.


Logarithm to trigonometry conversion I have an expression like
$$\frac{1}{2} i \log \left(\frac{a-b+i c}{a+b-i c}\right)$$
I was wondering if it is possible to write it in terms of trig functions. I guess it is possible to write all logs in some form of trig due to Euler's relation.
As an attempt to solve the problem, I was trying to play around the relation
$$\frac{1}{2} i \log \left(\frac{a+i c}{-a+i c}\right)=\tan^{-1}\left(\frac{a}{c}\right)$$ to change into arctan function but does not look like this alone is going to help in this case.
Any suggestions?

Assuming $a,c>0$, if you write $a+ic = re^{i\theta}$ where $r=\sqrt{a^2+c^2}$ and $\theta=\tan^{-1}(\frac{c}{a})$, then
$$-a+ic = re^{i(\pi - \theta)}$$
Since both numbers are positive the argument will be $\pi - \theta$ for $-a+ic$.
Now put this in your expression to get:
$$\frac{1}{2}i\log{\frac{a+ic}{a-ic}}=\frac{1}{2}i\log{\frac{re^{i\theta}}{re^{i(\pi -\theta)}}} = \frac{1}{2}i\log e^{i(2\theta-\pi)}=\frac{1}{2}i\times i(2\theta - \pi)$$
which becomes
$$-\frac{1}{2}(2\theta - \pi)=\frac{\pi}{2}-\theta=\frac{\pi}{2}-\tan^{-1}\frac{c}{a}=\tan^{-1}\frac{a}{c}$$
Use the fact $$\tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{\pi}{2}$$ for all $x>0$
.The expression might lead to different expressions depending on the values of $a,c$ I took the case where both are positive. It can be shown for rest of the cases as well. Hope this helps....


Angle between 2 vectors using the determinant I have a polygon like this:

I basically want to find the angles $\alpha$, inside the polygon, between  the vectors.
I'm using the determinant to calculate the angle alpha:

$det(\vec V2, \vec V2 ) = ||V2|| * ||V1|| * \sin \theta
$

So $\theta$ will be the arcsin. But my problem is that in this particular case, the angle that is calculated is not $\alpha$ but it's 180- $\alpha$ and I don't understand why.

This is my code in C:

TSMV_Point2D vecteur1, vecteur2;
vecteur1.x = p1.x - pC.x;
vecteur1.y = p1.y - pC.y;
vecteur2.x = p2.x - pC.x;
vecteur2.y = p2.y - pC.y;

float determinant = vecteur2.x * vecteur1.y - (vecteur1.x * vecteur2.y);
float normeVec1 = sqrt(vecteur1.x * vecteur1.x + vecteur1.y * vecteur1.y);
float normeVec2 = sqrt(vecteur2.x * vecteur2.x + vecteur2.y * vecteur2.y);

float division = determinant / (normeVec1 * normeVec2);
float angle = asin(division) * 180 / M_PI;

Can someone help me understand?

Well, usually angles between two vectors are considered to be lying in $[0,\pi]$ and so choosing a formula with $\arccos$ in it is a better choice, as @Badoe mentioned.

However, in your particular case, keep in mind that $\arcsin$ is defined over $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so, it is very possible that the angle you would like to get is actually larger than $90^\circ$ and, hence, $\arcsin$ moves is back into $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, resulting to giving you the complementary angle of what you've been looking for.

But, the above is just a speculation, it would be more helpful if someone could enlight us on how is $\arcsin$ implemented in $C$. Till then, you can always use the formula:
$$\theta=\arccos\frac{\vec{v}\cdot\vec{u}}{\lVert \vec{v}\rVert\lVert \vec{u}\rVert}$$

Because the formula you have is not correct. The dot product of two vectors equals the product of the norms of the vectors times cosine of the angle.

$$\mathbf{V_1} \cdot \mathbf{V_2}=||\mathbf{V_1}||\cdot ||\mathbf{V_2}|| \cdot \cos\theta$$


Lebesgue decomposition of an increasing function This problem asserts that any increasing function $f$ on $[a,b]$ can be decomposed as $$F=F_A+F_C+F_J,$$ with $F_A$ is absolutely continuous, $F_J$ is a jump function, and $F_C$ is a singular function, i.e., it is continuous and  $F'_C=0$, a.e.. Both $F_A$, $F_J$, $F_C$ are increasing on $[a,b]$. Moreover, the decomposition is unique upto a constant.

From Stein's book, I know that $F=F-J+J$ where $J$ is the jump function associated to $F$, and $F-J$ is increasing and continuous. But how can I proceed from here to prove the claim above?

Any hints would be greatly appreciated!

My suggestion is as follows: because $f$ is increasing on $I=[a,b]$, then it has bounded variation and hence $u\in BV(I)$. Thus, we have $Du$, as a Radon measure, is well defined. Now, by Lebesgue decomposition theorem, we could decompose $Du$ into two measures, namely the one $\mu_1<<\mathcal{L}^1$ and $\mu_2\bot \mathcal{L}^1$ such that $Du=\mu_1+\mu_2$. We future write $\mu_j:=\mu_2\lfloor J$ where $J$ is the jump set and $\mu_s:=\mu_2\lfloor(I-J)$. Now, $\mu_1$ is actually your $F_A$, $\mu_j$ is $F_J$ and $\mu_s$ is $F_C$.


Convergent/divergent series Is the following series divergent/convergent?

$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$

I think it is divergent since
$$
\begin{align}
S&>1-\frac{1}{2}-\frac{1}{2}+4\cdot\frac{1}{6}-\frac{4}{7}+\frac{5}{15}-\frac{6}{16}+...\\
&=1/3-4/7+1/3-6/16+...=\sum_{n=1}^\infty 1/3-\alpha_n
\end{align}
$$
where $\alpha_n=4/7, 6/16, 8/22$ which tends to 0, so the series on the right hand side is divergent. Is this the right answer? Thanks

After placing brackets around the terms with the same sign the series becomes
$$1+\sum_{n=2}^\infty(-1)^{n+1}(H_{n(n+1)/2}-H_{n(n-1)/2}),$$
where $H_n=1+\frac12+\frac13+\ldots+\frac1n$ is the $n$-th harmonic number. Now using well-known formula $$H_n=\log n+\gamma+O\left(\frac{1}{n}\right),$$
we get
$$a_n=H_{n(n+1)/2}-H_{n(n-1)/2}=\log\frac{n+1}{n-1}+O\left(\frac{1}{n^2}\right)=$$
$$=\log\left(1+\frac{2}{n-1}\right)+O\left(\frac{1}{n^2}\right)=\frac{2}{n}+O\left(\frac{1}{n^2}\right).$$
Therefore the main term of the series with braces is $$2\frac{(-1)^{n+1}}{n}+O\left(\frac{1}{n^2}\right),$$
which means that the series converges (by Leibniz's and comparison tests), and so does the original one.

S is convergent ( Using Leibnitz's Test for convergence of Alternating series)


Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist. I turned that into one fraction, then I factored the polynomial on the numerator as $-(x-1)(x+2)$ and the one on the denominator as $(x+1)(-x^2-x-1)$. What did I do wrong?

Using the identity
$A-B=\dfrac{1}{\dfrac{1}{A}}-\dfrac{1}{\dfrac{1}{B}}$ we get $$\dfrac{1}{x-1}-\dfrac{3}{1-x^3}=\dfrac{2-x^3-x}{-x^4+x^3+x-1}$$ so you go from the form $\infty-\infty$ to the form $\dfrac00$ and you can apply L'Hôspital so you have $$\lim_{x \to 1} \frac{1}{x-1}-\frac{3}{1-x^3}=\lim\dfrac{-3x^2-1}{-4x^3+3x^2+1}$$ and you can verify that the limits to the right and to the left are not equal, the first is $-\infty$ and the second is $\infty$.
Thus the limit does not exist.

I believe you messed up by looking at $$\lim_{x \to 1}\left(\frac{1}{x-1} \color{red}{+} \frac{3}{1-x^3}\right)$$ instead of $$\lim_{x \to 1} \left( \frac{1}{x-1}\color{red}{-}\frac{3}{1-x^3} \right)$$
The value of the first one is $1$, but the second one does not exist.
We can factor $1-x^3$ as $(x-1)(-x^{2}-x-1)$ so that the limit becomes $$\lim_{x \to 1}\frac{(-x^2-x-1)-3}{(x-1)(-x^2-x-1)}$$
Then it is clear that the limit does not exist since the numerator when plugging in $x = 1$ is $-6$ and the denominator is $0$.


$dX_t/X_t=\mu\, dt+\sigma \, dZ_t$, does this notation make sense? I understand that the notation

$$dX_t=\mu X_t \,dt + \sigma X_t \,dZ_t,$$

where $Z_t$ is Brownian Motion, is a shortcut to

$$X_t-X_0=\int_0^t\mu X_s \, ds+\int_0^t \sigma X_s \, dZ_s, \tag{*}$$

which has a precise meaning, since the Ito integral is defined.

But I don't get it when people write:

$$\frac{dX_t}{X_t}=\mu \, dt + \sigma \, dZ_t,$$

and later go on defining $\frac{dX_t}{X_t}$ as a variable. What does $\frac{dX_t}{X_t}$ even mean? To me, it looks like taking $X_s$ out of the integral in $(*)$, which does not make any sense.

For example, if $X_t$ is the price of an asset, $dX_t/X_t$ is sometimes defined as the rate of return the asset, but my point is that it is not even a mathematically defined object.

Am I missing something, or this is sloppy?

$$\frac{dx_t}{x_t}=\mu+\sigma dz_t\\$$let $f(x)=ln(x)$ apply ito formula to find $df(x)$ $$df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx+\frac{1}{2}\frac{\partial^2f }{\partial x^2}\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\df=0dt+\frac{1}{x}dx+\frac{1}{2}(\frac{-1}{x^2})dx^2\\$$now put $dx_t=\mu x_tdt+\sigma x_t dz_t$
$$df=0dt+\frac{1}{x_t}(\mu x_tdt+\sigma x_t dz_t)+\frac{1}{2}(\frac{-1}{x_t^2})(\mu x_tdt+\sigma x_t dz_t)^2\\$$
$$df=0dt+\frac{1}{x_t}(\mu x_tdt+\sigma x_t dz_t)+\frac{1}{2}(\frac{-1}{x_t^2})(\mu^2 x_t^2(dt)^2+\sigma^2 x^2_t (dz_t)^2+2\mu \sigma x^2_tdtdz_t)^2\\$$
apply $dt.dt \to 0\\dt.dz_t \to 0 \\dz_t.dz_t \to dt$
so
$$df=0dt+\mu dt +\sigma dz_t-\frac{1}{2}\sigma^2 dt$$apply integral for both sides
$$\int_{0}^{t}d(lnx_s)=\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s\\ \ln(x_t)-\ln(x_0)=\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s\\\frac{x_t}{x_0}=e^{\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s}\\x_t=x_0.e^{\int_{0}^{t}(\mu-\frac{1}{2}\sigma^2)ds+\int_{0}^{t}\sigma dz_s}$$


Partitions of $n$ with exactly 3 parts I'm learning about generating functions, and came across this question: find the generating function for the number of partitions of a number $n$, into exactly 3 parts. I just solved a problem where the condition was that each part is no greater than 3, but I'm kind of stuck on this one. Any ideas? Thanks!

Hint: The following can be found somewhat more detailed in this answer.

  A generating function for the number of partitions with exactly three parts is
  \begin{align*}
&\frac{1}{(1-x)(1-x^2)(1-x^3)}-\frac{1}{(1-x)(1-x^2)}\\
&\qquad=\frac{1}{(1-x)(1-x^2)}\left(\frac{1}{1-x^3}-1\right)\\
&\qquad=\frac{x^3}{(1-x)(1-x^2)(1-x^3)}\\
&\qquad=x^3+x^4+2x^5+3x^6+4x^7+\color{blue}{5}x^8+7x^9\cdots
\end{align*}

Example: There are $\color{blue}{5}$ partitions of $8$ with three summands
\begin{align*}
8&=1+1+6\\
&=1+2+5\\
&=1+3+4\\
&=2+2+4\\
&=2+3+3
\end{align*}

Other way at looking the problem.p(n,3) can be very useful.Idea behind is to put the answer as an arithmatic progression.p(n,3)=sum of all positive terms of (n-3-6t) and adding 1 if any term is zero,where t=0,1,2,3,4,5•••
Ex
p(11,3)=(11-3-6*0)+(11-3-6*1)=8+2=10;
p(12,3)=(12-3-6*0)+(12-3-6*1)=9+3=12;
p(15,3)=(15-3-6*0)+(15-3-6*1)+(15-3-6*2);
             In the last example we encounter the third term being zero.We have to add 1 for it.
Hence,
p(15,3)=12+6+0+1=19


Modular exponentiation WITHOUT modular exponentiation Given that $719$ is prime, find the least positive residue of $11^{721} \pmod {719}$, without using modular exponentiation.

So, I know how to use modular exponentiation and have done it to get the answer $612$, but I have no idea how to do this without modular exponentiation...

Any ideas for a method I can try?

This is a simple application of Fermat's Little theorem.

**
if M is prime, then to find ab % M,

r = b % (M-1)

c = a % M

ab % M = cr % M

**

721 % 718 = 3

11 % 719 = 11

Therefore, 11721 % 719 = 113 % 719 = 612

So, there is no need to perform modular exponentiation if Mod value is given prime.


Suppose M is a finitely generated non-zero R-module, where R is a commutative unital ring. Show that the tensor product of M with itself is non-zero. Suppose M is a finitely generated non-zero R-module, where R is a commutative unital ring. Show that the tensor product of M with itself is non-zero.
I know one way to show this is to find an R-bilinear map which is nonzero, but am not sure how to find it.

Not sure this is the best proof, but here goes. Note that $M\otimes_R M = 0$ if and only if $(M\otimes_R M)_{\mathfrak p} = 0$ for all primes $\mathfrak p\subset R$ (Atiyah-MacDonald 3.8). But
$$
(M\otimes_R M)_{\mathfrak p}\cong M_\mathfrak p\otimes_{R_{\mathfrak p}}M_{\mathfrak p}
$$
by Atiyah-MacDonald 3.7.
Hence we have reduced to the case where $R$ is a local ring. The statement now follows from the fact that tensor product is faithful for finitely-generated modules over local rings, e.g. exercise 3 in the second chapter of Atiyah-MacDonald. (This is a consequence of Nakayama's lemma, as far as I remember.)
Remark. So why doesn't the argument given here show that tensor products are faithful over any ring via reduction to the local case? Simply because a pair of distinct modules $M$ and $N$ might be supported on disjoint sets of primes. For instance, this is the case when $R = \mathbb Z$, $M = \mathbb Z/p$ and $N = \mathbb Z/q$ for distinct primes $p,q$. Then $\mathbb Z /p\otimes\mathbb Z/q = 0$ because, with $r$ ranging over primes, $(\mathbb Z/p)_{(r)} = 0$ for $r\neq p$ and similarly for $q$.

This is false. Let $R=\mathbb{C}[X]/(X^2)$. Let $x=\bar{X}$, so $x^2=0$, and let $M= R x$. Then $M$ is a finitely generated nonzero $R$-module, but $M\otimes M$ is generated by $x\otimes x=x^2\otimes 1=0\otimes 1=0$.
More generally, if $R$ is a commutative ring having a nonzero ideal $I$ satisfying $I^2=0$, then $I\otimes_R I=0$ and you get a counterexample.


How many integer solutions are there to the inequality $x_1 + x_2 + x_3 \leq 17$ if we require that $x_1 \geq 1$, $x_2 \geq 2$, $x_3 \geq 3$? How many integer solutions are there to the inequality
$x_1 + x_2 + x_3 \leq 17$,
if we require that
$x_1 \geq 1$, $x_2 \geq 2$, $x_3 \geq 3$?

The way I did it:

$x_4 = 17-x_1-x_2-x_3-x_4$

$x_1+x_2+x_3+x_4 = 17$

$x_1= x_i + 1$

$x_2= x_{ii}+2$

$x_3= x_{iii}+3$

$x_4= x_{iiii}$

So,

$x_i+1+x_{ii}+2+x_{iii}+3+x_{iiii}= 17$

$x_i+x_{ii}+x_{iii}+x_{iiii}=11$

$3$ bars, $\binom{14}{3} =364$ different ways?
Is this the right approach? correct answer?

for the inequality x1+x2+x3<=17,we may introduce a fourth variable say x4,so as to solve the number of integer solutions to;x1+x2+x3+x4=17 with x1>=1,x2>=2 and x3>=3.However, the three constraints implies x1-1>=o,x2-2>=o and x3-3>=o.so if we let x1-1=a,x2-2=b,x3-3=c, x4=d with x4>=o, we therefore solve the number of integer solutions to (a+1)+(b+2)+(c+3)+d=17 which implies a+b+c+d=11 with a,b,c,d>=o which is (11+4-1)Combination(4-1)=(14)combination(3)=364.thanks.


Finding a complex number So I have an illustrative picture of all of the roots of one complex number (the exponent is random). I have to find this number $z$, if we
know that these roots are on a circle, that's diameter is $3.4$ and degree between the real axis and the positive direction is $-61$ degrees.

Here is the illustrative picture that I have:

I got the answer that the complex number is $z=0.824+1.487i$.
but all I really need is $a$

Are my calculations right?

Assuming I understand the question correctly, we have:
$$w=z^{\frac{1}{6}}=\frac{3.4}{2} \cdot (\cos{(-61)}+i \cdot  \sin{(-61)}) = 0.82418-1.48685i$$
Which is what you have. But this is $w=z^{\frac{1}{6}}$ ;
$$\therefore z = w^6 = 24.00534-2.52306i$$

Change the $+$ to a $-$ and you're golden.

Presumably you have calculated
$$x = 1.7\cos(-61^\circ), y = 1.7\sin(-61^\circ),$$ which does indeed give the answer you suggest.


Null sets in $\mathbb{R}$ We know $A \subseteq \mathbb{R}$ is null if given $\epsilon > 0$, there exists intervals     $\{I_n\}_{n \geq 1}$ such that

$$ A \subseteq \bigcup_{n=1}^{\infty} I_n    \text{    and   }    \sum|I_n| < \epsilon $$

My question is that if we change the words interval in the above definition with open intervals, closed intervals ,and intervals of the form $(a,b]$, $[a,b)$, then we get an equivalent notion?

If you are dealing with Borel $\sigma$-field over $\mathbb{R}$ then for any property $P$ that depends only on the measure defined on that space, if $P$ holds for any one of the following class of sets, then it holds for any other class, and also for the entire Borel $\sigma$-field.

class of all intervals,
class of all open intervals,
class of all closed intervals,
class of all clopen intervals,
class of all half-lines etc.


conditional probability on Truth and lies Mike tells truth with probability 1/3 and lies with probability 2/3. Independently, David tells truth with probability 3/4 and lies with probability 1/4. Both watch a soccer match. David tells you that Spain won, Mike tells you that Spain lost. What probability will you assign to Spain's win?

Suppose there are 10 people instead of 2. Now, with each of them telling truth or lie, the probability of win will decrease? Or in the same way if i calculate P(loss) it will be 1-P(win)?

how should i approach for this question?

Mike tells truth with probability 1/3 and lies with probability 2/3. Independently, David tells truth with probability 3/4 and lies with probability 1/4. Both watch a soccer match. David tells you that Spain won, Mike tells you that Spain lost.

Edited: Full credit is due to joriki for this amended answer.

Correction: We're looking for Spain winning given what they said.

Letting $D_S, M_S$ be the events "David said Spain won" and "Mike said Spain won".  These will be a lie or truth depending on what the truth is.

$$\begin{align}\mathsf P(S \mid D_S, \neg M_S) & = \frac{\mathsf P(S, D_S, \neg M_S)}{\mathsf P(S, D_S, \neg M_S)+\mathsf P(\neg S, D_S, \neg M_S)}
\\ & = \frac{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)}{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)+\mathsf P(\neg S)\mathsf P(D_S\mid \neg S)\mathsf P(\neg M_S\mid \neg S)} \end{align}$$

We have   $\mathsf P(\neg M_S\mid S)=2/3 \\ \mathsf P(\neg M_S\mid\neg S)=1/3 \\ \mathsf P(D_S\mid S)=3/4 \\ \mathsf P(D_S\mid\neg S)=1/4$ .

What we don't have is: $\mathsf P(S)$, the prior probability that Spain won.

$$\begin{align}\mathsf P(S \mid D_S, \neg M_S) & = \frac{3\mathsf P(S)}{3\mathsf P(S)+2\mathsf P(\neg S)} \end{align}$$


Difference between inductive definition and recursive definition as used in "A concise introduction to mathematical logic" by W. Rautemberg. I'm trying to make sense of the definition of propositional formulas in "A concise introduction to mathematical logic" by W. Rautenberg. This is the definition given by Rautenberg (page 4):

"... we define a propositional language $F$ of formulas built up from the symbols $($, $)$, $\land$, $\lor$, $\lnot$, $p_1$, $p_2$, $\dots$    inductively as follows:

($F_1$) The atomic strings $p_1$, $p_2$, $\ldots$ are formulas, called prime formulas, also called atomic formulas, or simply prime.

($F_2$) If the strings $\alpha$, $\beta$ are formulas, then so too are the strings $(\alpha \land \beta)$, $(\alpha \lor \beta)$, and $\lnot \alpha$.

This is a recursive (somewhat sloppily also called inductive) definition
in the set of strings on the alphabet of the mentioned symbols..."

My question is, if he defines formulas inductively, why he then says that this is a recursive definition somewhat sloppily called inductive? Is that an inductive or recursive definition? What is the difference between defining inductively and a recursive definition?

Thanks,

At least some people feel that strictly speaking it's proofs that are inductive, while a definition may be recursive. If so the question is not the difference between an inductive definition and a recursive definition, the point is there's simply no such thing as an "inductive definition", strictly speaking.

I think the answer to your question is in the author's "somewhat sloppily called".

This definition is recursive because it depends on knowing about multiple predecessors. The set of formulas can be represented as a directed graph with sources the atomic formulas.

Strictly speaking, an inductive definition would depend on a single previously defined formula. The set of formulas would then be a chain starting with an atomic formula, or perhaps a union of disjoint chains.

Speaking sloppily, the author conflates the two ideas, since he probably never needs to distinguish and his readers may prefer "inductive".


Equivalence to properly discontinuous action Let $X$ be a metric space and let $G$ be a group of homeomorphisms $X \to X$ acting on $X$. We say $G$'s action is properly discontinuous in case for every $x \in X$ and compact $K \subseteq X$, there are at most finitely many $g \in G$ such that $g(x) \in K$. Equivalently (and this is not hard to show), $G \cdot x$ is discrete and $G_x$ finite for any $x$.

Why is it the case that $G$ acts properly discontinuously if and only if for any compact $K$, $g(K) \cap K \ne \emptyset$ for only finitely many $g$? One direction is relatively easy, but I just cannot seem to prove the "only if". The best I've been able to do is for finite $K$ (which is kind of the next best thing when you're stumped on proving something for compact sets, I guess).

I taken a look in Svetlana Katok's "Fuchsian groups", and I'm really confused.

For instance, it is proved in that book that $G$ acts properly discontinuously on $X$ iff every point $x \in X$ has a neighborhood $U_x$ such that $U_x \cap gU_x$ is not empty for only finitely many $g \in G$. (I don't understand the "only if" part, and I don't think that this holds in general).

Although, let be $G$ the group of all the (continuous) transformations $f_n$, $n \in \mathbb{Z}$,  of the plane without the origine $(0,0)$, with $f_n:(x,y) \mapsto (2^nx,2^{-n}y)$.

One can easily see that for each point $(x,y)$, a disc with this point as a center and a sufficiently small radius   intersects the orbit of $(x,y)$ only in one point, thus the orbit of each point is a discret subset, and clearly G acts freely on our set, thus the stabilizer of each point is finite.

On the other hand, consider the segment $K = \{(t, 1-t)| 0 \leq t \leq 1 \}$, clearly $K$ is compact, $K$ contains for each $n>0$ the element $P_n =(\frac{2^n-1}{2^{2n}-1}, \frac{2^n (2^n-1)}{2^{2n}-1}) $, and $f_n(P_n) \in K\cap f_nK$.  Therefore $K \cap f_nK$ is not empty for infinitely many elements $f_n$ of $G$.

Suppose there is a compact $K$ such that $g(K) \cap K \neq 0$  for infinitely many $g$. Denote by $L_k := g_k(K) \cap K$ where $g_k, k\in \mathbb N$ are among those elements of $G$ for which the inequality holds. Now pick any sequence $y_k \in L_k$. Because $K$ is compact, we have a convergent subsequence $y_{k_n} \to y$. But because every $y_{k_n} \in g_{k_n}(K)$ there must also be a sequence $x_{k_n}$ in $K$ such that $g_{k_n} \cdot x_{k_n} = y_{k_n}$ and so again by compactness there is a convergent subsequence $x_{k_{n_m}} \to x$ of $x_{k_n}$. But then $g_{k_{n_m}}\cdot x \to y$ is not discrete.


Inequality : $\sum_{k=1}^n x_k\cdot \sum_{k=1}^n \frac{1}{x_k} \geq n^2$ I have to show the inequality of $$\left(\sum_{i=1}^n x_i\right)*\left(\sum_{i=1}^n \frac{1}{x_i}\right) \geq n^2.$$For $x_1, ... x_n \in \mathbb{R_{>0}}$ and $ n \geq 1$. I wanted to show this inequality by induction. The basis is clear, but I am not sure how to do the inductive step. Can someone help me with that?

Expand it out as $\sum_{i,j} x_i/x_j$, and use $t + 1/t \ge 2$ for $t > 0$

By C-S
$$\sum_{i=1}^ns_i\sum_{i=1}^n\frac{1}{s_i}\geq\left(\sum_{i=1}^n\sqrt{s_i}\sqrt{\frac{1}{s_i}}\right)^2=n^2.$$

The Cauchy-Schwarz inequality it's the following.

For all reals $a_i$ and $b_i$ the following inequality holds.
$$\sum_{i=1}^na_i^2\sum_{i=1}^nb_n^2\geq\left(\sum_{i=1}^na_ib_i\right)^2.$$
In our case $a_i=\sqrt{s_i}$ and $b_i=\frac{1}{\sqrt{s_i}}.$

Good luck!


A family has three children. What is the probability that at least one of them is a boy? According to me there are $4$ possible outcomes:

$$GGG \ \
BBB \ \
BGG \ \
BBG $$

Out of these four outcomes, $3$ are favorable. So the probability should be $\frac{3}{4}$.

But should you take into account the order of their birth? Because in that case it would be $\frac{7}{8}$!

The complement of at least one boy is all three girls

So, $P($ at least one boy$)=1-P(GGG)$

$=\displaystyle1-\left(\frac12\right)^3$

This is the de facto way of solving problems of Probability of at least one in case of Binomial Distribution like tossing a coin etc.

The possibilities are

ggg ggb gbg bgg gbb bgb bbg bbb

at least one boy... 7/8


Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$? Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS?

$$\frac{d}{dt}\left(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{m \vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$

What is the differentiation rule used?

It's just the usual product and chain rules, but it helps to work with indices and sum implicitly when they repeat. I'll set $c=1$ to simplify notation:$$\frac{d}{dt}\left((1-u_ju_j)^{-1/2}mu_i\right)u_i=\left(u_k\dot{u}_k(1-u_ju_j)^{-3/2}mu_i+(1-u_ju_j)^{-1/2}m\dot{u}_i\right)u_i\\=(1-u_ju_j)^{-3/2}mu_i\dot{u}_i\left(u_iu_i+1-u_iu_i\right)=(1-u_ju_j)^{-3/2}mu_i\dot{u}_i.$$

The usual one. Try this: $u^2=(u,u)$ and then $\dfrac{d}{dt}u^2= 2(u',u).$


Let $f(n)$ be the number of prime factors of the positive integer $n$. Find $\lim_{n\to \infty}\frac{f(n)} n$ Let $f(n)$ be the number of prime factors of the positive integer $n$. Find $\displaystyle \lim_{n\to \infty}\frac{f(n)} n$.

I suspect it's equal to $0$, but how can I show this? Thank you.

Note that $2^{f(n)}\le n$, so $f(n)\le \log_2(n)$. Now the standard tools, such as L'Hospital's Rule, can be used to show that $\lim_{x\to\infty}\frac{\log_2(x)}{x}=0$.

Since we know infinite that there are infinitely many primes, we can take $n$ tending to be an infinitely large prime.

$\dfrac{f(n)}{n}=\dfrac{1}{p}$, here $f(p)=1$(Why?)


Predicate Logic and Sets (Review Questions) I have a couple questions I have tried. These are simple homework questions I am doing for review for my 3rd year course in university. These are not for marks.

Predicate Question

1)

Let the predicate E(x,y) represent "Person x eats food y"

Let the predicate M(y) represent (Food y is a meat product"

a) Express the statement in predicate logic: "Someone is a vegetarian."

My Answer: Ǝx∀y E(x,y) ¬M(y)

New Answer: Ǝx∀y (E(x,y) => ¬M(y))

b) Express the statement in predicate logic:"Nobody (except maybe John) eats lasagna."

My Answer: ¬∀xƎy E(John,Lasagna)

New Answer: ∀x(E(x,lasagna) => (x = John))

Sets Question

2) Determine which of the following sets are finite or infinite. If set is finite show all elements in the set. If infinite show 5 elements in the set.

N represents Natural Numbers

a) A = {n ∈ N: ∀m ≤ n, n ≠ m^2}

My Answer: Infinite Set, A = {1, 4, 9, 16, 25} (not sure about this one)

New Answer: Infinite Set, A = {3, 5, 7, 9 , 11}

b) B = {n ∈ N : ∀m ≤ n, n = m^2}

My Answer: Infinite Set, B = {1, 4, 9, 16, 25}

New Answer: Infinite Set, B = {2, 4, 6, 8, 10}

c) C = {n^2 : n ∈ N and 2 < n < 6}

My Answer: Finite Set, C = { 9, 16, 25 }

d) D = {(n,m)∈ N x N: 1 < n < m < 6}

My Answer: Finite Set, D = { (2,3), (3,4) , (4,5) }

New Answer: Finite Set, D = {(2,3),(2,4),(2,5),(3,4),(3,5),(4,5)}

I appreciate any sort of help, if the answer is correct just reply about the ones that are not correct, thanks in advance.

I find it helps to have an intermediate step that says the same thing as the predicate, but in words rather than symbols. So for your first one ("Someone is a vegetarian"), after some massaging, I'd work with the words to make it "There exists a person, such that for all food, if that person eats it then it's not meat". Which would then be written as $\exists{x}\forall{y}\left(E(x,y)\implies\neg M(y)\right)$. Similarly for your second one, I'd go from "Nobody (except maybe John) eats lasagne" to "If someone eats lasagne, that person must be John" to "For all people, if that person eats lasagne, then that person is John" which gives $\forall{x} \left(E(x, lasagne) \implies (x=John)\right)$. Although there are a few equivalent versions, such as $\forall{x} \left(\neg E(x,lasagne) \lor (x = John) \right)$, which roughly translates to "For all people, it's either true that that person doesn't eat lasagne, or that person is John".

Similarly, for the second set of questions, try to describe the predicate.

$A = \left\{n \in \mathbb{N}: \forall {m \leq n}, n \neq m^2\right\}$ says that "A is the set of all natural numbers n, such that for all m less than or equal to n, n is not equal to m squared". Translate it into marginally more natural English and you get "A is the set of natural numbers that are not equal to the square of another number", which then gets to "A is the set of non-square numbers". Which must mean that $A = \left\{2, 3, 5, 6, 7, 8, 10, \dots \right\}$ which is an infinite set. How would that work for the rest of the questions?

Question 1

I have a couple questions I have tried. These are simple homework questions I am doing for review for my 3rd year course in university. These are not for marks.

Predicate Question

1)

Let the predicate E(x,y) represent "Person x eats food y"

Let the predicate M(y) represent (Food y is a meat product"

a) Express the statement in predicate logic: "Someone is a vegetarian."

My Answer: Ǝx∀y E(x,y) ¬M(y)

b) Express the statement in predicate logic:"Nobody (except maybe John) eats lasagna."

My Answer: ¬∀xƎy E(John,Lasagna)

CHECKS:

1a  wrong (try if y is meaty then there is somebodty who won't eat it)

1b wrong again  (try if somebody is eating lasanga then ithat person is john)

btw the answers you gave also don't look welformed.

and in principle giving an answer that is not wellformed is even worse than giving a wrong answer.

Sets Question

2) Determine which of the following sets are finite or infinite. If set is finite show all elements in the set. If infinite show 5 elements in the set.

N represents Natural Numbers

a) A = {n ∈ N: ∀m ≤ n, n ≠ m^2}

My Answer: Infinite Set, A = {1, 4, 9, 16, 25} (not sure about this one)

b) B = {n ∈ N : ∀m ≤ n, n = m^2}

My Answer: Infinite Set, B = {1, 4, 9, 16, 25}

c) C = {n^2 : n ∈ N and 2 < n < 6}

My Answer: Finite Set, C = { 9, 16, 25 }

d) D = {(n,m)∈ N x N: 1 < n < m < 6}

My Answer: Finite Set, D = { (2,3), (3,4) , (4,5) }

CHECKS

2a what do you mean (try to describe it better, do you meean the infinite set of all quadrates , or the finite set containing  1, 4, 9, 16, and  25? , btw both answers are wrong)

2b again what do you mean (try to describe it better)

2c CORRECT  one point

2d you are missing some elements, hint (2,5) is one of them.

conclusion:

1 out of 6, you do need some more practice, Good luck

ps if you want to test new answers

mark in your question

Second round

and give new answers

Good luck

2nd Marking

1a 2nd answer   Correct

1b 2nd answer   Correct

2a  there is a bigger problem here you still say the set is infinite and then say it has a limited number of elements.

try to say what the elements of this set are in english, only after that translate ithat answer into a formula. DON't enumerate the set (do' t do something of the form: Infinite Set, A = {3, 5, 7, 9 , 11}

2b same problem, A = {a, b, c, d, e} is not an infinite set so where are you talking about?

Conclusion
Getting there

PS 1

Mauro's comment " We have to specify if $ 0 \in N$ or not. check your book different books have different conventions. so what is the convention in your class?


Find the Maximum Likelihood Estimator Let $X_1,X_2 ,X_3 ,..., X_n$ be iid with cdf $() = 1 − ^{−},  > 0,  > 0$.

What I have done so far,

$L(\theta)=\prod\limits_{i=1}^n\theta x_i^{\theta -1}=\theta^n\prod\limits_{i=1}^n\ x_i^{\theta -1}$

$ln(L(\theta))=nln(\theta)+(\theta-1)^{n}\prod\limits_{i=1}^n ln(x_i)$

$\frac{d}{d\theta}ln(L(\theta))=\frac{n}{\theta}+n(\theta-1)^{n-1}\prod\limits_{i=1}^nln(x_i)=0$

$\theta(\theta -1)^{n-1}=\frac{-1}{\prod\limits_{i=1}^nln(x_i)}$

How do I proceed?

You've got a number of errors.  First, the support should be $x > 1$, since $1 - x^{-\theta} < 0$ if $x < 1$.  Second, if $$F_X(x) = 1 - x^{-\theta}, \quad x > 1, \; \theta > 0,$$ then $$f_X(x) = F_X'(x) = \theta x^{-(\theta+1)}, \quad x > 1, \; \theta > 0$$ and the likelihood may be written as $$\mathcal L(\theta \mid \boldsymbol x) \propto \prod_{i=1}^n \theta x_i^{-(\theta+1)} \mathbb 1 (x_{(1)} > 1) \mathbb 1 (\theta > 0).$$  The log-likelihood is then $$\ell(\theta \mid \boldsymbol x) = n \log \theta - (\theta+1) \log \prod_{i=1}^n x_i + \log \left( \mathbb 1(x_{(1)} > 1) \mathbb 1 (\theta > 0) \right).$$  The derivative of the log-likelihood when the indicator functions are $1$ is $$\frac{\partial \ell}{\partial \theta} = \frac{n}{\theta} - \log \prod_{i=1}^n x_i,$$ which gives a critical point $$\hat \theta = \frac{n}{\log \prod_{i=1}^n x_i} = \frac{n}{\sum_{i=1}^n \log x_i} = \left( \overline{\log x_i} \right)^{-1};$$ that is to say, $\hat \theta$ is the reciprocal of the arithmetic mean of the log-transformed sample.  I leave it as an exercise to demonstrate this choice is maximal under the conditions $x_{(1)} > 1$ and $\theta > 0$.

To calculate the bias of $\hat \theta$, the nature of the MLE suggests that we should try to log-transform the distribution from which the sample is drawn.  That is to say, consider $Y = g(X) = \log X$.  Then $$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right| = \theta (e^y)^{-(\theta+1)} e^y = \theta e^{-\theta y}, \quad y > 0,$$ since $x > 1$ implies $y = \log x  > 0$.  Thus we recognize $Y$ as an exponential distribution with rate $\theta$, and we can express the MLE in terms of the log-transformed sample $\boldsymbol y = \log \boldsymbol x$:  $$\hat \theta = \frac{1}{\bar y}.$$  The log-sample total $n\bar y$ is gamma distributed with shape $n$ and rate $\theta$:  $$f_{n \bar Y}(y) = \frac{\theta^n y^{n-1} e^{-\theta y}}{\Gamma(n)}, \quad y > 0, \theta > 0,$$ being the sum of IID exponential variables; thus $\hat \theta /n = 1/(n \bar y)$ is inverse gamma distributed:  $$f_{\hat \theta/n}(y) = \frac{\theta^n e^{-\theta/y}}{y^{n+1} \Gamma(n)}, \quad y > 0, \theta > 0.$$  I leave it as an exercise to show this has expectation $$\operatorname{E}[\hat \theta/n] = \frac{\theta}{n-1}, \quad n > 1,$$ hence $$\operatorname{E}[\hat \theta] = \frac{n\theta}{n-1}, \quad n > 1.$$

I've already correct that mistake, I hope the solution would be usefull. By the way, in the first part you can use the exponents law's to do $a^nb^n=(ab)^n$, and then use the logarithm law's.


Converting a point to Cylindrical and Spherical Coordinates How is any point on the Cartesian coordinates converted to cylindrical and spherical coordinates. Taking as an example, how would you convert the point (1,1,1)? Thanks in advance.

Cylindrical coordinates: $(r, \theta, z)$ are given by the relationship $x = r \cos\theta$, $y= r\sin \theta$ and $z=z$. Using these we can solve for $r$ and $\theta$ explicitly:
$$
x^2 + y^2 = (r\cos\theta)^2+(r\sin\theta)^2= r^2(\cos^2\theta+\sin^2\theta)= r^2
$$
$$
\cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2+y^2}}$$

Spherical coordinates: $(\rho,\theta,\phi)$ can be given in a few different ways, I would check your textbook, but what I teach is $x = \rho\cos(\theta)\sin(\phi)$, $y= \rho\sin(\theta)\sin(\phi)$, $z = \rho \cos(\phi)$. You can use the same tricks as we used for cylindrical coordinates to isolate $\rho,\theta$ and $\phi$.

cylindrical:  $(r, \theta, z) = (\sqrt{2}, \pi/4, 1)$

spherical: $(\rho, \theta, \phi) = (\sqrt{3}, \pi/4, 0.955\text{ radians})$


Differential operator is not continuous between this metric spaces Let $\mathbb{D}$ be the set of functions $f:[0,1]\to \mathbb{R}$ of class $C^1$ (differentiable with continuous derivative). Let $\mathcal{C}[0,1]$ be the set of continuous functions in $[0,1]$ ($\to \mathbb{R}$).
Let $d_1,d_2$ be the metrics given by:

$d_1(f,g)=\displaystyle\int_0^1 |f(x)-g(x)| dx$

and

$d_2(f,g)=\sup_{x\in[0,1]} |f(x)-g(x)|$

Let $D:\mathbb{D}\to \mathcal{C}[0,1]$ be such that $D(f)=f'$.

a) Is $D:(\mathbb{D},d_1) \to (\mathcal{C}[0,1],d_1)$ a continuous function?

b) Is $D:(\mathbb{D},d_2) \to (\mathcal{C}[0,1],d_2)$ a continuous function?

DISCLAIMER: $D:(\mathbb{D},d_1)$ represents the topologic space given by the metric topology induced by $d_1$ (and so for $d_2$).

Well, this is a bit spoiler, but I'm quite sure that in both cases $D$ is not continuous. The idea would be to think of continuity in the "$\varepsilon-\delta$" sense. Some functions $F,G$ could be very "close", and however $F'$ and $G'$ be very far away. It's not hard to imagine examples of what i'm saying. However, it's a bit difficult to describe the problem precisely.

Hint: in both cases, you can find a sequence of functions which stay close to $0$, but whose derivatives go further and further away from $0$.

Look at the functions of the form $x^{-p}$ for various $p>0$.

(In fact, you may see that both of these metrics are induced by norms and $D$ is linear, so $D$ is continuous iff it is bounded.)

How about the sequence $(1/n)\sin 2n\pi x$? There is no $\delta$ that ensures $d_1(D[f],0)<\varepsilon<2\pi$


solving polynomial equation (showing non-negativity) I would like to show that for all $a,b\geq 0$, and integer $k\geq 1$, we have $(a+b)^k\leq 2^k(a^k+b^k)$. Is there an elegant way to prove this other than induction?

Thanks

This is equivalent as
$$\left(\frac{a+b}{2}\right)^k\leq a^k+b^k$$
Which is clearly true since $f(x)=x^k$ is strictly increasing, and
$$\min(a,b)\leq \frac{a+b}{2}\leq\max(a,b).$$

To explain further, suppose $a\leq b$, then because $f$ is increasing you have
$$f(a)\leq f\left(\frac{a+b}{2}\right)\leq f(b) \leq f(a)+f(b).$$

Thus it even holds if you replace the rhs of your original inequality by $2^k\max(a^k,b^k)$.

Note that
$$
a+b\leq2\max(a,b)\implies(a+b)^k\leq2^k(\max(a,b))^k=2^k(\max(a^k,b^k))\leq2^k(a^k+b^k).
$$
where we used $x^k$ is strictly increasing for $x\ge 0$.


To solve a Trig equation To solve for $\cos{x}$ of equation $2\cos{3x}-4\cos{2x}+10\cos{x}-3=0.\quad$ Now using usual identities i reduced this to a

$8\cos^3{x }-8\cos^2{x}+4\cos{x}+1=0$.

How do i proceed?

Thanks

Hint:

It can be shown the equation $p(t)=8t^3-8t^2+4t+1=0$ has only one real root between $-1$ and $0$, because

$p'(t)=24t^2-16t+44=4(6t^2-4t+1)$ has no real root, hence is positive for all $t$.
$p(-1)=-23$, $p(0)=1$, so we can apply the intermediate value theorem.

The rational root theorem yields no root (the candidates would be, taking into account the previous results, $-\frac12,-\frac14,-\frac18$).  We only can say the root is between $-\frac14$ and $-\frac18$.
 Thus the only exact method is Cardano's method:

First  eliminate the term in $t^2$, setting $z=2t-\frac23$, you should get the equation:
$$z^3+\frac23z+\frac{47}{27}=0 \qquad\text{(if I'm not mistaken)}.$$
Then set $z=u+v$ and add the condition $uv=-\frac29$ to simplify the resulting equation.
This leads to the (non-linear) system:
\begin{cases}u^3+v^3=-\frac{47}{27}\\u^3v^3=-\frac 8{243}\end{cases}
a classic problem on quadratic equations: $u^3$ and $v^3$ are the roots of
$$U^2+\frac{47}{27}U-\frac{8}{243}=0.$$

HINT

You can set for example $n=\cos x$, so then $\cos^2 x$ would change to $n^2$ and so on...

Your equation will look like $8n^3-8n^2+4n+1=0
$

After this, solve like you would do a cubic equation considering different methods like factoring, etc..

At the end don't forget to put back $\cos x$ in the place of $n$ !!


Is there any theoretical indication that this conjecture of Catalan could be true? Belgian mathematician Catalan  in $1876$ made next conjecture:
If we consider the following sequence of Mersenne prime numbers: $2^2-1=3 , 2^3-1=7 , 2^7-1=127 , 2^{127}-1$ then $$2^{2^{127}-1}-1$$ is also prime number. The last term has more than $10^{38}$digits and cannot be tested at present, so I would like to know is there  any theoretical indication that Catalan's conjecture could be true ?

EDIT:

At London Curt Noll's prime page I have found statement that this number has no prime divisors below $5*10^{51}.$

I don't know of any theoretical reason to think it would be true.  I would conjecture it to be false.

Standard heuristics suggest that the first unknown example would be prime with probability
$$e^\gamma\cdot2^{-120}\approx1.34\cdot10^{-34}\%$$
which is small.

It's possible that all numbers in that sequence are prime and there is an explanation. Bertrand's postulate states that there is no prime number that is more than double the previous prime number. It may seem like there's no possible way to prove it. It turns out that a not very long proof of Bertrand's postulate can be found in the Wikipedia article Proof of Bertrand's postulate. It derived some statements that don't appear to be useful but in the end, they all got combined to derive the final result.

The fact that $2^{2^{127} - 1} - 1$ has no prime divisors below $5 \times 10^{51}$ is not a very hopeful sign at all that $2^{2^{127} - 1} -1$ is prime. I believe that given that information, there is a 50% chance of there being a factor below $25 \times 10^{100}$. If one isn't found below $25 \times 10^{100}$, there's still a 50% chance that one will be found below $625 \times 10^{200}$. How many times do you have to square that in order to get past the square root of $2^{2^{127} - 1} - 1$?

However, people already managed to prove that $2^{82,589,933} - 1$ is prime. Maybe some day, they will prove that $2^{2^{127} - 1} - 1$ is prime.

I've noticed a pattern; 31, 89, and 127 are all Mersenne prime exponents. 31 and 127 are themselves Mersenne primes. 89 is a factor of $2^{11} - 1$.

For any positive integers, $x$, $y$, and $z$, we say that $x$ is congruent to $y$ modulo $z$ when the difference between $x$ and $y$ is a multiple of $z$. That statement is written $x \equiv y \mod z$. A group is an ordered pair of a set and a binary operation on that set, $(S, \cdot)$ that satisfies the following properties.

$\cdot$ is associative
There is an identity element $e$
Every element $x$ has an inverse $x^{-1}$ which means $x \cdot x^{-1} = e$ and $x^{-1} \cdot x = e$

It is a well known theorem that the multiplication modulo group of any prime number is cyclic. This shows that for any prime number $p$, all factors of $2^p - 1$ must be of the form $2kp + 1$. However, when a prime number is of the form $2kp + 1$, it's probability of being a factor of $2^p - 1$ is far higher than you would normally expect.

Maybe just like we could form many results and combine them into a proof of Bertrand's postulate, we can do a similar thing for proving that all numbers in the sequence you listed are prime.

There is a not very hard to prove already known theorem that for any odd prime number $p$, 2 is a square modulo $p$ if and only if $p$ is congruent to 1 or 7 modulo 8. I've noticed by observation that there is a pretty nice and simple way of determining of determining when ever any positive integer is a square modulo any prime number but failed to figure out how to prove it. It's possible that there is another simple test for determining for any larger prime number than 3 that's congruent to 1 modulo 6 whether 2 is a cube modulo that number. Maybe there's some special property of numbers in the sequence you listed. Maybe there is a nice easy way to determine when a number is a cube modulo any prime number congruent to 1 modulo 6, a fifth power modulo any prime number congruent to 1 modulo 10.

Also take the sequence $s_0 = 4$, $s_1 = 4^2 - 2 = 14$, $s_2 = 14^2 - 2 = 194$, $s_3 = 194^2 - 2 = 37634$ and so on. According to the Wikipedia article Lucas-Lehmer primality test, for any odd prime $p$, $2^p - 1$ is prime if and only if $s_{p - 2}$ is a multiple of $2^p - 1$. Maybe we could also derive other statements and combine them with those statements to derive the final result that all numbers in the sequence are prime.


Show that if $K \subset L$, then the separable closure of $K$ in $L$ is a field Let $K \subset L$ be a field extension. Consider the separable closure $K_s$ of $K$ in $L$ defined as
  $$
K_s = \left\{ {x \in L \mid x \text{ is algebraic and separable over } K} \right\}
$$
  Prove that $K_s$ is a field.

I know how to prove that the algebraic elements are closed under operations. If also the separable elements were closed under addition and multiplication, then I'm done (I think that this happens) but I don't know how to prove it.

One way to show this could be to first show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$ (that is, every element of $K(\alpha_1,\dots,\alpha_n)$ is separable over $K$). From this, it would follow that $$K_s = \bigcup K(\alpha_1,\dots,\alpha_n),$$ where the union is taken over all finite subsets $\{\alpha_1,\dots,\alpha_n\}$ of $L$ that consist of algebraic and separable elements over $F$. Then, one can show that $K_s$ is a field by a similar argument as used in showing that algebraic elements are closed under addition and multiplication.

One can use Galois theory to show that if $\alpha_1,\dots,\alpha_n \in L$ are algebraic and separable over $K$, then $K(\alpha_1,\dots,\alpha_n)$ is a separable extension of $K$. You can take a look at this answer of mine for a proof that uses Galois theory.

Let $\alpha,\beta\in K_{s}$, $\beta\neq0$.  Then by theorem, $K(\alpha,\beta)\subset L$ is a separable extension (adjoining a finite number separable elements gives a separable extension).  Then in particular, $\alpha-\beta$, $\alpha\cdot\beta^{-1}$ are separable so that both are in $K_{s}$.  Then you have it, closed under subtraction and multiplication by inverse, so $K_{s}\subset L$ is a subfield.

I believe that does it.


Is there a symbol for "taking a derivative of something"? When presented with an equation, say, $y=5x^3+7x^2+4x+9$, you can write on the second line, $\frac{dy}{dx}=15x^2+14x+4$. Similarly, $f(x)=5x^3+7x^2+4x+9$ and $f'(x)=15x^2+14x+4$. But is there a way to write "the derivative of $5x^3+7x^2+4x+9$ is $15x^2+14x+4$" in just one line?

What should l write, $\frac{dy}{d5x^3+7x^2+4x+9}=\cdots$? That fraction just gives me a headache trying to understand it.

What about $f'(5x^3+7x^2+4x+9)=\cdots$? For all the reader knows, $f(x)$ could be anything, and the writer wanted them to plug in $5x^3+7x^2+4x+9$ into the original $f(x)$ and then take the derivative.

So has anyone come up with a better way to write this that does not involve defining anything and then using the newly defined function/operator?

You would denote the derivative of $5x^3+7x^2+4x+9$ as
$$\frac{d}{dx}(5x^3+7x^2+4x+9)$$
That is the only notation I've ever seen unless the expression is expressed as a function.

Just as the symbol $$\int(\cdots)\;dx$$ denotes the antiderivative of something (the expression where the "$\cdots$" is), so the symbol $$\frac{d}{dx}(\cdots)$$ denotes the derivative of something (again, the expression where the "$\cdots$" is).

For example, you would have $$\frac{d}{dx}(13x^2-27x+1) = 26x-27$$
just as you would have $$\int(26x - 27)\;dx = 13x^2-27x + C$$

Occasionally "$D$" or "$D_x$" is seen in lieu of "$\frac{d}{dx}$", and it is very frequent to use an appended "prime" or "apostrophe" to mean the same thing, as $$(13x^2-27x+1)' = 26x-27$$

In other words, $$(\cdots)'\equiv \frac{d}{dx}(\cdots)$$


Is $0x_1+0x_2+0x_3=5$ a linear equation? Is the following equation regarded as a linear equation?

$$0x_1+0x_2+0x_3=5$$

The original question is as below:

Solve the linear system given by the following augmented matrix:

$\left(\begin{array}{ccc|c}2 & 2 & 3 & 1 \\2 & 5 & 3 & 0 \\0 & 0 & 0 & 5\end{array}\right)$

Note the words linear system in the original question. So, I was asking myself whether $0x_1+0x_2+0x_3=5$ is a linear equation. Can we call all of the equations given by the matrix collectively as a linear system?

It is a linear equation with no solutions, so the linear system has no solutions.

I would rather see $x$, $y$ and $z$ instead of $x_1$, $x_2$ and $x_3$... But it is still a linear equation because the degree of the polynomial is $1$. I know what you are stuck on, the fact that the coefficients are $0$, therefore making the left-hand-side evaluate to $0$, which is degree $0$. But because of the variables, we have to call it a linear equation. Just think of it as a linear equation with no solutions (because $0$ $\neq$ $5$).
EXTRA INFORMATION
Quadratic equations are of the form $Ax^2 + Bx + C = 0$, where $A$, $B$ and $C$ are all real numbers. The thing different about this equation is that the degree is $2$, not $1$. Similarly, cubic equations have degree $3$, quartic equations have degree $4$, and so on. Equations of degree $0$ have no special name.
Tip:
Many people get confused when they see things like $x_1$, $x_2$, and $x_3$. Whenever you see something like that, think of it as just being $x$, $y$ and $z$. They are different variables and are in no way related to each other.


Finding all real zeros of the polynomial? Okay, so I need to find all real zeros in this polynomial...

$$f(x) = 2x^3 + x^2 - 13x + 6$$

I know that the first step is to find the factors of 6 and 2, then see which when multiplied by the other coefficients have them add up to equal zero, but none of the factors I tried came out to zero. Is there an easier way to go about doing this???

Your method will in general not find all real, but only all rational zeroes.
If the leading coefficient were 1 instead of 2, all rational zeroes would have to be divisors of 6 (i.e. $\{\pm1, \pm2, \pm3, \pm6\}$).
However with a leading coefficient of 2, one should also check halves of these values (i.e. also {$\pm\frac12, \pm\frac32\}$).
Plugging in $x=2$, you will find that it is in fact a root. By polynomial division you thus obtain a quadratic for the other roots, which you can solve (or you will happen to find the remaining roots also by trying the above candidates).


How is the derivative truly, literally the "best linear approximation" near a point? I've read many times that the derivative of a function $f(x)$ for a certain $x$ is the best linear approximation of the function for values near $x$.

I always thought it was meant in a hand-waving approximate way, but I've recently read that:

"Some people call the derivative the “best linear approximator” because of how accurate this approximation is for $x$ near $0$ (as seen in the picture below). In fact, the derivative actually is the “best” in this sense – you can’t do better." (from http://davidlowryduda.com/?p=1520, where $0$ is a special case in the context of Taylor Series).

This seems to make it clear that the idea of "best linear approximation" is meant in a literal, mathematically rigorous way.

I'm confused because I believe that for a differentiable function, no matter how small you make the interval $\epsilon$ around $x$, there will always be for any $a$ near $x$ in that interval a line going through $x$ that is either as good an approximation of $f(a)$ as the one given by $f'(x)$ (in case the function is actually linear over that interval), or a better approximation (the case in which the line going through $(x, f(x))$ also goes through (a, f(a)) and any line between this line and the tangent at $x$).

What am I missing?

As some people on this site might be aware I don't always take downvotes well.  So here's my attempt to provide more context to my answer for whoever decided to downvote.

Note that I will confine my discussion to functions $f: D\subseteq \Bbb R \to \Bbb R$ and to ideas that should be simple enough for anyone who's taken a course in scalar calculus to understand.  Let me know if I haven't succeeded in some way.

First, it'll be convenient for us to define a new notation.  It's called "little oh" notation.

Definition: A function $f$ is called little oh of $g$ as $x\to a$, denoted $f\in o(g)$ as $x\to a$, if

$$\lim_{x\to a}\frac {f(x)}{g(x)}=0$$

Intuitively this means that $f(x)\to 0$ as $x\to a$ "faster" than $g$ does.

Here are some examples:

$x\in o(1)$ as $x\to 0$
$x^2 \in o(x)$ as $x\to 0$
$x\in o(x^2)$ as $x\to \infty$
$x-\sin(x)\in o(x)$ as $x\to 0$
$x-\sin(x)\in o(x^2)$ as $x\to 0$
$x-\sin(x)\not\in o(x^3)$ as $x\to 0$

Now what is an affine approximation? (Note: I prefer to call it affine rather than linear -- if you've taken linear algebra then you'll know why.)  It is simply a function $T(x) = A + Bx$ that approximates the function in question.

Intuitively it should be clear which affine function should best approximate the function $f$ very near $a$.  It should be $$L(x) = f(a) + f'(a)(x-a).$$  Why?  Well consider that any affine function really only carries two pieces of information: slope and some point on the line.  The function $L$ as I've defined it has the properties $L(a)=f(a)$ and $L'(a)=f'(a)$.  Thus $L$ is the unique line which passes through the point $(a,f(a))$ and has the slope $f'(a)$.

But we can be a little more rigorous.  Below I give a lemma and a theorem that tell us that $L(x) = f(a) + f'(a)(x-a)$ is the best affine approximation of the function $f$ at $a$.

Lemma: If a differentiable function $f$ can be written, for all $x$ in some neighborhood of $a$, as $$f(x) = A + B\cdot(x-a) + R(x-a)$$ where $A, B$ are constants and $R\in o(x-a)$, then $A=f(a)$ and $B=f'(a)$.

Proof: First notice that because $f$, $A$, and $B\cdot(x-a)$ are continuous at $x=a$, $R$ must be too. Then setting $x=a$ we immediately see that $f(a)=A$.

Then, rearranging the equation we get (for all $x\ne a$)

$$\frac{f(x)-f(a)}{x-a} = \frac{f(x)-A}{x-a} = \frac{B\cdot (x-a)+R(x-a)}{x-a} =  B + \frac{R(x-a)}{x-a}$$

Then taking the limit as $x\to a$ we see that $B=f'(a)$. $\ \ \ \square$

Theorem: A function $f$ is differentiable at $a$ iff, for all $x$ in some neighborhood of $a$, $f(x)$ can be written as
$$f(x) = f(a) + B\cdot(x-a) + R(x-a)$$ where $B \in \Bbb R$ and $R\in o(x-a)$.

Proof: "$\implies$": If $f$ is differentiable then $f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$ exists.  This can alternatively be written $$f'(a) = \frac{f(x)-f(a)}{x-a} + r(x-a)$$ where the "remainder function" $r$ has the property $\lim_{x \to a} r(x-a)=0$.  Rearranging this equation we get $$f(x) = f(a) + f'(a)(x-a) -r(x-a)(x-a).$$  Let $R(x-a):= -r(x-a)(x-a)$.  Then clearly $R\in o(x-a)$ (confirm this for yourself).  So $$f(x) = f(a) + f'(a)(x-a) + R(x-a)$$ as required.

"$\impliedby$": Simple rearrangement of this equation yields

$$B + \frac{R(x-a)}{x-a}= \frac{f(x)-f(a)}{x-a}.$$  The limit as $x\to a$ of the LHS exists and thus the limit also exists for the RHS.  This implies $f$ is differentiable by the standard definition of differentiability. $\ \ \ \square$

Taken together the above lemma and theorem tell us that not only is $L(x) = f(a) + f'(a)(x-a)$ the only affine function who's remainder tends to $0$ as $x\to a$ faster than $x-a$ itself (this is the sense in which this approximation is the best), but also that we can even define the concept differentiability by the existence of this best affine approximation.

First off, I'm surprised that of all the answers here, only one seems to point out an error (or at best an abuse of language), namely that the derivative of a function $f$ approximates it. This is flat out false. That is not what the derivative does. Perhaps this is where the main confusion comes from.

But what is the derivative, what does it do? Simply put, the derivative of a real-valued function $f$ of a real variable $x$ at a point $x_0\in\operatorname{Dom}(f)$, if it exists, is a number that quantifies how that function changes $\operatorname{wrt}$ the variable $x$ at that point $x_0$. You are no doubt aware of the many applications of this concept.

However, in trying to compute instantaneous rates of change, we come against the difficulty that the only functions we can directly do this for are linear functions, given that they have the same rate of change at every point (their graphs are nonvertical straight lines in the Cartesian plane). As usual then, we seek to understand any function in terms of linear functions, cumbersome as it might initially seem. However, this method is not applicable to all possible functions, but only to those that possess a remarkable property, namely that they behave like a linear function as you zoom in closer and closer on them at some point $x_0$ in their domain (obviously, if follows that such functions must be necessarily continuous); in other words, the graphs of such functions look like linear functions as you zoom in closer and closer on them about some fixed point. If this behaviour occurs at every point $x$ where these functions are defined, we say that they are differentiable or smooth.

So for these differentiable functions, how should we understand their derivatives in terms of that of linear functions? Well, we simply define the derivative of $f$ at the point $x_0$ to be the derivative of the linear function that $f$ resembles as $x\to x_0$. It is easy to see that this linear function, if it exists, must be unique, since there obviously cannot be a linear function that behaves like $f$ better than the limiting one at $x_0$. Now this is where the idea of linear approximations come from -- so that it is a linear function that is the approximation, and not its derivative.

Once again therefore, the derivative is not an approximation at all. We say that the derivative of $f$ at $x_0$ is the derivative of the best approximating linear function to $f$ at $x_0$. So that the relationship is between two derivatives at a point, two numbers; not between a derivative at a point, a number, and some linear function. Thus at the point $x_0$, the two derivatives are identified whereas the two functions are also identified (not function/derivative or derivative/function).