Rolling a dice $5$ times Rolling a dice $5$ times find the probability that we will get the continuum:

A. $12345$
B. $1234$
C. $434$

My attempt:

A: for each throwing there are $5$ possibilities so
$P(1)=P(2)=P(3)=P(4)=P(5)=\frac 1 6\Longrightarrow$ we are looking for $P(1)\cdot P(2)\cdot P(3)\cdot P(4)\cdot P(5)=(\frac{1}{6})^5=1/7776$
B: $P(1)=P(2)=P(3)=P(4)=\frac 1 6\Longrightarrow$ we are looking for $P(1)\cdot P(2)\cdot P(3)\cdot P(4)=(\frac{1}{6})^4=1/1296$ now we will multiply the result by two because the series could be here $\underline{1}\underline{2}\underline{3}\underline{4}\underline{}$ or here $\underline{}\underline{1}\underline{2}\underline{3}\underline{4}$ so the final result will be $2/1296$
C:$P(4)=P(3)=P(4)=\frac{1}{6}$ so $P(4)\cdot P(3)\cdot P(4)=(\frac{1}{6})^3=1/216$ now we will multiply the result by three because the series could be here $\underline{4}\underline{3}\underline{4}\underline{}\underline{}$ or here
$\underline{}\underline{4}\underline{3}\underline{4}\underline{}$ or here $\underline{}\underline{}\underline{4}\underline{3}\underline{4}$

This is correct?

In part C, you have overcounted because  $43434$ fits both $\underline{434\cdot\cdot}$ and $\underline{\cdot\cdot434}$. You should therefore subtract the probability of getting $\underline{43434}$, which is $1/6^5$, so the answer will be $3 (1/6^3) - 1/6^5$.


What's an example of a convergent, yet unbounded sequence? What's an example of a convergent, yet unbounded sequence?

I'm having trouble of thinking how to do this. I want to use a piecewise function, but I feel like there might be something easier than this.

Rise is correct. I can't believe I didn't see this sooner. all convergent sequences are bounded. I was confusing this with convergent functions, which are different. (i.e. $f(x)= \frac {1}{x}$)

Consider the sequence of functions on the interval $(0,1)$ given by $f_n(x) = nx^n$. Note that $f_n(x)$ converges to $0$ point wise on $(0,1)$. However, $f_n(x)$ is unbounded on $(0,1)$.


Why 2 distributions can not be multiplied? Possible Duplicate:
  what is product of delta function with itself ?

why $2$ or more dirac delta distributions can not be multiplied ?? i mean to define a coherent product of

$d(x)$  x $d'(x)$ with $d(x)$ being the dirac delta derivative

using the convolution theorem plus regularization i think we could.

If you mean $\delta (x) ^2$, it can be defined well as a very large positive number (infinity). Some people are shy of infinities and call the product "undetermined". It is undetermined only in one sense: it is larger than any finite number but it is not zero or any other finite number usually "ascribed" to the product in renormalizations.


What does this "\" mathematics symbol mean? I am reading a paper which has a mathematical argument: $X \subset V \setminus \{0\}$.

Can anyone explain this statement to me? Especially the "$\setminus \{0\}$".

Many thanks in advance!

This is set difference (or relative complement): $A\setminus B$ is the set of all things in $A$ but not $B$. That is, $$A\setminus B=\{x: x\in A, x\not\in B\}.$$ The latex code is "\setminus."

For example:

$A\setminus A=\emptyset$, for every $A$.
In your example, $V\setminus \{0\}$ is the set of all nonzero vectors in $V$.
Note that $A\setminus B$ makes sense even if $B\not\subseteq A$, e.g.:

$\{1,2\}\setminus \{2, 3\}=\{1\}$.
$\{1,2\}\setminus \{3\}=\{1, 2\}$.

In particular, "$X\subset V\setminus \{0\}$" means "$X$ is a subset of $V\setminus \{0\}$," or more succinctly "$X$ is a set of nonzero vectors (in $V$)." I'm guessing here that "$V$" is a vector space.

Note that unfortunately the symbol "$\subset$" is sometimes used to denote proper subset and sometimes general subset; context will indicate whether $X$ is meant to be a set of nonzero vectors, or a set of some but not all nonzero vectors (almost certainly the former).

The symbol $\setminus$ has several meanings, according to wiki, e.g., it sometimes also denotes double cosets
$$
 K\left\backslash G \right/ H
$$
Of course, in the context $V\setminus \{0\}$ it is the set difference. For a similar discussion, see

What does "\" mean in math


$\epsilon$-proof for the limit $f(x)=\frac{1}{\cosh{(x)}}+\log{\left ( \frac{\cosh{(x)}}{1+\cosh{(x)}} \right )}$ as $x$ goes to $\pm \infty$ I want to use a $\epsilon,\delta$-proof for the existence and value for the limit of $$f(x)=\frac{1}{\cosh{(x)}}+\log{\left ( \frac{\cosh{(x)}}{1+\cosh{(x)}} \right )}$$
for $x \rightarrow \pm\infty$.
Now, I know the definition for such proof to be:
1. Limit as $x \to +\infty$
$$\lim_{x \ \to \  +\infty} f(x) = L \Leftrightarrow
  \forall \ \epsilon>0\; (\exists \ \delta : (\;x>\delta\implies |f(x) - L|\leq\epsilon))
$$
2. Limit as $x \to -\infty$
$$\lim_{x \ \to \  -\infty} f(x) = L \Leftrightarrow
  \forall \ \epsilon>0\; (\exists \ \delta : (\;x<\delta\implies |f(x) - L|\leq\epsilon))
$$
3. Limit evaluating to +$\infty$
$$\lim_{x \ \to \  a} f(x) = +\infty \Leftrightarrow \forall M  > 0 \ (\exists \  \delta > 0  \ : \ (0<|x-a|<\delta \implies f(x) >M)$$
4. Limit evaluating to -$\infty$
$$\lim_{x \ \to \  a} f(x) = -\infty \Leftrightarrow \forall  N  < 0 \ (\exists \  \delta > 0  \ : \ (0<|x-a|<\delta \implies f(x)  < N)$$
I am struggeling to approach this problem as;

Wouldn't I need to know what the limit is before doing the actual proof and thus not being able to know which method to use?

This one seems rather complex and I am completly lost. Can someone explain to me how to identify and "attack" such a problem?

With $y:=\operatorname{sech}x>0$ this becomes $f=y-\ln(1+y)\sim\tfrac12y^2$, so the limit is $0$. We wish to prove $\forall\varepsilon>0\exists\delta>0\forall y(y<\delta\implies|f|<\varepsilon)$. Since $0<f<\tfrac12y^2$, we can choose $\delta=\sqrt{2\varepsilon}$. In particular, $\tfrac12y^2-y+\ln(1+y)=\int_0^y\tfrac{z^2}{1+z}dz>0$. The bound $y<\sqrt{2\varepsilon}$ is, of course, equivalent to $|x|>\operatorname{arcosh}\tfrac{1}{\sqrt{2\varepsilon}}$.

HINTS:
Note that the hyperbolic cosine function is even and so we need only examine the case $x\to +\infty$.
Then, apply the inequalities
$$\frac{y}{1+y}\le \log(y)\le y-1$$
and
$$0<\frac1{\cosh(x)}\le 2e^{-x}$$


How does this specific set look like? How does the set $  \begin{pmatrix}
        \cos(x) \\
      4\sin(x) \\
        \end{pmatrix} \in \mathbb R ^2$ with $x \in \mathbb R$ look like?

I guess it should be similar to $f(x) = \arccos(4\sin(x))$ , but none of the programs I know like to plot that

Consider that $\binom{\cos x}{\sin x}$ is the unit circle.  Multiplying the second element by $4$ stretches that circle in the vertical direction by a factor of $4$, so it becomes an ellipse with minor axis $2$ and major axis $8$.

You can note that $\sin(\arccos X) = \sqrt{1-X^2}$, then $4\sin(\arccos X) = 4\sqrt{1-X^2}$.  However, since $\arccos$ has range $[0, \pi]$, this only gives you the upper part of the ellipse.

The topological space is basically homeomorphic to $S^1$


Enlightening Mathematical Models What is your favourite Mathematical Model? What features make it intuitive or elegant?

This question is largely inspired by an example and a desire to find other's like it.

Suppose we have two military forces of sizes $x(t)$ and $y(t)$ respectively that are about to enter into a conflict with each other at time $t=0$. Then the aptitude for an army to kill should be proportional to its size. But if army $x$ is killing then army $y$ is dying so we have $$x'(t)=-ay$$ $$y'(t)=-bx$$ where $a$ and $b$ are two positive constants.

Consider the quantity $ay^2-bx^2$. By differentiating we see it is constant. So the trajectories of the system of equations form hyperbolas in the $x,y$ plane. So the strength of an army is directly proportional the square of its size. I love this example because it is effectively a mathematical proof of divide and conquer! (To see this, consider an army of 100 men. The strength of 1 100 man unit will be 10,000 but the strength 2 50 men units will be 5,000.) This is one of Lanchester's Laws

Has anyone got an example to match this?

If you like dynamical systems, there's a nice simple model for how romantic encounters evolve over time given the dating style of both people. There is a simplified version here. The short rules of thumb are, if both people like to give and receive attention, they fall madly in love :) . It also predicts cycles of love and hate for couples that have one person shying away from too much attention.

If you're interested in this further, John Gottman has developed a broader application of these kinds of models to the "mathematics of marriage."

I would argue that your "consider the quantity" comment in you OP is a consequence of a variables separable situation (which you show that this quantity is indeed conserved further down your thread). Thus (assume w.l.o.g that $a, b \neq 0$) thus,

\begin{eqnarray} \notag
\frac{y^{'}}{x^{'}} &=& \frac{dy}{dx} \\
                    &=& \frac{-bx}{-ay} \\
                    &=& \frac{b}{a}.\frac{x}{y} \\
\end{eqnarray}
Now, this is a first oder differential equation, which can be solved by separating out the vairables. Thus,
\begin{eqnarray} \notag
\frac{dy}{dx}  &=& \frac{b}{a}\frac{x}{y}     \\
\end{eqnarray}
\begin{eqnarray} \notag
\Rightarrow  ydy                &=& \left( \frac{b}{a} \right) xdx \\
\text{(Integrating)} \int ydy   &=& \left(\frac{b}{a}\right)\int xdx \\
\Rightarrow   \frac{1}{2}y^{2}  &=& \frac{1}{2} \left(\frac{b}{a} \right) x^{2} + C \\
 \end{eqnarray}
Assuming (w.l.o.g) that $C=0$ we have (upon re-arranging)
\begin{equation}
ay^{2} - bx^{2} = 0
\end{equation}
Which brings about your conserved quantity above.


Calcule of $\sum_{m=1}^\infty\sum_{n=1}^\infty{a_{nm}}$ and $\sum_{n=1}^\infty\sum_{m=1}^\infty{a_{nm}}.$ Let  for all $n,m\in\mathbb{N}$  $$a_{nm}= \left\{ \begin{array}{rc}
1, & n=m\\ -1, & m=n+1 \\ 0, & \text{in other case} \end{array} \right.$$

Calcule:

1) $\,\,\,\,\displaystyle\sum_{m=1}^\infty\sum_{n=1}^\infty{a_{nm}}$.

2) $\,\,\,\,\displaystyle\sum_{n=1}^\infty\sum_{m=1}^\infty{a_{nm}}.$

Intiutively I think that both sums are zero, but I don't can't to write this decently...

Let $s_m:=\sum_{n=1}^{\infty}a_{nm}$ and observe that $s_1=1$ and $s_m=0$ for $m>1$

That tells us that:  $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{nm}=\sum_{m=1}^{\infty}s_m=1+0+0+0+\cdots=1$$

On base of that conclude that $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{nm}=1$$

Let $t_n:=\sum_{m=1}^{\infty}a_{nm}$ and observe that $t_n=0$ for every $n\in\mathbb N$.

That tells us that:  $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}a_{nm}=\sum_{n=1}^{\infty}t_n=0+0+0+0+\cdots=0$$

Lets do part $1$ "formally":

If $m=1$, and study $n=1,2,\cdots$, the only non-zero value is at $n=1$ where $a_{nm}=1$.

For any other fixed $m>1$, if we run through the sum $\sum_{n=1}^\infty a_{nm}$, exactly two values of $n$ will have non-zero values, namely, $n=m-1$ and $n=m$ with values $-1$ and $1$ respectively so they cancel each other yielding a contribution of $0$.

So all these sums are $0$. Thus $$\sum_{m=1}^\infty \sum_{n=1}^\infty a_{nm}=\sum_{n=1}^\infty a_{n1} + \sum_{m=2}^\infty \sum_{n=1}^\infty a_{nm}=1+0=1$$

Now try doing the other part by yourself.


logic formalizing: The multiplication of every two positive numbers is also positive EDITED THE QUESTION WITH THE RIGHT SIGNS
I have to formalize the expression: "The multiplication of every two positive numbers is also positive" which I think is:
$$\forall a \forall b(ab>0 \to \left((a>0) \land (b>0)\right))$$
but I have to choose between two options:

$\forall a \forall b(ab \le 0  \to \left((a>0) \to (b
\le 0))\right)$

$\forall a \forall b (\left((a>0) \land (b<0)\right) \land ab<0))$

which one is more correct?

One (probably the most straighforward) correct logic expression can be: $$\forall a \forall b(((a\gt 0)\land (b\gt 0))\rightarrow ab \gt0) \tag 1$$
We shall compare others with this one to check their validity.
The expression $\forall a \forall b(ab\gt 0\rightarrow((a\gt 0)\land(b\gt 0)))$ means that if the product of two numbers is positive, then both of them are positive. This is the converse of what was asked. So, it's incorrect.
In option $2$, $\forall a \forall b (\left((a>0) \land (b>0)\right) \land ab<0))$ means that the all two numbers are positive and their product is negative. This is clearly incorrect. Thus, it doesn't generalize the statement either.
In option $1$,  $\forall a \forall b(ab\le 0  \to \left((a\gt0) \to (b\le 0))\right)$ is correct. It can be modified into $\forall a \forall b(ab\le 0  \to \left((a\le0) \lor (b\le 0))\right)$ which the contrapositive of $(1)$.
Thus, option $1$ is correct.

This “∀∀(>0→((>0)∨(>0)))” formalizes that if the product of two numbers a, b is positive then one of them is positive. This is simply false (as long as you include negative numbers in your universe)
I don’t see how 2. or 3. formalizes that the product of two positive numbers is positive. The easiest way to do it is $\forall a \forall b (a > 0 \land b > 0 \implies ab > 0)$


Struggling with connection between Clifford Algebra (/GA) and their matrix generators As I thought I understood things, the Gamma matricies behave as the 4 orthogonal unit vectors of the Clifford algebra $\mathcal{Cl}_{1,3}(\mathbb C)$, (also the Pauli matricies are for the 3 of $\mathcal{Cl}_{3}(\mathbb C)$??).

But, I'm not getting results that are intuitive when I perform algebra using multivectors built out of these.

The dot and wedge products are $A\cdot B=\frac{1}{2}(AB+BA)$ and $A\wedge B=\frac{1}{2}(AB-BA)$, respectively. But, I should be able to wedge multiply all 4 unit vectors in order to achieve the unit pseudovector, but the result is zero when I attempt to use these definitions to matrix multiply $\gamma_0\wedge\gamma_1\wedge\gamma_2\wedge\gamma_3$.

I think I'm missing something large...

Those formulas for dot and wedge product are only valid when $A$ and $B$ are vectors, not if they are bivectors, trivectors, etc.

You have to decide from the outset how many dimensions you want to work in.

In GA2 (2-D Geometric Algebra) A^B makes a 2-D bivector surface. There is no A^B^C that will "pop out" of the 2-D page, everything remains in 2-D.

In GA3 A^B^C makes a 3-D trivector volume, but you cannot "pop out" of 3-D into a fourth dimension, everything remains in 3-D.

And so on upward to higher dimensions.

It is part of the "closure" of GA, by the same principle that for a 1-D vectorA, A^A = zero and A.A = 1, so A*A = 1, collapsing a 1-D vector to a 0-D scalar.

Its like the "closure" of an odometer. Whenever it gets to 9999 +1 it rolls back around to 0000 to make sure your odometer does not "explode" if you exceed its limits, but simply rolls back to zero to start all over again, to make sure your calculation never "runs off the edge of the page" but remains within the dimensions of the algebra, like an "asteroids" spaceship that drifts off one edge of the screen only to reappear at the opposite edge.

It turns out that "closure" is one of the key features of GA seen in many different forms throughout the algebra.

Its explained here in simple intuitive terms.

https://slehar.wordpress.com/2014/03/18/clifford-algebra-a-visual-introduction/


Is the ideal $I = \langle x^2,x^3 \rangle $ in $\mathbb{R}[X]$ equal with $ \langle x^3 \rangle $? Are the two ideals in the title equal ?

What I think:

I'm not sure if it's true but I think the two ideals are equal, considering the elements in the ideal $I = \langle x^3 \rangle \{ \sum_{i=1} ^{n} f(x) \cdot x^3 \cdot h(x) | f(x),h(x) \in \mathbb{R}[X] \}$ if we take $f(x) = \frac{1}{x^3}$ and $h(x) = x^2$ we can get $x^2 = f(x) \cdot x^3 \cdot h(x)$ so that implies that $x^2 \in \langle x^3 \rangle$ and $ \langle x^2,x^3 \rangle =  \langle x^3 \rangle  $. I don't know if I can chose $f(x) = \frac 1 {x^3}$ since x can be 0.

Indeed, you cannot take $1/x^2$ as this is not a polynomial. Yet that $x$ can be $0$ is not the problem. You could not take, say, $1/(1+x^2)$ either. The elements of $\mathbb{R}[X]$ are the polynomials, that is expressions of the form $a_0 + a_1x + a_2 x^2 + \dots a_nx^n$.

If you restrict to polynomials correctly you can show that $x^2$ cannot be in the ideal generated by $x^3$. (For example consider the rank.) Thus the two ideals are not equal.

Would you allow expressions such as $1/x^2$ as elements of your ring, that is you consider rational functions, then indeed the ideals would be equal. Yet in the ring of rational functions $\mathbb{R}(X)$ – note the parenthesis instead of brackets – almost all ideals are equal as it is a field and the only ideals are $0$ and the ring itself.


$\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite How do I prove that $ \displaystyle\lim_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite?

$n! \geq (n/2)^{n/2}$ because half of the factors are at least $n/2$. Take $n$-th root.

Let's start out with a nice result expressed in harmonic numbers from one of the posted problems, see here. When n is very large we get that:

$$ (n+1) H_n - n \approx \ln n!\longrightarrow e^{(n+1) H_n - n} \approx  n!\longrightarrow e^\frac{{(n+1) H_n - n}}{n} \approx \sqrt[n]{n!} \longrightarrow \infty$$

REMARK: $\lim_{n\to\infty} \frac{(n+1) H_n - n}{\ln n!}=1$ may be proved by Cesaro-Stolz theorem such that we may avoid the use of Stirling's approximation.

The proof is complete.


Computing $\sum_{n = 1}^{+ \infty} \frac{1}{n^2 + 1}$ using Fourier series. $f$ is a $2 \pi$ periodic function on $ ] - \pi,\pi[$ defined as :
$$f(x) = e^{-x}$$
Using Fourier series, compute the sums :
$$ \sum_{n = 1}^{+ \infty} \frac{1}{n^2 + 1} , \sum_{- \infty}^{+ \infty} \frac{1}{n^2 + 1} $$

How do I compute $\sum_{n = 1}^{+ \infty} \frac{1}{n^2 + 1}$ ?
I have computed the fourier coefficient and found that:
$$a_0 = \frac{2}{\pi} (1 - e^{- \pi}) $$
$$a_n = \frac{2}{\pi(n^2 + 1)} (1 - (-1)^n . e^{- \pi} ) $$
$$b_n = \frac{2n}{\pi(n^2 + 1)} (1 - (-1)^n . e^{- \pi} ) $$
Using Dirichlet theorem and taking $x = 0$ to make the $b_n$ disappear, I get:
$$S_f (1) = \frac{2}{\pi} (1 - e^{- \pi}) + \sum_{n = 1}^{+ \infty} \frac{2}{\pi(n^2 + 1)} (1 - (-1)^n . e^{- \pi} ) . (-1)^n = 1 $$
I do not see how to proceed to get the value of the sum?
The same problem for the second sum.

With the correct coefficients (see comments), you have for $x \in]-\pi,\pi[$:

$$
e^{-x} = \frac{2}{\pi} \sinh(\pi) + 2 \sinh(\pi) \sum_{n=1}^{\infty}\frac{ (-1)^n  (n\sin (nx) + \cos(nx)) }{\pi(n^2 + 1)}
$$

so

$$
e^{x} + e^{-x} = \frac{4}{\pi} \sinh(\pi)  + 4 \sinh(\pi) \sum_{n=1}^{\infty}\frac{ (-1)^n \cos(nx) }{\pi(n^2 + 1)}
$$

Now we can evaluate this for $x \to \pm\pi$. To be more precise (see the answer by @kvantour), the evaluation $x \to \pm\pi$ corresponds to the average of the two values at $\pm\pi$ which is exactly what the Fourier series converges to at this point of discontinuity.

$$
\cosh (\pm \pi) = \frac{2}{\pi} \sinh(\pi)+ 2 \sinh(\pi) \sum_{n=1}^{\infty}\frac{1 }{\pi(n^2 + 1)}
$$

and in turn

$$
\sum_{n=1}^{\infty}\frac{1 }{(n^2 + 1)}  = \frac{\pi \coth(\pi) -1}{2  }
$$

Then the second question follows:
$$
\sum_{- \infty}^{+ \infty} \frac{1}{n^2 + 1} = 1 + 2 \sum_{1}^{+ \infty} \frac{1}{n^2 + 1} = \pi \coth(\pi) \simeq 1.0037 \; \pi
$$

Comment: for comparison,
$$
\int_{- \infty}^{+ \infty} \frac{1}{x^2 + 1} {\rm{dx}} = \pi
$$

Try using Parseval's theorem,

$$\frac{1}{\pi}\int_{-\pi}^\pi |f(x)|^2 dx = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n^2+b_n^2$$

If it comes out too nasty, then I think your coeffecients might be off.


A property for finite subsets of real numbers Let $B$ be a finite subset of real numbers. Then, is it true
that there exists $A\subseteq \mathbb{R}$ such that

(1) $(A-A)\cap (B-B)=\{ 0\}$,

(2) $(A-A)+B=\mathbb{R}$?

Not that if $B$ is a singleton, then it is true by choosing $A=\mathbb{R}$,
and if $|B|>1$ then $A-A$ can not be a subgroup (if $A-A$ is a subgroup, then $|\mathbb{R}:A-A|=|B|\geq 2$ that is a contradiction). Here, $B-B=\{ b-\beta:b,\beta\in B\}$, $A+B=\{a+b:a\in A,b\in B\}$.

Also see Two uncountable subsets of real numbers without any interval and two relations .

Take $A=\{0\}$ regardless of $B$, if $B$ is non-empty. $0 \in B-B$ always and $A-A=\{0\}$ trivially, so $(A-A) \cap (B-B) = \{0\}$.

If $A$ is finite and $B$ too, $A-A$ is finite and so is $(A-A) + B$ and so never equals $\Bbb R$.

$\Bbb R$ has no finite subgroups. I see no point in this question..


What is restriction of a measure to an algebra? I am reading Folland's Real analysis book and I saw following sentence in the Theorem 1.14:

  Let $A
\subset \mathcal{P}(X)$ be an algebra, $\mu_0$ a premeasure on $\mathcal{A}$, and $\mathcal{M}$ the $\sigma$-algebra generated by $\mathcal{A}$. There exists a measure $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$ -- namely, $\mu = \mu^* | \mathcal{M}$ (where $\mu^*$ is outer measure).

What I want to ask is, what book means by saying  There exists a measure $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$?

There exists a measure $\mu$ on $\mathcal M$ that gives you the same value for each element of $\mathcal A$ as $\mu_0$ would give you. That's all.

This means that if you define $ \mu(E) = \mu^*( E) $ for all $ E  \in  \mathcal{M} $, then $ \mu $ is a measure on $\mathcal{M}$.


Does there exist a formula for product of the primitive $ n $th roots of unity. I know there is a formula for the sum of the primitive $ n $th roots of unity which is the Mobius function of $ n $. See: The Möbius function is the sum of the primitive $n$th roots of unity.

I am curious about the existence of a formula for the product of the primitive $ n $th roots of unity.

If $g$ is a primitive $n$-th root of unity, then so is $g^{-1}$.

Unless $n\le2$, we have $g \ne g^{-1}$. When $n=2$, there is only one primitive $n$-th root of unity: $-1$.

Therefore, the product of all primitive $n$-th roots of unity is $1$ when $n\ne 2$ and $-1$ when $n=2$.

This is quite similar to Wilson's theorem for abelian groups:

  The product of all elements in a finite abelian group is $1$, except when there is exactly one element of order $2$, in which case the product is this element.

That's much easier than the sum!  The "primitive nth roots of unity" are $e^{\frac{ik\pi}{n}}$ for k from 1 to n.  Their product is $e^{(1+ 2+ ...+ n)(ik\pi)}= e^{\frac{n(n+1)ik\pi}{2}}$.


What is a characterisation function I am reading a paper: "Probability of Backtest Overfitting", and page 13 defines the relative frequency:

$$f(\lambda) = \sum_{c \in C_S} \frac{\chi_{\{\lambda\}} \left( \lambda_c\right)}{|\{C_S\}|}$$

the authors say that $\mathcal{X}$ is the characterization function. I would like to know what exactly that is since I have never come across that term before.

EDIT: I emailed the author of the paper, and according to him, $f(\lambda)$ is simply a PDF.

@i-squared-keep-it-real,

hello, I saw your edit but the fact $f(\lambda)$ is PDF still does not give a clue about what the $\chi$ is?

Should we solve the $\chi$ given the constraint $\int f(\lambda)d\lambda=1$ assuming $\chi$  is some type of function or should we fit something like Gaussian distribution to the frequency distribution ?

I mean without the $\chi$ how can I compute the $f(\lambda)$ to access to the POB?

Looking forward to an update!!

Thanks a lot!!

Sorry that I cannot comment...


Prove that $R$ is a ring with division. I'm having problems trying to know if my proof is wrong or not. The problem states:

  Let $R$ a ring with 1, not necessary commutative, such that for every $a\in R\setminus\{0\}$, there exists $b\in R\setminus\{0\}$ (which depends on $a$) such that $a\cdot b=1$. Prove that $R$ is a division ring.

I have almost everything to prove that is a ring with division, I think I only miss the part that $b\cdot a=1$, what I have done so far is the following:

We have $ab=1$ so multiplying $b$ on the left side we have
$$b\cdot a\cdot b=b\cdot 1$$
Since $1$ is the $1$ of $R$, we get:
$$b\cdot a\cdot b=1\cdot b$$
(this is the part I'm not sure about) since we are in a group, we can cancel $b$ on the left side
$$a\cdot b=1$$
and we are done?

If the only part you are missing is to show that $ab = 1 \Rightarrow ba = 1$ then proceed as follows:

$$ab = 1 \Rightarrow bab = b $$

Now because of the statement, $b$ has a right inverse, $c $:

$$babc = bc \Rightarrow ba = 1$$

And we are done.

@Alex Wertheim also provides us with a different approach in the comment section. Be sure to check it.

I think the point is that $a$ and $b$ are non zero, also not zero-divisors. Hence to cancel $b$, you simply multiply $a$ from the left for both sides. However this problem is asking to prove $R$ is a division ring. To me, for every nonzero element in $R$ it has a multiplicative inverse has already been the definition. I don't see what to prove.


How to prove that $\lim_{h\to 0} \, \frac{e^h-1}{h}$ when I try to show that the derivative of exp function is
$$\frac{\partial e^t}{\partial t}=e^t$$
I need to prove firstly that this limit is equal to 1.

$$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$

If you try to use L'Hospital's rule then you need firstly to know what is the derivative of exp function that we want to prove it.

If you want to use approximation of $e^\epsilon \simeq1+\epsilon$ you well found the answer '$1$' but I don't want to use this approximation because it based on the Taylor series that required also to know the derivative of exp function that we don't have it yet

The only way that I can prove it, is with the definition of $e$ number  $$ \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e $$
so I ask if there are another way to show this limit.

As I pointed out in my comment, the answer depends on which definition you are using.

Solution 1. Here we define $e^x = \lim_{n\to\infty}\left(1 + \frac{x}{n}\right)^n$. (We assume the existence of this limit is already established.) Using the binomial theorem, for $n \geq 2$ and $|x| < 1$ we obtain the following simple estimate:

$$ \left| \left(1 + \frac{x}{n}\right)^n - 1 - x \right|
= \left| \sum_{k=2}^{n} \binom{n}{k} \frac{x^k}{n^k} \right|
\leq \sum_{k=2}^{n} |x|^k
\leq \frac{|x|^2}{1-|x|}. $$

This bound remains true as we let $n\to\infty$, so it follows that

$$ \left| \frac{e^x - 1}{x} - 1\right| = \frac{\left|e^x - 1 - x\right|}{|x|} \leq \frac{|x|}{1-|x|} $$

and by the squeezing lemma as $x\to0$ the desired conclusion follows.

Solution 2. In case $e^x$ is defined as $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, the above argument carries over in almost identicaly way to yield a proof.

Solution 3. In case $e^x$ is defined as the inverse function of $x \mapsto \int_{1}^{x}\frac{dt}{t}$, it follows from the inverse function theorem.

Solution 4. Perhaps the most pesky case is where the exponentiation is defined by the process of extending rational exponents to reals and $e$ is simply defined as $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$. In this case, one can introduce two functions, $\log(x) := \int_{1}^{x}\frac{dt}{t}$ and its inverse, and observe that

$\log(xy) = \int_{1}^{x} \frac{dt}{t} + \int_{x}^{xy} \frac{dt}{t} = \log(x) + \log(y)$ and hence $\exp(x+y) = \exp(x)\exp(y)$.
$\log\left(\left(1+\frac{1}{n}\right)^n\right) = n\log\left(1+\frac{1}{n}\right) \to 1$ as $n\to\infty$ and $\log$ is continuous, so $\log(e) = 1$. In other words, $\exp(1) = e$.
Using this, it is routine to check that $\exp(k) = e^k$ for all integers $k$ and then $\exp(r) = e^r$ for all rationals $r$. Hence for any real $x$,

$$ e^x = \inf\{ e^r : r > x \text{ and } r \in \mathbb{Q} \} = \inf\{ \exp(r) : r > x \text{ and } r \in \mathbb{Q} \} = \exp(x) $$

by the continuity of $\exp$.

Now the conclusion follows by Solution 3.

Thank you all for your answers.

But my question is about if there are any other method without use any of the definition or derivative.

As I said in my Question I already know the "only way" to calculate this limit,
$$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$
"the only way" i mean :
using this definition

$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$$

by substituting $x=1/h$ you can rewrite it as

$$e=\lim_{h\to0}\left(1+h\right)^{1/h}$$

then :
$$\lim_{h\to0} e^h=\lim_{h\to0}(\left(1+h\right)^{1/h})^h =\lim_{h\to0}\left(1+h\right)     $$
so you can easily prove it the limit equal to one

$$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$

so we can say that : this limit is just redefine of the definition of the $e$ number


Definition of total and directional derivatives I'm reading Analysis II by Terence Tao and having a problem with solving one of the exercises. We are given two definitions:

Let E be a subset of R (the real numbers), $f:E \rightarrow R^m$ be a function, $x_{0}\in E$, and let $f'(x_{0}): R^n \rightarrow R^m$ be a linear transformation. We say that f is differentiable at $x_{0}$ with derivative $f'(x_{0})$ if

$\lim_{x \to x_{0}, x \in E- \{x_{0}\}} \frac{|f(x)-(f(x_{0})-f'(x_{0})(x-x_{0}))|}{|x-x_{0}|}=0$, where $|(x_{1}, ... x_{n})| = \sqrt{x_{1}^2+ ... + x_{n}^2}$.

Let $v \in R^n$, and let $x_{0}$ be an interior point of E.

If the limit $D_{v}f(x_{0})=\lim_{t \to 0; t>0, x_{0}+tv \in E}\frac{f(x_{0}+tv)-f(x_{0})} {t}$ exists, we say that
f is differentiable in the direction v at $x_{0}$.

Now I have to prove that if f is differentiable at $x_{0}$, then it is also differentiable in the direction v at $x_{0}$ and $D_{v}f(x_{0})=f'(x_{0})v$. Isn't, though, $f'(x_{0})$ a vector? How is that last equation defined?

Let's understand what the derivative really is.

Given a point $x_0\in E \subset \mathbb{R}^n$ we define de derivative of $f$ in $x_0$ as a linear transformation, then, the function that for each $x \in E$ outputs the derivative of $x$ is a function from $E$ to $\mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$ (the set of all linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$):

\begin{align}
f' : E \to & \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m) \\
x \mapsto & f'(x) : \mathbb{R}^n \to \mathbb{R}^m\\
\end{align}

Thus $f'(x_0)$ is a linear transformation and then $f'(x_0)v$ is a well-defined vector.

It is interesting to think about the following fact to fully understand this definition: we always listen that the derivative of a linear transformation $T:\mathbb{R}^n\to\mathbb{R}^m$ is $T$, but, in fact, the derivative of a linear transformation is a constant function $T': \mathbb{R}^n \to \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$ such that $T'(x) = T$ for all $x\in \mathbb{R}^n$.

We have

$$f(x_0+tv) - f(x_0) = f'(x_0)(tv) + o(tv)=tf'(x_0)(v) + o(t).$$

Dividing by $t$ then gives

$$\frac{f(x_0+tv) - f(x_0)}{t} = f'(x_0)(v) + o(1).$$

This gives the desired result, namely $D_v(x_0) = f'(x_0)(v).$


How to tell if $\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$ converges using Integral Test? So I have this problem:

Determine whether or not the following infinite series converge or diverge. State what test you used.

  $$\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$$

So I decided to do the Integral Test for $$\int^\infty_1\frac{ln(n)}{n^2}dx$$

I made $u = \ln(n)$ and $du = \frac{1}{n}dx$ and got:

$$\int^\infty_1\frac{u}{n}*du$$

which would become:

$$u[\ln(n)]|^\infty_1$$

This is incomplete, but does this look right so far? If not, how do I fix it?

By integral test we should obtain

$$\int_1^\infty \frac{\ln x}{x^2} dx=\left[-\frac{1+\log x}{x}\right]_1^\infty$$

or by $\ln x=u \implies \frac1x dx=du$

$$\int_0^\infty \frac{u}{e^u} du=\left[-\frac{u+1}{e^u}\right]_0^\infty$$

If you are not forced to use integral test, as an effective alternative, we can use limit comparison test with

$$\sum_{n=1}^\infty\frac{1}{n^p}$$

with $p>1$ such that

$$\frac{\frac{ln(n)}{n^2}}{\frac{1}{n^p}}\to 0$$

Since

$\begin{array}\\
\ln(n)
&=\int_1^n \frac{dt}{t}\\
&<\int_1^n \frac{dt}{t^{1/2}}\\
&=\int_1^n t^{-1/2}dt\\
&=2t^{1/2}|_1^n\\
&<2n^{1/2}\\
\end{array}
$,

$\dfrac{\ln(n)}{n^2}
\lt \dfrac{2n^{1/2}}{n^2}
= 2\dfrac1{n^{3/2}}
$
and the sum of this converges
by the integral test.

This can be modified
to show that
$\sum \frac{\ln(n)}{n^{1+c}}
$
converges for any $c > 0$.
Of course the sum diverges
for $c = 0$.


How do you denote the opposite of a range of numbers? What I want to do, is find the conventional standard to denote a range of numbers not included in a set, which is itself part of an infinate set.

For e.g:

A range of numbers, from say, -∞ to 0, and from 1,000,000 to ∞ i.e every number except, 1 through 1,000,000.

The most common notation would be $\mathbb Z\setminus\{1,2,3\}$ or $\mathbb Z-\{1,2,3\}$, where $\mathbb Z$ is the set of all integers. Notably, this would not include $\pm\infty$, since $\infty$ is not a number. Since this may get cumbersome if you wanted to leave out a lot of numbers-- say all the integers from one to one million--then you could say $\mathbb Z\setminus\{1,2,...,1{,}000{,}000\}$.

Sometimes sets of numbers are denoted $[a,b]$ for real numbers $a$ and $b$. This means the set of all numbers between $a$ and $b$ inclusive including decimals. Saying $\mathbb Z\setminus [1,10^6]$ still means the same thing, though because it indicates that we are removing all the numbers between $1$ and $10^6$ from $\mathbb Z$. If some of them weren't in $\mathbb Z$ to begin with, that is fine.

Another way is to reverse the
direction of the brackets
and exchange
open and closed.

For example,
the complement of
"(0, 1]"
would be
"]0, 1(".


a question about contractions on Hilbert spaces Let $\cal{H}$ be a Hilbert space, $T_1,T_2\in\cal{B(H)}$,

$\|T_1(h_1)+T_2(h_2)\|^2\leq\|h_1\|^2+\|h_2\|^2$ for all $h_1,h_2\in\cal{H}$.
$T_1T^\ast_1+T_2T^\ast_2\leq I$.

Then can we verify that 1 holds if and only if 2 holds ?

Consider the operator $T:\mathcal H\oplus \mathcal H\to\mathcal H$ defined by
$$T(h_1\oplus h_2):= T_1(h_1)+T_2(h_2)\, . $$

Then (1) says exactly that $\Vert T\Vert\leq 1$. Since $\Vert T\Vert=\Vert T^*\Vert$, this is equivalent to the condition $\Vert T^*\Vert\leq 1$, which is again equivalent to
$$TT^*\leq I\, .$$

Now, you will easily verify that $T^*:\mathcal H\to \mathcal H\oplus \mathcal H$ is given by the formula
$$T^*(u)=T_1^*(u)\oplus T_2^*(u)\, . $$

So we have $T(T^*(u))=T_1T_1^*(u)+T_2T_2^*(u)$, i.e.
$$TT^*=T_1T_1^*+T_2T_2^*\, .$$

Hence, (1) is equivalent to (2).

As Paul remarked above- this is not true. Take $h_1$ and $h_2$ to be two orthogonal vectors in $\mathcal{H}$, and $T_1=T_2=I$. Then, your condition 1) holds, but condition 2) doesn't.


Application of an exponential whose power is a second derivative Specifically I need to apply the exponential operator:
$$\exp\left[{\alpha\dfrac{\partial^2}{\partial x^2}}\right]$$
(where $\alpha$ is a purely complex number, so $\alpha = b i$, with $b$ real) to a Gaussian function of the form
$$\exp\left[ \frac{-(2\sigma x + C)^2}{4\sigma} \right]$$
where $\sigma$ and C are a complex numbers of the form $\beta= C_{1} + C_{2} i$, with $C_{1}$ and $C_{2}$ reals.

Any idea to correctly apply this operator?

Method 1. Under the unitary Fourier transform $ \mathcal{F}[f](\xi) = \int_{\mathbb{R}} f(x)e^{-2\pi i \xi x} \, \mathrm{d}x $ defined on the Schwarz space $\mathcal{S}(\mathbb{R})$,

$$ \mathcal{F}\big[e^{\alpha \partial_x^2} f(x)\big](\xi)
= \sum_{n=0}^{\infty} \frac{\alpha^n}{n!} \mathcal{F}[f^{(2n)}](\xi)
= \sum_{n=0}^{\infty} \frac{\alpha^n}{n!} (2\pi i \xi)^{2n} \mathcal{F}[f](\xi)
= e^{-4\pi^2 \alpha \xi^2} \mathcal{F}[f](\xi). $$

Now recall that, if $\operatorname{Re}(\sigma) > 0$ and $f(x) = \exp\left\{ -\frac{(2\sigma x+C)^2}{4\sigma} \right\}$, then

$$ \mathcal{F}\left[ f \right](\xi)
= \sqrt{\frac{\pi}{\sigma}} \exp\left\{ \frac{-\pi^2\xi^2 + iC\pi\xi}{\sigma }\right\}. $$

Plugging this back,

$$ \mathcal{F}\big[e^{\alpha \partial_x^2} f(x)\big](\xi)
= \sqrt{\frac{\pi}{\sigma}} \exp\left\{\frac{ -(1+4\alpha\sigma)\pi^2\xi^2 + iC\pi\xi}{\sigma }\right\}. $$

Taking inverse Fourier transform,

$$ e^{\alpha \partial_x^2} f(x)
= \frac{1}{\sqrt{4\alpha\sigma+1}} \exp\left\{ -\frac{(2\sigma x+C)^2}{4\sigma(1+4\alpha  \sigma)} \right\}. $$

Method 2. Assume for a moment that $\alpha$ is real. Since $\partial_x^2$ is the infinitesimal generator of the heat equation, we know that $u(t, x) = e^{t\partial_x^2} f(x)$ solves the equation

$$ \partial_t u = \partial_x^2 u \qquad \text{and} \qquad u(0, x) = f(x) $$

So the solution can be written as the convolution with the fundamental solution of the heat equation:

$$ u(t, x) = \int_{\mathbb{R}} f(y) \cdot \frac{1}{\sqrt{4\pi t}} e^{-(x-y)^2/4t} \, \mathrm{d}y. $$

Plugging $f(x) = \exp\Big\{ -\frac{(2\sigma x+C)^2}{4\sigma} \Big\}$ and computing the resulting gaussian integral, we obtain the same as before. Finally, we can check that the map $\alpha \mapsto e^{\alpha \partial_x^2}f(x)$ is analytic, and so, the same answer applies to complex $\alpha$ by the principle of analytic continuation.

Hint: express the operator as an infinite sum of derivatives, then work in Fourier space.


Bounding $L^{\infty}$ norm for elliptic PDE I'm trying to prove the following: If $L$ is a uniformly elliptic PDE, without costant or linear terms and with essentialy bounded coefficents $a^{ij}$. If $u\in H^1(B_2)$ satisfies $Lu = f$, then
$$||u||_{L^{\infty}(B_1)} \le C(||u||_{L^2(B_2)} + ||f||_{L^{\frac{n}{2}}(B_2)}),$$
at least as long $||f||_{L^{\frac{n}{2}}}$ is small enough.

I have seen a similar result for the Laplacian and I believe it should work but I'm having trouble:

The proof for the case of the Laplacian uses estimates that come from the inequality
$$\int\phi^2u^{p - 2}|\nabla u|^2 \le (\dfrac{2}{p - 1})^2 \int|\nabla \phi|^2 u^p - \dfrac{2}{p - 1}\int\phi^2u^{p - 1}\Delta u,$$

where $\phi$ is some cut off function. This inequality uses explicitly the laplacian and is proved via the divergence theorem, but in our case $u\in H^1$ so the Laplacian is not defined. I do not see how to change this inequality in some way that $L$ apears instead of $\nabla$, and without $L$ I do not know how to make $f$ appear suddenly.

In the cases I've seen for the Laplacian what they actually prove is that
$$||u||_{L^{\infty}(B_1)} \le  C(||u||_{L^2(B_2)}) $$
if $||f||_{L^{\frac{n}{2}}(B_2)}$ is small, so $f$ does not appear in the bounds. Can the inequality that I'm trying to prove be improved to an inequality of this form or there is a reason why $f$ has to appear in the right hand side?

If the inequality with the Laplacian can't be fixed is there any suggestion on how to make $L$ actually appear. Sobolev/Poincare inequalities seem to only make the gradient appear, not L (hence f), so I do not really know what to do.

Thanks!

You want to apply the standard bootstrapping getting $L^q$ for all $q$, followed by "Moser iterations" i.e. tracking the constants to push for $L^\infty$. I will sketch the argument in the simpler case of the Laplacian and where $u$ is zero on the boundary, leaving the details for you as an exercise.

Supposing first of all that $u\in H^1_0(B_2)$. Then
$$ - \Delta u = f $$
implies, multiplying by $u$ and itegrating by parts
$$\int_{B_2}|\nabla u|^2 dx =\int_{B_2} -\Delta u \cdot udx = \int_{B_2}f\cdot u dx \leq \frac{1}{2}(\int_{B_2} f^2 dx +\int_{B_2} u^2 dx ) $$
where we applied the Cauchy-Schwartz inequality. This is sometimes referred to as the Caccioppoli inequality. The key point is that it gives you higher integrability for $u$ when combined with the Sobolev inequality.

Going further, we can multiply the equation by $u^\alpha$ to get
$$ \int_{B_2} |\nabla (u^{\frac{\alpha+1}{2}})|^2dx \leq\frac{C}{\alpha+1}(\int_{B_2} f \cdot u^\alpha dx) \leq \frac{C}{\alpha+1} \left(\left(\int_{B_2} |u|^{\alpha \frac{n}{n-2}} dx\right)^\frac{2n}{n-2} + \left( \int_{B_2} |f|^{\frac{n}{2}} dx \right)^\frac{4}{n} \right)  $$
where we applied Holder to arrive at the crucial exponenet $\frac{n}{2}$. This then gives
$$\left( \int_{B_2} |u|^{(\alpha +1)\frac{n}{n-2} } \right)^\frac{n-2}{n} \leq \frac{C}{\alpha+1} \left(\left(\int_{B_2} |u|^{\alpha \frac{n}{n-2}} dx\right)^\frac{2n}{n-2} + \left( \int_{B_2} |f|^{\frac{n}{2}} dx \right)^\frac{4}{n} \right) $$
by applying the Sobolev inequality. This is a more general Caccioppoli inequality which holds for all $\alpha\geq1$, and applying it iteratively we arrive at $$\| u \|_{L^q(B_2)} \leq C(q) \left(\|u\|_{L^2(B_2)} + \|f\|_{L^{\frac{n}{2}}(B_2)}\right)^{R(q)} $$
for some $C(q)$ and exponent $R(q)$. Now it is for you to carefully go through this argument to estimate $C(q)$ and $R(q)$ explicitly and show that they are in fact uniformly bounded in $q$, hence we can pass to this limit as $q \to \infty$ and deduce
$$ \| u\|_{L^\infty(B_2)} \leq C\left( \|u\|_{L^2(B_2)} + \|f\|_{L^{\frac{n}{2}}(B_2)} \right)^R $$.

Once you have followed this argument, the next step is when $u\in H^1(B_2)$. You then have to introduce a test function $\chi$ which satisfies

$\chi\equiv 1 $ on $B_{R''}$, $\chi\equiv 0$ outside $B_{R''}$ and $|\nabla \chi| \leq \frac{C}{R'-R''}$

to cut off at the boundary. Now multiply through the equation instead by $u^\alpha \chi^2$. This is indeed a bit messy, but dealing with the lower order terms we eventually get a general Caccioppoli type inequality of the form
$$ \left( \int_{B''} |u|^{(\alpha +1)\frac{n}{n-2} } \right)^\frac{n-2}{n} \leq \frac{C}{(\alpha+1)(R'-R'')^2} \left(\left(\int_{B_{R'}} |u|^{\alpha \frac{n}{n-2}} dx\right)^\frac{2n}{n-2} + \left( \int_{B_{R'}} |f|^{\frac{n}{2}} dx \right)^\frac{4}{n} \right) $$

Then choosing a sequence of radii $R_2 > R^{\alpha_k} > R_1$ so that $R^{\alpha_k}$ is a decreasing sequence of radii, indexed by $\alpha_k \to \infty$, which tends to $R_1$ as $k\to \infty$. Applying the Caccioppoli inequality iteratively on this sequence of discs, we get $L^q(B_1)$ for all $q$, and track the constants to show that you can push to the limit as $q\to \infty$ and so get $L^\infty(B_1)$.

Once you have followed this argument, run the whole thing again for your uniformly elliptic operator. Everything goes the same as for the Laplacian except you will have to carry around your constants of ellipticity.

For constant coefficents, I recalled you can rotate the coordinate to get it into the standard form, then apply your result.


Del. $\partial, \delta, \nabla $: Correct enunciation I've come across various different symbols being pronounced as "del". What is the internationally accepted del? If not internationally, then what's the English/American(specify which one if they are different) one that most lecturers/&c use?

$\partial$: I have heard $\frac{\partial}{\partial x}$ being called "del by del x", and (rarely) "dou by dou x" and "der by der x". $\partial$ can be used without a fraction (einstein notation), in which case it gets confusing.
$\nabla$: Called Nabla or del. This has four different uses, which can be easily distinguished while reading out loud, but it gets confusing when the first and last uses (grad and covariant derivative) get mixed up with $\partial$ and $\delta$

Gradient/grad: $\vec{\nabla}\phi$ (phi is a scalar). Read as "nabla phi", or "del phi".
Divergence/div: $\vec{\nabla}\cdot\vec{v}$ Pretty clear, can be read as "nabla dot" or "del dot"
Curl/rot: $\vec{\nabla}\times\vec{v}$ Also clear, can be read as "nabla cross" or "del cross"
Covariant derivative: $\vec{\nabla}_{\vec{u}}\vec{v}$: Can be read as "del v" or "nabla v" . I've seen it called "del u v" also.

$\delta$ : Read out as "delta", but I've heard it used as "del" as well.

This entire thing has confused me. My questions are:

Which one can be correctly called "del"? I'm fine with div/curl being read out as del, as the dot/cross can be read out as well. The confusion is between the convariant derivative, grad, partial derivative, and lowercase delta. Or is it just a matter of context?
Where did this confusing terminology come from in the first place? Why name something del when we already have a bunch of other dels? A timeline of the dels would be appreciated, but not necessary :-).

For what it's worth, in the community I hang out with, we generally just say "partial ecks" for $\partial_x$, and when we are feeling even lazier and when the context is clear, we call the same operator "dee-dee ecks", as if it were the ordinary $\frac{d}{dx}$.

$\nabla$, however, is always "nabla", unless it is used for the gradient of a function, in which case we say "gradient of eff" for $\vec{\nabla} f$.

In class, however, if the expression is embedded in prose (say as part of a theorem statement), I would never read the symbol. I would instead say what it means. So while I may write

  Important, we always have $\partial_x\partial_y f = \partial_y\partial_x f$

I would say,

  Important, we always have that partial derivatives commute.

Or if I write

  Therefore $\partial_x f = 0$

I would say

  Therefore the partial derivative of eff with respect to ecks is zero.

Or if I write

  By the Maxwell's equations, $\nabla\cdot E = 0$

I would say

  By the Maxwell's equation, ee is divergence free.

The only time I might read the symbols as symbols is if I am performing a computation on the board and am just copying stuff directly from my notes. In those cases I honestly cannot remember what I would usually say.

There is no specific word you'll say  it.
So you can use anything.
BTW I personally like to say  'del del x of f' (seems fast and nice) for ∂f/∂x


Are $x \cdot 0 = 0$, $x \cdot 1 = x$, and $-(-x) = x$ axioms? Context: Rings.

  Are $x \cdot 0 = 0$ and $x \cdot 1 = x$ and $-(-x) = x$ axioms?

Arguably three questions in one, but since they all are properties of the multiplication, I'll try my luck...

I will assume this is in the context of rings (e.g., real numbers, integers, etc).  In this case, the axiom defining $0$ is that $x + 0 = x$ for all $x$.  $x*0 = 0$ is a result of this since we have $x*0 = x*(0+0) = x*0 + x*0$ which implies $x*0 = 0$ (canceling one of the $x*0$'s).

I am guessing that for the second one you mean $x*1 = x$.  This is a definition (axiom).

The third one is a consequence of the definition of $-x$ being the element such that $x + (-x) = 0$.  For then we have $(-x) + x$ is also zero so that $x$ is the negative of $-x$.

Answering what you asked, the question is poorly formed because you didn't specify the theory you are talking about. To do that, well, you must define symbols, deduction rules... and axioms.
Answering what you probably meant to ask, no. Those are properties that $-$ and $\times$ have in the $\mathbb Z$ set, not axioms. They can be deduced from the operations themselves.


Proving zero raised to the zeroth power is equal to one using the power series? Its been a while since I have done math, so I wondering if there was a proof for this, if so what?

Is zero to the zero power 1? Or is it not defined?.

Thank you

Limits don't apply to 0^0 any more than they do to 1^oo. Consider

lim(n->oo) (1+1/n)^n = e

lim(n->oo) (1+2/n)^n = e^2

Both are of the form 1^oo, but neither of these equals 1. And surely 1^oo = 1.

Also, various proposed limits for 0^0 come up with different answers:

lim(x->0) 0^x = 0

lim(x->0) x^0 = 1

So limits aren't helpful. But here's something that is helpful.

Consider

e^x = SUM(n=0->oo) x^n/n!

We are told that this is good for all real numbers, no? It's even good for complex numbers. Now, 0 is a real number. So this formula is good for e^0. But that only works if 0^0=1. (Oddly, the same books that tell us that 0^0 is "undefined" also tell us that the e^x formula above is good for all real numbers.)

Therefore, if you insist on calling 0^0 "undefined", you have to make an exception for using the above formula to calculate e^0. I doubt anyone is willing to do so! So 0^0=1.

This probably holds for any series in which you have the term a_0*x^0. We always equate this term as a_0, no matter what x is.

You get the same thing with the binomial theorem:

(a+b)^n = a^n + n * a^(n-1) * b + ... + b^n

In summation form you get

(a+b)^n = SUM(k=0->n) (n k) a^(n-k) b^k

        = SUM(k=0->n) n!/(k!*(n-k)!) a^(n-k) b^k

        = a^n b^0 + ... + a^0 b^n

        = a^n + ... + b^n

The summation version works because a^0 = 1 and b^0 = 1. Set a or b equal to zero and you get 0^0=1. And this is assuming n is a positive integer, where everyone agrees the theorem is true!

If you want to make the binomial theorem good for n=0, then 0^0=1 will do the job:

(a+b)^0 = 0!/(0!*0!) a^0 b^0 = a^0 b^0 = 1

0^0 = [a+(-a)]^0 = a^0 (-a)^0 = 1

No one has a problem defining 0! = 1. My high school book Modern Introductory Analysis additionally defines (n 0) = 1. So why not define (or conclude!) 0^0=1?

Some say you can't determine 0^0 because a^0 = a^1 / a^1 and with a=0 you can't do the division. But this need not be the defining condition! You can determine it otherhow, as I have above. The exponent rule used for this is

a^m * a^n = a^(m+n)

And a=0 satisfies it for integers m,n>=0. It satisfies the other exponent laws for non-negative exponents, too.

In conclusion, 0^0 = 1 unless you make x = 0 an exception to

e^x = SUM(n=0->oo) x^n/n!

and I really doubt anyone would.

One way you can arrive at this equality is by evaluating $$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{\ln(x^x)}  \\ = \lim_{x \to 0^+} e^{x\ln(x)}$$ Now using L'Hospital's rule on $x\ln(x) = \frac{ln(x)}{x^{-1}}$ we get $$\lim_{x\to 0^+} \frac{ln(x)}{x^{-1}} = \lim_{x\to 0^+} \frac{\frac{1}{x}}{-x^{-2}} \\ =\lim_{x\to 0^+} \frac{-x^2}{x} \\ = \lim_{x\to 0^+} =-x = 0$$ Hence $$\lim_{x \to 0^+} e^{x\ln(x)} = e^0 = 1$$ But the whole $0^0=1$ thing is pretty iffy. The right hand limit does not exist, for one thing.


Probability of a train and a bus will meet Would you explain the answer of the question in terms of independent events or other basic concepts in probability?

(Q) A train and a bus arrive at the station at random between 9 am and 10 am. The train stops for 10 minutes and the bus for x minutes. Find x so that the probability that the bus and the train will meet equals 0.5.

Many thanks in advance!

For any time the train arrives, the bus will meet it if it arrives between x minutes before the train arrives and 10 minutes after that.  That is a window of opportunity of $10+x$ minutes in the $60$ minute frame.  Find the $x$ that gives the required probability.


Simpson’s rule for monomial degree > 4 Does there exist a monic $x^n$, n > 4, for which Simpson’s rule is exact? If not, why?
$$ S(f) = \frac{b-a}{6}f(a) + \frac{2(b-a)}{3}f(\frac{a+b}{2})+ \frac{b-a}{6}f(b)$$

If you take, for sake of simplicity, $a=0, b=1$ (and $f(x)=x^n$) you would require that
$$
\frac 16 f(0) + \frac 23 f(1/2) + \frac 16 f(1) = \frac{1}{n+1}
$$
i.e.
$$
\frac 23 (1/2)^n+\frac 16 = \frac{1}{n+1}\Leftrightarrow \frac{1}{2^{n-2}}=\dfrac{5-n}{n+1}
$$
However, if $n>4$ this last equality is impossible.

Simpson's rule approximates a function, f(x), when integrating $\int_a^b f(x)dx$,  by dividing the interval from a to b an odd number of subintervals, then taking the quadratic that passes through each set of three points. If f is quadratic function the Simpson's rule would approximate it by itself so give an exact answer.
If f(x) is a function of "quadratic type", say $ax^4+ bx^2+ c$, where we can write $y= x^2$ to become $ay^2+ by+ C$, that will Simpson's rule would give an exact value for that.
As for monic polynomials, $x^8$ or $x^{12}$ will work.


What is the dimension of the set generated by $z \mapsto z + 1$ and $z \mapsto \frac{z}{1+z}$? Let's say I have subgroup of $SL_2(\mathbb{Z})$ generated by two elements, I wanted to compute the limit set.  So I wrote down two matrices:

$$
A =  \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) ,\;
B =  \left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right)
$$
These can be written as two maps of fractional linear transformations or Möbius transformations:
$$  z \mapsto z + 1\;\text{ and }z\mapsto  \frac{z}{1+z}$$
The plots of the orbits of these sets was surprising.  I represented a fraction $\frac{a}{b} = [a:b] \in \mathbb{R}P^2$ as a line in the Euclidean plane.

  What are the basic properties of this fractal?  E.g. it's dimension or that it's not connected.

a closer look:

an even closer look:

In fact, the set you've found is actually all of $SL_2(\mathbb{Z})$, not just a subgroup, because the two matrices $A$ and $B$ generate the group; the points you've plotted will eventually include all of the points on the circle that are on lines of rational slope.

The reason why you find a fractal-like structure is because of an interesting semi-universal property where simple solutions of an equation, simple points in a structure, etc. 'repel' other points.  In this case, it manifests as gaps around all of the points that are on rational lines $ax+by=0$ with particularly small values of $a$ and $b$; thought of another way, there are wider 'gaps' around rational numbers $\frac ab$ (in a Farey sequence) with small $a$ and $b$ than there are around more complicated numbers.

To see an example of this in action, look at https://en.wikipedia.org/wiki/Farey_sequence#/media/File:Farey_diagram_horizontal_arc_9.svg which demonstrates all the points in the Farey sequence of order 9 (that is, all rational numbers $\frac ab$ with $0\leq\frac ab\leq 1$ and $b\leq 9$).  Note that the nearest fraction to $0$ is $\frac19$, at distance $\frac19$, and a nearest fraction to $\frac12$ is $\frac49$, at distance $\frac1{18}$; by contrast, the nearest fraction to $\frac45$, for instance, is $\frac79$, which is only distance $\frac1{45}$ away from it!

The distribution of the Farey sequence is, arguably, understood both well and poorly at the same time.  It's known that the order-$N$ Farey sequence $F_N$ has $\theta(N^2)$ points in it (in fact, approximately $\frac6{\pi^2} N^2$, where the $\frac6{\pi^2}$ comes from $\frac1{\zeta(2)}$) and that it approaches a uniform distribution as $N\to\infty$; it's also known that the gaps between adjacent numbers range from $\theta(\frac1N)$ to $\theta(\frac1{N^2})$ in size. (For instance, the gap next to $\frac12$ is always approximately $\frac1{2N}$.)  But the question of how quickly it approaches a uniform distribution is much trickier:

Suppose that $f_n$ is the $n$th fraction in the order-$N$ sequence $F_N$, and consider the distance between $f_n$ and the point $\frac n{|F_N|}$ that corresponds to a uniform distribution with as many points as the Farey sequence.  Now, we can look at the 'mean square' size of these distances over all the sequence, $\displaystyle \Delta_N=\sum_{n=1}^{|F_N|}(f_n-\frac n{|F_N|})^2$; this is a measure of the total discrepancy of the sequence. It's known that $\Delta_N\in o(1)$ as $N\to\infty$; this is one way of formalizing the idea that the Farey sequence approaches a uniform distribution. But how does $\Delta_N$ go to zero as $N\to\infty$?  Well, the statement that $\Delta_N\in O(N^{-r})$ for all $r\lt 1$ would be nice; this says that $\Delta_N$ is 'essentially $1/N$', which is as good as we could hope for.  But this statement is equivalent to the Riemann Hypothesis!  (See the Wikipedia page on Farey fractions and this MathOverflow question for more on this)

It's a one dimensional object that is connected and path connected.  It is one dimensional because you can totally describe it using one variable

$$f(\theta) = (\cos(\theta),\sin(\theta))$$

Also, it is definitely connected but the image that you are seeing looks disconnected just because a computer has limits in presenting data.


Find examples of continuous functions $f:[0,1]\to[0,1]$ that satisfy $\forall x\in[0,1], f(f(x))=f(x)$ other than $f(x)=x$ I got this problem:

Find examples of continuous functions $f:[0,1]\to[0,1]$ that satisfy $\forall x\in[0,1], f(f(x))=f(x)$ other than $f(x)=x$.

I proved that $f([0,1])=\{x\in[0,1]|f(x)=x\}$,
But I couldn't find an example of a function other than $f(x)=x$ that satisfies the conditions.

Thanks on any examples.

The condition $f(f(x)) = f(x)$ is equivalent to $f(y)= y$ for all $y \in $ image of $f$.

Take an $f$ that is not surjective: $f(x) = x$ for $x\le 1/2$ and $f(x) = 1/2$ for $x\ge 1/2$.

I was thinking at first that you meant F(f(x))=x (ie involutary; or  a function that is its own inverse)

So I was going to mention F(x)=1-x as a possible candidate that is bi=jective nd continuous f:[0,1]\to[0,1]; until I read that you meant f(f(x))=f(x).

It would be interesting to note whether there are any functional equations that uniquely specify f(x)=x without explicitly mentioned a continuity (or continuity like condition); or the  the function, being almost already  in some sense itself F(F(x))=f(x), specified (at least given it being invertible).

I was wondering, whether if F(x)=x can be uniquely specified by
1.f(f(x))=x,
2.Where F is a one -one function from F:[0.1] to[0,1] closed and bounded.
3.That is Mononotoncally  Strictly Increasing in ;F(1)=1; F(0)=0, the end points of co-domain, / max points of domain, codomain etc being specified


Need help understanding graph theory question Suppose that $l$ lines are drawn through a circle and these lines form $p$ points of
intersection (involving exactly two lines at each intersection). How many regions
$r$ are formed inside the circle by these lines? Assume that the lines end at the
edge of the circle at $2l$ distinct points.
The answer in the book is $r-1=e-v+1=p+l+1$
But I believe that the correct answer is $p-l+1$
I was told that $v = p+2l$ and $e =\frac{1}{2} \text{(sum of degrees)} = 2p+3l$
But im not sure how that all connects to $r-1=e-v+1=p+l+1$
Im $99\%$ sure that this is a typo and that it's supposed to be $r-1=e-v+1=p-l+1$
Am I Wrong?

$$r-1=e-v+1=(2p+3l)-(p+2l)+1=p+l+1$$
You can tell $p-l+1$ is wrong if you draw multiple lines without intersection.
Number of regions cannot be $\le 0$.


How to prove $\displaystyle\lim_{x \to 0} \dfrac{\sin^{-1} x}{x} = 1$? How to prove this? Is there any geometrical proof? I have proved , btw, $\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x} = 1$ by Sandwich Theorem and little geometry.

you already have $$\lim_{t\to 0} \frac{\sin t}{t}  = 1.$$  we will make a change of variable $$\sin t = x, t = \sin^{-1} x $$  so that $$\lim_{x\to 0} \frac{\sin^{-1}(x)}{x} = \lim_{t\to 0}\frac{t}{\sin t} = 1.$$

I use l'Hôpital theorem. Here is a description


Check if segment crosses circle I have 2 points $p_1$ and $p_2$ in a $2D$ plane. There are $N$ balls of radius $r$ scattered in the plane. I want to know which balls will be crossed by the segment $p_1 \rightarrow p_2$.

I did it using line-point distance, but obviously, when checking the balls, some of them who are not crossed by the segment are giving me false-positives, since they are crossed by the line formed from the points $p_1$ and $p_2$.

Given $p_1$, $p_2$, $r$ and the position $(x,y)$ of a ball, how can I know if this ball is crossed by the segment?

  In the image below, the green balls are crossed while the reds are not.

Using coordinates, write $p_1=(x_{p_1},y_{p_2})$ and similarly for $p_2$. Now, you can parametrize the points of the segment between the two poins in this way:
\begin{equation}
p=\lambda(p_2-p_1)+p_1
\end{equation}

whith $\lambda\in[0,1]$ a parameter. This means that each point $p$ of the segment is written as
\begin{equation}
p=(\lambda(x_{p_2}-x_{p_1})+x_{p_1}, \lambda(y_{p_2}-y_{p_1})+y_{p_1}).
\end{equation}

Now, take a ball of center $(x,y)$ and radius $r$, compute its equation $(X-x)^2+(Y-y)^2=r^2$ and simply substitute the coordinates of the generic parametrized point $p$ inside the equation. The only variable now is the parameter $\lambda$, and the equation is quadratic. If this equation has some solution for $\lambda\in[0,1]$, the segment intersects the ball, otherwise not.

Checking the distance center-of-circle to line being less than radious is a good method.

To get only intersections with the segment, not the line, for those circles who intersect the line check also the distances from the center to both points $p1,p2$ to be less than radious.


What is the mathematical formula for Square Root? I wasn't at school when we were learning this, and I've forgot how to calculate a square root on paper using a formula?

Can anyone please help me? What is the formula?

I need this to write an algorithm for my college assignment home work. Thanks!

Let $ a \geq 0 $. If you want to find $ \sqrt{a} $, use the following iteration:
\begin{align}
x_{0}   &:= a. \\
x_{n+1} &:= \frac{1}{2} \left( x_{n} + \frac{a}{x_{n}} \right).
\end{align}
Then $ \displaystyle \lim_{n \rightarrow \infty} x_{n} = \sqrt{a} $. This is an example of the Newton-Raphson method, applied to the function $ f(x) = x^{2} - a $. This particular iteration exhibits quadratic convergence, which means that the number of correct digits about doubles in size with each successive step. Hence, on an ordinary scientific calculator, one can obtain a good approximation after just a few steps.

Observe that
$$\sin \left( {\frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}$$
so you can use Taylor series for either $\sin (\pi /4)$ or $\cos (\pi /4)$ to approximate ${\sqrt 2 }$ for example.

Or you could just use linear approximation from calculus.

I am not familiar with algorithms, but it seems like linear approximation should be easier.


Order of permutations A pack of 2n cards is shuffled by the "interlacing" method, in other words, if the original order is 1, 2, 3, 4,...,2n, the new order after the shuffle is 1, n+1, 2, n+2,... n, 2n. Work out how many times this shuffle must be repeated before the cards are again in the original order in the case of n = 10.

I know I have to use this proposition "The order of a permutation in cycle notation is equal to the least common multiple of the lengths of the cycles" but I can't find the order of the permutations.

We will consider $2n$ and $1$ a combined fixed position and the position before $2n$ the position index $0$ and $2n,1$ is at position index $1$.

Then the shuffle is defined by sequence $a_{x+1}=2a_x-1\pmod{2n-1}$

Solve the recurrence we have $a_x=c\cdot2^{x-1}+1\pmod{2n-1}$ where $c$ depends on which initial number you choose.

The period of the sequence is the smallest $y$ such that $2n-1|c(2^y-1)$ for all $c$ including $c=1$ ($c=1$ for the initial number $2$ case.) Hence it is basically the smallest $y$ that $2n-1|2^y-1$.

When $n=10$, $2n-1=19$ and hence $y=18$ by fermat little theorem. (We might still need to manually check if divisors of $18$ works and none of them work so $18$ is the smallest)


What are the differences between rings, groups, and fields? Rings, groups, and fields all feel similar. What are the differences between them, both in definition and in how they are used?

They should feel similar! In fact, every ring is a group, and every field is a ring. A ring is a group with an additional operation, where the second operation is associative and the distributive properties make the two operations "compatible".

A field is a ring such that the second operation also satisfies all the group properties (after throwing out the additive identity); i.e. it has multiplicative inverses, multiplicative identity, and is commutative.

Every field is a ring, and every ring is a group. A group has one operation which satisfies closure, associative property, commutive property, identity, and inverse property. A ring satisfies all properties of a group; it also has a second operation which has closure, associative, and distributive property between these two operations. A field has two operations and both satisfy all group properties.


Finding the standard deviation of a set of angles. My question is given a set of angles in the form of degrees minutes and seconds once finding the mean how do you find the standard deviation.

I know how to find the average or mean of a set (see below) but i'm not sure how to find the standard deviation.

For example say we have the following set of angles:

$$39^\circ 15'01''$$
$$39^\circ 14'15''$$
$$39^\circ 14'32''$$

The average is $39^\circ 14'36''$.  Now, how do I find the standard deviation.

I looked at the wiki page, but can't make since of it using degrees minutes seconds instead..

Your first number is $39 + \dfrac{15}{60} + \dfrac{1}{60^2} = 39.250277777\ldots$.  Deal similarly with the others.

And remember to do as little rounding as possible at least until the last step.  Rounding should always wait until the last step except when you know how much effect it will have on the bottom line.  One way to do avoid it is to work with exact fractions.  (Except that the seconds are of course probably rounded . . . .

This is a link to the Standard Deviation formula of a sample, applying it to your example:

√(((15′01″ - 14′36″)^2+(14′15″ - 14′36″)^2+(14′32″ - 14′36″)^2)/3-1)

= 23.26″


Matrix algebra linear transformation question Let $A = 4 \times 4$ matrix:
$\begin{bmatrix}
 3 & 2 &  10 &  -6 \\
 1  & 0  &  2 & -4 \\
 0 &  1  & 2  & 3 \\
 1  &  4 &  10 &  8
\end{bmatrix}$,
let $b = 4 \times 1$  matrix:
$\begin{bmatrix}
 -1  \\
 3  \\
 -1  \\
 4  \\
\end{bmatrix}$

Is $b$ in the range of linear transformation $x \rightarrow Ax$?
Why so or why not?
I'm not really sure what the question is asking. Any help would be greatly appreciated.

Sorry I don't know how to properly format the mathematical equations on this website yet, I did my best to make it legible.

This is a counterexample to the answer Hugh Mungus gave: consider the linear transformation $\mathbb{R}^3 \to \mathbb{R}^3: x \mapsto Ax$, where
$$A = \begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0
\end{pmatrix}.$$
Clearly the columns of $A$ do not span $\mathbb{R}^3$. What the map actually does is it sends $(x_1, x_2, x_3)^T$ to $(x_1, x_2, 0)^T$. Therefore, it is not difficult to see that the vector $(2,2,0)^T$ is an element of the range of the linear map.

The way to proceed is to check if the vector $b$ is a linear combination of the columns of $A$, which result in performing Gaussian Elimination on the augmented matrix (for example) and see if you can find an inconsistency (which you did, seeing your comment on Hugh Mungus' answer).

I just wanted to point out that it is not necessary for the columns to span the whole space in order to see if $b$ is in the range of not. Note however that if you would have found that the columns of $A$ do span the whole space, you are obviously done, but in the case it does not, you are still not sure if your vector is or isn't part of the range.

Hint: check to see if the columns of $A$ span $\mathbb{R}^4$.


Does this correctly interpret the "$\cong$" between groups as "$=$" between sets? I was trying to give the wording "isomorphic groups have the same structure" a precise interpretation, at least for the finite order case. A positive answer to my final question would reach the goal, as it would translate a "$\cong$" between groups into a "$=$" between sets.

For $n$ positive integer, let be:

$I_n:=\{1,\dots,n\}$;
$G$, $\overline G$ groups of order $n$;
$\psi\colon G \rightarrow \overline G$ isomorphism;
$f$, $\bar f$ bijections;
$\theta$, $\bar \theta$ embeddings;
in general, $\varphi^{(\alpha)}$ the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets;
$S_n$ the symmetric group of degree $n$.

Visually:

$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
   I_n     &   &    I_n      \\
    \da{f}         &           &     \da{\bar f}          \\
   G & \ras{\psi} &    \overline G  \\
    \da{\theta}         &           &     \da{\bar \theta}          \\
   \operatorname{Sym}(G) & \ras{\varphi^{(\psi)}} &    \operatorname{Sym}(\overline G)  \\
    \da{\varphi^{(f)}}         &           &     \da{\varphi^{(\bar f)}}          \\
   S_n &  &    S_n  \\
\end{array}
$$

  Is it $(\varphi^{(f)}\circ\theta)(G)=(\varphi^{(\bar f)}\circ\bar \theta)(\overline G)$?

No, those things need not be equal "for trivial reasons": you can take $G = G'$, and $\theta = \theta'$, and $\psi$ to be the identity (although it doesn't even factor into the equation you're asking about), but take $f \neq \bar f$. Then trivially $\theta(G) = \theta'(G')$, but the mappings $f, f'$ give different isomorphisms $\phi^{(f)}$ and $\phi^{(f')}$, so that you get different images in $S_n$. Concretely, you could take $G = C_4$ with generator $g$, and let $\theta$ be the Cayley embedding, and let $f(k) = g^k$, while $f'(1) = g, f'(2) = g^3, f'(3) = g^2, f'(4) = e$, so that $(1, 2, 3, 4)$ is in $(\phi^{(f)} \circ \theta)(G)$ but not in $(\phi^{(f')} \circ \theta')(G')$.

The problem with this is not that group isomorphism is the wrong notion, it's that you're asking the wrong question about it. It would work better with the following changes: suppose that $f, f'$ have the additional condition that $\psi \circ f = f'$, and that $\theta, \theta'$ are the Cayley embeddings for $G, G'$, then do the two mappings $I_n \to S_n$ given by
$$
\phi^{(f)} \circ \theta \circ f \qquad \text{and} \qquad \phi^{(f')} \circ \theta' \circ f'
$$
agree? And indeed the answer is that they do. For take $i, k \in I_n$, then
$$
\begin{align*}
\phi^{(f)}(\theta(f(i)))(k) &= f^{-1}(\theta(f(i))(f(k)))\\
&= f^{-1}(f(i)f(k))\\
&= f^{-1}\psi^{-1}\psi(f(i)f(k))\\
&= (f^{-1}\psi^{-1})(\psi(f(i))\psi(f(k)))\\
&= f'^{-1}(f'(i)f'(k))\\
&= f'^{-1}(\theta'(f'(i))(f'(k)))\\
&= \phi^{(f')}(\theta'(f'(i)))(k).
\end{align*}
$$

You cannot meaningfully have a bijection between a set and a group.

You can have a bijection between a set and the underlying set of a group, but the underlying set does not in general fix the group up to isomorphism, you can have non-isomorphic groups of the same order (same number of group elements). The simplest example being the cyclic group of order $4$ versus the Klein four-group:

The set $\{0,1,2,3\}$ together with addition modulo $4$ (that is, add the numbers, and then if you get beyond $3$, subtract $4$ to get back into the set) gives, up to isomorphism, the cyclic group of order $4$.
The set $\{0,1,2,3\}$ together with bitwise xor (that is write the number as two-bit binary, form the two-bit string that has an $1$ in positions where the bits of the two operands differ and a $0$ where they don't, and interpret the result as number) is isomorphic to the Klein four-group.

Those groups are very different. For example, in the cyclic group, $1+1=2\ne 0$, but in the Klein four-group, $n+n=0$ for all $n$ (nowhere do the bits of an element differ from the bits of itself!).

So your scheme fails at the very first step, defining a bijection between a set and a group.


Independence of linear combinations of Brownian motions Let $0<s\leq t\leq u\leq v$ and $W_x$ be a Brownian motion. Show that $aW_s+bW_t$ and $\frac{1}{v}W_v-\frac{1}{u}W_u$ are independent for $a,b$ satisfying $as+bt=0$.

The question seems easy but somehow I can't rearrange the terms to show the independence. I know that increments of Brownian motion are independent, so I tried to add, subtract, multiply, divide  but I didn't manage to make these increments appear. For example $aW_s+bW_t=aW_s+bW_t -(as+bt)W_s$ or $aW_s+bW_t=aW_s+bW_t -(as+bt)W_t$ don't lead me anywhere.

Thank you

Independence is equivalent to zero covariance when normals are jointly normal. So just compute the expectation of the product and see if it's zero.


Help with the Laplacian Formula Can someone please give me the formula of the Laplacian $∆$$_{x,y}$
where $x$ and $y$ are both in $R^n$?

The Laplacian operator is defined as :

$$\Delta = \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$

$$\Delta_x = \frac{\partial}{\partial x}(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2})$$
$$\implies \Delta_x = \frac{\partial^3}{\partial x^3} + \frac{\partial^3}{\partial x \partial y^2}$$

Similarly,

$$\implies \Delta_{x,y} = \frac{\partial^4}{\partial x^3 \partial y} + \frac{\partial^4}{\partial x \partial y^3}$$


GCD in multivariate polynomial ring and quotient ring 1. GCD in multivariate polynomial ring

I would like to prove the following but couldn't figure out how to.
Let $d$ and $h_1, h_2, \cdots, h_k$ be multivariate polynomials with $l$ number of indeterminates, i.e. $d, h_1, h_2, \cdots, h_k \in \mathbb{C}[x_1, x_2, \cdots, x_l]$. If
$$
    \tag 1 a \cdot d = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k
$$holds for arbitrary $a, b_1, b_2, \cdots, b_k \in \mathbb{C}[x_1, x_2, \cdots, x_l]$, then $d$ is a (greatest) common divisor (GCD) of $h_1, h_2, \cdots, h_k$.
 I think I can prove that the converse is true, i.e. if $d$ is a GCD
of $h_1, h_2, \cdots, h_k$, then $(1)$ holds, but I have no clue how to proceed with the original claim. Does Hilbert's Nullstellensatz somehow come into play in proving the above claim?
 Also, if $(1)$ holds, can I say that the principal ideal generated by $d$ equals the ideal generated by $h_1, h_2, \cdots, h_k$, i.e. $<d> = <h_1, h_2, \cdots, h_k>$? Can this be true even if the multivariate polynomial ring is not a principal ideal domain?

2. GCD in multivariate polynomial quotient ring

Now I have an ideal $I$ in $\mathbb{C}[x_1, x_2, \cdots, x_l]$. Let $d$ and $h_1, h_2, \cdots, h_k$ be multivariate polynomials  in the quotient ring $R=\mathbb{C}[x_1, x_2, \cdots, x_l]/I$. If
$$
\tag 2 a \odot d = b_1 \odot h_1 + b_2 \odot h_2 + \cdots + b_k \odot h_k
$$holds for arbitrary $a, b_1, b_2, \cdots, b_k \in R$
, where $\odot$ denotes the polynomial multiplication operation in the quotient ring $R$, can I call $d$ a (greatest) common divisor of $h_1, h_2, \cdots, h_k$ in $R$? How would I compute $d$ given $h_1, h_2, \cdots, h_k$ in this case?
The following is my initial thoughts. Since I know how to compute the Groebner basis $G=\{g_1, g_2, \cdots, g_m\}$ of $I$, I can lift $(2)$ to $\mathbb{C}[x_1, x_2, \cdots, x_l]$ and have
$$
\tag 3 a \cdot d + r_1 = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k + r_2
$$where $r_1$ and $r_2$ are unique polynomials in $<G>$. Collecting the terms, I get
$$
\tag 4 a \cdot d = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k + r_3
$$where $r_3 = r_2-r_1$. Since $r_3 \in <G>$, I can write the above into
$$
\tag 5 a \cdot d = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k + c_1 \cdot g_1 + c_2 \cdot g_2 + \cdots + c_m \cdot g_m
$$ for some $c_i$.
Now $(5)$ looks awfully like $(1)$ so I want to say $d$ is a GCD of $\{h_1, h_2, \cdots, h_k, g_1, g_2, \cdots, g_m\}$.
Am I on the right track?
I have only taken linear algebra and no other algebra course but I have access to most textbooks. A gentle nudge in the right direction and pointers to relevant materials will be greatly appreciated.

First, $d$ is a common divisor of $h_1,\ldots,h_k$.

Second, if $s$ is a common divisor of $h_1,\ldots, h_k$, then $s$ divides $d$.


What is the background for $\sum_{k=1}^n|f(x_k)-f(x_{k-1})|$? The question is from the following problem:

If $f$ is the function whose graph is indicated in the figure above, then the least upper bound (supremum) of
$$\big\{\sum_{k=1}^n|f(x_k)-f(x_{k-1})|:0=x_0<x_1<\cdots<x_{n-1}<x_n=12\big\}$$
appears to be

$A. 2\quad B. 7\quad C. 12\quad D. 16\quad E. 21$

I don't know what the set above means. And I am curious about the background of the set in real analysis.

If the supremum is finite, the function is called of bounded variation.

Function of bounded variation appear in some context. A simple example I like is the following: a function is of bounded variation if and only if it is  the sum of an increasing and a decreasing function.

Total variation sums up how much a function bobs up and down. Yours does this 16 units.  Therefore choose D.


If $f$ is Riemann integrable on $[a,b]$ , is $|f|$ Riemann integrable on $[a,b]$? If $f$ is Riemann integrable on $[a,b]$ , is $|f|$ Riemann integrable on $[a,b]$ ?
(The metric is $\mathbb R$ usual)

The other is question is $f$ is Riemann integrable on $[a,b]$ , can I claim $f$ is bounded on $[a,b]$ ? (I think the answer can be either yes or no that depend on considerating generalized function or not)

Update: I think I made a mistake on second question that both definitions of Riemann integral and generalized Riemann integral on [a,b] request the f being bounded on [a,b]. Sorry for my mistake.

Using the reverse triangle inequality we have

$$\begin{align}\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x) \\&=\sup_{x,y \in [x_{j-1},x_j]}| |f(x)|-|f(y)|| \\ &\leqslant \sup_{x,y \in [x_{j-1},x_j]}| f(x)-f(y)| \\ & = \sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x).\end{align}$$

For a given partition $P = (x_0,x_1, \ldots, x_n)$, the difference of upper and lower sums
is

$$U(P,|f|) - L(P,|f|) = \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x)\right)(x_j - x_{j-1})\\ \leqslant  \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x)\right)(x_j - x_{j-1})\\= U(P,f) - L(P,f)$$

Since $f$ is Riemann integrable, for any $\epsilon > 0$ there exists a partition $P$ such that the difference of upper and lower Darboux sums satisfies

$$U(P, |f|) - L(P, |f|) \leqslant U(P,f)-L(P,f) < \epsilon.$$

Hence, $|f|$ is Riemann integrable.


How to use Chebyshev's inequality to find a lower bound? I have that $\text{var}(X) = \frac13v^2$ and I want to find the lower bound for $P(|X| \le v)$

I tried doing the following:
$P(|X| \le v) = 1 - P(|X| \ge v)$ so using Chebyshev's inequality I have that

$1 - P(|X| \ge v) \le \frac13$

$\frac23 \le P(|X| \ge v)$,

but I don't know what can I say about $P(|X| \le v)$ since everything is now on terms of $P(|X| \ge v)$.

You are making a mistake in using Chebyshev's inequality. What you get $P(|X| \geq v) \leq 1/3$ (assuming that the mean is $0$) and not $1-P(|X| \geq v) \leq 1/3$


Throw 4 equal dice such that all of the faces show different numbers Imagine one throws 4 equal dice at the same time. How many combinations with different faces (in which no face appears twice) are possible? For the first die one has 6 possible faces, for the second 5, for the third 4 and for the fourth 3.

You have it if you think order matters.  You multiply the number of choices at each stage.  If you think 1234 is the same as 3142, you need to divide by the number of orders, here $4 \times 3 \times 2 \times 1=24$

Ross Millikan got the logic right, but did the arithmetic wrong. (6*5*4*3)/(1*2*3*4) = 15, not 24.

Another way to look at it is to see 6C4 choices for the four numbers that appear, and give them 4! permutations.   6C4 * 4! = 6*5*4*3.  (forgive the lousy notion for binomial coefficients; some day I will lurn to do it right.)

This problem is of interest to me because it is sort , of the complement of the problem I posted.   For example, suppose you roll the die ten times; in how many of the 6^10 outcomes do all six numbers appear?   10C6 * 6^4 is not correct because it counts 1234561111 five times.


Characterization of continuous one-parameter groups of linear operators on finite-dimensional spaces I'm looking for a nice textbook that proves this elementary result:

Theorem.  Suppose $V$ is a finite-dimensional real or complex vector space, and suppose that $\{U(t)\}_{t \in \mathbb{R}}$ is a continuous 1-parameter group of linear operators on $V$, meaning

$U(t) : V \to V$ is a linear operator for each $t \in \mathbb{R}$;
$U(t)$ depends continuously on $t$ (in the usual topology on the space of linear operators from $V$ to itself);
$U(0) = 1_V$;
$U(s+t) = U(s) U(t)$ for all $s, t \in \mathbb{R}$.

Then there exists a unique linear operator $H : V \to V$ such that $U(t) = \exp(tH)$.  Conversely, for any linear operator $H : V \to V$, if $U(t) = \exp(tH)$ then $\{U(t)\}_{t \in \mathbb{R}}$ is a continuous 1-parameter group of linear operators on $V$.

I'd really like a reference to a proof that only does the finite-dimensional case—and please give me the theorem number, since I need this for a paper I'm writing.  The referee is demanding a reference!

I find it sad that no reference to a proof of this result is given in the Wikipedia articles "Matrix exponential" and "One-parameter group".

I would accept a proof of the more general result where $V$ is a Banach space and $\{U(t)\}_{t \in \mathbb{R}}$ is a 1-parameter group of bounded linear operators that's continuous in the norm topology.  The proof is exactly the same!  But I'd rather not bring in these extra concepts.

I really don't want a theorem that considers the case where $V$ is infinite-dimensional and $\{U(t)\}_{t \in \mathbb{R}}$ is only continuous in the strong topology.  This is distractingly technical for what I need!

This result is proved in Chapter 1 (Linear Dynamical Systems), Section 2 (Finite-Dimensional Systems: Matrix Semigroups), of Engel's book (One-Parameter Semigroups for Linear Evolution Equations).

After Theorem 2.9 (p. 11) we read

  With this theorem we have characterized all continuous one-parameter groups on $\mathbb C^n$ as matrix-valued exponential functions

which is exactly what you want (for the complex case).

(Since the context of the book is primarily semigroups, the first paragraph in page 9 is relevant.)

Take your favorite book on Lie groups, a morphism $f:G\rightarrow H$ between Lie groups induces a morphism $Lie(f):Lie(G)\rightarrow Lie(H)$ between their Lie algebra which commutes with the exponential. Here take $G=\mathbb{R}$ and $H=Gl(n,\mathbb{R})$.


Largest rectangular area formed by origin and a point on a circle Lets say the circle has an equation of $(x-2)^2+(y-2)^2=4$.

How can I algebraically derive the largest possible rectangular area formed by the two corners (0, 0) and (x, y) where x and y are points on the circle?

Here are a couple of (educated) intuitions to explain why the answer must occur at a point where $x = y.$

Consider the area of a rectangle in the first quadrant with opposite vertices at $(0,0)$ and $(x,y)$ and edges parallel to the axes.
If the area is $1$ (for example), the point $(x,y)$ lies somewhere on the branch of the hyperbola $xy = 1$ in the first quadrant.
This curve is symmetric around the line $y = x,$ and lies above the line
$y = 2 - x$ except at the point of tangency, $(1,1).$
This is the level curve for area equal to $1.$

Every other level curve for the area of the rectangle has a similar shape:
a branch of a hyperbola in the first quadrant, symmetric around the line
$xy = k$ for some constant $k,$ tangent at a point of the form $(x,x)$
to a line $y = b - x$ for some constant $b.$

So find the points where the circle intersects the line $y = x,$ and
pick the one farthest from the origin; one of the level curves is
tangent to a line with slope $-1$ through that point, which is also
tangent to the circle.
Any other level curve that intersects the circle will lie below this one, and therefore any other point on the circle leads to a rectangle of smaller area.

Another approach:
Suppose the a rectangle has opposite vertices at $(0,0)$ and $(x,y),$
where $(x,y)$ is in the first quadrant.
If we do not make the restriction that the edges of the rectangle are parallel to the axes, then the largest possible area of the rectangle is the area of the square with diagonal from  $(0,0)$ to $(x,y).$

The area of that square is larger when $(x,y)$ is farther from the origin.
The area is maximized when $(x,y)$ is as far from the origin as it can get,
which is on a circle centered at the origin such that at least one point of the given circle is on the larger circle but all the other points of the given circle lie inside the larger circle.
The large circle is in fact the circle tangent to the given circle
at the farther point where the given circle intersects the line $y = x.$

So we choose the square with diagonal ending at that point.
No other square with diagonal ending on the circle is larger,
and for each of those squares, the rectangle with the same diagonal
but with sides parallel to the axes is not larger than the square,
so there is no rectangle with vertices at $(0,0)$ and on the circle
whose area is larger than the chosen square.

When you do this as an exercise, I would advise using the straightforward mechanics given in some other answers rather than relying on these intuitions, under the assumption that the purpose of the exercise is to learn to use those mechanics. But you can use these intuitions to check your answer: if the $x$ and $y$ values you get are not equal, look for the mistake in your calculations.

In the simplest form...
The circle has two points where x = y along the circumference of the circle.  The largest value that x = y forms the largest area for a rectangle, if the circle resides entirely in the first quadrant and the circumference touches both the x-axis and y-axis once and only once, hence the circle's center (x,y) both x and y are both equal to the value of r.

Largest area of a rectangle where one corner is the Origin (0,0) and the opposite corner is on the circumference of the circle.

A= ((r*(sqrt of 2)/2)+r) * ((r*(sqrt of 2)/2)+r)


Alternative proof of $\gcd(a,b)\mid\text{lcm}(a,b)$ I proved this problem using the theorem $\gcd(a,b)\text{lcm}(a,b)=ab$ Let $d=\gcd(a,b)$. Then $d\mid a$ and $d\mid b$. Thus, $a=dr$ and $b=ds$ for some integer $r$ and $s$. Then
$$d(\text{lcm}(a,b))=ab\Rightarrow \text{lcm}(a,b)=\frac{ab}{d}\Rightarrow \text{lcm}(a,b)=\frac{d^2rs}{d}\Rightarrow \text{lcm}(a,b)=drs\Rightarrow d\mid \text{lcm}(a,b)$$
as required.

Is my alternative proof for this problem correct?

Your alternate proof is correct, but you don't need to do all this, you can directly say d|a and a|lcm(a,b).

Hope it helps

$\gcd(a,b)\cdot\text{lcm}(a,b)=a\cdot b=>$
$$
\\\frac{\text{lcm}(a,b)}{\gcd(a,b)}=\frac{a}{\gcd(a,b)}\cdot\frac{b}{\gcd(a, b)}\in\mathbb N
$$


Maximum value of $\sin (x)\;\sin (2x)\;\sin (3x)$ To prove : $\sin (x)\;\sin (2x)\;\sin (3x) \lt 9/16$

With some transformations i was able to prove the above result using calculus but I am not getting the way to solve it without the use of calculus

I've tried grouping $\sin(x)\;\sin(3x)$ and then using product to sum transformations.

Also I've tried to find the maximum value of sinxsin3x and then using the fact that sin2x is a fraction. Although I was able to prove that sinxsin3x is less than 9/16 but the minimum value was less than -9/16 and since sin2x can be negative so i could not draw any conclusion.

Let $a= \sin x ,b = \cos x$ and $a^2+b^2=1$,

Since $\sin 2x = 2 \sin x \cos x = 2 a b$

And $\sin 3x = \sin x (2\cos x+1)(2\cos x-1) = a (2b+1)(2b-1)$

Thus we have $\sin x \sin 2x \sin 3x = 8a^3 b^3 -2a^3 b$

Substitute instead of $b = +\sqrt{1-a^2}$ and once $b=-\sqrt{1-a^2}$

Solving the inequality $8a^3 b^3 -2a^3 b < \frac{9}{16}$ will be easy from here.


Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.

$$z^{10}+\frac{1}{z^{10}}=?$$

For the solution:

the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
How do I proceed?


Algebra: What does "is defined for" mean? In algebra what does: "Is defined for" mean?

I have a question posted:

  $\sqrt{a+b}$ is defined for $-b \leq a$.

The question posed is: Is this true...

My question: WHAT DOES "Is Defined For" mean??

During philosophy class, and independent of that specific class, a math professor simply claimed that "is defined for" is best understood as an axiom and a bridge for us to understand conditions. We can't really use logic to define it, since we defined logic through the use of "is defined for".

Another way of putting it is to go the other way around; we can express the proposition "is defined for" through the use of logical symbols, and we define "is defined for" through those.

In this context, you could probably translate it to "is only true for the condition of".

Edit: it appears I missunderstood the question - yes, I'm trying to present the idea of how to define defined

It means, that the square root function $\sqrt{a+b}$ is properly defined, when you satisfy the given condition, in your example:
$$-b\leq a$$
E.g. for $b = 3$ and $a = -5$, then the inequality is not satisfied, because
$$
-b = -3 \nleq -5 = a.
$$
When we work in $\mathbb{R}$ then square root $f(x)=\sqrt{x}$is not defined for $x < 0$. So with  $b = 3$ and $a = -5$ the expression would be $\sqrt{a + b} =\sqrt{-5 + 3} = \sqrt{-2} $.


Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When is that last statement true?

You can use induction to prove
$$\frac{10^n-1}{9n} \tag{1}\label{eq1A}$$
is an integer when $n = 3^k$. The base case of $k = 0$ gives $3^k = 1$, with \eqref{eq1A} becoming $\frac{10-1}{9} = 1$. Next, assume \eqref{eq1A} is an integer for $k = m$ for some integer $m \ge 0$. Then, for $k = m + 1$,
$$\begin{equation}\begin{aligned}
\frac{10^{3^{m+1}} - 1}{9(3^{m+1})} & = \frac{\left(10^{3^{m}}\right)^3 - 1}{\left(9(3^{m})\right)3} \\
& = \frac{(10^{3^{m}} - 1)(10^{3^{2m}} + 10^{3^{m}} + 1)}{\left(9(3^{m})\right)3} \\
& = \left(\frac{10^{3^{m}} - 1}{9(3^{m})}\right)\left(\frac{10^{3^{2m}} + 10^{3^{m}} + 1}{3}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The first factor above is an integer based on the induction hypothesis, i.e., \eqref{eq1A} being an integer for $n = 3^{m}$. For the second factor, note
$$10 \equiv 1 \pmod{3} \implies 10^{3^{2m}} \equiv 10^{3^{m}} \equiv 1 \pmod{3} \tag{3}\label{eq3A}$$
This means
$$10^{3^{2m}} + 10^{3^{m}} + 1 \equiv 3 \equiv 0 \pmod{3} \implies 3 \mid 10^{3^{2m}} + 10^{3^{m}} + 1 \tag{4}\label{eq4A}$$
i.e., the second factor is also an integer. With the product of $2$ integers being an integer, \eqref{eq2A} shows \eqref{eq1A} is an integer for $n = 3^{m+1}$. This proves by induction that \eqref{eq1A} is an integer for $n = 3^k$ for all integers $k \ge 0$.

Recall a simple form (with simple proof)  of  LTE = Lifting The Exponent is
$$a\equiv b\!\!\! \pmod{kn}\,\Rightarrow\, a^k\equiv b^k\!\!\!\! \pmod{k^2n}$$
Applied inductively for $\,k=3\,$ (i.e. repeatedly cubing) yields the claim as follows
$$\begin{align}
&10^{\large 3^\color{#c00}0}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}0}}\\
\Rightarrow\ &10^{\large 3^\color{#c00}1}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}1}}\ \ \rm by\ \ LTE\\
\Rightarrow\ &10^{\large 3^\color{#c00}2}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}2}}\ \ \rm by\ \ LTE\\
&\ \ \ \ \ \ \ \ \ \vdots\\
\Rightarrow\ &10^{\large 3^\color{#c00}k}\! \equiv 1\!\!\! \pmod{3^{2+\color{#c00}k}}\ \ \rm by\ \ LTE\\[.1em]
\Rightarrow\ &9\cdot 3^k\mid 10^{3^k}\!\!-1\\[.1em]
\Rightarrow\ & 9\cdot n\ \mid 10^n-1
\end{align}\qquad$$


Getting stuck on simple logarithmic equation: $x \times \ln (x) = 1$ $$x \times \ln (x) = 1$$

I am trying to solve that equation. I used the theory $\ln(a) = \ln(b)$ being equivalent to $a = b$ and got stuck at

$$x = e^{\frac{1}{x}}$$

That's as far as I went and I know there's a solution (around 1.8 or 1.9), since I used my calculator, but I'd like to know how to do this by hand.

You can use the law of logarithms which states that for $a,b\in\mathbb{R}$: $a\ln{b}=\ln\left(b^{a}\right)$.

Therefore, you have:

$$x\ln{x}=1 \implies \ln{x^{x}}=1$$

You hence have:

$$x^{x}=e$$

Which does not have an elementary closed form, so you must use numerical methods (for instance Newton-Raphson iteration) to get an approximation (Mathematica gives $x\approx 1.76322$).

If you're interested, the closed form solution is: $$\frac{1}{W(1)}, \qquad \text{ where } W(z) \text{ is the LambertW function}$$

${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$See Example 4 here: http://en.wikipedia.org/wiki/Lambert_W_function


Inclusion exclusion principle questions i tried(doing it correct?) $x_1+x_2+x_3\le10$

how many natural numbers solve this problem if

$1\le x_1 \\ 2\le x_2 \\3\le x_3$

What i did:
i created $y_1,y_2 , y_3$ so $\\ y_1=x_1-1 \\y_2=x_2-2\\ y_3=x_3 -3$
and then added $y_4$ so it will be $=$ instead of $\le$

and got:

$y_1 +y_2 +y_3 +y_4 =4$

is it the correct way? thank you!

The key aspect that we need to consider is how to construct this problem.
Imagine a graph of 30 dots, arranged in 10 rows and 3 columns. As I go up a
column $i$, I add more integers to $x_i$, and as I move right from a column
$i$ to $i+1$, I stop adding integers to $x_i$ and I start adding integers to
$x_{i+1}$. The key idea is, how many ways can I get to row $10$, column $3$
with the given constraints? These give limits to how long a given column
up-path will be. Initially, the number of ways to get to row 10, column 3
unconstrained should be $\binom{10+3}{3}$ ways. Now remove the ways I can get
there when $x_1 < 1, x_2 < 2, \vee  x_2 < 3.$ This is the "exclusionary" step.


Factor of a product is product of factors I'm trying to prove the uniqueness of prime factorization of any natural number $n\ge 2$ (I take $0\in\mathbb N$). I've shown the existence of a prime factorization for any such number (by infinite descent on the statement that 'for each $n\ge 2$, there exists a greatest prime that divides $n$').
To show uniqueness, I realize that I will be done if I can show that
$$
\forall a\;\forall b\;\forall c\;\left(c\text{ divides } ab\implies\exists m\;\exists n\; (m\text{ divides } a\;\wedge\; n\text{ divides } b\;\wedge\; c=mn)\right).
$$
But after a lot of futile attempts, I'm not able to "see" any path leading me to this statement. Any help?

Note: I haven’t proven Euclid’s division lemma, and am not working with gcd’s. Hence I’ll appreciate a way that avoids these.

You can prove your displayed assertion by strong induction on $c$.  The assertion is clearly true for the base case $c=1$ (with $m=n=1$).  So assume it's true for all $c'\lt c$, where $c\gt1$. Let $p$ be any prime divisior of $c$ and write $c=pc'$, noting that $p\gt1$ implies $c'\lt c$.. Since $p\mid c\mid ab$, we know (by Euclid's lemma) that $p\mid a$ or $p\mid b$. Without loss of generality, let's assume it's $p\mid a$, and write $a=pa'$. We find now that $c'\mid a'b$, hence, by the strong induction hypothesis, there exist integers $m'$ and $n$ such that $m'\mid a'$, $n\mid b$, and $m'n=c'$. Now simply let $m=pm'$ and note that $m'\mid a'\implies m=pm'\mid pa'=a$ and $c=pc'=pm'n=mn$.

Proof of the uniqueness of prime factorization :
Let's suppose we have two factorizations : $p_1 ^{\alpha_1}...p_r ^{\alpha_r} = q_1 ^{\beta_1}...q_t ^{\beta_t}$.
Then for $1\leqslant i \leqslant r$, we have $p_i | q_1 ^{\beta_1}...q_t ^{\beta_t}$, and the $q_l$ are prime, so there exists one $1\leqslant j \leqslant t$ such that $p_i | q_j$ so $p_i = q_j$.
By doing the same with the $q_j$, we get that $\{p_1,...,p_r\} = \{q_1,...q_l\}$ and as all the $p_i$ and $q_j$ are different, we have that $r=l$.
We can then reorder the $(p_i)$ and $(q_j)$ to have $p_i = q_i$.
Let's prove we have $\alpha_i=\beta_i$ for $1\leqslant i\leqslant r$.
For $1\leqslant i\leqslant r$, $p_i^{\alpha_i} |q_1 ^{\beta_1}...q_r ^{\beta_r}$ and by Gauss Lemm, since $p_i$ is prime with all the $q_j$ for $j\geqslant 2$, we have $p_i^{\alpha_i} | q_i^{\beta_i} = p_i^{\beta_i}$, and we can do the same with $q_i$, so $p_i^{\alpha_i} = p_i^{\beta_i}$.
So $\alpha_i = \beta_i$, and this for all $i$.
Then the decomposition is unique.
For a proof of Euclid Lemma :
Let's suppose there exists $p$ prime and $a,b\in \mathbb{N}$ such that $p$ does not divide $a$ and $b$, but $p|ab$. Let's take, for $p$ and $a$ given, among all the possibilities of $b$, the smallest.
Then $0<b<p$ (because if not we could replace $b$ by a smaller one). We note $r$ the rest of Euclid's division of $p$ by $b$, which is not null since $p$ is prime.
Then $p=mb+r$ so $ar = ap -mab$ is a multiple of $p$.
But we know that $0<r<b<p$ so $ar$ is not a multiple of $p$ because of the minimality of $p$.
So we've got a contradiction, and that proves Euclid's Lemma.


Prove $\lim_{x\to a}\sqrt{x}=\sqrt{a}$ by definition Stewart solution error Let $a>0$. I want to show
$$\lim_{x\to a}\sqrt{x}=\sqrt{a}$$
Purpose of this question was to investigate Stewart's online solution, which turns out to be incorrect.
Their solution
I don't understand the solution on their website. Are they saying that
$1/x  < x $? Seems that way to me by the way they choose delta.
My solution :
$$|x-a|< a   \implies   \frac{1}{\sqrt{x} + \sqrt{a}} < \frac{1}{\sqrt{a}}$$
and choose delta accordingly. Is that correct?
Thanks for the help.

You have to discuss the inequality $|\sqrt{x}-\sqrt{a}|<\varepsilon$.

Now take $\delta=\min(\varepsilon\sqrt{a},a)$ and assume $|x-a|<\delta$. Since $\sqrt{x}+\sqrt{a}>\sqrt{a}$, we have
$$
|\sqrt{x}-\sqrt{a}|=
\frac{|x-a|}{\sqrt{x}+\sqrt{a}}<
\frac{|x-a|}{\sqrt{a}}<\frac{\delta}{\sqrt{a}}\le\varepsilon
$$

Your reasoning does not seem correct (honestly, I really can't follow it). Here's what I have.

Proof. Let $\epsilon > 0.$ Choose $\delta = \epsilon |\sqrt{x}+\sqrt{a}|.$ Then when $|x - a| < \delta,$

\begin{align*}
|\sqrt{x} - \sqrt{a} | &= \frac{x-a}{|\sqrt{x}+\sqrt{a}|} \\
&< \frac{\delta}{|\sqrt{x} + \sqrt{a} |} \\
&< \epsilon
\end{align*}

as required.

I welcome any refinements.