Is there a systematic way to calculate a correlated equilibrium? (in game theory) Consider the following game:

The pure Nash eq. are $( x_1,y_2 )$ and $( y_1,x_2 )$.

Now, in a correlated equilibria situation, you want to assign a probably of zero to playing $( y_1,y_2 )$.

So effectively choosing $p_1$ and $p_2$ in $U = 7 \cdot p_1 + 7 \cdot p_2 + 8 \cdot (1-p_1-p_2) = 8 - p_1-p_2$ to maximise $U$ but such that they are strictly positive.

But it is a constrained optimisation, but how do I work out what the constraints are?

Check out this paper on Correlated Equilibrium
Correlated-Q Learning
Amy Greenwald and Keith Hall
https://www.aaai.org/Papers/ICML/2003/ICML03-034.pdf


Linear Transformations - Linear Algebra I have a problem with this exercise.
I think it can be solved with the properties of linear transformations [ T(v+w) = T(v) + T(w) and T(cv) = cT(v) ] but I'm not sure if it's valid and how to use it in this context.

Does anyone have any idea how to solve it?

Exercise

//Here is the exercise if you can´t see the image//

Let V be a vector space and let T : V → V a linear transformation such that T($2v_1-3v_2$) = $5v_1 + 3v_2$ and T($−3v_1+5v_2$) = $4v_1 + 3v_2$. Write T($v_1$)  and  T($v_2$) in terms of $v_1$ and $v_2$.

The matrix $\begin{pmatrix}2 & -3\\-3 & 5\end{pmatrix}$ transforms coordinates with respect to basis $v_1, v_2$

to those with respect to basis $2v_1-3v_2, -3v_1+5v_2$.

The matrix $\begin{pmatrix}5&4\\3&3\end{pmatrix}$ represents $T$ with coordinates with respect to $2v_1-3v_2, -3v_1+5v_2$ in the domain

and $v_1,v_2$ for the range.

Therefore, the columns of $\begin{pmatrix}5&4\\3&3\end{pmatrix}\begin{pmatrix}2 & -3\\-3 & 5\end{pmatrix}^{-1}$ give the images of $v_1$ and $v_2$ under $T$.


Is ∑ with negative value solvable? Is it possible to have a negative value in sigma?

e.g.
$y = \Sigma_{k=0}^{k=-2} k \times 10$

Will this give the result $(0 \times 10) + (-1 \times 10) + (-2 \times 10) = -30 $?

Or will it be $\infty$ because $k$ will be increased with $1$ until it equals $-2$ (which is never).

Or something else?

There are several things to remark here:

First, $$\sum_{k=0}^{k=-2}\times 10$$ is not actually the correct notation.

You seem to mean $$\sum_{k=0}^{k=-2}k\times 10$$ which is correct notation (though usually the $k=$ part is not included in the top), but would be $0$. The reason is that the notation $$\sum_{k=i}^{k=j}k$$ means "take the sum of $k$ for each $k$ which is $\geq i$ and $\leq j$". In the case at hand, there are no such $k$, and by convention, this means the sum is $0$.

To get the desired result, you can do $$\sum_{k=-2}^{k=0}k\times 10$$ or, with the more common notation, $$\sum_{k=-2}^0k\times 10$$

Sigma notation is shorthand for iterated addition. The expression following the Sigma is the addend, and the operator for each iteration is addition.
As addition is commutative, it matters not the positive or negative direction of the iteration.
One can go from some initial index to a negative number. What may be forgotten in the notation is that there is an assumed increment. One may consider this assumption as the same as the assumed $+$ positive sign before a positive number and similarly to the trailing zeroes in decimal fraction of a Real number or Real part of a Complex number, i.e, $3.1$ is the same as $3.10,$ $3.100,$ etc..
The fact that the sign of a positive number is not always written out does not mean it does not exist. The same is true of the increment operator sometimes noted in Sigma notation.
In the convention of Sigma notation, the increment operation is defaulted as adding $+1$ to the index. When the decrement operator is used in the notation, it should not be seen as subtraction but as an increment of a negative amount.
If the upper limit of the index is lower than the initial index, one may logically assume that the operator is a negative increment. Some notationists will assist in the clarification by including the increment operator in the notation and, if not unary $(+1|-1),$ the step amount by which the index is to be positively or negatively incremented in the iteration.


Proof: How many digits does a number have? $\lfloor \log_{10} n \rfloor +1$ I read recently that you can find the number of digits in a number through the formula $\lfloor \log_{10} n \rfloor +1$ What's the logic behind this rather what's the proof?

Suppose that $n$ has $d$ digits; then $10^{d-1}\le n<10^d$, because $10^d$ is the smallest integer with $d+1$ digits. Now take logs base $10$: $\log_{10}10^k=k$, so the inequality becomes $d-1\le\log_{10}n<d$. If you now take the floor (= integer part) of $\log_{10}n$, throwing away everything to the right of the decimal point, you get $\lfloor\log_{10}n\rfloor=d-1$. Thus, $d=\lfloor\log_{10}n\rfloor+1$. This isn’t quite what you posted, but it’s the integer part of it, and clearly the number of digits must be an integer.

The number of decimal digits in $N$ would be $⌊\log_{10}N⌋+1$
Just remember that $⌊\log_{10}N⌋$ should always be in decimal form.


Are there some techniques for checking whether a statement implies another without truth tables? Are there some techniques for checking whether a statement implies another without truth tables?

For example, I was asked whether $P\Longrightarrow P_{1}$ given the following statements:

$$P: [p \land (q \land r)]\lor \neg[p \lor (q \land r)],$$

$$P_{1}: [p \land (q \lor r)]\lor \neg[p\lor(q\lor r)].$$

What I did was to find the logical equivalences of the negations of $\neg[p \lor (q \land r)]$ and $\neg[p\lor(q\lor r)]$, then tested some values, with the negations done, it was slightly easier to see what was happening as I drew 0's and 1's under $p,q,r$. I find that for $p:0,q:1,r:0$, $P$ and $P_{1}$ does not have the same truth statement, so $P\nRightarrow P_{1}$, yet it was still a tedious process.

Yes. I always think of using truth tables as a last resort; sometimes it actually pays off because the effort spent constructing the truth table can be considerably less than that using logical deductions, but there is not much at all learned by proving something with a truth table--it's very mechanical.

When trying to prove an implication of the form $\Omega\to\Psi$ without using a truth table, your job is to show that, given $\Omega$, the implication $\Omega\to\Psi$ is a tautology. I will give you two examples of how to check whether or not a statement implies another without a truth table (the first will not be a tautology but the second will be).

Example 1: Given $p$, does $p\to(p\land q)$?

Solution. Notice the following:
\begin{align}
p\to(p\land q) &\equiv \neg p\lor(p\land q)\tag{since $r\to s\equiv \neg r\lor s$}\\[0.5em]
  &\equiv \underbrace{(\neg p\lor p)}_{\text{Always true}}\;\;\land\underbrace{(\neg p\lor q)}_{\substack{\text{True if $q$ is true}\\\text{False if $q$ is false}}}\tag{distributivity}
\end{align}
Hence, the implication $p\to(p\land q)$ is not a tautology since it will be false when $q$ is false.

Example 2: Given $p\land q$, does $(p\land q)\to p$?

Solution. Notice the following:
\begin{align}
(p\land q)\to p &\equiv \neg(p\land q)\lor p\tag{since $r\to s\equiv \neg r\lor s$}\\[0.5em]
  &\equiv (\neg p\lor\neg q)\lor p\tag{DeMorgan}\\[0.5em]
  &\equiv \underbrace{(\neg p\lor p)}_{\text{Always true}}\lor\neg q.\tag{associativity of $\lor$}
\end{align}
Since $\neg p\lor p$ is always true, we have that $(p\land q)\to p$ is a tautology, and we proved this without using a truth table.

You could write a proof instead. Suppose $P$.

If $p\land q\land r$, then $p\land(q\lor r)$. Therefore $P_1$ tautologically.

Otherwise $\neg[p \lor (q \land r)]\Leftrightarrow \neg p \land \neg(q\land r)$. Then $p$ being true contradicts $P$, so $P\Rightarrow P_1$ in that case. In the case that $p$ is false, $P$ reduces to $\neg q \lor \neg r$ and $P_1$ reduces to $\neg q \land \neg r$. If $\neg q\not= \neg r$, then $P$ and not $P_1$, so $P\not\Rightarrow P_1$; otherwise $P$ and $P_1$ are both true or both false, so $P\Rightarrow P_1$.

Therefore $P\Rightarrow P_1$ exactly when $p$ is true or $q=r$, and $P\not\Rightarrow P_1$ exactly when $p$ is false and $q\not=r$.


An ellipse meets a hyperbola that the tangent lines at the points of intersection are perpendicular to each other . Show that $a^2-b^2=c^2+d^2$ Question: $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1 $ meets $\frac{x^2}{c^2} - \frac{y^2}{d^2}=1$ in such a way that the tangent lines at the points of intersection are perpendicular to each other. Show that $a^2-b^2=c^2+d^2$

I've been stuck on a while now , first I tried finding the intersection between the ellipse and the hyperbola and then creating tangent lines based on those specific $x_1 , y_1$ but that didn't turn out pretty. Second of all I tried taking the derivative of both the ellipse and hyperbola and using $m_1 \cdot m_2 =-1 $ but that didn't get me far either. What should I do?

By conformal mapping
\begin{align*}
  x+yi &= \lambda\cosh(u+vi) \\
  &= \lambda \cosh u \cos v+\lambda i\sinh u \sin v \\
  \frac{x^2}{\lambda^2 \cosh^2 u}+\frac{y^2}{\lambda^2 \sinh^2 u} &= 1 \\
  \frac{x^2}{\lambda^2 \cos^2 v}-\frac{y^2}{\lambda^2 \sin^2 v} &= 1 \\
  a^2-b^2 &= \lambda^2 (\cosh^2 u-\sinh^2 u) \\
  &= \lambda^2 \\
  c^2+d^2 &= \lambda^2 (\cos^2 v+\sin^2 v) \\
  &= \lambda^2 \\
  a^2-b^2 &= c^2+d^2
\end{align*}

Or just recognize that an ellipse and hyperbola that have a perpendicular intersection of tangent lines must be confocal (link below). Thus, the c value, or distance from center to focii for both ellipse and hyperbola must be equal. The focus of an ellipse is calculated by (a^2 - b^2)^0.5 while the focus of a hyperbola is (c^2 + d^2)^0.5. Set both these equal and you get a^2 - b^2 = c^2 + d^2.
https://en.wikipedia.org/wiki/Confocal_conic_sections#:~:text=In%20geometry%2C%20two%20conic%20sections,they%20have%20the%20same%20foci.&text=In%20the%20mixture%20of%20confocal,orthogonally%20(at%20right%20angles).


limit of a sequence $(-1)^n$ instead $1$ I should calculate the limit of a sequence. A friend told me, that the solution is $1$. But I got $(-1)^n$.

The exercise is: $\lim\limits_{n \to \infty} \frac{1}{n^2} + (-1)^n \cdot \frac{n^2}{n^2+1}$

I did following:
$$\begin{align*}
&=\frac{n^2 ((-1)^n n^2 + 1 + \frac{1}{n^2})}{n^2(n^2+1)}\\
&=\frac{(-1)^n n^2 + 1 + \frac{1}{n^2}}{(n^2+1)}\\
&=\frac{n^2(\frac{(-1)^n n^2}{n^2} + \frac{1}{n^2} + \frac{1}{n^4})}{n^2(1 + \frac{1}{n^2})}\\
&=\frac{(-1)^n + 0 +0}{1}\\
&=\lim\limits_{n \to \infty} (-1)^n
\end{align*}$$

What did I wrong?

Edit
Well, some answers confused me. Here the complete exercise.
I should check if the sequence is convergent for ${n \to \infty}$ and determine the limit if it exist. Also for a sequence which is $\infty$ or $-\infty$.

My friend got $1$ as limit. I got $(-1)^n$. I would say, that this sequence has no limit, just  limit points $1$ and $-1$.

The problem here is that the sequence does not converge at all!

To prove this, consider the subsequence where $n$ is an even number, and show that the limit is $1$. Then take the subsequence where $n$ is an odd number, and show that the limit is $-1$. Now, if a sequence converges, so do all its subsequences, and the limit is the same! Therefore it can't converge.

Note that, for the limit of a sequence to exist, it has to be unique. That means the sequence has to converge to a single value. In your case the sequence has subsequences which converge to two different values, namely
$ L = \left\{-1,1\right\}\,.$ So the limit does not exist. Here comes the notion of $\limsup$ and $\liminf$ which we use to define the existence of the limit. We say the sequence $a_{n}$ converges to $a$ $n\rightarrow \infty$, if

$$ \liminf_{n\rightarrow \infty} a_n =\limsup_{n \rightarrow \infty} a_n = \lim_{n \rightarrow\infty} a_n \,.$$

In your case $\liminf_{n\rightarrow \infty} a_n =-1$ and $ \limsup_{n \rightarrow \infty} a_n = 1 $, so they are not equal and hence the limit of the sequence does not exist.

You can compare this to the case of taking limits of functions when we require for the limit to exist at a point $x_0$ that
$$ \lim_{x\rightarrow x_0^{-}} f(x) = \lim_{x\rightarrow x_0^{+}} f(x) = \lim_{x\rightarrow x_0} f(x) \,.$$


Field extension of $\Bbb Q$ of degree $4$ has reals Suppose $L / \mathbb{Q}$ is a field extension with $[L : \mathbb{Q}]$ = 4, with $L \not\subset \mathbb{R}$. Is it true that $L \cap \mathbb{R} \neq \mathbb{Q}$? If not, is it true if $L / \mathbb{Q}$ is normal?
I have tried to suppose $a \in L \cap \mathbb{R} \implies a \in \mathbb{Q}$, then from $z = a + b i \in L$ we get, $z + \bar{z} = 2a$ ($\bar{z} \in L$ if $L$ is normal at least). So $a \in \mathbb{Q}$. But I don't know how to proceed from there. I guess one should be able to show that $L = \mathbb{Q}(i)$ which contradicts degree $4$.

Assume that $L/\Bbb Q$ is normal. Let $\sigma$ be the field automorphism given by complex conjugation (which is a field automorphism because the extension is normal). Then the subgroup $H$ of $\operatorname{Aut}(L)$ generated by $\sigma$ has order $2$, so $L$ has degree $2$ over the fixed field $L^H$. We get $[L^H:\Bbb Q]=4/2=2>1$ and $L^H\subset \Bbb R$, i.e. $L\cap \Bbb R\ne \Bbb Q$.
Without the normal assumption it is not necessarily true, take a polynomial of degree $4$ with Galois group $S_4$ and a nonreal root $\alpha$. Then $L=\Bbb Q(\alpha)$ doesn't have any non-trivial intermediate fields at all, so since $L\nsubseteq\Bbb R$ we get $L\cap \Bbb R=\Bbb Q$.

Hint: As an example, consider a primitive 8th root of unity $\xi$. It obeys the equation $\xi^4=-1$.
Then $L = \{a+b\xi+c\xi^2+d\xi^3\mid a,b,c,d\in\Bbb Q\}$.


Prove that a polynomial diverges to infinity. I would like to prove the following statement:

Let $P$ be a polynomial of degree $n$ where $n$ is an odd natural number and $x$ $\in$ $\mathbb{R}$. $P(x)=a_{0}+a_{1}x+ ... + a_{n}x^{n}$

If $a_{n} > 0$, then $\lim_{x\to\infty}P(x)=\infty$

I am thinking of three ideas in order to prove the statement above.

$1$. $\lim_{x\to\infty}P(x)=\infty$ if for every $M>0$, there exists $K \in \mathbb{R}$ such that if $x \in \mathbb{R}$ with $x>K$, then $P(x)>M$.

$2$. $P$ is continuous on $\mathbb{R}$.

$3$. Contradiction: $a_{n} > 0$ but $\lim_{x\to\infty}P(x)\neq\infty$

Although I'm thinking the ideas above might help me prove the statement, but I cannot prove it. Could anyone help me with this?

The fact that $n$ is odd is irrelevant. Rewrite the polynomial, over $(0,\infty)$ which is not restrictive, as
$$
x^n\left(a_n+\frac{a_{n-1}}{x}+\dots+\frac{a_0}{x^n}\right)
$$
Now it should be known that
$$
\lim_{x\to\infty}\left(a_n+\frac{a_{n-1}}{x}+\dots+\frac{a_0}{x^n}\right)
=a_n
$$
because each summand, except for the first one, has limit $0$.

Hence there is $x_0$ such that, for $x>x_0$,
$$
a_n+\frac{a_{n-1}}{x}+\dots+\frac{a_0}{x^n}>\frac{a_n}{2}
$$
Finally, for $x>x_0$,
$$
a_nx^n+\dots+a_0>\frac{a_n}{2}x^n
$$
and you should be able to finish.

The same strategy proves that if

$\lim_{x\to c}f(x)=\infty$
$\lim_{x\to c}g(x)=l>0$

then
$$
\lim_{x\to c}f(x)g(x)=\infty
$$
(here $c$ can be any real, $\infty$ or $-\infty$; $l$ can also be $\infty$).

Here's a pretty simple induction proof, done backwards (since it makes more sense that way):

Inductive Step: We can easily show that $\lim_{x\rightarrow\infty}P(x)=\infty$ if we can show that $\lim_{x\rightarrow\infty}P'(x)>0$. So, let's go about doing that. We differentiate, and get $P'(x)=nx^{n-1}+(n-1)x^{n-2}+(n-2)x^{n-3}...$. We don't seem to be any closer to proving the original statement, or our revised version. However, notice that if $\lim_{x\rightarrow\infty}P'(x)=\infty$, then $\lim_{x\rightarrow\infty}P'(x)>0$. So, we've now brought the whole problem down one derivative, by showing that if the $P'(x)$ case is true, then the $P'(x)$ must also be true.

Initial Case:
Notice also that in the case $n=1$, the statement is trivial to prove. So, if we can bring the problem all the way down to $n=1$, then we will proven the $n$th case. We can easily repeat the procedure already demonstrated, in total $n-1$ times, since $P^{n-1}(x)$ will be of degree $1$, which by the Principle of Math Induction means that the $n$th case is true.

Q.E.D.


Convergence of definite integrals given unknowns. Given the following equation,
$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx
$$
I am supposed to find the condition on the constant $a$ such that the above integral is convergent.
Substituting $u = \ln(x)$, we get
$$
\frac{du}{dx} = \frac{1}{x}
$$
Where the integral
$$
\int_{1}^{2} \frac{1}{x(\ln x)^{a}} dx = \int_{0}^{\ln 2} \frac{du}{dx}\frac{1}{u^{a}}dx = \int_{0}^{ln2} \frac{1}{u^{a}}du
$$
simplifies to
$$
\frac{(\ln 2)^{1-a}}{1-a}
$$
When $a = 1$, the above diverges. However, I am unsure of what values the above equation converges to, i.e when $a>1$ or $a<1$

Let's be a bit more formal here. First of all, it indeed suffices to consider the transformed integral
$$
\int_0^{\ln(2)} \frac 1{u^a}\,du = \int_0^{\ln(2)} u^{-a}\,du.
$$
Note that because of the possible division by zero, this is an improper integral. In other words, the quantity that we're after is the limit
$$
\lim_{b \to 0^+} \int_b^{\ln(2)} u^{-a}\,du.
$$
In the case that $a \neq 1$, this can be written as
$$
\lim_{b \to 0^+} \int_b^{\ln(2)} u^{-a}\,du = \lim_{b \to 0^+} \left(\frac{\ln(2)^{1-a}}{1-a} - \frac{b^{1-a}}{1-a}\right).
$$
What can we say about this limit when $a > 1$? When $a < 1$? What does the integral become in the case that $a = 1$?


Prove that f(x) is uniformly continuous Show that $f(x)=x^2cos(\frac{1}{x^2})$ on (0,1).

Let $\epsilon > 0$ and $x,y \epsilon (0,1)$. $| f(y) - f(x)| = |y^2cos(\frac{1}{y^2}) - x^2cos(\frac{1}{x^2})|$.

After this I do not know how to proceed.

A common way to show a function is uniformly continuous is to show it is Lipschitz. That is, to show that $|f(x)-f(y)|\leq L |x-y|$ on a set $A$, where $L$ is some constant independent of $x,y$. A common way to show a function is Lipschitz is the use the mean value theorem: $|f(x)-f(y)|=|f'(\xi)||x-y|$. For $1>\eta>0$, show that your function $f(x)$ is Lipschitz on $[\eta, 1)$.

Show also that for $x, y \in (0, \eta]$ that $|f(x)-f(y)|\leq 2 \eta^2$.

Combine both of these facts to show that $f$ is uniformly continuous. (That is, given $\epsilon>0$, pick $\eta>0$ and then pick $\delta > 0$ so that $|f(x)-f(y)|$ is small. You will have to find a way to "bridge the gap" between the intervals $(0, \eta]$ and $(\eta, 1)$. That is, you have to deal with the case $x\in (0, \eta]$ and $y \in (\eta, 1)$. Add and subtract and use the triangle inequality.)

[P.S. If this is homework, please remember to cite all help you receive.]

If you define $f(0) = 0$, you can prove that $f$ is continuous on $[0,1]$.


How prove this inequality $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}<6$ let $a,b,c\ge\dfrac{1}{3}$,and such $$a^2+b^2+c^2=a+b+c$$
show that
$$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}<6$$

I try:$$a-a^2=(b^2-b)+(c^2-c)\ge-\dfrac{1}{4}-\dfrac{1}{4}=-\dfrac{1}{2}$$
so we have
$$a^2-a-\dfrac{1}{2}\le 0\Longrightarrow a\le\dfrac{1+\sqrt{3}}{2}$$
so we have
$$a,b,c\in [\dfrac{1}{3},\dfrac{1+\sqrt{3}}{2}]$$
let $f(a)=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}$,and $f''(a)>0$. so $LHS$ is maximum when $a,b,c=\{\dfrac{1}{3},\dfrac{1+\sqrt{3}}{2}\}$.but I found all case this value is big $6$

Actually, this problem could be solved analytically or using Lagrange multipliers, but here the (Analytical solution).

show :

$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 6}$

s.t:

${a, b, c ≥ ⅓}$

${a² + b² + c² = a + b + c}$

Changing the problem a little

${a, b, c ≥ 1}$

$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$

rearranging the function

${a² = a + b + c - b² - c²}$

$\frac{a + b + c - b² - c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$

then

$\frac{a}{b}$ + ${1}$ + $\frac{c}{b}$  - ${b}$ - $\frac{c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$

${b ≥ 1}$ ⇒ ${1-b}$ ≤ 0
$\frac{c}{b}$ - $\frac{c²}{b}$ = $\frac{c-c²}{b}$   ⇒ $\frac{c-c²}{b}$ ${≤ 0}$
$\frac{a}{b}$ + $\frac{c²}{a}$ = $\frac{a² + bc²}{bc}$ =   $\frac{(a + b + c - b² - c² ) + bc² }{ba}$
$\frac{1}{b}$ + $\frac{b-b²}{ba}$ + $\frac{c - c²}{ba}$ + $\frac{c²}{a}$
$\frac{1}{b}$ ${≤1}$ and $\frac{b-b²}{ba}$ ${≤ 0}$ and $\frac{c - c²}{ ba}$ ${≤ 0}$

then we need to proof that $\frac{c²}{a}$ ${≤ 1}$

$\frac{c² + a² - a² }{a}$

= $\frac{c² + a²}{a}$ - ${a}$

= $\frac{c² + (a + b + c - b² - c²)}{a}$ - ${a}$

= ${1 - a}$ +  $\frac{2c - c²}{a}$ + $\frac{b-b²}{a}$

${1-a}$ ${≤ 0}$ and
$\frac{b-b²}{a}$ ${≤ 0}$

we need to proof that $\frac{2c - c²}{a}$ ${< 1}$  or  ${2c ≤ c²}$ but ${c≥1}$

then ${2c ≤ c²}$ for all ${c≥1}$


Calculate the limit: $\lim_{n\rightarrow 0}\frac{\sin(2n)}{\ln(1+n)}$ Calculate the limit with L'Hôpital:

  $$\lim_{n\rightarrow 0}\frac{\sin(2n)}{\ln(1+n)}$$

I get $2$ as limit but not sure because I tested with calculator and I get some small thing like $0.032\ldots$

$$(\sin(2n))'=2\cos(2n)$$

$$(\ln(1+n))'=\frac{1}{1+n}$$

$$\lim_{n\rightarrow 0}\frac{2\cos(2n)}{\frac{1}{1+n}}= \lim_{n\rightarrow 0}2\cos(2n) \cdot(1+n)=\lim_{n\rightarrow 0}2\cos(2n)+2n\cdot \cos(2n)=$$

$$2\cdot \cos(0)+0\cdot \cos(0)=2\cdot1+0\cdot1=2$$

Yeah, you're doing right. When you check the result, please make sure that you calculate the angle using radian mode.

$$\lim_{n\rightarrow 0}{\sin(2n)\over \ln(1+n)}$$
$$=\lim_{n\rightarrow 0}{\sin(2n)\over \ln(1+n)}{2n\over 2n}$$
$$=\lim_{n\rightarrow 0}{2n\over \ln(1+n)}$$
$$=\lim_{n\rightarrow 0}\large{2\over {\ln(1+n)\over n}} = 2$$


Prove that the least upper bound of $A$ cannot be rational Let $A\subseteq\mathbb{Q}$ be defined by
$$
A=\{x\in\mathbb{Q}:x^2<2\}.
$$

Prove that $A$ cannot have a least upper bound in $\mathbb{Q}$.

I know I could prove that $\sqrt2$ is the least upper bound of $A$ and prove that $\sqrt2$ is irrational. However, I'm looking to prove this using an alternative method because it might not always be obvious what the least upper bound should be.

Initial thoughts

My initial thinking is to prove this by contradiction.
So, suppose $y\in\mathbb{Q}$ is the least upper bound of $A$. Then for any $z\in\mathbb{R}$ with $z<y$, we can find $a\in A$ such that $a>z$.
Since $a\in A$, we have that $a=m/n$ for some $m,n\in\mathbb{Z}$ and that $a^2<2$. Similarly, since $y\in\mathbb{Q}$, we have that $y=p/q$ for some $p,q\in\mathbb{Z}$.

Problem

I'm not sure how to combine these facts to construct a contradiction. Any hints/insights would be great.

There exists a function $f: [0, \infty) \to [0, \infty)$ with the following properties:

$f(x) \in \mathbb{Q}$ if $x \in \mathbb{Q}$,
$f$ is increasing,
$x < f(x) < 2$ if $x < 2$.

Thus for any proposed $a \in A$, we have

$f(a) \in A$ and
$f(a) > a$.

This shows that $a$ cannot be the supremum of $A$.
Can you find such a function? Spoiler below.

 The rational function
 $$f(x) = \frac{4 + 3x}{3 + 2x}$$ works.

 It is constructed as
 $f(x) = g(g(x))$, where     $$g(x) = \frac{2 + x}{1 + x}$$
 which also converges to $\sqrt{2}$ upon
 iteration, but in an alternating manner.

Suppose that $c = \max{A}$. So $\forall x \in A, x\le c$. Take $\varepsilon > 0$ such that $0 < \varepsilon < \min\{\frac{2 - c^2}{2c + 1},1\}$ (*). It implies that
\begin{align}
\varepsilon < \frac{2 - c^2}{2c + 1} &\Rightarrow (2c + 1)\varepsilon < 2 - c^2\\
&\Rightarrow c^2 + \varepsilon(2c + 1) < 2\\
&\Rightarrow c^2 + 2\varepsilon c + \varepsilon < 2\\
&\Rightarrow c^2 + 2\varepsilon c + \varepsilon^2 < c^2 + 2\varepsilon c + \varepsilon < 2\\
&\Rightarrow (c+\varepsilon)^2 < 2.
\end{align}
Since $c>0$, it means that $(c+\varepsilon)>0$ and  $(c+\varepsilon)^2<0$. Now, taking $\varepsilon \in \mathbb{Q}$ and satisfacting (*), $c + \varepsilon \in A$, and this contradicts the fact that $c = \max{A}$. So, the set $A$ doesn't have a maximum.


Question from Real Analysis -- making proof cleaner I'm taking my first real analysis class after about maybe 7 years from the last time I took one. I have a question about how to make my proof nicer. The question is: Let $f : X \rightarrow Y$ be a mapping. Let $Y_\alpha,{\alpha \in A}$ be a family of subsets of Y. I need to prove $f^{-1}(\bigcup_{\alpha \in A} Y_\alpha) = \bigcup_{\alpha \in A} f^{-1}(Y_\alpha)$

So here's how I tried it.

Fix $a \in f^{-1}(\bigcup_{\alpha \in A} Y_\alpha) = \left\lbrace x \in X : f(x) \in \bigcup_{\alpha \in A} Y_\alpha \right\rbrace$. Then $f(a) \in \bigcup_{\alpha \in A}Y_\alpha$, so there exists some $\beta \in A$ so that $f(a) \in Y_\beta$. This implies that $a \in f^{-1}(Y_\beta)$, and hence $a \in \bigcup_{\alpha \in A} f^{-1}(Y_\alpha)$

This shows containment one way. I haven't written it out but I assume to show the other containment is similar. It seems however, that there should be a bit more elegant way to write this than how I did. I think the work I did was right, but is there a way to make it cleaner?

You can write in this way

$a \in f^{-1}(\bigcup_{\alpha \in A} Y_\alpha) = \left\lbrace x \in X : f(x) \in \bigcup_{\alpha \in A} Y_\alpha \right\rbrace$ $\iff$ $f(a) \in \bigcup_{\alpha \in A}Y_\alpha$ $\iff$ $ \exists \beta$ s.t $ f(a) \in Y_\beta$   $\iff$ $\exists \beta$ s.t $a \in f^{-1}(Y_{\beta})$$\iff$ $a \in \bigcup_{\alpha \in A} f^{-1}(Y_\alpha)$


What if the intermediate point is $c=(0,0)$ in the MVT? Let $f: \Bbb R ^2 \to \Bbb R$ be differentiable everywhere. Assume $f(-\sqrt 2, -\sqrt 2) = 0$ and also that

  $$ \left| \frac {\partial f} {\partial x} (x,y) \right| \le | \sin (x^2 + y^2) | \\
\left| \frac {\partial f} {\partial y} (x,y) \right| \le | \cos (x^2 + y^2) |$$

  for each $(x,y) \in \Bbb R ^2 \setminus \{ (0,0) \}$. Prove that $|f(\sqrt 2, \sqrt 2)| \le 4$.

So by Mean Value Theorem there is a point $c$ in the straight line segment that joins $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$ such that,

$\|f(\sqrt{2},\sqrt{2})-f(-\sqrt{2},-\sqrt{2})\|\leq |\nabla f(c)| \;\|(\sqrt{2},\sqrt{2})-(-\sqrt{2},-\sqrt{2})\|$.

This implies $\|f(\sqrt{2},\sqrt{2})\|\leq |\nabla f(c)| 4$. Now if $c\neq (0,0)$ then by the given inequalities we can show the result. But, what if $c=(0,0)$? Because $(0,0)$ is a point on the line segment.

Write

$$f(\sqrt 2,\sqrt 2) = f(\sqrt 2,\sqrt 2) - f(0,0) + f(0,0) - f(-\sqrt 2,-\sqrt 2).$$

Use the mean value theorem on each segment; recall the MVT doesn't require differentiability at the end points.

The problem is $f$ may be discontinuous at $0$.

However we can evade $(0,0)$ and follow two line segments.

Call A = $(\sqrt{2},\sqrt{2})$.

Fix $\epsilon>0$ and call E = $(0, \epsilon)$.

Take the path -A to E to +A. Then by making $\epsilon$ arbitrarily small we can approach 4 as closely as needed. Hence $f(A)$ must be $\le 4$.


Maximal free submodule over a PID Let M be a f.g. module over a p.i.d. and T(M) be its torsion submodule. Then M is the direct sum of T(M) and of a free submodule F, unique up to isomorphism, and in addition which is a maximal free submodule. Conversely, let F be a maximal free submodule of M (which is a stronger condition than being of maximal rank). Is it true that F is a direct summand, so that M is the direct sum of T(M) and F?

The claim is false in general, as we shall illustrate below by a counterexample. Before moving on two key observations are needed:

as I passingly mentioned in a comment to the original question, there is identity between the collection of submodules of $M$ which intersect $\mathrm{T}(M)$ trivially and the collection of free submodules. Clearly, if $N \leqslant_A M$ is free then $N \cap \mathrm{T}(M)=\mathrm{T}(N)=\{0_M\}$, since free modules over (commutative) integral domains are torsionless. On the other hand, by the same token any submodule $N$ that intersects the torsion submodule trivially must itself be torsionless. $M$ being finitely generated over the Noetherian ring $A$ means that $M$ itself is a Noetherian module, with the consequence that in particular $N$ is finitely generated. Since any torsionless finitely generated $A$-module is free, $N$ is in particular free. The problem can therefore be reformulated as analysing whether any submodule $N$ maximal among those which intersect $\mathrm{T}(M)$ trivially is a direct summand or not.
assuming that the submodule $L \leqslant_A M$ is maximal free and at the same time a direct summand, let us remark that it has a unique supplementary, namely the torsion submodule. Indeed, consider an arbitrary supplementary $P$ of $L$. Without proof we mention a very general result:

Lemma. Let $A$ be an arbitrary ring, $S \leqslant_{\mathrm{Mon}} \mathrm{Z}(A)$ a central multiplicative system of $A$, $M$ a left $A$-module, $I$ an arbitrary index set and $N \in \mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(M)^I$ a family of submodules whose (internal) sum is direct. Given any family $P \in \displaystyle\prod_{i \in I}\mathscr{S}_{\operatorname{\mathit{A}-\mathrm{Mod}}}(N_i)$ of submodules, the following relation holds between $S$-saturations:
$$\mathrm{Sat}_S\left(\sum_{i \in I}P_i\right)=\sum_{i \in I}\left(\mathrm{Sat}_S(P_i) \cap N_i\right)$$

and apply this result to the particular case of submodules $\{0_M\} \leqslant_A L$ respectively $\{0_M\} \leqslant_A P$ and multiplicative system $A^{\times}\colon=A \setminus \{0_A\}$, yielding the conclusion that $\mathrm{T}(M)=\mathrm{Sat}_{A^{\times}}\left(\{0_M\}\right)=\left(\mathrm{Sat}_{A^{\times}}\left(\{0_M\}\right) \cap L\right)+\left(\mathrm{Sat}_{A^{\times}}\left(\{0_M\}\right) \cap P\right)=\mathrm{T}(L)+\mathrm{T}(P)=\mathrm{T}(P)$. Since $P$ is finitely generated over a principal ideal domain, its torsion submodule must admit a (free) supplementary $Q$ in $P$, and it follows that $L+Q$ and $\mathrm{T}(P)=\mathrm{T}(M)$ are supplementary in $M$. By maximality of $L$, it further follows that $Q \subseteq L+Q=L$, whence $Q \subseteq L \cap P=\{0_M\}$ and thus finally $P=\mathrm{T}(M)$.
Now on to the promised counterexample: consider $A=\mathbb{Z}$ and $M=\mathbb{Z} \times \mathbb{Z}_4$, where I am using the abbreviation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for any $r \in \mathbb{N}$. It is clear that $\mathrm{T}(M)=\{0\} \times \mathbb{Z}_4$. We argue that the submodule $N\colon=\mathbb{Z}\left(2, \overline{1}\right)$ is maximal free yet not a direct summand (the bar notation $\overline{m}$ denotes the residue class of $m \in \mathbb{Z}$ modulo $4$). The condition $N \cap \mathrm{T}(M)=\{0_M\}$ is easily verified. Consider a submodule $P \leqslant_{\mathbb{Z}} M$ such that $N \leqslant_{\mathbb{Z}} P$ and $P \cap \mathrm{T}(M)=\{0_M\}$. If $p \colon \mathbb{Z} \times \mathbb{Z}_4 \to \mathbb{Z}$ denotes the canonical projection onto the left factor, we have $2\mathbb{Z}=p[N] \leqslant_{\mathbb{Z}} p[P] \leqslant_{\mathbb{Z}} \mathbb{Z}$, which gives the only options $p[P]=\mathbb{Z}$ or $p[P]=2\mathbb{Z}$. The former option leads to a contradiction, since $1 \in p[P]$ meant there existed $a \in \mathbb{Z}_4$ such that $(1, a) \in P$ and therefore $(2, 2a) \in P$. Since we also have $\left(2, \overline{1}\right) \in N \subseteq P$, it follows that $0_M \neq \left(0, 2a-\overline{1}\right) \in P \cap \mathrm{T}(M)$, which constitutes the contradiction (the fact that $\overline{1} \notin 2\mathbb{Z}_4=\left\{\overline{0}, \overline{2}\right\}$ is crucial here). We thus necessarily have $p[P]=2\mathbb{Z}$, so for any $x \in P$ there exist $n \in \mathbb{Z}$ and $t \in \mathbb{Z}_4$ such that $x=(2n, t)$. As we also have $n\left(2, \overline{1}\right)=\left(2n, \overline{n}\right) \in P$, we infer that $\left(0, t-\overline{n}\right) \in P \cap \mathrm{T}(M)=\{0_M\}$ whence $t=\overline{n}$ and thus $x \in N$. This proves the equality $P=N$ and establishes $N$ as maximal among those submodules which intersect $\mathrm{T}(M)$ trivially. However, the second observation tells us that $N$ cannot be a direct summand: indeed, if this were the case it would necessarily follow that $N+\mathrm{T}(M)=M$, whereas in our case $N+\mathrm{T}(M)=2\mathbb{Z} \times \mathbb{Z}_4<_{\mathbb{Z}} M$ is a proper submodule.

Consider $M=\mathbb{Z}$ over $\mathbb{Z}$ and $F=2\mathbb{Z}$, the sequence $0\rightarrow 2\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}/2\mathbb{Z}\rightarrow 0$ does not split and $2\mathbb{Z}$ is maximal.


Please help me with this trigonometric limit without using L'Hopital's rule I need to solve the following limit without using L'Hopital's rule:

$$\lim _{x\to 0}\left(1+\sin\left(x\right)\right)^{\frac{1}{x}}$$

The thing is that I can not figure out what to do. One of my ideas was to apply this rule: $a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$, getting this:

$$\lim _{x\to 0}e^{\frac{1}{x}\ln \left(1+\sin \left(x\right)\right)}$$

I already know that the answer is $e$, so the exponent is definitely 1. However, I tried everything I could but have no idea how to solve $\lim _{x\to 0}\left(\frac{1}{x}\ln \left(1+\sin \left(x\right)\right)\right)$ which needs to be 1.

I would  really appreciate your help, and if you find a totally different way to solve the limit without using L'Hospital's rule it will be good as well.

Consider the function $f\colon(-\frac{\pi}{2},\frac{\pi}{2})\to(-1,1)$ defined by
$$
f(x) = \ln(1+\sin x)\,.
$$

We have that $f(0)=0$, $f$ is continuously differentiable, and $f'(x) = \frac{\cos x}{1+\sin x}$ by the chain rule; so that
$$
\lim_{x\to 0} \frac{\ln(1+\sin x)}{x}=\lim_{x\to 0} \frac{f(x)-f(0)}{x} =  f'(0) = \frac{\cos 0}{1+\sin 0} = 1
$$
so that by continuity of $\exp$
$$
\lim_{x\to 0} e^{\frac{1}{x}\ln(1+\sin x)} =  e^{\lim_{x\to 0} \frac{1}{x}\ln(1+\sin x)} = e^1 = e\,.
$$

We have

$$\left(1+\sin x\right)^{\frac{1}{x}}=\left[\left(1+\sin x\right)^{\frac{1}{\sin x}}\right]^{\frac{\sin x}{x}}\to e^1=e$$


Is mathematics invented or discovered? In physics for example, and in science in general, facts are "discovered" in the sense that they arise from observing nature. A particle is discovered if we can measure its existence in nature. A law is discovered if the predictions it makes are observed in nature.

Is this the case with mathematics? Is mathematics "invented", in the sense that we think up concepts and ways to relate them and all mathematical derivations are just toying with those base elements we made up, or "discovered", in the sense that there is some underlying fundamental "natural" framework we shed light on?

On one hand, one could think that mathematics is an invention, as it usually derived from a set of axioms and all we do is take those axioms as given and creatively work from there, but on the other hand, there seem to be truths akin to physical laws or constants. Take $\pi$ for example. Mathematicians in India computed the same number as Archimedes. If mathematics was just an invention by man, how could there be an agreement on this as there is with any other fundamental physical law? Same with the binomial theorem, for example. This suggests that mathematical truths are "discovered", not merely "invented".

In fact, does it even make sense to ask this? Is this an open philosophical question?

I would say, personally, that axioms are invented, and theorems are discovered (as a result of those axioms). Different people will probably have different opinions.

An analogy may be that by inventing axioms, we are planting the seed of a tree. The resulting tree that grows from that seed is the mathematical framework that is a result of our planting of the tree, but we didn't have a part in shaping each branch.

America was discovered.It was there when people lived in Europe but they didn't know about it's existence.When I was 6 years old I could certainly count up to 20.I surely counted 8 and 9.I did not know of course that 8 and 9 are the only consecutive powers but this was and is actually true.I know this is now a fact (thanks to Preda Mihailescu, 2004).It is the same thing with America.Mathematics is discovered.

One more thing:You may destroy an "invention" such as the radio but you cannot destroy the fact that 8 and 9 are consecutive powers.Even if you destroy the radio the ability to construct one from the beginning will remain alive.
Certainly Mathematics is discovered.


Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.

My methods:

$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
Let it be $0$ for some $x=k$. Then $x^2+x+1=0$ has a real solution. But since $1^2\not\gt4$, this has no real solution. Therefore it is more than $0$.

We note that:

$$x^2+x+1=\frac{x^3-1}{x-1}$$

and the sign of the RHS numerator and denominator are always equal, except for $x=1$.

We handle $x=1$ separately, but this is trivial on the LHS.


How to get the short time asymptotics of this integral? The integral is like this:

$$
\int_0^\infty \mathrm{d} x \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2}
$$

The short time asymptotics is like this (some constant maybe missing):

$$
\sim \frac{1}{\ln t}
$$

I don't know how to get this...

Edit: what's the meaning of short time?

The source I am reading says:

  At short times the integral rises or falls sharply as $\sim \frac{1}{\ln t}$

My guess that the meaning of short time is the time smaller than the first maximum or minimum of the integral.

For $x$ real we have

$$
\left| \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} \right| \leq \frac{1}{1+x^2},
$$

we know that

$$
\frac{1}{1+x^2} \in L^1([0,\infty)),
$$

and for fixed $x \geq 0$ we calculate

$$
\lim_{t \to 0^+} \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} = \frac{1}{1+x^2}.
$$

Therefore we may conclude that

$$
\lim_{t \to 0^+} \int_0^\infty dx \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} = \int_0^\infty \frac{dx}{1+x^2} = \frac{\pi}{2}.
$$

by the Dominated Convergence Theorem.

here is an idea:  i will look at the asymptotics of the simpler $$I= \int_0^\infty \frac{\cos(te^x)}{1+x^2}\, dx  \text{ for small } t. $$

observe that the graph of $y = \frac{\cos(te^x)}{1+x^2}$ oscillates between the envelopes $y = \pm \frac 1{1+x^2}$ starts at $(0,1)$ and the first $x$ -intercept is $\ln(\pi/2) - \ln(t).$ there after the intercepts are spaced by $\ln 2, \ln 3. \cdots.$

therefore $$I= \int_0^\infty \frac{\cos(te^x)}{1+x^2}\, dx=\int_0^{\ln(\pi/(2t)}\frac{\cos(te^x)}{1+x^2}\, dx +\cdots = \frac12 \times 1 \times \ln\left(\frac{\pi}{2t}\right) + \cdots$$ the estimate being the area of the triangle with base $ \ln\left(\frac{\pi}{2t}\right)$ and height $1.$


Is there a symbol for "Time Displacement" Does the "Time Displacement" property of a wave function have an associated symbol?

  Amplitude = $\hat U$

  Period = $T$

  Ordinary frequency = $f$

  Angular frequency = $ω$

  Phase = $ϕ$

  Time Displacement = ?

The term appears in my college lecture notes:

For this analogy assume $\alpha > 0$.

The author of that text is simply using the term time displacement to mean the time it would take the lagging function to achieve the same $y$ value (displacement, field magnitude etc.) as the leading function.

In that sense $\Delta t$ is not a bad choice for a symbol. Especially considering that distance displacement is commonly represented by the symbol $\Delta x$.

Physicists use $t$ for time, the variable, units second.


Two questions on continuous functions I am stuck on two questions :

If $f,g\in C[0,1]$ where $C[0,1]$ is the set of all continuous functions in $[0,1]$ then is the mapping $id:(C[0,1],d_2)\to (C[0,1],d_1)$ continuous ? where $id$ denotes the identity mapping.

where  $d_2(f,g)=(\int _0^1 |f(t)-g(t)|^2dt )^{\frac{1}{2}} $ and $d_1(f,g)=(\int _0^1 |f(t)-g(t)|dt )$
?

2.If $f\in L^2(\Bbb R)$ does it imply that $f\in L^1(\Bbb R)$.

My try:

1.The first question reduces to proving the fact that if $(\int _0^1 |f(t)-g(t)|^2dt )^{\frac{1}{2}} <\infty$ does it imply that  $(\int _0^1 |f(t)-g(t)|dt )<\infty$ which I am unable to prove.

I am also unable to conclude anything for the 2nd one.

Please give some hints

@marwalix got the first question.

For the second, consider the function $f(x) = \min\left\{1,\frac{1}{x}\right\}$.

$$\int_{\Bbb R} |f|^2 = \int_{-\infty}^{-1} \frac{dx}{x^2} + \int_{-1}^1 dx + \int_{1}^{\infty} \frac{dx}{x^2}= 1+2+1 = 4$$

but $\int_{\Bbb R}|f|$ doesn't converge.

Take $f(x)=\sqrt{x}$ one has $f\in L^2([0,1])$ and obviously $f\notin L^1([0,1])$ for the integral diverges at $0$.

Now consider a sequence of functions in $L^2$ that converges to the above $f$. If the identity were continuous the sequence should converge in $L^1$ but we have seen that $f\notin L^1$ so the identity is not continuous.


Limit of $\left(1-\frac{c f(n)}{n}\right)^n$ I know that $(1-1/n)^n \approx 1/e$, but what is the result for $\left(1-\frac{c f(n)}{n}\right)^n$, where $c$ is an arbitrary constant and $f(n)$ is an arbitarry function of $n$?

I am asking because I am having trouble understanding these derivations:
$E(x)=\binom{n}{2} (1-p)(1-p^2)^{n-2}$

Let $p=c \sqrt{\ln n / n}$

$E(x) \approx n^2/2 (1-c \sqrt{\ln n / n})(1-c^2 \ln n / n)^n$

$\approx n^2/2 e^{-c^2\ln n} \approx 1/2 n^{2-c^2}$

For $c>\sqrt 2$, then $E(x) \to 0$

Source: https://www.cs.cmu.edu/~avrim/598/chap4only.pdf, page 11

You cannot say much about an arbitrary function of $n$: the result will depend on the function.

For instance, for $f(n)=1/n$
$$
\left(1+\frac{1}{n^2}\right)^n \xrightarrow[n\to\infty]{} 1
$$
while for $f(n) = n$, of course
$$
\left(1+1\right)^n \xrightarrow[n\to\infty]{} \infty\,.
$$
and for say $f(n) = -\sqrt{n}$
$$
\left(1-\frac{1}{\sqrt{n}}\right)^n \xrightarrow[n\to\infty]{} 0\,.
$$

In your specific case, however, for $f(n) = -c^2 \ln n$,
$$
\left(1-\frac{c^2\ln n}{n}\right)^n = e^{n\ln\left(1-\frac{c^2\ln n}{n}\right)} = e^{-c^2\ln n + o(1)} = \frac{1+o(1)}{n^{c^2}} \xrightarrow[n\to\infty]{} 0\,,
$$
using that $\ln(1+u)=u+O(u^2)$ when $u\to 0$.

You have to set the following change of variable x= -n/c. Then your equation become  :
((1+1/x)^x)^-c. Because c is a constant you can pass the limit into the exponential and when x goes to infinity, you find e^-c


Quadratic non-residues While reading the book What is Mathematics? by Courant and Robbins, I've found a statement that I don't know how to prove, although it seems that it shouldn't be really difficult. Literally, they write:

  We have just seen that every quadratic residue $a$ of $p$ satisfies the congruence $a^{(p-1)/2} \equiv 1$ (mod $p$). Whithout serious difficulty it can be proved that for every non-residue $b$ we have the congruence $b^{(p-1)/2} \equiv -1$ (mod $p$).

Here, $p$ is a prime number and $a$ and $b$ are any integers not multiples of $p$. The case of quadratic residues is quite easy, you only have to apply Fermat's little theorem, but although they say that it can be proved whithout serious difficulty, I can't see how to do the case of non-residues. I've been looking on the Internet for some ideas, but my knolewdge of the theory of numbers is quite elementary (in fact, I only know some facts which are mentioned in the book: congruences, Fermat's little theorem, ...) and all the proofs I've found are out of my reach. Is there any elementary proof of the fact that $b^{(p-1)/2} \equiv -1$ (mod $p$)?

Let $b$ be a quadratic non-residue modulo a prime $p$. Let $c=b^{(p-1)/2}$. By Fermat, $c^2\equiv1\pmod p$. It follows that either $c\equiv1\pmod p$ or else $c\equiv-1\pmod p$. Now, the congruence $x^{(p-1)/2}\equiv1\pmod p$ is of degree $(p-1)/2$ and has among its solutions all the quadratic residues, of which there are exactly $(p-1)/2$; therefore, its only solutions are the quadratic residues. So, we can't have $c\equiv1\pmod p$, and we must have $c\equiv-1\pmod p$.

We have used the fact that the integers modulo a prime form a field, and over a field the number of zeros of a polynomial can't exceed the degree of the polynomial.

Note that $b^{p-1} \equiv 1 \pmod{p}$, assuming $p$ is a prime which does not divide $b$. Taking the square root of both sides, we get $b^{(p-1)/2} \equiv \pm 1 \pmod{p}$. Of course, it cannot be $1$, because it would then be a quadratic residue, so it has to be $-1$.


Found artwork probabilities A stolen piece of art is found. The probability of it being the real one is 0.70. To find out if it is the real one, two experts are asked to give their verdict. Expert 1 is correct in 80% of the cases. Expert 2 is correct in 90% of the cases. What is the probability of

expert 1 saying it is real and expert 2 saying it is fake;
the art being real if expert 1 says it is real and expert 2 says it is fake ?

My attempts

$A$: it is real
$B$:  expert 1 is correct
$C$: expert 2 is correct

So

$P(A) = 0.70$
$P(B) = 0.80$
$P(C) = 0.90$

I would say the asked probability is $P(B \cap C^c) = P(B)P(C^c) = 0.80 \times 0.10 = 0.08.$
I used this because I'm pretty sure $B$ and $C$ are independent.
I'm just gonna be honest and say I don't have a clue for this one. I would maybe try $P(A|B\cap C^c) =\frac{ P(B \cap C^c|A)}{P(A)} = \frac{0.8 \times 0.9}{0.7} = 0.80.$

I also thought about making a Venn diagram, but that didn't work. Can you guys please give some hints? I'd be very grateful!

Let's perform $1000$ trials.  The art is real in $700$ of those trials and fake in the other $300$.

If the art is real, the first expert will say it's real in $560$ of the cases.  Of those, the second expert will agree that it's real in $504$ of the cases and will declare it fake the remaining $56$ times.  In the remaining $140$ "real" cases (where the first expert is wrong), the second expert will declare the art real $126$ times and also will be fooled $14$ times.

In the $300$ cases where the art is fake, the first expert will be fooled $60$ times and will declare it fake the other $240$ times.  Of the $60$ times the first expert is fooled, the second expert also will be fooled $6$ times and will declare the art fake the other $54$ times.  The second expert will be fooled in $24$ of the "fake" cases where the first expert gets it right, and the second expert will agree with the first expert that the art is fake the other $216$ times.

So the first expert will say the art is real and the second expert will say it's fake on $56+54=110$ occasions (out of $1000$), or with probability $0.11$.  If that is their collective verdict, the art will be real $56$ times and fake $54$ times, so the likelihood that it's real in that case is $0.50909...$

Assume Expert 1 and Expert 2 do their analysis independently and give their verdicts.

Let E1 be Expert 1 and E2 be Expert 2.

P(E1 saying real) = P(A).P(E1 saying real|A) + P($A^c$).P(E1 saying real|$A^c$) = (0.7*0.8) + (0.3*0.2) = 0.62

P(E2 saying fake) = P(A).P(E2 saying fake|A) + P($A^c$).P(E2 saying fake|$A^c$) = (0.7*0.1) + (0.3*0.9) = 0.34

P(E1 saying real and E2 saying fake) = P(E1 saying real) * P(E2 saying fake) = 0.62*0.34 = 0.21

P(E1 saying real and the artwork is real) = P(A).P(E1 saying real|A)/P(E1 saying real) = (0.7*0.8)/0.62 = 0.9032

P(E2 saying fake and the artwork is real) = P(A).P(E2 saying fake|A)/P(E2 saying fake) = (0.7*0.1)/0.34 = 0.2059

P(The artwork is real and E1 saying real and E2 saying fake) = P(E1 saying real and the artwork is real) * P(E2 saying fake and the artwork is real) = 0.9032*0.2059 = 0.186


Proving $\sum_{i=1}^n 2^i = 2^{n+1} - 2$ using strong induction I just started learning proof by induction in class, but got a problem requiring proof by strong induction.

Here is the problem.

  Prove by strong induction: $$\sum_{i=1}^n 2^i = 2^{n+1} - 2$$

I've done the base, showing that the statement holds for $n=1$, $n=2$, and $n=3$. (I won't show the simple math here). For $n=k$, the statement would be $2^{k+1}-2$. But that's where I get stuck, as I'm still trying to grasp the concept of strong induction.

For $n=k+1$, do I do the following and simplify?

$$\sum_{i=1}^{k+1} 2^i = \sum_{i=1}^k 2^i + 2^{(k+1)+1} - 2$$
$$=[2^{k+1}-2]+[2^{k+2}-2]$$
$$=\text{etc}\ldots?$$

For strong induction, a proof goes something like this:

Proof of a base case: here $n= 1$ will do:

$2 = 4 - 2$.

The only thing different between "strong" and "regular" induction is how we state the inductive step:

We assume that for ALL $n_0 \leq k < n$, the theorem holds, and then use that to show it holds for $k = n$. This means we have access to ANY previous non-negative integer $k$ in that range, not just "the previous integer", $n-1$.

In this case, that means we can assume the result for all $1 \leq k < n$.

Of course, since $k = n - 1 < n$, we can use that case, too.

So we have:

$\sum\limits_{i = 1}^n 2^i = \sum\limits_{i = 1}^{n-1} 2^i + 2^n = (2^n - 2) + 2^n = \dots ?$

We have to prove, $$\sum_{i=1}^{n}2^{i}=2^{n+1}-2$$
Step 1: Setting $n=1$ in the above equality, we get $$\sum_{i=1}^{1}2^{i}=2^{1+1}-2\iff 2=4-2\iff 2=2$$ hence it holds for $n=1$

Step 2: Assuming that it holds for $n=k$, we get $$\sum_{i=1}^{k}2^{i}=2^{k+1}-2$$

Step 3: Now, setting $n=k+1$ in the given equality, we get $$\sum_{i=1}^{k+1}2^{i}=2^{k+1+1}-2 $$$$\sum_{i=1}^{k}2^{i}+2^{k+1}=2^{k+2}-2$$ $$\sum_{i=1}^{k}2^{i}=2.2^{k+1}-2-2^{k+1}$$ $$\sum_{i=1}^{k}2^{i}=2^{k+1}-2$$
Which is true from (2), hence it holds for $n=k+1$

It is clear from (1), (2) & (3) that the given equality holds for all positive integers $\color{blue}{n\geq 1}$


Arc Length of an Ellipse using integration I was thinking about what the arc length of an ellipse is, but throughout my calculations I got stuck. Here is how I approached the problem:$$$$We have an ellipse in the form:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\pm \frac{b}{a}\sqrt{a^2-x^2}$$By applying the formula of the arc length of a function, we get:$$L=4\int_0^a\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx=4\int_0^a\sqrt{\frac{a^4+(b^2-a^2)x^2}{a^2(a^2-x^2)}}dx$$Now I made a little subsitution recalling trigonometry: $$x=a\sin(u)\\dx=a\cos(u)du$$So the Integral now can be expressed as:$$L=4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2\cos^2(u)+a^2\sin^2(u)-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}\sqrt{a^2+(b^2-a^2)\sin^2(u)}du$$So we have:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+\frac{(b^2-a^2)}{a^2}\sin^2(u)}du$$
Letting $m=\frac{(b^2-a^2)}{a^2}$ we finally get:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}du$$
However, at this point I do not know any way on how to integrate this function because the $m$ is 'in the way'. Does anyone have any hints?

I suppose that you missed an $a$ somewhere and have in fact, for your last expression,$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}\,du=4a\,E(-m)$$ where appears  the complete elliptic integral.

This cannot be expressed in terms of elementary functions but some approximations are available by the great Ramanujan. In particular
$$L=\pi  \left(3 (a+b)-\sqrt{(3 a+b) (a+3 b)}\right)\tag 1$$ and  $$L=\pi  (a+b) \left(1+\frac{3\frac{
   (a-b)^2}{(a+b)^2}}{10  \sqrt{4-3\frac{
   (a-b)^2}{(a+b)^2}}}\right)\tag 2$$

Appliead to your example $(a=4,b=8)$, $(1)$ would give $4\pi \left(9-\sqrt{35}\right)\approx 38.7537 $ and  $(2)$ would give $\frac{2\pi}{55} \left(330+\sqrt{33}\right)\approx 38.3554$ while the exact value should be $16 E(-3)\approx 38.7538$.

You also could use an infinite sum formulation $$L=\pi(a+b)\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}^2 h^n \qquad \qquad \text{where} \qquad h=\frac{
   (a-b)^2}{(a+b)^2}$$ Limiting to $p$ terms, the convergence is quite fast
$$\left(
\begin{array}{cc}
 p & L_p \\
 0 & 37.69911184 \\
 1 & 38.74630939 \\
 2 & 38.75358160 \\
 3 & 38.75378361 \\
 4 & 38.75379237 \\
 5 & 38.75379285 \\
 6 & 38.75379288
\end{array}
\right)$$

This is an elliptical integral.
You can use the elliptical integral tables to look up a value for the integral- much like the standard normal distribution.


Determine whether the set $\{v_1 + v_2 - v_3, 2v_1 + 2v_3, -v_1 + v_2 - 3v_3\}$ is linearly dependent or independent. We had a question on our last test that was very similar to this and I only got $2$ points of $6$ and I want to make sure I do it right this time. Here's my solution to that one:

Let $v_1, v_2,$ and $v_3$ be three linearly independent vectors.

My teacher told me that to qualify for full credit every detail of my suloution must be presented and the logical steps that lead to my conclusion must be clear, justified, and readable.

Here's what my solution would be to the question, but i don't think it's enough.

Here's my updated solution, this should be enough right?

Thanks guys

i think you can explain what you are doing in words. you have not made clear that you understand what linear dependence of vectors mean. true, you are showing some row reduction, but why and how does it relate to the question.

one way to do is to explain how you determine linear independence by supposing $$a(v_1+v_2-v_3) + b(2v_1+2v_3) + c(-v_1 + v_2 - 3v_3) = 0 $$ rewrite as $$v_1(a+2b-c)+v_2(a+2c)+v_3(-a+2b-3c)= 0 $$ now, use the linear independence of $v_1, v_2, v_3$ to get three equations.

do you see the point?

You can find if a set is linearly dependent or independent by calculating the determinant.

If it's zero, then the set is dependent.
If it's non-zero, then the set is independent.


let $a,b,c\in \mathbb{N} $ and Quadratic equations $x^2-2ax+b=0,x^2-2bx+c=0,x^2-2cx+a=0$ has positive integer roots. find $a,b,c$ let $a,b,c\in \mathbb{N} $ and Quadratic equations $x^2-2ax+b=0,x^2-2bx+c=0,x^2-2cx+a=0\,\,$  has positive integer roots.  Find $a,b,c$

My working:
WLOG $a\ge b\ge c$
and $a^2-b={\lambda}^2,\,\,b^2-c=\mu^2, \,\,c^2-a=\gamma^2; \lambda,\mu,\gamma\in \mathbb{N}$

This is what I got.

Consider the discriminants of the quadratic equations. These are:

$4a^2-4b$

$4b^2-4c$

$4c^2-4a$

Or you know what, let's just use the quadratic formula to lay down the roots of $x$ from each equation.

$(1) \ldots x=\dfrac{2a \pm \sqrt{4a^2-4b}}{2}=a \pm \sqrt {a^2-b}$

Similarly for the rest:

$(2) \ldots x=b \pm \sqrt {b^2-c}$

$(3) \ldots x=c \pm \sqrt {c^2-a}$

So here's what we know. Those constant terms are positive. Furthermore, in order for us to talk about integer solutions at all, the discriminants must be $0$ or perfect squares. Or at least the reduced expressions in the $\sqrt{}$

$\text{Case 1: Discriminants are null}$

$a^2-b=0 \implies b=a^2$

$b^2-c=0 \implies c=b^2=a^4$

$c^2-a=0 \implies a^8-a=0$

$a(a^7-1)=0 \implies a=\{0,1\}$

So either everything is $0$ and the equations don't exist or everything is $1$ and at least we have something. Hence here's a solution:

$\{a,b,c\}=\{1,1,1\}$

$\text{Case 2: Discriminants are slightly bigger}$

In all honesty, I'm stuck here. I'll update if I can get any more insight. But I guess that's half the work anyway.

Hint. Subtract the quadratic equations from each other. You get three linear equations in $x.$ Eliminate this from the system. Now you have a system for $a,b,c.$


Find a unitary matrix given its action on a class of vectors Im studying quantum computation and have encountered an interesting problem.

I need to find an unitary matrix $U$, such that

$$U\begin{bmatrix}1 \\ 1 \\ 1\\ \vdots \\ -1 \\ \vdots \\ 1\end{bmatrix} = \frac1k\begin{bmatrix}0 \\ 0 \\ 0\\ \vdots \\ 1 \\ \vdots \\ 0\end{bmatrix}$$

where the vector is $n\times1$ and $k$ is a constant (as necessary). Essentially, every $1$ element goes to $0$, and $-1$ goes to $1$. The $-1$ element can be present anywhere in any row the column vector, and the corresponding row in the image vector should have $1$ at that place and $0$ everywhere else.

Can we solve for $U$? If yes,how? (It need not be unique)

The constraints on $U$ can be collected into the single equation $$U(\mathbb 1_n-2I_n)=\frac1kI_n$$ where $\mathbb 1_n$ is an $n\times n$ matrix of all ones. This has the unique solution $$U=\frac1k(\mathbb 1_n-2I_n)^{-1}=\frac1{2k}\left(\frac1{n-2}\mathbb1_n-I_n\right)$$ for $n\gt2$. This $U$ is real and symmetric, so $UU^*=U^2$, which is diagonal only for $n=4$, so it doesn’t seem possible in general to construct a unitary matrix that will perform this mapping.


A strictly convex polynomial is coercive if and only if it has a positive definite Hessian I have some difficulties in the following problem.
Thank you for all comments and helping.

Let $f:\mathbb{R}^n\rightarrow \mathbb{R} (n\in \mathbb{N})$ be a polynomial.
Suppose that $f$ is strictly convex, i.e., for all $x,y \in\mathbb{R}^n, \lambda \in (0,1)$ we have
$$
f(\lambda x+(1-\lambda)y)<\lambda f(x)+ (1-\lambda) f(y).
$$
Then the following statements are equivalent

(i) $f$ is coercive, i.e.,
$$
\lim_{\|x\|\rightarrow\infty}f(x)=+\infty;
$$

(ii) There exists $x^*\in \mathbb{R}^n$ such that $\nabla^2f(x^*)$ is positive definite. Moreover, the set of such points $x^*$ is a set of full measure.

This is clearly not true: p(x1,x2)=x1^4+x2^4 is strictly convex (as it is a convex and positive definite form) and coercive (as it is a positive definite form), but its Hessian at (x1,x2)=(1,0) is not positive definite.

(i) Is not true.  Counter example.  I can't think of a better one but this works. Take $ F(x) $ is the standard normal distribution,  $ F _2 (t) = \int _{-\infty} ^t F(x) dx $ is strictly convex and the $ \lim _{t \rightarrow -\infty} F _2(t) \neq + \infty $.  Although this is one dimension the absolute value should be good enough.  You could always have a random vector.

(ii)  Is True.  I'm not quite sure how the proof would go though.  Positive definite is equivalent to $ < y, \nabla ^2 f(x) y> \; \; > 0, \; \forall y \in \mathbb{R}^n $.  You may also have to use the equivalent definition of convexity $ f(x) > \; \;  < \nabla f(y), x-y> + f(y) $.  I think you would have to prove that locally in is convex and then because you let your location be any where it proves the whole system is convex.


Bump Functions in Dynamical Systems. Define

$B(x) = e^{-1/x^2}$ for $x > 0, B(x) = 0 $  otherwise.

sketch the graph of B(x);

prove that B'(0) = 0.

When $x = 0$, $B(x) = 0$. It follows that the rate of change of a constant function is 0 and hence B'(x) = 0.

Inductively prove that $B^{(n)}(0) = 0$ for all n. Conclude that $B(x)$ is a $C^{\infty}$ - function.

The base case is trivially true from the previous part of the question.
Assume that $B^{(n)}(0) = 0$ for some n > 0.
We must show that $B^{(n +1 )}(0) = 0$ is true to complete the inductive phase.

$B^{(n+1)}(0) = (B^{(n)})' (x)$

since  $B^{(n)}(0) = 0$ we can conclude by saying that the derivative of a constant function is 0. (?)

by the induction hypothesis,$B^{(n+1)}(0) = 0 $  is true. Hence by mathematical induction the result follows.

That last part i think is wrong, but my REAL issue is what comes next.

3.Modify $B(x)$ to construct a Bump function $D(x)$ on the interval $[a,b]$ that is:

$D(x) = 1 , x \in [a,b]$

$D(x) = 0 $for$ x < \alpha $and$ x > \beta,$ where $ \alpha < a $and$ \beta > b.$

$D'(x) \ne 0$ on the intervals $(\alpha,a)$ and $(b,\beta)$.

I have sketched something that resembles a bump and satisfies all of these conditions. and it looks like a bump!.

$D(x)= e^0 $for $x \in [a,b]$

$D(x) = 0 $for$ x < \alpha $and$ x > \beta,$ where $ \alpha < a $and$ \beta > b.$

$D(x) = e^{-1/x^2} $ for $x\in[\alpha,a),$

but i am unsure how to cover the interval $(b,\beta)$

Then there is the next part.

4.Use a bump function to construct a diffeomorphism $f:[a,b] \rightarrow [c,d]$ which satisfies $f'(a) = f'(b) = 1$ and $f(a) = c, f(b) = d$.

this part I have not attempted because I'm really stuck. I know it doesn't specify that I use the previous "bump" function, but I don't really know what a bump function is anyway

Thanks in advance.

Alright... let's start with part 2.

  When x=0, B(x)=0. It follows that the rate of change of a constant function is 0 and hence B'(x) = 0.

I'll try to explain why this reasoning is flawed with a different example. Consider the following function:

$$f(x)=\begin{cases}
    0 & x >0\\
    x & x = 0\\
    0 & x <0\\
  \end{cases}$$

Now, if you think about this for a second, you should realize that this is nothing more than a convoluted way of writing $f(x)=0$. If this is not immediately obvious, try drawing the graph. So, we know $f'(x)=0$, and, in particular, $f'(0)=0$.

But what happens if we try to apply your previous logic?

When $x=0$, $f(x)=x$. It follows that the rate of a change of a linear function is equal to the slope, and hence $f'(0)=1$.

The reason this doesn't work is that you are only using the information at a single point, but derivatives are defined by the limiting behavior of the function in a neighborhood around a point. To compute the derivative, we need to use this definition.

$$B'(0)=\lim_{h\rightarrow0}\frac{B(h)-B(0)}{h}=\lim_{h\rightarrow0}\frac{e^{-1/h^2}}{h}=\lim_{h\rightarrow0}\frac{1/h}{e^{1/h^2}}$$

We apply Hôpital's rule to get

$$B'(0)=\lim_{h\rightarrow0}\frac{-1/h^2}{-2(1/h^3)e^{1/h^2}}=\frac{1}{2}\lim_{h\rightarrow0}\frac{h}{e^{1/h^2}}=0$$

For the sake of completeness, we should also note that the limit approaching from the left is also zero, so the derivative is well defined.

Now, on to part 3. We again apply the definition of a limit.

$$B^{(n+1)}(0)=\lim_{h\rightarrow0}\frac{B^{(n)}(h)-B^{(n)}(0)}{h}=\lim_{h\rightarrow0}\frac{B^{(n)}(h)}{h}$$

So we need some sort of expression for $B^{(n)}$, preferably in terms of $B$.

Suppose $f(x)$ is a rational function, and suppose $B'(x)=g(x)B(x)$, for some rational function $g(x)$. (We derived $g$ in step 2, so we know this is true). We can obtain the following.

$$\frac{d}{dx}[f(x)B(x)]=f'(x)B(x)+f(x)B'(x)=[f'(x)+f(x)g(x)]B(x)$$

Since rational functions are closed under derivation, addition, and multiplication, we know $f'(x)+f(x)g(x)$ is also a rational function. Thus, in general, we know $B^{(n)}(x)=q(x)B(x)$ for some rational function $q$.

And so, for some rational function $R_1$., we know $B^{(n+1)}(0)=\lim_{h\rightarrow0}\frac{R_1(h)}{e^{1/h^2}}$

Define a new rational function $R_2(h)=R_1(1/h)$, to get $B^{(n+1)}(0)=\lim_{h\rightarrow \infty}\frac{R_2(h)}{e^{h^2}}$.

As the argument tends to infinity, any rational function is necessarily bounded by some sufficiently high power of $x$. I.e. for some $H>0$, $h>H \implies -h^{2m}< R_2(h)<h^{2m}$, for some sufficiently large $m$.

Making the substitution $\delta=h^2$, we get $\lim_{\delta\rightarrow \infty}-\frac{\delta^m}{e^\delta}\leq B^{(n+1)}(0)\leq\lim_{\delta\rightarrow \infty}\frac{\delta^m}{e^\delta}$

Repeated application of Hôpital's rule shows both bounds go to zero, as desired.

On to part 3! (your question has two part 3s. How cool is that?)

This might not be the most elegant way to solve this part, but it works.

Before worrying about arbitrary intervals, let's try to construct a bump function on $[-1,1]$.

$B(x)$ is $0$ for $x\leq0$, so $B(x-1)$ is zero for $x\leq1$, and positive everywhere else.

Thus, $B(x^2-1)$ is $0$ for $-1\leq x\leq1$, and positive everywhere else.

We know $B((2)^2-1)\approx 0.89$, so $0.5-B(x^2-1)$ is positive on some interval $[-\delta,\delta]$, $\delta>1$, negative outside $[-\delta,\delta]$, and $0.5$ on $[-1,1]$.

Therefore, we can conclude $\frac{B(0.5-B(x^2-1))}{B(0.5)}$ is $1$ on $[-1,1]$, $0$ outside some larger interval $[-\delta,\delta]$. Moreover, we know $\frac{d}{dx}B(f(x))=0$ if and only if $f(x)<0$ or $f'(x)=0$. By construction, $0.5-B(x^2-1)$ only satisfies these conditions on the interval $[-1,1]$ and outside $[-\delta,\delta]$, so we can conclude that $\frac{B(0.5-B(x^2-1))}{B(0.5)}$ has a non-zero derivative everywhere else.

Let's define a new function. $F(x)=\frac{B(0.5-B(x^2-1))}{B(0.5)}$.

Since $F(x)=1$ iff $x\in[-1,1]$, we conclude that $F(\frac{x-b}{a})=1$ iff $x\in [-a+b,a+b]$.

To find the correct pair $(a,b)$, we just need to solve the system of equations $\alpha = b-a$, $\beta = b+a$, which is trivial.

Part 4!

I don't have a full proof for this bit. I'm not sure that my construction in part 3 is necessarily the one you want.

However, $x-kF(x)$ takes $[0,2]$ to $[-k,2]$. It has slope $1$ at both endpoints. You can probably apply some suitable transformation to map arbitrary intervals to each other in the desired way.

Is this a diffeomorphism? I don't know. I suspect it might be. Maybe someone else can tell you.

I am new to M.SE and degree levels maths but i will have an attempt at this.

The induction proof seems ok.
$B^{(n+1)}(0) = (B^{(n)})'(0) $ makes sense. You should add a bit more detail in your inductive phase.

However, a more simple attempt at modifying $B(x)$, would give:

\begin{equation}
  f(x)=\begin{cases}
    1, & x \in [a,b] \text{ }.\\
    0, & x < \alpha. \text{ and  } x > \beta\\
    x, & x \in (\alpha,a)\\
    -x,& x \in (b, \beta)\\
  \end{cases}
\end{equation}

where $ \alpha < a$ and $\beta > b$

This would satisfy the conditions.

for the last part you have to know what a diffeomorphism is.

It is a function that is bijective, continuous such that its inverse is also continuous. Also the function and its inverse has to be a $C^r$. This means that the r-th derivative exists and is continuous.

Now there are some added requirements in the question. that require $f'(a) = f'(b) = 1$ and that $f(a) = c, f(b) = d$.

I would suggest thinking in terms of the exponential function. Anyone please feel free to destroy my answer.


Set of digits modulo m A set S of integers $\{s_1, s_2, . . . , s_k\}$ such that 0 ∈ S is called a set of digits
modulo m if and only if any integer b can be written as
$$b = f_nm^n + f_{n−1}m^{n−1} + · · · + f_1m + f_0$$
for some $n \geq 0$ and some $f_i$ ∈ S, moreover such expression is unique assuming
that $f_n \neq 0$.
Show that:

a) $k = m$ and the set of $s_i$ is a complete set of representatives modulo m

b) S = {0, 1, −1} is a set of digits modulo 3

c) S = {0, 2, −2} is NOT a set of digits modulo 3

I can't really seem to get anywhere with this.  I've tried representing b in a few ways using the division algorithm and then trying to convert to the expression above but this doesn't really seem to get me anywhere.  I'd appreciate any help, thanks!

1) $S$ must contain a complete representative of modulo $m$ and $k \ge n$.

Suppose not: Let $a\in \mathbb Z$ be such that $a \not \equiv s_i \mod m$ for any $s_i$ in $S$.  Then  $a = \sum_{i=0}^n s_i m^i; s_i \in S$ is impossible as $\sum_{i=0}^n s_i m^i \equiv s_0 \mod m$ but $a \not \equiv s_0 \mod m$.  So $a$ can not be represented in terms of $S$.

2) $S$ may not contain any two integers, $s \ne t$ but $s \equiv t \mod m$.

Suppose $S$ could: If $s, t \in S$ and $s \ne t$ and $s \equiv t \mod m$.  So $s = t + km$ for some k.  Let $k = \sum k_i m^i;k_i \in S$ be the unique representation of $k$ using terms of $S$.  Then $s = t + \sum k_i m^{i+1}$ is one representation of $s$ using terms of $S$.  And $s = s$ is another.  That contradicts that all representations are unique.

So 1) and 2) mean $S$ must be a completer modulo $m$ representation.

b)  Will show by induction.

We want to find $n = \sum s_i 3^i; s_i \in \{0,1,-1\}$.

So we will need $n \equiv \sum s_i 3^i \mod 3 \equiv s_0 \mod 3$.

Now $n \equiv 0, -1, 1$ so we can and we must choose $s_0 = 0, -1, 1$ so that $n \equiv s_0 \mod 3$.

Now $n - s_0 \equiv 0 \mod 3$.  Let $n_1 = n/3$.  We can, and must, repeat these steps inductively. To get $n = \sum_{i = 0}^N s_i 3^i$ being a unique representation.

c) Note if $n = \sum_{i=0}^m s_i 3^i; s_i = 0, -2,2$ then $n$ is even.  No odd number can be thusly represented.  So this fails to be a set of digits modulo $3$.

I feel we are asking the exact same problem and the rest of it goes this way

d) S = {0,1,2} is a set of digits modulo -3

e) S = {0,1,-1,2,-2} is a set of digits modulo 5

f) S = {0,1,-1,8,-8} is NOT a set of digits modulo 5

g) S={0,1,-1,17,-17} is a set of digits modulo 5

h) S={0,1,-2,(-2)^2, (-2)^3, (-2)^4, (-2)^5} is a set of digits modulo 7

I am still kind of confused about how to proceed for the rest of the question. Can someone give a hint? Thank you so much!


Equality of two binomial coefficient containing expressions Why is

  $$ \begin{align}
 &\sum_{k=0}^n(-1)^k\left[\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\right]2^{n-2k}\\
 &=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2}\\[6pt] & \end{align} $$ ?

Should'nt it be $$\begin{align}\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k}-\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k-1}2^{n-2k}\\[6pt] & \end{align} $$
I know this must be a silly question but I can't understand this...please help me out!


Show that $A$ is an invertible matrix if $\left(A+I\right)^3=0$ and find $A^{-1}$ If $A\in M_{n\times n}\left(\mathbb{R}\right)$ is such that $ (A+I)^3=0$, show that $A$ is an invertible matrix and find the inverse of $A$.

My idea was:
\begin{eqnarray*}
0&=&\left(A+I\right)^3\\
&=&A^3+3A^2I+3AI^2+I^3\\
&=&A^3+3A^2+3A+I,
\end{eqnarray*}
then
$$I=-A^3-3A^2-3A,$$
so
$$I=A\left(-A^2-3A-3I\right).$$
It follows that
$$A^{-1}=-A^2-3A-3I.$$
Now I found the inverse matrix, but how does this show that an inverse actually exists?

Your idea is good, it works. You have shown that
$$A(-A^2-3A-3I)=I,$$
which means that $-A^2-3A-3I=A^{-1}$ by definition of the inverse. This also shows that the inverse exists, because you have constructed it already.

All of you are right actually but for one minor overlook perhaps... the given relationship between A and I cannot be interpreted as characterstic polynomial but rather MINIMAL POLYNOMIAL because A is of order n rather than three. Keeping that in mind, proving invertibility of A is simple: all eigenvalues of A are non-zero; each being -1, and therefore, det(A)= Product of eigenvalues is non-zero as well and hence, inverse of A exists. As far as, construction of the inverse goes, may be we require more information, just like that in this case A has been assumed to be a 3 x 3 matrix.


If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?

So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.

We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$

$n = 9, k = 0$

$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$

$n = 9, k = 2$

$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$

$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$

$n = 9, k = 6$

$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$

$n = 9, k = 8$

$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$

Add all of these up:

$$=.64$$ so there's a 64% chance of probability?

The probability is $\frac{1}{2}$ because the last flip determines it.

Here's an analytical answer with greater emphasis on reasoning than anything specific to probability which might lend greater insight into the problem.

Consider if there was only one coin. The probability to have an even number of heads is $1\over2$, since there are two possible outcomes and we are only interested in one of them.

Now, let there be $N \gt 1$ coin tosses. The $N^{\text{th}}$ coin toss will either give a heads or tails. If the $N-1$ tosses resulted in an even number of heads the probability of the $N$ coin tosses resulting in an even number of heads is $1\over2$ since the $N^{\text{th}}$ coin toss will add either $0$ or $1$ to the number of heads from the $N-1$ tosses, and we are only concerned with the parity of the count.

The conclusion is the same for the $N-1$ coin tosses resulting in an odd number of heads.

This approach is valid as this reasoning applies to all possible values of $N$ in our given domain.

$\therefore$ The probability of $N$ coin tosses resulting in an even number of heads is $1\over2$, with $N \in \mathbb{N}$.


The probability of no cars passing within a certain time interval Let's say there is an induction loop in a road capable of counting the number of cars passing over it. By keeping a list of moments for when a car passed the detection loop, I am able to determine the average interval $\mu$ and variance  $\sigma^2$ there over. I also know the amount of intervals $N$ measured and the last moment a car passed the induction loop.

I wish to know the chance of no cars passing for a time interval since the last moment a car passed the loop. This question seems to ask for a Poisson-distribution, but these seem no to keep into account the variance of time intervals, nor do they seem to use the amount of measurements. Which method should I use to determine the probability of no car passing the detection loop within a certain interval?

known values:

$\mu$: The average interval between two cars passing the induction loop
$\sigma^2$: The variance in these intervals
$N$: the amount of interval measurements

unknown:

$P(t,c)$: The chance of $c$ cars passing the induction loop in interval $t$

where:

$t$: the interval for which the probability is sought
$c$: the amount of cars which should pass within the interval. 0 in my case.

Contrary to the existing answer, the probability does depend on the variance.

If you kept the entire information from the previous measurements, you could estimate the cumulative distribution function $F(x)$ of the time interval between car passages using the empirical distribution function. Since you only recorded the first two moments, you can approximate it using a normal distribution. The cumulative distribution function of a normal distribution with mean $\mu$ and variance $\sigma^2$ is

$$
F(x)=\frac{1}{2}\left[1 + \operatorname{erf}\left( \frac{x-\mu}{\sigma\sqrt{2}}\right)\right]\;.
$$

Given $F(x)$, the probability that no cars will pass during the interval $t$ is $1-F(t)$.

Variance is not needed, and it is perfectly fine to use Poisson formula. So your answer would be just

$$P(X=0)=e^{-\mu}$$


Why do $x^{x^{x^{\dots}}}=2$ and $x^{x^{x^{\dots}}}=4$ have the same positive root $\sqrt 2$? What is  $x$ when is satisfies $x^{x^{x^{\dots}}}=2$ ?

I am really confused with this; the root is $\sqrt{2}$, but why does the equation $x^{x^{x^{\dots}}}=4$ have the same root?

Let's first start with defining what we mean by the expression $a = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$: the infinite power tower is the limit of the recursion

$$a_{n+1} = x^{a_n},~~~~~~~ a_0 = x$$

If $x>1$ then $x^x > x$ so $a_{n}$ is monotonically increasing and converges iff it's bounded above (this is the case only if $e^{-e} \leq x \leq e^{\frac{1}{e}}\approx 1.44$; see e.g. Wiki:Tetration).  If $x=\sqrt{2}$ then it follows by induction that $a_n \leq 2$ since $a_0 = \sqrt{2} < 2$ and

$$a_{n+1} = \sqrt{2}^{a_n} \leq \sqrt{2}^2 = 2$$

Thus $a_n$ is bounded above by $2$ (and the recursion therefore converges). Taking the limit of the recursion we get that it satsify the equation $a=x^a$. As you have noted this equation has the two solutions $a=2$ and $a=4$ (which are also the only real solutions). Since the limit is well defined only one of these two solutions are valid and since $a\leq 2$ we have that the solution $a=4$ does not describe the limit of the recursion above.

Why is there an extra solution? Note that

$$\lim\limits_{n\to\infty} a_n = a \implies a = x^a$$

is a one way implication and it's not true that $a = x^a \implies \lim\limits_{n\to\infty} a_n = a$. This is similar to

$$x=1 \implies x^2 = 1 \implies x = \pm 1$$

In this simple example it's squaring that introduces an extra solution. This happens quite a lot when manipulating equations: if not all of the steps are two-way implications then we might introduce extra solutions and we must use some other means to determine which one is the right one to choose.

In general when the infinite tower converges the limit is given by

$$a = -\frac{W(-\log(x))}{\log(x)}$$

where $W$ is the (principal branch of the) Lambert function. The maximum value (for $x$ where it converges) is found for $x = e^{\frac{1}{e}}$ where we have $a = e < 4$. Thus $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = 4$ has no solutions at all in real numbers so the manipulations you do to get $x = \sqrt{2}$ are not justified.

Assuming the limit exists, then since $$x^{x^{x^{\cdots }}} = 2$$ then it follows that $$x^2 = 2$$ which has two solutions: $x = \sqrt{2} $ and $x = -\sqrt{2}$. If you're working over the reals, then you'll want to ignore $-\sqrt{2}$.

Do note that all this operates under the assumption that the limit exists, you need to prove it does before the working above is valid.


Why do actions become commutative as we reduce the 'size' of it? If we have a physical object, then infinitesimal rotations have commutative behaviour while finite rotations have anti-commutative behaviour. I find this very intriguing as to why as we take smaller and smaller actions the actions start becomes commutative whilst for large actions it's not. So, due to this phenomena rotations become vectors as when we take differential rotations.
Reference from a physics book 'Fundamentals of physics' by Resnick, Halliday walker:

This is true not just of rotations but any Lie group actions.  'Small' actions, meaning those that are near the identity, can be written as exponentials, $R=e^{X}$. For example, for rotations, $X$ is a skew-symmetric matrix.
The Baker-Campbell-Hausdorff formula states $$e^{X}e^{Y}=e^{X+Y+[X,Y]/2+\cdots}=e^{Y}e^{X}E $$ If $X$, $Y$ are small then $E=e^{[X,Y]+\cdots}$ is almost the identity, because it consists of second order terms, so the actions related to $X$, $Y$, almost commute.

This is because small rotations are relatively close to being the identity function which commutes with everything. Lets look at two dimensional rotations to get a feel for why this is.
A rotation matrix with respect to the canonical basis is of the form $\pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} = \cos \theta I + \sin \theta J$ with $J=\pmatrix{0 & -1 \\ 1 & 0}$. In this case, because $\theta$ is small, $\sin \theta$ is approximately $x$ near zero, the contribution of $J$ to the rotation is also small. We can also see that $\cos \theta$ should be close to $1$ when $\theta$ is near zero so the rotation as a whole is approximately $I$.


If $a, b$ are positive integers such that $a|b^2, b^2|a^3,....$ then $a=b$. The following is a well known problem

If $a|b^2; b^2|a^3; a^3|b^4; ....$ then $a=b$.

I started with $a$ is a prime $p$ and the conclusion immediately holds.
So I was comparing powers of primes but can't get anything useful.
 Please help.

Hint $\quad\ \forall n\in\mathbb N: \ \color{#c00}a\:\!\left(\dfrac{a}b\right)^{2n}\!\!\in\mathbb Z\,\Rightarrow \color{#c00}a\,$ is a denominator for unbounded powers of $\,\dfrac{a}b\,$ so $\,\dfrac{a}b\in\Bbb Z$

Likewise $\forall n\in\mathbb N:\ b\:\!\left(\dfrac{b}a\right)^{2n+1}\!\!\in \mathbb Z\ \ \Rightarrow\ \dfrac{b}a\in \Bbb Z.\ $ Thus $\ b\mid a\mid b\ $ so $\,a = b.$

Remark $ $ This generalizes to any Noetherian integrally-closed domain $D,\,$ e.g. any PID, since if unbounded powers of a fraction $f$ over $D$ have a common denominator then $\,f\in D\,$ (Proof)

Hint:  that is a good start.  Now can you prove it for $a=p^k$ with any prime $p$?  $b$ has to again be a power of $p$.  After that, claim that it is multiplicative, so when you factor $a$ you get the two cases you have proved.


Need to find $\lim_{n\to\infty}$ $\frac 1{(n!)^\frac 1{n}}$ I've attempted to solve this  $\lim_{n\to\infty}$ $\frac 1{(n!)^\frac 1{n}}$ via converging 1/n to zero, but somehow it's incorrect. Can someone explain what i'm doing wrong?

Hint:
$$0<\frac 1{(n!)^\frac 1{n}}=\sqrt[n]{\frac{1}{n!}}<\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}\to 0.$$
In general, we have: if $a_n\to a$, then
$$\lim_{n\to\infty}\frac{a_1+a_2+\cdots+a_n}{n}=a.$$

Use the common limit
$$\lim_{n \to \infty} \left( \frac{n^n}{n!}\right)^\frac 1n = \lim_{n \to \infty} n \times \frac{1}{(n!)^\frac1n} = e$$
It's easy to see that $$\lim_{n \to \infty} \frac{1}{(n!)^\frac1n} = \lim_{n \to \infty} \frac{e}{n} = 0$$
And that's the final answer.

It's easier to do this via Stirling's approximation, as $n! = \sqrt{2\pi n}\left( \frac ne\right)^n$ and the nth root of $1/n!$ is equal to $e/n$, which tends to 0.


How to prove $n^5 - n$ is divisible by 30 without reduction How can I prove that prove $n^5 - n$ is divisible by 30?

I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$

Now, $n(n-1)(n+1)$ is divisible by 6.

Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by 5.

My guess is using Fermat little theory but I don't know how..

By the Fermat little theorem, $n^{5} \equiv n \mod 5$ i.e. $n^{5} - n \equiv 0 \mod 5$.


Number of prime divisors of element orders from character table. From wikipedia:

  It follows, using some results of Richard Brauer from modular representation theory, that the prime divisors of the orders of the elements of each conjugacy class of a finite group can be deduced from its character table (an observation of Graham Higman).

Precisely how is this done?  I would be happy with a reference.  (In particular, I am looking for the location of the elements in a solvable group with which have the maximum number of prime divisors in their order.  If somebody has any extra information on that particular situation that would also be appreciated.)

I think I probably wrote the quoted passage in Wikipedia. If we let $\pi$ be a  prime ideal of $\mathbb{Z}[\omega]$ containing $p,$ where $\omega$ is a primitive complex $|G|$-th root of unity, then it is the case that two elements $x$ and $y$ of $G$ have conjugate $p^{\prime}$-part if and only if we have $\chi(x) \equiv \chi(y)$ (mod $\pi$) for each irreducible character $\chi$ of $G$. This is because the $\mathbb{Z}$-module spanned by the restrictions of irreducible characters to $p$-regular elements is the $\mathbb{Z}$-span of irreducible Brauer characters. The rows of the Brauer character table remain linearly independent (mod $\pi$), as
Brauer showed. Furthermore, the value (mod $\pi$) of $\chi(x)$ only depends on the $p^{\prime}$-part of $x,$ since $\eta- 1 \in \pi$ whenever $\eta$ is a $p$-power root of unity.
  We now work inductively: for any prime $p$, we can recognise elements $g \in G$ which have $p$-power order, since $g$ has $p$-power order if and only if $\chi(g) \equiv \chi(1)$ (mod $\pi$) for all irreducible characters $\chi.$ If we choose a different prime $q,$ we can recognise $q$-elements by the same procedure. But then we can recognise the elements whose orders have the form $p^{a}q^{b}.$ Such an element $h$ must have $p^{\prime}$-part $z$ which has order $q^{b},$ so we have previously identified the possible $p^{\prime}$-parts. Furthermore, $h$ has $p^{\prime}$-part $z$ if and only if $\chi(h) \equiv \chi(z)$ (mod $\pi$) for all irreducible characters $\chi.$ And then, given a different prime $r,$
we can identify all elements whose orders have the form $p^{a}q^{b}r^{c}$ in a similar manner, etc.

A formula for the order of each conjugacy class of a finite group is given in Exercise 2.21 of Fulton and Harris Representation Theory: A first Course; namely $c(g) = {|G| \over \sum_{\chi} \overline{\chi(g)}\chi(g)}$ where the sum is over the irreducible representations, and $c(g)$ is the number of elements conjugate to $g$.

This isn't exactly what you asked for, but in any particular case you could calculate the prime divisors of the orders of the elements of each conjugacy class from this by factorisation.


Unable to Understand the meaning of a term in question in book Richard Brualdi Introductory Combinatorics in Chapter Polya Counting I am self studying Combinatorics from Richard Brualdi and I am unable to understand  a term in the following exercise on page 577 of 5th edition of the book:

  Determine the symmetry group and corner symmetry group of an equilateral triangle .

I am unable to understand what author means by symmetry group and unfortunately unable to find it in the text.

While the meaning of term corner symmetry is clear that author means symmetries obtained by naming corners 1,2,3 and then rotational and reflection symmetry.

Edit 1-> Notation of Will Orrick

Can someone please tell what author means here?

I shall be really thankful.

I have used Brualdi's book in the past, but do not currently have access to a copy.  Google Books shows me three relevant excerpts from the 2nd edition.  The first is on page 554:

As a result a set of permutations which results by considering all the symmetries of a figure is automatically a permutation group.  Thus we have a corner-symmetry group, an edge-symmetry group, a face-symmetry group and so on.

Servaes has now posted a very useful answer that includes a longer excerpt containing the same text.
The second is a pair of exercises on page 590:

Determine the symmetry group and corner-symmetry group of a triangle which is neither equilateral nor isosceles.

Determine the symmetry group of a regular tetrahedron.  (Hint: There are 12 symmetries.)

The third, on page 609, is an apparent answer to an exercise:

The symmetry group contains only the identity motion.  The corner symmetry group contains only the identity permutation of the three corners.

This looks like the answer to Exercise 5 mentioned above, although the number of the problem does not correspond.
The quotation in Servaes's answer confirms that the symmetry group of a figure is the set of geometric motions that map the figure to itself.  You will have to find the relevant passages in the book that define whether these motions include rotations and reflections or rotations only.  (I would start by looking for the text quoted above.  The surrounding text may contain useful information.)  My guess, based on the hint to Exercise 6, is that it is rotations only.
I will discuss the tetrahedron because it provides a good example of why it is necessary to distinguish the symmetry group, corner-symmetry group, edge-symmetry group, and face-symmetry group, even though all four groups are isomorphic in this case.  Assuming we are concerned with rotations only,  the symmetry group of the regular tetrahedron consists of the identity motion, eight rotations of either $120^\circ$ or $240^\circ$ about the four axes joining a corner to the center of the opposite face, and three $180^\circ$ rotations about the axes joining the centers of opposite edges.  To fully describe the group, the rules for composing these motions need to be stated.  Note that a set of motions forms a group in the algebraic sense when it is closed under composition of motions and contains inverses of all motions.
One typically gives names to each motion.  Label the tetrahedron vertices with the numbers $1$ through $4$, with $2$ $3$, and $4$ in counterclockwise order as viewed from vertex $1$.  You might use $r_1$ to mean a $120^\circ$ clockwise rotation about the axis through vertex $1$ (as viewed from above).  Then the $240^\circ$ rotation about this axis would be $r_1^2$.  Similarly, you might use $f_1$ for the $180^\circ$ rotation about the axis through the center of edge $1$-$4$ and the center of edge $2$-$3$.  The relations $r_1^3=e$ and $f_1^2=e$ hold, where $e$ is the identity motion (no motion at all).  Also $r_1f_1$ (the operation of first doing $r_1$, then doing $f_1$) equals $r_2$ (the $120^\circ$ clockwise rotation about the axis through vertex $2$, viewed from a vantage point above vertex $2$).  Two other $120^\circ$ rotations, $r_3$ and $r_4$ are defined similarly, as are two other $180^\circ$ rotations, $f_2$, about the axis joining the centers of edges $2$-$4$ and $1$-$3$, and $f_3$, about the axis going the centers of edges $3$-$4$ and $1$-$2$.  Hence the symmetry group is the set of motions
$$
\{e,r_1,r_1^2,r_2,r_2^2,r_3,r_3^2,r_4,r_4^2,f_1,f_2,f_3\}
$$
with composition of motions the group operation.  By a detailed analysis, one can work out the "multiplication table" for the group, which shows how all $12$ motions compose with each other.
The corner-symmetry group would be obtained by writing down the permutations of the vertices that correspond to each of the motions listed above.  The permutations corresponding to $r_1$, $f_1$, and $r_2$ are $(23)(24)$, $(14)(23)$, and $(14)(13)$.  The symmetry group and the corner symmetry group are clearly isomorphic.  The corner-symmetry group acts on a four-element set, the set $\{1,2,3,4\}$ of vertex labels.  The symmetry group, in contrast, acts on the geometrical object itself.
One can define the edge symmetry group in a similar way by labelling edges and determining how each of the motions permutes the edges.  Let edges $1$-$4$, $2$-$4$, $3$-$4$, $2$-$3$, $1$-$3$, and $1$-$2$ be labeled $a$, $b$, $c$, $x$, $y$, and $z$.  Then $r_1$ corresponds to the permutation $(az)(ay)(bx)(bc)$, $f_1$ to the permutation $(by)(cz)$, and $r_2$ to the permutation $(ac)(ay)(bx)(bz)$.  The edge-symmetry group acts on a six-element set, the set $\{a,b,c,x,y,z\}$ of edge labels.
Likewise for the face-symmetry group, which acts on the four-element set of face labels.  If each face is labeled with the same label as the vertex opposite, the face-symmetry group is identical to the corner-symmetry group and both are identical to the alternating group $A_4$ (consisting of all even permutations) on the set $\{1,2,3,4\}$.
All four groups are, in fact, isomorphic to each other, meaning, for example, that if each element in the multiplication table of the symmetry group is replaced with the corresponding element of the corner-symmetry group, the result is the correct multiplication table for the corner-symmetry group.  This raises the question: will the permutation group associated with any set of geometrical features of the tetrahedron will be isomorphic to the symmetry group?  The answer is no, as a trivial example shows: the body-symmetry group consists only of the identity motion since the tetrahedron has only one three-dimensional "facet", which always maps to itself.  A more interesting example comes from the set of three $180^\circ$ rotation axes mentioned in connection with the rotations $f_1$, $f_2$, and $f_3$.  If these axes are called $I$, $J$, and $K$, we can ask how they permute among themselves under the $12$ geometric moves in the symmetry group.  You can check that the three $180^\circ$ rotations, as well as the identity motion, all produce the identity permutation.  The rotations $r_1$, $r_2$, $r_3$, and $r_4$ all produce the permutation $(IK)(IJ)$, while their squares all produce the permutation $(IJ)(IK)$.  These are the cyclic permutations of $I$, $J$, and $K$ and therefore the "$180^\circ$-axis-symmetry group" is isomorphic to the cyclic group of order $3$.  There is a homomorphism of the symmetry group of the tetrahedron to this group, but not an isomorphism.
I'll leave it for you to work out the symmetry group and corresponding corner symmetry group for the equilateral triangle.  Assuming only rotations are to be considered, this is a group with three elements.  The symmetry group, corner symmetry group, and edge symmetry group are all isomorphic in this case, but the face symmetry group is not, as it's the group containing only the identity permutation.

From your description, this is what I think the author means:

Symmetric group (of order 3):  All possible permutations or 3 elements.  For example if you have elements a, b, and c, then the possible permutations are abc, acb, bac, bca, cab, and cba.  Standard notation for this group is $S_3$.

ref: https://en.wikipedia.org/wiki/Symmetric_group

"Corner symmetry group" (aka Dihedral Group of a regular triangle):  All reflections and rotations of polygon (triangle) with uniquely labelled corners.  The Dihedral Group of a regular triangle would generally be known as the Dihedral Group of order 6, since there are 6 distinct results for any combination of rotation and reflection about an axis of symmetry.  Standard notation for this group is $D_3$.

ref: https://en.wikipedia.org/wiki/Dihedral_group


Negative exponent Fourier Transform of Sequences Why the exponent must be a negative in the Fourier transform of any sequence?
What happens with expressions

$$x(m)=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}X(w)\exp(jmw)dw$$
if we define the Fourier transform of sequences as:

$$X(w)=\sum_{-\infty}^{\infty}x(m)\exp(jmw)$$ tal que $-\pi\leq w \leq \pi.$

recall that Fourier Transfrom of sequence $x(n)$ is $X(w)=\sum_{-\infty}^{\infty}x(n)\exp(-jnw)$

I am found that $x(m) = x(-m)$ this is true?

The reason comes from what we want to end product to be: an approximation to the given function $x$ in the form of a trigonometric polynomial: $$x(t)\sim \sum c_k \exp(ikt) \tag1$$
The formula for coefficients $c_k$ is derived from (1), and necessarily involves the minus sign. In other words, we have the minus sign in the transform formula because we don't want to have it in  (1).

That said, there is no real  difference between $i$ and $-i$ -- the difference is  purely imaginary :). If you go through all Fourier analysis formulas changing the sign of $i$ everywhere, all statements will remain true.

As for your other question: the Fourier coefficients of a real-variable function $f$ satisfy $c_{-k}=\overline{c_k}$, as you can find by conjugating the integral that gives $c_k$.

Because Fourier wanted to make a convergent transform. If a function is convergent then it has a lot of nice proprieties very useful in math. And if you make the exponent positive, then the integral would not be convergent and you can't find those proprieties that Fourier found.


Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$. Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$

I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$
So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.
Something i’m not sure whether it’s true:
$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$
Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?

In respect of the stated comment, I add our second case (and so the whole solution) I left for OP to solve.
First, check $m=1$.
If $m≥2, m-2=k$ and $n=2n_1$, then
$$3×2^{k}+1=n_1^2, n_1≥2$$
Case $-1.$
$n_1=3a+1$, where $a≥1.$
$$3×2^k=9a^2+6a$$
$$2^k=3a^2+2a$$
$$a≥1 \Longrightarrow \begin{cases} a=2b \\ k≥3 \end{cases}$$
$$2^k=12b^2+4b$$
$$2^{k-2}=3b^2+b=b(3b+1)$$
$$b=2^x, 3b+1=2^y$$
If $x≥1$, then we get $3b+1≥7$ and $3b+1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Then, we can check $n_1=3a+2$ by the same way.
Of course, $n_1=3a$ is impossible.

The method of looking at our second case is the same as the first one, as I mentioned earlier.
Case $-2.$
$n_1=3a+2$, where $a≥0.$
First, check $a=0$. If, $a≥1$ then
$$3×2^k+1=9a^2+12a+4$$
$$2^k=3a^2+4a+1, a≥1$$
$$a≥1 \Longrightarrow \begin{cases} a=2b-1 \\ k≥3 \end{cases}$$
We have,
$$2^k=3(2b-1)^2+4(2b-1)+1, b≥1$$
$$2^k=12b^2-4b$$
$$2^{k-2}=3b^2-b=b(3b-1)$$
$$b=2^x, 3b-1=2^y$$
If $x≥1$, then we get $3b-1≥5$ and $3b-1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Finally, backward all the steps and you will find all required integer values of $m$ and $n.$

hint
It is easy to check that $ m $ satisfies
$$m\ge 2.$$
but
$$3.2^m=(n+2)(n-2)$$
with $ n $ even.
So
$$(n+2)|3.2^m $$
and by Gauss Theorem
$$(n+2)|2^m$$


Approximate the probability An immortal snail is at one end of a perfect rubber band with a length of 1km. Every day, it travels 10cm in a random direction, forwards or backwards on the rubber band. Every night, the rubber band gets stretched uniformly by 1km. As an example, during the first day the snail might advance to x=10cm, then the rubber band gets stretched by a factor of 2, so the snail is now at x=20cm on a rubber band of 2km.

The question: Approximate the probability that it will reach the other side at some point (better approximations are obviously preferred, but any bounds are acceptable as long as they are found by doing something interesting)

With the doubling interpretation, as you have written it, the snail can never reach the end:

Fix the end at $1$, start at $0$

$$X_0=0$$

$$\mbox{suppose} \ \forall n \ X_n=X_{n-1}+2^{1-n} 10^{-5} $$

This is a probability $0$ case, but moving forwards every time is clearly the best case for reaching the end anyway. In this case:

$$\lim_{n \rightarrow \infty}X_n=\frac{2}{10^5}\sum_{n=1}^{\infty} 2^{-n} = \frac{2}{10^5} \ll 1$$

I do not believe the snail will reach the end in finite time, but it may reach the end in the limit, Assuming the extreme case, the snail always moves forward, at time 0, the snail moves 10cm to the right, and is at 10cm on a 1km band. The band then stretches, so the snail is now at 20cm on a 2km band. The snail moves 10cm again and is now at 30cm on a 2km band, but then the band stretches and the snail is at 45 cm on a 3km band.

Let $i$ be a given 'time'. What we have here is that the snail moves $\frac{10}{i\cdot 100,000}$ of the bands length at time $n$, after which the band gets stretched by $\frac{i+1}{i}$, so $\textrm{Pos}(n)$, the snails position at time $n$ (more precisely, before time $n+1$: after movement and expansion) should be:
$$
\begin{align}
\textrm{Pos}(n) &= \sum_{i=1}^n \frac{i+1}{i}\frac{10}{i\cdot 100,000}\\
\textrm{Pos}(n) &= \frac{1}{10,000}\sum_{i=1}^n \frac{i+1}{i^2}
\end{align}
$$
Basically, we have a harmonic series here, which does diverge, but only at infinity, which means that the snail will reach the end with probability 1, but will never reach it at any finite observation.

Once moving backwards is included, it takes on the properties of a 1-dimensional random walk, which, given infinite time, I believe reaches the end, but given finite time won't since even moving forward always doesn't.

Edit
Ross is correct, $\textrm{Pos}(n)$ is the fraction of the band at which the snail sits, so once $H_n$ is greater than 10,000, the snail will hit the end of the band, which is about $e^{10000}$ or a very big number. I am still not sure if it ever hits in finite time when there is positive probability for moving backwards.


Problems in Algebraic Number Theory if $z$ is an element of $Q(\zeta)$ ,where $\zeta$ is some $k$ th root of unity then $z^{(1/2)}$ is an element of $Q(\zeta^{(1/2)})$ ?

This is false.  Let $\zeta = e^{2\pi i/p}$ for any prime $p$.  Since the $p$th cyclotomic field is the same as the $2p$th ($\zeta^{1/2} = -\zeta^{(p+1)/2}$), we have $\mathbb{Q}(\zeta) = \mathbb{Q}(\zeta^{1/2})$.  Taking any integer $n \not= \pm p$, $\sqrt{n} \notin \mathbb{Q}(\zeta)$ since one of $\mathbb{Q}(\sqrt{p})$ or $\mathbb{Q}(\sqrt{-p})$ is the unique quadratic subfield of $\mathbb{Q}(\zeta)$.