How big does a rectangle have to be so that while rotating it, another smaller rectangle behind it is never visible? Imagine I'm drawing a rectangle (rectSmall) of the size 1280x720.

This rectangle is static and never changes its position, size, rotation, etc.

Now I want to draw another, bigger rectangle on top of it (rectBig) that is continuously rotating around its center.

If rectBig has the same size as rectSmall, rectSmall will naturally be partly visible behind rectBig while rectBig is rotating (unless rectBig's rotation angle is 0 or 360).

My question is: How big does rectBig have to be so that rectSmall is never visible behind rectBig?

(Sorry for my English :D)

The diagonal of rectSmall must be the minimal side length of rectBig.

If the big rectangle has same centre as small rectangle,then big rectangle with minimum area to completely cover(during rotation) the small rectangle will be a square. The length of side of this square is the diameter of circle circumscribing the smaller rectangle. If you draw it, it would be clearer to you.


How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.

I have managed to solve the base case, which gives 9, which is a multiple of 3.

From here on,

I have $(n+1)((n+1)^2 + 8)$

$n^3 + 3n^2 + 11n + 9$

$n(n^2 + 8) + 3n^2 + 3n + 9$

How can I show that $3n^2 + 3n + 9$ is a multiple of 3?

You've already solved the case $n=1$, so I'll not repeat that there.

Assuming as the induction hypothesis that $n$ has the property that $3|n(n^2+8)$, we can rewrite $(n+1)((n+1)^2+8)$ to obtain
$$
\begin{split}
(n+1)((n+1)^2+8) & = (n+1)(n^2+2n+1+8)\\
& =n(n^2+8)+n(2n+1)+((n+1)^2+8)\\
& = n(n^2+8)+3n^2+3n+9
\end{split}
$$
In the latter, all the terms are divisible by $3$, hence it follows that $(n+1)((n+1)^2+8)$ is also divisible by $3$. This finishes the induction proof, so we may conclude that $n(n^2+8)$ is divisible by $3$ for all $n\geq 1$.

Quote: How can I show that $3n^2 + 3n + 9$ is a multiple of $3$? End of quote

I'm surprised at the complexity of some of the answers.

$$
3n^2 + 3n + 9 = 3(n^2 + n + 3) = (3\cdot\text{something}).
$$

A "multiple of $3$" is anything that is $3$ times something (where "something" means of course an integer).


Maximum clique in intersection graph of $3$-element subsets of a $9$-element set? How big is the largest collection of $3$-element subsets of $\{1,\ldots,9\}$ such that every pair of sets intersects nontrivially?

I have a hard time visualizing the problem, or getting a grip on it. I
have found a few lower and upper bounds in different ways, the
tightest I can get is that it is between $28$ and $44$, but every slight sharpening requires increasingly tedious work. Is there a better approach than brute-forcing
this with a computer or many hours of paperwork?

Bonus question: Is there any nice construction of such a maximal collection of subsets?

The general theorem you are looking for here is the Erdős–Ko–Rado theorem, which states that for $n \geq 2r$, the largest size of an intersecting family of $r$-element subsets of $\{1, \ldots, n\}$ is ${n-1 \choose r-1}$.


Help with a proof that a countable union of measure zero sets is measure zero. I don't know anything about measure theory, I'm studying real analysis and this showed up in the book I'm reading as a way to characterize integrable functions. The author defined that a subset $X \subset \mathbb{R}$ has measure zero if for each $\epsilon > 0$ we can find infinitely countable open intervals $I_n$ such that $X \subset \bigcup_{n=1}^{\infty}I_n$ and $\sum_{n=1}^{\infty} |I_n| < \epsilon $ where $|I|$ is the size of $I$, as in, if $I = (a,b)$, then $|I| = b - a$.

Now, the author gives the following proof that the countable union of measure-zero sets has measure zero:

"Let $Y =\bigcup_{i=1}^{\infty} X_i $, where each $X_i$ has measure zero. Now, given $\epsilon > 0 $ we can, for each $n$, write $X_n \subset \bigcup_{i=1}^{\infty} I_{n_i}$ where each $I_{n_i}$ is an open subset and $\sum_{i=1}^{\infty}|I_{n_i}| < \epsilon / 2^n$. Therefore, $Y \subset \bigcup_{n,j=1}^{\infty} I_{n_J}$ where $\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n = \epsilon$. Therefore, $m(Y) = 0$"

I'm really confused about the ending. It is very intuitive, but it's not rigorous enough for me, I wanna see formally why this holds:

$\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n$

Like maybe looking at the definition of a series, the limit of the sequence of partial sums. Any help?

Given two sequences $x_n$ and $y_n$, we have the relation
$$
x_n\leq y_n\Rightarrow \sum_n x_n\leq \sum_n y_n.
$$
Therefore, if we set $x_n=\sum^{\infty}_{j=1}|I_{n_j}|$ and $y_n=\frac{\epsilon }{2^n}$, applying the above gives
$$
\sum_n\sum^{\infty}_{j=1}|I_{n_j}|<\sum_n\frac{\epsilon}{2^n}
$$

In general, if $(X,\mathcal M,\mu)$ is a measure space and $\{E_n\}$ is a sequence of sets in $\mathcal M$, let
$$F_n=E_n\setminus\bigcup_{j=1}^{n-1}E_j. $$
Then for any $n$, $$\bigcup_{j=1}^n F_n=\bigcup_{j=1}^n E_n$$
and $F_n\cap F_m=\varnothing$ for $m\ne n$. So by countable additivity,
$$\mu\left(\bigcup_{n=1}^\infty E_n\right) = \mu\left(\bigcup_{n=1}^\infty F_n\right) = \sum_{n=1}^\infty \mu(F_n). $$

Now, as $E_n$ is the disjoint union of $E_n\cap F_n$ and $E_n\setminus F_n$, we have
$$\mu(E_n) = \mu(E_n\cap F_n) +\mu (E_n\setminus F_n) = \mu(F_n)+\mu(E_n\setminus F_n)\geqslant \mu(F_n). $$
It follows that
$$\sum_{n=1}^\infty \mu(F_n)\leqslant \sum_{n=1}^\infty \mu(E_n) $$
and since $\mu(E_n)=0$ for all $n$, that
$$\sum_{n=1}^\infty \mu(E_n)=0. $$


Why do we add 6 in BCD addition? When performing addition in BCD, if we get an invalid BCD value, we remedy this by adding 6 to our sum.

For example:

$0101 + 0110 = 1011$ (Invalid in BCD)

So, we add 6 to fix this.

$1011 + 0110 = 0001 0001$

I've been told that it is to skip the 6 invalid BCD stages, but I have a problem with this. But, if our result was 14 we would only need to skip 1 invalid BCD combination So, why do we need to add 6 specifically?

$$10_{bcd}=0001\ 0000_b=16_d$$
$$11_{bcd}=0001\ 0001_b=17_d$$
$$12_{bcd}=0001\ 0010_b=18_d$$
$$13_{bcd}=0001\ 0011_b=19_d$$
$$14_{bcd}=0001\ 0100_b=20_d$$
$$...$$
As you see, the discrepancy is always $6$. A carry from a BCD digit to the next converts ten units to a single tenth, but a BCD tenth has weight $16$ instead of $10_d$, hence the correction $6$.

Whenever you add 2 numbers (0-9), if the sum>=10, the required bcd should be:
0001, followed by 4 bits representing the last decimal digit.

Eg:-
8+8=16
1000+1000= 10000, but required BCD: 0001 0110

To obtain the binary representation of the last digit, you need to subtract ten
(i.e, 1010)

This is the same thing as adding the 2’s complement of 1010 which is, 0110( 6 )

In binary subtraction of unsigned numbers using 2’s complement method, you will always have an end carry when the result is positive.

Continuing the previous example,

10000

+0110

Ans:  1(end carry) 0110

Upon adding leading zeroes:

0001 0110, which is the required BCD


Solving a Smullyan style knight and knave problem using natural deduction. How to shorten this proof? [ EDITED, there were 2 useless lines in the deduction below]

The Smullyan style problem :

John: If (and only if) Bill is a knave, then I am a knave.

Bill: We are of different kinds.

With the understanding that knights only tell true statements and knaves only tell false statements, identify John and Bill's types.

(It therefore follows that if $J$ is the proposition that John is a knight, and John says a logical statement $X$, that $J\leftrightarrow X$ is a tautology.)

My attempt uses natural deduction; but the reasoning is rather long.

Can you think of a shorter natural deduction style proof?

What would be the quickest way to solve the problem according to you?

Below , "J" means " John is a knight" and "B" means " Bill is a knight".

Note: this solution makes Bill's statement true and John's statement false; which is coherent with their being a knight and a knave respectively.

Indeed if we have ( B & ~J) ( that is our alledged solution) John's statement is false, since this statement is equivalent to ( B <-> J) , which, in turn, is equivalent to ~ (B&~J) & ~ ( J&~B).

Here's a proof I made on http://proofs.openlogicproject.org/
  Mine is 13 statements versus your 39 (since your proof didn't go to my statement 14).  On the other hand, I stuck with a bidirectional formulation and I gather different systems do different things with $\leftrightarrow$E.  Still, make of this what you will.

I understand that this is not a natural deduction proof, but in response to benrg's comment above, I read somewhere else that you can use algebra in the Galois field of two elements GF(2) to solve things like these, and in that case if you take Z="A is a knave" then you would formalize "A asserts p" as Z + p = 1, since either Z or p are true, but not both . I don't know the details of how to scale this for difficult problems but for the OP's problem, a solution is pretty short:

Take: A = "A is a knight", B = "B is a knight", Z = "A is a knave" and Y = "B is a knave"
By definition: A+Z = B+Y = 1
Inclusive p or q is "p + q - pq = 1"
The claims give us the equations: Z + AB + YZ - ABYZ = 1 and Y + AY + BZ - ABYZ = 1
ABYZ = 0 so Z + AB + YZ = Z(1 + Y) + AB = 1 and Y + AY + BZ = Y(1 + A) + BZ = 1
Using (2) (1+Y) = B and (1+A)=Z, substituting in (5) we end up with: B(Z+A)=1 and Z(Y+B)=1
Then B=1 and Z=1, that is, A is a knave and B is a knight.


Show $\sum_{n\le x}\frac1{\sqrt n}=2\sqrt x+c+O(x^{-1/2})$ I am trying to show the asymptotic expansion for $$\sum_{n\le x}\frac1{\sqrt n}=2\sqrt x+\zeta(1/2)+O(x^{-1/2}).$$

(The exact identity of the zeta term is not important, it need only be some $c$.) To that end, I am attempting to prove the following slightly stronger theorem, which is supported by numerical evidence:

  Let $F(x)=2\sqrt x-\sum_{n\le x}\frac1{\sqrt n}$. If $0<a\le b$, then $|F(a)-F(b)|\le\frac1{\sqrt a}$.

So far, I have only been able to show the weaker statement $|F(a)-F(b)|\le\dfrac2{\sqrt{\lfloor a\rfloor}}$, which is good enough for the asymptotic statement but can clearly be improved to work for all positive reals. My method:

$$\sum_{a<n\le b}\int_{n-1}^{n}(t^{-1/2}-n^{-1/2})\,dt=\sum_{a<n\le b}\left[2\sqrt n-2\sqrt{n-1}-\frac1{\sqrt n}\right]=F(\lfloor b\rfloor)-F(\lfloor a\rfloor)$$

Since the integrand is less than $\frac1{\sqrt{n-1}}-\frac1{\sqrt{n}}$ on its domain, we have:

$$|F(\lfloor b\rfloor)-F(\lfloor a\rfloor)|\le\sum_{a<n\le b}\frac1{\sqrt{n-1}}-\frac1{\sqrt{n}}=\frac1{\sqrt{\lfloor a\rfloor}}-\frac1{\sqrt{\lfloor b\rfloor}}$$

Finally, $F$ is differentiable on $(\lfloor x\rfloor,x)$ with $F'(y)=y^{-1/2}\le\lfloor x\rfloor^{-1/2}$, hence $|F(x)-F(\lfloor x\rfloor)|\le\frac1{\sqrt{\lfloor x\rfloor}}$. Putting it all together, we have

\begin{align}
|F(b)-F(a)|&\le|F(a)-F(\lfloor a\rfloor)|+|F(b)-F(\lfloor b\rfloor)|+|F(\lfloor b\rfloor)-F(\lfloor a\rfloor)|\\
&\le\frac1{\sqrt{\lfloor a\rfloor}}+\frac1{\sqrt{\lfloor b\rfloor}}+\frac1{\sqrt{\lfloor a\rfloor}}-\frac1{\sqrt{\lfloor b\rfloor}}=\frac2{\sqrt{\lfloor a\rfloor}}.
\end{align}

How can I avoid the use of $\frac1{\sqrt{\lfloor a\rfloor}}$, which is not even defined for $a<1$?

Using Abel's summation formula we have $$\sum_{n\leq x}\frac{1}{\sqrt{n}}=\sum_{n\leq x}1\cdot\frac{1}{\sqrt{n}}=\sqrt{x}-\frac{\left\{ x\right\} }{\sqrt{x}}+\frac{1}{2}\int_{1}^{x}\frac{\left\lfloor t\right\rfloor }{t^{3/2}}dt=2\sqrt{x}-\frac{\left\{ x\right\} }{\sqrt{x}}-1-\frac{1}{2}\int_{1}^{x}\frac{\left\{ t\right\} }{t^{3/2}}dt
 $$ where $\left\lfloor t\right\rfloor
 $ is the floor function and $\left\{ t\right\}
 $ is the fractional part of $t$. So $$\left|F\left(b\right)-F\left(a\right)\right|=\left|\frac{\left\{ b\right\} }{\sqrt{b}}-\frac{\left\{ a\right\} }{\sqrt{a}}+\frac{1}{2}\int_{a}^{b}\frac{\left\{ t\right\} }{t^{3/2}}dt\right|
 \leq\left|\frac{\left\{ b\right\} }{\sqrt{b}}-\frac{\left\{ a\right\} }{\sqrt{a}}\right|+\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\leq\frac{1}{\sqrt{a}}+\frac{\left\{ a\right\} }{\sqrt{a}}\leq\frac{2}{\sqrt{a}}
 $$ since $0\leq\left\{ t\right\} <1
 $. It is not the bound that we want but we have eliminated $1/\left\lfloor a\right\rfloor
 $.

This is a special case of the following theorem:

  If $a:\Bbb R^{>0}\to\Bbb R^{\ge 0}$ is a decreasing function, and $A$ is an antiderivative of $a$, then letting $F(x)=A(x)-\sum_{n\le x}a(n)$, we have $$x\le y\implies |F(x)-F(y)|\le a(x).$$ Thus in particular, if $a\to 0$, then $F$ converges to some $c$, and $|F(x)-c|\le a(x)$.

The OP is the special case of this theorem for $a(x)=\frac1{\sqrt x}$ and $A(x)=2\sqrt x$.

Proof: Define the functions $$G_-(x)=F(x)-\{x\}a(x)=A(x)-\{x\}a(x)-\sum_{n\le x}a(n)$$ and $$G_+(x)=G_-(x)+a(x)=A(x)+(1-\{x\})a(x)-\sum_{n\le x}a(n).$$

We will show that $G_-$ is increasing and $G_+$ is decreasing, and $G_-(x)\le F(x)\le G_+(x)$. From this it follows that

$$F(y)-F(x)\le G_+(y)-G_-(x)\le G_+(x)-G_-(x)=a(x)$$
$$F(x)-F(y)\le G_+(x)-G_-(y)\le G_+(x)-G_-(x)=a(x)$$
so $|F(x)-F(y)|\le a(x)$.

Clearly $F(x)-G_-(x)=\{x\}a(x)\ge 0$, and $G_+(x)-F(x)=(1-\{x\})a(x)\ge 0$.

For fixed $k\in\Bbb N$, we note that $G_-(x)=A(x)-(x-k)a(x)-\sum_{n\le k}a(n)$ on $[k,k+1]$. On $[k,k+1)$ this is by definition, and at $k+1$ we have $$G_-(k+1)=A(k+1)-\sum_{n\le k+1}a(n)=A(k+1)-a(x)-\sum_{n\le k}a(n)$$. Thus given $k\le x\le y\le k+1$ we have
\begin{align}
G_-(y)-G_-(x)&=A(y)-A(x)-(y-k)a(y)+(x-k)a(x)\\
&\ge A(y)-A(x)-(y-x)a(y)\\
&=\int_x^y[a(t)-a(y)]\,dt\ge 0.
\end{align}

Similarly,
\begin{align}
G_+(x)-G_+(y)&=A(x)-A(y)+(k+1-x)a(x)-(k+1-y)a(y)\\
&\ge A(x)-A(y)+(y-x)a(x)\\
&=\int_x^y[a(x)-a(t)]\,dt\ge 0.
\end{align}

Thus $G_-$ is increasing and $G_+$ is decreasing on $[k,k+1]$, and by pasting together the intervals at the common points, we conclude that $G_-$ is increasing and $G_+$ is decreasing on $\Bbb R^+$.


How many 7-digit even numbers less than 3000000 can be formed using all the digits 1,2,2,3,5,5,6? I've somewhat got this question down but I'm only half way.

  How many $7$-digit even numbers less than $3,000,000$ can be formed using all the digits $1,2,2,3,5,5,6$?

So I figured that there's about $4$ possible approaches

$1$ _ _ _ _ _ $6$
$1$ _ _ _ _ _ $2$
$2$ _ _ _ _ _ $6$
$2$ _ _ _ _ _ $2$

How do I fill in the middle? I tried $5!$ and dividing out the similar factorials but I didn't get the right answer .

It appears that both previous answers missed that the number was to be even.  So the real count is 5!/2! + 5!/2!/2! + 5!/2! + 5!/2! = 210

Lets think that first digit is fixed due to which we get x __ __ __ __ __ __

Now these 6 places can be filled with 6 digits therefore total number of possibilities for the other 6 digits can be given as 6!

but here X has 2 possibilities so total no of possibilities = $$2*6!  = 1440$$

At last we have two 2's and two 5's so we get $$\frac{1440}{2!*2!} = 360$$


Find all real number satisfying $10^x+11^x+12^x = 13^x+14^x$ Find all real number $x$ satisfying $10^x+11^x+12^x = 13^x+14^x$

My Work
Dividing by $13^x$ we get
$$\left( \frac{10}{13} \right)^x +
\left( \frac{11}{13} \right)^x +
\left( \frac{12}{13} \right)^x
= 1 + \left( \frac{14}{13} \right)^x$$
The LHS is a decreasing function of $x$ and the RHS is an increasing function of $x$. So there is only one intersection in their graph. I am looking for a formal way to find the root. I know that $x=2$ works. But how to formally find this root?

As you have done, $$\left( \frac{10}{13} \right)^x +
\left( \frac{11}{13} \right)^x +
\left( \frac{12}{13} \right)^x
= 1 + \left( \frac{14}{13} \right)^x$$
has a decreasing function on the left and an increasing function on the right. So there is only one solution.

If you want to find them, just plug in some values. Plug in $x=1,2, \dots$. Now, we see $x=2$ works. $$10^2+11^2+12^2=13^2+14^2=365$$
Problems like this usually have a solution over the integers, or don't have a closed form solution at all. The OP pointed out that one might make a mistake if doing it by hand. However, plugging in certain values is helpful in most cases. This is true since it would help approximate the solution using IVT.

I doubt that there is a truly formal/analytical way to solve this equation. If there was, we could express the roots of $$2^x+3^x+4^x=5^x+6^x$$
using some well known functions. However, this kind of equation is known to have no closed form solution. We just know by plugging the values that $x$ lies somewhere between $0$ and $1$. So this is impossible.

However, you could approximate the roots using Newton's Method or a similar method.

I just solved my problem. A well known fact that

$$a^2 +(a+1)^2 + \cdots + (a+k)^2 = (a+k+1)^2 +(a+k+2)^2+ \cdots (a+2k)^2$$
Satisfies when $a = k(2k+1)$.
So in my question $a = 2(2\cdot2+1)$, So without any thinking $x=2$ is an answer.
This helps suppose the equation was -
$$55^x+56^x+57^x+58^x+59^x+60^x=61^x+62^x+63^x+64^x+65^x$$
Then it is hard to just substitute $x = 1,2,3,..$ and check? But using this fact without any thinking we can say $x=2$, as $55 = 5(5\cdot2+1)$.


check that $\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2sin^2\theta}}d\theta$ is equivalent to $-\log\sqrt{1-k^2}$ for $k\to1$. Please help check that $\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2sin^2\theta}}d\theta$ is equivalent to $-\log\sqrt{1-k^2}$ for $k\to1$.

I really have no clue how to do this. Could anyone kindly help or provide some hint?
Thanks so much!

Since $k \rightarrow 1$ try working backwards simplifying the given expression so that $k$ is a value. To check it, use the method of solving improper integrals.

I could be wrong. I would change the expression so that the denominator would be
$\sqrt{1-1{\sin{x}}^2}$. Then solve the integral using whatever method is necessary. Then compare the results?

Sorry this should be a comment rather than an answer.

P.S. I only took first year courses for Calc, so sorry in advance, if you think my explanation is not clear and lacking some other information..


different ways to arrange the word ARRANGEMENT in 5 words. The total different ways that can the letters in the word

"ARRANGEMENT"

to be arranged in 11 letters is:

$$\frac{11!}{2!\times2!\times2!\times2!}=2494800\space ways $$

But how many different ways can the letters be arranged in only 5 letters?

The $$\frac{11!}{2!^4}$$ part is already right.

To divide it in 5 words, you have to multiply by:

$${{6 + 4}\choose{4}}$$

Given that words have at least 1 letter, you have $11 - 5 = 6$ letters left to divide over 5 buckets for 0 or more letters.

This is typically done with an eggs in baskets calculation with bars: you have to place 4 bars between the eggs (letters in this case). Out of 10 positions, you have to select 4 bars.


Questions for understanding fiber bundle definition From Wikipedia:

A fiber bundle consists of the data $(E, B, π, F)$, where $E, B, $and
$F$ are topological spaces and $π : E → B$ is a continuous surjection
satisfying a local triviality condition outlined below:
for every $x$ in $E$, there is an open neighborhood $U ⊂ B$ of $π(x)$
(which will be called a trivializing neighborhood) such that
$π^{-1}(U)$ is homeomorphic to the product space $U × F$, in such a
way that $π$ carries over to the projection onto the first factor.

I was wondering why the local triviality condition (the second
paragraph) is initiated from "every $x$ in $E$"? In other words, can
it be instead initiated from $B$ as follows:

there is an open cover of $B$ such that
each open subset in the cover is homeomorphic to the product space $U × F$, in such a
way that $π$ carries over to the projection onto the first factor.

What does "the first factor" mean?

Generally, what does "a mapping carries over to another mapping onto
another thing" mean?

Thanks and regards!

(1) Yes, your formulation is easily seen to imply and be implied by (hence be equivalent to) Wikipedia's one.

(2) In this case "the first factor of $U\times F$" means simply $U$ -- the first of the two factors that were multiplied to get the product space. So the condition is that there is a $\psi: \pi^{-1}(U) \to U \times F$ which is a homeomorphism and satisfies that if $\psi(e)=\langle u,f\rangle$ then $\pi(e)=u$. One might also phrase the condition as: "... there exists $\phi: \pi^{-1}(U) \to F$ such that the map $e\mapsto \langle\pi(e),\phi(e)\rangle$ is a homeomorphism $\pi^{-1}(U) \to U \times F$.

(3) I don't think "carries over" is a technical term with a distinct meaning here, apart from "such that the obvious diagram one might draw commutes".

Tim's suggested definition of a fibre bundle is in fact not equivalent to the standard definition. For a counterexample -

Let $C_0$ be the standard Cantor discontinuum.
Let $C_n=C_0+n$ - The Cantor discontinuum translated by $n$.

Consider the following subspaces of $\mathbb{R}^2$

$$A_0=\{0\} \times C_0$$
$$A_{n,(n\geq1)}=\left\{\frac{1}{n}\right\} \times C_0$$
$$D_{n,(n\geq1)}=\left\{\frac{1}{n}\right\} \times C_{n+1}$$

Let $E$ be the union of $A_0$ and all of the $A_n$ and $D_n$.

Let $B = \{(0,0)\} \cup \left\{\left(\frac{1}{n},0\right)|n\geq1\right\}$

$E$ and $B$ are taken with the subspace topologies from $\mathbb{R}^2$.

Let $F$ be the standard Cantor discontinuum.

Now define $\pi: E \to B$ as the projection $(x,y) \mapsto x$

Note that the disjoint union of two copies of the Cantor discontinuum is
homemorphic to the Cantor discontinuum. So all fibres of $\pi$ are homeomorphic.

Then $E$,$B$,$\pi$,$F$ satisfy Tim's suggested definition of a fibre bundle.

But $\pi$ in fact is not locally trivial at $(0,0)$.


find all $ a \in \mathbb N $ such that $2^a − 1 \equiv 0 \pmod 3$ and $ 4x^2 + 1 \equiv 0 \pmod {\frac{2^a -1}{3}}$ has a solution I want to find all $ a \in \mathbb N $ such that $2^a − 1 \equiv 0 \pmod 3$  and $ 4x^2 + 1 \equiv 0 \pmod {\frac{2^a -1}{3}}$ has a solution.
I know the approach is to first prove that $a$ should be a power of 2. But I am not able to solve this question.

Partial answer, showing that the two conditions hold true when $a$ is a power of $2$.
The first requirement implies that $a$ is even. To see why, notice that $2^a \equiv (-1)^a \pmod{3}$.
For the second requirement, let $N_a = \frac{2^a-1}{3}$. We can rewrite $4x^2+1\equiv0\pmod{N_a}$ as $(2x)^2\equiv-1\pmod{N_a}$. Since $N_a$ is always odd, this is equivalent to finding values of $N_a$ for which $-1$ is a quadratic residue. We'll show that, if $a = 2^k$ for some positive integer $k$, then $-1$ is a quadratic residue mod $N_a$.
We proceed by induction. Verify that our base case, $k = 1$, is true. Now, let $k > 1$. Then $$ N_{2^k} = \frac{2^{2^k} -1}{3} = \frac{(2^{2^{k - 1}} - 1)(2^{2^{k-1}}+1)}{3} = N_{2^{k-1}}(2^{2^{k-1}} + 1). $$
By our induction hypothesis, $-1$ is a quadratic residue mod $N_{2^{k-1}}$. Furthermore, $-1$ is also a quadratic residue mod $2^{2^{k-1}} + 1$, since $$ (2^{2^{k-2}})^2 \equiv -1 \pmod{2^{2^{k-1}} + 1}. $$
By CRT, $-1$ is a quadratic residue mod $N_{2^k}$, so there is always a solution to the second equation when $a$ is a power of $2$.

COMMENT.- Noting $2^{a'}-1\equiv0\pmod3$ it is clear that $a'=2a$ so we have $\dfrac{4^a-1}{3}=\dfrac{(3+1)^a-1}{3}=3^{a-1}+a\cdot3^{a-2}+\cdots+a$ from which
$$4x^2+1\equiv0\mod{(3^{a-1}+a\cdot3^{a-2}+\cdots+a)}$$ so we have the diophantine equation $$4x^2+1=(3^{a-1}+a\cdot3^{a-2}+\cdots+a)y$$ Helped by Wolfram we have the following results till $a=8=2^3$.
$$a=2\Rightarrow 4x^2+1=5y\Rightarrow \color{red}{(x,y)=(1,1)\text{ is solution}}\\a=3\Rightarrow 4x^2+1=21y\Rightarrow \text{there are not  solution}\\a=4\Rightarrow 4x^2+1=85y\Rightarrow \color{red}{(x,y)=(19,17) \text{ is  solution}}\\a=5\Rightarrow 4x^2+1=341y\Rightarrow \text{there are not  solution}\\a=6\Rightarrow 4x^2+1=1365y\Rightarrow \text{there are not  solution}\\a=7\Rightarrow 4x^2+1=5461y\Rightarrow \text{there are not  solution}\\a=8\Rightarrow 4x^2+1=21845y\Rightarrow\color{red}{ (x,y)=(3606,2381)\text{ is   solution}}$$ The conclusion of this comment is that it is true what OP says: that $a'$ is a power of $2$. I have no disposable time to try to prove this.


Singular value decomposition of product of matrices Given SVD(A) and SVD(B) and B is a diagonal matrix, is there a way or method to construct SVD(AB) ?

Actually there is an answer, but somewhat complex.
Denote the SVD decomposition of $A$ by $U^*S^*V'$, denote $\operatorname{qr}(BU) = QR$, and denote SVD decomposition of $RS$ by $U_1^*S_1^*V_1'$ then the SVD decomposition of $BA$ is $(QU_1)S_1(VV_1)'$.

Proof: $BA= BUSV'=QRSV'=(QU_1)S_1(V_1'V') = (QU_1)S_1(VV_1)'$. $QU_1$ is unitary and $VV_1$ is unitary and $S_1$ is diagonal.

Interpretation: the singular values of $BA$ are the singular values of $RS$. So the original singular values are multiplied by the upper triangular $R$ of the qr decomposition of $BU$, i.e., $B$ after unitary transformation by right bases $U$ of $A$.

This won't be a complete answer but maybe a step in the right direction.

Consider the equivalent problem of the SVD of $BA$, where $B$ is diagonal.  This is without loss of generality because if we know the SVD of $BA$, then we know the SVD of $(AB)^T=A^TB^T$, which is of the form you mention. We will assume that $B$ has non-zero diagonal elements, such that it has full rank.

Let the SVD of A be $USV^*$, where $U$ and $V$ are unitary, $S$ non-negative diagonal, and $V^*$ is the conjugate transpose of $V$.  Further, let $B=QR$ be the unique QR decomposition of $B$ obtained by the Gram-Schmidt method such that the diagonal elements of R are positive, where $Q$ is unitary and $R$ is upper triangular.  The positivity of the diagonal elements of $R$ will be useful in what follows.

Since $R$ is upper triangular, and $S$ is diagonal, $RS$ will also be diagonal. Namely, $RS$ will be a diagonal matrix formed of the products of the diagonal elements of $R$ with the corresponding elements of $S$. Since $R$ and $S$ both have positive diagonal elements, so does their product, which we will call $D$. Thus an SVD of $BA$ is given by:

$$BA=QDV^*$$


Is it smarter to defend with one army or with two in the game Risk? I've recently played Risk with the following rules:

One player can attack another player with at most 3 armies.
The defender can defend with at most 2 armies. This is independent of the number of attacking armies.
Each player (attacker and defender) gets to throw a die for each army they have. So the attacker throws at most 3 dice and the defender throws at most two dice.
When an attack is launched and the dice are thrown, the question is which armies get defeated. To do so, the result of the attacker's dice $a_1, a_2, a_3$ and the defender's dice $d_1, d_2$ are ordered descending. They are compared pairwise ($a_1$ vs $d_1$; $a_2$ vs $d_2$; $a_3$ is ignored). The attacker wins if $a_i > d_i$, otherwise the defender wins. There will be a loss of exactly $k$ armies, where $k = \max(\text{number of attacking armies}, \text{number of defending armies})$. So $k \in \{1, 2\}$. Any combination is possible (the attacker loses 2 armies, the defender loses 2 armies, both lose 1 army for the case with $k=2$).

For the attacker, I think it always makes sense to attack with 3 armies if he can. However, suppose the attacker has $n$ armies and will keep attacking until the defender has lost or he has no armies to attack with any more. The interesting question is:

Should the defender always use two armies?

Okay, I assume:
 (1) the defender chooses # of dice before seeing what the attacker rolls,
 (2) the attacker and defender each have "lots" of armies, and
 (3) the attacker always attacks with 3 armies.

Then the defender can either defend with two or defend with one.  If the latter, we take the attacks in pairs.  The data are from my link in the comments.

Defending with two dice, the outcomes are  +2 (attacker loses 2) with probability $.29$, 0 (each loses one army) with probability $.34$, and -2 (defender loses 2) with probability $.37$.

Defending with one die twice, the outcomes are +2 with probability $.34^2=.12$, 0 with probability $.34\cdot .66+.66\cdot .34=.45$, and -2 with probability $.66^2=.44$.

Hence, it is generally better to defend with two dice.  Circumstances when it is better to defend with one are hard to construct.  One example is if you're trying to delay the game by causing lots of die rolls.  Another is if you're trying to lose on purpose (e.g. to an ally).

Some more mathy stuff on Risk can be found here.

According to
Lanchester's laws
(see https://en.wikipedia.org/wiki/Lanchester%27s_laws),
if
the forces fight
with firearms engaging each other directly
with aimed shooting from a distance,
the effectiveness of a force
is proportional to
the square of its size.

Therefore
it would be better to
have the armies as large as possible,
so the number of armies should
be as small as possible
for a fixed number
of fighters.


Dice Auction Optimal Strategy Suppose two dice are rolled, and Alice and Bob are playing an auction-style game on the sum of the two dice. The first dice roll is shown to Alice, and the second one is shown to Bob. Then, Alice proposes a bid. Bob, after hearing Alice's bid, proposes his bid. The winner of the auction (whoever bids higher) must pay the loser the value of the winning bid, and the loser pays the winner the sum of the two dice.
Example Game: Suppose Alice is shown a die roll of 4, and bids $\$10$. Bob sees a $5$, and in response, bids $\$11$. Then, Bob, the "winner" of the auction, pays Alice $\$11$ and Alice pays Bob $\$4 + \$5 = \$9$. The net gain for Alice is $2, and Bob loses the same amount in a zero-sum fashion.
I'm looking to find the Nash Equilibrium of this game.
Per the comments, I've edited background and my own thoughts on the problem so far.
Background - I've been trying to self-study game theory, and I've been reading about auction types and different types of payouts. Auction game payouts tend to have a "punishment" feature for betting excessively high, so that participants are incentivized to keep their bids relatively small. This had me thinking--what if we designed a zero-sum payout so that the winner pays the loser, but the loser pays the value of the item. I proposed this to a colleague, and while he wasn't able to solve it, he helped me reformulate the problem so that issues of randomness and incomplete information are also relevant--making this an even more interesting problem.
Progress - As mentioned in my original post, I've taken to looking at Bob's perspective. Bob is missing only the value of Alice's die, but even without that information, Bob decides if he wants to win the auction or lose the auction by betting higher or lower than Alice's bid.
If Bob decides that he wants to win the auction, then he will be paying his bid to Alice, so it's optimal to bet $a + 1$, where $a$ is Alice's bid. On the other hand, if he thinks it'd be more optimal to lose the auction, it doesn't matter what he bets--either way, he bets lower than Alice, so he gets Alice's bid, but pays Alice the sum of the two dice. For convention, we can simply say he bets $\$0$. Thus Bob only has two choices.
But, thinking about Alice's strategy is harder, and I've formulated it into a minimax problem. Define $E(a, y)$ to be the expected value of the sum of the dice given Alice's bid $a$ and the roll on Bob's die $y$. Then, we can expect Bob to bet $a + 1$ if $E(a, y) - (a + 1) > a - E(a, y)$, and 0 otherwise. Bob will always choose the larger of the LHS and the RHS, so we want to choice a strategy for Alice such that the max between the LHS and RHS is minimized.
Edit:
Here's the best solution I could come up with. Consider Bob's perspective,  where he essentially has two options: betting $\$a + 1$ and winning optimally, or betting $\$0$ and losing the auction. Note that the expected value of the sum of the two dice from Bob's perspective is $E[X+Y | a] = E[X | a] + E[Y | a] = E[X | a] + y$.
As the payouts are  structured, Bob will bid $a + 1$ iff $E[X | a] + y - (a + 1) > a - (y + E[X | a])$, and 0 otherwise. The expected value of Bob's winnings will be the max of the LHS and RHS of the inequality.
Next, we define $k = a - E[X | a]$. All 6 values of Bob's dice are equally likely, so we can say that Bob's expected earnings will be $$E[B] = \frac{1}{6}\sum_{y=1}^{6}max(k - y, -k + y - 1).$$
Alice wants to minimize $E[B]$ wrt to k, so we have a minimax problem. With some analysis, I was able to show that this value is minimized when $3 \leq k \leq 3.5$, with $E[B] = 1$. Take $k = 3$ as one strategy. Then, recalling that $k = a - E[X|a]$, one simple strategy that yields $k = 3$ is for Alice to simply bet $\$3$ more than the dice roll she sees (assuming that Bob knows this is her strategy.)
This yields an expected loss of $\$1$ for Alice each game, which seems to be supported by a Python simulation.

I'm not sure there is a "nice" solution to this problem, as even such a simple-looking problem can actually have many parameters and considerations.
First, let's assume that the bids must be integers and that if Bob and Alice bid the same, one of them is chosen to be the winner by a flip of a coin (other tie-breaking rules can be considered, it doesn't matter much).
Clearly, Alice's strategy shouldn't be pure.
If she plays pure, she reveals her information, and Bob, who has all the information now, can always make sure to bid the correct amount w.r.t to her bid and $X+Y$. Her payoff is strictly negative in this case.
Thus, Alice should play a mixed strategy, which means that for each $x\in\{1,\ldots,6\}$ she needs to decide on a distribution $p_x$ over $\{2,\ldots,12\}$, then randomize and bid accordingly.
Bob, knowing the second roll $Y$ and Alice's choice $a$, needs to estimate the first dice (Probability of $i$ is $\tfrac{p_i(a)}{p_1(a)+p_2(a)+\ldots+p_6(a)}$) and then he can calculate the expected payoff of winning the auction, the expected payoff of losing and chose the best.
Given the $p$s his strategy is quite simple, but Alice has $60$ variables to work with and optimize her payoff. I'm not sure how to do it nicely and no "obvious" strategy comes up (part of the problem: Alice's bid is both payoff relevant AND tries to conceal information from Bob).

Very nice editing of the query.  The following long-winded reaction can not in any way be
construed as an answer.  However, it is so long-winded, that it would be ridiculous to
present it as a series of comments,
Look at the situation from Alice's viewpoint, and assume that Alice rolls an
$n$, where $n \in \{1,2,3,4,5,6\}.$  My initial (blind) instinct is that Alice's
optimal strategy is to bid $(n+3)$ 50% of the time and $(n+4)$ 50% of the time, with
Bob (presumably) assuming that she will follow this strategy.
This means that at any given point, Bob doesn't know whether her bid represents
$(n+3)$ or $(n+4)$, but he assumes that it must be one of those two.  I think
the first issue to consider is whether Alice should practice deception, and if so,
how often.  An example of this would be Alice rolling a 6 and bidding 7, rather than
bidding 9 or 10.
A different
deception strategy would be that 10% of the time, Alice will pretend that her roll is
2 less than it was (unless she rolled a 1 or 2) and similarly 10% of the time, Alice will
pretend that her roll is
2 more than it was (unless she rolled a 5 or 6).  My (blind) conjecture is that
Bob can defeat any attempts at deception by pretending that Alice never tries to deceive.
This blind conjecture may well be wrong.
Remember, I have no knowledge of Nash equilibriums.  From this (ignorant) perspective, you would
have to calculate Bob's expected profit assuming that Alice never tries to deceive.
You would then have to conjure various strategies that Alice might adopt to lessen
Bob's profit.  If you can actually find a way to alter Alice's strategy so that Bob's
profit is lessened (assuming Bob sticks with the "Alice is not deceiving idea"), you then
have to consider how Bob might alter his strategy to react to a possible alteration in
Alice's strategy.
Further, you have to consider the possibility that Bob will start with one strategy,
then (perhaps) vary it based on Bob's analysis of what Alice is doing.  Similarly, you
have to consider that Alice will start with one strategy, and then (perhaps) vary it
(perhaps temporarily) based on Alice's analysis of what Bob is doing.
To get (and prove) a definitive answer (absent any knowledge of Nash equilibriums),
you may have to write a computer program to test out various strategies that Alice
&/or Bob may adopt.


Uniqueness of exponential function To my knowledge, the exponential function is the unique function satisfying

$f'=f$ and
$f(0)=1$

however, unless I've made a mistake, we have

$$\frac{\partial}{\partial x} (ax)^x = x (ax)^{x-1} a = ax (ax)^{x-1} = (ax)^x$$

and

$$(a0)^0 = 0^0 =1$$

so I feel like I must be missing something special about $e^x$.  Any pointers would be greatly appreciated.

You differentiated $x^x$ wrong. In fact,

$$  (x^x)' = (e^{x \log x})' \overset{\text{chain rule}}{=} [x \log x]' e^{x\log x} = (\log x +1 )x^x$$

The rule $[x^n]' = n x^{n-1}$ only applies when $n$ is a fixed constant.

The simplest of the mistakes in your derivation is $0^0 = 1$. That is incorrect.
$0^0$ is undefined (it is the same as $0/0$).


Find an example such that $X_n\to 0$ in probability but $X_{N(n)}\to 1$ a.s. with $N(n)\to \infty$ It is an exercise from Durrett's probability.

Find an example with $X_n\in \{0,1\}$,$X_n\to 0$ in probability, $N(n)$ increases to infinity a.s. and $X_{N(n)}\to 1$ a.s.

I was thinking that first we divide $[0,1]$ into two equal pieces and let $X_1=1_{[0,\frac{1}{2}]}$ and $X_2=1_{[\frac{1}{2},1]}$. Then we divide $[0,1]$ into four equal pieces and define $X_3=1_{[0,\frac{1}{4}]}$ and $X_4=1_{[\frac{1}{4},\frac{1}{2}]}$ and so on. So we have $$X_{2^n+m}=1_{[\frac{m}{2^n},\frac{m+1}{2^n}]} \text{ for } 1\leq m < n$$

$X_n\to 0$ in probability, but I could not find a subsequence $N(n)$ which satisfies the above condition.

If you have a convergent sequence it implies that all sub-sequences must converge at the same value, because having a sequence with sub-sequences approaching two values would imply that it isn't caunchy (i.e. the points don't get infinitesimally closer as $n\to \infty$) which means it isn't convergent. So either $X_n\to 1$ or $X_{N(n)}\to 0$ not both.


Handling a sequence in a series: $\sum_{n=1}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}$ for $a_n \rightarrow \tfrac{1}{2}$ The question is whether the series $\sum_{n=1}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}$, where  $\{ a_n \}$ is a sequence of positive numbers that converges to ½, converges absolutely or not.

My hesitation is that we know nothing about the behaviour of $a_n$ (whether it is strictly increasing/decreasing or oscillatory). Here is what I am thinking, but I'm uncertain if it is sufficiently attentive to details:

$\displaystyle \sum_{n=1}^{\infty} \frac{(a_n)^n\cos(n\pi)}{n} = \sum_{n=1}^{\infty} \frac{(a_n)^n(-1)^n}{n}$

We're interested in absolute convergence, so we consider the absolute value of that series, viz:

$\displaystyle \sum_{n=1}^{\infty} \frac{(a_n)^n}{n}$

Case I: Assume $a_n \rightarrow ½$ from above, then there exists some $n > N$ such that $a_n < r$, where $r<1$, for all $n > N$. As such,

$\displaystyle \sum_{n=N}^{\infty} \frac{(a_n)^n}{n} < \sum_{n=N}^{\infty} \frac{(r)^n}{n} <   \sum_{n=N}^{\infty} r^n$,

which is a convergent geometric series because $r<1$.

&nd then similarly for for $a_n \rightarrow ½$ from below. Now that I think of it, we could probably handle the entire thing in one case by using absolute values, right?

An approach.

Let $0<\epsilon<1$. Since $\left\{a_n\right\}_{n\geq1}$ converges to $\frac 12$, then
$$
\exists \,N\geq0, \forall n\geq N, \left|a_n-\frac 12\right|\leq \epsilon.
$$
Taking for example $\epsilon:=\frac14$,
$$
|a_n|\leq\left|a_n-\frac 12\right|+\frac 12\leq\frac 34, \quad n\geq N,
$$
and
$$
|a_n|^n\leq\left(\frac{3}{4}\right)^{n}, \quad n\geq N.
$$
Then you may write
$$
\left|\sum_{n=N}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}\right|\leq \sum_{n=N}^{\infty} \frac{1}{n}\left(\frac{3}{4}\right)^{n}
$$
But on the right handside we have the rest of a convergent series, thus on the left handside the rest of our series may be as small as we want as $N$ is great.

The initial series is absolutely convergent.

We know that $\{a_n\}$ is a sequence of strictly positive numbers that converges to $\frac{1}{2}$, so we can take the absolute value of the whole thing to get rid of the cosine term, and perform the Comparison Test with the series $${\sum_{n=1}^\infty} {a_n}^n$$ which is a geometric series for all $n > N $ for some $N$ and hence converges. We know that:
$${\sum_{n=1}^\infty} \frac{{a_n}^n}{n} \leq {\sum_{n=1}^\infty} {a_n}^n$$
and the result should fall out. Obviously not the most rigorous proof, and you would need to elaborate on this argument, but that should be the general gist of it.


Giving randomly 5 red balls and 5 blue balls to 5 kids, each kid get 2 balls. What is the expected value of number of kids got 2 different balls? Giving randomly 5 red balls and 5 blue balls to 5 kids, each kid get 2 balls. What is the expected value of number of kids got 2 different balls?

I solve it by:

Let $X_i$ be random variable that holds whether kid $i$ got different balls (i.e. red and blue), or same balls (i.e. red and red or blue and blue).
$X_i = 1 \iff$ got red and blue. $X_i = 0 \iff$ got blue and blue or red and red.
Get probabilities: $P(X_i = 0) =  red\cdot red + blue\cdot blue = \frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{1}2{} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.$

From here, $P(X_i = 1) = 1 - P(X_i = 0) = \frac{1}{2}$.
Find $E(X_i = 1) =\sum_1^5 X_i\cdot P(X_i = 1) = 5\cdot \frac{1}{2} = \frac{5}{2}$.
It's not the answer (got 0/9 on the exam sheet).
Why? What is the correct answer? Thanks in advance.

Your approach with indicator variables using linearity of expectation went exactly in the right direction. You just used the wrong probabilities. Give the first child a ball; no matter which colour it is, $5$ of the remaining $9$ balls are of the other colour, so the probability that the second ball you give them is of the other colour is $\frac59$. Thus the expected number of kids with different colours is $5\cdot\frac59=\frac{25}9$.

I solved it with some brute force but I found it to be the easiest way. If there were a higher number of balls and kids It would be better to use some algebra.

You can have $5$, $3$ or $1$ kid(s) with different colored balls. You only need to count in how many ways each can happen.

For {{R,B},{R,B},{R,B},{R,B},{R,B}} there's only one possibility.

For {{R,B},{R,B},{R,B},{R,R},{B,B}} there's $5\times 4=20$ options (5 possible kids who have the red balls, then 4 possible kids who have the blue balls).

For {{R,B},{R,R},{R,R},{B,B},{B,B}} there's $5\times 6=30$ options (5 possible kids who have different balls and $4\choose 2$ to order the other 4).

All together you have $51$ possibilities. Now let's calculate the mean of kids with different boys:

$$\mu = 5\times\frac{1}{51}+3\times\frac{20}{51}+1\times\frac{30}{51}=
\frac{5+3\times 20+30}{51}=\frac{95}{51}\\
\mu \approx 1.863
$$

Hope it helped you.


Number of $Z$ homomorphisms from $Z_{n}$ to $Z_{m}$ using tensor product It is a standard result that $\textrm{Hom}_{\mathbb{Z}}(\mathbb{Z}_{n},\mathbb{Z}_{m})=\mathbb{Z}_{(n,m)}$. Is it possible to show this using the following result:

$\mathbb{Z}_{n} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = \mathbb{Z}_{(n,m)}$

This is the answer for the "opposite direction". All tensor products and homs are over $\mathbb{Z}$.

The basic argument is based on the adjointness of Hom and tensor ( Adjointness of Hom and Tensor). For abelian groups (aka $\mathbb{Z}$-modules) we get

$Hom(M\otimes N,Q) \cong Hom(M,Hom(N,Q))$.

This is pretty clear from the universal property of tensor product: maps from $M\otimes N$  to any group are bilinear maps from $M \times N$, and those are nothing but maps from $M$ to $Hom(N,-)$.

So $Hom(\mathbb{Z}_{n} \otimes \mathbb{Z}_{m}, \mathbb{Z}_t)=Hom(\mathbb{Z}_{n}, Hom(\mathbb{Z}_{m}, \mathbb{Z}_t))=Hom(\mathbb{Z}_{n}, \mathbb{Z}_{(m,t)})=\mathbb{Z}_{(n,m,t)}$,
which is also $Hom( \mathbb{Z}_{(n,m)}, \mathbb{Z}_t)$.

Now we want to apply Yoneda embedding (see first paragraph of the accepted answer to
Can someone explain the Yoneda Lemma to an applied mathematician? ).

The question is to which category. That depends on what we are willing to assume known about tensor product of two cyclic groups. If we assume it is cyclic, we apply Yoneda to the category of cyclic groups and are done. If we assume it known that it is finite abelian (which is easy to see from the construction of tensor product as a quotient of an abelian group generated by $m\otimes n$ with each such symbol of finite order), and know that every finite abelian group is a product of cyclic ones, then using the obvious fact $Hom(X, A\oplus B)=Hom(X, A)\oplus Hom(X, B)$ we apply Yoneda embedding to the category of finite abelian groups and we are done. Finally, if we only know the tensor product is finitely generated abelian and know structure theorem for finitely generated abelian groups, then we need only additional equality $Hom(\mathbb{Z}_{n} \otimes \mathbb{Z}_{m}, \mathbb{Z}) = Hom(\mathbb{Z}_{n}, Hom(\mathbb{Z}_{m}, \mathbb{Z}))=0=Hom( \mathbb{Z}_{(n,m)}, \mathbb{Z})$, to apply Yoneda to the category of finitely generated abelian groups.

I think I know a direct answer
If you prove that R/I⊗R/J ≅ R/I+J , then u can easily reach that Zn⊗ZZm=Z(n,m) by the fact that you can write Zn as Z/nZ and Z(n,m) as Z/nZ+mZ
If you don't know how to prove R/I⊗R/J ≅ R/I+J just ask again and I'll answer... It's a little bit long and you have to use nakayama's lemma twice in it.


Does the unit sphere $S$ consist of its boundary only? I want to prove that a normed space is separable iff its unit sphere is separable. I am trying to build up the proof by analyzing unit sphere and its density.

I would like to know if unit sphere $S$ consist of its boundary only?
Because the definition of separable is given as follows:

A normed space $X$ is separable if it contains dense, countable set $A$.

Now density of set is defined as $\bar A = \partial A \cup A$.

I am not sure if the sphere contains only its boundary $ S =\partial S$?

If so, then $\partial S \cup S = S = \bar S.$

Yes, it holds $\partial S = S$.

We have $$\partial S = \overline{S} \cap \overline{X\setminus S} = S \cap \overline{X\setminus S} \subseteq S$$

since $\overline{S} = S$, because $S$ is closed.

For the converse, take $x \in S$ and $r > 0$. Note that $x \in B(x,r) \cap S$, and $$\left(1 + \frac{r}2\right)x \in B(x,r) \cap S^c$$

Therefore, every ball around $x$ intersects both $S$ and $S^c$, hence $x \in \partial S$.

We conclude $\partial S = S$.


How to calculate arc length I forgot my secondary school maths, so I need to ask to confirm.

Arc Length = Radius*(Angle In Radian)

Is it correct?

Yes. This works because $C = 2\pi R$ and, coincidentally, there are $2\pi$ radians in one full rotation. Clearly, a fraction of a full rotation produces a fraction of circumference.

To be fair, your equation is in fact the definition of radian.

It's approximately correct only because you assume that the arc is the base of a triangle... It only works where you can say that sin(θ) is approximately θ. For small angles only.
General solution using $sin^{-1}$ is also an approximation tho because of the curvature.
If it was me, I would calculate the perimeter with the given Radius and calculate the ratio corresponding to the ratio $\frac{360}{\theta}$.


How prove this inequality for $a,b,c,d$ are real numbers let $a,b,c,d$ are real numbers,show that
$$2\sqrt{a^2+c^2}+\sqrt{a^2+c^2+3(b^2+d^2)-2\sqrt{3}(ab+cd)}+\sqrt{a^2+c^2+3(b^2+d^2)+2\sqrt{3}(ab+cd)}\ge6\sqrt{|ad-bc|}$$

This problem  is creat by  China's famous mathematician hua luogeng,http://en.wikipedia.org/wiki/Hua_Luogeng
when he is child,and  Now this problem has some ugly methods,I think this inequality has nice methods,Thank you

It is the isoperimetric inequality for a triangle with vertices $0$, $x+y$, and $2x$, where $x=(a,c)$ and $y = \sqrt{3} (b,d)$.

If $P$ is the perimeter of the triangle and $A$ its area, Hua's inequality says $$|2x| + |x-y| + |x+y|  \geq 6\sqrt{\frac{A}{\sqrt{3}}} .$$ The sum on the left is $P$ and the squared inequality is $ P^2 \geq (12 \sqrt{3}) A$.

This is the correct lower bound on $P^2/A$, because equality is attained at an equilateral triangle.

In Hua's parameterization, the triangle is equilateral when $(a,c)$ is a 90 degree rotation of $(b,d)$, or $(a,c) = \pm (-d,b)$.  So it could be an inequality discovered by trying to prove the isoperimetric inequality (for triangles) in coordinates and choosing variables in which the algebra is simplified. Which is something that a very clever child might do.  But did he rediscover the isoperimetric principle on his own?


Do $L^2$ convergence and continuity imply pointwise convergence? It is said here that $L^2$ convergence and continuity imply pointwise convergence (just before paragraph $5.2$) but I can't find how to prove it. Does anyone see how ?

It's wrong : $f_n(x) = \log(n)e^{-nx}$ over $[0,1]$ is a sequence of continuous functions which converges in $L_2$ toward the zero function.
However, $f_n(0) \rightarrow +\infty$

Convergence in $L^2$ does not imply convergence everywhere but it does imply point wise convergence almost everywhere. See http://en.m.wikipedia.org/wiki/Convergence_of_Fourier_series


Well-definedness of uncomputable functions. I have been reading about Rayo's function and uncomputable functions in general, and have gotten very confused. There is apparently concern over the well-definedness of Rayo's function, but I never had any doubt that things such as the busy beaver function were well-defined. Then I read this article where it was stated that two numbers described by uncomputable functions like BB might be incomparable within a certain system of axioms - it could be independent of the system which one was larger. This was taken to mean that one might not be larger than the other at all. I took this to suggest that even a simple uncomputable function like BB(n) could be undefined in a sense.

However, it seems to me that, given two numbers, it should always be possible to decide which is bigger. It also seems like the busy beaver function should be well-defined since it is making a precise statement about machines which could theoretically be implemented in the physical world.

So, are uncomputable functions well-defined in general? What characteristics would make one well-defined or not, and is there something I'm conceptually missing in the way that I am approaching these questions? Thank you for helping me understand!

The busy-beaver-function is well-defined because every halting turing machine can only write a finite number of ones on the tape, hence there must be a champion, and $bb(n)$ is the number of ones this champion writes down.

But the busy-beaver-function is not computable because we cannot determine which turing machines halt and which do not. This (and only this) prevents us from determining the busy-beaver-champion and therefore determining $bb(n)$ in general.

"Computable" means that there is an algorithm being able for every n to either determine $f(n)$ or determine that $f(n)$ is not defined.

"Well-defined" means that there is a description guaranteeing that for every $n$ there is either only one possible value $f(n)$ or that the function is not defined.  (It is not necessary that this $f(n)$ can actually be determined , if $f(n)$ is defined.)


Calculate rectangle vertices I have a programming problem and I've forgotten some of my math rectangle formulas. I have to make a program that creates a rectangle that isn't parallel to the X and Y axis, I assume this means a diagonal rectangle in the first quadrant of a Cartesian graph.

My inputs are the following:

coordinates of the vertex closest to the X axis and the origin
angle the long side makes with relation to the X axis.
area of the rectangle
length of the long side

The outputs expected are the following:

length of the short side,
coordinates of all the rest of the rectangle's vertices.

Given these input how do I calculate the output mentioned?

Hope the following can be of some help:

For example, $x_S = h - (\dfrac {A}{L}) \sin\theta$. All other co-ordinates can be done in the similar fashion.

All the lengths of the line segments can be found and hence the x-coordinates of S, Q and R (in terms of $h, k, A, L, \theta$). The y-coordinates can also be found by drawing appropriate dotted vertical lines.

Well, to calculate the length of the rectangle's short side, you can do the area divided by the length of the rectangle's long side. Then, if you use the distance formula, you can say if the starter coordinate is (0,0) and the length of the short side is, say, 3, you can do $3 = \sqrt{(0-x_2)^2+(0-y_2)^2}$ and then do, say, $9 = (0-x_2)^2+(0-y_2)^2$. I'm not sure how to proceed from there, I'll continue thinking about it.


When does the equality hold for norm equivalence We know that for a vector $x\in \mathbb{R}^n$, its 1-norm and 2-norm satisfy that
$$\frac{1}{n}\|x\|_1\le\|x\|_2\le \|x\|_1,$$
could anyone please give me some hints that on what condition these equality holds?

Hint:
If you are given a proof for those estimates, look at them.
The proofs contain some other inequalities that you may be more familiar with.
When does equality hold in them?

A different kind of hint:
What special kind of vectors can you come up with in $\mathbb R^n$?
What are some simple nonzero vectors?

The first inequality is not optimal and thus never reached.
If you can show that $\frac{1}{\sqrt n}\|x\|_1\le\|x\|_2$, you can conclude that equality only holds for $x=0$ (or when $n=1$).
To prove this, find a vector $y$ so that $x\cdot y=\|x\|_1$ and use Cauchy-Schwarz.

The null vector $x=\vec{0}_n$ yields the equality of norms.


Taylor series expansion of log function. Hi please help me with Taylor series expansion of $\log\{\frac{a*(1-x)}{x}\}$ upto order=3.

This function has no Taylor expansion around $x_0=0$ because $f(0)$ is not defined.

Hint: the Taylor expansion formula with Lagrangian remainder is:
$$\frac {d^{n+1}f}{(n+1)!df^{n+1}} \vert _{x=\zeta}(x-x_0)^{n+1}+\sum _{i=0} ^n \frac {d^if}{i!df^i} \vert_{x=x_0}(x-x_0)^i$$
where $\zeta$ is a value between $x_0$ and $x$.


In Linear Algebra, what is a vector? I understand that a vector space is a collection of vectors that can be added and scalar multiplied and satisfies the 8 axioms, however, I do not know what a vector is.

I know in physics a vector is a geometric object that has a magnitude and a direction and it computer science a vector is a container that holds elements, expand, or shrink, but in linear algebra the definition of a vector isn't too clear.

As a result, what is a vector in Linear Algebra?

In modern mathematics, there's a tendency to define things in terms of what they do rather than in terms of what they are.

As an example, suppose that I claim that there are objects called "pizkwats" that obey the following laws:

$\forall x. \forall y. \exists z. x + y = z$
$\exists x. x = 0$
$\forall x. x + 0 = 0 + x = x$
$\forall x. \forall y. \forall z. (x + y) + z = x + (y + z)$
$\forall x. x + x = 0$

These rules specify what pizkwats do by saying what rules they obey, but they don't say anything about what pizkwats are. We can find all sorts of things that we could call pizkwats. For example, we could imagine that pizkwats are the numbers 0 and 1, with addition being done modulo 2. They could also be bitstrings of length 137, with "addition" meaning "bitwise XOR." Or they could be sets, with “addition” meaning “symmetric difference.” Each of these groups of objects obey the rules for what pizkwats do, but neither of them "are" pizkwats.

The advantage of this approach is that we can prove results about pizkwats knowing purely how they behave rather than what they fundamentally are. For example, as a fun exercise, see if you can use the above rules to prove that

  $\forall x. \forall y. x + y = y + x$.

This means that anything that "acts like a pizkwat" must support a commutative addition operator. Similarly, we could prove that

  $\forall x. \forall y. (x + y = 0 \rightarrow x = y)$.

The advantage of setting things up this way is that any time we find something that "looks like a pizkwat" in the sense that it obeys the rules given above, we're guaranteed that it must have some other properties, namely, that it's commutative and that every element has its own and unique inverse. We could develop a whole elaborate theory about how pizkwats behave and what pizkwats do purely based on the rules of how they work, and since we specifically never actually said what a pizkwat is, anything that we find that looks like a pizkwat instantly falls into our theory.

In your case, you're asking about what a vector is. In a sense, there is no single thing called "a vector," because a vector is just something that obeys a bunch of rules. But any time you find something that looks like a vector, you immediately get a bunch of interesting facts about it - you can ask questions about spans, about changing basis, etc. - regardless of whether that thing you're looking at is a vector in the classical sense (a list of numbers, or an arrow pointing somewhere) or a vector in a more abstract sense (say, a function acting as a vector in a "vector space" made of functions.)

As a concluding remark, Grant Sanderson of 3blue1brown has an excellent video talking about what vectors are that explores this in more depth.

The YouTube channel 3Blue1Brown has recently put out an amazing short series on the "Essence of linear algebra". As it so happens, the first chapter is called "Vectors, what even are they?" and is an outstanding explanation, far simpler than any of the answers above: https://www.youtube.com/watch?v=fNk_zzaMoSs

While I highly recommend just watching the video (since vectors are really best understood visually), I'll try to summarize here: vectors are simply lists of numbers--that's really it. They can be used in a geometric sense (similar to the physics sense you're already familiar with on a grid) where each number represents coordinates relative to some "axes" (formally called "basis vectors"). In the most common case with basis vectors $\hat{i}$, a 1-unit-long vector pointing directly right along the x-axis (represented as $[1, 0]$), and $\hat{j}$, a 1-unit-long vector orthogonal to $\hat{j}$ pointing up along the the y-axis (represented as $[0, 1]$), vectors are just coordinates on the plane. So in that case $[1, 1]$ is a vector pointing up and to the right from the origin to coordinate $(1, 1)$.

The video also goes into how vectors can also be seen as geometric transformations of the plane (e.g. squashing, stretching, shearing, or rotating), but that's something you really need to see to understand.


How to solve the integral $\int\frac{x-1}{\sqrt{ x^2-2x}}dx $ How to calculate  $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx $$

I have no idea how to calculate it. Please help.

we have,

$$I=\int \frac{x-1}{\sqrt(x^2 -2x)}dx----------(1)
    $$

put ,              $$ (x^2 -2x)=t $$
                      on differentiating, we get,    $$ 2xdx-2dx=dt$$

$$2dx(x-1)=dt$$
$$dx(x-1)=\frac{dt}{2}$$
substitute on equation (1) we get,

$$I=\int \frac{\frac{1}{2}dt}{\sqrt(t)} $$

$$I=1/2 \int t^-(\frac{1}{2}) dt $$

$$I=1/2 \frac{t^(-\frac{1}{2} +1)}{(-\frac{1}{2} +1)} + C $$

$$I=\frac{1}{2} \frac{t^\frac{1}{2}}{\frac{1}{2}} + C$$

now substitute the value of t,

$$I=\sqrt(x^2-2x) + C-------(2)$$

equation (2) is the required integration of equation (1)


calculating number of boolean functions I would just like to clarify if I am on the right track or not.  I have these questions:

Consider the Boolean functions $f(x,y,z)$ in three variables such that the table of values of $f$ contains exactly four $1$’s.

Calculate the total number of such functions.
We apply the Karnaugh map method to such a function $f$. Suppose that
the map does not contain any blocks of four $1$’s, and all four $1$’s
are covered by three blocks of two $1$’s. Moreover, we find that it is
not possible to cover all $1$’s by fewer than three blocks. Calculate
the number of the functions with this property.

1a: I have answered $70 = \binom{8}{4}$.

1b: I have manually drawn up Karnaugh maps and have obtained the answer $12$, but my friend has $24$.  Is there another way to do this?

Thank you

A "block of two" on a Karnaugh map corresponds to a Boolean function of the form $xy$ (complements of variables allowed).  So we are looking for Boolean functions of weight $4$ ($4$ ONEs in the truth table or on the Karnaugh map) that can be covered by $3$ "blocks of two" but not by $2$ "blocks of two", that is, the minimum sum of products expression has three terms, and not two.  The simplest form is thus
$$xy \vee xz \vee yz$$
which is the T shape referred to (see also my comment on the main question), and complementing variables gives us $2^3 = 8$ different T shapes, some of which are "wrapped around" the edges (which is perfectly acceptable).  Are there any others?  Well, let's try counting.  The principle of inclusion-exclusion tells us that

  the total weight of $4$ equals the sum of the weights of the three "blocks of two" minus the sum of the weights of the pairwise
  intersections of "blocks of two" plus the weight of the intersection of all
  three "blocks of two".

Well, two different "blocks of two" must intersect in one position because
if they intersect in both positions, they are identical, and if they
don't intersect at all, they cover all $4$ ONEs in the function. So,
all three "blocks of two" must also intersect in the same position giving
$$4 = (2+2+2) - (1 + 1 + 1) + 1$$
as the total weight, and also showing that the T shape is the only one
possible.

L shapes: --. or --' or '-- or .-- count: 8

T shapes: -'- or -.- count: 4

Z shapes: -.. or ..- count: 4

Each of these shapes is 3-across on the map, so you can shift each across by one to get another boolean function.

total: 16

If you treat the side edges as wrap-arounds you get 32


Is every even number greater than 4 the sum of a prime number with another number that belongs to the set of twin primes? Thinking about Goldbach conjecture, I have the following question:

  Is every even number greater than 4 the sum of a prime number with another number that belongs to the set of twin primes?

For example, as 31 and 17 belong to the set of twin primes,  38=31+7 and 40=17+23.

Assume the Twin Prime conjecture is True, It's equivalent to there being infinitely many numbers of form $12m$ that have a Goldbach Partition $(6m-1,6m+1)$. Now consider what happens to the values of 3 times the sum of the values of $m$,  they get 2 Goldbach partitions for $6m$ equidistant for the higher value. Only when the sums of $m$ values are even, can this continue with 3 times the sum of all $m$ used (with duplication possible) over 2 when it uses 4 indices. So the conjecture, relies on the density of the indices $m$ that sum to even values, being dense enough.

Analyzing this conjecture was really complicated task, but I will pose some constructive arguments and discuss it's consequences.

The proposed conjecture states that Every even number greater than $4$ is the sum of a prime number and another prime number that belongs to the set of twin primes

First of all, it heavily assumes the truth of Goldbach's Conjecture and then the Infinitude of twin primes.

2n = a + b partition form is the mother of two famous conjectures namely The Goldbach's Conjecture and The Twin Prime Conjecture
This article will be followed by my humble apologies for not being able to give complete exposition of my above statement, but I'll assure that we won't include it much in our further arguments.

Consider representing every Positive even Integer $2n$ as sum of $a$ and $b$ where $a,b \in{N}$. For further arguments, we will assume $a\leq{b}$ and $n>2$

Consider $a+b=44$
Define a set $T = \{3,5,7,11,13,17,19\}$ now what if $b$ is non prime for every $a\in{T}$ ?

Obviously the Conjecture will fail. For significantly larger values of $n$, prime numbers are spaced out and if we target the gap length of $2$,  they are even more spaced out.

So for very large values of $n$, we could expect that for every $a$ which assumes the value of all twin primes less than $n$ won't get $b$ such that $b$ is any prime and $a+b=2n$

Note that $a$ and $b$ can easily assume all the values of twin primes which are available in the range from $1$ to $\leq{2n-1}$

A close miss
Consider $38=a+b$ if we keep $19+19$ aside for a while, we are left with $T= \{3,5,7,11,13,17,19\}$ Except for $7$ all the other values of $a$ have assumed non prime $b$ , so it could be a really really close miss at Infinitude too, nobody knows!

Aspects of numerical evidences

Let $f(2n)$ denote the total number of ways in which we can represent $2n$ as sum of two primes. Then we have $f(12)=1 , f(18)=2, f(22)=3$ and so on.

We have very compelling numerical evidences for the Goldbach's Conjecture and roughly speaking, the value of $f(2n)$ grows as $n$ grows significantly larger. So it's a good news for all those $a$ which assumes values of twin primes less than $n$

Observe that if there are finite twin primes, then for significantly larger values of $n$ we will fail to get $b$ which is any prime, this is what our arguments and evidences suggests. So the truth of just any one of the Conjecture won't be sufficient to prove the main proposed conjecture.

In Conclusion, there are $3$ essential things which keeps the proposed conjecture alive.

Truth of Goldbach's Conjecture and the Twin Prime Conjecture
The growth of $f(2n)$
$a$ assuming the twin primes, which follows from point $1$ itself.

The most beautiful Aspect is that, in some way, both the Conjectures ( the Goldbach's and the twin prime ) suggests and support our heuristics. But even more interesting would be a Counterexample to it!


I need help with defining the integral of a marginal pdf Given: $f(x,y)=1/\pi a^2,$ when $x^2+y^2\le a^2\,(a>0)$
In order to find the marginal pdfs, I have to set up the integral. I don't understand how to go about it as the expression is composed of constants only.


Problem on Contraction mapping Here is my question :

Define $T :C[0,1] \rightarrow C[0,1] $ by $T(x)(t) = 1+\int_0 ^t
x(s)ds$. Where the metric in $C[0,1]$ is defined as $d(f,g) = max_{x \in[0.1]}|f(x) - g(x)|$.

a) Is T a contraction?

b) Now we change the space to $C[0,\frac{1}{2}]$ . Is T a contraction?

Any help will be appreciated.

Thank you.

for the first one answer is no. note that for $f=1$ we have $T(f)=1+t$ and for $g=0, T(g)=1$ so $d(f,g)=1=d(T(f),T(g))$

for the second one yes.
\begin{align}
d(T(f),T(g))&=max_{x \in[0.1]}|1+\int_0 ^t
f(s)ds- (1+\int_0 ^t
g(s)ds)|\\
 & =max_{x \in[0.1]}|\int_0 ^t
f(s)ds-
g(s)ds)| \\
 & \leq \frac{1}{2}max_{x \in[0.1]}|f(x)-g(x)| \\
 & =\frac{1}{2}d(f,g)
\end{align}


Expected number of coin tosses question To build on my understanding from this question, I was wondering what is the expected number of tosses required to obtain the sequence H-T-H-T-T-H, given that the coin is fair and independent?

Edit: The way I was thinking about it, which I am not sure makes sense. Let $e$ be the expected number of coin tosses. Then, if we get a Tail immediately, expected number is $e+1$. If we get a H and another H, then the expected number is $e+2$. For the third toss, if we have a H,T,T then it is $e+3$ and for the fourth, fifth and sixth it is $e+4$, $e+5$ and $e+6$ respectively.

So, we have the equation

$$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{64}(e+6)+\frac{1}{64}(6).$$

I get that $e = 126$ which appears to be wrong compared to @Did's Answer.

If you start with $H-H$, you don't go all the way back to the start - you have already begun your second try with the second $H$.
I think it is a little messy:
$$e=\frac12(e+1)+\frac12(f+1)\\
f=\frac12(g+1)+\frac12(f+1)\\
g=\frac12(h+1)+\frac12(e+1)\\
...
$$
then solve the collection of equations.

Probability is (1/2)^6. Approximately 64 attempts (of a sequence of 6 tosses in a row) would be required to gain the desired result.


Translating "Such that" when writing statements in symbolic form. (Discrete math) So im confused as to how to do the following question:

  A: “If c is a prime number such that c divides ab, then c divides a or
  c divides b”

  B: “If c is a prime number such that c divides ab, and c does not
  divide b, then c divides a”

  (a) Write the statements A and B in symbolic form and then show that
  they are logically equivalent.

Here is what I have tried to do:

Let p be "C is a prime number such that C divides ab"

Let q be "C divides a"

Let r be "C divides b"

So I get the equation:

$$
p \rightarrow (q \vee r) \Longleftrightarrow p \wedge (\neg q \rightarrow r)
$$

But these equations dont appear to be logically equivalent.  I believe that I might be interpreting "such that" incorrectly?

Thanks

We have $A: p\rightarrow(q\vee r)$ and $B: (p\wedge\neg r)\rightarrow q$. Therefore $$p\rightarrow(q\vee r)\Leftrightarrow (p\wedge\neg r)\rightarrow q$$

$\bullet$1:I would prefer to see more brackets. Give $p\to (q\lor r)\iff p \land (\neg p\to r)$ to a computer. It might not interpret it as $(p\to (q\lor r))\iff (p\land ((\neg p)\to r))). $ $\bullet$2:By definition, $x\to y$ is equivalent to $(\neg x)\lor y.$ So $((\neg q)\to r))\iff ((\neg \neg q)\lor r)\iff (q\lor r).$ So your final "equation" (you should say "formula" ) is true. $\bullet$3: Sometimes "such that" is written as  brackets, especially when there is a preceding logical quantifier "$\exists$" or :"$\forall$". E.g.: "There exists a  real $r$ such that $r^2=2$" is $\exists r\in R\;(r^2=2).$ This can also be re-written so that "such that" becomes "and". Consider that $\exists r\in R\; (r^2=2)$ is short for $\exists x\;(x\in R \land x^2=2).$


If $L$ is a normal subgroup of $G$ and if $L=HK$ (where $H$ is normal in $G$ and $K$ is any subgroup), then must $K$ also be normal? I think the answer is yes, but I just want to ask in case I'm missing something...

We know that $aKH=KHa$ for all a in G. But we also know that $aH = Ha$, so $KHa = KaH$.

So $aKH = KaH \implies aK=Ka$.

Thanks in advance

This is false. Take G=L=H to be nonabelian of order 6, and K of order 2.

Notice that aKH = aKG = G, and that KaH = KaG = G, so aKH = KaH only tells you G = G, nothing about K or a. You cannot generally cancel subsets.

So $aKH=KaH  \ then  K^{-1}a^{-1}Ka\in H$\
It seems this way cannot be Ok. you can use this way :
$K=H^{-1}L$ is normal subgroup ,because $H\  L$ are both normal .


Transition function of the generalized nondeterministic finite automata Why does the transition function of a GNFA map from two states to a regular expression
$$
\delta: (Q \setminus \{q_{start}\}) \times (Q \setminus \{q_{accept}\}) \to R $$
instead of mapping from the current state and the regular expression to the next state?

It seems to be part of a normalization for the sake of convenience.
The traditional definition would work as well.

Citing from [HanWood2004] "The Generalization of Generalized Automata:
Expression Automata"

  Definition 2. An expression  automaton A is  specified  by  a  tuple
  $(Q, \Sigma, \delta, s, f )$ , where $Q$ is a finite set of states, $\Sigma$ is an input
  alphabet, $\delta \subseteq Q \times R_\Sigma \times Q$ is a finite set of expression transitions,
  where $R_\Sigma$ is the set of all regular expressions over $\Sigma$,
  $s \in Q$ is the
  start state and $f \in Q$ is the final state. (Note that we need only have
  one final state.) We require that, for every pair $p$ and $q$ of states,
  there is exactly one expression transition $(p, \alpha, q ) \in δ$ , where $\alpha$ is a regular expression over $\Sigma$ .

So this early GNFA definition starts with the traditional specification of the transition function as relation from state $p$ via regular expression $\alpha$ to state $q$.

The next paragraph in that paper is (boldface added by me):

  We can also use the functional notation $\delta : Q × Q \to R_\Sigma$ that gives the equivalent representation.
  An expression transition $( p, α, q )$
  gives $\delta( p, q )= \alpha$ . Note that $\delta$ contains
  exactly $|Q|^2$
  transitions, one transition for each pair of states, and whenever
  $( p,\emptyset,q )$ is in $\delta$, for some $p$ and $q$ in $Q$ , $A$ cannot move from $p$ to $q$ directly.

We note that this functional notation ($\delta$ is a function, not a relation) only allows for one expression transition from $p$ to $q$ via the regular expression $\alpha$ simply because functions are single-valued.
Thus the requirement at the end of definition 2 above can be dropped.

This convenience might explain the use of the functional notation.
To cite [Sipser2006] p. 70:

  For convenience we require that GNFAs always have a special form that meets the following conditions.

  The start state has transition arrows going to every other state but    no arrows coming in form any other state.
  There is only a single accept state, and it has arrows coming in from every other state but no arrows going to any other state.
  Furthermore, the accept state is not the same as the start state.
  Except for the start and accept states, one arrow goes from every state to every other state and also from each state to itself.

References:

[HanWood2004] Yo-Sub Han and Derick Wood. "The Generalization of Generalized Automata: Expression Automata." In: 9th International Conference on Implementation and Application of Automata, CIAA 2004, Kingston, Canada, July 22–24, 2004, Revised Selected Papers, LNCS 3317, pp. 156–166.

[Sipser2006] Michael Sipser. 2006. Introduction to the Theory of Computation (2nd ed.). International Thomson Publishing.


Context Free Grammar How can a context free grammar be generated for the following language:

$$\{a^ib^jc^k : i = j + k\}$$

I assume that any production rule that places a $b$ or $c$ must also place an $a$ but I don't know how to do this while maintaining order.

Start generating $c$s and $a$s, by say the rule $S \to aSc$ ($S$ the start variable), we need a possibility to switch to $b$s, by say $S \to T$. The rule $T \to aTb$ will generate $a$s and $b$s, and $T\to \epsilon$ (the empty word) will allow us to stop. So our grammar is $\{S \to aSc|T, T \to aTb|\epsilon\}$.

You can just reduce that to a single production as S -> aSc | aSb | ab

I hope I am correct.


A matrix involving distances of $n$ points in $\mathbb{R}^3$ Let $x_1,\ldots ,x_n$ be $n$ distinct points in $\mathbb{R}^3$. Consider the $n\times n$ real symmetric matrix $A$ defined by $A_{ij}:=|x_i-x_j|$. I would like to show that

$$Ker\,A\;\cap\,\{v\in\mathbb{R}^n\,:\, v_1+v_2 +\ldots +v_n=0\}=\{0\}$$

Thank you for any suggestions.

$\def\RR{\mathbb{R}}$Numerical experiments suggest that this matrix is negative semidefinite on the plane $\sum v_i=0$. Specifically, I generated 20 sets of 10 points each, drawn uniformly from the unit cube, and this was true every time. I repeated the experiment with points in 2, 4 and 5 dimensions, and the same held.

I am reminded of this answer by Noam Elkies but can't make a precise connection.

Switching this answer to CW to write up darij's proof from the comments. We will show that:

If $x_i$ are any $n$ points in $\RR^d$ and $v_i$ are any scalars with $\sum v_i=0$, then $\sum v_i v_j |x_i-x_j| \leq 0$ and
If the $x_i$ are distinct and the $v_i$ are not all $0$, we have strict inequality.

The latter shows that the matrix $\left[ |x_i-x_j| \right]$ times the vector $\left[ v_i \right]$ is nonzero. We start, however, by proving the former. We turn to the issue of when zero occurs after the horizontal line.

We start with the averaging trick: By rotational and scaling invariance, we can see that there is some positive $c$ such that
$$\int_{|w|=1} \left| \langle w \cdot x \rangle \right| = c |x|.$$
So
$$\sum v_i v_j |x_i-x_j| = c^{-1} \int_{|w|=1} \sum v_i v_j \left| \langle w \cdot (x_i-x_j) \rangle \right|$$
and thus it is sufficient to show $\sum v_i v_j \left| \langle w \cdot (x_i-x_j) \rangle \right|\leq 0$ for a particular vector $w$. Now, $w \cdot (x_i-x_j)$ only depends on the orthogonal projections of $x_i$ and $x_j$ onto the line $\RR w$, so we may (and do) assume all the $x_i$ lie on a line. Our goal is now to show, for any $n$ values $x_i \in \RR$, that $\sum v_i v_j |x_i-x_j| \leq 0$.

We have $|z| = \max(z,0) + \max(-z,0)$, so $\sum v_i v_j |x_i-x_j|=2 \sum v_i v_j \max(x_i-x_j,0)$.
We use the notation $\left[ \mbox{statement} \right]$ to mean $1$ if the statement is true and $0$ if it is false. So
$$\max(x_i-x_j,0) = \int_{t \in \RR} \left[x_j < t < x_i \right] dt$$
and
$$\sum_{i,j} v_i v_j \max(x_i-x_j,0) = \int_{t \in \RR} \sum_{i,j} v_i v_j \left[x_j < t < x_i \right] dt.$$ So it is enough to show that, for any $t$, we have
$$\sum_{x_i < t < x_j} v_i v_j \leq 0 . $$
Let $I = \{ i : x_i < t \}$ and $J = \{ i : x_j > t \}$. (For almost all $t$, none of the $x_i$ equal $t$, so we can disregard the boundary case.) Then
$$\sum_{x_i < t < x_j} v_i v_j  = \sum_{i \in I,\ j \in J} v_i v_j = \left( \sum_{i \in I} v_i \right) \left( \sum_{j \in J} v_j \right) = - \left(\sum_{i \in I} v_i \right)^2 \leq 0 .$$
In the final equality, we have finally used the hypothesis $\sum v_k=0$.

Now, let's consider what happens for distinct $x_i$. As long as the $x_i$ as distinct, for almost as $w$, the orthogonal projections of the $x_i$ onto $\RR w$ will stay distinct. In order to get $0$, we must get $0$ from all these choices of $w$. Let's say the orthogonal projections are ordered as $x_1 < x_2 < \cdots < x_n$. Once again, we must get $0$ from every choice of $t$. So we must have $\left( \sum_{i=1}^k v_i \right)^2 =0$ for all $k$, and this means that all $v_i$ are zero.


Check if line intersects with circles perimeter Say I have this circle here.

What would be an algorithm to check if lines like the green or blue lines intersect the edge of the circle, but not the red line.

So lets say

if(amountOfPointsHit(line, circle) > 1) then return true
else return false

I only know the start and finish points of the lines, and not where the line intersects the circle. So the pseudo code would more be like

if(pointBetweenXYHit(x1, y1, x2, y2, circle) > 1) then return true
else return false

Here is the code I came up with based on the answers...

/**
*@param l1  Line point 1, containing latitude and longitude
*@param l2  Line point 2, containing latitude and longitude
*@param c   Center of circle, containing latitude and longitud
*@param r   Radius of the circle
**/
Maps.ui.inCircle = function(l1, l2, c, r){
    var a = l1.lat() - l2.lat()
    var b = l1.lng() = l2.lng()
    var x = Math.sqrt(a*a + b*b)
    return (Math.abs((c.lat() - l1.lat()) * (l2.lng() - l1.lng()) -
           (c.lng() -  l1.lng()) * (l2.lat() - l1.lat())) / x <= r);
}

pseudo code for that ^^ is

method inCircle (line1point, line2point, center, radius)
    let l1 = line1point
    let l2 = line2point
    let c = center
    let r = radius
    x is length between points line1point and line2point
    return true if ((c.x - l1.x) * (l2.y - l1.y) - (c.y - l1.y) * (l2.x - l1.x)) / x)
    is less then or eqaul to r

You can find the shortest distance from a point to a line using the formula
$$\operatorname{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. $$
Put $(x_0,y_0)$ = center of circle. If this distance is smaller (or equal) than radius of circle, then your line and circle intersects.

Since you know start  point $(x_1,y_1) $ and end point $(x_2, y_2) $, you can get the equation of line using formula
$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x-x_1)$$
Simplifying, we get
$$ (x_2 - x_1) y + (y_1 - y_2)x +(x_1-x_2)y_1 + x_1(y_2-y_1) = 0$$
It would be nice to store $a = y_1 - y_2, b = x_2 - x_1, c =  (x_1-x_2)y_1 + x_1(y_2-y_1)$

It would be something like

return (Math.abs((l2.lat() - l1.lat())*c.lng() +  c.lat()*(l1.lng() -
       l2.lng()) + (l1.lat() - l2.lat())*l1.lng() +
       (l1.lng() - l2.lng())*l1.lat())/ Math.sqrt((l2.lat() - l1.lat())^2 +
       (l1.lng() - l2.lng())^2) <= r)

something like
$$ \frac{\left | (x_2 - x_1)x_0 + (y_1 - y_2)y_0 + (x_1-x_2)y_1 + x_1(y_2-y_1) \right |}{\sqrt{(x_2 - x_1)^2 + (y_1 - y_2)^2}} \le  r$$

ryu jin is correct, but his pseudo code is a tad off, as observed by morteza khosravi. For anyone lazy or unsure how to correct it:

c = ax^2 + ay^2 - r^2;
b = 2*(ax*(bx - ax) + ay*(by - ay));
a = (bx - ax)^2 + (by - ay)^2;

not:

a = ax^2 + ay^2 - r^2;
b = 2*(ax*(bx - ax) + ay*(by - ay));
c = (bx - ax)^2 + (by - ay)^2;

(notice the change in a and c).
Also note, that this solution works for line segments, not infinite lines.


Simple generators with a complex Gröbner basis It's known that finding a Gröbner basis of a polynomial ideal has a worst-case space complexity of $O(2^{2^{c\cdot n}})$, where c is constant and n is the number of variables $k[x_1,\ldots,x_n]$.

However, in practice it seems that most ideals have a simple Gröbner basis.

Can anyone give some concrete examples of small generators whose ideal has a large Gröbner basis?
How would I go about searching for such examples (besides a brute-force approach of trying random ideals)?

I just found the following example, from this paper:

GroebnerBasis[{x^5 + y^5 + z^5 - 1, x^3 + y^3 + z^2 - 1}, {x, y, z}] (* Is long *)


Countability and uncountability of a set $A$ and the set of equivalence classes $A / R$ Let $A$ be a set and $R$ an equivalence relation on $A$. Prove or disprove:

  If $A$ is countable then all the equivalence classes of $R$ are countable.
  If $A$ isn't countable then the quotient group $A/R$ isn't countable.
  If $A$ isn't countable and $A/R$ is countable, then there is an equivalence class in $R$ that isn't countable.

Well, I thought of going from the definitions and say that two equivalence class is either identical or foreign, also, the equivalence relation (that is transitivity, reflexivity and symmetry) can add only a finite number of elements so at most it will be countable. So it's true.
False: $A=\Bbb R , \ R=mod (2) $ so there are only two equivalence class and the quotient group consists of only $1$ and $0$.
I'm pretty sure it's true,  $A/R$ at most has the cardinality $\aleph_0$ and it's possible to have an equivalence class that is in one to one correspondence with $A$. But I don't really have an idea on how to prove it.

Please share your thoughts on how to solve this.

Thanks.

Note: This is from set theory intro course so I probably won't understand solutions that utilize knowledge from abstract algebra, rings or group theory.

First note that equivalence classes $R$ are subsets of $A$. Because the size of any subset of a countable set is countable, all equivalence classes are countable.

For part $3$, suppose all equivalence classes are countable. The order of a union countable sets is also countable, so the order of $A$ (which is the order of the union of all equivalence classes) is countable. This is a contradiction, so there is at least one non-countable equivalence class.

I have prove for 3. If A isn't countable and A/R is countable then there is an equivalence class that isn't countable.
We have example for it is-
If A= the set of real numbers
Relation between a and b is either both are rational or irrational
Then Q/R is countable having two equivalence  classes one of rationals and second is of irrationals. So here equivalence class of irrationals is uncountable.


Find all $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy $ f(m-n+f(n))=f(m)+f(n) $ Question -

Find all $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the relation -

$$
f(m-n+f(n))=f(m)+f(n)
$$
for all $m, n \in \mathbb{N}$ where  $N={1,2,3....}$

solution -

Observe $f(n) \geq n .$ Consider $F(n)=f(n)-n .$ Show that $F$ satisfies
$$
F(F(n)+m)=F(m)+n
$$
Using this, conclude that $F(1)=1$

and $F(n+1)=F(n)+$
$F(1)$ for all $n \geq 1 .$ Thus $F(n)=n F(1) .$ It follows that $F(n)=n$ and $f(n)=2 n$

Now i did not understand how they proved $F(1)=1$ ???

Any help will be appreciated

thankyou

I don't understand how he proved it, but here's my solution:
$$f(m-n+f(n))=f(m)+f(n) \implies P(m,n)$$
$$P(0,0) \implies f(f(0))=2f(0)$$
Let $f(0)=c$  giving $f(c)=2c$, so
$$P(m,c)\implies f(m+c)=f(m)+2c $$
$$P(m,0) \implies f(m+c)=f(m)+c $$
So, $c=2c \implies c=0 \implies f(0)=0$
$$P(n,n) \implies f(f(n))=2f(n)$$
Let $f(1)=k$
$$P(1,1) \implies f(k)=2k$$
$$P(m,1) \implies f(m-1+k)=f(m)+k$$
$$P(m-1,k) \implies f(m-1-k+2k)=f(m-1+k)=f(m-1)+2k$$
So, combining the previous 2 equations, we get
$$f(m)+k=f(m-1)+2k \Leftrightarrow f(m)-f(m-1)=k \implies H(m) $$
$$H(2)\implies f(2)-f(1)=k \implies f(2)=2k$$
$$H(3)\implies f(3)-f(2)=k \implies f(3)=k+f(2)=k+2k=3k$$
and so on, by simple induction, and the fact that $f(0)=0$
$$f(x)=kx \text{ } \forall \text{ } x \in \mathbb{N}$$
Substituting in the original equation,
$$km-kn+{k^2}n=km+kn \Leftrightarrow {k^2}n=2kn \Leftrightarrow k^2=2k \Leftrightarrow k^2-2k=0 $$
$$ \Leftrightarrow k(k-2)=0$$
This gives $$(1) \text{  } k=0 \implies f(x)=0 \text{ } \forall \text{ } x \in \mathbb{N}$$
$$(2) \text{   } k=2 \implies f(x)=2x \text{ } \forall \text{ } x \in \mathbb{N}$$
the only 2 solutions $\Box$.

Note that when $0$ doesn't belong to $\mathbb{N}$, you can discard all the steps of relating to $0$ and the proof would still hold true.

let $F(n) = f(n) - n$. $F$ verify the functional equation:
$$F(m + F(n)) = F(m) + n$$
By choosing $m=0$, we have $F(F(n)) = F(0) + n$, so $F$ is injective.
By letting $n=0$, we have $F(m+F(0)) = F(m)$ and by injectivity we deduce
$F(0) = 0$.
Now we have $F(F(n)) = n$ so now $F$ is surjective. Let $n_0$ s.t. $F(n_0) = 1$.
We have now:
$$F(m + 1) = F(m) + n_0$$
so:
$$F(m) = n_0 m$$
by inserting this into last equation we deduce $n_0^2 = 1$ so $n_0 = 1$.
I have prouved $F(1) = 1$. Now by induction we deduce $F(n) = n$, i.e. $f(n) = 2n$


Dimension is preserved by normalization How do I show that the dimension of a scheme is the same as its normalization's?

Let $v: \tilde X\rightarrow X$ be the normalization of $X$. I want to show that dim $\tilde X$ = dim $X$.

At first, I wanted to say that the dimensions should be equal because the normalization is a birational map that induces a bijection on the irreducible components. But this doesn't seem to be sufficient because of the following example I thought of: let $\pi: \Bbb A^1 \rightarrow Y$, where $Y$ is a line with a second line glued to its origin, so that the origin is the generic point of the second line. Then we have a birational map sending $\Bbb A^1 \setminus \{0\}$ to the equivalent subset of $Y$, which is dense there as well. However, dim $Y = 3$, so birational maps don't preserve dimension.

For the normalization, we have in addition that $\forall p\in \tilde X$, dim $\mathcal O_{\tilde X, p} =$ dim $\mathcal O_{X, v(p)}$, so dim $\tilde X \le$ dim $X$ But how do I get the other inequality?

I guess a rigorous solution is this:

a normalization gives a bijection between the irreducible closed subsets. Since dimension is determined locally on any irreducible component that maximizes 'the dimension' (more precisely, that maximizes the maximal length of a chain of irreducible closed subsets), we can pick an affine open Spec $A \subset X$ that has that maximal dimension and look at its preimage Spec $B \subset \tilde X$. Then $A$ and $B$ have the same dimension, which gives us the other inequality.


$u$-substitution integral I have the following integral:

$$\int_0^{y_c} e^{-y} e^{-a e^{-y}}y^2 \operatorname dy$$

My first attempt at solving in Mathematica was fruitless (I did not try using any assumptions though).

However, I think I can do it by hand when I make the substitution:
$u = e^{-y}$ which means $\operatorname du = -e^{-y}dy = -u\operatorname dy$, and $y = -\ln u$.
The integral then becomes

$$-\int_1^{e^{-y_c}} u e^{-a u}(\ln u)^2 \frac{\operatorname du}{u} = -\int_1^{e^{-y_c}} e^{-a u}(\ln u)^2 \operatorname du$$
Mathematica can then perform the integral in terms of exponential integrals and hypergeometric function.

Does it look like I've performed the $u$-substitution correctly?

Yes, the substitution have been correctly done.

HINT: first of all I think that your substitution isn't correct.

$ e^{e^{-y}}=t$ and if we apply ln on both sides we get

$e^{-y}=\ln(t)\Rightarrow -y=\ln(\ln(t)) \Rightarrow dy=\frac{-1}{t\ln(t)}dt$ and $y^2=(\ln(\ln(t)))^2$

If we substitute this into integral we get $I=-\int t^{a-1}(\ln(\ln(t)))^2 dt$ ,you can calculate bounds....

Now make substitution $\ln(t)=p \Rightarrow \frac{1}{t}dt=dp \Rightarrow I=-\int e^
{ap}(\ln(p))^2dp$

The last integral we can solve by partial integration method with $u=e^{ap}\ln(p)$ and $dv=\ln(p)dp$


A property of irreducible and aperiodic Markov chains Let $P$ denote the $s\times s$ Markov transition matrix. We know that irreducibility and aperiodicity implies the following:

There exists an integer $N\geq 1$, such that $[P^n]_{ij}>0$ for all $i,j$ and all $n\geq N$.

Is the following property true for irreducible and aperiodic chains?

Fix $i,j$. Suppose $[P^n]_{ij}>0$ for some integer $n$, then $[P^{m}]_{ij}>0$ for all $m\geq n$.

Nevermind, the statement is incorrect. I found a counterexample:

Suppose $P$ is $3\times 3$. Let $P_{11}=P_{13}=P_{22}=0$ and $P_{12}=1$. All other entries are positive. This chain is aperiodic and irreducible. You can check that $[P^2]_{12}=0$ even though $P_{12}>0$.

If the chain is irreducible, and we have $[P^n]_{ij}>0$ for all $i,j$ and all $n\geq N$, it means we communicate states $i$ and $j$ with some probability.

Your assumption is false, since the value $P_{ij}^n$ doesn't make sense (in the probabilistic mean.). for $n=1$, this represents the probability of going from state $i$ to state $j$, but to power it, means what?.

On the other hand, what you may want to do, it to factorize you matrix, using it's eigenvalues and eigenvectors, so you can operate matrix power much faster. We have $$P^m =U^{-1} \Sigma^m U,$$ where the matrix $U$ is composed by the eigenvectors of $P$ and $\Sigma$ is the diagonal matrix with the eigenvalues of $P$. And the nice thing about $\Sigma$ is that $\Sigma^m= \left(\sigma_{i,j}^m \;;\; (i,j) \in |s|^2\right),$ where $\sigma_{i,j} = 0$ if $i \neq j$.


If $a$ in $R$ is prime, then $(a+P)$ is prime in $R/P$. Let $R$ be a UFD and $P$ a prime ideal. Here we are defining a UFD with primes and not irreducibles.

Is the following true and what is the justification?

  If $a$ in $R$ is prime, then $(a+P)$ is prime in $R/P$.

The bijection between ideals of $\,R/I\,$ and ideals of $R$ containing $I$ restricts to prime ideals, hence $\,(a+P)\,$ is prime in $\,R/P\,$ iff $\,I = (a)+P\,$ is prime in $\,R.\,$ But generally this is not true, e.g. $\,I=1\,$ for $\,a\nmid P\,$ primes in $\,\Bbb Z,\,$ or $\,I = (4\!-\!x)+(x) = (4,x)\,$ is not prime in $\,\Bbb  Z[x],\,$ by $\,2^2\in I\,$ but $\,2\not\in I.$

I think thats wrong. If $ a \in P $ holds, then $ a + P = 0 + P \in R/P$ and therefore $a+P$ not prime.


Solving extremal value problems with boundaries using lagrange-multipliers Find the max and/or min of $f=x+2y$ on $x^2+y^2=1$
So, I'd like to do it with Lagrange-multipliers. I'd like to demonstrate two approaches for that.
Approach 1:
We first write $g=x^2+y^2-1$
Lagrange-Function: $L=f-\lambda g = x + 2y - \lambda(x^2 + y^2 -1)$
We solve:
$\frac{\partial L}{\partial x}=1-2\lambda x = 0 \quad \Rightarrow \quad \lambda=\frac{1}{2x}$
$\frac{\partial L}{\partial y}=2-2\lambda y = 0 \quad \Rightarrow \quad \lambda=\frac{1}{y}$
Using the results of the above we get: $y=2x$
By using $x^2+y^2=1$ we get $x^2+4x^2-1=5x^2-1=0 \quad\Rightarrow\quad x=\pm\frac{1}{\sqrt{5}}$ and using $y=2x$ we get $y=\pm \frac{2}{\sqrt{5}}$
So we found the two candidates: $p_1=\big( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \big), p_2=\big( \frac{-1}{\sqrt{5}}, \frac{-2}{\sqrt{5}}\big)$
Using $\operatorname{Hess}(L)(x,y,\lambda)=\begin{pmatrix}-2\lambda & 0 \\ 0 & -2\lambda\end{pmatrix}$
By evaluating the hessian matrix in the two points ,we find that $p_1$ is a maximum and $p_2$ is a minimum.
Approach 2:
Again we have the same $f,g$ as above.
$\nabla f = (1, 2), \quad \nabla g=( 2x, 2y)$
We get:
I)   $1=\lambda 2x \quad\Rightarrow\quad x=\frac{1}{2\lambda}$
II)  $2=\lambda 2y \quad\Rightarrow\quad y=\frac{1}{\lambda}$
III) $x^2+y^2 = 1$
From I and II we also get $ x=2y$
We solve the system of equations for $\lambda$, getting: $\lambda=\pm\frac{\sqrt{5}}{2}$
Using the results from I and II we get $p_1=\big( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \big), p_2=\big( \frac{-1}{\sqrt{5}}, \frac{-2}{\sqrt{5}}\big)$
We evaluate $f(p_1)=\sqrt{5}$ and $f(p_2)=-\sqrt{5}$ so $p_1$ is maximum and $p_2$ is minimum.
Question:
Is there any significant difference here? I really think the second one is way more intuitive. I really think actualy using the gradient just makes more sense. I think in the first one we kind of create a new vectorfield wich depends on the lagrangian-variable and then just apply the extremal problem on that. But I'm not 100% sure about the first approach.
This really is a question about details, so I know that the basic concept of both approaches is the same. But I'm e.g. wondering why, in the first approach, they didn't just put the value into the function.
If going with the 2nd approach, how would I know if a point is a saddle point? Which one is the more general approach?

In the first approach, if there is more than one local maximum then how do you consider?

The second approach is more efficient than the first approach.


$f$ is Lipschitz if and only if there exists $L\geq0$ such that $|f'(x)|\leq L$ Let $f:[a,b]\to\mathbb{R}$ be absolutely continuous.
Prove that $f$ is Lipschitz if and only if there exists $L\geq0$ and a set $E\subset[a,b]$ such that, $m(E)=0$ and $f$ is differentiable at each $x\in[a,b]\setminus E\quad$  with $|f'(x)|\leq L$.
My attempt:
For $\Rightarrow$ direction:
Since $f$ is absolutely continuous on $[a,b]$, it is differentiable a.e on $[a,b]$
And notice:
$\begin{align}
|f'(x)|&=\left|\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\\
&\leq\lim\limits_{h\to0}\left|\frac{f(x+h)-f(x)}{h}\right|\\
&\leq \lim\limits_{h\to0}\left| \frac{L|(x+h)-x|}{h}\right|\\
&=L \quad\text{ Where, $L$ is the Lipschitz constant}
\end{align}$
Is that correct?
And for the other side of the implication I would really appreciate your help

Since $f$ is absolutely continuous, we have that
$$ f(x) - f(y) = \int_x^y f'(t) \, dt .$$
Then
\begin{align}
\text{$f$ is Lipschitz with constant $L$} &\Leftrightarrow
-L (y-x) \le \int_x^y f'(t) \, dt \le L (y-x) \text{ for all $x < y$} \\
&\Leftrightarrow  \int_x^y (L - f'(t)) \, dt \ge 0 \text{ and } \int_x^y (L + f'(t)) \, dt \ge 0\text{ for all $x < y$}
\end{align}
Also
$$ \int_x^y g(t) \, dt \ge 0 \text{ for all $x < y$} \Leftrightarrow g \ge 0 \text{ a.e.} $$
Hence $f$ is Lipschitz with constant $L$ if and only if $L - f'(t)$ and $L + f'(t)$ are greater than or equal to zero for almost every $t$.

Important: this proof only works when $f$ is differentiable in $\mathbb{R}$.
The reverse implication can easily be seen with the Mean value theorem. Assuming $|f'(x)|\leq L$ (being $L\geq 0$) for any $x\in\mathbb{R}$, using the mean value theorem we know that, given $x,y\in\mathbb{R}$ (we will assume $x<y$), there exists $c\in(x,y)$ such that
$$\frac{f(x)-f(y)}{x-y}=f'(c).$$
Using absolute values, and using that $|f'|\leq L$ we get that
$$\frac{|f(x)-f(y)|}{|x-y|}\leq |f'(c)|\leq L,$$
so we end up with
$$\boxed{|f(x)-f(y)|\leq L\cdot|x-y|}$$
and we conclude $f$ is Lipschitz


$4^{x}+2^{x+1}=18$ Please help me solve? I tried using logs on both sides or tried treating it as a quadratic but didn't manage to simplify it, Help?:D

$$(2^x)^2+2\cdot (2^x)-18=0$$

Same as $x^2+2x-18=0$ (after substitution).

Use the quadratic formula to find the roots. Set a root equal to $2^x$ then do the log.


Does $A^2 \geq B^2 > 0$ imply $ACA \geq BCB$ for square positive definite matrices? Assume we have two $n \times n$ real nondegenerate matrices $ A^2 $ and $B^2$, such that
$$
A^2 \geq B^2 > 0,
$$
where "$\geq$" means positive semidefinite (Loewner) ordering. Does the following inequality holds for any real matrix $C$
$$
ACA \geq BCB \ ?
$$
If not, under which conditions on $C$ (or additional conditions on $A$ and $B$) does it holds?

I would appreciate any ideas, suggestions, counterexamples.
Thanks!

I don't know if there are any nice (i.e. not-too-strong) conditions for the inequality to hold, but I'm sure that it doesn't always hold, even when $C$ is positive definite. Counterexample:
\begin{align}
A&=A^2=I,\\
B&=B^2=\frac12\pmatrix{1&1\\ 1&1},\\
C&=\operatorname{diag}(1,4).
\end{align}
In this case, we have $A^2\ge B^2\ge 0$ but $ACA-BCB=\frac14\pmatrix{-1&-5\\ -5&11}$ is indefinite. While $B$ is not positive definite here, by continuity, we can obtain a valid counterexample by adding a small positive multiple of $I$ to both $A$ and $B$.

Edit.

Note that if $ACA\ge BCB$ for all real symmetric $C$, we must have $A=B$ because $AIA\ge BIB$ and $A(-I)A\ge B(-I)B$ imply that $A^2=B^2$.
It isn't quite meaningful to consider $ACA\ge BCB$ for all $C\ge0$ either. In particular, if $A(vv^\ast)A\ge B(vv^\ast)B$ for every vector $v$, then $Bv$ must be equal to $\lambda_v Av$ for some $0\le\lambda_v\le1$. Therefore, by linearity, $B=\lambda A$ for some $0\le\lambda\le1$.
It is interesting to ask, however, if $A\ge B>0$ and $A^2\ge B^2$, what class of $C$ (under perhaps some additional conditions on $A$ and $B$) would satisfy the inequality $ACA\ge BCB$.

Here I posted a related question. My (preliminary) conclusion to this question might be a valuable idea/suggestion to your answer:
I strongly assume (!) when your matrices commute (i.e. $(AB)^T AB = AB (AB)^T$) then $ACA \geq BCB$. This could be the case when $A$, $B$, and $C$ are diagonal matrices (and of course in the trivial cases of $C$ or $A$ and $B$ being multiples of $I$).