Let $X$ be an infinite set with a topology $T$, such that every infinite subset of $X$ is closed. Prove that $T$ is the discrete topology. Let $X$ be an infinite set with a topology $T$, such that every infinite subset of $X$ is closed. Prove that $T$ is the discrete topology.

I have somewhat of an answer but I don't think it's enough to prove it, especially with respect to the subsets being infinite.

Let $S$ be contained in $X$, then $X \setminus S$ is also contained in $X$. Therefore we can say that $X \setminus S$ is closed, therefore $S$ is open for any $S$ contained in $X$. Hence $T$ is the discrete topology.

Thanks

If $X$ is infinite, let $x \in X$ ; it follows that since $\{x\}$ is finite, $X \backslash \{x\}$ is infinite, hence closed. Therefore, for any subset $S \subseteq X$,
$$
S = \bigcup_{x \in S} \{x\}
$$
is a union of open sets, thus open. This means $T$ is all subsets of $X$. You have the discrete topology.

Hope that helps,

Every infinite subset of X closed means every finite subset of X is open. In particular, the singleton sets {x} for all x in X are open. This is sufficient to conclude that T is discrete!

Hope it helps.


Calculate sum of a series using telescopy $$A=\sum_{n=1}^\infty \frac{2^n n} {(n+2)!} $$

By using induction, one can show that $A = 1$ but I don't know how to solve this problem by using other methods.


closure = union of the set and the set of limit points Let $S \subseteq \mathbb{R^n}$ and denote the set of limit points of
  $S$ as $S'$.

  Show that $S \cup S' = \bar{S}$ where $\bar{S}$ is the closure.

(i) I want to show that $\bar{S} \subseteq S \cup S'$ .

Let $x \in \bar{S}$ then either $x \in S$ or $x \in \bar{S}\backslash S $.

If $x \in S$ then $x \in S \cup S'$

If $x \in \bar{S}\backslash S $ ...?

(ii) I want to show that $S \cup S' \subseteq \bar{S}$.

Let $x \in S \cup S'$. Then, $x \in S$ or $x \in S'$.

If $x\in S$ then $x \in \bar{S}$.

If $x \in \bar{S}$ ... ?

Thank you

Looking at (i) you're almost done.  If $x \in \overline{S} \backslash S$ then $x \notin S$ and since $x \in \overline{S}$ then all neighborhoods $U$ of $x$ intersect $S$.  But since $x$ is not in $S$ then $U$ must intersect $S$ at some point other point that is not $x$.  Thus $x \in S'$.  Therefore you can conclude that $\overline{S} \underline{\subset} S \cup S'$.

For (ii) note that if $x \in S \cup S'$ then all neighborhoods $U$ of $x$ intersect $S$ at some point, doesn't matter where.  Thus $x \in \overline{S}$.  Therefore $S \cup S' \underline{\subset} \overline{S}$.

The final conclusion being that $S \cup S' = \overline{S}$

(This is a general result in topology that you can reformulate to real euclidean space if you wish)

Not sure what you are using as your definition of closed, but one very common definition is: A set S is closed iff it contains all its limit points. If that is a definition you are using or can use, this proof should be easier.

$\underline{Show: S\cup S' \subseteq \bar S}$: If $x \in S$ then we know $x \in \bar S$ as $S \subseteq \bar S$. If $x \in S'$ then $x \in \bar S$ because $\bar S$ contains all its limit points. Thus $S\cup S' \subseteq \bar S$.

$\underline{Show: \bar S \subseteq S\cup S'}$: Let $x \in \bar S$. We know either $x \in S$ or $x \in \bar S - S$. The first case is trivial, so we will focus on any $x \in \bar S - S$. Since $x\notin S$ but $x \in \bar S$, $x$ must be a limit point of $S$. Thus $x \in S'$. This establishes that for any $x \in \bar S$ either $x\in S$ or $x \in S'$, so $\bar S \subseteq S\cup S' $.


Find all integer solutions to the equation $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$ I was going through one of my Mathematics books and I came to this problem:

Find all integer solutions to the equation: $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$

I tried a few things to begin with but none went in the right direction, any suggestions?

Another thing, if we have the left side of the equation to be equal to 9, there are supposed to be 16 solutions in integers. What way can they be found?

First, let us suppose that $x,y,z$ are positive integers satisfying the proposed equality. Then
$$ xyz(15-x-y-z)=\frac{1}{2}\left((xy-yz)^2+(yz-zx)^2+(zx-xy)^2\right)\ge0$$
So we have $x+y+z\le 15$.
Now, note that squares modulo 5 are $\{0,-1,1\}$. Thus, if the sum of three squares modulo $5 $ is $0$ then either all three of them are multiples of $5$, (the case $0+0+0$) or just exactly one of them is a multiple of $5$ and the other two are equal to $+1$ and $-1$ modulo $5$, ( the case $0+1-1$). In both cases $5$ divides at least one of the three numbers.
The equality
$$(xy)^2+(yz)^2+(zx)^2=15xyz\equiv 0\mod 5\tag1$$
shows then that one of the numbers $xy,yz,zx$ is a multiple of $5$. So, we may and will suppose that $x=5a$.  But then $(1)$ proves that $5$ must divide $yz$ and again we may suppose that $ y=5 b$.
Now, $5+5+1\le 5a+5b+z\le15$ implies that $ a=b=1$, that is $x=y=5$ and the equation implies then that $z=5$.
We conclude that $(5,5,5)$ is the only solution consisting of positive integers.
The general case. Suppose that $x=sx'$,$y=ty'$ and $z=uz'$ is a solution to the proposed equation where
$x',y',z'$ are positive integers, and $s,t,u$ are $\pm1$. It follows from
$$xyz\left(\frac{1}{x^2}+ \frac{1}{y^2}+\frac{1}{z^2}\right)=15$$
That
$$stu x'y'z'\left(\frac{1}{x'^2}+ \frac{1}{y'^2}+\frac{1}{z'^2}\right)=15$$
This implies that $stu=1$ and
$$ x'y'z'\left(\frac{1}{x'^2}+ \frac{1}{y'^2}+\frac{1}{z'^2}\right)=15$$
So, $x'=y'=z'=5$ and exactly two of $s,t,u$ are equal $-1$, or they are all equal to $+1$. This gives the following set of solutions $$\{(5,5,5),(-5,-5,5),(-5,5,-5),(5,-5,-5)\}.$$

Translating the equation to form
$$(xy)^2+(yz)^2+(zx)^2=15xyz,\qquad(1)$$
easy to see that $xyz>0.$
That means that all solutions with different signs can be obtained from positive solutions by sign replacing of two unknowns at the same time.
So we can search for positive solutions,
$$x,y,z>0.$$

All rational roots of equation with integer coefficients
$$\left(\dfrac{yz}x\right)^2-15\dfrac{yz}x+y^2+z^2=0$$
must be integer, so $x\ |\ yz$ and similarly $y\ |\ zx,\ z\ |\ xy.$
Let
$$u=\dfrac{xy}z,\quad v=\dfrac{yz}x,\quad w=\dfrac{zx}y,$$
then we get
$$u+v+w=15,\quad u,v,w,\sqrt{uv},\sqrt{vw},\sqrt{wu}\in\mathbb N$$
with the single solution
$$u=v=w=5,$$
$$x=\sqrt{uw}=5,\quad y=\sqrt{uv}=5,\quad z=\sqrt{vw}=5.$$
So the solution set is
$$\boxed{(x,y,z)\in\{(5,5,5),(5,-5,-5),(-5,5,-5),(-5,-5,5)\}.}$$

Note

For the equation $LHS=9$ we have
$$u+v+w=9,\quad u,v,w,\sqrt{uv},\sqrt{vw},\sqrt{wu}\in\mathbb N$$
with the solutions
$$(u,v,w)=\{(4,4,1), (4,1,4), (1,4,4), (3,3,3)\},$$
so the positive solutions of $LHS=9$ are
$$(x,y,z)=\{(2,4,2),(4,2,2),(2,2,4),(3,3,3)\}$$
and with account to pair signs permutations we have 16 solutions.


Puzzling solution for $1/x = 3 - 2\sqrt{x}$ What is the solution for
$$\frac{1}{x} = 3 - 2\sqrt{x}$$
When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.
However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.
What am I missing?

If we let $y = \sqrt{x}$, we get
\begin{align*}
\frac{1}{y^{2}} = 3 - 2y & \Longleftrightarrow 1 = 3y^{2} - 2y^{3}\\
& \Longleftrightarrow 2y^{3} - 3y^{2} + 1 = 0\\\\
& \Longleftrightarrow (2y^{3} - 2y^{2}) - (y^{2} - 1) = 0\\\\
& \Longleftrightarrow 2y^{2}(y-1) - (y-1)(y+1) = 0\\\\
& \Longleftrightarrow (2y^{2} - y - 1)(y-1) = 0\\\\
& \Longleftrightarrow \left(y + \frac{1}{2}\right)(y-1)^{2} = 0\\\\
& \Longleftrightarrow \left(y = -\frac{1}{2}\right)\vee(y = 1)
\end{align*}
Since $y = \sqrt{x} > 0$, there is only one possible solution, which is $x = 1$.
Hopefully this helps!

But $x$ cannot be negative, given we must have $\sqrt x >0$, right?
You solved the equation by squaring it, and your solutions solve the squared equation.  But when squaring both sides of an equation, please realize that can introduce an extraneous solution.
Hence you need to check which solution(s) is/are valid: only the solutions for which the original question is defined are valid.  But given the original equation, $x=-1/4$ makes the equation undefined.


Semisimple Lie algebras are perfect. Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ?

The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not perfect would be expected to be solvable. But it doesn't seem to work.

Let $\mathfrak{g}$ be a semi simple Lie algebra. Then $\mathfrak{g}$ can be written as direct sum of simple ideals.

For simplicity, we can write $\mathfrak{g}=\mathfrak{g}_1\oplus\mathfrak{g}_2$ where $\mathfrak{g}_1,\mathfrak{g}_2$ are simple ideals.

We then have $$[\mathfrak{g},\mathfrak{g}]=[\mathfrak{g}_1\oplus \mathfrak{g}_2, \mathfrak{g}_1\oplus \mathfrak{g}_2]= [\mathfrak{g}_1,\mathfrak{g}_1]\oplus [\mathfrak{g}_2,\mathfrak{g}_2]=\mathfrak{g}_1\oplus \mathfrak{g}_2=\mathfrak{g}$$

See that $\mathfrak{g}_1=[\mathfrak{g}_1, \mathfrak{g}_1], \mathfrak{g}_2=[\mathfrak{g}_2,\mathfrak{g}_2]$ as $\mathfrak{g}_1,\mathfrak{g}_2$ are simple ideals. Thus, we have $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$.

here you can find some relevant discussion.

https://mathoverflow.net/questions/60498/lie-algebra-semisimple


How does one prove that if $\cos(t) = \cos(t')$ and $\sin(t) = \sin(t')$ then $t = t' + 2k\pi$? How does one prove that if $\cos(t) = \cos(t')$ and $\sin(t) = \sin(t')$ then $t = t' + 2k\pi$ ?

I've tried proving the above statement, which I think is valid.

I know $\sin(t)$ is injective on $[-\pi/2; \pi/2]$ and $\cos(t)$ is injective on $[0; \pi]$, but until now I've not been able to use this to prove the statement rigorously.

also can
$$\cos{x}-\cos{y}=0\Longrightarrow 2\sin{\dfrac{x-y}{2}}\sin{\dfrac{x+y}{2}}=0\tag{1}$$
$$\sin{x}-\sin{y}=0\Longrightarrow 2\sin{\dfrac{x-y}{2}}\cos{\dfrac{x+y}{2}}=0\tag{2}$$
with $(1),(2)$ must
$$\sin{\dfrac{x-y}{2}}=0$$
since we kown
$$\sin{(k\pi)}=0,k\in Z$$
so
$$\dfrac{x-y}{2}=k\pi,\Longrightarrow x=y+2k\pi,k\in Z$$

Cos(x) and sin(x) are periodic functions of period 2k(pi).hence sin(x)=sin(x') and cos(x)=cos(x').


Count the divisors of n with particular property Take $n = \prod_{i=1}^r {p_i}^{\alpha_i}$, where each $p_i$ is a prime and $\alpha_i\geq 1$.

How many divisors of $n$, not equal to $n$, contain at least one $p_i$ with the corresponding multiplicity $\alpha_i$?

I'm not sure if your posted answer is correct, but here's one way to think about the problem. How many divisors are there where no prime occurs with maximal multiplicity? This is easy to compute as $\prod_m \alpha_m$ since you can choose exponents between $0$ and $\alpha_m-1$ for each prime divisor, each giving rise to a unique product for each choice. Then how many total divisors are there? This is $\prod_m (\alpha_m + 1)$. The difference between these two quantities is the answer you want, and subtract $1$ if you want to exclude $n$ as a divisor of itself.


Math Piecewise style, which one is correct/best? I am writing my Computer Science Master Thesis. I am wondering what is the correct or in your opinion best looking style for piecewises:

See examples below:

Do I use a comma at the end of each line (ex last line)?
Do I end the last line with a dot?
Do I use a separator, and if yes which (in this example a colon) to seperate condition and formula?
What other things are stylewise wrong in the example below?

$$f(x) = \left\{
     \begin{array}{lr}
       0 & : x \leq 0, \\
       1 & : x > 0.
     \end{array}
   \right.$$

I use

$$f(x)=\begin{cases}0, & \text{if $x<0$;} \\
    1, & \text{if $x\geq 0$.}\end{cases}$$

resulting in

$$f(x)=\begin{cases}0, & \text{if $x<0$;} \\
1, & \text{if $x\geq 0$.}\end{cases}$$

IIRC this was also suggested by the authors of AMSmath.

Best practice is to ask your master's thesis supervisor if your Computer Science department has a required/recommended LaTeX style/class file that you should/could use for writing up your master's thesis as a whole (and that's not just for piecewise definitions of functions).  If said file is not available, ask your CS department for the thesis layout guidelines, if there are any.


If $f$ is even and $y'=f(y)$ then $y$ is odd Let $f\in C^1(\mathbb{R}, \mathbb{R})$ be an even function.
Consider the maximal solution $y\colon\left]\alpha ,\beta\right[\to \mathbb{R}$ of the IVP $$y'=f(y),\ y(0)=0$$
Prove that $y$ is an odd function and $\beta =-\alpha$.

To be able to prove that $y$ is odd, I first need its domain to be symmetric ($x\in \left]\alpha ,\beta\right[\implies -x\in \left]\alpha ,\beta\right[$), from here I can conclude that $\alpha=-\beta$. But how to prove the domain is symmetric?
And how to prove that $y(-x)=-y(x)$ for all $x\in \left]\alpha,\beta\right[$? I suspect it has something to do with the fact that the derivative of an even function is odd and $y'=f(y)$, but I can't see how to get the desired result.

Clearly, $\alpha<0<\beta$.
Let $r=\min\{-\alpha,\beta\}$ so that we at least have a solution on $\left]-r,r\right[$.

Then show: If $y$ is a solution on $]\alpha,\beta[$, then $x\mapsto-y(-x)$ is a solution on $\left]-\beta,-\alpha\right[$.

By uniqueness, these coincide on $\left]-r,r\right[$.

Then you can combine $y$ with $x\mapsto- y(-x)$ to find a solution on $\left]-\max\{-\alpha,\beta\},\max\{-\alpha,\beta\}\right[$. By maximality, we conclude $\alpha=-\max\{\alpha,\beta\}=-\beta$.

I'll leave this here, for someone that has a similar question.

Since f is odd, y' = f(y)=f(-y) — This means that not only the function is odd, but f[y(t)]=f[-y(t)] (we'll get to this later)

From this, we come to the point that y(t) has a known point (0,0) , from the inicial condition.

Let us consider a point α :

f[y(α)] = y'(α) = f[-y(α)]

Considering now α and -α=β, we come to :

f[y(α)] = y' = f[-y(-α)] =f[-y(β)]

This means that f[y(α)] = f[-y(β)] = f[-y(-α)]

This implies that y(α) = -y(-α) , condition for a odd function f(x) = -f(-x) , so y(t) is odd, and it's done.


Infimum and supremum of finite ordered subsets I am currently taking an introductory proofs course, and I have come across this problem.  It's asking to prove the following:

  Let $S$ be an ordered set. Let $A$ be a non-empty finite subset. Then $A$ is bounded. Furthermore, $\inf A$ exists and is in $A$ and $\sup A$ exists and is in $A$. Hint: Use induction.

It's finite, so it's got $n$ elements, say. If $n=1$, you're done (why?).

If $n > 1$, then try removing some element $a$. The resulting set (by inductive hypothesis) has an inf and a sup; now put the element $a$ back in.


Which order to perform integration after coordinate change? Let $A$ be the unit circle. I know that:

$\iint_A (x^2 + y^2) \,dx\,dy$ = $\int_{0}^{1} \int_{0}^{2 \pi} r^2\ d\theta \ dr $

I'd like to know why wouldn't it be:

$\iint_A (x^2 + y^2) \,dx\,dy$ = $\int_{0}^{1} \int_{2 \pi}^{0} r^2\ d\theta \ dr $

My thinking is that the notation $\iint_A $ comes with an implicit assumption about the order of integration, though I fail to see how this carries over when we transform to polar coordinates or more generally $(u,v)$-coordinates.

Any pointers would be great. Thanks :)

When you change to polar coordinates, you replace the integral over the region $A$ in the $xy$-plane with an integral over a corresponding region $B$ in the $r \theta$-plane, namely the rectangle given by $0 \le r \le 1$ and $0 \le \theta \le 2\pi$.

And when you integrate over a rectangle in the $xy$-plane, I don't think you would ever dream of putting the bounds of integration “backwards” (from the greater to the smaller), so why would you do it in the $r \theta$-plane?

The correct set up, by definition and convention, is the following one

$$\iint_A (x^2 + y^2) \,dx\,dy = \int_{0}^{1} \int_{0}^{2 \pi} r^3\ d\theta \ dr $$

Reversing the order for $\theta$ is equivalent to make a change of variable

$\alpha=2\pi-\theta\implies d\theta=-d\alpha$

then

$$\int_{0}^{1} \int_0^{2 \pi} r^3\ d\theta \ dr=-\int_{0}^{1} \int_{2 \pi}^0 r^3\ d\alpha \ dr$$


Can you answer my son's fourth-grade homework question: Which numbers are prime, have digits adding to ten and have a three in the tens place? My son Horatio (nine years old, fourth grade) came home with some fun math homework exercises today. One of his problems was the following little question:

  I am thinking of a number...

  It is prime.
  The digits add up to $10.$
  It has a $3$ in the tens place.

  What is my number?

Let us assume that the problem refers to digits in decimal notation. Horatio came up with $37,$ of course, and asked me whether there might be larger solutions with more digits. We observed together that $433$ is another solution, and also $631$ and $1531.$ But also notice that $10333$ solves the problem, based on the list of the first $10000$ primes, and also $100333$, and presumably many others.

My question is: How many solutions does the problem have? In particular, are there infinitely many solutions?

How could one prove or refute such a thing? I could imagine that there are very large prime numbers of the decimal form $10000000000000\cdots00000333$, but don't know how to prove or refute this.

Can you provide a satisfactory answer this fourth-grade homework question?

This question was found while searching for mathematical recreational problems to use for didactic reasons in programming. More exactly in sage (computer algebra system = CAS used for structural mathematical questions, based on python) for different aspects of using or not using (python native) "iter tools". The idea was to illustrate instances of the classes like Partitions and Combinations. Since sage also comes with a quick detection of prime number if it has a "decent size", i tried to write a short code snippet for this problem. I will insert first some dialog with the python interpreter, showing why (and how) the constructors Partitions and Combinations can be useful, then the compact code to search for prime numbers satisfying the conditions in the OP.

First of all, finding partitions of ten with parts $\ge 1$ and at least one piece equal to $3$ is simple:
partitions = [part for part in Partitions(10) if 3 in part]

Let us show them in the interpreter:
sage: for part in partitions:
....:     print(part)
....:
[7, 3]
[6, 3, 1]
[5, 3, 2]
[5, 3, 1, 1]
[4, 3, 3]
[4, 3, 2, 1]
[4, 3, 1, 1, 1]
[3, 3, 3, 1]
[3, 3, 2, 2]
[3, 3, 2, 1, 1]
[3, 3, 1, 1, 1, 1]
[3, 2, 2, 2, 1]
[3, 2, 2, 1, 1, 1]
[3, 2, 1, 1, 1, 1, 1]
[3, 1, 1, 1, 1, 1, 1, 1]

Now i would like to construct - just as an illustration, all prime numbers having $25$ digits that are using the non-zero digits from the last entry, say. We have to only generate the places for the many ones. The numbers to be generated
(and then tested for primality) are thus of the shape
$$
\underbrace{1}_{24}0\dots0
\underbrace{1}_{e}0\dots0
\underbrace{1}_{d}0\dots0
\underbrace{1}_{c}0\dots0
\underbrace{1}_{b}0\dots0
\underbrace{1}_{a}0\dots0
\underbrace{3}_1
\underbrace{1}_0
\ .
$$
The positions $a,b,c,d,e$ in the interval $[2,23]$ can be generated by asking sage for the combinations of five elements in the interval.
So the code to collect the primes is...
candidates = [ 10^24 + 31 + sum([10^a for a in pos])
               for pos in Combinations([2..23], 5) ]
solutions = [N for N in candidates if N.is_prime()]
solutions.sort()
print(f"There are {len(solutions)} solutions 10...031\n"
       "with 25 digits with zeros and ones in the ellipsis.")
print("The 10 biggest solutions among them are:")
for N in solutions[-10:]:
    print(N)

We obtain after some seconds:
There are 1332 solutions 10...031
with 25 digits with zeros and ones in the ellipsis.
The 10 biggest solutions among them are:
1111000000000100001000031
1111000000010000000000131
1111000000010000010000031
1111000000010000100000031
1111000000010010000000031
1111000100000000000001031
1111010000000010000000031
1111010000001000000000031
1111010000100000000000031
1111011000000000000000031

Now we are in position to search all solutions with a given (decent) amount of digits. Here is the code for numbers with LENGTH=$33$ digits, say:
LENGTH = 33
solutions = []    # so far

for units_digit in [1, 3]:
    N0 = 30 + units_digit
    for main_digit in [1 .. 10 - 3 - units_digit]:
        N1 = main_digit * 10^(LENGTH-1)
        for part in Partitions(10 - 3 - units_digit - main_digit):
            print(f"Candidates for {main_digit}{part}3{units_digit}")
            digits = list(set(part))
            count_dic = dict([(d, list(part).count(d)) for d in digits])
            places_ranges = [[ tuple(c)
                               for c in Combinations([2..LENGTH-2], count_dic[d])]
                             for d in digits ]
            places_list = cartesian_product(places_ranges)

            for places in places_list:
                 used_places = [place for d_places in places for place in d_places]
                 if len(set(used_places)) < len(used_places):
                     continue
                 N = N0 + N1
                 for k in range(len(digits)):
                     d = digits[k]
                     d_places = places[k]
                     N += d*sum([10^a for a in d_places])

                 if N.is_prime():
                     solutions.append(N)

print(f"There are {len(solutions)} solutions "
      f"with {LENGTH} digits")
print("The 10 biggest solutions among them are:")
for N in solutions[-10:]:
    print(N)

The above delivers the following ten biggest solutions with $33$ digits:
There are 13008 solutions with 33 digits
The 10 biggest solutions among them are:
200000100000000001000000000000033
200000000000010010000000000000033
200000000010000100000000000000033
200000010000001000000000000000033
200000000011000000000000000000033
200000100010000000000000000000033
200001000010000000000000000000033
200000001100000000000000000000033
200010100000000000000000000000033
300000000001000000000000000000033

Some other random solutions among the computed ones are:
sage: import random
sage: for _ in range(10):
....:     print(random.choice(solutions))
....:
100200000100000000100000001000031
200100000101000000000000000010031
100030000011000000000000000000031
200000000100010000100000010000031
100010000000020000010000001000031
110020000000000000000000010100031
100000000000001000001010002000031
100000000000000011000000010110031
100000000010000000021000010000031
100003000000000110000000000000031

Working with 40 instead, we get:
There are 31202 solutions with 40 digits
The 10 biggest solutions among them are:
2000000000000010000000001000000000000033
2000000100000000000010000000000000000033
2000100000000000000010000000000000000033
2000000000100000010000000000000000000033
2000000000000100100000000000000000000033
2000000000001010000000000000000000000033
2100000000000010000000000000000000000033
2000000110000000000000000000000000000033
2110000000000000000000000000000000000033
3000010000000000000000000000000000000033

One can ask for solutions with "special properties", for instance all solutions with $40$ digits and three times the digit 3 are...
sage: [N for N in solutions if N.digits().count(3) == 3]
[3000000000000000000000000000030000000031,
 1000000000030000000000000000000000000033,
 1000000030000000000000000000000000000033,
 3000010000000000000000000000000000000033]

Some words on the search method, so that the code can be digested without problems. Let us consider for instance the solution
100000000000000000000000002100231

its units_digit is 1. It has position zero in the number. (Counting from right.)
its main_digit is 1. It has position $32$ in the number. (Counting from right.)
the search is done now for number of the shape $1\dots 31$ with $33$ digits. "Inside" the dots we have many occurrences of zero, but some few digits are not zero and have to add to $5=10-1-3-1$. For this reason we consider all partitions of $5$. At some point we consider also the partition part = [2, 2, 1]. The digits inside this part are building the list [2, 1]. (We take the set associated to the list part, then make it a list for programmatic reasons.)
it is handy to record how many digits are in the part for each encountered digit, this gives the dictionary {2: 2, 1: 1}, so two comes twice, the one once.
we are searching now for possible positions for the two digits 2, and for the one digit 1. First we consider the places for the 2. We build the Combinations([2..31], 2). An element in these combinations object is [2, 6]. (It leads to the above solution.)
Then we consider the place"s" for the 1. We build the Combinations([2..31], 1). An element in these combinations object is [5]. (It leads to the above solution.) We consider the cartesian product. The tuple $(\ $[2, 6]$\ ,\ $[5]$\ )$ is somewhere in the cartesian product. The used places are [2, 6, 5], and we have the full length, i.e. there is no collision of the places dedicated for the 1 digit and the 2 digits.
then we consider the corresponding number
$N=1\cdot 10^{32}+2\cdot (10^6+10^2)+1\cdot(10^5)+30 +1$ and check if it is a prime number. Yes, it is. So we append it to the list of (already found) solutions (and search further and possibly append further...)

Some words on the true mathematical problem behind the puzzle. The "digits of numbers" do not have a structural meaning. Methods to find infinitely many solutions are a matter of human inventive force. A more concrete question would be if there are infinitely many prime numbers in a specific infinite subset like the one mentioned in the OP:
$$
N(a) :=
10^a + 333\ .
$$
Well, this is also a complicated problem. Some cheap exclusions would be...

considering $N(a)$ modulo $7$, we have $N(a)=0$ mod seven for $a=1$ mod six. (So $N(a)$ has the factor $7$ for $a=7, 14, 21,$...
considering $N(a)$ modulo $p=31$, we have $N(a)=0$ mod $p$ for $a=3$ mod $(p-1)$. In fact even for $a=3$ mod $15$.

and so on. Here are the factorizations of the first few numbers in the sequence:
sage: for a in [3..60]:
....:     N = 10^a + 333
....:     print('PRIME' if N.is_prime() else '     ', N, '=', N.factor())
....:

      1333 = 31 * 43
PRIME 10333 = 10333
PRIME 100333 = 100333
PRIME 1000333 = 1000333
      10000333 = 7 * 857 * 1667
      100000333 = 79 * 1265827
      1000000333 = 17 * 139 * 423191
      10000000333 = 19 * 526315807
      100000000333 = 347 * 288184439
PRIME 1000000000333 = 1000000000333
      10000000000333 = 7 * 1428571428619
      100000000000333 = 23 * 103 * 42211903757
      1000000000000333 = 74761 * 13375958053
      10000000000000333 = 2383 * 31727 * 132265613
      100000000000000333 = 29 * 3448275862068977
      1000000000000000333 = 31 * 59 * 1573679 * 347432263
      10000000000000000333 = 7 * 113 * 157 * 9049 * 8898632591
      100000000000000000333 = 167265053 * 597853515761
      1000000000000000000333 = 79 * 402859 * 31420988107753
      10000000000000000000333 = 439 * 22779043280182232347
      100000000000000000000333 = 2350339 * 42547053850529647
      1000000000000000000000333 = 43 * 11083 * 2098332035864691157
      10000000000000000000000333 = 7 * 17 * 84033613445378151260507
      100000000000000000000000333 = 244129 * 5233003 * 78276183759559
      1000000000000000000000000333 = 61 * 267139 * 13269967 * 4624481284981
      10000000000000000000000000333 = 19 * 5387 * 6550519 * 14915015630858219
      100000000000000000000000000333 = 4270499 * 23416467255934259673167
      1000000000000000000000000000333 = 4517 * 1159027 * 191010110705909287187
      10000000000000000000000000000333 = 7 * 431 * 3314550878355982764335432549
      100000000000000000000000000000333 = 10729 * 86117699364577 * 108230168748901
      1000000000000000000000000000000333 = 31 * 32258064516129032258064516129043
      10000000000000000000000000000000333 = 79 * 179 * 15443 * 45791851773167882549468491
      100000000000000000000000000000000333 = 3058170639611 * 32699287183242339516503
      1000000000000000000000000000000000333 = 23^2 * 4129417913867017 * 457778604072487381
      10000000000000000000000000000000000333 = 7 * 47 * 1579 * 19249611639085181456464115836463
      100000000000000000000000000000000000333 = 191 * 1223 * 26683 * 59000101 * 133534237 * 2036386152511
      1000000000000000000000000000000000000333 = 5183468525235985007 * 192921013242667183619
      10000000000000000000000000000000000000333 = 163 * 148250309 * 413825061582392670299853092099
      100000000000000000000000000000000000000333 = 17 * 41809 * 140695853552499954273847595437514861
      1000000000000000000000000000000000000000333 = 1049011 * 57358632811 * 16619622950403176530634173
      10000000000000000000000000000000000000000333 = 7^2 * 877 * 1699 * 2731 * 50152114413015514081748431875409
      100000000000000000000000000000000000000000333 = 1303 * 461596713173393557 * 166261952579593717268623
      1000000000000000000000000000000000000000000333 = 29 * 43 * 75344436707208992173 * 10643448330526537714943
      10000000000000000000000000000000000000000000333 = 19 * 526315789473684210526315789473684210526315807
      100000000000000000000000000000000000000000000333 = 79 * 21493 * 2140363 * 7291466811954023 * 3773753526074490611
      1000000000000000000000000000000000000000000000333 = 31 * 103 * 313185092389602254932665205136235515189476981
      10000000000000000000000000000000000000000000000333 = 7 * 669998127191476652301479 * 2132202122056322773969661
      100000000000000000000000000000000000000000000000333 = 173909 * 84555100781 * 6800457497533245880372446409581277
      1000000000000000000000000000000000000000000000000333 = 44319249889 * 863620780381437833 * 26126697381028729821509
      10000000000000000000000000000000000000000000000000333 = 467 * 919 * 12457 * 2145379 * 871866755809868463760709656795160507
PRIME 100000000000000000000000000000000000000000000000000333 = 100000000000000000000000000000000000000000000000000333
      1000000000000000000000000000000000000000000000000000333 = 2087 * 2423 * 15013 * 2594461187 * 6381048309061 * 795641525842044878663
      10000000000000000000000000000000000000000000000000000333 = 7 * 139 * 192106702699 * 923265594035567 * 57945269915555160600013037
      100000000000000000000000000000000000000000000000000000333 = 4603 * 803977 * 34682531183 * 779120470267938849650272081658642321
      1000000000000000000000000000000000000000000000000000000333 = 17 * 823 * 971 * 1658571172243 * 7209852591239633 * 6155615937870395875787
      10000000000000000000000000000000000000000000000000000000333 = 23 * 405720054989371 * 1071632036299123913967888194095887739225601
      100000000000000000000000000000000000000000000000000000000333 = 816271 * 122508333629395139604371587377231336161642395723969123
      1000000000000000000000000000000000000000000000000000000000333 = 79 * 35423 * 357344884625843825276933351962913318493991424437458749

The "easy prime" mentioned last as PRIME in the above has $54$ digits. The next ones are for $223$, then $232$ digits...
sage: for a in [3..2000]:
....:     N = 10^a + 333
....:     if not N.is_prime():
....:         continue
....:     print(f"The number 10...0333 with {len(str(N))} digits is a PRIME")
....:
The number 10...0333 with 5 digits is a PRIME
The number 10...0333 with 6 digits is a PRIME
The number 10...0333 with 7 digits is a PRIME
The number 10...0333 with 13 digits is a PRIME
The number 10...0333 with 54 digits is a PRIME
The number 10...0333 with 223 digits is a PRIME
The number 10...0333 with 232 digits is a PRIME
The number 10...0333 with 417 digits is a PRIME

We have a hard question behind the puzzle, i have no method to attack it.
But experimenting with such questions may be a good idea. Didactically, or as a programming exercise. Even "isolated questions" that are not quickly answered by "cheap tricks" or computer power can be hard. For instance, letting the above code go a little further, and printing the value of $a$ shows that the computer does not find an "easy factor" in case of $a=2111$. The isolated question is then:
$$
\text{Is }
10^{2111}+333
\text{ a prime?}
$$
(I will maybe try to use elliptic curves to have some more insight for this "special $a$" and for the next hard ones...)


largest group that fulfills a certain equality Let $S_n$ be the symmetric group of degree $n$ and let $G,H$ be subgroups of $S_n$. Is there a way to find the largest group $G'$ such that $GH = G'H$? What can we say about $G'$?

In particular im interested in the case $H=S_{M_1}\times S_{M_2} \times ... \times S_{M_k} \subseteq S_n$ for $M_1 \sqcup M_2 \sqcup ... \sqcup M_k = \{1, 2, ..., n\}$.

The largest subgroup $G'$ of $S_n$ for which $GH=G'H$ is just $GH$. Indeed, the equation $GH=GHH$ trivially holds, and if $K$ is another subgroup of $S_n$ for which $GH=KH$, then $K \leq KH=GH$ because $e \in H$.

Actually, the above argument would work for any group in place of $S_n$.


Dirichlet Domain of a Fuchsian Group Recall that a Fuchsian group is a discrete group $\Gamma \leq PSL(2, \mathbb{R})$ and that a Dirichlet domain for $\Gamma$ is a set $D \subset \mathbb{H}^2$ of the form $$\{z \in \mathbb{H}^2: d(z, p) \leq d(z, \gamma p) \forall \gamma \in \Gamma \}$$

where $p \in \mathbb{H}^2$ is a point not fixed by any element of $\Gamma - e$.

Every book I read very casually declares that because $D$ is an intersection of hyperbolic half-planes $\{z : d(z,p) \leq d(z, \gamma p)\}$, D is "bounded by segments of the geodesics $\{z :d(z,p) = d(z, \gamma p)\}$."  While this is somewhat clear if this intersection is finite, its not at all clear to me that this is true (or what this even means!) when D cannot be written as a finite intersection of halfplanes. Presumably you want to say something like: "D is the region bounded by a simple closed curve in $\mathbb{H}^2$ which is comprised of geodesic arcs and arcs along the x-axis" but I'm not at all sure if thats true or how to prove it.

Can someone give me a very explicit and technical statement (ie without using words like "is comrpised of") about the boundary of Dirichlet regions and some indication of how its proved? Thanks!

The key is that the collection if bisectirs defining Dirichlet domain is locally finite (this follows immediately from discreetness if the group). Therefore, the same proof as in the case if finitely many bisectors goes through.

There is some online software to generate data for 3-manifolds which might be worth looking at before going further. Here is a link to the screenshots: http://www.math.uic.edu/t3m/SnapPy/doc/screenshots.html


MST uniqueness: Is there a necessary and sufficient condition? Is there a necessary and sufficient condition for a minimum spanning tree to be unique?

I've searched and found a sufficient condition like this:

  If each edge has a distinct weight then there will be only one, unique minimum spanning tree.

But I can't find a necessary condition!

For any graph $G$, let $E'$ be the set of edges such that $e$ is the lightest (not necessarily unique) edge across some partition. $G$ has a unique MST $\iff$ $E'$ forms a spanning tree.

Proof:

If $G$ has a unique MST, then there are exactly $|V|-1$ edges which appear in every MST. Those edges must be unique lightest edges. Suppose not, and that there is an edge $e$ and $e'$ which are lightest edges across a cut. Then the MST in $G$ is not unique. This means that the edges in the MST are also all of the edges in $E'$, so $E'$ forms a spanning tree.

If $E'$ is a spanning tree, then it has $|V|-1$ edges which connect the graph. This means that every edge must be a unique lightest edge. Suppose not. Then across some edge, there exists $e$ and $e'$ which are both lightest edges, so $E'$ contains at least $|V|$ edges, and is no longer a spanning tree. By the Unique Cut Property, every lightest unique edge must appear in every MST.


Why can we prove mathematically that a formula to solve an (n+5) order polynomial does not exist? I understand that the quadratic equation can solve any second order polynomial. Furthermore, equations exist for polynomials up to fourth order. However, without a graduate level degree and a deep understanding of mathematics, is there an explanation of how we can prove that the equations for solutions to a fifth order and above don't exist?

Thanks in advance.

Here's a very vague intuition for why such a proof might be possible.

Many people are shocked when they hear that the general quintic cannot be solved.  "What does that mean?", they say. "Just because many people have tried and failed does not mean that's impossible.  Flight was impossible until the Wright Brothers took off at Kitty Hawk...", etc. etc.  But when we talk about solving the quintic, we are not talking about  as some Herculean task that only a genius can accomplish. We're just talking about something that, as it turns out, is not allowed according to the rules of the algebra game.

One way of proving that a given task is impossible is to find something that stays the same when you perform any of the possible "moves" that are allowed.  For example, suppose you were teaching someone the rules of checkers, you play a couple games, and then you point out that a piece on a black square can never find its way to a white square.  "What?!", your friend bellows.  "I don't see why not!  Flight was impossible until...", etc.  But all you need to do to prove this simple fact is to observe that at any given time, none of the rules allow you to move from black to white.

In the algebra game, you're given a polynomial $p(x) = a + bx + cx^2 + dx^3 + ex^4+fx^5$ and you're allowed to manipulate the symbols $a,b,c,d,e,f$ and numbers $1,2, \ldots$ in various ways.  For example, you can add $a$ and $b$ to get $a+b$; you're allowed to multiply, divide, etc.  This gives you what mathematicians call a field.

You're also allowed to use the symbol $\sqrt{}$, so to take square roots, cube roots, etc..  This lets you make complicated-looking expressions like $$\sqrt[8]{\frac{a + b \sqrt{d +e^3}}{e-\sqrt[4]{5}}}.$$  But to do this, you have to make your field bigger.  This is called a field extension.
As it turns out, there's a certain property that's always the same when you make field bigger by taking roots.  It's a little complicated to explain what - for that you must learn Galois theory.  But it's a little bit like the way a checker can't change colors.

It turns out that for certain polynomials $p(x)$, you can prove that any field that contains any root of $p(x)$ doesn't have that property.  That means that you can spend all day playing with roots, etc., and you will never solve the general quintic.  A clever mathematician like Évariste Galois can see the whole game at once and realizes that your task is not possible.

At one time I had the idea that if there wasn't one formula perhaps there was a variety of formula that could be selected from; even infinity one of an infinite.

The class I took explained it very carefully.
First it started with a straight edge and compass and computed the domain of calculations that could be done with them; and then showed trisecting the angle was not in that domain.  Then the book went on to construct the domain of radical solutions and showed it didn't include certain roots of fifth order polynomials.

Now there are certainly solutions; but the point is they are not even in the domain of radical algorithms.  No collection of formulas work because none of them can even reach those roots.
And as mentioned above Galois showed clearly which formulas could be solved and which couldn't.

It's like expressing pi as a rational number; you might get close but are never going to get there for a finite rational.


Minimum number of 'moves' required to re-order a list of numbers Consider the following 'game'. You have a list of $N$ integers ordered in descending order from $N$ to $1$. The aim of the game is to sort the list back into ascending order.
Rules:

You can only compare 2 numbers at any one time.

When comparing two numbers in the list, if $n_i$ is larger than $n_j$ (where $i$<$j$) then $n_j$ is moved 'up' the list to $n_{j-1}$.

Otherwise, there is no change.

My intuition is telling me that the minimum number of moves required to re-order the list in ascending order is the same as the number of moves a bubble sort sorting algorithm requires. This list of integers from $N$ to $1$ is the worst possible starting position for a bubble sort, so it would require $N^2$ moves. Am I correct?
When the comparisons are chosen randomly
The previous question about the minimum number of moves applies when you can choose which 2 numbers to compare and the order in which you choose the pairs of numbers to compare. But what if I extend the question to say that you have no choice over how the comparisons are made. That is to say, $i$ and $j$ (i.e. the indices we are comparing) are chosen randomly each move (although $i$<$j$ still applies). Is there any way to calculate how long on average it would take to return to an ascending ordered list? Is there a way to prove it would take a finite number of moves? Also, how does the time it takes depend on $N$? This seems a really interesting problem, has it been studied before?
Simulation results
I've had a go at writing some code to investigate this. For example, for $N=10$, it takes my script (which uses a random number generator to choose the indices $i$ and $j$ to compare) around 350 moves to reach ascending order. For reference, I calculated 'orderedness' using the formula from this answer - https://math.stackexchange.com/a/3201037/668220.

But for $N=30$ it takes around 10,000 moves

For $N=100$ it seems to take longer than 100,000 moves

I would be really interested to know if there is a way to describe this behaviour mathematically.

Optimal order will require $\frac{N \cdot (N - 1)}{2}$ steps, not $N^2$. To prove it we can count number of pairs $(i, j)$ such that $i < j$ but $n_i > n_j$. Initially this number is $\frac{N \cdot (N - 1)}{2}$, at the end it is $0$, and every step decreases it by at most $1$.

If we choose elements to compare by random, then it can take arbitrary time, as there can be cycle: initial order $3, 2, 1$, take $i = 1$, $j = 3$ - we get $3, 1, 2$. Then take again $i = 1$, $j = 3$ - we get back to initial order.

However the array will be sorted with probability $1$ after finite number of moves. For example, number $n$ never moves to left, so it will eventually get to the last position. After it we have to sort remaining $n - 1$ elements (steps when $j = n$ will do nothing), and it will be done in finite number of steps almost surely.

This also gives some upper bound on average of moves we need: if $f(n)$ is average number of steps we need to sort the worst (requiring maximum number of steps on average) permutation of $n$ elements, then $f(n) \leqslant n^3 + \frac{n}{n - 2} \cdot f(n - 1)$: less then $n^3$ steps involving $n$ on average (as probability of a step to move $n$ to right is at least $\frac{1}{n^2}$, and we need at most $n - 1$ such steps), and after that we need $f(n - 1)$ steps not involving $n$ - and probability of step to not involve $n$ is $\frac{n - 2}{n}$. This bound is, of course, very bad - we likely will not end up with the worst sequence after moving $n$ to the last position.


Calculation with Hyperbolic Cosine Could you please check my work?

  $\cosh \left(\ln \sqrt{5}\right) =\ ?$

\begin{align*}\cosh(x) &= \frac{e^x + e^{-x}}{2} \\ \\
\frac{e^{\ln \sqrt{5}} + e^{-\ln \sqrt{5}}}{2} &=
\frac{\sqrt{5} + \frac{1}{\sqrt{5}}}{2}\\
&=\frac{\sqrt{5}}{2}+\frac{1}{2\sqrt{5}} \\
&= \frac{3}{\sqrt{5}}
\end{align*}

Thanks.

Umm...this seems like it has an error to me, unless I misread something.  (There appear to be some LaTeX problems, so it's entirely possible I've misread.)

But

\begin{align}
\cosh \left( \ln \sqrt{5} \right)
    & = \frac{1}{2} \left( e^{\ln \sqrt{5}}+e^{-\ln \sqrt{5}} \right) \\
    & = \frac{1}{2} \left( \sqrt{5} + \frac{1}{\sqrt{5}} \right) \\
    & = \frac{1}{2} \left( \frac{5+1}{\sqrt{5}} \right) \\
    & = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}
\end{align}

That's correct, no errors.

As a note, its well worth learning how to present questions well (like using the LaTex capabilities of this website, as all the editors have been doing).
It makes it much easier for viewers to read and understand your question. It also encourages you to think about what's important in your question, as this determines a lot about how you present it.


Local Truncation error We have error formulas for Gaussian Quadrature and polynomial interpolation with 10 Gaussian points, use them to derive formulas for the local truncation error, where $$P = \frac{f^n(\xi)}{n!}\prod_{i=0}^{n}(t-t_i)$$ is the error for Polynomial interpolation,
and error of Gaussian Quadrature:
$$Q = \frac{(b-a)^{2n+1}(n!)^4}{(2n+1)![(2n)!]^3}$$
How can I derive the formula for local error truncation using error formula above?

Are you in Math 128B with Professor Strain?

If I'm not mistaken, I think this is the problem he's trying to solve:
https://math.berkeley.edu/~strain/128b.S15/p10.pdf

Here are some notes on the spectral integration method that must be used:
https://math.berkeley.edu/~strain/128b.S15/bvp.pdf

Some thoughts in the right direction: You're integrating over $[a, a + Ih]$ and $[a + (I+1)h, b]$ using Gaussian Quadrature, so you know the error for that exactly. How do you find the error of the spectral integration method that works over $[a + Ih, s_{Im + i}]$?


Are matrices rank 2 tensors? I know that this is sometimes the case, but that some matrices are not tensors.
So what is the intuitive and specific demands of a matrix to also be a tensor?
Does it need to be quadratic, singular or something else?

Some sources I read seem to suggest that all rank 2 matrices are tensors while other just claims that "some" matrices are rank 2 tensors.

What's the connection between tensors and matrices?

This question doesn't have a single good answer, because there isn't a universally agreed upon definition of "tensor" in mathematics.  In particular:

Tensors are sometimes defined as multidimensional arrays, in the same way that a matrix is a two-dimensional array.  From this point of view, a matrix is certainly a special case of a tensor.
In differential geometry and physics, "tensor" refers to a certain kind of object that can be described at a point on a manifold (though the word "tensor" is often used to refer to a tensor field, in which one tensor is chosen for every point).  From this point of view, a matrix can be used to describe a rank-two tensor in local coordinates, but a rank-two tensor is not itself a matrix.
In linear algebra, "tensor" sometimes refers to an element of a tensor product, and sometimes refers to a certain kind of multilinear map.  Again, neither of these is a generalization of "matrix", though you can get a matrix from a rank-two tensor if you choose a basis for your vector space.

You run into the same problem if you ask a question like "Is a vector just a tuple of numbers?"  Sometimes a vector is defined as a tuple of numbers, in which case the answer is yes.  However, in differential geometry and physics, the word "vector" refers to an element of the tangent space to a manifold, while in linear algebra, a "vector" may be any element of a vector space.

On a basic level, the statement "a vector is a rank 1 tensor, and a matrix is a rank 2 tensor" is roughly correct.  This is certainly the simplest way of thinking about tensors, and is reflected in the Einstein notation.  However, it is important to appreciate the subtleties of this identification, and to realize that "tensor" often means something slightly different and more abstract than a multidimensional array.

All matrices are not tensors, although all tensors of rank 2 are matrices.

Example: Matrix T (T11=x , T12=-y , T21=x^2 , T22=-y^2) .This matrix is not tensor rank 2. We test matrix T to rotation matrix A: a11=cos(theta),a12=sin(theta),a21=-sin(theta),a22=cos(theta). Now Expand tensor equation rank2,for example T'11=Σ(a1ia1jTij) (1)

Now calculate T'11=x'=xcos(theta)+ysin(theta) (2)

You see (1) unequal to (2) then conclude Matrix T isn't a tensor rank2!!

Tensor must follow the conversion(transformation) rules, but matrices generally are not.


Loopy integral $\int xe^x\sin(x)$ How to find integral of: $$\int xe^x\sin(x)$$
I know it should be repeated integration by parts, but i do not know how many times i should do this and when to stop.

You had better use the formula $\sin x = \dfrac{e^{ix}-e^{-ix}}{2}$. In this way you obtain an integral like $\int xe^{ax}dx$ and it is easy to reach your goal by integration by parts.

Also, another ways are possible.

Treat x*sinx as first term and e^x as the 2nd term. Then use integration by parts.


Matrix factorization as a product of triangular matrices Is it true that any square matrix $A $ can be factorized as a product of only triangular matrices?
That is, can we write $A $ as $\prod_{i=1}^k B_i$, where every $B_i$ is either a lower or an upper triangular matrix (for some natural $ k $)?

Note $ k $ is not assumed to be 2 above. So the question is not (directly) about the $LU$ decomposition.

Comment: We know that for any square matrix $A$ we have: $A=PLU$ where $P$
is a permutation matrix, $L$ is a lower triangular matrix and $U$ is an upper triangular matrix. So the question possibly boils down to whether any permutation matrix can be factorized to triangular matrices.

The answer to your question is yes. A permutation matrix is in fact a product of permutation matrices associated to transpositions, which means a matrix obtained from the identity matrix by interchanging two rows. Reduce the problem to this case and notice that this case can be easily deduced from the case of the $2\times 2$ matrix $$\left(\begin{array}10 & 1 \\ 1 & 0\end{array}\right).$$
This matrix can be transformed into the identity matrix by using the following elementary transformations (which correspond to triangular matrices): add the second row to the first, substract the first column from the second, substract the first row from the second, and last multiply the second row by $-1$.

Moreover, in the paper of Nagarajan et al., Products of three triangular matrices, Linear Algebra and its Applications, 292(1999), 61-71 it is proved that any matrix $n\times n$ over a field is a product of at most three triangular matrices.

Any square matrix if diagonalizable can be written as the product of as many matrices as you want. For instance, $A=SDS^{-1}$ for some invertible matrix $S$, then the decomposition $A=B^{2}$ where $B=SD^{1/2}S^{-1}$ is equally valid. For our peace of mind, if we want distinct matrices, Any $B_{i}=SD^{\theta_{i}}S^{-1}$ such that $\sum_{i=1}^{k}\theta_{i}=1$ will give you $A=\prod_{i=1}^{k}B_{i}$. Now each of this $B_{i}=L_{i}U_{i}^{T}$, Hence you get your square matrix as product of as many triangular matrices as you want.


Does T3 spaces implies every open set contains a closure of an open set I have two questions regarding this spaces:

First Question:  Suppose $(X,\tau)$ is T3. Does it implies that for every open set $U \subset X$ and for every $x \in U$ there exist an open set $V$, s.t. $x \in V \subseteq U$ and $V \subseteq \overline{V} \subseteq U$?
It seems to me that this is true since:
Take $U$ open and $x \in U$. Then $U^c$ is closed. $X$ is T3 so there are two open disjoint sets $V_1,V_2$ s.t. $x \in V_1$ , $U^c \subset V_2$. So $V_1$ is the set we are looking for. since  $\overline{V_1} \subseteq {V_2}^c \subseteq U$

Am I right?

Second question: I am looking for an example for a space which is T3 but not T0,T1,T2.
I have looked in "Counter examples in topology" examp. 90, but I don't know what Tychonoff Corkskrew is..

Thank you!! Shir

The answer to your first question is yes, and your proof is correct. (This is sometimes even taken as the definition of regularity.)

As for your second question, if you use what I consider the standard definition of $T_3$, every $T_3$ space is automatically $T_0,T_1$, and $T_2$. To me a $T_3$ space is one that is regular (points can be separated from closed sets not containing them) and $T_1$, and such spaces are automatically $T_2$. Some people use $T_3$ to mean what I call regular, which is just the property that points can be separated from closed sets not containing them. In that case you can let $X=\Bbb R\cup\{p\}$, where $p$ is some point not in $\Bbb R$, with the following topology $\tau$:

$$\tau=\{U\subseteq\Bbb R:0\notin U\in\mathscr{E}\}\cup\{U\cup\{p\}:0\in U\in\mathscr{E}\}\;,$$

where $\mathscr{E}$ is the usual Euclidean topology on $\Bbb R$. This simply makes $p$ a second copy of $0$, so that $X$ is obtained from $\Bbb R$ by splitting $0$ into $0$ and $p$ and keeping both of these in the same open sets. Clearly $X$ is not $T_0,T_1$, or $T_2$, since neither of the points $p$ and $0$ can be separated from the other, but the space is still regular.

Your mention of the Tikhonov corkscrew suggests that you may instead be looking for an example of a $T_3$ space (in my sense) that is not Tikhonov. If so, you may find this answer helpful: it explains in more detail part of the construction of the corkscrew given by Steen & Seebach and presents fully a simpler example of a $T_3$ space that isn’t Tikhonov.

A simple answer to your question is any Set X with at least two elements with the indiscrete topology T={empty set, X}, then it is clear that (X, T) is T3 (with respect to defintion of T3 in the book "Counter examples in topology") but it is neither To nor T1 nor T2
best regards
Dr. Halgwrd M. Darwesh
Assistant Professor
in General Topology and Dimensions Theory
Math. Dep. , College of Sci. University of Sulaimani


How to solve $(x^2 + 2xy)y' = 2(xy + y^2)$, with $y(1) = 1$ I've just started learning differential equations(so far I've learnt about separate variables and homogeneous equations) and I'm trying to solve an exercise that is listed as an example for a homogeneous equations. I substituted $\cfrac yx$ for $u$ and after I integrated the equation I still can't get the value of $u$.
This is the exercise:

$(x^2  + 2xy)y' = 2(xy + y^2)$

$y(1) = 1$

$y' = 2(xy+y^2)/(x^2 + 2xy)$

$\cfrac yx = u => y = xu $

$u +xu' = 2(u + u^2)/(1+2u) $

$xu' = u/(1+2u)$

$u' = u/((1+2u)x)$

Knowing that u' = du/dx

$(1 + 2u)/u du = 1/x dx$

After integration I got the following:

$\ln(|u|) + 2u = \ln(|x|) + c$

$y(1) = 2 => c = \ln2 + 4$

I may have messed up somewhere. What is the easiest way to solve this problem?

$$(x^2  + 2xy)y' = 2(xy + y^2)$$
I cannot see an easier way that what you did. Your solution $\quad \ln|u|+2u=\ln|x|+c\quad$ is correct.

And with condition $y(1)=1\quad\implies\quad c=2\quad\implies\quad ue^{2u}=e^2x$

Nevertheless it is possible to go a bit further, thanks to the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html

$$2ue^{2u}=2e^2x \quad\implies\quad 2u=W\left(2e^2x  \right)\quad\implies\quad u=\frac12\: W\left(2e^2x  \right)$$
$$y(x)=\frac{x}{2}\: W\left(2e^2x  \right)$$


First Order Logic - exactly one predicate is true I am working on a question for an assignment and I am to declare a clause in Conjunctive Normal Form that says exactly one of three predicates is true. Or given a bit of context - given three suspects and exactly one is guilty.

My first attempt to capture this in FOL (where A, B, and C are constants) Guilty$(A) \lor \text{Guilty}(B) \lor \text{Guilty}(C)$ but I do not think that this restricts to exactly one.

Attempts to better capture as only one can be guilty is Guilty$(x) \land \lnot \text{Guilty}(y) \land \lnot \text{Guilty}(z)$ thought I do not think this is quite right either as it maybe wouldn't play nice when using resolution proofs?

I know I mentioned this is for homework but this an isolated clause - the actual question is to write an answer extraction proof which I am not asking about as I want to figure it out on my own but this clause is causing a bit of trouble with that.

Thanks in advance.

If you prefer a more systematic approach, the CNF of a statement can be found easily with the following 3 steps:

$1.$ Negate the statement.
$2.$ Construct the DNF of the negated statement.
$3.$ Negate the statement again (once it is in DNF), and apply DeMorgan's Laws.

An example below...

Given propositions $G(a),G(b),$ and $G(c)$, consider each possible scenario:
scenario $1$: $G(a)$ is true while $G(b),G(c)$ are false
scenario $2$: $G(b)$ is true while $G(a),G(c)$ are false
scenario $3$: $G(c)$ is true while $G(a),G(b)$ are false
Now we can translate this rather intuitively into DNF as follows:
$(G(a) \wedge \neg G(b) \wedge \neg G(c)) \vee (G(b) \wedge \neg G(a) \wedge \neg G(c)) \vee (G(c) \wedge \neg G(a) \wedge \neg G(b))$
But we have a problem here: we have our statement in DNF, but what we really want is the DNF of the negation. So, let's begin step $1$ above by negating the statement and then use DeMorgan's Law to find...
$\neg \bigg[ (G(a) \wedge \neg G(b) \wedge \neg G(c)) \vee (G(b) \wedge \neg G(a) \wedge \neg G(c)) \vee (G(c) \wedge \neg G(a) \wedge \neg G(b)) \bigg]$
$\Leftrightarrow \bigg[ \neg G(a) \vee G(b) \vee G(c) \bigg] \wedge \bigg[ \neg G(b) \vee G(a) \vee G(c) \bigg] \wedge \bigg[ \neg G(c) \vee G(a) \vee G(b) \bigg]$
We now have the negated statement in CNF. My suggestion is to convert this into DNF by applying the distribution law numerous times and then simplifying with other applicable equivalencies. Once you have done this, you have completed step $2$, and you may complete the final step - step $3$ above - rather straightforwardly.

Thinking about it more - would the clause (assuming x, y, and z are variables and can be substituted)  $Guilty(x) \lor Guilty(y) \lor Guilty(z)$ capture the sentence exactly only one is guilty? Pardon my ignorance, I am quite new to this.


How to show that $x_n=-\sqrt{n} + n\ln\Big(1+\frac{1}{\sqrt{n}}\Big)$ is decreasing? I am a non-mathematician who knows some elemententary calculus ans I want to prove that the sequence $(x_n)$ given by

$$
x_n=-\sqrt{n} + n\ln\Big(1+\frac{1}{\sqrt{n}}\Big)
$$

is decreasing. Is there an elegant way to show this?

This does not answer your question but might be helpful. Let $x>0$ be a real number.  Then it follows from $$\log(x) = \int_1^x \frac{dt}{t}$$ and the estimates $$2-t <\frac{1}{t}<1-\frac{(t-1)x}{x+1}$$ for $t\in(1,1+1/x)$ that

$$
x-\frac{1}{2} < x^2\log\left(1+\frac{1}{x}\right)<x-\frac{1}{2}+\frac{1}{2(x+1)}.
$$

In particular $$\lim_{x\to\infty}x^2\log\left(1+\frac{1}{x}\right)-x=-\frac{1}{2}$$ which might or might not be what you are after.

as @user130512 said, you could prove that $x_{n+1}$ is greater than $x_n$. Sometimes, it can help to look at the ratio between the two, that is $x_{n+1}/x_n$. If it is greater than $0$, $x_{n+1}$ is greater than $x_n$.
I think you also should be able to look at the derivative of $x_n$ to deduce whether it is positive or negative. So,  positve $ \frac{d(x_n)}{dn}$ means it is increasing while negative would mean it is decreasing. Note that it can be increasing for one period but decreasing for another, so, as in your case, you would need to prove that $ \frac{d(x_n)}{dn}$ is smaller than $0$ for all values of $n$ to show that it is negative since the derivative will not be a constant.

The derivative in your case is:

$$\dfrac{2(n+\sqrt{n})\log\left(\tfrac1{\sqrt{n}}+1\right)+1}{2(n+\sqrt{n})}$$

As you can see, we only need to find what values would make this function equal to zero. Because of this, we can work with the nominator only, $$(n+\sqrt{n})\times \log\left( \frac{1}{\sqrt{n}} +1\right)+1 =0$$

You can also see it graphically in the derivative:


Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:

  What is the blue area?

  Hint: Think outside the box

My answer:

I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:

This then gave me the following equation to solve:

\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}

And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}

I solved these to obtain:

$$a=2,
b=2,
c=3,
d=1,
y=4$$

And thus $$x=4^2-9=7$$

My question:

Is there another way I could have solved this, using the hint to think outside the box?

The hint may assume two things: to draw extra lines 1) outside; 2) inside the box so that the solution is simple to understand.

Here is yet another method: draw the lines to divide the square into four rectangles indicated by $A, B, C$ and $3+3$:

$$\begin{cases}
A+B=8 \\
A+C=4 \\
CB=6A
\end{cases} \stackrel{(1)-(2)}\Rightarrow \begin{cases}
B=C+4 \\
A=4-C \\
C(C+4)=6(4-C)
\end{cases} \Rightarrow $$
$$C^2+10C-24=0 \Rightarrow C=2, A=2, B=6$$

Hence: Blue area = $(A+B+C+6)-(2+3+4)=16-9=7.$

P.S. I wonder what brilliant method the proposer (on the brilliant.org) has.

Assume that the hypotenuses of the yellow and green triangles bisect the right side of the quadrilateral, which is assumed to be a square with side s.  Area formula is (1/2)bh where b=s and h=s/2, so the area of the yellow triangle is equal to ss/4, which is equal to 4.  ss is thus equal to 4*4=16, so s=4.  The area of the square is also equal to s*s, which is 16.  The blue triangle is thus equal to 16-4-3-2 = 7.


Is a discrete set inside a compact space necessarily finite? Is it true that if $A$ is discrete as a subspace of $X$, and $X \;$ is compact, then $A$ is finite?

If this doesn't hold, then does it hold for $X\;$ manifold?

It is not true in general. Let $X = \{0\}\cup\{2^{-n}:n\in\mathbb{N}\}$ with the topology inherited from $\mathbb{R}$; then $X$ is compact, and $X\setminus \{0\}$ is an infinite discrete subset of $X$. Of course every closed discrete subset of a compact space is finite, so infinite discrete subsets won’t be closed, but in general they will exist. For instance, the space $X$ just described can be embedded in any infinite compact metric space.

Singleton sets are open .there union is a cover for X.so this is the cover which do not have a subcover hence X is not compact


The relationship between matrix rank and its characteristic polynomial coefficients Given the matrix characteristic polynomial coefficients. Is there a quick way to determine the rank of the matrix?

In general nothing can be said about rank of the matrix by merely looking at char polynomial. Take $$A=\left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right]$$ Rank$A=1$, though char polynomial is $x^2=0$. But if your matrix is diagonalizable, ''effective degree'' of the characteristic polynomial is equal to the matrix rank, since for a diagonalizable matrix, rank turns out to be number of non zero eigenvalues.

For a $n\times n$ matrix, if the characteristic polynomial is $X^mP(X)$ where $P$ is not divided by $X$, the rank is $n-m$.


Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$ So i was given this question. Show by combinatorial argument that ${2n\choose 2} = 2{n \choose 2} + n^2$

Here is my solution:

Given $2n$ objects, split them into $2$ groups of $n$, $A$ and $B$. $2$-combinations can either be assembled both from $A$, both from $B$ or one from each. There are ${n \choose 2}$ from $A$ and ${n \choose 2}$ from $B$. For the mixed pair, each choice from $A$ can be coupled with $n$ choices from $B$ so the total is $n^2$.

Therefore ${2n\choose 2} = 2{n \choose 2} + n^2$

Is this correct. Is there any other combinatoric proof to solve this?

Just another way to think about it:

Let's find the coefficient of $x^2$ in the expansion of $(1+x)^{2n}$. On one side, it's equal to ${2n}\choose2$. On the other hand, it's equal to the coefficient of $x^2$ in the expansion of $(1+x)^n(1+x)^n$ which is equal to the sum of products of coefficients of $x^0$ and $x^2$, $x^1$ and $x^1$, and $x^2$ and $x^0$ in the expansion of $(1+x)^n$. That's equal to ${n\choose2}\cdot{ n\choose0}+{n\choose1}\cdot{ n\choose1}+{n\choose0}\cdot{ n\choose2}$ and the result follows.

A more direct (and, in my opinion, simple) approach is to expand and simplify until both sides are equivalent.

$${{2n}\choose{2}}=2{{n}\choose{2}}+n^2$$

First, expand both sides using ${{n}\choose{k}}=\frac{n!}{k!(n-k)!}$:

$$\frac{(2n)!}{2!(2n-2)!}=2\left[{\frac{n!}{2!(n-2)!}}\right]+n^2$$

Expand again, this time focusing on the factorials:

$$\frac{(1)(2)(3)...(2n-2)(2n-1)(2n)}{(1)(2)*(1)(2)(3)...(2n-2)}=2\left[\frac{(1)(2)(3)...(n-2)(n-1)(n)}{(1)(2)*(1)(2)(3)...(n-2)}\right]+n^2$$

In this format, it's easy to see that some variables of the factorials on each side cancel each other out:

$$\frac{(2n-1)(2n)}{2}=(n-1)(n)+n^2$$

Factor in the monomials on each side:

$$\frac{4n^2-2n}{2}=n^2-n+n^2$$

Simplify both sides for the last time:

$$2n^2-n=2n^2-n$$

This method takes a little longer but I feel it's more intuitive and solid.


Writing vectors in the $\hat{r}$ and $\hat{\theta}$ basis I am having a bit of confusion about writing down vectors in the $\hat{r}$,  $\hat{\theta}$ basis. Let's say I want to write $-1$ $\hat{i}$ $+$ $1$ $\hat{j}$ in the $\hat{r}$,  $\hat{\theta}$ basis. Could I write $1$ $\hat{r}$ $+$ $1$ $\hat{\theta}$ so long as I specify that $\hat{r}$ is "at" $\pi/2$ radians? (...and in the above question $\hat{\theta}$ would be "at" $\pi$ radians)

Could I have described the exact same point with $\sqrt2\hat{r}$ $+$ $0$ $\hat{\theta}$ so long as I specify this time that $\hat{r}$ is "at" $3\pi/4$ radians? (...and $\hat{\theta}$ would be "at" $5\pi/4$ radians, but we wouldn't move along the $\hat{\theta}$  direction this time).

Yes, you can do so.

Remember the vectors $\hat r,\hat\theta$ are defined by the equations
$$\begin{cases}\hat r=\cos\theta\hat i+\sin\theta\hat j\\ \hat\theta=-\sin\theta\hat i+\cos\theta\hat j.\end{cases}\tag{1}$$

This is equivalent to $$\begin{cases}\hat i=\cos\theta\hat r-\sin\theta\hat\theta\\ \hat j=\sin\theta\hat r+\cos\theta\hat\theta.\end{cases}\tag{2}$$

To see why this is true, remember that each of $\hat r,\hat\theta$ are obtained by rotating each of $\hat i,\hat j$ by $\theta$ radian anti-clockwise (which is what $(1)$ does). So we can rotate each of $\hat r,\hat\theta$ by $-\theta$ radian anti-clockwise to obtain $\hat i,\hat j$ (which is what $(2)$ does).

So $-1\hat i+1\hat j=-(\cos\theta\hat r-\sin\theta\hat\theta)+(\sin\theta\hat r+\cos\theta\hat\theta)=(\sin\theta-\cos\theta)\hat r+(\sin\theta+\cos\theta)\hat\theta$. Plug in the value $\theta$ and you can express $-1\hat i+1\hat j$ in terms of $\hat r,\hat\theta$.

For case 1, you put $\theta=\frac{\pi}{2}$, giving $(\sin\theta-\cos\theta)\hat r+(\sin\theta+\cos\theta)\hat\theta=\hat r+\hat\theta$. For case 2, you put $\theta=\frac{3\pi}{4}$, giving $(\sin\theta-\cos\theta)\hat r+(\sin\theta+\cos\theta)\hat\theta=\sqrt2\hat r$.

You need a combination. You are correct in saying that your value of $\hat{r}$ needs to be $\sqrt{2}$, but by saying $0\hat{\theta}$ you are saying that your point lies in the $\hat{i}$ direction. Saying that it is "at" $3\pi /4$ radians is precisely what $3\pi /4 \hat{\theta}$ means.

Your answer in polar coordinates $(r, \theta)$ should be $\left(\sqrt2 , \frac{3\pi}{4}\right).$

Think of it in terms of $re^{i\theta} = r\cos\theta\hat{i} + r\sin\theta\hat{j}$


Why can we use the minimal polynomial to generate the kernel in this case too? This might be a really easy question, but I am confused. Further more, I am not satisfied with the title of my question. Please feel free to improve it.

This is kind of a standard setting when looking at extensions of Dedekind rings (in algebraic number theory):

  Let $A$ be a Dedekind ring and $K$ its field of fractions. Let further $L/K$ be a finite separable field extension and $B$ the integral closure of $A$ in $L$.

In this situation we always have a primitive element $\beta \in B$ such that $L = K(\beta)$. And of course $\beta $ has a minimal polynomial — let's call it $m_{\beta,K} \in K[X]$. Since $\beta$ is integral over $A$, we even know $m_{\beta,K} \in A[X]$.

If we look at the homomorphism

$$ \begin{align}\Psi \colon K[X] &\longrightarrow K[\beta] \\
 f &\longmapsto f(\beta) \end{align}$$
which substitutes $\beta$ for $X$, we have $\ker(\Psi)=(m_{\beta,K})$, i.e. the principal ideal $K[X]m_{\beta,K}$. So far so good.

But now I often read that for the homomorphism

$$ \begin{align}\varphi \colon A[X] &\longrightarrow A[\beta] \\
 f &\longmapsto f(\beta) \end{align}$$

we also have $\ker(\varphi)=(m_{\beta,K})$, where $(m_{\beta,K})$ now of course means the principal ideal $A[X]m_{\beta,K}$.

It is clear to me that $A[X]m_{\beta,K} \subseteq \ker(\varphi)$. But why exactly do we have $A[X]m_{\beta,K} \supseteq \ker(\varphi)$? I am also confused because $A[X]$ is not necessarily a PID (right?), which would explain $\ker(\varphi)=A[X]m_{\beta,K}$.

In general, when you have a field extension $L/K$ and $\beta \in L$ algebraic over $K$, the minimal polynomial $m_\beta \in K[X]$ of $\beta$ over $K$ divides any polynomial that has $\beta$ as a root i.e. is in the kernel of the evaluation map. To see this remember that since $K$ is a field, $K[X]$ is a PID. Now in your case with the evaluation map $\phi: A[X] \rightarrow A[\beta]$, $\ker(\phi)$ may not be principal but we have $(m_\beta) \subset \ker(\phi)$ because $\beta$ is integral over $A$. Now since $m_\beta$ divides any polynomial in $K[X]$ (including $A[X]$) that vanishes at $\beta$, it divides everything in $\ker(\phi)$. Here I made a mistake and assumed it was obvious that after division by $m_\beta$, the remaining polynomial would lie in $A[X]$. One way to see that this is true because $A$ is integrally closed and $K$ is its fraction field (Gauss's Lemma). Knowing this, we have $\ker(\phi) \subset (m_\beta)$.

Another way to see that division by $m_\beta$ is valid in $A[X]$ is to note that in any polynomial ring the division algorithm works when dividing by a monic polynomial. This only works because we already know $m_\beta \in A[X]$.

  Monic Polynomial Division (Algorithm):
  Let $R$ be a commutative ring with $1$ and $f,g \in R[X]$ with $g \neq 0$ and monic. Then there exists $h,r \in R[X]$ such that $f=gh +r$ with $\deg(r)<\deg(g)$ or $r=0$.

Proof: The cases when $f=0$ and $\deg(g)>\deg(f)$ are obvious so assume otherwise. Inductively assume that for polynomials of degree less than $f$, this is true. Then simply get rid of the highest order term. If $f=a_n x^n+a_{n-1} x^{n-1}+\cdots$ and $g$ has degree $m \leq n$, then $f-a_n x^{n-m}g$ has lower degree so we can find $h,r$ with $f-a_n x^{n-m} g=gh+r$. Then $f=(a_n x^{n-m}+h)g+r$, as required.

I wanted to write a short answer to my own question for future reference.

Let $f \in A[X]$ such that $f(\beta)=0$ and let $a \neq 0$ be the leading coefficient of $f$ (w.l.o.g. $f \not\equiv 0$).

Proof I We have $\left(\frac{1}{a}f\right)(\beta) = 0 $ and therefore $f \in \ker(\Psi) = K[X] m_{\beta,K}$. Hence there is a $g \in K[X]$ such that
  $$ \frac{1}{a}f = g \cdot m_{\beta,K}. $$
Since both $\frac{1}{a}f$ and $m_{\beta,K}$ are monic, $g$ has to be monic as well. Now Gauss's Lemma grants $g \in A[X]$ and we get
  $$f = \underbrace{ag}_{\in A[X]} \cdot m_{\beta,K}.  $$
This shows $f \in A[x]m_{\beta,K}$ and further $\ker(\varphi) \subseteq A[x]m_{\beta,K}$.

Proof II By polynomial division in $A[X]$ (note $m_{\beta,K} \in A[X]$ is monic) we find $g,r \in a[X]$ with
  $$ f = g \cdot m_{\beta,K} + r$$
and $\deg r \lt \deg m_{\beta,K}$. Since $f(\beta)=0$ we must have $r(\beta)=0$, which means $r \in \ker(\Psi)=K[x]m_{\beta,K}$. Because of $\deg r \lt \deg m_{\beta,K}$ it follows that $r \equiv 0$ and thus
  $$ f = g \cdot m_{\beta,K} $$
This shows $f \in A[x]m_{\beta,K}$ and further $\ker(\varphi) \subseteq A[x]m_{\beta,K}$.


Rudin's proof that the Cantor set has no segments In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(\alpha,\beta)$ contains a segment of the form $(\frac{3k+1}{3^{n}},\frac{3k+2}{3^{n}})$ if $3^{-n}\lt\frac{\beta-\alpha}{6}$. I don't know how $\frac{\beta-\alpha}{6}$ is obtained.

We want to show that there is an integer $k$ such that $\alpha\le\frac{3k+1}{3^n}$ and $\frac{3k+2}{3^n}\le\beta$,

so we want to have $\frac{\alpha(3^n)-1}{3}\le k\le\frac{\beta(3^n)-2}{3}$.

Such an integer $k$ will exist if $\frac{\beta(3^n)-2}{3}-\frac{\alpha(3^n)-1}{3}\ge1$, so

$(\beta-\alpha)3^n\ge4$ gives $3^{-n}\le\frac{\beta-\alpha}{4}$.

If Cantor set contains a segment $(\alpha,\beta)$, then it contains an interval $$\left(\frac {3k+1}{3^m}, \frac {3k+2}{3^m}\right)$$ for some $k$ and $m$. This surely contradicts the claim that the Cantor set has no common points with any such intervals.

From the above, we can see that $$\beta-\alpha >3^{-m}$$ is needed. If we have $$\frac {\beta-\alpha}{6}>3^{-m},$$
then it is more than needed.

-------Edited by the author: to see why we just need $3^{-m}<\beta-\alpha$. ----

Suppose that Cantor set $P$ contains an interval $(\alpha,\beta). $ We expand $\alpha$ in base-3.
$$\alpha = \frac {a_1}{3}+\frac {a_2}{3^2}+\cdots+ \frac {a_n}{3^n}+\cdots,a_i=0,1,2,$$
We want to find two rational numbers, one is strictly larger than $\alpha,$ the other is strictly less than $\beta$. Let $m$ to be determined. We construct the first one from the $m$-th position of the decimal expansion of $\alpha$: $a_m$ is at most $2$; moreover the tail of the series expansion is at most $3^{-m}$; we sum the first $m-1$ terms using common denominators to see that there exists an integer $k$. To be strictly larger than $\alpha,$, adding one more $\frac {1}{3^m}$, we obtain the first rational is $$\frac {3k+2+1+1}{3^m}.$$
For the second rational, in order for the numerator of the above adding $1$ so that $(\frac {3k+2+1+1}{3^m}, \frac {3k+2+1+1+1}{3^m})$ is in $(\alpha,\beta)$, we need
$$\frac {1}{3^m}<\beta-\alpha.$$
This alternatively gives $$3^{-m}<\beta-\alpha.$$


Math Major: How to read textbooks in better style or method ? And how to select best books? I'm a bachelor student of Math major. The question is about the better way to select and read maths text- or non-text- books, without considering the course and exam. (Because we follow the lecture notes during the course, and the exam is totally covered in lecture notes, so it's not an issue, but just about how to really learn)

Question:

(1) Reading the book one by one, or just focus and re-read lots of times on one certain book, and then continue with little others ? Because recently I've tried the first way, for example, to learn Calculus, I took one book, the read it from first page to the last and finish some exercises, and then immediately read another calculus book, doing the same thing, and then the third one, and so forth. After this, I realize it is really bad result, it seems that I don't master the calculus pretty well, just memorize lots of things. So, I'm thinking about whether the second way is better. Because some friends told me before, said that there're lots of good maths textbooks, I should read as much as possible.

(2): If the second way is better, could I just use the UTM and GTM series(published by Springer, Undergraduate Texts in Mathematics) for all the maths courses ? Because there're too many, tons of books in the market, I'm not sure how to select the perfect books. But lots of people said that the UTM & GTM are two perfect series in Maths. And my teacher told me that it is better to spend more time on the books written by the famous mathematicians who are the masters in their areas. Less time on the so-called popular textbooks, like thousand-page widely-used textbooks. He's not saying that the latter one is bad, it just means that the latter one is too specified for the general course in university, passing exam , or something like that. Although it's organized well, ton's of exercises, examples and so on,  the former one is better to grab the deep insight of the field and focus more on thinking, not just knowledge. So, briefly, is it good to just use to the whole UTM series for every course in math bachelor program?

In my opinion it's much better to pick some book and read it in depth, solving many of the exercises while reading. If you read a 100 books without actually concentrating on what you're doing, it will be of no help. Of course, reading a book thoroughly and attentively is slow, so don't expect it to be quick.

You seem to be somewhat biased by your experience so far. Apart from the first few courses, there are no expensive, shiny, popular textbooks - only the dense, terse, substantial ones. If you really want to understand math (rather than to be just able to apply it), then the shiny books don't help you at all.

Finally, I have no idea whether UTM/GTM books are "enough", but if for some reason they're all that you have access to, you're probably fine. But why limit yourself? It might happen that in a particular subject, none of the "standard" books are UTM/GTM, e.g. Kunen's Set Theory is Elsevier and Jech's is Springer but neither UTM/GTM. Instead of committing to UTM/GTM, just pick whatever friends or professors recommend and is actually available.

You can sometimes download books for free to try them out: try googling "ebooks title".  If you have access to a university library then go their and pick out the one(s) you like best.  I've picked out at least one book using the first method.  Sometimes google books will have it listed and sometimes amazon's look inside feature is available.


problem 19, section 3.2 book Chung - a course in probability path I can´t prove this problem. Can you help me? The problem says:

If $\{X_n\}$ is a sequence of identically distributed random variables with finite mean, then
$$lim_{n\to\infty}\frac{1}{n}\mathbb{E}\Big[\max_{1\leq j\leq n} |Xj|\Big] = 0$$
[HINT: Use Exercise 17 to express the mean of the maximum.]

In problem 17 show that if X is a positive random variable, then we have
$$\mathbb{E}[X]=\int_0^\infty \!\! P(X > x)\, dx=\int_0^\infty \!\! P(X \geq x)\, dx$$

I prove the problem 19 assuming independence and for the general case I find some upper bounds for the maximum of dependent variables on the internet but none worked.

Thank you for help me.

Use Dominated convergence theorem after applying Exercise 17. The dominating function is P(|X_1|>x).

since h(x)=|f(x)| for any real valued function f, is a convex function(why?) then max(|Xi|) is also convex. by using Jensen's inequality you can prove this. be happy


Integrate $\int \sqrt{a + x^3}dx$ How does one go about integrating:

$$\int\sqrt{a + x^3}dx$$

Wolframalpha seems to be giving a result that seems slightly too elaborate.

why not use u-substitution?  u = a+x^3 then du=3x^2 dx.  Should be a good start.


Finding a basis that allows the matrix of a linear transformation to have [1,0,0,...,0] as its first row I'd like some help with problem 4 from section 3C of Linear Algebra Done Right:

Suppose $w_1,⋯⋯,w_n$ is a basis of $W$ and $V$ is finite-dimensional. Suppose $T∈L(V,W)$. Prove that there exists a basis $v_1, ⋯⋯,v_m$ of $V$ such that all the entries in the first row of $M(T)$ (with respect to the bases $v_1, ⋯⋯, v_m$ and $w_1, ⋯⋯, w_n$) are $0$ except for possibly a $1$ in the first row, first column.

I've found a solution here, but it only provides a formulation of the basis vectors and doesn't explain why they lead to the desired first row of $M(T)$.

I am finding it hard to believe such a basis exists.

Take for instance $T:\Bbb R^2\to \Bbb R^2$ such that $T(v_1)=v_2;T(v_2)=v_1$

The matrix will be given by \begin{bmatrix} 0 &1\\1&0\end{bmatrix}

Now suppose that (as claimed in the problem) there exists  a  basis $\{v_3,v_4\}$ such that the matrix becomes \begin{bmatrix} 1 &0\\0&0\end{bmatrix}.

Then $T(v_3)=v_4;T(v_4)=v_3$.

By the matrix given $T(v_3)= v_3;T(v_4)=0\implies v_3=0$ which is false as $\{v_3,v_4\}$ is a basis


How can I index 2 dimensional points, starting at the origin and going outwards? Is there any way that I can mathematically determine a unique index number for 2D points that increases the further away I get from the origin? I do not know how far out that this coordinate system extends to.

I am working on a system where I need random access to data indexed by a 2D point.

In other words, I need a function that will always return a unique integer given the same two integers, but not conflict with other sets of numbers. Think of it like a hash. However, I need the indices generated to increase as the distance from the origin increases - this is so that my index growth remains constant as I add more data.

EDIT:

Here is some procedurally generated data that would work as a solution. I simply need a function that I can execute in constant time that produces something to the effect of:

   0,   0 = 0
  -1,   0 = 1
   0,  -1 = 2
   0,   1 = 3
   1,   0 = 4
  -1,  -1 = 5
  -1,   1 = 6
   1,  -1 = 7
   1,   1 = 8
  -2,   0 = 9
   0,  -2 = 10
   0,   2 = 11
   2,   0 = 12
  -2,  -1 = 13
  -2,   1 = 14
  -1,  -2 = 15
  -1,   2 = 16
   1,  -2 = 17
   1,   2 = 18
   2,  -1 = 19
   2,   1 = 20
  -2,  -2 = 21
  -2,   2 = 22
   2,  -2 = 23
   2,   2 = 24
  -3,   0 = 25
   0,  -3 = 26
  -3,  -1 = 27
  -3,   1 = 28
  -1,  -3 = 29
   1,  -3 = 30
  -3,  -2 = 31
  -3,   2 = 32
  -2,  -3 = 33
   2,  -3 = 34
  -3,  -3 = 35

If you're using points with integer coordinates, then try a spiral like this one: http://upload.wikimedia.org/wikipedia/commons/1/1d/Ulam_spiral_howto_all_numbers.svg.

If the coordinates of the points can have any real value, then I don't think such a function is possible because you are looking for an injective function from $\mathbb{R}^2$ to $\mathbb{R}$.

EDIT: The examples came up after I posted this. If you're just using integers, then such a function is possible.


Finding a vector given its dot products with given vectors Given vectors $\vec{A} = -4.7\hat{\imath}  + 8.2\hat{\jmath}$  and $\vec{B}= 9.6\hat{\imath} + 6.7\hat{\jmath}$, determine the vector C that lies in the $xy$-plane perpendicular to $\vec{B}$ and whose dot product with $\vec{A}$ is $20.0$.

Can someone walk me through the steps of solving this problem? I need to solve for $\vec{C} = ?\hat{\imath} + ?\hat{\jmath}$

Instead of "?", you can fill in the missing components with letters, say $x$ and $y$.  Suppose that $\vec C=x\hat i+y\hat j$.  Then $\vec C\cdot \vec A=-4.7 x+8.2y$.  What must the last expression equal? Similarly, write down an expression for $\vec C\cdot \vec B$ in terms of $x$ and $y$ to get an equation based on the assumptions in the problem.  Then try to solve the system of equations.

Hint:
Vectors perpendicular to $\vec B = 9.6 i + 6.7j$ are of form $k (6.7 i - 9.6 j) $ for some scalar $k$.
Can you work out why and find $k$ from the condition on dot product with $\vec A$?


distribution of prime powers Let $f(x)$ be the number of prime powers less than $x$
(by prime power I mean any number of the form $p^n$ such that $p$ is prime and $n \ge 1$.)

Consider the limit: lim x->∞ f(x)/x
I would like to know the value of L
In particular, this article seems to suggest that the limit tend neither towards 0 nor 1 however I do not have the knowledge needed to understand the it:

http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160270602/abstract

Thanks

Let's define $\Pi_n(x)$ as the number of prime powers $p^n$ with exponent $n$ that are less than $x$. Clearly $\Pi_1(x) = \Pi(x)$, the usual prime counting function.

Since $p^n < x$ implies that $p < x^{1/n}$, we have
$$
\Pi_n(x) = \Pi\left(x^{1/n}\right)\, .
$$
Your $f(x)$ is:
\begin{align}
f(x) &= \sum_{n=1}^{\infty} \Pi_n(x)\\
&= \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)\, .
\end{align}
For fixed $x$, $x^{1/n}$ will eventually go to 1 as $n\rightarrow\infty$, so this sum will actually terminate at some point. (Actually the sum need only go up to the point where $x^{1/n} < 2$, i.e. $n_{max} = \left\lfloor \ln x / \ln 2 \right \rfloor$.)

Beyond that, at present, all I have are numerics (done in Mathematica) which loosely suggest that
$$
\lim_{x\rightarrow\infty} \frac{f(x)}{x} = 0\, ,
$$
although convergence seems to be very, very slow. (Perhaps going as $1/\ln(x)$?)

Edit:

Numerically, it looks like the limit
$$
\lim_{x\rightarrow\infty} \frac{f(x)}{x/\ln(x)}
$$
may exist, and have a value $\ge 1$. (I suspect the actual limit is 1 itself, but the convergence is very slow.)

Edit 2:

For those down-voting this, note that the original questioner asked me (in the comments above) to post the proof of $f(x) = \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)$. I'm just trying to be helpful. (Unless I've made an error here, in which case let me know.)


Diminishing upper limit on Rubik's Cube solutions - why so long? I distinctly remember the news in 2007 about how researchers at Northwestern University had established the upper bound on moves needed to solve a Rubik's Cube at 26.  (I haven't been paying attention since, but it's obviously a lot lower now.)

Given:  The number of possible states of a proper Rubik's Cube is

$$
11! \cdot 8! \cdot 3^{8} \cdot 2^{12} = 43,252,003,274,489,856,000 ~~ (4.3252 \times 10^{19})
$$

Using the quarter-turn metric, there are only six (6) possible moves available for any cube at any time (using Singmaster notation):

$$
\text{L, R, U, D, F, and B}
$$

Assuming that:

only the quarter-turn metric is used (e.g. $\text{F}2$ is considered two moves, F F),
all moves are in one direction only (e.g. $\text{R'}$ would be accomplished by $\text{R R R}$ [or $\text{R}3$]),
applying a move and subsequently its inverse would be ignored stipulated by (2) above, and also ignoring the same move 4 times (to yield the same original state prior to the 4 moves),

then:  a decision tree can be made where each node represents a unique (valid) state.  This would be a 6-ary decision tree, with each branch representing each of the 6 basic moves above.

From my CS education, I know that any decision tree with $n$ nodes and $b$ daughter branches per node will have at most $\lceil\textbf{log}_{b} \textbf{n}\rceil$ levels.  For states of a Rubik's Cube, this is

$$
\log_{6} (11! \cdot 8! \cdot 3^{8} \cdot 2^{12}) \cong \lceil 24.8473 \rceil = 25 ~\text{levels}
$$

What this means is that one can shuffle a Rubik's Cube from any state to any other state (e.g. the solved cube) in not more than 25 moves.

Even though many symmetries and optimizations noted in code20.org show that the max number of moves is lower, my question is regarding the Northwestern Univ. discovery in particular.

Given that

the Rubik's Cube was invented in 1974 and publicly introduced in 1980,
decision trees in computer science have been around since at least the 1970s, and
the Northwestern University research (2007) gave 26 moves as a newly-established upper bound (in agreement with my calculation above),

then my question is:

  What took so long?

Is there something lacking in my math above?  Or was the complexity of solving the cube not fully understood?

One last side note:  I distinctly remember this research announcement in particular as I was teaching a CS lab in grad school at the time, and I discussed this topic with my students.  The Northwestern press release seemed to imply that their discovery was a "really big deal", yet the the mathematical analysis seemed obvious to me at the time.  Strange...

For the record, I have read the various Rubik's cube questions (and answers) here on math SE (and also CS SE and Stack Overflow), but I couldn't find anything about this topic in particular.

The nodes in your tree you described are not unique.  Sure you can fit 43 quintillion nodes in 25 levels, but because your tree allows for duplicates, you can't be sure that you got all unique positions of the cube.

For example consider the node reached by turning RL compared to the node reached by turning LR.

I think that's what you were getting at with your Rule #3: trying to delete obvious duplicates, but in that case each node no longer has 6 children, therefore you no longer can do $\lceil\log_b n\rceil$.  For example how many children does the node reached by RRR have?

FWIW you don't need to define a one-direction-only metric to do this (unfortunately flawed) analysis.  In regular half-turn metric, each node will have 18 children.  So $\lceil\log_{18} n\rceil=\lceil15.6428\rceil=16$, which is way less than what we know is required, due to so many duplicate nodes being in our tree.

(Credit where credit is due: this is basically @Arthur's answer he gave in the comments.)

Yes, I've given @timidpueo the credit for answering this, and @Arthur made some great comments, but still I feel a need to clarify the graph topology model I was claiming in reference to solving the Rubik's Cube in 26 or less moves.

First:  My question may appear borderline rhetorical (gee, I hope the SE community doesn't kick me out for that!), but I was sincerely curious what took so long for the math to catch up with proving solution move sequences in a finite number of moves.

Second:  The topology I had described in the 6-ary tree I mentioned above was substantially lacking in what I had envisioned about 10 years ago (around the time I saw the Northwestern press release).  This is mainly carelessness on my part in asking the question above, so I'd like to re-make the claim that the Cube can be solved in 26 moves or less.

Given a (valid) $3 \times 3$ Rubik's Cube, with the same 3 restrictions above in representing a solution, but with the additional restrictions below, I still claim that a solution sequence could be modeled by a 6-ary tree:

With all moves in one direction only, then the decision tree becomes a directed graph.  For example,  If one were to apply a move $R$ to a cube, then the $R'$ cannot be made simply to return to the original state.  (You'd have to make moves $\{R R R\}$ to return to the first state.)
Each node in the 6-ary tree represents one unique state of the cube.  This means that there are NO DUPLICATES in the tree.  More on that later...
In a theoretical situation where someone considered applying the move sequence $\{R R\}$ twice (returning to the original state), then this represents a philosophical breakdown in what defines solving the Cube.  Thus, being a directed graph with each state being unique, then this means that you cannot simply go from $S \rightarrow \{F F\} \rightarrow S' \rightarrow \{F F\} \rightarrow S$.  Thus, no cycles, and no duplicate nodes.  If you were trying to go from e.g. a solved state to a different state, then using the 6-ary tree model implies that the destination state is the top (root) node.
This also means that the desired (root node) state is arbitrary.  My claim is that one can go from any state to any other state in 26 moves (or less), regardless of what those states be.  (Of course, most people want to solve the cube, so it goes without reason that the solution state is also the root node.)
If two different solutions were to exist from the same initial state to the same final state, then each would take a different route up the tree, but each intermediate state along each path would be unique from any of the others.  But this can't happen!  Proof by contradiction:  If the same state existed along two different solution paths, then only one solution could exist from that state to the solved state.  Inductively, this could be shown all the way down to just two moves, but as I've stated above in (4), that can't happen.
One last point, but a very important one:  Despite the limitations I've described, it is still possible to apply a move sequence from any possible state to any other possible state with just the six moves $\{L R F B U D\}$.  For those familiar with Singmaster Notation, then think of the counterclockwise "prime" moves, (e.g. $\{L', R', etc.\}$) as the forward move three times, and the inner-layer moves (e.g. $\{.L, etc.\}$) as a combination of $\{L L L R\}$ which accomplishes the same.

A final note:  In answering my "What took so long" question, I'm guessing that basing a tree model mathematically to solving the Rubik's Cube took some time to develop properly, and seeing how all the above restrictions in modeling the solution decision tree made things a little complicated, I can now understand why it took a while.

I'd also like to thank the cube20.org web page as it mentions some of the reasons why "God's Number" is only 20 for the complete set of Singmaster moves--but note that their site still mentions 26 moves as a minimum upper bound on solving the Cube using the quarter-turn metric.

Comments are certainly welcome.  Thanks!


Proving that $64$ divides $3^{2n+2}+56n+55$ by induction Let $n ≥ 0$ be an integer. Prove by induction: 64 divides $3^{2n+2} + 56n + 55$

general expression: $3^{2n+2} + 56n + 55 = 64m$

1st I substitute $P(0)$ and it gives me true:
$9+55 = 64$ (if m = 1 the condition is true)

2nd I assume $n = k$ and I substitute saying $3^{2k+2} + 56k + 55$

and then I don't know what to do to prove the claim.

Let $F(n) = 3^{(2n+2)} + 56n + 55$
Then $F(n+1) = 3^{(2(n+1)+2)} + 56(n+1) + 55$
Then we get:
$F(n+1) = 3^{(2n+4)} + 56n + 56 + 55 = \\
9.3^{(2n+2)} + (9.56.n - 8.56.n) + 56 + (9.55-8.55) = \\
(9.3^{(2n+2)} + 9.56.n + 9.55.n) - 8.56.n - 8.55 + 56 = \\
9.F(n) - 448.n - 384 = \\
9.F(n) - 64.7.n - 64.6$

Now from the last form of F(n+1) it's already obvious that: if 64/F(n) then 64/F(n+1).
And this is exactly the induction step you have to make to finalize your proof.

In general: usually such problems are solved by expressing $F(n+1)$ in terms of $F(n)$. This way you can use the induction hypothesis (which you assume it's true). In our case that's the fact that $64 / F(n)$. And then what's left will be usually something which is obviously divisible by that number which they ask you to prove that it's divisible by (in our case that leftover is the value $64.7.n + 64.6$ which is obviously divisible by 64).


If $k$ divides $n$ then there's only one subgroup of $G$ of order $k$ Let $G$ be a cyclic group of order $n$ generated by, say, $g$. Show
  that if $k$ divides $n$, then there is one, and only one, subgroup of
  $G$ of order $k$, and that this is generated by $g^{n/k}$.

I can see it being true if we take $n$ to be prime. The $k$ is either $1$ or $n$, thus if $k=1$, then the subgroup of order $1$ will be generated by $e$ and if $k=n$, then the subgroup will be $G$ itself generated by $g$ which is true since $n$ is prime.

How about in a case where $n$ is not prime?

Suppose $\,k\mid n\implies n=dk\;$:

If $\;G=\langle g\rangle\;$ is ciclyc  then clearly $\,K:=|\langle g^d\rangle|=k\;$ .

Suppose now we have $\;1\ne H\le G\;,\;\;|H|=k \;$ and let $\,r>0\;$ be the minimal natural number s.t. $\,g^r\in H\;$ , then

$$1=(g^r)^k=g^{rk}\implies n\mid rk\implies rk=ns=dks\implies r=ds\implies s\mid d$$

and from here we have

$$g^r=(g^d)^s\in K$$

and from $\,|H|=||K|\;$ we get equality and thus uniqueness of $\,K\,$.

Suppose $h\in H \leq G$, for $|H|=k$. What can the order of $h$ be? Is it in the group generated by $g^{g/k}$?


Prove $\gcd(n, n + 1) = 1$ for any $n$ Let $n \in \mathbb Z$ be even. Then $n + 1$ is odd. So, $2$ doesn't divide $n + 1$. Thus there's no even number for which $\gcd(n, n+1)$ is not $1$. I am not sure how to show it for odd numbers. Is there a better way to prove the statement?

Let $a$ and $b$ be integers.

Any integer that divides both $a$ and $b$ must also divide their difference (can you see why this is?).

Take $n$ and $n-1$, if they have a common prime $p$ then from $n-(n-1)$ we can write
$$n-(n-1)=p\cdot\left(\dfrac{n}{p}-\dfrac{n-1}{p}\right)=p\cdot\dfrac{n-n+1}{p}=p\cdot\dfrac{1}{p}$$ and that implies p divides 1, absurd, because p is prime. $quod$ $erat$ $demonstrandum$


The derivative of $ \tan x$ is $ \sec^2 x$. Why? I understand why the derivative of $\sin x$ is $\cos x$, and why the derivative of $\cos x$ is $-\sin x$ in a geometric way.

But I can not understand why the derivative of $\tan x$ is $\sec^2 x$.

Can someone please explain this in a visual, geometric way using the unit circle?

The following is maybe a little bit hand-wavy, but the idea is hopefully clear.

If $\theta$ is the angle $EAH$ in the figure, then $\tan \theta$ is the lenght of the segment $BC$. If we increase that angle slightly (or “infinitesimally”), then the segment $EF$ is approximately $d\theta$, and we would like to know the length of $CD$, since that's the change in $\tan \theta$.

Since the triangle $AHE$ is $\cos \theta$ times smaller than the triangle $ABC$, the segment $EF$ is $\cos \theta$ times shorter than $CG$. And $CG$ is in turn $\cos \theta$ times shorter than $CD$ (similar triangles; the angle $GCD$ is $\theta$ and the angle $CGD$ is $\pi/2$ up to an infinitesimal correction). So $EF$ is $\cos^2 \theta$ times shorter than $CD$, and thus
$$
d( \tan\theta ) = CD = \frac{EF}{\cos^2\theta} = \frac{d\theta}{\cos^2 \theta}
.
$$

I can't explain it geometrically, but I can prove it algebraically:

$$\begin{aligned}
\frac{d}{dx} [\tan x] &= \frac{d}{dx} \left[\frac{\sin x}{\cos x} \right] &\quad \text{Definition of } \tan x\\ &= \frac{\frac{d}{dx}[\sin x]\cdot (\cos x) - (\sin x) \cdot \frac{d}{dx} \left[\cos x \right]}{\cos^2 x} &\quad \text{Quotient Rule}\\ &= \frac{\cos^2 x + \sin^2x }{\cos^2 x} &\quad \text{Derivatives of } \sin x \text{ and } \cos x \\ &= {\frac{1}{\cos^2x}} &\quad \text{Pythagorean Identity} \\ &= {\sec^2 x}  &
\end{aligned}$$


Why is $i^2$ equal to $-1$? In this KhanAcademy link at 2:25, Sal (the narrator) says that $i^2$ is negative 1 and he didn't explain why.

Why is this so? What is the intuition behind it?

It is defined that way.

I believe the proper definition of $i$ is $i^2 = -1$ and not $i = \sqrt{-1}$ as is commonly stated.

It is useful in many deep areas of mathematics, but as a starting point, it serves as a solution to $x^2 + 1 = 0$.

I recommend looking deeper into the theory of complex numbers, particularly the polar form.

To help you understand further. Realize that we have many sets of numbers such as the natural numbers, the rationals, and the reals. The complex number system is just another set and each complex number can be defined as an ordered pair $a + ib$ or $(a,b)$ where $a,b \in \mathbb{R}$.

A note:
Since $i^2 = -1$,
there is no way to tell the difference
between
$i$ and $-i$.


How do I change this parametric equation: $x=t+1/t, y=t^2 + 1/t^2$ into a Cartesian equation? I've just started parametric equations on my own & I am a bit confused on how to convert this parametric equation into a Cartesian equation.

$$\begin{array}{rcl}
x=t + \frac{1}{t},
y= t^{2} + \frac{1}{t^{2}}
\end{array}\qquad$$