Quotient rules and fractions! I'm doing a problem in my textbook that I have the answer too thanks to it being an odd problem in the back. The problem is that I can't get to the answer of this problem.

Problem by the way is shown to be "Differentiate: (1 / y^2 - 3 / y^4)(y + 5y^3)".

The answer in the book happens to be: "5 + 14/ y^2 + 9/Y^4".

What I've tried thus far is use (fg)' = fg' + gf'

With F being the first parenthesis and G being the second parentheses respectably.

But doing so gets me (y^2 - 3)(5y^2 + 1) / y^3 + (y^2 - 3)(5y^2 + 1). I stop at this point because I know theirs no way to get the answer from here.

Can someone point me to the right way?Also how can I make fractions like others are doing on this site? I can't find the button on here..

$\frac{1}{y^2}$

You can expand what you have to give:
$$\left(\frac{1}{y^2}-\frac{3}{y^4}\right)\left(y+5y^3\right)=5y-14y^{-1}-3y^{-3}$$
Therefore, the derivative will be given by:
$$\frac{d}{dy} (5y-14y^{-1}-3y^{-3})=5\cdot\frac{d}{dy}(y)-14\cdot\frac{d}{dy}(y^{-1})-3\cdot\frac{d}{dy}(y^{-3})$$
Now, use the fact that $\frac{d}{dy}(y^n)=ny^{n-1}$.

Alternatively, you can use the product rule or the quotient rule, as you have mentioned. I personally prefer to expand it in this case.

I see that you asked how to format questions using LaTeX. Here is a nice guide showing how to use the formatting.

So, you are taking the derivative with respect to y. What you have to do is to first expand your expression by multiplying it out. You will get

$\frac{1+5y^2}{y}-\frac{3+15y^2}{y^3}$ after some simplification.

Then you simply take the derivative using the quotient rule: $\frac{f'g-fg'}{g^2}$ and you'll have it!


Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that

$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?

Let $A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$, then $A^2 = \begin{bmatrix} a & b\\ c & d\end{bmatrix}\cdot \begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd\\ ac + cd & bc + d^2\end{bmatrix}$. Thus:

$A - A^2 = \begin{bmatrix} a-a^2-bc & b-ab-bd\\ c-ac-cd & d-bc-d^2\end{bmatrix} = \begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$. Thus:

$a-a^2-bc = 3$

$d-bc-d^2 = 4$

$b-ab-bd = 1$

$c-ac-cd = 1$.

Subtract the last $2$ equations to each other:

$b-c - a(b-c) - d(b-c) = 0 \to (b-c)(1-a-d) = 0$.

And subtract the first $2$ equations to obtain:

$a-d -(a^2-d^2) = -1 \to (a-d)(1-a-d) = -1$. Thus: $b-c = 0$, and $b = c$. Let's rewrite the system:

$a-a^2-b^2 = 3$

$d-b^2-d^2 = 4$

$b-ab-bd = 1$. Using Wolframalpha:

$a \approx 0.5 - 0.479i$

$b \approx 1.588i$

$d \approx 0.5 + 1.109i$.

Thus all roots are imaginary numbers, which means there is no matrix $A$ with real entries for the equation.

Assume such an $A$ exists. Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$. Then, the eigenvalues of $A-A^2$ are $\lambda_1-\lambda_1^2$ and $\lambda_2-\lambda_2^2$.

The eigenvalues of $\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$ are $\dfrac{7 \pm \sqrt{5}}{2}$. Hence $\lambda_1-\lambda_1^2 = \dfrac{7 - \sqrt{5}}{2}$ and $\lambda_2-\lambda_2^2 = \dfrac{7 + \sqrt{5}}{2}$.

Solving gives us two possibilities for each eigenvalue:

$\lambda_1 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13-2\sqrt{5}}$ and $\lambda_2 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13+2\sqrt{5}}$.

If the entries of $A$ are all real, then either $\lambda_1, \lambda_2 \in \mathbb{R}$ or $\lambda_1 = \overline{\lambda_2}$, which is not the case here.

Hence, there is no real $2 \times 2$ matrix $A$ such that $A-A^2 = \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$.


Probability that a head eventually turns up (From Grimmett and Stirzaker) This problem is from the exercise section of Grimmett and Stirzaker. I have also listed my proof.
Question.

A fair coin is tossed repeatedly.
(a) Show that a head turns up sooner or later with probability one.
(b) Show similarly that any given finite sequence of heads and tails occurs eventually with probability one.
(c) Explain the connection with Murphy's law.

Proof.

Let $A_{1},A_{2},A_{3},...,A_{n}$ be the events that a head turns up on the $(1,2,3,...,n)$-th toss after a sequence of $n-1$ tails. This can be visualized in the form of binomial tree. The sample space $\Omega=\{H,TH,TTH,TTTH,...\}$ is a countably infinite set.
The probability measure $P$ is countably additive over disjoint sets.
$P(\bigcup\limits_{i=1}^{\infty}{A_{i}})=\sum_{i=1}^{\infty}P(A_{i})$
In this example, all the members of $\mathscr{F}$ are disjoint.
Hence, $P(\bigcup\limits_{i=1}^{\infty}{A_{i}})=P(A_{1})+P(A_{2})+...=(1/2)+(1/2^2)+(1/2^3)+...\infty=1$.
Thus, a head will turn up with a probability 1.

I would like to know, if this proof is correct and sound. Are there alternative approaches to it? Also, I was wondering, how I can extend it to a sequence of heads and tails of length $m$. What is Murphy's law? How do I define the $\sigma$-field $\mathscr{F}$?

Your proof is correct, and extending it is probably possible, although I don't immediately see a way. An alternate way, which is easier to extend to the second case is the following.

The probability that no heads occur in $N$ tosses is $2^{-N}$. As $N \to \infty$, this goes to $0$. Thus, the probability that at least one head occurs in an infinite experiment is $1$.

This is easier to extend this as follows:

Pick any sequence of $m$ digits. The probability that it doesn't occur in $N$ tosses is upper bounded by the probability that it doesn't occur as sequence $(x_{jm - m}, \dots, x_{jm})$ for $1 \le j \le \lfloor N/m \rfloor$. The latter probability is $\left(\frac{2^m - 1}{2^m}\right)^{\lfloor N/m \rfloor} \to 0$ as $N \to \infty$.

Murphy's law is a tongue in cheek folk-saying that states that anything that can go wrong, will go wrong. Through the above, this is true for, well, coins, assuming some finite sequence makes things wrong for you, and if you're willing to hang around for a while.

I don't think your proof is right because of the definitions of $A_n$ and $\mathscr F$. Are the flips independent?

Consider a probability space $(\Omega, \mathscr F, \mathbb P)$ where

$\Omega = \{H,T\}^{\mathbb N}$

So we have $\omega = (\omega_1, \omega_2, ...)$ where $\omega_n \in \{H, T\} \ \forall n \in \mathbb N$

$\mathscr{F} = \sigma(\omega_n = W | \ W \in \{H, T\})$ like here (because I guess $\mathscr{F} = 2^{\Omega}$ doesn't work)
$P(\omega_n = H) = P(\omega_n = T) = 1/2$

where

$(\omega_n = H) = (\omega_1, \omega_2, ..., \omega_n = H, ...)$

$(\omega_n = T) = (\omega_1, \omega_2, ..., \omega_n = T, ...)$

In your case, $(\omega_n = H) = A_n$

So $P(A_n) = 1/2$ not $\frac{1}{2^n}$

Let $H_1, H_2, ...$ be events where $H_n$ = {nth flip is heads and 1st, ..., (n-1)th flips are tails}.

Thus, we have $$H_n = A_1^C \cap A_2^C \cap ... \cap A_{n-1}^C \cap A_n$$

$$=\bigcap_{k=1}^{n-1} [A_k^C] \cap A_n$$

Assuming independence of the flips i.e. independence of the $A_n$'s, $$P(H_n) = [\prod_{k=1}^{n-1} P(A_k^C)] \times P(A_n) = \frac{1}{2^n}$$

Now for Q1, we want to show that at least one of the flips will be heads

Or:

$$P(\bigcup_{n=1}^{\infty} A_n) = 1$$

Or:

Almost surely, $\forall \omega \in \Omega$,

$$\omega \in \bigcup_{n=1}^{\infty} A_n$$

Or:

$$\exists z \ge 1 s.t. P(A_z) = 1$$

Or:

Almost surely, $\forall \omega \in \Omega$,

$$\exists z \ge 1 s.t. \omega \in A_z$$

One route: Now we can do what you attempted earlier because the $H_n$'s are pairwise disjoint (*): $$P(\bigcup_{n=1}^{\infty} H_n) = \sum_{n=1}^{\infty} P(H_n) = 1 \ \text{if the flips are independent}$$

Hence almost surely, $\forall \omega \in \Omega$,

$$\omega \in \bigcup_{n=1}^{\infty} H_n$$

$\to \exists! \ q \in \mathbb N$ s.t. $\omega \in H_q$

Observe that $H_q \subseteq A_q$.

Hence, $$\omega \in A_q \subseteq \bigcup_{n=1}^{\infty} A_n \ QED$$

Or prove that $$\bigcup_{n=1}^{\infty} H_n = \bigcup_{n=1}^{\infty} A_n$$

(*) They are pairwise disjoint because:

$\forall m > n$,

$$\omega \in H_n \cap H_m$$

$$\to \omega \in H_n \cap \omega \in H_m$$

So, $$\omega \in H_n \to \omega \in A_n$$

However, $$\omega \in H_m \to \omega \in A_n^c ↯  \ QED$$

Another route:

$$P(\bigcup_{n=1}^{\infty} A_n) = 1 - P(\bigcap_{n=1}^{\infty} A_n^C)$$

$$= 1 - \prod_{n=1}^{\infty} P(A_n^C) \ \text{if the flips are independent}$$

$$= 1 - \prod_{n=1}^{\infty} (1/2)$$

$$= 1 - \lim_{m \to \infty} \prod_{n=1}^{m} (1/2)$$

$$= 1 - \lim_{m \to \infty} (1/2)^m = 1 - 0 = 1 \ QED$$

Yet another route:

$$\because \sum_{n=1}^{\infty} P(A_n) = \infty,$$

by Borel-Cantelli Lemma 2, if the flips are independent, we have $P(\limsup A_n) = 1$

Observe that $$\limsup A_n \subseteq \bigcup_{n=1}^{\infty} A_n$$

Hence, by monotonicity of probability, $$P(\bigcup_{n=1}^{\infty} A_n) = 1 \ QED$$

Or:

Hence, almost surely, $\forall \omega \in \Omega, \forall m \ge 1, \exists n \ge m$ s.t.

$$\omega \in A_n \subseteq \bigcup_{n=1}^{\infty} A_n$$

For Q2, let $B_{n,r}$ be a block of length r where

$$B_{n,r} = \bigcap_{i=n}^{n+r-1} A_i*$$

where $A_i* = A_i$ or $A_i^C$

$$\because \sum_{n=1}^{\infty} P(B_{n,r}) = \sum_{n=1}^{\infty} (\frac{1}{2^r}) = \infty,$$

using BCL2 again gives us

$$P(\limsup B_{n,r}) = 1$$

This means that almost surely $\forall \omega \in \Omega, \forall m \ge 1, \exists n \ge m$ s.t. $\omega \in B_{n,r} \ QED$

To see why the statement doesn't hold for a block of infinite length, define $$B_{n, \infty} := \lim_{r \to \infty} B_{n, r}$$

By the continuity of probability, $P(B_{n, \infty}) = \lim_{r \to \infty} P(B_{n, r}) = \lim_{r \to \infty} \frac{1}{2^r} = 0$

$$\because \sum_{n=1}^{\infty} P(B_{n,\infty}) = \sum_{n=1}^{\infty} 0 < \infty,$$

BCL1 gives us

$$P(\limsup B_{n,\infty}) = 0$$

For Q3, Murphy's Law is 'Anything that can go wrong, will go wrong', w/c is technically false:

Flipping 10 coins is a 'thing'. If we define 'go wrong' to be 'at least one head', then we may have 10 tails.

Mathematically,

$P(\bigcup_{n=1}^{10} E_n) \ne 1$ even if $P(E_n) > 0$

or

$P(\bigcup_{n=1}^{\infty} E_n) \ne 1$ where $E_k = \emptyset$ for $k \ge 11$ even if $P(E_n) > 0$ for $n = 1, 2, ..., 10$

Even if we flip infinitely, but $P(E_k) = 0$ for $k \ge 11$, we still may not have $P(\bigcup_{n=1}^{\infty} E_n) = 1$

Let us add the condition that it is not the case that all but a finite number of the $E_n$'s have zero probability (the $E_n$'s have positive probability infinitely often) to have Murphy's Law #2.

To put Murphy's Law #2 mathematically,

$$P(E_n) > 0 \text{i.o} \to P(\bigcup_n E_n) = 1$$

Is Murphy's Law #2 true? If not, what are some sufficient conditions for Murphy's Law #2?

Case 0: $\exists z \in \mathbb N s.t. P(E_z) = 1$

Obviously, Murphy's Law #2 holds.

Case 1: $E_n$'s are independent, $P(E_n) < 1$

$$P(\bigcup_{n=1}^{\infty} E_n) = 1 - P(\bigcap_{n=1}^{\infty} E_n^C)$$

$$= 1 - \prod_{n=1}^{\infty} P(E_n^C) = 1$$

Case 2: $E_n$'s are not independent but disjoint and $\sum_n P(E_n) = 1$, $P(E_n) < 1$

$$P(\bigcup_{n=1}^{\infty} E_n) = \sum_n P(E_n) = 1$$

Case 3: $E_n$'s are disjoint, not independent but $\sum_n P(E_n) = \infty$, $P(E_n) < 1$

Impossible.

Case 4: $E_n$'s are not independent but $\sum_n P(E_n) = \infty$, $P(E_n) < 1$

$$\sum_n P(E_n) = \infty to P(\limsup E_n) = 1 \to P(\bigcup_{n=1}^{\infty} E_n) = 1$$

Just kidding. BCL2 needs independence.

Case 5: $E_n$'s are not independent but $1 < \sum_n P(E_n) < \infty$, $P(E_n) < 1$

Here, we have $P(\liminf E_n^C) = 1$. So for some m, $\omega \in E_m^C, E_{m+1}^C, ...$.

$\omega$ may or may not be in $\bigcup_{n=1}^{m-1} E_n$.

So Murphy's Law #2 does not hold.

To sum up:

Case 0 is obvious. Case 1 corresponds to Q1. Case 2 corresponds to Q2. Case 3 is impossible. Cases 4 and 5 suggest counterexamples.


Functional equation: $f\left(\frac{x-1}{x}\right)+ f\left(\frac{1}{1-x}\right)= 2- 2x$ There is a function given $f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x
,f\colon \Bbb R\setminus\{0,1\}\to \Bbb R$ How many fuction exist? I have no idea how to start

$f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x\tag{1}$

Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(1)$

$\displaystyle f\left(\frac{1}{1-x}\right)+f\left(x\right) = \frac{2}{x}\tag{2}$

Similarly Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(2)$

$\displaystyle f(x)+f\left(\frac{x-1}{x}\right) = \frac{2x}{x-1}\tag{3}$

Now $(1)-(2)-(3),$ we get

$$\displaystyle -2f(x) = 2-2x-\frac{2}{x}-\frac{2x}{x-1} = -2\left(\frac{x^3+x-1}{x^2-x}\right)$$

So $\displaystyle f(x) = \frac{x^3+x-1}{x^2-x}$ for all $x\in \mathbb{R}\setminus\{0,1\}$


Projective Space vs. Vector Space In my geometry course we defined the projective space $\mathbb{P}(V)$ of a vector space $V$ as

  $\mathbb{P}(V) := \{ U \subseteq V \, | \, U \text{ is a vector
 subspace of dimension } 1 \}$.

Let's observe $\mathbb{P}(\mathbb{R}^2)$. This is the set of all lines passing through the origin. In my imagination these lines cover $\mathbb{R^2}$. So instead of thinking about $(x, y)^T \in \mathbb{R}^2$ as a point only, I can think about them as a generator, of one of the lines in $\mathbb{P}(\mathbb{R}^2)$. [Beside the origin itself...]

Why can I not just say that $\mathbb{R}^2$ "is the same" as $\mathbb{P}(\mathbb{R^2})$? [Or at least a subset/superset]

I would like to understand the differences between the projective space and the vector space itself. First of all, what is the difference?

Bonus-Question:

Does the answer to the above question depends on the vector space? What happens if $V$ would be $\mathbb{C}^n$, or $\mathbb{F}_q^n$ [Field with $q$ elements], or  a set of linear functions, etc. ?

You can't identify $\mathbb{P}(\mathbb{R}^2)$ with $\mathbb{R}^2$, but there's something like you'd like to think.
$\def\vv#1#2{\left(\begin{smallmatrix}#1\\#2\end{smallmatrix}\right)}
\def\ww#1#2{\begin{pmatrix}#1\\#2\end{pmatrix}}$

Consider, over the set $\mathbb{R}^2\setminus\{\vv{0}{0}\}$ the relation
$$
\ww{a}{b}\sim\ww{c}{d}
\text{ if and only if there exists $\rho\ne0$ such that }
\ww{c}{d}=\rho\ww{a}{b}
$$
This is an equivalence relation over the set. Note that each equivalence class corresponds to a single vector subspace of $\mathbb{R}^2$, the one generate by any element in the equivalence class.

Then you can “identify” $\vv{x}{y}$ with a point in the projective space, provided you “don't distinguish” between, say, $\vv{1}{2}$ and $\vv{\smash-1}{\smash-2}$ or $\vv{3}{6}$.

These are indeed the homogeneous coordinates, often written $(1:2)$ to mean the equivalence class of $\vv{1}{2}$. Thus $(1:2)=(-1:-2)=(3:6)$.

You can do similarly when dealing with any vector space. If you fix a basis of $V$, then a one-dimensional subspace is identified by any of its nonzero vectors; if you take the coordinates of such a nonzero vector with respect to the chosen basis, you get the homogeneous coordinates with respect to the coordinate system induced by the chosen basis.

What you get is actually a bijection from $\mathbb{P}(V)$ to the quotient set $(\mathbb{R}^n\setminus\{0\})/{\sim}$, where the equivalence relation is defined as before.

In the simple case of $\mathbb{P}(\mathbb{R}^2)$, there is a basis involved, namely the canonical basis. Choosing a different basis will induce a different coordinate system on the projective line.

Is $\mathbb{R}$ relevant? No, everything works the same over any base field.

Why can I not just say that $\mathbb{R}^2$ "is the same" as
   $\mathbb{P}(\mathbb{R^2})$?

Simple, because it is not the same. For one, $\mathbb R^2$ has the zero vector which doesn't generate any line (or rather, it generates too many of them). Also, $(1,1)$ and $(2,2)$ are two points in $\mathbb R^2$, but they generate only one line.


If there are 50 notes whose total value is 100 rupees but 2 rupee note should not be there in the count of those50 notes How many such notes can be? If there are 50 notes whose total value is 100 rupees but 2 rupee note should not be there in the count of those 50 notes.How many such notes can be ?

Notes available are
$1$ Rupee
$2$ Rupees ( but NOT to use as per question )
$5$ Rupees
$10$ Rupees
$20$ Rupees
$50$ Rupees
$100$ Rupees ( obviously, this won't be needed!)
$500$ Rupees ( obviously, this won't be needed!)
$1000$ Rupees ( obviously, this won't be needed!)

Putting it other way

let us assume $a$,$b$,$c$,$d$ & $e$ as variables satisfying following condition :

1. $a+b+c+d+e = 50$
2. $a + 5b + 10c + 20d + 50e = 100$

Then what can be values of $a,b,c,d$ & $e$ ? ( Given that they must be non-negative integers!)

Thanks!

One way to think about this is that the average note has value 2 rupees and only one note is any lower.  So each large note has to be paired with enough 1 notes to bring the average down to 2, and the group can be viewed as a "note".  A 5 rupee note has to be paired with 3 1 rupee notes making a package that is worth 8 rupees and has 4 notes.  Similarly a 10 rupee note must be paired with 8 1 rupee notes to make a package worth 18.
Your new notes have value 8, 18, 38, 98 and you need to make 100 (the count will take care of itself).  You can't use 98 at all.  Working mod 8 can make it easy.  If you use two 38's, you need 24 more, which can only be done with 3 8's.  So this gives 2*20+3*5+45*1 (in the original notes).  If you use one 38, that is 6 mod 8, so to get to 4 mod 8 we need 3 18's and 1 8, giving 20+3*10+5+45*1.  Finally, if we use no 38's, we need 2 18's and 8 8's, which leads to 2*10+8*5+40*1.


A metric defined in terms of the minimum distance to a subset of metric space For a set $S$ in a metric space $(M,d)$ and $x \in M$, let

$$
d(x,S):=\inf_{s \in S}d(x,s)
$$

Show that the distance for points in the closure of a set is zero:

$$
d(x,S)=0 \Leftrightarrow x \in \bar{S}
$$

Hint: $x \in \bar{S}$ if and only if $\forall \epsilon > 0$, $\exists s \in S$ such that $d(x,s)<\epsilon$.


Probability of birthday chain Let $B(k,n)$ be the probability that a consecutive birthday chain of length $k$ can be found in $n$ people, excluding leap days and assuming that December 31 and January 1 are consecutive. After some calculations, I found that (formulae marked with * are found by myself after original question asked; formulae marked with ** are found in other answers):

$B(n,n)=\dfrac{n!}{365^{n-1}}, B(n-1,n)^*=\dfrac{n!(-n+729)}{2\times365^{n-1}},\\ B(n-2,n)^*=\dfrac{n!(3n^2-4379n+1598698)}{24\times365^{n-1}},\\ B(2,n)^{**}=1-\dfrac{1}{365^{n-1}}\displaystyle{\sum_{k=1}^{n}\dfrac{1}{k!}(\prod_{i=1}^{k-1}(365-k-i))(\sum_{j=0}^{k-1}(-1)^j\dfrac{k!}{j!(k-j)!}(k-j)^n})$
$B(k,n)=0$ if $k>n$
$B(k,n)*365^{n-1}=\begin{bmatrix}1&365&133225&48627125&17748900625&6478348728125&2364597285765625\\0&2&2178&1576910&948848970&512466548222&257645853120138\\0&0&6&8700&7884390&5716075620&3625913148606\\0&0&0&24&43440&47175960&39848163600\\0&0&0&0&120&260280&329320320\\0&0&0&0&0&720&1819440\\0&0&0&0&0&0&5040\\\end{bmatrix}$

However, I wasn't able to find the general form of $B(k,n)$. How can I generalize this probability?
Reference:

Consecutive birthdays probability

(Choose $k$ people out of $n$ people)$\times$(choose a particular day $d$ from $365$ days)$\times$(number of ways these $k$ people can have birthdays on days $d,d+1,\cdots,d+k-1$)$\times$(number of possible birthdays for the other $n-k$ people, with either no birthdays on $d-1$ or no birthdays on $d+k$)
$$=\binom{n}{k}\times\binom{365}{1}\times k!\times\left(\binom{2}{1}364^{n-k}-363^{n-k}\right)$$
Hence $$P(k,n)=\frac{\frac{n!}{(n-k)!}\times\left(2\times364^{n-k}-363^{n-k}\right)}{365^{n-1}}$$
Obviously here $k\le365$.

The first result is more than doubtful, since $\frac{n!}{365^{n-1}}>1$ for $n\gg1$.


Normal Subgroups and Internal Direct Products Why are Normal Subgroups important?

Why are Internal Direct Products important?

I'm studying abstract algebra and I have always wondered about its relevance and usefulness.
Does anyone could help me please?

Normal subgroups are important for the same reason that factor groups are important, since normal subgroups correspond to factor groups and vice-versa.

Direct products (whether internal or external; they correspond to one another in a natural way) give you both ways of producing new groups from old, and of (sometimes) understanding more complicated groups in terms of simpler ones. A classical example of the latter is the Fundamental Theorem of Finitely Generated Abelian Groups (which is later generalized to any finitely generated module over a PID), which tells you that any finitely generated abelian group is a direct product of cyclic groups that, in addition, have orders satisfying certain restricting relations. This makes understanding finitely generated abelian groups very easy.

It is important because all those studies lead e.g. to Galois Theory. Galois Theory is important e.g. in solving polynomial equation. Solution to polynomial equation is of course important as it solves no one knows how many problems in real world from theoretical physics to engineering and economy.

I feel your tone though. So, YES, most of the teachings are bad in a sense that they:

Ignore the intention of a theorem that we are left in the dark and missing the excitement;
Skip too many critical arguments by assuming the students somehow could been suddenly skilled in tens of theorems just presented earlier;
Give too many puzzles to be forgotten later that the sole purpose seems to just slow down the learning.

Self-learning is almost impossible as they don't seem to care the learning process of a student.


Are the assertions "$2 + 2$ equals $4$" and "$2 +2$ is $4$" identical Are the assertions "$2 + 2$ equals $4$" and "$2 +2$ is $4$" identical? Or is this a linguistic, psychological or murky philosophical thing rather than a mathematical thing

I think that we should first ask "In what language are these assertions made?". The assertion does not belong to the mathematical language, where it should be written "2+2=4" or more basically, using the language of the Peano axioms: "$s(s(0))+s(s(0))=s(s(s(s(0))))$". If it's in the English language, then one must ask if "equals" and "is" mean the same thing all the time or only when they are used between two numbers or numeric expression.

Clearly, 2+2 is not identical to 4. 2+2 is the sum of two twos but 4 is a single number. What we mean by either way of saying it, is that 2+2 is quantitatively equivalent to 4. And I wish we would use the identity symbol and reserve the equals sign for conditional quantitative equivalence.

This also applies to 0.9999... is quantitatively equivalent to 1. They are obviously not identical. Mathematics should not rule out common sense.


Alternative proof of Taylor's formula by only using the linear approximation property So a function $f: E \to F$ between the normed spaces $E,F$ is called differentiable in $x \in E$ if there exists a bounded linear map $Df(x): E \to F$ such that for every $h \in E$ we have $$f(x+h)=f(x)+Df(x)h + o(||h||). \tag{1}$$
If $f$ is differentiable for every $x \in E$ and $Df: x \mapsto Df(x)$ is differentiable for every $x \in E$ too we get analogously
$$Df(x+e)=Df(x)+D^2f(x)e+o(||e||). \tag{2}$$
Then $f$ is called twice differentiable and for every $h\in E$ we have the "Taylor expansion of second degree"
$$f(x+h)=f(x)+Df(x)h+\frac{1}{2}D^2f(x)[h] + o(||h||^2), \tag{3}$$
where $D^2f(x)[h]:=(D^2f(x)h)h$ for better readability.
I have two questions:

How can $(3)$ be proven without resorting to the "standard proof" of using integrals? I want to show it by only using the linear approximations  given in $(1)$ and $(2)$. Inserting $(2)$ in $(1)$ doesn't result in something useful though. Can this be done?
Can $(3)$  be used as an alternative definition off twice-differentiability? Analogously what about  the general case of $n$-times differentiability:
$$ f(x+h) = f(x) + \sum_{j=1}^{n} \frac{1}{j!} D^jf(x)[h] + o(\|h\|^n) \tag{4}$$

A few days back I wrote an answer with some detail about Taylor polynomials for maps between Banach spaces. You can see my answer here. What I proved is that (I'm sorry about the differences in notation)

  Taylor Expansion Theorem:

  Let $V$ and $W$ be Banach spaces over the field $\Bbb{R}$, let $U$ be an open subset of $V$, and fix a point $a \in U$. Let $f:U \to W$ be a given function which is $n$ times differentiable at $a$ (in the Frechet differentiable sense). Define the Taylor polynomial $T_{n,f}:V \to W$ by
  \begin{equation}
T_{n,f}(h) = f(a) + \dfrac{df_a(h)}{1!} + \dfrac{d^2f_a(h)^2}{2!} + \dots + \dfrac{d^nf_a(h)^n}{n!}
\end{equation}
  Then, $f(a+h) - T_{n,f}(h) = o(\lVert h \rVert^n)$.

  Explicitly, the claim is that for every $\varepsilon > 0$, there is a $\delta > 0$ such that for all $h \in V$, if $\lVert h \rVert < \delta$, then
  \begin{equation}
\lVert f(a+h) - T_{n,f}(h) \rVert \leq \varepsilon \lVert h \rVert^{n}.
\end{equation}

The proof is pretty short if you know what you're doing.
The idea is to use induction, and most importantly, the mean-value inequality for maps between Banach spaces. I don't think it is possible to derive $(3)$ directly from $(1)$ and $(2)$ alone, because $(2)$ talks about how the derivative $Df$ changes, while $(1)$ talks about how the function $f$ changes, and ultimately $(3)$ talks about how much $f$ changes. So, you somehow have to relate changes in $Df$ to changes in $f$... this is roughly speaking, what the mean-value inequality does.

The proof I showed in my other answer is pretty much from Henri Cartan's excellent book Differential Calculus. Also, Henri Cartan's book has a proof of the mean-value inequality which doesn't rely on integrals. Alternatively, you can take a look at Loomis and Sternberg's book Advanced Calculus. Here, they prove the mean value inequality in a rather elementary way without integrals, and it's also a relatively short proof. It is proven in Chapter 3, Theorem 7.4 (which uses theorem 7.3); this is on page 148-149 of the book (I prefer this proof to Cartan's proof).

For your other question, I assume you mean the following:

  Does the existence of a polynomial $P$, which equals $f$ up to order $2$ at $x$ imply that $f$ is twice differentiable at $x$? Or more precisely, does the existence of a continuous linear map $A_1:E \to F$, and a symmetric continuous bilinear map $A_2:E \times E \to F$ such that for all $h \in E$,
  \begin{equation}
f(x+h) = f(x) + A_1(h) + A_2(h,h) + o(\lVert h \rVert^2)
\end{equation}
  imply that $f$ is twice Frechet differentiable at $x$?

The answer to this question is no. We can see this even in the single variable case (this following example is from Spivak's Calculus, page 413, 3rd edition). Take $E=F=\Bbb{R}$, and define $f: \Bbb{R} \to \Bbb{R}$ by
\begin{equation}
f(x) =
\begin{cases}
x^{n+1}& \text{if $x$ irrational} \\
0 & \text{if $x$ rational}
\end{cases}
\end{equation}
($n\geq 2$). Then choose $x=0$ and the zero polynomial $P \equiv 0$. It is easy to verify that
\begin{equation}
f(0 + h) = 0 + o(|h|^n)
\end{equation}
However, if $a \neq 0$, then $f'(a)$ doesn't exist so $f''(0)$ is not even defined. Hence, what this shows is that the existence of a well-approximating polynomial does not guarantee that the function is sufficiently differentable.

The differential Df(x) is a linear map that approximates your function f, according to your relation (1).  And that linear map depends on x (the point around which you consider the Taylor expansion).  The map Df , which associates a vector to a linear map doesn't have to be linear indeed.  The second differential  DDf(x)   is a linear map that associates a vector to a linear map , and so on. Your relation (1) tells us that the differential of f exists (definition). Your relation (2) tells us that the differential of Df exists (definition). Your relation (3) tells us that there is a valid finite second order (degree) Taylor expansion of f at x (also a definition). From (1) and (2) you cannot derive (3). That's not always possible  even for real function , even more so in this abstract setting.  In a valid, finite Taylor expansion,  the remainder must grow slower than the last term of the Taylor expansion.  And no, (3) is not equivalent to second order differentiability of f.


Laurent series coefficient $a_{-1} $ for $f(z)=\frac{1}{2-e^z} $ I am still a beginner in complex analysis. I am trying to compute the coefficient $a_{-1}$ of the Laurent series $\sum_{m=-1}^{\infty}a_n(z-z_0) $ of the function $$f(z)=\frac{1}{2-e^z} $$ near the singularity $z_0=log(2)$.

Here is what I did so far, since the singularity $log(2)$ is a pole of order 1 and the fact that there are no other singularities in the unit circle centered at z=0 the other nearest singularities are at $log(2) \pm 2i\pi$ and their modulus is $\gt 1 $.

I tried to compute $a_{-1} $ using contour integration around the unit circle centered at z=0 $$a_{-1}=\frac{1}{2i\pi} \int_{0}^{2\pi}\frac{ie^{it}}{2-e^{e^{it}}}dt $$

Then I get $$ \frac{1}{2i\pi}[\frac{1}{2}(e^{it}-log(2-e^{it})) ]_{0}^{2\pi} $$

Which is equal to 0. From what I know, I should get 0 if there was no singularity inside my contour but $z_0=log(2) $ has a modulus 0.301.

I already know that the expansion of $f(z)$ near $z_0=log(2) $ is of the form
$$ R(z)=-\frac{1}{2}\frac{1}{z-log(2)}+...$$

Then I should get $a_{-1}=-\frac{1}{2} $. I did not find my mistake..

$a_{-1}=\operatorname {Res} (f,\log 2)=\lim _{z\to \log 2}\, \dfrac{z-\log 2}{2-e^z}$

Using L'Hopital rule

$a_{-1}=\lim _{z\to \log 2}\, \dfrac{1}{-e^z}=-\dfrac{1}{2}$

You're right, we're looking for the coefficient, but you have to delimit an outline or you have at least one pole inside like |z-\ln{2}|\leq\epsilon after you make your parametrisation with z = \epsilon exp{(i\theta)}


"expectation of sum is sum of expectation", is this claim true? if yes, how to justify this claim? this post is saying

  linearity of expectation gives following equation
  $$\mathbb{E} [\sum_{j\neq i} Y_i Y_j] = \sum_{j\neq i} \mathbb{E} [Y_i Y_j]$$

per wiki, Linearity of Expected_value is saying

$$
{\displaystyle {\begin{aligned}\operatorname {E} [X+Y]&=\operatorname {E} [X]+\operatorname {E} [Y],\\[6pt]\operatorname {E} [aX]&=a\operatorname {E} [X],\end{aligned}}}$$

where $X$ and $Y$ are arbitrary random variables, and $a$ is a constant.

is not saying some rules like "expectation of sum is sum of expectation"

this post is saying

  expectation of sum of something = sum of expectation of something even
  when the items involved are not independent.

how to justify this claim?

When $E[X_1], E[X_2], \dots, E[X_n]$ exist
$${\displaystyle {\begin{aligned}
E [\sum_{i=1}^n X_i] &= E[X_1+X_2+\dotsb+X_n] \\
&= E[X_1]+E[X_2]+\dotsb+ E[X_n] \\
&= \sum_{i=1}^n E [X_i]
\end{aligned}}}$$


Evaluate $\lim_{x\to\infty}(1 + \frac{2}{x})^x$ So I have to evaluate: $$\lim_{x\to\infty}\left(1 + \frac{2}{x}\right)^x$$

This is:

$$\lim_{x\to\infty} e^{x\ln\left(1 + \frac{2}{x}\right)}$$

And:

$$x\ln\left( 1 + \frac{2}{x} \right) = \frac{\ln\left( 1 + \frac{2}{x} \right)}{\frac{1}{x}} $$

So I should apply L'Hospital rule and calculate:  $$\lim_{x\to\infty}\frac{\ln\left( 1 + \frac{2}{x} \right)}{\frac{1}{x}}$$

However I'm stuck at this step and  I don't know how to move further and land at the final result ( that is, $\displaystyle{\lim_{x\to\infty}}\left(1 + \frac{2}{x}\right)^x = e^2$ )

( Alternative options are also welcome as long as they're more simple and straightforward to apply than L'Hospital rule ).

The key trick for computing the last limit is to set $y=\frac{1}x$ then compute the limit $\lim_{y\to 0^+} \frac{\ln(1+2y)}{y}$, which is much easier using L'Hopital.

You can apply L'Hopital because this limit is of the indeterminate form $\frac{0}{0}$.


Showing that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$. Problem: Show that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
  Solution. Upon the given condition, the quadratic equation $ax^2+bx+c$ can be written as $(px+q)(rx+s)$ and we have:
   $$\overline{abc}=100a+10b+c=(10p+q)(10r+s)$$
  i.e. $\overline{abc}$ is not a prime which is a contradiction, so there is no such prime available.

I don't get the reasoning completely.

We show a more general fact:

  There is no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=d^2$ where $d$ is a non-negative integer.

We first note that $d\leq b\leq 9$. Since $b^2-4ac=d^2$, we have that
$$P(x):=ax^2+bx+c=a\left(x-\frac{-b+d}{2a}\right)\left(x-\frac{-b-d}{2a}\right)$$
Assume that $P(10)=\overline{abc}=p$ where $p$ is some prime of three digits. Then
$$4ap=\left(20a+b-d\right)\left(20a +b+d\right)$$
which implies that $p$ divides $\left(20a+b-d\right)$ or $\left(20a+b+d\right)$.

Now $1\leq a\leq 9$, $0\leq b\leq 9$, and
$$0<\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 9+9+9\leq 198.$$
Hence $a=1$, and
$$\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 1+9+9\leq 38.$$
Therefore $p>100$ can not divide any of them and we have a  contradiction!

$c$ can take only $1, 3, 7, 9$ . [$abc$ is prime]
$b$ can take only $1, 2, 3, 4, 5, 6, 7, 8, 9$. [$b\neq 0$ as $-4ac$ can't be a square of a number]
$a$ can take $1, 2, 3, 4, 5, 6, 7, 8, 9$.
$b^2-k^2=4ac$.
Thus $b$ and $k$ both should be both odd or both even and $b>k$ as $c$ can't be $<0$.
$4ac=\text{Multiple of 4,12,28 or 36}$ as $c$ can take only $1, 3, 7, 9$.
$$(b,k)=(2,0),(4,0),(6,0),(8,0),(4,2),(6,2),(8,2),(6,4),(8,4),(8,6)(3,1),(5,1),(7,1),(9,1),(5,3),(7,3),(9,3),(7,5),(9,5),(9,7).$$
For all this values of $b$ and $k$, $a$ and $c$ can't assume such values such that $abc$ becomes a prime number. Thus, $b^2-4ac$ is never a perfect square when $abc$ is prime.


How do you solve: $y'=c_1e^{-\frac{2}{3}x^{\frac{3}{2}}}$? How do you solve that ODE? I understand it has gamma functions but I have no clue where to start. Thanks
$$y'=c_1e^{-\frac{2}{3}x^{\frac{3}{2}}}$$

As Amzoti commented, just write $$y=c_1 \int e^{-\frac{2}{3}x^{\frac{3}{2}}}dx$$ Life could be easier if you change variable and do $$\frac{2}{3}x^{\frac{3}{2}}=z$$ $$x=\left(\frac{3}{2}\right)^{2/3} z^{2/3}$$ $$dx=\frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{z}}dz$$ which then leads to $$y=\sqrt[3]{\frac{2}{3}}c_1\int \frac{ e^{-z}}{\sqrt[3]{z}}dz$$

Since you mention gamma function, I am sure that you can take from here.

You can start by finding that $$\begin{array}{rrcl} & c_{1}\cdot \mathsf{e}^{-\frac{2}{3}\cdot\sqrt{x^{3}}} & = & y^{\prime} \\ & & = & \frac{\mathrm{d}y}{\mathrm{d}x} \\ \Rightarrow & \mathrm{d}y & = & c_{1}\cdot\mathsf{e}^{-\frac{2}{3}\cdot\sqrt{x^{3}}}\,\mathrm{d}x \\ \Rightarrow & y & = & \int c_{1}\cdot\mathsf{e}^{-\frac{2}{3}\cdot\sqrt{x^{3}}}\,\mathrm{d}x \end{array}$$


Calculate intercept location and time I want to calculate the coordinates and time where object B2 can intercept object B1.

Asuming the following:

All coordinates are in pixels

B1: Moves from coordinates A1 to A2, it has a speed of 2 pixels per hour and starts moving at 15:00.

B2: Moves from A3 to B1 and has a max speed of 1,5 pixels per hour and starts moving when it can intercept B1

A1: has the coordinates 20,45 (X,Y)

A2: has the coordinates 60,86 (X,Y)

A3: has the coordinates 55,78 (X,Y)

Results of the calculations I would like to achieve:

The time (00:00 - 24:00) and coordinates (X,Y) B2 is able to intercept B1 while both moving on a straight line, keeping in mind that both objects have a different movement speed.

B2 can intercept B1 as long as A3-A2 distance <= 75% of (15,0) to A2 passing by A1

Let's call the point where B1 start A0 :

B2 intercept B1 <=> dist(A3,A2) <= 75/100 (dist(A0,A1) + dist(A1,A2))

(Since 1.5 = 75% of 2)


Proof of Lindelöf Theorem I have been surfing the net to read the proof of the Lindelöf Theorem:

  Let $U\in \mathbb{R}^n$ be open and $U=\bigcup_{\lambda \in \Lambda}
 U_{\lambda}$where $\Lambda$ is an index set, $\{U_{\lambda}\}$ is a
  collection of open sets. Then, ther eis a countable subcollection
  $\{U_i\}$ of $\{U_{\lambda}\}$ so that $U=\bigcup_{i=1}^\infty U_i$.

I found out that most of the proof in the internet are from complex analysis. The level does not fit me as a beginner of real analysis, so I hereby ask for a more detailed and suitable proof. (Again, please don't lead me to some websites which talk about complex analysis!)

Thanks in advance.

Check that the collection $\mathcal{B}$ of open balls $B(x, r)$ of $\mathbb{R}^n$ where $x$ ranges over the points with all rational coordinates (of which there are only countably many) and $r>0$ ranges over the rational numbers, is a countable base of $\mathbb{R}^n$, which means that every open set $O$ of $\mathbb{R}^n$ can be written as a union of members from $\mathcal{B}$. Enumerate  $\mathbb{B}$ as $\{B_n: n \in \mathbb{N}\}$.

For every $x \in U$, pick $\lambda(x) \in \Lambda$ such that $x \in U_{\lambda(x)}$. Then pick $n(x) \in \mathbb{N}$ such that $x \in B_{n(x)} \subseteq U_{\lambda(x)}$. For every $n$ in $I = \{n_x: x \in U \}$ (which is countable) pick one $\lambda(n)$ such that $\lambda(n) = \lambda(n_x)$ where $n_x = n$. Then $\{ U_{\lambda(n)} : n \in I \}$ is the required countable subfamily: pick $p \in U$, then we have $n(p)$ and $\lambda(p)$ such that $p \in B_{n(p)} \subseteq U_{\lambda(p)}$, so $n = n(p) \in I$, and we have $\lambda(n)$ to witness that $p \in U_{\lambda(n)}$, as required.

This could all be made explicit if we start with a concrete enumeration of the base, and take minimal index elements whenever we make choices above. The above is just a general proof of the fact that a second countable space is hereditarily Lindelöf...

Choose one open set $U_\lambda$ for every $n$-tuple with all rational coordinates.


Rewriting sylvester's sequence in a closed form Sylvester's sequence is defined as

http://upload.wikimedia.org/math/1/6/f/16feba8ab6368dc9d965dbec35e445bb.png

but according to wikipedia and wolfram mathword, this can be rewritten in closed form as

http://upload.wikimedia.org/math/e/d/b/edb03103fbb72767dd6e3ab262ed9e66.png

How did they reach that equation?

(sorry about the links, I'm a new user and can't post images)

To recap, Sylvester's sequence is defined by
$$
s_0=2, \qquad s_{n+1}=1+s_0\dots s_n\ \ (n\ge 0)
$$
but can also be computed by the formula
$$
s_n=\lfloor E^{2^{n+1}}+\frac12\rfloor\qquad (*)
$$
for some constant $E$.  The question is how (*) can be derived.

The first step in the derivation will be to observe that each term in Sylvester's sequence is approximately the square of the previous one.  The second step will be, for each $n$, to find bounds for $E$ so that (*) will work for that $n$.  The third step will be to prove that all these bounds are consistent and allow some single $E$ to work for all $n$.

First, notice that, if $n\ge 1$,
$$
s_{n+1}=1+(s_0\dots s_{n-1})s_n
=1+(s_n-1)s_n
$$
so for all $n\ge 1$ (and in fact also for $n=0$)
$$
s_{n+1}-\frac12=(s_n-\frac12)^2+\frac14. \qquad (+)
$$
For (*) to work for a given $n$, it is sufficient to have the following bounds on $E$:
$$
E_n^-:=(s_n-\frac12)^{2^{-(n+1)}}\le E \le E_n^+:=(s_n)^{2^{-(n+1)}}.
$$
Here, the lower bound for $E$ is $E_n^-$ and the upper bound $E_n^+$.
Now by (+),
$$
E_{n+1}^-=(s_{n+1}-\frac12)^{2^{-(n+2)}}
=((s_n-\frac12)^2+\frac14)^{2^{-(n+2)}}> (s_n-\frac12)^{2^{-(n+1)}}=E_n^-
$$
and, since $s_n>1$ for all $n$,
$$
E_{n+1}^+=(s_{n+1})^{2^{-(n+2)}}
=((s_n-\frac12)^2+\frac34)^{2^{-(n+2)}}< (s_n)^{2^{-(n+1)}}=E_n^+.
$$
Therefore,
$$
E_0^-< E_1^-< E_2^-< \dots < E_2^+< E_1^+ < E_0^+.
$$
Take $E:=\sup_n E_n^-$.  Then $E\ge E_n^-$ for all $n$.  Also, for any $m$ and $n$,
$$
E_m^-\le E^-_{\max(m,n)}<E^+_{\max(m,n)}\le E_n^+. \qquad (++)
$$
Now, if $E>E_n^+$ for some $n$, then, by the definition of the supremum, there would be some $m$ such that $E_m^->E_n^+$, contradicting (++).  Therefore,
$E$ is in every interval $[E_n^-, E_n^+]$, so (*) will be satisfied for all $n$.

This derivation is constructive in the sense that it gives an algorithm to compute $E$ to any desired accuracy (by computing the bounds $E_n^{\pm}$), although it does not give a good algorithm for computing Sylvester's sequence since, to find an $E$ that will work for any given $n$, you must first compute $s_n$.

For a general method, see this paper:

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fib. Quart., 11 (1973), 429-437.

In that paper, the constant E is written as the exponential of an infinite sum, but the sum involves the terms of the sequence.

For an equivalent expression for E as a product, see

On Certain Nonlinear Recurring Sequences
Solomon W. Golomb
The American Mathematical Monthly
Vol. 70, No. 4 (Apr., 1963), pp. 403-405.


Simplify this expression fully How would i simplify fully the following expression?

$\dfrac{{\sqrt 2}({x^3})}{\sqrt{\frac {32}{x^2}}}$

So far i have got this

$\dfrac{{\sqrt 2}{x^3}}{{\frac{\sqrt 32}{\sqrt x^2}}}$ = $\dfrac{{\sqrt 2}{x^3}}{{\frac{4\sqrt 2}{x}}}$

Am not quite sure if this is correct however, could someone help explain how i would simplify this expression?

There is a mistake in the OP.  Recall that $\sqrt{x^2}=|x|\ne x$ when $x<0$.  To simplify, we can write

$$\frac{\sqrt 2 x^3}{\sqrt{\frac{32}x}}=\frac{\sqrt 2 x^3}{\frac{4\sqrt 2}{|x|}}=\frac{x^3|x|}{4}$$


Why is there no theory of $G$-ic varieties, for linear algebraic groups $G$? A toric variety is an algebraic variety $X$ with an embedding $T \hookrightarrow X$ of an algebraic torus $T$ as a dense open set, such that $T$ acts on $X$ and the embedding is equivariant.
It turns out that, given this setup, essentially all the algebraic-geometric information of $X$ can be encoded in a finite combinatorial structure (the fan of $X$). This makes toric varieties very appealing to algebraic geometers, since they're much easier to calculate with than arbitrary varieties. And indeed, toric varieties have been intensively studied since they were introduced in the early 1970s.
It seems natural to generalise this by replacing $T$ by some other linear algebraic group $G$, and studying the resulting class of varieties. To reiterate, that means we are looking at $G$-varieties $X$ with a dense equivariant embedding $G \hookrightarrow X$.
But I have never seen anyone mention this more general setup. So my questions are:

Q 1. Is there a clear conceptual reason that, for general $G$, this setup does not lead to a nice theory?
Q 2. If so, can we put some restrictions on $G$ so that we do get a nice theory?
Q 3. Finally, are there any references that discuss this general setup?

Actually there is a concept of Spherical varieties. It is a variety that admits an action of a reductive algebraic group G wherein a Borel subgroup has an open dense orbit. So some kind of  generalization from torus to connected solvable group is there. The names of people who have contributed to this study that I remember are Michel Brion, Luna, Vust.

Lakshmibai also has done some work on Schubert Varieties X in G/P that are spherical varieties for some Levi subgroup of G.

It turn out that torus are the only compact abelian groups.

So both the algebra and the topology are especially simple and well mastered.


Prove continuity by the epsilon-delta definition Let $a>0$, $b>0$ and $f:\mathbb{R} \to \mathbb{R}$ define by
\begin{align}
f(x) = \left\{
\begin{array}{ll}
x^a \sin{x^{-b}} \quad &\text{if } x \ne 0, \\
0 \quad &\text{if } x = 0
\end{array}
\right.
\end{align}
I have to find all values of $a$ and $b$ when $f$ is continuous.

We know that $f$ is continues on $(\infty, 0)$ if $\forall x<0$, $\forall \epsilon > 0$,  $\exists \delta > 0$ and $\forall y<0$ such as
$$0 < |x-y| < \delta \implies |x^a\sin{x^{-b}} - y^a\sin{y^{-b}}| < \epsilon$$

But I don't know how to approach the right side of the implication to isolate $|x-y|$. Can you help me?

The question is unclear, but it sounds like you want to show that the function is continuous on $(-\infty, 0)$, which does not require epsilons and deltas. Just use the facts that:

Compositions of continuous functions are continuous
Quotients of continuous functions are continuous if the divisor doesn't attain $0$ on the interval.

Then $\dfrac{1}{x^b}$, $x^a$, and $\sin{x}$ are continuous on $(-\infty, 0)$, so $x^a \sin{\dfrac{1}{x^b}}$ is also continuous.


C-language binary Math - Convert bytes to hex and to decimal I am having concerns on a homework assignment.

  All problems are in C-Language Syntax (<--What does that imply? Am I supposed to write my answers in a certain format?)

  Given two bytes x=0x23 and y=0xA2, fill in the following table indicating the byte values of the different C expressions: (ONE BYTE
  ANSWERS) (<--What does that imply? What is a "one-byte" answer?)

 --------------------------------------------
| Expression    Hex Value    Decimal Value   |
|   x & y                                    |
|  ~x | y                                    |
| !x || y                                    |
| x && ~y                                    |
 --------------------------------------------

I have looked up the binary equivalencies of 0x23 and 0xA2. I found that:

0x23=0010 0011=23=x

and

0xA2=1010 0010=162=y

I know that the & symbol is the AND operator and | is the OR operator. Knowing this, I ANDed the binary values of x and y together to find x & y:

x:     0010 0011
y:     1010 0010
x & y: 0010 0010

0010 0010=0x22=34=x & y

...and OR'd the vaues of ~x and y together to find ~x | y:

~x:     1101 1100
y:      1010 0010
~x | y: 1111 1110

1111 1110=0xfe=254=~x | y

Are those correct?

I asked an online tutor to assist me, and he got the following results:

 -------------------------------------------
| Expression    Hex Value    Decimal Value  |
|   x & y         0022          34          |
|  ~x | y       ffff fffe       -2          |
| !x || y          0001          1          |
| x && ~y          0001          1          |
 -------------------------------------------

How did he get -2 as the decimal value for ~x | y? Is getting negative values even possible? Also, are his hex values formatted correctly? In the instructions, it says to give "ONE BYTE ANSWERS". Doesn't that mean they should be formatted like they are in the question? (i.e. 0x23 and 0xA2).

Finally, that leaves the && and || operators. On this page, it says that && and || are "logic operators", while & and | are bitwise operators. I'm not sure what that means or how to apply that to the evaluation of the binary expressions.

By "C-Language Syntax", the problem presumably means that all the symbols are to be interpreted as given in the C standard. This helps to clarify, for example, the distinction between | and ||.

You're right, the first row is 0x22 = 34 and the second row is 0xfe. There are two ways to interpret 0xfe, either as a uint8_t, in which case its decimal value is 254, or an int8_t, in which case its decimal value is -2. The problem is ambiguous! It describes the data types of x and y as simply "bytes". I think you're justified in assuming an unsigned type, so your 254 is correct. If possible, I would explicitly write this assumption in your homework. You can even mention that -2 is the other option, just to cover your bases in case the grader makes the opposite assumption.

0xfffffffe is incorrect because it occupies 4 bytes, which violates the "ONE BYTE ANSWERS" part of the question.

As for the logical operators, their behavior is not something that you can divine from mathematical principles; it really depends on how C represents booleans as integers, which is not universal among programming languages. Your class really ought to cover that stuff!

First, in your example, ~x|y=0xffff fffe instead of 0xfe -- the leading 1's make all the difference.

Are you familiar with the two's complement representation of numbers? Basically, 0xffff ffff is -1 in decimal. 0xffff fffe is 1 less than 0xffff ffff and is therefore -2.

Read more about two's complement here:
https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html

Regarding && and ||. C treats non-zero integers as True (1) and zero as False (0). So !x || y becomes (not True) OR True which evaluates to True (1).

Since y is non-zero, ~y is non-zero too. So x && ~y becomes True AND True which evaluates to True(1).


What is correct approach for test of divergence/convergence for $\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n+1}}{n^4}$ What is the correct approach for determining if the following series is convergent or divergent?

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n+1}}{n^4}$$

My thought process was to use the limit comparison test where:

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n+1}}{n^4} \leq \frac{1}{n^4}$$

According the p-series test this series converges. Is this correct?

After some experience with such problems, the answer is obvious: the terms approximate $n^{-5/2}$, so the series converges. But how do we approach the problem from first principles?

The first step is to make a crude estimate of the magnitudes involved. In this case, we have numerator $\sqrt{n^3+n+1}\approx n^{3/2}$, and denominator $n^4$. This gives a crude estimate of $\dfrac{n^{3/2}}{n^4}=n^{-5/2}$.

As I expect you know, $\sum n^\alpha$ converges if $\alpha<-1$, so this crude estimate converges.

The next step is to show that your crude estimate of the numerator is a good enough approximation. That means that it is less than the true value by a constant factor $C$, i.e. $\sqrt{n^3+n+1}<Cn^{3/2}$ for some $C$. Then you have $\sum \dfrac{\sqrt{n^3+n+1}}{n^4}<C\sum n^{-5/2}$, which converges. So look for $C$ such that $n^3+n+1<C^2n^3$ for all $n$, and you are done. In fact you only need to find $C$ such that this inequality holds for all sufficiently large $n$, which often makes the problem much simpler.

In the previous paragraph, I implicitly used the fact that all the terms are positive. If this is not the case, you might have a little more work to do.


If every element of $H$ and $G/H$ is a square, then prove that so is every element of $G$. Let $H$ be a subgroup of an abelian group $G$ such that every element of $H$ can be written as $b^2,\, b \in G$ and similarly for $G/H$. Then how to prove that every element of $G$ can also be written as a square ?

Alternative proof. Apply the Snake Lemma to
$$\begin{array}{c} 0 & \rightarrow & H & \rightarrow & G & \rightarrow & G/H & \rightarrow & 0  \\ & & 2 \downarrow ~~& & 2 \downarrow ~~ & & 2 \downarrow ~~ \\0 & \rightarrow & H & \rightarrow & G & \rightarrow & G/H & \rightarrow & 0 \end{array}$$

Hint: Note that $gH=aHaH=a^2H$ so elements of the coset $gH$ have the form...what?


Boundary and Interior Points of the set: Rational Numbers While I was studying metric spaces, I saw this question on the book:

  Let $Q$ be the set of all rational numbers. What are $intQ$ and $∂Q$?

I think the answer should be: $intQ$ is all the rational numbers, and the $∂Q$ is all the real numbers. But I do not have the answer and do not feel confident with my answer. Is my answer true?

edit: To make it clear why I though $intQ$ is all the rational numbers: if I take $X_0$ = 3/2 and r=1, $(X_0-r,X_0+r)$ is a subset of $Q$ and as long as I choose $X_0$ a rational number, for all values of r, I can get a rational number.

You have to specify what metric space you are working in. If you say that $\mathbb{Q}$ is a subset of the metric space $(\mathbb{R},d)$ where $d$ is the usual metric given by the usual absolute value then we get that $\operatorname{int}{\mathbb{Q}} = \emptyset$ since if we take any $x \in \mathbb{Q}$ and $r \in \mathbb{R}$, $r>0$, then any open ball (i.e. open interval) centered at $x$ will contain irrationals, and hence it will not lie entirely in $\mathbb{Q}$.

Also we have $\partial \mathbb{Q} = \mathbb{R}$ since the boundary is defined to be the closure minus the interior, and the closure if simply $\mathbb{R}$ since if $x \in \mathbb{R}$ then $x$ ``adheres'' to $\mathbb{Q}$, i.e. there will be some sequence of rational numbers tending to $x$ (for instance, a decimal expansion).

Ok, here is a explanation why Int($\mathbb{Q}$) is empty.
We know that a point $p$ is an interior point of any set $S$ if there is a neighborhood of $p$ contained entirely in the set $S$. Now suppose that any real number $r$ is interior point of set $\mathbb{Q}$, then there should be a neighborhood of $r$ contained entirely in $\mathbb{Q}$, but we can't find any such neighborhood, because the set of irrationals is dense in $\mathbb{R}$.
Hence int($\mathbb{Q}$) is empty.


Determine the least upper and the greatest lower bounds of the following sets Determine the least upper and the greatest lower bounds of the following sets:

$a)$ A={${\frac{m} {n} : m,n \in N ,m<2n}$}

$b) $B={${n^{1/2} -[n^{1/2}] : n \in N}$}

my attempts :  for option a) if m= n=1  then greatest lower bound will be  1 and if m = 2n ,the the least upper bound will be 2

for option b) here greatest lower bound will be 0... that is $n^{1/2} -n^{1/2}$=0 here i don't know the what is least upper bounds

Pliz verified its  and tell me the solution

thanks in advance

Consider (a). All elements of $A$ are between $0$ and $2$. Now think about limits.

a) Supremum will be 2. As choose any $\epsilon >0$ there exist a rational number such that $m-\epsilon <m/n<2$
And Infimum will be 0. As by Archimedean property for any $\epsilon>0$ there exist $m\in N$ such that $0<1/m<\epsilon$

b)note that $0\leq b<1$ for all $b\in B$ so Infimum is 0 as 0 is the element of B for any square number.
Now Supremum is 1(we show that to you)
Now choose $\epsilon>0$ and let $y=1-\epsilon$ and $1>y>0$ consider any integer $m$ such that $m>\frac{y^2}{2(1-y)}$ so then $y<\sqrt{m^2+2m}-m$ now note that, $\sqrt{m^2+2m}-m =\sqrt{m^2+2m}-[\sqrt{m^2+2m}]<1$

so given any $y>0$ we find a number $m'$ in $B$ such that $1-\epsilon<m'<1$ so Supremum is 1.

To understand the solution, you must know the definition of Supremum and Infimum.


Two coupled second order differential equations I have two coupled equations in the form:

$$f''(x) + g'(x) + f(x) = 0$$
$$g''(x) + f'(x) + g(x) = 0$$

Looking at the form, i can guess a relation of the form $g(x) = \lambda f(x)$.
where $\lambda$ is some constant. I can find the constant by replacing $g(x)$ in the above equations and comparing the coefficients of every derivative. Finally i'm left with a single equation which is easily solvable.

The question is: is this procedure legal? and is the solution that i get is the most general one?

Thanks

Hint

Make a substitute
$$
f' = p \\
g' = q
$$
then you get a system
\begin{align}
f' &= p \\
g' &= q \\
p' &= -f - q \\
q' &= -g - p
\end{align}
or in matrix form
$$
\left[ \begin{array}{c}
f \\ g \\ p \\ q
\end{array}\right ]' = \left [ \begin{array}{cccc}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 0 & 0 & -1 \\
0 & -1 & -1 & 0
\end{array}\right ] \cdot
\left[ \begin{array}{c}
f \\ g \\ p \\ q
\end{array}\right ]
$$
which is in form of
$$
\mathbf y' = \mathbf A \cdot \mathbf y
$$
Can you take it from here?


Is the Empty set an orientable manifold? The empty set can be regarded as an object in the category of smooth manifolds, at least for technical considerations.

Is the empty set an orientable manifold?

Contradicting Espen's answer: The empty map does (vacuously) provide a nonzero differential form at every point on the manifold, and does so continuously. Therefore the empty manifold is orientable in every dimension.

(But really this will depend on the exact definition you're working with, and among the various usually-assumed-equivalent definitions for "orientable" there are some that are only really equivalent when the manifold is assumed to be nonempty).

EDIT: This is wrong. See Henning Makholm's answer below.

An $n$-manifold is orientable if and only if it has a nonzero differential form taking $n$ arguments. The empty manifold has no nonzero forms whatsoever, so it is not orientable in any dimension.


$x_{n+1}=x_n\cos(y_n)-y_n\sin(y_n), y_{n+1}= x_n\sin(y_n)+y_n\cos(y_n) \implies \lim_{x \to \infty}x_n, \lim_{y \to \infty}y_n$ Question_

  The sequences $\{x_n\}$ and $\{y_n\}$ are defined as $(x_1, y_1)=(0.8, 0.6)$,

  $x_{n+1}=x_n\cos(y_n)-y_n\sin(y_n)$ and

  $y_{n+1}= x_n\sin(y_n)+y_n\cos(y_n)$.

  Then, if $$\lim_{x \to \infty}x_n, \space\space\space \lim_{y \to \infty}y_n$$ exist, find the value of them. If they does not exist, prove it.

I'm not sure whether they have values or not. Here's the procedure I've made:

My Own Procedure_

  Let $\{z_n\}$ be a sequence given by $\tan\left(z_n\right)=y_n/x_n$.

  We can drive:
  $$x_{n+1} = \sqrt{x_n^2+y_n^2}\cos(y_n+z_n)$$
  $$y_{n+1} = \sqrt{x_n^2+y_n^2}\sin(y_n+z_n)$$

  Using two equations, we can make a new equation about $y_n$ and $z_n$:
  $$\tan(z_{n+1}) = \tan(y_n+z_n)$$
  Subsequently,
  $$z_{n+1} = y_n+z_n+k\pi$$

And I'm stuck here. Do you think that the procedure is going in the right way? Could you give me other key ideas about the problem? Thanks very much.

If $x_0=y_0=0$, then $x_n=y_n=0$. Assume that $x_n^2+y_n^2>0$. Set
$$
x_n+iy_n=r_n\mathrm{e}^{i\vartheta_n},
$$
then the recursion provides
$$
r_{n+1}\mathrm{e}^{i\vartheta_{n+1}}=x_{n+1}+iy_{n+1}=\mathrm{e}^{iy_n}(x_{n}+iy_{n})=
r_n\mathrm{e}^{iy_n}\mathrm{e}^{i\vartheta_n}
$$
Hence, $r_{n+1}=r_n=\cdots=r_0$, and
$$
\vartheta_{n+1}=\vartheta_n+y_n=\vartheta_n+r_0\sin \vartheta_n,
$$
modulo $2\pi$.

I have run a program and it does not always converge.

Convergence can be proved if $r_0<1/6$ and $\vartheta_0$ near $2k\pi$, in which case
$$
x_n\to r_0\quad\text{and}\quad y_n\to 0.
$$

It's easy to notice that the sum of the squares of the $2$ sequences is constant (i.e. $x_1^2+y_1^2$) here $x_n^2+y_n^2 = x_1^2+y_1^2 = 1$
Now,
$|x_{n+1}| = |x_n\cdot\cos(y_n) - y_n\cdot\sin(y_n)|$ , which is bounded above at $(x_n^2+y_n^2)^{1/2} = 1$
Also,
$x_{n+1} = x_n\cdot\cos((1-x_n^2)^{1/2}) - ((1-x_n^2)^{1/2})\cdot\sin(y_n) \le x_n - (1-x_n^2)^{1/2} < x_n$.
So by monotonic convergence theorem we can assert that $x_n$ is convergent
Similar arguments provide that , $|y_n| \le1$. Due to the nature of sine function when $y_n$ is less than $1$ we also conclude that $y_n$ is non-negative .
We also get that $y_n$ is increasing owing to the fact that $x_n$ is increasing and sum of the $2$ sequences is $1$ .
So we can again assert by monotone convergence theorem that $y_n$ converges .
Now passing over the lim gives $\lim y_n = 0$ and $\lim x_n^2=1$
Now notice that after a certain cutoff say $N$ belonging to set of Natural numbers $x_n$ is negative ($n>N$) so $\lim x_n < 0 \to \lim x_n = -1$.


Prove for every $a_n>0$ if $\lim _{n\to \infty }\:\:\left(\sqrt[n]{a_n}\right)=L$ then $\lim _{n\to \infty }\:\:\left(\frac{a_{n+1}}{a_n}\right)=L$ Prove for every $a_n>0$ if $\lim _{n\to \infty }\:\:\left(\sqrt[n]{a_n}\right)=L$ then $\lim _{n\to \infty }\:\:\left(\frac{a_{n+1}}{a_n}\right)=L$  is true or false

  i know that's its true if $\lim _{n\to \infty }\:\:\left(\frac{a_{n+1}}{a_n}\right)=L$ then $\lim _{n\to \infty }\:\:\left(\sqrt[n]{a_n}\right)=L$

but in this case i think that this is false and my example is if $a_n=\sqrt[n]{n^n}$ then $\lim _{n\to \infty }\:\:\sqrt[n]{n^n}=\infty \:$ but $\lim _{n\to \infty \:}\left(\frac{\left(n+1\right)^n}{n^n}\right)=e$

is that correct?

Your first statement does not hold. Consider the sequence $a_n = 1, 2, 1, 2, \ldots$. Then
$\lim_{n \to \infty} \sqrt[n]{a_n} = 1$, but $\lim_{n \to \infty}
\frac{a_{n+1}}{a_n}$ does not exist.

One always has (compare
Relationship Between Ratio Test and Power Series Radius of Convergence or Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$)
$$
 \liminf _{n \to \infty}
\frac{a_{n+1}}{a_n} \le \liminf_{n \to \infty} \sqrt[n]{a_n}
 \le \limsup_{n \to \infty} \sqrt[n]{a_n}
 \le  \limsup_{n \to \infty}
\frac{a_{n+1}}{a_n}
$$
so that the existence of $\lim_{n \to \infty}
\frac{a_{n+1}}{a_n}$ implies the existence of
$\lim_{n \to \infty} \sqrt[n]{a_n}$ (with the same limit, which can
be finite, $+\infty$, or $-\infty$),
but not the other way around.

The theorem says that if both sequences has finite limit, then the two limits are equal.


Is there any way to calculate the simple product of binomial coefficients Given the sum $$ \sum_{k=0}^{m} {n \choose k} {m \choose k}, $$ where $ n > m$. Could it be somehow calculated into a shorter an nicer expression which doesn't contain the sum?

Thanks in advance!

We use the  coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ of a series. So, we can write e.g.
 $$\binom{m}{k}=[x^k](1+x)^m$$

  We obtain for $a\geq 0$
  \begin{align*}
\sum_{k=0}^{m}&\binom{m+a}{k}\binom{m}{k}\\
&=\sum_{k=0}^{\infty}[x^k](1+x)^{m+a}[y^k](1+y)^m\tag{1}\\
&=[x^0](1+x)^{m+a}\sum_{k=0}^{\infty}x^{-k}[y^k](1+y)^m\tag{2}\\
&=[x^0](1+x)^{m+a}(1+\frac{1}{x})^m\tag{3}\\
&=[x^0](1+x)^{m+a}\frac{1}{x^m}(1+x)^m\\
&=[x^m](1+x)^{2m+a}\\
&=\binom{2m+a}{m}
\end{align*}

Comment:

In (1) we use the coefficient of Operator and change the limit to $\infty$ without changing the sum, since we add only zero.
In (2) we use the linearity of the coefficient of operator and $[x^{n+k}]A(x)=[x^n]x^{-k}A(x)$
In (3) we use the substitution rule
\begin{align*}
A(x)=\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^{\infty}x^k[y^k]A(y)
\end{align*}

it is related to a question I answered yesterday :

$$\sum_{k=0}^m {n \choose k} {m \choose k} = \frac1{2\pi}\int_0^{2\pi} (e^{ix}+1)^n (e^{-ix}+1)^m dx =  \frac{2^{n+m}}{2\pi} \int_0^{2\pi}e^{ix(n-m)/2}\cos^{m+n}(x/2) dx $$


How to solve for $x$ in the equation $y = (x + 25)^x$ where you know the value of $y$ I am trying to figure out how to solve for $x$ in the equation $y = (x + 25)^x$ for any known $y$.
For example, I know that in the example $2758547353515625 = (x + 25)^x$, the value of $x$ is 10.
Is there a way to solve for $x$ for any known $y$?

Making the problem more general, you want to find the zero of function
$$f(x)=(x+a)^x-y$$
I shall assume $a>0$ and suppose that $y$ could be a huge positive number.
$f(x)$ varies very quickly and this is not good for any root-finding algorithm.
At the opposite, consider instead
$$g(x)=x \log(x+a)-\log(y)$$ which is much more linear.
I am very lazy and I shall expand as series to have
$$g(x)=x \log (a)+O\left(x^2\right)-\log(y)$$ which gives as an estimate
$$x_0=\frac{\log (y)}{\log (a)}$$ which will allow to start Newton method.
Let us take the case of $a=25$ and $y=1234567898765432123456789$. The iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 17.233582865360804139 \\
 1 & 15.056838056149540045 \\
 2 & 15.034341020407527174 \\
 3 & 15.034338495863187264 \\
 4 & 15.034338495863155457
\end{array}
\right)$$

With $y\neq0$ (trivial), using logarithms
$$\ln(y)=x\ln(x+25)\implies \frac{\ln(y)}{\ln(x+25)}=x\implies\frac{1}{\ln(x+25)/\ln(y)}=x$$
and using a property of logarithms
$$\frac{1}{\log_y(x+25)}=x\implies \frac{1}{x}=\log_y(x+25)$$
which is a transcendental equation, therefore requires numerical solutions on most cases.
If you want to invert transcendental equations involving the exponential function, sometimes it useful to consider the Lambert function


Does a random variable come from a probability distribution or is it vice-versa? I am curious to know which statement is correct:

a random variable come from a probability distribution
OR
a probability distribution is created from observing the behavior of a random variable

Interesting question! The answer is that it can go both ways.

For any probability distribution F, i.e. a cadlag (right continuous with left limits) function $F: (-\infty, \infty) \to [0,1]$ satisfying

\begin{equation} \lim_{x \to -\infty} F(x) = 0, \lim_{x \to \infty} F(x) = 1,
\end{equation}

there exists a random variable $X$ with $\mathbb{P}(X \leq x) = F(x)$. Specifically, this means there exists a probability space $(\Omega, \mathcal{A}, \mathbb{P})$, and a measurable function $X: \Omega \to \mathbb{R}$ with $\mathbb{P}(\{\omega: X(\omega) \leq x\}) = F(x)$ for every $x \in \mathbb{R}$. Depending on the distribution function $F$, there may be many possible probability spaces that work, but usually $\Omega$ just sits in the background, and we don't think much about it. (It turns out that the space $\Omega = [0,1]$ with the Borel sigma-field and Lebesgue measure always works, so you can assume that's where $X$ lives if you like.)

Conversely, if $X$ is a measurable map from a probability space $\Omega$ into $\mathbb{R}$, then the function $F_X: \mathbb{R} \to [0,1]$ is called the distribution function of $X$, defined as $F_X(x) = \mathbb{P}(\{\omega: X(\omega) \leq x\})$. One can use the axioms of a probability space to show that $F_X$ satisfies all the properties mentioned above.

Philosophically, what is happening? Roughly speaking, a statistician would like the first definition. The idea is: we observe data from some process, and form an empirical distribution from it. We assume there is a random variable $X$ generating the data, and we can try to understand how the empirical distribution relates to the true distribution of $X$.

For a mathematician, the other order is more typical. We define processes or variables abstractly using random variables on probability spaces, and prove theorems about them, or about the corresponding distribution functions.

Any random variable is a real-measurable function on the given space, equipped with some $\sigma$-algebra. It induces the probability mass function, in case the image is a discrete set, etc.


Substructure of $\mathbb Q$ that is not a field We have a  structure: $\mathbb Q=\langle \mathbb Q,\,+,\,-,\,*,\,0,\,1 \rangle$

Question: Does $\mathbb Q$ contain a substructure which is not a field?

I have in my notes that there is such a substructure and it is $\mathbb Z$ because $\mathbb Z \subseteq \mathbb Q$

But I don't think that I fully understand it.

We wrote the axioms of field and that

$x \neq 0 \rightarrow (\exists y)(x*y=y*x=1)$  does not satisfy rest of the theory.

but I don't know exactly why.

The answer to your question depends on what kind of substructure you're looking for. (That might have been explained in class but didn't get to your notes.)

There is no subset of $\mathbb{Q}$ that is itself a field with the usual addition and multiplication.

$\mathbb{Z}$ is a subring of $\mathbb{Q}$ when you think of the latter as a ring.

Any subset of $\mathbb{Q}$ is a "substructure" if "being a set" is the only structure you care about.

A possible "substructure" here is "subring" in this context. The integers are a subring of the field $\mathbb{Q}$. But they are not a subfield, because not all nonzero elements are invertible, e.g. the equation $2x=1$ has no solution in the integers.


Why should the sign of remainder be same as dividend? I was reading binary division from this site and this statement came across that "remainder will have the same sign as the dividend". But why? Like from the utmost conceptual level, the very basic idea of remainders and negative numbers.

I'm going to hazard a guess at why someone might adopt a convention that the remainder should have the same sign as the dividend. (As others have noted, the usual convention, when writing $n=qd+r$ is that $0\le r\lt|d|$, where, to fix terminology, $n$ is called the dividend, $q$ the quotient, $d\not=0$ the divisor, and $r$ the remainder. This convention is convenient for most number-theoretic purposes, although occasionally one wants to use the convention $-|d|/2\lt r\le |d|/2$.)

Let's start with a non-controversial example: $17$ divided by $5$. No one, I hope, objects to writing

$$17=3\cdot5+2$$

for the answer.  Now the easiest thing to do, if we put a negative sign in front of the $17$, is to write

$$-17=-3\cdot5-2$$

which we can interpret as either

$$-17=(-3)\cdot5-2$$

if we are dividing $-17$ by $5$, or as

$$-17=3\cdot(-5)-2$$

if we are dividing $-17$ by $-5$.  (The final combination of signs, $17$ divided by $-5$, is $17=(-3)\cdot(-5)+2$.)

In other words, the convention used in the OP's link begins by taking absolute values of both the dividend and the divisor, does the non-controversial calculation to get absolute values for the quotient and remainder, and then inserts signs in the appropriate places to turn $|n|=|q||d|+|r|$ into $n=qd+r$. The key here is that the final adjustment doesn't affect the digits of any of the numbers, only the numbers' signs. Given the link's computer-sciencey nature, that makes sense as a desirable feature for a convention.

It is a very important convention since we usually take our dividends to be positive. This is also in other contexts called the modulus, and the remainder the residue. Thus given a modulus $d> 0$ and an integer $n,$ we have that there are always integers $q$ and $r\ge 0$ such that $$n=qd+r.$$ whatever negativity is absorbed by the first part, in particular with the quotient $q.$


If P1, P2, P3 are altitudes of a triangle ABC from the vertices A, B, C and is the area of the triangle, then P^-1+P^-2+P^-3 is equal to If P1, P2, P3 are altitudes of a triangle ABC from the vertices A, B, C and  is the area of the triangle , then P^-1+P^-2+P^-3  is equal to

Using Area of Triangle in Terms of Side and Altitude formula,

$$\triangle=\frac12a\cdot P_1\implies\frac1{P_1}=\frac a{2\triangle}$$

$$\implies \sum\frac1{P_1}=\frac{a+b+c}{2\triangle}$$

Using Area of Triangle in Terms of Inradius formula, $$\triangle=r\cdot s$$

where $r$ is the inradius  and $2s=a+b+c$

Using Area of triangle in terms of side and altitude formula,
Triangle= 1/2 a. P1=>1/p1=a/2triangle
        => £1/p1=a+b+c/2triangle

Using Area of triangle in terms of inradius formula,
Triangle= r.s
Where r is inradius and
2s=a+b+c
Hope it is useful......


Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit

$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$

My approach :

If I divide numerator and denominator by $n^2$ I get :

$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$

but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.

For each $1\leq i\leq n$, $\frac{1}{n^2+i}\leq \frac{1}{n^2}$ and $\frac{1}{n^2+i}\geq \frac{1}{n^2+n}$, and so we may bound the sum from above and below by $$\sum_{i=1}^{n}\frac{i}{n+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{n^{2}}.$$ Since $\sum_{i=1}^{n}i=\frac{n(n+1)}{2},$ this becomes $$\frac{1}{2}=\frac{n(n+1)}{2n(n+1)}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\frac{n(n+1)}{2n^{2}}=\frac{1}{2}\left(1+\frac{1}{n}\right),$$ and so it follows from the squeeze theorem that the limit is $\frac{1}{2}$.


Trigonometric integrals. Integrate $\int\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}\mathrm{d}x$ I tried factoring the denominator but does not seem to suit the factorization of quadratic. Is there another way?

\begin{align*}
 J &= \int \frac{1}{\cos^4 - \cos^2 x \sin^2 x + \sin^4 x} dx \\
   &= \int \frac{\sec^4}{1 - \tan^2 x + \tan^4 x} dx
\end{align*}

Let $u = \tan x$ and $\sec^2 x = \tan^2 x + 1$

$$ J = \int \frac{(1+\tan^2 x) \sec^2 x}{1- \tan^2 x +\tan^4 x} dx $$

again $ u = \tan x$

$$ J = \int \frac{u^2 + 1}{u^4 - u^2 + 1} du $$

Since $ \int \frac{u^2 + 1}{u^4 - u^2 + 1} du = \arctan(\frac{u}{1-u^2}) + c $

$$ J = -\arctan(2\cot(2x)) + c$$

Yeah from $J=\int\frac{u^2+1}{u^4-u^2+1}\mathrm{d}u$       i manipulated it to get $\int\frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}-1}\mathrm{d}u$  now we can rewrite the the denominator as $u^2+\frac{1}{u^2}=\left(u-\frac{1}{u}\right)^2+2$ now putting it back in give me                           $\int\frac{1+\frac{1}{u^2}}{\left(u-\frac{1}{u}\right)^2+1}\mathrm{d}u$ by doing substitution letting $\left(u-\frac{1}{u}\right)=w$   i got the result $\tan^{-1}\left(\frac{\tan^2x-1}{\tan x}\right)+c$


Need help finding the series in order to prove by induction I need to prove that for all natural numbers $n$, $\frac{n(n+1)(n+2)}{6}$ is a natural number. The problem is that I can't seem to figure out what the series is, the LHS. Any help would be appreciated.

Hint: The numerator is a product of $3$ consecutive numbers, so we can say something about the prime factors of at least two of the numbers (think of the prime factors for $6$).

You seem to be completely misunderstanding the question, thinking, apparently, that the fraction you are given is the sum of some series.  Whether it is or not is irrelevant to the question.  You have n(n+1)(n+ 2).

Now, either n is even or n is odd.  If n is even, it is divisible by 2- it can be written a n= 2k and so is n(n+ 1)(n+ 2)= 2k(2k+1)(2k+ 2) which is divisible by 2.  If n is odd, it can be written n= 2k+ 1. n(n+1)(n+2)= (2k+1)(2k+1+1)(2k+1+2)= (2k+1)(2k+2)(2k+3)= 2(2k+1)(k+1)(2k+3).  So what ever n is, n(n+ 1)(n+2) is divisible by 2.

Now, "3" is just a little harder.  Either n is divisible by 3 or it has remainder 1 when divided by 3 or it has remainder 2 when divided by 3.
If n is divisible by 3, n= 3k so n(n+1)(n+2)= 3k(3k+1)(3k+2) and is divisible by 3.
If n has remainder 1 when divided by 3, it is of the form n= 3k+1 so n(n+1)(n+2)= (3k+1)(3k+1+1)(3k+1+2)= (3k+1)(3k+2)(3k+3)= 3(3k+1)(3k+2)(k+1) and is divisible by 3.
If n has remainder 2 when divided by 3, it is of the form n= 3k+2 so n(n+1)(n+2)= (3k+2)(3k+2+1)(3k+2+2)= (3k+2)(3k+3)(3k+4)= 3(3k+2)(k+1)(3k+4) and is divisible by 3.

So for any integer n, n(n+1)(n+3) is divisible by both 2 and 3 and so divisible by 6.


Inverse Bessel Process as continuous local martingale Let $B$ be a $n$-dimensional brownian motion. This question shows, that
$$\Big(\sum_{i=1}^n\int_0^t\frac{B^i_s}{||Bs||}dB^i_s,||B_t||\Big)$$
is a weak solution of $dX=\frac{n-1}{2X}dt+dW$.
Now I would like to show, that $||B||^{-1}$ is a continuous local martingale iff $n=3$.
Let $n=3$. Using the SDE, $$||B_t||^{-1}=\int_0^t||B_s||ds+W_t$$
Therefore $||B||^{-1}$ is a continuous local martingale, since this is sum of two continous local martingales? But wouldn't this apply to any other dimension as well?
How could I deal with the other direction?
I'd appreciate some hint or help on this problem. Thanks for your attention, again!

The function $$h:(0,\infty)\rightarrow(0,\infty),\quad h(x):=x^{-1}$$ has derivatives
$$\begin{align}
\frac{\partial h}{\partial x}(x)&=-x^{-2}\\
\quad\frac{\partial^2 h}{\partial^2 x}(x)&=2x^{-3}
\end{align}$$
Using both Ito's lemma and the SDE, one finds
$$\begin{align}
||B_t||^{-1}=h(||B_t||)&=h(||B_0||)-\int_0^t||B_s||^{-2}d||B_s||+\frac{1}{2}\int_0^t2||B_s||^{-3}d[||B||]_s\\
&=h(||B_0||)-\int_0^t\frac{n-3}{2}||B_s||^{-3}ds-\int_0^t ||B_s||^{-2} dW_s
\end{align}$$
The first and last term are local martingales. The middle term is a local martingale iff $n=3$.


Probability that both 2 already born children are girls given that one of them is a girl Question:You have asked a woman if she has any children. She answered that she has two children. Then you have asked her whether if she has daughter out of those two. She said yes. What is the probability that both of her child are girls?

One answered 1/3 because one thinks that there are 3 outcomes:

boy and girl

girl and boy

girl and girl

What is the correct answer?
And if you think the answer isn't 1/3, how do you explain that the above explanation is wrong?

you are quite right, $S=\{gg, gb,bg\}$(since she has said she has atleast one girl then bb doesnt happen,both are girls =$A=\{gg\}$ thus answer is $p(A)=\frac{n(A)}{n(S)}=\frac{1}{3}$.

Your answer is correct. You can also work it out using the formula for conditional probability:

\begin{eqnarray}
P(\textrm{two girls}\ |\ \textrm{at least one is a girl}) & = & \frac{P(\textrm{two girls } \textbf{and} \textrm{ at least one is a girl})}{ P(\textrm{at least one girl})} \\
& = & \frac{\frac{\left|\{GG\}\right|}{\left|\{BB, BG, GB, GG\}\right|}}{\frac{\left|\{GB, BG, GG\}\right|}{\left|\{BB, BG, GB, GG\}\right|}}\\
& = & \frac{\frac{1}{4}}{\frac{3}{4}} \\
& = & \frac{1}{3}
\end{eqnarray}


Arguing away/bounding an integral in contour integration basically I have a contour integration problem where I am trying to determine the value of
 $$\int \frac {(z^2-1)^{1/2}} {(z^2+1)} dz$$ around a dumbbell contour. I am computing this integral with the ultimate aim of finding  $$\int_{-1}^{1} \frac {(1-x^2)^{1/2}} {(x^2+1)} dx$$

In my problem I need to "argue away" the components of my integral around two circles of small radius approaching zero, i.e. show these components tend to $0$ as the radius tends to $0$

We have on a circle $\gamma$ of radius $\epsilon$ centred at 1

$$\vert \int \frac {(z^2-1)^{1/2}} {(z^2+1)} dz \vert \le 2\pi\epsilon .\max \vert \frac {(z^2-1)^{1/2}} {(z^2+1)} \vert$$

It is the $\epsilon\max \vert \frac {(z^2-1)^{1/2}} {(z^2+1)} \vert = \epsilon\frac {\max \vert (z^2-1)^{1/2} \vert}{\min \vert z^2+1 \vert}$ on $\gamma$ that I'm struggling to find. I need to show this is of order $\epsilon$ to a positive power and hence as $\epsilon \to 0$ so too does the integral. I was wondering if anyone could help?

You have to imagine that, attached to your dumbbell is a circle of radius $R$ centered at the origin.  Thus, letting all of the radii of the small detours about $z=\pm1$ go to zero, the contour integral is equal to

$$\oint_C dz \frac{(z^2-1)^{1/2}}{z^2+1} = i 2 \int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{(R^2 e^{i 2 \theta}-1)^{1/2}}{R^2 e^{i 2 \theta}+1}$$

As $R \to \infty$, the latter integral about the big circle approaches $i 2 \pi$.

By the residue theorem, the contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand at $z=\pm i$.  Thus,

$$i 2 \int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} + i 2 \pi = i 2 \pi \left (\frac{\sqrt{2} e^{i \pi/2}}{2 e^{i \pi/2}} + \frac{\sqrt{2} e^{-i \pi/2}}{2 e^{-i \pi/2}} \right )$$

or,

  $$\int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} = (\sqrt{2}-1) \pi$$

ADDENDUM

In my haste I did not see that the OP's concern is those small circles.  Consider $z=-1$.  Above the real axis, we parametrize by $z=e^{i \pi}+\epsilon \, e^{i \phi}$ and have the contribution to the contour integral being

$$i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{(2 e^{i \pi} \epsilon e^{i \phi})^{1/2}}{2} $$

which vanishes as $\epsilon \to 0$.  Same goes below the axis.  Similar analysis for $z=-1$, except we only need one integral from $\pi$ to $-\pi$.


Showing that the given sequence is bounded as follows Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.
Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows:
To show $x_{n+1}<x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that is s, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$
Back to the problem. It's clear that $A$ is bounded above by $\frac{1}{3^6}$ (Should I show this first?).
Then, to show that $A$ is bounded, it's suffices to show that $A$ is bounded below by $0$.
I show this one again by induction. Indeed, it's true for $n=1$.
Assume that it's true for $n=k$; that is $0 < x_k$. Then, $x_{k+1} = \frac{1}{3^{(k+1)+5}}
= \frac{1}{3^{k+5}} \cdot \frac{1}{3} > 0$. Hence, $0< x_n$ for all $n \in \Bbb Z^+$. Therefore, $A$ is bounded below by $0$. Thus, $A$ is bounded, as desired.
Now, I claim that $\sup A = \frac{1}{3^6}$ and $\inf(A) = 0$.
For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.
For the supremum, let $M$ be an another upper bound of $A$. To show: $\frac{1}{3^6} \le M$. Suppose $M < \frac{1}{3^6}$. Then, by the Density Theorem, there exists $s \in \Bbb Q$ such that
$M < s < \frac{1}{3^6}$. Hence, $s \in A$, contradiction with the fact that $M$ is an upper bound of $A$.
Therefore, $\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.
Does those approach true?

Then, by the Density Theorem, there exists $r \in \Bbb Q$ such that $0
< r < m$.

This is true, but irrelevant.

Hence, $r \in A$. A contradiction, since $m$ is a lower
bound of $A$.

No, you haven't show that $r \in A,$ so no, you haven't shown a contradiction. You correctly showed that there is a rational number between $0$ and $m$, but you haven't (and you won't be able to) get from "$r$ is a rational number" to "$r$ is a member of $A$".
In summary, you haven't shown that $\ r \in A.$ Think about what it means to show that $\ r \in A $ [Hint: what is $A$?]
You could, however, amend your argument as follows:
For the proof of infimum, let $m$ be an another lower bound of $A$. To show: $m \le 0$. Suppose
$m > 0$. Then, since [$\color{blue}{\text{*insert reasoning here*}}$], there exists $\color{blue}{r \in A}$ such that
$0 < r < m$. Hence, $r \in A$. A contradiction, since $m$ is a lower bound of $A$.
Thus, $m \le 0$ and therefore, $\inf A = 0$.


Compute $P(\{6|xy\})$ and $P(\{8|xy\})$ if $x$ and $y$ are independent random positive integers (Please edit this post if it is ill-formed) If $x$ and $y$ are independent random positive integers,
 1. How to compute the probability of $(2*3)|xy$ ?
 2. How to compute the probability of $(2^3)|xy$ ?

For both problems it is necessary to set up a probabilistic environment.

The divisibility of $xy$ by $6$ depends only on the remainders of $x$ and $y$ modulo $6$. It is natural to assume that both $x$ and $y$ are independently and uniformly distributed modulo $6$. In this case they are as well independently and uniformly distributed modulo $2$ and $3$. The probability that $x$ is not divisible by $2$ is ${1\over2}$. Therefore with probability $1-{1\over2}\cdot{1\over2}={3\over4}$ at least one of $x$ or $y$ is divisible by $2$. Similarly, the probability that $x$ is not divisible by $3$ is ${2\over3}$.
Therefore with probability $1-{2\over3}\cdot{2\over3}={5\over9}$ at least one of $x$ or $y$ is divisible by $3$. It follows that with probability $p={3\over4}\cdot{5\over9}={15\over36}$ the product $xy$ is divisible by $6$.

The divisibility of $xy$ by $8$ depends only on the remainders of $x$ and $y$ modulo $8$. It is natural to assume that both $x$ and $y$ are independently and uniformly distributed modulo $8$. The following cases lead to divisibility of $xy$ by $8$:
$$\eqalign{x\in\{1,3,5,7\}\quad&\wedge\quad y=0,\cr
x\in\{2,6\}\quad&\wedge\quad y\in\{0,4\},\cr
x=4\quad&\wedge\quad y\in\{0,2,4,6\},\cr
x=0\quad&\wedge\quad y\in\{0,1,2,3,4,5,6,7\}\ .\cr}$$
Therefore the probability that $xy$ is divisible by $8$ is given by
$$q={1\over2}\cdot{1\over8}+{1\over4}\cdot{1\over4}+{1\over8}\cdot{1\over2}+{1\over8}\cdot1={5\over16}\ .$$

Things might get more easy to compute by realizing that:$$\{6\mid xy\}=\{6\mid x\}\cup\{6\mid y\}\cup(\{2\mid x\}\cap\{3\mid y\})\cup(\{3\mid x\}\cap\{2\mid y\})$$
combined with the application of the principle of inclusion/exclusion.

This approach might be fruitful especially if $x$ and $y$ are independent.

Success not guaranteed, though.


I'm actually lost regarding inductions Basically, it's what it says in the title, could someone solve step by step this induction? It would be even better with an expanation for said steps, but it's not all that needed, I want to read the solution mostly to decipher how to solve it, since I have a few questions regarding Inductive Base, demonstration through hipothesis, and thesis... Esentially everything.
Here's the problem:
0+2+4+...+2n = n(n+1), whereas n is a Natural.

So you want to prove the statement by the mathematical induction, right?

start with $ n = 1 $.

then you get $ 2 = 1 \times (1+1) $

so the statement holds.

now suppose that the statement holds for $ n = k $

claim that the statement also holds for $ n = k + 1 $.

since we assumed that $ 0 + 2 + ... + 2k = k(k+1) $, add $ 2n = 2(k+1) $ on both sides.

then $ 0 + 2 + ... + 2k + 2(k+1) = k(k+1) + 2(k+1) = (k+1)(k+2) $

therefore the statement holds for $ n = k+1 $

due to the mathematical induction, the statement holds for every natural number.


Why can the determinant be assumed to be 0? I'm trying to work through how to calculate eigenvalues and eigenvectors.

I start with

$$Ax=\lambda x$$

Where $A$ is a $p \times p$ matrix, $\lambda$ is the eigenvalue and $x$ is the eigenvector.

This is the same as:

$$Ax=I\lambda x$$

$$Ax-I\lambda x=0$$

$$(A-I\lambda) x=0$$

We define the matrix $A$ as a $2 \times 2$ matrix:

$\begin{bmatrix}4 & -2\\-3 & 6\end{bmatrix}$

Thus this -$I\lambda$ equals

$\begin{bmatrix}4-\lambda & -2\\-3 & 6-\lambda\end{bmatrix}$

$$Det(A-I\lambda)=(4-\lambda(6-\lambda)-(-3)*-2)$$

$$Det(A-I\lambda)=24-10\lambda +\lambda^2 -6$$
$$Det(A-I\lambda)=18 - 10\lambda + \lambda^2 $$

Then, out of the blue my textbook claims that

$$0=30 - 10\lambda + \lambda^2 $$

How do I justify setting the determinant to $0$?

(I do "not" have an advanced knowledge in linear algebraic analysis, I only know how the determinant is used to calculate the inverse matrix)

The text is not claiming that the determinant is $0$. The text is saying "Let's find out for which values of lambda the determinant is $0$!"

So the determinant is $\lambda^2 - 10\lambda + 30$, and you want to find the $\lambda$ such that it is equal to zero. What do you do? You set it equal to zero and solve for $\lambda$. That is, you solve the equation

$$\lambda^2 - 10\lambda + 30 = 0$$

As for why you are interested in the values of $\lambda$ that make the determinant equal to $0$, remember that

$$rank(A-\lambda I) = n \iff det(A - \lambda I) \neq 0$$

So, if $det(A-\lambda I) \neq 0$, you will find that the only solution to $(A - \lambda I)x = 0$ is $x = 0$ (due to the fact that the rank of the matrix is full, hence the kernel only contains the $0$ vector). This means that the only $x$ such that $Ax = \lambda x$ is $x=0$, which means that $x$ is not an eigenvector.

So the only way to have eigenvectors is to have the determinant of $A - \lambda I$ be equal to zero, so that's why to find eigenvalues you look for the values of $\lambda$ that make $det(A - \lambda I) = 0$

As per the Eigen vectors for a transformation matrix , the transformation on such vector will result in scaling it to some values λ. Let X be the Eigen vector and A be the transformation matrix of X.
Thus,
AX = λX (λ being the scaling factor)
AX-λX=0
(A-λ*I)*X = 0

Now this (A-λI) can be considered as another transformation matrix on X which needs to be 0.
According to a algebraic rule if a transformation matrix transforms the vector X to zero , then its determinant must be zero. Which is exactly the condition we need to satisfy by finding λ(We call it Eigen values from here on) using det(A-λI)=0.
Thus we make the condition det(A-λ*I) =0 and find λ (Eigen value) from it.
Then supplying back Eigen values we get its corresponding Eigen vectors.

This is something I understood from why we need to use determinant .


Discontinuous derivative. Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.

Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example.  From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $\mathbb R$?  Here's a closer look at the Volterra-type functions referred to in Haskell's answer, together with a little indication as to how it might be extended.

Basic example

The basic example of a differentiable function with discontinuous derivative is
$$
f(x) = \begin{cases}
  x^2 \sin(1/x) &\mbox{if } x \neq 0 \\
0 & \mbox{if } x=0.
\end{cases}
$$
The differentiation rules show that this function is differentiable away from the origin and the difference quotient can be used to show that it is differentiable at the origin with value $f'(0)=0$.  A graph is illuminating as well as it shows how $\pm x^2$ forms an envelope for the function forcing differentiablity.

The derivative of $f$ is
$$
f'(x) = \begin{cases}
  2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)&\mbox{if } x \neq 0 \\
0 & \mbox{if } x=0,
\end{cases}
$$
which is discontinuous at $x=0$.  Its graph looks something like so

Two points

The next step is to modify this example to obtain a function that is everywhere differentiable with a derivative that is continuous on all of $\mathbb R$, except for two points.  To this end, consider
$$
f(x) = \begin{cases}
  x^2 (1-x)^2 \sin \left(\frac{1}{\pi  x (1-x)}\right)&\mbox{if } 0<x<1 \\
0 & \mbox{else}.
\end{cases}
$$
The graph of $f$ and its derivative look like so.

A cantor set of discontinuties

Now that we have a way to construct a differentiable function whose derivative is discontinuous exactly at the endpoints of an interval, it should be clear how to construct a differentiable function whose derivative is discontinous on a Cantor set constructed in the interval.  For $n\in\mathbb N$ and $m=1,2,\ldots,2^n$, let $I_{m,n}$ denote one of the $2^n$ intervals removed during the $n^{th}$ stage of construction of the Cantor set.  Then let $f_{m,n}$ be scaled to have support $I_{m,n}$ and to have maximum value $4^{-n}$.  The function
$$F(x) = \sum_{n=0}^{\infty} \sum_{m=1}^{2^n} f_{m,n}(x)$$
will be everywhere differentiable but its derivative will be discontinuous on the given Cantor set.  Assuming we do this with Cantors standard ternary set, we get a picture that looks something like so:

Of course, there's really a sequence of functions here and care needs to be taken to show that the limit is truly differentiable.  Let
$$F_N(x) = \sum_{n=1}^{N} \sum_{m=1}^{2^n} f_{m,n}(x).$$
The standard theorem then states that, as long as $F_N$ converges and $F_N'$ converges uniformly, then the limit of $F_N(x)$ will be differentiable.  This is guaranteed by the choice of $4^{-n}$ as the max for $f_{m,n}$.

Increasing the measure

Again, the last example refers to the standard Cantor ternary set but there's no reason this can't be done with any Cantor set.  In particular, it can be done with a so-called fat Cantor set, which can have positive measure arbitrarily close to the measure of the interval containing it.  We immediately produce an everywhere differentiable function whose derivative is discontinuous on a nowhere dense set of positive measure.  (Of course, care must again be taken to scale the heights of the functions go to zero quickly enough to guarantee differentiability.)

Finally, we can fill the holes of the removed intervals with more Cantor sets (and their corresponding functions) in such a way that the union of all of them is of full measure.  This allows us to construct an everywhere differentiable function with derivative that is discontinuous on the union of those Cantor sets, which is a set of full measure.

I guess that you are looking for a continuous function $ f: \mathbb{R} \to \mathbb{R} $ such that $ f $ is differentiable everywhere but $ f' $ is ‘as discontinuous as possible’.
We have the following theorem in real analysis.

Theorem 1 If $ f: \mathbb{R} \to \mathbb{R} $ is differentiable everywhere, then the set of points in $ \mathbb{R} $ where $ f' $ is continuous is non-empty. More precisely, the set of all such points is a dense $ G_{\delta} $-subset of $ \mathbb{R} $.

Note: A $ G_{\delta} $-subset of $ \mathbb{R} $ is just the intersection of a countable collection of open subsets of $ \mathbb{R} $.
The proof of Theorem 1 is an application of the Baire Category Theorem, and it can be found in Munkres’ Topology or Measure and Category by Oxtoby. By this theorem, it is therefore impossible to find an $ f: \mathbb{R} \to \mathbb{R} $ whose derivative exists but is discontinuous everywhere.
There is another theorem that provides a necessary and sufficient condition for a set $E$ to be the set of discontinuities of some derivative.

Theorem 2 A set $E$ is the set of discontinuities of some derivative if and only if $E$ is a meagre $ F_{\sigma} $-subset of $ \mathbb{R} $.

Note: An $ F_{\sigma} $-subset of $ \mathbb{R} $ is just the union of a countable collection of closed subsets of $ \mathbb{R} $.
Let me end off with a non-trivial example to add to yours. Volterra’s Function is differentiable everywhere, but its derivative is discontinuous on a set of positive measure, not just at a single point.


Matrix of a Transformation Let $\psi$ be a linear operator on $V$, a vector space of dimension two. How do I show, given $\psi$ is not a scalar multiple of the identity, that there is a $v$ such that {$v$,$\psi$(v)$} forms a basis, and further how do I write the transformation as a matrix with respect to that matrix? For the matrix, the first column would be 0,1, but I don't know what the second column would be.

The key observation here seems to be: let $V$ be a vector space over a field $F$ [not necessarily finite-dimensional!], and let $\psi: V \rightarrow V$ be a linear operator on $V$.  Then exactly one of the following occurs:

(i) There is $\lambda \in F$ such that $\psi(v) = \lambda v$ for all $v \in V$.
(ii) There is $v \in V$ such that $\psi(v)$ is not in the span of $v$.

It is clear that the conditions are mutually exclusive.  Seeking a contradiction we suppose that neither (i) nor (ii) holds.  Then there are $\lambda_1 \neq \lambda_2$ in $F$ and $v_1,v_2 \in V$ with $\psi(v_1) = \lambda_1 v_1$ and $\psi(v_2) = \lambda_2 v_2$.  Observe that $v_1$ and $v_2$ must be linearly independent.  Let $v = v_1 + v_2$.  Then $\psi(v) = \psi(v_1) + \psi(v_2) = \lambda_1 v_1 + \lambda_2 v_2$.  Now suppose there is $\lambda_3 \in F$ such that

$\lambda_1 v_1 + \lambda_2 v_2 = \psi(v) = \lambda_3 v = \lambda_3 v_1 + \lambda_3 v_2$,

so

$(\lambda_1 -\lambda_3)v_1 = (\lambda_3 - \lambda_2) v_2$.

Since $v_1$ and $v_2$ are linearly independent, this equation forces $\lambda_1= \lambda_3 = \lambda_2$.  This is a contradiction.

In other words, the only endomorphisms which preserve every one-dimensional subspace are the scalar linear transformations.  The OP's first question is a special case of this.

As for the second question: yes, the first column of the matrix is $(0,1)$.  In fact the second column of the matrix could be anything whatsoever: since $\psi(v)$ is linearly indpendent from $v$, for any $w \in V$ (here I am assuming, as the OP is, that $V$ is two-dimensional) there is a unique linear transformation $v$ such that $v \mapsto \psi(v)$ and $\psi(v) \mapsto w$.

What you need is a v such that
$\psi(v)$ is not a multiple of v, and not zero; (1)
and then you have your basis.

Is there such a v?  There must be a v such that $\psi(v) \ne 0$ (v is not in the nullspace of $\psi$) or else $\psi$ is identically 0.  We will assume this is not the case.

So let v be any vector which is not in the nullspace of $\psi$. There are two cases:

I) v is not an eigenvector of $\psi;$ that is $ \psi(v) \ne \lambda v$.  Then this v  satisfies (1).

II) v is an eigenvector of $\psi$.  Let u be any other non-zero vector which is not a multiple of v.  There must be such a u because this is a 2 dimensional space.  Let w = av + bu where a,b are not zero.

First, $\psi(w) \ne 0$ because $\psi(v) = a\lambda v \ne 0$.  Next, there must be a choice of a and b such that w is not an eigenvector of $\psi$; if a particular choice of a,b, causes w to be an eigenvector, choose a different b. Then av + bu cannot be a multiple of w; and it cannot be a multiple of v.  Since there are at most two eigenvectors in a 2 dimensional space, w cannot be one.

Then w and $\psi (w)$ are independent and w satisfies (1).


probability on countable infinite sets My question relates to probabilities on countable infinite sets.  For example, what is the probability of choosing an even number from the positive integers.  Believe it or not I am interested in this question from a practical standpoint.  I am writing a paper on Boltzmann's brains (Brains that occur spontaneously from the vacuum in De Sitter space). Specifically the problem is, is it more likely that I am a Boltzmann brain or a regular brain if space is infinite.  There are of course no good calculations about what the odds of a Boltzmann brain identical to mine are vs. the odds of an identical brain evolving on a twin earth (my money would be that a copy on a twin earth is much more likely).  But would the specific odds matter if space is infinite and there are an infinite number of both, regardless of which is more common?  I know the answer would of course depend on your philosophy of probability (frequency, Bayesian, etc.) so I would be interested in the answer for each of the main theories in probability.

In conventional countably additive probability, in order to assign a probability to the set of even positive integers, you would need to have a probability assigned to each integer, and then you add up the ones for the even integers.

You can't assign equal probabilities to all positive integers, since then either they're all $0$ and their sum is $0$, whereas the sum should be $1$, or they're all some positive number as the sum is $\infty$, whereas it should be $1$.

However, there is also the idea of letting the probability of each set $A\subseteq\{1,2,3,\ldots\}$ be the "density" of $A$, which is
$$
\lim_{n\to\infty} \frac{|A\cap\{1,\ldots,n\}|} n.
$$
If you do that, then probabilities are finitely additive but not countably additive, and then not all of the theorems of conventional probability hold.

What do you think of the answer of Anubhav Balodhi here ? I have copied/pasted it in the following. He proposes a simple mathematical proof that leads to 0.5 probability to pick an even number among all natural numbers. His answer makes much more common sense than getting a 1/3 probability to pick an even number...


Minimizing a function with summation I'm looking to minimize:
$$F(m,b) = \sum_{i=1}^n (mx_i + b -y_i)^2 $$
By taking first order conditions $w.r.t$  m, I find:
$$\begin{align*} \frac{\partial F}{\partial m}= 2 \sum_{i=1}^n x_i(mx_i+b-y_i) &=0 \\
\ 2m\sum_{i=1}^nx_i^2+2b\sum_{i=1}^n x_i -2\sum_{i=1}^n y_ix_i &=0\\
\end{align*}$$
By dividing everything by 2 and rearranging:
$$m\sum_{i=1}^nx_i^2+b\sum_{i=1}^n x_i = \sum_{i=1}^n y_ix_i$$
Finally I find that
$$ m = \frac{\sum_{i=1}^n y_ix_i - b\sum_{i=1}^nx_i}{\sum_{i=1}^nx_i^2} $$
However, this answer seems strange to me as I am unsure about the math.
Can this be simplified further?

Yes the Math is correct. You minimize the equation wrt m to get the minimum point equation for m. Since the equation wrt to m is the form (f(m))^2 or g(f(m)) where g(x)=x^2, we use the chain rule to get 2 * (f(m)) * df/dm or dg/dm * (f(m)) * df/dm