On the definition of Markov chains A Markov chain with discrete time dependence and stationary transition probabilities is defined as follows. Let $S$ be a countable set, $p_{ij}$ be a nonnegative number for each $i,j\in S$ and assume that these numbers satisfy $\sum_{j\in S}p_{ij}=1$ for each $i$. A sequence $\{X_j\}$ of random variables with ranges in $S$ is defined to be a Markov chain if
$$
P[X_{n+1}=j|X_0=i_0,\ldots,X_n=i_n]=p_{i_nj}\quad(1)
$$
for every $n$ and every sequence $i_0,\ldots,i_n$ in $S$ for which
$P[X_0=i_0,\ldots,X_n=i_n]>0$.

The property
$$
P[X_{n+1}=j|X_n=i]=p_{ij}\quad(2),
$$
is a consequence of the definition (1), but I don't think that (2) implies (1). So are there any interesting examples which satisfy (2) but are not Markov chains?

(Of course (1) determines $P[X_{n}=i_n,X_{n-1}=i_{n-1},\ldots,X_{n-k}=i_{n-k}]$ in terms of the transition probabilities and the initial probabilities for any $n$ and all $k\leq n$, whereas (2) does so for any $n$ and for $k=0,1$.)

One can consider primitives of stochastic processes, in the following sense. Assume that $(U_n)$ is i.i.d. and define $(X_n)$ and $(S_n)$ recursively by $X_0=S_0=0$ and,  for every $n\geqslant0$, $$X_{n+1}=X_n+U_{n+1},\qquad S_{n+1}=S_n+X_{n+1}.
$$
Then, the process $(U_n)$ is i.i.d. by definition (hence a Markov chain), the process $(X_n)$ is a (homogenous) Markov chain with transition probabilities $p_{i,j}=P(U_1=j-i)$ for every states $(i,j)$, but, in general, the process $(S_n)$ is not a Markov chain.

For example, assume that every $U_n$ is uniform on $\{-1,1\}$, then $(X_n)$ is the standard random walk on $\mathbb Z$ with transitions $p_{i,i+1}=p_{i,i-1}=\frac12$ for every $i$ in $\mathbb Z$. Using the notation $\xi_{i:j}$ for $(\xi_i,\xi_{i+1},\ldots,\xi_j)$ for various symbols $\xi$ and integers $i\leqslant j$, one can check that
$$P(S_{1:4}=(1,1,0,2))=P(U_{1:4}=(1,-1,-1,3))=0,$$ while
$$P(S_{1:4}=(-1,-1,0,2))=P(U_{1:4}=(-1,1,1,1))\ne0,$$ hence the distribution of $S_4$ conditionally on $S_{1:3}$ does not depend only on $S_3$, that is, $(S_n)$ is not a Markov chain.

...However, the process $(S_n)$ is "Markov of order $2$", in the sense that the distribution of $S_{n+2}$ conditionally on $S_{1:n+1}$ depends on $(S_{n},S_{n+1})$ only. Specifically, for every integers $(i,j,k)$, $$P(S_{n+2}=k\mid S_{1:n-1},S_n=i,S_{n+1}=j)=P(U_1=k-2j+i).$$
Iterating the primitivation mechanism described above produces "Markov processes of higher orders".

To get stationary processes, that is, processes $(S_n)$ whose transitions do not depend on time, consider again $(U_n)$ i.i.d. uniform on $\{-1,1\}$ but consider some processes $(X_n)$ and $(S_n)$ on the state space $\mathbb Z/N\mathbb Z$ for some large $N$, defined recursively by $X_0=S_0$ uniform on $\mathbb Z/N\mathbb Z$, and $$X_{n+1}=X_n+U_{n+1}\pmod{N},\qquad S_{n+1}=S_n+X_{n+1}\pmod{N}.
$$
Then, every $X_n$ and every $S_n$ is uniform on $\mathbb Z/N\mathbb Z$ and $X_{n+1}$ is independent of $S_n$ (this one demands a little work). In particular, the process $(S_n)$ is not a Markov chain since, for every $(i,j)$ in $\mathbb Z/N\mathbb Z$,
$$
P(S_{n+1}=j\mid S_n=i)=\frac1N,\qquad P(S_{n+1}=j\mid S_{n-1}=2i-j,S_n=i)=0.$$

Any probability distribution at all will satisfy your equation 2. The point of a Markov chain is that in the equation
$$
P[X_{n+1}=j|X_0=i_0,\ldots,X_n=i_n]=p_{i_nj},
$$
the probability only depends on the previous element:
$$
P[X_{n+1}=j|X_0=i_0,\ldots,X_n=i_n]=P[X_{n+1}=j|X_n=i_n].
$$


How to prove that if $a^x \equiv a^y \pmod n$ then $x \equiv y \pmod{\operatorname{ord}_{n}(a)}$? I can easily prove the inverse, however I cannot prove this.
We know that $\operatorname{ord}_{n}(a)\mid\phi(n)$, so I have been trying to prove that $x \equiv y \pmod{\phi(n)}$ and it follows that $x \equiv y \pmod{\operatorname{ord}_{n}(a)}$.
I have also noted some relevant properties that might be of help:
$$a^{\operatorname{ord}_{n}(a)}\equiv 1 \pmod n $$
$$\gcd(a,n)=1$$
$$\operatorname{ord}_{n}(a)\mid\phi(n)$$
$$a^{\phi(n)} \equiv 1\pmod n$$

If $\gcd(a,n)=1$ then $a$ is invertible modulo $n$, so we actually have (multiplying by $(a^{-1})^y$) that $a^{x-y}\equiv 1 \mod n$. Now it is clear that $x-y\equiv 0 \mod \text{ord}_n(a)$.

You only need to use the definition of the order of $a$ modulo $n$:
It is the least positive integer $k$ such that $a^k\equiv 1\pmod n$. The set of such integers is an ideal in $\mathbf Z$, and  $\operatorname{ord}_n(a)$ is the positive generator of this ideal.
Now, $a^x\equiv a^y\mod n\iff a^{x-y}\equiv 1\mod n$, which means  $\operatorname{ord}_n(a)\mid x-y$ .


Why can't you solve $0 = -5\sqrt{x+2}-2$ So i did this,
\begin{aligned}
2&=-5\sqrt{x+2},\\
\tfrac{2}{-5}&=\sqrt{x+2},\\
\tfrac{4}{25}&=x+2,\\
x&=-\tfrac{46}{25}.
\end{aligned}
Is there something wrong with my algebra?

Remember that, $\sqrt{x+2}≥0$ and multiplying each side by $(-1)$  then you get,
$$-5\sqrt{x+2}-2=0 \Longleftrightarrow 5\sqrt{x+2}+2=0$$ which is impossible.
The sum of  non-negative and positive  numbers can not equal to zero.
EDIT:
Your argument is not generally correct. Because, $$a=b \not \Longleftrightarrow a^2=b^2.$$ For example,
$$-1=1 \\  (-1)^2=1^2 \\  1=1 $$ which is invalid.

You've incorrectly assumed $f(x)=f(y)\implies x = y$ (line 2 to line 3) which is not true when $f$ is non injective.


Combinatorics question involving distributing 9 different candies to three different kids Nine different chocolate bars are to be distributed to 3 different kids.

a) In how many ways can this be done if there are no restrictions?

b) In how many ways can this be done if the child A receives exactly four chocolate bars?

c) In how many ways can this be done if each kid gets at least one chocolate bar?

I'm pretty confident with the first two parts (a and b)

Here is my work:

a) Since there are no restrictions we can assign the kids to the chocolate bars in $3^9$ ways.

b) Since the first child must get four candies we can assign the candies to him in ${9\choose 4}$ ways. Then we can assign the rest of the candies using the same logic from part a in $2^5$ ways. So there are $2^5$${9\choose 4}$ ways.

c) I'm confused here because I did this problem in two distinct methods and did not obtain the same answer in the slightest...

$\textbf{Method 1:}$ Assign 3 candies to three kids: This can be done in 9×8×7 ways but we also have to consider that there are 3! arrangements of the kids. Then we need to assign the remaining candies to the kids, which can be done in $3^6$ ways. So in this first method I obtained $3^6$×3!×(9×8×7) ways = 2,204,496 ways.

$\textbf{Method 2:}$ I then tried using the inclusion exclusion principle. First I saw that there were $3^9$ ways to assign the kids to the candies in total. And then I took the cases in which one of the three kids received no candies. For each of the three kids this would be $2^9$ and thus would be $3×2^9$ for the kids collectively. However, we would have to add back the case in which none of the three kids received a candy, which would be 3. Thus the answer would be $3^9 - 3×2^9 + 3$ ways.

I'm really not sure which of part c or if either are correct at all...but could someone also check my answers for part a and part b?

Your answers to the first two questions are correct, as is your second method for the third problem.

Your first method for the third problem over counts.  To see this, suppose the candy bars the children receive are $A, B, C, D, E, F, G, H, I$.  You count each distribution multiple times.  For instance, if we give $A, B$ to the first child, $C, D, E$ to the second child, and $F, G, H, I$ to the third child, your method counts this case $2 \cdot 3 \cdot 4 = 24$ times, once for each of the two ways of designating one of the candy bars the first child receives as the candy bar reserved for that child, once for each of the three ways of designating one of the three candy bars received by the second child as the candy bar reserved for that child, and once for each of the four ways of designating one of the four candy bars the third child receives as the one reserved for that child.  To illustrate a few ways you count this distribution, consider the following table.

\begin{array}{c | c |c | c | c | c}
R_1 & R_2 & R_3 & A_1 & A_2 & A_3\\ \hline
A & C & F & B & D, E & G, H, I\\
B & C & F & A & D, E & G, H, I\\
A & D & F & B & C, E & G, H, I\\
B & D & F & A & C, E & G, H, I\\
A & E & F & B & C, D & G, H, I\\
B & E & F & A & C, D & G, H, I
\end{array}
where $R_i$, $1 \leq i \leq 3$, denotes the candy bar reserved for the $i$th child, and $A_i$, $1 \leq i \leq 3$, represents the additional candy bars received by the $i$th child.

Your method $1$ with restrictions far exceeds $3^9$ with no restrictions.

Your method $2$ is close. You didn't consider the case where $2$ children  receive no candy and also, there isn't a case where all $3$ receive no candy (that wouldn't be a valid distribution of the candy).

Like you said, there are $3$ ways for one child to receive no candy and for each of those there are $2^9$ ways for the other $2$ children to receive the candy.

There are $3$ ways for $2$ children to have zero candy and for each of those there is only one way for the other child to have all $9$ candies.

c) $3^9 - 3\cdot 2^9 - 3 = 18144$


What's the intuition behind the sin of an obtuse angle? The definition of sin of an angle is opposite / hypotenuse. Obviously, this definition is easy to use for acute angles within a right triangle, but it's hard to see how it carries over for obtuse angles.

I know there are trig identities and unit circle concepts that allow one to get to the sin of an obtuse angle, but I'm looking for a more intuitive explanation for what exactly the sin of an obtuse angle means.

"The definition of sin of and angle is opposite/hypotenuse" - well that only works for a right-angled triangle, and is the beginning of the definition of the sine function.

In order to extend the definition draw a unit circle with centre at the origin. Measure the angle counterclockwise from the positive $x$-axis and take a point $(x,y)$ on the circle. The sine of the angle (equivalent to opposite/hypotenuse in the first quadrant, with the hypotenuse made equal to $1$) is $y$ and the cosine of the angle is $x$. You can go round the circle more than once, so you can see that the functions are periodic.

This is why the functions are sometimes known as circular functions and underlies why they come in so surprisingly useful.

In case you are interested ...

In more advanced work the sine function is sometimes defined very differently, with the angle measured in radians ($2\pi$ radians $=360^{\circ}$). Then $$\sin x = \sum_{r=1}^\infty (-1)^{r-1}\frac{x^{2r-1}}{(2r-1)!}=\frac {e^{ix}-e^{-ix}}{2i}$$This can be applied in more general circumstances still, and the series is convenient because it converges rapidly and enables the sine function to be computed accurately for practical use. The first few terms are $$x-\frac {x^3}6+\frac {x^5}{120}-\frac {x^7}{5040}+\dots$$

Let's use the definition of the slope or tangent of the terminal line for each angle ....
For the terminal line of an acute angle the slope is y/x....
For the terminal line of an obtuse angle the slope is also y/x.....(though negative in sign)....
Now we multiply each slope by x to get us the sin values for the two angles (in the unit circle ) as y and -y which are equal in absolute values.....


Stating a suitable domain (composite functions) One of the questions was:

Given $f(x)=\frac{1}{x^2-4}$    &    $g(x)=\frac{1}{x} -2$

a) find $g(f(x))$ (Which i think is $\frac{1}{2}$?)

b) State a suitable domain

I dont know how to write about domains - any help would be appreciated

To compute $g(f(x))$ (let's call it $h(x)$) you take

$$
y = \frac{1}{x^2-4}
$$

and replace that on $g$, yelding

$$
g(y) = \frac{1}{\frac{1}{x^2-4}} - 2 = x^2 - 4 -2 = x^2 - 6
$$

As for the domain, this new function $h$ is defined for all $x \in \mathbb{R}$.


Showing two random variables independent despite seemingly looking dependent I just met this in probability and it got me completely stumped:

  We define an i.i.d sequence of normally distributed random variables
  $ \{ X_n \}_{n=1}^{\infty} $ such that $ X_n \sim \mathcal{N} (0,\sigma ^2) $ and we define a random variable $ N \sim Pois(\Lambda) $ independent of the $ X_n $'s

  We are asked to show that $ X_N $ and N are independent random variables

I cannot seem to get past the simple fact that these seem clearly dependent as N clearly appears as a subscript for the X random variable so I do not really know what is wrong or how to prove that these are independent random variables so this is where I need help. Thanks all.

Independence follows from the fact that for every fixed $k\in\mathbb N$:$$P(X_N\in A)=\sum_{n=0}^{\infty} P(X_N\in A\mid N=n)P(N=n)=$$$$\sum_{n=0}^{\infty} P(X_n\in A)P(N=n)=\sum_{n=0}^{\infty} P(X_k\in A)P(N=n)=$$$$P(X_k\in A)=P(X_N\in A\mid N=k)$$

This is not different from rolling a green and a red die ans using a coin flip to decide which of the dice you use and which you ignore. You still have a coin flip and a die.

All that matters is that the $X_n$ are identically distributed (they need not be normally distributed and not independent of one anaother) and that $N$ is independent of all $(X_1,X_2,\ldots)$.

For any measurable $A$ and $n\in\Bbb N$, we have (letting $Y=X_N$ for clarity)
$$\begin{align} P(Y\in A\mid N=n)&=P(X_n\in A\mid N=n)&\text{def of $Y$}\\&=\frac{P(X_n\in A\wedge N=n)}{P(N=n)}&\text{def of conditional prob.}\\&=P(X_n\in A)&\text{independent}\\&=P(X_1\in A)&\text{identically distributed}\end{align}$$
so that (as one might have guessed)
$$ \begin{align}P(Y\in A)&=\sum_n P(Y\in A\mid N=n)P(N=n)\\&=\sum_nP(X_1\in A)P(N=n)\\&=P(X_1\in A)\sum_nP(N=n)\\&=P(X_1\in A)\end{align}$$
and hence
$$P(Y\in A\mid N=n)=P(Y\in A).$$


Writing a permutation group in 2 row notation I have a permutation group in $S_7$, namely: $$(12345)(137)(56)$$

How do I write this in two row notation? I am to write it as disjoint cycles and then as transpositions but I feel better working in two row notation.

Just compute where each element goes:

$1\mapsto3\mapsto4$

$2\mapsto3$

$3\mapsto7$

$4\mapsto5$

$5\mapsto6$

$6\mapsto5\mapsto1$

$7\mapsto1\mapsto2$.

So you have
$$\begin{pmatrix}
1&2&3&4&5&6&7\\
4&3&7&5&6&1&2
\end{pmatrix}$$

It should be $$\binom{1234567}{4375612}$$
And you compose from right to left


Show that every Euclidean ring is a principal ring Question: Show that every Euclidean ring is a principal ring.

My attempt: Let $(R, +, \cdot)$ be a Euclidean domain. Let $I \subseteq R$ be a nonzero ideal of $R$. Let $d \in I$ with $d \neq 0$ be any element such that $\delta (d)$ is minimum in $I$. We claim that $I = dR = \{ dx : x \in R \}$. Since $d \in I$ we have that:
$$\begin{align} \quad dR \subseteq I \quad (*) \end{align}$$
Now let $a \in I$ with $a \ne 0$. Since $R$ is a Euclidean domain there exists elements $q,r \in R$ such that $a=dq+r$ and $r=0$ or $\delta (r) < \delta (d)$. Now we have that $r=a−dq$ where $a \in I$ and $−dq \in I$, so $r=a−dq \in I$. From the minimality of $\delta(d)$ we have that:
$$\begin{align} \quad \delta (r) \geq \delta (d) \end{align}$$
Hence, since $\delta(r) ≮  \delta(d)$ we must have that $r=0$. So $0=a−dq$. Hence $a=dq$. So $a \in dR$. Thus:
$$\begin{align} \quad dR \supseteq I \quad (**) \end{align}$$
From $(∗)$ and $(∗∗)$ we conclude that $I=dR$. In other words, $I=<d>$. So every Euclidean domain is a principal ideal domain.

From here, how to show that every Euclidean ring is a principal ring???

No, the relation $0\leq \delta(r)<\delta(d)$ implies that $\delta(r)=0$ and so $r=0$. Thus each element of $I$ is a multiple of $d$. On the other hand, the multiplies of $d$ form an ideal. Hence, the result follows.


Example of complete formula on constants. I have question considering example of complete formula, that was presented in Keisler/Chang - Model Theory:

2.3.1. EXAMPLES
(1). Let $T$ be a complete theory and let $c_0,c_1,c_2,\ldots$ be constant symbols of $\mathscr{L}$. Then any formula of $\mathscr{L}$ of the form
$$x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$$
is complete in $T$. If $\mathfrak{U}$ is a model of $T$ such that every element of $A$ is a constant, then $\mathfrak{U}$ is an atomic model.

Why so? Am I right, when I think, that if theory is complete than -> every of it's formulas is complete too? So that formula with conjunction too?
Also, I pin basic definitions about it:

Consider a complete theory $T$ in $\mathcal{L}$. A formula $\varphi(x_1\ldots x_n)$ is said to be complete (in $T$) iff for every formula $\psi(x_1,\ldots x_n)$ exactly one of
$$T\vDash\varphi\rightarrow\psi,\quad T\vDash\varphi\rightarrow\neg\psi$$
holds. A formula $\theta(x_1\ldots x_n)$ is said to be completable (in T) iff there is a complete formula $\varphi(x_1\ldots x_n)$ with $T\vDash\varphi\rightarrow\theta$. If $\theta(x_1\ldots x_n)$ is not completable it is said to be incompletable.
A theory $T$ is said to be atomic iff every formula of $\mathcal{L}$ which is consistent with $T$ is completable in $T$. A model $\mathfrak{U}$ if is said to be an atomic model iff every $n$-tuple $a_1,\ldots,a_n\in A$ satisfies a complete formula in $\mathrm{Th}(\mathfrak{U})$.

$T$ is complete iff for every sentence $\phi$ either $T \models \phi$ or $T \models \lnot \phi$.  This requirement only applies to sentences, i.e., closed formulas: if $\phi = \phi(x_1, \ldots, x_n)$ has free variables then $\phi$ might hold under some interpretations of $\phi$ but not under others. E.g., consider $\phi(x, y) = x < y$ in the theory of a dense linear order without endpoints.
The formula $x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ is complete in the sense of the definition you give, because for any $\phi$,
$T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$ iff $T \models \phi(c_1, \ldots, c_n)$. Now, $\phi(c_1, \ldots, c_n)$ is a sentence, so if $T$ is complete either $T \models \phi(c_1, \ldots, c_n)$ or $T \models \lnot\phi(c_1, \ldots, c_n)$ and so either $T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$ or $T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \lnot\phi$.
If every element of a model $\mathfrak{A}$ is a constant, then every $n$-tuple, $a_1, \ldots a_n$ satisfies the formula $x_1\equiv a_1\wedge x_1\equiv a_2\wedge\cdots\wedge x_n\equiv a_n$, which we have just shown is complete. Hence such an $\mathfrak{A}$ is an atomic model according to the definition you have quoted.

I changed my answer to myself:
Thus $T$ is complete, therefore it's consistent therefore it has model. So, it's sufficient to show, that
$T \cup x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ also will be consistent (and from that complete).
Suppose it isn't true.
Then, $T \models \forall x_0...\forall x_n \neg(x_0 \equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n)$
or $T \models \forall x_0...\forall x_n (\neg x_0 \equiv c_0\vee \neg x_1\equiv c_1 \vee \cdots\vee \neg x_n\equiv c_n)$
Interpretation $I^{\mathfrak{A}}_{}$ of $c_{i} := t_{2i}$, as a term from signature in the model $\mathfrak{A} = <A,I^{\mathfrak{A}}_{}>$ is $I^{\mathfrak{A}}_{}[c_i] = a_{i}$.
Some labeling $\gamma_{j}$ of $x_{i} := t_{1i}$, as a term, in model $\mathfrak{A} =<A,I^{\mathfrak{A}}_{}>$ is $t^{\mathfrak{A}}_{1i}[\gamma_{j}] = \gamma_{j} x_{i} = a_{1i}$.
So, thus in the last formula with all variables bounded, it's must be true on all labelings $\gamma_{j}$ and on all of our variables.
But for all $i \in {0,...,n}$, $x_{i}$ takes all values from $A$, and eventually it takes $a_i$ in some labeling $\gamma_{j}$.
The resulting contradiction completes the proof.


Determine monic and degree 3 polynomial in $\mathbb Z_p$ I stumbled upon this kind of problem and I really can't get the hang of it. Will anyone please outline the way to solve it?

Determine for which of the first $p > 0$ values the polynomial $f = 42x^4+21x^3-x+1 \in \mathbb Z_p$ is monic and has degree 3. Then factor it as product of irreducibile polynomials in the polynomial rings found.

Hint $\ $ If $\rm\:f\ mod\ p\:$ is cubic then $\rm\:p\:|\:42,\ p\nmid 21\:$ hence $\rm\:p = \ldots$  Further, mod this $\rm\:p\:$ we see $\rm\:f\:$ has no roots, so $\rm\:f\:$ is an irreducible cubic. The point of this is that it implies that if $\rm\:f\:$ factors over $\Bbb Q\,$ then it must split as a linear times a cubic. Thus to show $\rm\:f\:$ is irreducible over $\Bbb Q\,$ it suffices to show it has no root, e.g. by using the Rational Root Test.

Edit $\ $ The OP later clarified that factorization over $\Bbb Q$ is not needed, so the second half of my answer is not needed. But I'll leave it since it may still prove of interest. In fact this is the way some polynomial factorization algorithms work: by deducing constraints on the degrees of possible factors from factorizations mod $\rm p$ for various primes $\rm p.$

$\,f\,$ has degree $<4\,$ iff
$$2\cdot 3\cdot 7=42=0\Longleftrightarrow p=2,3,7$$

Furthermore, it will be a monic cubic if also $\,21=1\pmod p\Longrightarrow p=2\,$ , since $\,21=0\pmod 3\,,\,\pmod 7\,$

So only in $\,\Bbb Z_2[x]\,$ is$\,f\,$ a monic cubic polynomial, namely: $\,f=x^3+x+1\,$


Regular ring not UFD I have to prove that the ring $R=K[x,y]/(x^2-y^3+y)$ is not a UFD showing that the prime ideal $(x,y)R$ has height 1, but it's not principal.

Do someone know a simple way to prove it? I know there are others way to solve the problem, for example to consider the Picard group to the elliptic curve, but I am interested to solve it in the way I explained. Thanks!

Here is an elementary algebraic solution. I prefer $R = K[x,y]/(y^2-x^3+x)$.  To show that $R$ is not a UFD, it's enough to
show that $y$ is irreducible in $R$ (it's not prime, as $R/(y)$ is not a domain; the same reasoning shows it's not a unit).
This incidentally also implies that the ideal $(x,y)$ isn't principal, as any generator has to divide $y$ and $(x,y) \ne (y)$.

Suppose $y$ factors in $R$, so $y = fg + h(y^2-x^3+x)$ in $K[x,y]$ for some $f,g,h \in K[x,y]$.
We now use division in $K[x,y]$ by the polynomial $y^2-x^3+x$ with unique remainder of degree at most 1 in $y$.

So WLOG $f = f_1(x)y+f_2(x)$ and $g = g_1(x)y+g_2(x)$, where $f_i(x), g_i(x) \in K[x]$ and we get

$$(f_1(x)y+f_2(x))(g_1(x)y+g_2(x)) \equiv y \pmod{y^2-x^3+x},$$

i.e.

$$f_2 g_1 + f_1 g_2 = 1, \ f_1 g_1 (x^3-x) = - f_2 g_2$$

in $K[x]$. We see that $f_1 = 0$ iff $g_2 = 0$ iff $f \in K^\times$. Similarly, $g_1 = 0$ iff $f_2 = 0$ iff $g \in K^\times$.
so if our factorisation above is non-trivial, then $f_1,f_2,g_1,g_2$ are all non-zero. Moreover,
$\gcd(f_1,f_2) = \gcd(g_1,g_2) = 1$. Hence $f_1$ divides $g_2$ and $g_1$ divides $f_2$ in $K[x]$. writing $g_2 = f_1 v$ and
$f_2 = g_1 w$, we get

$$g_1^2 w + f_1^2 v = 1, \ x^3-x = - vw.$$

In particular, $\deg(v) \not\equiv \deg(w) \pmod 2$, so $\deg(g_1^2 w) \not\equiv \deg(f_1^2 v) \pmod 2$.
This implies that either $g_1^2 w$ or $f_1^2 v$ is of degree 0 and the other vanishes, contradiction.

The ring $K[x,y]$ has Krull dimension $2$. So the ring $R$ has Krull dimension $1$, hence the maximal height of a prime ideal is $1$. And your prime ideal does not have height $0$ because it strictly contains the prime ideal $(x^2-y^3+y)R$.


Finding the bounds of a triple integral (spherical coordinates) I'm currently learning how to calculate the volume of a 3D surface expressed in spherical coordinates using triple integrals.
There was this exercice (from here) which asked me to find the volume of the region described by those two equations:
$x^2+y^2=1$
$x^2+y^2+z^2=4$
(The image of the region from the website)
From this I've found that:
$csc\phi\le \rho\le2$ and $0\le \theta\le 2\pi$
I was also able to find a value of $\phi$ which was $\frac{\pi}{6}$ (I wasn't able to deduce if it corresponded to the lower or higher bound).
And from here I didn't knew what to do next, so I assumed that $0\le \phi\le \frac{\pi}{6}$ because I was stucked.
And the actual right setup of the integral in spherical coordinates was:
$\int_{\frac{\pi}{6}}^{5\frac{\pi}{6}}\int_{0}^{2\pi}\int_{csc(\phi)}^{2}\rho^2sin(\phi) d\rho d\theta d\phi$
So my brain died.
I couldn't figure why first of all $d\theta$ was located before $d\phi$ (what constraint would have caused this ?).
And also why the other bound of $\phi$ was $5\frac{\pi}{6}$, how was I supposed to find this value ?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\underline{"Outside"}\,\,\, Volume\
{\Large\cal V}_{\mrm{out}} \pars{~\mbox{Cylindrical Coordintaes}~}}$.
Hereafter, $\ds{\bracks{\cdots}}$ is an
Iverson Bracket.
Namely, $\ds{\bracks{P} =
\left\{\begin{array}{rcl}
\ds{1} & \mbox{if}\ P\ \mbox{is}\ true.
\\[1mm]
\ds{0} & \,\,\,\,\,\mbox{if}\ P\ \mbox{is}\ false.
\end{array}\right.}$

\begin{align}
{\Large\cal V}_{\mrm{out}} & \equiv
\bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}\bracks{x^{2} + y^{2} > 1}\bracks{x^{2} + y^{2} + z^{2} < 4}
\dd^{3}\vec{r}}
\\[5mm] & =
\iiint_{\mathbb{R}^{3}}\bracks{\rho^{2} > 1}
\bracks{\rho^{2} + z^{2} < 4}
\rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] & =
\iiint_{\mathbb{R}^{3}}
\bracks{1 < \rho^{2} < 4 - z^{2}}
\rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] & =
\iiint_{\mathbb{R}^{3}}
\bracks{1 < 4 - z^{2}}
\bracks{1 < \rho < \root{4 - z^{2}}}
\\[2mm] &\ \rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] & =
\iiint_{\mathbb{R}^{3}}
\bracks{\verts{z} < \root{3}}
\bracks{1 < \rho < \root{4 - z^{2}}}
\\[2mm] & \rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] & =
2\pi\int_{-\root{3}}^{\root{3}}
\int_{1}^{\root{4 - z^{2}}}\rho\,\dd\rho\,\dd z
\\[5mm] & =
\pi\int_{-\root{3}}^{\root{3}}\pars{3 - z^{2}}\dd z
\\[5mm] & =
\bbx{4\root{3}\pi} \approx 21.7656 \\ &
\end{align}

It does not matter what order $\rho, \theta, \phi$ are integrated in, as long as the limits on each are expressed only in terms of variables yet to be integrated. Since $\theta, \phi$ have constant limits, they can be interchanged freely. Since the limits for $\rho$ and $\theta$ are independent of each other, they can also be exchanged. But to exchange the order of the integrations of $\rho$ and $\phi$, you would need to express the limits of $\phi$ in terms of $\rho$ instead of $\rho$ in terms of $\phi$. However, between the two, your approach is simpler. That is:
$$\begin{align}\int_{\frac{\pi}6}^{\frac{5\pi}6}\int_0^{2\pi}\int_{\csc \phi}^2\rho^2\sin \phi\, d\rho\, d\theta\, d\phi &= \int_{\frac{\pi}6}^{\frac{5\pi}6}\int_{\csc \phi}^2\int_0^{2\pi}\rho^2\sin \phi\,d\theta\,d\rho\,d\phi \\
&= \int_0^{2\pi}\int_{\frac{\pi}6}^{\frac{5\pi}6}\int_{\csc \phi}^2\rho^2\sin \phi\,d\rho\,d\phi\,d\theta\\
&=\int_0^{2\pi}\,d\theta\int_{\frac{\pi}6}^{\frac{5\pi}6}\int_{\csc \phi}^2\rho^2\sin \phi\,d\rho\,d\phi\\
&=2\pi\int_{\frac{\pi}6}^{\frac{5\pi}6}\left(\sin \phi\int_{\csc \phi}^2\rho^2\,d\rho\right)\,d\phi\end{align}$$
since the integrand and limits are constant with respect to $\theta$, the $\sin \phi$ factor of the integrand does not depend on $\rho$.
Now $\int_{\csc \phi}^2\rho^2\,d\rho$ is not hard to figure out. And once you do, the integrand for $\phi$ may not be as bad as you think.


If $3x^{2}-2(a-d)x+(a^{2}+2(b^{2}+c^{2})+d^{2})=2(ab+bc+cd)$, then If $3x^{2}-2(a-d)x+(a^{2}+2(b^{2}+c^{2})+d^{2})=2(ab+bc+cd)$, then

$A.$ a,b,c,d are in G.P.
$B.$ a,b,c,d are in H.P.
$C.$ a,b,c,d are in A.P.
$D.$ None of the above

Tried writing the expression as a sum of squares but it was of no use.

NOTE: G.P.- Geometric progression, H.P.-Harmonic Progression and A.P.-Arithmetic Progression. $x,a,b,c,d \in \mathbb{R}$

I wrote and put it in the jpg format.

I hope it helps!


Example of an easy non-pseudotopological convergence Recently, I have had an introduction to convergence spaces and I was wondering if there is an easy example of a non pseudotopological convergence space.

Thank you for your help.

I think the following is the most basic example of a convergence space which is not (equivalent to) a pseudotopological space:

Let $X$ be an infinite set, and fix some $\infty \in X$.  Define a convergence on $X$ so that

$\dot{x} \rightarrow x$ for each $x \in X$;
if $\mathcal{U}$ is a free (nonprincipal) ultrafilter on $X$, then $\mathcal{U} \rightarrow \infty$.

(And there are no other convergences.)

It is fairly straightforward to show that this defines a convergence on $X$.  However, if $\mathcal{F}$ is a (free) filter (but not an ultrafilter) on $X$ extending the co-finite filter, then $\mathcal{F} \not\rightarrow \infty$ even though $\mathcal{U} \rightarrow \infty$ for each ultrafilter $\mathcal{U} \supseteq \mathcal{F}$.


Problems with understanding a given Sum Identity In a textbook I found the following equation without any explaination:

\begin{align}
\sum_{\substack{j, k=1,\\ j\neq k}}^n\frac{1}{(x-x_j)(x-x_k)} - \frac{3}{2}\left( \sum_{j=1}^n \frac{1}{x-x_j}\right)^2 \\
 = -\frac{1}{2}\sum_{j=1}^n \left( \frac{1}{x-x_j}\right)^2 - \left( \sum_{j=1}^n \frac{1}{x-x_j} \right)^2
\end{align}

I was wondering why this is true and tried to proof it but somehow failed. Can anyone give a hint?

There is a misrpint here. The factor $\frac 1 2$ should apply to the second term on the right side, not the first.

This modified identity follows immediately from the fact that the first term on the left side is $ \sum\limits_{j,k=1}^{n} \frac 1 {(x-x_j)(x-x_k)}-\sum\limits_{k=1}^{n} \frac 1 {(x-x_j)^{2}}$ which is same as $ (\sum\limits_{k=1}^{n} \frac 1 {(x-x_j)})^{2}-\sum\limits_{k=1}^{n} \frac 1 {(x-x_j)^{2}}$

Consider the following expression
\begin{eqnarray*}
\left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right)^2 = \left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right) \left( \sum_{k=1}^{n} \frac{1}{x-x_k} \right).
\end{eqnarray*}
Now when you calculate the product we have either $j \neq k$ or $j=k$, so we have
\begin{eqnarray*}
\left( \sum_{j=1}^{n} \frac{1}{x-x_j} \right)^2 =  \sum_{\substack{j, k=1,\\ j\neq k}}^n \frac{1}{x-x_j} + \sum_{j=1}^{n} \frac{1}{(x-x_j)^2} .
\end{eqnarray*}
So the first term in the stated formula needs a minus sign!


Eigenvalue $0$ iff $\ker(A) \neq \{0\}$ I read that a Matrix $A$ has the eigenvalue $0$ if and only if $\ker(A) \neq \{0\}$.

Why so?

Edit Okay actually I figured it out myself. If $0$ is an eigenvalue of a matrix $A$ then $\det(A)=0$ and then $A$ is not invertible, therefore  the rows are not linearly independent: $\ker(A) \neq0$.

$A$ has eigenvalue $0$ $\iff$ $\exists v \neq0$ such that $Av=0v=0$ $\Rightarrow 0\neq v\in \ker (A) \Rightarrow \ker (A) \neq\{0\} $ .

On the other hand : $\ker (A) \neq\{0\} \Rightarrow \exists v\neq0$ such that $v \in \ker (A)\Rightarrow \exists v\neq0$ such that $Av=0=0v$. So $0$ is eigenvalue of $A$.

Because there exists non-trivial vector $x$ such that $Ax = 0$, so then $Ax = 0 *x$ and $0$ is an eigenvalue.


Is there an efficient method to find all the self-inverse matrices with integers in a given range? Given $n$ and a range, for example $[-10,10]$, is there an efficient method to find
 all $n \times n$-matrices $A$ with integers in the given range, which are self-inverse (that means the equation $A=A^{-1}$ holds)?

Some necessary conditions for $A$:

$\det(A)=-1$ or $\det(A)=1$
$A$ has no eigenvalues other than $-1$ and $1$
The minimal polynomial of $A$ divides $x^2-1$

With $A$, the matrices $-A$ , $A^T$ and $B^{-1}AB$ for any invertible matrix B
are also self-inverse.

So, is there a method to find the matrices systematically without checking
all possible matrices, which would be infeasible for, lets say $n = 4$ and
range $[-10,10]$?

This doesn't seem to have been studied much in the literature. The latest paper at MathSciNet was by Robert Hanson, titled "Self-Inverse Integer Matrices" (College Mathematics Journal, Vol 16, No 3 (Jun 1985), pp. 198-198). He proves that you can generate all self-inverse integer matrices by starting with a matrix of the form $\left[\begin{array}{c}I&A\\ 0&-I\end{array}\right]$ ($A$ is a rectangular matrix), and calculating $BAB^{-1}$, where $B$ ranges over all matrices you get from the identity matrix by doing the following row operations:

Swapping two rows
Adding a multiple of one row to another row
Multiplying all the entries in a row by -1

I'm not sure this helps much. Maybe you can limit the row operations once you know the matrix $A$.

I think the number is infinite...
because all possible rotation angles (a) are possible (analogy with spin in Quantum mechanics) for n=2 we can write in general the matrix [cos(a), sin(a)](line 1) [sin(a), -cos(a)] (line 2)

Zeno Toffano (France)


Confusion regarding conditional probability Let's take an example. I am throwing a fair dice many times. I want to get a $6$. Every time I throw the dice , probability of getting $6$ is $\frac{1}{6}$. Suppose that I haven't got a $6$ for $20$ consecutive times, then what is the probability of getting $6$ on $21^{st}$ roll? Is it $1/6$ or $(5/6)^{20}*1/6$ or anything else?

I am confused because the dice doesn't know it did not get $6$ for so many times

No dice roll affects any future roll or previous roll - the die has no memory of its previous rolls when it rolls for you. That means, that if you do 20 rolls and they all flip up 6 (or all not 6, or 1 not 6 and 19 6s, etc.), then on your next roll there will always be a 1/6 chance of getting any of the 6 possible values.

Let's dnote $\delta(n)$ the probability of getting 6 after $n$ shots.

$\delta(0)=1/6$ obviously

$\delta(1)=5/6*1/6=5/36$  consequently?? no! false! I will explain.

$\delta(1)$ is the chance to hit $6$ regarded failing all the other cases, because the sequence {$62$} is not taken as an option, the tosser should halt at a specific stage, when he makes his first trial. which means $\delta(1)=P(6\ after\ all\ failings)=\frac{combinations(succeeding\ a\ 2nd\ toss)}{combinations(All\ \ available\ tosses)}=\frac{5}{1+6*5}=5/31$

Extrapolating along the basis leads to ....

$\delta(20)=\frac{5^{20}}{1+5+5^2+...+5^{20}6}=\frac{5^{20}}{(5^{20}-1)/4+5^{20}6}$


Question in probability and statistics I'm terrible at questions involving probability, so I'm hoping you can help me!

Suppose there are eight people in a group, each person with a distinct first name.  Each person has a card with their first name written on it.

If the eight cards are randomized and each person selects a card at random, what is the probability that at least one of the first SIX people draws the card with their name on it.

I'm completely baffled by this problem, thanks for the help!

We want to count how many permutations of $\{1,2,\ldots,8\}$ fix at least one of $1,2,3,4,5$ or $6$. Note that the number of permutations that fix one of these points is $7!$, that fix two is $6!$, etc, and that there are $\binom61=6$ elements in $\{1,2,3,4,5,6\}$, $\binom62$ pairs, $\binom63$ triples, etc. Thus by the inclusion-exclusion principle, the number of permutations that fix an element of $\{1,2,3,4,5,6\}$ is
$$\binom61\cdot 7!-\binom62\cdot 6!+\binom63 5!-\binom64 4!+\binom65 3!-\binom662!=21514$$
Since the total number of permutations of $\{1,2,\ldots,8\}$ is $8!=40320$, the probability that one of the first six people draws a card with their name on it is
$$\frac{21514}{40320}\approx .534$$


Condition for which $ \text{dist}(x,A) = d(x,y_0)$ If $(X,d)$ is metric space, $x\in X$ and $A \subset X$. Let's mark $$\text{dist}(x,A) = \text{inf}\; \Big\{ d(x,y) : y \in A \Big\}$$

Does there always exists $y_0 \in A$ that $ \text{dist}(x,A) = d(x,y_0)$ ? Will the answer change, if set $A$ is closed?
Can someone help to solve this exercise?

First Consider a finite dimensional vector space like $\mathbb R$. Take for $x$ the origin.

An example of a bounded subset for which distance is not attained below:

   $A = (1,2]$

If $A$ is compact, then the distance is always attained. For the proof consider a sequence $(x_n)$ such that $d(x_n ,A) \to d(x,A)$ and extract a converging sub-sequence.

If $A$ is closed, then the distance is also always attained. For the proof, consider a point $a \in A$. Then $\overline A = \{y \in A \colon d(x,y) \le d(x,a)\}$ is compact and you can use point above.

Second. Case of an infinite dimensional vector space.

In that case, the distance may not be attained even if $A$ is closed. Look here for a counterexample.


What does it mean to eliminate $\theta$ from these equations? How should I do it? Eliminate $\theta$ from the following pairs of equations:
A) $x=\sin \theta$, $y=\sin 2\theta$
B) $x=3\cos 2\theta +1$, $y=2\sin \theta$

My problem is I really don't understand what the question is asking? The answer in the book seems like it is asking me to put everything in terms of $x$ and $y$? And if it is, how do I go about that?
Thank you!
EDIT: My attempt:
$$y = 2\sin \theta \cos \theta $$
$${y \over {2\sin \theta }} = \cos \theta $$
so as $$x = \sin \theta $$
$$\cos \theta  = {y \over {2x}}$$
what do i do from here?

You've done the hard part. Of A):

  $$
\begin{align}
\sin\theta & = x \\ \\
\cos\theta  & = {\frac y{2x}}
\end{align}$$

This is where the Pythagorean Identity (as you call it) comes in very handy, indeed:

$$\quad\quad\quad\sin^2(\theta) + \cos^2(\theta) = 1$$

$$\iff \left(\sin\theta \right)^2 + \left(\cos\theta \right)^2 = 1$$

$$x^2 + \left(\dfrac{y}{2x}\right)^2 = 1$$

Then just simplify!

Part B:

  $$x = 3 \cos 2\theta, \quad x = 2\sin \theta$$

$$x = 3(1-2\sin^2\theta) \tag{$\cos(2\theta) = 1 - 2\sin^2\theta$}$$
$$x= \frac 32(2 - (1 - \cos^2\theta)) \tag{$\sin^2\theta = 1 - \cos^2 \theta$}$$
$$x = \frac 32 (\cos^2 \theta + 1) \iff \frac 23 x - 1 = \cos^2\theta$$
$$ $$
$$ y = 2 \sin \theta \iff \sin \theta = y/2$$
$$ $$
$$\sin^2 \theta + \cos ^2 \theta = 1\tag{Pythagorean identity again}$$
$$\left(\frac y2\right)^2 + \left(\frac 23 x - 1\right) = 1$$

Simplify.

A) $y = \sin 2\theta$, $y = 2\sin\theta\cos\theta$, $y = 2x\sqrt{1-x^2}$
B) Similarly you will find the answer as $3-3y^2+1$


Why does $\frac{d}{d\theta} \left(\theta\prod\limits_{i=1}^nx_i\right) = \sum\limits_{i=1}^n\ln x_i$ Is this just the product rule? I have this in my notes but I didn't think anything of it and now I'm wondering how this happens?

Edit:
Im working with maximum likelihood estimation and in my notes I have that the likelihood funciton $=L(x;\theta)=\prod_{i=1}^nf(x;\theta)$ where $x$ is the variable and $\theta$ is the parameter of a probability distribution. To estimate I was told that we take the log of the likelihood function, i.e. $\ln(L)$, then take its derivative to estimate the parameter. The function I'm working with is $f(x;\theta)=(\theta +1)x^{\theta}$. So
$$
  L(x;\theta)=\prod_{i=1}^n(\theta +1)x_i^{\theta}=(\theta+1)^n\prod_{i=1}^nx_i^{\theta}.
$$
Now
$$
  \ln(L(x;\theta))=n*\ln(\theta+1)+\theta \ln\left(\prod_{i=1}^nx_i\right).
$$
Here's where I'm confused, I have in my notes that
$$
  \frac{d(\ln L)}{d\theta}=\frac{n}{\theta+1}+\sum_{i=1}^n\ln(x_i).
$$
Why does the product of $x_i$ become the summation of $x_i$?

$$
\ln(a\cdot b) = \ln(a) + \ln(b)
$$
This is a fundamental property of logarithms.
so lets use $$\prod_{i=0}^n x_i = P_n$$
then we have
$$
\ln\left(\prod_{i=0}^n x_i\right) = \ln\left(x_n\prod_{i=0}^{n-1} x_i\right) = \ln(x_n) + \ln\left(\prod_{i=0}^{n-1} x_i\right) = \ln(x_n) + \ln(P_{n-1})
$$
using induction or just seeing a pattern you find that we have
$$
\ln\left(\prod_{i=0}^n x_i\right)  = \sum_{i=0}^n\ln x_i
$$


Binary Division Vs Decimal Divison As I was working with a program for my CRC code calculation, I hit upon a situation where I had to divide two binary strings. Now, what I did was just convert the binary string to decimal format, did a modulo operation to find out the remainder.

But as it turns out, I am wrong in every single sample cases.
For example, in this post, the dividend is 3664 and the divisor is 27.

Now, 3664 mod 27 is 19, which does not agree with the remainder presented in the figure which says it's 4.

Please help me out here.

Thanks in advance :)

Short answer: don't do that. The CRC calculation is not division of
one number by another, and if you try to treat it as such
(by converting to decimal and doing ordinary division on the decimal numbers)
you will usually get the wrong answer.

Long answer:

The CRC calculation represents the incoming data as a polynomial
whose coefficients are in $\mathbb Z/2\mathbb Z$,
that is, the coefficients are integers modulo $2$,
which have the subtraction rules
\begin{align}
0 - 0 \equiv 0 \pmod 2 \\
1 - 0 \equiv 1 \pmod 2 \\
1 - 1 \equiv 0 \pmod 2 \\
0 - 1 \equiv 1 \pmod 2 \\
\end{align}
The last equation is the one that makes it impossible to treat the
XOR "division" algorithm the same as division of one multiple-digit
binary number by another.

So a string of binary bits in this algorithm is never treated as a
binary number, that is, as a sum of powers of $2$.
Instead it is treated as a polynomial, a sum of powers of $x$.
The rightmost bit is the constant term of our polynomial; the
bit to the left of that is the coefficient of $x$, the next bit
to the left is the coefficient of $x^2$, and so forth.

For example, suppose the bits of our divisor are $1011$ as in the
example on the Wikipedia page.
We do not treat these bits as a number (whose value would be eleven);
instead, we treat the bits $1,1,0,1$ as coefficients of the polynomial
$1x^3+0x^2+1x+1 = x^3+x+1$ modulo $2$.

Now to make this a very simple example,
suppose the incoming message is not $11010011101100,$ as
in Wikipedia's example, but instead is  $00000000000001.$

The first step of the algorithm is to pad the message
to $00000000000001000,$ which will be treated as the polynomial
$1x^3+0x^2+0x+0 = x^3$ modulo $2.$
Now the synthetic division of $x^3$ by $x^3+x+1$
looks like this modulo $2$:
\begin{align}
1 &\\
1x^3+0x^2+1x+1\ \overline{)\ 1x^3 +0x^2+0x+0} &\\
\underline{1x^3+0x^2+1x+1} \\
1x + 1
\end{align}

Notice that we used the rule $0 - 1 \equiv 1 \pmod 2$
for the last two terms on the right,
that is, we used the fact that $-1\equiv1 \pmod2.$
Another way to write this is:
$$
x^3 - (x^3+x+1) = x^3 - x^3 - x - 1 = -x - 1 \equiv x + 1 \pmod 2.
$$
The remainder of the division is the polynomial $x+1$,
which gives us the CRC bits $011.$

But notice that if you try to view the dividend and divisor as
numbers and the CRC bits as the remainder when dividing one
number by another, the result above makes no sense at all.
The dividend is the number eight, the divisor is eleven,
and therefore the remainder should just be eight again
(since the divisor is greater than the dividend).
But the CRC bits appear to be the number three!
(Notice that this result is just as nonsensical for binary numbers
as for decimal numbers, which is one reason I wrote the names of
the numbers in English rather than using their decimal
representations.)

Incidentally, there is a very good reason not to use an algorithm
that would be equivalent to division of numbers:
division of numbers is a very expensive operation compared to
the XOR "division" algorithm, even if all we want is the remainder
and we ignore the quotient.
In fact, division of one number by another is one of the most
time-consuming of the "simple" arithmetic operations on a computer.
(The exception is division by a power of $2,$ which is very fast
but would be useless in a CRC-like algorithm.)

By using the XOR algorithm, a computer can process a long stream
of incoming data while using a lot less of its resources than would
otherwise be required, leaving those resources free to do something else.
If the CRC check is performed by specialized hardware,
that hardware can be far simpler than it would have had to be
if it needed to find the remainder of a numerical division.

Aha!

Here is the part you mean:

11010011101100 000 <--- input right padded by 3 bits

1011               <--- divisor (4 bits) = x³ + x + 1

If you convert 110100111011000
from binary,
you get
27096.

If you convert 110100111011
from binary,
you get
3387,
not
3664,
which is
11100101000 in binary.


$\operatorname{Supp}(\mathbb{Q}/\mathbb{Z})$ as a $\mathbb{Z}$-module This question provides the def of $\operatorname{Supp}(M)$ and a simple example

I want to find $\operatorname{Supp}(\mathbb{Q}/\mathbb{Z})$ as a $\mathbb{Z}$-module.

My attempt. $\mathbb{Z}$ is a PID. $\operatorname{Ann}(\mathbb{Q}/\mathbb{Z})=(0)$, because there is no number $x \neq 0$ such that $\mathbb{Q}(x) \subset \mathbb{Z}$. All prime ideals contain (0). This means that $\operatorname{Supp}(\mathbb{Q}/\mathbb{Z}) = \operatorname{Spec}(\mathbb{Z})$

Is it right?

UPD: the answer is $\operatorname{Spec}(\mathbb{Z}) \setminus (0)$ as the zero ideal does not belong to the support.

Definition: For an $R$-module $M$, the set $\{P\ \mid $ $P$ prime ideal of $R$, $M_P\not=0\}$ is called the support of $M$ and is denoted by $\mathrm{Supp}(M)$.

  Theorem: Let $R$ be a Noetherian ring and $M$ a finitely generated $R$-module. For any prime ideal $P$ of $R$, the following conditions are equivalent
  $(i)$ $P \in \mathrm{Supp}(M)$.
  $(ii)$ $P'\subseteq P$ for some $P' \in \mathrm{Ass}(M)$.
  $(iii)$ $\mathrm{Ann}_R(M)\subseteq P$.

But $\mathbb{Q}/\mathbb{Z}$ is not a finitely generated as a $\mathbb{Z}$-module.

Let $P$ be a prime ideal  of $\mathbb{Z}$. We have two cases:

1) If $P=0$, then $(\mathbb{Q}/\mathbb{Z})_P\cong\mathbb{Q}_P/\mathbb{Z}_P\cong\mathbb{Q}_0/\mathbb{Z}_0\cong\mathbb{Q}/\mathbb{Q}=0$. Thus $0\not\in \mathrm{Supp}(\mathbb{Q}/\mathbb{Z})$.

2) If $P\not=0$, then $(\mathbb{Q}/\mathbb{Z})_P\cong\mathbb{Q}_P/\mathbb{Z}_P$. Now since $\mathbb{Q}_P\not=\mathbb{Z}_P$, we have $P\in \mathrm{Supp}(\mathbb{Q}/\mathbb{Z})$.

Therefore, $\mathrm{Supp}(\mathbb{Q}/\mathbb{Z})=\mathrm{Max}(\mathbb{Z})=\mathrm{Spec}(\mathbb{Z})\setminus\{0\}$.

Yes, it is correct that $\operatorname{Ann}(\Bbb{Q}/\Bbb{Z})=0$ and $\operatorname{Supp}(\mathbb{Q}/\mathbb{Z}) = \operatorname{Spec}(\mathbb{Z})$.

In general, for a $\Bbb{Z}$-module $M$ it is true that $\operatorname{Ann}(M)=\operatorname{exp}(M)$, the exponent of $M$, and hence that
$$\operatorname{Supp}(M)=\{p\Bbb{Z}:\ p\mid\operatorname{exp}(M)\}=\operatorname{Spec}(\Bbb{Z}/\operatorname{exp}(M)\Bbb{Z}),$$
and in particular, if $M$ is a ring then $\operatorname{exp}(M)=\operatorname{char}(M)$.


New Year 2017 Riddle (3 quadratic Diophanitine equations) What is the smallest prime number n which can be formulated with all of the
three following integer equations?
$$u^2 +  4^2 \cdot v^2 = n$$
$$w^2 +  9^2 \cdot x^2 = n$$
$$y^2 + 11^2 \cdot z^2 = n$$

Using the Theorem:

  If a prime can be expressed as sum of two squares then the representation is unique.
  Proof

Then we have that the three expressions are the same.

Then for minimize $n$ we need find a prime  that can be expressed as any of this forms:
$$4^2\cdot 11^2 r^2+9^2 j^2$$ or $$4^2\cdot 9^2 l^2+11^2 f^2$$
But $36^2+11^2=1417$ isn't a prime and by coincidences of the life $44^2+9^2=2017$ is a prime number. We are done.

$$ax_1^2+bx_2^2=ax_3^2+bx_4^2=ax_5^2+bx_6^2$$

The solution is easy to write.

$$x_1=t(bp^2+ak^2)(bs^2+ak^2)$$

$$x_2=y(bp^2+ak^2)(bs^2+ak^2)$$

$$x_3=((bp^2-ak^2)t-2bpky)(bs^2+ak^2)$$

$$x_4=((bp^2-ak^2)y+2apkt)(bs^2+ak^2)$$

$$x_5=((bs^2-ak^2)t-2bsky)(bp^2+ak^2)$$

$$x_6=((bs^2-ak^2)y+2askt)(bp^2+ak^2)$$

Here the representation of 3 options, but it is easy to see that can be written in the form of a combination with any number of options.


Find $\lim_{x\to 0}x^{\frac{1}{x}}$ I tried to solve $$\lim_{x\to 0} x^{\frac{1}{x}}$$

I tried doing it like this:

$$\lim_{x\to 0} x^{\frac{1}{x}}= \lim_{x\to 0} e^{\ln(x^{\frac{1}{x}})} = \lim_{x\to 0} e^{\frac{\ln(x)}{x}} = \exp\left(\lim_{x\to 0} \frac{\ln(x)}{x}\right)$$

Then solving $\lim_{x\to 0} \frac{\ln(x)}{x}$ with l'Hospital's Rule I get

$$\lim_{x\to 0} \frac{\ln(x)}{x}=\lim_{x\to 0} \frac{1}{x}$$

Which remains indeterminate as it has no two-sided limit.

So is that the answer, that it remains indeterminate?  Or would I say that it has two limits depending on the side you approach?  How do I phrase this result?

The expression $\lim_{x\to 0^-}x^{\frac{1}{x}}$ makes no sense, for in general $x^y$ is undefined if $x$ is negative. As a consequence, $\lim_{x\to 0}x^{\frac{1}{x}}$ does not exist.

As was pointed out by Lubin, L'Hospital's Rule is not appropriate here, since after taking the logarithm we are looking at $\frac{\ln x}{x}$. For $x$ near $0$, the numerator $\ln x$ is very large negative, but the denominator is close to $0$, so we do not have a suitable form. But luckily, the rule is not needed, since raising a number close to $0$ to a large power gives a number very close to $0$.


What's your favorite proof accessible to a general audience? What math statement with proof do you  find most beautiful and   elegant, where such is  accessible to a general audience, meaning you could state, prove, and explain it to a general audience in roughly $5 \pm\epsilon$  minutes.
Let's define 'general audience' as approximately  an average adult with education and experience comparable to someone holding a bachelor's degree in any non science major (e.g. history) from an average  North American university.

I really like the proof of $$\sum_{i=1}^n i = \dfrac{n(n+1)}{2}$$ in which $1 + 2 + \cdots + (n-1) + n$ is written forwards then backwards and summed. It is claimed that Gauss had come up with this when he was just a child, although contested.

The proof

Let $$s = 1 + 2 + \cdots + (n-1) + n.$$
Clearly,
$$ s = n + (n-1) + \cdots + 2 + 1.$$

Sum to get $$2s = \underbrace{(n+1) + (n+1) + \cdots  + (n+1)  + (n+1)}_{n \text{ times}}.$$

Hence, $$2s = n(n+1),$$ and $$s = \dfrac{n(n+1)}{2}.$$

One of the first proofs I was shown was in a discrete structures class.  I don't think this would survive the rigors of formal proof, but I really enjoyed it.

Definitions:

$N$ is the set of numbers starting with 0, 1, 2, $\dots $

A number in our set $N$ is odd if and only if there exists another number in our set $N$ that when multiplied by two and added to one yields the number in question.  Or more formally

$\forall a \in N$, a is odd if $\exists b \in N \ni a = 2b + 1$

To Prove:

If $n$ is odd then $n^2$ is odd

Proof: Direct

Let us assume that we take an "odd" number, $n$.  Then "n is odd" is true.  Let us prove that $n^2$ is true.

By our assumptions $n$ is odd

This is the critical part of the proof, that might need some convincing

Since $n$ is odd we can use the definition of an odd number above and state that, $ \exists x \in N \ni n=2x+1$.

Now we can substitute our number $n$ with $2x+1$ and state $n=2x+1$

Further to prove $n^2$ is odd, we need only show that $(2x+1)^2$ is odd

This also might need some examples to convince

By distributive property, $(2x+1)^2=4x^2+4x+1$

Now we can refactor and rewrite $4x^2+4x+1=2(2x^2+2x)+1$

We know from rules of arithimetic that $2x^2+2x$ is a number which we can call $z$ and $z=2x^2+2x$

By substitution $2(2x^2+2x)+1=$2z+1$

$\therefore$ by definition of odd number $2(2x^2+2x)+1$ is odd


Value of $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n!}}$? I was reading this old question and fascinated by the second infinite sum $$\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n!}}.$$ This is clearly convergent (by comparison or ratio test) and, we can obtain some crude approximations of this using inequalities like $n^4\le n!\le (n!)^2,$ here the first inequality holds for all $n\ge7.$ But, I wonder if we can find the exact value of this series using (familiar) special functions/constants. How would you attack to this series?


S-shaped Reverse Logistic Curve Is there any curve that grows very slow at the beginning then growth picks up exponentially before hitting the wall.
I need sort of reverse behavior of the logistic curve.

Here's raw idea how it should look like  (drawn with paint)

And here it is compared with logistic curve

The function
\begin{align*}
f: (-6,6) &\to \mathbb{R} \\
x &\mapsto \frac{\log(\frac{1}{2}\left(1+x/6)\right)}{1-\frac{1}{2}\left(1+x/6)\right)},
\end{align*}

which is just an adaptation from the one found here, seems to satisfy your needs. (The adaptation is done in order to conform with the interval in your image.) From this you may be able to make further modifications which are relevant to you, if needed.

Looking for the same thing too. This looks promising.

y = 0.5 + log(x / (1-x)) / 4

Where 4 is a number that determines the "sharpness" of the curve.
Of couse this never actually touches 0 and/or 1, so there should be edge cases to protect against INF and -INF.


Measure the length of a wrapping I'm interested in learning how to find the length of a wrapping.

Let's say I'm going to be wrapping some flat fabric webbing around a pole. I'd like to find the amount (length of fabric) i'll need to buy to wrap a portion of the pole.

Desired length of wrap: 30 in.
Outer diameter of pole: 2 in.
Width of webbing: 0.5 in.
Angle of spiral: 15° (arbitrary value I came up with)

Thanks!

Basically, you want to cover a cylinder with $h = 30\text{ in}$ and $\Phi = 2R = 2\text{ in}$. The area of a cylinder without the top and bottom is $A = 2R\pi h = \Phi \pi h$.

Imagine that you cut this "open" cylinder after you have wrapped it: you obtain a rectangle with sides $h$ and $2R\pi$ and this rectangle is covered by slightly tilted "$2D$ strips" of fabric.

The width of each strip is $w = 0.5\text{ in}$ and the area covered by a fabirc of length $\ell$ equals $w\ell$ (which must equal $A$). Note that the angle doesn't matter. So you need
$$\ell = \frac{\Phi \pi h}{ w}  \doteq \frac{2\cdot3.14 \cdot 30}{0.5}\text{ in}\doteq 377\text{ in}$$
of fabric.

Consider unwrapping the wrapped strip, keeping the angle at $15°$. The final result is a right triangle with acute angle of $15°$ where the edge of the strip is the hypotenuse, and the side opposite the angle is the $30$-inch segment of the pole. Thus the length of the cloth strip is the length of the hypotenuse of that triangle:
 $30\,/\left(\sin 15°\right)$  inches,
somewhat close to $115.9$ inches. As long as the $"\!15°\!"$ is fixed, the width of the strip and the diameter of the pole determine the number of times you loop around the pole but do not affect the length in the question.


Can be $f_n:[0,1]\rightarrow\mathbb{R},\:f_{_n}(x)=x^n \cdot\ e^x$ uniform convergence? We have $f_n:[0,1]\rightarrow\mathbb{R},\:f_{_n}(x)=x^n \cdot\ e^x$.

I don't know how we can find the pointwise convergence...This sequence can be a uniform convergence? and explain your argument.

Pointwise convergence is very simple.

In order for a series of functions $f_n$ to converge pointwise towards a function $f$, the only condition is:

  For all values $x$, the limit $$\lim_{n\to\infty} f_n(x)$$ is equal to $f(x)$

Did you try to calculate the limit in your case? Since $x$ is fixed, it should be easy!

Pointwise convergence for any $x\in [0,1)$ we have $x^ne^x\to 0$ and $f_n(1)\to e$ so the sequence has a non continuous pointwise limit and therefore the sequence cannot be uniformly convergent


trouble understanding when to use $b^{\log_b(x)} = x$ I've been learning logarithm properties in a finite mathematics course and I am confused about a particular example in the text book. They are using the property $b^{log_b(x)}=x$, $x>0$, to solve the following problem:

$\frac{log_e(x)}{log_e(b)}$

The first thing they do is replace x in the equation with $b^{log_b(x)}$ giving the result:

$\frac{log_e(b^{log_b(x)})}{log_e(b)}$

Then use another property ($log_bM^p=p*log_bM$)to get:

$\frac{log_b(x)*log_e(b)}{log_e(b)}$

The two "$log_e(b)$" cancel out and we are left with the answer

$log_b(x)$

My question is that before #3(or at any point really) I could replace the only "x" in the equation with blogb(x) again could I not? And I could continue replacing "x" with blogb(x) and get a massive equation. What I need to understand is why do they replace x with that logarithm and why can't I continue to replace it over and over(as pointless as that would be). I put in some numbers and constantly changing "x" to that log changes the result so it obviously can't be done continuously. Also in the property it states x>0. Does that mean that after using this property x now is greater than zero. Or can this property only be applied if x is greater than zero. In which case how can I know x=0 if no value has been assigned to x at this point?

I am not great at math and my teachers have told me this textbook has many typos and errors so I end up questioning everything. So any clarification on when I can and cannot use blogb(x)=x would be great thanks.

When $x$ and $b$ are positive, you can always say that $x = b^{\log_b(x)}$. (When $x$ is negative, this doesn't hold up since the logarithm of a negative number isn't well-defined in the reals).

Your book uses this because it is often useful to replace $x$ with $b^{\log_b(x)}$. One example of this property being useful is the proof you stated in your question. However, while you can always use this property (when $x$ and $b$ are both positive), it is not always useful.

Your textbook is doing something weird, namely working from a specific application of the change of base formula to the "more frequently encountered expression" $\log_b x$. It is easy to derive the change of base formula to see what it is I am talking about.

Let $y=\log_b x$, whereby it must be true that $b^y=x$ by the definition of a logarithm. Now take the logarithm of base $a$ of both sides:
$$
y=\log_b x\Leftrightarrow b^y=x\Leftrightarrow \log_a(b^y)=\log_a(x)\Leftrightarrow y\log_a(b)=\log_a(x).
$$
From above, you can see that $y\log_a(b)=\log_a(x)$ or simply $y=\frac{\log_a(x)}{\log_a(b)}$. Remember that we defined $y=\log_b x$. That is, we have shown that
$$
\log_b x=\frac{\log_a x}{\log_a b}
$$
for any suitable base value $a$. Your specific problem is handled by letting $a=e$, where $e$ is Euler's constant.

Note that when $f(x)=a^x$ and $f^{-1}(x)=\log_a x$ we have
$$
\log_a(a^x)=x,\qquad x\in\mathbb{R}\\[0.5em]
a^{\log_a x}=x\qquad x>0.
$$
Why? [Hint: This is because of the Inverse Function Property, and also consider the domain and range of $f$ as opposed to $f^{-1}\ldots$]


Differentiate complex function? $$f(z)=3z^2+\bar{z}$$

I want to show the function is either differentiable or not so I can state if it is holomorphic or not. What is the method for this ?

Edit - Can some give an example of how to apply cauchy riemann equation to this question ?

If $f(z)$ were differentiable, then $f(z)-3z^2=\bar{z}$ would be differentiable since $3z^2$ is. However, $\bar{z}$ is not complex differentiable. Check this with the Cauchy-Riemann equations.

This function is differentiable everywhere except at z=0.  As with real differentiation, complex differentiation is still a linear operation, meaning that d/dz( $3z^2 + \bar{z}$) = d/dz ($3z^2$) + d/dz ($\bar{z}$).  The first term is differentiable everywhere.  The second term is not differentiable at zero because $\bar{z}'(0) = \lim_{z->0} (\bar{z}-0)/(z-0)$, and $\bar{z}/z$ can take on different values as $z->0$ for different directions (e.g. if you approach z from the positive real axis, $\bar{z}/z -> 1$; if you approach from the positive imaginary axis, $\bar{z}/z->-1$.  Here is a good page detailing this: http://en.wikipedia.org/wiki/Holomorphic_function


How to define a morphism $\varphi:L\otimes_K\hat K\to\prod_i \hat L_i$ from completion of finite field extensions Let $L/K$ be a degree $n$ extension of fields, where $K$ has discrete valuation $v$, which can be prolonged to the discrete valuations $w_i$ on $L$. We can therefore define the completion of $K$ w.r.t. $v$ to be $\hat K$, and the completion of $L$ w.r.t. $w_i$ to be $\hat L_i$, then in Theorem II.3.1 of Serre's Local Fields, we have a homomorphism $$\varphi:L\otimes_K\hat K\to\prod_i\hat L_i$$which we then show to be an isomorphism. However, I don't see how this morphism is defined in the first place.

The basic idea is to take $L$ to be $K[\alpha]\cong K[x]/(f(x))$ where $f(x)$ is the minimum polynomial of $\alpha$.  Then $L\otimes_K\hat K \cong K[x]/(f(x))\otimes_K\hat K \cong \hat K[x]/(f(x))$ and factor $f(x)$ in $\hat K[x]$.

In order to define the canonical morphism $\varphi:L\otimes_K\hat K\to\prod_i\hat L_i$, we will make use of the completion functor $\mathrm{comp}:\mathbf{DVF}\to\mathbf{DVF}$ acting on the category of discretely valued fields, with continuous field embeddings (that is to say, field extensions that prolong the discrete valuation) as morphisms. Then since we have a natural transformation $\mathrm{id}_{\mathbf{DVF}}\to\mathrm{comp}$ given by inclusion in the completion, we obtain the following commutative diagram:

$$
\require{AMScd}
\begin{CD}
(K,v) @>>> (\hat{K},\hat{v})\\
@VVV @VVV \\
(L,w_i) @>>> (\hat{L}_i,\hat{w}_i)
\end{CD}
$$
which allows us to define a ring morphism $\varphi:L\otimes_K \hat{K}\to\hat{L}_i$ for all $i$, which is $\hat{K}$-linear by construction. This gives us the morphism $$\varphi:L\otimes_K\hat K\to\prod_i\hat L_i.$$


Construct a multiplication table for $\mathbb{Z}_2[i]$. Construct a multiplication table for $$\mathbb{Z}_2[i]=\{a+bi|a,b \in \mathbb{Z}_2\}$$.
I have figured this table out for the most part. I know that each side has $0,1,i,1+i$, but $(1+i)(i)=i-1$, and what would that become in $\mathbb{Z}_2[i]$? I have the same question for $(1+i)(1+i)$, I don't know how to reduce these in terms of the mod.

As others also mentioned , we can generate the field ${Z_2[x]}$ with irreducible polynomial ${x^2+x+1}$ ,and the elements are ${\{0,1,x,x+1\}}$  , here ${x^2=x+1}$,and  ${-1=1}$ , ${0=(1+i)^2 = i+i}$

Here ${(x+1)^{-1}=x}$, cayley table may looks like..


Why is negative times negative = positive? Someone recently asked me why a negative $\times$ a negative is positive, and why a negative $\times$ a positive is negative, etc.

I went ahead and gave them a proof by contradiction like so:

Assume $(-x) \cdot (-y) = -xy$

Then divide both sides by $(-x)$ and you get $(-y) = y$

Since we have a contradiction, then our first assumption must be incorrect.

I'm guessing I did something wrong here. Since the conclusion of $(-x) \cdot (-y) = (xy)$ is hard to derive from what I wrote.

Is there a better way to explain this?  Is my proof incorrect?  Also, what would be an intuitive way to explain the negation concept, if there is one?

This is pretty soft, but I saw an analogy online to explain this once.

If you film a man running forwards ($+$) and then play the film forward ($+$) he is still running forward ($+$). If you play the film backward ($-$) he appears to be running backwards ($-$) so the result of multiplying a positive and a negative is negative. Same goes for if you film a man running backwards ($-$) and play it normally ($+$) he appears to be still running backwards ($-$). Now, if you film a man running backwards ($-$) and play it backwards ($-$) he appears to be running forward ($+$). The level to which you speed up the rewind doesn't matter ($-3x$ or $-4x$) these results hold true.
$$\text{backward} \times \text{backward} = \text{forward}$$
$$ \text{negative} \times \text{negative} = \text{positive}$$
It's not perfect, but it introduces the notion of the number line having directions at least.

Why a negative times a negative can be reduced to the question of why
-1 x -1 = 1.
The reason for that is because it is forced upon you by the other rules of arithmetic.

1 + (-1) = 0 because of the definition of -1 as the additive inverse of 1

Now multiple both sides by -1 to get

-1(1+(-1)) = 0 because 0 times anything is 0

Use distributive law to get:

-1* 1 + (-1)x(-1) = 0

Now -1 * 1 = -1 because 1 is multiplicative identity.

So we have
-1 + (-1)x (-1) = 0

Put -1 on the other side by adding 1 to both sides to get

(-1) x (-1) = 1

So -1 x -1 = 1.

Now for any other negative numbers x, y we have

x = (-1) |x| and y= (-1) |y|

So x * y = (-1) |x| * (-1) |y| = (-1) *(-1) * |x| * |y| = |x * y| is positive.

Now that you know the reason it really doesn't make much difference in understanding. This question is not really that important.
It's like asking why is 1 raised to the 0 power equal to 1?
Because that's forced upon you by other rules of exponents,etc.

A lot of time is wasted on this. This is not the kind of problem kids should be thinking about.


What is the difference between Newton integrable function and Riemann integrable function? I am supposed to determime the difference between Newton integrable function and Riemann integrable function.

I know that  A function $f : (a, b) \to \Bbb R$ is said to be Newton
integrable if $f$ has a primitive $F$ in $(a, b)$, and if the one sided limits $F(a_+)$ and $F(b_−)$ exist and are finite.

I also understand what is Riemann integral: I have to make partition on the graph and the Riemann integral is the limit of the Riemann sums of a function as the partitions get finer.

But somehow I am unable to distinguish the difference between those two, althougtht there are the definition.

Is somehow the area below the graph different or what is the difference?

Thanks.

Take for example $f$ from $[-1,1]$ to $\Bbb R$ given by
$$f(x)=-1 \text{ if } x<0$$
and
$$f(x)=1 \text{ if } x\ge 0$$

then $f$ is Riemann integrable since it has only one discontiuity point,

But

It is not Newton integrable since it has no primitive at $(-1,1)$.


Closed $\iff$ weakly closed subspace On this link
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2004;task=show_msg;msg=1414.0001
is the argument that a linear subspace in a normed space is closed w.r.t. norm iff it is weakly closed.

On the other hand, $c_0$ (sequences convergent to $0$) is a norm-closed linear subspace of $l_\infty$ (bounded sequences), but it is not weakly closed, since the base vectors $e_i$ are weakly dense in $l_\infty$.

Since I studied func.an. quite a while ago, my question is - what am I missing?

Edit: the previous answer was wrong. Below is a new version.

Weak convergence of a net $\{x_n\}$ to $1\in\ell^\infty$ means that, for any functional $\varphi$ in the dual of $\ell^\infty$, $\varphi(x_n-1)\to0$.

We can see $\ell^\infty$ as $C(\beta\mathbb N)$, the continuous functions on the Stone-Čech compactification of $\mathbb N$. Let $\omega\in\beta\mathbb N\setminus\mathbb N$ (in other words, $\omega$ is a free ultrafilter). Then $1(\omega)=1$ and $x(\omega)=0$ for all $x\in c_0$. So we have, letting $\varphi_\omega$ be the point-evaluation at $\omega$,
$$
\varphi_\omega(x-1)=\varphi_\omega(x)-\varphi_\omega(1)=0-1=-1,
$$
and so no net in $c_0$ will make the limit go to zero. This means that $c_0$ is not weakly dense in $\ell^\infty$. Weak-star density works because one has to deal with less functionals.

As was mentioned, the Hahn-Banach theorem guarantees that $c_0$ (being convex) is both norm and weakly closed.

There is a fundamental fact about convex sets in locally convex spaces (and, in particular normed linear spaces) that a convex set is closed if and only if it is closed in the weak topology (the topology of convergence on the (continuous)
dual.  WARNING: convergence is of nets, NOT sequences.  Also, every space has a weak topology since every space has  continuous dual (albeit possibly only the zero vector).  There is SOMETIMES a weak*-topology if the space in question is itself a dual but only then.


An easy way to remember PEMDAS I'm having trouble remembering PEMDAS (in regards to precedence in mathematical equations), ie:

Parentheses
Exponentiation
Multiplication & Division
Addition & Subtraction

I understand what all of the above mean, but I am having trouble keeping this in my head. Can you recommend any tricks or tips you use to remember this.

Thanks

PEMDAS is an acronym to help you remember. try different forms of mnemonic devices, like acrostics:
Please Excuse My Dear Aunt Sally;
Pancake Explosion Many Deaths Are Suspected;
Purple Egglants Make Dinner Alot Sickening;
Pink Elephants March, Dance, And Sing;
Pizza ended my donuts addiction Saturday


Convex dense subset of $\Bbb{R}^n$ is the entire space Say we have a convex dense set $X\subset\Bbb{R}^n$, does it follow that $X=\Bbb{R}^n$ ?

For $n=1$ it's true because convex set of real numbers are intervals, and if it's dense then it's $\Bbb{R}$.
Anyway, it seems more difficult in the general case, perhaps it's false, I don't know.

Any ideas ?

This question (Why does a convex set have the same interior points as its closure?) shows that the interior of $X $ is the same as the interior of the closure of $X $ (which is all of $\Bbb{R}^n $). Since a set contains it's interior,  we are done.

Let $x \in X$ so we can choose three points that $x$ is inside the triangle(Or in hider dimentions $n+1$ points). We can proxsimate dots from $X$ that $x$ is still in the "triangle". So because $X$ is convex $x \in X$. We proved that $X= \mathbb{R}^n$. Q.E.D


Why can a Venn diagram for $4+$ sets not be constructed using circles? This page gives a few examples of Venn diagrams for $4$ sets. Some examples:

Thinking about it for a little, it is impossible to partition the plane into the $16$ segments required for a complete $4$-set Venn diagram using only circles as we could do for $<4$ sets. Yet it is doable with ellipses or rectangles, so we don't require non-convex shapes as Edwards uses.

So what properties of a shape determine its suitability for $n$-set Venn diagrams? Specifically, why are circles not good enough for the case $n=4$?

The short answer, from a paper by Frank Ruskey, Carla D. Savage, and Stan Wagon is as follows:

... it is impossible to draw a Venn diagram with circles that will represent all the possible intersections of four (or more) sets. This is a simple consequence of the fact that circles can finitely intersect in at most two points and Euler’s relation F − E + V = 2 for the number of faces, edges, and vertices in a plane graph.

The same paper goes on in quite some detail about the process of creating Venn diagrams for higher values of n, especially for simple diagrams with rotational symmetry.
For a simple summary, the best answer I could find was on WikiAnswers:

Two circles intersect in at most two points, and each intersection creates one new region. (Going clockwise around the circle, the curve from each intersection to the next divides an existing region into two.)
Since the fourth circle intersects the first three in at most 6 places, it creates at most 6 new regions; that's 14 total, but you need 2^4 = 16 regions to represent all possible relationships between four sets.
But you can create a Venn diagram for four sets with four ellipses, because two ellipses can intersect in more than two points.

Both of these sources indicate that the critical property of a shape that would make it suitable or unsuitable for higher-order Venn diagrams is the number of possible intersections (and therefore, sub-regions) that can be made using two of the same shape.
To illustrate further, consider some of the complex shapes used for n=5, n=7 and n=11 (from Wolfram Mathworld):

The structure of these shapes is chosen such that they can intersect with each-other in as many different ways as required to produce the number of unique regions required for a given n.
See also: Are Venn Diagrams Limited to Three or Fewer Sets?

I'm quite late to the party, but I'd like to add a very simple but much less rigorous answer for variety.

Each region of a Venn diagram must represent some combination of true or false for each category. For instance, the two circle Venn diagram has 4 regions: the intersection, the outside, and the last two circle parts. The intersection contains everything true to both categories, the outside neither category, and the last two only one of the two categories. (You could list them like TT, TF, FT, FF)

In order to fully cover all cases of true/false combinations, we need $2^n$ sections. Each new category we add doubles the number of regions as it must split each existing region into 2 smaller regions. (One for the true case of the new category, the other for false, so the TT region is now split into TTT and TTF)

We can see how the third circle intersects all 4 regions, but there's no way to draw a 4th circle that intersects all 8 of the new regions, making it impossible to draw a 4 circle Venn diagram.


Can I compute this integral analytically? I will give a small background and explain the variables and the system first. I have two images which are observed and are constant and we can treat them as continuous functions and I will call them $r$ and $f$. In my problem, I am trying to find a continuous transform (which is very non-linear) that makes $f$ looks like $r$ according to some similarity criteria or cost function. I will call this transformation function $t$ and I am trying to estimate its parameters $w$.

So, the integral I need to compute turns out to be

$$
Z = \int_{-\infty}^{\infty} \exp-{\frac{\left( r(i) - f\left(t(w)\right)\right)^2}{2\sigma^2}} \, dw
$$

where $\sigma$ is a constant. Now, given a constant linear function $A$, $f(t(w))$ is computed as:

$$
f(t(i, w)) = (\lceil{Aw}\rceil - Aw) * f(\lfloor{Ax}\rfloor) + (Aw - \lfloor{Aw}\rfloor) * f(\lceil{Ax}\rceil)
$$

where $\lceil \rceil$ gives the ceiling function and $\lfloor \rfloor$ is the floor function. This basically means that I am using linear interpolation to make the transformation function continuous. This is because the images and the transformation are defined in the digital domain and are computed only on a uniform grid (corresponding to the pixel locations) and the transformation $t$ is telling me what the location of a pixel $i$ in image $r$ is in image $f$ through $w$.

Can someone tell me if I can compute such an integral? My first instinct was to use Taylor series to linearise $t(w)$ but then I realised it is not a good idea as $t(w)$ is in the integral and we are integrating over $w$. So the higher order terms will not cancel out and I cannot justify that approximation.

Yes, it can be solved using double integration. For simplicity, we integrate $\int_{ - \infty }^{\infty}  {{e^{ - {x^2}}}dx}$. Consider the circular disc ${D_b}:{x^2} + {y^2} \le {b^2}$ with polar coordinates $(r,\theta)$ in the set $\Gamma :0 \le \theta  \le 2\pi ,0 \le r \le b
$. Therefore,
\begin{align}
\int_{{D_b}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }  &= \int_\Gamma  {\int {{e^{ - {r^2}}}rdrd\theta } } \\
 &= \int_0^{2\pi } {\int_0^b {{e^{ - {r^2}}}rdrd\theta } }  = \int_0^{2\pi } {\frac{1}{2}\left( {1 - {e^{ - {b^2}}}} \right)d\theta }  = \pi \left( {1 - {e^{ - {b^2}}}} \right)
\end{align}
Let $S_a$ be the square $-a\le x \le a$, $-a\le y\le a$. Since $D_a \subseteq S_a \subseteq S_{2a}$ and ${{e^{ - \left( {{x^2} + {y^2}} \right)}}}$ is positive,
\begin{align}
\int_{{D_b}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }  \le \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }  \le \int_{{D_{2a}}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }
\end{align}
It follows that
\begin{align}
\pi \left( {1 - {e^{ - {a^2}}}} \right) \le \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }  \le \pi \left( {1 - {e^{ - 4{a^2}}}} \right)
\end{align}
As $a \to \infty$, $\pi \left( {1 - {e^{ - {a^2}}}} \right) \to \pi$ and $\pi \left( {1 - {e^{ - 4{a^2}}}} \right) \to \pi$. Therefore,
\begin{align}
\mathop {\lim }\limits_{a \to \infty } \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }  = \pi.
\end{align}
But
\begin{align}
\int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }  = \int_{ - a}^a {\int_{ - a}^a {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} }  &= \left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)\left( {\int_{ - a}^a {{e^{ - {y^2}}}dy} } \right)
\\
&= {\left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)^2}.
\end{align}
Thus,
\begin{align}
\mathop {\lim }\limits_{a \to \infty } \int_{ - a}^a {{e^{ - {x^2}}}dx}  = \mathop {\lim }\limits_{a \to \infty } {\left( {\int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } } \right)^{1/2}} = \sqrt \pi.
\end{align}


Show that $x^3+x^2+x+1 \mid x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}$ in $\Bbb Z[x]$ Problem:

Show that $x^3+x^2+x+1 \mid x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}$ in $\Bbb Z[x]$ for all $m,n,q,s$ nonnegative integers

So my algebra sucks.
I'm reviewing some algebra and found this problem. I was able to play with it and get a trivial solution that I'm sure is missing the point of the problem.
My (dumb) solution:

We prove by induction.
For $m=n=q=s=0$ we have that $x^3+x^2+x+1 \mid x^3+x^2+x+1$.
Now assume $x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}=p(x)(x^3+x^2+x+1)$.
Then: $$(p(x)+x^{4m+3}(x-1))(x^3+x^2+x+1)=x^{4(m+1)+3}+x^{4n+2}+x^{4q+1}+x^{4s}$$
$$(p(x)+x^{4n+2}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4(n+1)+2}+x^{4q+1}+x^{4s}$$
$$(p(x)+x^{4q+1}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4n+2}+x^{4(q+1)+1}+x^{4s}$$
$$(p(x)+x^{4s}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4(s+1)}$$

However this is very unenlightening. What is the point of the problem? Is there more enlightening solution?

$i,-1,-i$ are all roots of your polynomial. It is divisible by $(x-i)(x+1)(x+i)=x^3 + x^2 + x + 1.$ I suppose I should add that it  is so divisible in $\mathbb Q [x],$ while Gauss theorem on content says it is then divisible in  $\mathbb Z [x] \; .$

The algebraic answer by @Will Jagy is excellent (of course). I'll add a naive solution by induction, especially that induction was mentioned in the OP post.

Let's do induction over $\ E\ := m+n+q+s,\ $ where $\ m\ n\ q\ s\ $ are arbitrary non-negative integers (see the OP post). The theorem holds for $\ E=0.\ $ Now, assume that $\ E>0.\ $ Then there exists a non-negative integer $\ t\ $ and $\ r\in\{0\ 1\ 2\ 3\}\ $ such that
$\ x^{4\cdot(t+1)+r}\ $ is one of the four summands of
$\ g(x)\ :=\ x^{4\cdot m+3} + x^{4\cdot n+2} + x^{4\cdot q+1} + x^{4\cdot s}. $ Define polynomial $\ f\ $ obtained by replacing the said summand by monomial $\ x^{4\cdot t+r}.\ $ Then

$$ g(x) - f(x) = (x^4-1)\cdot x^{4\cdot t+r}  $$

Thus, polynomial $\ x^3+x^2+x^1+1\ $ divides $\ g-f,\ $ and it
divides $\ f\ $ by induction, hence it divides $\ g.\ $ END of (inductive) proof

EDIT Actually, I've virtually rewritten the OP's solution. I think that it is as good as any. I'll keep this version here  too since it may be clearer this way when you have both of these renditions.


Big Theta of this modification of the secondary branch of the Lambert W function I am looking to find the big-$\Theta$ of $-W_{-1}(-\frac{a}{n})$ in terms of elementary functions where $a$ is a constant. Looking around and I find that this should be  $O(\log(n))$ and with maxima I found that the ratio seems to decrease towards $1$ as you increase $n$ implying that it is also $\Omega(\log(n))$

I get this from the fact that $W_{-1}(x) = -\ln(-x) - \ln(-(\ln(-x) - \ln(-(\ln(-x)-\cdots))))$

Is there any proof that this is correct or not?

Maple has
$$
W_{-1}(x)=
\ln  \left( -x \right) -\ln  \left( -\ln  \left( -x \right)  \right) +
{\frac {\ln  \left( -\ln  \left( -x \right)  \right) }{\ln  \left( -x
 \right) }}+{\frac {-\ln  \left( -\ln  \left( -x \right)  \right) +1/2
\, \left( \ln  \left( -\ln  \left( -x \right)  \right)  \right) ^{2}}{
 \left( \ln  \left( -x \right)  \right) ^{2}}}+O \left( -{\frac {
 \left( \ln  \left( -\ln  \left( -x \right)  \right)  \right) ^{3}}{
 \left( \ln  \left( -x \right)  \right) ^{3}}} \right)
$$
as $x \to 0-$.  Since $W_{-1}(x) < 0$ near $0$, at least my sign is better than yours.


An isomorphism between fields of fractions Let $R$ be an integral domain, let $K$ be its field of fractions, and let $S$ be a ring between $R$ and $K$.  Prove that the field of fractions of $S$ is (isomorphic to) $K$.

My attempt:

Since $S \subset K',$ define $f:K' \rightarrow S$ to be epimorphism, $g:S \rightarrow R$ to be epimorphism since $S$ between $K,R$, and $h:R \rightarrow K.$ I say if we take $t= h \circ g \circ f: K' \rightarrow K$, then this function is isomorphism?

I am not sure if my argument true or not, so I appreciate any help.

Here is another attempt:

Let us denote the field of fraction of $R$ by $K$, and the field of fraction of $S$ by $K'$. So the functions $f_1:R \rightarrow K,\ f_2:S \rightarrow K'$ defined by $f_1(r)=\frac{r}{1},\ f_2(s)=\frac{s}{1}$ is a ring monomorphism. Next, define $f_3:R \rightarrow S$ by $f_3(r)=r$ since $R \subset S$, so it is a ring monomprphism. Now, we can define $g_1:R \rightarrow K'$ by $g_1=f_2 \circ f_3$, so this function is clearly a ring monomorphism. Hence there exists a unique ring homomorphism $\alpha: K \rightarrow K'$ such that $\alpha \circ f_1=g_1$. Now I am getting stuck here because I am not sure why this function $\alpha$ must be injective and surjective.

Thanks for any further help.

Let $Q(S)$ be the field of fractions of $S$ and let $\iota_R\colon S\to Q(S)$ be the canonical embedding. Let $\iota_R=\mathrm{id}_K\Bigm|_{R}$ be the restriction of the identity map of $K$ to $R$ (so $\iota_R$ is the embedding of $R$ into $K$) and let $j_S=\mathrm{id}_K\Bigm|_{S}$ be the restriction to $S$ (so $j_S$ is the embedding of $S$ into $K$. Finally, let $j_R\colon R\to S$ be the inclusion of $R$ into $S$.

The universal property of the field of fractions guarantees that:

If $f\colon S\to F$ is any ring homomorphism with the property that $f(s)$ is invertible for every $s\neq 0$, then there exists a unique ring homomorphism $\mathfrak{f}\colon Q(S)\to F$ such that $f=\mathfrak{f}\circ \iota_S$. Moreover, if $F$ is no the trivial ring, then $f$ is one-to-one (since $0$ is not invertible in $F$, so only $0$ can map to $0$), and hence $\mathfrak{f}$, whose domain is a field, must be one-to-one.
If $g\colon R\to F$ is any ring homomorphism with the property that $g(r)$ is invertible for every $r\neq 0$, then there exists a unique ring homomorphism $\mathfrak{g}\colon K\to F$ such that $g=\mathfrak{g}\circ \iota_R$. Moreover if $F$ is not the trivial ring, then $g$ must be one-to-one and hence $\mathfrak{g}$ must also be one-to-one.

Now, the map $\iota_S\circ j_R\colon R\to Q(S)$ induces a map $g\colon K\to Q(S)$. And the map $j_S\colon S\to K$ induces a map $f\colon Q(S)\to K$.

I claim that $f$ and $g$ are inverses of each other.

Indeed, consider the map from $R$ to $K$ to $Q(S)$ to $K$ given by $f\circ g\circ\iota_R$. There must exist a unique morphism $h\colon K\to K$ such that $f\circ g\circ\iota_R = h\circ\iota_R$. Now, the identity map of $K$ certainly works, but so does $f\circ g$. By the uniqueness clause of the universal property, it follows that $f\circ g$ is the identity of $K$.

Symmetrically, consider the map from $S$ to $Q(S)$ to $K$ to $Q(S)$ given by $g\circ f\circ \iota_S$. There exists a unique morphism $h\colon Q(S)\to Q(S)$ such that $g\circ f\circ \iota_S = h\circ\iota_S$. Now, certainly $g\circ f$ is one such function, but so is the identity map of $Q(S)$. By the uniqueness clause, we have that $g\circ f= \mathrm{id}_{Q(S)}$.

Thus, $g\circ f = \mathrm{id}_{Q(S)}$ and $f\circ g = \mathrm{id}_{K}$, hence $f$ and $g$ are isomorphisms, so $Q(S)\cong K$, as claimed.

The field of fractions $K'$ of $S$ is equivalence classes $\frac{a}{b}$ for $a, b\in S$, $b\ne 0$. But $S\subset K$, so $\frac{a}{b}\in K$ since $K$ is a field. Thus $K'\subset K$. It is clear that $K\subset K'$ since $R\subset S$.


Is substitution a good strategy when I have to integrate a natural log? For example, I have the integral ∫xln(x^2)dx. Would it be better to use parts?

So far, I have tried: u= x^2

I came up with the new integral of 1/2∫ln(u)du...
I am not sure what the next step is

To solve
$$\int\ln(u) \,du$$
a common trick is to treat the integral as
$$\int 1 \cdot \ln(u) \, du$$
and use integration by part.
So
$$\int 1 \cdot \ln(u) \, du = u \ln(u)+\int 1\cdot \,du = u\ln(u)+u +C. $$


Meaning of Rays in Polar Plot of Prime Numbers I recently began experimenting with gnuplot and I quickly made an interesting discovery. I plotted all of the prime numbers beneath 1 million in polar coordinates such that for every prime $p$, $(r,\theta) = (p,p)$. I was not expecting anything in particular, I was simply trying it out. The results are fascinating.

When looking at the primes beneath 30000, a spiral pattern can be seen:

For comparison, here is the same graph with the multiples of 3 and 7 superimposed on it. Primes are in yellow, multiples of 3 and 7 in green and red respectively.

What is really interesting to me, though, is the behavior when the range is increased. Multiples of a given number appear to spiral out in the same pattern into infinity, but the primes begin to form rays in groups of 3 or 4. See below:

Compared to multiples of 3 and 7 again:

Now, I must admit that I am very much a novice mathematician with little experience beyond trigonometry. I am just going into Calculus and Discrete Mathematics this upcoming fall.

I know that there is something called the Prime Number theorem - are these patterns related to it? Are these rays the same phenomenon as the diagonal lines found in Ulam Spirals?

EDIT:
In response to Greg Martin's explanation, I decided to add a couple more graphs. To see why they are relevant, read his answer.

$(r,\theta)=(n,n), n \in  \mathbb{N}$

What we're seeing is arithmetic progressions (not prime-producing polynomials) of primes, combined with a classical phenomenon about rational approximations.

When the integers (or any subset of them) are represented by the polar points $(n,n)$, of course integers that are close to a multiple of $2\pi$ apart from each other will lie close to the same central ray. Figuring out when integers are close to a multiple of $2\pi$ apart is a perfect job for continued fractions. The continued fraction of $2\pi$ is $\langle 6; 3,1,1,7,2,146,\dots\rangle$, giving the convergents
$$
\left\{6,\frac{19}{3},\frac{25}{4},\frac{44}{7},\frac{333}{53},\frac{710}{113},\frac{103993}{16551},\dots\right\},
$$
which are the rational approximations of $2\pi$ that will dominate the picture on different scales.

For example, if you plot the polar points $(n,n)$ for $1\le n\le 25000$, you will notice the points aligning themselves into $44$ spirals: jumping ahead from $n$ to $n+44$ is almost the same as going around the circle $7$ times (note the convergent $\frac{44}7$ showing up); moving from $n$ to $n+1$ jumps ahead $7$ spirals. Each spiral corresponds to an arithmetic progression $a\pmod{44}$;  going from one spiral to the next one counterclockwise corresponding to changing the arithmetic progression from $a\pmod{44}$ to $a+19\pmod{44}$ (note that $19\equiv7^{-1}\pmod{44}$).

If instead you plot only the primes $(p,p)$, you will get reasonable representation in the $\phi(44)=20$ spirals corresponding to arithmetic progressions $a\pmod{44}$ where $\gcd(a,44)=1$, and no primes in the other $24$ spirals. That's what we're seeing in the top two pictures.

As the scale moves farther out, these particular spirals become more tightly wound and harder to see (go from the 1st picture to the 5th, then the 3rd, then the 4th), and the next convergent takes over. In this case, the convergent $\frac{710}{113}$ is an extremely good rational approximation to $2\pi$ (as we know from the large partial quotient $146$). Therefore the integer points $(n,n)$ will group themselves into $710$ spirals, but these spirals are so close to straight lines at the beginning that they almost don't look like spirals, and persist for a large interval of possible scales. Each ray thus corresponds to an arithmetic progression $a\pmod{710}$.

When we plot only prime points $(p,p)$ (the 4th picture is best here), we will only see the $\phi(710)=280$ arithmetic progressions $a\pmod{710}$ where $\gcd(a,710)=1$. The fact that the visible rays are mostly grouped in fours is a consequence of the fact that $5\mid710$ and so every fifth ray doesn't contain primes. Really, though, we are seeing four out of every ten rays rather than four out of every five; the arithmetic progressions $a\pmod{710}$ with $a$ even have no primes at all and are thus invisible. There are four exceptional groups containing only three rays instead of four; these correspond to the four arithmetic progressions $a\pmod{710}$ where $a$ is a multiple of $71$ but not a multiple of $2$ or $5$.

$3$Blue$1$Brown has given a beautiful explanation to the spiral and ray patterns. Please watch his video titled "Why do prime numbers make these spirals?"


Can anyone help showing why my calculation of $\int x\ln x$ $dx$ is wrong? Suppose $\frac{dy}{dx}=x\ln x,$ my teacher asks me to find $y$.

So I assume I got to integrate the right hand side:

$$\int x\ln x\, dx$$

The result I got is
$$
\int x\ln x\, dx=x\ln x-x+C\tag{1}
$$
But, apparently, it is wrong since taking the derivative gives:
$$
(x\ln x-x+C)'=\ln x+1-1=\ln x.
$$

Can you please give me a hand?

The answe you got is wrong indeed, since

$$(x\ln x - x)' = 1\cdot \ln x + x\cdot \frac{1}{x} - 1 = \ln x + 1 - 1 = \ln x\neq x\ln x$$

To calculate the integral, use the per partes method. Since $$\int u(x)v'(x)dx = u(x)v(x) - \int u'(x) v(x)dx,$$ you should select $u$ such that you can calculate its derivative, while $v'$ should be such that you can calculate its antiderivative. In your case, it should be clear which of the functions $x$ and $\ln x$ is easier to find the antiderivative of.

You have to integrate the product $x\ln x$, the antiderivative of which is not immediate, and there is no obvious substitution to try.

So let us consider integration by parts. We can integrate the factor $x$ (but not $\ln x$), giving

$$\int x\ln x\,dx=\frac{x^2}2\ln x-\int\frac x2dx.$$

Bingo !


Can Hilbert (style) prove my tautology? Can Hilbert (style) axioms prove the following tautology?

$$A\wedge(C\rightarrow B)\oplus(A\wedge B\leftrightarrow C)\rightarrow(C\rightarrow(A\rightarrow B))\qquad\text{algebraic style}
$$$$
(((A\wedge(C\rightarrow B))\oplus((A\wedge B)\leftrightarrow C)))\rightarrow(C\rightarrow(A\rightarrow B))\qquad\text{formal style}
$$

Here $\wedge$ means logical and, $\oplus$ means exclusive-or, and $\leftrightarrow$ means logical equivalence.

Due to Wikipedia/Hilbert style the axioms below describe classical propositional logic:

P1. $\phi \to \phi$

P2. $\phi \to \left( \psi \to \phi \right)$

P3. $\left( \phi \to \left( \psi \rightarrow \xi \right) \right) \to \left( \left( \phi \to \psi \right) \to  \left( \phi \to \xi \right) \right)$

P4. $\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$

How to go? Is it necessary to add definitions of connectives?

You can see Elliott Mendelson, Introduction to mathematical logic (4ed - 1997) : PROPOSITION 1.14 (COMPLETENESS THEOREM).

The proof (due to Kalmar,1935) allows us to build a proof of a tautology $\mathcal B$ whatever.

Assume $\mathcal B$ is a tautology, and let $B_1, \ldots, B_k$ be the statement letetrs in $\mathcal B$. For any truth value assignment to $B_1, \ldots, B_k$ we have :

  $B_1', \ldots, B_k' \vdash \mathcal B$

where let $B_i'$ be $B_i$ if $B_i$ takes the value $T$, and let $B_i'$ be $\lnot B_i$ if $B_i$ takes the value $F$.

Thus, due to the fact that your formula $\mathcal F$ has only three statement letters : $A,B,C$, it is enough to rewrite it in "primitive" notation, i.e. using only $\rightarrow$ and $\lnot$ (call it : $\mathcal F'$) and then apply the procedure described in the above Proposition.

Starting from example from the truth value assignment $v(A)=v(B)=v(C)=T$, we have :

    $A,B,C \vdash \mathcal F'$.

To do this you would want to first define the other connectives.  I will use Lukasiewicz/Polish notation.  For your "∧" I write "K".  For your "→" I write "C".  For your "⊕", following A. N. Prior, I will write "J".  For "$\lnot$", "N".  For "↔" I will write E.

The axioms become

P2 CpCqp.

P3 CCpCqrCCpqCpr.

P4 CCNpNqCqp.

I will note that P1 is NOT an axiom given that a requirement for axioms is that they are independent, and I simply fail to understand this new standard where axioms don't have to be independent.

Now we'll need definitions for the other connectives, which in effect will end up behaving like axioms.  The symbol "$\delta$" indicated a variable function of one argument.  This means we can make substitutions for $\delta$.

PK: C$\delta$Kpq$\delta$NCpNq.

PE: C$\delta$Epq$\delta$NCCpqNCqp.

PJ: C$\delta$Jpq$\delta$CCpqNCqp.

Your formula is CJKaCcbEKabcCcCab.

Using definition PK we can tell that  CJKaCcbEKabcCcCab is equivalent to

CJNCaNCcbENCaNbcCcCab.

Using definition PE we can tell that CJNCaNCcbENCaNbcCcCab is equivalent to

CJNCaNCcbNCCNCaNbcNCcNCaNbCcCab.

Using definiton PJ we can tell that CJNCaNCcbNCCNCaNbcNCcNCaNbCcCab is equivalent to

CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcbCcCab.

To try to prove it using Prover9, or OTTER you'll only need to parenthesize all of the axioms and use a unary predicate (which you may think of as meaning "provable" or "is formally provable"), and rewrite the variables as "x", "y" and "z" respectively for "a", "c", and "b" or "p", "q" and "r"... something like this:

P2: P(C(x,C(y,x))).

P3: P(C(C(x,C(y,z)),C(C(x,y),C(x,z)))).

P4: P(C(C(N(x),N(y)),C(y,x))).

Parenthesize P(CCCNCxNCzyNCCNCxNyzNCzNCxNyNCNCCNCxNyzNCzNCxNyNCxNCzyCzCxy), and add this as the goal.  This formula is:

 P(C(C(C(N(C(x,N(C(z,y)))),N(C(C(N(C(x,N(y))),z),N(C(z,N(C(x,N(y)))))))),N(C(N(C(C(N(C(x,N(y))),z),N(C(z,N(C(x,N(y))))))),N(C(x,N(C(z,y))))))),C(z,C(x,y)))).

You'll also want an assumption which enables Prover9 to prove things like it were using condensed detachment:

-P(C(x,y)) | -P(x) | P(y).  (it is not the case that either it is provable that x implies y, or it is not the case that x is provable, or y is provable).  Then you might find a proof equivalent to a proof using condensed detachment.  You'll want to prove that since you have

PK: C$\delta$Kpq$\delta$NCpNq.

PE: C$\delta$Epq$\delta$NCCpqNCqp.

PJ: C$\delta$Jpq$\delta$CCpqNCqp.

you can prove

PK reversed: C $\delta$NCpNq $\delta$Kpq.

PE reversed: C $\delta$NCCpqNCqp $\delta$Epq.

PJ reversed: C $\delta$CCpqNCqp $\delta$Jpq.

And it does come as possible to prove each of those, since it has gotten proven that from

C $\delta$ P $\delta$ R

we can derive

C $\delta$ R $\delta$ P

where "R" and "P" are constants, and we only have substitution and detachment as the rules of inference.

Another approach could involve proving CCNpqCCNpNqp (or CCNpNqCCNpqp).  Then one could set up a natural-deduction like demonstration of the C-N fomrula you've referred to, and then use the proof of the deduction meta-theorem to produce a formal proof.


Dependencies of various of areas of mathematics Specifically, I want to learn category theory to improve my programming ability in Haskell, but it seems like that would be difficult without a knowledge of abstract algebra, and it seems that a solid understanding of linear algebra is useful to understand abstract algebra.

I'm therefore looking for something showing how these areas of maths depend on each other, but I think that a more general table or graph probably exists (or would be useful).

This question is the same as this one on Stack Overflow, but it seems like a good question for this site, and likely to get much better answers here.

I can't give a fully authoritative answer to this, but I can try to give some indication. The key issue with category theory is that whilst you could probably understand the theory without too many pre-requisites, examples of categories come from all branches mathematics and in order to fully appreciate it, you'd need to have seen some of these examples before. You'd also need a good deal of experience in other disciplines, especially abstract algebra. As such at my university in England, category theory is not offered as an undergraduate course.

I'm going to assume that you have a basic high school proficiency in mathematics - as I have to start somewhere.

First thing you'll need is some elementary set theory - by which I mean a knowledge of what we mean by sets (unions, intersections etc), functions (injections, bijections, surjections), basic notions of countability etc. In fact this is one of the simplest examples of a category.

After this, you'll need  a first course in linear algebra, and a first, and possibly second course in abstract algebra covering the rudiments of group theory, ring theory, and perhaps module theory and field theory. This will give you a large bank of examples of categories, as well as some familiarity with some of the ideas that will be explored and their significance.

Other useful things to know about would be posets, which can often be found in a first course on logic. It'd also be handy to know some basic topology for some more examples.

I'd say the raw essentials are:
 - Elementary set theory
 - Linear algebra
 - Group theory
 - Ring theory

But the more experience you have in mathematics, the more accessible it will become.

Not sure if this is what you're looking for, but the book Mathematics - Form and Function by Saunders Mac Lane contains several figures with "dependency trees" connecting different mathematical areas or different concepts within one area.  See pages 149, 184, 306, 408, 423, and 425-428.


Prove that $\det(A)=\det(A^T)$ algebraically If we use row operations to turn matrix $A$ into an upper triangular matrix then the $\det(A)$ is equal to the product of the entries on its main diagonal. So if we transpose $A$, then those row operations can be made column operations and we would have the same upper triangular matrix where $\det(A^T)$ is equal to the product of the entries on its main diagonal. So $\det(A)=\det(A^T)$.

Is there an algebraic way to show this.

If you use the definition of determinant where you take the sum of all signed products of n elements where each element is chosen from a distinct row and distinct column, then you get that $\det(A) = \det(A^T)$ for free, and I guess you could call it an algebraic proof because you get the same algebraic formula for both $\det(A)$ and $\det(A^T)$.

There are at least three ways to show this, all very good exercise: (i) SVD, which takes one line (ii) show it for diagonalizable matrices and use density of diagonalizable matrices (iii) show this for invertible and singular matrices where the former is by decomposition to elementary matrices and latter is just 0.