Mistake proving that $3 = 0$ I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:

Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.

Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $(2)$

From eq. $(1)$ we conclude that $x +1 = -x^2$ and if we substitude that in eq. $(2)$ we get: $-x^2 + \frac{1}{x} = 0$ $(3)$

Simplifying eq $(3)$ we get: $x^2 = \frac{1}{x} \iff x^3 = 1$, so we get that $x = 1$ is a solution.

If we substitude the solution $x = 1$ back into eq. $(1)$ we get $1^2 + 1 + 1 = 0 \iff 3 = 0$

I'll leave the answer and the rest of my question as a spoiler if you first want to try this yourself:

 He explains that the mistake is when we substitute $x + 1=-x^2$ into eq. $(2)$, because when we do so we are actually adding the solution $x = 1$. The thing is that in the video he only says that and that got my wondering, why is that so?

Consider the equation $x=x^2$. Its solutions are easy to find: they are $0$ and $1$, right?!
However, if $x$ is a solution of that equation, then $x=x^2$, and therefore $x=(x^2)^2=x^4$. But the equation $x=x^4$ has other solutions, besides $0$ and $1$, namely $-\frac12+\frac{\sqrt3}2i$.
The problem lies in the if-then clause of the first sentence of the previous paragraph: we are saying (correctly) that if $x$ is a solution of the given equation, then $x$ is also a solution of another equation. But there is no reason for you to think that the new equation only has the solutions of the original one!
That's what happens in your example: if $x$ is a solution of $x^2+x+1=0$, then $x$ is a solution of $x^2=\frac1x$, but you have no reason to suppose that every solution of $x^2=\frac1x$ is also a solution of $x^2+x+1=0$.

Call $f(x)=x^2 + x + 1$, then his equation is $f(x)=0$.
From there, he derives the equation $(2)$, $\frac{f(x)}{x}=0$.
To reach equation $(3)$, he substracts the second equation to the first one, obtaining the equation $\left(\frac{1}{x}-1\right)f(x)=0$.
Thus, the equation $(3)$ will be true when either $f(x)=0$, which are the solutions of the original problem, or when $\frac{1}{x}-1=0$, that is, $x=1$. That´s what he means when he says he is adding the solution $x=1$.


Is there a way to find expected value of equation? If the random variable $X$ is binomially distributed with parameters $n=6$ and $p=0.3$, what is $$E(4+3X^2)$$

I know $E(X) = np = 1.8$. I solved this problem by finding $P(X)$ of all $X$ using $(^6_x)(0.3)^x(0.7)^{6-x}$ and then apply them with $4+3X^2$ to solve for $E(4+3X^2)$. But this takes too long and I think there should be a formula or another way to solve this.

Can anyone tell help me out with this? Thank you in advance!

Since $X$ is binomial you also know that $Var(X)=np(1-p)$. So, use that (this holds in general) $$Var(X)=E[X^2]-E[X]^2\implies E[X^2]=Var(X)+E[X]^2$$ and linearity of expectation to obtain that
\begin{align}E[4+3X^2]&=E[4]+E[3X^2]=4+3E[X^2]=4+3(Var(X)+E[X]^2)\\[0.2cm]&=4+3(np(1-p)+(np)^2)\end{align}

Another way of solving this is you compute second derivative of MGF and put $t=0$.


The identity of a domain and the identity of its field of fractions Let D be a domain and let F be its field of fractions. If e is the identity of the domain D and e' is the identity of its field of fractions F, can we say that e=e'?

Would you help me, please?
Thank you in advance.

This has nothing to do with domains and field of fractions:

Lemma. If $G$ and $H$ are groups such that $H \subset G$, then $e_H=e_G$.

Proof. $e_H=e_H * e_G=e_G.$

The answer to your question follows from setting $H=D$ and $G=F$.


Proving that I cannot factor a cubic polynomial with Algebra alone How do I prove a given polynomial is not factor-able using only radicals, basic arithmetic operations, and real, rational numbers?  For example, how can I prove that

$$0=x^3-3x+1$$

Has no solution with the given restraints?

  Persistently, I want an algebraic (pre-calc. type of algebra) proof.

An idea I've had is to assume it does have one solution expressible with the given and then disproving using contradictions, but it seems be fruitless.

Let $a,b,c$ be the three real roots of $p(x) = x^3-3x+1$.

Consider the interpolating polynomials $f,g$ such that $f(a)=b, f(b)=c, f(c)=a$ and $g(a)=c, g(b)=a, g(c)=b$.
With some Galois theory, one can show that they are defined over $\Bbb Q(\sqrt \Delta)$ (and since all roots are real, the polynomials are real, so $\Delta > 0$). In fact since here $\Delta$ is a square, they have rational coefficients.

It turns out that $f,g$ are the polynomials $x \mapsto x^2-2$ and $x^2 \mapsto -x^2-x+2$. It's a bit hard to find, but easy to check because a straightforward computations shows that if $a$ is a root, then so are $f(a)$ and $g(a)$.

Well this tell us that if $K$ is any subfield of $\Bbb R$ then if $p(x)$ has a root in $K$ then the other two roots are also in $K$. So either $p(x)$ is irreducible or it has three linear factors.

Now, if $K \subset L = K(\sqrt[n] y) \subset \Bbb R$ with $y \in K$ and $n$ prime, then if $p(x)$ splits in $L$ then it was already split in $K$ :

Since $n$ is prime, there is no intermediate subfield between $K$ and $L$, so if $p$ splits in $L$ and is irreducible over $K$ then $L = K(a)$ and $n=3$.

Then $L$ is Galois over $K$ which is impossible because cube root extensions are not Galois :

If $q(a)$ is a root of $Y^3-y$ for some polynomial $q \in K[X]$, then $Y^3-y$ splits into $(Y-q(a))(Y-q(b))(Y-q(c))$ over $K(a)$ (remember that $b=f(a)$ and $c=g(a)$), and we get $(q(a) -q(b))^2 + 3q(c)^2 = 0$, from which we get $q(c)=0$ then $y=0$, and $K=L$, contradiction.

Now if you had a way to obtain $a$ from rationals with the usual operations and taking real roots, then there would be a finite tower of radical extensions $\Bbb Q \subset K_1 \subset \ldots \subset K_n \subset \Bbb R$ with $a \in K_n$. With the above result, if $a \in K_n$ then $a \in K_{n-1}$ and so on, which implies that $a \in \Bbb Q$ in the first place.

Since the roots are not rationals, it is impossible to obtain them from rationals with those operations.

Let us define our cubic as:

$$f(x)=ax^3+bx^2+cx+d=0$$

We consider the polynomial:

$$D(x)=3ax^2+2bx+c$$

We test for when $D(x)=0$, say $x_1,x_2$.

If $x_1,x_2$ don't exist, then $f$ has only one real root.

If $x_1,x_2$ both exist, and $f(x_1),f(x_2)$ have different signs, then $f$ has three real roots.

If either or both of $f(x_1),f(x_2)$ are zero, then $f$ has three real roots, two (or all three if both) being the same.

Otherwise, i.e. $\operatorname{sign}(f(x_1))=\operatorname{sign}(f(x_2))$, there is only one real root.


Elementary question about real analysis Suppose $A \subseteq \mathbb{R}$, and let $x \in [a,b]$. Put

$$ F(x) = \int_{[a,x]} \chi_A dm $$

Then of course we know that $F $ is differentiable for almost all $x \in [a.b]$ and
$$ F'(x) = 1  \; \; \text{for almost all } x \in [a,b] \cap A $$

Can someone explain to me why does it follow that $F'(x) = 1 $ implies that there exists $\epsilon$ and $\delta >0$ such that

$$ 1 \geq \frac{ m( [x , x +h] \cap A )}{h} > 1 - \epsilon \; \; for\; \; 0 < h < \delta$$

and

$$ 1 \geq \frac{ m( [x , x +k] \cap A )}{k} > 1 - \epsilon \; \; for \; \; 0 < k < \delta$$

This is not true. If $A$ is the Cantor set, then $F=0$ and so $F'=0$ everywhere.


Probability Concepts Which of these numbers cannot be a probability?

a) $-0.00001$

b) $0.5$

c) $1.001$

d) $0$

e) $1$

f) $20\%$

Probability can either be expressed in percentages, decimals, proper fractions or numbers between 0 to 1, that is, not likely to happen or most likely to happen. Therefore a and c are not probability. What are the concepts behind this? Thanks in advance!

(Thank you. I found the defining properties of probability which sums up your comments)

Probability has two defining properties:

1.The probability of any event is a number between 0 and 1, or 0 < P(E) < 1. A P followed by parentheses is the probability of (event E) occurring. Probabilities fall on a scale between 0, or 0%, (impossible) and 1, or 100%, (certain). There is no such thing as a negative probability (less than impossible) or a probability greater than 1 (more certain than certain).

2.The sum of all probabilities of all events equals 1, provided the events are both mutually exclusive and exhaustive. If events are not mutually exclusive, the probabilities would add up to a number greater than 1, and if they were not exhaustive, the sum of probabilities would be less than 1.

You have just given the answer! Probability can be $0$ between it and $1$ or $1$. So a) and c) are the answers.
Laws in probability state that sum of all probability equals $1$.


Why won't a series converge if the limit of the sequence is 0? Just thinking about it in terms of logic, shouldn't the series of a sequence whose limit as $n$ approaches infinity is 0 converge?

I know that the $n$th term test for divergence says that if a series is convergent, then the limit of its sequence is 0 and I also know there are some sequences for which it has been "proven" that their series does not converge even though the sequence converges to 0, but I just don't believe these tests. If we stretch $n$ out to infinity and the terms are approaching 0, then how is it possible for the sum of the terms to "overflow" and diverge if the terms are becoming negligibly small?

A very easy counterexample would be
$$ 1, \underbrace{\frac12, \frac12}_{2\text{ halves}},
\underbrace{\frac13, \frac13, \frac13}_{3\text{ thirds}},
\underbrace{\frac14, \frac14, \frac14, \frac14}_{4\text{ fourths}},
\underbrace{\frac15, \frac15, \frac15, \frac15, \frac15}_{5\text{ fifths}},
\ldots $$
This sequence clearly converges to $0$, but if you try to sum it, it should be obvious that it has partial sums as large as you'd like them to be -- so the series diverges.

Try whichever argument you have in mind for believing that the series should converge, and attempt to figure out why it doesn't work for this one.

Just look at the problem from the other end.

Take a function like $f(x) = \sqrt{x}$ or $\log x$ - they grow infinitely with the $x$ growth, though their growth ratio decreases (their first derivatives tend to zero).

Consider the function increments for equal steps of argument, say, for arguments natural:
$$\Delta f = f(n+1) - f(n)\quad \text{ for } n\in\mathbb N^+$$
As the $f$ function grows slower and slower, these deltas get smaller and smaller for growing $n$, and they actually tend to zero: for arbitrarily small $\varepsilon > 0$ there exists $n$ big enough, for which $\Delta f < \varepsilon$.

Now $\Delta f$ defines a sequence convergent to zero, whose partial sums are values of $\sqrt n$, that is a function growing with no bound, hence a divergent series.


In $k[X,Y,Z,W]/(XW−YZ)$, there is no GCD of $xw$ and $xy$ I wanted to find an integral domain in which we do not have the property that any two elements have a greatest common divisor (gcd).

Somewhere on this forum someone claimed that the following quotient ring of the polynomial ring in four variables over a field $k$ is such a ring: $A:=k[X,Y,Z,W]/(XW−YZ)$. I would appreciate some help in proving this. We denote the classes of $X,Y,Z,W$ in $A$ by $x,y,z,w$.

1) $A$ is an integral domain
It suffices to prove that $XW−YZ$ is irreducible in $k[X,Y,Z,W]$. But this holds with the Eisenstein criterion if we consider $XW−YZ$ as an element in $k[Y,Z,W][X]$.

2) The elements $xw$ and $xy$ have no gcd in $A$.
Obviously the idea is to use that (because of the relation $xw=yz$), both $x$ and $y$ are common divisors of $xw$ and $xy$. But I'm having a hard time proving rigorously that there cannot exist a gcd.

Here is a proof I put together using the above arguments of @user26857.

First I state all the lemmas we will use in the proof. We use the notation $(a,b)$ for $\gcd(a,b)$.

Lemma $1$: Let $R$ be an integral domain and $a,b\in R$ two irreducible elements that aren't associates. Then $(a,b)=1$.
Euclid's Lemma: Let $R$ be an integral domain and $a,b,c\in R$. If $(ac,bc)$  exists, then $(a,b) = (ac,bc)/c$.
Claim $2$: $y\nmid x$ in our Ring $A$. In particular, $y$ and $x$ aren't associates.
Claim $3$: $x,y,z,w$ are irreducible in our ring $A$.

Proofs for Euclid's Lemma and Claim $3$ were given by @Bill Dubuque and @user26857 in the comments to my question. Lemma $1$ is trivial.
Proof of Claim $2$:
Suppose $y\mid x$, i.e. that there exists a polynomial $f$ such that $y\cdot f(x,y,z,w)=x$. Then $Y\cdot f(X,Y,Z,W)-X\in (XW-YZ)$. By plugging in $0$ for $Y,Z,W$ we get $0\cdot f(X,0,0,0)-X=0 $, a contradiction.

Finally, we prove that $d=(xw,xy)$ doesn't exist. If $d=(xw,xy)$
then $d=xd′$ (since $x \mid d$) and therefore by Euclid's Lemma $d′=(w,y)$. Now, $y$ and $w$ are both irreducible by Claim $3$ and not associates by Claim $2$ (just replace $x$ with $w$ in proof of Claim $2$), so by Lemma $1$ we get $d′=1$. This means $d=x$. On the other side, $y∣xw(=yz)$ and $y∣xy$, so $y∣x$, a contradiction to Claim $2$ and we are done.

I would suggest you denote the classes of $X,Y,Z,W$ by $x,y,z,w.$ Otherwise, it's difficult to know in which ring you perform computations.

I didn't check the details, but it seems that $A$ may be identified to a subring of $B=k(Y)[X,W]$ (namely $k[X,Y,XW/Y,W]$), by sending $Z$ to $XW/Y$. The latter ring  is a UFD. If $xw$ and $xy$ have a gcd $d=P(x,y,z,w)$, then $D=P(X,Y,XW/Y,W)$ would be a gcd of $XW$ and $XY$ in $B$. Hence, $D$ would be of the form $F(Y)X$, and I'm sure you can derive a contradiction from here.

Otherwise, I can provide you another kind of example: take $A=\mathbb{Z}[i\sqrt{5}]$. This is an integral domain.

One can show that the common divisors of $9$ in $A$ are $\pm 1,\pm 3,\pm 9,\pm (2+i\sqrt{5}), \pm (2-i\sqrt{5})$.

(Hint: solve the equations $\vert z\vert^2=1,3$ or $9$ , for $z\in A.$)

I let you deduce from the latter that $9$ and $3(2+i\sqrt{5})$ have no gcd in $A.$

(HInt: find the possible common divisors among the list above, and check that none of them satisfy the definition of a gcd)


How many $7$- letter words can be formed from the letters in bookkeeper? I was thinking that I would write it out like $\dfrac {10} 3 \dfrac 8 2 \dfrac 6 2 5!$, but I realized that would just show the total possible ways the word can be rearranged. If I were to guess as to what to do I would say that $\dfrac {10} 7$, which translates to $\dfrac {10!} {7!} 3!$.

BOOKKEEPER: one each B,P,R, two each K,O, and three E.

To form a seven letter word (assuming not limited to actual English words) we may select

a triple and two doubles: $\binom{1}{1}\binom{2}{2}\frac{7!}{3!2!^2}$
a triple, double, and two singles: $\binom{1}{1}\binom{2}{1}\binom{4}{2}\frac{7!}{3!\,2!\,1!}$
a triple, four singles: $\binom{1}{1}\binom{5}{4}\frac{7!}{3!\,1!^4}$
three doubles, one single: $\binom{3}{3}\binom{3}{1}\frac{7!}{2!^3\,1!}$
two doubles, three singles: $\binom{3}{2}\binom{4}{3}\frac{7!}{2!^2\,1!^3}$
one double, five singles $\binom{3}{1}\binom{5}{5}\frac{7!}{2!\,1!^5}$
seven singles: $\binom{6}{7}\frac{7!}{1!^7}$  (whut?)

$$\frac{7!}{3!2!^2}+\frac{7!\,12}{3!\,2!}+\frac{7!\,5}{3!}+\frac{7!\,3}{2!^3}+\frac{7!\,12}{2!^2}+\frac{7!\,3}{2!}+0$$

you have a multiset {${1\cdot B, 3\cdot E, 2\cdot K, 2\cdot O, 1\cdot R, 1\cdot P}$} There are 6 letters in this set.

First look at taking one of each letter on then one double (E,K or O) this can be done in $3\choose 1$ ways and the chosen letters can be arranged in $\frac{7!}{2!}$ ways.This give a total of (${3\choose 1}\cdot \frac{7!}{2!}$) ways. Follow on with this by looking at taking 2 doubles, 3 doubles, a double and a triple and 2 doubles and a triple to get your final answer. Note the only possible triple is 'E'. All of these are straight forward with the exception of 2 doubles which will require some thought.


Conservative Measures under a group action (reference request) I was reading a paper and the author define the concept of conservative measure:

Let $(X,\mathcal{B})$ a measurable space and $G$ a group that acts on $X$ by $$G\times X:(g,x)\mapsto T_g(x)$$ where all the functions $T_g$ are measurable.

A measure $\mu$ on $(X,\mathcal{B})$ is called conservative under $G$ if for every $A\in \mathcal{B}$ with $\mu(A) > 0$ exists $g \in G\setminus\{1\}$ with $$\mu(A\cap T_g^{-1}A) > 0$$
I was wondering if this definition is known in the literature or if it is specific to this paper?.

One can notice that if the action of $G$ is given by an invertible measurable function $T$
and the measure $\mu$ is ergodic, then by the von Neumann ergodic theorem (on $L^2(\mu)$) we have that for every $A \in \mathcal{B}$ with $\mu(A) > 0$
$$\frac{1}{n} \sum_{k=0}^{n-1} \mu(A \cap T^{-k}A) = \left\langle 1_A , \frac{1}{n} \sum_{k=0}^{n-1} 1_A \circ T^k \right\rangle_{L^2(\mu)} \stackrel{n \to \infty}{\longrightarrow} \mu(A)^2 > 0$$ So it must exists $k \in \mathbb{N}$ such that
$$\mu(A \cap T^{-k}A) > 0$$
Hence $\mu$ is conservative by the group action $(T^n)_n$.

Edit: More generally if the measure is only invariant by the action $(T^n)_n$ we have the same result thanks to the Poincare Recurrence Theorem .

There is a more general connection between conservative measures and ergodic measures?

Any comment or reference will be appreciated.

Conservativity of the action is not necessarily immediate for infinite measure spaces, and this is where this definition is useful (see Aaronson's book), for example $x\mapsto x+1$ is not conservative on $\mathbb{R}$ with the Haar measure (but certianly measure-preserving).

Maharam's theorem relates conservativity and recurrence in general.

Conservative measures are a well known (old) concept coming from physics, where the system is conservative if it does not "loose" energy (e.g. frictionless). In this case a corresponding volume of the phase space is preserved and if one formalizes this notion we arrive at a conservative measure.
Nevertheless this notion is a bit old-fashioned as people nowadays mostly talk about ergodic measures. Conservative measures have been kind of the predecessors of the notion of ergodic measures.

For a not that old reference (in the case of Z-actions), see the book Invitation to ergodic theory by Silva. Older books (from the 60s and 70s) on ergodic theory usually have this as well.


When is a definition via properties considered valid? How one would define the validity of a definition of an object by its properties?

Little background: I'm trying to implement a kind of framework of mathematics in which the user is not restricted to any formal system, but can define anything at will just like in english but imposing logical constraints on the steps possible during a proof preventing mistakes. This obviously requires to prove that the definition is "valid".

Example: If we consider $G$ to be a group and $+$ its group operation, we can define $0\in G$ s.t. $\forall x\in G : x+0=0+x=x$, we can prove that two sets with such property are equal ($0_1 = 0_1 + 0_2=0_2$) so the definition is solid. Here equality is by definition $x=y :\iff (\forall z:(x\in z \iff y\in z)\land \forall w:(w\in x \iff w\in y)$, so basically in any formula of set theory we can freely replace $x$ with $y$ and vice versa. But what if we don't want to be restricted to set theory? Then what seems reasonable is to require for every definition a list of formulas(not necessarily of set theory) which constitutes the notion of "equality" and then proving that in these formulas they are indeed interchangeable. This issue does not arise when considering objects defined through expressions($3:=2+1$) as they are just directly replaceable short-hand notations.

A partial answer that works in first-order logic is to have a definitional principle that says that if you can prove $\forall x_1, \ldots x_n\exists_{1}y\phi(y, x_1, \ldots, x_n)$, then you can introduce a new $n$-place function symbol $f$, whose defining property is $\forall x_1, \ldots, x_n\phi(f(x_1, \ldots, x_n), x_1, \ldots, x_n)$. The special case when $n = 0$ lets you introduce new constant symbols. For a proof that this definitional principle is conservative (i.e., that it doesn't increase the logical strength of the theory), see Mendelson's Introduction to Mathematical Logic section 2.9.

Definitional principles along these lines are implemented in interactive proof systems such as the various members of the HOL family. The definitional principles are formulated as inference rules (they have one or more antecedent theorems) that make theory extensions (the signature of the language is extended in the theorem that is the succedent of the rule).

If we go into the mathematical realm a "thing which does not exist" is just a set of properties which cannot be satisfied.


Measures: Atom Definitions Let $\Omega$ be a measure space with measure $\mu$.
(Here, a measure is only meant to be countable additive!)

Consider a subset $A\in\Sigma$.

Then according to the wikipedia article it is an atom if:
$$(1)\quad\forall E\subseteq A:\quad\mu(E)<\mu(A)\implies\mu(E)=0\quad(\mu(A)>0)$$
and according to the paper by Johnson it is an atom if:
$$(2)\quad\forall E\in\Sigma:\quad\mu(E\cap A)=0\lor\mu(E^c\cap A)=0\quad(\mu(A)>0)$$

Now, these definitions agree for the really trivial cases:
$$\mu\equiv0$$
$$\mu\equiv\infty$$
but they differ for the less trivial case:
$$\quad\mu(E\neq\varnothing):=\infty,\,\mu(\varnothing):=0$$
namely the atoms are w.r.t. (1) all nonempty subsets whereas w.r.t. (2) only the singletons.

Excluding this pathological case, are the definitions equivalent?
(More precisely, assume there exists a measurable subset $0<\mu(F)<\infty$.)

So far I checked that:
$$(2)\implies(1):\quad\mu(E)<\mu(A)\implies\mu(A\setminus E)>0\implies\mu(E)=0$$
but what about the other direction?

First, as mentioned in the comments excluding the pathological case is not enough:
(Thanks alot @tomasz!!)

Consider the measure space:
$$\mu:\mathcal{P}(\mathbb{N}\cup{\infty})\to\overline{\mathbb{R}}_+:\quad\mu(\varnothing):=0,\;\mu(\{\infty\}):=1,\;\mu(E):=\infty\text{ else}$$
Then the atoms are w.r.t. (1) all nonempty subsets whereas w.r.t. (2) only the singletons
and for both (1) and (2) additionally the point at infinity.

Next, for atoms of finite mass the proof goes as follows:
(Thanks alot @asaf karagila!!)

Assume, it holds (1). Let $E\in\Sigma$ arbitrary.

If $\mu(A\cap E)=0$ we're done.
Otherwise $\mu(A\cap E)=\mu(A)$ by (1). But then $\mu(A\cap E^c)=0$ as $\mu(A)<\infty$ and we're done, too.

Concluding that (2) holds.

Finally, for the sigma-finite case atoms have finite mass and the preceding applies:

Assume $A$ is an atom w.r.t. (1) with $\mu(A)=\infty$.

By sigma-finiteness, there exists a sequence $A_n\uparrow A$ with $\mu(A_n)<\infty$.
By continuity, there exists a natural $n_0\in\mathbb{N}$ with $\mu(A_{n_0})>0$.
By defintion of an atom, this implies $\mu(A_n)=\mu(A)=\infty$.
Contradiction!

Concluding that $\mu(A)<\infty$.


Formal proof that $N$ is the smallest infinite set I wish to write a formal proof of the following statement: For any infinite set $X$, there exists an injection $f:\mathbb{N}\to X$.

I'd like the proof to explicitly use the full axiom of choice (for every family of sets $\{S_\alpha\}$ there exists a family of elements $\{x_\alpha\}$ such that each $x_\alpha\in S_\alpha$). When this was asked before, none of the answers were explicit about where choice is invoked.

Motivation: I'm TAing a course in discrete math and was embarrassed to find that I can't prove this homework question.

A set $X$ is infinite if and only if there is an injection $\phi:X\to X$ that is not a surjection.  In other words, $\phi$ misses out a nonempty subset $S_1$ of $X$. Let, $X_1:=X\setminus S_1$.  $X_1$ must be infinite (because its bijectively related to $X$), hence there is a map $\phi_1:X_1\to X_1$ that is injective but not surjective, misses out nonempty $S_2\in X_1$, etc.  (Induction will make the argument rigorous, and shows that $X_k\cap X_j=\emptyset$ for all $j\leq k$.) Axiom of choice helps "picking" $f(i):=x_i\in S_i$.


If irrational numbers are uncountable, then why did I find this? I understand that irrational numbers are uncountable.  I've seen the proof and it makes perfect sense.  However, I came up with this (most likely false) proof that says that they're countable.  Chances are, I made a mistake and the proof doesn't mean anything, but I just want to be sure.  Here is the proof:

You can't find a solid chunk (range of numbers) that does not contain a rational number.  Which means that irrational numbers are just points on the number line, not lines.  Which means you just need to name all the points to count the irrationals.

What is the mistake here?

In mathematical language, the claim that you can't find a solid chunk of the line (meaning an interval) that does not contain a rational is correct, and it is usually stated that the rationals are dense in the line.

I think your intuition is then that there is one irrational between "neighboring" rationals, which is the sense of saying that the irrationals are points on the line instead of lines.  This is the problem.  Between any two rationals there are an uncountable infinite number of irrationals and a countable infinite number of rationals. When you think of dense sets you need to banish the thought of "neighboring elements" from your mind.  They do not exist because between any two there is another.

The difference between the rationals and the reals as a field is that the reals are complete.  If you think of the interval $[0,1]$ you can make a binary expansion of a number.  The expansion will converge, but without the completeness axiom you don't know that it will converge to a number in your field.  A classic example is $\sqrt 2 -1$.  If you make its binary expansion the approximation by increasing numbers of bits is a Cauchy sequence.  The reals have an axiom that says it converges to a real, while the rationals do not.  Now the classic Cantor proof applies to say the reals are not countable.

Q: "What is the mistake here?"
A: I believe the mistake is assuming that there is a countable number of segments.

In your 'proof', you say each segment (which contained only rational numbers) is terminated at an irrational number.  Since there are an uncountable number of segments, there are an uncountable number of irrational numbers (each one being at the end of its 'companion' segment).

[EDIT: Removed trailing garbage from earlier edit]


The product of any number and zero is zero I don't understand why this proof is valid. The proof just swaps the left and right hand side of the equality. Is this a valid method of proof for equalities? If so, what's the logical intuition behind it?

Theorem: $0⋅a = 0$ for any integer $a$.

Proof: Add zero!

$$\begin{align} 0 + \color{Green}{0⋅a} = 0⋅a & = (0 + 0)⋅a  \\ & = 0⋅a + \color{Green}{0⋅a} \end{align}$$ by using the identity property
twice and then the distributive law. Use cancellation on the first and last piece to get
$0 = 0⋅a$

The problem is not algebra. I don't understand why transforming $0\cdot a =0$ to $0=0\cdot a$ proves this. I mean to prove $a=b$, we transform it to $b=a$, why is it a proof method. What logic does it depend on?

you want to prove

$$0\cdot a = 0$$

assume, where x is any number

$$0\cdot a = x \tag{1}$$

We know that

$$0\cdot a = (0+0)\cdot a$$

so

$$(0+0)\cdot a = x$$

$$0\cdot a + 0\cdot a = x\tag{2}$$

subtracting $(2)-(1)$

$$(0\cdot a + 0\cdot a) - (0\cdot a) = x-x = 0$$

that is

$$0\cdot a + 0\cdot a = 0\cdot a$$

so

$$0\cdot a = 0\cdot a - 0\cdot a$$

as a result

$$0\cdot a = 0$$

if you want to prove 0.a=0 you need not to prove it through algebra.
consider following 2.4=8(dot means multiply)
what actually we have done is 2+2+2+2=8
multiplication itself do not have any significance
it can be said as short form of addition or shortcut of addition for adding large no.
so what will be value of 0.4=?
it is 0+0+0+0+0=0
similarly 0.a is 0+0+0...a


Why multiply a matrix with its transpose? This might be a very stupid question, but I do not seem to understand why I would multiple a matrix with its transpose. I am not a mathematician, but I am very interested in understanding the practical usage of equations:

Imagine I have three products sales Apple, Orange and Pear for the last 3 days in a matrix form called A:
$$ A=
\begin{bmatrix}
        Apple & Orange & Pear \\
        10 & 2 & 5 \\
        5 & 3 & 10 \\
        4 & 3 & 2 \\
        5 & 10 & 5 \\
        \end{bmatrix}$$

What will $AA^{\rm T}$ tell me?

I have seen this long answer link: Is a matrix multiplied with its transpose something special?, but I did not get it at all.

I see that a lot of equations use the product $AA^{\rm T}$ and I really hope that someone will give a very simple answer.

In your case, $AA^T$ just sitting on a park bench doesn't tell you anything of great interest.

Hyprfrcb's answer talks about units. One of the elements has units of squared apples. Another has units of pear-oranges. Another has units of orange-pears! This by itself should be a red flag that values don't mean anything by themselves.

It can be a means to get to a least-squares solution if you were looking to model, say, how much of each fruit you'd expect to sell on a particular day. (This was one of the answers in the linked question.)

But by itself? It's just a fruit hybridization experiment gone terribly wrong.

I've been trying to figure this out my myself recently, and I think I understand it.

John is right, in your example, it doesn't make sense.  But I'll attempt to explain it in a simpler example than what Brethlosze said... (bare with me and please tell me where I am wrong).

Let's say you are measuring something like acceleration.  But your measurement of acceleration always has an error.  Each time you measure acceleration 'a', you have an error of +- 0.1*a.

This error propagates as you continue your calculations.  If you want to calculate velocity from that acceleration, then that acceleration error is going to propagate into the velocity calculations.  If you continue to calculate distance, than that acceleration error is going to also continue to propagate into the distance calculation.

Now this is a very simple example, but it shows the relationship between this measurement (and its error) and all the calculates values (state variables).  Now imagine a more complicated scenario where you need more than 1 measurement to get the desired state variable that you are looking for.  You will have an equation with 2 variables to show the relationship between the measurements and the state variable.  What if you are looking for 4 state variable that each are composed of 2 other state variables that those in itself take 2 or 3 or 4 or however many measurements to calculate.  Now you can use a matrix to show the relationships between all these measurements and state variables.

So now, if we transpose the matrix and multiply it by the original matrix, look at how those equations in the matrix are being multiplied with all the other variables (and itself).  Try the math of a simple 2x2 times the transpose of the 2x2.  This is the covariance.  Wikipedia: In probability theory and statistics, covariance is a measure of the joint variability of two random variables.  Too me, its like the covariance matrix mixes all the variables and measurements in every way possible to show how all of them vary against and with each other.

This is very important in inertial measurement units (IMUs).  There are many error states (sometimes more than a dozen) in an IMU in order to calculate the state variables (position, velocity, body orientation, etc.).  These error states and state variables are defined in a matrix.  When you find the covariance of the matrix, you can define the joint variability between all these error states and state variables.  This is important in order to know how these errors and measurements can affect your accuracy of position, velocity, and body orientation.  By knowing your covariance, you pretty know how accurate your IMU is at defining your position, velocity, orientation, etc.


How to remove divergence from any vector field? I am trying to simulate fluids in 2D. I have a vector field which represents the flux; a continuous function that maps 2D positions to 2D velocity. I do some simulation steps and the flux changes. The new flux may have non-zero divergence, which is undesired, because I want to simulate non-compressible fluid. Therefore I want to have zero divergence at all points by changing the flux as little as possible. How can I calculate this most optimally and accurately?
My failed attempts:

I found Helmholtz decomposition which can be useful, but I don't think it tries to stay as close as possible to the original vector field with non-zero divergence.

I could not get this to work. The formula there does not even make sense to me.

The solution that worked for me has been Hodge decomposition.
I have found exactly what I was looking for here. It even has an example implementation.
In hindsight, I should have asked this question in gamedev. What I have been trying to achieve was similar to plasma pong.


Convex set definition Munkres defines thus: Given an ordered set $X$ a subset $Y$ of $X$ is convex in $X$ if for each pair of points $a<b$ of $Y$ the entire interval $(a,b)$ of points of $X$ lies in $Y$. But again in Connected subspaces of real line section he recalls the definition of convex but now using a closed interval $[a,b]$. Is it $(a,b)$ in the definition or $[a,b]$?

As noted in the comments, when $a$ and $b$ are distinct points in $Y$ so knowing that $a,b\in Y$ tells you that both $(a,b)$ and $[a,b]$ lie in $Y$ and so there's no particular need to distinguish between the two.
There is one subtlety though: if $a=b$ in $Y$ then $(a,b)=\emptyset$ and $[a,b]=\{a\}$ (or $\{b\}$ if you prefer) and then it may make a difference whether you talk about open or closed intervals.  In this case $a$ is referred to as an extreme point of $Y$ and it becomes interesting to consider if a set if the closed convex hull of all of its extreme points (for example).
Speaking of examples:  the unit ball in $\ell_2$ is convex and possesses no extreme points, whereas the unit ball in $\ell_1$ is convex and does possess extreme points as you should be able to easily verify.


How many solutions to this under-determined equation I have a system of equations that looks as such:

$$A=\begin{pmatrix}-3&-3&2\\ \:-9&-9&6\end{pmatrix}$$
$$b=\begin{pmatrix}2\\ \:-4\end{pmatrix}$$

How many solutions exist to this matrix? I tried to answer this using RREF, which gives me:

=\begin{pmatrix}-9&-9&6\\ 0&0&0\end{pmatrix}

Thereby, since I have 2 free variables, I have a plane of infinitely many solutions. Is this fair?

Your notation is vague. If you mean "How many solutions are there to
$$A\begin{bmatrix}x\\y\\z\end{bmatrix}=b?"$$ the answer is "none". Sophisticated reason: rank of coefficient matrix=1 $\ne$2=rank of augmented matrix. Basic reason: multiplication of 1st given equation by 3 gives $$-9x-9y+6z=6$$ a contradiction to the second given equation which is $$-9x-9y+6z=-4.$$

No, there is no solution. Let $A\in\mathbb R^{m\times n}$ and $b\in\mathbb R^m$. The main idea is to reduce the augmented coefficient matrix $(A\vert b)$ to row echolon form via Gaussian elimination. This does not change the solution set. We get
$$
\begin{pmatrix}
a_{11} & a_{12} & \dots & a_{1k} & \dots & a_{1n} & b_1\\
0 & a_{22} & \dots & a_{2k} & \dots & a_{2n} & b_2\\
\vdots & \ddots & \ddots & \vdots & \ddots & \vdots & \vdots\\
0 & \dots & 0 & a_{kk} & \dots & a_{kn} & b_k\\
\vdots & \ddots & \vdots & 0 & \dots & 0 & \vdots\\
0 & \dots & 0 & 0 & \dots & 0 & b_m
\end{pmatrix}
$$
Now, if $b_i$, $k+1\leq i\leq m$ is non-zero, then the linear equation system has no solution. This is equivalent to saying that the rank of $A$ is smaller than the rank of $(A\vert b)$. So, we have at least one solution if
$$\operatorname{rank}(A) = \operatorname{rank}(A\vert b).$$
It is exactly one solution if
$$\operatorname{rank}(A) = n$$
and infinitely many solutions if
$$\operatorname{rank}(A) = r < n.$$
In the latter case the solution set has the dimension $n-r$.


how does expectation maximization work? I'm reading a tutorial on expectation maximization which gives an example of a coin flipping experiment (the description is at http://www.nature.com/nbt/journal/v26/n8/full/nbt1406.html?pagewanted=all). Could you please help me understand where the probabilities in step 2 of the process (i.e. in the middle of part b in the below illustration) come from? Thank you.

EM starts with an initial guess of the parameters. 2. In the E-step, a probability distribution over possible completions is computed using the current parameters. The counts shown in the table are the expected numbers of heads and tails according to this distribution. 3. In the M-step, new parameters are determined using the current completions. 4. After several repetitions of the E-step and M-step, the algorithm converges.

These are the likelihoods of the corresponding set of $10$ coin tosses having been produced by the two coins (using the current estimate for their biases) normalized to add up to $1$. The estimated probability of $k$ out of $10$ tosses of coin $i$ ($i\in\{A,B\}$) yielding heads is

$$p_i(k)=\left({10\atop k}\right) \theta_i^k (1-\theta_i)^{10-k}\;.$$

The binomial coefficient is the same for both coins, so it drops out in the normalization, and only the ratio of the remaining factors determines the result.

For instance, in the second row, we have $9$ heads and $1$ tails. Given the current bias estimates $\theta_A=0.6$ and $\theta_B=0.5$, the factors are

$$\theta_A^9 (1-\theta_A)^{10-9}\simeq0.004$$

and

$$\theta_B^9 (1-\theta_B)^{10-9}\simeq0.001\;,$$

resulting in the numbers

$$\frac{0.004}{0.004+0.001}=0.8$$

and

$$\frac{0.001}{0.004+0.001}=0.2$$

in the second row.

Let o1 <5H,5T>,o2 <9H,1T>,o3 <8H,2T>,o4 <4H,6T>,o5 <7H,3T> be five observation respectively.
So each of these observation can either come from coin A with probability 0.5 or from coin B with probability 0.5
If we select coin A , pA_Head denote the probability of Head from coin A , which is initialized as 0.60
If we select coin B , pB_Head denote the probability of Head from coin B , which is initialized as 0.50

Now we have to find out weather these five observation has come from Coin A or Coin B.
i.e we have to find P(A | o2) and P (B | o2)
Now, P(A|o1) = P(A,o2) / P(o2)

o2 can come from either Coin A or Coin B, So P(o2) = P(o2,A) + P(o2,B) [ Sum rule of Probability]
P(A,o2) = P(A) * P(o2|A)
P(B,o2) = P(B) * P(o2|B)
P(o2|A) = (9+1) C 9 * pA_head^(9) * (1-pA_head)^1
P(o2|B) = (9+1) C 9 * pB_head^(9) * (1-pB_head)^1
P(A|o2) = [P(A) * 10 C 9 * pA_head^(9) * (1-pA_head)^1 ] / [ {P(A) * 10 C 9 * pA_head^(9) * (1-pA_head)^1 } +  {P(B) * 10 C 9 * pB_head^(9) * (1-pB_head)^1  } ]
Since P(A) = P(B) = 0.5 , Hence P(A|o2) =  pA_head^(9) * (1-pA_head)^1  / [ pA_head^(9) * (1-pA_head)^1 + pB_head^(9) * (1-pB_head)^1  } ] = (0.6^9 *0.4 ) / [ (0.6^9 * 0.4^1)  + ( 0.50^9 * 0.50^1)]


finding all composition series of ${\rm sym} \ (4)$ How can I find all composition series of ${\rm sym}\ (4)$ ? I think first I have to find all maximal normal subgroups. But how ? Thanks.

this isn't correct, since {(1),(12)(34)} is only a subgroup of A4, but isn't normal,
since a normal subgroup is a union of complete conjugacy classes, and elt (12)(34)
isn't a complete c.class.

or, try conjugation of {(1),(12)(34)} by elt (123) (which is even) and you get
{(1),(14)(23)} not ={(1),(12)(34)}.

of course {(1),(12)(34)}⊲{(1),(12)(34),(13)(24),(14)(23)}
is true, since {(1),(12)(34),(13)(24),(14)(23)} is abelian, and every subgroup of
an abelian group is Normal, but {(1),(12)(34)} isn't normal in A4


$\lim_{n \to \infty}f_n(x_n)=f(x)$ if $f_n \to f$ and $x_n \to x$? $f_1,f_2,...$ is a sequence of continuous functions defined on $\mathbb R$ such that $\forall x \in \mathbb R$ $\lim_{n \to \infty}f_n(x)=f(x)$, and $\{x_n\}\subseteq \mathbb R$ is a sequence converging to $x$.

Can we prove that $\lim_{n \to \infty}f_n(x_n)=f(x)$?

My attempt (motivated by Show that $f_n(x_n) \to f(x).$)

\begin{align*}|f_n(x_n)-f(x)|&=|f_n(x_n)-f(x_n)+f(x_n)-f(x)|\\
& \leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|\\
& \leq \sup_{t \in \mathbb{R}}  |f_n(t)-f(t)|+|f(x_n)-f(x)|\\
& \to 0\end{align*}

To complete the proof we need to show that $\sup_{t \in \mathbb{R}}  |f_n(t)-f(t)|  \to 0$.

Using the definition of convergence, we know that $$\forall \epsilon\, \forall t\, \exists n^*(t) \text{ such that }|f_n(t)-f(t)|<\epsilon\quad \forall n \ge n^*(t)$$
We could complete the proof by showing that the $n*(t)$ are bounded above; that is, we need to show that there exists an integer $n^*$ such that $$n^*=\max_{t \in \mathbb R}n^*(t)$$ since then we will have $$\forall t\, \forall n \ge n^*\quad|f_n(t)-f(t)|< \epsilon$$ and therefore $$\forall n \ge n^*\quad\sup_{t \in \mathbb{R}}  |f_n(t)-f(t)| \le \epsilon$$

How can we show that $n^*$ exists?

This is only true if $f_n(x)$ converges uniformly to $f(x)$. In fact, it will break for any situation where $f_n(x)\rightarrow f(x)$ pointwise but not uniform on an interval $[L,R]$. In other words, there exists $\epsilon>0$ s.t. for all $N>0$, there exists a point $a$ such that $|f_n(a)-f(a)|>\epsilon$. Pick a sequence of $\epsilon_n$ and increasing $N_n$. This gives a sequence of $a_n$. By Bolzano-Weierstrass, pick a convergent subsequence of $a_n$ that converges, and call it $x_n$. Now follow through the proof.

Example: $f_n(x)=2^nx$ when $0\leq x\leq 2^{-n}$, $f_n(x)=2-2^nx$ when $2^{-n}\leq x\leq 2^{-n+1}$ and 0 elsewhere. Then $f_n(x)\rightarrow 0$. Yet pick $x_n=1/2^n$. Then $x_n\rightarrow 0$ but $f_n(x_n)=1$. Essentially you have a bump which is clamped to 0 on either side of $[0,1]$, it's squishing to the left but there's always a sequences which rides on the value 1.

Alternatively, we may assume that the family $\{f_n\}_{n=1}^\infty$ is equicontinuous at $x$.


What do angle brackets ($\langle\rangle$ ) mean in mathematics/statistics (autocorrelations)? Okay, so the logarithmic return on a stock is given by:

$$r_τ (t) = \ln P(t+τ) - \ln P(t),$$ where τ is the interval of time.

I have no problem calculating that. My question comes to the following formula:

$$ρ(T) \sim 〈r_τ (t+T) \cdot r_τ (t)〉$$

This is supposedly the autocorrelation function of log-returns. What's the deal with the brackets?

In Bra-ket notation, we denote an inner product of two vectors $|u\rangle,\,|v\rangle$ as $\langle u|v\rangle$, where the map from vectors $|v\rangle$ in the inner product space to scalars $\langle u|v\rangle$ is a linear map labelled $\langle u|$. (The set of such linear maps is a vector space called the dual space; its elements are bras $\langle u|$, whereas the original space has ket elements $|v\rangle$.)

If the matrix $M$ keeps $|v\rangle$ in the same vector space by mapping it to $M|v\rangle$, $\langle u|M|v\rangle$ is not only an inner product; if $|u\rangle,\,|v\rangle$ are elements of some basis, this inner product is a matrix element in that basis.

In quantum mechanics, to each observable $O$ there corresponds an $\widehat{O}$ such that in the pure state $|\psi\rangle$ the expectation of $O$ is $\langle\psi|\widehat{O}|\psi\rangle$. If the state can be taken for granted, we abbreviate this as $\langle\widehat{O}\rangle$.

Even with classical probability, we can express the covariance of two random variables as an inner product on a vector space: namely, a quotient space of equivalence classes of finite-variance random variables. I go through the details here. Then $A,\,B$ has covariance $\langle AB\rangle-\langle A\rangle\langle B\rangle$, but the second term can be deleted if at least one of $A,\,B$ has mean $0$.

If variables have variance $1$, their covariance is also their correlation. Autocorrelation is one special case.

A clear and concise definition I found in Wolfram MathWorld which states a below,

Its history is discussed by Professor Brailsford in Computerphile youtube channel here.

Another question was asked here which is more in the context of statistics and can be helpful to understand.


derivative of cost function for Logistic Regression I am going over the lectures on Machine Learning at Coursera.

I am struggling with the following. How can the partial derivative of

$$J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}y^{i}\log(h_\theta(x^{i}))+(1-y^{i})\log(1-h_\theta(x^{i}))$$

where $h_{\theta}(x)$ is defined as follows

$$h_{\theta}(x)=g(\theta^{T}x)$$
$$g(z)=\frac{1}{1+e^{-z}}$$

be $$ \frac{\partial}{\partial\theta_{j}}J(\theta) =\sum_{i=1}^{m}(h_\theta(x^{i})-y^i)x_j^i$$

In other words, how would we go about calculating the partial derivative with respect to $\theta$ of the cost function (the logs are natural logarithms):

$$J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}y^{i}\log(h_\theta(x^{i}))+(1-y^{i})\log(1-h_\theta(x^{i}))$$

The reason is the following. We use the notation:

$$\theta x^i:=\theta_0+\theta_1 x^i_1+\dots+\theta_p x^i_p.$$

Then

$$\log h_\theta(x^i)=\log\frac{1}{1+e^{-\theta x^i} }=-\log ( 1+e^{-\theta x^i} ),$$ $$\log(1- h_\theta(x^i))=\log(1-\frac{1}{1+e^{-\theta x^i} })=\log (e^{-\theta x^i} )-\log ( 1+e^{-\theta x^i} )=-\theta x^i-\log ( 1+e^{-\theta x^i} ),$$ [ this used: $ 1 = \frac{(1+e^{-\theta x^i})}{(1+e^{-\theta x^i})},$ the 1's in numerator cancel, then we used: $\log(x/y) = \log(x) - \log(y)$]

Since our original cost function is the form of:

$$J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}y^{i}\log(h_\theta(x^{i}))+(1-y^{i})\log(1-h_\theta(x^{i}))$$

Plugging in the two simplified expressions above, we obtain
$$J(\theta)=-\frac{1}{m}\sum_{i=1}^m \left[-y^i(\log ( 1+e^{-\theta x^i})) + (1-y^i)(-\theta x^i-\log ( 1+e^{-\theta x^i} ))\right]$$, which can be simplified to:
$$J(\theta)=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\theta x^i-\log(1+e^{-\theta x^i})\right]=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\log(1+e^{\theta x^i})\right],~~(*)$$

where the second equality follows from

$$-\theta x^i-\log(1+e^{-\theta x^i})=
-\left[ \log e^{\theta x^i}+
\log(1+e^{-\theta x^i} )
\right]=-\log(1+e^{\theta x^i}). $$ [ we used $ \log(x) + \log(y) = log(x y) $ ]

All you need now is to compute the partial derivatives of $(*)$ w.r.t. $\theta_j$. As
$$\frac{\partial}{\partial \theta_j}y_i\theta x^i=y_ix^i_j, $$
$$\frac{\partial}{\partial \theta_j}\log(1+e^{\theta x^i})=\frac{x^i_je^{\theta x^i}}{1+e^{\theta x^i}}=x^i_jh_\theta(x^i),$$

the thesis follows.

Notice that,

$$\frac{\partial}{\partial\theta_j}y_i\theta x^i = \frac{\partial}{\partial\theta_j}y_i(\theta_0 + \theta_1x^i_1 + ... + \theta_jx^i_j)= $$
in this $\partial\theta_j$ order derivative, $y_i$ is a constant, so
$$=y_i\frac{\partial}{\partial\theta_j}(\theta_0 + \theta_1x^i_1 + ... + \theta_jx^i_j)=$$
because it is a linear model ($\frac{\partial}{\partial \theta}k\theta = k$), so
$$=y_i(0 + x^i_1 + ... + x^i_j)=$$
$$=y_ix^i_j$$
Finally,
$$\frac{\partial}{\partial\theta_j}y_i\theta x^i = y_ix^i_j$$


Find a particular solution $y_p$ for $5y''+8y'+8y=\cos^{2}(x)$ I'm getting $A = \frac{-3}{68}$ and $B = \frac{1}{34}$ for the guess $y_p = A\cos(2x)+B\sin(2x)$ using the method of undetermined coefficients.  Apparently this answer is wrong (with $A$ and $B$ plugged in).  Can anyone help me find the right one?

Thanks!

Note that the right hand side is ${\cos ^2}(x)$, so want to choose something similar as our particular solution. But $\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x) = 2{\cos ^2}(x) - 1$ so
$${\cos ^2}(x) = \frac{1}{2} + \frac{1}{2}\cos (2x)$$
and we can look for a particular solution of the form
$${y_p} = A + B\sin (2x) + C\cos (2x).$$

Then you need to use the initial conditions to determine the coefficients or plug ${y_p}$ into the differential equation, collect like terms, and equate the coefficients.

See below for the details. Your equation is

$$\left( {5{D^2} + 8D + 8} \right)y = {\cos ^2}\left( x \right)$$

Make that into

$$\left( {5{D^2} + 8D + 8} \right)y = \frac{{1 + \cos \left( {2x} \right)}}{2}$$

and apply $D$

$$\eqalign{
  & \left( {5{D^2} + 8D + 8} \right)y = \frac{{1 + \cos \left( {2x} \right)}}{2}  \cr
  & \left( {5{D^2} + 8D + 8} \right)y' =  - \sin \left( {2x} \right)  \cr
  & \left( { - 5{D^2} - 8D - 8} \right)y' = \sin \left( {2x} \right) \cr} $$

Thus, we have $$\eqalign{
  &  - 5{D^2} - 8D - 8 = \left( { - 5{D^2} - 8} \right) - 8D  \cr
  & \varphi \left( D \right) =  - 5D - 8  \cr
  & \xi \left( D \right) =  - 8 \cr} $$

and then

$$y' = \frac{{3\sin 2x + 4\cos 2x}}{{100}}{\text{ }}$$

and you solution is $$y = \frac{{ - 3\cos 2x + 4\sin 2x}}{{200}} + C{ _1}$$

I leave it to you to find $C_1$.

I know the following is not completely rigorous, but it has its justifications.

As a generalization of Tapu's solution, suppose we're given

$$\phi(D)=\sum_{k=0}^m a_kD^k$$

where $D=\dfrac d {dx}$. Since $D^2\sin(ax)=-a^2\sin (ax)$, we have that

$$\phi(D^2)\sin(ax)=\phi(-a^2)\sin(ax)$$

But we may write

$$\phi (D) = \sum\limits_{k = 0}^{m'} {{a_{2k}}{D^{2k}}}  + \sum\limits_{k = 0}^{m'} {{a_{2k + 1}}{D^{2k + 1}}} $$

whence

$$\phi (D) = \varphi \left( {{D^2}} \right) + D\xi \left( {{D^2}} \right)$$

thus an equation of the form

$$\phi (D)y = \sin ax$$

becomes $$\left[ {\varphi \left( {{D^2}} \right) + D\xi \left( {{D^2}} \right)} \right]y = \sin ax$$

We apply the "conjugate" of the operator to the equation, to get $$\eqalign{
  & \left[ {{\varphi ^2}\left( {{D^2}} \right) - {D^2}{\xi ^2}\left( {{D^2}} \right)} \right]y = \left[ {\varphi \left( {{D^2}} \right) - D\xi \left( {{D^2}} \right)} \right]\sin ax  \cr
  & \left[ {{\varphi ^2}\left( {{D^2}} \right) - {D^2}{\xi ^2}\left( {{D^2}} \right)} \right]y = \varphi \left( { - {a^2}} \right)\sin ax - a\xi \left( { - {a^2}} \right)\cos ax  \cr
  & \left[ {{\varphi ^2}\left( {{D^2}} \right) - {D^2}{\xi ^2}\left( {{D^2}} \right)} \right]y = \alpha \sin ax + \beta \cos ax  \cr
  & \Phi \left( {{D^2}} \right)y = \alpha \sin ax + \beta \cos ax \cr} $$

Assuming  $$\Phi \left( { - {a^2}} \right) \ne 0$$ we may write $$\frac{{\sin ax}}{{\Phi \left( { - {a^2}} \right)}} = \frac{{\sin ax}}{{\Phi \left( {{D^2}} \right)}}$$ so that

$$\eqalign{
  & y = \frac{{\alpha \sin ax + \beta \cos ax}}{{\Phi \left( { - {a^2}} \right)}}  \cr
  & y = \frac{{\varphi \left( { - {a^2}} \right)\sin ax - a\xi \left( { - {a^2}} \right)\cos ax}}{{{\varphi ^2}\left( { - {a^2}} \right) + {a^2}{\xi ^2}\left( { - {a^2}} \right)}} \cr} $$

For the cosine, you get $$y = \frac{{\varphi \left( { - {a^2}} \right)\cos ax + a\xi \left( { - {a^2}} \right)\sin ax}}{{{\varphi ^2}\left( { - {a^2}} \right) + {a^2}{\xi ^2}\left( { - {a^2}} \right)}}$$

As an example, consider

$$(D^2-3D+2)y=\sin 3x$$

Then $\phi(D)=D^2-3D+2=D^2+2-3D=\varphi(D^2)-D \xi(D^2)$ where $\varphi(D)=D+2$ and $\xi(D)=-3$. Thus $$y = \frac{{ - 7\sin 3x - 3\left( { - 3} \right)\cos 3x}}{{49 + 9{{\left( { - 3} \right)}^2}}} = \frac{{9\cos 3x - 7\sin 3x}}{{130}}$$


Calculate probability of an infinite geometric series Suppose in a single round, I win with probability p, and you win with probability
  1 − p. We play repeatedly and keep score until one of us is two ahead (for example
  a score of 2 − 0, 3 − 1, 2 − 4, etc.). What is the probability that I win overall?

I have been working through this probability problem. But I am not sure if my method is the most efficient or even may not be even correct and was wondering if someone could advise if I am carrying this out correctly, and if there more effective way of approaching this type of question?

My method for calculation

If I have understood the above question correctly, the only way to win is if you I am 2 ahead?

So one way I could win two time in a row would if I have a probability of $p^2$.

But if I get a $p$ then a $q$ or the other persons throws and $q$ then $p$ we have drawn, so to me this is like starting again i.e $0-0$ again so for me two win now I would need the following $pqpp=p^3 q$ or if I draw again then I would need to win the following $pqpqpp=p^4 q^2$ and so fourth.

As probabilities are independent (each future try is not effected by the previous) then I can add the following probabilities to find my overall chance of winning.

$P(Over \:all \:win)=p^2+p^3q+p^4q^2+.....=p^2+p^3(1-p)+p^3(1-p)^2+....$

so now using the infite geometric sum $S=\frac{p^2}{1-p(1-p)}=\frac{p^2}{1-p+p^2}$.

So my probability for winning over all is:

$$P(Over \:all \:win)=\frac{p^2}{1-p+p^2}$$

I am bit rusty with probability as it something I don’t do too often. But as mentioned is there more efficient way of approaching this type of question, as I was thinking of extending this question and exploring if I had to be 3 ahead what would the over all win be.

The case is random walk start at 0, reach 2 before -2. Recall Gambler's ruin
. The proof if you learned martingale is trivial, otherwises you can use $P_k=pP_{k+1}+(1-p)P_{k-1}$ and bound condition prob  0,1 to derive.


Existence of Nested Countable Neighborhood Basis I'm stuck on an intro to point-set topology homework problem. I'm given that $(X, \mathcal T)$ is a topological space and $p\in X$ has a countable neighborhood basis. I need to then show that $p$ has a nested countable neighborhood basis. The definition of "nested" we are using is: If $\{U_i\}_{i \in \mathbb N}$ is a collection of subsets of a space $X$, we say that $\{U_i\}_{i \in \mathbb N}$ is nested if $U_{i+1} \subset U_i$ for all $i \in \mathbb N$.

Here is my work so far:

$p$ has a countable neighborhood basis $\implies \exists \{ U_\alpha \}_{\alpha \in \lambda}: \lambda \subseteq \mathbb N,$ $p \in U_\alpha \in \mathcal T$ $\forall \alpha \in \lambda$, and $\forall U \in \mathcal T$ with $p \in U$, $\exists \alpha \in \lambda: U_\alpha \subseteq U$. I am now trying to define a nested countable neighborhood basis for $p$ as the sequence of open sets $\{V_n\}_{n \in \mathbb N}$. Any tips on how to do this? Here's what I tried to do:

Define $\{V_n\}_{n \in \mathbb N}$ recursively. First case: $V_1 = U_\alpha$ for some $\alpha \in \lambda$.

Then for $k \geq 1,$ we define $V_{k+1}$. We know that $V_k$ is an open set containing $p$. Let $W$ be an open set containing $p$ such that $W \subset V_k$. Then $W\cap V_k$ is another open set containing $p$, so $\exists \beta \in \lambda: U_\beta \subseteq W \cap V_k \subset V_k$. Define $V_{k+1} = U_\beta$.

We have that $\{V_n\}_{n \in \mathbb N}$ is countable and nested as defined, but is it a neighborhood basis for $p$? We know that $p \in V_n \in \mathcal T$ $\forall n \in \mathbb N$, but do we know that $\forall U \in \mathcal T$ with $p \in U$, $\exists n \in \mathbb N: V_n \subseteq U$? Or maybe I have defined $\{V_n\}_{n \in \mathbb N}$ incorrectly?

While this approach will mostly work (there is some difficulty in defining $V_{k+1}$ from $V_k$ since you really don't say how the open set $W$ comes into being), you are working much too hard.  You are not asked to show that some subcollection of $\{ U_\alpha \}_{\alpha \in \lambda}$ is a countable nested neighbourhood basis; you are only asked to show that one exists, perhaps by constructing a new countable neighbourhood basis (which happens to be nested) from the given one.

I'll leave you with one final hint: the intersection of finitely many open sets is open.

Actually, if a topological space X has a countable neighborhood base at a point x in X for example, say {U(n)} n in N, then it is easy to construct a collection of neighborhoods with desired properties. We can take V(1)=U(1), V(2)=U(1)∩U(2), V(3)=U(1)∩U(2)∩U(3), ..., V(n)=U(1)∩....∩U(n), and so on.


arranging 2 blue balls, 2 red balls and 1 green ball How many ways are there to arrange 2 blue balls, 2 red balls and 1 green ball?

My Answer: $$\frac{5!}{2!*2!}$$

If this is incorrect then please help me understand where I am wrong.
But if this is correct then please help me with this:

"A bag holds $4$ red marbles, $5$ blue marbles, and $2$ green marbles. If $5$ marbles are selected one after another without replacement, what is the probability of drawing $2$ red marbles, $2$ blue marbles, and $1$ green marble?"

Correct Answer:
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}\left(\frac{4}{11}\right)\left(\frac{3}{10}\right)\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)\left(\frac{2}{7}\right) = \frac{20}{77}$$

why does it have this?
$$\binom{5}{2}\binom{3}{2}\binom{1}{1}$$
instead of this:
$$\frac{5!}{2!*2!}$$

for more info, you can refer to my pervious question here:
Basic combinations logic doubt in probability

It is because they are in fact equal.$$\begin{align}\binom 5 2\binom 32\binom 11&=\dfrac{5!}{2!~3!}\dfrac{3!}{2!~1!}\dfrac{1!}{1!~0!}\\[1ex]&=\dfrac{5!}{2!~2!~1!}\end{align}$$

Some just prefer presenting such expressions as binomial coefficients as it reflects the logic used to derive the expression.

I rather the multinomial form. $$\binom 5{2,2,1}$$Use whatever you find most comfortable.

This is called The Mississippi Formula and yes you are correct! Basically you are getting rid of the "overcounts" by dividing by 2! 2!


Motivation for different mathematics foundations I've been studying an introductory book on set theory that uses the ZFC set of axioms, and it's been really exciting to construct and define numbers and operations in terms of sets. But now I've seen that there are other alternatives to this approach to construct mathematics like New Foundations (NF), category theory and others.

My question is, what was the motivation to look for an alternative for ZFC? Does ZFC have any weakneses that served as motivation? I know there are some concepts that don't exist in ZFC like the universal set or a set that belongs to itself. So I thought that maybe some branches of mathematics needed these concepts in order to develop them further, is that true?

Well. That depends on whom you might ask this.

Set theory might be inconsistent. In particular $\sf ZFC$ and its extension by large cardinal axioms. It's a nontrivial thing, to feel safe with these theories, and it takes a lot of practice and time until you understand that $\sf ZFC$ is self-evident to some extent, and [some] large cardinal axioms are somewhat self-evident as well.

But of course, we might be tricked. Crooks have been known to seem honest, until you're left with a big heap of nothing in your hands. That is why Nigerian royalty is going to have a very hard time emailing people around the world.

If that happens, we need to ask ourselves where the problem lies. Is it in one of the axioms, or maybe specifically in the existence of infinite sets? Will a slightly weaker set theory (e.g. $\sf ZC$) work better, or maybe we have to resort to arithmetic theories to fix things?
Our grasp of things is usually largely inconsistent,1 but we do like to think in "types". So working inside different categories when needed, or working with some type theory or another. A lot, and I mean a lot, of people will twitch when you ask them what are the elements of $\pi$. Or whether or not $\frac13$ is a subset of $e$.

Of course, those who have a firm understanding of this know that this is a question of implementation, and this is like asking whether or not the machine code of one implementation of an algorithm is the same or different of another. But people don't think about it this way, although in a way they kinda do.

Instead, people focus on their math, and they just remember (or more accurately: they don't) that you can formalize this in terms of set inside set theory.
$\sf ZFC$ is incomplete. That doesn't sound like a big deal. But it is, in a deep sense. We might half expect a foundational theory to be able and decide all sort of things about the mathematical universe. Enough so most, or any, questions we might have can be answered there. Again, those who study foundations long enough should realize this is not the goal of a foundational theory, but it is somewhat expected.

And some people feel very awkward with this incompleteness. They will also have a problem with arithmetic based foundations, since that is an incomplete foundation also. But it's true that in $\sf ZFC$ those incompleteness phenomenon are far more gaping, and as soon as you leave the term "countable" and enter the "uncountable", more or less everything becomes open and independent.

This poses an actual problem. Imagine that half the functional analysts would work in one set theory (say $V=L$), and the other half in another (say $\sf ZFC+PFA$). It would create some strange discrepancies which will eventually tear the field up. And the same goes to everything else, except set theory, whose main occupation is these different axioms.

But that's just counting three strikes against $\sf ZFC$. And I cannot, in good conscience finish my post like this. So let's balance things as to why $\sf ZFC$ is a good foundation.

Set theory, and in particular $\sf ZFC$ is quite self evident. When I was a starting masters student, I attended a course by one of the generation's greatest mathematician and asked him about the axioms of $\sf ZFC$ and he said that a good foundational theory is one whose axioms you don't feel you're using. It might be that we're trained to think in "set theory ways", but it's also true that the axioms give you exactly the expressive power to talk about everything you care about and not care about one specific implementation over another (looking your way, Replacement axioms!), which is quite great.

Can you imagine a mathematical universe where the reals constructed with Dedekind cuts and with Cauchy sequences are different? (Those mathematical worlds exists in other foundations, by the way, and at least to me, that is weird.)
The previous point leads us exactly to this argument. Humans take an abstract algorithm, implement it in various languages, on various processors, using various data structures and types. But the code and data all turn into electronic signals. So working with high level objects like $\Bbb N$ and $\Bbb R$, and function spaces and so on is our algorithms. And we can turn that into electronic signals, or sets and first-order structures in this case.
We like sets more than we are willing to admit. One of the problems of second-order logic is that the logic is incomplete.2 And if you have an algorithm for a list of inference rules, to verify a proof, then there will be statements which are true (even valid) that these inference rules cannot prove. So your proof theory is quite insufficient in this aspect. Because at least the valid sentences should be provable.

So we can, instead, use a first-order based foundation to fix this. Instead of proving something about $\Bbb R$, we prove that $\sf ZFC$ proves that thing about $\Bbb R$; and this we can verify mechanically. So one option is to resort to some arithmetic foundation. But this causes a different problem. We cannot "take objects" anymore. We don't have sets of reals, or rings, or groups. So when you prove something about groups, you can't say "Let $G$ be an abelian group, then bla bla bla" anymore. You have to say "The theory of abelian groups proves that bla bla bla".

And this is quite a big issue, since we think about mathematics in some material sense. The objects exists somewhere. They are not just axiomatic consequences. And set theory, in particular $\sf ZFC$ with its implementation agnosticism, provides us with the means to do exactly this.

Footnotes.

(1) Is $\Bbb N\subseteq\Bbb R$? Often the answer is yes, many other times the answer is no. And it really depends on what you want to do, or how you mostly use these two objects. But since we know that either method can be replaced by the other we're not worried about it too much. But it is a concrete question that has convenient answer either way, and we exploit it. And that, in a nutshell, a "global inconsistency" in our thinking.

You can argue about this, but it's going to be besides the point.

(2) Incomplete here means in the sense of the completeness theorem. Something true in every model need not be provable. This is a different kind of incompleteness than the one referred to earlier, where $\sf ZFC$ is incomplete in the sense that it does not prove or refute every statement.

Yes. Category theory is one example of a theory than transcends ZFC. Some other examples include Hyperreals and Surreals. There are many more.


Is getting a random integer even possible? On googling I got

  Random numbers are numbers that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is impossible to predict future values based on past or present ones. Random numbers are important in statistical analysis and probability theory.

What if I want to pick a random integer from the number line ?
The number line consists of infinitely many integers.
My favorable event is to get a number (random) and the number of events possible is the complete number line i.e. infinity.
I won't be considering any limits here as our number line doesn't tend to infinity but does contain infinitely many elements.

Now the probability of getting a random number

P(E) = 1 / ∞ = 0

So I won't be able to get a random number, ever.
This implies, never would I ever be able to get a random number.
That's absurd as many of our science topics do depends on things being random.

Most typically you choose a random natural from a finite range.  If you toss a fair coin, you get a random number that is either $0$ or $1$.  If you call a random number generator on your computer, you might get a number in the range $[0,2^{32}-1]$.  There is no problem having equal probability in these cases.  If you want to choose a random natural, you can't have the probability of all numbers be equal, but you can use a non-uniform distribution that sums to $1$.  If you don't think $0$ is a natural, you can have $p(n)=2^{-n}$, for example.

Progress through the integers beginning with the first element in a well-founded wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element, given that you land on it, where $f(x)$ satisfies $\forall x:0<f(x)<1$.

Then by this means you can not only do so, but given an appropriate function, infinitely many integer outcomes (including possibly every integer) can have non zero probability.

An appropriate function for the natural numbers $\Bbb N^+$ is $f(x)=1/2^x$


"Marginalizing out" a parameter in a PDF Say I have a variable $x$ whose PDF is a von Mises distribution with a single parameter, $\kappa$, that determines the distribution's concentration: $VM(x; \kappa)$.

But suppose $\kappa$ itself is not fixed, but drawn from a Gamma distribution: $G(\kappa; \bar{\kappa}, \tau)$, where the mean is $\bar{\kappa}$ and the variance is $\bar{\kappa}\tau$.

An author whose work I'm replicating has this to say about this situation: "The predictions [for each $x$ value's probability] have to be averaged over all possible values of precision $\kappa$ (mathematically: we need to 'marginalize out $\kappa$').  We do this by discretizing the (gamma) distribution over $\kappa$ into 50 bins with equal probability masses, computing the model [von Mises] prediction at each bin center, and then averaging the predictions.  This solution is accurate as long as the number of bins is not too small (in the limit of an infinite amount of bins, the solution is equal to an analytical marginalization)" (emphasis mine).

I've read elsewhere that probabilities should be summed, not averaged, in marginalization.  Another confusion I have regards this author's statement that the average should be "over all possible values of $\kappa$."  This seems to imply that each possible value of $\kappa$ should be treated equally -- but then we're talking about a uniform distribution, not a Gamma one.  Clearly, the author was using language loosely.

Can someone shed some light on this, maybe with language that is more precise?

Suppose you  have:
\begin{align}
\Pr(X = 0 \mid k=0) & = 2/3 & & & \Pr(X=1\mid k=0) & = 1/3 \\[10pt]
\Pr(X = 0 \mid k=1) & = 1/2 & & & \Pr(X=1\mid k=1) & = 1/2
\end{align}
If you want the marginal distribution of $X$, you average the two rows above, with weight equal to $\Pr(k=0)$ for the first row and to $\Pr(k=1)$ for the second row.  Let's suppose those weights are each $1/2$.  Then the joint distribution is:
\begin{align}
\Pr(X=0\ \&\ k=0) & = 1/3 & & & \Pr(X=1\ \&\ k=0) & 1/6 \\[10pt]
\Pr(X=0\ \&\ k=1) & = 1/4 & & & \Pr(X=1\ \&\ k=1) & 1/4
\end{align}
Then rather than averaging, you sum, to get
\begin{align}
\Pr(X=0) = & \frac 7 {12}, & & & \Pr(X=1) & = \frac 5 {12}.
\end{align}

In one case you have a conditional distribution given $k$, and you find an average; in the other case you have a joint distribution, and you find the sum.  Either way you get the same marginal distribution.

In the situation you describe, the von Mises distribution is a conditional distribution given the value of $\kappa$.

The average should be a weighted average with weights given by the distribution of $\kappa$.

You have a conditional density distribution: $f_{X\mid K}(x\mid k) = VM(x;\kappa)$

By "analytic marginalisation" the author means:

$$\begin{align}f_X(x) ~=~ & \mathsf E\big(f_{X\mid K}(x\mid K)\big) \\[1ex] ~=~ & \int_0^\infty f_{X\mid K}(x\mid \kappa)~f_K(\kappa) ~\operatorname d \kappa\end{align}$$

However the author claims that as long as the number of bins is not too small, this can approximated by "averaging out".


Curl of Cross Product of Two Vectors I want to prove the following identity

$$\text{curl } \left(\textbf{F}\times \textbf{G}\right) = \textbf{F}\text{ div}\textbf{ G}- \textbf{G}\text{ div}\textbf{ F}+ \left(\textbf{G}\cdot \nabla \right)\textbf{F}- \left(\textbf{F}\cdot \nabla \right)\textbf{G}$$

But I do not know how! Also, what does $\textbf{F}\cdot \nabla $ mean, isn't it the divergence of $\textbf{F}$!

You only need two things to prove this.  First, the BAC-CAB rule:

$$A \times (B \times C) = B(A \cdot C) - C(A \cdot B)$$

And the product rule.  Let $\dot \nabla \times (\dot F \times G)$ mean "differentiate $F$ only; pretend $G$ is constant here".  So the product rule would read

$$\nabla \times (F \times G) = \dot \nabla \times (\dot F \times G) + \dot \nabla \times (F \times \dot G)$$

Now, apply the BAC-CAB rule.  I'll do this for just one term for brevity:

$$\dot \nabla \times (\dot F \times G) = \dot F (\dot \nabla \cdot G) - G(\dot \nabla \cdot \dot F)$$

Now, here's where the dots become important:  since $G$ is not differentiated in this whole equation, $\dot \nabla \cdot G$ is a directional derivative, conventionally written $G \cdot \nabla$.  Indeed, we have

$$\dot F(\dot \nabla \cdot G) = (G \cdot \nabla) F$$

On the other hand, the $G(\dot \nabla \cdot \dot F)$ term can just drop the dots to get something that looks like a divergence:

$$G (\dot \nabla \cdot \dot F) = G(\nabla \cdot F)$$

Carry out the same expansion for the $\dot \nabla \times (F \cdot \dot G)$ term, and you're done.

It's easiest to see by writing it out in components:

[ ( ∇ ⋅ a ) b ]i = ( ∂x ax + ∂y ay + ∂z az )bi = ( ∂x ax )bi + ( ∂y ay )bi + ( ∂z az )bi

Whereas

[(∇⋅a)b]i=(∂x ax +∂y ay +∂z az)bi =(∂x ax)bi +(∂y ay)bi +(∂z az)bi

and clearly these are not the same. So while a ⋅ b = b ⋅ a a⋅b=b⋅a holds when a and b are really vectors, it is not necessarily true when one of them is a vector operator. This is one of the cases where the convenience of considering ∇ ∇ as a vector satisfying all the rules for vectors does not apply.


How many 8-bit binary numbers have a 1 as the first bit, or third bit, or both? I tried doing C(8,1) for when 1 is the first bit, and third bit, and C(8,2) when it's both and multiplying them but I didn't get the right answer.

$C(8,1)$ is the number of 8-bit binary numbers with exactly one 1, and $C(8,2)$ is the number of binary numbers with exactly two $1$'s. I don't think that's what you want to count.

Hint: How would you most easily describe a number which doesn't fulfill that criterion? How many such numbers are there? And how many 8-bit binary numbers are there in total?

There are total $2^8$=256 8 bit binary numbers.
You want those having 1 at $1^{st}$ or $3^{rd}$ bit.
So, go for those who do not have; i.e. numbers having 0 at their $1^{st}$ and $3^{rd}$ bit

Fix them as 0, and fill remaining 6 bits in either of 0 and 1
There are $$2\cdot2\cdot2\cdot2\cdot2\cdot2=2^6=64$$
64 numbers do not have 1 at $1^{st}$ and $3^{rd}$ bit.
So, total-not having 1= those having 1

  $$256-64=192$$ is your answer

You can also go for counting individually numbers having 1 at $1^{st}$ (0 at $3^{rd}$), 1 at $3^{rd}$ bit (0 at $1^{st}$), and 1 at both $1^{st}$ and $3^{rd}$ bit, and adding all.    (Try!)


How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$ Let $n$ be a given positive integer,and let $a_{1},a_{2},\cdots,a_{n}\ge 0$,such that

$$a_{1}+a_{2}+\cdots+a_{n}=1$$

Find this  minimum value

$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}$$

I think maybe we can use this well known

$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}\ge a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}?$$

But this problem is only
$$-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}$$
so I can't it.Thank you for help.
By the way:
I don't know this problem have Someone research?if No,I think this is interesting problem.

You almost did the problem yourself, just left out a few bits:

Assume, without loss of generality,

$$a_1 \geq a_2 \geq \dots \geq a_n$$

By the rearrangement inequality,

$$\implies a_1^2 + a_2^2 \dots a_n^2 \geq a_1a_2 + a_2a_3 \dots a_{n-1}a_n + a_na_1$$

Clearly, the required will be minimum if equality holds true above.

This will happen when

$$a_1 = a_2 = a_3 = \dots  = a_n = a$$

This means, the maximum value of the inequality is:

$$\sum_{i = 1}^n a^2 - 2a^2 = -na^2$$

Since, from the first equation,

$$na = 1$$

$$a = \frac{1}{n}$$

Therefore, the minimum is,

$$-n\cdot\left(\frac{1}{n^2}\right) = -\frac{1}{n}$$


Number of solutions to infinite sum of polynomial equation How many solutions (at most) are there in the following equation?

$a_0+a_1x+a_2x^2+a_3x^3+\cdots=0$, where each $a_i\in\mathbb{R}$, and $x\in (0,1)$?

Is it possible to find some $\underline{x}\in (0,1) $, such that all $x\geq \underline{x}$ are the solutions to the above equation?

I know that the fundamental theorem of algebra tells us that there are $n$ solutions for the $n$ degree polynomial, but don't know that happens when $n$ is infinity.

The theorem only says there can be ATMOST n solutions for n degree polynomial. There could be much less. As your equation ends with $=0$, the no. of solutions can be given by the number of  indivisible factors of your polynomial; if there are no factors, there is 1 solution (not necessary a real no.).


Do we have negative prime numbers? Do we have negative prime numbers?

$..., -7, -5, -3, -2, ...$

I don't know why this question has a down vote, because it identifies a subtle point about arithmetic which becomes particularly significant when the notion of "prime" is extended to other contexts, and is relevant so far as the integers are concerned when looking at issues like unique factorisation.

When we first encounter prime numbers, we do so in the context of the positive integers. The significant point about this context is that $+1$ is the only unit (the only positive integer with a multiplicative inverse). So the question here does not really arise. Often, when the main focus of work is the positive integers, the word prime will be used to imply a positive integer.

As soon as we start to extend this to the integers, and in particular, to consider the integers as having the structure of a ring, we add in a second unit $-1$ with $(-1)^2=1$. Even in this context it is possible to define the prime numbers as positive integers without too much inconvenience.

But if we extend further and add $i$ with $i^2=-1$ as another unit - note that $i\cdot -i=1$, we are in a different world. For example, $2=(1+i)(1-i)$ and $(1+i)=i(1-i)$ so that $2=i(1-i)^2$. Are these factorisations of $2$ to be taken as the same or different?

So very soon, in the context of ring theory and the theory of algebraic integers, we start talking about prime ideals (initially thought of as all the multiples of some prime $p$ - but extended beyond that idea too - an ideal which consists of all the multiples of a single element is called principal). And it is somewhat natural, if the ideal is principal, to call the generator a prime element of the ring. However, the primes are then only identified up to multiplication by units - $1+i$ generates the same ideal as $1-i$. One of the reasons for using ideals is that the uniqueness of factorisation can be maintained in this larger context. In $\mathbb Z$ both $2$ and $-2$ generate the same ideal.

Of course there are negative prime numbers. For a number $p$ in a given domain to be prime, it needs to satisfy $p|ab$ always means $p|a$ or $p|b$ (maybe both). For example, if $p = -3$, $a = -2$, $b = 30$, we see the condition is satisfied with $p|b$. But if $p = -4$, neither condition is satisfied. This means that $-3$ is a negative prime number and $-4$ is a negative composite number.

It also needs to be said that there is a negative unit, $-1$. A unit divides every number in the domain. If $p$ is prime and $p = ab$, either $a$ or $b$ (but not both) must be a unit. For example, $-3 = -1 \times 3 = 1 \times -3$.

This is no problem whatsoever for unique factorization (if it exists in the domain at hand). We simply say that factorization is unique regardless of ordering (e.g., $5 \times 3$ is not a distinct factorization of $15$) and regardless of multiplication by units.

However, for the prime counting function $\pi(x)$, we're generally only concerned with positive primes. After all, $\pi(7) = \infty$ would be kind of useless. Or we can just say that $\pi(x)$ counts how many primes there are between $0$ and $x$. Then it turns out that $\pi(x) = \pi(-x)$, e.g., $\pi(-10) = 4$ just as $\pi(10) = 4$.


What is the maximum number of regions produced, i.e. $f(n)$, by joining all vertexes with line segments of a convex polygon with $n$ sides? What is the maximum number of regions produced, i.e. $f(n)$, by joining all vertexes with line segments of a convex polygon with $n$ sides?

For example, for the hexagon on the left, number of regions is 24, but the one on right is 25, which I think is in fact $f(n)$ for the case $n=6$.

So is there any way to find the general equation for $f(n)$?

Thank you.

Let $v$ be the number of vertices, $e$ the number of edges and $f$ the number of faces (not counting the exterior of the given $n$-gon $P$) in the resulting configuration. Then by Euler's formula
$$v+f = e+1\ .$$
On the other hand the number of vertices is given by
$$v=n+{n\choose 4}\ ,$$
because any four vertices of $P$ determine a unique vertex in the interior of $P$. Finally the number of edges satisfies
$$2e=n(n-1)+4{n\choose 4}\ ,$$
because each vertex of $P$ has degree $n-1$ and each inner vertex has degree $4$.

Solving these three equations for $f$ we get
$$f={n(n-1)\over2}+2{n\choose 4}-n-{n\choose 4}+1={1\over24}(n^4-6n^3+23n^2-42n+24)\ .$$

The most convenient expression for f(n) is
f(n) = n(n-1)(n-2)(n-3)/24 + (n-1)(n-2)/2 which, of course, is f(n) for any irregular polygon with n sides (or any regular 'odd-sided' polygon}. For regular 'even-sided' polygons it is necessary to reduce the value given by f(n)
by counting the number of 6,8,10,12 ... nodes produced by the intersecting diagonals. For each 6-node the number of regions is reduced by one, for each
8-node by 3, each 10 node by 6, each 12 -node by 10 and so on in a triangle number sequence. For example, in an 18-sided regular polygon there are 216 6-nodes, 54 8-nodes, 54 10 nodes and one central 18-node ('hides' 28 regions). The f(n) value of 3196 is thus reduced by 730 to give the 'observed'(!) value of 2466.


Differential Equations: Time-Reversal Symmetry In Strogatz's Nonlinear Dynamics and Chaos (page: 163-164), he claims that any mechanical system of the form mx" = F(x) is T-R symmetric. He considers a system that has T-R symmetry to be one that is invariant under the change of variable

$ t -> -t $

First, I see that if we reverse time along a trajectory, that the first derivative would inherit a negative sign. Second, I'm not positive I see why the second derivative, x", would remain unchanged. Is this due to the following?

$ x''= d^2x/dt^2 $  ... which since t is squared would yield $t^2$ = $(-t)^2$ under the mapping of t -> -t.

However, it seems a bit odd that you can manipulate the $dx/dt$ notation like that. If someone could show why $dt^2$ is able to obey the above I would appreciate it. Namely, why is $d(t)^2$ considered a product or function where we can first compute $t^2$ then apply d?

Thanks!

This is the chain rule. If you have a function
$$
f(t)
$$
and you compose with the function $g(t)=-t$, then
$$
\frac{d}{dt}f(-t)=\frac{d}{dt}f(g(t))=g'(t)f'(g(t))=-f'(-t)\\
\implies \frac{d^2}{dt^2}f(-t)=\frac{d}{dt}-f'(-t)=-\frac{d}{dt}f'(-t)=f''(-t)
$$

Due to the power rule, if the first derivative is odd such that $F(-x)=-F(x)$, than the second derivative is even. That is why it is invariant to the transformation $t\to-t$


Divide set of numbers into twosub sets with equal totals Given a finite sequence of natural numbers. Determine wether it is possible to divide the numbers into two sets such as totals of both sets are equal. Show one variant of such distribution. Is there any subset of initial set with total of 100.

Now I only see a bruteforce approach to this problem - check  totals of all of S(n,2) (Stirling number of the second kind) combinations for equality and show one such combination. And also check all possible combinations of initial set for equality to 100. Is there more elegant solution?

This is the partition problem, a well-known NP-complete problem.

This means that no "elegant" solution is known assuming that we require as a necessary condition for "elegance" that it can be proven to require only a polynomial amount of work.

However, if the numbers themselves (or the target sum) are small, there is a good dynamic programming algorithm; see the above linked Wikipedia article.

This is an NP problem, which means that it can only be solved in non-polynomial time, in this case if the set is called $S$, then the NP approach will cost:
$$(|S|-1)!$$
steps (at least that's the fastest I method know of).
If you find a way to do it in polynomial time, you'll have proven that P=NP and you'll win a millennium prize (also if you prove that such an approach does not exist).


Inverse of $y= x^{3} +x+1$ Can you help me find the inverse function for  $y= x^3 +x+1$? This question proposed by one of my friends and I don't know the real source of the problem.
We can't use Cardano's method for solve this problem.

Thanks!

Why won't Cardano work?

I'll tackle this problem the algebraic way because it's the most straightforward, despite the intimidating algebra.

Replacing $x$ with $y$, we get$$y^3+y+1=x\quad\implies\quad y^3+y+1-x=0$$Now recall Cardano's formula for the cubic $x^3+qx+r=0$ who has the real root at$$x_1=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}$$Substituting $q=1$ and $r=1-x$ gives$$\begin{align*}y=\sqrt[3]{-\frac {1-x}2+\sqrt{\frac {27x^2-54x+31}{108}}}-\sqrt[3]{\frac {1-x}2+\sqrt{\frac {27x^2-54x+31}{108}}}\end{align*}$$And that's your inverse! Here's a visual diagram in Desmos

Given

$$ y = x^3+x+1 $$

Interchange $x,y$ to see inverse function graph.

$$ x= y^3+y+1 $$

Using Cardano it is quite possible to find the solution $f(x)$ in terms of $y$ analytically. There is one real root at $x=1$ coinciding at its inflection point.


Quotient of a locally compact Hausdorff space by a proper action is Hausdorff I am trying to prove the following:

  Let $G$ be a topological group acting properly on a Hausdorff locally
  compact space $X$, i.e. preimages of compacts sets by the map
  $$G\times X\to X\times X$$ $$(g,x)\mapsto (gx,x)$$ are compacts. Then
  the quotient $X/G$ is Hausdorff.

Take $x,x'\in X$ such that $Gx\not=Gx'$.

By using that X is Hausdorff locally compact and property of the action, I found
$V_1,V_2$ disjoint open set with compact closure and $H\subset G$ compact, such
that $x\in V_1,y'\in V_2$ and $$(G-H)\overline{V_1}\cap
    \overline{V_2}=\varnothing.$$
So now we need to work on $H$. Since
    $x'\in (Hx)^c\cap V_2$ open, there exists $x'A_2\subset V_2$ such
    that $$A_2\cap Hx=\varnothing.$$ It implies that $x\in
        (H^{-1}A_2)^c\cap V_1$.
Now, I'd like to find as before $x\in A_1\subset V_1$ open such that
$$A_1\cap H^{-1}A_2=\varnothing,$$
which would be enough to conclude. But $H^{-1}A_2$ is not closed like $Hx$, so I'm stuck...

Are my first ideas correct ? Is there a way to exit this dead end ?

So using that locally compact spaces are compactly generated, it is easy to show that a proper map $f: A\to B$ with $B$ locally compact is closed.

Therefore, $f: G\times X\to X\times X$ as defined above is closed. But the quotient map $\pi: X\to X/G$ is closed, so
$$\tilde f=\pi\circ f: G\times X\to X/G\times X/G$$
is also closed. Note that $\tilde f(g,x)=([gx],[x])=([x],[x])$, so the diagonal
$$\Delta(X/G)=\tilde f(G\times X)$$
is closed, i.e. $X/G$ is Hausdorff.

Is that correct ?


Magic Squares with Random Numbers I'm trying to solve a problem related to Magic Squares. The thing is:

Given a list of n numbers, I need to answer if it is possible to create a magic square with them. These numbers are random (no order at all) and may/can be repeated. For example:

-1 -11 -6 19 -21 4 14 29 9 (Which in fact works as a Magic Square 3x3)

I know I can calculate the size of the matrix by sqrt() the lenght of the array, but I cannot find any hint on the way I can solve this efficiently (No, I cannot afford trying every single combination as my matrix can be as big as 29x29)

Any help??

Thanks !!

Preconditions:

$a_i$: the sequence of input numbers, $i \in \{1, 2, \dots, m\}$.

$n$: the number of row/columns. Obviously, $n=\sqrt{m}$. If there is no integer solution for $n$, then there is no magic square.

$s$: the sum of numbers in each row, column and diagonal.
$$s \cdot n=\sum{a_i}$$
$$s=\frac{\sum{a_i}}{n}$$
Like $n$, there has to be an integer solution for $s$, or there is no magic square.

Solution:
Except for the conditions above, I don't see any other way than trying out possible solutions. However, we do not have to try each and every permutation; we can do that a bit more systematically, e.g. by modelling the problem as a contraint satisfaction problem (CSP) with $n^2$ variables and $2n+2$ constraints (one per row/column/diagonal: $\sum=s)$

For $n=29$ it's still quite hard, though.

For 29x29 that is a lot of work.  I am here looking for similar info. Where I am at problem solving-wise is you can find all sums of n numbers adding to the magic sum (s above): order the list, test the sum of the smaller numbers until the sum is too large, get a list of possible rows, which is also a list of possible columns.  Search through the rows to find n that are disjoint (have all the numbers in them), then search through the columns (which are also the rows) to find columns that intersect each of your n rows just once.  I think you still have to test (n!)^2 possible row-column permutations then test the diagonal sums.


Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$ $$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$

I was doing some questions on limits, I saw one in which there is $\arccos x$. I am stuck there, not able to proceed.

Can you give me some hint?

Set $t=\arccos(2x\sqrt{1-x^2})$, so $2x\sqrt{1-x^2}=\cos t$ and, by definition, $0\le t\le\pi$. Then
$$
\sin t=\sqrt{1-\cos^2t}=\sqrt{1-4x^2+4x^4}=|2x^2-1|
$$
and so $t=\arcsin|2x^2-1|$. Thus your limit (from the right) can be written
$$
\lim_{x\to(1/\sqrt2)^+}\sqrt{2}\frac{\arcsin|2x^2-1|}{\sqrt{2}x-1}=
\lim_{x\to(1/\sqrt2)^+}\sqrt{2}
\frac{(\sqrt{2}x+1)\arcsin(2x^2-1)}{2x^2-1}
$$
which you should be able to compute.

Note then that, for $x<1/\sqrt{2}$, $|2x^2-1|=-(2x^2-1)$ and you can go similarly for the limit as $x\to(1/\sqrt{2})^-$

That $\sqrt{1-x^2}$ shouts "Trigonometric functions!!!". So:

Put $x=\sin\theta$, which means that $\sqrt{1-x^2}=\cos\theta$. Then $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\arccos \left(\sin 2\theta \right)}{\sin \theta-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\frac{\pi}{2}-2\theta}{\sin \theta-\frac{1}{\sqrt{2}}}$$

You can use l'Hôpital from here. Or...

Put $\phi=\theta-\frac{\pi}{4}$. The limit becomes $$\lim_{\phi \to 0} \frac{-2\phi}{\sin \phi \cos\frac{\pi}{4} + \cos \phi \sin\frac{\pi}{4}-\frac{1}{\sqrt{2}}} \equiv \lim_{\phi \to 0} \frac{-2\phi}{\frac{1}{\sqrt{2}}(\sin \phi + \cos \phi -1)}$$

…which you can l'Hôpital again, or note that for small $\phi$, $\sin\phi$ is practically $\phi$ and $\cos\phi$ is practically 1, so that the limit becomes $$\frac{-2}{\frac{1}{\sqrt{2}}}$$
…which I leave to you to work out. This is essentially reproducing what l'Hôpital does, but by doing it yourself you can actually see what you are doing.


Find a possible pair of numbers given HCF and a factor of the LCM I was just going through a GCSE paper with a student and I came across a question that I'm struggling to find a good method for. The question was this:

  Martin thinks of two numbers.

  The Highest Common Factor of the two numbers is 6   The Lowest Common
  Multiple is a multiple of 15

  Write a pair of numbers that Martin could be thinking of.

It took me an embarrassingly long time to come up with a solution but I still don't have a method that will work quickly for different numbers. What would be the best way to come up with a solution?

Both numbes must be multiples of $6$ and at least one must be a multiple of $5$ (because $5$ is a prime dividing $15$). So the first attempt is $a=30$ and $b=6$ and works

this question is incorrect!

To have an LCM of 15 the numbers can only be 1 and 15 or 3 and 5

The given answer of 6 and 30 would give a HCF of 6 and and LCM of 30


What is $\left\lfloor0.\overline{9}\right\rfloor$? We know that $0.\overline{9} = 1$ but then what is $\left\lfloor0.\overline{9}\right\rfloor$?

My thought process went:
$0.\overline{9} = 1$ so therefore $\left\lfloor0.\overline{9}\right\rfloor = \left\lfloor1\right\rfloor$

But also $\left\lfloor0.9\right\rfloor = 0$ and it shouldn't change no matter how many numbers you add on to the back of it.

So what is the right answer, if there is one, and why?

As you say, $0.\overline9=1$. Thus, $0.\overline9$, like $2-1$, $\frac33$, etc., is just a different notation for the number $1$, and consequently $\left\lfloor0.\overline9\right\rfloor=\lfloor1\rfloor=1$.

Added: The fact that $\lfloor 0.9\ldots9\rfloor=0$ for any finite string of $9$s is irrelevant: the floor function is not continuous from the left at integers. In effect you’re saying that $\lfloor 1\rfloor$ ought to be $0$ simply because $\left\lfloor 1-10^{-n}\right\rfloor=0$ for each $n\in\Bbb Z^+$.

To me the intuitive explanation is that if you put any number of 9s after the 0, $\lfloor 0.999...99 \rfloor =0$. But $0.\overline{9}$ has infinite 9s after the 0, and infinity isn't a number. So the pattern doesn't hold.

Cameron Williams pointed out the rigorous answer, which is that you're implicitly assuming that because a sequence of numbers (0.9,0.99,0.999...) gets closer to 1, then if you apply a function to them (in this case, $\lfloor x \rfloor$), the the resulting sequence ($\lfloor 0.9\rfloor$,$\lfloor 0.99\rfloor$,$\lfloor 0.999\rfloor$...) should also get closer to  $\lfloor 1\rfloor$. But functions don't always do this. If a certain function does have this property, we say that the function is continuous at that point, but $\lfloor x\rfloor$ is not continuous at 1. In fact, you've more or less given a proof that $\lfloor x\rfloor$ is discontinuous at 1.


Computing $ \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2}$ I would like to show that

$$ \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2}=\frac{\pi^2}{15}-\ln^2(\phi) $$

where $$ \phi = \frac{1+\sqrt{5}}{2}$$

How can this result be derived from the integral representation

$$  \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2} = -\int_0^{1/\phi^2}\frac{\ln(1-x)}{x} \mathrm dx $$

?

To justify the integral representation, observe that
$$
\frac{-\ln(1-x)}{x}=\sum_{n\geq 1} \frac{x^{n-1}}{n}
$$
converges normally (hence uniformly absolutely) on $[0,1/\phi^2]$.
So term by term integration is possible and yields the desired formula.

Now to compute the integral, we need to know that an antiderivative of $-\ln(1-x)/x$ is the so-called dilogarithm. According to Wikipedia, Don Zagier once said that the dilogarithm is the only mathematical function with a sense of humor.

So
$$
\sum_{n\geq 1} \frac{1}{\phi^{2n}n^2}=\mbox{Li}_2(1/\phi^2)=\mbox{Li}_2\left( \frac{3-\sqrt{5}}{2}\right).
$$

I found here: http://en.wikipedia.org/wiki/Polylogarithm that
$$
\mbox{Li}_2\left( \frac{3-\sqrt{5}}{2}\right)=\frac{\pi^2}{15}-\ln^2\left(\frac{\sqrt{5}-1}{2}\right).
$$

That's your formula, since $\ln^2(1/\phi)=\ln^2(\phi)$.

The dilogarithm formula used above is most likely proved here: http://books.google.fr/books?id=u_UVn_iquj0C&pg=PA141&lpg=PA141&dq=trilogarithm+identities&source=bl&ots=5lp7VWoN1p&sig=xkvQm5jZFj_bcb0CpPKqsmIa1Fw&hl=fr&sa=X&ei=d5EOUd_iPLK-0QH8k4CAAQ&ved=0CDoQ6AEwAQ#v=onepage&q=trilogarithm%20identities&f=false

Use the fact that $- \ln (1-x) = \sum \frac{x^n}{n}$. Then prove you can swap sum and integral.


What does the star symbol means in the binary operation a*b .... Ex: x*y=x+y+xy what does the star symbol means in binary operation algebra (logic)
Does it refer to a certain rule

It means the operation defined on the RHS.

For example
$$2*3= 2+ 3+2\times 3=11$$

In binary logic $+$ refers to OR and $\cdot$ refers to AND.

You could also use alternate notation $x\lor y\lor(x\land y)$ but it is equivalent to simple $x\lor y$ or $x+y$ in the origninal notation.

Assuming it is really a logical operation, I do not see the purpose of defining $*$ if it is the same thing as $+$ or maybe this is th epurpose of the exercise to show they are the same rules.

Anyway, I sometimes saw $*$ being used for NAND i.e. $\lnot(x\land y)$ or $(xy)'$ in both notations.

But this is not the case here, are you certain about the definition of $x*y$ ?

https://en.wikipedia.org/wiki/Logic_gate