Trouble finding probabilities. I am working through some problems about probability and seem to be having trouble working through this one in particular. I'd love some help learning how to go about solving problems such as this.

A website estimates that 19% of people have a phobia regarding public speaking. If three students are assigned to a project group, what's the probability...

a.) That all 3 students have a fear of public speaking.

b.) That none have a fear of public speaking

c.) That at least one of the students has a fear of public speaking.

Use the fact that these students are chosen independently, and so $$\Bbb P(A\text{ and }B\text{ and }C)=\Bbb P(A)\Bbb P(B)\Bbb P(C)$$ For your problem, you could say $A$ is the event that the first student has a fear of public speaking, etc. You can also use this property for part $B$.

For questions where they say something like "at least", think of the opposite - the opposite of at least one student having this fear is for none of them to have this fear. This is easier to calculate, and in fact you have already worked it out in another question. Then the probability you'd want is $1-\Bbb P(\text{the opposite})$.

For this question,:

A) If one kid has a fear then probability of picking him will be $0.19$ while picking one more reduces the probability to $(0.19)*(0.19)$ . For three people: it becomes $(0.19)^3$
 This is from selection via independent events.

B)None have fear: $(1-0.19)^3$
This is via noting that 81% population won't be fearful as such. and following the above principal again.

C) At least one have fear: $1-(1-0.19)^3$
This one is by asking in account that for at least one to be fearful, we may take it as Anti-event when none is fearful. Thus 1-P(part B)


$a^m=b^m$ and $a^n=b^n$ imply $a=b$ Let $D$ be an integral domain and let $a^m=b^m$ and $a^n=b^n$ where $m$ and $n$ are relatively prime integers, $a,b \in D$.

How do I show $a=b$?

I came across a similar problem but with the condition that $m$ and $n$ are positive coprime integers. The following is my solution.

If $a=0$, then $b^m=a^m=0$. It follows that $b=0=a$ since $D$ is an integral domain.

If $a \ne 0$, since gcd(m,n)=1, $\exists s, t\in \mathbb z $ such that $1=sm+tn$. Because $ m, n>1$, $s,t \in \mathbb z$ and $1 = sm + tn$, we have either $s>0, t<0$ or $t>0, s<0$ (Otherwise we cannot have $1=sm+tn$).

Let's assume $s>0, t<0$. Then $-t>0$. From $a^m = b^m$, we have $(a^m)^s=(b^m)^s$. From $a^n=b^n$, we have $(a^n)^{-t}=(b^n)^{-t}$. We can factor out as follows.

$$a^{-tn}(a^{sm+tn})=b^{-tn}(b^{sm+tn})$$

$$a^{-tn}(a^{sm+tn})=a^{-tn}(b^{sm+tn})$$
$a^{-tn} \ne 0$ since $a \ne 0$ and $a$ is in an integral domain. We can cancel $a^{-tn}$ on both sides.Then we get $a^{sm+tn}=b^{sm+tn}$, i.e. $a^1=b^1$.

The proof for the case where $s<0, t>0$ is similar.

What's the necessity of $D$ to be a domain?
As $(m,n) = 1$ we have two integers $x$ and $y$ such that $mx+ny = 1$

so, $a^{mx} = b^{mx}$ and $a^{ny} = b^{ny}$ which implies, $a^{mx+ny} = b^{mx+ny} = a = b$


Why are groups more important than semigroups? This is an open-ended question, as is probably obvious from the title. I understand that it may not be appreciated and I will try not to ask too many such questions. But this one has been bothering me for quite some time and I'm not entirely certain that it's completely worthless, which is why I've decided to ask it here.

Why are groups so immensely more important in mathematics and its applications than semigroups are? I know little of group theory, mathematics and its applications, so I cannot understand how much more important they are. But I know they are because I see how many more people are interested in groups than in semigroups.

I wouldn't ask this question if this fact didn't seem a little strange to me. I know that groups are associated with symmetries, or with automorphisms of structures. Cayley's Theorem tells me that every group can be seen as a set (closed with respect to taking compositions and inverses) of automorphisms of a structure with no constants, functions or relations, i.e. a plain set. I know that the automorphisms of a vector space, a module or a group form a group. Obviously, automorphisms are important.

Then we have inverse semigroups. These are less popular but this I can understand. They are associated with partial symmetries, by the Wagner-Preston Theorem. Partial functions do seem much less used than functions.

But then come semigroups. Just like in the two previous cases, there is an "embedding theorem", which says that every semigroup can be embedded in a semigroup of maps from a certain set into itself. And, analogously to the case of groups, the endomorphisms of a vector space, a module or a group form a semigroup.

This seems to say that semigroups are to endomorphisms what groups are to automorphisms. The "conclusion" would be that

$\frac{\mbox{the importance of semigroups}}{\mbox{the importance of endomorphisms}}=\frac{\mbox{the importance of groups}}{\mbox{the importance of automorphisms}}$

Assuming that endomorphisms are about as important as automorphisms, we get that semigroups are about as important as groups. My feeling is that the assumption is correct.

I understand that I must be oversimplifying something at some point. But what am I oversimplifying and where?

I realize that this is possibly a very dumb question, but my confusion is genuine.

Edit: I have realized, after reading your answers and comments, that I made the mistake of using a very vague term without even attempting to define it. The problem seems to be what importance is. Is it popularity or usefulness, or being used a lot, or interesting or being interesting, or something else, or a mixture of many traits? I'm not going to force my understanding of the word now. But if someone still wants to answer this question, perhaps it would be good idea if the answer contained an attempt at clearing this up. (Or maybe not. I'm not sure!)

Edit: A question similar to mine is dealt with here.

To add a remark related to Jim Belk's answer and the OP's comments on that answer:

In many naturally occurring situations, including some of those where group theory is particularly useful, endomorphisms are automatically automorphisms.

For example, if $E/F$ is a finite extension of fields, any endomorphism of $E$ which is the identity on $F$ is automatically an automorphism of $E$.

As another example, if $C$ is a Riemann surface of genus at least $2$, then
any (nonconstant) endomorphism of $C$ is necessarily an automorphism.

Any endomorphism of a Euclidean space which preserves lengths is necessarily an automorphism.

Another point to bear in mind is that the groups that arise in practice in geometry are often Lie groups (i.e. have a compatible topological, even smooth manifold, structure).  One can define a more general notion of Lie semigroup,
but if your Lie semigroup has an identity (so is a Lie monoid) and the semigroup
structure is non-degenerate in some n.h. of the identity, then Lie semigroup
will automatically be a Lie group (at least in a n.h. of the identity). A related remark: in the definition of a formal group, there is no need to include an
explicit axiom about the existence of inverses.

To make a point related to Qiaochu Yuan's answer: in some contexts semigroups
do appear naturally.

For example, studying the rings of endomorphisms of an object is a very common technique in lots of areas of mathematics.  (E.g., just
to make a connection to my first point, for genus $1$ Riemann surfaces, there can be endomorphisms that aren't automorphisms, but then genus $1$ Riemann surfaces can also be naturally made into abelian groups --- so-called elliptic curves --- and there is a whole theory, the theory of complex multiplication, devoted to studying their endomorphisms rings.)

As another example, any ring of char. $p > 0$ has a Frobenius endomorphism, which is not an automorphism in general; but the semigroup of endomorphisms
that it generates is typically an important thing to consider in char. $p$ algebra and geometry.  (Of course, this semigroup is just a quotient of $\mathbb N$.)

One thing to bear in mind is what you hope to achieve by considering the group/semigroup of automorphisms/endomorphisms.

A typical advantage of groups is that they admit a surprisingly rigid theory (e.g. semisimple Lie groups can be completely classified; finite simple groups can be completely classified), and so if you discover a group lurking in your particular mathematical context, it might be an already well-known object, or at least there might be a lot of known theory that you can apply to it to obtain greater insight into your particular situation.

Semigroups are much less rigid, and there is often correspondingly less that can be leveraged out of discovering a semigroup lurking in your particular context.  But this is not always true; rings are certainly well-studied, and the appearance of a given ring in some context can often be leveraged to much advantage.

A dynamical system involving just one process can be thought of as an action of the semigroup $\mathbb N$.  Here there is not that much to be obtained from
the general theory of semigroups, but this is a frequently studied context. (Just to give a perhaps non-standard example, the Frobenius endomorphism of a char. $p$ ring is such a dynamical system.)
But, in such contexts, precisely because general semigroup theory doesn't help much, the tools used will be different.

E.g. in topology, the Lefschetz fixed point theorem is a typical tool that is used to study an endomorphism of (i.e. discrete dynamical system on) a
topological space.  Interestingly, the same formula is used to study the action of Frobenius in char. $p$ geometry (see the Weil conjectures).  So even in contexts
such as action of the semigroup $\mathbb N$, there is some coherent philosophy that can be discerned --- it is just that it is supplied by topology rather than algebra, since the algebra doesn't have all that much to say.

I think the conclusion to be drawn is not to be too doctrinaire, and to be sensitive to the actual mathematical contexts in which and from which the various notions of group, semigroup, automorphism, and endomorphism arise and have arisen.

An arbitrary semigroup may lack an identity (or left or right element) element and/or inverse elemments for every element in the semigroup. This prevents such a rich theory without other constraints on the semigroup.


Automorphisms of transcendental extensions. I'm trying to solve the following two exercises from chapter 18 of Garling's A Course in Galois Theory.

I haven't been able to get started (only thing I've done is verify that $\sigma^3$ is the identity, and calculated some examples that me guess the fixed field of $\sigma$ is K). The chapter of the book where this is taken from gives few hints on where to start to tackle these as it only mentions some theorems about algebraic independence (which won't be aplicable here as the extension is simple) and the fact that $K(\alpha)$ is $K(x)$ if $\alpha$ is transcendetal over $K$ which I already knew and its the way I think about $K(\alpha)$ but I haven't been able to put it to good use.

Any help would be greatly appreciated.

I use the variable $t$ instead of $\alpha$ since its easier type.

Clearly the elementary  symmetric functions in $t,\sigma(t), \sigma^2(t)$ belong to the fixed field. And further $t$ is the root of a cubic polynomial in these functions. If you calculate them you get only one quantity,
$$s=\frac{t^3-3t+1}{t(1-t)}$$

Thus $k(t)/k(s)$ is an extension of degree $3$ with $\sigma$ as automorphism.
Thus the fixed field is $k(s)$.


If $A^*x_n \to y$, there exists a sequence such that $A^*Ay_n \to y$. I'm struggling with a problem from Young's Introduction to Hilbert Space (7.30 to be more specific)

Let $\mathbb{H}$ be a Hilbert space, $A\in B(\mathbb{H})$ (ie a bounded operator) and $(x_n)_{n=1}^{\infty}$ a sequence in $\mathbb{H}$.
Prove that if $A^*x_n\rightarrow y$, there exists a sequence $(y_n)_{n=1}^{\infty}$ such that $A^*Ay_n \rightarrow y$.

I'm 99% sure this has something to do with the relationship between the kernel and the image of $A$ and $A^*$ ($\ker A^*$ being the orthogonal of the image of $A$, etc), but I haven't been able to make much progress. Could I get a hint on how to proceed?

Recall that $\ker A = \ker A^*A$. Namely, clearly $\ker A \subseteq \ker A^*A$. Conversely

$$x \in \ker A^*A \implies A^*Ax = 0 \implies 0 = \langle A^*Ax, x\rangle = \langle Ax, Ax\rangle \implies Ax = 0\implies x \in \ker A$$

Since $(\ker T)^\perp  = \overline{\operatorname{Im} T^*}$, taking orthogonal complements in the above relation gives

$$\overline{\operatorname{Im}A^*} = \overline{\operatorname{Im}A^*A}$$

Therefore, $A^*x_n \to y$ implies that $y \in \overline{\operatorname{Im}A^*}$, so $y \in \overline{\operatorname{Im}A^*A}$ implying that there exists a sequence $(y_n)$ such that $A^*Ay_n \to y$.

The image of a linear operator is closed, so $y$ is a limit of elements of an image iff it is an element of that image.  So this is equivalent to showing that given any $v$, there is some $u$ such that $A^* Au = A^*v$.

$v$ can be written as $h+p$, where $h$ is in $ker A^*$ and $p$ is orthogonal to $ker A^*$, and $A^*v =A^*p$. So we need, given arbitrary $p$, to find $u$ such that $Au = p$.


Endomorphism $T : V \to V$ $ \Leftrightarrow \det T \neq 0$ Prove that $T$ is bijective $\Leftrightarrow \det T \neq 0$, where $T$ is a linear map $T : V \to V$ of a finite-dimensional vector space $V$.

Note that the determinant of the square matrix $A$ of the linear map $T$ is denoted by $\det T$ because $\det A$ depends only on $T$.(Not sure if this is common notation or not so I put this here)

I'm not sure what I can do to prove this. I don't think I've yet developed an intuition for linear algebra but I sadly don't see the connection between the two facts we need to prove.

Recall three  facts:

1) Let $A$ be a square matrix, and let $B$ be another matrix, obtained from $A$ via an elementary operation. Then $\det A=0\Leftrightarrow\det B=0.$

2) If the rows of $A$ are linearly dependent, then it is possible to obtain from $A$ a matrix with a $0$ row via elementary operations.

3) If $A$ represents a transformation which is not bijective, then the rows of $A$ are linearly dependent.

On the other hand, if $A$ represents a bijective transformation, then $A$ is invertible. By multiplicativity of the determinant we have$$\det A\cdot\det(A^{-1})=1,$$which yields$$\det A\neq0.$$

If $A$ is invertible then $AA^{-1} = I$ and then $\det(A) \det(A^{-1}) = 1$ and then $\det(A)$ is invertible and therefore not zero.

Now, since $A A' = A'A = \det(A) I$ where $A'$ is the classical adjoint of $A$, $\det(A)\neq0$, implies that $\frac{1}{\det(A)} A'$ is the inverse of $A$


Polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers) Let the polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers). If $f(p) = f(q) = 0$, where $ 0 < p < q < 1$, then find the minimum possible value of $a$.
The vertex is $-\frac{-b}{2a}=\frac{b}{2a}>0$ and lies between $0$ & $1$.
$f(0)>0$ and also $f(1)>0$, hence $c>0$ and $a-b+c>0$, also $b^2-4ac>0$. Even after proceeding up to these steps I am not able to find the minimum value of $a$.

Let the quadratic be $a(x-u)^2-v$, where $u=(p+q)/2$ and $-v=f(u)\lt0$.
Then $f(0)=au^2-v\ge1$ so $au^2\gt1$ and likewise $a(1-u)^2\gt1$.  Either $u\le\frac12$ or $1-u\le\frac12$ so $\frac a4\gt1$, and $a\ge5$.
An example is $f(x)=5x^2-5x+1$

You have $ax^2-bx+c = a(x-p)(x-q)=ax^2-a(p+q)x+apq$. So $a(p+q)=b\geq 1$ and $apq=c\geq 1$. This implies $a\geq 1/(p+q)$ and $a\geq 1/pq$. But your constraints on $pq$ force $1/pq>1/(p+q)$. So really the bound is:
$$
a\geq 1/pq.
$$
Without more information there is no improvement.
For example let $p=1/3$ and $q=1/2$. Then You can take $a=6$, $b=5$, and $c=1$.


When $\min \max = \max \min$? Let $X \subset \mathbb{R}^n$ and $Y \subset \mathbb{R}^m$ be compact sets.

Consider a continuous function $f : X \times Y \rightarrow \mathbb{R}$.

Say under which condition we have

$$ \min_{x \in X}  \max_{y \in Y} f(x,y) = \max_{y \in Y} \min_{x \in X} f(x,y). $$

From this we have that $\max_{y \in Y} \min_{x \in X} f(x,y) \leq \min_{x \in X}  \max_{y \in Y} f(x,y)$. So here we are looking for conditions on $f$ such that we have the equality.

If you look at this problem as a Primal Problem and its Dual Problem then you're basically asking when Strong Duality Holds.

Basically when $ f\left( x, y \right) $ is Convex in $ x $ and Concave in $ y $.

take two examples:

$$f(x,y) = \cos(x+y)$$

and

$$f(x,y)=2xy(x-y)$$

In the first case $\min_x \max_y f(x,y)=1 $ and $\max_y\min_x f(x,y)=-1$ for all $x,y$.

In the second one we have $y=0.5x$ and $x=0.5y$ since we have $f(x,y)^{''}>0$ for $\frac{d^2}{dx}f(x,y)$ and $f(x,y)^{''}<0$ for $\frac{d^2}{dy}f(x,y)$, they are minima and maxima respectively. As a result one gets $$\min_x \max_y f(x,y) = \max_y \min_x f(x,y)$$

for the second case.


Growth of Derivative Assume that $f \in C^1(\mathbb{R})$ such that $\lim_{x \rightarrow 0} x^{-(k+1)} f(x) = 0$ for some $k \in \mathbb{N} = \{1, 2, 3, \ldots \}$. Do we have $\lim_{x \rightarrow 0} x^{-k} f'(x) = 0$?

I suppose this question is really simple, but I'm struggeling with finding a counterexample or proof for this statement even if $k = 1$. Any suggestions?

I think I found the counterexample just after I posted the question: Take $f(x) := x^3 \sin(1/x)$. We have $f(x) / x^2 = x \sin(1/x) \rightarrow 0$ as $x \rightarrow 0$, but $f'(x) = 3x^2 \sin(1/x) - x \cos(1/x)$ and hence $f'(x) / x$ doesn't converge as $x \rightarrow 0$.


Find $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4x} dx$. I was wondering, how to calculate the following definite integral that appeared in an integration bee (preferably the quickest way possible).

$$\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4x} dx$$

Hint:

$$\sin^4x + \cos^4x = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x$$

And again

$$2\sin^2x\cos^2x = 2(\sin x \cos x)^2 = 2\left(\frac{\sin2x}{2}\right)^2 = \frac{1}{2}\sin^22x$$

Hence in the end

$$\sin^4x + \cos^4x = 1 - \frac{1}{2}\sin^22x$$

You can treat this.


Schurs Lemma for endomorphisms Schur's lemma states that for an irrep $(\varphi,V)$ any endomorphism $\phi: V \mapsto V$ is given by a scalar mapping.

Lets say we are in the complex case, then this would mean:
$\phi = \lambda I$

Since any homomorphism between two representations is defined by its property $\varphi_1 = \phi^{-1} \varphi_2 \phi$, this would mean:

$\varphi_1 = (\lambda I)^{-1} \varphi_2 (\lambda I) = \varphi_2$

Therefore only a single irrep over $V$ exists, since $\varphi_1 = \varphi_2$.
What am I missing?

This should be false, since I found for the group $D_4$ at least two different irreps:
$$1:  \langle \{ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \}\rangle$$
$$2:  \langle \{ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \}\rangle$$

where $\phi$ should equal $\lambda I$, but for all $\lambda$
$$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \ne (\lambda I)^{-1} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} (\lambda I).$$

Where goes everything wrong? Thanks for any help!

In the first two sentences you state versions of Schur's lemma involving an endomorphism $\phi$ (although the first statement is not correct unless you already assume yourself to be over an algebraically closed field). Starting from the third sentence you take $\phi$ to be a morphism, not an endomorphism. But what does it mean for a morphism between two representations to be multiplication by a scalar?

The correct version of Schur's lemma for a morphism between two irreducible repesentations is that the morphism must be either zero or invertible. It is zero if they are not isomorphic and invertible if they are. Note that in your second equation you assume that $\lambda \neq 0$.

Unitary has nothing to do with any of this.

Doesn't the condition for two irreps to be equivalent require the transformation ($\phi$ in your case) to be unitary? If $|\lambda|\neq 1$, the corresonding transformation is non-unitary.


If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$ Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
  Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.

(problem composed by Laurentiu Panaitopol)

So far no idea.

For any positive integer x this is true: $x \leqslant x^2$ (From $1 \leqslant x$ for any positive ineger x ). So $a + b + c \leqslant a^2 + b^2 + c^2$. But for 2 positive integers $x$, $y$ $x$ is divisible by $y$ only if $x \geqslant y$. So $a + b + c \geqslant a^2 + b^2 + c^2$ if $a + b + c \mid a^2 + b^2 + c^2$.
From these 2 inequalities:
$a + b + c = a^2 + b^2 + c^2$

So  $a + b - a^2 - b^2 = c^2 - c$

Evaluation for the right part $c^2 - c \geqslant 0$ (because $x \leqslant x^2$ for any positive integer x).
Evaluation for the left part $a + b - a^2 - b^2 \leqslant 0$ (adding 2 inequalities $a - a^2 \leqslant 0$ and $b - b^2 \leqslant 0$)

So the left part is $\leqslant 0$ and the right part $\geqslant 0$. But they are equal so $a + b - a^2 - b^2 = c^2 - c = 0$ and $c^2 = c$. c is positive so we can divide both part of the last equation by c and get $c = 1$.

Similarly $b = 1$ and $a = 1$. So $a + b + c = 3$ and $a^n + b^n + c^n = 3$ for any positive $n$. 3 is divisible by 3 so $a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer n.


Sum of divisors is power of 2, is n square-free? We were given this question on one of our recent exams, and we can't seem to generate a proof, nor find a counter-example.

If we let $\sigma(n)$ be the sum of the divisors of $n$, the question is, if

$$\sigma(n)=2^k$$

for some $k$, does this imply $n$ is square free?

This is clearly true if we consider just the number of divisors being a power of two, but we cannot extend it to the sum of the divisors. We have tried numerous $n$ that are not square-free, and none have given a counter-example. We made some progress in noting that if

$$n=p_1^{k_1}p_2^{k_2}...p_j^{k_j}$$

is the prime factorization of $n$, then

$$2^k=\sigma(n)=(1+p_1+p_1^2+...+p_1^{k_1})...(1+p_j+p_j^2+...+p_j^{k_j})$$

This then implies that each factor must be a power of two itself, and thus for each $p_i$,

$$p_i+p_i^2+...+p_i^{k_i}=2^t-1$$

for some $t$. Even with this, we are unable to find a counter-example. There does not seem to be a non-trivial solution.

Is there a proof we can't see, or is there a counter-example we can't find?

EDIT: From a comment by @lulu $\sigma(n)=2^k$ when $n$ is a product of distinct Mersenne primes, and is thus square-free. Since we have not covered these in our class, I'm still open to other solutions.

Mersenne prime is a  prime number that can be written in the form Mn = 2^n − 1 for some integer  n. mersenne numbers are all integers of that same form including the ones that fail primality test .


Why are they called "Isothermal" Coordinates? If I understand correctly, Gauss proved that given any oriented Riemannian surface, one can find a complex structure on the surface so that the metric on the charts is just $f|dz|$, where $f>0$.

I've heard these coordinates referred as "conformal coordinates," which makes sense, but I've also heard of them referred to as "isothermal coordinates."

Why are these coordinates called "isothermal"? Who named it "isothermal"? Does it have anything to do with physics, or is it kind of like "Inertia"?

If $(u,v)$ are isothermal coordinates, then $u$ and $v$ are harmonic functions with respect to the Laplace-Beltrami operator on your Riemannian manifold, that is $\Delta u=\Delta v=0$.

Now, the equation $\Delta f=0$ characterizes the stationary states for the heat equation. The level curves for a harmonic function are therefore the isothermal curves for some heat distribution.

It follows that if $(u,v)$ is an isothermal system of coordinates, then the level curves, i.e., the coordinate lines, are isothermal curves.

I can recommend you the book "Modern Differential Geometry of Curves and Surfaces with Mathematica" Alfred Gray. This would be useful.


Is there any way to define arithmetical multiplication as other thing than repeated addition? Is there any way to define arithmetical multiplication as other thing than repeated addition?
For example, how could you define $a\cdot b$ as other thing than $\underbrace{a+a+\cdots+a}_{b \text{-times}}$ or $\underbrace{b+b+\cdots+b}_{a \text{-times}}$?

Given two sets $A$ and $B$ of cardinality $a$ and  $b$, respectively, the cardinality of the cartesian product $A\times B$ is called the product of $a$ and $b$, and is denoted by $a\cdot b$.

Update

When I wrote this answer I didn't have infinite sets in mind. I just wanted to convey a mental picture of multiplication that does not involve repeated addition.

No. Because this is the fundamental definition of arithmetical multiplication, that, repeated addition is known as multiplication.

If you try to generate another definition of multiplication then you have to use this definition either directly or indirectly.


How to show $\{f_n\}_{n=1}^\infty$ has uniformly convergent subsequence on [0,1]? Let $\{f_n\}_{n=1}^\infty$    a sequence of second order differentiable functions on the interval [0,1].

If $\forall n\in \Bbb N$ $f_n(0)=f_n'(0)=0$ and for all $n\in \Bbb N$ and $x \in [0,1]$  ,      $|f_n''(x)|\le 1$  then show $\{f_n\}_{n=1}^\infty$ has uniformly convergent subsequence on [0,1].

Do we use arzela-ascoli which says if A is compact and $B\subset C(A;\Bbb R^m)$ is bounded and equicontinous, every sequence in B has a convergent subsequence at $C(A;\Bbb R^m).$ ?

Yes, you can use Arzela-Ascoli's theorem. To prove boundedness of $B = \{f_n \mid n \in \mathbb N\}$ recall that
$$ f_n(x) = \int_0^x f_n'(\xi)\, d\xi = \int_0^x \int_0^\xi f_n''(\eta) \,d\eta\, d\xi $$
as $f_n(0) = f_n'(0)= 0$. For equicontinuity, prove that the $f_n$ are uniformly Lipschitz, using the mean value theorem and boundedness of $(f_n')$.

By Lagrange Theorem we have $$|f'_n (x) |=|f'_n (x) -f'_n (0) | =|f''_n (\xi ) ||x|\leq 1 \mbox{ for } 0\leq x\leq 1.$$
Analogously $$|f_n (x) |=|f_n (x) -f_n (0) | =|f'_n (\xi_1 ) ||x|\leq 1 \mbox{ for } 0\leq x\leq 1.$$
Hence the family $\{ f_n :n\in\mathbb{N}\}$ is uniformly bounded.
Moreover $$ |f_n (u) -f_n (v) |=|f'_n (\xi_2 ) ||u-v|\leq |u-v|$$ hence the family $\{ f_n :n\in\mathbb{N}\}$ is equicontinous. So by the Arzela - Ascoli's Theorem the sequence $(f_n )$ has convergent subsequence.


Why do we write 'finite' as an adjective to proposition when defining an axiomatic system? At 58:54, Professor Frederic Schuller defines an axiomatic system in the following way:

An axiomatic system is a finite sequence of propositions $a_1,a_2....,a_n$ which are called axioms

Why do we have the word finite before the sequence? Precisely speaking into what I want, What problems would be run into if we had an infinite sequence of axioms?

This isn't actually something we do in general! Both $\mathsf{ZFC}$ and (first-order) $\mathsf{PA}$, two of the most important axiom systems in mathematics, are not finitely axiomatizable. So Schuller is just wrong here. It's been a while but my recollection is that in general he's pretty sloppy about the treatment of foundations - see e.g. this older MSE question (and I recall another question along similar lines, but I can't find it right now). I recommend finding a different source on this topic.

If we have an infinite sequence of axioms, then we need some sort of method to define how the sequence of axioms could get produced given an "infinite amount" of time.  Of course though, there is no such thing as an infinite amount of time that anyone has ever known from experience at least.  I read also one person analyzing Galielo a while back also saying that he never used the term "infinite number", and regarded that phrase as a contradiction ( the concept of "infinite number" probably could get contested at least somewhat successfully by an ultrafinist argument).  Also, no one has ever been, nor will be ever able to, produce an infinite sequence.  And again, we need some method to define all of those axioms.
A finite sequence of axioms doesn't have these (potential) issues.  At least for some small finite sequences, we can list the entire sequence, and thus determine at a glance whether or not some well-formed formula belongs to the sequence of axioms.  We don't need some method to define it.  There is no need, ever, to get into addressing perhaps a very well-founded objection of there not existing an "infinite amount" or "infinite number", since those concepts are irrelevant to what any given axiomatic system consists of.  Finite sequences of well-formed formulas exist.  Infinite sequences of well-formed formulas existing is at least potentially false.  Therefore, using a finite sequence of axioms has at least a few advantages, and there can exist strong reasons to prefer them.
Additionally, a finite sequence of axioms can get shown to have an independent set of axioms by a finite process.  At the very least, the process of showing such seems potentially more complicated for an infinite set of axioms.
A computer can process some or many finite sequences of axioms.  A computer can't process an infinite sequence of axioms, since there is no computer that can perform an "infinite number" of computations and will only for a finite amount of time, correct?


Is every composite number the average of two primes? I'm interested in this question because it relates to a bad joke about people in their prime.

It seems to work for the first 20 numbers:

4 is the average of 5 and 3.
6 is the average of 5 and 7.
8 is the average of 5 and 11.
$\vdots$
16 is the average of 13 and 19.
$\vdots$

It would not be hard to write a program that checks more cases, but I suspect the hypothesis is true. If that is the case, then a more mathematical approach is needed.

So you are asking whether for every composite number $n$ there exist primes $p,q$ such that
$$n=\frac{p+q}{2}$$
That is, $2n=p+q$, so you are asking whether $2n$ can be written as the sum of two primes.

The question whether every even integer greater than $2$ is a sum of two primes is a famous open problem known as the Goldbach conjecture.

I think that it is all about prime gap.

If all natural numbers larger than 4 can be proved that it is the average of primes, the Goldbach's Conjecture will be solved, immediately


$\lim_{n \to \infty} x_n = 0$ and $\lim_{n \to \infty} f(x_n) = a$ Let $f(x)$ be a continuous function defined on $(0,1]$. Define a  subset $I \subseteq \Bbb R$ as follows: $a\in I$ if and only if there exists a sequence in the interval $(0,1]$ such that $\lim_{n \to \infty} x_n =0$ and $\lim_{n \to \infty} f(x_n) = a.$

I want to show that in general, if $I$ is nonempty, then it is a connected closed set.

When $f(x)$ is $\sin1/x$, I think $I=[-1,1]$ by seeing graph. (I was taught this by Elliot G.). More precisely,by taking $x_n＝1/2nπ ,y_n＝1/（２nπ＋1/2）$,we can check $1,-1∈I$.In the same way, for arcitrary a∈[-1,1],we just need to let $x_n＝1/（２πn＋arcsina）$.
But in general, I cannot show $I$ is connected closed set in real line.
According to Michael, we just need to show $ let a,b∈I,a＜b,c∈（a,b）,
then, c∈I$. I understand that, but I don't know how to apply intermediate value theorem in this situation. Thank you.

For connectivity, apply the intermediate value theorem. Suppose that $x, y \in I$ and without loss of generality take $x < y$. Then there exist monotonically decreasing sequences $x_n$, $y_n$ converging to $0$ so $\lim\limits_{n \to \infty} f(x_n) = x$, $\lim\limits_{n \to \infty}f(y_n) = y$. Using terms from these sequences, one can form a decreasing sequence $z_n \to 0$ so that $\lim\limits_{k \to \infty}f(z_{2k}) = x$ and $\lim\limits_{k \to \infty}f(z_{2k+1}) = y$. Now fix $\alpha$ so $x < \alpha < y$. We can find a sequence of real numbers $\alpha_k$ so that for all $k$, $f(z_{2k}) \leq \alpha_k \leq f(z_{2k+1})$ and $\lim\limits_{k \to \infty}\alpha_k = \alpha$. Continuity of $f$ and the intermediate value theorem provide us a decreasing sequence $\tilde{z}_k \to 0$ so $f(\tilde{z}_k) = \alpha_k$. We conclude that $\alpha \in I$, so $I$ is connected.
To see that $I$ is closed, take $\alpha \in \mathbb{R}$ and a sequence $\alpha_k$ in $ I$ converging to $\alpha$. By the definition, we get decreasing sequences $x_{k, n}$ so that $\lim\limits_{n \to \infty} x_{k,n} = 0$ for all $k$ and $\lim\limits_{k \to \infty } f(x_{k,n}) = \alpha_k$ for all $k$. With this one can inductively construct a decreasing sequence $x_k$ converging to $0$ so that $|f(x_k) - \alpha_k| < \frac{1}{k}$. One can then see that $\lim\limits_{k \to \infty}f(x_k) = \alpha$, so $\alpha \in I$ completing the proof.
One should note that $I$ need not be a closed interval. The function $f(x) = \frac{1}{x} \sin(1/x)$ is an example where $I = \mathbb{R}$.

Assume $a \notin I$.  Then $\exists \varepsilon \gt 0$ such that $\{ x \mid \vert f(x) - a \vert \lt \varepsilon \}$ is finite.  But then there must be an open ball around $a$ for which the same statement is true:  If $\vert y - a \vert \lt \frac {\varepsilon}{2}$, then $\{ x \mid \vert f(x) - y \vert \lt \frac {\varepsilon}{2} \} \subseteq \{ x \mid \vert f(x) - a \vert \lt \varepsilon \}$ is finite (because it's a subset of a finite set), so also $y \notin I$.  Therefore $\Bbb R \setminus I$ is open, and $I$ is closed.
Alex Nolte's proof demonstrates that $I$ must be connected.


Testing properties of mapping given graph Suppose we are given a graph of a mapping (in $\Bbb R^2$ or $\Bbb R^3$) and we want to test graphically if it is a graph of a function, if it is one-to-one and if it is onto.

$\Bbb R^2$ : $$\begin{align}
\text{Function} & \iff \text{all lines parallel to the y-axis intersect the graph no more than once}.\\
\text{one-to-one}& \iff \text{all lines parallel to the x-axis intersect the graph no more than once.}\\
\text{onto}& \iff \text{all lines parallel to the x-axis intersect the graph at least once.}\\
\end{align}$$

$\Bbb R^3$ : $$\begin{align}
\text{Function} & \iff \text{all lines parallel to the z-axis intersect the graph no more than once}.\\
\text{one-to-one}& \iff \text{all planes parallel to the xy-plane intersect the graph no more than once.}\\
\text{onto}& \iff \text{all planes parallel to the xy-plane intersect the graph at least once.}\\
\end{align}$$

Are these methods correct?

In the case of Function, change "no more than once" to "exactly once" (assuming the function is defined on all of $\mathbb R$ or $\mathbb R^2$ respectively.


Let $G$ a group of order $6$. Prove that $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$. Let $G$ a group of order $6$.  Prove that:
i) $G$ contains 1 or 3 elements of order 2.
ii) $G \cong \Bbb Z /6 \Bbb Z$ or $G \cong S_3$.

I haven´t covered Sylow groups and normal groups. This is an exercise from the chapter about group actions. I have covered Lagrange and cosets.

Surely given $n\in \mathbb N, n\geq 1$, there exists a cyclic group of order $n$: In our case, there exists a cyclic group $G\cong \mathbb Z /\mathbb 6Z \cong \mathbb Z_6$.

So that covers one option: Now, $\mathbb Z_6$ has exactly how many elements of order $2$?   Use Lagrange, and determine the possible orders of subgroups. Any element of order $2$ will generate a subgroup of order $2$. $Z_6$ has only one subgroup of order $2$: What subgroup of $\mathbb Z$ has order $2$ ,and which element necessarily has order $2$?
For any group of even order, there can exist only an odd number of elements of order $2$. Why?
We can rule out $5$ such elements for a group of order $6\;\ldots\quad$WHY?
Hence, that leaves us with $G$ having $1$ or $3$ elements of order $2$.  $\mathbb Z_6$ covers one possibility. If a group $G$ of order $6$ has $3$ elements of order $2$, how does, how does this fully determine the corresponding group $G$? (For any $G\not\cong \mathbb Z_6$, all it's elements must of order $1$,$2$, or $3$.)

Now, use these facts to justify that the only groups of order $6$ must be isomorphic to $\mathbb Z_6$ or to $S_3$.

If $G$ isn't $\mathbb Z/6\mathbb Z$, then it isn't cyclic. Hence, the maximal order is 3 (see Lagrange). It easily follows that an element of order 2 exists as well. From here to show that it's actually $S_3$ should be trivial. This solves the second part.

The first part follows directly from the second.


Is the inverse of a symmetric matrix also symmetric? Let $A$ be a symmetric invertible matrix, $A^T=A$, $A^{-1}A = A A^{-1} = I$ Can it be shown that $A^{-1}$ is also symmetric?

I seem to remember a proof similar to this from my linear algebra class, but it has been a long time, and I can't find it in my text book.

You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more complete proof. Given A is nonsingular and symmetric, show that $ A^{-1} = (A^{-1})^T $:

$$ I = I^T $$

since $ AA^{-1} = I $,

$$ AA^{-1} = (AA^{-1})^T $$

since $ (AB)^T = B^TA^T $,

$$ AA^{-1} = (A^{-1})^TA^T $$

since $ AA^{-1} = A^{-1}A = I $, we rearrange the left side

$$ A^{-1}A = (A^{-1})^TA^T $$

since $ A = A^T $, we substitute the right side

$$ A^{-1}A = (A^{-1})^TA $$
$$ A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$
$$ A^{-1}I = (A^{-1})^TI $$
$$ A^{-1} = (A^{-1})^T $$

and we are done.

Inverse of $A$ can be expressed as a polynomial $p(A)$ of $A$ (from Cayley-Hamilton theorem).

So it is sufficient to prove that if $A$ is symmetric then power $A^k$ is symmetric, sum of symmetric matrices is symmetric and multiply by scalar is symmetric.


What can we say about the convergence of this series $$\sum {z^n\over n!} $$

I used Alembert's Ration test and get $$\lim_{n \infty}{u_n\over u_{n+1}}={n+1\over z}$$ As this tends to $\infty>1$ can i say that the given series is convergent for all values of $z$ ?

Note : z is a complex number

Yes. A good thing to remind is the proof of this property, which in your case writes:

$$
\frac{u_{n+1}}{u_n} = \frac z{n+1}
\implies
\left|\frac{u_{n+1}}{u_n}\right| \le 1/2 <1
$$when $n$ is big enough, and then
$$
|u_n| \le C2^{-n}
$$for every $n$, for a certain $C>0$. Hence the series is (absolutely) convergent.

The summation of the given series can be given as $e^z$ and hence the given series is convergent.


Why is the range of $D(x) = \frac{(5x+8)^2}{(x-15)^2}$ all non-negative numbers? $$D(x) = \frac{(5x+8)^2}{(x-15)^2}$$
The domain = all real numbers excluding $15$, because we cannot divide by $0$.
But, why is the range of this all real numbers greater than or equal to 0?
I've tried setting $D(x) = y$ and solving for $x$, but it's quite a tricky thing to solve for $x$. Any easier methods that you guys can suggest?

It is clear that for all $x \in \mathbb{R} \setminus \lbrace 15 \rbrace$, you have $D(x) \geq 0$. So the range is included in $\mathbb{R}_+$.
Moreover, $$\lim_{x \rightarrow 15^-} D(x) = +\infty \quad \text{ and } \quad D \left( \frac{-8}{5}\right) = 0$$ By continuity on $(-\infty, 15)$, the intermediate value theorem tells you that $\mathbb{R}_+$ is included in the range.
So the range of $D$ is $\mathbb{R}_+$.

Write
$$
D(x) = \frac{(5x+8)^2}{(x-15)^2} = \left(\frac{5x+8}{x-15}\right)^2 = f(x)^2
$$
The image of $f$ is $\mathbb R \setminus \{5\}$ because it is easy to solve $f(x)=a$ when $a\ne 5$. Therefore, image of $D$ is $\mathbb [0,\infty)$ except possibly for $25$. Since the image $f$ contains $-5$, the image of $D$ contains $(-5)^2=25$ and so is actually $[0,\infty)$.


When is the double cover of a smooth curve reducible Let $C$ be a smooth projective plane curve. Suppose $f:C'\rightarrow C$ is a double cover of $C$ ramified at $d$ points. Suppose $C'$ is irreducible, then the Riemann Hurwitz formula says that $d$ has to be even, is that correct?

Consequently, if I know that $d$ is odd, does that mean that the double cover $C'$ is reducible?

Yes, I think you are correct. The RH-formula for a branched double cover $f:S' \rightarrow S$ gives:$$ \chi(S') = 2\chi(S) - \sum_{p \in S'}(E_{p}-1) $$ where $e(p)$ is the ramification index at $p\in S'$. Since $f$ is a ramified cover of degree 2 the map is $2:1$ on the complement of the ramification subset, implying that $e(p) \leq 2$ for all $p \in S'$, since otherwise we get a point in $S$ (over which the map is regular) whose pre-image has more than $3$ preimages: a contradiction.

Hence, $\sum_{p \in S'}(E_{p}-1)$ is equal to the number of ramification points in $S'$, and is clearly even by the formula.

The definition of ramified cover implies easily that a ramified cover of a Riemann surface has to be a (possibly disconnected) Riemann surface. So, if a curve is branch covered by some reducible curve, then the irreducible components are pairwise disjoint smooth curves. Then we can apply the above argument to each component of the domain, giving in general that the number of ramification points is always even. Although I should mention that if $S'$ has more than one component, then we have that $S'$ is just two copies of $C$, in particular it is an unramified cover.


If two parabolas don't intersect, is there a line that doesn't intersect either of them? Two parabolas in a plane are given, such that they don't intersect. Is it true that there is a line in plane such that doesn't intersect any of them?

Parabolas are convex, so they'll always be on the same side of their tangent lines. Find both tangent lines at the points of closest approach. These lines are parallel. Intuitively, if they were not then taking a 'small step' along the parabola in the right direction would bring you closer. (Slightly) more rigorously, if you know how to derive the method of lagrangian multipliers, this is a similar idea.  so you can pick any line between these parallel lines to satisfy the problem.

If one is contained inside the other the above won't work, because even when a point of closest approach does exist, the parabolas will both be on the same side of the tangent line. It's a simple case though, just use the directrix of the larger parabola as your answer.

Here's a simple approach, which I hope will suit.

Here is a parabola, courtesy of WolframAlpha:

Here is another parabola.

Do the parabolas intersect? (No.) Is there a line, say y = -0.25, that won't intersect either parabola? (Yes, and there are others.)


A Hard Geometry Problem on circle $$\angle(ABC) = 30°\\
\angle(BCO) = 20°\\
\angle(OCD) = 20°$$

How do i find $\angle(ODC)$? so i wanted to show my teacher this but he is not available yet. Can someone help me to solve? geometry problems on circle seems hard to me. Thanks!

Without loss of generality, we can consider $R=OC=OB=1$.

From $\Delta OBC: \frac{OB}{\sin20}=\frac{BC}{\sin140}\Rightarrow BC=\frac{\sin140}{\sin20}=2\cos20.$

From $\Delta BCD: \frac{BC}{\sin110}=\frac{CD}{\sin30}\Rightarrow CD=\frac{\cos20}{\sin110}=1.$

Finally from $\Delta OCD: \frac{OC}{\sin{x}}=\frac{CD}{\sin{(x+20)}}\Rightarrow \sin x=\sin{(x+20)}\Rightarrow x=80.$

First join $AC$ and $AO$. Now, $\angle ABC$ is $30^\circ$,so AOC is 60*.AO=OC.so,OAC=OCA=(180*-60*)÷2=60*. Triangle AOC is equilateral Triangle.so,AC=OC=OA.angleACD=(ACO-DCO)=(60*-20*)=40*.In triangle ACB ,CAB=70*,ACD=40*.So,ADC=70*.So,AC=CD. Now AC=CO=CD,So,OC=CD.in triangle CDO,DCO=20*,and OC=CD.So,CDO=COD=(180*-20*)÷2=80*.HENCE,CDO=80*.(ANS.).


Conditional expectation of X given X+Y I suppose this is a trivial question, but sometimes you´re just stuck.

Let $X,Y$ be integrable i.i.d. Random Variables. Then $E(X|\sigma(X+Y)) = E(Y|\sigma(X+Y))$.

To show this I obviously have to verify $\int XZ \; dP = \int YZ \; dP $ for any $\sigma(X+Y)$-measurable $Z$. But I just can´t formulate a rigorous general proof...

Hint: $X$ and $Y$ may be realized on a probability space $\Omega = \Omega_1 \times \Omega_2$.   Consider the symmetry $(x,y) \mapsto (y,x)$ of $\Omega$.


I'm trying to find the longest consecutive set of composite numbers Hello and I'm quite new to Math SE.

I am trying to find the largest consecutive sequence of composite numbers. The largest I know is:

$$90, 91, 92, 93, 94, 95, 96$$

I can't make this series any longer because $97$ is prime unfortunately.

I can however, see a certain relation, if suppose we take the numbers like (let $a_1, a_2, a_3,...,a_n$denote digits and not multiplication):

$$a_1a_2a_3...a_n1,\ a_1a_2a_3...a_n2,\ a_1a_2a_3...a_n3,\ a_1a_2a_3...a_n4,\ a_1a_2a_3...a_n5,\ a_1a_2a_3...a_n6,\ a_1a_2a_3...a_n7,\ a_1a_2a_3...a_n8,\ a_1a_2a_3...a_n9,\ a_1a_2a_3...(a_n+1)0$$

The entire list of consecutive natural numbers I showed above can be made composite if:

The number formed by digits $a_1a_2a_3...a_n$ should be a multiple of 3
The numbers $a_1a_2a_3...a_n1$ and $a_1a_2a_3...a_n7$ should be composite numbers

If I didn't clearly convey what I'm trying to say, I mean like, say I want the two numbers (eg: ($121$, $127$) or ($151$, $157$) or ($181$, $187$)) to be both composite.

I'm still quite not equipped with enough knowledge to identify if a random large number is prime or not, so I believe you guys at Math SE can help me out.

You can have a sequence as long as you wish. Consider $n\in\Bbb{N}$ then the set

$$S_n=\{n!+2,n!+3,\cdots,n!+n\}$$

is made of composite consecutive numbers and is of length $n-1$

I wish to point out 2 ideas which arent as good as the primorial idea nor as good as some of the other ideas, but which are enhancements of the idea mentioned that $n!+2, n!+3, n!+4, .... n!+n$ are composite, which is that
$n!-2, n!-3, n!-4, ... n!-n$ will be composite provided $n!-m=(n!/m-1)*m$ has both factors greater than 1 for $m=2,3,...n$.  clearly the second factor $m$ is ok,
for the other factor we need $n!/m-1>=2$ ie $n!/m>=3$ ie $n!>=3m$, this will be true for all the $m$ if $n!>=3n$ ie $(n-1)!>=3$ which is true if $n-1>=3$, ie $n>=4$
so for $n>=4$, the consecutive numbers $n!-2, n!-3, ...., n!-n$ are composite!
The other enhancement is that if $N=lcm(2,3,4,5,6,...,n)$, then $N-2,N-3,...,N-n$ are composite provided $N-m=(N/m-1)*m$ has both factors greater than 1. The second factor is clearly alright.
so we need $N/m-1>=2$ ie $N/m>=3$ ie $N>=3*m$. Now $(m-1)$ and $m$ are coprime, so $N>=(m-1)*m$. Thus for $m-1>=3$ ie $m>=4$ this is true. We thus need to check $m<4$ namely $m=2,3$. for $m=2$ we need $N>=3*2=6$ which will be true for $n>=3$ because 2 and 3 are coprime and thus $N>=2*3=6$. And for $m=3$ we need $N>=3*3=9$ which will be true for $n>=4$ because 3 and 4 are coprime, hence $N>=3*4=12>9$.
so for $n>=4$, $N-2, N-3, ...., N-n$ are composite!
I agree entirely that this is not as good a result as some of the other ones given, but I give this as the idea simply because it is different and somewhere in between the results given.
Here follows an enhancement of the primorial argument:
if we consider $n\#-2, n\#-3,...,n\#-n$, and choose $2<=m<=n$, and let $p$ be a prime factor of $m$, then $p$ is a factor of both $n\#$ and $m$. So $n\#-m=p*((n\#-m)/p)$, and this is composite if the second factor is greater than 1, ie $(n\#-m)/p>1$, ie $n\#>p+m$. If we can show that $n\#>=3n$ for $n>=5$, then $n\#>=3n>p+m$ and thus $n\#-m$ is composite.
In fact via another post, $n\#>=3m$ for $n>=5$, namely:
Prove by elementary means that $n\#\geq 3n$ for $n\geq 5$, where $n\#$ is the primorial function.
so for $n>=5$, $n\#-2, n\#-3,...,n\#-n$ are consecutive composite numbers in descending order. The result is false for $n=4$ because $4\#-3=3$ and $4\#-4=2$ which are primes.


A committee of 8 people consisting of 3 men and 5 women are lining up next to each other for a photograph. Please help me with the following questions:

A committee of 8 people consisting of 3 men and 5 women are lining up next to each other for a photograph.

i)   In how many different ways can they be arranged for the photo?

ii)  How many arrangements are possible if a male has to stand at both ends
     of the line.

iii) In how many ways can a subcommittee of four people be chosen if it has
     to contain at least one male.

My attempt:

i)   8!/(3!5!) = 56 ways

ii)  6!/5! = 6 ways

iii) 8!/(5!3!) - 5!/(2!4!) = 54 ways

But I am not confident with my answers at all.

When dealing with these sort of counting problems, you should try to figure out "what makes one arrangement a 'different' arrangement than another?"

In general, problems about arranging "people" implies that the people are distinct.  They each have their own unique name, their own unique life experiences, maybe even their own unique heights or ages.  (It is helpful to assume in some cases that their ages are all distinct so that we can have an arbitrary way of ordering them for certain counting arguments).

As such, an arrangement where person $A$ is in one location is considered to be a different arrangement than one where person $B$ is in that location instead, regardless of whether they are the same gender or wearing the same color clothing, etc...

Compare this to a problem about arranging balls where here we instead commonly assume the balls are indistinguishable from one another.  Whether it was this red ball in the first position or a different red ball in that position, we can't tell the balls apart so it does not matter.

For your problem, since we are dealing with arranging people, treat each person as being different.

How many ways to arrange the eight people in a line for a photograph

There are $8!$ ways.  This is seen easily by application of the rule of product and remembering that each person is different.  An answer of $\binom{8}{3}$ would have been correct if we couldn't tell people apart by anything except gender.

How many ways to arrange the eight people in a line for a photograph if at each end there must be a male?

First pick who the male is on the far left.  There are $3$ such choices.  Then pick the male who stands on the far right.  There are $2$ such choices.  Then, from left to right in the remaining available spaces select which person stands there.  This gives a final total of $3\times 2\times 6!$ number of arrangements.

An answer of $6$ would have been correct if we couldn't tell people apart in any way except gender.

How many ways are there to form a subcommittee if it must contain at least one male?

Here, since we started talking about "committees" and "subcommittees" we now traditionally assume that the order within the committee does not matter (E.g., we don't care who the secretary is, who the leader is, who the subordinates are, etc...).

To count the number of ways to select a subcommittee here where we don't care about there being a male or not, this would simply be $\binom{8}{4}$ number of ways.  Since we want at least one male, we can subtract away the "bad" arrangements which are those consisting only of women.  There are $\binom{5}{4}$ "bad" arrangements, so the final total number of committees with at least one male is $\binom{8}{4}-\binom{5}{4}$.

As an aside, you could have done this directly rather than indirectly as $\binom{3}{1}\binom{5}{3}+\binom{3}{2}\binom{5}{2}+\binom{3}{3}\binom{5}{1}$ and arrived at the same answer.  (A now deleted answer incorrectly claimed your attempt for the third problem was incorrect, despite giving the same answer just written differently)

if everyone is distinct, then
1. Simply permute everyone i.e. $8!$
2.Now for the second part, you have to select two men out of three which is ${3 \choose 2}2!$ to place at ends with permutation, for the arrangement part you have 1 man and 5 women after fixing two men. So, their arrangement is ${3 \choose 2}2!6!$.
3. For the third part,
 choose 1 man out of three and 3 women out of 5 i.e $\binom{3}{1}\binom{5}{3}$

choose 2 men out of three and 2 women out of 5 i.e $\binom{3}{2}\binom{5}{2}$

choose 3 men out of three and 1 woman out of 5 i.e. $\binom{3}{3}\binom{5}{1}$
Adding this will give the solution which is $\binom{3}{1}\binom{5}{3}+\binom{3}{2}\binom{5}{2}+\binom{3}{3}\binom{5}{1}$ as posted by  JMoravitz.


Delta function equation difficulty 1) Can we solve if we have two delta functions with $a=b$? Can we solve this? If not why?

$$\int\delta (x−a)\delta(a−x)dx$$

2) Can we solve this

$$\int f(x)\delta (x)dx$$

...where $f(x)=1/x$ i.e. $f(x)$ is infinite at $x=0$ and also delta function is infinite at $x=0$. We can have many such similar $f(x)$ which may face same difficulty.

The problem you are facing with this lies in the fact that the Dirac-delta distribution is not actually a function, it's a type of measure.  So you can't treat it formally in the same sense you would treat a function $f:\mathbb{R} \to \mathbb{R}$.  When you integrate a function with respect to normal integration

$$
\int f(x) dx
$$

you are implicitly integrating the function $f$ with respect to a type of measure called Lebesgue measure, and Lebesgue measure is for the most part consistent with Riemann integration (they are exactly the same in the case of continuous functions).  When you write  $\int f(x) \delta(x-a) dx$ it looks like you're doing Riemann integration, but you are in fact not.  Instead you are integrating the function $f$ with respect to a different measure.

Another way of thinking about measure, is to see measure as assigning "mass" or importance to certain parts of the real line.  In Lebesgue measure, all points are treated equally and mass is assigned to intervals and corresponds to length as we normally would interpret it.  For the Dirac measure given by $\delta(x-a)$, one point is considered to contain all of the mass, namely the point $a$.

So when we think of integrating with respect to Dirac measure, it doesn't really make sense to treat $\delta(x-a)$ as a function.  Therefore writing $\int \delta(x-a) \delta(a-x)dx$ doesn't really make sense.  Another important point is that given a measure defined on a set, we can describe functions as being measurable with respect to that certain measure.  It is only when a function is measurable with respect to a certain measure that we are then allowed to integrate it with respect to that measure.  It so turns out that $f(x) = \frac{1}{x}$ is not Dirac-measurable, therefore we can't compute the integral $\int f(x) \delta(x)$.

The papers in Canadian Journal of Physics: A relook at radiation by a point charge- I and II are based on above discussion. In conventional Poynting’s theorem analysis, the rate of change of work done on a system of charges is written as addition of two terms, rate of change of stored energy and surface integral of Poynting vector. For a delta function source, the first two terms of this equation are either non-integrable or difficult to evaluate.


Which is the best English term for "the result of a mathematical integration"? I am writing an academic paper. And I wonder which is the best English term for "the result of a mathematical integration".
For example, I have a mathematical integration as below.
$$F = \int f(u) du$$
Which is a more suitable name for $F$? "integral $f(u)$" or "integrated $f(u)$"? I must name $F$ in a way expressing its relationship with $f(u)$.

Thanks for your comments! As it is the "integral of $f(u)$", can I name it "integral $f(u)$" for short? I think I have to omit the preposition when defining a new academic word. ($f(u)$ is an already-defined physical term.)

I agree with some comments.  The term "integral" is used both for the problem and for the answer.  This is like a lot of other words in mathematics:

5+3 is an easy sum
What is 5+3? The sum is 8.

5! is the product of the numbers from 1 to 5
120 is the product of the numbers from 1 to 5

The notation $\int f$ is generally used in one of two contexts:  either it represents a number (which we generally interpret as the area under the graph of $f$, but it is actually more general than that), or it represents a function.
A number
If $\int f$ represents a number, then it is a definite integral.  More commonly, this is written in the form
$$ \int_{a}^{b} f(x) \,\mathrm{d}x
\qquad\text{or}\qquad
\int_E f(x) \,\mathrm{d}x.$$
Sometimes, a definite integral will be written without giving explicit bounds for the integration, or without specifying a domain of integration.  In such a context where a number is still meant, this notation is generally understood to indicate integration over the entire domain of the function.  For example,
$$ \int \mathrm{e}^{-x^2}\,\mathrm{d}x = \int_{-\infty}^{\infty} \mathrm{e}^{-x^2}\, \mathrm{d}x = \sqrt{\pi}.$$
In this context, the number which is obtained by integration is called the integral.  More precisely, one might say that $\int_E f(x)\,\mathrm{d}x$ is called the "definite integral of $f$ over $E$".
A function
The same notation is also used to denote a function.  For example, if $f$ is a "sufficiently nice" read-valued function defined on the real numbers, and there is a function $F$ with the property that $F'(x) = f(x)$ for all $x$, then we say that $F$ is an antiderivative or primitive of $f$.  This is often written as
$$ F(x) = \int_{a}^{t} f(t)\,\mathrm{d}t
\qquad\text{or}\qquad
F = \int f. $$
It should be noted that $F$ is not unique—a function may have many antiderivatives, though these antiderivatives differ by only a constant, so it is not hard to specify the entire family of antiderivatives.  The notation $\int f$ is, perhaps, confusing, but it is justified by the Fundamental Theorem of Calculus, which demonstrates that definite integrals and antiderivatives are related, e.g. in the setting of Riemann integration, $F$ is an antiderivative of $f$, then
$$ \int_{a}^{b} f(x)\,\mathrm{d}x = F(b) - F(a). $$
This function might also be called the indefinite integral or inverse derivative. Wikipedia gives another couple of terms.
A note on notation
The notation $f$ denotes a function, while $f(u)$ or $f(x)$ denotes a value of that function—this latter notation represents a number (or dependent variable), not a function.  Thus it is a little strange to talk about "the integral of $f(u)$".  Thus it is fine to say that something is the "integral of $f$", but it is not quite right to say that it is the "integral of $f(u)$".
TL;DR
If $\int f$ denotes a number, that number is the integral or definite integral of $f$ (over some interval or domain).  If $\int f$ denotes a function, that function is an antiderivative or primitive of $f$.  In either case, I would not elide the preposition—keep that "of" in there.


Name of an inference rule that I can't find anywhere As a homework question, I must prove, using inference rules, that:
p ∨ q
(¬p ∧ q) → r

∴ ¬r → p
I have done the following progress:

p ∨ q

(¬p ∧ q) → r

¬r → ¬(¬p ∧ q)

¬r → (¬(¬p) ∨ ¬q)

¬r → (p ∨ ¬q)

1 premisse
2 premisse
3 contrapositive from 2
4 morgan law from 3
5 double negation from 4
And I can't figure out how to finish it. I can easily prove the whole expression using a truth-table, but I simply can't find a formal way of merging 1 and 5. Any help would be hugely appreciated.

Here's a way to do this directly:
$p \to p,$ LEM as an implication
$\neg q \to p,$ $1$ as an implication
$(p \vee \neg q) \to (p \vee p),$ constructive syllogism
$(p \vee \neg q) \to p$
$\neg r \to p,$ hypothetical syllogism from the above and $5$

hint
You want to prove that $\lnot r \to p$.
you proved that $ \lnot r \to (p\vee \lnot q)$.
Assume  $ \lnot p$.
then by DS, $ \lnot q$.
but the first premise is $ p\vee q$
so, we can't have $ \lnot p \wedge \lnot q$. this is the contradiction.


Let $f :[0,\pi] \rightarrow \mathbb R$ be a continuous function such that $f(0) = 0$ . Which of the following statements are true? 1.If $$\int_0^{\pi} f(t)\cos(nt)~dt=0$$

for all $n\in\{0\}\cup\mathbb N$ , then $f\equiv0$ .

2.If $$\int_0^{\pi} f(t)\sin(nt)~dt=0$$

for all $n\in\{0\}\cup\mathbb N$ , then $f\equiv0$ .

A hint: Extend your function $f:\>[0,\pi]\to{\mathbb R}$ to a continuous function $g:\>[{-\pi},\pi]\to{\mathbb R}$ in two different ways: For 1. let $g$ be even, and for 2. let $g$ be odd.What can you say about the Fourier series of $g$ in the two cases?

For (2)

take f(t) = Sin(t)

$\int_0^{\pi}$ sint sin(nt)dt = $\int_0^{\pi}$ (cos(n-1)t - cos(n+1)t) dt = $[\frac{sin(n-1)t}{n-1} - \frac{sin(n+1)t}{n+1}]_0^{\pi} = 0$ for all n $\in \mathbb N$

please tell me I am right or not

Thank you


How to check if a ray intersects a circle Given a ray (Origin(x,y) and Direction(x,y)) and a circle (Center(x,y) and Radius) determine if the ray intersects the circle (touches one or two points in the circle).
I've followed this solution (both geometic and analityc) but I can't get it to work on my python code. Anyone can see my mistake?
The code is the following:
def dot(x1, y1, x2, y2):
    return (x1*x2) + (y1*y2)

def intersects(x, y, vx, vy, cx, cy, r):
# r is radius, x and y are the ray origin, vx and vy are the ray direction and cx and cy are the circle center

    dirx = vx-x
    diry = vy-y

    Lx = cx - x
    Ly = cy - y

    tca = dot(Lx,Ly,dirx,diry)

    if tca < 0:
        return False

    Lsq = dot(Lx,Ly,Lx,Ly)
    tcasq = tca*tca

    dsq = Lsq - tcasq

    rsq = r*r

    if dsq>rsq:
        return False

    thc = math.sqrt(rsq-dsq)

    t0 = tca - thc

    t1 = tca + thc

    if t0 < 0:
        t0 = t1
        if t0 < 0:
            return False

    return True

I would suggest using this more simple approach rather than the one you're using. You can do this by trying to analytically find the minimum distance between the points on the line and the centre of the circle.
Define the variable $\lambda$ by:
$$ \lambda = - \dfrac{(x-c_x)v_x + (y-c_y)v_y}{v_x^2 + v_y^2} $$
Now use lambda to calculate the variable below:
$$ r^* = \sqrt{(x + \lambda v_x - c_x)^2 + (y + \lambda v_y - c_y)^2} $$
$r^{*}$ is the minimum distance between the "ray" and the centre of the circle. Clearly if $r^* > r$, the line and the circle do not intersect.(You can derive the relation I provided for $\lambda$ by differentiating this equation and setting it to $0$.) Otherwise they intersect. If $r^* < r$ they intersect at two points. Since you're using Python the chances of $r^* = r$ is really low because of rounding errors, so the downside to this method is you likely won't be able to tell if the line is exactly tangent to the circle or not.

I think this might be your mistake
dirx = vx-x
diry = vy-y
needs to be
dirx = vx
diry = vy
Since vx,vy is already a direction vector.


How to show that the following summation is true for matrix multiplication? I have following matrices $$\mathbf{b}\in R^n,~\mathbf{W}\in R^{n\times n},~ A\in R^{n\times m}.$$ Where $\mathbf{W}$ is a diagonal matrix and let us assume that the $i$-th row of $\mathbf{A}$ is written as $\mathbf{a}_i$ and the $i,i$-th entry of $\mathbf{W}$ is $w_i$. In this how to show that the following equality is true $$\mathbf{b^TWA(A^TWA)^{-1}A^TWb}=\sum_{i=1}^{n}w_i^2b_i^2\mathbf{a}_i^T\left(\sum_{j=1}^nw_j\mathbf{a}_j\mathbf{a}_j^T\right)^{-1}\mathbf{a}_i$$ where $b_i$ is the $i$-th entry of $\mathbf{b}$. Thanks in advance.

Isn't simply
$$\mathbf b^T\mathbf W\mathbf A(\mathbf A^T\mathbf W\mathbf A)^{-1}\mathbf A^T\mathbf W\mathbf b =\mathbf b^T\mathbf W\mathbf A \mathbf A^{-1}\mathbf W^{-1}(\mathbf A^T)^{-1}\mathbf A^T\mathbf W\mathbf b=\mathbf b^T\mathbf W\mathbf b $$
?


General solution of $\sin 2x+\cos x=0$ Solve the trigonometric equation $\sin 2x+\cos x=0$

My Attempt
$$
2\sin x\cos x+\cos x=0\implies\cos x=0 \text{ or }\sin x=\frac{-1}{2}=\sin\frac{-\pi}{6}\\
x=(2n+1)\frac{\pi}{2} \text{ or }x=n\pi+(-1)^n\frac{-\pi}{6}\\
x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}
$$
Reference
$$
\cos x=-\sin 2x=\cos\Big(\frac{\pi}{2}+2x\Big)\implies x=2n\pi\pm\Big(\frac{\pi}{2}+2x\Big)\\
-x=2n\pi+\frac{\pi}{2}\text{ or }3x=2n\pi-\frac{\pi}{2}\implies x=2m\pi-\frac{\pi}{2}\text{ or }x=\frac{2m\pi}{3}-\frac{\pi}{6}
$$
But my reference gives the solution $x=2n\pi-\dfrac{\pi}{2}$ or $x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}$. I understand how it is reached and both represent the same solutions. But, how do I derive the solution in my reference from what I found in my attempt ? i.e.,

  How to derive
  $$
\bigg[x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}\bigg]\\
\implies \bigg[x=2n\pi-\dfrac{\pi}{2}\text{ or }x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}\bigg]
$$

There are two ways to solve the equation.

Your method: $2\sin x\cos x+\cos x=0$, so $\cos x(2\sin x+1)=0$. Thus we have either $\cos x=0$ or $\sin x=-1/2$. Thus
\begin{align}
x&=\frac{\pi}{2}+2n\pi &\text{or}&&  x&=-\frac{\pi}{2}+2n\pi && \text{(from $\cos x=0$)} \\[4px]
x&=-\frac{\pi}{6}+2n\pi  &\text{or}&& x&=\frac{7\pi}{6}+2n\pi && \text{(from $\sin x=-1/2$)}
\end{align}
(you grouped together the families of solutions of $\cos x$ and of $\sin x=-1/2$, but I prefer to keep them separated). Good job.

Alternative method: $\cos x=-\sin2x=\cos(\frac{\pi}{2}+2x)$. Therefore either
$$
x=\frac{\pi}{2}+2x+2n\pi \to x=-\frac{\pi}{2}-2n\pi
$$
or
$$
x=-\frac{\pi}{2}-2x+2n\pi \to 3x=-\frac{\pi}{2}+2n\pi \to x=\frac{2n\pi}{3}-\frac{\pi}{6}
$$

How do you recover the previous sets of solutions?

One set is already present. For the other three, consider the cases when $n=3k$, $n=3k+1$ or $n=3k+2$, with integer $k$. Then
\begin{align}
n&=3k & x&=\frac{6k\pi}{3}-\frac{\pi}{6}=2k\pi-\frac{\pi}{6} \\[4px]
n&=3k+1 & x&=\frac{(6k+2)\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{2\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{\pi}{2}\\[4px]
n&=3k+2 & x&=\frac{(6k+4)\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{4\pi}{3}-\frac{\pi}{6}=2k\pi+\frac{7\pi}{6}
\end{align}

$$\sin(2x)+\cos(x)=0$$
$$2\sin(x)\cos(x)+\cos(x)=0$$
$$\cos(x)\left(2\sin(x)+1\right)=0$$
so you have two sets of solutions:
$$\cos(x)=0,\,\sin(x)=-\frac{1}{2}$$
EDIT:

firstly they have:
$$x=2n\pi-\pi/2=\pi(2n-1/2)$$
and you have:
$$x=(2n+1)\pi/2$$
let:
$$\pi/2(4n-1)=\pi/2(2m+1)$$
so:
$$4n-1=2m+1$$
$$m=(4n-2)/2=2n-1$$
so for all integer values of $n$, $m$ is also an integer and so they are equivalent?


Game theory: Given a number N, decide the winner. Given a number $N$:

Two people A and B subtracts a number k
$\left(~0 < k < N,\ k\mid N~\right)$ alternatively from $N$.
The one who cannot make a move loses the game.Provided that A always starts first, decide the winner.

Initially, I tried solving this problem using dynamic programming considering the fact that a state can be a winning one if it leads to at least one losing state. Also, a state is a losing one if all the states it goes to are winning. Checking for all the states didn't prove to be time efficient. After looking at the solutions I found that this problem boils down to checking whether N is odd or even. What I think is, one can say that if a state X is losing then $\mathbf{X + 1}$ $\left(\,\mbox{as}\ \mathbf{1\mid\left(\,X + 1\,\right)}\,\right)$ will always be a winning state but what if $\mathbf{X}$ is a winning state ?.

How can one prove that even states will always be winning and vice versa ?.

Note a few things:

1) If you are passed an even number, you can not be made to lose in that turn.

This is because $N = 2m > 1$ and $1|N$ so you can pass $N -1$.

2) If you are passed an odd number, then either a) you are forced to lose if $N =1$ or b) you are forced to pass an even number. And your opponent cannot be made to lose in the next turn.

a) if $N =1$ you can't do anything.

b) if $N > 1$ but $N$ is odd so if $k|N$ then $k$ is odd and you must pass $N -k$ which is an odd minus an odd must be even.

3) if you are passed an even you can make it so that you are passed an even the next turn (if there is a next turn).

Just pass the $N-1$.  That will be odd and your opponent must pass you an even by 2). (or the opponent must lose if you passed a $1$).  [If you have a larger odd $k$ that divides $N$ you can pass $N -k$ which will also be odd.  That will speed the game up.]

4) The game is finite and must end.

You are passed a finite $N$ and you pass an $1 \le N - k < N$ that is smaller.  There are only a finite number (maximum $N$) turns that can be taken.

...... so ......

So if you are passed an even you need not lose that turn or, by induction, any other turn.  If you are passed an odd you might lose but you can not make your opponent lose.  All games must end.

Therefore (if both players play the best strategy) player $A$ will lose if the first $N$ is odd.  Player $A$ will win if the first $N$ is even.

If $N=2$, then the second player, B wins the game.

Lets set some notations. Given $N$ as described above.
Let $a_1=N$.

First state:

Assume that A chooses in the first state $\alpha_1$.
Then let $b_1:=a_1-\alpha_1$.
Assume that B chooses in the first state $\beta_1$.
Then let $a_2:=b_1-\beta_1$.

Second state:

Assume that A chooses in the second state $\alpha_2$.
Then let $b_2:=a_2-\alpha_2$.
Assume that B chooses in the second state $\beta_2$.
Then let $a_3:=b_2-\beta_2$.

.......
.......

$i$-th state:

Assume that A chooses in the $i$-th state $\alpha_i$.
Then let $b_i:=a_i-\alpha_i$.
Assume that B chooses in the $i$-th state $\beta_i$.
Then let $a_{i+1}:=b_i-\beta_i$.

In all other cases, i.e., $2 \leq N$, the first player, A wins the game.

Let $k$ be the least natural number, which does'nt divides $N$, i.e.

$k:=min\{n \in \mathbb{N} : N-1 \geq n \nmid N  \}$.

  The strategy for the first player in the $i$-th state, is to choose the unique element $\alpha_i \in \{1, 2, 3, ..., k-2, k-1\}$ such that $b_i:=a_i-\alpha_i$ is divisible by $k$.

Let $k$ be the least natural number, which does'nt divides $N$, i.e.

$k:=min\{n \in \mathbb{N} : N-1 \geq n \nmid N  \}$.

[Notice that the set $\{n \in \mathbb{N} : n \nmid N \}$ is not empty.

Notice that $N-1 \nmid N$, since N is greater than $1$.]

Also notice that for every $t \in \{1, 2, 3, ..., k-2, k-1\}$ we have that:

$t\mid N$, by definition.

In $i$-th state the first player, A subtracts

the unique element $\alpha_i \in \{1, 2, 3, ..., k-2, k-1\}$;

such that $b_i:=a_i-\alpha_i$ is divisible by $k$.

[By the euclid's algorithm, A can be able to do such a move.]

Now it's the second player turn.

We claim that B could not win.

Suppose on contrary that B could win, so he/she must choose the whole remaining number; i.e. $b-i$

which means that the remaining number $b_i$ divides $N$.

On the other hand, $k \mid b_i$;
since A makes the remaining number divisible by $k$.

So we have: $k \mid b_i \mid N$,

so by the property of transitivity of division we have

that $k \mid N$, which contradicts the definition of $k$.

  Example

$N=60$.

In this case $k:=7$ is the least number which does not divides $60$.

The remainder of $a_1=60$ by $7$; is equal to $4$, so A subtracts $4$, and $56=b_1=a_1-\alpha_1=60-4$ is divisible by $7$.

Now it is the second player's turn for example he chooses $15$,
$41=a_2=b_1-\beta_1=56-15$ has the remainder equal to $6$ module $7$, so A subtracts $6$, and $35=b_2=a_2-\alpha_2=41-6$ is divisible by $7$.

Now again, it is the second player's turn for example he chooses $5$,
$30=a_3=b_2-\beta_2=35-5$ has the remainder equal to $2$ module $7$, so A subtracts $2$, and $28=b_3=a_3-\alpha_3=30-2$ is divisible by $7$.

Now again, it is the second player's turn for example he chooses $20$,
$8=a_4=b_3-\beta_3=28-20$ has the remainder equal to $1$ module $7$, so A subtracts 1, and $7=b_4=a_4-\alpha_4=8-1$ is divisible by $7$.

Now again, it is the second player's turn for example he chooses $2$,
$5=a_5=b_4-\beta_4=7-2$ has the remainder equal to $5$ module $7$, so A subtracts $5$; $0=b_5=a_5-\alpha_5=5-5$, and he wins the game.

  Notice that every integer like $a$ which is divisible by $7$ could not divide $60$, since $7$ does not divides $60$.


Small proximity of important points of a function Let $a,b,c$ be coprime integers with c greater than b and a,  $a^2 + b^2 \gt c^2$ and consider the function $f(x) = a^x + b^x - c^x$. It is easy to verify that there exist $r$ and $s$ such that $f(r) \ge f(x)$ for all $x$ and $f(s) = 0$. Prove the inequality below (in other words, when $c$ is big $r$ and $s$ are very close to each other  or even f-graph appears vertically from its peak).

$$ 0 < s-r < \frac{\ln(\ln c)}{\ln c}$$


Alternative interpretation of ball and urns problem Suppose an urn has r red balls and b black balls. They are withdrawn one at a time at random until a total of k, k $\leq$ r, red balls have been withdrawn. Find the probability that a total of n balls are withdrawn.

A first approach would be to consider the probability of drawing k-1 red balls in the first n-1 withdrawals, then multiply it by the probability of the final ball drawn being red. This gives

$$\frac{\binom{r}{k-1}\binom{b}{n-k}}{\binom{b+r}{n-1}}\cdot\frac{r-k+1}{b+r-n+1}\;.$$

All this is well. However, I wanted to try solving the problem not by considering probabilities in each step, but by counting outcomes as a whole. This is where I run into problems and wish that someone can enlighten me.

This is one of the futile attempts that I made:

$$P = \frac{\text{#ways to draw k red in n attempts}}{\text{#ways to draw k red in any number of attempts}} = \frac{\text{#ways for red ball to be nth position and exactly k-1 red balls preceding it in any of the n-1 slots}}{\text{#number of ways for any number to choose k red balls and insert any number of blue balls in their way}} = \frac{\binom{r}{k-1}\binom{b}{n-k}(r-k+1)}{\binom{r}{k}2^b}\;.$$

Clearly, the numerator is the same. Since in my second method i'm not involving the variable in the denominator, this already hints to disaster as the former answer does have an n in the answer. After a few more tries I'm no longer sure what objects I'm supposed be counting exactly. Please shed some light

You can only obtain a probability as the quotient of the number of favourable outcomes over the total number of outcomes if all outcomes are equiprobable. The way you're trying to count, there's no reason to expect the outcomes to be equiprobable, as they involve different numbers of draws.

The equiprobable outcomes that you'd need in order to do it this way are all possible outcomes of $b+k$ draws (including the ones with more than $k$ red balls drawn). But counting those is a bit complicated; it's easier if you go all the way and consider all possible outcomes of $b+r$ draws. That yields

$$
\frac{\binom{n-1}{k-1}\binom11\binom{b+r-n}{r-k}}{\binom{b+r}r}\;,
$$

which coincides with your first result.

I think the question should be - Find the probability that a total of n "red" balls are withdrawn.

Then let p = r/(r+b) and q = 1 - p = b/(r+b), and let X be the number of red ball. So X has a binomial distribution.

Thus

\begin{align}
\ P(X=n)= \binom{r+b}{n} p^{n}q^{r+b-n}
\\ =\frac{(r+b)!}{n!(r+b-n)!} p^{n}q^{r+b-n}
\end{align}


Throw a dice 4 times. What is the probability `6` be up at-least one time? First time I approach a probability question (:

  Throw a dice 4 times. What is the probability 6 be up at-least one time?

Intuitively, I would say: $\frac{1}{6}\times4$.

I would explain as: If you throw one time, probability is $\frac{1}{6}$.

If you do it 4 times, then multiply by 4.

According to the answers I'm wrong.

Can you explain please? thanks in advance.

Expanding the comment of Vladimir Vargas about binomial distribution you can, alternatively, use the complementary probability to take zero 6's in the four throws:

$$P(X\ge 1)=1-P(X=0)=1-\binom{4}{0}\left(\frac16\right)^0\left(1-\frac16\right)^4=1-\left(\frac56\right)^4=\frac{671}{1296}\approx52\%$$

Consider the complement problem, there is a 5/6 probability of not rolling a six
for any given die, and since the four dice are independent, the probability of not rolling a six is (5/6)^4 = 5^4/6^4 = 625/1296. The probability of rolling at least one six is therefore
1 − 625/1296 = 671/1296 ≈ .517.


Explain the proof behind maximal ideals $\iff $ quotient is a field. $R$ is a commutative ring with unity and $M$ is a maximal ideal $\iff$ $R/M$ is a field.

Proof is here

Can someone explain the inclusion $J \subset \ker \psi$?

Can someone also explain the hypothesis that

To see this note that if for all $r \in R$, there is some $r_1 \in J$ such that $$\psi(r) = \psi(r_1)$$

Sorry but where did this hypothesis come from and why are we assuming this to prove $J = R$?
Proof is here

Can someoene explain the statement

This ideal $\psi^{-1}(r + M)$ contains $M$ because $r \in \psi^{-1}(r+M)$.

Isn't the definition of the preimage of $r+M$ just $$\psi^{-1}(r+M) = \{r \in R : \psi(r) = r +M \}$$
I mean I could just prove the inclusion, but such an attempt won't really reveal what the statement's intent.
($x \in M$, $\psi(x) = x + M = M$)

$1.$ $0+M$ is the zero of $R/M$. So $\psi(J)=0+M$ implies $J\subset ker(\psi)$.

$2.$ Since $\psi(J)=R/M$, then for all $r\in R$, we have $$\psi(r)=r+M\in\psi(J),$$thus there exists $r_1\in J$ such that $$\psi(r_1)=r+M=\psi(r).$$ Here, $\psi(r-r_1)=0+M$ and so $r-r_1\in M$, i.e., $r-r_1=m$, where $m\in M$. Therefore $$r=r_1+m\in J+M\subseteq J$$ Since $r$ is arbitrary, it follows that $R\subseteq J$.

$3.$ $\psi^{-1}(r+M)=\{s\in R:\psi(s)=s+M=r+M\}$.

I'll try to go in order here.

The first point that's somewhat brushed under the rug here is that if $R, S$ are commutative rings, $\phi: R \to S$ is surjective, and $I$ is an ideal of $R$, then $\phi(I)$ is an ideal of $S$. To see this, note that if $s \in S$, then $s = \phi(r)$ for some $r \in R$, so
\begin{equation}
\phi(r')s = \phi(r')\phi(r) = \phi(rr') \subset \phi(R)
\end{equation}

Then, one uses the fact that a field has only two ideals, and we examine each case individually; $\phi(J) = 0 \subset R/M$, and $\phi(J) = R/M$.

It seems like you feel comfortable with the first case, $\phi(J) = 0$.

For the second case, we $\textbf{assume}$ the map is surjective (as we have handled the other case in which it is not), and so the statement that $\phi(r_1) = r + M$ for some $r_1 \in J$ is precisely this surjectivity.

I think the proof of the converse is perhaps too complicated. One can show that if $R$ is a commutative ring with unity, $R$ is a field $\textbf{if and only if}$ $R$ has precisely two ideals. It turns out that the ideals of $R/M$ are precisely the images of the ideals of $R$ containing $M$ under the quotient map. To prove this, start with the fact that the pre-images of ideals are ideals and examine ideals of $R/M$.

Once you've established that, the result follows easily. Since $M$ is maximal, there are only two such ideals, $\phi(R)$ and $\phi(M)$.


Ordering compositions of integers, where cyclic shifts are not considered distinct A composition of an integer $n$ of length $k$, is a tuple (ordered list) of $k$ positive integers whose elements sum to $n$, (sometimes also called integer composition, ordered partition or ordered integer partition). For instance, $[1,1,2,4], [1,2,1,4], [1,2,4,1], [4,4],$ and $[1,1,1,3,1,1]$ are examples of distinct compositions of $8$. These have lengths $4,4,4,2$, and $6$, respectively.

My question is: if I consider standard lexicographical ordering of all compositions of a fixed length $k$ of an integer $n$ (see example below for $k=4$, $n=8$), which is the most efficient way to check whether a given composition is the smallest (with respect to this ordering) among all its cyclic shifts?

I'm demonstrating with an example: Below I enumerate the partitions of 8 of length 4. For example, the composition $[1,1,5,1]$ has a smaller cyclic shift composition $[1,1,1,5]$; the composition $[2,3,2,1]$ has a smaller cyclic shift $[1,2,3,2]$; $[2,2,2,2]$, and $[1,3,2,2]$ are the smallest cyclic shifts of such a composition. In the case of $[2,3,2,1]$, I observed that there's a $1$ in there, so I shifted the composition until I got $[1,2,3,2]$, but I'm not sure how I do a similar check in the general case.

\begin{array}{llll}
    1 & 1 & 1 & 5 \\
    1 & 1 & 2 & 4 \\
    1 & 1 & 3 & 3 \\
    1 & 1 & 4 & 2 \\
    1 & 1 & 5 & 1 \\
    1 & 2 & 1 & 4 \\
    1 & 2 & 2 & 3 \\
    1 & 2 & 3 & 2 \\
    1 & 2 & 4 & 1 \\
    1 & 3 & 1 & 3 \\
    1 & 3 & 2 & 2 \\
    1 & 3 & 3 & 1 \\
    1 & 4 & 1 & 2 \\
    1 & 4 & 2 & 1 \\
    1 & 5 & 1 & 1 \\
    2 & 1 & 1 & 4 \\
    2 & 1 & 2 & 3 \\
    2 & 1 & 3 & 2 \\
    2 & 1 & 4 & 1 \\
    2 & 2 & 1 & 3 \\
    2 & 2 & 2 & 2 \\
    2 & 2 & 3 & 1 \\
    2 & 3 & 1 & 2 \\
    2 & 3 & 2 & 1 \\
    2 & 4 & 1 & 1 \\
    3 & 1 & 1 & 3 \\
    3 & 1 & 2 & 2 \\
    3 & 1 & 3 & 1 \\
    3 & 2 & 1 & 2 \\
    3 & 2 & 2 & 1 \\
    3 & 3 & 1 & 1 \\
    4 & 1 & 1 & 2 \\
    4 & 1 & 2 & 1 \\
    4 & 2 & 1 & 1 \\
    5 & 1 & 1 & 1 \\
\end{array}

You have to start with the smallest part. If there is only one of that size, you are done. Otherwise find the smallest number following one of the smallest parts and so on.

A rule for the order of all compositions of a positive integer n is derived in the paper, “Relationships between Ordered Compositions and Fibonacci Numbers” published in the Journal of Mathematics research $(Vol-7, No-3, 2015)$ under Canadian Center of Science and Education.
        ‘Rule for Significant Order of Compositions of n’ or ‘SOC(n)’ : Under SOC(n), the summands of the 1st C(n) are all 1; the last C(n) is n itself; and  for n ≥ r ≥ 2,  if any kth C(n) is: $x_1 + …  + x_r$  then $(k + 1)$-th  C(n) is: $x_1 + …+ x_{r – 2}  + (x_{r – 1 + 1)} +$ the  sum of $x_{r – 1}$ summands which are all 1 such that if $r \ge 3$ then the first $r – 2$ summands of k-th C(n) appear also in (k + 1)th C(n) in the same order, but if r = 2 then such common summands of k-th C(n) and (k + 1)-th C(n) cannot exist. The number of summands of k-th C(n) and (k + 1)th C(n) under SOC(n) are $r$ and $r + x_{r – 2}$ respectively.
Example: SOC(4)  ≡ $4, 1 + 1 + 1 + 1,   1 + 1 + 2,   1 + 2 + 1,   1 + 3,   2 + 1 + 1,   2 + 2,   3 + 1,   4$.
        In the connection of your enquiry, you can download and read the paper.


Number of ways to write n as a sum of k nonnegative integers How many ways can I write a positive integer $n$ as a sum of $k$ nonnegative integers up to commutativity?

For example, I can write $4$ as $0+0+4$, $0+1+3$, $0+2+2$, and $1+1+2$.

I know how to find the number of noncommutative ways to form the sum: Imagine a line of $n+k-1$ positions, where each position can contain either a cat or a divider. If you have $n$ (nameless) cats and $k-1$ dividers, you can split the cats in to $k$ groups by choosing positions for the dividers: $\binom{n+k-1}{k-1}$. The size of each group of cats corresponds to one of the nonnegative integers in the sum.

Note : I Can't Use MathJax
First, think of the number n as the sum of n ones.
Now think of the k non-negative integers
So now consider this row of n+k-1 objects where there are the n ones and k-1 walls.
We Arrange these (k-1) walls in this row .
Now the k-1 walls cordon off k numbers who shall add up to the number n.
Note that the walls can be adjacent as they shall cordon off zero and zero is non-negative.
So to summarize the total number of ways to do this is the number of arrangements of this row. This is obviously (n+k-1) choose (k-1) as the selection as the selection of the (k-1) places where the walls must go completely determines the rest of the row .
Hope this helps!
P.S ( This Is My First answer on math s.e)


Relation between points of inflection and saddle points Let $I$ be an interval and $f\colon I \to \mathbb{R}$ a differentiable function. Suppose the following definitions:

For $x_0 \in I$ the point $(x_0,f(x_0))$ is called saddle point if $f'(x_0) = 0$ but $x_0$ is not a local  extremum of $f$.

For $x_W \in I$ the point $(x_W,f(x_W))$ is called point of inflection if there is a neighborhood $U$ from $x_W$ in $I$ such that $f'$ is strictly monotonic increasing (resp. decreasing) for $x < x_W$ on $U$ and strictly monotinic decreasing (resp. increasing) for $x > x_W$ on $U$.

What is the logical relation between saddle points and points of inflection?

My first intuitive guess was that a point $(x,f(x))$ is a saddle point iff it is a point of inflection and $f'(x) = 0$. However the implication "$\implies$" seems to be wrong.  Consider the following counterexample:
$$
f(x) =
\begin{cases} x^4 \cdot \sin\left(\frac{1}{x}\right) & x \neq 0 \\
0 & x = 0
\end{cases}
$$

Then $(0,0)$ is a saddle point but not a point of inflection because the derivativative oscillates on every neighborhood of $0$.

Is this correct so far? Is the other implication true? If so, how to prove it?

Your example does indeed show that a saddle point need not be an inflection point. (The function $x^2\sin(1/x)$ also works, but your example has the virtue of being continuously differentiable.)

In the other direction, if $(a,f(a))$ is a point of inflection and $f'(a) = 0,$ then $(a,f(a))$ is a saddle point. To see this, suppose WLOG that for some small $\delta > 0$ that $f'$ strictly increases in $[a-\delta,a]$ and strictly decreases in $[a,a+\delta].$ In $[a-\delta,a)$ we have $f'(x)<0,$ because these values must be less than $f'(0)=0.$ The same reasoning shows that $f'(x) < 0$ for $x\in (a,a+\delta].$ The mean value theorem then shows $f$ strictly decreases on both $[a-\delta,a]$ and $[a,a+\delta].$ Hence $f$ strictly decreases on $[a-\delta,a+\delta].$ It follows that $f(a)$ is neither a local max. nor min. for $f$ at $a.$

Can only be brief, sorry  you can fill in the gaps. Information available in Wiki.. For z = f(x,y) ; second derivative test. <0 for max, >0 for min, test fails but at saddle points the both signs prevail. E.g., monkey saddle.$ f(x,y) = x^3 - 3 x y^2 $. when considering inflection points along certain directions ( 3 of 6 directions). Like a Col point in mountainous range, one direction upward, one downward, one is neither, topography negative Gauss curvature, inflection along asymptotic direction, normal curvature vanishes..  See also Minimax, Nash equilibrium.


Find $a$ so that $f$ has a specific local extreme point We are given a function $f : \mathbb{R} \rightarrow \mathbb{R}$ with:

$$f(x) = \frac{x^2 - ax}{\sqrt{x^2 + 1}}, a \in \mathbb{R}$$

Find $a$ so that the function $f$ has a point of local extreme situated at a distance of $2$ from the $Oy$ axis.

The derivative of a function in a local extreme point is $0$. Calculating the derivative of $f$ and then equalizing with $0$, I got $x^3 + 2x - a = 0$.

At this point I got stuck.

Thank you in advance!

let $$h(x)=x^3+2x-a$$

$$h(2)=0 \rightarrow a=13$$

$$h(-2)=0 \rightarrow a=-13$$

When $-13 \lt a \lt 13$,  $h(2) \gt 0$ and $h(-2) \lt 0$
Thus by intermediate value theorem, there is a $c \in (-2, 2)$ such that $h(c)=0$. We can conclude there is an critical point within (-2, 2).


One divided by infinity is not zero? I know that $\frac{1}{\infty}$ is undefined. But my question is - can we say that $\frac{1}{\infty}\neq0$ ?

I've got some idea how to explain that: Let's say we have a random-number generator that generates numbers in interval $(0, \infty)$. What is the probability that it generates 5? $\frac{1}{\infty}$. So it cannot equals zero because if it was zero, it wouldn't generate any number (every number has the same probability - $\frac{1}{\infty}$).

Am I right? Or is that wrong idea?

Thanks

It is worth noting that since $\infty$ is not a number, the expression $\frac{1}{\infty}$ is not meaningful. That is, you cannot evaluate this expression. It is not a number. You can think of it in much the same way as $\frac{1}{0}$ --- an expression without any interpretation.

The only means that we have to talk about expressions involving $\infty$ is through the concept of limits. The shorthand
$$ \lim_{x \to \infty} \frac{1}{x} = \frac{1}{\infty} = 0 $$
is sometimes used in a careless way by students, but the middle expression in this string of equations isn't strictly meaningful. It is only a device used to remind you that we are looking at what happens as we divide 1 by increasingly large numbers.

So, I suppose that in some sense, you can say that $\frac{1}{\infty} \neq 0$, as the left-hand side here isn't any number at all.

If we are to understand "$=$" as a binary relation of identity and "$\neq$ as a binary relation of non-identity, then in most contexts, formally, $1/\infty$ is not identical to zero. Therefore it would make logical sense to write $1/\infty \neq 0$ as long as we "define what we mean by $1/\infty$". This is more in a strictly logical sense than in a mathematical sense.

In order to interpret $1/\infty \neq 0$ mathematically, you have to carefully describe the underlying structure of the statement.

If we are to only consider "$=$" and "$\neq$" for numbers or algebraic expressions involving numbers, then $1/\infty \neq 0$ doesn't make sense because the left side isn't a number. However, if you define $1/\infty$ appropriately (e.g. as a limiting procedure), then the statement makes sense and can be evaluated as false.


Let $f$ be such that $f(0)=0$ and $f'(0)=1$. Find iff conditions for normality of successive iterations $f\circ f\circ\ldots$ Let $f$ be an entire function normalized by the conditions $f(0)=0$ and $f'(0)=1$. Find necessary and sufficient conditions on $f$ for normality of the family of successive iterates $\{f\circ f,f\circ f\circ f, f\circ f\circ f\circ f,\ldots \}$.

I know that if $|f'(0)|>1$, it certainly does not. But for $|f'(0)|=1$ I'm having trouble coming with conditions so strong they are necessary and sufficient. Would a condition on the second derivative of $f$ help?

Any help is greatly appreciated. Since I haven't made much progress on this I'd prefer if I were just given hints, else I feel like I'm asking for answers. Thank you.

We have the following relevant fact:

  If $\mathscr{F}\subset \mathscr{O}(U)$ is a normal family, then $$\mathscr{F}' = \{ f' : f \in \mathscr{F}\}$$ is also a normal family.

The proof is simple, for any sequence $(g_n)$ in $\mathscr{F}'$ select a sequence $(f_n)$ in $\mathscr{F}$ such that $f_n' = g_n$ for all $n$. By the normality of $\mathscr{F}$, we can extract a compactly convergent subsequence. Without loss of generality - to save subscripting - assume the whole sequence $(f_n)$ is compactly convergent with limit $f$. For every $z_0 \in U$ choose $r > 0$ such that $D_{3r}(z_0)\subset U$. For $z \in D_r(z_0)$ we then have

$$g_n(z) = f_n'(z) = \frac{1}{2\pi i} \int_{\lvert\zeta-z_0\rvert = 2r} \frac{f_n(\zeta)}{(\zeta-z)^2}\,d\zeta.$$

Since the circle $C = \{\zeta : \lvert\zeta-z_0\rvert = 2r\}$ is compact, $f_n$ converges uniformly to $f$ on $C$, and since $\lvert \zeta-z\rvert > r$ for $(\zeta,z) \in C \times D_r(z_0)$, we have

$$\frac{f_n(\zeta)}{(\zeta-z)^2} \to \frac{f(\zeta)}{(\zeta-z)^2}$$

uniformly on $C\times D_r(z_0)$, whence

$$g_n(z) \to g(z) = \frac{1}{2\pi i} \int_{\lvert\zeta-z_0\rvert = 2r} \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta = f'(z)$$

uniformly on $D_r(z_0)$. Since $z_0$ was arbitrary, it follows that $(g_n)$ converges locally uniformly, hence compactly, on $U$.

This can be iterated, so for a normal family $\mathscr{F}$, the family $\mathscr{F}^{(k)} = \{ f^{(k)} : f \in \mathscr{F}\}$ of $k^{\text{th}}$ derivatives is again normal.

For our setting, where $\mathscr{F} = \{ f^n : n \in \mathbb{N}\}$, we need only look at the derivatives at $0$. If $\mathscr{F}$ is normal, then the sequence $\bigl((f^n)^{(k)}(0)\bigr)_{n\in \mathbb{N}}$ must be bounded for every $k$. Investigating that will lead you to a necessary and sufficient condition.

I think you donʼt have to consider so precisely. The answer is the family ${\frac{d^3 f^n}{dz^3}}$ is normal.


Is the symmetric definition of the derivative equivalent? Is the symmetric definition of the derivative (below) equivalent to the usual one?

\begin{equation}
\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}
\end{equation}

I've seen it used before in my computational physics class. I assumed it was equivalent but it seems like it wouldn't matter if there were a hole at $x=h$ in the symmetric derivative, whereas with the usual one it wouldn't be defined. Which is kinda interesting...

If they're not equivalent - is there a good reason as to why we should use the common one? Or is the symmetric one actually more useful in some sense because it "doesn't care" about holes?

As I noted in a comment to the other answer, Milly's computation is incorrect.  I am posting this answer to rectify the situation.  The symmetric derivative is defined to be
\begin{align}
\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}.
\end{align}
If $f$ happens to be differentiable, then the symmetric derivative reduces to the usual derivative:
\begin{align}
\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}
&= \lim_{h\to 0} \frac{f(x+h)-f(x)+f(x)-f(x-h)}{2h} && (\text{add zero}) \\
&= \frac{1}{2} \left( \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} + \lim_{h\to 0} \frac{f(x) - f(x-h)}{h}\right) \\
&= \frac{1}{2} \left( \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} - \lim_{-h\to 0} \frac{f(x+h) - f(x)}{-h}\right) && (\ast)\\
&= \frac{1}{2} \left( f'(x) + f'(x) \right) && (\text{since $f'(x)$ exists}) \\
&= f'(x).
\end{align}
I'm begin a little trixy at $(\ast)$.  Notice that we can think about the difference quotient as being the slope of a secant line through the points $(x,f(x))$ and $(x+h,f(x+h))$.  This slope is given by
\begin{equation*}
\frac{f(x+h)-f(x)}{(x+h)-x}.
\end{equation*}
Multiplying through by $-1$, this becomes
\begin{equation*}
-\frac{f(x+h)-f(x)}{(x+h)-x}
= \frac{f(x+h)-f(x)}{x-(x+h)}
= \frac{f(x+h)-f(x)}{-h}
\end{equation*}
Taking $h$ to zero (which is the same as taking $-h$ to zero) on the left gives $-f'(x)$, justifying $(\ast)$.  Again, this proves the key statement:

  Proposition: If $f$ is differentiable at $x$ (in the usual sense), then $f$ is symmetric differentiable at $f$, and the symmetric derivative agrees with the usual derivative.

The converse does not hold.  The usual example is the absolute value function which is not differentiable at zero, but which is symmetric differentiable at zero (with derivative zero):
\begin{equation*}
\lim_{h\to 0} \frac{|0+h|-|0-h|}{2h}
= \lim_{h\to 0} \frac{|h|-|h|}{2h}
= \lim_{h\to 0} \frac{0}{2h}
= 0.
\end{equation*}
In particular, this demonstrates that the symmetric derivative is a generalization of the usual derivative.  It cannot be equivalent, because it can meaningfully define the derivative of a larger class of functions.

As to the why? of the symmetric derivative (and why we don't use it instead of the usual derivative), I think that adequate answers can be found attached to this question and in comments elsewhere on MSE.

$$\lim_h \frac{f(x-h)-f(x+h)}{2h}= \frac12 \lim_h \frac{f(x-h)-f(x)+f(x)-f(x+h)}{h}$$
$$=\frac12(\lim_h \frac{f(x-h)-f(x)}h+\lim_h \frac{f(x)-f(x+h)}h)=\frac{1}{2}(-f'(x)-f'(x))=-f'(x)$$
However this definition can be "strange" sometimes (in the sense it does not catch the idea of "derivative as slope"): let $f(x):=1/x^2$, and
$$\lim_h \frac{f(-h)-f(h)}{2h}=0.$$
Even without asymptotes, you can have something like $g(x):=|x|$, and
$$\lim_h \frac{g(-h)-g(h)}{2h}=0.$$


Colored egg carriage problem Bob has lots of painted eggs to $50$ different colors. He wants to carry one egg of each color from point $A$ to $B$. The probability that he breaks an egg is $P$ (same for all colors). I want to formulate cost of carrying all $50$ eggs for two different conditions:

If there is any broken egg when he reaches point $B$, he goes back to point $A$ and tries to bring eggs of the same color he broke (only broken eggs will be brought and he can take one egg of each color).
If there is any broken egg when he reaches point $B$, he goes back to point $A$ and tries to bring all eggs from scratch (all $50$ eggs will be brought again).

Going from point $A$ to $B$ count as $1$ cost. Going back to from $B$ to $A$ does not count.

[Edit: The question has been clarified. I'll leave my original interpretation and answer because it is beneficial as stepping stone to the new question.]

Let $n$ be the number of eggs and $p$ be the probability that each egg breaks, independently of other eggs.

[My original interpretation of the second condition: All eggs are discarded at B if at least one is broken, and he goes back to A and restarts the whole procedure.]

Let $x$ be the expected number of trips and find an equation for $x$ by considering the expected number of trips after one trip. There are two cases corresponding to whether no eggs break or at least one egg breaks. I'm sure you can compute the probability $q$ that the first case occurs, in terms of $n$ and $p$. In that case we need to restart the whole procedure and so the expected number of trips from that point is $x$. In the other case we are done and the expected number of trips from that point is $0$. So we get $x = 1 + q \cdot x + (1-q) \cdot 0$ and then we can find $x$.

[The new question: Unbroken eggs are kept at B and he stops if he has at least 1 unbroken egg of each colour at B.]

Note that the two conditions are equivalent because bringing a egg of a colour that he already has (unbroken) at B makes no difference. Let $f(k)$ be the expected number of trips where $k$ eggs need to be brought. Clearly $f(0)=0$. For each $k>0$, there are now $k+1$ cases corresponding to how many eggs break on the first trip, and you can compute the probability that each case occurs, which gives as before a linear equation involving $f(k)$ and $\{ f(m) : 0 \le m \le k \}$ with coefficients depending on $p$ and $n$, which is $f(k) = 1 + \sum_{m=0}^k P(m\text{ eggs broke}) f(m)$. You can now see that you can find $f(k)$ for any $k$ from $1$ to $n$.

Let's divide this by steps. Let step 1 be the travel between point $A$ and point $B$.

You have $P(x)$ as a chance to break one single egg on arrival to $B$. As the probability is equal for all other eggs, and assuming the events are independent, the probability of $n$ eggs breaking is given by $P(x)^n$.

This being said, one can calculate the chance all eggs arrive safely to the other side.

$$
P = 1 - \sum_{i=1}^{n} P(x)^n
$$

The cost for the 1st step is therefore given by the following:

$$
Cost = P \times C
$$

Where $C$ is the cost of the travel. However, this doesn't account the fact that you have to get back and get any egg that broke. To get the total cost one would have to consider all hypothesis iteratively. Assuming that no egg broke on the way back, we would have a repetition of the thought process in the first step. The cost for the 2nd step would be the following.

$$
Cost = P \times C + (1-P)(P \times 2C)
$$

Note the $2C$, as you had to travel two times.

For the 3rd step:

$$
Cost = P \times C + (1-P)(P \times 2C) + (1-P)(1-P)(P \times 3C)
$$

The full series (and as such, the full cost) is the following:

$$
Cost = \sum_{i=1}^{n \rightarrow \infty} iPC (1-P)^{i-1}
$$

The answer is valid for both situations, as replacing one or every egg has no cost or effect on probability, the only thing that matters are the rides from $A$ to $B$.