Why $(x)$ is a prime non-maximal ideal in $K[x,y]$ where $K$ is a field? The ring of poynomials $K[x,y]$, $K$ a field is not a principal ideal domain. I want to show that in this ring non-zero prime ideals need not be maximal.

I am having difficulties in showing that $(x)$ is prime but it is not maximal.

Let $f(x)$, $g(x)$ be two elements of $K[x,y]$ such that $f(x)g(x)$ is in $(x)$. Then there exists $h(x)$ in $K[x,y]$ such that $f(x)g(x)=xh(x)$. So, $x$ divides the product $f(x)g(x)$.

How can I arrive in the conclusion that either $f(x)$ or $g(x)$ is in $(x)$?

Would you help me, please?

The ring homomorphism $K[x,y] \to K[y], f(x,y) \mapsto f(0,y)$ induces an isomorphism $K[x,y]/(x) \cong K[y]$. Now $K[y]$ is a domain, but not a field, thus $(x)$ is prime, but not maximal.

The answer of MatheiBoulomenos exploits the equivalence "$P$ prime $\iff$ $A/P$ domain" and is certainly correct.

But you can prove that $(x)$ is prime directly from the definition as follows.

Let $P(x,y)$ and $Q(x,y)$ be two polynomials with
$$
P(x,y)Q(x,y)\in(x).\qquad\qquad(*)
$$

Write $P(x,y)=p(y)+xP_1(x,y)$ and $Q(x,y)=q(y)+xQ_1(x,y)$ separating the monomials that do not contain $x$.
Then
$$
P(x,y)Q(x,y)=p(y)q(y)+x[P_1(x,y)q(y)+Q_1(x,y)p(y)+xP_1(x,y)Q_1(x,y)]
$$
and the hypothesis $(*)$ implies that  $p(y)q(y)=0$. But since $K[y]$ is a domain this implies that, for instance, $p(y)=0$ which means that $P(x,y)\in(x)$.


What is this problem asking? (Arc length) All of the previous questions are asking me to find the distance of a line in between a set of x-values, for example the distance of $y=x^2$ when $-1\leq x\leq 2$. The problem I don’t understand is asking for the length of $y=\int_0^x \tan t\,dt$ when $0\leq x\leq \frac{\pi}{6}$. How do I go about this?

The arc length of the curve given by  $y=\int_0^x \tan t\,dt$ when $0\leq x\leq \frac{\pi}{6}$ is
$$
\begin{align*}
\text{ArcLength}(y)
&= \int_0^{\pi/6} \sqrt{1+(y'(x))^2}dx \\
&= \int_0^{\pi/6} \sqrt{1+\tan^2 x} \: dx \hspace{4mm} \mbox{ by the Fundamental Theorem of Calculus}\\
&= \int_0^{\pi/6} \sqrt{\sec^2 x} \: dx \\
&= \int_0^{\pi/6} |\sec x| \: dx \\
&= \int_0^{\pi/6} \sec x \: dx   \hspace{4mm} \mbox{ since }\sec x \geq 0 \mbox{ on } [0,\pi/6]  \\
&= \ln|\sec x + \tan x | \: \bigg|_0^{\pi/6} \\
&= \ln \left|\sec \left(\frac{\pi}{6}\right)
 + \tan \left(\frac{\pi}{6}\right) \right| - \ln|\sec 0 + \tan 0| \\
&= \ln \left(\frac{2}{\sqrt{3}}+ \frac{1}{\sqrt{3}}\right) - \ln (1+0) \\
&= \ln \left( \frac{3}{\sqrt{3}}\right) \\
&= \ln\left( \sqrt{3} \right) \\
&= \boxed{\frac{\ln\left( 3\right)}{2}}. \\
\end{align*}
$$

Hint By the Fundamental Theorem of Calculus,
$$y'= \tan(x)$$


When will $\operatorname{det}\left(A\cdot A^{\top}\right)=0$? I am writing a small computer program to solve certain linear algebra equations as part of a larger program.  For two of my functions I need to evaluate $\left(A\cdot A^{\top}\right)^{-1}$.  This got me wondering.  Is there any significance to $\operatorname{det}\left(A\cdot A^{\top}\right)=0$ ?  When will this happen, and is it common in the real world?  In my particular case it is unlikely that $A$ itself would ever be a square matrix.  Just throwing that out there.

If $A$ is $m\times n$ real matrix and $\operatorname{rank}A = k\le\min(m,n)$, then $AA^T$ is $m\times m$ matrix of rank $k$. Thus $AA^T$ is invertible if and only if $\operatorname{rank}A = m$ (in particular: if $m > n$, then $AA^T$ must be singular.)

For complex matrices it does not hold in general, as baharampuri suggests. However for complex matrix $A$, $\operatorname{rank} A = \operatorname{rank} AA^*$ holds where $A^*$ is the conjugate transpose of $A$.

For A being a n x n matrix :
det(A * A^T)=det(A)*det(A^T)=det(A)^2
and det(A)=0 implies that A is not invertible.


Why does $\sum_{n=0}^\infty \frac{n!}{(2n+1)!!}$ converge to $\frac{\pi}{2}$ Why does:

$$\sum_{n=0}^\infty \frac{n!}{(2n+1)!!}$$

converge to $\frac{\pi}{2}$?

I'm honestly not sure where to start.

$$\sum_{n\geq 0}\frac{n!}{(2n+1)!!}=\sum_{n\geq 0}\frac{2^n n!^2}{(2n+1)\cdot (2n)!}=\sum_{n\geq 0}\frac{2^n}{(2n+1)\binom{2n}{n}}=\sum_{n\geq 0}\frac{2^n \Gamma(n+1)^2}{\Gamma(2n+2)}$$
can be written, through Euler's Beta function, also as
$$ \sum_{n\geq 0}2^n\int_{0}^{1}x^n(1-x)^n\,dx = \int_{0}^{1}\frac{dx}{1-2x(1-x)}\stackrel{\text{symmetry}}{=}2\int_{0}^{1/2}\frac{2\,dx}{1+(2x-1)^2} $$
or, through the substitution $x=\frac{1-z}{2}$, as
$$ \int_{0}^{1}\frac{2\,dz}{1+z^2}=2\arctan(1)=\color{red}{\frac{\pi}{2}}.$$

As an alternative, you may notice your series (related to the Taylor series of the arcsine function) is half the Euler transform of Gregory series. See pages 20-21 of my notes, for instance.

What about finding a function $f$ for which there exists an $x$ such that $f(x)=\frac{\pi}{2}$, calculating its Taylor series and using the Euler Transform for series?

Start with $a\tan(1)=\frac{\pi}{4}$ and search for the Taylor series of $a\tan(x)$:
$$a\tan(x)=f(x) = \sum _{n=0}^{\infty } \frac{(-1)^n}{(2n+1)}x^n.$$

Looking at the Euler transform  gives us $$ \left(\frac{1}{1-x}\right)f\left(\frac{x}{1-x}\right) = \sum _{n=0}^{\infty } \left(\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)}\right)x^n.$$

Using that  $$\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{(2k+1)} = \frac{(2n)!!}{(2n+1)!!}$$

We see that the series $$ \sum_{n=0}^{\infty} \frac{n!}{(2n+1)!!}  =\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^n =
 2f(1)=\frac{\pi}{2}$$


Mathematics for Graduate Political Science I'm preparing to attend graduate school for political science here in Canada and I'm having something of a crisis. Midway through my degree program I chose to drop my first love (English) to focus on what I was more interested in professionally. I'm now completing a honours degree in political science and international relations. Thing is, I've lost much of my quantitative ability (not that I had much to begin with). I was able to do some differential calculus a couple years ago, but I fear I've forgotten much of what I knew how to do.

I've become fascinated with quantitative methodology and want to practice quantitative research in graduate school. I'm currently on track to study theory. So here's my question: If I want to be able to use quantitative methods in grad school (say negative binomial regression, for example) how much math should I know prior to entering? Multivariable calculus? Linear algebra? Differential equations? When it comes to eventual knowledge, the sky's the limit, but I'm immediately concerned with applied maths. Pardon any ignorance inherent in this post.

For quantitative social science, the more math you can learn before you get to grad school, the better. A minimal preparation would be something like:

Three semesters of calculus, which is usually two of single-variable and one of multi
One semester of linear algebra
One semester of probability and statistics

If you have time, the following courses would look good:

Differential equations
Advanced probability or mathematical statistics courses
Real analysis, which might be called advanced calculus

Topology also couldn't hurt, in case you want to learn functional analysis later on. You should certainly take a couple of computer science courses.

The standard list of maths needed are as follows:

1) A "good" calculus class this is at the level of MITOCW calculus or edx.org calc 1A, calc 1B, calc 1C etc.   Free or 49 dollars if you want certificate

2) Linear Algebra say MITOCW linear algebra by Gilbert Strang  free

3) Econometrics try MITOCW econometrics or coursera.org econometrics  free or 49 dollars

4) Bayesian Methods try coursera.org Bayesian Methods UC Santa Cruz  free or 49 or more if you take 5 course series

5) Software Learn how to use R and Stata at a minimum

6) Learn how to clean a data set

7) take the survey methodology series on Coursera at least 4 courses 49 each or free

8) Don't take any more than this without doing some serious reading of political science journals.   If you can't read the journal with
 understanding as in being able to replicate the analysis

9) Work for a politician or sit and watch at local senate/house meetings
10) read some partisan mags./radio shows/  etc.

11) Learn the language of political sci don't apply to you know the area break down of all the areas of political science American, Political Economics, Methodology, etc.

12) more important to know the language of political science and to have an imagination of how to write about it than to be have math phd so don't overdo the math.

13) Don't forget to sit in on some classes at your local phd mill.

14) coursera.org Game theory I & 2 stanford and Tokyo Uni Intro game theory

15) Network Theory class coursera.org or edx.org

16) Model Thinking Coursera.org

17) You shouldn't need Grad real analysis, Measure theory, Functional Analysis, Topology, etc.

18) You might need Control theory if dynamic models is an interest.  But read the journal first because if you don't like those types of papers or if no one really does them why should you.


Do vectors need to work in translated coordinate systems as well as rotated coordinate systems? In the Feynman Lectures on Physics, Chapter 11, section 5 (http://www.feynmanlectures.caltech.edu/I_11.html#Ch11-S5) Feynman states: "In order for it to be a vector, not only must there be three numbers, but these must be associated with a coordinate system in such a way that if we turn the coordinate system, the three numbers “revolve” on each other, get “mixed up” in each other, by the precise laws we have already described."

Is there not an equivalent requirement for when we translate the coordinate system?

If we ignore y and z components and just focus on x components, so we can describe each point in our coordinate system using only one number, we have the points:

A(a), B(b), and C(a + b).

In a new coordinate system, translated by α to the right, we have the points:

A(a'), B(b'), C(a' + b').

Substituting into the equation linking the two coordinate systems (x' = x - α), we get:

a' = a - α, b' = b - α, a' + b' = a + b - α

However, it is clear that adding a' and b' gives a' + b' = a + b - 2α, not a' + b' = a + b - α - they don't match.

Is my reasoning here wrong or is there simply no requirement for translations of coordinate systems as there is with rotations?

In that context, that is as in linear algebra, vectors are considered "applied" to the origin, therefore we are not interested in translations. In particular the text is referring to a rotation around the $z$ axis.


Cantor set--nowhere dense, complete I can't figure out this out.

Cantor set is closed in $\mathbb{R}$.

$\mathbb{R}$ is a complete metric space.

Every closed subset of a complete space is also complete; thus, so is the Cantor set.

Complete space can't be written as a countable union of nowhere dense sets; thus, the Cantor set has no such representation. (1)

Cantor set is nowhere dense (its interior is empty, and no intervals are contained in it)

Thus, the Cantor set is a finite (then countable) intersection of nowhere dense sets.

Contradiction with statement (1).

Where is a mistake?

Does representation in (1) refer only to infinite countable intersections?

the assumption that since its finite thus countable is false.  Countable means a bijection with N.  It can be infinite (countably infinite, or simply just stated countable)


Numerical range of a matrix contains the convex hull of the eigenvalues. I am stuck with the following question.

Question: Let $A \in \mathbb{C}^{m \times m}$ be arbitrary. Let $W(A)$ be the numerical range i.e. the set of all Rayleigh quotients of $A$ corresponding to a all nonzero vectors $x \in \mathbb{C}^m$. Show that $W(A)$ contains the convex hull of the eigenvalues of $A$.

Progress: We can write $W(A) = \{ x^TAx : \| x \| = 1 \}$. If $(\lambda_i, q_i)$ is an eigenpair of $A$, we know that the Rayleigh quotient is $r(q_i) = \frac{q_i^TAq_i}{q_i^Tq_i} = \lambda_i$. It is enough to show that every convex combination $\alpha_1\lambda_1 + \alpha_2\lambda_2 \in W(A)$ (i.e. $\alpha_1 + \alpha_2 = 1$).

Because $A \in \mathbb{C}^{m \times m}$ we don't have an orthogonal basis of eigenvectors. Since we can, I thought it would be helpfull to choose $\| q_i \| = 1$. I tried taking $x = \sqrt{\alpha_1}q_1 + \sqrt{\alpha_2}q_2$, then
$$
r(x) = \frac{\alpha_1\lambda_1 + \alpha_2\lambda_2 + \sqrt{\alpha_1\alpha_2}(\lambda_1 + \lambda_2)q_1^Tq_2}{\| \text{norm} \|^2},
$$
but then the problem with the $q$'s not being orthogonal arises.

Can somebody help me solve this problem?

The numerical range $W(A)$ is convex. This is a non-trivial fact, as far as I know, so maybe you are supposed to use this fact and not prove it.

According to Wikipedia page for Rayleigh quotient, Rayleigh quotient is usually defined for Hermitian Matrix, so you can assume that eigenvectors are orthonormal and hence the last part of your equation is equal to zero.


Is the set $\{\langle \varnothing, a \rangle ,\langle \{ \varnothing \}, b \rangle \}$ considered as function? The set defined as
$$F=\{\langle\varnothing,a \rangle, \langle \{\varnothing\},b\rangle\}$$
a function?

Assuming you're using the "functions-as-sets-of-ordered-pairs" approach, then yes, it is: its domain is $\{\emptyset,\{\emptyset\}\}$, it sends $\emptyset$ to $a$, and it sends $\{\emptyset\}$ to $b$. Crucially $\emptyset\not=\{\emptyset\}$, so there's no inconsistency here.
(The other standard approach to functions views a function as a set of ordered pairs together with a specific mention of domain and codomain. Of course the domain is determined by the set of ordered pairs (at least, if we don't allow partial functions), but the codomain is not, so this is a genuinely different approach. If we take this stance then what you've written is not a function.)
EDIT: note that the parenthetical remark is exactly the point Arturo Magidin makes in his comment above.

Yes, it is a function, and if you adopt the usual definition that $0 = \varnothing$, $1 = \{ \varnothing \}$, then this function maps from $\{ 0, 1 \}$ to real (or complex) numbers, and its explicit form is: $F(0) = a, F(1) = b$.


Calculate distance between two points on concentric circles I am trying to find the shortest distance between two concentric circles. I already know the angle between the two points and radii of the circles, but I am not sure how to calculate the distance between the two.

Imagine the example below, where $R_1 = 2$ and $R_2 = 3$ and the points are separated by 45 degrees. I would like to calculate the distance between these two points. Any help is much appreciated. Thanks!

EDIT:

Let me be more clear. I do not need the linear distance, I can do that no problem. I need the distance to be of the arc that connects these two points. The arc should "rise" at a constant rate between $R_1$ and $R_2$. This connection should also not intersect with the innermost circle.

Basically I am trying to simplify my problem. I currently have the great circle distance of two points on a globe, but this assumes that the two points are at the same radius. This is not always true when you are trying find the distance between a ground station and a satellite or something like that. I know the latitude, longitude, and altitude of both objects, so I can find this their angular separation, but don't know how to calculate the distance of the arcs that connect them.

Assume: $R_1 < R_2$

Since the radius increases at a constant rate relative to the change of the angle $\theta$ the polar equation would be $r = r_1 + \frac{r_2-r_1}{\theta_1 - \theta_0}\theta$ where $r_2 > r_1$.  To find the distance you need the integral $\int_{\theta_0}^{\theta_1}\sqrt{1+(\frac{\text{d}r}{\text{d}\theta})^2}\text{d}\theta$, where $\frac{\text{d}r}{\text{d}\theta}=\frac{r_2-r_1}{\theta_1 - \theta_0}$.  Which would yield a length of $(\theta_1 - \theta_0)\sqrt{1 + (\frac{r_2-r_1}{\theta_1 - \theta_0})^2}$.  Even simpler it would be $\sqrt{(\theta_1 - \theta_0)^2 + (r_2-r_1)^2}$.

Let $C_2$ and $C_3$ denote the two circles. Any point on a circle is determined by the angle $\theta$. That is any point on $C_r$ is given by $(r\cos\theta,r\sin\theta)$. Therefore if you know the difference between the two angles, lets call this $\theta$ again, we have distance
$$
||(3\cos\theta,3\sin\theta) - (2\cos 0,2\sin 0)|| = \sqrt{(3\cos\theta - 2)^2 + (3\sin\theta)^2}.
$$


Linearization of a product of two decision variables I am trying to solve a problem that involves constraints in which products of two decision variables appear.
So far, I read that such products can be reformulated to a difference of two quadratic terms:

$x_1\cdot x_2=y_1^2-y_2^2$

Where $y_1 = 0.5 \cdot \left( x_1+x_2 \right)$ and $y_2 = 0.5 \cdot \left( x_1-x_2 \right)$

As stated in "Model building in mathematical programming" by H.P. Williams, I tried to linearize $y_1^2$ and $y_2^2$ by piecewise approximation. When I only use two node points (one interval) for the piecewise approximation, my solvers (Gurobi 5.6.3 and CPLEX 12.5.1) are able to solve the problem, but when I introduce more node points, both conclude that the problem becomes infeasible.

I already tried SOS2 variables as well as a binary approach for the linearization of $y_1^2$ and $y_2^2$.

I also used Taylor series to directly approximate $x_1\cdot x_2$ which resulted in similar results as the approximation using two node points. Since this method succeeded, I suppose that the rest of my model is feasible and only this product makes things complicated.

So my questions are:

What can cause the infeasibility? I thought that adding more node points would be beneficial to the problem.
Are there better ways to approximate such products? I have also experimented with a reformulation using logarithms, but this caused more computational effort and did not solve the infeasibility problem.

The AIMMS Modelling Guide in section 7.7 provides the following hint for bounded variables:

Have a look at this paper. This article introduces 10 methods for your problem. on page 5, the authors introduce SOS procedure for bilinear terms.


Show that every principal ideal domain is a unique factorisation domain I have to show that every principal ideal domain is a unique factorisation domain.

I can show that every non invertible element that is not equal to $0$ can be written as product of irreducible elements.

However, I have troubles showing that if

$$p_1 \dots p_n = q_1 \dots q_m$$ where $q_i, p_i$ are irreducibles elements, then $n = m$ (I already showed that every $p_j = u_iq_i$ with $u_i$ invertible)

Any hints?

You know that in a PID a element is irreducible if and only if it is prime. Then, if $p_1$ on the left divides $q_1 \cdot \ldots \cdot q_m$ then it must divide one of this factors, lets say $q_i$. But $q_i$ is irreducible, thus

$$ p_1 = uq_i $$

where $u$ is invertible. By induction, you can then show that $m=n$

after $k$ times, if $m$ is less than $k$ (which happen only when $m<n$)
$$ p_k \cdot \ldots \cdot p_n = u_k $$
Which is a contraddiction (the product of prime elements cannot be invertible). So $m \geq k$ and
$$ p_k \cdot \ldots \cdot p_n = u_k q_k \cdots q_m $$
We can then apply the same argument I explained on top.

If otherwise m > n eventually you will have
$$ 1 = u_n q_n \cdot \ldots \cdot q_m $$
which is a contraddiction, as shown before

According to JayTuma you get (w.l.o.g. $i=1$):

$q_1\ldots q_m=p_1\ldots p_n=u_1q_1p_2\ldots p_n$

so $q_2\ldots q_m=u_1p_2\ldots p_n$ since every PID is an integral domain. Now substitute $p_2$ by (w.l.o.g.) $u_2q_2$ to get $q_3\ldots q_m=u_1u_2p_3\ldots p_n$ and so on. From this you easily deduce that $n$ has to be equal to $m$.


Functions in a Reproducing Kernel Hilbert Space are Lipschitz continuous I would like to show that all the functions in a Reproducing Kernel Hilbert Space (RKHS) are Lipschitz continuous. So that, I take two points in the domain $\vec{x}_{1} ,\vec{x}_{2} \in X$ then from the Riez Representation Theorem it follows that,

\begin{equation}  \begin{array}{l} {\left|f\left(\vec{x}_{1} \right)-f\left(\vec{x}_{2} \right)\right|=\left|\left\langle f,K_{\vec{x}_{1} } \right\rangle _{H_{K} } -\left\langle f,K_{\vec{x}_{2} } \right\rangle _{H_{K} } \right|} \\ {\quad \quad \quad \quad \quad \quad \; =\left|\left\langle f,K_{\vec{x}_{1} } -K_{\vec{x}_{2} } \right\rangle _{H_{K} } \right|} \\ {\quad \quad \quad \quad \quad \quad \; \le \left\| f\right\| _{H_{K} } \left\| K_{\vec{x}_{1} } -K_{\vec{x}_{2} } \right\| _{H_{K} } } \end{array} \end{equation}

Here $H_{K}$ is a RKHS and the inequality comes from Cauchy-Schwarz inequality.
In a document online, it is that $ \left\| K_{\vec{x}_{1} } -K_{\vec{x}_{2} } \right\|=(K_{\vec{x}_{1} }-K_{\vec{x}_{2} })^{2}$. But I can not show it. I am pretty sure that it equals  to difference of their kernelized positions, i.e., $ \left\| K_{\vec{x}_{1} } -K_{\vec{x}_{2}} \right\|=d(K_{\vec{x}_{1}},K_{\vec{x}_{2}})=\sqrt{\left\langle(K_{\vec{x}_{1}}-K_{\vec{x}_{2}}),(K_{\vec{x}_{1}}-K_{\vec{x}_{2}})\right\rangle}$

Much obliged for your help in advance.

Pag.28 of these slides may be useful to you.


order isomorphism maps $f < g$ to $Tf < Tg$? Suppose $C(X)$ is the set of functions $f:X \rightarrow \mathbb{R}$.

A bijection $T:C(X) \rightarrow C(Y)$ is called order isomorphism if $f \leq g$ if and only if $Tf \leq Tg$.

$f \leq g$ means that $f(x) \leq g(x)$ for all $x \in X$. Similarly, $f < g $ means that $f(x) < g(x)$ for all $x \in X$.

Question: If $f < g$, then $Tf < Tg$?

The statement seems correct but I don't know how to prove it.

If the statement above is not correct, then is it possible to obtain $f \geq 0 \Rightarrow Tf \geq 0$?

First of all, one usually defines $f < g$ to mean $f \leq g$ and $f \neq g$. This is different from your definition, which is a little confusing. For the sake of clarity, I will write $f \prec g$ whenever $f(x) < g(x)$ holds for all $x \in X$. (Now we have $f \preceq g$ if and only if $f \prec g$ or $f = g$. Note that $f \preceq g$ implies $f \leq g$, but not the other way around.)

For your first question, the answer is yes. Note that $C(X)$ is a lattice and that we have
$$ (f \wedge g)(x) = \min(f(x), g(x)),\qquad\text{for all $x\in X$}. $$
Furthermore, note that an order isomorphism between lattices automatically is a lattice isomorphism. In other words, we have $T(f \wedge g) = Tf \wedge Tg$ for all $f,g\in C(X)$. Now suppose that $f,g\in C(X)$ are given with $f \prec g$ but $Tf \not\prec Tg$, then there is some $y_0\in Y$ such that $Tf(y_0) = Tg(y_0)$ holds. We define $h \in C(Y)$ by
$$ h(y) = \begin{cases} Tf(y) + 1,&\quad\text{if $y = y_0$}; \\ Tf(y),&\quad\text{if $y \neq y_0$}. \end{cases} $$
Now set $h' = T^{-1}(h)$, then we have
$$ T(h' \wedge g) = h \wedge Tg = Tf. $$
Since $T$ is injective, it follows that $h' \wedge g = f$ holds. Since we have $f(x) = \min(h'(x), g(x)) < g(x)$ for all $x\in X$, it follows that $h' = f$ must hold. But now we have $T(h') = h \neq Tf$, which is a contradiction.

For the second question, the answer is no. Suppose that $X = Y$ holds and consider the function $T : C(X) \to C(Y)$ given by $f \mapsto f - 1$, that is:
$$ (Tf)(x) = f(x) - 1. $$
Then clearly $T$ is an order isomorphism, but we have $T(0) \not\geq 0$.

You are correct:
$$
\def\imp{\quad\Longrightarrow\quad}
f < g \imp f \le g \imp Tf \le Tg.
$$
But, if $Tf = Tg$, then $f=g$ (since $T$ is an injection), which contradicts the fact that $f < g$.


Def. of lin independence for subsets I use following definition:

  let be $v:\Bbb{N} \to T$ with $T\subseteq V $ and $V $ a $K-$vector space: $$ (v_i)_{i \in \Bbb{N}} \text{ is lin. ind. if } \\ \forall \alpha \in K^\Bbb{N}: \sum_{i \in \Bbb{N}} \alpha_i \, v_i=0 \wedge \exists m \in \Bbb{N}:\forall n>m: \alpha_n=0 \to \forall i \in \Bbb{N}: \alpha_i=0$$

Now I need a definition not for family of vectors but for subsets of $V $ . I thinked $T \subseteq V$ is lin ind if $\forall x\in T: x \notin <T\setminus \{x\}> $ but I have following problem:

  the family of vectors  $((1,2), (1,2)) $ ist not lin ind, but the set $\{(1,2)\} $ ist lin ind...

Are there therefore two definitions of lin ind, one for families and one for sets?!

By definition singleton  sets are always linearly independent.

Let $\{\alpha \neq 0\}$ be a singleton  set. $a_1 \alpha = 0 \implies a_1 = 0 $.

Now $a_1(1,2) + a_2 (1,2) = (0,0) \implies (a_1 +a_2, 2a_1+2a_2) = (0,0) \implies a_1 = -a_2 $. Thus the family of vectors  $((1,2), (1,2)) $ is not linearly independent.


Question about proving existence of a function $f$ such that $f \circ f = g$ for an odd function $g$ I have been reading a book on elementary mathematics and have come to a problem where I don't understand where the solution comes from. The problem goes :
"Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be an odd function, such that $g(x) > 0$
whenever $x > 0$. Show that there exists a function $f : \mathbb{R} \rightarrow \mathbb{R}$
for which $g = f \circ f$."
In the answer key they just say :
"Take $f(x) = -g(x)$ for $x \leq 0$ and $f(x) = -x$ for $x \geq 0$."
I am wondering what the reasoning is behind this answer. The truth is that I just want to see a more detailed solution to the problem. Can anyone help with this ?

Interesting question, as an example consider the identity function for $g$, then $f$ is either trivially the identity function or more interestingly the negative the identity function (or more generally Babbage's functional equation, but that takes us off topic). What about $f(x)=x^3$? Well it's just as easy to solve the general case.
Now consider $f\circ f(x)=g(x)$ the naive attempt would be to let $f(x)=g(x)$ but then things wouldn't workout, so to fix the issue we need to define $f(x)=x$ sometimes, and $f(x)=g(x)$ other times. If we let $f(x)=x$ if $x>0$ then we get stuck because $f\circ f(x)=x\neq g(x)$ if $x>0$, so to avoid getting stuck we let $f(x)=-x$ if $x>0$, and once we do that we know that $f(x)$ must equal $g(-x)$ if $x<0$. so that $f\circ f(x)=g(-(-x))=g(x)$ if $x>0$, or $f\circ f(x)=-g(-x)=g(x)$ if $x<0$.(Since $g$ is odd)
Notice the solution is not unique, since we could have defined $f(x)=-x$ if $x<0$ then we would define $f(x)=-g(x)$ if $x<0$. Are there more?

Here's the reasoning: Since $g$ is odd, we really only need to know how it maps positive numbers to know all of $g$. We set up $f$ so that the first application gets the correct $g$ value, if you ignore the sign, and the second application fixes the sign. At least that's what it does for negative numbers. For positive numbers, it does roughly the same two steps, but in the other order.
It's pretty clever -- it's as if $f$ uses the sign to both encode and decide whether or not $g$ has applied yet.


How to show $\lim_{\epsilon \rightarrow 0} \frac{\epsilon}{x^2+\epsilon ^2} = \pi \delta (x)$? Show that :

$\lim_{\epsilon \rightarrow  0} \frac{\epsilon}{x^2+\epsilon ^2} = \pi  \delta (x)$

Where $\delta (x)$ is the dirac-delta function.

I can't show that the integral of this over all $x$ is $\pi$

Since you tagged distribution theory, I think the following should work.

For any test function $\varphi(x)\in\mathcal{D}(\mathbb{R})$, where $\mathcal{D}(\mathbb{R})$ consists of $C^{\infty}$-smooth functions with compact support (which indicates that $\lim_{x\to\pm\infty}\varphi(x)=0$). We have the following
$$
\begin{aligned}
\int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&=\int_{-\infty}^{\infty}\varphi(x) d\tan^{-1}
\left(\frac{x}{\varepsilon}\right) \\
&=
\left[
  \varphi(x)\tan^{-1}
\left(\frac{x}{\varepsilon}\right)
\right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\tan^{-1}
\left(\frac{x}{\varepsilon}\right)d\varphi
\end{aligned}
$$
then we have
$$
\begin{aligned}
\lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\frac{\varepsilon\varphi(x)}{x^{2}+\varepsilon^{2}}dx&=
-\lim_{\varepsilon\to0}\int_{-\infty}^{\infty}\tan^{-1}
\left(\frac{x}{\varepsilon}\right)d\varphi \\
&=-\frac{\pi}{2}\int_{0}^{\infty}d\varphi+\frac{\pi}{2}\int_{-\infty}^{0}d\varphi \\
&=-\frac{\pi}{2}
\Big[
  \varphi(x)
\Big]_{0}^{\infty}+
\frac{\pi}{2}
\Big[
  \varphi(x)
\Big]_{-\infty}^{0} \\
&=\pi\varphi(0)
\end{aligned}
$$
then
$$\lim_{\varepsilon\to0}\frac{\varepsilon}{x^{2}+\varepsilon^{2}}=\pi\delta(x)$$
in the distributional sense.

For $x\ne 0$,
$$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=0$$
For $x=0$,
$$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\infty$$
Besides,
$$\int_a^b \frac{\epsilon}{x^2+\epsilon^2}dx=\int \frac{1/\epsilon}{1+(x^2/\epsilon^2)}dx$$
$$=\int_a^b \frac{1}{1+(x^2/\epsilon^2)}d(x/\epsilon)$$
$$=\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}$$
Now if $a>0, b>0$, then
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - \pi/2 =0$$
Similary, for $a<0, b<0$, we have
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=(-\pi/2) - (-\pi/2) =0$$
For $a<0, b>0$, we have
$$\lim_{\epsilon\to infty}\arctan(x)\bigg|_{a/\epsilon}^{b/\epsilon}=\pi/2 - (-\pi/2) =\pi$$
Hence
$$\lim_{\epsilon\to 0}\frac{\epsilon}{x^2+\epsilon^2}=\pi\delta(x)$$


Proving completeness of modal systems In reading through Chellas' Modal Logic I'm trying to understand his proof of completeness.  Here is my best outline of his proof:

If $\Sigma$ is any normal system of modal logic then it has a proper canonical standard model, and every proper canonical standard model is a canonical standard model.  So every system has at least one canonical standard model.  If $\mathcal{M}$ is a proper canonical standard model and $\mathcal{M}\in\mathrm{C}$ is a class of modals containing it, and $A$ any sentence, then $\vDash_{C}A \quad \Rightarrow \quad \vDash^{M}A \quad \Rightarrow\quad \vdash_{\Sigma}A$ so that $\Sigma$ is complete with respect to $C$.

Now of course my question may be premised on a misunderstanding of his proof, but if my understanding is correct, then this seems to prove that every modal logic system is complete, since every one has a proper canonical standard model.

Is it that we shouldn't be talking about the completeness of a system per se, but completeness only with respect to some class?  So would that mean that we have to first pick a class and then, even though we know the system has a proper canonical model, we need to show that the proper canonical standard model that we come up with is actually in the class that we have chosen?

If that's right, then how would we for instance prove the completeness of $K5$, assuming it's complete?  What class would we choose?  I feel like I have some kind of fundamental misunderstanding about the goal or the method or both.

I'm trying to summarize the issue without too many details.

We have to start from the language of modal logic, with the "operator" $\Box$, and its semantical interpretation, following Kripke.

Of course, we want that the calculus for modal logic turns to be sound and complete with respect to semantics; and so it is.

The first step is to prove that every tautology is valid. The set of tautologies (taken as axioms) and the rule of inference modus ponens will constitute the basis of all modal logics. Thus, we have to show that all the tautologies
are valid, and that the rule of modus ponens preserves validity.

The next step is to consider the system K (the "basic" system of modal logic, i.e. the weakest normal modal logic system). We prove that the logic K is sound, that is, that every theorem of K is valid. We do this by showing that the axiom $K$ is valid and that the rule of necessitation
preserves validity. This, together with the previous demonstration that the tautologies are valid and that mp preserves validity, guarantees
the soundness of K.

With Kripke models, we can also show the invalidity of some schema, like $\Box \varphi \to \Diamond \varphi$.

Also others schemata turn to be invalid :

  $\varphi \to \Box \Diamond \varphi$ (B)

or

  $\Box \varphi \to \Box \Box \varphi$ (4)

or

  $\Box (\varphi \lor \psi) \to (\Box \varphi \lor \Box \psi)$.

But some of those schemata are still "interesting" from a modal point of view; thus, we have a further step. We look at some invalid schemata that nevertheless prove to be valid with respect to a certain class of models, in the sense of being true in all the models of a certain class. Some of these formula will be
valid with respect to all models in which the accessibility relation meets a certain condition.

The schema $4$ proves to be valid with respect to the class of models having
a transitive accessibility relation. Once we have seen that a schema $S$ is valid with respect to a certain interesting class $\mathsf C$ of models, we are able to show that the normal modal logic based on $S$ (i.e., having $S$ as an axiom schema) is sound with respect to $\mathsf C$, i.e., that every theorem
of the logic based on S is valid with respect to $\mathsf C$.

For example, the modal logic K4 has the tautologies, instances of $K$, and instances of the $4$ schema as axioms, and has as theorems all of the formulas derivable from these by modus ponens and necessitation. We can prove that every theorem of the normal modal logic K4 is valid with respect to the class of transitive models. Then we show that K4 is complete with respect to the class of transitive models, showing that the formulas which are valid in the class of transitive models are theorems of K4.

Intuitively, adding axioms to the "basic" modal logic K, will restrict the "numbers" of models. The issue is to find a suitable "characterization" of this logic in terms of a class of models, i.e., the class of all modles with a "special" feature with regard to the accessibility relation.

See Modal Axioms and Conditions on Frames :

  Each of the modal logic axioms we have discussed corresponds to a condition on frames in the same way. The relationship between conditions on frames and corresponding axioms is one of the central topics in the study of modal logics. Once an interpretation of the intensional operator $\Box$ has been decided on, the appropriate conditions on $R$ [the accessibility relation] can be determined to fix the corresponding notion of validity. This, in turn, allows us to select the right set of axioms for that logic.

i haven't read Chellas but I refer to the new edition of 'Hughes and Cresswell's Introduction to Modal Logic'  for an extensive treatment of your question. Completeness of a modal logic also depends onto the axioms chosen.


What does it mean to describe an adjunction of functors in terms of bimodules? Suppose $M$ and $N$ are $(A,B)$- and $(B,A)$-bimodules, respectively, where $A$ and $B$ are symmetric $R$-algebras, such that the functors $M\otimes_B - $ and $N\otimes_A - $ are both left and right adjoint via some isomorphisms $M\simeq\hom_B(N,B)$ and $N\simeq \hom_A(M,A)$. (I don't think the specific details are relevant for what follows.)

What does it mean to describe these adjunctions in terms of morphisms of bimodules? I read that these adjunctions can be described by
$\epsilon_{M,N}\colon M\otimes_B N\to A$, $\epsilon_{N,M}\colon  N\otimes_A M\to B$, $\eta_{M,N}\colon B\to N\otimes_A M$, and $\eta_{N,M}\colon A\to M\otimes_B N$.

This notation seems to suggest various units and counits. However, I'm used to be a counit/unit as being a natural transformation, say $\epsilon\colon M\otimes_B -\circ N\otimes_A -\to \text{Id}_A$, so it would be described in terms of the component morphisms, $\epsilon_X\colon M\otimes_B N\otimes_A X\to X$, where $X$ is an $A$-module. Now $X\simeq A\otimes_A X$, so viewing a component as a map of modules
$$
\epsilon_X\colon M\otimes_B N\otimes_A X\to A\otimes_A X
$$
seems like it would closely related to the bimodule map $\epsilon_{M,N}\colon M\otimes_B N\to A$. Is it saying something like $\epsilon_X=\epsilon_{M,N}\otimes 1_X$? And likewise for all the other three cases?

Remark: The classical Eilenberg-Watts theorem says the following: Let $R,S$ be commutative $k$-algebras. If
$$F: Mod(R) \rightarrow Mod(S)$$
is a $k$-linear right exact functor commuting with direct limits it follows there is an
$R\otimes_k S$-module $M$ and an isomorphism of functors
$$F(-) \cong -\otimes_R M.$$
Note: This result has "recently" been generalized to schemes and you will find all details in the online paper.


AP stat multiple choice need help AP Statistic

A simple random sample produces a sample mean, x(bar), of 15. A 95 percent confidence interval for the corresponding population mean is 15 +/- 3. Which of the following statements must be true?

(A) Ninety-five percent of the population measurements fall between 12 and 18.

(B) Ninety-Five of the sample measurements fall between 12 and 18.

(C) If 100 samples were taken, 95 of the sample means would fall between 12 and 18.

(D) $P(12\leq x(bar) \leq 18) = 0.95$

(E) If mu = 19, this x(bar) of 15 would be unlikely to occur.

Can you guys help me with this one? I thought the answer was B, but the answer is actually E. I don't get how come B isn't right.

A 95 percent confidence interval for the population mean says that 95 percent of times the interval will "catch" the true population mean (which is not known).
We have to take numerous samples and 95 % of the times, our confidence interval will contain the true population mean.

A is incorrect as we dont everything about the population.

B is incorrect as there can be samples where all measurements are way off.

C is incorrect as we have to take numerous samples, say in millions.

D is incorrect as Confidence intervals are specified for population means and not sample mean, x(bar).

E is correct as mu is population mean. If it were known to be 19, it is unlikely that x(bar) would be 15. This is the basis of hypothesis testing.

First thing wrong with A has to do with wording. You would say something like I am or We are 95% confident..... Or at the 95% confidence interval.....
B would need to say after taking many samples of size  whatever the confidence interval was calculated using a method that will capture the population parameter in 95% of the samples.


How do I find integers that solve the following inequalities? Give integers $a,b,c,x,y,z$ such that:

$$|a|+|b|+|c| < |x|+|y|+|z|$$ and $$\sqrt{x^2 + y^2 + z^2} \le \sqrt{a^2+b^2+c^2}.$$

I have no idea when this would ever be true, it seems like it would never be true. Any help is appreciated.

Because of the strict inequality involved and the constraint to integer values only, the problem has no solution. To see this, let the ordered tuples of integers $(x,y,z)$ and $(a,b,c)$ represent points in the integer grid $\mathrm R^3.$ Then we can show that the problem $$\sqrt{x^2+y^2+z^2}\le \sqrt{a^2+b^2+c^2}\le |a|+|b|+|c|< |x|+|y|+|z|$$ has no such solutions, by showing that if the sum of coordinate lengths of one of the points, say $A,$ is less than that of the other, say $B,$ then $A$ is strictly closer to the origin than $B,$ contradicting the given conditions.

Thus, without loss of generality let all coordinates be nonnegative (we're in the first octant), and let $p=a+b+c<x+y+z=q.$ Then since we're dealing with integers, assume again without loss of generality that $p=q-1.$ Then any nonnegative three-partition of $p$ will have at least one coordinate less than those of $q.$ It then follows that $A$ is on a sphere of smaller radius than $B.$


Hint: Compute $(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n+1}+\sqrt[4]{n})(\sqrt[4]{n+1}-\sqrt[4]{n})$.

Warm up: Simplify the expression $(a^2+b^2)(a+b)(a-b)$.
Pre-warm up: Simplify the expression $(a+b)(a-b)$ and deduce that $\sum\limits_{n=1}^{99}\frac1{\sqrt{n+1}+\sqrt{n}}=9$.


Explaining the definition about greatest common divisor, what do they mean up to the equivalence relation? The definition of gcd is provided below:

Let R be an principal ideal domain. Let a,b ∈ R. Then there is a greatest
common divisor of a and b, that is, an element d that divides both a and
b and such that if c divides both a and b, then c divides d. The element d is
unique up to the equivalence relation ∼ (a ∼ b if a and b are associate).

Here, a ~ b means a|b and b|a which was proved to be an equivalence relation.

My question is what do they mean by "up to the equivalence relation ~"?

Thanks in advance.

When you have an equivalence relation you separate the set into equivalence classes.  Just as $d$ divides $a$ and $b$, so does any associate of $d$.  Any of the associates of $d$ would work as well.  As an example, in the Gaussian integers, the units are $1,-1,i,-i$ and the associates of a number are the number multiplied by these.  Once you know, for example, that the GCD of $8$ and $3+3i$ is $1+i$, you know that $-1-i, -1+i, 1-i$ are also GCDs and that is all there are.  All the elements of the equivalence class are GCDs.

It means that this statement is true as long as a and b are not equivalent.


Schauder basis and Eigenbases There are several question in this site comparing different basis functions including Schauder basis and others, but I could not connect the difference between the Schauder basis and Eigenbasis clearly. Would someone please point the best way to contrast between the two?

An eigenbasis is a basis consisting entirely of eigenvectors (vectors associated with a linear system of equations).

A Schauder basis is like a basis except they allow for infinite sums.


Greatest common divisor 2 Find all integers m such that $0 < m < 1008$ and $\gcd(m, 2016-m) = 1$.

The following is my solution:
Note that $1 = \gcd(m, 2016-m) = \gcd(m, 2016)$.
Since $2016$ is even and divisible by $3, 7, 9, 21, 63$ and $\gcd(m, 2016) = 1$, so $m$ is odd and $m\neq 3k, 7l$, where $k, l\in \mathbb Z,$ and $k, l$ are odd numbers.
Since $0 < m< 1008,$ we deduce that $1\leq k\leq 335, \ 1\leq l\leq 143$ and $k, l$ are odd.
We set $$A = \{3k|\ k\in\mathbb Z,\ 1\leq k\leq 335,\ k \mbox{ odd}\}$$
and $$B = \{7l|\ l\in\mathbb Z,\ 1\leq l\leq 143,\ l \mbox{ odd}\}.$$
Then, we have $|(A\cup B)\setminus (A\cap B)| = 168 + 72 - 48 = 192.$
On the other hand, from $1$ to $1008$ we have $\frac{1007-1}{2} + 1 = 504$ odd numbers. Combining these data, we conclude that there are $504 -192 -24 = 288$ positive integers $m$.

Commenting on your solution process, I think you are making life a little more difficult than it needs to be in this instance by separating out odd numbers. You also end up with a wrong answer.
Here it makes more sense to simply take the three prime factors of $2016$ - $2,3,$ and $7$ - and count all the multiples of these numbers up to $1008$, and then by subtraction find how many numbers in the range are coprime to $2016$.

It's true here that $\gcd(m, 2016{-}m) = \gcd(m, 2016)$, but as observed it might be worth another line of justification for a well-paved proof. For solving the challenge it's not so important.

You correctly identify that we need to avoid multiples of $2,3,$ and $7$ in the range

You miss that you have twice subtracted numbers that are multiples of $3\cdot 7 = 21$, so your solution comes out low.

It's perhaps easier to leave $1008$ in the range because it will be eliminated anyway as not coprime to $2016$.

As a hand-worked alternative method of getting the count, you could note that the numbers coprime to $(2,3,7)$ will repeat on a block of $2\cdot3\cdot 7=42$ and in fact that pattern is itself symmetric about $21$.


Numerically solve differential equation that has no solution I am not really experienced in the field of numerical differential equations. I am only studying existence and uniqueness results about PDEs at the moment. Now I was wondering what would happen, if one tries to solve a PDE numerically, although the PDE 1) has no solution or 2) the solution is not unique?

The Navier-Stokes-Equation in 3 dimensions comes to mind. How do physicists approach this equation? Do they work with numerical solutions although there is no proof, that there is a solution at all?

I really appreciate your help!

Best,

Luke

What happens depends strongly on the software to compute the solution, but in general if you don't have a proof of existence and uniqueness as well as a proof of convergence to the unique solution, you are on very shaky ground.

Let's consider $y' = f$ with $y(0) = 0$. This can be solved by an ODE solver, under some assumptions on $f$ (say, continuity). Now let's say $f$ is a delta function at a random location in the domain. A numerical solver will never find it, and the answer will always be wrong.

The next situation is non-uniqueness. I wrote a few solvers for a PDE which had many solutions. What we attempted was to compute them all, and then select a subset of the solutions based on some property we thought was relevant. (To be fair, what actually happened is we got into fights about what was relevant, and eventually we just modeled the phenomenon with a different PDE that did have a unique solution. I can honestly say that non-uniqueness was one of the biggest headaches I've had solving PDEs.)

I've never attempted to solve an equation which known to not have a solution, but I cannot imagine it turning out well. The best you can hope for is an error message, but you probably won't get it. My suspicion is that your numerical method will compute the solution to an equation which is in some sense "close" to the original, and then you'll be in a truly devastating situation where you have a bunch of garbage which you think is a solution.

Hopefully someone else can offer insight into Navier-Stokes; I've never worked on it so I can't help you there.

I am engineer and I made quite a lot of numerical simulations on PDEs. My experience does not result form mathematical insight, but by practical experince.
Engineers are interested in dayly work problems. We can assume that some kind of solutions exists: No such thin like a Dirac impuls somewhere. You need experience, some kind of feeling, which problems tend to fail and which are not problematic. The usual necessary mesh play an important role.
If I am unsure whether the solution is correct I tend to change something slightly: The geometry, the mesh, the boundary conditions. It may be happen that I got a very different solution. That means I am on a wrong way.
There are other situations where I know that I solutions seems to exist, lets say turbulent hydrodynamics. But I know that cannot come close to that solution with my algorithms. Then it is a useful idea to calibrate the solution against a real measuring. Then you may change something (say optimzing a geometry) an need only to take into account the changes against the calibrated situation.


Prove differential equation solution is even function Let $0\leq c<1$,$u,v,w \in C^1(\mathbb{R})$. Suppose $u'=vw ,\; v'=-uw, \; w'=-c^2 uv$ and $u(0)=0, v(0)=1, w(0)=1$
Then, for all $x\in \mathbb{R} $,$v(x)=v(-x)$ and there is $x_0$ with $v(x_0)=0$

I know that $u^2+v^2=1,\; c^2 u^2+w^2=1$ and I think I use this property in order to prove $v$ is even and has zero point.

If $x\mapsto \bigl(u_0(x),v_0(x),w_0(x)\bigr)$ is the solution of the original IVP then the triple $x\mapsto \bigl(u_1(x),v_1(x),w_1(x)\bigr)$ defined by
$$u_1(x):=-u_0(-x),\quad v_1(x):=v_0(-x),\quad w_1(x):=w_0(-x)$$
is also a solution of the same IVP (check this!). The general existence and uniqueness theorem for systems of ODEs then implies that in fact
$$\bigl(u_0(x),v_0(x),w_0(x)\bigr)\equiv\bigl(u_1(x),v_1(x),w_1(x)\bigr)\ ;$$
in particular $x\mapsto v_0(x)$ is an even function.

First equality:
Multiply the first equation by $u$ and the second by $v$. Adding them together yields $uu' + v v' = 0$, which is equivalent to $\frac{1}{2} \frac{d}{dx}(u^2 + v^2) = 0$. Hence $u^2 + v^2 = C$. Plugging in the initial condition implies that $u^2 + v^2 = 2$.

Second equality:
Hint: Multiply first equation by $c^2 u$ and third equation by $w$.


What is the most frequent number of edges of Voronoi cells of a large set of random points? Consider a large set of points with coordinates that are uniformly distributed within a unit-length segment. Consider a Voronoi diagram built on these points. If we consider only non-infinite cells, what would be (if any) the typical (most frequent) number of edges (that is, neighbors) for those cells? Is there a limit for this number when number of points goes to infinity? Does it have anything common with kissing number? If so, doest it generalize to higher dimensions, that is, 6 for 2D, 13 for 3D etc?

I don't have access to the book that @lhf referenced, but here is a nice topological argument for the expected number of edges of a Voronoi cell in $2$D. Unfortunately, it does not generalize to more than two dimensions.

Consider the dual of the Voronoi diagram, which is the Delaunay triangulation of the given points. This is a planar connected graph, so its Euler characteristic is $\chi = 2$. The Euler characteristic is also given by $\chi = V - E + F$, where $V$, $E$, and $F$ are the number of vertices, edges, and faces in the Delaunay triangulation respectively. Now every face has $3$ edges, while all non-boundary edges are adjacent to $2$ faces. Under reasonable conditions*, the proportion of boundary edges tends to zero, so let us ignore them. This means that $3F \approx 2E$, and plugging this into $V - E + F = 2$ gives $V \approx \frac13E + 2$, where by "$\approx$" I mean the ratio tends to $1$ as $V \to \infty$. So there are about three times as many edges as vertices, and since each edge is incident on $2$ vertices, the average degree of the vertices approaches $6$. As the degree of a vertex in the Delaunay triangulation is precisely the number of edges of the corresponding Voronoi cell, this agrees with your intuition for the $2$D case. (Although, strictly speaking, the expected number of edges is not the same as the most frequent number of edges).

In three dimensions, the Euler characteristic is still $2$, but is now given by $V - E + F - C$, where $C$ is the number of tetrahedral cells in the triangulation. A similar argument as above gives $4C \approx 2F$, but we have no control over $E$. Indeed, there are triangulations on the same point set with the same boundary (the convex hull) but which have different numbers of edges. In the $2$D case, every such triangulation had exactly the same number of edges! So in $3$D one will have to think about the geometry of the Voronoi diagram and/or Delaunay triangulation, and cannot get a result purely via its topology.

* I believe it is sufficient for the points to be drawn from a uniform distribution on a strictly convex area, but I don't know the details.

I think the book Spatial tessellations : concepts and applications of Voronoi diagrams by
OKABE, BOOTS,and SUGIHARA discusses this.


The behaviour of the composition of the polynom and a function I need to prove

  Suppose $f(x)$ is defined and continuous on $[0, 1]$ and it attains its the largest value at three points. Then $g(x) = f^3(x) - f^2(x) + 5 f(x) - 14$ attains its the largest value also at three points

$(f^3(x) - f^2(x) + 5f(x) - 14)' = f'(x) (3 f^2(x) - 2f(x) + 5) = 0 \iff f'(x) = 0$.

Derivatives of functions attain zero at the same points, but can it be necessarily condition to say that $g(x)$ attains the greatest value at three points such as $f(x)$ ? I mean, that the function can haven't got a derivative at the extrema points, can it? $|x|$ for example. So, as I understand I can't use the equality of derivatives in order to prove the statement.

Also, I've noticed that $g(x) = (f(x) - 2)(f^2(x) + f(x) + 7)$ but have no idea, how that fact can help me...

Could you please give me any hints, how to prove the statement?

Thanks in advance!

Note that, if $p(x)=x^3-x^2+5x-14$, then $g(x)=p\bigl(f(x)\bigr)$. On the other hand, $p$ is strictly increasing. Therefore, $p\circ f$ attains its largest value exactly at the points at which $f$ attains its largest value.

Assuming $f $ differentiable, The equation $$g'(x)=f'(x)(3f^2 (x)-2f (x)+5)=0$$

has only three roots ( those of $f'(x)=0$), since the discriminant of the second equation $3X^2-2X+5=0$ is

$2^2-60 <0$.


Is a Borel finite measure on $\mathbb{R}^2$ bounded by "width" absolutely continuous? Consider a finite Borel measure $\mu$ on $\mathbb{R}^2$ such that for each stripe* $S$ of width $\leq \varepsilon$ (with $\varepsilon\geq 0$) one has $\mu(S)\leq \varepsilon$. Show that $\mu$ is absolutely continuous with respect to the Lebesgue measure, or find a counter example.

So far I've only managed to show that $\mu(A) = 0$ for numerable sets or for sets whose projection in one of the two components is zero.

*A set $S$ is a stripe $S$ of width $\leq \varepsilon$ if there exists a continuous function $f:
\mathbb{R} \rightarrow \mathbb{R} $ such that $S \subseteq \{(x,y) \in \mathbb{R}^2 | f(x)\leq y \leq f(x)+\varepsilon\} $ or  $ S \subseteq \{(x,y) \in \mathbb{R}^2 | f(y)\leq x \leq f(y)+\varepsilon\}$.

Show that a square with side $\epsilon$ has measure less than $\epsilon$. Then, the bounded circle with radius $\frac{\epsilon}{\sqrt{2}}$ has also measure less than $\epsilon$.

For each set with Lebesgue measure zero, cover it with a finite union of balls such that the sum of areas is less than some small number- the measure $\mu$ of the cover would be bounded by that small number times $\sqrt{2}$. So by the exterior measure property, the measure $\mu$ of the Lebesgue measure zero set, is also zero.


Prove that: $\vdash \forall x \exists !y(y=x)$ Prove that: $$\vdash \forall x \exists !y(y=x)$$ in first order logic.

The first thing to do would be to write this as $$\forall x (\exists y(y=x) \land \forall y\forall z (y=x \land z=x \to y=z))$$

However, I cannot proceed from here. The equality axioms of first order theories don't seem to give me anything.

This seems like a formal matter, thus I will describe how you do this formally using proof rules.

In order to prove
$$\forall x (\exists y(y=x) \land \forall y\forall z (y=x \land z=x \to y=z))$$
We show that
$$\exists y(y=a) \land \forall y\forall z (y=a \land z=a \to y=z))$$
hold for a general constant $a$, and then if we want to be formal use $\forall$-introduction.

To prove $\exists y(y=a)$ we may just notice that $a=a$ hold by $=-$introduction and then use $\exists$-intrudction to prove $\exists y(y=a)$. If we want to use the axioms for equality instead of the $=-$introduction rule, we may use $\forall x(x=x)$ and use $\forall-$elimination to conclude that $a=a$ hold and then continue as above.

In order to prove $\forall y\forall z (y=a \land z=a \to y=z))$ We assume that $b,c$ are general constants such that $b=a\land c=a$ hold. Now we may use $=$-elimination (or equality axioms) to conclude that $b=c$ hold. If we want to use the axioms for equality instead we may use the axiom $\forall x\forall y (y=x \wedge P(x)\rightarrow P(y))$, which should hold for any predicate $P$ Thus it also hold if P is the predicate $c=x$. In this case we get the formula
$$\forall x\forall y (y=x \wedge x=c\rightarrow y=c)$$
If we do $\forall-$eliminationputting in $a$ and $c$ we get $a=b\wedge b=c\rightarrow a = c$. which we may use to prove what we wanted.

To finnish the proof we use $\forall-introduction to conclude the formla we want to prove.

That was tricky! This isn't standard FOL, but you should be able to translate it if you are familiar with all the "subtleties" of standard FOL (I am not). I introduce the unary predicate $U$ to make the domain of discussion explicit. (This is not the usual practice in standard FOL.) Using this convention, we want to prove:

$$\forall x:[U(x) \implies \exists y:[U(y) \land [x=y \land \forall z:[U(z)\implies [x=z \implies y=z]]]]]$$

Proof:

$U(a)$  (Premise)
$a=a$  (Reflexivity)
$U(c)$  (Premise)
$a=c$  (Premise)
$a=c \implies a=c$  (Conclusion, 4)
$\forall z:[U(z) \implies [a=z \implies a=z]]$ (Conclusion, 3)
$a=a \land \forall z:[U(z) \implies [a=z \implies a=z]]$  (Join, 2,6)
$U(a) \land [a=a \land \forall z:[U(z) \implies [a=z \implies a=z]]]$  (Join, 1, 7)
$\exists y:[U(y) \land [a=y \land \forall z:[U(z) \implies [a=z \implies y=z]]]$ (E Gen, 8)
$\forall x:[U(x) \implies \exists y:[U(y) \land [x=y \land \forall z:[U(z)\implies [x=z \implies y=z]]]]]$ (Conclusion, 1)


Partial order relation aRb is true when b contains a, 001R0101 is true - why? Note: I'm taking a course in Polish, where the word "ciąg" is used, which translates to "sequence", but may be translated as a "word" or a "string".

We have a set of binary sequences: A = { 0, 00, 01, 10, 11, 000, 001, 010, 100, 111, ... }.

We define a partial order relation $R \subseteq A^2$ so that $aRb$ is true if, and only if, sequence $b$ contains subsequence $a$. Examples: $001R0101$ is true, $110R0101$ is false.

...but 0101 doesn't contain 001 "as is", but rather after "skipping" the first 1 from it. I thought that if if a sequence contains a subsequence, it must contain it "in one block, as is" - without manipulating $b$ in any way. Why is this true/allowed?

In English math and computer science, there is a common notion of "subsequence" (not "substring") that does not have to be a contiguous block. Under this notion, if we have a string, say $a_1 a_2 a_3 \ldots a_n$, a subsequence is defined to be $a_{i_1} a_{i_2} a_{i_3} \ldots a_{i_k}$, where $1 \le i_1 < i_2 < \cdots < i_k \le n$.
I would suspect that this is the definition intended in the course you are taking, and its a perfectly reasonable and useful definition. For, under this definition, $001$ is absolutely a subsequence of $0101$.

Note that the word "substring" is usually reserved for contiguous blocks, where "subsequence" may not have to be contiguous. This usage agrees with Wikipedia, see substring and subsequence.

Caveat: Of course, this is only speculation. This question is technically impossible to answer, because it's a matter of the way your course defined things. If your course makes a mistake, then how are we to know? And if not, there is nothing to answer definitively, only speculation. So I would avoid asking this type of question on mathSE, i.e. a question that does not have a real answer.

There must be a typo somewhere,
because it looks to me
that it should be false,
not true.


Calculating a complex integral How can I calculate $$\int_{|z-i|=1}\frac{1}{z+i}\;dz$$

It seems like the non-continuous point is: $z=-i$, and it is not inside the contour, would that mean that the integral is simply zero?

Thanks

Cauchy Integral formula gives us
\begin{align}
f(a) = \oint_\gamma \dfrac{f(z)}{z-a} \,dz
\end{align}

Let $f(z) = 1 \; \forall z \in \mathbb{C}$, so, you are evaluating $f(-i)$, which by our definition is 1.


Is $\frac{x}{e^x-1}$ a monotonically decreasing function? On the first sight, the function $f(x)=\frac{x}{e^x-1}$ is a monotonically decreasing function (https://i.stack.imgur.com/ctArv.png).
But when I zoom the graph to the neighborhood of the point $(0,f(0))$, the function kind of "oscillates":
(https://i.stack.imgur.com/Wpd2T.png)
Either it's not a monotonically decreasing function, or it's an error of the calculator (desmos.com/calculator).

Formally, your question makes no sense, because $f$ is undefined at $0$. But suppose that you extend the domain of $f$ to $\mathbb R$, defining$$f(0)=\lim_{x\to0}\frac x{e^x-1}=1.$$Then, yes, $f$ is decreasing near $0$. That's so because $f(x)=\frac1{g(x)}$, where\begin{align}g(x)&=\begin{cases}\frac{e^x-1}x&\text{ if }x\neq0\\1&\text{ otherwise}\end{cases}\\&=1+\frac x2+\frac{x^2}{3!}+\cdots\end{align}Then $g'(0)=1$ and, since $g$ is increasing near $0$, $f$ is decreasing near $0$.

$$
f(x)=\frac{x}{e^x-1} $$
   Find its first derivative   it is zero at x=0 only and function is not defined at and x= 0 is not the vertical asymptot  ; as x approach towards 0       , y approaches towards 1 ;
f'(x)<0 for all x in its domain hence decreasing function.


Dimension of irreducible component of reduced ring Let $X= SpecA$ denote the spectrum of a reduced ring $A$. Is there any way to tell the dimension of an irreducible component $Y$ of this variety?
Each irreducible component corresponds to a minimal prime ideal, does this mean that the dimension is zero? (or one?)

It is not zero or $1$, consider the quotient   $A=C[x,y,z]/x(x-1)$ where $C$ is the field of complex numbers, $Spec(A)$ has two irreducible components corresponding to the ideal generated respectively by the classes of $x$ and $x-1$ in $A$. The dimension of these components is 2.

Let $k$ be a field and consider the rings $A_0=k, A_j=k[x_1,...,x_j]$ and  $A=\Pi^n_{i=0} A_i$.
The scheme $X=\operatorname {Spec}(A)$ has dimension $n$ but its irreducible components $X_i=\operatorname {Spec}(A_i)=\mathbb A^i_k$ have dimension  $\dim X_i=i$.
So, no,  the dimension of an affine scheme $X$ won't tell you the dimension of its irreducible components, which can be any number up to $\dim X$.


what is $\lim\limits_{x\to 0} \lceil x\rceil/x$? What is $\displaystyle \lim_{x \to 0} \frac{\lceil x \rceil}{x}$ ?  Here, $\lceil x \rceil$ is the ceiling function at $x$.

For left limit and right limit as $x\to 0$.

Assuming that [x] is the floor of x, then look at this graph.

Assuming that [x] is the ceiling of x, then look at this similar looking graph.

Assuming that [x] is the "nearest integer" function, then consider what the nearest integer is on the interval [-0.49, 0.49].

If this is something else, please specify.

$\lim \limits_{x \to 0} \frac{\lceil x \rceil}{x}$

This is what the Alpha Wolf says:

  $\lim \limits_{x \to 0^-} \frac{\lceil x \rceil}{x} = 0$

  $\lim \limits_{x \to 0^+} \frac{\lceil x \rceil}{x} = \infty$

The two limits aren't equal. And what could this possible mean?

How did we get this? Look here:

  $\lim \limits_{x \to 0^-} \lceil x \rceil = 0$

  $\lim \limits_{x \to 0^+} \lceil x \rceil = 1$

And:

  $\lim \limits_{x \to 0^-} x = 0$

  $\lim \limits_{x \to 0^+} x = 0$


Finding $\sum \frac{1}{n^2+7n+9}$ How do we prove that $$\sum_{n=0}^{\infty} \dfrac{1}{n^2+7n+9}=1+\dfrac{\pi}{\sqrt {13}}\tan\left(\dfrac{\sqrt{13}\pi}{2}\right)$$

I tried partial fraction decomposition, but it didn't work out after that. Please help me out. Hints and answers appreciated. Thank you.

Start with the infinite product expansion of $\cos x$

$$\cos x = \prod_{k=0}^\infty \left(1 - \frac{x^2}{(k+\frac12)^2\pi^2}\right)$$

Taking logarithm of $\cos(\pi x)$ and differentiate, we have

$$-\pi\tan(\pi x) = \sum_{k=0}^\infty \frac{2x}{x^2-(k+\frac12)^2}
\quad\implies\quad
\sum_{k=0}^\infty\frac{1}{(k+\frac12)^2-x^2} = \frac{\pi}{2x}\tan(\pi x)
$$
This lead to
$$
\sum_{k=0}^\infty \frac{1}{k^2+7k+9}
= \sum_{k=0}^\infty \frac{1}{(k+\frac72)^2 - \frac{13}{4}}
= \sum_{k=3}^\infty \frac{1}{(k+\frac12)^2 - \frac{13}{4}}\\
= \frac{\pi}{2\cdot\frac{\sqrt{13}}{2}}\tan\left(\frac{\pi\sqrt{13}}{2}\right) - \sum_{k=0}^2 \frac{1}{(k+\frac12)^2 - \frac{13}{4}}
= 1 + \frac{\pi}{\sqrt{13}}\tan\left(\frac{\pi\sqrt{13}}{2}\right)
$$

Can't you use the first comparison test? Your series is always less than $\sum_{n=0}^\infty \frac{1}{n^2}$ which converges.


Overview of basic results about images and preimages Are there some good overviews of basic facts about images and inverse images of sets under functions?

I have no doubt that you there are many useful online resources for these, but many such results are available here at MSE, together with their proofs.
I'll give a list of some basic results about images and preimages links to the posts, which have proofs here at MSE. I am making this CW, so feel free to add more identities and pointers to further useful questions and answers.
$\newcommand{\Map}[3]{{#1}\colon{#2}\to{#3}}\newcommand{\Img}[2]{{#1}[#2]}\newcommand{\Pre}[2]{{#1}^{-1}[#2]}$
If $\Map fXY$ is a function and $A\subseteq X$ and $B\subseteq Y$ are some set then the set
$$\Img fA=\{f(x); x\in A\}$$
is called the image of the subset $A$ and the set
$$\Pre fB=\{x; f(x)\in B\}$$
is called the preimage or inverse image of the subset $B$.
In the other words, we have
$$y\in\Img fA \Leftrightarrow (\exists x\in A)f(x)=y$$
and
$x\in\Pre fB \Leftrightarrow f(x)\in B$.
In the notation below we always assume $A,A_i\subseteq X$ and $B,B_i\subseteq Y$.

$A\subseteq \Pre f{\Img fA}$ and, if  $f$ is injective, then $A=\Pre f{\Img fA}$.

see Proving that $C$ is a subset of $f^{-1}[f(C)]$
In fact, the equality is equivalent to the fact that $f$ is injective: Show $S = f^{-1}(f(S))$ for all subsets $S$ iff $f$ is injective or Is $f^{-1}(f(A))=A$ always true? Some counterexamples to the equality can be found in answers to Why $f^{-1}(f(A)) \not= A$

$\Img f{\Pre fB}\subseteq B$ and, if $f$ is surjective, then $\Img f{\Pre fB}=B$.

For the first part see: Need help for proving that: $f(f^{-1}(A)) ⊆ A$ For the second part, see Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$ or Prove $F(F^{-1}(B)) = B$ for onto function.

The operation of taking images and preimages preserves inclusion, i.e. $A_1\subseteq A_2$ implies $\Img f{A_1}\subseteq \Img f{A_2}$ and $B_1\subseteq B_2$ implies $\Pre f{B_1}\subseteq\Pre f{B_2}$

Image of union of two sets is the union of their images, i.e. $\Img f{A_1\cup A_2}=\Img f{A_1}\cup\Img f{A_2}$.

See, for example, Prove $f(S \cup T) = f(S) \cup f(T)$ or
Functions-Set Theory Proof that $f(C \cup D) = f(C) \cup f(D)$ or How do I prove the following: $f(S\cup T) = f(S) \cup f(T)$ or Maps - question about $f(A \cup B)=f(A) \cup f(B)$ and $ f(A \cap B)=f(A) \cap f(B)$ or Unions and Functions on Sets or The preimage of a subset

The same is true for arbitrary union: $$\Img f{\bigcup_{i\in I} A_i} = \bigcup_{i\in I} \Img f{A_i}.$$

See Show that $\bigcup_i f(A_i) = f(\bigcup_i A_i)$

In the case of intersection, we only have one inclusion: $$\Img f{A_1\cap A_2}\subseteq \Img f{A_1} \cap \Img f{A_2}.$$ But if $f$ is injective, then $\Img f{A_1\cap A_2} = \Img f{A_1} \cap \Img f{A_2}$.

See, for example Is this a valid proof of $f(S \cap T) \subseteq f(S) \cap f(T)$?
or Is this proof correct for : Does $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$?
or Prove $f(S \cap T) \subseteq f(S) \cap f(T)$
or Do we have always $f(A \cap B) = f(A) \cap f(B)$?
or Maps - question about $f(A \cup B)=f(A) \cup f(B)$ and $ f(A \cap B)=f(A) \cap f(B)$
or Verifying a proposition on image and preimage: $f(A\cap B)\subseteq f(A)\cap f(B)$ and $f^{-1}(C\cap D)=f^{-1}(C)\cap f^{-1}(D)$ or How to prove this? "For all sets A,B⊆D and functions f:D→R, we have f(A∩B)⊆(f(A)∩f(B))." or $f(A \cap B)\subset f(A)\cap f(B)$, and otherwise?
The part about injective functions can be found in
Conditions Equivalent to Injectivity or in
Proving: $f$ is injective $\Leftrightarrow f(X \cap Y) = f(X) \cap f(Y)$
A counterexample showing that equality is not true in general can be found here: Do we have always $f(A \cap B) = f(A) \cap f(B)$?
In fact, if this equality is true for all subsets $A_1,A_2\subseteq X$, then $f$ must be injective, see To prove mapping f is injective and the other f is bijective

Again, the same claims hold for arbitrary intersection: $\Img f{\bigcap_{i\in I} A_i} \subseteq \bigcap_{i\in I} \Img f{A_i}$ and if $f$ is injective, then $\Img f{\bigcap_{i\in I} A_i} = \bigcap_{i\in I} \Img f{A_i}$.

See Proof for Image of Indexed Collection of Sets?

Preimages preserve intersection: $\Pre f{B_1\cap B_2}=\Pre f{B_1} \cap \Pre f{B_2}$.

See, for example: What are the strategies I can use to prove $f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$? or how to prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$ or Verifying a proposition on image and preimage: $f(A\cap B)\subseteq f(A)\cap f(B)$ and $f^{-1}(C\cap D)=f^{-1}(C)\cap f^{-1}(D)$

The same is true for arbitrary intersections: $\Pre f{\bigcap_{i\in I}B_i} = \bigcap_{i\in I} \Pre f{B_i}$.

Preimages preserve union: $\Pre f{B_1\cup B_2}=\Pre f{B_1} \cup \Pre f{B_2}$.

See: Show that $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$

Again, this is also true for union of more than just two sets: $$\Pre f{\bigcup_{i\in I}B_i} = \bigcup_{i\in I} \Pre f{B_i}.$$

See: Union of preimages and preimage of union
We can also ask whether image and preimage preserve difference:

$\Img fA\setminus\Img fB \subseteq \Img f{A\setminus B}$, but the equality does not hold in general

See Proving $f(C) \setminus f(D) \subseteq f(C \setminus D)$ and disproving equality. Equality holds for injective functions, see If $f$ is 1-1, prove that $f(A\setminus B) = f(A)\setminus f(B)$. In fact, validity of the equality $\Img fA\setminus\Img fB = \Img f{A\setminus B}$
characterizes injectivity: Does $f(X \setminus A)\subseteq Y\setminus f(A), \forall A\subseteq X$ imply $f$ is injective ?.

$\Pre fA\setminus\Pre fB=\Pre f{A\setminus B}$

See Proof of $f^{-1}(B_{1}\setminus B_{2}) = f^{-1}(B_{1})\setminus f^{-1}(B_{2})$. As a consequence of this we get that the preimage of complement is complement of the preimage, see here: How to approach proving $f^{-1}(B\setminus C)=A\setminus f^{-1}(C)$?, Show that for any subset $C\subseteq Y$, one has $f^{-1}(Y\setminus C) = X \setminus f^{-1}(C)$ and Show $f^{-1}(A^c)=(f^{-1}(A))^c$
The image and inverse image for composition of maps can be expressed in a very simple way. For $\Map fXY$, $\Map gYZ$ and $A\subseteq X$, $C\subseteq Z$ we have

$\Img g{\Img fA}=\Img{g\circ f}A$

$\Pre f{\Pre gC}=\Pre{(g\circ f)}C$

See Prove that if $Z\subseteq Y$, then $(g\circ f)^{-1}(Z)=f^{-1}(g^{-1}(Z)).$

The "online compendium of mathematical proofs" known as ProofWiki has most of these results -- with one or more proofs. (For the record, I am somewhat affiliated with that site.)

Most results regarding images and preimages should be in the Mapping Theory category (look under the "M").


Center of fundamental group Let $f_t: X \rightarrow X$ be a homotopy of maps such that $f_0 = f_1 = \mathrm{id}_X$. For any $x_0 \in X$, the map $t \mapsto f_t(x_0)$ is a loop based at $x_0$.

To prove: $[f_t(x_0)]$ is contained in the center of $\pi(X, x_0)$.

Attempt:

If $y$ is an element of $\pi(X, x_0)$, then $y$ commutes with $[f_t(x_0)]$ because $[f_t(x_0)]$ is homotopic to the identity. That is, we can replace $[f_t(x_0)]$ with the identity loop, since they are in the same class, and the identity loop is in the center of $\pi(X, x_0)$.

Thoughts?

Let $p:[0,1] \to X$ be a loop based at $x_0$.

The map $h(s,t) = f_t(p(s))$ is a continuous mapping $[0,1] \times [0,1] \to X$.

On the boundary of the square, the following is true:

$h(0,t) = f_t(x_0) = h(1,t) \quad$ and $\quad h(s,0) = p(s) = h(s,1)$.

In other words, $h$ restricts to $p(s)$ along the bottom and top sides and restricts to $f_t$ along the left and right sides.  The concatenation of two paths, $f_t*p(s)$, is the left side followed by the top side.  The concatenation of two paths, $p(s)*f_t$, is the bottom side followed by the right side.  That $h$ extends this pair of concatenated paths over the square shows that they are homotopic relative to their endpoints.

If you sketch this square and then draw a diagonal from the upper left to the lower left, then you can visualize the homotopy (rel endpoints) by restricting $h$ to the family of line segments which join the lower left vertex to a point on the diagonal and then to the upper right vertex.

Let $C(X)$ be the set of curves in $X$ and $Hom(X)$ the homeomorphisms $X \to X$.

For $F : id_X\to id_X \in C(Hom(X))$ and some $x_0 \in X$ let $f_t = F_t(x_0)$ so that $f \in C(X)$.

For any curve $g:x_0 \to x_0 \in C(X)$ let $h_t = ( f_\tau,\tau \in [0,t]) \cup F_t(g) \cup ( f_{t-\tau},\tau \in [0,t])$ so that $h \in C(C(X))$.

$h_0 = g$ and $h_1 = f  \cup g \cup f_{-}$ and $h_t$ is an homotopy between the two.

Thus we obtained that $f$ commutes with $g$ in $\pi_1(X)$.


Prove that $XY$ crosses the midpoints $\triangle ABC$ has altitudes $AD$, $BE$, $CF$. The reflections of $E$, $F$ in $H$ are $E'$, $F'$. The circle $DE'F'$ intersects $BE$, $CF$ at $X$, $Y$. Prove that $XY$ goes through the midpoints of $AB$, $AC$.

I can show that $XY$ is parallel to $BC$ by simply angle-chasing. $EYFX$ is cyclic as well as $APFY$.

I also tried showing that $AP$=$PH+HD$

Take point $C$ as the origin of a Cartesian coordinate system. Then let $|PD|=a$. It suffices to note that since the line $XPY$ is parallel to the base of the triangle, its equation is $y=a$. Now since the midpoints of $AC$ and $AB$ have coordinates $(x_1,a)$ and $(x_2,a)$ respectively, it's clear that they lie on the line of interest. $\square$


Determine the minimal polynomial of $\sqrt 3+\sqrt 5$ I am struggling in finding the minimal polynomial of $\sqrt{3}+\sqrt{5}\in \mathbb C$ over $\mathbb Q$

Any ideas? I tried to consider its square but it did not helped..

As was mentioned in the other answers, $x^4-16x^2+4$ is a polynomial with root $\sqrt{3}+\sqrt{5}$, so it suffices to prove that this polynomial is irreducible. There are several methods:

1) Calculation of the quadratic factors, see Lubin's answer.

2) Field theory: $\mathbb{Q}(\sqrt{3},\sqrt{5}) = \mathbb{Q}(\sqrt{3})(\sqrt{5})$ has degree $2 \cdot 2=4$ over $\mathbb{Q}$ essentially because $\sqrt{5} \notin \mathbb{Q}(\sqrt{3})$ (easy computation!) and hence $X^2-5$ stays irreducible over $\mathbb{Q}(\sqrt{3})$. So it suffices to prove that $\mathbb{Q}(\sqrt{3},\sqrt{5})=\mathbb{Q}(\sqrt{3}+\sqrt{5})$ (because then $\sqrt{3}+\sqrt{5}$ has degree $4$ over $\mathbb{Q}$). Since $\supseteq$ is trivial, it suffices to check $\subseteq$. Notice that rationalisiation yields
$$\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}-\sqrt{5}}{3-5}.$$
Hence, $\sqrt{3}-\sqrt{5} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. But then we also have $2 \sqrt{3} = (\sqrt{3}-\sqrt{5}) + (\sqrt{3}+\sqrt{5}) \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. This shows $\sqrt{3} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$ and then we also get $\sqrt{5} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$. $\checkmark$

The same proof shows: If $p,q$ are distinct prime numbers, then $\mathbb{Q}(\sqrt{p},\sqrt{q})=\mathbb{Q}(\sqrt{p}+\sqrt{q})$ has degree $4$ over $\mathbb{Q}$. The minimal polynomial of $\sqrt{p}+\sqrt{q}$ is $(x^2-p-q)^2 - 4 pq$.

3) Kummer theory: It is a special case of Kummer theory that for distinct prime numbers $p_1,\dotsc,p_n$ (actually pairwise coprime square-free integers also work) $\mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_n})$ has degree $2^n$ over $\mathbb{Q}$, has primitive element $\alpha=\sqrt{p_1} + \dotsc + \sqrt{p_n}$ and that its minimal polynomial therefore has degree $2^n$. It is given by the product of all $X-\beta$, where $\beta$ runs through the conjugates $\pm \sqrt{p_1} \pm \dotsc \pm \sqrt{p_n}$. (It is quite obvious that this polynomial is rational and has $\alpha$ as a root, but its irreducibility is not so easy to prove directly, and this is where Kummer theory helps.)

$\mathbb Q(\sqrt 3 + \sqrt 5)/\mathbb Q$ is a finite separable normal extension whose Galois group consists of the four $\mathbb Q$-automorphisms
$$
\sqrt 3 + \sqrt 5 \mapsto \pm\sqrt 3 \pm \sqrt 5.
$$
Since there are precisely $4$ such $\mathbb Q$-automorphisms, by the Fundamental Theorem of Galois Theory, $[\mathbb Q(\sqrt 3 + \sqrt 5):\mathbb Q] = 4$, so the minimal polynomial has degree $4$.

The elements $\pm\sqrt 3 \pm \sqrt 5$ are all distinct and $\sqrt 3 + \sqrt 5$ is one root of the minimal polynomial, so the minimal polynomial is simply the product of the linear factors of Galois conjugates
$$
[x-(\sqrt 3 + \sqrt 5)][x - (\sqrt 3 - \sqrt 5)][x - (-\sqrt 3 + \sqrt 5)][x - (-\sqrt 3 - \sqrt 5)].
$$
This is computed to be $x^4 - 16x^2 + 4$.


Why is calculating the area under a curve required or rather what usage it would provide I understand Integration and Differentiation and see a lot of Physics / Electrical Theory using them.

Take for example a sine wave. So area for me means the space any object would occupy. So what's usage it comes to find the area of a sine curve?

There are so a many formulas that calculates the area by Integration - but why calculation is required - I mean what information we can get (isn't it just space occupied) or rather what data we can find by calculating area of a curve via Integration?

The point is that in science and in real life, you come up against situations where you need to totalize or to aggregate or to accumulate some kind of growing quantity. Say you know how fast the snow is falling at any particular time, and want to know the accumulation after five hours. Say you have pollutants flowing into a lake at varying rates during the day or week, and want to know how much crud is in the lake after a few weeks. Both these examples are quantities that are accumulated from a varying contribution, and both are measured by an integral. And there’s no area in sight.

A function, differentiate it represent how fast it changes, integrate it represent how much change had accumulated.

If we know the derivative of the function (how fast it changes), by integrate it, we find function itself. If we know how much change had accumulated, by differentiate it, we can find of the function of this thing.


Example when the torsion of a module is not a submodule. Can any one suggest me an example of a ring $R$ and an $R$-module $A$ s.t. torsion of module $A$ is not a sub-module?

Torsion of module $A$, i.e. $\operatorname{Tor}(A)$, denotes all torsion elements in module $A$.

*(And we already know that if $R$ is a commutative integral domain then $\operatorname{Tor}(A)$ is a submodule of $A$).

An example for a commutative ring which is not a domain: $R=\{0,a,1-a,1\}$ with $a^2=a,a+a=1+1=0$ and $A=R$. The elements $a,1-a$ are torsion, but $a+(1-a)=1$ isn't.

actually, the Torsion subset is a submodule for all $R$-modules $M$.
Recall that $m\in M$ is called torsion, if there is $r\in R$ which is regular (i.e. not a zero divisor) such that $r.m=0$.

Assume $m$ and $m'$ are torsion with corresponding regular elements $r$ and $r'$. Then $rr'$ is non-zero and still not a zero divisor and we have $rr'.(m+m')=0$. Hence $m+m'$ is torsion.
It is trivial to see that if $m$ is torsion then the same is true for $s.m$ for all $s\in R$.

Note also that over a finite ring every module is torsion.
To see this, notice that an element of a finite ring is either a unit or a zero-divisor. Hence all regular elements act invertibly on modules and hence cannot kill any non-zero elements.


Show that if $x$ is a limit point of $A\subset X$ and $f:X \rightarrow Y$ is continuous then $f(x)$ is a limit point of $f(A)$ Suppose $f:X\rightarrow Y$ is a continuous function between the topological spaces $X$ and $Y$. Suppose that: $A \subset X$ and that  $x$ is a limit point of $A$.

Show that $f(x)$ is a limit point of $f(A)$

By a limit point $c$ of $B$ I mean that every neighborhood of $c$ intersects $B$ in a point $d$ such that $c\not = d$

What I have done is that: I supposed that $x$ is a limit point of $A$ but $x\not\in A$ and showed the required. but in the case when $x \in A$. My trial goes as follows:

Suppose $W$ was a neighborhood of $f(x)$ so $f^{-1}(W) \text{is a neighborhood of } x$. So $ f^{-1}(W) \cap A \not=\emptyset$ because $x$ is a limit point of $A$. Say $y\in f^{-1}(W) \cap A$. So $f(y)\in A$ and $f(y)\in W$. Now, if  $x\not\in A$ then $f(x) \not\in f(A)$ but $f(y)\in A$ so $f(y)\not = f(x)$. So $W$ intersects $f(A)$ in a point $f(y)$ which is different of $f(x)$

If $x\in A$, it may happen that $f$ is not injective and happen that  $f(x)=f(y)$ in which case I can't show that f(x) is a limit point of $f(A)$.

I've tried to came with a counter-example using constant functions on real functions. but it didn't works

What to do? any hints?

As the question was answered in comments and there is no one who added a correct answer to close the question. I'm adding this answer (I got the idea of the  answer from This comment)

The statement is false, here is a concrete example:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ where, $f(x)=2$ for all $x\in \mathbb{R}$. So $f$ is a constant function.

Note that we denote the set of limit points of $H$ by $H'$.

Let $A = [0,1]$.
So, $A' = [0,1]$.

$f(A)=f([0,1])=\{2\}$, but $f(A)'=\emptyset$ because {$2$} has no limit points as if $x\not = 2$  is a real number then there is always an interval $C$ containing $x$ but not contaning $2$.

So Although $1$ is a limit point of $A$, $f(1)=2$ is not a limit point of $f(A)=\{2\}$.

As the question was answered in comments and there is no one who added a correct answer to close the question. I'm adding this answer (I got the idea of the  answer from comments).

$f(x)$ is a limit point of $f(A)$ if and only if every open neighborhood $V\subseteq Y$ of $x$ intersects $f(A)$.

We show the contrapositive. Assume that $f(x)$ is not a limit point of $f(A)$. Then there is an open neighborhood $V_0$ of $f(x)$ such that $V_0\cap f(A) = \emptyset$. Since $f:X\to Y$ is continuous, the preimage $f^{-1}(V_0)$ is an open neighborhood of $x$.

Claim:  $A \cap f^{-1}(V_0) = \emptyset$. Proof of claim: Suppose $z \in A \cap f^{-1}(V_0)$, then $f(z) \in f(A) \cap V_0$ contrary to assumption.

Now the claim implies that $x$ is not a limit point of $A$ and we're done.


Why is my logical statement wrong? "Express the following sentence symbollically, using only quantifiers for real numbers, logical connectives, the order relation < and the symbol Q having the meaning 'x is rational'"

I have to translate the sentence "There is a rational number between any two unequal real numbers". I worked a bit on it and eventually deduced the following:

$$(\forall x,y\in \mathbb{R})[x> y](\exists q\in \mathbb{Q})[q>y \wedge q< x]$$

In light of some comments a correct version of my incorrect statement should be:
$$(\forall x,y\in \mathbb{R})[x≠ y \Rightarrow (\exists q\in \mathbb{Q})[q>y \wedge x> q]\vee[y>q \; \wedge \;q>x]]$$

Can you help me understand why my answer is wrong?

In your first formula, following @Danul G's advise, I assume $\mathbb R$ as the domain of the variables, and I'll rewrite it with the predicate $Q(x)$ (for $x$ is rational) to stay with the original formulation of the problem :

  $\forall x \forall y ( x > y ) \exists q (Q(q) \land [q>y \land q<x])$

there are basically two mistakes :

(i) one substantial: as pointed in the above comments, in this way you are assuming that $x > y$, but your problem says $x \ne y$; so you must correct the formula accordingly.

(ii) the other one is connected to the way you have written the formula; juxtaposition of the two sub-formulas is not formally correct: at most you can read it as a conjunction.

But in this way you are saying that :

  $\forall x \forall y ( x > y )$ and $\exists q (Q(q) \land [q>y \land q<x])$

i.e.that for two real numbers whatever, the first is greater than the second (false ! : take $0$ as $x$ and $1$ as $y$) and ... and the resulting statement is false, being the conjunction of two formulas, one of which is false.

$$\forall x,y\in\mathbb{R} x\not=y \implies \exists q\in\mathbb{Q} (y-q)(q-x) > 0 $$


Why is a reflection followed by another reflection is a rotation? I just started abstract algebra and we are working with dihedral groups. I've made Cayley tables for D3 and D4 but I can't explain why two reflections are the same as a rotation

Consider the dihedral group $D_5$, and consider its action on the pentagon. In particular, every element of the group can be thought of as some combination of rotations and reflections of a pentagon whose corners are labeled $1,2,3,4,5$ going clockwise.

First, notice that no matter what we do, the numbers will be in the order $1,2,3,4,5$ in either the clockwise (cw) or counterclockwise (ccw) direction.

If our change switches the order from ccw to cw (or vice versa), then we must have reflected the image.  On the other hand, if no such change occurs, then we must have rotated the image.

Note that reflecting twice results in switching from ccw to cw, then to ccw.  So, the numbers still go $1,2,3,4,5$ in the ccw direction.  So, we must have rotated the image.

A non-identity rotation leaves only one point fixed-the center of rotation. A reflection leaves only the axis of rotation fixed, while a reflection followed by a different reflection leaves only one point fixed-the intersection of the two axes of reflection , so it must be a rotation since only a rotation leaves a point fixed.


Let $f$ be continuous. If for every open $A$, $f(A)$ is also open, then $f$ is injective Question:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. For every $A\subset\mathbb{R}$ open, prove that if $f(A)$ is open, then $f$ is injective.

My attempt:

We prove the contrapositve. We exibit an open set $A$ such that $f(A)$ is not open. We have that $f$ is not injective. Hence, there exists $x,y\in\mathbb{R},x\neq y$, such that $f(x)=f(y)$. Take $A=(x,y)$. We notice that A is an open set. Since $f$ is continuous, it follows that $f(A)$ is an interval. We argue that $f(A)$ cannot be an open interval since, this would force $f(A)=(f(x),f(y))$ or $f(A)=(f(y),(f(x))$, both of them being the empty set (which is not an interval).

Hence, since an interval is an open set iff it is an open interval, we conclude that $f(A)$ cannot be open.

Why i posted this question: since $x$ and $y$ are not in $A$, the only way to the image of $f(A)$ to be open, that is, for the extremes of this interval don't be part of it, is: $f(A)$ must have $f(x)$ and $f(y)$ as its extremes. Is this reasoning correct? Any counter example?

I think the idea could work, you just need to change stuff a bit. I will assume $x<y$ and your setting.

Your wrong assumption:

I agree $f(A)$ is an interval (image of connected sets trough continuous functions is connected), but it has not necessarily $f(x)$ and $f(y)$ as extreme points, but the maximum and minimum of the image of the closed interval $[x,y]$ (that exist because $[x,y]$ is compact).

A way to end the proof (spoiler):

If they max and min are met at the borders $x,y$ is all done: $(x,y)$ is mapped to a point $max=min=f(x)=f(y)$ because $[x,y]$ is mapped there that is not open.

If not, I notice at least one between $min, max$ are image of interior points, they have to be met inside: that leads us to $min, max \in f(A)$, so $f((x,y))=(min,max]$ or $[min,max)$ or $[min,max]$, that is not open.

We contradicted that $f(A)$ has to be open.

Suppose $f$ is not injective, then there is some $x_1<x_2$ such that
$f(x_1) = f(x_2)$.

Let $M = \max_{x \in [x_1,x_2]} f(x) \ge f(x_1)$, $m = \min_{x \in [x_1,x_2]} f(x) \le f(x_1)$.

If $m=M$, then $f((x_1,x_2)) = \{f(x_1)\}$.

If $M > f(x_1)$, then either $f((x_1,x_2)) = [m,M]$, or $f((x_1,x_2)) = (m,M]$.

If $m< f(x_1)$, then the same sort of reasoning applies.


Net gain on dice rolling Suppose you have to pay $15$ to roll a dice repeatedly until the first $6$ comes up. Then you will get paid as many dollars as the square of the no of rolls $X^2$.

I have a equation as follows:
$$G = \text{Gain} = X^2 -15 \\
E(G) = E(X^2 - 15) = E(X^2) -15 $$

Since this is a geometric distribution
I found the mean to be $E(X) = 1/p = 1/(1/6) = 6$ and variance $= (1-1/6)/(1/6)^2 = 30$

Therefore I found $E(X^2)$ to be $66$ using $V(X) = E(X^2) − (EX)^2$

The overall net gain was $66-15 = 51$

I am just wondering whether I am on the right track. I thought the net gain would be very low for this $Q$ as you need to roll at least 4+ times to get a gain.

Also can someone direct me how to find the probability that the users net gain will be positive. I was thinking of calculating the probability of rolling at least 4 times.

Your calculation of $E(X^2)$ is absolutely correct, it indeed equals $66$. You indeed need to be able to roll the die at least four times to make profit, which means that you need to not throw a $6$ three times in a row. The probability of not throwing a $6$ on a given roll equals $\frac{5}{6}$, so the probability $P(X > 3)$ of making a profit equals:

$$P(X > 3) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \approx 0.579$$

You are on the right track. Intuitively a large reward is plausible because of the squared increase in rewards. Very unlikely events have a huge reward which in this case outweigh the small losses.

Regarding your last question: Calculating the probability of rolling at least 4 times is correct, however, rather difficult. It is easier to calculate the probability of a loss and then using $P(\text{gain}) = 1 - P(\text{loss})$ since we know the two probabilities must sum up to 1.

(I get $P(\text{gain}) \approx 0.58$)


Using Pumping Lemma prove that the language $L = \{a^ib^j \mid i,j \in N \}$ is Not Regular. Using Pumping Lemma prove that the language $L = \{a^ib^j \mid i,j \in N \}$ is Not Regular.

Proof:

Assume that $L$ is Regular.
Pumping Length = $P$.
 We choose $w = a^{P-2}b^{P+2} \in L$

We divide $w$ in $xyz$.
$x= a, y = a^{P-3}bb, z = b^P$

Now we Pump with $i = 2$. $xy^iz = xy^2z$ = $aa^{P-3}bba^{P-3}bbb^P$ = $a^{P-2}bba^{P-3}b^{P+2} \notin L$. QED

Did I do it right?

Let's suppose that $w$ is regular.

We can take the string $s = a^ib^i$. Like s $\in w$ and $|s| \ge p$, the pumping lemma states that we can separate $s$ in 3 parts that are: $xyz$. With $n \gt 0, xy^iz \in w$.
Because one of the lemma condition says that $|xy| \le p$, $y$ must consist only of $0s$ otherwise $|xy| \ge p$.

If we take the string $xyyz$ (which is supposed to be regular), we notice that we have more $0s$ than $1s$. So it's a contradiction.

Your language is regular and equal to $a^*b^*$. That being said, your application of the pumping lemma is faulty.

  Assume that $L$ is Regular. Pumping Length = $p$. We choose $w = a^{p-2}b^{p+2} \in L$.

Apart from the telegraphic style, so far, so good.

  We divide $w$ in $xyz$. $x= a, y = a^{p-3}bb, z = b^p$.

Wrong! The pumping lemma just tells you there must be some decomposition $w=xyz$ with $|xy| \leqslant p$ and $|y| \geqslant 1$ such that $xy^iz \in L$ for every $i \geqslant 0$. The decomposition you propose may not work, but there might be another decomposition that works. And indeed, the decomposition $x = a^{p-3}bb$, $y = b^p$ and $z=1$ will work since $a^{p-3}bb(b^p)^n \in L$ for all $n$. So you didn't find a contradiction.


What can you say about a number with remainder 1 and 2 when divided by 3 and 4 respectively? I was trying to solve a problem which states:

  How many two-digit numbers have remainder 1 when divided by 3 and remainder 2 when divided by 4?

and solved it by writing down individual numbers...

Looking at the solution, I was baffled by the explanation :

  The lowest common multiple of 3 and 4 is 12. Hence both of the required
  conditions are satisfied only by numbers that are 2 less than multiples of 12
  and also less than 100, ie: 10, 22, 34, 46, 58, 70, 82 and 94.

Why on earth does the number have to be 2 less than multiples of 12? Could someone offer a better explanation please?

remainder 1 when divided by 3
$x \equiv 1 \pmod 3$
$x \equiv -2 \pmod 3$

remainder 2 when divided by 4
$x \equiv 2 \pmod 4$
$x \equiv -2 \pmod 4$

And so
$x \equiv -2 \pmod 3$
$x \equiv -2 \pmod 4$

Since 3 and 4 are relatively prime, this implies that we can combine the two equations to
$x \equiv -2 \pmod {12}$

Another way of looking at this is

$x \equiv -2 \pmod 3$ means that $x = 3n - 2$ for some $n$, or $x + 2 = 3n$ for some n.

Similarly, $x + 2 = 4m$ for some m.

Hence $3n = x + 2 = 4m$.
Since $\gcd(3,4) = 1, $ then m must be a multiple of 3.

It follow that $x + 2 = 4m = 12r$ for some $ m = 3r.$

finally, $ x = 12r - 2$.

This is a consequence of the Chinese Remainder Theorem: https://en.wikipedia.org/wiki/Chinese_remainder_theorem


Can the derivative be defined game-theoretically? It is often said, by way of intuitive explanation, that the derivative of a function at a point is the slope of the line that “best fits” the function through that point. Can this be pressed into a mathematically-rigorous definition, somewhat along the lines of the prisoners’ dilemma?

That is, two parties, who cannot communicate with each other, are given the same function and the same point, and told that if they both come up with the same line through that point that is best relative to the function, they will be set free, otherwise they will both have to serve time.

Would it then be some kind of theorem in Probability/Logic/Game Theory, that they would, if acting rationally and intelligently, come up with the tangent line “almost surely”?

Regards,

Mike Jones

2.Jun.2011 (Beijing time)

As I stated in my comment I don't think your solution quite works. However, I think I just thought of a nice game-theoretical way of defining the derivative of a function:

Two players are both given a function $f$ and a point $(x,f(x))$. Both players then independently come up with a line through $(x,f(x))$, say $f_1$ and $f_2$.

Now the players announce which line they chose and both come up with some $\epsilon>0$ and the minimum of these is selected.

Then the final stage of our game begins: if $|f(x+\epsilon) - f_1(x+\epsilon)| = |f(x+\epsilon) - f_2(x+\epsilon)|$, the game ties and nothing happens. If $|f(x+\epsilon) - f_1(x+\epsilon)| < |f(x+\epsilon) - f_2(x+\epsilon)|$ player 1 wins and player 2 loses. If $|f(x+\epsilon) - f_1(x+\epsilon)| > |f(x+\epsilon) - f_2(x+\epsilon)|$ player 2 wins and player 1 loses.

To spice the game up a little one should of course introduce prison sentences or death penalties for the loser at this point, but I will leave that as an exercise to the reader.

Any proper game theorist would now wonder about the optimal strategy for both players, and unless I am gravely mistaken that would be choosing the line corresponding to the right-derivative of $f$ at $(x,f(x))$.

You've only talked about the 1st derivative of a function, and the geometric way of looking at "the derivative" as a tangent line only works for the 1st derivative.  It does come as possible to talk about the 2nd, 2.5th, pi-th, the -1th derivative (the integral), and other derivatives via [fractional calculus].1  I don't see how a game-theoretic definition would allow for anything else than the 1st derivative.


What is the practical difference between convolution and cross correlation? Good morning,

I am coming from learning machine learning convolution for neural nets and was wondering about cross-correlation vs convolution.

I referenced this answer here: What's the difference between convolution and crosscorrelation?

But I fail to understand the practical difference that a mirrored 'filter' (not sure if that is the correct term in this context) produces when using convolution rather than cross-correlation. It seems that either method contains different representations of the same data. Whether (as in the link above) it is X+Y or Y-X, they both contain similar, albeit opposite, data.

Is this simply used to adjust the direction of the vector, as seemingly the magnitude would remain unchanged? Or am I missing some subtleties?

Thank you!

From more research I performed, per a Coursera course in which I am enrolled (DeepLearning.Ai), in "signal processing or in certain branches of mathematics, doing the flipping in the definition of convolution causes the convolution operator to enjoy [the associativity property]" - Professor Andrew Ng

The 'flipping' referring to using convolution rather than cross-validation, both terms used in the mathematical context.


Prove that $a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}$ Question -

  Let $a, b, c$ be non-negative real numbers. Prove that
  $$
a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}
$$

my doubt -

author writes

Applying AM-GM inequality, we have

$\sum_{c y c} a^{2} \sqrt{b^{2}-b c+c^{2}}=a \sqrt{a^{2}\left(b^{2}-b c+c^{2}\right)} \leq \frac{1}{2} \sum_{c y c} a\left(a^{2}+b^{2}+c^{2}-b c\right)$

but how we can apply am-gm if there is negative term inside ??? can someone clear this step that how they apply am-gm and got this result..

thankyou

As pointed out by others, why do you even assume $b^2-bc+c^2$ can be negative?

Simply write it as $(b-c)^2+bc$. The first term, being a perfect square, is clearly always non-negative, while the second term, being a product of two non-negative numbers ( as per the question), must also be non-negative. Thus their sum is also non-negative.


Calculating the area of an irregular polygon Given the length of the sides of an irregular polygon (no coordinates provided) how do you compute the area of the maximum area of the polygon?

Thanks in advance

The polygon must be cyclic. For details, check this out.

because we know the sides So the angles formed by these sides at the center of the circle will be in the same ratio as the sides
for example if the sides of the polygon are a,b,c,d and e then the sum of all the angles at the center will be = a*x+b*x+c*x+d*x+e*x and this sum will be equal to 360.Sonow we know all the angles made by the sides at the centre. Now we can find the area of each triangle and sum of these areas is the area of the polygon


Finding the derivative of $x$ tetrated to the $x$ Differentiating the functions $x^x$, $x^{x^x}$ (or ${^2{x}}$ and ${^3{x}}$), etc., although somewhat tedious, is pretty straightforward. I've even seen in a couple of books (and even on a post on this forum) a general expression for the derivative of ${^n{x}}$. However, I would like to find the derivative ${^x{x}}$, or $x$ tetrated to the $x$. I seem to have absolutely no tools to even approach this problem, as I am not at all familiar with the maths associated with tetration.