Is it more accurate to use the term Geometric Growth or Exponential Growth? On Wikipedia, the terms Exponential Growth and Geometric Growth are listed as synonymous, and defined as when the growth rate of the value of a mathematical function is proportional to the function's current value but I question whether one term is more mathematically correct than the other?  For example, there is a Geometric Progression but no Exponential Progression article on Wikipedia, so perhaps the term Geometric is a bit more accurate, mathematically speaking?  Why are there two terms for this type of growth?  Perhaps exponential growth is more popular in common parlance, and geometric in mathematical circles?

In my 50 or so years of studying mathematics, I've never encountered "geometric growth", but often have met "exponential growth". So that's one small bit of evidence that if you want to sound like most mathematicians, you should use "exponential growth." I wouldn't say either is "more mathematically correct," for the assignment of words as names for concepts is not one that's subject to mathematical evaluation. (Is "continuous" really the right word for the thing we call continuity? Perhaps not...consider a function like
$$
f(x)= \begin{cases}
\frac{1}{2^k} & \text{ if $x = \frac{p}{2^k}$ in lowest terms} \\
0 & \text{otherwise}
\end{cases}
$$
which is continuous at every irrational, and at no rational. But if you used any term other than "continuous" for this thing, mathematicians would regard it as somewhere between "peculiar" and "redundant" and "wrong" :) . )

Wow! I can't believe I'm in this discussion. I am NOT any kind of mathematician. But I thought that it worked like this:

Geometric: a number is multiplied by a fixed factor, and then the product is multiplied by that same fixed factor... etc. For example:
$$1\times1.2=1.2$$
$$1.2\times1.2=1.44$$
$$1.44\times1.2=1.728$$
$$1.728\times1.2=2.0736$$

Exponential: a number is multiplied by a factor that, instead of being fixed, grows  as the succeeding products grow; i.e., the factor is in proportion to each product. For example:
$$1\times1.2=1.2$$
$$1.2\times(1.2\times1.2)=1.728$$
$$1.728\times(1.728\times1.2)=3.583$$
$$3.583\times(3.583\times1.2)=15.405$$


Quantifier Logic : Translation of sentence "All students have already done the exam for all course except course X" As the title said, what is the best translation for "All students have already done the exam for all course except course X"?
Me and my fellow students debated for quite a long time about this question. We use x is a domain of students and y is a domain of the courses, and Predicate P(x,y) is x already done exam course X. My answer is $$\forall x \forall y (P(x,y) \iff (y \neq Course X))$$
One of my friend answer is $$\forall x \forall y(P(x,y) \implies (y \neq Course X))$$
and the other answer is $$\forall x \forall y (P(x,y) \implies (y \neq Course X)) \land \exists x(\lnot P(x, Course X))$$
Which one do you think is right? And if none are right, what is the translation and why?

Consider $S$ be the domain of students.

And $C:$ be the domain of course tests.

Let  $\text{X-Course}$ be the "name" of the course to be excluded (err, the course that no student has yet taken a test in).

Let $P(x, y)$ denote "$x$ has taken $y$."

$$\forall x \in S, \forall y \in C \Big((y \neq \text{X-Course})\to P(x, y) \Big)$$

Or we can say, $$\forall x \in S, \forall y \in (C\setminus \{\text{X-Course}\}) \Big(P(x, y)\Big)$$

where $C\setminus \{\text{X-Course}\}$ is the set of all course tests, except for the test for Course $X$.

$\Huge($ And if you insist on the fact that we are to assume that the domain for $x$ is "All students", and the domain for $y$ is "All courses", with $P(x, y)$ meaning $x$ has taken the test in course $y$, and $X$ the name for the course in which no test has yet been taken,

Then at its simplest, we have: $$\forall x, \forall y\Big(\big(y\neq X\big) \ \rightarrow P(x, y)\Big)\Huge)$$

Alternatively, let the domain $D$ of x, y, consist of all persons, places, things.

Let $S(x)$ denote that element $x\in D$ is a student.

Let $C(x)$ denote that element $x \in D$ is a test for an academic course.

Let $P(x, y)$ denote "x has taken y".

Let $\text{X-Course }$ name the "not-yet-if-ever tested academic course X"

Then the sentence you are given can be translated as follows:

$$ \forall x, \forall y\Big(\big(S(x) \land C(y) \land (y \neq \text{X-course})\big) \to P(x, y)\Big)$$

$\forall x,y (P(x,y) \iff (y \neq Course X))$ can also be accepted, meaning no one has done exam for $X$ but all did other exams

EDIT

$$\forall x \forall y ((y \neq Course X) \implies P(x,y)) \land \exists x(\lnot P(x, Course X))$$
So the other answer you provided is close to being right only it has to switch places in the implication. This means each exam not being $X$ has been done but also fact is that not everyone has done $X$, so there is someone who didn't do it


Prove two line segments are perpendicular to each other in a triangle with orthocenter See below chart:

$H$ is the orthocenter of $\triangle ABC$. $AD, BE$ are altitudes of the triangle. $F$ is on $AD$. $\angle BFD = \angle BAC$. $\frac{FH}{HD} = \frac{KF}{FD}$. $KL \perp BE$. Prove that $FL \perp AB$

My thoughts: obviously there are a lot of similar triangles..such as $\triangle BFD \sim \triangle BAE $,  $\triangle HKL \sim \triangle HAE \sim \triangle HBD$ etc. But $FL$ and $AB$ seem hard to relate.

I think we can also extend $LK$ to intersect $AB$ at $J$, then we just need to prove that $JL^2 - JF^2 = BL^2 - BF^2$.. that might be closer to what we want to prove.

We will show that $ \angle FLH = \alpha \, ( = \angle BAC)$.
If so, then since $\angle LHC = \angle EHC = 90^ \circ - \angle HCE = \alpha$, so $FL \parallel HC \perp AB$.

Let $ \angle FLH = \theta$.

From the given condition, $ \frac{ FH}{ FK } = \frac{ HD } { FD } = \frac{ DH / BD } { FD / BD }  = \frac{ \tan ( 90^\circ - \gamma) } { \tan (90^ \circ - \alpha } = \frac{ \tan \alpha } { \tan \gamma} $.

By sine rule on $ \Delta FHL $, $\frac{ FH}{\sin \theta } = \frac{ FL } { \sin \gamma}$.
By sine rule on $\Delta KFL$, $ \frac{ KF} { \cos \theta } = \frac{ FL } { \cos \gamma}$.
Hence $ \frac{ FH } { FK } = \frac{ FH / FL } { KH / FL } = \frac{ \sin \theta / \sin \gamma } { \cos \theta / \cos \gamma } = \frac{ \tan \theta } { \tan \gamma}$.

Comparing these equations, $ \tan \alpha = \tan \theta$. Since these angles are $ < 180^\circ$, hence $ \theta = \alpha$.

Other interesting results:
1.  $\Delta LHF \sim \Delta ABC$
2. Circumcircle of $LHF$ is tangential to $BF$

Note: I made the assumption that $\alpha < 90 ^ \circ$ as hinted in the diagram. For completeness (in an olympiad), this proof could benefit from checking cases / using directed angles.

Experimental approach:

Extend LK to touch AB at Q.

In this optimized drawing we have:

$\frac{FH}{HD}≈ \frac{KF}{FD}$

$\angle FBD=\angle BFD=\angle BAC=\angle BQL=45^o$

Triangles HBD and HQL are similar so:

$\angle HBD=\angle HQL$

$\angle FQL +\angle BQF=45^o$

$\angle HBD +\angle FBH=45^o$

$\angle FBH=\angle BQF$

$\angle BQL=\angle QBL=45^0$

⇒$\angle FBQ=\angle FQB=\angle HBD$

Therefore LF must be a part of altitude of triangle QLB. that is LF is perpendicular to AB.


Summation of combination of binomial coefficient Is there any way to find:

$$\sum_{i=0}^n {\binom{n}{i}i^k}$$

I know that we can find it for small k by using binomial theorem by differentiating both sides and then multiplying both sides by x and repeating till the form presents itself and then putting x =1, but is there a way for any general power( k here) of i?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{i = 0}^{n}{n \choose i}i^{k} & =
\sum_{i = 0}^{n}{n \choose i}\bracks{z^{k}}\pars{k!\expo{iz}} =
k!\bracks{z^{k}}\sum_{i = 0}^{n}{n \choose i}\pars{\expo{z}}^{i} \\[5mm] & =
\bbx{k!\bracks{z^{k}}\pars{1 + \expo{z}}^{n}}
\end{align}


Solve $ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$ Solve $$ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$$

I first tried to use Eulers formula

$$ \frac{1}{2i} \sum_{k = 1}^{ \infty} \frac{1}{k} \left( e^{2ik} - e^{-2ik}  \right)$$

However to use the geometric formula here, I must subtract the $k=0$ term and that term is undefinted  since $1/k$. I also end up with something that diverges in my calculations and since $\sin 2k$ is limited the serie should not diverge.

The summation is as follows:
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{1}{2i} \, \sum_{n=1}^{\infty} \frac{ e^{2ai n} - e^{- 2ai n}}{n} \\
&= - \frac{1}{2i} \left( \ln(1 - e^{2ai}) - \ln(1 - e^{-2ai}) \right) \\
&= - \frac{1}{2i} \ln\left( \frac{1 - e^{2ai}}{1 - e^{- 2ai}} \right) = - \frac{1}{2i} \ln\left( - \frac{e^{ai}}{e^{-ai}} \cdot \frac{\sin(a)}{\sin(a)} \right) \\
&= - \frac{1}{2i} \ln\left( - e^{2ai} \right) = - \frac{1}{2i} \left( \ln(e^{- \pi i}) + \ln(e^{2ai}) \right) \\
&= - \frac{1}{2i} \left( - \pi i + 2ai \right) = \frac{ \pi - 2a}{2}.
\end{align}
This provides
\begin{align}
\sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{ \pi - 2a}{2}.
\end{align}

For finite non-zero $a,$

$$\sum_{k=1}^\infty\frac{e^{2iak}}k=-\ln(1-e^{2ia})$$

$$1-e^{2ia}=-e^{ia}(e^{ia}-e^{-ia})=-e^{ia}[2i\sin(a)]$$

$$\ln(1-e^{2ia})=\ln(e^{i(a+2m\pi)})+\ln2+\ln(-i)+\ln[\sin(a)]$$
$$=i(a+2m\pi)+\ln2-\frac{i\pi}2+\ln[\sin(a)]$$

as $-i=\cos\left(\frac\pi2\right)+i\sin\left(\frac\pi2\right)=e^{i\dfrac\pi2}$ and where $m$ is any integer

Hope you can take it home from here setting $a=1,-1$ one by one


Conditions for Linear Independence for functions defined by integration Given that the set of strictly positive and continuous functions $$f_i(x,y) >0, \quad i=1,\dots,n$$  are defined on $[0,1]^2$  and
 $\mathbb{R}$-linearly  independent for $(x,y)  \in [0,1]^2$. That is if $c_1, \ldots, c_n \in \mathbb{R}$ and  $\sum_{i=1}^n c_i f_i(x,y)=0$ for all $(x,y) \in [0,1]^2$ then each $c_i$ is zero.
 Let $g_i$ be defined by
$$
g_i(x) = \int_{y\in [0,1] } f_i(x,y) d y, \quad  i=1,\dots,n
$$
what are other conditions are needed on $f_i(x,y)$ so that the set of functions $g_i(x)$ are  $\mathbb{R}$-linearly  independent for $(x,y)  \in [0,1]$.

I'm having difficulties on this problem and any suggestions or references to read would be greatly appreciated. Some thoughts or attempts that I have done are

If $f_i(x,y)= h_i(x) k_i(y)$ where the set of $h_i(x)$ is  $\mathbb{R}$-linearly  independent for $x  \in [0,1]$ and if the set of $k_i(y)$ is  $\mathbb{R}$-linearly  independent for $y  \in [0,1]$ then both $f_i(x,y)$ and $g_i$ are $\mathbb{R}$-linearly  independent?
If $f_i(x,y)$ is also $\mathbb{R}$-linearly independent when $x \in [0,1]$ for every fixed $y  \in [0,1]$. I'm just not sure how the proof would work in this case.

$\renewcommand{\p}{\partial}$
HINT:
Recall that any set of linearly independent functions has non zero Wronskian.

\begin{align}
\forall c_{1}, c_{2} \ldots, c_n \in \mathbb R  \quad
c_{1} g_{1}\left(x\right) + c_{2} g_{2}\left(x\right) + \ldots + c_{n} g_{n}\left(x\right)  \not\equiv 0
 \impliedby
W\big( g_{1}, g_{2}, \ldots, g_{n} \big)
\not \equiv 0
\end{align}

Recall also  Leibniz formula  for differentiation under the integral with variable limits:

$$
\frac{d}{dx}\left(
\int_{a\left(x\right)}^{b\left(x\right)} f\big(x,t\big)\, dt  \right)
= f\big(x,b\left(x\right)\big) \cdot b'\left(x\right)
- f\big(x,b\left(x\right)\big) \cdot a'\left(x\right)
+ \int_{a\left(x\right)}^{b\left(x\right)}
\frac{\partial}{\partial x}\,f\left(x,t\right)\, dt
$$

In your particular case

\begin{align}
g_{i} \left(x\right) = \int_{0}^{1} f_{i} \left(x,y\right) \,d y
\implies
\frac{d\,g_i}{dx} = \int_{0}^{1} \frac{\partial\,}{\partial x}\Big( f_i\left(x,y\right) \Big)\,d y
\implies
\frac{d^{k}g_i}{dx} = \int_{0}^{1} \frac{\partial^{k}\,}{\partial x^{k}} \Big( f_i\left(x,y\right) \Big)\,d y,
\end{align}
for $\,k = 1, \,\ldots,\, n-1\,$ under assumption that functions $\,f_{i}\,$ have $\,n-1\,$ derivatives.

In order for a set of functions $\,g_{1}, \ldots, g_{n}\,$ we require the non-zero Wronskian

\begin{align}
W =
\begin{vmatrix}
g_{1} & g_{2} & \cdots & g_{n} \\
g_{1}^{\left(1\right)} & g_{2}^{\left(1\right)} &\cdots& g_{n}^{\left(1\right)}\\
g_{1}^{\left(2\right)} & g_{2}^{\left(2\right)} &\cdots& g_{n}^{\left(2\right)}\\
\vdots & \vdots & \ddots & \vdots\\
g_{1}^{\left(n-1\right)} & g_{2}^{\left(n-1\right)} & \cdots & g_{n}^{\left(n-1\right)}
\end{vmatrix} =
\begin{vmatrix}
\int_{0}^{1} f_{1} \,dy
& \int_{0}^{1} f_{2}\,dy
& \cdots
& \int_{0}^{1} f_{n}\,dy
\\
\int_{0}^{1} \frac{\partial \,f_{1} }{\partial x }\,dy
& \int_{0}^{1} \frac{\partial \,f_{2} }{\partial x}\,dy
& \cdots
& \int_{0}^{1} \frac{\partial \,f_{n} }{\partial x}\,dy
\\
\vdots & \vdots & \ddots & \vdots\\
\int_{0}^{1} \frac{\partial^{n-2}\,f_{1} }{\partial x^{n-2}}\,dy
& \int_{0}^{1} \frac{\partial^{n-2}\,f_{2} }{\partial x^{n-2}}\,dy
& \cdots
& \int_{0}^{1} \frac{\partial^{n-2}\,f_{n} }{\partial x^{n-2}}\,dy
\end{vmatrix}
\not \equiv 0
\end{align}

Nonzero Wronski determinant  is nonzero means that the column vectors of the matrix  are linearly independent.
Hope you can pick it from here.


Can we rotate a 3D lattice of deformed spheres? EDIT by EricStucky: The full text of original post below for reference, but I have talked with the OP in chat and believe that this is the mathematical core of the question.
Suppose that we have objects with the following shape:

(In the comments, Alan suggests $r = 10 + .2(cos(5\theta) + cos \frac{4\phi}{2})$ as a possible parametrization for this surface.)
(Edit:  Feb 2021  I have found out the formula to create the shape...I do not know if it matches the equation given)

If we create a face-centered cubic arrangement of these objects, the ridges will "settle into" one another.
If we now think of them as physical objects, so that they are never allowed to intersect, is it possible to rotate them?
(For example, if we have infinitely many gears in a line, they can be rotated. But if we have three arranged in a triangle, it is not possible to rotate them.)
Yes.  The question I seek to answer, is... In a friction-free environment, what might naturally emerge when each of the components is rotating...

And how to calculate it.
I realize it is much more complicated (the math) than I had originally thought.  I am not really any longer expecting a concrete answer to this question as it stands.

Hello mathematicians of advanced skills and intuition.
I'm a visual guy, who needs a little help describing what I see.
I see the ball presented below, as a sort of "2-sided" sphere. But also a a 12 sided sphere, because I think that two overlapped tetrahedrons inside of it, could spin and rotate in a toroidal pattern, about a common fixed barycenter, to create this shaped sphere.  But my visual math exceeds my written math abilities, I think.
Maybe has something to do with two paired equally overlapped tetrahedrons, of different size, by a factor of 12/13, with their different volumes, describing a common empty barycenter for all of the enclosed space, with the two paired tetrahedrons rotating in toroidal and counter toroidal directions, until their tips have painted the patterns seen with red for the larger, and blue for the smaller, and purple where they balance.  Or something like that??
Could this shape infinitely co-exist in a pattern of balanced counter-rotation, in which gaps accumulate, and dissipate in a balanced fashion where there are changes in speed in any of three paired intersecting coil like balances, patterns of energy transfer, and creation of balance points in the accumulation of matter, from a mathematical perspective.
Is this shape stackable and expandable "geodesically"?
Please help me decipher what I see.

Added:  It is my understanding that the source of the shape is derived from this pattern of oscillation I think called e mode and b mode polarization.  The source pattern is from the oscillation in gravity as derived through detection of magnetism in the cosmic microwave background.

I am trying to determine if it CAN stack tetrahedrally, and counter-rotate in unison, in a field, mathematically, in 3 pairs of offsetting orientation.
Pattern of potential axis orientation if counter-rotated in a field, if the intersection point of 12 of them is always balanced:

I don't know how to answer the question I have asked, but I think I can answer it by analogy, in the hopes that someone will see my pathetic attempt to explain a mathematical concept on a board where math is a specialty, take pity on me, and try to provide an answer that is more in keeping with the nature of this boards math section.

If one was to imagine two boys with their backs to one another, using their hands and feet to press outwards on the silo, and trying to crabwalk up it, but "mis-programmed their feet" to work in slightly opposite "phases of movement" where two limbs would purposely "short step" by the smallest touch, while their other two limbs would "longstep", then they would oscillate and the alternate limb do the "short step" or "long step".  Then if in one boy, an  cross mixed pair was also given "mis-programming" to "step up" adjust, and across the diagonal, the other pair was given a "step down".

Then because in a 3D array, there would be the relative motion of the walls, which are other balls in counter-rotation.  The combined "up and down" and "left and right" and the cross mixed reversal of those actions, and the appropriate counter action from the "limbs" of virtual people inside the spheres, will create a 3 dimensional array of "square dancers".  The interference pattern of their hands and feet, will equal the interaction pattern, of where "pressure" will be felt, and "released, while the spheres counter-rotate.


Inconsequent use of coordinates in function evaluation and differentiation. Assume $X = (0,\infty)$, $Y = \mathbb{R}$ and that $f : X \to Y$ is a function with some rule, for example
$$
f(x) = ax; \qquad a\in \mathbb{R} := A.
$$
These symbols provide the following inference:

($i$) the point $x$ in the domain of $f$ is associated to the point $ax \in Y$ by the relation $f$.

($ii$) the point $x$ in the domain of $f$ is associated to the point
$$
\lim_{t \to 0} \frac{1}{t} \left[ a(x + t) - a(x) \right] \in Y
$$
by the relation $\partial f/\partial x$.

Two concerns:

I have seen authors switch between the notation $f(x)$ and $f(b)$ carelessly. That is no problem for inference provided above, because in the first instance I know that I should associate $x \mapsto ax$ and in the second $b \mapsto ab$.  However, some authors seem to switch between «$f(x)$» and «$f(a)$» as to claim that «the symbol $x$ appears in the rule $f$» and «the symbol $a$ appears in the rule $f$» respectively. Every time this happens, I wonder if the picture in ($i$) and ($ii$) is indeed correct and/or if it is reconcilable with that meaning.

However, this dilemma becomes vivid if you consider the derivative:

Suppose (to my horror) that an author wrote $\partial f/ \partial b = 0$ and I wanted to give meaning to this expression. It seems to me that the formula for the derivative no longer applies because we were never told how to define the limit (the derivative) for any other coordinate than $x$ (the coordinate on the domain of $f$). At best, I can give meaning to $\partial f/ \partial a$ if I redefine the function $f : A \to Y$ with a different domain by $f(a) = ax$ because the symbol $a$ appears in the rule $f$ above. However, I would not know how to define the limit with respect to the symbol $b$, because it does not appear in the product $ax$.

If the view in ($i$) and ($ii$) is indeed correct, can you explain what happens when authors write $f(a)$ «seemingly to say that $f$ contains the symbol $a$» and $\partial f/\partial b = 0$ «seemingly to say that $f$ does not contain the symbol $b$» such that these notions can be reconciled? If ($i$) or ($ii$) is false, please let me know what is correct.

Some authors highlight the function and pay less attention to using the same designation for the independent variable. In essence, they are implicitly using a function notation similar to the following behind-the-scenes:

$$f(~~) = a(~~)$$

I have even seen some that intentionally do this to ensure that readers are breaking away from the comfort zone of always having $x$ as the independent variable or even $f$ as the function.

For instance, they might define a particular function with $f(x)$ and then in some later text all of a sudden switch to $g(x)$ or $h(k)$.

In the end, it comes down to the trade-off between crystal clear consistency (where $f$ always means the same $f$ and $x$ always meaning the same $x$) and the attempt to write mathematics texts that challenge the reader and not always spelling things out in great detail (thereby opening up for the "personal discovery" aspect of math learning).

Of course, it is always possible that it is due to sheer laziness or that some error has crept in that has not been fixed during editing. After all, many mathematics textbooks have errata lists seemingly no matter how many editions of it that have been published.

Also, not all math texts are equally rigorous. Just look at Thomas versus Spivak for calculus, for instance.

My personal preference is for consistency.


Function with the symmetry of a hyperbolic regular tiling I'm looking for a meromorphic complex function

$$f: \mathbb{D} \to \mathbb{C}$$

where $\mathbb{D} = \{z\in \mathbb{C} : \vert z \vert \le 1\}$ is the unit disk identified with the Poincaré disk model of the hyperbolic plane, which has a zero of multiplicity 7 at each vertex of the order-7 triangular tiling of the disk:

(with one of the vertices at the origin and one of the central edges in $\mathbb{R}^+$, for example, so that the function is unique up to scale), and a pole of multiplicity 3 at each vertex of its dual tiling.

Has this function been studied somewhere? Is there an explicit expression for it, maybe in terms of special functions?

I know that similar functions exist for regular spherical tilings in the Riemann sphere (see for example this question), and powers of Weierstrass P functions give the corresponding functions for Euclidean tilings of the complex plane.

First of all, the proper description of your map is as a holomorphic map ${\mathbb D}\to CP^1= {\mathbb C}\cup \{\infty\}$ (this is what it means to have poles) which is automorphic with respect to a certain group $\Gamma$ of linear fractional transformations of ${\mathbb D}$ preserving the order order 7 tiling:
$$
f(\gamma z)= f(z), \gamma\in \Gamma, z\in {\mathbb D}.
$$
Such $\Gamma$ is not unique and neither is $f$ (even after your normalization), unless you impose further restrictions. Firstly, you want to have $\Gamma$ which is the $(2,3,7)$ von Dyck group. To describe this group geometrically, pick one of the triangles $T$ in your tiling. Let $v$ be one of its vertices, $o$ the center of $T$ and $p$ the midpoint of an $e$ edge containing $v$. The group $\Gamma$ is discrete, generated by order 7 primitive elliptic rotation around $v$, order 2 elliptic rotation around $p$ and order $3$ rotation around $o$. (Rotations are understood here as linear fractional transformations conjugate to the euclidean rotations.) The quotient $X={\mathbb D}/\Gamma$ is topologically a 2-sphere with three distinguished points $[v], [o], [p]$, projections of $v, o, p$. The complex structure from  ${\mathbb D}$ descends to a complex structure on $X$ making it a Riemann surface. As any compact Riemann surface of genus $0$, it is biholomorphic to the extended complex plane, $CP^1$. Such a biholomorphic map $h$ is not unique, but it becomes unique if you require
$$
h([o])=\infty, h([v])=0, h([p])=1.
$$
The composition of the (holomorphic) projection $\pi: {\mathbb D}\to X$ with $h$ is your automorphic function $f$. It will have a triple pole at $o$, order 7 zero at $v$ and also will send the edge $e$ to  the unit segment $01$ in the real line. By automorphicity, the same will be true for all other vertices are centers of triangles.

This description is not explicit of course. However, the function $f$ (or its multivalued inverse) can be found as a solution of some ODE of the form $S(F)=\varphi$, where $\varphi$ is a certain meromorphic quadratic differential and $S$ is the Schwarzian derivative. Equivalently, $F$ is the ratio of two independent solutions of Fuchsian linear 2nd order ODE, of the type: $y'' + \varphi y=0$, where $y$ is a function of one complex variable.

This ODE can be made very explicit if you look at the target space (the extended complex plane); I can write it down if you are interested and when I have time. The differential $\varphi$ can be made very explicit on the target space but less so in the unit disk (one can find it as an infinite sum, a Poincare series). All of what I describe was known to Poincare and can be found in books on automorphic functions, like Ford's "Automorphic functions".

Edit. The inverse to the function $f$ is the ratio $F_1/F_2$, where $F_1(z), F_2(z)$ are two linearly independent solutions of the following holomorphic 2nd order ODE
$$
z(z-1) F'' + ((a+b+1)z-c)F' + abF=0
$$
on the complex plane, where
$$
|1-c|=1/7, |c-a-b|=1/2, |a-b|=1/3.
$$
One has to further fiddle with the choice of $F_1, F_2$ to make sure that $f^{-1}$ sends the upper half plane $H=\{z: Im(z)>0\}$ in the target complex plane to your favorite triangle $T$ (as above) in the unit disk.

See for instance Notes on differential equations and hypergeometric functions page 29. I do not have access to any math books right now and I got this reference just by googling "monodromy of hypergeometric functions".

Once you know $f: T\to H$, you get $f: {\mathbb D}\to CP^1$ by using the Schwarz reflection principle (to define $f$ on the image of $T$ under the hyperbolic reflection in the edge $e$) and then the automorphic equation $f(\gamma z)= f(z)$ on the rest of the unit disk.

I have not heard of this.

A lot of people who play with hyperbolic tilings check in with me to seek my opinion on this or that.

The ring structures grow very fast, this series https://oeis.org/A265041 shows the first few rings of {3,3,3,5} as found by my process.

The equiangular or steriographic projection of hyperbolic geometry is bounded.  The only unbounded one i know is the equiperimeter version, which has rings that run as the cosh(nr).

But i can't ever recall anyone ever putting the h2 or e2 onto the complex plane.  A number of people have tried to suggest some sort of coordinate system, but i point out these turn very dense very quickly.


Conditional expectation $E(X\mid XY)$ when $(X,Y)$ is standard normal I have to calculate
$\mathbb{E}(X|X*Y)$ with X,Y being independent and standard normal distributed. I got at tip in this post (Conditional expectation on components of gaussian vector), that I should use the definition and Bayes Thm to solve the problem. I played around a bit, but I just don't get it :(

May anyone give me another hint?

As every conditional expectation, $E[X\mid XY]=w(XY)$ for some measurable function $w$. Recall that:

  Conditional expectations depend only on the joint distribution of the random variables considered, in the sense that, if $E[X\mid XY]=w(XY)$, then $E[X'\mid X'Y']=w(X'Y')$ for every $(X',Y')$ distributed like $(X,Y)$.

Choosing $(X',Y')=(-X,-Y)$ above, one gets $X'Y'=XY$ hence $$w(XY)=E[-X\mid XY]=-E[X\mid XY]=-w(XY).$$ Thus, $$E[X\mid XY]=0.
$$
One sees that $E[X\mid XY]=0$ for every centered gaussian vector $(X,Y)$, neither necessarily independent nor standard.

Still more generally:

  Let $\Xi$ denote any centered gaussian vector, $u$ an odd measurable function such that $u(\Xi)$ is integrable and $v$ an even measurable function. Then, $$E[u(\Xi)\mid v(\Xi)]=0.$$

you need to compute the integral :
$$ \int_{-\infty}^{\infty} (x f_{x|xy}) dx $$
where $f_{x|xy}$ is the conditional pdf. To find this pdf, $$f_{x|xy} = \frac{f_{x,xy}}{f_{xy}}$$
Try to do a transformation with $U=XY$ and $V=.....$ that can help you to compute the expectation


Motivation behind Dedekind's cut set I want to know the motivation behind Dedekind's real number construction. The motivation of such properties of the cut sets is not clear to me.
BTW, I am new to real analysis and just have started reading the first chapter of Rudin.

what made Dedekind to thought of sets with some nice properties (what are the motivations behind such properties ? For example, why cut set should not have any greatest element ?). I read somewhere that initially Dedekind thought that any real number can be uniquely identified by rational numbers less than that. So, this was his starting point. From that the first and second property of cut set is understandable, but the third property (no greatest element) is not clear to me.

Here's my best guess. Dedekind:

Hmmmmmm.

The rational number line is full of holes. At $\sqrt{2}$, there should be a number, but there isn't.

What does that even mean?

Hm.

Okay, there is no hole at $2$. We can make this precise by observing that $(2-\epsilon,2)_\mathbb{Q}$ is distinct from $(2-\epsilon,2]_\mathbb{Q},$ for all strictly positive $\epsilon$.

But there is a hole at $\sqrt{2}.$ We can make this more precise by observing that $(\sqrt{2}-\epsilon,\sqrt{2})_\mathbb{Q}$ has exactly the same elements as $(\sqrt{2}-\epsilon,\sqrt{2}]_\mathbb{Q}$ for all strictly positive $\epsilon$.

But wait. If we're trying to build the real numbers, then the notation $\sqrt{2}$ isn't 'allowed' yet. Okay, so lets make it precise like this.

There is a hole at $\sqrt{2},$ which means that $\{x \in \mathbb{Q} \mid x^2 < 2\}$ has exactly the same elements as $\{x \in \mathbb{Q} \mid x^2 \leq 2\}.$

But no, this is too symmetrical; the above statement could equally well be seen as the claim that there's a hole at $-\sqrt{2}$. Okay, lets go ahead and destroy that symmetry.

There is a hole at $\sqrt{2},$ which means that the downward closure of $\{x \in \mathbb{Q} \mid x^2 < 2\}$ has exactly the same elements as the downward closure of $\{x \in \mathbb{Q} \mid x^2 \leq 2\}.$

Aha! So maybe we should define that $\sqrt{2}$ is the downward closure of $\{x \in \mathbb{Q} \mid x^2 < 2\}.$ But wait, why not the downward closure of the non-strict version? Okay, lets just disregard that possibility for the moment.

So, a real number is the downward closure of a set of rational numbers that is bounded above, but has no greatest element.

Wait. How about (Eureka!):

A real number is a downward-closed set of rational numbers that is bounded above, but has no greatest element.

One way to approach mathematics is to see it as building up more and more complex structures on top of simple foundations. One commonly used foundation (but not the only one) is set theory of one sort or another. Once you have set theory, you can build on it to form the natural numbers, the the integers, then the rational numbers, and then the reals. Dedekind cuts are one of the simpler ways to construct reals.

Added after question edit:

It's important that every real number be represented by exactly one Dedekind cut. Without the "no greatest element" bit, every rational number could be represented by two different cuts.


Show that for any $0<y<1$ there is $c\in(0,1)$ with $f(c)=f(y+c)$ Suppose $f:[0,1]\to\mathbb{R}$ is continuous and $\min_{x\in[0,1]}f(x)=f(0)=f(1)$. Show that for any  $y\in(0,1)$, there is $c\in(0,1)$ s.t. $f(c)=f(y+c)$.

Intuitively I can see what is going on by drawing a graph, but I am having difficulty writing the proof down formally using mathematical languages. Can some one help me?

The magic words are: intermediate value theorem, applied to the function $f(x+y)-f(x).$

Hint:

Since f is continuous in $[0,1]$ and f(0)=f(1), hence there exists at least one x$\in~(0,1)$ such that $$f\prime(x)=0 \text { (Rolle's theorem)}$$

If you have drawn the graph you will notice that the curve will be symmetric about x


Is horizontal line always parallel to the x-axis? I am having an argument about the answer of the following question with my friend:

Determine if the following sentence is always, sometimes or never true:
- A horizontal line is parallel to the x-axis.

I think it's 'always', but my friend thinks it's 'sometimes'.

My friend thinks that the answer is 'sometimes' because a horizontal line at x-axis is not parallel to x-axis, it is the x-axis.

We are unable to find any resources to prove it one way or other, and hoping this community would help.

Thanks.

A horizontal line is always parallel to the $x$-axis.  Note that a line is considered to be parallel to itself.

Sometimes; a horizontal line has the same slope as the x-axis. If the horizontal line is not y=0, then it is parallel to x-axis.


Give a formula for the volume of the solid under a surface $z=xy$ and a triangle? Give a formula for the volume of the solid with unit density lying under the surface $z = xy$ and above the triangle in the $xy$-plane with vertices $(0, 1, 0)$, $(1, 1, 0)$ and $(0, 2, 0)$.

What is the correct way to do this?

Attempt:

The volume $V$ lying beneath the surface $z = f(x,y)$ and above the region

$a\leq x\leq b,g_{1}(x)\leq y\leq g_{2}(x)$

$V=\iint_{D}f(x,y)dA=\int_{a}^{b}\int_{g_1(x)}^{g_2(x)}f(x,y)dydx$

In the triangle with vertices $(0, 1, 0)$, $(1, 1, 0)$ and $(0, 2, 0)$, $x$ lies between

$0$ and $1$

The $y$ component could lie between $y = 1$ and the line joining the points $(1, 1)$ and $(0,2)$. And it could have the equation $y = -x + 2$.

$D= \left \{ (x,y):0\leq x\leq 1,1\leq y\leq -x+2 \right \}$

$V=\iint_{D} (xy)dA=\int_{0}^{1}\int_{1}^{ -x+2}xydydx$

Here's another way to get answer to this question:

Rename the vertices: $A_1 = (0,1,0), A_2 = (1,1,0), A_3 = (0,2,0)$.
Find area $$A_x = [\min{(A_{1x},A_{2x},A_{3x})} .. \max{(A_{1x},A_{2x},A_{3x})}] = [\min(0,1,0) .. \max(0,1,0)] = [0..1]$$
$$A_y = [\min{(A_{1y},A_{2y},A_{3y})} .. \max{(A_{1y},A_{2y},A_{3y})}] = [\min(1,1,2) .. \max(1,1,2)] = [1..2]$$

$A=[0..1]\times[1..2]$. (bounding rectangle)
Find subset of A which is inside the triangle, $K \subset A$:
$$ K = \{ (x,y) | isInsideTriangle(x,y) \wedge (x,y) \in A \}$$, where isInsideTriangle can be found from url https://stackoverflow.com/questions/2049582/how-to-determine-if-a-point-is-in-a-2d-triangle
Note that $z=xy$ can be written in form $z-xy=0$ which is of form $f(x,y,z)=0$.
Use implicit function and heightmap substitution $z=f(x,y)$: $$f(x,y,z) = f(x,y,f(x,y)) = 0$$
$$ f(x,y)-xy = 0$$, $$ f(x,y) = xy $$
Calculate area of the triangle $$ A_{tri} = \frac{1}{2} A_x \cdot A_y = \frac{1}{2} 1 \cdot 1 = \frac{1}{2} $$
Calculate average of f(x,y) by random sampling, utilizing the area $K$ and a sum $av_f(K) = \frac{\sum_{i=1}^{k} f(x_i,y_i)}{k}$, where $f(x_i,y_i)$ is random sample taken from $(x_i,y_i) \in K, i \in [1..k], k>10000$.
Note that $\int_K{f(x,y)dK} = A_{tri} \times av_f(K) $ where $av_f(K)$ is the average of $f(x,y)$ in area $K$.

What you did looks good providing some light modifications:

The volume is a 3d integral. So I would start saying that your aim is to compute the volume V of the domain $\mathcal{V}=\{(x,y,z) \in \mathbb{R}^3 \ | 0\leq x\leq 1,1\leq y\leq -x+2, 0 \le z \le xy \}$.
Then $V= \int \int \int_\mathcal{V} dx dy dz=\int \int_Dxy dx dy$.
Then you have a small issue of using $x$ as both a bound of integration and a variable in the function to be integrated which is confusing.


Positive ordered pair of $(10x)^2+(5y)^2<1,000,000$ How can I find the number of ordered pair $(x,y)\in\mathbb{Z}_{>0}$ where $(10x)^2+(5y)^2<1,000,000$?

Thanks

To get an approximate answer, note that the ellipse with axes [-100,100] in $x$ and $[-200,200]$ in $y$ has area $20000\pi\approx 62864$.  This is the approximate number of lattice points in the ellipse.  $600$ are on the axes, and a quarter of what is left is in the first quadrant, so there are approximately $\frac 14 \cdot 62264=15566$ solutions

I'm trying to answer this myself, check my work if I'm correct:

Since x can't be more than or equal to 100, and for y can't be more than or equal to 200, so there's 99*199 combinations possible?


Double integral on domain $x^2+y^2 \leq 1$ and $0 \leq x \leq 1/2$ I need to compute the following double integral. I am stuck.

  Let $$\mathbb D = \left\{ (x,y)\in\mathbb R^2 : 0 \leq x \leq \frac12, x^2 + y^2 \leq 1 \right\}$$ Find the following integral $$I = \iint_{\mathbb D}\sqrt{x^2+y^2} \, \mathrm d x \, \mathrm d y$$

By symmetry, the integral equals $2$ times the integral of $r$ over the sector $S=\{(r,\theta),r\in[0,1], \theta\in[\pi/3,\pi/2]\}$ and the triangle $R=\{(r,\theta),r\in[0,1/(2\cos\theta)], \theta\in[0,\pi/3]\}$.
$$I=2\int_{\pi/3}^{\pi/2}\int_0^1r\cdot rdrd\theta + 2\int_{0}^{\pi/3}\int_0^{1/(2\cos\theta)}r\cdot rdrd\theta$$


Flat connection over a simply connectied manifold I am looking for a proof of the following result.

  Let $p:P\to M$ be a $G$-principal bundle over a connected and simply connected manifold.
  Suppose there is a flat connection $A\in\Omega^1(P,\mathfrak{g})$, that is $F^A\equiv 0$.
  Then $P$ is trivial and there exists a global section $s:M\to P$ such that $s^*A \equiv 0$.

That a global section on $P$ implies that it is trivial is well known, so the first task is showing that $P$ is trivial.
It seems that this is a "well known fact", but I could find no concise proofs.
Wikipedia suggests that this is closely related to the holonomy of $A$.

Even then, I don't know how do we hit the $s^*A$ condition.

A connection in a principal $G$-bundle $P$ is said to be flat at $u\in P$ if there is a local trivialization of the bundle in a neighborhood of u, $(U_{\alpha},\psi_{\alpha})$,  under which the given horizontal subspaces map to the horizontal subspaces of the trivial connection on the corresponding trivial bundle.
Consider the local section $s(m)=\psi_{\alpha}(m,e)$, being $e$ the identity element in $G$. Since the horizontal space at $s(m)$ consists of those $V\in T_{s(m)}P$ such that $A_{s(m)}(V)=0$  it is clear that
$$
s^*(A)_{m}(X)=A_{s(m)}(ds_m(X))=0
$$
and so $s^*(A)=0$.
On the other hand, I don't know how to solve the part that $P$ being trivial, that is, that $s$ is global. Sorry.

A connection on a principal bundle is defined by a distribution ${\cal H}$ transverse to the fibers. Saying that the connection is flat is equivalent to saying that ${\cal H}$ is integrable and define a foliation transverse to the fibres this is equivalent to saying that $P$ is a suspension bundle, that is a quotient  $(x,y)\rightarrow (\gamma(x),h(\gamma)(y))$ where $h:\pi_1(M)\rightarrow Diff(G)$ is a representation of $\pi_1(M)$ and $\hat M$ the universal cover of $M$, if $\pi_1(M)=1$, then $P$ is isomorphic to $M\times G$.


$C^\infty(R^n)$ is a Banach Space when equipped with topology of uniform convergence Prove $C^\infty(\Bbb R^n)$ is a Banach Space when equipped with topology of uniform convergence.
$C^\infty(\Bbb R^n)$ is space of all continuous functions that converge to $0$ at $\infty$.
And, the topology of uniform convergence is defined by the norm
$f \mapsto \sup_{x\in\Bbb  R^n}|f(x)|$.

The statement is false. Norms have to be finite, but if we let $n=1$ and take $$f(x)=\begin{cases} 1/x & x \neq 0 \\ 0 & x=0\end{cases}$$ then $f$ is in your space but has infinite norm.

Norms also have to be positive, but if we take $f(x)=-1$ then the norm is $-1$.

It's not difficult to show $\|f\|=\sup_{x\in \mathbb{R}^n}|f(x)|$ is a norm. We need to show the space is complete.

Suppose $\{f_n\}_{n=1}^\infty$ is a Cauchy sequence in the space $i.e.\ \|f_n-f_m\|\to0$ as $m,n\to\infty$. For any fixed $x\in \mathbb{R}^n,\ \{f_n(x)\}$ is Cauchy in $\mathbb{R}$(I hope I didn't misunderstand the map $f: \mathbb{R}^n\to\mathbb{R}$).

Define $f(x):=\lim_{n\to\infty}f(x)$, this is a uniform convergence $i.e. \|f_n-f\|\to 0$ as $n\to\infty$. Hence, $f$ is continuous and moreover,

$$ |f(x)|\le |f_n(x)-f(x)|+|f_n(x)|\le \|f_n-f\|+|f_n(x)|
$$
$\forall \epsilon>0,\ \exists\ M>0\ s.t.\ \|f_n-f\|\le\epsilon$ when $n>M$. For a fixed $n>M,\ \exists\ N>0,\ s.t.|f_n(x)|<\epsilon$ if $|x|>N$. Therefore, there is an $N$ such that for any $|x|>N$
$$ |f(x)|\le \|f_n-f\|+|f_n(x)|<2\epsilon\  \Rightarrow\ f\ converges\ to\ 0\ at\ \infty
$$
So $f$ is an element in your space.


Calculating maximum velocity in simple harmonic motion I'm a bit confused about simple harmonic motion...

If a particle is in simple harmonic motion, to calculate the maximum velocity can I use either displacement = 0 or acceleration = 0, since i know in both graphs a zero in displacement and acceleration corresponds to maximum velocity?

You're correct.

Physically, the maximum speed (velocity is a vector! Speed is its magnitude, which is what you're talking about) happens when the object that's moving passes by the origin and that is the point of zero displacement. It's also the point of zero acceleration because the restoring force behind the motion is zero at that point (and acceleration is force divided by mass).

Incidentally, the minimum speed happens at the opposite situation, namely, at the extreme ends of the motion. Since those are the extreme ends, the particle can't move beyond those and has to momentarily stop there (hence, zero speed). Those points are also where the acceleration is largest in magnitude, pulling the particle back towards the origin.


In simple English, what does it mean to be transcendental? From Wikipedia

A transcendental number is a real or complex number that is not algebraic
A transcendental function is an analytic function that does not satisfy a polynomial equation

However these definitions are arguably rather cryptic to those who are not familiar with the literature of higher mathematics.

So in layman's terms, what exactly does it mean to be transcendental? How would a transcendental number be different from an ordinary number, say 5. And respectively, how would a transcendental function be different from an ordinary function, say $f(x) = x^2$

We will play a game. Suppose you have some number $x$.  You start with $x$ and then you can add, subtract, multiply, or divide by any integer, except zero.  You can also multiply by $x$.  You can do these things as many times as you want. If the total becomes zero, you win.
For example, suppose $x$ is $\frac23$. Multiply by $3$, then subtract $2$. The result is zero. You win!
Suppose $x$ is $\sqrt[3] 7$. Multiply by $x$, then by $x$ again, then subtract $7$. You win!
Suppose $x$ is $\sqrt2 +\sqrt3$. Here it's not easy to see how to win. But it turns out that if you multiply by $x$, subtract 10, multiply by $x$ twice, and add $1$, then you win. (This is not supposed to be obvious; you can try it with your calculator.)
But if you start with $x=\pi$, you cannot win. There is no way to get from $\pi$ to $0$  if you add, subtract, multiply, or divide by integers, or multiply by $\pi$, no matter how many steps you take. (This is also not supposed to be obvious. It is a very tricky thing!)

Numbers like $\sqrt 2+ \sqrt 3$ from which you can win are called algebraic. Numbers like $\pi$ with which you can't win are called transcendental.

Why is this interesting?  Each algebraic number is related arithmetically to the integers, and the winning moves in the game show you how so.  The path to zero might be long and complicated, but each step is simple and there is a path. But transcendental numbers are fundamentally different: they are not arithmetically related to the integers via simple steps.

Well, in simple english which isn't rigorous, "transcendental" means it can not be expressed in terms of whole numbers and roots.  $\sqrt[3]{2 + 1/\sqrt{3}}, \sqrt[7]{43} + 5\sqrt[3]{49},$ etc (or simply $\sqrt[4]{3}$ or even $39/47$ or $5$) can be and so are not transcendental.  $\pi$ can not be expressed in any such way (take my word for it) so it is transcendental.

But you can see that is a terrible, poorly defined, ambiguous definition that is useless in any formal mathematical sense.

More formally a real number $m$ is algebraic if there is some polynomial $P(x) = a_nx^n + a_{n-1}x^{n-1} + .... + a_2x^2 + a_1x + a_0$ where each of the $a_i$s are integers and each of the $x^i$ are raised to a non-negative integer power-- if there is such a polynomial where if you plug in your number $m$ and get $P(m) = 0$-- if that is possible, the number is algebraic.  If it is not possible for there to be any such polynomial, the number is *transcendental.

So take my number $m =\sqrt[3]{2 + 1/\sqrt{3}}$ it is the solution to ... $P(x) = 3x^6-12x^3 + 11 = 0$[*].  There is no finite polynomial with integer coefficients where $P(\pi) = 0$.  (Take my word for that).

===

$3(\sqrt[3]{2 + 1/\sqrt{3}})^6 - 12 (\sqrt[3]{2 + 1/\sqrt{3}})^3 + 11 =$

$3({2 + 1/\sqrt{3}})^2 - 12({2 + 1/\sqrt{3}}) + 11 =$

$3(4 + 4/\sqrt{3} + 1/3) - 12(2 + 1/\sqrt{3}) + 11 =$

$12 + 12/\sqrt{3} + 1 - 24 - 12/\sqrt{3} + 11 = 0$

My other number  $\sqrt[7]{43} + 5\sqrt[3]{49}$ is also a solution to  polynomials but it'd be a real pain for me to figure out which ones.  (It'd be a 21 degree polynomial.  If I set $P(x) = a_{21}x^{21} + ... + a_1x + a_0$, plug in $ \sqrt[7]{43} + 5\sqrt[3]{49}$, set equal to 0, I will get 21 equations to solve for 22 of the $a_i$s in terms of the 22nd $a_i$ which I would choose just to make them all integers.  Very tedious but doable.  Always doable for any finite expression in terms of roots and whole numbers.)

$\sqrt[4]{3}$ is the solution to $x^4 - 3 = 0$.

$39/47$ is the solution to $47x - 39 = 0$ (which is a single degree polynomial) and $5$ is the solution to $x - 5 = 0$.


Is arrow notation for vectors "not mathematically mature"? Assuming that we can't bold our variables (say, we're writing math as opposed to typing it), is it "not mathematically mature" to put an arrow over a vector?

I ask this because in my linear algebra class, my professor never used arrow notation, so sometimes it wasn't obvious between distinguishing a scalar and a vector. (Granted, he did reserve $u$, $v$, and $w$ to mean vectors.) At the same time, my machine learning class used arrows to denote vectors, but I know some other machine learning literature chooses not to put arrows on top of their vectors.

Ultimately, I just want a yes or no answer, so at least I do not seem like an immature writer when writing my own papers someday.

A sign of mathematical maturity is the awareness that truth is invariant under changes of notation.

There's a reason I avoid it, which is that I hand write almost all my mathematics, writing in bulk, and it slows me down to write the arrows overhead. It helps me get down detailed thoughts quickly to use the same notation for writing those notes that I would use when I'm writing a full proof or reasoning from a prepared vantage point. Boldface isn't better, at least for speed (Maybe clarity? I have scribbly handwriting.), but knowing some calligraphy helps especially to cheat and have distinctions clearly hold within the text. Either way, if I don't have two sets of things using the same letters, I'll just use unadorned, pen-weight letters, and just draw them kind of formally clean. I have a feeling I'm not alone in this, and that the adoption of this practice would influence those exposed to its effect on an author's style, which would be another explanation for the large amount of authors there are that don't use arrows or boldface to distinguish vectors from other things. It's also normal if someone doesn't do that just because they don't feel the need to distinguish vectors' names from scalars' names, or have chosen some other notation sufficient to distinguish vectors and scalars in a formula, like a dedicated symbol and order of arguments for the scalar multiplication operator (useful for talking about rings as a module over themselves). The hat over unit vectors or just normalizations of other vectors, on the other hand, is useful for keeping your number of symbols small and showing by the name when things are related.


How to calculate Vapnik-Chervonenkis dimension it's my first post here, so I apologize if I broke a rule!

I'm reading Introduction to Machine Learning and got stuck on VC dimension. Here's a quote from the book:

"...we see that an axis-aligned rectangle can shatter four points in two dimensions. Then VC(H), when H is the hypothesis class of axis-aligned rectangles in two dimensions, is four. In calculating the VC dimension, it is enough that we find four points that can be shattered; it is not necessary that we be able to shatter any four points..."

And I don't understand that. If it's enough to find some separable combinations, why can't we just choose a "rectangle with positive examples" from the image above, put another $n$ positive ones therein, and then say $VC(H)$ increased by $n$? And if all cases must be separable then why we don't consider 4 points placed on a line - which is in general not possible to shatter by a rectangle?

The same with the linear classifier example on wikipedia VC article - on their image four points are impossible to shatter, but we can come up with a layout where it is possible. And conversely, we can put 3 points (as "+", "-", "+") on a line and it won't be possible to separate the positives from the negatives by a linear classifier.

Can anyone explain where's my mistake?

I think you just need to think carefully about how your examples relate to the definition. The VC dimension of H is the maximum h such that there exists a set of cardinality h shattered by H. To show that VC(H)=4 you must show:

There is a set of cardinality 4 shattered by H, and
Any set of cardinality greater than 4 is not shattered by H

In the picture, they are doing the first thing - giving a lower bound on the VC dimension by giving an example of a set that is shattered. To show VC(H)<5 they should also show that no set of five points is shattered.

There will in general be lots of sets of various sizes that are not shattered, but that doesn't matter, essentially because VC dimension is a maximum over sets that are shattered. Your example of $n$ points on a line does not imply anything about the VC dimension. I hope this helps.

@andreister, just adding some color:

The confusion arises because the definition of VC dimension involves BOTH an existential qualifier ("there exists"), and a universal qualifier ("for all"):

  Note that, if the VC dimension is h, then there exists at least one
  set of h points that can be shattered, but it in general it will not
  be true that every set of h points can be shattered Burgess (1998)

Thus, one such set of points has to exist, but their shattering == for all labelings == a universal assertion.


What is the result of $\infty - \infty$? I would say $\infty - \infty=0$ because even though $\infty$ is an undetermined number, $\infty = \infty$. So $\infty-\infty=0$.

From a layman's perspective, imagine that I have an infinite number of hotel rooms, each numbered 1, 2, 3, 4, ...
Then I give you all of them. I would have none left, so $\infty - \infty = 0$
On the other hand, if I give you all of the odd-numbered ones, then I still have an infinite number left. So $\infty - \infty = \infty$.
Now suppose that I give you all of them except for the first seven. Then $\infty - \infty = 7$.
While this doesn't explain why this is indeterminate, hopefully you can agree that it is indeterminate!

Infinity does not lead to contradiction, but we can not conceptualize $\infty$ as a number. The issue is similar to, what is $ + - \times$, where $-$ is the operator. The answer is undefined, because $+$ and $\times$ are not the kind of mathematical objects that $-$ acts upon. (We could of course define something, but then we are no longer talking about the usual operations, we are changing the meaning of the symbols.)

We can not conceptualize $\infty$ as a number to a large part because $\infty - \infty$ does not have a good interpretations. In particular for any finite set, the size of the set, its cardinality, is the same as the place of its last member, its ordinal. So if I have a concrete set $\{a, 57, \text{apple}\}$, then one of the members is at the place $3$, because there are $3$ members of the set. That is no longer true for infinity.

This is shown by the Ross–Littlewood paradox, you receive $10$ coins at each time step, and then you give a single coin back. The question is, how many coins do you have after infinite many steps. The problem is called a paradox because it has a simple and intuitive answer, unfortunately not many people agree which answer is simple and intuitive.

The first answer is, you have $\infty$ coins, which is a notational shorthand for, the total number of coins in your possession growth without limit. Proponents of this answer parse the problem in terms of total value where individual coins loose their identity. One could interpret that as taking the ten coins to the bank, then wiring the value of a single coin back. The bank updates your balance and the operation at a time step $t$ is $b_{t+1}=b_{t} + 10 - 1$.

The other answer is $0$, because each coin only spends a finite time in your possession and so after an infinite amount of steps you gave each and every coin back. In this case, one puts the coin into a line, one by one, adds $10$ coins to the end of the line, and gives $1$ coin back from the front of the line. That means that the $n$-th coin spends the timesteps $n/10$ until $n$ in your possession, and you give it away in a finite amount of time. (You get the $65$-th coin at time step $6$ and give it away during the time step $65$.) In "the end," infinity does not have an end" you have given all coins away and so the balance is $0$.

The difference between the two scenarios is, in the second the order of the coins is preserved, in the first the coins are piled and one only looks at the total value of the coins. And if we modify the second scenario a little bit, we end up with different answers. If you give back always the last coin you received, then you keep coins  $1,\dots,9$ from the first step, coins $11, \dots, 19$ from the second and so on and you get to keep an infinite amount. (I use the order of received here to number the coins, not the place the coins lie in at the end.) If you take the ten coins, put $1$ on a pile, and you put $9$ in a line, and you do that for the first $n$ steps and after that you get lazy and just put all the coins into the line, then you will give all the coins in the line away, but you end up with a pile of $n$ coins.

These different procedures could all be plausibly called $\infty - \infty$, and we just constructed ways to parse that expression in a way to get any answer between $0$ and $\infty$. This is the issue with infinity, we have to be precise how to conceptualize $\infty$ and we can not just treat $\infty$ like a kind of number.


If $\lim_{x\to n}f(x) = 0$ and $f(n)=0$, does $\lim_{x\to n}\frac{\sin(f(x))}{f(x)} = 1$ always? If $\lim_{x\to n}f(x) = 0$ and $f(n)=0$, does $\lim_{x\to n}\frac{\sin(f(x))}{f(x)} = 1$ always?
I have been playing around with some graphs on desmos and there's always the indication that the limit equals $1$. I know that $\sin x$ becomes very linear, is there any function with the potential to "un-linearize" it?

It is always true if $f(n) \neq 0$ in a punctured neighborhood of $x=n$ otherwise we can't take the limit.

hint
Let $ \epsilon>0$.
we know that
$$\lim_{X\to 0,\ne}\frac{\sin(X)}{X}=1$$
Thus
$\exists \eta>0$ such that
$$\color{red}{0<}|X|<\eta \implies |\frac{\sin(X)}{X}-1|<\epsilon$$
But
$$\lim_{x\to 0}f(x)=0$$
then
$\exists \alpha>0 $ such that
$$|x|<\alpha \implies |f(x)|<\eta$$
The red condition $ \color{red}{0<|f(x)|}$ is not always satisfied to insure the existence of $\frac{\sin(f(x))}{f(x)}$.
Your conclusion will be correct if you put$$\frac{\sin(0)}{0}=1$$


Proving $|\sin x| \leqslant |x |$ Is it possible to prove $|\sin x| \leqslant  |x |$ with only  trigonometric identities? (not with well-known calculus or geometry proofs)

If we have the following definitions

$$\cos(x) = \frac{z + z^{-1}}{2}$$
$$\sin(x) = \frac{z - z^{-1}}{2i}$$

where $z = e^{ix}$, then all finitary trig identities are reduced to rational polynomial identities with this definition. Let me show an example (of the strongest inequality possible that is close to what we want):

$$\begin{array}{rcl}
\sin(x)^2 + \cos(x)^2
&=& \left(\frac{z + z^{-1}}{2}\right)^2 + \left(\frac{z - z^{-1}}{2i}\right)^2 \\
&=& \frac{z^2 + 2 + z^{-2}}{4} - \frac{z^2 - 2 + z^{-2}}{4} \\
&=& \tfrac{4}{4} \\
&=& 1
\end{array}$$

The hidden reason that it is impossible to prove $|\sin(x)| < |x|$ with trig identities lies in the complex plane. All these trig identities hold uniformly for real and complex $x$. There are obstructions to this inequality in the complex plane:

$$|\sin(\sqrt{-1})| = 1.1752\ldots \not < 1 = |\sqrt{-1}|$$

This tells us that the inequality is something special about trigonometry on the real line, and because trig identities are unable to distinguish this we need a new tool (such as geometry, calculus for example).

Lastly just let me note there are infinitary trig identities such as Euler's product $$ \frac{\sin(x)}{x} = \prod_{n = 1}^\infty\cos\left(\frac{x}{2^n}\right)$$ which do prove that $|\sin(x)| < |x|$ (to see this, just take the absolute value of both sides multiply up $|x|$ and note the infinite product is bounded by $1$ which is a consequence of the triangle identity applied to the Pythagorean trig identity)

it's prove from that:

$cos(x)\leq \frac{sin(x)}{x} \leq 1$
then
$|\frac{sin(x)}{x}|\leq 1 \Rightarrow |sin(x)|\leq |x|, \quad for \quad x\neq 0$

and as sin(0)=0 then

$|sin(x)|\leq |x|$.

the complete answer is in calulus silverman.


Explain $du = g'(x) dx$ when $u=g(x)$ I have seen the following multiple times in the book Calculus by James Stewart.

For instance, it appears in the proof of the substitution rule (or change of variable) for integral.

Suppose $u = g(x)$ and $\dfrac{du}{dx} = g'(x)$.

Then, $du = g'(x) dx$.

$\dfrac{du}{dx}$ is the derivative of $u$ with respect to $x$ not it is definitely not a fraction.

For instance, the $\dfrac{dy}{dx} = \dfrac{dy}{dt}\times \dfrac{dt}{dx}$ holds true as proven as the Chain Rule.

So why does $du = g'(x) dx$ hold  true? And what does this mean?

I need a clear explanation on this concept please. Appreciate your help, thanks :)

The Leibniz notation does not denote a fraction, as you know, so it is not technically correct to treat it as one and 'separate' the top and bottom. Also, given $u = g(x)$ we never need to say things like $du = g'(x) dx$. After all, simple analysis suffices to get what you actually want, in this case the change of variable theorem.
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

Take any differentiable function $g : D \to U$, and any real/complex parameter $x$ varying in $D$.

Let $u = g(x)$. Then $\lfrac{du}{dx} = g'(x)$.

Take any continuous function $f$ on $U$. Then $\int f(u)\ du$ exists (on the path traced by $x$).

Thus $\lfrac{d(\int f(u)\ du)}{dx} = \lfrac{d(\int f(u)\ du)}{du}·\lfrac{du}{dx} = f(u)\lfrac{du}{dx}$, by chain rule and definition of anti-derivative.

Thus $\int f(u)\ du = \int f(u) \lfrac{du}{dx}\ dx + k$ for some constant $k$. Note that here we are using a basic theorem that if $\lfrac{dy}{dx} = z$ everywhere (on the path traced by $x$) then $y = \int z\ dx + k$ for some constant $k$.

You can see that nowhere do we break apart the derivative notation.

To look at the little bits, recall that as $Δx \to 0$ we have $\lfrac{Δu}{Δx} \approx g'(x)$, which implies much more than $Δu \approx g'(x)·Δx$. It in fact implies that $Δu - g'(x)·Δx \ll Δx$.

Intuitively, as $x$ changes by $Δx$, $u$ will change by $Δu$, and the change in $\int f(u)\ du$ divided by $Δu$ is approximately $f(u)$. Hence $Δ(\int f(u)\ du)$ divided by $Δx$ is approximately $f(u)·\lfrac{Δu}{Δx}$. If you add up the little bits over the entire path of $x$, then the total change of $\int f(u)\ du$ is approximately the sum of $f(u)·\lfrac{Δu}{Δx}$ over all the little changes in $x$.

When you are working with differentials you actually zoom so close to point on the curve that the curve behaves as a straight line.

Coefficient of the line is derivative of $g(x)$ in that point.

So yes, it is actually a fraction.


Show that $f$ is Lebesgue integrable I am given a function defined on $(\mathbb R,$ Borelians, Lebesgue measure)
such that
$$f= \begin{cases}  + \infty &\text{if }  x=0 \\
\ln(|x|) &\text{if } 0\lt |x|\lt1 \\
0 &\text{if } |x|\ge1 \end{cases}
$$

  How can I show that this function is Lebesgue-integrable ?

Since $f$ is even it is enough to show that it is integrable on $(0,1)$. But $$\int_{(0,1)} f(u)du =ulnu-u|_0^1 =1$$


Modular equation with exponents on the exponents I want to solve this modular equation:

$ 7x + 5 \equiv 2^{11^{2017}} \pmod {31}$

As far as I know, dividing the exponent by 31 and substituting it with the remainder is not allowed.

I've looked here and here but I don't see how to apply that to my equation.
I'm sorry for this short question but I don't really know how to approach this equation.

$2$ has order $5\bmod 31$, so
$$2^{11^{2017}}\equiv2^{11^{2017}\bmod 5}\pmod{31}.$$
Now $11\equiv 1\mod 5$, so $11^{2017}\equiv 1\mod 5$ and eventually $2^{11^{2017}}\equiv2^1=2\mod{31}$, so the congruence equation becomes
$$7x+5\equiv 2\iff7x\equiv -3\iff x\equiv 9(-3)=-27\equiv \color{red}{4}\mod31$$
since $7^{-1}\equiv 9\mod 31$.

To solve such question, I would simplify the right hand side first. Notice that ${(2^{11})}^{2017} \equiv 2^{2017} \pmod{31}$. Further more, $2017 = 67 \times 30 + 7$. Therefore, By Fermat's theorem, $2^{2017} \equiv {(2^{30})}^{67} \times 2^7 \equiv 1 \times 2^7 \equiv 128 \equiv 4 \pmod{31}$.

The equation finally becomes

$$7x+5 \equiv 4 \pmod{31} \to 7x \equiv -1 \pmod{31}$$

Notice that I divided by $30$ instead of $31$ when using Fermat little theorem.


Is $\pi$ periodic in any numeral system? Is $\pi$ periodic in any base-k numeral system, where k is integer ? And what is the status of this problem?

$\pi$ is irrational - that was settled hundreds of years ago. That implies that the expression of $\pi$ to any integer base $b$ will be aperiodic. If you have some other kind of numeral system in mind, please edit your question accordingly.

Not for an integer base, but it just happens that $\pi=0.1111111...=0.\bar{1}$ in base $\kappa=\frac{1}{\pi}+1 \approx 1.31831$. :)


Show that $f(x)=\begin{cases}1/b& x=\frac{a}{b}\in [0,1],a,b\in\mathbb Z\\ 0&x\in \mathbb R\backslash \mathbb Q \cap [0,1]\end{cases}$ is integrable. Show that $$f(x)=\begin{cases}\frac{1}{b}& x=\frac{a}{b}\in [0,1],a,b\in\mathbb Z\\ 0&x\in \mathbb R\backslash \mathbb Q \cap [0,1]\end{cases}$$
is integrable on $[0,1]$.

Let $$S_\sigma =\sum_{i=1}^n m_i(x_{i+1}-x_i)\quad \text{and}\quad S^{\sigma }=\sum_{i=1}^n M_i(x_{i+1}-x_i),$$
where $$M_i=\max_{ [x_{i+1},x_i]}f,\quad \text{and}\quad m_i=\min_{[x_{i+1},x_i]}f.$$

I have to show that $\overline{S}=\underline{S}$ where $$\overline{S}=\sup_{\sigma }S_\sigma  \quad \text{and}\quad \underline{S}=\inf_\sigma  S^\sigma .$$

Obviously $S_\sigma =0$ for all partition $\sigma $, and thus $\overline{S}=0$. But for $\underline{S}$ I have some problem. I just can't find $M_i$. But maybe something as :

I consider $ \sigma _n : 0<\frac{1}{n}<...<\frac{n-1}{n}<1$. Then I would say that $M_i\leq \frac{1}{i}$, and thus

$$S^{\sigma _n}\leq \frac{1}{n}\sum_{k=1}^n\frac{1}{k}\underset{n\to \infty }{\longrightarrow }0,$$
and thus, the claim follow. It work ?

$f$ is integrable iff $\exists P$ for every $\epsilon >0$ such that $S^{\sigma}-S_{\sigma}<\epsilon$.

Choose $P=\{x_0, ....,x_n\}$ such that $x_i - x_{i-1} = \frac{b-a}{n}$.

$S_{\sigma} = 0$ on $P$ because for every interval $[x_i, x_{i+1}]$, $m_i=0$.

Also, since $f(x)<1$ for $x<1$, we also have $M_i < 1$ and so $$\sum_{i=1}^n M_i < n$$

Put $\sum_{i=1}^n M_i = M <n$

Therefore $S^{\sigma}-S_{\sigma} = \frac{b-a}{n}M - 0<\epsilon$ on $P$ for large enough $n$.


How difficult exactly is $\int\tan(x^2)\ dx$? How difficult exactly is $\int\tan(x^2)\ dx$ ?

Is it possible to express this integral in terms of elementary functions?

If not, is there anything one could say about it, that would be in some way helpful?

I have not done anything to answer this question myself. (Well, I googled it,
Wolfram alpha tells me no result found in terms of standard mathematical functions, so it seems safe to assume that no such result exist.)

This integral looks somewhat similar to $\int e^{x^2} dx$ (which cannot be expressed in terms of elementary functions) but I just need some reassurance (possibly with a link or an explanation) specifically for $\int\tan(x^2)\ dx$ .

Just in case, here is the Taylor series expansion
$\tan(x) = x+x^3/3+2x^5/15+17x^7/315+62x^9/2835+O(x^{11})$ and
$\tan(x) = \sum_{n=0}^\infty \dfrac{(-1)^{(n-1)}2^{2n}(2^{2n}-1) B(2n)}{(2n)!} x^{2n-1}$, where $B(n)$ are the Bernoulli numbers.
Someone asked me about this integral and I realized I couldn't say much about it.

If the question is : "Is it possible to express the integral $\int \tan(x^2)dx$  in terms of elementary functions ?" the answer is : Yes, on the form of infinite series of elementary functions.

If the question is : "Is it possible to express the integral $\int \tan(x^2)dx$  in terms of the combination of a finite number of elementary functions ?" the answer is : No. (as it was already pointed out in a preceeding answer).

If the question is : "How difficult exactly is $\int \tan(x^2)dx$ ?" the answer is : No more difficut than the integrals : $$\int \sin(x^2)dx=\sqrt{\frac{\pi}{2}}\  S
\left( \sqrt{\frac{2}{\pi}}\ x\right)+constant$$
where $S(X)$ is defined as a special function, namely the Fresnel S integral : http://mathworld.wolfram.com/SineIntegral.html

and no more difficult than the integral : $$\int \cos(x^2)dx=\sqrt{\frac{\pi}{2}}\  C
\left( \sqrt{\frac{2}{\pi}}\ x\right)+constant$$
where $C(X)$ is defined as a special function, namely the Fresnel C integral: http://mathworld.wolfram.com/CosineIntegral.html

The only difference is that in : $$\int \tan(x^2)dx=\sqrt{\frac{\pi}{2}}\  T
\left( \sqrt{\frac{2}{\pi}}\ x\right)+constant$$
the special function $T(X)$ is not referenced among the standard special functions, doesn't appear in the handbooks of special functions and is not implemented in the maths softwares.

One could say that just giving a name to an integral is no more than a cleaver trick. Nevertheless, one should think about it. A paper for general public on the subject : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales

Impossible to express in terms of elementary functions. The proof is quite clever, and you should give it a shot. If you can't figure it out, let me know and I'll post the solution.

Here's a hint: http://en.wikipedia.org/wiki/Combinatorial_species#Further_operations


If $\sum_{n=1}^{\infty} a_{n}$ converges, then$\sum_{n=1}^{\infty} \sin(a_{n})$ also converges If series $\sum_{n=1}^{\infty} a_{n}$ converges, prove series $\sum_{n=1}^{\infty} \sin(a_{n})$ converges  too.

Is this a series function problem or something related to?

I tried this:

If $a_{n} \ge 0 $ for all $n$, then $| \sin (a_{n})| \le a_{n}$, on the other hand $\sin(x)$ is continuous function in $[0,x]$ and differentiable in $(0,x)$ then exist $c \in (0,x)$ and
$$(x-0)\cos (c)= \sin(x)-\sin(0)$$
then
$$x\cos(c)= \sin(x)$$

but, $|\sin(x)|=|x\cos(c)|= |x| |\cos(c)| \le |x|$, thus $|\sin (x)| \le |x|$.

We know, $|\sin (a_{n})| \le |a_{n}| $ and $\sum_{n=1}^{\infty} a_{n}$ converges, then $\sum_{n=1}^{\infty} |\sin(a_{n})|$ converges. Therefore $\sum_{n=1}^{\infty} \sin(a_{n})$ converges.

But this proof use ${a_{n}}$ positive and in the original problem I don't have this hypotesis.

This is a "Community Wiki" answer recording a comment by Daniel Fischer under the question.  The comment provides a link The set of functions which map convergent series to convergent series to a proof that the result in question is false in general, though certainly true when the $a_n$ are non-negative. My reason for writing this answer is that comments can vanish more easily than answers and can probably also be more easily overlooked.

Write $ \sum_n \sin(a_n) = \sum_n a_n + \sum_n a_n(\frac{\sin(a_n)-a_n}{a_n}) $. Since $ \sum_n a_n $ converges, $ \lim_n a_n = 0 $. Thus, for $ n $ large enough, by a Taylor series expansion, $ |\frac{\sin(a_n)-a_n}{a_n}| = O(a_n^2) $. Hence, by the triangle inequality, $ |\sum_n \sin(a_n)| \leq | \sum_n a_n | + O(\sum_n |a_n|^3) $. Finally, we use the fact that if $ a_n $ is nonnegative, then $ \sum_n |a_n|^3 $ converges whenever $ \sum_n a_n $ converges.


Limit of $\frac{\sin(\theta)}{\theta}$ in degrees What does $\lim \limits_{\theta\to0}\dfrac{\sin(\theta)}{\theta}$ equal when $\theta$ is expressed in degrees?

I know that theta in degrees is $\frac{\pi}{180}$ theta radians, but I don't get the final answer of $0.01745$.

$$
\lim_{\theta\to0}\frac{\sin\theta^\circ}{\theta^\circ} = \lim_{\theta\to0} \frac{\sin\left( \dfrac{\pi\theta}{180}\text{ radians} \right)}{\theta} = \lim_{\eta\to0} \frac{\sin(\eta\text{ radians})}{\left( \dfrac{180\eta}{\pi} \right)} = \frac\pi{180}\lim_{\eta\to0} \frac{\sin\eta}{\eta}.
$$

This is why radians are used: When radians are used then $\lim\limits_{\eta\to0} \dfrac{\sin \eta} \eta=1$.

Postscript in response to comments below:

The question is how to find
\begin{align}
& \lim_{\theta\to0} \frac{\text{sine of $\theta$ degrees}}\theta = \lim_{\theta\to0}\frac{\text{sine function in radians}\left(\dfrac{\pi\theta}{180}\right)} \theta \\[12pt]
= {} & \frac\pi{180}\lim_{\theta\to0} \frac{\text{sine function in radians}\left(\dfrac{\pi\theta}{180}\right)} {\dfrac{\pi\theta}{180}} = \frac\pi{180}\lim_{\eta\to0} \frac{\text{sine function in radians}(\eta)}\eta.
\end{align}

In other words the notation "$\sin$" means a particular function: the one that maps a number $\eta$ to the sine of $\eta$ radians.

End of postscript

It's the same as the reason why $e$ is the "natural" base for exponential functions:
\begin{align}
\frac{d}{dx} 2^x & = (2^x\cdot\text{constant}) \\[10pt]
\frac{d}{dx} 3^x & = (3^x\cdot\text{a different consant}) \\[10pt]
\frac{d}{dx} 20^x & = (20^x\cdot\text{yet another constant})
\end{align}
etc.  Only when the base is $e$ is the "constant" equal to $1$.

The question is not properly posed.
In whatever units the is angle chosen the result is the same $(=1).$
For all units (degree, radian)
$$ lim_{X \rightarrow 0}\dfrac{\sin X}{X}$$
equals unity.
You should get that answer independently  when asked " how many radians are there in a degree?"


If a solution exists, how can I construct another solution with coprime integers? Suppose , $n$ is an odd positive integer and positive integers $a,b$ not being coprime satisfy the equation $$a^2-2b^2=n$$

  Can I always find a solution of $$c^2-2d^2=n$$ with coprime integers $c,d$ ?

I know that I can construct infinite many solutions from one solution of this pell-like equation, but the problem is that the usual procedure gives integers not being coprime , if I start with a solution with integers not being coprime. What can I do ?

No. Consider $n=9$.

There are non-coprime solutions
gotten by taking solutions to
$x^2-2y^2=1$
and multiplying them by $3$.

Suppose
$c^2-2d^2 = 9$.
Mod 3,
$d^2 = 0$ or $1$.
Therefore
$2d^2 = 0$
or $2$,
so $c^2 = 0$ or $2$.
This can only hold
if $c^2 = 0$,
so $c = 0$
mod 3.
Therefore
$d = 0$ mod 3
so $c$ and $d$
are not relatively prime.

Knowing the first solution $(x_0 ; y_0)$ for the equation. $$x^2-qy^2=n$$

It is necessary to use any solution of the Pell equation.  $$p^2-qs^2=1$$

Then knowing what some solution $(x_0 ; y_0)$ we can find the following for equation $x^2-qy^2=n$

$$x=px_0+qsy_0$$

$$y=sx_0+py_0$$


Do these two equalities hold? $$\mathbb P (x+y > N) = \mathbb P \left(x > \frac N2 \right) + \mathbb P \left(y > \frac N2 \right) \tag{1}$$

$$\mathbb P (\mid x+y-z \mid> N) = \mathbb P \left(\mid x-z \mid > \frac N2 \right) + \mathbb P \left(\mid y \mid > \frac N2 \right) \tag{2}$$

They seem to me unreasonable since for neither of them the two events on the left hand side of the equality are mutually exclusive. If there is one theorem similar to the above expressions, could anyone kindly show me? Many thanks in advance!

No, it doesn't hold. You can easily visualize that this is wrong. The region R1 = {(x,y)|x+y>N} is being incorrectly broken down into R2 = {(x,y)|x>N/2} and R3 = (x,y)|y>N/2}. For probability to be equal, R1 should be the union of R2 and R3 ( which is false, (0,3N/4) belongs to R3 but doesn't belong to R1. Further, R2 and R3 should be disjoint, clearly not true, (N,N) belongs to both. In fact RHS > LHS if you think about it.

Let $x$ and $y$ be uniformly distributed on $[N,N]$. (Yes, one point).  Then your equation says $1 = 1 + 1$, which seems unlikely.

Now let $x$ be distributed uniformly on $[N/2+1, N/2+1]$ and $y$ be distributed uniformly on $[0,0]$.  Then your equation is $0 = 0+1$.

Now suppose that $x$ and $y$ are both uniformly distributed on $[0,N]$.  The two probabilities on the right are $1/2$, as is the probability on the left.  (Draw a $[0,N] \times [0,N]$ square.  The probability on the left hand side is the fraction of the square above the line $x+y = N$, which is half the area.  The probabilities on the right hand side are the right half and upper half of the area of the square, respectively.)  Then your equation says $1/2 = 1$.

In fact, the diagram above suggests the truth : the set of outcomes with $x + y > N$ is a subset of the union $U = \{(x,y): x > N/2\} \cup \{(x,y): y > N/2\}$.  Therefore \begin{align*}
&P(x+y > N)  \\
&\leq  P(x > N/2) + P(y > N/2) - P(x > N/2 \text{ and } y > N/2)  \\
&\leq P(x > N/2) + P(y > N/2)  \text{,}
\end{align*}
where the middle uses inclusion-exclusion to only count the contributions from $U$ once.


Proof verification : $X$ and $Y=g(X)$ are independent random variables $\implies$ $Y$ is degenerate THE PROBLEM :

Source : Alan F. Karr (Probability), p.$96$, problem $3.10.(b)$

MY SOLUTION :

Suppose $B \subset \mathbb{R}$ such that $P\left\{Y \in B\right\}=1$. We want to show that $B$ is singleton. First of all $B \neq \phi$ since $g(0) \in B$. Assume, for sake of contradiction, $B$ is not singleton. Let $A \subset B$ such that $P\left\{Y \in A\right\}>0$ and $P\left\{Y \in B \setminus A\right\}>0$. Such a set $A$ can always be constructed because of the assumption. Note that

$$P\left\{X \in g^{-1}(A), Y \in A^c\right\} = 0 \neq P\left\{X \in g^{-1}(A)\right\} \cdot P\left\{Y \in B \setminus A\right\}$$

Both the terms in the right side are strictly positive by assumption $(P\left\{X \in g^{-1}(A)\right\}=P\left\{Y \in A\right\}>0)$. Hence, $X$ and $Y$ are not independent. Thus, we have arrived at a contradiction.

Please verify whether or not my solution is technically okay and if it can be improved in any regards. Thanks in advance.

For any measurable set $B$, we have
$$P(Y \in B) = P(\{Y \in B\} \cap \{X \in g^{-1}(B)\}) = P(Y \in B) P(X \in g^{-1}(B)) = P(Y \in B)^2,$$
so $P(Y \in B) \in \{0,1\}$, regardless of what $B$ is. Now consider $B$ of the form $(-\infty, y]$.

I may be using facts above what you currently know* but here goes:

Y is a distraction. We have that $X$ & $g(X)$ are independent. By definition, $$\sigma(X) \ \text{&} \ \sigma(g(X)) \ \text{are independent.}$$

$$\to \sigma(X) \ \text{&} \ \sigma(X) \ \text{are independent} \tag{Why? Hint: subset}$$

This means that $X$ is independent of itself and thus is constant or at least almost surely constant! Why?

Let $B \in \mathscr B$.

$$P(X \in B, X \in B) = P(X \in B)P(X \in B)$$
$$P(X \in B, X \in B) = P(X \in B)$$

Equating the RHS's, we have $P(X \in B) = 0,1$

I think this means $X$ is constant, but this certainly means that $X$ is almost surely constant i.e. $\exists d \ \in \ \mathbb R$, s.t.

$$P(X=d)=1$$

$$\to P(g(X)=g(d))=1 \tag{Why? Hint: subset}$$

$$\to P(Y=g(d))=1$$

Now choose $c=g(d)$. Then

$$P(Y=c)=1 \ \text{QED}$$

*Do you know Kolmogorov 0-1 Law?


How to prove this function to be identically zero I need help in this particular question due to the reason that I am not very comfortable in questions when integrals are to be proved identically zero.

Question Let $f:[0,1]\times[0,1]\longrightarrow[0,\infty)$ be a continuous function . Suppose that $$\int_{0}^{1}\bigg(\int_{0}^{1} f(x,y)dy\bigg)dx=0.$$ Then prove that $f$ is an identically zero function.

I think this would be due to the fact that range of $f$ is non-negative. But In such kind of questions I am not able to rigorously prove these. Actually, I was taught analysis by a really poor instructor and problem solving/assignments in our university are minimal. So, I use to try exercises by myself and ask questions here.
Can anyone please tell on how should I attempt this particular problem.
I shall be really thankful.

Hint :

Prove the result in the one-dimensional case : that is, if $\varphi : [0,1] \rightarrow \mathbb{R}$ is a non-negative and continuous function such that
$$\int_0^1 \varphi(x) \mathrm{dx} = 0$$
then $\varphi = 0$ (maybe you already know how to prove that).

Apply this result to the function $x \mapsto \int_0^1 f(x,y) \mathrm{dy}$ (check all the hypothesis correctly !) to deduce that for all $x \in [0,1]$, you have $\int_0^1 f(x,y) \mathrm{dy} = 0$.

Re-apply the result to the function $y \mapsto f(x,y)$ with $x$ fixed to deduce finally that for all $x,y$, you have $f(x,y)=0$.

Since the function is strictly non-negative we can interpret this integral as the volume under the graph of the function on the square $[0,1]\times [0,1]$. Now assume there exists some point $(x_0,y_0)$ such that $f(x_0,y_0) \neq 0$. Since the function is continuous there exists some small neighborhood of $(x_0,y_0)$ where all $f(x,y) > 0$ in that neighborhood, by the intermediate value theorem. Since this neighborhood is strictly positive, it must have some positive, non-zero volume. This is impossible because the volume is given as zero.


Is taking the derivative inside this integral acceptable? I have run into an integral where I take the partial derivative of x when the upper limit of the integral is x.  This is how I solve it but I am not sure if this is correct.  Can someone please check?  Thank you!
$$ \frac{\partial}{\partial x}\int_0^xf(t)dt=\int_0^x\frac{\partial f(t)}{\partial x}dt=0 $$

No.
Assuming that $x$ does not appear free in $f(t)$, then by the fundamental theorem of calculus:
$$\dfrac{\partial ~~}{\partial x}\int_0^x f(t)\mathrm d t=f(x)$$


What Does Homogenisation Of An Equation Actually Mean? For example, if we have a conic:

$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$

What does homogenising this equation with another line (say $ax + by + c = 0$ ) actually mean? As in, what are the graphical consequences of this? I've been told that it represents a pair of straight lines passing through the origin and the points of intersection (of the conic and the line), but how did this come about?

Update: I'm sorry for the lack of context, but I've just started algebraic geometry and I like understanding what things actually mean instead of mugging up a formula.

A function $f:\mathbb{R}^n \to \mathbb{R}$ is homogeneous of degree $r$ when it has the following property:

$$f(\lambda x_1,\lambda x_2,\ldots,\lambda x_n) = \lambda^r f(x_1,x_2,\ldots,x_n)$$

Examples of such functions include:

Linear functions, they are of degree 1. If you scale the graph of the function by a factor $\lambda$, you still get the same graph, except that all points have coordinates scaled up by the factor $\lambda$.
The monomials in one variable: $f:\mathbb{R} \to \mathbb{R}: x \mapsto x^n$ is homogeneous of degree $n$, scale $x$ by a factor $\lambda$ and the function will scale by a factor $\lambda^n$.
Any polynomial (function) in $n$ variables such that each term appearing in the polynomial is of degree $k$ is homogeneous of degree $k$. For instance $x^2 y + z^3$ is homogeneous of degree $3$.

The importance of homogeneity is the scale invariance of the functions. Which implies that the graphs of the functions will be scale invariant. Indeed, imagine a homogeneous function is used to define a geometrical object implicitly:

$$f(x_1,x_2,\ldots,x_n)=0$$

meaning all points with coordinates $(x_1,\ldots,x_n)$ that satisfy this equation will belong to the geometrical figure defined by $f$. If $f$ is homogeneous, it immediately follows that any multiple of these coordinates also satisfies the equations. In other words, any point that satisfies the equation immediately implies the entire ray going through that point and the origin of the space belong to the geometrical object.

Homogenizing an implicit polynomial equation means adding an extra variable $z$ and multiply any term by $z^k$ with $k$ such that the resulting polynomial is homogeneous. Of course, since any $z$-multiple of the polynomial will also be homogeneous, you choose the resulting homogeneous polynomial with smallest possible degree.

In your example, this would become

$$ax^2 + 2hxy + by^2 + 2gxz + 2fyz + cz^2 = 0$$

If you take the intersection of the geometrical figure defined by above equation with the plane $z=1$, you get back the original figure. If you choose another plane parallel with the $z=1$ plane, you get a scaled up or scaled down version of the figure.

Homogeneous equation should have [all] terms of equal powers..
For Example: $$x^2 + y^2 + 2xy = 0$$

the degree of $x^2$, $y^2$, and $xy$ is 2, hence the equation is the homogeneous equation of 2 degree.


Is most of the GM-AM Inequality in its codicil? Let’s define the codicil of the Geometric Mean – Arithmetic Mean Inequality to be the statement that if the means are equal, then all the terms are equal. Then: I conjecture that most of the GM-AM Inequality is actually in its codicil, in the sense that if you can use the codicil (that is, assume the codicil as part of the hypothesis), then it is a hundred times easier to prove the AM-GM Inequality.

 Rephrased by Y.F.

Is a proof along the following lines possible? Show that as long as not all variables are equal, there is a way to make them "more equal" which decreases the difference $\mathrm{AM}-\mathrm{GM}$. Deduce somehow that "the worst case" is when all variables are equal, in which case we know that $\mathrm{AM}=\mathrm{GM}$.

 Addendum by OP Mike Jones

I thought it was obvious that although the setting of this question is real analysis (specifically, inequalities), the germane issue is proof theory. I want to explore the interrelationships between the various parts of the hypothesis and conclusion of this celebrated theorem. I would have tagged it as “proof theory” in the first place, but didn’t see “proof theory” in my cursory glance at the tags. Now, I have found it buried under the tag “logic”. I was attempting to add the “logic” tag when I got thrown into the comment box, so, could someone who knows how please add this tag to the question? Also, if the answer to the question is affirmative, then it should also be tagged with “education” and “teaching”. About 80% of learning consists of simply acquiring (deep) familiarity, and so the teacher could pose the problem to the students, or to an especially talented/eager younger student: “Prove the GM-AM Inequality, but use its codicil as part of the hypothesis this first time around.”. Indeed, that is the original motivation for my question: how to make this celebrated result much more accessible to students. (I teach mathematics in high school.) It occurred to me that perhaps a great deal, even most, of the difficulty is tied up in the codicil, precisely as user “Moron” has described:
(GM = AM implies values equal) implies GM <= AM.

I found that this is too long for a comment, so I bailed out of the comment box and am putting this in as an addendum to the original question.

Oh, I see now how to do addtional tags:-)

Regards,
Mike Jones
American expatriate in Beijing
5.May.2011

One way to prove the AM-GM inequality along the lines suggested by my rephrasing is as follows. Denote the variables by $x_1,\ldots,x_n$, all positive. Suppose that it is not true that all variables are equal. There are thus variables $x_i,x_j$ which are different, say $x_i < x_j$. Suppose we brought them closer together by letting $x'_i = x_i + \epsilon, x'_j = x_j - \epsilon$ for some $0 < \epsilon < x_j - x_i$. This doesn't change the arithmetic mean but increases the geometric mean:
$$
x'_i x'_j = (x_i + \epsilon)(x_j - \epsilon) = x_i x_j + \epsilon (x_j - x_i - \epsilon) > x_i x_j.
$$
Note that any $\epsilon$ for which $x'_i \leq x'_j$ will do: such an $\epsilon$ satisfies the stronger requirement $\epsilon \leq (x_j-x_i)/2$.

Using $n-1$ such operations we can go from the original assignment to the constant assignment, in which the two means are the same. This shows that the geometric mean is always at most the arithmetic mean, with equality only if all variables have the same value.

The dual proof maintains the geometric mean while decreasing the arithmetic mean. Details left to the reader. A similar proof should also work for the harmonic mean - geometric mean inequality.

A different proof along similar lines runs as follows. Given $x_i \neq x_j$, modify the variables by averaging $x_i,x_j$. Since
$$ x_i x_j = \left(\frac{x_i + x_j}{2} + \frac{x_i - x_j}{2}\right) \left(\frac{x_i + x_j}{2} - \frac{x_i - x_j}{2} \right) = \left(\frac{x_i + x_j}{2}\right)^2 - \left(\frac{x_i - x_j}{2}\right)^2, $$
this operation increases the geometric mean while not changing the arithmetic mean. Intuitively, it's obvious that if we apply these operations successively then each variable tends to the arithmetic mean (this can be proven formally using a potential function argument). Since the geometric mean is continuous, the AM-GM inequality follows.

Affirmative.

It is indeed very easy to prove the “main part” of the celebrated GM-AM Inequality from its codicil (and by appealing to the continuity of the functions involved).

Note that if the GM-AM Inequality holds for positive integers, then it is trivially easy to show that it also holds for positive rational numbers (-for if AM < GM, then w*AM < w*GM, where w is the product of the denominators. w can then be pulled into each expression, yielding positive integers for which the GM-AM Inequality fails, a contradiction). Furthermore, if the GM-AM Inequality holds for all positive rational numbers, the by continuity (of the functions involved), it holds for all positive real numbers. Hence, the whole edifice can be made to rest on whether the GM-AM Inequality holds for positive integers. But if we can assume its codicil, proving that it holds for positive integers is very easy to do, as shown below.

Codicil Axiom. If n is a positive integer and x(1), …, x(n) are positive real numbers such that GM(x(1),…,x(n)) = AM(x(1), …,x(n)), then x(1) = x(2) = … = x(n).

Theorem. If n is a positive integer and x(1), …, x(n) are positive integers, then GM(x(1),…,x(n)) <= AM(x(1),…,x(n)).

Proof: By contradiction. Suppose that there exists a positive integer n and positive integers x(1),…, x(n) such that AM(x(1),…,x(n)) < GM(x(1),…,x(n)). Since it is trivially easy to establish the GM-AM Inequality for n = 1 and n = 2 we can assume, wlog, that n > = 3. Furthermore, in any such case, we can assume, wlog, that x(1) <= … <= x(n).
Let M(0) be the set {(x(1),…,x(n)) | x(1), …, x(n) are positive integers such that x(1) <= … <= x(n) and AM(x(1),…,x(n)) < GM(x(1),…,x(n)).
By hypothesis, M(0) is non-empty. Since its members are positive-integer-based, M(0) has a (lexicographically) “smallest” element. We want to find it. To that end, we squeeze down on it by means of the following definition.
For each positive integer k <= n, let M(k) be the set {(x(1),…,x(n) | (x(1),…,x(n))  M(k – 1) and for each (y(1),…y(n))  M(k – 1), x(k) <= y(k)).
Then M(n) contains exactly one element, which we will denote by (x(1),…,x(n)). (It is the lexicographically smallest element of M(0) that we were looking for.) And, of course, it is the case that x(1) < x(n). (It is important to bear in mind throughout that x(1) <= … <= x(n).)

Case 1. x(1) > 1.
Change the value of x(1) to x(1) – 1 in a manner that is monotone and continuous. Then the relation AM < GM will have changed to GM <= AM by, or at, the time x(1) – 1 is reached (by the minimality of x(1), and by the fact that x(1) – 1 is a positive integer, since x(1) is a positive integer > 1). But for this to happen, the relation GM = AM must (by continuity of the functions involved) be the case somewhere along the way. However, it cannot happen in the open interval (x(1) – 1,x(1)), because that is a non-integer value, and, by the Codicil Axiom, x(2), …, x(n), which are all integers (with n > 2) would have to be equal to it, which would be a contradiction. Thus equality of means must occur for x(1) – 1. That is, GM(x(1) – 1,…,x(n)) = AM(x(1) – 1,…,x(n)). Then by the Codicil Axiom, x(n) = x(1) – 1. Then x(n) < x(1), a contradiction.

Case 2. x(1) = 1.

Case 2.1. x(n) > 2.
Change the value of x(n) to x(n) – 1 in a manner that is monotone and continuous. Then the relation AM < GM will have changed to GM < AM by, or at, the time x(n) – 1 is reached. But for this to happen, the relation GM = AM must (by the continuity of the functions involved) be the case somewhere along the way. However, it cannot happen in the open interval (x(n) – 1,x(n)), because that is a non-integer value, and, by the Codicil Axiom, x(1), …, x(n – 1) (with n > 2), which are all integers, would have to be equal to it, which would be a contradiction. Thus equality of means must occur for x(n) – 1. That is, GM(x(1),…,x(n – 1),x(n) – 1) = AM(x(1),…x(n – 1),x(n) – 1). Then by the Codicil Axiom, x(1) = x(n) – 1. But x(n) – 1 > 1, since x(n) > 2. Thus x(1) > 1, a contradiction.

Case 2.2. x(n) = 2 and x(n – 1) = 2.
Moving x(n) from 2 to 1 forces x(n – 1) to be 1, a contradiction.

Case 2.3. x(n) = 2 and x(n – 1) = 1.
In this case, remembering that x(1) <= … <= x(n – 1) <= x(n), the relation GM < AM can easily be established by direct calculation.

This covers all the cases, thereby establishing the result.

(end of proof)

Now, assuming I haven’t blundered somewhere in this attempt. I ask you: Isn’t this a hundred times easier than any of the proofs of the full result?

Regards,
Mike Jones


Solution of $\frac{d^2y}{dx^2} - \frac{H(x) y}{b} = H(-x)$ Does the equation

$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(x)$$

have a solution where $H(x)$ is the Heaviside step function and $b$ and $c$ are constant?

Update: What about the second step function be $H(-x)$:
$$\frac{d^2y}{dx^2} - \frac{H(x)}{b} y = c H(-x)$$

For $x<0$, $y''=0$, i.e. $$y=C_1x+C_0.$$

For $x\ge0$, $y''-\frac1by=1$.

Assuming $b>0$, the solution of the homogenous equation is
$$y=A\exp(\frac x{\sqrt b})+B\exp(-\frac x{\sqrt b}),$$
and a particular solution is
$$y=-b.$$
Hence the general solution,
$$y=A\exp(\frac x{\sqrt b})+B\exp(-\frac x{\sqrt b})-b.$$

You will ensure continuity of the function by equating the function and first derivative at $x=0$:
$$C_0=A+B-b\\C_1=\frac A{\sqrt b}-\frac B{\sqrt b}.$$

In particular, starting from the steady state, $A=B=\frac b2$, and
$$y=b\left(\cosh(\frac x{\sqrt b})-1\right)H(x).$$

I am not sure wether this answer is correct or not, but as far as the step function is only a sign, I brought it out of the integration.

Please note that, for finding the constants, you need the boundary values and compare the value of the function at the point $x=0$.

$$\begin{array}{l}\frac{{{d^2}y}}{{d{x^2}}} - \frac{{H(x)y}}{b} = H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \frac{{H(x)y}}{b} + H(x)\\\frac{{{d^2}y}}{{d{x^2}}} = \left( {\frac{y}{b} + 1} \right)H(x)\\\frac{{{d^2}y}}{{\left( {\frac{y}{b} + 1} \right)}} = H(x){d^2}x\\\int {\frac{b}{{y + b}}{d^2}y}  = H(x)\int {{d^2}x} \\\int {\left\{ {bLn\left( {y + b} \right) + {c_1}} \right\}dy}  = H(x)\int {x + {c_3}dx} \\b\left\{ {\left( {y + b} \right)Ln\left( {y + b} \right) - \left( {y + b} \right)} \right\} + {c_1}y + {c_2} = H(x)\left\{ {\frac{1}{2}{x^2} + {c_3}x + {c_4}} \right\}\end{array}$$


A subset of an infinite-dimensional simplex where no element is in the convex hull of other elements of the subset. Let
$$S=\{x\in R_{+}^\infty \mid \sum_{n=1}^\infty x_n=1\}$$
be an infinite simplex.
Suppose $\hat{S}$ is a countable subset of $S$ such that

for each $x\in \hat{S}$ only at most finitely many coordinates are non-zero, denote the set of those coordinates $N(x)$;
for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\neq N(x’)$.

How could one prove that no $x$ is in a convex hull of other elements of $\hat{S}$, i.e., no $x\in \hat{S}$ can be represented as a convex combination of other elements of $\hat{S}$?
EDIT: Condition (2) should be read: "for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\triangle (\cup_{x'\in \hat{S}}N(x’)) \neq \emptyset$". Here $\triangle$ stands for a symmetric difference of sets.
Thank you! (These two infinities intimidate me...)

No countable set is convex [unless it is empty] (since in a convex set, an interval between two points is contained in the set, and an interval has uncountably many points. However, your statement after the i.e. has nothing to do with convexity.


What is $\mathbb R^\omega$? I have seen  $\mathbb R^\omega$ mentioned in my topology texts but cannot find where $\omega$ is defined. Could someone please tell me what it means in comparison to $\mathbb R^n$?

tl; dr;

$$X^\omega = X\times X\times\cdots$$

is an infinite countable product of $X$, i.e. the set of all countable sequences over $X$.

Longer explanation: so $\omega$ is an ordinal number defined as the ordinal number of naturals $\mathbb{N}$. If we use the Von Neumann definition of ordinals then for an ordinal number $k$ we have a very straight forward generalization of finite Cartesian product:

$$X^k=\{f:k\to X\ |\ f\text{ is a function}\}$$

and so you can think about $X^k$ as the set of all ordered "sequences" with values in $X$. For finite $n$ this coincides with classical finite product because $n$ as a Von Neumann ordinal is recursively defined as $\{0,1,2,\ldots,n-1\}$ with $0=\emptyset$. For $\omega$ (which in Von Neumann case is simply $\mathbb{N}$) this coincides with the usual notion of sequences. For uncountable ordinals this probably goes beyond any intuition. Nevertheless typically by $X^\omega$ people understand the countable infinite product $X\times X\times\cdots$.

Note that there are alternative (and arguably simplier) definitions of an infinite Cartesian product, e.g. the set of all choice functions.

The cardinality of $\Bbb N$, which is usually written $\aleph_0$ (the smallest infinite cardinal), is sometimes/often also written $\omega$. In other words

$$|\Bbb N| = \aleph_0 = \omega$$

Hence $\Bbb R^{\omega} = \Bbb R \times \Bbb R \times \Bbb R \times \dots$ is the cartesian product of countably infinitely many $\Bbb R$.


Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even. Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even. And if $n = 2k$, then $A$ is similar over the field of real numbers to a matrix of the block form

$$\begin{bmatrix}
    0      & -I \\
    I & 0 \\
\end{bmatrix}$$
where $I$ is the $k \times k$ identity matrix.

I have done the first part. Since $\det(A^2) = \det (-I)$ we must have $\det (-I)$ non negative and hence $n$ must be even.

Need Help in the second part.

Thank You.

You can solve both parts of the question at the same time as follows: Given $A$, define a map $\mathbb C\times\mathbb R^n\to\mathbb R^n$ by $(u+iv,x)\mapsto (u+iv)\cdot x:=ux+vAx$. Evidently, this is linear over $\mathbb R$ in both components and satisfies $(1,x)\mapsto x$. Using that $A^2=-I$, one easily verifies that it is multiplicative, i.e. $z\cdot (w\cdot x)=(zw)\cdot x$. Hence is satisfies all properties required from scalar multiplication, and hence makes $\mathbb R^n$ into a complex vector space, so $n=2m$ for some $m\in\mathbb N$. Moreover, choosing a complex basis $\{x_1,\dots,x_m\}$ for this vector space, $\{x_1,\dots,x_m,i\cdot x_1,\dots,i\cdot x_m\}$ is a real basis for $\mathbb R^n$. By definition the matrix representation of $A$ with respect to this real basis is the block matrix you have indicated.

I solve it another method : $det(A^2)=det(-I)=(-1)^n ....(1)$.

Now $\lambda$ is an eigenvalue of $A$ then ${\lambda}^2+I^2=0\Rightarrow \lambda =i,-i.$ So $detA=i(-i)=1\Rightarrow det(A^2)=(detA)^2=1 ....(2)$

So from $(1)$ and $(2)$ we get $n=2m$ for dome $m\in \mathbb{N} $