Do users of RTL languages adopt an LTR standard for mathematics (in the same way they often do when using LTR words or phrases in RTL text)? Non-mathematician here. There is a discussion on this forum titled "Is “applying similar operations from left to right” a convention or a rule that forces us to mark one answer wrong?" I found it trying to answer a question I have. I could not comment as I am new here (trolling protection I guess)
My interest is software localisation. My question is whether mathematics is globally written Left to Right (LTR). i.e. do those substantial countries that use a RTL languages adopt an LTR standard for mathematics (in the same way they often do when using LTR words or phrases).

Note that I am not asking what is mathematically correct (i.e. use parenthesis properly) - I am asking what is commonly actually done?
Thanks

Most university-level math education has conformed to a left-to-right standard, regardless of how the native language is written. However, pre-university education differs, and it depends on the region.

For instance, see here: http://en.wikipedia.org/wiki/Modern_Arabic_mathematical_notation.

One of the reasons for the predominance of left-to-right mathematical writing is that a majority of mathematical papers are written in left-to-right languages. Furthermore, it is difficult to find equivalent texts for some (advanced) topics written in a right-to-left sense. Even translation is particularly difficult. While there might be a translation for "limit" or "derivative" in some languages, there often isn't a direct translation for something like "cotangent bundle" or "Hom functor." How does one translate an advanced text when the nomenclature is so-far removed from the native language? (One may even argue that some of the nomenclature is pretty far removed from English, as well. "Eigenvector" is a horrible Frankenstein's monster of a word, grammatically speaking. And don't even get me started on "homomorphism" vs. "homeomorphism").

In fact, this phenomenon has led to English becoming almost mandatory for university-level technical education in many countries.

Just put in some parentheses, and then you don't have to worry about the LTR-vs-RTL issue. The expression $(a-b)-c$ means the same thing everywhere.


Examples of patterns that eventually fail Often, when I try to describe mathematics to the layman, I find myself struggling to convince them of the importance and consequence of "proof". I receive responses like: "surely if Collatz is true up to $20×2^{58}$, then it must always be true?"; and "the sequence of number of edges on a complete graph starts $0,1,3,6,10$, so the next term must be 15 etc."

Granted, this second statement is less logically unsound than the first since it's not difficult to see the reason why the sequence must continue as such; nevertheless, the statement was made on a premise that boils down to "interesting patterns must always continue".

I try to counter this logic by creating a ridiculous argument like "the numbers $1,2,3,4,5$ are less than $100$, so surely all numbers are", but this usually fails to be convincing.

So, are there any examples of non-trivial patterns that appear to be true for a large number of small cases, but then fail for some larger case? A good answer to this question should:

be one which could be explained to the layman without having to subject them to a 24 lecture course of background material, and
have as a minimal counterexample a case which cannot (feasibly) be checked without the use of a computer.

I believe conditions 1. and 2. make my question specific enough to have in some sense a "right" (or at least a "not wrong") answer; but I'd be happy to clarify if this is not the case. I suppose I'm expecting an answer to come from number theory, but can see that areas like graph theory, combinatorics more generally and set theory could potentially offer suitable answers.

I'll translate an entry in the blog Gaussianos ("Gaussians") about Polya's conjecture, titled:

A BELIEF IS NOT A PROOF.

  We'll say a number is of even kind if in its prime factorization, an even number of primes appear. For example $6 = 2\cdot 3$ is a number of even kind. And we'll say a number is of odd kind if the number of primes in its factorization is odd. For example, $18 = 2·3·3$ is of odd kind. ($1$ is considered of even kind).

  Let $n$ be any natural number. We'll consider the following numbers:

  $E(n) =$ number of positive integers less or equal to $n$ that are of even kind.
  $O(n) =$  number of positive integers less or equal to $n$ that are of odd kind.

  Let's consider $n=7$. In this case $O(7) = 4$ (number 2, 3, 5 and 7 itself) and $E(7) = 3$ ( 1,  4 and 6). So $O(7) >E(7)$.

  For $n = 6$: $O(6) = 3$ and $E(6) = 3$. Thus $O(6) = E(6)$.

  In 1919 George Polya proposed the following result, know as Polya's Conjecture:

  For all $n > 2$, $O(n)$ is greater than or equal to $E(n)$.

  Polya had checked this for $n < 1500$. In the following years this was tested up to $n=1000000$, which is a reason why the conjecture might be thought to be true. But that is wrong.

  In 1962, Lehman found an explicit counterexample: for $n = 906180359$, we have $O(n) = E(n) – 1$, so:

  $$O(906180359) < E(906180359).$$

  By an exhaustive search, the smallest counterexample is $n = 906150257$, found by Tanaka in 1980.

  Thus Polya's Conjecture is false.

  What do we learn from this? Well, it is simple: unfortunately in mathematics we cannot trust intuition or what happens for a finite number of cases, no matter how large the number is. Until the result is proved for the general case, we have no certainty that it is true.

The partition numbers https://oeis.org/A000041 2,3,5,7,11, and then 15 instead of 13.


How to write $\pi$ as a set in ZF? I know that from ZF we can construct some sets in a beautiful form obtaining the desired properties that we expect to have these sets. In ZF all is a set (including numbers, elements, functions, relations, etc...).

For example we can define a copy of $\mathbb{N}$, using the empty set axiom and others, called sometimes $\omega$ and defining,

\begin{align}
0_\mathbb{N}&=\{\}\\
1_\mathbb{N}&=\{\{\}\}\\
2_\mathbb{N}&=\{\{\},\{\{\}\}\}\\
3_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\\
4_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\\
&\ \vdots
\end{align}

Well, ZF allow us to build this type of sets that are kind of arrangements of brackets and commas. We can follow with $\mathbb{Z}$ since we can define $(a,b)=\{\{a\},\{a,b\}\}$:

$\mathbb{Z}$ is defined as the set of equivalence classes $\mathbb{Z}=(\mathbb{N}\times\mathbb{N})\big/\sim$ where

$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{N}\times\mathbb{N})\times (\mathbb{N}\times\mathbb{N}):(m+_\mathbb{N} k)= (h+_\mathbb{N} n)\}$$

here, the integers are sets more complicated than natural numbers. For example,

\begin{align}
-2_\mathbb{Z}&=\{(1_\mathbb{N},3_\mathbb{N}),(2_\mathbb{N},4_\mathbb{N}),(3_\mathbb{N},5_\mathbb{N}),\ldots,(n_\mathbb{N},(n+2)_\mathbb{N}),\ldots\}\\
&=\{\{\{1_\mathbb{N}\},\{1_\mathbb{N},3_\mathbb{N}\}\},\{\{2_\mathbb{N}\},\{2_\mathbb{N},4_\mathbb{N}\}\},\ldots\}\\
-2_\mathbb{Z}&=\{\{\{\{\{\}\}\},\{\{\{\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\},\\
&\quad\{\{\{\{\},\{\{\}\}\}\},\{\{\{\},\{\{\}\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\}\},\ldots\}\\
\end{align}

Here we note the importance of notations. We continue with  $\mathbb{Q}=(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\big/\sim$ where

$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\times (\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\}):m\odot_\mathbb{Z}k=h\odot_\mathbb{Z}n\}$$

And for example:

$$(0.2)_\mathbb{Q}=\{(1_\mathbb{Z},5_\mathbb{Z}),(2_\mathbb{Z},10_\mathbb{Z}),(3_\mathbb{Z},15_\mathbb{Z}),\ldots,(n_\mathbb{Z},(5n)_\mathbb{Z}),\ldots\}$$

Imagine if we write the integer numbers as before and write the ordered pairs in unabridged form ($(0.2)_\mathbb{Q}$ is a nice abbreviation for this monster). However we can.

Finally we define $\mathbb{R}$ as the set of all Dedekind cuts, for example:

$$(0.2)_\mathbb{R}=\{x\in\mathbb{Q}:x<_\mathbb{Q} (0.2)_\mathbb{Q}\}$$

Note that $(0.2)_\mathbb{R}$ is even more monstrous than $(0.2)_\mathbb{Q}$. Also I can write $(\sqrt{2})_\mathbb{R}$ showing its elements in a simple form,

$$(\sqrt{2})_\mathbb{R}=\{x\in\mathbb{Q}:(x^2<_\mathbb{Q} 2_\mathbb{Q}) \lor (x<_\mathbb{Q} 0_\mathbb{Q})\}$$

But I don't know how to do it with $\pi_\mathbb{R}$, since

$$\pi=\lim_{k\to\infty}\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}$$

I only know that

$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{R}$$

Since monotonically converge to $\pi$ we have

$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left\{x\in\mathbb{Q}:x<_\mathbb{Q} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{Q}\right\}$$

Is there any way to avoid infinite union (and
the choice of a particular convergent sequence
) as the case of $(\sqrt{2})_\mathbb{R}$? If not, why?

Can we write the set representing $\pi$ listing its elements as we do with integers or rationals (as $\mathbb{Q}$ is countable I guess that should be able to do, but I don't know how to do)?

If in ZF all is a set, is so surprising the fact that so many things can be defined, then my last question is

How many more things can be built using ZF?, ZF could define us what is a derivative, an integral, a limit or a measure?

Thanks in advance.

Note that when we define the natural numbers we have a good sense of addition and multiplication (ordinal arithmetics), and from those we can define the operations on $\Bbb Z$ and $\Bbb Q$ and then by using Dedekind cuts construction we can extend these to $\Bbb R$ as well.

So we have that $\Bbb R$ has the operations $+,\cdot$ and they all satisfy all the things we know they do from the times we did mathematics without writing all the sets explicitly.

Now we can use these things to start and define anything else that we desire using the $+$ and $\cdot$ and whatnot as our stones. For example you can define $\pi$ to be the length of the semi-circle of radius $1$.

How do we do that? We define what is an integral, and a path integral, and so on. All from the sets which are addition and multiplication and so on, and then we can define $\pi$ in a painfully tedious way.

The whole point of using set theory, and in this case $\sf ZF$, as our foundation is that we can do things, once we can define the real numbers with their basic properties we have formulas which define things from that structure, and we don't have to write everything in set-form explicitly.

Once we have the real numbers (with the order) it is easy to define the collection of open intervals, and then it is easy to define the standard topology (the smallest collection containing the intervals and having certain properties), from there we can define the Borel sets, the Lebesgue sets, and the Lebesgue measure (being the unique function from the Lebesgue sets into the real numbers which satisfies certain properties), then we can define integration and with respect to the measure, and we can define derivation.

All these things end up being immensely long and complicated formulas, but the point is that we can write them up. And all this with just $\in$ and the axioms of $\sf ZF$. (Although we may want to add $\sf DC$ or even $\sf AC$ if we discuss measure theory.)

But if you do want to insist on $\pi$ being written in set form:

$$\pi = \left\{x\in\Bbb Q\mathrel{}\middle|\mathrel{} x<_\Bbb Q0\lor \left(x\geq_\Bbb Q0\land\exists k\in\Bbb N:\frac{x^2}6<_\Bbb Q\sum_{n=1}^k\frac1{n^2}\right)\right\}$$

I wrote the original equation in Word and added it here


Calculate the degree of convergence I was studying about numerical methods to find root of functions. In the False Position Method, I came to the following claim:
$$\text{Degree of convergence } = P = \frac{1+\sqrt5}{2}$$
I cannot find any proof about it. Please help.

The claim is wrong. This rate of convergence applies only to the secant method. The pure vanilla regula falsi applied to monotone convex functions will develop a one-sided behavior, where only one side of the guard interval is moved. This can be easily checked on a diagram with paper and pencil. Convergence in this case is linear, order 1.

That is why the active point modifications were invented, the most simple being the Illinois variant. They restore superlinear convergence, but I'd think that a proof would be rather technical.

Here's a proof of the rate of convergence for a similar method.  Perhaps you can use a similar proof for your purposes.


Limit Question: a better way to solve this limit? where $x \to -\infty$ Hey guys so I have this limit:

$$\lim_{x \to -∞} f(x) = {(x+\sqrt{x^2+2x})}$$

I solved it by multiplying numerator and denominator by $$x-\sqrt{x^2+2x}$$ and got $-1$ as my answer, but I really don't like how I solved it; any better way to solve it?

You are looking, setting $t=|x|$, at
$$
\lim_{t\to\infty} \sqrt{t^2-2t} - t
$$
For the sake of it, here is a solution using the Taylor expansion of $\sqrt{1+u}$ around $0$: for $t>0$,
$$
\sqrt{t^2-2t} - t
= t\sqrt{1-\frac{2}{t}} - t
= t\left(\sqrt{1-\frac{2}{t}} - 1\right)
= t\left(1-\frac{1}{t} + o\left(\frac{1}{t}\right) - 1\right)
= -1+o(1)
$$
showing the limit is indeed $-1$.

I suggest to flip the sign by $y=-x\to \infty$ then

$$\lim_{x \to -∞} f(x) = {x+\sqrt{x^2+2x}}=\lim_{y \to \infty} f(x) = {-y+\sqrt{y^2-2y}}$$

and then

$$\sqrt{y^2-2y}-y=\left(\sqrt{y^2-2y}-y\right)\cdot\frac{\sqrt{y^2-2y}+y}{\sqrt{y^2-2y}+y}$$

Edit As another alternative we can use binomial series $t\to 0, \,(1+t)^n=1+nt+O(t^2)$, are you aware about that?

We proceed as follows

$$\sqrt{y^2-2y}=\sqrt{y^2}\cdot \sqrt{1-2/y}=y\cdot(1-2/y)^\frac12=y\cdot(1-1/y+O(1/y^2))=y-1+O(1/y),$$

then

$$\sqrt{y^2-2y}-y=y-1+O(1/y)-y=-1+O(1/y) \to -1$$

Indeed the $O(1/y)$ notation collects all the terms in absolute value eventually smaller or equal than $c/y$ with $c$ some positive constant then

$$|O(1/y)|\le \frac c y \to 0 \implies O(1/y)\to 0$$


Proof that some inverse limit is connected Consider the inverse limit
$$\varprojlim \mathbb{R}/n\mathbb{Z}$$
with respect to the system $(\mathbb{R}/n\mathbb{Z})_{n \geq 1}$ of additive groups togehter with the projections $\pi_{m,n} \colon \mathbb{R}/m\mathbb{Z} \to \mathbb{R}/n\mathbb{Z}$ whenever $n \mid m$.

We equip each $\mathbb{R}/n\mathbb{Z}$ with the quotient topology
 and $\varprojlim \mathbb{R}/n\mathbb{Z}$ with the induced topology from
$\prod_{n \geq 1} \mathbb{R}/n\mathbb{Z}.$

My question: How can one prove that $\varprojlim \mathbb{R}/n\mathbb{Z}$ is a connected topological space?

I see that each $\mathbb{R}/n \mathbb{Z}$ is connected, hence the product $\prod_{n \geq 1} \mathbb{R}/n\mathbb{Z}$ is connected. But I do not know how to proceed from here.

The connectedness can be proved as follows. By a continuum we mean a connected compact Hausdorff space.

Every intersection of a downwards directed family of continua is a continuum. For simplicity, imagine just a sequence of continua $X_0 ⊇ X_1 ⊇ \cdots$ and put $X := ⋂_{n ∈ ω} X_n$. Clearly, $X$ is Hausdorff compact. It is also connected. Otherwise $X = A ∪ B$ for some nonempty disjoint subsets $A$, $B$ that are clopen in $X$ and so closed in $X_0$. Since $A$ and $B$ are compact and $X_0$ is Hausdorff, there are disjoint open sets $U, V ⊆ X_0$ such that $A ⊆ U$ and $B ⊆ V$. We have that $(X_n)_n$ is a decreasing sequence of compacta such that $⋂_n X_n ⊆ U ∪ V$. Hence, for some $n$ we have $X_n ⊆ U ∪ V$. Since $X_n ∩ U ⊇ A ≠ ∅$ and similarly $X_n ∩ V ≠ ∅$, we have that $X_n$ is disconnected.
Every limit of a directed inverse system is homeomoprhic to the intersection of subspaces of the product, and the subspaces may be taken homeomorphic to products of some of the original spaces. Again, imagine a sequence as before. We have $\varprojlim (X_n, f_{n,m}) = \{x ∈ ∏_n X_n: x_n = f_{n,m}(x_m)$ for every $n ≤ m\} = ⋂_k Y_k$ where $Y_k= \{x ∈ ∏_n X_n: x_n = f_{n,m}(x_m)$ for every $n ≤ m ≤ k\} \cong ∏_{n ≥ k} X_k$.

By combining the two facts we obtain the connectedness since our spaces are circles. Also note that we don't need a general proof for directed systems since our system has a cofinal subsequence.

Consider $C$ the subset of $\mathbb{N}^{\mathbb{N}}$ the set of maps $\mathbb{N}\rightarrow \mathbb{N}$ such that for every $u\in C$, $u(n)= u(m)$ mod $n$ if $n$ divides $m$ endowed with the discrete topology,  there exists a continuous map $f:C\times \mathbb{R}\rightarrow \mathbb{R}^{\mathbb{N}}$ defined by $f(u,x)=(x+u(1),..,x+u(n),..)$ the image of the composition of $f$ with $\mathbb{R}^{\mathbb{N}}\rightarrow \Pi_n\mathbb{R}/n\mathbb{Z}$ is $\varprojlim\mathbb{R}/n\mathbb{Z}$. We deduce that $\varprojlim\mathbb{R}/n\mathbb{Z}$ is connected since $C\times \mathbb{R}$ is connected.


Proof that every polygon with an inscribed circle is convex? In many elementary (and not-so-elementary) Euclidean geometry texts, a (simple) polygon is said to be tangential  if it is convex and has an inscribed circle (i.e., a circle that intersects and is tangent to each side of the polygon).  The assumption of convexity is not needed: I've come up with a rather laborious proof that every polygon with an inscribed circle is convex.  But I'd like to find either a simple elementary proof or a reference to a proof in the literature.  (By "elementary," I mean using only standard facts of axiomatic Euclidean geometry.)

Does anyone know of a reference for a proof of this fact (elementary or not)?  Or can anyone think of a straightforward elementary proof?  You can use any definition of "convex polygon" that you like, but the easiest one to work with is that for each edge, the vertices not on that edge lie on one side of the line through that edge.

(Interestingly, the corresponding fact for circumscribed circles--i.e., that every polygon with a circumscribed circle is convex--is quite easy to prove: If P has a circumscribed circle, any two nonadjacent sides of P are non-intersecting chords of the circle; it is easy to show that both endpoints of each chord lie on the same side of the line through the other, and from there it is an easy matter to prove that P is convex.)

Thanks to everyone who suggested approaches to this problem. In the end, none of the suggested approaches fit into the axiomatic framework that I was working in, so I had to write up my own rather laborious proof. It's a bit long to post here, but in order to close this question, I just want to post the reference and a quick summary of the approach.

You can find the complete proof in my textbook Axiomatic Geometry (Theorem 14.31). The basic idea is first to prove the following lemma:

  Lemma. Let $\mathscr P$ be a polygon circumscribed about a circle $\mathscr C$. Suppose $A$ is any vertex of $\mathscr P$, and $E$ and $F$ are the points of tangency of the two edges containing $A$. Then there are no points of $\mathscr P$ in the interior of $\triangle AEF$.

To prove that a tangential polygon $\mathscr P$ must be convex, the basic idea is to show that if $\ell$ is any edge line of $\mathscr P$, then there can't be any vertices of $\mathscr P$ on the "wrong" side of $\ell$ (the side not containing the inscribed circle), because that would violate the lemma.

Here is the sketch of a much shorter proof.

If a (simple) polygon is tangential, then it is convex (all its angles are each less than two right angles). Let the polygon be $n$-sided. Then draw radii from the centre of the circle to the $n$ points of tangency (by hypothesis). This splits the $n$-gon into $n$ quadrilaterals which are kites (that they are kites follows from the $RHS$ congruency test, after having connected each vertex of the $n$-gon to the centre of the circle). Each of these kites has a vertex of the $n$-gon as a vertex and (by a well-known result of Euclidean Geometry about tangent straight lines to circles and radii thereof) two right angles. The last angle of each kite is formed at the centre of the circle by two radii. Since the sum of angles in a quadrilateral is four right angles, it is clear that the two non-right angles in each quadrilateral must be supplementary, so that either one of them is always less than two right angles. The result follows.


What's a proof that the angles of a triangle add up to 180°? Back in grade school, I had a solution involving "folding the triangle" into a rectangle half the area, and seeing that all the angles met at a point:

However, now that I'm in university, I'm not convinced that this proof is the best one (although it's still my favourite non-rigorous demonstration). Is there a proof in, say, linear algebra, that the sum of the angles of a triangle is 180 degrees? Or any other Euclidean proofs that I'm not aware of?

Here's a decent Euclidean proof:

Let $x$ be the line parallel to side $AB$ of $\triangle ABC$ that goes through point $C$ (the line is unique because of the fifth postulate). $AC$ cuts $x$ and $AB$ at the same angle, $\angle BAC$ (corollary of the fifth postulate). $BC$ cuts $x$ and $AB$ at the same angle, $\angle ABC$. These two angles and the final angle $\angle ACB$ form a straight angle on $x$, which is always  $180^\circ$ (corollary of the third postulate).

If we have a Triangle with vertices A,B,C

As we can see in the Triangle, if a man starts from Vertices A and goes in the direction of the line extended from Vertices A, suddenly decides to go towards Vertices B, Meaning he has to turn in Anti-Clockwise Direction for Angle(PI - A). now he walks to Vertices B, then suddenly he wants to go towards C and turns  in Anti Clockwise Direction for Angle(PI - B) and moves along C, finally he does the same and turn towards A with a Rotation of Angle(PI - C), So completes One Revolution or 2*PI Radians, or the sum of all the Three Rotations the man took to complete one Perimeter walk on the Triangle Walk or an equivalent of Angle(3*PI -(A+B+C)).

Meaning we Reach an Equation :
Angle(2*PI) = Angle(3*PI -(A+B+C))
=> Angle(A+B+C) = Angle(2*PI)

Assumptions Taken to Calculate This result :

A rotation of Angle(PI) is required to move in an exactly opposite direction.
Obviously to come to the same line of sight we need to rotate an angle of Angle(2*PI).

I think i can safely assume these and get to the result.


Let $T: V \to V$ be a linear map such that $T^2-3T+2I=0$. Let $T: V \to V$ be a linear map such that $T^2-3T+2I=0$, where $I$ is the identity map.

question:

a) Prove that $V=\ker(T-2I) \oplus\ker(T-I)$

b) let $A$ be an $n \times n$ matrix such that $A^2-3A+2I_n=0$ Where $I_n$ is the $n\times n$ identity matrix. True or false: $A$ is diagonalizable

I attempted $\ker(T-2I)=(T-2I)V=0$

$\ker(T-I)=(T-I)V=0$

I know the direct sum should be the join of $\ker(T-2T)$ and $\ker(T-I)$ is $0$, and I am not sure how to prove it

I am a first year student from McGill U. I am doing linear mapping on linear algebra. The textbook I am using is Linear Algebre edition sixth by SEYMOUR LIPAXHUTZ

Some hints: To show $V= \ker(T-2I) \oplus \ker(T-I)$, you need to show that

(i) $\ker(T - 2I) \cap \ker(T - I) = \{0\}$ and (ii) $\ker(T - 2I) + \ker(T - I) = V$.

(i) Given $v \in \ker(T - 2I) \cap \ker(T - I)$, what can you say about $T(v)$? What does $v \in \ker(T - I)$ imply? What does $v \in \ker(T - 2I)$ imply?

(ii) Given $v \in V$, then
$$
0 = (T^2 - 3T + 2I)(v) = (T - 2I)((T - I)(v)) = (T-2I)(T(v) - v) \, .
$$
What does this say about $T(v) - v$? Similarly,
$$
0 = (T^2 - 3T + 2I)(v) = (T - I)((T - 2I)(v)) = (T-I)(T(v) - 2v) \, .
$$
What does this say about $T(v) - 2v$?

Spanning part is already proved. Now, say $v\in ker (T-2I)\cap ker (T-I)$. Then $(T-2I)v=0$

$\Rightarrow (T-I)v=v $

$\Rightarrow v=0$


I need help with chapter 5 of Spivak calculus I am currently going through chapter 5 of Spivak's book on calculus. That chapter is on limits. I have done the entire chapter, however, I feel I have not completely grasped the concept of limits. In particular, my problem is in proving that the limit of a function does not exist. Moreover, another problem I am facing is understanding the concept of the very definition of limits. That is: if $f$ tends towards the limit $l$ close to \alpha, then for every $\epsilon>0$ there is a $\delta>0$  such that, for every x, if $0<|x - \alpha|<δ$, then $0<|f(x) - l|<\epsilon$ .Can someone please suggest some further reading I should do?

Awaiting for your reply

Andrew

You can consult other books then return to it later.

But I'll give you some ideas on limits that I hope you'll find helpful.

Consider the function $f$ defined by $x + 1$, when $x$ gets closer to $1$, $f$ gets closer to 2.
But getting $x$ closer to 1 actually means that the difference between $x$ and $1$ is  as small as possible. Succinctly, it means
  $0<|x - 1|<\delta$
Also, $f$ being made close to $2$ actually means we want
$0<|f(x) - 2|<\epsilon$
Here, epsilon and delta are positive real numbers that may or may not be less than 1( but as Spivak would put it, it's only of small epsilons that are of interest).

I only explained the absolute value terminology. You can visit YouTube and good textbooks(Mir Titles too) for further explanations.

One way of thinking about limit is imagining you have measuring devices for $x$ and $y$. A limit means that given any measuring device for $y$ with a particular amount of precision, there is a measuring device for $x$ with a precision for which for every $x$ close to $x_0$, $y$ is close to $y_0$. For instance, suppose $x$ is time and $y$ is distance. Then you have a watch for $x$ and a measuring tape for $y$. Each of those is going to have some degree of precision. For instance, the watch might measure times to a hundredth of a second, and the measuring tape measures distances to a millimeter. And maybe for every $x$ within a hundredth of second within $x_0$, $y$ is within a millimeter of $y_0$. Then we can say that for a watch with precision .01 s and tape with precision 1mm, if $x$ is close to $x_0$ then $y$ is close to $y_0$.

Saying that $f(x)$ goes to $y_0$ as $x$ goes to $x_0$ means that given any measuring tape $T$, there is some watch $W$ such that whenever $W$ measures $x$ as being indistinguishable from $x_0$ (that is, the difference between $x$ and $x_0$ is less than the precision of $W$), $T$ measures $y$ as being indistinguishable from $y_0$.


Is there an elegant way to prove a function is linear? I'm reading Hoffman and Kunze's linear algebra book and on page 73 in the exercise 7, they ask to verify this function

  $$T(x_1,x_2,x_3)=(x_1-x_2+2x_3,2x_1+x_2,-x_1-2x_2+2x_3)$$
  is a linear transformation.

This exercise is really simple, but a little bit tedious. We have to define arbitrary $u=(x_u,y_u,z_u)$ and $v=(x_v,y_v,z_v)$ elements of $F^3$ and show $T(u+v)=T(u)+T(v)$ and $T(ku)=kT(u)$ for $k\in F$. (we can see $F$ as $\mathbb R$ or $\mathbb C$)

Is there a way more elegant to prove this function is linear?

You can observe that the function is composed out of smaller linear functions. You know that $f(x) = x$ is linear and so is $f(x,y,z)=x$. Also that the sum of linear functions $f+g$ is linear, as well as a scalar multiplier $2f$. Finally, for linear $f,g,h$, the function $x \mapsto (f(x), g(x), h(x))$ is also linear.

When you know all of the above, you can just say that $F$ is made out of these linearity preserving compositions and a few obviously linear functions. That should be a good enough proof.

For any function, if $f(ax + by) = af(x) + bf(y)$, then the function is linear.

It's easy to show (just write it out) that your T satisfies this.


What is the order of operations in trig functions? Is $\sin(x)^2$ the same as $\sin^2(x)$ or $\sin(x^2)$? I thought it would mean the former interpretation, $\sin^2(x)$, rather than the latter, but my teacher and I had a long argument on this and in the end I found that Casio calculators have a space between before the parenthesis, so it would look like $\sin\text{ }(x)^2$ and what the calculator would do is calculate $x^2$, and then take the $\sin$ of that whereas on Texas calculators, there is no space, so it would look like $\sin(x)^2$ and it will calculate $\sin$ of $x$ first, and then take the result and square it.

$$\Large \sin(x)^2 \equiv [\sin(x)]^2 \equiv \sin^2(x).$$

Think about it: if $\sin(x)^2$ were equivalent to $\sin(x^2),$ then the exponent $\quad ^2 \quad$ would be inside the brackets.

It is not only in trigonometric functions, all functions fall under this notation. $$(f(x))^2 \equiv f(x)^2 \equiv f^2(x)$$
It is commonly used for trigonometric functions and logarithmic ones, but it doesn't mean it is wrong for the rest of functions.


Which of the following real numbers are algebraic? A real number is said to be algebraic if it is root of a non-zero polynomial with integer coeﬃcients.

Which of the following real numbers are algebraic?
(a) $\cos (\pi/5)$
(b) $e^{\frac{1}{2}\log2}$
(c) $5^{1/7}+7^{1/5}$

(a) I am not sure.
(b) it is algebraic since this is $\sqrt{2}$ and $x^2-2$ is the polynomial.
(c) not sure but I think this is also algebraic and a polynomial og order 35 may be the answer.but I cant prove this.
can I get some help please.

You are correct on (b).

For (a) and (c), the key facts are that a sum, quotient and product of algebraic numbers are algebraic. (You need to prove this if you haven't yet!)

$$\cos (\pi/5)=\frac{ \zeta_{10}+\zeta_{10}^{-1}}2$$ where $\zeta_{10}$ is a $10$th root of unity, and (c) is plainly obvious from the key facts.

(c) this is a solution of 35x7 deg polynomial
when we solve :  x-5^(1/7)=7^(1/5)
raising 35 power
x^35 - 5^5 - 7^7 = 5^(1/7)*(polynomial of degree 34)

again raising 7th power we get desired polynomial.

(a) refer wolfram alpha.


A Probabilistic Drinking Problem This question was asked by a fellow MSE user in chat. Motivational credits to @Quintec.

Question:

  Bob goes to a bar and drinks a drink. On drink $n$, Bob has a $1-\left(\frac12\right)^n$ probability of getting drunk. If he doesn't get drunk that drink, then he will add $n$ drinks to the amount of drinks he drinks. Even if he gets drunk before finishing his "queue" of drinks, he will still finish the rest of the drinks. He will stop drinking after that. What is his expected number of drinks?

Attempt:

We have the following relationships:

1

drunk -> 1 (probability 1/2)

not drunk -> 1 2 (adds on one drink - probability 1/2)

1 2

drunk -> 1 2 (probability 3/4)

not drunk -> 1 2 3 4 (adds on two drinks - probability 1/4)

1 2 3 4

drunk -> 1 2 3 4 (probability 7/8)

not drunk -> 1 2 3 4 5 6 7 (adds on three drinks - probability 1/8)

1 2 3 4 5 6 7 and so on

Therefore the expectation is given by $$\frac12\times1+\frac12\left[\frac34\times2+\frac14\left[\frac78\times4+\frac18\left[\cdots\right]\right]\right]=\sum_{i=1}^\infty\left[\frac1{2^{i(i-1)/2}}\times\frac{2^i-1}{2^i}\times\left(1+\frac{i(i-1)}2\right)\right]$$ on noticing that $1,2,4,7,\cdots$ are the sequence of one plus the triangular numbers, and this can be written as $$\sum_{i=1}^\infty \frac{2^i-1}{2^{(i^2+i+2)/2}}\left(i^2-i+2\right)\approx1.800$$ Unfortunately, W|A does not return a closed form. So is there any methods to find such a form, if any exists?

Let $S$ be our sum. We have:
$$S=\sum_{i=1}^\infty \frac{2^i-1}{2^{(i^2+i+2)/2}} \cdot (i^2-i+2) = \sum_{i=1}^\infty \frac{2^i}{2^{(i^2+i+2)/2}} \cdot (i^2-i+2) - \sum_{i=1}^\infty \frac{1}{2^{(i^2+i+2)/2}} \cdot (i^2-i+2)$$
$$S=\sum_{i=1}^\infty \frac{1}{2^{(i^2-i+2)/2}} \cdot (i^2-i+2)- \sum_{i=1}^\infty \frac{1}{2^{(i^2+i+2)/2}} \cdot (i^2-i+2)$$
$$S=\sum_{i=1}^\infty \frac{1}{2^{(i^2-i+2)/2}} \cdot (i^2-i+2)- \sum_{i=1}^\infty \frac{1}{2^{(i^2+i+2)/2}} \cdot (i^2+i+2)+ \sum_{i=1}^\infty \frac{2i}{2^{(i^2+i+2)/2}}$$
Note that $(i-1)^2+(i-1)+2 = i^2-i+2$
$$S=\sum_{i=1}^\infty \frac{1}{2^{(i^2-i+2)/2}} \cdot (i^2-i+2)- \sum_{i=2}^\infty \frac{1}{2^{(i^2-i+2)/2}} \cdot (i^2-i+2)+ \sum_{i=1}^\infty \frac{2i}{2^{(i^2+i+2)/2}}$$
$$S=\frac{1^2-1+2}{2^{(1^2-1+2)/2}}+\sum_{i=1}^\infty \frac{i}{2^{(i^2+i)/2}}$$
$$\bbox[5px,border:2px solid red] {S=1+\sum_{i=1}^\infty \frac{i}{2^{i(i+1)/2}}}$$

Now, our sum is simplified to a simpler infinite sum. However, I am unsure whether the sum can actually have any closed form, as the numerator is quite small compared to the denominator, and the exponents are triangular numbers. If this sum does have any closed form, a good idea to find it would be to first evaluate the sum:
$$S(k)=\sum_{i=k}^\infty \frac{1}{2^{i(i+1)/2}}$$
for $k \in \mathbb N$. Then, we can express:
$$S=\sum_{i=1}^\infty S(i)$$
As you stated, it seems that the answer tends to the value of $S=1.8009367251072776$. I would say that this is quite good an approximation since the terms get small really fast. Thus, $S \approx 1.8$.

This is really a comment, but I can't fit it in the comment box.

We can calculate the expected number of drinks by adding up the probabilities of Bob's taking each drink, by linearity of expectation.

The probability that Bob takes the first drink is $1$ and the probability that he takes the second is $\frac12$ which gives us $\frac32$ so far.  The probability that he takes the third and forth drinks is the probability that he takes the second drink, $\frac12,$ times the probability that it doesn't make him drunk, $\frac14$ or $\frac18,$ so we have $\frac32+\frac28=\frac74.$  He take drinks $5,6,7$ if drink $3$ doesn't make him drunk.  The probability that he takes drink $3$ is $\frac18$ and it makes him drunk with probability $\frac18$ so we now have $\frac74+{3\over64}= {115\over64}.$  Now if he takes the fourth drink (probability $\frac18$) and it doesn't make him drunk (probability ${1\over16}$), Bob will drink another $4$ drinks, so we're up to ${115\over64}+{1\over32}={117\over64}=1.828125.$

Since this is already greater than the sum of your series, I feel there must be something wrong with your calculations.  I must admit that I wasn't able to follow your calculations in detail, so I don't know what the problem is.  Of course, it's also possible that I don't understand the question.


Show that every graph can be embedded in $\mathbb{R}^3$ Show that every graph can be embedded in  $\mathbb{R}^3$ with all
  edges straight.

(Hint: Embed the vertices inductively, where should you
 not put the new vertex?)

I've also received a tip about using the curve ${(t, t^2 , t^3 : t \in \mathbb{R} )}$ but I'm not sure how to do that and how I should go about proving this rigorously but without using any pure topology.

Any hints or useful suggestions?

Thanks a lot!

Let $\vec{p} : \mathbb{R} \to \mathbb{R}^3$ be the function
$\vec{p}(t) = (t, t^2, t^3)$ and $\mathscr{C}$ be the curve
$\Big\{\;\vec{p}(t) : t \in \mathbb{R}\;\Big\}$.

For any four distinct $t_1, t_2, t_3, t_4$ in $\mathbb{R}$, the volume of the tetrahedron $\mathscr{T}$ formed by
$\vec{p}(t_i) \in \mathscr{C}$ is proportional to a Vandermonde determinant:

$$\begin{align}
6 \text{Volume}(\mathscr{T})
& = \Big|
(\vec{p}(t_2)-\vec{p}(t_1)) \cdot  \left[
(\vec{p}(t_3) - \vec{p}(t_1)) \times (\vec{p}(t_4) - \vec{p}(t_1)) \right]
\Big|\\
& = \det \begin{bmatrix}
1 & t_1 & t_1^2 & t_1^3\\
1 & t_2 & t_2^2 & t_2^3\\
1 & t_3 & t_3^2 & t_3^3\\
1 & t_4 & t_4^2 & t_4^3
\end{bmatrix} \ne 0
\end{align}$$
The implies any four distinct points on $\mathscr{C}$ are not coplanar. As a result,
the edges of the tetrahedron $\mathscr{T}$ intersect at and only at the appropriate
vertices.

Now take arbitrary $n$ distinct points $\vec{p}_i$ from $\mathscr{C}$.
Above argument shows that if we form a complete graph $K_n$ from them,
the edges intersect at and only at appropriate vertices. This gives us
an embedding of the complete graph $K_n$ into $\mathbb{R}^3$.
Since every graph of $n$ vertices is a sub-graph of $K_n$, we are done.

A graph can be seen as a manifold; there is a result that the graph of a smooth function is a manifold; in this case, it is a 1-manifold. Then use Whitney's theorem that guarantees  that a smooth m-manifold can be smoothly-embedded in $\mathbb R^{2m}$.


Probability of Event Occurring in Time-Series So I'm trying to model email openings over the course of a day.

For example, I have a data set for one individual that simply has a bunch of time-stamps of when he has opened an email. I don't care about the date, just the time of day.

So over the course of a 24 hour day, there will be events along the time-series (X) axis. I want to model this so that I can plug in a time and it will give me the probability of the event (email opening) occurring at that specific time given the historical data. So the Y axis has no magnitude, it's just an event happened at that specific time or it didn't.

Not sure how to represent this mathematically. Any help would be appreciated!

You may be interested in this article about Steven Wolfram's analysis of 2 decades worth of timestamped emails: http://www.nytimes.com/2012/04/08/business/mining-our-personal-data-for-our-own-good.html.

The state space is a cylinder, since the X axis representing time of day is a (periodic) phase variable, but I disagree that the Y axis has no magnitude. Rather, the Y axis will initially be a count of the # of emails that were opened/sent occurring within a narrow time window, say, every hour or every minute.

This count histogram by time of day can then be normalized (by the total count) to obtain the rate (ie fraction of all emails opened/sent in any time window), which has a direct interpretation as a probability distribution.

First off, if you consider time to be continuous then the answer is trivial. The chance of opening an email at a particular instance is precisely 0. That is one of the fundamental properties of continuous random variables.

What I think you are interested in is the probability of an email being opened during the time between two instants.

You need to refine your question to consider:

the duration you are interested in
are you interested in one and only one or one and more than one


What made you choose your research field? I was wondering how come modern mathematicians do not seem to discover as many theorems as the older mathematicians seem to have done. Have we reached some kind of saturation limit where all commonly needed mathematics has already been discovered or is it just that the standard common textbooks do not get updated with the newly discovered theorems ? Are mathematicians of our generation only left with research topics in super specialized sub-fields ? I am wondering what made you choose your research field, what's so beautiful about it ? I understand that this may be a little personal question and i do not mean to intrude on your privacy. I guess if you could just shed light on the field you are familiar with and why u would or would n't recommend me to conduct research in it then i'd be great full to you. This would help me make an informed decision about picking a field.

PS:I know "commonly needed mathematics" is subjective to interpretation but i was thinking of defining that as anything one is taught in an undergraduate level.

Edit: As per the request my educational level is i have a bachelor's degree in computer science, a graduate diploma in mathematics. I am currently a honours student and would be starting a PHD next year. I have taken  mostly non-rigrous undergrad level math courses, mostly because that's all the uni was offering at the time. These courses were on financial maths, Dynamics, ODEs, Mathematical modelling with multiple ODEs, linear algebra, basic Probability, Statistcial modelling, statistical inference, vector calculus, Time series. Out of rigrous fields i have only self studied basic abstract algebra and some basic mathematical analysis. I guess i am in mathematical infancy and being made to choose which seems very scary.

I began graduate school wanting to be an analyst. My first year I did the standard introductory algebra, analysis, and topology sequences. In the second semester, the algebra course did a lot of module theory and homological/categorical things, and I was hooked. I work mainly in module theory because I find it interesting seeing what does and doesn't generalize from vector spaces and Abelian groups, and what I can learn about rings from modules. I enjoy what I do. I'm not at a research university so I don't have to worry if my work isn't fashionable or high-powered. I do it for me, in my own time. I didn't get in the business to get rich, or famous, but to pursue what I consider some beautiful ideas, and maybe make a contribution here or there. It's been a good ride so far.

I think mathematics is always establishing new results. In my experience, the basic graduate texts provide an introduction to the various fields. For the most part, you won't find in them the advanced results you need to do serious research. That's where advanced texts, papers, etc. become very valuable.

What made me to choice my field of studies?

Passion and amusement.


Trains describing a parabola From the train station – the point S – originante two tracks, i.e. rays, which do not lie on a common straight line.
Along these move two trains, which are line segments.
On the first track a train is moving toward the station S,
while on the other track the second train leaves from the station.
We shall consider the motion of train-segments only for such period of time during which they are leaving the track-rays.

At any fixed time consider a convex quadrilateral whose vertices are the endpoints of the train-segments. In particular, consider the point where its diagonals intersect.

Find out the necessary and sufficient conditions for all these points of intersection to lie on some parabola.

The whole setup is invariant under affine transformations. Therefore you can (without loss of generality) assume a coordinate system where the incoming train rides on the positive $x$ axis and the outgoing on the positive $y$ axis, and both travel with unit speed. So you'd have the heads of the trains at $(x_1-t,0)$ and $(0,y_1+t)$ and the tails at $(x_2-t,0)$ and $(0,y_2+t)$. Then the point of intersection is

$$
P(t)=
\frac1{(x_2y_1-x_1y_2)-t(x_1-x_2+y_1-y_2)}
\begin{pmatrix}
(t-x_1)(t-x_2)(y_1-y_2) \\
(t+y_1)(t+y_2)(x_2-x_1)
\end{pmatrix}
$$

So this is a rational curve of degree $2$ in $t$. If you compute its implicit form, you will find that it is indeed a conic section. So you want to know when this conic section will be a parabola. For that you care for the limit cases, $\lVert P\rVert\to\infty$. There are two ways how this could happen: either the denominator of the scalar coefficient becomes zero, or you have $t\to\infty$.

\begin{align}
t&\to\frac{x_2y_1-x_1y_2}{x_1-x_2+y_1-y_2} &
t &\to\infty \\
P(t)&\to\lambda\begin{pmatrix}x_1-x_2\\y_2-y_1\end{pmatrix} &
P(t)&\to\mu\begin{pmatrix}y_1-y_2\\x_2-x_1\end{pmatrix}
\end{align}

Your curve is a parabola if and only if these two directions agree, i.e. if these two vectors are linearily dependent.

$$
\begin{vmatrix}
x_1-x_2 & y_1-y_2 \\
y_2-y_1 & x_2-x_1
\end{vmatrix} = (y_1-y_2)^2 - (x_1-x_2)^2 = 0
$$

which is simply asking that both trains have the same length – in this coordinate system. Taking our special choice of coordinate system into account, this means that the ratio between length and speed must be the same for both trains:

$$\frac{\ell_1}{v_1} = \frac{\ell_2}{v_2} $$

Some experimentation suggested that the curve (in general a conic) is symmetric about the angular bisector of the two tracks, so that is how I set up my model in Geometry Expressions.

One train is located at distance t from the station, the other at u*t+v (so u is the ratio of speeds of the two trains)

The complicated formula is the implicit equation of the curve (most of it is off screen, but here it is in all its glory:

$X^{2}\cdot \left (a^{2}\cdot sin(\theta)^{2}-a\cdot b\cdot sin(\theta)^{2}-a\cdot b\cdot u^{2}\cdot sin(\theta)^{2}+b^{2}\cdot u^{2}\cdot sin(\theta)^{2}\right )+X\cdot \left (2\cdot Y\cdot a\cdot b\cdot sin(\theta)\cdot cos(\theta)-2\cdot Y\cdot a\cdot b\cdot u^{2}\cdot sin(\theta)\cdot cos(\theta)+2\cdot a^{2}\cdot b\cdot sin(\theta)^{2}\cdot cos(\theta)-2\cdot a^{2}\cdot b\cdot u\cdot sin(\theta)^{2}\cdot cos(\theta)-2\cdot a\cdot b^{2}\cdot u\cdot sin(\theta)^{2}\cdot cos(\theta)+2\cdot a\cdot b^{2}\cdot u^{2}\cdot sin(\theta)^{2}\cdot cos(\theta)+4\cdot a\cdot b\cdot v\cdot sin(\theta)^{2}\cdot cos(\theta)-4\cdot a\cdot b\cdot u\cdot v\cdot sin(\theta)^{2}\cdot cos(\theta)\right )-Y^{2}\cdot a^{2}\cdot cos(\theta)^{2}-Y^{2}\cdot a\cdot b\cdot cos(\theta)^{2}-Y^{2}\cdot a\cdot b\cdot u^{2}\cdot cos(\theta)^{2}-Y^{2}\cdot b^{2}\cdot u^{2}\cdot cos(\theta)^{2}-2\cdot Y\cdot a^{2}\cdot b\cdot sin(\theta)\cdot cos(\theta)^{2}-2\cdot Y\cdot a^{2}\cdot b\cdot u\cdot sin(\theta)\cdot cos(\theta)^{2}+2\cdot Y\cdot a\cdot b^{2}\cdot u\cdot sin(\theta)\cdot cos(\theta)^{2}+2\cdot Y\cdot a\cdot b^{2}\cdot u^{2}\cdot sin(\theta)\cdot cos(\theta)^{2}-4\cdot Y\cdot a\cdot b\cdot v\cdot sin(\theta)\cdot cos(\theta)^{2}-4\cdot Y\cdot a\cdot b\cdot u\cdot v\cdot sin(\theta)\cdot cos(\theta)^{2}-4\cdot a^{2}\cdot b\cdot v\cdot sin(\theta)^{2}\cdot cos(\theta)^{2}+4\cdot a\cdot b^{2}\cdot u\cdot v\cdot sin(\theta)^{2}\cdot cos(\theta)^{2}-4\cdot a\cdot b\cdot v^{2}\cdot sin(\theta)^{2}\cdot cos(\theta)^{2}=0$

As there is no term in $X \cdot Y$, we can tell the conic is aligned with the axes.

You can see that this will be a horizontal parabola when the coefficient of $X^{2}$ is 0.

This happens when $\theta=0$ (degenerate situation disallowed by the problem statement) or $a=b$ or $a=b \cdot u^{2}$


How do I prove that $\lnot(q \rightarrow (p \rightarrow q)) \not\equiv p$ using logical equivalences? I was able to prove that
$\lnot(q \rightarrow (p \rightarrow q)) \not\equiv p$
using truth tables, but when it comes to propositional equivalences, I'm a little stuck. Do I need to show that the left hand side contradicts the right hand side? Any help is appreciated!

We say that $A ≡ B$ iff $A,B$ have the same truth-value for every truth-assignment (i.e. no matter what truth-values the propositional atoms have). So $A ≢ B$ iff $A,B$ have different truth-value for some truth-assignment. Such statements cannot be proven using propositional equivalences! For example, $p ≢ p∧q$, but we can prove neither $p ⇔ p∧q$ nor $¬( p ⇔ p∧q )$, because they are each neither a tautology nor a contradiction.
The only way to prove this kind of statement is therefore by finding some truth-assignment that differentiates the truth-values of the two expressions. As you noted, a truth-table is one way to do that. You can first simplify each expression via equivalences such as peter.petrov showed, but you eventually still need to find a truth-assignment to finish.
Hence the implicit assumption in your question that you can prove the desired claim using merely propositional equivalences is incorrect, because propositional equivalences only allow you to prove equivalences (i.e. "≡") and nothing else. To prove anything else, you have to step beyond boolean algebra.

Two propositions are logically equivalent if they are both true or both false.
Let $$R : q \rightarrow (p \rightarrow q)$$
If $ q $ is false, then $ R $ will be true no matter what  $ p $ is.
So, if $ p $  and $ q $ are true, $ \lnot R$ is false.
we conclude that $ p $ and $ \lnot R $ are not equivalent.


What does it mean if a sequence is not eventually constant From What I understand a sequence $(a_k)$ is eventually constant such that $a_k=a$ whenever $k>N$. Does that mean a sequence is not eventually constant such that $a_k=a$ whenever $k<N$?

No, it means that the points keep moving around.  No matter what point $a$ and number $N$ you pick there is a point $a_k$ with $k \gt N$ and $a_k \neq a$

Yes, it means that for any k, there exist m and n> k such that $a_k\ne a_m$.


Coefficients of a minimal polynomial are homogeneous polynomials. If we have an endomorphism $\varphi$ in a $n$-dimensional vector space, and the expression of his characteristic polynomial $P_{\varphi}(X)$ is as follows :
\begin{equation}
P_{\varphi}(X)= X^n + \alpha_1X^{n-1} + \dots + \alpha_{n-1}X + \alpha_n
\end{equation}
then, we know the $\alpha_i$'s are all homogeneous polynomials of degree $i$ of the coefficients of the matrix representation of $\varphi$.
We also know that the minimal polynomial $\mu_\varphi$ of $\varphi$ divides $P_\varphi(X)$.

My question is : does the coefficients of $\mu_\varphi$ also follows the same rules as the characteristic polynomials? i.e. if we have
\begin{equation}
\mu_\varphi(X)= X^r + \beta_1X^{r-1} + \dots + \beta_{r-1}X + \beta_{r}
\end{equation}

Are the $\beta_i$'s homogeneous polynomials of degree $i$ of the coefficients of the matrix representation of $\varphi$ ? If yes, how do we find that ?

No: indeed the $\beta_i$ are not even continuous functions of those coefficients, essentially because $r$ itself is not a continuous function of those coefficients. Consider for example the family of matrices
$$\bigg\{\begin{pmatrix}1&t\\0&1\end{pmatrix}\colon t\in\Bbb R\bigg\}.$$
Here the minimal polynomial is $X^2-2X+1$ for all $t\in\Bbb R\setminus\{0\}$, but $X-1$ for $t=0$.

Let $t$ be trancendental/indeterminate.

Then
$$
\mu_{t\phi}(X)\ |\ \mu_{\phi}(tX)\ |\ \mu_{t\phi}(X)
$$
and so these minimal polynomials are equal.

Now this gives us the characteristic property of homogeneity:
$$
\beta_j(t a_{11}, \dots, t a_{nn})=t^j\beta_j( a_{11}, \dots,  a_{nn})
$$
where the $a_{ij}$ are the matrix entries.


Can a piece of A4 paper be folded so that it's thick enough to reach the moon? While procrastinating around the web I stumbled on a page that contained  the image below, from cracked.com.

I can't help but believe that this is false… Even though the article header says:

  22 Statistics That Will Change The Way You See the World

My question: is what the image below implies a mathematical impossibility? (…Just for procrastination's sake…)

  If you could fold a piece of A4 paper just 42 times it would be thick enough to reach the moon

Even if the sheet of paper were infinitely foldable, the answer is that no, you can't reach the moon by folding a sheet of A4 paper any number of times, for a reason that bears calling out (and in fact explains why a sheet of paper that size can only be folded a certain number of times — that is, why it's impossible to fold it 42 times in the first place): consider the last fold and imagine looking at the sheet in a cross-section perpendicular to this fold.  The 'faces' of the folded paper that are at the top and at the bottom after the last fold must be connected along the fold edge, since they were part of a single 'face' before the fold — but this means that the distance along the paper between the top and bottom must be  at least as long as the distance 'through' the paper on a straight line between them.  In other words, you need to start with a sheet of paper that's at least 385,000km along at least one direction (using Sabyasachi's numbers) to be able to reach that far, regardless of what sequence of folds you use.

Mathematically, it is possible. however realistically you will loss paper every time you fold, because you have to count the width of the paper being folded.


How can I study the convergence of the improper integral $\int_{0}^{ \infty} \frac{\sin(x)}{x+1} \, \mathrm dx\,$? I need to study the convergence of the following improper integral:
$$\int_{0}^{\infty} \dfrac{\sin(x)}{x+1}\, \mathrm dx$$
I did the following:
$$ -1 \leq \sin(x)  \leq 1  \\
\implies \dfrac{-1}{x+1}  \leq \dfrac{\sin(x)}{x+1}  \leq \dfrac{1}{x+1} \\
\implies \left|\dfrac{\sin(x)}{x+1}\right|  \leq \dfrac{1}{x+1} \\
 \implies \int_{0}^{\infty} \left|\dfrac{\sin(x)}{x+1}\right| \, \mathrm dx   \leq \int_{0}^{\infty}\dfrac{1}{x+1}\, \mathrm dx = \infty  $$
I planned to use the comparison criterion and then the absolute convergence criterion. However, the idea did not work for me.

Notice that
$$\int_0^\infty \frac{\sin x}{x+1}\,dx = \frac{-\cos x}{x+1}\Bigg|_0^\infty - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx = 1 - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx$$
and the last integral converges absolutely since
$$\int_0^\infty \frac{\left|\cos x\right|}{(x+1)^2}\,dx \le \int_0^\infty \frac{dx}{(x+1)^2} = \int_1^\infty \frac{dx}{x^2} < +\infty.$$
However the original integral does not converge absolutely. Namely, we have $$x \in \bigcup_{k \in \mathbb{N}_0} \left[\frac\pi6+k\pi,\frac{5\pi}6+k\pi\right] \implies \left|\sin x\right| \ge \frac12$$
so
$$\int_0^\infty \frac{\left|\sin x\right|}{x+1}\,dx \ge \frac12\sum_{k=0}^\infty \int_{\frac\pi6+k\pi}^{\frac{5\pi}6+k\pi} \frac{dx}{x+1} = \frac12\sum_{k=0}^\infty \ln \frac{\frac{5\pi}6+k\pi+1}{\frac\pi6+k\pi+1} = +\infty.$$

Let
$$
a_n = \int_{\pi n}^{\pi(n+1)}\frac{|\sin x|}{x+1}dx.
$$
Note that
$$
\int_0^\infty \frac{\sin x}{x+1}dx = \sum_{n=0}^\infty (-1)^n a_n.
$$
So, if the series converges, the integral must also converge. For any positive integer $n$, we can see that $a_n$ is positive, and we can rewrite it the following way:
$$
\begin{align}
a_n = \int_{\pi n}^{\pi(n+1)}\frac{|\sin x|}{x+1}dx &= \int_{\pi n}^{\pi(n+1)}\frac{\sin (x - \pi n)}{x+1}dx\\
&= \int_0^\pi\frac{\sin x}{x+1+\pi n}dx.
\end{align}
$$
This makes it clear that the denominator of $a_{n+1}$ is larger than the denominator of $a_n$ over the entire interval integrated, so $a_n$ must be decreasing. Furthermore, it is easy to see that $\lim_{n\to\infty} a_n=0$. Therefore, by the alternating series test, the integral converges.


$\sup\{||Ax|| : x\in E, A \in \Bbb A\} < \infty$ for every bounded set $E \subset X$ Let $X$ be a Banach Space, $Y$ be NLS and $\Bbb A\subset B(X,Y)$ be such that $\{Ax:A\in \Bbb A\}$ is bounded in $Y$ for every $x\in X$. The $\sup\{||Ax|| : x\in E, A \in \Bbb A\} < \infty$ for every bounded set $E \subset X$.

Why is the problem? IS it not direct consequence from Uniform Boundedness Principle?

$\{Ax:A\in \Bbb A\}$ is bounded in $Y$ for every $x\in X$  means $A$ is pointwise bounded and by UBP $A$ is uniformly bounded.

you can use the UFP and $\exists M,\sup A＜M$.

So $$\sup\{||Ax|| x\in E\}＜M\sup\{||x||,x\in E\}<\infty$$


Prove existence of $n$ such that $x>(1-\epsilon)^n$ for some $x,\epsilon$ I am stuck in an analysis problem where I have to prove this inequality-
$$x>(1-\epsilon)^n$$ for some $n\in\mathbb{N}$ where $x\le1$ and $\epsilon\in(0,1)$ are specified.
Background: I am solving an analysis problem where I'm supposed to show, $\sup S_x=1$ where $S_x=\{x,x^{(1/2)},...,x^{(1/n)},...\}$ when $0<x\leq 1$ and $x\in\mathbb{R}$.
I have no idea how to proceed. Please help me.

We have $0\lt 1-\epsilon \lt 1$
Then $\frac 1{1-\epsilon} \gt 1$
Let $\frac 1{1-\epsilon}=1+a $  where $a\gt 0$
Then $(1+a)^n \gt na $ which implies
$\frac 1{(1+a)^n}=(1-\epsilon)^n \lt \frac 1{na} \to 0$ as $n\to \infty$
So whatever $x$ is given in $(0,1)$ , there exist some $n\in \mathbb{N}$ such that
$(1-\epsilon)^n \lt x$
Edit : If you don't want to use limits, note that by Archimdean Property of Real numbers, there is $n_o\in \mathbb{N} $ such that
$n_o(xa)\gt 1$ which implies
$(1-\epsilon)^{n_o} \lt \frac 1{n_oa} \lt x$

$\lambda=1-\epsilon$ is certainly between $0$ an $1$. The integral
$$
\int_{0}^{1}\lambda^n d \lambda=\frac{1}{n+1}
$$
Converges, hence the integrand converges to $0$. Your result follows.


Elementary motivations for free resolutions Let $M$ be a finitely generated module over a Noetherian ring $R$ which admits a finite free resolution $0 \to F_n \to \dots \to F_0 \to M \to 0$. There is no doubt that knowing such a resolution is very useful in practice since it allows us to compute many invariants of $M$, such as homology groups, the rank of $M$, Betti numbers in the local case, or the Hilbert polynomial in the graded case. More generally, we can compute any invariant which is easily determined for free modules and behaves well under exact sequences.

While this all is clear for me, I am not able to explain the usefulness of finite free resolutions at an elementary level. Of course it always contains a finite presentation $F_1 \to F_0 \to M \to 0$ which I can motivate: With this data we are able to describe all elements of $M$ and to prove any equation which holds in $M$. In short: a finite presentation allows us to do calculations in $M$. But this explanation creates the impression that a finite presentation is all we need to understand the module $M$, doesn't it? Are there elementary reasons why we should be interested in higher syzygies?

Maybe this is a related question: What were the historical reasons to consider free resolutions?

Thank you in advance!

Intuitively, $M$ is an alternating sum of free modules.


Prove that $n^n>\left(\dfrac{n+1}{2}\right)^{n+1}$ for all positive integer $n>1$. Question: Prove that $n^n>\left(\dfrac{n+1}{2}\right)^{n+1}$ for all positive integer $n>1$.

I could not understand what should be the initial approach.

We want to show the $2$s at the bottom are enough to defeat the $+1$s at the top. The effect of the $2$s is easy. Just write

$$ \left(\dfrac{n+1}{2}\right)^{n+1} = \frac{1}{2^{n+1}}(n+1)^{n+1}$$
We'd like a bound like  $(n+1)^{n+1} < Cn^n$ for some $C < 2^{n+1}$.

$$(n+1)^{n+1}= (n+1)(n+1)^{n}$$

To bound $(n+1)^n$ use the binomial expansion, round each $n^i$ up to $n^n$, then use how the coefficients add up to $2^n$. This should complete the proof.


Prove this trig identity $$\sin^2(x) - \cos^2(x) - \tan^2(x) = \frac{2\sin^2(x) - 2\sin^4(x) - 1 }{ 1-\sin^2(x)}.$$

I tried this but I can't figure out how they got $-2\sin^4(x)$.

\begin{align*}
\sin^2\theta -\cos^2\theta-\tan^2\theta & = \sin^2\theta -\cos^2\theta-\frac{\sin^2\theta}{\cos^2\theta}\\
& = \frac{\sin^2\theta \cos^2\theta -\cos^4\theta-\sin^2\theta}{\cos^2\theta}\\
& = \frac{\sin^2\theta (1-\sin^2\theta) -(1-\sin^2\theta)^2-\sin^2\theta}{\cos^2\theta}\\
\end{align*}
Now simplify it.

you can factor sin^2 theta from 2sin^2-2sin^4

then put 1-sin^2 theta as cos ^2 theta

write tan^2 theta as sin^2/cos^2

then put cos ^2 as 1-sin^2 theta


Find a norm so that its closed unit ball is the area between $y=x^2-1$ and $y=1-x^2$ As the title specifies, I need to find an explicit formula for a norm $|||\cdot|||$ so that:
$$B_{|||\cdot|||}=\{\mathbf{x} : ||| \mathbf{x}|||\le1  \}$$ where $\mathbf{x}=(x,y)\in\mathbb R^2$,  is the area between $y=x^2-1$ and $y=1-x^2$.
I have already found many norms but none of them satisfies that $|||\alpha \mathbf{x}|||=\alpha |||\mathbf{x}|||$ for $ \alpha \in R$

Any help will be appreciated.

A necessary and sufficient condition for a subset $B$ of a vector space $V$ to the unit ball of a norm on $V$ is that $B$ is non-empty, convex and symmetric about the origin ($-B = B$). If $B = \{(x, y) : x^2-1 \le y \le 1-x^2\}\subseteq\mathbb{R}^2$, then $B$ is non-empty, convex and symmetric about the origin and so it is the unit ball of a norm on $\mathbb{R}^2$.

Given the unit ball $B$ of a normed space there is only one possible definition of the norm, namely $\|v\| = \inf\{x > 0 : v/x \in B\}$. (Because, for $x > 0$, $v/x \in B$ iff $\|v/x\| \le 1$ iff $\|v\| \le x$.)

Balls in general have to be convex. I dont think its possible.


Use Fourier series for computing $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$ I need to compute Fourier series for the following function: $f(x)=\frac{-\pi}{4}
$ for $-\pi \leq x <0$, and $\frac{\pi}{4} $ for $ 0 \leq x \leq \pi$, and then to use it and compute $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$

I tried to use Parseval equality:

$$\widehat{f(n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}=\frac{1}{4in}-\frac{(-1)^n}{4in}, \sum_{-\infty}^{\infty}|\widehat{f(n)}|^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2.$$

$$\sum_{-\infty}^{\infty}|\frac{1}{4in}-\frac{(-1)^n}{4in}|=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)=\frac{\pi^2}{16}.$$

Does someone see how can I compute form that the requsted sum?

Thanks!


Proving $\,2\sqrt 2 + 1\,$ is irrational by contradiction I am working on some review questions for my Discrete Structures final and I needed some assistance for the problem "Prove by contradiction that $\,2\sqrt 2 +1$ is irrational".

Now I know how prove that $\sqrt 2$ is irrational on its own, by showing that $\sqrt 2 = a/b$ and showing they are not in lowest terms. My issue is that other stuff is messing up my proof and I am not sure how to solve it.

Here is what I have so far. Assume that $\,2\sqrt 2 +1$ is rational, therefore $\,2\sqrt 2 +1 =$ $\frac {a}{b}$. Then:
$$2 \sqrt 2 +1 = \frac {a}{b}$$
$$ \sqrt 2 = \frac {a-b}{2b}$$
$$ a = 2(b \sqrt 2 + \frac{b}{2})$$
$$ a= 2k$$
$$\sqrt 2 = \frac{2k-b}{2b}$$

And since they are not in lowest terms therefore $\sqrt 2$ is irrational.

Any help would greatly appreciated. Thanks.

If you already know that $\sqrt{2}$ is irrational, because you know how to prove this fact, then having written
$$
\sqrt{2}=\frac{a-b}{2b}
$$
is sufficient to end the proof, because the right hand side is rational by construction.

There's nothing wrong in showing something by reducing it to the proof of another fact that you already know how to do. Indeed, this is a normal way of doing things in mathematics. Since proving $\sqrt{2}$ is irrational is easy, you can use this for proving that, whenever $x$ and $y$ are rational and $x\ne0$, then $x\sqrt{2}+y$ is irrational too: from $x\sqrt{2}+y=r$ you get
$$
\sqrt{2}=\frac{r-y}{x}
$$
and, if $r$ is rational, also $(r-y)/x$ is rational.

You can prove it in other ways, though. If $u=2\sqrt{2}+1$, then $u-1=2\sqrt{2}$, so $u^2-2u+1=8$ and so
$$
u^2-2u-7=0
$$
A rational root of the polynomial $X^2-2X-7$ has to be among $1$, $-1$, $7$ and $-7$, but none of these numbers is a root:
$$
1^2-2\cdot1-7=-8\\
(-1)^2-2\cdot(-1)-7=-4\\
7^2-2\cdot 7-7=28\\
(-7)^2-2\cdot(-7)-7=56
$$

Let $\mathbb{Q}$ denote the set of rational numbers. If $2\sqrt2+1 = x$ (say) is in $\mathbb{Q}$, then $(x-1)/2$ is in $\mathbb{Q}$, and so $\sqrt2$ is in $\mathbb{Q}$, which is a contradiction.


Negative binomial expansion problem So here's the problem:-

Show that if x is small, the expression
$$\sqrt{(1+x)(1-x)^{-1}}\approx 1+x+0.5x^2$$


Finding the equation of a tangent line that passes through a curve at point P. I'm not sure how to long the question in the title, I have a curve $y = lnx^{3}$ and the point $P=(0,1)$. I have the derivative of the function $y' = \frac{3}{x}$, but I'm not sure where to go from here. Can anyone point me in the right direction?

Let's find the general equation of the tangent to this curve.
Let's write $ln(x^3)$ as $3ln(x)$.

You know the slope. You know one point through which it passes,say,$(h,3ln(h))$.
You are done. Equation of line:

$y-3ln(h)=\frac 3 h (x-h)$

It passes through $(0,1)$

Lets try plugging that in $(x.y)$ in line equation.

$1-3ln(h)=\frac 3 h (0-h)$

As h is not zero as function is not defined there, we can cancel it.

$1-3ln(h)=-3$

$ln(h)=4/3$

$h=e^{\frac4 3}$

Lets plug it in equation of line

$y-3 \frac 4 3 =\frac {3} {h}(x-h)$

$y-4=\frac {3x} h - 3 $

$y-1=\frac {3x}{e^{\frac 4 3 }}$

Congos...You did it.


Uniformly distributed probabiity density over $\mathbb{R}^n$ I came across the following phrase in my reading:

  Let $\rho(x)$ be a uniformly distributed probability density defined on $\mathbb{R}^n$.

I know that $\rho(x)$ being a probability density function means that $\displaystyle \int_{\mathbb{R}^n} \rho(x) dx = 1$, but doesn't uniformly distributed mean that $\rho(x) = c$ for some positive constant $c$?

But, then $$\displaystyle \int_{\mathbb{R}^n} \rho(x) dx = \displaystyle \int_{\mathbb{R}^n} c dx = \infty$$, so how is it possible for a probability density to be uniformly distributed over $\mathbb{R}^n$?

Does it implicitly assume for instance that $\rho$ vanishes outsie a set of finite measure?

Uniformly distributed means it follows a Uniform distribution.

As you correctly said, $\displaystyle \int_{\mathbb{R}^n} \rho(x) dx = 1$.

But the density function is:
$
\rho(x)=\begin{cases}
  \frac{1}{b - a} & \mathrm{for}\ a \le x \le b, \\[8pt]
  0 & \mathrm{for}\ x<a\ \mathrm{or}\ x>b
  \end{cases} $

$\begin{align}\displaystyle \int_{\mathbb{R}^n} \rho(x) dx &=\displaystyle \int_{-\infty}^a \rho(x) dx+\displaystyle \int_{a}^b \rho(x) dx+\displaystyle \int_{b}^\infty \rho(x) dx\\ &=0+\displaystyle \int_{a}^b \frac{1}{b - a} dx+0\\ &= \left.\frac{x}{b - a}\right |_a^b\\ &=1\end{align}$


Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$ I am trying to show the followng:

Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ converges uniformly on $\mathbb{R}$.

I have shown it for a compact subset of $\mathbb{R}$ however, do not know how to extend it to the reals. Below is my proof for convergence on a compact subset.

I break up the sum into two parts and attack each individually

$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1 + \frac{x}{n})$ =
$\underbrace{\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1)}_{(1)}$ +
$\underbrace{\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}[\sin(1 + \frac{x}{n})-\sin(1)]}_{(2)}$.

$\textbf{Equation (1)}$

$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\sin(1)$ converges because $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}$ converges by the alternating series test.

$\textbf{Equation (2)}$

we will first bound $\sin(1 + \frac{x}{n})-\sin(1)$ as follows:

\begin{align*}
|\sin(1 + \frac{x}{n})-\sin(1)| &= \Big|\int_{1}^{1+\frac{x}{n}} \cos(t)dt\Big| \\
                                &\leq \Big|\int_{1}^{1+\frac{x}{n}} |\cos(t)| dt\Big| \\
                                &\leq \Big|\int_{1}^{1+\frac{x}{n}} dt\Big| \\
                                &\leq \Big|1+\frac{x}{n} - 1 \Big| \\
                                &= \frac{|x|}{n} \\
\end{align*}
Therefore, on any compact interval $ x \in [-M, M]$
\begin{align*}
|\sin(1 + \frac{x}{n})-\sin(1)| \leq \frac{M}{n}
\end{align*}

It follows that

\begin{align*}
\Big|\frac{(-1)^{n}}{\sqrt{n}} \Big( \sin(1 + \frac{x}{n})-\sin(1)\Big)\Big| &= \Big|\frac{(-1)^{n}}{\sqrt{n}}\Big|\Big|\sin(1 + \frac{x}{n})-\sin(1)\Big| \\
&= \frac{1}{\sqrt{n}}\Big|\sin(1 + \frac{x}{n})-\sin(1)\Big| \\
&\leq \frac{1}{\sqrt{n}}\frac{M}{n} \\
&= \frac{M}{n^{\frac{3}{2}}}
\end{align*}

Hence,

\begin{align*}
\sum_{n}^{\infty}\frac{1}{n^{\frac{3}{2}}} < \infty &\implies \sum_{n}^{\infty}\frac{M}{n^{\frac{3}{2}}} < \infty \\
&\implies \sum_{n=1}^{\infty}\Big|\frac{(-1)^{n}}{\sqrt{n}} \Big[ \sin(1 + \frac{x}{n})-\sin(1)\Big]\Big| < \infty \\
&\implies \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}} \Big[ \sin(1 + \frac{x}{n})-\sin(1)\Big] < \infty
\end{align*}

Hence, we have that both $(1)$ and $(2)$ converge on a compact interval $[-M,M]$ which implies our original equation of interest does also.

I want to extend this proof to all $\mathbb{R}$ but I do not know how to. Could someone please help me with this. Anything is appreciated.

EDIT: Since this is now a bountied question, I will explain the major sticking point in my proof where I began to hand-wave. For a fixed $N$, I have to show that

$$\inf_{x\in\mathbb{R}} \left \{\sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{x}{n}\right)+(-1)^n\right)^2\right\}=0$$

I proved this fact by showing that a subsequence was equidistributed over $[0,1]$. Unfortunately, this portion of the proof is hand-wavy at best. However, the rest of the proof is solid.

We will attempt to prove that

$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}\sin\left(1+\frac{x}{n}\right)$$

does not converge uniformly on the reals. Unfortunately, some details of the proof require more careful attention to the analysis of special functions in the complex plain. While these details might be lacking in a formal proof, I feel that enough progress has been made here to write everything down.

Suppose that it converges uniformly. Let $N_0\in\mathbb{N}$ such that for all $M\geq N\geq N_0$

$$\left|\sum_{n=1}^{M}\frac{(-1)^n}{\sqrt{n}}\sin\left(1+\frac{x}{n}\right)-\sum_{n=1}^{N}\frac{(-1)^n}{\sqrt{n}}\sin\left(1+\frac{x}{n}\right) \right|=\left|\sum_{n=N+1}^{M}\frac{(-1)^n}{\sqrt{n}}\sin\left(1+\frac{x}{n}\right)\right|<1$$

(this is the Cauchy criteria for uniform convergence with $\epsilon=1$). In fact, as long as $M\geq N\geq N_0$, we are free to choose any natural numbers. As such, choose $N= \text{max}(2,N_0)$ and $M=2N$ to get

$$\left|\sum_{n=N+1}^{2N}\frac{(-1)^n}{\sqrt{n}}\sin\left(1+\frac{x}{n}\right)\right|<1$$

Now, consider the function

$$f(x)=\sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{x}{n}\right)+(-1)^n\right)^2$$

Clearly, this function is defined for all $x\in\mathbb{R}$ and we know

$$0\leq \sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{x}{n}\right)+(-1)^n\right)^2\leq \sum_{n=N+1}^{2N}4=4(2N-N)=4N$$

Thus, we know that the infimum of $f(x)$ over the reals exists (as well as the supremum) and is well defined. We shall show that this infimum is $0$. To do this, we will show that a subsequence of real numbers is equidistributed over $[0,4N]$. First normalize by $4N$ so that

$$0\leq \frac{f(x)}{4N}\leq 1$$

Next, consider the sequence $x_k=k(2N)!$ for $k\in\mathbb{N}$. If we can show that

$$a_k=\frac{1}{4N}\sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{k(2N)!}{n}\right)+(-1)^n\right)^2$$

is equidistributed over $[0,1]$ then we are done. To do this, we will use Weyl's criterion. That is, a sequence $a_k$ is equidistributed over $[0,1]$ if and only if for all $l\in\mathbb{N}$ the limit

$$\lim_{m\to\infty}\frac{1}{m}\sum_{k=1}^m e^{2\pi i l a_k}=0$$

Now, plugging our sequence into the test we get

$$\lim_{m\to\infty}\left|\frac{1}{m}\sum_{k=1}^m e^{2\pi i l a_k}\right|=\lim_{m\to\infty}\left|\frac{1}{m}\sum_{k=1}^m \exp\left[ \frac{2\pi i l}{4N}\sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{k(2N)!}{n}\right)+(-1)^n\right)^2\right]\right|$$

$$\lim_{m\to\infty}\left|\frac{1}{m}\sum_{k=1}^m \prod_{n=N+1}^{2N}\exp\left[ \frac{2\pi i l}{4N}\left(\sin\left(1+\frac{k(2N)!}{n}\right)+(-1)^n\right)^2\right]\right|$$

Since these are finite sums and products, there is no issue with reversing their order:

$$=\lim_{m\to\infty}\left| \prod_{n=N+1}^{2N}\frac{1}{m}\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\sin\left(1+\frac{k(2N)!}{n}\right)+(-1)^n\right)^2\right]\right|$$

$$=\prod_{n=N+1}^{2N}\lim_{m\to\infty}\left|\frac{1}{m}\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\sin\left(1+\frac{k(2N)!}{n}\right)+(-1)^n\right)^2\right]\right|$$

Thus, if we can show that

$$\lim_{m\to\infty}\left|\frac{1}{m}\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\sin\left(1+\frac{k(2N)!}{n}\right)+(-1)^n\right)^2\right]\right|=0$$

for arbitrary $n\leq 2N$ then we are done. To begin, since $n\leq 2N$, it is clear that

$$1+\frac{k(2N)!}{n}=1+k\frac{(2N)!}{n}$$

is an arithmetic sequence that increases by $(2N)!/n$ at each step. To simplify notation, we may as well call this natural number $b$. Then the problem simplifies to showing that

$$\lim_{m\to\infty}\left|\frac{1}{m}\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\sin\left(bk+1\right)+(-1)^n\right)^2\right]\right|=0$$

for some $b\in\mathbb{N}$ and all $l\in\mathbb{N}$. Next, we will separate the $\sin(x)$ function

$$\sin(kb+1)=\cos(1)\sin(kb)+\sin(1)\cos(kb)$$

Now,

$$\left|\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\sin\left(bk+1\right)+(-1)^n\right)^2\right]\right|=\left|\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\cos(1)\sin(kb)+\sin(1)\cos(kb)+(-1)^n\right)^2\right]\right|$$

$$=\left|\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\cos^2(1)\sin^2(kb) + 2 \cos(1)\sin(kb) \sin(1)\cos(kb) + \sin(1)^2\cos^2(kb) + 2 \cos(1)\sin(kb) (-1)^n + 2 \sin(1)\cos(kb) (-1)^n + 1\right)\right]\right|$$

As this is extremely difficult to read, let us rename these terms $s_1$ through $s_5$ (where $s_i$ is real):

$$=\left|\sum_{k=1}^m\exp\left[ i\left(s_1\sin^2(kb) + s_2\sin(kb) \cos(kb) + s_3\cos^2(kb) + s_4\sin(kb) + s_5\cos(kb) + 1\right)\right]\right|$$

Now, we can rewrite the exponential as

$$\exp\left[ i\left(s_1\sin^2(kb) + s_2\sin(kb) \cos(kb) + s_3\cos^2(kb) + s_4\sin(kb) + s_5\cos(kb) + 1\right)\right]$$
$$=e^{i}\exp\left[ i\left(s_1\sin^2(kb) + s_2\sin(kb) \cos(kb) + s_3\cos^2(kb) + s_4\sin(kb) + s_5\cos(kb) \right)\right]$$

$$=e^{i}\exp\left[ i\left((s_3+s_5)+(s_2+s_4)kb+\frac{1}{2}(2s_1-2s_3-s_5)k^2b^2+\cdots \right)\right]$$

where the infinite series is the Taylor series expansion for the trigonometric functions from the previous equality. What is important to note is that each power of $k$ will have an imaginary coefficient (as the addition of real numbers $s_i$ is multiplied by $i$) which go to zero (as sums and products of sinusoids are analytic, the coefficients must go to zero). Thus, we can rewrite this exponential as

$$=e^{i+g_0}\exp\left({\sum_{q=1}^\infty g_q k^q}\right)$$

where $g_q$ is a purely imaginary number. Having thoroughly run roughshod over all of our math at this point, we can put this series back into our original sum from $k=1$ to $m$:

$$\left|\sum_{k=1}^m\exp\left[ \frac{2\pi i l}{4N}\left(\sin\left(bk+1\right)+(-1)^n\right)^2\right]\right|=\left|\sum_{k=1}^me^{i+g_0}\exp\left({\sum_{q=1}^\infty g_q k^q}\right)\right|$$

$$\left|e^{i(1+\text{Im}(g_0)}\right|\left|\sum_{k=1}^m\prod_{q=1}^\infty\exp\left( g_q k^q\right)\right|$$

As the infinite product converges absolutely, we can switch it with the finite sum (and discard $e^{i(1+\text{Im}(g_0)}$)

$$=\left|\prod_{q=1}^\infty\sum_{k=1}^m\exp\left( g_q k^q\right)\right|=\prod_{q=1}^\infty\left|\sum_{k=1}^m\exp\left( g_q k^q\right)\right|$$

At this point, we are leaving what I at least am familiar with and instead journey into the land of special functions and limits in mathematica. That is, if anyone can fill in more details than what I will put down in this next section, please feel free.

First, note that we can approximate the finite sum with its integral counterpart to get

$$\sum_{k=1}^m\exp\left( g_q k^q\right)\approx\int_1^m\exp\left( g_q x^q\right)dx=\frac{E_{\frac{q-1}{q}}(-g_q)-m E_{\frac{q-1}{q}}\left(-g_q m^q\right)}{q}$$

Now, note that the next detail is slightly ambiguous, but it seems that since $Re(g_q)=0$ and $q>0$

$$\lim_{m\to\infty}m E_{\frac{q-1}{q}}\left(-g_q m^q\right)=0$$

Again, I do not have a proof for this and am only able to submit the following graph produced in mathematica as evidence

As you can see, for $1\leq q\leq 10$ and $-10\leq \text{Im}(g_q)\leq 10$ (with Re$(g_q)=0$) the limit is always $0$ (the artifact at the origin is simply because I set $m=1000$ to get this plot). Then for large $m$, the expression above becomes

$$\frac{E_{\frac{q-1}{q}}(-g_q)-m E_{\frac{q-1}{q}}\left(-g_q m^q\right)}{q}\approx \frac{E_{\frac{q-1}{q}}(-g_q)}{q}$$

In much the same way as above, I conjecture (and numerical evidence supports), that

$$\left|E_{\frac{q-1}{q}}(-g_q)\right|$$

goes to $0$ for large $q$ (mainly since $|g_q|\to 0$). Especially, it should be the case that this expression is bounded above by some constant (let us call it $Q$). Perhaps someone with more experience with complex analysis could investigate? With this, we get

$$\lim_{m\to\infty}\frac{1}{m}\prod_{q=1}^\infty\left|\sum_{k=1}^m\exp\left( g_q k^q\right)\right|\leq \lim_{m\to\infty}\frac{1}{m}\prod_{q=1}^\infty\left|\frac{Q}{q}\right|=0$$

After all that handwaving and limits by inspection, we have finally arrived at our original goal: namely that

$$a_k=\frac{1}{4N}\sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{k(2N)!}{n}\right)+(-1)^n\right)^2$$

is equidistributed on $[0,1]$. Now, we can return to more solid ground. By definition, there exist infinite $k$ such that $a_k<1/(64\cdot 4N)$. Define $x=k_0(2N)!$ where $k_0$ is the smallest such $k$ where the inequality holds. Then we have

$$\frac{1}{64\cdot 4N}>\frac{f(x)}{4N}=\frac{1}{4N}\sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{x}{n}\right)+(-1)^n\right)^2$$

$$\sum_{n=N+1}^{2N}\left(\sin\left(1+\frac{x}{n}\right)+(-1)^n\right)^2<\frac{1}{64}$$

Since this must hold as we sum from $N+1\leq n\leq 2N$, it must hold for the individual $n$. That is, for $N+1\leq n\leq 2N$

$$\left(\sin\left(1+\frac{x}{n}\right)+(-1)^n\right)^2<\frac{1}{64}$$
$$\left|\sin\left(1+\frac{x}{n}\right)+(-1)^n\right|<\frac{1}{8}$$

This implies that

$$\frac{7}{8} <\sin\left(1+\frac{x}{n}\right)\leq 1$$

for even $n$ and

$$-1 \leq\sin\left(1+\frac{x}{n}\right)<-\frac{7}{8}$$

for odd $n$. We can use these bounds in our original equation

$$\left|\sum_{n=N+1}^{2N}\frac{(-1)^n}{\sqrt{n}}\sin\left(1+\frac{x}{n}\right)\right|>\frac{7}{8}\left|\sum_{n=N+1}^{2N}\frac{1}{\sqrt{n}}\right|$$

Now, one bound for this sum is

$$\frac{7}{8}\left|\sum_{n=N+1}^{2N}\frac{1}{\sqrt{n}}\right|=\frac{7}{8}\sum_{n=N+1}^{2N}\frac{1}{\sqrt{n}}>\frac{7}{8}\int_N^{2N+1}\frac{1}{\sqrt{x}}dx$$

$$=\frac{7}{4}(\sqrt{2N+1}-\sqrt{N})>\frac{7}{4}\sqrt{2N}-\sqrt{N})=\frac{7}{4}\sqrt{N}(\sqrt{2}-1)$$

$$\geq\frac{7}{4}\sqrt{2}(\sqrt{2}-1)=\frac{7}{2}-\frac{7}{2 \sqrt{2}}=1.02513>1$$

This is a contradiction. Thus, the series in question does not uniformly converge.

In conclusion, we have proved that

$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}\sin\left(1+\frac{x}{n}\right)$$

does not converge uniformly on the reals. Unfortunately, there are a few details in the middle that need to be cleaned up by someone other than me as I do not trust my knowledge/expertise to do so.


Boundary operator: what does it mean? I'm getting some confusion in simplicial homology...Take a very simple example, a (solid) tetrahedron:

Following the well known property that "the bounday of a boundary is zero", we would end up with $\partial\partial=0$.

Instead, using the "chain complex" concept, the boundary operator seems to map the solid tetrahedron to the "void tetrahedron", and the latter to its edges.

So it seems to me that "taking the boundary" has at leat two different meanings...where am I wrong?

In Homology, we formally add and subtract directed simplices.

0-simplices are the vertices. They have boundary $0$ by definition.
1-simplices are directed segments, so that $BA=-AB$. $\ \partial(AB) =B-A$.
The $n$-simplex $A_1\dots A_n$ is considered to be the same as $A_{\sigma(1)}\dots A_{\sigma(n)}$ if $\sigma$ is an even permutation, and its negative if $\sigma$ is odd.
Define $\partial(A_1\dots A_n) =(A_2\dots A_n)\, - \, (A_1A_3\dots A_n) \, +\, \dots\,\pm\, (A_1\dots A_{n-1})$.
(the $i$th term omits $A_i$, with alternate signs)

Now, in particular for the $ABCD$ simplex, we get
$$\partial(ABCD)=BCD - ACD + ABD - ABC\\
\partial(XYZ)=YZ+ZX+XY\\
\partial(\partial(ABCD)) = BC+CD+DB\ - \ AC-CD-DA\ +\ AB+BD+DA\ -\ AB-BC-CA\ =\ 0$$

Simply because when you take one of these triangle and you apply once again the boundary operator you obtain the (oriented) segments of the triangle.
To put it "simply" the boundary operator decompose a figure into pieces that compose him.
The accuracy of my answer is maybe not too high, because I also begin in the study of such object.


Link between two gradient definitions Let $M$ a $n$ dimensional embedded Riemannian manifold of $\mathbb{R}^{n+1}$. There are two definitions of $\nabla_M f$ that seem different, but I think they should be related:
Def 1: $\nabla_M f = \nabla f - \langle\nabla f, n\rangle n$, where $n$ is the outer unit normal.
Def 2: Definition through parametrization.
By Def 1, one sees that $\nabla_M f \in \mathbb{R}^{n+1}$, while using Def 2 (in practice), we get $\nabla_M f \in \mathbb{R}^{n}$.
How to relate this two definitions? (Maybe, behind my confusion there is some hidden identification).
Edit 1:
Take, e.g., $f(x,y,z)=\frac{1}{2}x^2$ on $\mathbb{R}^3$. Then
by Def 1: $\nabla_{S^2} f(x,y,z)=(x-x^3, -x^2y,-x^2z)$ and by Def 2: $\nabla_{S^2} f(\phi(\theta,\varphi))=(-\frac{1}{2} \sin(2\theta),\frac{1}{2} \sin(2\varphi) \cos^2 \theta)$.
But in Riemannian geometry, they said that they are the same, they even prove one from the other.

Mind that $Def. 1$ is like all derivatives and gradients only define at a given point on $M$ where $f$ is sufficiently steady otherwise there is need to derive from directions towards the selected points. If there is set in $M$ where all the gradients exist then the gradient can be written in the abstracted form of Your question set. This inherent in the definition of gradients.
There very same is true is the definition of $f$ is some kind of path on $M$. The point matters not any paths along which the gradient is taken. The gradient exists if it is independent of the path of differentiation.
Your example is not so suitable since $f:\Bbb R^{3} \rightarrow \Bbb R$. This is not like in Your definition $\Bbb R^{3}$ with $n=2$ and Your $n$ is $1$.
Your $S^{2}$ is meant the surface of the unit sphere in $\Bbb R^{3}$ which has the dimension 2 but You forgot that $n$ from the $Def.1 $s is a unit vector. It has to have the form
$$\vec n=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$$
So
$\vec \nabla f(x,y,z)=(x,0,0)$
without any restriction to the sphere $S^{2}$.
An example normal on the $S^{2}$ is:
$\vec n=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$
$$\vec \nabla_{S^{2}} f(x,y,z)=(x,0,0)-<\vec \nabla f,\vec n>\vec n$$
$$=(x,0,0)-<(x,0,0),\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}>\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$$
$$=(x,0,0)-\frac{x^2}{\sqrt{x^2+y^2+z^2}}\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$$
$$=\frac{(x(x^2+y^2+z^2)-x^2,-x^{2}y,-x^2z)}{x^2+y^2+z^2}$$
Your results is rather different!
$\vec n$ is directed outwards. The same is true is $-\vec n$ is used but oriented inwards. There are other normals on $S^{2}$.
This shows up that the normal has to be specified in the definition before calculating the gradient. And the normal has to be assigned to a point on the unid sphere centered around zero for example the point (0,0,1) where the gradient becomes (0,0,-1).
I in the second example the same $f$ is used why are there only two components of the gradient? $r=1$ is only the definition of the unit sphere in spherical coordinates is not allowed to ignore the $r$ derivatives.
As a check transform the function $f$ properly. $x=r sin(\theta) cos(\phi)$ so
$$f(r,\phi,\theta)=\frac{r^2 sin(\theta)^2 cos(\phi)^2}{2}$$
The gradient is now with the projection on the normal:

Form this point with use the unit sphere restriction for our example normal. The example normal is the $\vec e_{r}$ unit vector of the coordinate system. $r=1$.
So the second term is
$$-(cos(\phi)cos(\theta))^2\vec e_{r}$$
The $\vec e_{r}$ is simply $cos(\theta)^2$. The others remain unchanged at $r=1$.


Convergence of $\lim_{n \to \infty} \left(\sum_{k=1}^{n}f(k)-\int_{1}^{n+1}f\left(x\right)dx\right)$ I have seen similar questions about specific cases (see here for $f(x) = \frac{1}{\sqrt{x}}$ and here for $f(x) = \frac{1}{x^a}$). I would like to find when $$\lim_{n \to \infty} \left(\sum_{k=1}^{n}f(k)-\int_{1}^{n+1}f\left(x\right)dx\right)$$

converges. I know that one possible way for this to converge is if $\int_1^{\infty}f(x)dx \leftrightarrow \sum_{k=1}^{\infty} f(k)$ converges. However, this is not the only way for it to converge such as with $f(x)= \frac{1}{x^a}, 0 \le a \le 1$. We can split the integral into parts so that the limit becomes $$\sum_{k=1}^{\infty}\left(f(k)-\int_k^{k+1} f(x)dx \right)$$

The summand must approach $0$ for the sum to converge. I believe this only happens if $f(x)$ is a function approaching $0$. However, this probably is not sufficient for the sum to converge. This brings me to my main question:

What conditions on $f(x)$ are necessary and sufficient for $\lim_{n \to \infty} \left(\sum_{k=1}^{n}f(k)-\int_{1}^{n+1}f\left(x\right)dx\right)$ to exist?

Edit: as jvc pointed in the comments, it is not necessary for $f(x)$ to approach zero, such as with $f(x)$ a constant. I now think it is necessary (perhaps even sufficient) for $f'(x)$ to approach $0$, not necessarily $f(x)$.

A sufficient condition is that $f$ is differentiable and $\int_1^\infty |f'(x) | \, dx < \infty$.

Integrating by parts we get

$$\int_k^{k+1} (x - k - 1)f'(x) \, dx = \left.(x- k- 1)f(x)\right|_k^{k+1} - \int_k^{k+1} f(x) \, dx  = f(k) -   \int_k^{k+1} f(x) \, dx$$

Thus,

$$\left|f(k) -   \int_k^{k+1} f(x) \, dx\right| = \left|\int_k^{k+1} (x - k - 1)f'(x) \, dx \right|\leqslant \int_k^{k+1} |x - k - 1|\,|f'(x)|\, dx \\ \leqslant \int_k^{k+1} |f'(x)|\, dx , $$

and we have absolute convergence of the series

$$\sum_{k=1}^{\infty}\left|f(k)-\int_k^{k+1} f(x)dx \right|\leqslant \sum_{k=1}^{\infty}\int_k^{k+1} |f'(x)|\, dx = \int_1^\infty |f'(x)| \, dx,$$

which implies convergence of

$$\sum_{k=1}^{\infty}\left(f(k)-\int_k^{k+1} f(x)dx \right) = \lim_{n \to \infty}\left(\sum_{k=1}^{n}f(k)-\int_{1}^{n+1}f\left(x\right)dx\right)$$

can we use the fact that:
$$\int_1^{n+1}f(x)dx=\lim_{N\to\infty}\sum_{i=0}^Nf\left(1+\frac{in}{N}\right)\frac nN$$
and compare the two sums?


Are there intractable problems? Obviously, there is no problem in NP which can be shown to be intractable (by which I mean: not in P).

Is there a problem (outside of NP) which can be shown to be intractable (not lying in P)?

What you want is the Time Hierarchy Theorem, which broadly speaking states that for any "reasonable" function $f$ there are problems that take about $f(n)$ time to decide.

Concretely, each of the known EXPTIME-complete problems would be an example of something that has been proved not to lie in P.

Here is a related question on compsci.

NP-hard / NP. A problem is in NP-hard when every problem L in NP can be reduced in polynomial time to H. Under the assumption of $P \neq NP $ ,NP-hard is intractable. Therefore NP-hard / NP is intractable.

https://en.wikipedia.org/wiki/NP-hardness


Basic prove that boolean function is self-dual I'm tring to prove this function:
   $$ f(x,y,z) = x'y'z'+x'yz+xyz'+xy'z $$

is self-dual, I've tried some basic manipulations like using double not on the function with de-morgan rules but got no results.
I also looked at Kerno's map and it's really suppose to be self-dual.
I would really appreciate if someone would guide me how to prove this basic thing...

thanks :).

Since $f(x,y,z)$ is true if and only if an even number (2 or 0) of $x,y$, and $z$ are true, the dual of $f(x,y,z)$ will be true if and only if it is not the case that an even number of $\neg x,\neg y$, and $\neg z$ are true, i.e., if and only if an odd number (1 or 3) of $x,y$, and $z$ are false. But that's equivalent to the original criterion for $f(x,y,z)$ being true, so the dual of $f(x,y,z)$ is the same Boolean function as $f(x,y,z)$.

See here, a function is self-dual if it is equal to its own dual.
So you simply take the dual of this function, and if it gives you the original function, then its self-dual. Otherwise, its not.

For what its worth, I did the calculation, and I didn't find it to be dual.


Singularities of $f(z)=z/\cos(z)$ Regarding complex functions (in complex variables), I was wondering why the function $g(z)= \cos(z)$ has a singularity at $z = \infty$ but $f(z)= \dfrac{z}{\cos(z)}$ does not.

I am a bit confused about the concept of singularities at $\infty$.

Thank you.

So, to sum up after all these comments:

$g(z)$ indeed has an essential singularity at $z=\infty$ because the function $h(z)=g\big(\frac1z\big)$ has an essential singularity at $z=0$:
$$h(z) = \cos\left(\frac1z\right) = 1- \frac1{2!}\left(\frac1z\right)^2 + \frac1{4!}\left(\frac1z\right)^4+ \dots $$
has infinitely many terms in its Laurent expansion at $0$.

On the other hand, $f(z)$, all appearances to the contrary, does not have an essential singularity at $\infty$: $f(z)$ has poles at $z=z_n=n\pi + \pi/2$ and $z_n\to\infty$, and so $\infty$ is not an isolated singular point.

For $\,\frac1z\,$ "close enough to zero" ( say, $\;\frac1{|z|}<\sqrt[3]6\;$ ) , we have

$$f\left(\frac1z\right)=\frac1{z\cos\frac1z}=\frac1z\frac1{\left(1-\frac1{2z^2}+\frac1{24z^4}-\ldots\right)}=$$

$$=\frac1z\left(1+\frac1{2z^2}+\frac1{4z^4}+\ldots\right)=\frac1z+\frac1{2z^3}+\ldots$$

which shows $\,f(z)\;$ has a singularity at $\,z=\infty\;$


Proving properties of the images and inverse images of functions Let $f : X\to  Y; \;\;g : Y\to Z;\,$ and $\;g \circ f : X\to Z.$

Prove or disprove

a) For all subsets $\,A \subseteq X,\;\; f^{-1}(f(A)) = A$.

b) For all subsets $\,B \subseteq Y,\;\; f(f^{-1}(B)) = B$.

c) For all subsets $\,E \subseteq Z,\; \; (g \circ f)^{-1}(E) = f^{-1}(g^{-1}(E))$

Well for any element $x\in X$, $f^{-1}(f(x))=x$,and since $A\subseteq X$,$x \in A \implies x\in X$,
$y\in Y$$$f(f^{-1}(y))=y$$
Since $B\subseteq Y$,$y \in B \implies y\in Y,$
$e\in E$
$$(g^{-1}(f^{-1}(e))=(f^{-1}(g^{-1}(e))$$
Since $E\subseteq Z$,and $f^{-1}:Y\to X$ means that e might not be in $Y$ therefore it's false


perfect coin is tossed n times. Let Sn denotes the number of heads obtained. What is the expectation of Sn? The Problem is: A perfect coin is tossed n times. Let Sn denotes the number of heads obtained. What is the expectation of Sn?

I got to E of $S_n$ = $\sum_{n=1}^{+\infty} \space\space\space Sn (\frac{1}{2})^n$

Let's say a coin is tossed once: we get the Expectation of number of heads as

E[R] = no.of heads*prob(no.of heads) = 1*1/2 = 1/2.

Now let n=2, E[R2] = 2*1/4 + 1*2/4 + 0*1/4 = 1.

Now let n=3, E[R3] = 3*1/8 + 2*3/8 + 1*3/8 = 12/8 = 3/2.

As we keep continuing this we observe that E[Rn] = n*1/2.

This is the linearity of Expectation. That is, E[R] = E[R1] + E[R2] + .. . + E[Rn].


map of graded rings and projective schemes Let $\phi$ a morphism between $S$ and $R$, two graded rings (there is a $d>0$ such that $S_{n}$ maps to $R_{dn}$ for all $n$). How could I show that this induces a morphism of schemes
$$\operatorname{Proj} R - V(\phi(S^+))\to \operatorname{Proj} S,$$
where $S^+$ is the irrelevant ideal?

As Cantlog says, this can be found in every introduction to algebraic geometry. For example EGA II, 2.8. Or Görtz-Wedhorn, Remark 13.7.


Polar Coordinate function of a Straight Line I was having some problem when trying to come out a polar coordinate function with straight line equation.

I know it is not good to post images here, but please bear with me as the question requires us to solve the equation from the straight line in the image.

What I have done is I tried to come out with an implicit function for the straight line L.

2x + 3y - 6 = 0

But then I not sure how can I continue from there to come out with a polar coordinate function. Any ideas?

Thanks!

$x = r\cos \theta\\
y = r\sin\theta\\
2x + 3y - 6 = 0\\
2r\cos\theta + 3r\sin\theta = 6\\
r(2\cos\theta + 3\sin\theta) = 6$

Now I could say:

$r = \frac {6}{3\cos\theta + 2\sin\theta}$

and be done.

But I think that this is a little bit more informative:

$\sin (\arctan \frac ab) = \frac {a}{\sqrt {a^2 + b^2}}\\
\cos (\arctan \frac ab) = \frac {b}{\sqrt {a^2 + b^2}}$

$r\sqrt {2^2 + 3^2}(\sin (\arctan \frac{3}{2})\sin\theta + \cos(\arctan \frac 32)\cos\theta) = 6\\
r\sqrt {13}(\cos(\theta - \arctan \frac 32)) = 6\\
r\sqrt {13}(\cos(\theta - \arctan \frac 32)) = 6\\
r = \frac {6}{\sqrt {13}} \sec (\theta - \arctan \frac 32)$

As it gives the angle of rotation and the distance to the line.

Given the linear equation in two intercept form, x/a+y/b=1
The equivalent polar form is r=(absec(theta))/(b+tan(theta))


Work on Springs using Hooke's Law I'm currently stuck on parts c and d of this problem. The problem says Suppose a force of 20 N is required to stretch and hold a spring 0.4 m from its equilibrium position (0). I found k constant to be 50 and part b required using integration on 0 to -0.3 for spring compression on the and I found the work to be 2.25 J. Here's part c and d...

c) does it take twice as much work to compress the spring 0.6 meters from equilibrium as it does to compress 0.3 meters from equilibrium?

d)Does it take twice as much force to compress the spring 0.6 meters from equilibrium as it does to compress 0.3 meters from equilibrium?

I'm assuming you just solve the work for 0.6 compression and compare it to 2.25 J for c but im not sure about d. Please let know your thoughts and ways to approach this.

Hint: From Hooke's Law
$$
F=kx
$$
we can derive the formula for spring potential energy by integrating with respect to position
$$
E=\frac12kx^2
$$

For any future people who'd like to know the answer to this question.

For c) you can just calculate the work for compressing the spring 0.6 meters and compare it to the work for 0.3.

And for d) all you have to do is plug in -0.3 and -0.6 and compare the result. Hooke's law assumes a linear spring ( F(x) = kx ) which is an odd function so stretching or compressing x amount either direction takes the same amount of force and since F(x) = kx doubling or tripling the x-value will do the same to the result, as shown below.

F(x)  = kx

F(2x) = k(2x) = 2kx

or

F(-x) = -kx

F(-2x) = k(-2x) = -2kx