TITLE: Question About Irreductibility of an element in a ring
QUESTION [1 upvotes]: I $A$ is a principal ideal domain  and let N  the set of non zero elements, that do not have an inverse and that cannot be decomposed in a product of irreductibles.
If $a \in N$
Let's create the set $I = aA$
I am trying to understand why a is not irreductible
(This is part of the proof of the fundamental theorem of arithmetics our professor gave us)
Is it just that if $a$ was irreductible then $a = a$ so it can be decomposed in a product of irreductibles?

REPLY [0 votes]: if A  a factorial domain , for $a\in A$, a not 0 and not invertible, we have equivalence :
a irreductible
a prime
aA irreductible ideal
aA maximal ideal in the set of principal ideal and propre
aA prime ideal


TITLE: How does the form affect sound traveling thru solid material
QUESTION [0 upvotes]: Take for example a boom pole: does its diameter affect how sound (like handling noise) is traveling towards the mic? Would a short section of smaller diameter (say 10% of the regular diameter) reduce the transmitted energy? And if so, would it affect all frequencies or would it act maybe as a hp or lp filter?

REPLY [1 votes]: Elastic systems (like your boom pole) have a resonant frequency which is related to the form of the system. Because a boom pole involves fairly complex junctions between solids it's considerably more complex than a hollow cylinder.  Still, you might start in your estimation using the equation for the natural frequency of a simply-supported hollow cylinder:
$$ f = 2 \pi  L^2 \ A \; \sqrt{E \;  \frac{I}{m}} $$
where:

$A$ = 9.87 for first mode
$L$ = length of pole (m)
$E$ = Young's modulus (N/m^2)
$I$ = area moment of inertia (m^4)
$m$ = mass per unit length of the pole (kg/m)

Your question about adding a short section of smaller diameter seems to me to be a question of insertion loss. Essentially what you're doing is breaking the pipe into three coupled systems - the first (now, shorter) tube of the original diameter, a tube of smaller diameter in the middle, and a second (also shorter) tube of the original diameter at the end.  I suspect it would act to make the transfer of energy less efficient and would reduce the levels at the end of the boom.
I don't know how to perform the calculations, but you'd expect - for the same material and wall thickness - the smaller diameter tube to be less elastic.  Thus, a higher Young's modulus, thus a higher natural frequency.

The last part of your question involves impedance.

Would it affect all frequencies or would it act maybe as a hp or lp filter?

You'd expect your smaller section to be driven by the natural frequency of the larger diameter section near your hands.  That is, a driving frequency < the natural frequency.  As above, we expect quite a bit of dampening of the lower pipe's resonant frequency.

But we should expect the impulsive sound of handling noise to be broadband in nature.  In that case the center section would do a good job of transmitting energy at it's natural frequency, which would be dampened again somewhat by the larger section of pipe at the end.  Again, I'm not sure of the calculations but intuitively that strikes me as more of a bandpass filter than either a high or lowpass filter.


TITLE: How prove this there exist positive integer sets $A$ such $|A+A|>|A-A|$
QUESTION [4 upvotes]: let positive integer $n\ge 8$ is given, show that: there exist sets $A$ with the set of is positive integer.such $|A|=n$,and such
$$|A+A|>|A-A|$$
where
$$A-A=\{a-b|a\in A,b\in A\},A+A=\{a+b|a\in A,b\in A\}$$
This problem seem like this How prove this set inequality $|B|\ge 2|A|^2-1$
I search some paper, Found this problem background is :
The following are called Minkowski sum and difference of sets:
$$A-A=\{a-b|a\in A,b\in A\},A+A=\{a+b|a\in A,b\in A\}$$
The other notation is the cardinality $ |X|$ of a set $ X$
But I can't prove this, maybe $n\ge 8$ is useful

REPLY [3 votes]: You will find much discussion of this question if you type "more sums than differences" into the web. For example, here's a paper by Greg Martin and Kevin O'Bryant that should answer your question, and more.


TITLE: Why there is a relation among the second-order minors of a symmetric $4\times 4$ matrix?
QUESTION [35 upvotes]: A $4\times 4$ symmetric matrix
$$
\left(
\begin{array}{cccc}
 a_{11} & a_{12} & a_{13} & a_{14} \\
 a_{12} & a_{22} & a_{23} & a_{24} \\
 a_{13} & a_{23} & a_{33} & a_{34} \\
 a_{14} & a_{24} & a_{34} & a_{44} \\
\end{array}
\right)
$$
contains exactly 21 minors of order 2, but there is a linear combination of them, namely
$$
-(a_{13} a_{24}-a_{14} a_{23})+(a_{12} a_{34}-a_{14} a_{23})-(a_{12} a_{34}-a_{13} a_{24})\, ,
$$
which vanishes, as opposed as to the case of symmetric $3\times 3$ symmetric matrices, whose 6 minors of order two are linearly independent.

BIG QUESTION: Why is that? I mean, what is the theory behind such a phenomenon?
PHILOSOPHICAL QUESTION: What is the "true number" of minors (from the example above), 20 or 21?

I already gave myself an explanation, but I still cannot see the big picture. I would be grateful if anyone pointed out a reference, sparing me the efforts of reinventing the wheel.
If an $n\times n$ matrix $A$ is regarded as an element of $V\otimes_{\mathbb{K}} V^\ast$, with $V\equiv \mathbb{K}^n$, then there is an obvious way to extend $A$ to a $\mathbb{K}$-linear map
$$
A^{(k)}:\bigwedge^kV\longrightarrow\bigwedge^kV^\ast\, .
$$
If $A$ is symmetric, then so is $A^{(k)}$, i.e., there is a (polynomial of degree $k$) map
$$
S^2(V^\ast)\ni A\stackrel{p}{\longmapsto} A^{(k)}\in W^k:=S^2\left( \bigwedge^kV ^\ast \right)\, .
$$
Observe that $W_k$ has dimension $\frac{{n\choose k}\left({n\choose k}+1\right)}{2}$, which is 21 for $n=4$ and $k=2$. So, I was led to identify $W^k$ with the space of "formal minors" of order $k$ of a $n\times n$ matrix (indeed, the entries of $A^{(k)}$ are precisely such minors).
How to explain now the dependency of three of them?
There is a linear map
$$
S^2\left( \bigwedge^2V ^\ast \right)\ni\rho\odot\eta\stackrel{\epsilon}{\longmapsto}\rho\wedge\eta\in \bigwedge^4V ^\ast\equiv\mathbb{K}\, ,
$$
and it can be proved that $\epsilon\circ p=0$, i.e., that $p$ takes its values in the subspace
$$
W^2_0:=\ker\epsilon\, ,
$$
which has dimension 20 (vanishing of the above linear combination corresponds precisely to the equation $\epsilon(p(A))=0$).

SIDE QUESTION: How to define this $W_0^k$ for arbitrary $n$ and $k$? It should be the "linear envelope" of the image of $p$: but how to exhibit it explicitly?

Probably, the theory of representations may answer this: $W^k$ is not always an irreducible $\mathfrak{gl}(n,\mathbb{K})$-module, and $W_0^k$ is the only irreducible component which contains the image of $p$. Provided this guess is true, it doesn't help me finding the expression of $W_0^k$ in terms of tensors on $V$.

REPLY [15 votes]: This is a comment regarding Robert Bryant's conjecture: It is indeed true and not hard if you know the representation theory of $GL(V)$ well enough. In fact, the following stronger statement is true:

The only common constituent of $(S^2 V)^{\otimes k}$ and $(\bigwedge^k V)^{\otimes 2}$ is the representation with highest weight $2 \alpha_k$.

One must then check that this constituent lies in the symmetric parts of the tensor powers, which is probably most easily done by pointing out that the $k \times k$ minors of a symmetric matrix give a nonzero $GL(V)$ equivariant map $(\bigwedge^k V)^{\otimes 2} \longrightarrow (S^2 V)^{\otimes k}$.
Write weights as partitions $(\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n)$, so $S^2 V$ has height weight $(2,0,0,\ldots,0)$ and $\bigwedge^k V$ has highest weight $(\overbrace{1,1,\ldots,1}^k,0,0,\ldots,0)$. Tensoring with $S^k V$ and with $\bigwedge^k V$ is described by the Pieri rule and its transpose. From the Pieri rule, any irreducible constituent $V_{\lambda}$ of $(S^2 V)^{\otimes k}$ has $\sum \lambda_i=2k$ and $\lambda_{k+1} = \lambda_{k+2} = \cdots = \lambda_n=0$. From the transpose of the Pieri rule, any $V_{\lambda}$ in $(\bigwedge^k V)^{\otimes 2}$ has all $\lambda_i \leq 2$. The only way to satisfy these conditions is $\lambda = (\overbrace{2,2,\ldots,2}^k,0,0,\ldots,0)$.
More generally, the only common constituent of $(S^{\ell} V)^{\otimes k}$ and $(\bigwedge^k V)^{\otimes \ell}$ is the representation of highest weight $\ell \alpha_k$.
It is definitely in $S^{\ell} (\bigwedge^k V)$. If $\ell$ is even, then it is in $S^k(S^{\ell} V)$, otherwise it is in $\bigwedge^k (S^\ell V)$.


TITLE: Rings with exactly 3 ideals
QUESTION [1 upvotes]: So I am reading some notes which are a bit confusing so I’m trying to make sense of everything.
The statement is
A commutative ring $R$ has exactly 3 ideals if and only if:

$I := R \setminus R^* $ is a non zero ideal.
$I=\langle r \rangle $ for some non zero $r \in I $ .
$r^2=0 $ for all $r\in I$.

Here $R^*$ is the unit group.
I can show that if $R$ has exactly 3 ideals then 1, 2 and 3 hold. But for the converse, do we need all 3 conditions to hold for $R$ to have 3 ideals or are the conditions s equivalent so we just need one of them?

REPLY [1 votes]: You only need conditions 2) and 3), since 2) implies 1). Let $R$ be a commutative ring.
The first/second condition tells us that $R$ contains at least three ideals $(0) \subset I \subset R$.
Now let $J$ be any non-zero, proper ideal of $R$. By definition of $I$ we have that $J \subset I$. Our goal is to show that $J = I$. Now, certainly $J$ contains some non-zero non-unit element $s \in J$, so we have the chain of inclusions $$
(s) \subset J \subset I
$$
Thus if we can show $(s) = I$ we are done.
Now, by condition 2) we can write $I = (r)$ for some non-zero $r \in I$. We claim that for any $a \in R$, if $a \notin R^*$ then $ar = 0$.
Assume that $a \notin R^*$. Then $a \in I$. Since $I = (r)$, you have $a = br$ for some $b \in R$. But then $ar = br^2 = 0$ by condition 3), as desired.
Now, since $s \in I$, we can write $s = ar$ for some $a \in R$. Since $s$ is nonzero, by the above discussion $a$ is a unit, thus $a^{-1}s = r$. So $I = (r) \subset (s)$.
Thus we have $I \subset J \subset I$, hence $J = I$ and $R$ has exactly three ideals.
Condition wise, it seems to be enough to say that $I = R\setminus R^* = (r)$ for some nonzero $r$ with $r^2 = 0$.


TITLE: Fourier Transform of $f(t)=4te^{-t^2}$
QUESTION [0 upvotes]: I am trying to find Fourier Transform of:
$$f(t)=4te^{-t^2}$$.
I found in MatLab that $\mathfrak{F}\left \{ f(t) \right \}=i\sqrt{2}e^{- \frac{w^2}{4}}w$ .So is this possible to come to same result by using shifting theorem since the function is Gaussian.
What I was thing was: Suppose $g(t)=-2e^{-t^2}$ and it's first derivative is $g'(t)=4te^{-t^2}$ so by using these facts I am trying to use properties of Fourier Transforms to come to same result.

REPLY [1 votes]: we know that $f'(x) = -2 a x f(x) \implies \ln f(x) = -a x^2 + cste \implies f(x) = C e^{-a x^2}$.
now if $g(t) = e^{-a t^2}$ then $g'(t) = -2 a t g(t)$ and if $a > 0$, $|g(t)|$ and $|g'(t)|$ fastly  decreases when $t \to \infty$ so :
$$i \omega \int_{-\infty}^\infty g(t) e^{- i \omega t }dt =  \int_{-\infty}^\infty g'(t) e^{- i \omega t }dt = -\int_{-\infty}^\infty 2 a t g(t) e^{- i \omega t }dt =  -2 a i \frac{d}{d\omega} \left(\int_{-\infty}^\infty g(t) e^{- i \omega t }dt\right) $$
hence $$\int_{-\infty}^\infty g(t) e^{- i \omega t }dt = C e^{-\omega^2/(4 a)}$$ and $C = 1/\sqrt{2 a}$ is obtained from the Parseval theorem


TITLE: Is it correct to say the field of complex numbers is contained in the field of quaternions?
QUESTION [5 upvotes]: I believe it is correct to refer to the complex numbers and their 'native algebra' as a field.  See for example Linear Algebra and Matrix Theory, by Evar Nering.  I assume the same may be said for the set of quaternions.  Please correct me if that is wrong.
Based on that assumption, I ask: is the field of complex numbers a subset (sub-field?) of the quaternions in the same sense that the real numbers are 'contained' in the complex numbers?
Is there a term for such a subordination of number fields?
I admit that I am reaching well beyond my realm of familiarity, and that my question may be nonsensical to those who are more knowledgeable in these matters.

REPLY [1 votes]: The Hurwitz-Frobenious theorem is your friend here.  It states that the only normed division algebras over the real numbers are the reals, the complex numbers which are obtained by doubling the reals, the quaternions that are obtained by doubling the complex numbers, and the octonions that are obtained by doubling the quaternions.
With each doubling, you lose something.  You lose ordering moving to the complex numbers.  You lose commutativity moving to the quaternions, and you lose associativity moving to the octonions.
You can continue the process to get the Clifford algebras with dimension $2^n$, but you loose the existence of inverses for all non-zero elements, and so these are no longer normed division algebras.
The doubling process is mechanically isomorphic to constructing $2\times2$ matrices from two elements of the previous algebra with with the diagonal duplicating one value 'a' and the cross diagonal conjugating the other value 'b'.
Geometrically, the quaternions are a four dimensional vector space with Euclidean norm that admits a well defined multiplication with inverses.  This multiplication is a representation of the four dimensional rotations (which rotate a plane through the origin and fix the two dimensions orthogonal to that plane).  The three dimensional rotations are a proper subset of the four dimensional rotations.
Any plane through the origin within the quaternions is algebraically isomorphic to the complex numbers, which in turn may be thought of as a representation of the two dimensional rotations.
These are the basic facts, consideration of which should serve to clarify your terminology and answer your questions.


TITLE: Poisson structure on the cotangent bundle
QUESTION [5 upvotes]: Let $X$ be a smooth variety over $\mathbb{C}$ and $\mathscr{A}$ a sheaf of twisted differential operators on $X$. The latter comes equipped with a natural filtration and the associated graded algebra $\text{gr } \mathscr{A}$ is identified with $\text{Sym } \mathscr{T}_X$, the sheaf of functions on the cotangent bundle $T^*X$.
Both of these algebras have a Poisson structure: $\text{gr } \mathscr{A}$ the one induced by the commutator, and $\text{Sym } \mathscr{T}_X$ the one arising from the canonical symplectic structure on $T^*X$. Can anyone explain to me why these Poisson structures coincide under the aforementioned isomorphism?
Edit: When I say "sheaf of twisted differential operators" I just mean a quasicoherent sheaf of filtered associative algebras $\mathscr{A}$ on $X$ together with an isomorphism $\text{gr } \mathscr{A} \cong \text{Sym } \mathscr{T}_X$. The answer probably doesn't involve any difficult computations but I'm pretty murky about what's happening on the cotangent bundle. Any general remarks in that direction, especially from an algebraic perspective, would be helpful.

REPLY [1 votes]: Both of these constructions are étale local on $X$.  I'm using étale since you seem to want to work in algebraic geometry; it's probably better to say it's local in the classical/analytic topology.
In either case, I only need to show this for $\mathbb{C}^n$.  In that case, it's simply a matter of writing things down; my TDO is trivial now, so it's just algebraic differential operators on C^n.  It's associated graded is the polynomial ring $\mathbb C[x_1,\dots, x_n,\partial_1,\dots, \partial_n]$ with the rule that $\{\partial_i,x_j\}=\delta_{ij}$.
On the other hand, if you trace through the definition of the Poisson bivector on $T^*\mathbb{C}^n$, then what happens?  If you want to compute $\{w_i,x_j\}$, where $w_i$ is the dual coordinate to $x_i$ on $T^*\mathbb{C}^n$: you turn these into vector fields using the symplectic form (you get $d/dx_i$ and $-d/dw_j$), and you pair these using the symplectic form.  If you chose the symplectic form $\sum_i dw_i\wedge dx_i$ then you get a cookie, because you exactly get $\{w_i,x_j\}=\delta_{ij}$, and all is right.  You'll note, I ignored the Poisson brackets of "like" coordinates, but these follow easily from the same principle.
If you want to do this in a coordinate-free way, what you should do is realize the computations you need to do are:

$\{f,g\}=0$ for functions on $X$ in both cases (easy)
$\{Y,f\}=Y(f)$ for $f$ a function on $X$ and $Y$ a vector field on $X$, thought of as a function on $T^*X$ linear on the fibers.
$\{Y,Z\}=[Y,Z]$ for $Y$ and $Z$ both vector fields (here $[Y,Z]$ is Lie bracket); at that point you'll be set, since the symmetric powers of the tangent bundle are generated by these.

These are essentially defining properties for a TDO, so we just need to check them for the canonical structure.
The latter 2 equations follow from a basic property: that if you think of a vector field as a function on the contingent bundle, and turn it into Hamiltonian vector field using the symplectic form, you'll get the vector field itself on the base and a natural extension of it to the cotangent bundle which you can think of as something like geodesic flow. (In any set of coordinates, it's clear what to do, but the coordinate free version isn't coming to mind).  Since we can think of $\{Y,f\}$ or $\{Y,Z\}$ as Lie derivative with respect to this vector field, the desired formulas follow.


TITLE: To solve a PDE by separation of variables $\displaystyle 4\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=3u$
QUESTION [1 upvotes]: To Solve: $\displaystyle 4\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=3u$
My attempt:
Let $\displaystyle u=X(x)Y(y)$. So, $\displaystyle 4X'Y +XY'=3XY$
Separating the variables, $\displaystyle \frac{4X'-3X}{X}=\frac{-Y'}{Y}=k$(say)
$\displaystyle X'-\frac{(3+k)}{4}X=0$
$\displaystyle Y'+kY=0$
Integrating, $\displaystyle X=c_1e^{(k+3)x/4}$
and $\displaystyle Y=c_2e^{-ky}$
So, $\displaystyle u=c_1c_2e^{(k+3)x/4-ky}$
Now I check this against the initial value given as $\displaystyle u=3e^{-y}-e^{5y}$ when $x=0$
The final answer also (obviously) has two components $u=3e^{x-y}-e^{2x-5y}$
How come there are two components here? What did I miss?

REPLY [2 votes]: I don't have time to type it out in Mathjax, I have taken an image of the solution and please pardon me if it is not clear.  Thanks

REPLY [1 votes]: As the PDE is linear and homogeneous, a linear combination of two solutions is another solution. So, even though your assumed solution is of the form $X(x)Y(y)$, you can get a solution of the form $X_1(x)Y_1(y) + X_2(x)Y_2(y)$.
The two terms in the solution are each separate solutions.
Why is the sum of two solutions again a solution? Suppose $u_1$ and $u_2$ two solutions. Then $4\dfrac{\partial (u_1 + u_2)}{\partial x} + \dfrac{\partial (u_1 + u_2)}{\partial y} = 4\dfrac{\partial u_1}{\partial x} + \dfrac{\partial u_1}{\partial y} + 4\dfrac{\partial u_2}{\partial x} + \dfrac{\partial u_2}{\partial y} = 3u_1 + 3u_2 = 3(u_1 + u_2)$, so $u_1 + u_2$ is also a solution.


TITLE: Continuity for absolute value function
QUESTION [0 upvotes]: These are true/false questions.

If f is continuous on the open interval I, then |f| is continuous on I.
If |f| is continuous on the open interval I, then f is continuous on I.

I need help picturing these, any feedback is greatly appreciated.

REPLY [1 votes]: For the first question, note that
$$
||f(x)| - |f(c)|| \leq |f(x) - f(c)|;
$$
so $f$ continuous implies $|f|$ is continuous.
For the second question, consider the function $f: x \mapsto 1$ on $]0,1[$ and $\mapsto -1$ on $\{ 1 \}$ and $\mapsto 1$ on $]1, 2[$. Then the domain of $f$ is the open interval $]0, 2[$ and $|f|$, being a constant function, is continuous on $]0,2[$; but $f$ is discontinuous at $x = 1$. You can see the discontinuity by taking $\varepsilon := 1/2$, say.


TITLE: Sum of continuous functions in non-metrizable space
QUESTION [2 upvotes]: I found the following problem in Brezis, Functional Analysis. It seems obvious, but I can't find a start point to solve it.

Let $X$ be a topological space and $E$ be a Banach space. Let $u,v : X \to E$ be two continuous maps from $X$ with values in $E$ equipped with the weak topology $\sigma(E,E^*)$.

Prove that the map $x \mapsto u(x)+v(x)$ is continuous from $X$ to $E$ equipped with $\sigma(E,E^*)$.

Let $a : X \to \Bbb{R}$ be a continuous function. Prove that the map $x \mapsto a(x)u(x)$ is continuous from $X$ into $E$ equipped with $\sigma(E,E^*)$.

Since the weak topology is not metrizable, I cannot use an approach similar to metric spaces case. Please give me a hint. Thank you.

REPLY [4 votes]: The $\sigma(E,E^*)$ topology on $E$ is an example of an
initial topology; that is, it is the coarsest topology on
$E$ that makes $\ell : E\to\mathbb{R}$ continuous for all
$\ell\in E^*$.
Consequently, a map $f:X\to E$ from another topological
space $X$ is continuous iff every composition
$\ell\circ f$ is continuous from $X$ to $\mathbb{R}$.
For instance, if $u,v:X\to E$ are continuous, then so are
$\ell\circ u$ and $\ell\circ v$. Therefore,
 $x\mapsto\ell(u(x)+v(x))=\ell(u(x))+\ell(v(x))$ is
continuous into $\mathbb{R}$, and so $x\mapsto u(x)+v(x)$
is continuous into $E$.
Similarly,  $x\mapsto\ell(a(x)u(x))=a(x)\ell(u(x))$ is
continuous into $\mathbb{R}$, and so $x\mapsto a(x)u(x)$
is continuous into $E$.


TITLE: How many ways are there of tying a tie?
QUESTION [3 upvotes]: I am sorry if this is useless.
I have read in a newspaper that mathematicians have found the number of ways a tie can be tied. How could such a problem be solved?
I'm asking out of curiosity.

REPLY [3 votes]: You can read the solution to your question here http://en.wikipedia.org/wiki/The_85_Ways_to_Tie_a_Tie. [solution of 1999, seems outdated]


TITLE: Evaluate $\lim_{n \to \infty} \sum_{j=0}^{n} \sum_{i=0}^j \frac{i^2+j^2}{n^4+ijn^2}$
QUESTION [1 upvotes]: I am asked to evaluate: $$\lim_{n \to \infty} \sum_{j=0}^{n} \sum_{i=0}^j \frac{i^2+j^2}{n^4+ijn^2}$$
I am not experienced with double summations, but I tried simplifying the expression above into:
$$\lim_{n \to \infty}\frac{1}{n^2} \sum_{j=0}^{n} \sum_{i=0}^j \frac{{\left(\frac{i}{n}\right)}^2+{\left(\frac{j}{n}\right)}^2}{1+\left(\frac{i}{n}\right) \left(\frac{j}{n}\right)}$$
Perhaps this is Riemann integral (double)?

REPLY [1 votes]: To evaluate the integral we use symmetries and geometric and telescoping series viz.$$\begin{align}\int_0^1\int_0^x\frac{x^2+y^2}{1+xy}dydx&=\frac12\int_0^1\int_0^1\frac{x^2+y^2}{1+xy}dxdy\\&=\int_0^1\int_0^1\frac{x^2dxdy}{1+xy}\\&=\sum_{n\ge0}(-1)^n\int_0^1x^{n+2}dx\int_0^1y^ndy\\&=\sum_{n\ge0}\frac{(-1)^n}{(n+1)(n+3)}\\&=\frac12\sum_{n\ge0}\left(\frac{(-1)^n}{n+1}-\frac{(-1)^n}{n+3}\right)\\&=\frac12\left(1-\frac12\right)\\&=\frac14.\end{align}$$


TITLE: Understanding Proof that Interval $[0,1]$ is Compact.
QUESTION [5 upvotes]: On page 44-45 of the notes from http://www.msc.uky.edu/droyster/courses/fall99/math4181/classnotes/notes5.pdf, the author writes in the proof that the  interval $[0,1]$ is compact that supposing the interval isn't compact, that this implies either $[0,1/2]$ or $[1/2,1]$ isn't compact. How do we know that both $[0,1/2]$ and $[1/2,1]$ are not compact?

REPLY [3 votes]: Suppose $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ are compacts. Now take any open cover of $[0,1]$. This open cover is an open cover of $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$. Now you use the compactness of $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ and conclude that $[0,1]$ is compact.


TITLE: Fundamental group of the n-fold torus is nonabelian for n > 1
QUESTION [1 upvotes]: I'm attempting to show that the $n$-fold torus (i.e. the connected sum of $n$ tori, $T_1 \# T_2 \# \ldots \# T_n$) has non-abelian fundamental group when $n > 1$.
It has been suggested that I consider mapping the free group on the $2n$ letters $\alpha_1, \beta_1, \ldots, \alpha_n, \beta_n$ (where $\alpha_i, \beta_i$ are generators of $\pi_1(T_i, b)$) to the free group on $2$ letters $\gamma, \delta$, such that $\alpha_1, \beta_1 \to \gamma$ and $\alpha_i, \beta_i \to \delta$ for $i \neq 1$.
I am able to show that this map is a homomorphism. And I can show that if there were a homomorphism from $\pi_1(T\#\ldots\# T, b)$ to a non-abelian group (such as the free group on $\gamma, \delta$), then $\pi_1(T\#\ldots\# T, b)$ would necessarily be non-abelian itself. But $\pi_1(T\#\ldots\# T, b)$ is not the free group on 2n letters, but rather this group mod the least normal subgroup containing $[\alpha_1, \beta_1] \cdots [\alpha_n, \beta_n]$.
I've also attempted another approach, via considering the $n$-fold torus as a $4n$-gon with appropriate identifications of edges, but wasn't able to draw anything concrete from that. Perhaps I simply lacked the appropriate geometric intuition there?

REPLY [4 votes]: As you point out, $\pi_1(\Sigma_g) \cong \langle \alpha_1, \beta_1, \dots, \alpha_g, \beta_g \mid [\alpha_1, \beta_1] \cdots [\alpha_g, \beta_g]\rangle$. Define a homomorphism from this to the free group on two letters by $\alpha_1 \mapsto x_1$, $\beta_1 \mapsto x_2$, $\alpha_2 \mapsto x_2, \beta_2 \mapsto x_1$, and all other generators map to $1$. Because $[x_1,x_2][x_2,x_1] = 1$, this does indeed define a homomorphism $\pi_1(\Sigma_g) \to F_2$, and because each of the generators are mapped to, it's surjective, so $\pi_1(\Sigma_g)$ is nonabelian.
You can see explicitly why this fails for $g=1$.


TITLE: Find the infinite sum $\sum^\infty_{n=2}\frac{\left(-1\right)^{n}}{\log n}$, speed up its convergence
QUESTION [1 upvotes]: $\displaystyle\sum_{n\in\mathbb{Z}_{\geqslant{2}}}\frac{\left(-1\right)^{n}}{\log n}$
Since $\log x$ grows more slowly than any positive power, the sum above converges extremely slowly. Is there a series that converges faster and converges to the same value as this sum? Or an integral?

REPLY [2 votes]: A first transformation that gives an equivalent series which converges a it more rapidly is to break the series into even and odd terms, and let $n = 2k$ or $2k+1$.
then you get
$$
\sum_{k=1}^\infty \left( \frac{1}{\log (2k)} -  \frac{1}{\log (2k+1)} \right) =
\sum_{k=1}^\infty \frac{\log\left( 1+\frac 1{2k}\right)}{\log(2k)\log(2k+1)}
$$
And now you can work with that because  $\log\left( 1+\frac 1{2k}\right) = \sum_1^\infty \frac{(-1)^{m+1}}{(m+1)(2k)^m} $
$$
\sum_{k=1}^\infty \frac{\log\left( 1+\frac 1{2k}\right)}{\log(2k)\log(2k+1)} =
\sum_{k=1}^\infty \frac{\sum_{m=1}^\infty \frac{(-1)^{m+1}}{(m+1)(2k)^m}}{\log(2k)\log(2k+1)} = \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m+1}\sum_{k=1}^\infty \frac{ 1}{(2k)^m\log(2k)\log(2k+1)}
$$
Now we see another alternating sum, in $m$, and we can apply the same sort of technique, but the algebra gets a bit messy if we do. However, when $m > 1$ the sum over $k$ converges reasonably rapidly due to the power of $k^m$ in the denominator. So we are left with just one headache, that is how to quickly approximate
$$
\sum_{k=1}^\infty \frac{ 1}{(2k)\log(2k)\log(2k+1)}
$$
That is predictably slow to converge, since with9out the logs it would diverge and the logs grow slowly. However, it does converge much more rapidly than the original sum.


TITLE: How long will it take to complete Shannon's project?
QUESTION [1 upvotes]: If Shannon has a project that will take 550 hours to complete and she only works on the project 3 days a week for only 3-4 hours a day, about  how long will it take Shannon to complete her project?

REPLY [0 votes]: We could solve your problem with quite classical methods, but it is sometimes interesting to look at some problems differently to develop your intuition for other problems. Sometimes, it is useful to compare the quantities of the things involved in the wording of a problem.
Suppose she had a work requiring $h$ hours and she works $x$ hours a day in average, $y$ days a week. It takes $h/x$ days to complete the work and it takes $(h/x)/y$ weeks to complete it, i.e. $7\cdot(h/x)/y=\frac{7h}{xy}$ days in total.
Justifications: if she works $x$ hours per day of work, it would take her $h/x$ consecutive days to complete it. But she only works $y$ days per week, so it will take her $(h/x)/y$ weeks to complete it. There are $7$ days in a week, so it will take her a total amount of $7\cdot(h/x)/y$ days to complete the job.
The "most difficult" part of the reasoning may be the one from $(h/x)$ consecutive days to $(h/x)/y$ weeks. But look at the units: $h$ is in hours (unit: $\text{hours}$), $x$ in hours a day (unit: $\frac{\text{hours}}{\text{days}}=\text{hours}\cdot\text{days}^{-1}$) and $y$ in days per week (unit:$\frac{\text{days}}{\text{week}}$):
$$\frac{h}{x}\cdot\frac{1}{y}\Rightarrow\underbrace{\frac{\text{hours}}{\text{hours}\cdot \text{days}^{-1}}}_{(h/x)}\cdot\underbrace{\frac{\text{weeks}}{\text{days}}}_{1/y}=\text{weeks}$$
Then, you just have to know that a $1\,\text{week}$ is equivalent to $7\,\text{days}$ so that there are $7$ times more $\text{days}$ than weeks required to complete the job, i.e. $7\cdot\frac{h}{x}\cdot\frac{1}{y}=\frac{7h}{xy}$.
For your particular example, it takes her $7\cdot(550/3.5)/3\approx 336.67$ days in total.