TITLE: An example of a divisible ideal
QUESTION [0 upvotes]: Let $R$ be a commutative ring (other than a field) with identity. I am looking for an example of a divisible ideal of $R$.

REPLY [1 votes]: Not sure what notion of divisibility you're interested in.
If you are interested in it only being a divisible abelian group, then you can take any divisible abelian group $G$ and look at the matrix ring
$$
R=\left\{\begin{bmatrix}n&g\\0&n\end{bmatrix}\mid n\in \Bbb Z, g\in G\right\}
$$
The group $G$ is embedded as an ideal consisting of the set of strictly upper triangular matrices. (Actually this is a relative construction of Hurkyl's example. This is a subring of a "split-null extension", whereas his is a Dorroh extension.)

If you are interested in the definition Thomas Andrews mentions in the comments, (where you are checking for divisibility over non-zero divisors) then my matrix ring still works, since the zero divisors are exactly the elements with $n=0$. (This version of divisibility coincides with Lam's definition below when the ring is a domain.)

There is yet another good definition in Lam's Lectures on modules and rings:

An $R$ module $M$ is divisible if for any $m\in M$ and $r\in R$ such that $r.ann(r)\subseteq ann(m)$, there exists $n\in M$ such that $m=nr$.

You can also find it proven there that a ring is von Neumann regular iff all right $R$ modules are divisible (in this sense). In particular, all of its ideals would be divisible.


TITLE: What Are The Physics and Air Pressure Effects of Extremely Large Bells
QUESTION [2 upvotes]: How would one go about figuring out the physical properties of when a bell is hit.
Can I parameterize the diameter of the bell, its height, and the kinetic force of the knock to get the volume in decibels?
ultimately, I want to figure out its shockwave effects. Assuming that this bell is gargantuanly massive, like maybe the size of a large building or skyscraper being hit by 5 kN of bus sized hammer.
( (0.5)(8000kg)(5 m/s)2 ) / 20 m = 5 kN of force
.
I was reading about this Bell of Lost Souls, a bell as big as a cathedral, whose toll forced the populace into bunkers to escape death. Could the implications of such a massive bell be that bad?

REPLY [1 votes]: First, some background on sound intensity and its relation to pressure etc. can be found at:
https://physics.stackexchange.com/a/266046/59023
https://physics.info/intensity/
Second, this question turned out to be much more interesting and complicated than I had initially anticipated.  Some great discussions/background on bell acoustics can be found in the following references:

Fletcher, N.H., et al., "Bell clapper impact dynamics and the voicing of a carillon," J. Acoust. Soc. Am. 111(3), pp. 1437--1444, doi:10.1121/1.1448517, 2002.
Woodhouse, J., et al., "The Dynamics of a Ringing Church Bell," Adv. Acoust. Vib. 2012, pp. 19, doi:10.1155/2012/681787, 2012.

Can I parameterize the diameter of the bell, its height, and the kinetic force of the knock to get the volume in decibels?

Yes, but it's not trivial, it turns out (see discussion in references above).  We can, however, do some simple order of magnitude estimates.
Let's start with an absolute limit to get an idea of upper/lower bounds on values.  Suppose the sound pressure intensity generated by this hypothetical, Bell of Lost Souls is the maximum possible in Earth's atmosphere, namely ~194 dB.  This would correspond to a maximum gauge pressure, $\Delta P_{max}$ ~ 100,237.5 Pa.  We know that the gauge pressure can be defined as:
$$
\Delta P = 2 \pi \ \rho \ f \ C_{s} \ \Delta s \tag{0}
$$
where $\rho$ is the mass density of air, $f$ is the frequency, $C_{s}$ is the speed of sound (because we assume sound waves), and $\Delta s$ is the displacement the particles in the sound wave from equilibrium.  We know that at standard temperature and pressure (STP) that $\rho$ ~ 1.225 kg m-3 and $C_{s}$ ~ 331.3 m s-1.  If we assume the sound emitted occurs at f = 1000 Hz, then we find that $\Delta s_{max}$ ~ 0.039 m.  This displacement may look small, but normal sound waves that don't cause your head to explode only displace air particles by, at most, a few 10s of micrometers.
Note that while the sound wave propagates at $C_{s}$, the particles oscillating in the longitudinal mode have velocities that vary above and below this value.  The magnitude of the velocity oscillations is given by:
$$
\Delta V = 2 \pi \ f \ \Delta s \tag{1}
$$
which then results in $\Delta V_{max}$ ~ 247 m s-1.  As an aside, at this point you should start to be concerned about the applicability of our linear assumptions to this point since the velocity fluctuations are approaching the magnitude of the mean, i.e., $\Delta V_{max}/C_{s}$ > 70%.
The next part requires that we make some assumptions about the bell size and composition.  Suppose the local area of the bell of interest is ~100 m by 100 m (or an area $A$ ~ 104 m2) with a thickness, $d$ ~ 1 m.  We also assume the bell is made of steel so that we know its Young's modulus is $E$ ~ 200 GPa.  We can estimate a rough potential energy per unit area of the bell by treating it as a thin plate such that:
$$
\frac{ U_{s} }{ A } \approx \frac{ E \ \Delta s^{2} }{ 2 \ s_{o} } \tag{2}
$$
where we assume $s_{o}$ ~ 1 m then we find (assuming 100% radiation efficiency) to be $U_{s}/A$ ~ 154 MJ m-2.  If we multiply by the area we assumed above, then $U_{s}$ ~ 1.54 TJ.
As discussed by Woodhouse et al. [2012], the conversion of kinetic energy of the clapper (i.e., the object striking the bell surface) to vibrational energy is smaller than 100% (not by a tremendous amount, but definitely smaller).  So we can get a lower bound on the impact speed of the clapper if we assume a mass.  Let's assume the clapper has a mass of ~1000 kg then its speed must satisfy $V_{c}$ $\gtrsim$ 56 km s-1 (or $\gtrsim$125,000 mph).  This is over five times the escape velocity of Earth's gravity, so not very realistic.  So we increase the clapper mass to ~1 Mkg and then $V_{c}$ $\gtrsim$ 1.8 km s-1 (or $\gtrsim$4,026 mph), which is still absurd.
Note that even though the energy conversion efficiency is ~90%, there are numerous other issues that will alter these estimates (e.g., bell shape, shape of clapper surface at impact, effective masses associated with different excited normal modes, etc.).

Assuming that this bell is gargantuanly massive, like maybe the size of a large building or skyscraper being hit by 5 kN of bus sized hammer.

So we can estimate the force on the bell surface (very handwavy) as just $U_{s}/\Delta s_{max}$ ~ 39.3 TN.  This is very likely a bad estimate as there is some deformation and displacement of the bell during impact, so the displacement value of $\Delta s_{max}$ is likely too large here.  That is, the impacting force is almost certainly much larger than 39 TN.  For a reference, Fletcher et al. [2002] provides some typical values on a normal sized bell in the area of ~20--150 kN (i.e., ~200 million times smaller).

I was reading about this Bell of Lost Souls, a bell as big as a cathedral, whose toll forced the populace into bunkers to escape death. Could the implications of such a massive bell be that bad?

Well, I think the human ear drum ruptures somewhere just north of ~155--160 dB.  Every ~3 dB of intensity increase is basically doubling the intensity amplitude so 194 dB has an intensity ~2512 times larger.  Note that it is very possible to kill a living organism with sound if it has sufficient intensity at certain frequencies.  Shock waves are a nonlinear type of sound wave and the abrupt change in pressure can exert lethal forces on a human body.


TITLE: Ellipse in Quadratic Form: Finding Intercepts with Principal Axes
QUESTION [3 upvotes]: Where an ellipse is expressed in quadratic form (e.g. $ax^2 + bxy + cy^2 = k$ is expressed as $x^TQx = k$), the principal axes are in the directions given by the eigenvectors of Q. I understand this.
Now, if we want to find the intercepts of the ellipse and the principal axes, apparently they can be found as intercepts = $\pm \sqrt{k/\lambda_i}$ (where $\lambda_i$ is the eigenvalue corresponding to the eigenvector which describes the axis we are looking for the intercept with).
What I don't understand is how to derive this. I can see how to derive the intercepts in terms of $a, b, c, k$ but not in terms of the eigenvaluesand $k$.
Can someone please help shed some light on this - perhaps provide a derivation?
Thanks in advance for your help.

REPLY [5 votes]: Suppose $(\hat x , \hat y)$ is an eigenvector corresponding to the eigenvalue $\lambda_i$. In this case $Q$ is given by:
$$Q = \left( \begin{array}{cc}
a & \frac{1}{2} b \\
\frac{1}{2} b & c \end{array} \right).$$
Then we have $Q \binom{\hat x}{\hat y} = \lambda_i \binom{\hat x}{\hat y}$, and so
$\binom{\hat x}{\hat y}^T Q \binom{\hat x}{\hat y} = a\hat x^2 + b\hat x \hat y + c \hat y^2 = \lambda_i (\hat x^2 + \hat y^2)$. To find out where the ellipse intersects the principal axis (ie, the line going through $0$ and $(\hat x, \hat y)$), we need to scale the eigenvector so it lies on the ellipse. In other words, we need to choose a $\sigma$ such that $\binom{\sigma \hat x}{\sigma \hat y}^T Q \binom{\sigma \hat x}{\sigma \hat y} = \lambda_i ((\sigma \hat x)^2 + (\sigma \hat y)^2)$ = k. Thus the distance from the origin to the intercept is given by:
$$\sqrt{(\sigma \hat x)^2 + (\sigma \hat y)^2} = \sqrt{\frac{k}{\lambda_i}}$$


TITLE: Is the function $f(x)=x^TAx$ convex when $A \in S^n,A\geq0$,$x\in R^n$?
QUESTION [4 upvotes]: Is the function $f(x)=x^TAx$ convex  when $A \in S^n,A\geq0$,$x\in R^n$?
Notation
$S^n$: Symmetric $n$ x $n$ matrix.
$R^n$: Column vector $n$x$1$
$A \geq 0$: $A$ is positive semi-definite matrix

I know that $x^TAx$ is  convex  when $A \in S^n,S\geq0$,$x\in R^n$ but I not sure why exactly this is happening. Can somebody elaborate on it?
Is it safe to say that when $A\geq 0$ and $x$ is non negative the function
$x^TAx$ is a parabola in $R^n$ towards $+\infty$. I mean in $R^2$  $f(x)=ax^2+bx+c$ is a parabola. But is there parabola in $R^n$ for example in $R^3$, is $f(x,y)=ax^2+by^2+cxy +d$  a parabola on $R^3$ or there is no such
thing?

REPLY [3 votes]: We can check the convexity of $f$ directly. First, we observe that for all $p\in[0,1]$ and $q=1-p$ and $r,s\in\mathbb{R}$, we have
$$
p^2r^2+2pqrs+q^2s^2\leq pr^2+qs^2
$$
because $\text{RHS}-\text{LHS}=pq(r-s)^2$. Now let's use this: for $x,y\in\mathbb{R}^n$, $p\in[0,1]$, and $q=1-p$, we can write $s^2=x'Ax$, $r^2=y'Ay$ and have the following:
$$
f(px+qy)\leq p^2r^2+2pqrs+q^2s^2\leq pr^2+qs^2=pf(x)+qf(y).
$$
The first inequality above is a simple application of the Cauchy-Schwarz inequality. The claim follows.


TITLE: Weights of BGG dual of Verma module
QUESTION [4 upvotes]: Consider the Verma module $M(\lambda)=\oplus_{\mu \in \mathfrak{h}^{\star}} M(\lambda)_{\mu}$. Denote its BGG dual by $M(\lambda)^{\vee}=\oplus_{\mu \in \mathfrak{h}^{\star}} M(\lambda)^{\star}_{\mu}$ as in section $3.2$ of Humphreys' book.
When $\lambda$ is anti-dominant, $M(\lambda)$ is simple and $M(\lambda) \cong M(\lambda)^{\vee}$. My question is whether
$$\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{-\mu}?$$
I have doubts because on one hand, I feel that since $M(\lambda) \cong M(\lambda)^{\vee}$ we should have $\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{\mu}$. On the other hand, I feel that the weights of $M(\lambda)^{\vee}$ should be negative of the weights of $M(\lambda)$. Hence, it might be true that $\dim M(\lambda)^{\vee}_{\mu}=\dim M(\lambda)_{-\mu}$.
Thank you for explaining.

REPLY [4 votes]: If you compute the dual of a weight module $M$ using the usual notion of dual of a module over a Lie algebra, then you get a module $M^{\ast}$ such that $M^{\ast}_{-\mu}$ is the vector space dual of $M_{\mu}$ (and thus has the same dimension).  This why you think the displayed equation is correct.
But that's not the notion of dual that anyone uses in category $\mathcal{O}$, because as you can see it doesn't send objects in category $\mathcal{O}$ to objects in category $\mathcal{O}$.  Thus, $M^\vee$ is something different: it's the usual vector space dual (EDIT: not actually the full one; as the OP writes, it's the sum of the vector space duals of the individual weight spaces, which is a version of graded dual), with the action twisted by the Cartan involution $E_i\mapsto -F_i, F_i\mapsto -E_i, H_i\mapsto -H_i$ (for the classical groups, this is transpose and negate).  Because of the negative sign in front of $H_i$, this flips the sign of weights, and $M^\vee_\mu$ is the vector space dual of $M_\mu$ (and thus the same dimension).


TITLE: The application of Nimbers to Nim strategy
QUESTION [2 upvotes]: I've been reading about combinatorial game theory, and some works start with the game of Nim.  After that, they introduce Nimbers, which are numbers that represent Nim games.  So far so good.  I get that, and I understand Nimber addition.
But I don't understand how any of this helps to play Nim.
So I'm interested in knowing how to use Nimbers to play Nim better.  Specifically..

How do Nimbers represent Nim games?
What does Nimber addition mean in the context of Nim?
What does Nimber multiplication mean in the context of Nim?

If there's a worked-through example of playing Nim via Nimbers, that would be awesome.

REPLY [4 votes]: It's been a while since I've read up on Nimbers (introduced to me by the immortal Martin Gardner), but if I remember correctly, nimbers are basically regular numbers representing the number of stones in a Nim heap. The nimber "sum" is the XOR of the nimbers, and represents a sort of "combined" heap, except that removing stones can either increase or decrease the resulting total nimber, but will always change the total. In particular, since nimbers are always nonnegative, if the total nimber is $0$, there is no place to go but up. Note that the case of all heaps empty has nimber $0$.
So this is the strategy: at the start of your turn, calculate the total nimber. If it is not $0$, determine a change that will make it $0$ and make that move (nim-add each heap to the total you've already calculated (start with the largest heap) and when you find one where the result is less than the heap size, reduce that heap to the result). If the total is $0$, hope and pray that your opponent makes a mistake.
Because it is always possible to reduce a total nimber to $0$ in one move, and since no move will keep a total nimber of $0$ at $0$, as long as you are the one reducing the nimber to $0$, your opponent can't win. They must always move in a way that increases the nimber, meaning that it is never $0$ at the end of their turn. Which means that they cannot reduce all heaps to $0$, since that position has a total nimber of $0$.


TITLE: A function and its derivative chasing tails
QUESTION [2 upvotes]: For which $t\ge0$ does there exist a differentiable function $f$ with $f(0)=0$, $f'(x)>f(x)$ for all $x>0$ and with $f'(0)=t$?
This question was inspired by (and is a variation of) the following question and its answer.
Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $f'(x)>f(x)$ for all $x>0$.
It could be shown then, as in the answer referred to above,
that $f(x)>0$ for all $x>0$.
The solution goes as follows (with some details added).
Let $g(x)=e^{-x} f(x)$. Because $g'(x)=e^{-x}\Bigl(f'(x)-f(x\Bigr)>0$ we have that $g(x)$ is strictly increasing on the interval $(0,\infty)$. Since $g(0)=0$ we have $g(x)>0$ for all $x\in(0,\infty)$, hence $f(x)=e^x g(x)>0$ whenever $x\in(0,\infty)$.
On close inspection this argument works even if we add to the assumptions the requirement that $f'(0)<0$. This seems contradictory, since then on one hand $f(x)>0$ for all $x>0$, but on the other hand $f(x)<0$ for all $x\in(0,\varepsilon)$, for some small enough $\varepsilon>0$. This contradiction only shows that no such $f$ exist that satisfy all these conditions (including
$f'(0)<0$).
Thus the following:
Question.  For which $t\ge0$ does there exist an $f$ with $f'(0)=t$,
and with $f(0)=0$, $f'(x)>f(x)$ for all $x>0$?
Easily $t≥1$ works, e.g. $f(x)=e^{tx}−1$. I wonder if any $t<1$ might also work, in particular if $t=0$ works.
I feel the solution may have something to do with the notion of blow up as understood in this answer, but I don't seem to be able to figure the details. Tried to get a modification of $e^{-\frac1{x^2}}$ but it only works on a bounded interval of positive reals, not on all of $(0,\infty)$. Thank you for your help.
(I included tag integration only because I thought the answer may have also something to do with the average value of $f$ on any interval $[a,b]$, defined as $\frac1{b-a}\int_a^b f(x) dx$. In particular the average value of $f'(x)$ on $[0,b]$ is $\frac1b\int_0^b f'(x) dx=f(b)$.)

REPLY [1 votes]: You could go with a function of the form
$$f(x) = g(x)e^x,$$
for some nonnegative $g(x)$ with positive derivative. Then
$$f'(x) = g'(x)e^x + f(x) > f(x).$$
Also, I think this is related to the question I asked here:
Effect of differentiation on function growth rate
about function growth rates.

Edit:
Also, if you are only interested in the existence of such a function and not a simple formula, one could use the existence and uniqueness theorem for ODEs. One sets up the following ODE:
$$f'(x) = f(x) + w(x),$$
with the boundary condition
$$f(0)=0.$$
Then for any "gap" function $w$ with
$$w(0)=t,$$
the other "boundary condition"
$$f'(0)=0 + w(0) = t$$
is also guaranteed to be satisfied as well.


TITLE: Termination of Bellman-Ford algorithm
QUESTION [0 upvotes]: Good day,
I am currently going over some problems from my optimisation class and I can not seem to figure this one out.

Assume that in the Shortest Path Problem, Bellman-Ford terminates but some vertices still remain unchanged i.e. $y_v=\infty$ for some $v\in V$
Show that $\exists  \ X\subset V $ such that $c(uv)=\infty \ \forall \ uv \in A, \ u \in X, \ v\in V\setminus X $ (there is a subset $X$ such that the cost of each edge with one vertex in $X$ and one outside of it is $\infty$).

I am not really sure how to get started on this one.

REPLY [1 votes]: I think what might help you is the following:
The Bellman-Ford algorithm computes one-to-all shortest paths by repeated relaxation. If at the end of the algorithm some vertices still have value $\infty$, then those vertices must be in a different connected component than the starting vertex $a$. Otherwise they would have been reached over some path starting from $a$ and then they wouldn't have value $\infty$ anymore.
Now to solve this you have to somehow define your set $X$. I suggest you try with $X = \{ v \in V \text{ such that v has value } \infty \text{ when Bellman-Ford has terminated} \}$.
(The cost of an edge between to vertices $u, v$ is considered $\infty$ if in the original graph there is no such edge. At least that's what I assume. The transformation from no edge to an edge with value $\infty$ is made for Bellman-Ford to operate on the graph.)


TITLE: Distribution Theory. How to solve $x^{m}{u}=\delta$?
QUESTION [2 upvotes]: I just got to study this topic and trying to get familiar with new sources.
I want to find all $u$ in $D'(\mathbb R)$ such :
$$ {x}^{m}{u}=\delta.$$
where $m$ is a positive integer.
My attempt was to try $u=\operatorname{p.v.} \frac{1}{x^m}$.
Any hint?

REPLY [0 votes]: Paul Garrett's comment gives a particular solution, then you'll need all the solutions to $x^m w=0$ that you'll find from the same method as in your previous question (take $\psi$ constant $=1$ on $[-1/2,1/2]$ to simplify the computation).


TITLE: Quick Volume with Vertical Lines Question
QUESTION [0 upvotes]: So I was given the following prompt with the graph:
"Let $M$ be the region in the first quadrant bounded by the graphs of $y=\arctan(x)$ and $y=\frac{\pi}{4}x$, as shown in the figure below. What is the volume of the solid generated when region $M$ is revolved around the vertical line $x=2$?"

So I guess I'm just a bit confused about what my volume equation might look like here. I understand that the equations here would both have to be expressed in terms of $y$, but I'm a bit lost about where to go after that. Any help would be appreciated!

REPLY [3 votes]: You have to find volume of the area bound between curves $C1$ and $C2$ when rotated around $x = 2$
$C1: y=\frac{\pi}{4} x$ and $C2: y=\arctan x$ $(0 \leq x \leq 1, 0 \leq y \leq \frac{\pi}{4}$)
When you rotate it around $x = 2$, please note the radius of the rotation of both curves at any given $y$.
For $C1$, $r = 2 - \frac{4y}{\pi}$
For $C2$, $r = 2 - \tan y$
So your triple integral to find volume should be $\displaystyle \int_0^{2 \pi} \int_0^{\pi/4} \int_{2 - \frac{4y}{\pi}}^{2 - \tan y} r \ dr \ dy \ d\theta$


TITLE: Muon neutrino and electron neutrino discoveries and their Nobel prizes
QUESTION [3 upvotes]: In this lecture, around 44:00
https://youtu.be/KU94cMZif8Y?t=2635
The lecturer mentioned quickly that the 2nd neutrino (of muon) results in a Nobel prize earlier than the 1st neutrino (of electron's).

Puzzle: Could some experts please explain why and how the research results happened? Naively
the 2nd neutrino (of muon) is heavier than the 1st neutrino (of electron's). So why the 1st neutrino (of electron's) appears to be awarded or confirmed later?

REPLY [1 votes]: The 1956 discovery of electron antineutrino from nuclear reactions is credited to Clyde Cowan and Frederick Reines in their famous Savannah River experiment. They based the expected number of neutrino events to a parity-conserving weak interaction theory. Cowan and Reines reported the observed neutrino events consistent with parity-conserving theory.
However, the parity violation was confirmed in Wu experiment the following year. A parity-violating theory (so-called V$-$A theory) predicts the number of neutrinos to be twice the amount Cowan and Reines claimed. After these developments on the theory side, Reines insisted that he overestimated the detector efficiency. In their renewed experiment in 1960, Cowan and Reines reported that the number of neutrinos is consistent with parity-violating theory. One could interpret this inconsistency suspicious, and could be reason for the 39-year-delayed Nobel prize for the neutrino discovery.

Naively the 2nd neutrino (of muon) is heavier than the 1st neutrino (of electron's).

We have not been able to measure the masses of the neutrinos, only the upper limits. We also do not now which one of these two neutrinos is heavier, in a statistical sense.


TITLE: How do I know if this equation can be solved symbolically?
QUESTION [0 upvotes]: Can these equations be solved symbolically for $x$?
$$
\begin{align}
  x &= \frac{p - p_m(x)}{p_m(x) - p_m(x)^2} \\
  \\
  p_m(x) &= \frac{e^x}{e^x + e^y}           \\
\end{align}
$$
If not (which I assume), how do I know?

REPLY [0 votes]: If it could be done, it would be most likely done with the Lambert W function.  However, with the amount of addition in there, it is unlikely you could solve it like so.
So putting this all in, we have
$$x=\frac{p-p_m(x)}{p_m(x)[1-p_m(x)]}$$
$$=\frac{(e^x+e^y)\left(p-\frac{e^x}{e^x+e^y}\right)}{e^x\left(1-\frac{e^x}{e^x+e^y}\right)}$$
$$=\frac{(e^x+e^y)[p(e^x+e^y)-e^x]}{e^{x+y}}$$
$$=e^{-x-y}[p(e^{2x}+2e^{x+y}+e^{2y})-e^{2x}-e^{x+y}]$$
$$=e^{-x-y}[(p-1)e^{2x}+(2p-1)e^{x+y}+2e^{2y}]$$
$$x=(p-1)e^{x-y}+(2p-1)+2e^{y-x}\tag1$$
At this point, I am fairly confident there is no way to solve for $x$.  We can, however, approach the problem numerically to get close values for $x$.
For Newton's method, we have
$$a_{n+1}=a_n-\frac{(p-1)e^{a_n-y}+(2p-1)+2e^{y-a_n}-a_n}{(p-1)e^{a_n-y}-2e^{y-a_n}+1}$$
And $x=\lim_{n\to\infty}a_n$.
Perhaps if you make $p$ or $y$ known, the problem could be handled better, but I am unsure if there is much you can do here.

Furthermore, if we assume the answer is some very large $x$, then we have
$$\lim_{x\to\infty}p_m(x)=1$$
We would also have
$$\lim_{x\to\infty}x=\lim_{x\to\infty}\frac{p-p_m(x)}{p_m(x)[1-p_m(x)]}\approx\lim_{p_m(x)\to1}\frac1{1-p_m(x)}$$
Upon further inspection, we find that $p_m$ will approach $1$ from the negative side, and thus it can be seen that for $x$ to be large, we must have $p<p_m=1$ for both sides to be positive and large.
Assume negative and large $x$, and we will find
$$\lim_{x\to-\infty}p_m(x)=0$$
$$\lim_{x\to-\infty}x=\lim_{p_m\to0^+}\frac{p-p_m(x)}{p_m(x)[1-p_m(x)]}$$
For that to be true, we must have $p<0$.


TITLE: Understanding distances and unit vectors in this problem
QUESTION [0 upvotes]: A charge $-q$ is at $x = -a$ and a second charge $+q$ is at $x = a$

Between the charges, the contribution from each charge is in the negative direction.
An expression for $${\overrightarrow{E}} = \frac {kq}{(x-a)^2} \hat{{e_+}} + \frac{k(-q)}{(x+a)^2} \hat{{e_-}}$$ for $-a< x< a$
Where $\hat{{e_+}}$ is a unit vector that points away from the point $x=a$ for all values of $x$ (except $x=a$) and $\hat{{e_-}}$ is a unit vector that points away from the point $x=-a$ for all values of $x$.
The first thing I don't understand is that if you take any point between the origin of coordinates and $a$, the distance from $a$ to the point is $x-a$. But if you  place that point to the left of the origin of coordinates  the distance is not $x-a$ but $x+a$.
Then you can't have the expression $\frac {kq}{(x-a)^2} \hat{{e_+}}$ because if the point is to the right of the origin you have $x-a$, but if it is to left you have $x+a$.
Also I don't understand the explanation about the unit vectors here, $\hat{{e_+}}$ and $\hat{{e_-}}$

REPLY [1 votes]: To try and explain how the equation was arrived at for $-a< x<a$, ie the centre region, I have drawn all the relevant unit vectors including those that have to do with the directions of the electric fields produced by the two charges $\hat E_+$ and $\hat E_-$.

Adding the two electric fields between the two charges with the individual fields being shown as a magnitude plus a direction.
$$\vec E_{\rm centre}= \frac{kq}{(a-x)^2}\, (-\hat x)\, + \,  \frac{kq}{(a+x)^2}\, (-\hat x)$$
Noting that in the centre region $-\hat x = \hat e_+ = - \hat e_-$  and $(a-x)^2 = (x-a)^2$   you obtain the equation given in your question.


TITLE: Why is the proof of Gödel's first incompleteness theorem no contradiction?
QUESTION [2 upvotes]: I consider the following version of Gödel's first incompleteness theorem:
Assume $F$ is a formalized system which contains Robinson arithmetic $ Q$. Then a sentence $G_F$ of the language of $F$ can be mechanically constructed from $ F$ such that:
- If $F$ is consistent, then $F$ $ ⊬$ $G_F$
- If $F$ is $1$-consistent, then $F$ $⊬$ $¬G_F$
In the proof we first use the Diagonalization Lemma and apply it to the negated provability predicate $¬Prov_F(x)$: this gives a sentence $G_F$ such that
(A) $F ⊢ G_F ↔ ¬Prov_F(⌈G_F⌉)$.
Where $⌈G_F⌉$ is the Gödel number of $G_F$. Thus, it can be shown, even inside $F$, that $G_F$ is true if and only if it is not provable in $F$.
It is not difficult to show that $G_F$ is neither provable nor disprovable in $F$, if $F $ only is $1$-consistent by assuming $F$ is provable, ... and so on and so forth.
But when we prove $G_F$ is unprovable didn't we also prove $G_F$, since A holds? That is, we have proved $G_F$ and at the same time we have shown we cannot prove $G_F$.
This seems contradictory to me, or am I missing something? I guess the answer must have to do with Tarski's undefinability of truth.

REPLY [3 votes]: This is a great question, and highlights a common issue with Godel's theorem: where does the proof take place?
You're quite right that if the entire proof took place inside $F$, then this would be a contradiction. But it doesn't! It takes place in the slightly stronger theory $F+1Con(F)$ (actually, Rosser later improved this to $F+Con(F)$). That is:

$F$ knows that, if $F$ is 1-consistent, then $G_F$ is true and not $F$-decidable.
$F+1Con(F)$ knows that $G_F$ is true and not $F$-decidable.
In particular, $G_F$ is $F+1Con(F)$-decidable, but not $F$-decidable.
So there's no contradiction.

A couple comments:

Rosser's improvement was the following. Instead of the Godel sentence $$\mbox{$G_F$="I am $F$-unprovable",}$$ Rosser used the sentence $$\mbox{$R_F$="For any $F$-proof of me, there is a shorter $F$-disproof of me."}$$ (Just like for $G_F$, the formulation of $R_F$ in the language of arithmetic uses Godel numbering and the fixed point theorem.) It's a good exercise to show that $R_F$ let's us bring the consistency assumption down to just $Con(F)$.
Note that we can iterate the Godel (or Rosser) process! We can consider a sequence of stronger and stronger theories, starting with some $F$ and repeatedly adding the Godel (or Rosser) sentence. It turns out this leads to some deep ideas in computability theory; see e.g. https://mathoverflow.net/questions/12865/using-consistency-to-create-new-axioms-in-set-theory.


TITLE: Intuition behind $x^T A x$
QUESTION [2 upvotes]: I have been encountering the expression $x^T A x$ (where $x$ is a vector and $A$ is a matrix) in many topics related to linear algebra. However, I fail to grasp the essence; what is happening here? For example, I know (or more like, have been told) that if the expression is greater than zero for all vectors $x$, then we claim that $A$ is positive-definite. I would like to understand why such property - and many others - matter by understanding why should we care about $x^T A x$.
I understand that $x^T x$ is a dot product and that $x^T A$ and $A x$ are vectors of linear combinations of $x$ defined by columns and rows of $A$ respectively.

REPLY [1 votes]: To me most obvious use of $x^TAx$ is that it can represent lot's of quadratic functions.
For example, if:
$x=\begin{pmatrix}
x \\
y \\
\end{pmatrix}$ and $A=\begin{pmatrix}
a & b \\
b & c \\
\end{pmatrix}$ the expression will result in:
$$x^TAx=ax^2 + byx + cy^2$$
So by choosing the right values for $(a,b,c)$ you can encode a variety of quadratics.
Whether the matrix $A$ is positive-definite matters a lot in optimization problems, because if the matrix is positive-definite the resulting function will be convex, and will have a minimum.
You can improve on the reach of quadratic functions you can represent by adding the terms $x^Tb + c$ where $b$ is a n-vector and $c$ is a real a constant. This way you have $x^TAx + x^Tb + c$, which can represent any quadratic. And again, it will have a minimum vector, if and only if the matrix $A$ is positive-definite.
I hope you have gained some insight as to why the properties of the matrix $A$ are important and why $x^TAx$ is too. Am happy to answer any doubts about what I've written.


TITLE: In how many ways can 2k people be seated on a table such that there are n couples, and each couple must sit together?
QUESTION [0 upvotes]: Edit: The table is ROUND.
By sitting together, I mean that the husband and wife must sit together. I think we should first fix one husband, giving us two options for the wife. We should then consider the remaining couples as blocks and the singles as elements( giving us $2k - 2n + n - 1$ ?) elements and then for each couple block, there are 2 arrangements inside the block so $2^n * (2k-n-1)!$ is the ans???

REPLY [0 votes]: A broad hint, not a complete solution
I could not follow your answer, but it might be worth checking it with an example like $k = 3, n = 1$ to see whether it works.
Rather than placing the husband first, I'd say that you have $n$ items of one type (couples) $C$ and $2(k-n)$ items of another type, singles, $S$.
Label the seats of the table $1, 2, 3, \ldots, 2k$. A seating plan then consists of

a sequence of $n$ C's and $2(k-n)$ S's.
An allocation of the couples to the $C$s and the singles to the $S$s
For each couple, a choice of who gets the first of the pair of seats.

How many items of type $1$ are there? [That's the hard part to work out]
Assigning the couples to the $C$s can be done in $n!$ ways. What about assigning the singles to the $S$s?
Picking a first member for each of $n$ couples amounts to choosing a sequence of $n$ $0$s and $1$s; how many of those are there?

For the first bullet, a stars-and-bars kind of solution is probably what you want to think about.


TITLE: Checking differentiability at the origin
QUESTION [4 upvotes]: Let $f: \mathbb{R}^2 \to \mathbb{R} $ be given as
$$
f(x,y) =
\begin{cases}
e^{\frac{-1}{x^2+y^2}}, & \text{if }\text{ $(x,y) \neq (0,0)$} \\
0, & \text{ }\text{ $(x,y)=(0,0)$}
\end{cases}
$$
I want to check $f$ is not differentiable at the origin. My attempt would be to consider the sequence $\mathbf{x_n} = (\frac{1}{n}, \frac{1}{n} ) \to (0,0) $. But, notice
$$ f( \mathbf{x_n}) = e^{\frac{-1}{2/n^2}}= e^{-\frac{n^2}{2}} \to e^0 = 1 \neq f(0,0) = 0$$
Hence, $f$ is not continuous at $(0,0)$. In particular, it cannot be differentiable at the origin. Is this a correct approach?

REPLY [1 votes]: The definition of differentiability can be stated in many ways.  The way that I prefer (and usually teach my students) is this:
Definition: We say that $f(x,y)$ is differentiable at $(x,y)=(a,b)$ if
$$
\lim_{(x,y)\to(a,b)}\frac{f(x,y)-L(x,y)}{\sqrt{(x-a)^2+(y-b)^2}}=0,
$$
where $L(x,y)$ is the linear approximation to $f$ near $(a,b)$:
$$
L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).
$$
In your particular case, $(a,b)=(0,0)$. Using the definition of the partial derivative, you can show that $f_x(0,0)=f_y(0,0)=0$. This means that the linearization is
$$
L(x,y)=0+0(x-0)+0(y-0)=0.
$$
So, you need to check whether or not
$$
\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}\overset{?}{=}0.
$$
Since in the limit as $(x,y)\to(0,0)$, we can assume $(x,y)\neq(0,0)$, this is equivalent to checking whether or not
$$
\lim_{(x,y)\to(0,0)}\frac{e^{-1/(x^2+y^2)}}{\sqrt{x^2+y^2}}\overset{?}{=}0.
$$
Now, this limit is a perfect candidate for checking in polar coordinates:
$$
\lim_{(x,y)\to(0,0)}\frac{e^{-1/(x^2+y^2}}{\sqrt{x^2+y^2}}=\lim_{r\to0^+}\frac{e^{-1/r^2}}{r}\overset{?}{=}0.
$$
Can you see how to do this?


TITLE: Question about Real and Complex primes (Algebraic number theory)
QUESTION [2 upvotes]: I posted same question ago, but there was no answer.
I'm reading Jurgen Neukirch, Algebraic number theory, p.183

Now I can't understand the underlined statement.
Why that  $\mathfrak{p}$ is real or complex embedding depends whether the $K_{\mathfrak{p}}$ is isomorphic to $\mathbb{R}$ or to $\mathbb{C}$? (By the Ostrowski theorem, $K_{\mathfrak{p}}$ is isomorphic to $\mathbb{R}$ or $\mathbb{C}$)

REPLY [1 votes]: The completions of an embedding is either $\mathbb R$ or $\mathbb C$, depending on wheter the embedding is real or not (this is trivial!).
Now we know that if $v$ is an archimedian valuation on $K$ then up to equivalence $v$ comes from an (archimedic) embedding of $K$. This means that the bijection into the embedding is isometric under the valuation. Note that in $\mathbb R/\mathbb C$ there is only one valuation up to equivalence.
So if we take the completion $K_v$ it must be isomorphic to the completion of the embedding, which is either $\mathbb R$ or $\mathbb C$, depending on whether the embedding is real.


TITLE: "Vectors", i.e. (1,0)-tensors, their definition and motivation for relativity
QUESTION [4 upvotes]: I'm reading Einstein Gravity in a Nutshell (by Zee) and here he defines a vector as an object which is invariant under coordinate representation; concretely, if in one coordinate representation, $V$, $p_V=(p^1,p^2)$ then when we transform it via  a rotation $p_W = R(\theta)p_V$ then we do not violate any physical laws.
To motivate the question further, I understand that individual terms after a transformation are not preserved but the law as a whole is: For example if $ma_V = F_V$ then $ma_W=F_W$ and although $F_W \neq F_V$ it is the case that the scalar (tensor) is preserved
$$ma_W-F_w=0=ma_V-F_V$$
The question that motivated this post is the following:  Prove that if $p_V$ is a vector then $p'=(ap^1,bp^2)$ cannot be a vector unless $a \equiv b$.  This seems easy enough to show
$$Rp'= (p_1a\cos(\theta)-p_2b\sin(\theta), p_1 a\sin(\theta) + p_2b\cos(\theta))^t$$
And clearly, we cannot just factor out the $a$ and $b$ unless $a\equiv b$.  But what does this really mean?  The converse must be true, namely if $p'$ is a vector then surely $p_V$ cannot be a vector.
I suspect if I keep reading there will be ample examples and it will click better (for example the force-mass-acceleration equation makes sense in how it transforms).  I suppose my confusion rests in the idea that "well of course a vector is just a tuple and if rotate it that's another tuple and that old-new tuple pair was prescribed by a rotation so of course all tuples are vectors!"
EDIT:  Here's another example to make the point (this is an actual exercise):
Suppose we are given two vectors ${p}$ and $q$ in ordinary 3-dimensional space.  Consider this array of three numbers: $(p^2q^3,p^3q^1,p^1q^2)^t$. Prove that is not a vector, even though it looks like a vector. (Check how it transforms under rotation!) In contrast, $(p^2q^3-p^3q^2, p^3q^1-p^1q^3,p^1q^2-p^2q^1)^t$ does transform like a vector (so it's a vector).  It is in fact the vector cross product $p \times q$.

REPLY [1 votes]: I just wanted to comment on what I think the question from Zee's book is getting at. I don't think that this question from Zee is a particularly good question, and I think ACuriousMind's answer is excellent and gets at the real points that you should be trying to learn. So this is being written purely for the sake of trying to clarify Zee's book, rather than trying to give a competing answer to this question.
Let's say we have a vector $\vec{p}$ which is expressed in some basis (let's call it the 'unprimed basis') as $\vec{p}=p_1 \hat{e}_1 + p_2 \hat{e}_2$. We can express $\vec{p}$ in a new basis formed by rotating by an angle $\theta$ (this is a 'passive' view of rotations)
\begin{equation}
\vec{p} = p_1'\hat{e}_1' + p_2' \hat{e}_2' = \left(p_1\cos\theta+ p_2 \sin \theta \right)\hat{e}_1' + \left(-p_1 \sin \theta + p_2 \cos \theta \right) \hat{e}_2'
\end{equation}
So far, so good.
Now let's define an object $\vec{q}$, which has the following components in the unprime basis:
\begin{equation}
\vec{q}=q_1 \hat{e}_1 + q_2 \hat{e}_2 = a p_1 \hat{e}_1 + b p_2 \hat{e}_2
\end{equation}
Now in order for $\vec{q}$ to be a vector it must be that expressed in the prime basis, it has the form
\begin{equation}
\vec{q} = \left(a p_1 \cos \theta + b p_2 \sin \theta\right)\hat{e}_1' + \left(- a p_1 \sin \theta + b p_2 \cos \theta \right)\hat{e}_2'
\end{equation}
So far there is no problem for any value of $a$ or $b$. In a fixed basis, the components of  a vector can have any value they want, so I might as well just call the components of $\vec{q}$ in the unprimed basis $a p_1$ and $b p_2$. It's just a strange looking parameterization of an arbitrary vector in the unprimed basis. So long as, after a rotation, $\vec{q}$ has the form above, then everything is fine.
I think the point that Zee is trying to get at with his question can be phrased like this: In the prime basis,
\begin{equation}
\vec{q} \neq a p_1' \hat{e}_1' + bp_2' \hat{e}_2'
\end{equation}
unless $a=b$.
In other words, even though in the original frame it is true that $q_1 = a p_1$ and $q_2= b p_2$, in the rotated frame it is certainly not true that $q_1' = a p_1'$ and $q_2' = b p_2'$.
Of course this is completely obvious it you just rotate $\vec{q}$, then you clearly see that $q_1' = ap_1 \cos \theta + b p_2 \sin \theta \neq a (p_1 \cos \theta + p_2 \sin \theta) = a p_1'$.
I think the larger point that Zee is trying to get at is that you should only trust vector relationships, not relationships between components: $q_1 = a p_1$ is not a "good" relationship because it is only true in a specific frame.
The only exception to this is if $a=b$; in other words, if it turns out that if all the components of two vectors are proportional in some frame, then you actually do have a vector statement $\vec{q}=a\vec{p}$ which will then be true in any frame. This fact is used frequently in GR--to show two tensors are equal you only need to show they are equal in some frame, so you often choose the simplest possible frame in which to evaluate the tensors.
However in my opinion, this is a bad question (or at least poorly phrased). There is nothing wrong with $\vec{q}$ being a vector for any values of $a$ and $b$, the real problem is that $q_1=ap_1$ is not a covariant statement, and that was pretty unclear from the original question.


TITLE: Points on a plane and triangle
QUESTION [2 upvotes]: Given any three non-collinear points on a plane. Can there be an equilateral triangle such that the points lie on the perimeter of the equilateral triangle ?

REPLY [1 votes]: Think in reverse.
Take an arbitrary equilateral triangle.
From an outer point A', draw lines through two vertices, that form the same angle as CAB.
Then draw a third line that forms the angles ABC and BCA with the former, and translate it to the third vertex, giving the intersections B' and C'.
Now you have a triangle which is similar to the desired triangle and circumscribed to an equilateral triangle. Apply the similarity transform that maps A'B'C' to ABC, and you have it.


TITLE: Does A∪C=B∪C imply A = B? What about intersection instead of union?
QUESTION [2 upvotes]: if A, B and C are sets;
for (u = union, n = intersection)

$A \cup C = B \cup C$, would $A = B$ ?

Similar for

$A \cap C = B \cap C$, would $A = B$ ?

For my conclusion I have got for the first one, $A \neq B$ because $A$ can contain values differently from $B$ and for the second one $A$ does equal to $B$. Is this correct?

REPLY [1 votes]: That depends on what quantification you do on $C$. If the premise is true for all sets $C$ then your conclusion holds, not if it's a specific unfortunate set.
To see why this is the case in the first interpretation you can choose $C$ to be a fortunate set. If we have $C=A\cap B$ in the first equation and $C=A\cup B$ in the second:
$A = A \cup (A \cap B) = B \cup (A \cap B) = B$
$A = A \cap (A \cup B) = B \cap (A \cup B) = B$
However if the premises don't hold for any set $C$ we may be unfortunate with the value $C$ has. In the first equation we could have that $C = A\cup B$ for the first equation and $C = A \cap B$ for the second and we would have regardles of $A$ and $B$ (which could be unequal):
$A \cup C = A \cup (A \cup B) = A\cup B = B \cup (A\cup B) = B \cup C$
$A \cap C = A \cap (A \cap B) = A \cap B = B \cap (A \cap B) = B \cap C$


TITLE: Showing independence of $Y/X$ and $X$ where $(X,Y)$ is uniform on a triangle.
QUESTION [0 upvotes]: Let $X$, $Y$ be r.v.s with the uniform distribution on the triangle $A$ = $\{(x,y) \in \mathbb{R^2}:0 \leq y \leq x \leq 1\}$.
So $d\mu_{(X,Y)}(x,y)= 2\mathbf{1}_A(x,y)dxdy$.
I found that $P(Y/X\leq k)=k$ when $k \in (0,1)$. How do I use this to show that $X$ and $Y/X$ are independent?

REPLY [1 votes]: It is not strictly necessary to find the distribution of $Y/X$ to show independence, and in fact the distribution will be obvious from the proof. Fix $j$ and $k$, and we want to calculate
$$P\Big( X\le j, \, Y/X \le k\ \Big) = \iint_{A_{j,k}} 2 \, dxdy,$$
where $A_{j,k}:= \big\{ (x,y) \,:\, x\le j, \, y/x \le k \big\}$. Use the change of variables $(x,y) = (u,uv)$ (which we could rewrite as $(u,v) = (x,y/x)$). One has
$$ \begin{bmatrix}
\frac{dx}{du} & \frac{dx}{dv} \\
\frac{dy}{du} & \frac{dy}{dv}
\end{bmatrix} = \begin{bmatrix}
1 & 0 \\
v & u
\end{bmatrix}, $$
so the Jacobian of the transformation is $u$. Note that $(x,y)\in A_{j,k}$ if and only if $0 \le u \le j$, $0 \le v \le k$. Hence,
$$P\Big( X\le j, \, Y/X \le k\ \Big) = 2\int_0^j \int_0^{k/u} u\,dvdu = 2k\int_0^j u\, du = j^2k.$$
Since this is a function of $j$ times a function of $k$, it follows that $X$ and $Y/X$ are independent.


TITLE: Where can I find visualizations of quantum field theory equations?
QUESTION [8 upvotes]: As I work through A. Zee's "Quantum Field Theory in a Nutshell," I'd love to be able to visualize the different equations so I can explore the effects of different terms.  Any suggestions? Python script, Matlab, Mathematica, or are there any tools or videos already out there that cover this subject?

REPLY [0 votes]: The difficulty of visualizing a quantum field comes from the large number of degrees of freedom it has: Even if you discretize the field to a 2-dimensional mattress of, e.g., $100\times100$ points each of which can only oscillate in one dimension (similar to how Zee does it in one of the entry examples of his book), you already need a state function $\Psi: \mathbb R^{10\,000} \to \mathbb C$ (i.e. from a 10 000 dimensional real space to the complex plane).
One way of visualizing such a high dimensional function is to use a parallel axes diagram and plot only a relatively small number of randomly chosen points.
You can see how the result looks like in this paper:
https://www.researchgate.net/publication/321684743_A_new_way_of_visualizing_quantum_fields
Best regards
Helmut


TITLE: Tilings of n-board
QUESTION [1 upvotes]: Apparently it is equivalent to talk about a tiling of an n-board with squares and dominoes and a tiling of an (n+1)-board with odd length tiles (of any odd length).
Can someone explain to me how this is true?

REPLY [1 votes]: There is a bijection between the two tilings.  If you start with squares and dominoes on an $n$-board, add one more square at the right hand end and then glue each domino to whatever is to its right.  Conversely, if you start with odd-length tiles on an $(n+1)$-board, chop each $(2k+1)$-tile into $k$ dominoes followed by a square; this necessarily leaves a square at the right hand end, so delete it.


TITLE: Does this product have analytic continuation?
QUESTION [11 upvotes]: The product
$$
F(s)=\prod_{p}\frac1{(1-p^{-s})^p},
$$
converges for $\mathrm{Re}(s)>2$, when $p$ runs over all primes. Does it admit analytic continuation beyond the line $\mathrm{Re}(s)=2$? Any papers where it has been studied?

REPLY [8 votes]: $$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$

$P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$
so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(ns)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$ :
$P(s)$ has a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.
Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.
For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.


TITLE: Superstring vacuum amplitude on the torus
QUESTION [1 upvotes]: My question is how to obtain the superstring (Type II A and B) vacuum amplitudes on a torus. They are given in Polchinski's String Theory Vol. 2 equation (10.7.9):
$$Z_\psi^{\pm}=\frac{1}{2}[Z^0_0(\tau)^4-Z^0_1(\tau)^4-Z^1_0(\tau)^4\mp Z^0_0(\tau)^4].$$
I understand how each individual $Z^\alpha_\beta$ is obtained but do not understand how they are put together to get $Z_\psi^{\pm}$.

REPLY [1 votes]: One way to think about this is the following. In general, the partition function (which is the integrand of the vacuum amplitude and not the vacuum amplitude itself) will be of the form
$Z_\psi^{\pm}\propto\sum_{a,b}C[^a_b]Z^a_b(\tau)$
where $a$ and $b$ sum over the different sectors as given in the text and the $C$s are some phases. Many of these phases are fixed by modular invariance, ie by the requirement that $Z$ is invariant under $\tau\rightarrow\tau+1$ and $\tau\rightarrow-\frac{1}{\tau}$. Given that $C[^0_0]=1$ (it's only the relative phases that matter, so we can fix this one), modular invariance demands that $C[^0_1]=C[^1_0]=-1$. However, both choices $C[^0_0]=\pm$ give modular invariant theories, so we end up with two different viable theories.