TITLE: Calculation of angles in regular polyhedrons
QUESTION [1 upvotes]: For any regular polyhedron (tetrahedron, cube, octahedron, dodecahedron, icosahedron, ...), how do I calculate an angle from the center of the regular polyhedron to the center of each face? Obviously, the angle decreases with the increasing number of faces. What is the formula, when the number of faces n is the variable?

For cube, n= 6, the angle is 90 degrees:

REPLY [0 votes]: A full outline on corner angles of polygons, dihedral angles of polyhedra, and dihedral angles of polychora can be found here: https://bendwavy.org/klitzing/explain/dihedral.htm
It also includes the asked for supplements as well as an explicite listing of all relevant cases, even the acrons.
--- rk


TITLE: Integer solutions using stars and bars with bounds on integers
QUESTION [0 upvotes]: In this page of brilliant , an example problem is as follows

$$\sum_{i=0}^6 a_i \leq 100$$ How many positive integers $(a_1,a_2...,a_6)$ with $ i \leq a_i$ for $i=1,2,3,4,5,6$ satisfy the sum?

In the solution they start of with introducing an extra variable $c$ into the equation making it:
$$ c+ \sum_{i=0}^6 a_i  \leq 100$$
With, $ 0 \leq c$
But I don't understand why?  They call this addition of 'c' is some kind of bijection but I'm not sure what purpose it is serving..

REPLY [3 votes]: The term $c$ has been introduced to the expression as a dummy variable, which saves you a lot of work in the long run.
Via the introduction of this new dummy variable '$c$', the below problem has been transformed from
$$\sum_{i=0}^6 a_i  \leq 100$$
to
$$\sum_{i=0}^6 a_i + c = 100$$
Why this works out so well is that each non-negative value taken up by this dummy variable $c$ corresponds to a particular case of the inequality. For example, when $c=10$, $\sum_{i=0}^6 a_i = 90$. To reiterate, taking this extra variable saves you the effort of having to calculate so many different cases.
The number of positive integral solutions of the above equation can be calculated easily via the well known formula for the same. I think you can take it from here!


TITLE: How to solve the equation $\cos(\alpha-c)=a$ for a given pair of numbers $(c,a)$
QUESTION [0 upvotes]: I am trying to solve the equation $$\cos(\alpha-c)=a$$
where the values of $c$ and $a$ are known, for example:
$$\cos(\alpha - 200^\circ) = -0.49;$$
Could you please explain to me how to find the value of $\alpha$ in degrees?
Thanks for any advice!
Best wishes,
Gendryn

REPLY [1 votes]: Taking your example:
We have the equation $$\cos(a-200^{\circ})=-0.49.$$
Let $\cos\alpha=0.49\Rightarrow\alpha=\arccos0.49$.  Since we want a negative cosine ratio, we know the angle we want is in quadrant $2$ (given by $180^\circ-\alpha$) or quadrant $3$ (given by $180^\circ+\alpha$).
So we have
$$a-200^\circ=180^\circ-\arccos0.49+360^\circ n\implies a=20^\circ-\arccos0.49+360^\circ n,\ n\in\Bbb Z$$
or
$$a-200^\circ=180^\circ+\arccos0.49+360^\circ n\implies a=20^\circ+\arccos0.49+360^\circ n,\ n\in\Bbb Z$$


TITLE: A question relating to exponent laws
QUESTION [3 upvotes]: Here is the question:
If:
$$125*(3^x) = 27*(5^x)$$
Then find the value of $x$.
Here is what I have done so far:
I found that I can manipulate the numerical values to get the same bases:
$\to 125 = 5^3$ and $27 = 3^3$
$\to 5^{3-x} = 3^{3-x}$
Here is where I got stuck. I do not have similar bases anymore so where can I go from here? Should I change it to a logarithm or should I do a trial and error to see what value(s) work or is there a way to get one of the sides into the same base as the other?

REPLY [3 votes]: $$\frac{3^{3-x}}{5^{3-x}} = 1 \implies \left(\frac{3}{5} \right)^{3-x} = 1$$
and you know that $a^0 = 1$ for any real number $a$.

REPLY [3 votes]: You have $5^{3-x}=3^{3-x}$. Divide both sides by $3^{3-x}$, then $\left(\frac{5}{3}\right)^{3-x}=1\Rightarrow 3-x=0\Rightarrow x =3$.
(Remember $\frac{5^{3-x}}{3^{3-x}}=\left(\frac{5}{3}\right)^{3-x}$ using the index law $\frac{a^m}{b^m}=\left(\frac{a}{b}\right)^{m}$.)


TITLE: Is there any relation between Poisson Brackets and the Jacobian Matrix?
QUESTION [7 upvotes]: The Poisson brackets for $u,v$ can be written as,
$$
\frac{\partial u}{\partial q} \frac{\partial v}{\partial p} - \frac{\partial u}{\partial p}\frac{\partial v}{\partial q}.
$$
We can write this as determinant of this matrix
$$
 \begin{bmatrix} \frac{\partial u}{\partial q} & \frac{\partial u}{\partial p} \\\frac{\partial v}{\partial q} & \frac{\partial v}{\partial p} \end{bmatrix}
$$
Which is the Jacobian matrix.
Is there any relation between them?

REPLY [9 votes]: For a 2-dimensional phase space, they are the same.
More generally, for a $2n$-dimensional symplectic manifold $(M;\omega)$ with symplectic two-form
$$\tag{1} \omega~=~\frac{1}{2} dz^I ~\omega_{IJ} \wedge dz^J, \qquad \omega_{IJ}~=~-\omega_{JI}, $$
the Poisson bracket is given by
$$ \tag{2} \{f,g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ} \frac{\partial g}{\partial z^J}, \qquad \pi^{IJ} ~:=~(\omega^{-1})^{IJ} . $$
See also e.g. this and this Phys.SE posts.
The canonical volume form on $(M;\omega)$
$$\tag{3} \Omega~=~\omega^n~=~\rho~ dz^1\wedge \ldots \wedge dz^{2n},$$
is the $n$'th exterior power of the symplectic two-form $\omega$, with volume density
$$\tag{4} \rho~=~{\rm Pf}(\omega_{IJ}) $$
given by the Pfaffian, which is a square root of the determinant. This is closely related to Liouville's theorem.


TITLE: The precise meaning of the limit that defines the integral
QUESTION [1 upvotes]: I'm having some trouble understanding what the following definition is saying, i think i have a basic understanding but would appreciate some clarification on the technical aspects. The precise meaning of the limit that defines the integral follows as:

for every number $\epsilon$ there is an integer N such that
$$\vert \; \int_a ^b f(x)dx - \sum_{i=1}^nf(x_i^*)\Delta x \;\vert
< \epsilon $$
for every integer $n > N$ and for every choice of $x_i^*$ in $[x_{i-1},
x_i] $

I know that $\int_a^bf(x)dx$ is the integrand and that $ \sum_{i=1}^nf(x_i^*)\Delta x$ is a Riemann Sum (Approximation of the area under a curve by rectangles). Judging by the absolute value sign I would take this precise definition to be formalizing the idea of "closeness", or saying that the difference between the actual area, the integrand, and the approximation, the Riemann sum is extremely small (i.e highly accurate).
I don't understand how the integer N comes into it, in particular the last sentence "for every integer $n > N$ and for every choice of $x_i^*$ in $[x_{i-1}, x_i]$. I'd appreciate any clarification on this and confirmation of whether i have the basic idea of what this is saying, i unfortunately couldn't find any explanation of this elsewhere.

REPLY [0 votes]: $N$ comes from the step $h_n$ of a regular ( equidistant) partition of the interval $[a,b]$ containing $n$ sub-intervals,  defined by
$$h_n=\frac{b-a}{n}$$.
to say that $h_n$ becomes small means $n$ is greater than a certain $N$ or
$$\lim_{h_n \to 0} \equiv \lim_{n\to +\infty}.$$


TITLE: Why does the dependent product type need "forall"?
QUESTION [3 upvotes]: I feel stupid asking this question because it is so fundamental to logic and math. However, in my starting to learn proof theory and now type theory, I have not seen an explanation on why you need the ∀ quantifier for dependent products. It seems like it can be left out / assumed. Let me explain.
Background
In section 4.2: Typing Rules and Dependent Products in Coq'Art: The Calculus of Inductive Constructions, they provide this definition of the dependent product type:

∀v:A,B is the type of the functions that map any v of type A to a term of type B where v may occur in B.

Then they provide the example:
Check le_n.
le_n : ∀n:nat, n ≤ n

So that example is saying

forall n of type natural number, n is less than or equal to n.

Problem
That makes sense, and it comes from how you would write it in classical logic proofs. However, I keep wondering why can't it just be written like this:
Check le_n.
le_n n:nat : n ≤ n

That seems like it means the same thing to me. It means that

for all n of type natural number, n is less than or equal to n.

Or it feels to me it could mean "for any n", which may be different.
In both cases, you need some input to create the type, so it seems like they both mean "dependent type".
Basically, I'm just wondering why you have to include the forall ∀ in the definition (in Coq), when the fact that it's a definition means that it will apply to all cases anyways. Any thoughts?

REPLY [2 votes]: Well, it turns out the type of le_n is indeed a dependent product:
le_n : ∀ (n : nat), n ≤ n

The fact that it is displayed as a function with named arguments is syntactic sugar. In fact, when I check le_n in my version of Coq it displays this:
le_n
     : forall n : nat, n <= n

And whenever you write something like this:
Definition f (a1 : T1) (a2 : T2) : T3 -> T4 := ...

Your are really declaring:
Definition f : ∀ (a1 : T1) (a2 : T2), T3 -> T4 := λ a1 a2 → ...


text:
/-
Copyright (c) 2022 Michael Stoll. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Michael Stoll
-/
import ring_theory.trace
import field_theory.finite.basic
import field_theory.finite.galois_field

/-!
# The trace map for finite fields

We state the fact that the trace map from a finite field of
characteristic `p` to `zmod p` is nondegenerate.

## Tags
finite field, trace
-/

namespace finite_field

/-- The trace map from a finite field to its prime field is nongedenerate. -/
lemma trace_to_zmod_nondegenerate (F : Type*) [field F] [fintype F] {a : F}
 (ha : a ≠ 0) : ∃ b : F, algebra.trace (zmod (ring_char F)) F (a * b) ≠ 0 :=
begin
  haveI : fact (ring_char F).prime := ⟨char_p.char_is_prime F _⟩,
  have htr := trace_form_nondegenerate (zmod (ring_char F)) F a,
  simp_rw [algebra.trace_form_apply] at htr,
  by_contra' hf,
  exact ha (htr hf),
end

end finite_field

subset_name: curated


TITLE: Are associativity and solvability for $x$ in all $x+a=b$) sufficient laws for a group?
QUESTION [1 upvotes]: Typically a group is defined as a set with a single binary operation mapping pairs of elements to elements of the set.  The operation is associative, and there exists a neutral element $0$ such that $0+a=a$.
It occurs to me that if we have a set with an associative binary operation defined in it, and declare that for every equation of the form $x+a=b$ there exists an element $x$ satisfying the equation, we can show the existence of both a neutral element, as well as an inverse to every element.
That is, the solution to $x+a=a$ will be $0$.  Once we demonstrate the existence of $0$, we require solutions to all equations $x+a=0$.
Are these two assumption sufficient to characterize a group?

REPLY [2 votes]: Unfortunately, it isn't, even if we add the assumption that the underlying set is non-empty. The simplest counterexample I can think of is to take any set $S$ with more than one element and define an operation $*$ on $S$ by $a*b=a$ for all $a,b\in S.$ Then $$a*(b*c)=a=a*c=(a*b)*c,$$ so $*$ is associative. Moreover, for any $a,b\in S,$ we have $x=b$ as a solution to $x*a=b,$ as you desired--in fact, $x=b$ is the unique solution! However, this operation cannot have an identity. If there were such, say $e,$ then taking $a$ and $b$ to be distinct elements of $S,$ we obtain the absurdity $$a=e*a=e=e*b=b.$$
On the other hand, let's suppose we imposed the following conditions:

$S$ is a non-empty set.
$\star$ is an associative operation on $S.$
For any $a,b\in S,$ there exists some $x\in S$ such that $x\star a=b.$
For any $a,b\in S,$ there exists some $y\in S$ such that $a\star y=b.$

Then $S$ is a group under the operation $\star.$
To prove it, we begin by proving that the identity exists. That is, we must show that there is some $e\in S$ such that for all $a\in S,$ $a\star e=a=e\star a.$
Take any $a\in S.$ We know that there exists $x\in S$ such that $x\star a=a.$ We'd like to show that $x\star b=b$ for all $b\in S.$ Take any $b\in S.$ We know there is some $y\in S$ such that $a\star y=b.$ Then $$x\star b=x\star(a\star y)=(x\star a)\star y=a\star y=b,$$ as desired. Thus, we've shown that there is at least one left-identity element in $S.$ A similar proof shows that there is a right-identity element in $S.$
Suppose that $l,r\in S$ are left- and right-identity elements. That is, for all $x\in S,$ we have $l\star x=x=x\star r.$ Then with $x=r$ we see that $l\star r=r,$ and with $x=l,$ we see that $l\star r=l.$ Thus, every left-identity element is a right-identity element, and vice versa, so we've proved that there is at least one identity element. We can (and should) further prove that there is exactly one identity element, but I leave that to you.
Finally, we prove the existence of inverses. Letting $s\in S$ be arbitrary and letting $e$ be the identity element, we know that there exists $x\in S$ such that $x\star s=e.$ It remains only to show that $s\star x=e.$ We know that there is some $y\in S$ such that $(s\star x)\star y=e.$ Since $e$ is the identity element, then $$s=s\star e=s\star(x\star s)=(s\star x)\star s,$$ so we see that $$s\star x=\bigl((s\star x)\star s\bigr)\star x= (s\star x)\star (s\star x),$$ and so
\begin{eqnarray}e &=& (s\star x)\star y\\ &=& \bigl((s\star x)\star(s\star x)\bigr)\star y)\\ &=& (s\star x)\star\bigl((s\star x)\bigr)\star y)\\ &=& (s\star x)\star e\\ &=& s\star x,\end{eqnarray} as desired.


TITLE: Mertens Theorem
QUESTION [0 upvotes]: If (i) $\sum a_n$ converges absolutely
(ii)$\sum a_n = A$
(iii)$\sum b_n = B$
(iv)$c_n = \sum_{k=0}^n a_k b_{n-k}$,
Then $\sum c_n = AB$.
Here, what if (iii) is changed to ' $\sum b_n$ diverges'? If it is inconclusive, please give me an example.
Plus, if $\sum a_n$ and $\sum b_n$ both diverge, does $\sum c_n$ diverge?

REPLY [2 votes]: Observe that $(c_n)$ is the Cauchy product (or discrete convolution) of $(a_n)$ and $(b_n)$. That is, it satisfies (at least as a formal identity)
$$ \sum c_n x^n = \left( \sum a_n x^n \right)\left( \sum b_n x^n \right). \tag{1} $$
This shows that it is natural to consider power series to test if your new condition

(iii') $\sum b_n$ diverges

leads to an inconclusive statement.

First, let us consider the following example:
$$ \sum a_n x^n = 1 - x, \qquad \sum b_n x^n = \frac{1}{1-x}. $$
Then $a_n = (1, -1, 0, 0, \cdots)$ and $b_n = (1, 1, 1, 1, \cdots )$ satisfies (i), (ii) and (iii'). Now, $(c_n)$ defined by (iv) satisfies
$$ \sum c_n x^n = \left( \sum a_n x^n \right)\left( \sum b_n x^n \right) = (1+x)\left(\frac{1}{1+x}\right) = 1, $$
so that $c_n = (1, 0, 0, 0, \cdots)$. Therefore in this case, $\sum c_n$ converges.
However, if we change $(b_n)$ so that
$$ \sum b_n = \frac{1}{(1-x)^2},$$
then we have $ (b_n) = (1, 2, 3, 4, \cdots)$ instead, which again satsifies (iii'), but now the application of $(1)$ again gives $c_n = (1, 1, 1, 1, \cdots)$. Therefore $\sum c_n$ diverges.
Finally, consider the following example
$$ \begin{align*}
\sum a_n x^n &= \frac{1-x}{1+x} = 1 - 2x + 2x^2 - 2x^3 + \cdots, \\
\sum b_n x^n &= \frac{1+x}{1-x} = 1 + 2x + 2x^2 + 2x^3 + \cdots.
\end{align*} $$
Then both $\sum a_n$ and $\sum b_n$ diverge, but $\sum c_n$ converges since from $(1)$,
$$ \sum c_n x^n = \left( \sum a_n x^n \right)\left( \sum b_n x^n \right) = 1, $$
so that $c_n = (1, 0, 0, 0, \cdots)$.


TITLE: Algorithm for testing a sequence for convergence
QUESTION [1 upvotes]: I have this exam problem which states the following:
A sequence of real numbers ${\{x_n\}_{n=1}^\infty}$ is given with $x_1=1$, $x_{n+1}=\sqrt{20+x_n}$, $n \geq 1$. Test for convergence of this sequence, and if it is convergent evaluate its limit.
So after evaluating a few values of $x_n$, I realized that the sequence is increasing and that its bounded from above with $5$. What implies that the sequence is converging(by the Monotone Convergence Theorem).
My question is, what would be the algorithmic way to solve this and to prove it properly ? How should the limit be evaluated in this case later on?
Also, would we say for such stated sequence that it is given recursively ?

REPLY [1 votes]: Picking up from where you left off in the comment:
Suppose x_n (defined above) is monotonically increasing.
The upper bound case is easy to show!
$$x_1=1<21$$
Suppose $x_k < 21$ for some $k \in \mathbb{N}$. Then $x_{k+1} = \sqrt{20 + x_{k}} \leq \sqrt{41} \lt 21$, hence we have shown $x_n \lt 21$ for all $n \in \mathbb{N}.$
By the monotone convergence theorem, there is a limit. Find it now:
Remember that $\lim s_{n+1} = \lim s_{n}$.
Then $\lim x_{n+1} = \lim x_n$ satisfies $ x = \sqrt{20 + x}$ for a grand result of $x=5$. ($x$ here is the limit of the $x_n$ sequence)
As a whole, the process here is to show the monotonic increasing-ness of the sequence, use induction to show an upper bound with some constant substitution, and then use a recursive algebra trick to understand the limit's actual form.
I hope this supports your exam prep.


TITLE: The probability of rolling a 5 is four times the probability of rolling a 3. What is the probability that a 2 or a 5 is rolled?
QUESTION [0 upvotes]: A 6-sided die is weighted so that the probabilities of rolling a 1, 2, or 3 are equally likely; the probabilities of rolling a 4, 5, or 6 are equally likely; and the probability of rolling a 5 is four times the probability of rolling a 3. What is the probability that a 2 or a 5 is rolled?
How I solved it:
$P(\text{rolling a } 1, 2, 3) = x$

$P(\text{rolling a } 4, 5, 6) = 4x$ (because rolling a 4, 5, 6 is 4x the probability of rolling a 1, 2, 3)
$3(x)+3(4x)=15x$ (because 1,2,3 each has $x$ probability of being rolled and same with 4,5,6 except each of them has a $4x$ chance)

$15x = 1 ,  x=\frac{1}{15}$
Therefore, there's a $\frac{1}{15}$ chance of rolling a 1, rolling a 2, rolling a 3 and a $\frac{4}{15}$ chance of rolling a 4, rolling a 5, rolling a 6.
So to roll a 2 or 5, that would be $\frac{1}{15} + \frac{4}{15} = \boxed{\frac{1}{3}}$.
My teacher told me I got the answer correct, but my solution wasn't right and she said the part where it mentions that rolling a 5 is four times the probability of rolling a 3 is irrelevant. Can someone tell me what I'm doing wrong? Thanks!

REPLY [1 votes]: Your solution was correct but there was an easier way to see it.
The probability of a $1$ or $4$ is the same as the probability of a $2$ or $5$, which is the same as the probability of a $3$ or $6$.  Thus, the answer has to be $\frac 13$ no matter what the relative probability of the two groups are.


TITLE: Writing up the solution for a nonhomogeneous differential equations system with complex Eigenvalues
QUESTION [0 upvotes]: The nonhomogeneous linear system of differential equations is given as:
$$ x'(t)=Ax(t)+b $$
It has the following Eigenvalues and Eigenvectors:
$$ \lambda=-1+i, \sqrt{2}, -1-i $$
$$ v_1 =
        \begin{bmatrix}
        -i \\
        0 \\
        1 \\
        \end{bmatrix}, \qquad
v_2=
 \begin{bmatrix}
        0 \\
        1 \\
        0 \\
        \end{bmatrix},\qquad v_3=
 \begin{bmatrix}
        i \\
        0 \\
        1 \\
        \end{bmatrix}$$
A particular solution is:
$$ x_p(t)=
        \begin{bmatrix}
        2 \\
        \sqrt{2} \\
        -1 \\
        \end{bmatrix}
 $$
How do I find the general real solution for this system? I got:
$$ x(t)=c_1Re\left(\begin{bmatrix}
        -i \\
        0 \\
        1 \\
        \end{bmatrix}e^{-t}(\cos(t)+i\sin(t))\right)+c_2\begin{bmatrix}
        0 \\
        1 \\
        0 \\
        \end{bmatrix}e^{\sqrt{2}t}+c_3 Re\left(\begin{bmatrix}
        i \\
        0 \\
        1 \\
        \end{bmatrix}e^{-t}(\cos(t)+i\sin(t))\right)+\begin{bmatrix}
        2 \\
        \sqrt{2} \\
        -1 \\
        \end{bmatrix} $$
But the answer is to take the imaginary part of the last term. I don't see how that makes sense.

REPLY [1 votes]: If $\lambda$ is an eigenvalue of the matrix, and $\vec v$ is an eigenvector for that function, then $e^{\lambda t}\vec v$ is a solution to the system of differential equations. In your case, you get $\pmatrix{-i\cr 0\cr 1\cr}\cdot e^{(-1+i)t}$ and $\pmatrix{i\cr 0\cr 1\cr}\cdot e^{(-1-i)t}$.
These solutions have complex number in them; you want to find two new functions which are real-valued, and which span the same subspace (in the linear algebra sense). These two new functions are the real and imaginary parts of either of these solutions. (They are complex conjugates of each other, which also explains why you can't take the real part of both solutions: The real part of $f(t)+g(t)\,i$ equals the real part of $f(t)-g(t)\,i$, so you don't get as many solutions.)


TITLE: If $A$ and $B$ are $n \times n$ diagonal matrices, prove that so also is $A^pB^q$ for all positive integers $p, q$.
QUESTION [0 upvotes]: If $A$ and $B$ are $n \times n$ diagonal matrices, prove that so also is $A^pB^q$ for all positive integers $p, q$.

My Proof
Let $A = \begin{cases}
      a_{i,j} & i = j \\
      0 & i \not= j
   \end{cases}$
and
$B = \begin{cases}
      b_{i,j} & i = j \\
      0 & i \not= j
   \end{cases}$
be $n \times n$ matrices and $p, q$ be positive integers.
$\therefore [A^pB^q]_{i,j} = \sum_{k = 1}^n a^p_{i,k}b^q_{k,j} = a^p_{i,j}b^q_{i,j}$, where $i = j \ \forall n \ge i,j \ge 1$.
I would greatly appreciate it if people could please take the time to review my proof and provide feedback.

REPLY [2 votes]: I think you are making it more complicated than needed.

If $C$ and $D$ are diagonal, then (prove directly that) $CD$ is diagonal.
Then it follows inductively (which you may prove) that $A^p$ is diagonal (with $C= D = A$) for any positive integer $p$. (If $A^{p-1}$ is diagonal, then let $C= A^{p-1}$ and $D = A$, and use 1. concluding that $CD = A^p$ is diagonal.)
Hence, $B^q$ must also be diagonal for any positive integer $q$, from 2.
Use 1. with $C = A^p$ and $D = B^q$ which are diagonal by 2. and 3. to conclude that $CD = A^p B^q$ is diagonal.

Note that 3. and 4. are direct consequences of 1. and 2.


TITLE: Expected number of questions when the student knows 50 out of 250 questions and the teacher selects 25
QUESTION [3 upvotes]: A teacher has 250 sample test questions and he selects 25 of which to put on the test. Johnny only practices 50 out of the 250 questions.
a) What's the expected number of questions that appear on the exam that Johnny have practiced on?
My answer:
I'm quite confused so I don't think it's the right answer. I know that the expected value is just
$$\sum_{k=1}^{25} \left[kp(k)\right]$$
However, I'm not sure what $p(k)$ is. I think it's $$ \frac{25}{250} \frac{50}{250}  = 0.02 $$ which would make $E(X) = 6.5$
b) What's the probability that Johnny hasn't practiced any of the exam questions?
My answer:
Here I tried using the hypergeometric distribution and set N = 250, r = 25, n = 50, and x = 0. Obviously assuming here that if Johnny sees a question he's practiced he'll be able to solve it.
This means the final answer will be
$$\frac {{50 \choose 0}{225 \choose 50}} {{250 \choose 50}}$$
which is
$$\frac {{225 \choose 50}} {{250 \choose 50}}$$
Are those answers correct? Any help would be appreciated! Thank you.

REPLY [3 votes]: What's the expected number of questions that appear on the exam that Johnny have practiced on?

Johnny practices on $50$ out of $250$ questions.  This implies that at random, any question has a $(1/5)$th probability of being a question that Johnny has practiced on.
Linearity of Expectation does not require that the events be independent.
Consider the following $2$ events:

$E_1$ : Johnny practiced on the 1st question chosen by the teacher.
$E_2$ : Johnny practiced on the 2nd question chosen by the teacher.

When computing the expected number of the questions, out of the $(25)$ chosen by the teacher, that Johnny will have practiced on, it is irrelevant (for example) that events $E_1$ and $E_2$ above are not independent events.
So, the expected number of questions, out of the $(25)$ questions chosen by the teacher, that Johnny will have practiced on is
$$(1/5) \times 25 = 5.$$

What's the probability that Johnny hasn't practiced any of the exam questions?

The probability may be expressed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
$N$ will denote the number of ways of selecting $25$ questions from the $200$ that Johnny did not practice on.
$D$ will denote the number of ways of selecting $25$ questions from any of the $250$ questions.
So, the probability is
$$\frac{\binom{200}{25}}{\binom{250}{25}}.$$
Edit
See the comment of JMoravitz, following my answer.  In fact, I did think that the OP's (i.e. original poster's) answer of
$$\frac{\binom{225}{50}}{\binom{250}{50}}$$
was automatically wrong because it involved Combinations different from mine.  It never occurred to me that the $2$ answers are equivalent.

For what it's worth, my approach was to focus on the $25$ questions selected, reasoning that they had to be part of the $200$ questions that Johnny did not practice on.
The OP took the valid (but opposite) viewpoint.  He reasoned that the $50$ questions that Johnny did practice on had to be contained in the $225$ questions that were not selected.


TITLE: Antisymmetry of Electromagnetic tensor
QUESTION [0 upvotes]: From Mueller-Kirsten's Electrodynamics, the electromagnetic tensor is constructed by writing out all components of the field strengths $\mathbf{E}=-\nabla\phi-\frac{\partial\mathbf{A}}{\partial t}=-c\left(\boldsymbol{\nabla}A^0+\frac{\partial \mathbf{A}}{\partial x_0}\right)$ and $\mathbf{B}=\boldsymbol{\nabla}\times\mathbf{A}.$ I am seeking any physical implication that the antisymmetry of this tensor would imply. In other words, apart from the mathematical structure of electromagnetic theory, what is the physical requirement that the $F_{\mu\nu}$ is antisymmetry? My priority is the physical aspect of this property, but any mathematical explanation (ex. from the perspective of gauge theory) is welcome too.

REPLY [1 votes]: Unable to comment due to reputation. But check out this possible duplicate: What does it mean that the electromagnetic tensor is anti-symmetric?
There is a detailed accepted answer therein.


TITLE: What does the colons in $:\hat{P}:$ mean?
QUESTION [2 upvotes]: When going through some past papers I came across the following exercise:

For the momentum operator
  $$\hat{P}^\mu =\frac{1}{2}\int \frac{d^3 k'}{(2\pi^3)2E(\textbf{k'})}k'^\mu (\hat{a}^\dagger(\textbf{k'})\hat{a}(\textbf{k')} + \hat{a}(\textbf{k'}) \hat{a}^\dagger(\textbf{k}'))$$
  prove the commutator relation
  $$[:\hat{P}^\mu:, \hat{a}^\dagger(\textbf{k})]=k^\mu\hat{a}^\dagger(\textbf{k})$$

Does the ": :" symbol around the momentum operator represent anything? If so would it affect the calculations or would I just prove the commutation by doing:
$$\hat{P}^\mu\hat{a}^\dagger(\textbf{k})-\hat{a}^\dagger(\textbf{k})\hat{P}^\mu$$

REPLY [1 votes]: The "::" symbols represent normal ordering, a very common operation in Quantum Field Theory. In practical terms, it effectively means that the operator inside the colons (in this case the momentum operator) is the same operator but with all annihilation operators placed at the end:
$$
:\hat { P } ^ { \mu }: = \frac { 1 } { 2 } \int \frac { d ^ { 3 } k ^ { \prime } } { \left( 2 \pi ^ { 3 } \right) 2 E \left( \mathbf { k } ^ { \prime } \right) } k ^ { \prime \mu } \left( \hat { a } ^ { \dagger } \left( \mathbf { k } ^ { \prime } \right) \hat { a } \left( \mathbf { k } ^ { \prime } \right) + \hat { a } ^ { \dagger } \left( \mathbf { k } ^ { \prime } \right) \hat { a } \left( \mathbf { k } ^ { \prime } \right) \right) = \int \frac { d ^ { 3 } k ^ { \prime } } { \left( 2 \pi ^ { 3 } \right) 2 E \left( \mathbf { k } ^ { \prime } \right) } k ^ { \prime \mu } \left( \hat { a } ^ { \dagger } \left( \mathbf { k } ^ { \prime } \right) \hat { a } \left( \mathbf { k } ^ { \prime } \right) \right)
$$
And you can then proceed as usual from here.

REPLY [1 votes]: ::  means normal ordering of the creation and destruction operators inside the colons. The normal order happens when all destruction operators are placed to the right of the creation operators. So $:{a}(\textbf{k}){a}^\dagger(\textbf{k}):={a}^\dagger(\textbf{k}){a}(\textbf{k})$


TITLE: What is the expected number of rolls to get a subset of all possibilities?
QUESTION [1 upvotes]: I need to know how many times I should expect to need to roll an $n$-sided fair die to get each of $m$ (where $m < n$) outcomes at least once.
In my specific example I have a $7$ possible outcomes, of which I want to achieve $3$, with order being irrelevant.
Please excuse if this has been answered before, I searched but all I could find were answers for how many rolls you need to get one specific outcome, or every possible one, nothing for a subset of the possible outcomes and I couldn't figure out how to generalize those answers to apply to my problem.

REPLY [0 votes]: Method I (Markov):  Let $E_i$ be the expected number assuming you have already seen $i$ of your list (so the answer you want is $E=E_0$.  Then We have the recursion $$E_i=\frac {n-i}m\times \left(E_{i+1}+1\right)+\frac {m-(n-i)}m\times \left(E_{i}+1\right)$$  Thus $$ E_i=\frac m{n-i}+E_{i+1}$$
In your case $(m,n)=(7,3)$, that gives $E=12.8\overline 3$
Method II: (Coupon Collector Problem) If you only had the $n$ desired coupons the answer would be $nH_n$, where $H_n$ denotes the $n^{th}$ Harmonic Number.  Here, only $\frac nm$ of the tosses count, so the answer would be $$\frac mn \times nH_n=mH_n$$  In your case that gives $$E=7\times \frac {11}{6}=12.8\overline 3$$


TITLE: Do relativistic phenomena affect communication with satellites?
QUESTION [0 upvotes]: Communication with satellites certainly requires knowing the speed of light.  Beyond having that magic number, do relativistic phenomena affect communication with satellites?  For example, would one be able to take an accurate photograph of a specific spot on a remote planet without knowledge of relativity?
A similar question can be raised in the context of GPS: would one get accurate transmissions based on knowledge of the speed of light alone, or is knowledge of the details of relativity required to get accurate transmissions?
In other words, consider the stage of physics at the end of the 19th century when the speed of light was known already, but the formalism of relativity did not exist yet.  Would people have been able to communicate accurately and in time with space probes, and get accurate transmissions via GPS?
GE was confirmed experimentally long ago by Mercury precession (see links below) but beyond such fairly esoteric phenomena in astronomy I was wondering about the specific practical manifestations this has, as described above.

REPLY [1 votes]: Don't forget stellar aberrations--the (special) relativistic treatment is certainly required for star trackers, and would be required for anything that requires accurate pointing--perhaps COM.