Why is it possible that $f(x)$ has real outputs, but $\exp(\ln(f(x))$ has complex outputs? This is probably a trivial question, but I can't seem to wrap my head around it.

Say I have a function: $f(x)=x(x-6)^{1/5}$. Is the output of the function just the real numbers?

Desmos tells me yes, but my friend told me that it isn't, because $f(x)=\exp(\ln(x(x-6)^{1/5})=\exp(\ln(x)+\frac15\ln(x-6))$, and because $\ln$ of a negative number will yield a complex output, $f(x)$ should give complex answers as well.

Plotting $f(x)=\exp(\ln(x)+\frac15\ln(x-6))$, I see that it only gives real values from 6 to infinity.

Or does this mean that $x(x-6)^{1/5}$and $\exp(\ln(x)+\frac15\ln(x-6))$ are actually different functions after all?

I thought about when $(-1)^n$ is imaginary and real. When $n=0.2$, $(-1)^n=-1$ and so my function $f(x)$ should be able to handle negative $x$'s perfectly well.

However, I found that $f(x)=x(x-6)^{1/10}$ outputs real values only from 6 to infinity. This is probably because $(-1)^n$ is imaginary when $n=0.1$.

This is my question. Why is my friend's logic sound when $f(x)=x(x-6)^{1/10}$ but it breaks down when $f(x)=x(x-6)^{1/5}$? Does it have anything to do with when $(-1)^n$ is real or is there something else about the exponential/logarithmic function that I don't know about?

I feel like this might be in the realm of complex analysis, but I haven't studied that topic yet.

Edit: Let's say the domain of $f$ is $\mathbb{R}$.

In the complex field the logarithm is a multivalued function, let's designate it by $Log$:
$$
Log(z) = \ln (\left| z \right|) + i\,\arg \left( z \right) + i2k\pi
$$

When you take the principal branch of that function
$$
\log (z) = \ln (\left| z \right|) + i\,\arg \left( z \right)
$$
and apply it to a real $z$, you get
$$
\log (x) = \left\{ {\matrix{
   {\ln (\left| x \right|) + i\,\pi } & {x < 0}  \cr
   {\ln (\left| x \right|)} & {0 < x}  \cr
 } } \right.
$$

Understanding by $e^z$ the exponential function $e^{|z|} (\cos (\arg(z))+i \sin(\arg(z))$, then you have for instance
$$
\eqalign{
  &  - 1 = e^{\,\log ( - 1)}  = e^{\,0} \left( {\cos \pi  + i\sin \pi } \right) =  - 1  \cr
  & \left( { - 1} \right)^{\,1/5}  = e^{\,\,1/5\,\log ( - 1)}  = \cos \pi /5 + i\sin \pi /5 = i^{\,2/5}  \cr}
$$
That, keeping on the principal branch of the logarithm.

But $ \left( { - 1} \right)^{\,1/5} $ has five different solutions in total, which correspond to the various branches of the multivalued $Log$.
If you take the branch corresponding to $k=2$ you get
$$
\left( { - 1} \right)^{\,1/5}  = e^{\,\,1/5\,\ln 1 + 1/5i5\pi }  = e^{\,\,i\pi }  =  - 1
$$

But you do not have any branch that can provide $\left( { - 1} \right)^{\,1/10} =-1$.

The matter is that if you choose a branch of the logarithm and remain within it, then you can reverse - to a certain extent-
the exponentiation keeping the results congruent, while if you "jump among the branches" you end with incongruencies.

The answer (to the titular question) is simple. Whereas the equation $$e^{\log x}=x$$ is true when $x>0$ (and therefore $\log x$ is well-defined and real), it fails to be true when $x<0.$

In more detail, we provide answers to the body of your question. You say,

  Say I have a function: $f(x)=x(x-6)^{1/5}.$ Is the output of the function just the real numbers?

Yes, $f(x)$ always takes a real value whenever $x$ is real, since we can always extract the fifth root of any real number (which is again always real, by definition), and the product of any two real numbers is again a real number.

Then your friend's redefinition of $f(x),$ and the attendant consequences, is explained by my answer above to your titular question.


what is the volume of a tilted glass I was wondering about the volume of a tilted glass (60$^o$).
Not sure, but my attempt:
$$y=\sqrt{x}, \space 0<x<9$$
$$x=y^2,\space 0<y<3$$
$$A(y)=\pi x^2=\pi y^4$$
$$V(y)=\int_0^3 \pi y^4\,dy$$
$$V=\frac{243 \pi}{5}$$
$$V(\text{adjusted})=\sin(60^o)*V=\frac{243\sqrt{3 \pi}}{10}$$
Some help to get?
Image:

Notice that if we reposition our axis to be parallel and perpendicular to the water level this is the equaivalent of performing the following:
$$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos(60)&-\sin(60)\\\sin(60)&\cos(60)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$
so you can substituted these into the equation to get new formulas for the lines and then work out from there. Alternatively you could imagine the current axis at normal positions and notice the water is at a 60 degree angle and form new limits to the integral for this

The line of the $60$ degree angle can be described by $x = \frac{1}{\sqrt{3}}y + 9 - \sqrt{3}$
The area enclosed by the curve and the line can be calculated with an integral.
$\int{}{}{\frac{1}{\sqrt{3}}y + 9 - \sqrt{3}} - y^2$
You need to find the limits of integration. For the right side, it will be 3 since the limit of the glass is 9. For the left side, it will be $-\sqrt{9-2\sqrt{3}}$.
$$\int_{-\sqrt{9-2\sqrt{3}}}^3{\frac{1}{\sqrt{3}}y + 9 - \sqrt{3}} - y^2 dy$$
Find the value of the integral and that is the two dimensional area of the glass tilted.
To find the volume for a three dimensional version you will have to evaluate an integral of the area of a circle from zero to the liquid level.
$$f_r(x) = \sqrt{x}$$
$$\int_0^{9 - \sqrt{3}} \pi f_r(x)^2 dx$$
$$\int_0^{9 - \sqrt{3}} \pi x \ dx$$
$$ \pi [\frac{x^2}{2} ]^{9 - \sqrt{3}}_0 $$
$$ \pi * \frac{(9-\sqrt{3})^2}{2} $$
$$=~ 82.974...$$


Is there a one-word term for "grows by square root"? We can describe a growth relationship $y = x$ as linear (e.g. "linear growth").
We can describe $y = \log(x)$ as logarithmic (e.g. "logarithmic growth").
We can describe $y = x^2$ as quadratic (e.g. "quadratic growth").
We can describe $y = 2^x$ as exponential (e.g. "exponential growth").
Is there a one-word term to describe a growth relationship where $y = \sqrt{x}$ ?
The best suggestion I've seen is "radical growth", which does not seem to be a standard term and I think would cause confusion if used without explanation.

The asymptotic growth of a root is known as "sublinear growth". Another term is "fractional-power" growth but it sounds odd imo.
The sublinear term originates from linear algerba. It can be used not only for square-root but for any $n-th$ root complexity.
In strict terms as sublinear function is a function that satisfies the following properties:

Positive Homogeneity: $f(kx) = k^n f(x) \quad n\in\mathbb{R^+_0}$ and
Subadditivity: $f(x+y) \leq f(x) + f(y)$

Roots satisfy both. The proof of positive homogeneity is trivial. As far as the proof of subadditivity is concerned you can check this post
Other seemingly sublinear functions (such as log) do not satisfy these properties therefore, in a sense you can separate them from sublinears.

Well since $y=\sqrt x \implies y^2=x$, I would call it inverse quadratic growth, i.e. the inverse of the function has quadratic growth.


Is there any solution such that.. Is there any positive integer solutions of $x$, $y$, $z$ such that both $$\frac{x^2+y^2+z^2}{x+y+z}$$ and $$\frac{x^2+y^2+z^2}{xyz}$$ are integers?

Edit: Forgot to mention that $x$, $y$, $z$ are distinct positive integers.

A remark, how to start: The second condition can be written as Diophantine equation
$$
x^2+y^2+z^2=nxyz,
$$
and it is easy to see that we must have $n=1$ or $n=3$. This is already very helpful. Compare also with this question.

For $n=3$ the triples $(x,y,z)$ with $\frac{x^2+y^2+z^2}{xyz}=3$ are known as Markov numbers, and we can produce a Markov tree. Now it gets easier if we also require that $(x,y,z)$ is a solution of the Diophantine equation $x^2+y^2+z^2=m(x+y+z)$.

Special case

Well, if $x = y = z$ then your equations become

$$\frac{3x^2}{3x} = x$$

$$\frac{3x^2}{x^3} = \frac{3}{x}$$

If they are coupled, then you notice that $x = 1$ is the simplest result. There is also $x = 3$: $3/x$ is an integer if and only if $x = 1$ or $x = 3$.

Unless you count negative integers too, in that case you have also $x = y = z = -1$ and $x = y = z = -3$


A lady and a monster A famous problem:

A lady is in the center of the circular lake and a monster is on the boundary of the lake. The speed of the monster is $v_m$, and the speed of the swimming lady is $v_l$. The goal of the lady is to come to the ground without meeting the monster, and the goal of the monster is to meet the lady.  Under some conditions on $v_m,v_l$ the lady can always win. What if these conditions are not satisfied?

Edited: the monster cannot swim.
If the conditions are not satisfied, then monster can always perform a strategy such that the lady will not escape the lake. On the other hand this strategy is not desirable for both of them because they do not reach their goals.
As there was mentioned, this deals with undecidability of the problem. On the other hand, if you imagine yourself to be this lady/monster, you can be interested in the strategy which is not optimal. What is it? If there are such strategies in the game theory?

Edited2:
My question is more general in fact. If we have a game with one parameter $v$ when two players $P_1, P_2$ are enemies and if $v>0$ then for any strategy of $P_2$ the player $P_1$ wins.
If $v\leq 0$ then for any strategy of $P_2$ there is a strategy of $P_1$ such that $P_2$ does not win and vice versa. I am interested in this case. From the mathematical point of view as I have understood the problem is undecidable since there is no an ultimate strategy neither for $P_1$ nor for $P_2$. But we are solving somehow these problem IRL.
Imagine that you are a lady in this game - then you would like to win anyway even while knowing that your strategy can be covered by the strategy of the monster. On the other hand, the monster knows that if he will cover all strategies of the lady she will never reach the shore and he will never catch her. I mean they have to develop some non-optimal strategies. I hope now it's more clear.

Since you seem to know the answer, I will give it here.

Suppose that $v_l = v_m / k $ and the radius of the lake is $r$.  Then the lady can reach a distance $\frac{r}{k}$ from the centre and keep the monster directly behind her, a distance $r\left(1 + \frac{1}{k}\right)$ away.  One way would be to swim in a spiral gradually edging outwards as the monster runs trying to close the distance; another would be to swim in a semi-circle of radius $\frac{r}{2k}$ away from the monster once it starts to run.  And the lady can sustain this distance by going round in a circle as the monster tries in vain to close the distance.

The next stage is for the lady to try to swim direct to shore at some point away from the direction the monster is running.  If the monster starts at the point $(-r,0)$ running anti-clockwise and the lady starts at the point $\left(\frac{r}{k},0\right)$ her best strategy is to head off in a straight line initially at right angles to the line between her and the monster: a less steep angle and the monster has proportionately less far to run than the lady has to swim, but a steeper angle and it is worth the monster changing direction.  (If the monster changes direction in this right-angle case, the lady changes too but now starts closer to shore.)   As they are both trying to get to the point $\left(\frac{r}{k},r \sqrt{1-\frac{1}{k^2}}\right)$ then they will arrive at the same time if  $ \pi + \cos^{-1}(1/k) = k \sqrt{1 -1/k^2}$ which by numerical methods gives $k \approx 4.6033$.

So if the monster is less than 4.6033 times as fast as the lady, the lady can escape; if not then she stays in the lake and the monster stays on the edge and they live unhappily ever after.

I was thinking about this problem and it seems to me that in a given formulation this game is somehow equaivalent to the following game.

If $v_m/v_l>k$ then on any action of the lady monster can either move (M) and follow her an catch her if she will come to the shore or stay (S) on the same place and wait (and thus let the lady run away). The lady also has to possibilities - either to move (M) and get out of the lake - or to stay (S) always inside the lake.

From these arguments it's easy to see that M is a dominating strategy for the monster (if we don't consider a penalty for the moving), hence it's optimal for the monster to follow the lady. Hence the lady on her side have to stay inside the lake.

For a pair (Lady, Monster) if $v_m/v_l>k$ there is a Nash equilibrium (Stay, Move).


Derivation of Backward Kolmogorov Equation I'm following Kallianpur-Gopinath's textbook "Stochastic analysis and diffusion processes" to study Kolmogorov equations and I got stuck in a step of the derivation of the backward equation.

In particular the authors (guess it's a pretty standard construction actually) start from the solution of a d-dimensional SDE

$$dX_t = b(t,X_t)dt + \sigma(t,X_t)dB_t$$

and, after computing the Taylor expansion of the probability transition function, arrive to write something like

$$ -\big(p(s,x;t,y) -p(s-h,x;t,y)\big) = \sum_{i=1}^d \frac{\partial}{\partial x_i}p(s,x;t,y) \int_{\mathbb{R}^d}(z_i-x_i)p(s-h,x;s,z)dz + \\
\frac{1}{2} \sum_{i=1}^d\sum_{j=1}^d \frac{\partial^2}{\partial x_i \partial x_j}p(s,x;t,y) \int_{\mathbb{R}^d}(z_i-x_i)(z_j-x_j)p(s-h,x;s,z)dz +\\
\int_{\mathbb{R}^d} o\big(|z-x|^3\big) p(s-h,x;s,z)dz\\$$

By dividing by $h$ and letting $h \to 0$ we should get the backward equation

$$ -\frac{ \partial p(s,x;t,y) }{ \partial s } = b(s,x) \cdot \nabla p(s,x;t,y) + \frac{1}{2} (\sigma(s,x), \nabla^2 p(s,x;t,y) \sigma(s,x) ) $$

what I don't understand is the step

$$ \lim_{h\to 0} \frac{1}{h} \int_{\mathbb{R}^d} (z_i-x_i) p(s-h,x;s,z)dz \stackrel{?}{=} b_i(s,x)$$

I understand this should be related with the fact that the solutions of SDE are diffusion processes, but the definition of diffusion process requires

$$ \lim_{h\to 0} \frac{1}{h} \int_{U_x} (z_i-x_i) p(s-h,x;s,z)dz = b_i(s,x)$$

for any bounded neighborhood $U_x$ of $x$, but instead we are considering an integral over $\mathbb{R}^d$. Even if I split the integral in two, one over a bounded set and one over its complement I don't see how to prove that the second goes to $0$. Any help is appreciated.

EDIT:

Quoting from the same book:

  A Markov process $X_t$ with transition function $p(s,x;t,A)$ is called a diffusion process if it has continuous paths and if there exists an $\mathbb{R}^d$-valued function $b(s,x)$ and a $d \times d$-matrix-valued function $a(t,x)$ s.t. for every bounded open neighborhood $U_x$ of $x$

  $\lim_{h \downarrow 0} \frac{1}{h} \int_{U_x^c} p(t,x;t+h,dy) = 0$
  $\lim_{h \downarrow 0} \frac{1}{h} \int_{U_x} (y_i - x_i) p(t,x;t+h,dy) = b_i(t,x) \ \ \ \ i \in \{1...d\}$
  $\lim_{h \downarrow 0} \frac{1}{h} \int_{U_x} (y_i - x_i)(y_j - x_j) p(t,x;t+h,dy) = a_{i,j}(t,x) \ \ \ \ i,j \in \{1...d\}$

and immediately after there is a theorem that states that the solution of the SDE written at the top of this post, under the usual hypotheses of Lipschitzianity of $b$ and $\sigma$, is a diffusion process with drift $b$ and diffusion coefficient $\sigma \sigma^*$.

The authors say that 'we will formally derive a pair of parabolic pdes known as the Kolmogorov backward and forward equations' at the beginning of Section 8.3. It's easy to miss, but the use of expansion indicates a heuristic presentation. A rigorous derivation is provided in Section 8.6.


ODE with change of variable I'm trying to solve
$$y'=\frac{x+y}{y}.$$
I then tried the change of variable
$$ v=\frac{y}{x},$$
and -I think-
$$y'=xu'+u$$
I went on few lines but then I got struck since the thing was clearly not going in the direction of the solution. What did I do wrong?

With your Substitution we get
$$u+u'x=1+\frac{1}{u}$$ this is $$xu'=\frac{1+u-u^2}{u}$$

Can you finish?


Show that $A$ is countable if and only if $A=\emptyset$ or there exists a surjective function $\mathbb N\rightarrow A$. Show that $A$ is countable if and only if $A=\emptyset$ or there exists a surjective function $\mathbb N\rightarrow A$.

I am having some trouble with this. This is what I have so far. Can someone verify it or help lead me to the correct path? Thank you.

Proof:

In the first direction, if $A=\emptyset$, then it is clear. If not, let there be a function $f:A\rightarrow \mathbb{N}\quad f(x)=2x$. Then $f$ is bijective, so it has an inverse: $g:\mathbb{N}\rightarrow A\quad g(x)=\frac{1}{2}x$.

$f\circ g=2(\frac{1}{2}x)=x=id_\mathbb{N}$ and $g\circ f=\frac{1}{2}(2x)=x=id_A$, so $g$
is bijective. Thus, $A\sim\mathbb{N}$, so $A$ is countable.

In the second direction, if $A$ is countable, then $A$ is equinumerous to $\mathbb{N}$ so there exists a bijective function from $A$ to $\mathbb{N}$.

Definition: A set is countable if it is finite, $\emptyset$ or equinumerous to $\mathbb{N}$

The left to right implication is the easiest :

Let $A$ be countable.

if $A = \emptyset$, we are done
if $A$ is finite, say $A = \{a_0, \dots, a_k\}$ then the following map is a surjection $f : \mathbb{N} \mapsto A$ :
$$f : n \mapsto
\left\{ \begin{array}{ll} a_n & \textrm{if } n \leqslant k \\
a_0 & \textrm{otherwise} \end{array} \right. $$
finally, if $A$ is equinumerous with $\mathbb{N}$ then you have a bijection $f : \mathbb{N} \mapsto A$, which is hence a surjection.

The right to left implication is the hard part :

If $A=\emptyset$ or $A$ is finite we are done. So let's assume that $A$ is infinite and we have a surjective function $f : \mathbb{N} \mapsto A$. We need to find a bijection $g : \mathbb{N} \mapsto A$.
By induction, for each $n \in \mathbb{N}$, we construct the family of functions  $(g_n)_{n\in \mathbb{N}}$ satisfying :

for each $n$, $g_n$ is an injective function $\{0, \dots n\} \mapsto A$.
for each $m \leqslant n$, $g_n$ extends $g_m$, ie $\forall k \leqslant m$, $g_n(k) = g_m(k)$.
for each  $k \leqslant n$ , $f(k) \in \operatorname{Im}(g_n)$

If we can do that we are done since $g : n \mapsto g_n(n)$ is a bijection from $\mathbb{N}$ to $A$.

induction start We let $g_0 : \{ 0 \} \mapsto A$ be defined by $g_0(0) := f(0)$.

Induction step Let $n \in \mathbb{N}$ Suppose we have already constructed $g_0, \dots, g_n$ satisfying points 1. 2. and 3. above.
 As $A$ is infinite, $g_n$ can't be a surjection (for otherwise $A$ would be finite), hence the set $X_n := \{k \in \mathbb{N} \ | \ f(k) \notin \operatorname{Im}(g_n) \}$ isn't empty. This allows us to define $g_{n + 1}(n + 1):= f(\min X_n)$.
For $k \leqslant n$, one sets $g_{n + 1}(k) := g_n(k)$.

Clearly, $g_{n+1}$ is injective and extends $g_m$ for $m \leqslant n + 1$. Let's check point 3. :
let $k \leqslant n + 1$ , then

if $k \leqslant n$ we are done since $f(k) \in \operatorname{Im}(g_n) \subseteq \operatorname{Im}(g_{n+1})$.
if $k = n+1$ then either $f(k) \in \operatorname{Im}(g_n)$ and we are done or we have $n + 1 \in X_n$, and since $0, \dots n, \notin X_n$ we have $n+1 = \min X_n$ hence $a= g_{n+1}(n+1)$.

For any element $a$ in $A$ consider the set $\phi(a) = \{n|f(n)=a \}$ where $f$ is your surjection.Now $\phi(a)$ is non-empty for all $a$ in $A$ so,we define $g(x)= \min.\phi(x)$ ,now it is easy to check that $g$ is bijection from $A$ to a subset of natural numbers.(Note that min. exists by Well Ordering Principle)


Can every orthogonal matrix be written as a product of Givens rotations? I'd like to know whether every orthogonal matrix

$$ A \in \mathcal{O}_n(\mathbb{R})$$

can be written as a product of givens-rotations. I know that when we do QR-decomposition of matrix $A$ we get

$$ A = Q R $$

So my idea was to prove that $R$ must be the identity $I_n$, however I'm stuck at that. Can somebody give me a hint on how I could prove this?

Givens rotations are... rotations, they preserve orientation ($det(M)=+1$), however $\mathcal{O}_n(\mathbb{R})$ has two components, one component is rotations ($det(M)=+1$), the other is reflections ($det(M)=-1$).

Example of $M\in\mathcal{O}_2(\mathbb{R})$ that can not be written as a Givens rotation (its determinant is $-1$)
$$
M=\left(\begin{array}{cc}
-1 & 0 \\
0 & 1
\end{array}\right)
$$
You can not find Givens rotation $G_\theta=\left(\begin{array}{cc}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{array}\right)$ such that $M=G_\theta$

If you want to have a Givens rotation decomposition you must restrict yourself to $SO_n(\mathbb{R})$, the special orthogonal group, which is defined by $M^tM=I_d$ and $\det{M}=+1$.

If you write the matrix $A=QR$ then there are two ways you can perform this.

$$ A \underbrace{R_{1} R_{2} \cdots R_{n}}_{\hat{R}^{-1}} = \hat{Q} \tag{1} $$

or the following, which is what you want

$$ \underbrace{Q_{n} Q_{n-1} \cdots Q_{2}Q_{1}}_{{Q}^{*}}A = R \tag{2} $$

In $1$ we see that Gram-Schmidt applies a sequence of elementary triangular matrices $R_{k}$ on the right of $A$. For $2$ we see a sequence of elementary unitary matrices $Q_{k}$ on the left of $A$. The matrices $Q_{k}$ can be givens matrices.

$$ \underbrace{G_{n} G_{n-1} \cdots G_{2}G_{1}}_{{G}^{*}}A = R \tag{3} $$

All you need to prove is that $G_{k}$ is a unitary matrix. It is. There are notes provided here.


Why is Euclid's proof on the infinitude of primes considered a proof? I've expressed Euclid's proof on the infinitude of primes on Mathematica:

f[x_] := Product[Prime[n], {n, 1, x}] + 1
TableForm[Table[{f[x], PrimeQ[f[x]]}, {x, 1, 20}]]

Which results in:

$\begin{array}{ll}
 3 & \text{True} \\
 7 & \text{True} \\
 31 & \text{True} \\
 211 & \text{True} \\
 2311 & \text{True} \\
 30031 & \text{False} \\
 510511 & \text{False} \\
 9699691 & \text{False} \\
 223092871 & \text{False} \\
 6469693231 & \text{False} \\
 200560490131 & \text{True} \\
 7420738134811 & \text{False} \\
 304250263527211 & \text{False} \\
 13082761331670031 & \text{False} \\
 614889782588491411 & \text{False} \\
 32589158477190044731 & \text{False} \\
 1922760350154212639071 & \text{False} \\
 117288381359406970983271 & \text{False} \\
 7858321551080267055879091 & \text{False} \\
 557940830126698960967415391 & \text{False} \\
\end{array}$

The proof flaws for all those values, how is it considered a proof then? I guess that there might be infinite prime numbers according to the proof, but what is the guarantee that at some point it won't fail indefinitely?

Euclid’s proof differs from what many mathematicians tell you it is.  He said this:

Take any finite set of primes.  (Don’t assume it’s the set of all primes; don’t make it a proof by contradiction; don’t assume it’s the first $n$ primes; for example it could be $\{2,7,31\}$.)

Multiply them and add $1$.  Then show (and this part was done by contradiction) that the prime factors of the resulting number are not in the finite set you started with.

Thus every finite set of primes can be extended to a larger finite set of primes.

Nothing in that argument gives you any reason to think that if you multiply the first $n$ primes and add $1$, the result is prime.  That’s a confusion resulting from inattentiveness to what Euclid actually wrote.

I had a joint paper with Catherine Woodgold about this in the Mathematical Intelligencer in autumn 2009. “Prime Simplicity”

An excerpt from our paper:

  Only the premise that a set contains all prime numbers could make anyone conclude that if a number is not divisible by any primes in that set, then it is not divisible by any primes.

  Only the statement that $p_1\dots p_n+1$ is not divisible by any primes makes anyone conclude that that number "is therefore itself prime", to quote no less a number theorist than G. H. Hardy [who] actually attributed that conclusion to Euclid!  (Euclid's statement "Certainly [that number] is prime, or not" [...] clearly shows that Euclid's reasoning did not follow that path.)

  The mistake of thinking that $p_1\dots p_n+1$ has been proved to be prime is made all the more tempting by the very obvious fact that that would entail the result to be proved.

  [ . . . ]

  In any proof by contradiction, once the contradiction is reached, one can wonder which of the statements asserted to have been proved along the way can really be proved in just the manner given (since the argument supporting them does not rely on the initial assumption later proved false), which ones are correct but must be proved in some other way (since the argument supporting them does rely on the initial assumption) and which ones are false.  It is easy to neglect that task.  One's consequent ignorance of the answers to those questions can lead to confusion: after all, when one remembers reading a proof of a proposition, might one not think the proposition has been proved and is therefore known to be true?  G. H. Hardy probably was aware that because the conclusion that $p_1\dots p_n+1$ "is therefore itself prime" was contingent on a hypothesis later proved false, it could not be taken to be proved.  But he did not say that explicitly.  It seems hard to justify a similar confidence that all of his readers avoided the error into which he inadvertently invited them.

The proper way to express Euclid's idea in Mathematica would be something like this:

LimitedPrimeQ[x_, y_] := Not[Or@@(Divisible[y,#]&/@Prime/@Range[x])]
f[x_] := Product[Prime[n], {n, 1, x}] + 1
TableForm[Table[{f[x], LimitedPrimeQ[x, f[x]]}, {x, 1, 20}]]

Here LimitedPrimeQ checks whether y is divisible by the first x primes. If there were only x primes in total, as the assumption of the proof by contradiction states, then this would be equivalent to PrimeQ. But the above will print True for every single row, and you can proove that it does so for any row, just as Euclid did.


If $x$ and $y$ are linearly independent, there is a separating linear functional Let $E$ be a $\mathbb R$-vector space and $x,y\in E$ be linearly independent. How can we show that there is a $\varphi\in E^\ast$ with $\varphi(x)=0$ and $\varphi(y)=1$?

I guess we need to construct $\varphi$ in terms of the coordinate functionals $(\varphi_e)_{e\in B}$ related to a basis $B$ of $E$.

EDIT 1: If it's easier to prove, I'd specifically interested in the cases where $E$ is a Hilbert space or finite-dimensional (or both).

EDIT 2: Maybe I'm missing something, but shouldn't the claim trivially follow in the following way: Let $x_1:=x$ and $x_2:=y$. Since $x_1$ and $x_2$ are linearly independent, $U:=\mathbb Rx_1+\mathbb Rx_2$ is a $\mathbb R$-vector space of dimension $2$ and, by construction, $(x_1,x_2)$ is a basis of $U$. So, there are $\varphi_1,\varphi_2\in U^\ast$ with $\varphi_i(x_j)=\delta_{ij}$ for all $i,j\in\{1,2\}$. So, all we need to do is showing that $\varphi_2$ has a (not necessarily unique) linear extension to $E$.

If you assume that every vector space has a basis, you can also prove the corollary that every linearly independent set can be extended to a basis.

Indeed, if $\mathscr{A}$ is linearly independent and $U$ is the span of $\mathscr{A}$, the vector space $E/U$ has a basis $\mathscr{C}=\{x+U:x\in\mathscr{B}\}$ for some set $\mathscr{B}$. Then it's easy to prove that $\mathscr{A}\cup\mathscr{B}$ is a basis of $E$.

In your case take $\mathscr{A}=\{x,y\}$; once you get $\mathscr{B}$ as above, define $\varphi\colon E\to \mathbb{R}$ by
$$
\varphi(x)=0,\quad \varphi(y)=1,\quad \varphi(z)=0\ (z\in\mathscr{B})
$$
This is nothing else than the functional corresponding to $y$ in the dual basis of $\{x,y\}\cup\mathscr{B}$.

So you see that the trick is to choose a suitable basis.

In the case of a Hilbert space, you want $\varphi$ to be continuous. In this case you can use the fact that the span of $\{x,y\}$ is closed and you can take the orthogonal complement $K$ thereof. Then you can consider $\operatorname{span}(\{x\})\oplus K$ and apply Hahn-Banach.

Addition 1

How do you prove that every linearly independent set can be extended to a basis? One needs Zorn's lemma, as existence of bases for all vector spaces is equivalent to the axiom of choice (and to Zorn's lemma).

Let's fix a vector space $E$.

Let $\mathscr{A}$ be a linearly independent set in $E$. Consider the set $\mathfrak{I}$ consisting of all linearly independent sets $\mathscr{B}$ in $E$ such that $\mathscr{B}\supseteq\mathscr{A}$, ordered by set inclusion. The union of a chain of members of $\mathfrak{I}$ is again linearly independent, because linear independence is checked on finite subsets.

By Zorn's lemma, the set $\mathfrak{I}$ has a maximal element $\mathscr{C}$. I claim that $\mathscr{C}$ spans $E$. Otherwise there would exist $v\in E$ not in the span of $\mathscr{C}$; but then $\mathscr{C}\cup\{v\}$ would be linearly independent, contradicting maximality of $\mathscr{C}$.

Addition 2

Suppose you have as an axiom that every vector space has a basis. How can we prove the theorem above? The easiest way is to consider the subspace $U$ spanned by $\mathscr{A}$. The map
$$
\pi\colon E\to E/U,\qquad \pi(v)=v+U
$$
is linear. If $\mathscr{C}$ is a basis of $E/U$, whose existence is guaranteed by the axiom, then you can write $\mathscr{C}=\{x+U:x\in\mathscr{B}\}$ (choose a representative for every element of $\mathscr{C}$). The set $\mathscr{A}\cup\mathscr{B}$ is then a basis of $E$.

Indeed, suppose we have
$$
\alpha_1v_1+\dots+\alpha_mv_m+\beta_1x_1+\dots+\beta_nx_n=0
$$
with $v_i\in\mathscr{A}$ and $x_i\in\mathscr{B}$. Then applying $\pi$ we obtain
$$
\beta_1(x_1+U)+\dots+\beta_n(x_n+U)=0+U
$$
By linear independence of $\mathscr{C}$, we get $\beta_1=\dots=\beta_n=0$. Now, since $\mathscr{A}$ is linearly independent, also $\alpha_1=\dots=\alpha_m=0$.

Let's prove that $\mathscr{A}\cup\mathscr{B}$ is a spanning set. If $v\in E$, then $\pi(v)=v+U=\beta_1(x_1+U)+\dots+\beta_n(x_n+U)$, with $x_1,\dots,x_n\in\mathscr{B}$, because $\mathscr{C}$ is a spanning set of $E/U$. This implies that
$$
v-\beta_1x_1+\dots+\beta_nx_n\in U
$$
so we can write $v-\beta_1x_1+\dots+\beta_nx_n=\alpha_1v_1+\dots+\alpha_mv_m$, with $v_i\in\mathscr{A}$ and we're done.

If $E$ is infinite dimensional, it may not be "such an elementary linear algebra result".

In finite dimension $\geq 2$, take any $x_0$ in the hyperplane orthogonal to $\mathbb{R}x$ which is not also orthogonal to $\mathbb Ry$, then define $\varphi(y_0) = \frac{ \langle x_0, y_0\rangle}{\langle x_0, y \rangle}$.


Twin primes of the form $n^2+1$ and $N^2+3$? Assume that there are infinity many primes of the form  $n^2+1$ and there are infinity many primes of the form  $N^2+3$ , Then could we show that there are infinity primes of the form $n^2+1$  and   $N^2+3$ (twin primes ) ?

Edit: I have edit the question just to show that $n$ and $N$ are not the same

This is certainly not sufficient.  For example, it could be that all the primes of the form $n^2+1$ are of the form $(4k)^2+1$ and all the primes of the form $n^2+3$ are of the form $(4k+2)^2+3$.  Then there would be no twin primes of this type, even if there were infinitely many of each.  Yes, $17$ and $19$ is an example, so I could say that all the primes above $10^{1000}$ satisfy my restriction.  There could then be many twin primes of your type, but not an infinite number.

A (trivial) insight...It's a lot easier to show which primes $n^2+1$ can't produce. Consider: $n^2+1 = n^2-1+2 = (n-1)(n+1)+2$; if $n^2+1 = p$, $p$ prime, then $(n-1)(n+1)+2 = p$, or $(n-1)(n+1) = p - 2$. Clearly $p$ can never be the larger of any two twin prime pairs because $p-2$ has factors $(n-1)$ and $(n+1)$.  Further, $n^2+1 = (n^2-a^2)+(a^2+1)$. If $(n^2+1)$ is prime, then $p-a^2-1$ cannot be for all $0<a<n-1$. For a even, $a^2+1$ is odd and $p-a^2-1$ is even, so only $a$ odd is of interest. Conclude: Given two primes $p_1$ and $p_2$, where $p_2>p_1$, then if $p_2=n^2+1$ for certain $n>0$, then $p_2-p_1$ cannot equal $a^2+1$, $0<a<n-1$.

So the primes $n^2+1$ can produce are somewhat reduced in number because of the growing number of gap conditions that need to be satisfied.  And the further one goes out, the more conditions there are to satisfy. So another form of the $n^2+1$ conjecture might be to ask whether it is possible to satisfy the conditions as $n$ increases.  -FW


Index of a Symmetric Matrix In Hansjorg Geiges' introductory textbook on Symplectic Geometry, is defined a projective conic given by $q^tAq=0$ where $A$ is a symmetric matrix of rank 3 and index 2. What does "index 2" mean? I googled it in vain. I am seeing the term "index of a matrix" for the very first time. It was not taught to me either. The reason I want to know is that I would like to plug some value for A and visualise the projective conic given in the text. Any help is greatly appreciated.

Index of real quadratic form x'Axmeans number of 1  s in canonical reduction .It  is same as number of  positive eigen values of real symmetric matrix A.


What are the odds of hitting exactly 100 rolling a fair die I roll a fair die and sequentially sum the numbers the die shows. What are the odds the summation will hit exactly 100?

More generally, what are the odds of hitting an exact target number t while summing the results of numbers sequentially drawn uniformly from a set S.

I have experimentally tested this and the result is (warning - spoilers ahead):

   1 over the expectation of the drawn numbers. In the case of the fair die 1/((1+2+3+4+5+6)/6) = 6/21 = 2/7

However I do not have a strong intuition, let alone a formal proof, why this is the case.

I'll be happy to get your thoughts!

The probability of hitting $n$ on the nose is a function of $n$, call it $f(n)$. It satisfies the recurrence
$$f(n)=\frac{f(n-1)+f(n-2)+f(n-3)+f(n-4)+f(n-5)+f(n-6)}6$$
with boundary values $f(0)=1$ and $f(n)=0$ for $n\lt0$.
I didn't try to solve the recurrence, but my numerical calculations agree with yours, it looks like $f(n)\to2/7$.

I would say since the possible outcomes are $1,2,3,4,5,6$ on the die that the chances of getting exactly $100$ are the same as that of getting any single desired digit on the die once you are in the $94$ to $99$ range. Eventually you will get into that range such that you need either a $1,2,3,4,5$ or $6$ to make a $100$ sum.  Then you must get that number to achieve the $100$ sum.  So I would say $1$ in $6$ chance at that point.  For example, suppose the partial sum is $94$, then you would need a $6$.  If you instead roll a $2$ and get $96$ then you would need a $4$.  If instead you roll a $3$ then you need a $1$.  So anytime you get into the $94$ to $99$ range, you have a $1$ in $6$ chance of hitting exactly $100$ on a single final roll.  If you instead consider the probability from the beginning when the partial sum is $0$ or when the partial sum is in the range $94$ to $99$ and considering multiple "final" rolls, then the answer will be different.

So the answer depends on when you start counting the probability but in my case, I simplified and only care about the one final roll that would give you a $100$ sum.

One can also argue/claim that the probability is $0$ if the number of die rolls for the $100$ sum case is less than $17$ rolls.  The original question didn't specify how many rolls are allowed or will take place.

Another answer is to interpret the question to mean only 1 die roll and sum (as many times as needed) only the numbers showing.  This could be interpreted many ways such as only the top number (in which case $1, 2, 4$, and $5$ would "always" sum to exactly $100$), or all sides except the "hidden" bottom side (assuming it wasn't rolled on a see thru table or equivalent), or even summing all $6$ sides if they are all visible.

Also the question doesn't state only base $10$ of the sum $100$ so if I go past that, I could change to base $11$ and keep rolling (and go for $121$ in base $10$ which is $100$ in base $11$).  If I miss that I can go for $144$ which is $100$ in base $12$....

How about if you interpret the question as strictly $1$ die roll and all $6$ faces are visible thus adding up to $21$ and then expressing that sum in base $21^{1/2}$ which would be written as $100$?  Then it is $100$% chance of getting it.

I don't know the answer to the more general question yet but I will think about it some.


Finitely generated $R$-module in the product of quotient field. Let $R$ be an integral domain with the quotient field $K$.

Let $M$ be a finitely generated $R$-submodule in $K^n$.

Is it true that $M$ is free $R$-module?

Non, it is almost never true. E.g. an finitely generated ideal $M$ in $R$ is free only if it is generated by one element because two elements $x_1, x_2\in M$ are always related by the linear relation $x_1.x_2+(-x_2).x_1=0$, thus any basis of $M$ can only have one element.

Example: $R=\mathbb C[X, Y]$, $M=XR+YR$. If $M$ was free, it would be generated by one element $P(X,Y)$. Then $P(X,Y)$ divides $X$ and $Y$,so $P(X,Y)$ is constant and generates the unit ideal $R$. But $M\ne R$.

Let $\,M=\langle\,m_1,...,m_s\,\rangle_R\leq K^n\,$ be a finitely generated $\,R-\,$submodule of $\,K^n\,$. Assuming  the above is a minimal generating set for $\,M\,$, we thus clearly have $(1)\,\,\,s\leq n\,$ (why?) and also $(2)\,\,\,\{m_1,...,m_s\}\,$ are $\,K-\,$ linearly independent (why?). Thus

$$r_1m_1+...+r_sm_s=0\,\,,\,\,r_i\in R\Longrightarrow r_i=0\,\,\forall\,i$$

since otherwise we get a contradiction to (2) above.


Prove a function is a norm? I want to show that $${||x||}_{\infty} = \max{\{|x_1|,|x_1|,...,|x_n|\}}$$
is a norm.

Where the properties of a norm are

${||x||} \ge 0$ for all $x$ in${\mathbb R}^n$
${||x||} = 0$ when $x=0$
${||ax||} = |a|{||x||}$ for all $a$ in ${\mathbb R}$
$||x+y|| \le ||x||+||y||$

I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?

To prove 4., notice that

$$ ||x+y||_{\infty} = |x_l+y_l | $$

for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| \leq |x| + |y| $ and since for example $|x_l| \leq \max |x_i| = ||x|| $, the result follows

For 1. $\max = |x_i| \geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component.
For 2. let $||x||=0 \rightarrow 0 = \max \geq |x_i| \geq 0 \forall x_i \rightarrow x=0$. Now let $x=0 \rightarrow \forall x_i, x_i = 0 \rightarrow \max = 0 \rightarrow ||x||=0$.
For 3. $||ax|| = \max |ax| = |a||x_i| = |a|\max |x| = |a|||x||$, where $|x_i|$ is the max element.
For 4. $||x+y|| = |x_i+y_i| \leq |x_i| + |y_i| \leq \max x + \max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).


If $y'(t)=.2y(t)(1000-y(t))$, what are the values of $y$ such that $y(t)$ is increasing at a decreasing rate? A population rate changes at a rate modeled by the differential equation
$ \frac{dy}{dt} = .2y(1000-y),   $ where t is measured in years. What are all y values for which y is increasing at a decreasing rate?

An answer online claims that the second derivative of this would be:

$ 200 - .4y  $, however, isn't this answer leaving out dy/dt? In other words, should not the second derivative be:

$200 - .4y\frac{dy}{dt}   $?

On this site, the solution does not have the last dy/dt: http://www.rtmsd.org/cms/lib9/PA01000204/Centricity/Domain/211/2013%20BC%20Multiple%20Choice%20SOLUTIONS.pdf

Isn't that incorrect?

We have $\frac{dy}{dt} = .2y(1000-y)$

$\frac{dy}{dt} = 200y - .2y^2$

1.) If finding derivative with respect to t, then it is -

$\frac{d^2y}{dt^2} = 200 \frac{dy}{dt} - 0.4y\frac{dy}{dt}$

2.) If finding derivative with respect to y, then it is -

$= 200 - 0.4y$

Yes it seems that the person who posted that answer took the derivative with respect to y and did not use the chain rule to get the derivative with respect to t.


Derivatives of equations Assume that $x$ and $y$ are related by the equation $y\ln x=e^{1−x}+y^3$. Compute $dy/dx$ evaluated at $x=1$.

I do not understand how to compute the derivative of an equation. Please explain.

Here are the first few steps you need to do. As others have noted, it is called implicit differentiation. It is very similar to "normal" differentiation, but the difference is that you consider $y$ to be a function of $x$. When you want to differentiate it with respect to $x$, you get that $\frac{d}{dx}(y) = \frac{dy}{dx}$ (very simply) . Also, you need to remember to apply to $y$ all of the rules you normally apply to functions of $x$. For example, you need to apply the product rule to $\frac{d}{dx}(xy)$, among other things.

So you have that

$$y\ln x = e^{1-x} + y^3$$
thus
$$\frac{d}{dx} (y\ln x) =  \frac{d}{dx} ( e^{1-x} + y^3 )$$
so
$$\frac{dy}{dx}\ln x + \frac{y}{x} = -e^{1-x} + 3y^2 \frac{dy}{dx}$$

Now you should be able to solve for $\frac{dy}{dx}$, which will be dependent of both $x$ and $y$.

Hint: You're not computing the derivative of an equation. You're computing the derivative of $y$ with respect to $x$. Since you aren't given $y$ in terms of $x$, your answer will most likely be in terms of both $y$ and $x$, not just $x$.


$(f) = (c)(g)$, where $R$ is a UFD, $c \in R$, $f,g \in R[x]$, and $g$ is primitive Let $R$ be a ring, and let $I$ be an ideal of $R$. Then $IR[x] = \{a_0 + \cdots + a_dx^d \in R[x] \mid a_i \in I \}$ is an ideal of $R[x]$.
Let $R$ be a commutative ring, and let $f = a_0 + a_1x + \cdots + a_dx^d \in R[x]$ be a polynomial.
$f$ is a primitive if for all prime ideals $\rho$ of $R$, $f \notin \rho R[x]$.

Let $R$ be a UFD. The content of a nonzero polynomial $f \in R[x]$, denoted $\operatorname{cont}_f$, is the $\gcd$ of its coefficients. As any $\gcd$ in an integral domain, it's defined only up to associate relation.
We also know that if $R$ is a UFD, then $f$ is primitive if and only if $(\operatorname{cont}_f) = (1)$.

Now, what I need to prove is that:

Let $R$ be a UFD, and let $f \in R[x]$. Then if $(f) = (c)(g)$ (where $(c)(g) = (cg)$), with $c \in R$, and $g$ primitive, then $(c) = (\operatorname{cont}_f)$.

I tried some things, but I to no avail (like reducing ideals the equality to $f = ucg$ for a unit $u$ in $R$ and then seeing what can we extract from the equality $a_i = ucb_i$). Right now I don't have any ideas on approaching a proof.

Suppose $(f)=(c)(g)$. Then $(f)=(cg)$. So, $f =ucg$ for some unit $u \in R$.

What is the $\gcd$ of the coefficients of $f$?

It is the $\gcd$ of the coefficients of $ucg$. Since $g$ is primitive, the $\gcd$ of the coefficients of $ucg$ is $uc$. Thus, the $\gcd$ of the coefficients of $f$ is $uc$.

But since $\gcd$ is defined up to units, then we can say the $\gcd$ of the coefficients of $f$ is just $c$.

So, $(\mathrm{cont}_f) =(c)$.

$\newcommand{\cont}{\operatorname{cont}}$Hint:$f=\cont_f{f\over {\cont_f}}$ and ${f\over {\cont_f}}$ is a primitive polynomial, $f=cg=\cont_f{f\over {\cont_f}}$ where $g$ is a primitive polynomial implies that $c|\cont_f$ and $\cont_f|c$.


How does factoring a quadratic equation relate to the parabola on the graph and to common sense? I just learned how to factor quadratic equations (at least, how to factor them when they're simple and easy; I don't know how to factor when $a$ isn't equal to $1$). I'm fine at doing this, and I get that the outcome gives me the roots of the parabola. However, I would love to know how the process itself (as in the factoring and the creation of two linear equations) relates to graphing parabolas and to the general behavior of quadratic equations. Is there any visual representation of what's happening?

By the way, I am somewhat familiar with other ways to solve quadratic equations such as completing the square and using the quadratic formula, and I know how to find the vertex; I would love to know how factoring relates to these things (conceptually) as well.

Thank you, and I'm sorry for the vagueness of my question.

It's not entirely clear how you use your method to factor a quadratic polynomial,
but the usual goal of factorizing a general quadratic,

$$ ax^2 + bx + c, $$

is that you end up with something like this:

$$ ax^2 + bx + c = a(x - r_1)(x - r_2). $$

(That is, to deal with the case where $a \neq 1,$ you can factor $a$ out of the polynomial as the very first step, and then you just have to factor something where the $x^2$ term has coefficient $1$.)

The numbers $r_0$ and $r_1$ are the roots of the quadratic, that is, they are the $x$ coordinates where the graph of $y = ax^2 + bx + c$ crosses the $x$ axis.

The $x$ coordinate of the vertex is midway between $r_0$ and $r_1$.
The $y$ coordinate is not obvious from the factorization; basically I would find the $x$ coordinate and then plug it into either the original polynomial or the factorization in order to compute the $y$ coordinate.

There is also something called the vertex form of a quadratic equation,
which is somewhat related to factorization (and can help you with the factorization), but it is not actually a factorization:

$$ ax^2 + bx + c = a(x - h)^2 + k. $$

Now the coordinates of the vertex are $(h,k),$ but the $x$ coordinates where the graph crosses the $x$ axis are not as easy to find as in the factorization.

The vertex form is not a factorization in general because it is not the product of some expressions multiplied together; it is the sum of
$a(x-h)^2$ and $k.$
In the case where $k = 0$ you are left with just $a(x-h)^2,$ which is a factorization, but in other cases the $k$ prevents you from using $x - h$
as a factor of the quadratic.

Suppose a function $F: R -> R$ has n-roots ${r1, r2, r3 ... rn}$ then the function F can be written as $F(X) = (X - r1)*(X - r2)*(X - r3) ... *(X - rn)$ .

So basically when we factor a function $F(x)$ in simpler degree-1 terms like $(X - ri)$ we are actually finding root ri that contributes to the function F.

But why is thing true ?

Lets see what remainder theorem is :
Suppose :
$F(x) = G(x) + rem$
Now dividing by $(X - r1)$ on both sides gives:
$F(x)/(x - r1) = Q(x) + rem/(x - r1)$
$=> F(x) = Q(x)(X - r1) + rem $
If $X = r1$, then $F(r1) = rem$.

So we can see if $rem = 0$ or $F(r1) = 0$, then $r1$ is a root of F(x). Similarly $Q(x)$ can be factored sown to get all the remaining roots.

References : https://brilliant.org/wiki/polynomials/#factoring-polynomials
Try this video. Will surely help. https://www.youtube.com/watch?v=MHXO86wKeDY.


"Recursively" expressing continuous-time trajectories I am interested in functions of time I call "trajectories" $h$, each in an arbitrary codomain $D$. $D$ is a set.

I wish to express that these trajectories are determined by an initial condition $h_0$ and an evolution/determining function $f$. (almost-)Like a dynamical system. $f$ can take any required form/signature, even complicated. And $f$ is supposed to be given.
I also wish to express that the universe hosting these trajectories, represented by $f$, is gifted with a memory so any value taken by $h$ depends on all past values taken by $h$, and not on future values.

If time were discrete, I would express it this way:

\begin{equation}
\newcommand{\funmap}[4]{{%
        \left\{%
            \begin{array}{ccl}
            #1 & \to     & \ #2 \\
            #3 & \mapsto & \ #4
            \end{array}
        \right.
    }}
    h : \funmap{\mathbb{N}}{D}{n}{h(n)}
\end{equation}
then
\begin{align}
h(0) &= h_0\\
\forall n \in \mathbb{N}^{*},\ h(n) &= f\left(h_{\big|[0,n[}\right) \text{ or } f\left(h_0, ..., h_{n-1}\right) \text{ ..}
\end{align}
.. depending on how we define $f$. Anyway, I repeat I'm mostly interested in expriming that:

1. given $f$ and $h_0$, $h$ is (constructively) completely determined (exists + is unique + we can compute it).
2. any value of $h$ depends on all its past values.

In continuous time however, I spontaneously try to express 2. this way:

\begin{equation}
h : \funmap{\mathbb{R^+}}{D}{t}{h(t)}
\end{equation}

then
\begin{align}
h(0) &= h_0\\
\forall t \in \mathbb{R}^{+*},\ h(t) &= f\left(h_{\big|[0,t[}\right)
\end{align}

.. but I'm not sure that it makes 1. hold. Is 1. true then?
If not, how to express both 1. and 2. in continuous time?

I am aware that this sounds like a recursive definition over $\mathbb{R}^+$, which I think is impossible.. unless we accept the axiom of choice and make it a transfinite recursion by forcing $\mathbb{R}^+$ to be well-ordered? Which is not constructive..

Is there a better way to write this?

Tracks I could think of:

discretizing $\mathbb{R}^+$, defining $h$ recursively on small bits.. then limit-reducing the size of the bits to zero. But I am afraid that everything would collapse at the limit resulting in $h$ being undefined.. would it be the case?
generalizing the concept of "derivative" and "local variation" to any arbitrary set $D$. Then expressing 1. and 2. like a generalized version of an ODE. I like this option best, and I am aware that it necessitate some kind of continuity in $h$ that cannot be defined for any codomain $D$. But I have a couple of hypotheses we could make us of for this track:

Here are two interesting hypothese on $h$, that may help us:

if $D$ topological, then $h$ is piecewise-continuous, and every piece interval has positive lenght.
if not, then $h$ is piecewise-constant, and every piece interval has positive length as well.

Am I looking for something impossible to express?
Can I exploit these hypotheses in a useful way to express 1. and 2.?

This is what I came up with. I'll accept my own answer once someone makes me confident it's valid.

First, property "2.: $h$ depends on its past values" is a dumb property since it is always verified. I.e. for any given $h$, one may build the function $f$ so that $h(t) = f\left(h_{\big|[0,\ t[}\right)$.

However, we may approximate $h$ by a sequence $u$ for an arbitrarily accurate precision $\varepsilon$, then state than "$f$ determines $h$ from its past" if and only if the series it determines is the approximation of $h$ for all $\varepsilon$.

For any precison $\varepsilon\in\mathbb{R}^{+*}$, let ${}^\varepsilon h$ be the sequence approximating $h$ in the sense that:

\begin{equation}
\forall n \in \mathbb{N},\quad {}^\varepsilon h_n = h(n\varepsilon)
\end{equation}

Now, let's consider $F:\left\{\begin{array}{lcl}
\mathbb{R}^{+*} &\to& \mathcal{F}\!\left(\bigcup_{i\in\mathbb{N}}D^i\to D\right)\\
\varepsilon &\mapsto& F(\varepsilon) = f_\varepsilon
\end{array}\right.$

Then, once given $h_0 = h(0) \in D$, any $F(\varepsilon)$ defines a sequence $f_\varepsilon\times h_0 = (u_n)_{n\in\mathbb{N}}$ by recursion with:

\begin{align}
u_0 &= h_0\\
\forall n \in \mathbb{N}^*,\ u_n &= f_\varepsilon(u_0,\ \dots,\ u_{n-1})\\
\end{align}

So I can finally state than

\begin{equation}
«\ F \text{ determines } h \text{ from its past } » \iff \forall \varepsilon \in \mathbb{R}^{+*},\ f_\varepsilon\times h_0 = {}^\varepsilon h
\end{equation}

This way, I do have:

1. Given $f_\varepsilon$ and $h_0$, ${}^\varepsilon h$ is entirely defined. So given "$F$ determining $h$ from its past", I can approximate $h$ with arbitrary precision.

2. Any "present" approximation ${}^\varepsilon h_n$ is constructed from its "past" ${}^\varepsilon h_n = f_\varepsilon({}^\varepsilon h_0,\ \dots,\ {}^\varepsilon h_{n-1})$.


Dihedral group for arbitrary polygon Is a dihedral group only considered for shapes that when reflected or rotated fit exactly back into place of the original image?

My confusion arises from this wikipedia article and specifically from this picture:

Are these symmetries considered a dihedral group?

Those four actions form a dihedral group of order $4$. They are not symmetries of the letter "F" though, they are symmetries of a line segment (2-gon). The rotations are applied to the letter "F" in the picture to help you see what they do. If they are symmetries of anything in the picture, they are symmetries of the $y$-axis (or the $x$-axis).

Look, Michael, dihedral groups are easy, they are the next best thing after a cyclic group. While in a cyclic group you turn things around in a circle, see your dihedral group as a prism, like two identical cycles above each other, forming a kind of coin. Then your movements consist in turning the coin as if it were cyclic, and the extra moves being the flips of the coin and consider the "F" as a symbol stamped through the coin.


Is a basin of attraction necessarily an open set? Definition:
The basin of attraction is the defined as the set of all initial conditions $x_{0}$ such that $x(t$) tends to an attracting fixed point  $x^{\ast}$ as time $t$ tends to $\infty$.

Is this basin of attraction necessarily an open set?
My text mentioned nothing about the basin of attraction being an open set-Of course this could imply that the audience is meant to think on a deeper level about the said properties of it being an open set. It is in a given example that I concluded that the author implicitly claimed that the basin of attraction is an open set.
I would like to know if it is indeed true that the basin of attraction is an open set and if it is how can it be shown on a heuristic level.
Thanks in advance.

As pointed out by Did, for non-attracting fixed points $x_0$ the set of points $x$ with $\lim_{t\to \infty} x(t)=x_0$ does not have to be open.

For an attractive fixed point, if the dynamical system is continuous then the attraction basin is indeed open (Note: For discontinuous dynamics this is clearly no true; for a counterexample, just pick any set $S$ containing a neighbourhood of $x_0$ and send it to $x_0$ and send the rest to some point $y_0$ not in $S$; the basin of attraction is now $S$).

To see this for continuous dynamics, argue as follows:

Start with $x_1$ in the attraction basin. We will find a neighborhood of it also in the attraction basin.

By definition of attractive fixed point, there exists  open $U$ around $x_0$ such that for any $x\in U$ $\lim_{t\to \infty} x(t)=x_0$. The fact that $x_1$ is attracted of $x_0$ means that $\lim_{t\to \infty} x_1(t)=x_0$. In particular, for some large $T$ we have $x_1(T)\in U$. Since the dynamics is continuous, the preimage of $U$ under $\phi_T$ (lets call it $V$) is open, and contains $x_1$.

I claim that all of $V$ is also in the basin of attraction -- which is obvious because any $x\in V$ ends up in $U$ after time $T$ and  then attracts to $x_0$; in formulas $\lim_{t\to \infty} x(t)=\lim_{(s=t-T)\to \infty} x(T)(s) =x_0$.

So there you have it: any $x_1$ in the basin has an open neighbourhood $V$ around it which is also entirely in the basin.

I'm pretty sure it's an open set and can give a heuristic proof (note: this is excluding basins of attraction for unstable steady states since they are trivial (the point itself) which is closed). In 2D if you apply the Poincure-Bendixon theorem (basically, just that there's no chaos in 2D), then every flow goes to a fixed point or a periodic orbit. But between any of these, you must have an unstable steady state / periodic orbit, and so the boundary between basins of attraction are unstable states/orbits.

In higher dimensions, this is more difficult because of the possibility of chaos. However, it seems like it's always the case that between any region where there is an attractor (either a chaotic attractor, attracting orbit, or attracting steady state) there are unstable fixed points on the boundary. I wouldn't be able to prove it (and it could be wrong in some crazy example).


creating a balanced gray code with digits of different parity I have some code that generates all combinations from something like this:

[the or a] and [angry or mad or furious] and [cat or feline]

to this:

the angry cat
the mad cat
the furious cat
the angry feline
the mad feline
the furious feline
a angry cat
a mad cat
a furious cat
a angry feline
a mad feline
a furious feline

The problem is that this cross product may become too large, in which case I would like to fairly sample as many combinations as possible in a given time limit.  This is very similar to balanced gray codes but the parity is not the same for each "digit".

The obvious algorithms (e.g. Breadth-First search and Depth-First search) will oversample some combinations and completely ignore others which probabilistic sampling like monte-carlo will evenly sample all combinations very well but it cannot efficiently enumerate all combinations.  Does anyone know of any algorithm for solving this problem?

Say you have $n_i$ options for slot $i$. Find the total number $n=\prod_in_i$ of possibilities, and find the next higher prime $p$. Find a primitive root $a$ modulo $p$, and traverse its powers $a^k$ modulo $p$. Skip the very few cases where the residue is greater than $n$. In all other cases, subtract $1$, successively divide by the $n_i$ and use the remainders as indices for the options.

In your example $n=12$, $p=13$, and e.g. $6$ is a primitive root modulo $13$. Its powers are $6$, $10$, $8$, $9$, $2$, $12$, $7$, $3$, $5$, $4$, $11$, $1$. If we subtract $1$, then divide by $2$ and use the remainder to index $\{\text{the},\text{a}\}$, then divide by $3$ and use the remainder to index $\{\text{angry},\text{mad},\text{furious}\}$, and then use what's left to index $\{\text{cat},\text{feline}\}$, this yields the following enumeration:

a furious cat
a mad feline
a angry feline
the mad feline
a angry cat
a furious feline
the angry feline
the mad cat
the furious cat
a mad cat
the furious feline
the angry cat

Here's a very similar approach to the answer by @joriki, which is in may ways simpler to implement (it requires neither primes nor the discovery of primitive roots), and which has been analyzed extensively:

As above, find $n=\prod_in_i$ and then find $m=2^{\lceil{log_2 n}\rceil}$. Finally, generate a random $a$ and $c$ such that $0<a,c<m, a\equiv1\pmod 4, c\equiv1\pmod2$. Then use a Linear Congruential Generator (LCG) starting with some randomly-selected $x_0$, producing at each step $x_{i+1} = a x_i + c\pmod m$. Skip over any values of $x_i$ which are greater than or equal to $m$ (which will be fewer than half the generated values) and converted the remaining ones to option sequences using a mixed-base representation.

The LCG produced in this way is guaranteed to cycle through all $m$ values, and for particular values of $a$ and $c$ could be used as a cheap pseudo-random number generator. It's performance as such has been analysed extensively. Not all values of $a$ are equally good; in general, you should use values somewhere in the vicinity of $\sqrt m$. One weakness of an LCG is that the low-order bits of successive entries are highly correlated. However, that will not bias the composition of a sample.


Does re-ordering a shuffled pile while drawing from it affect the randomness of the draw? In the game "Settlers of Catan", there is a stack of fourteen tiles, four of which are "water", two "gold", and the other eight an arbitrary mixture of "sheep", "wheat", "ore", "brick", and "lumber".

I shuffled these tiles, and then placed the tiles to create one to four islands.

I thought about this as a loop:
 1. I pulled the top tile and placed it in the first spot.
 2. Then I took the new top tile and placed it on the bottom of the pile.
 3. Repeat until all tiles placed.

The other players argued that my moving the top tile after every draw affected the randomness of the placement -- of the water tiles specifically -- and created a unrandom setup. They thought the usual procedure of placing the top one into the first spot, then the next tile into the second slot, and so on, would be better.

It seemed to me that my procedure was completely random and would not affect the location of the four water tiles in any way, as I had no knowledge of their location in the stack.

How do I prove to my friends that I didn't affect the outcome of the tiles other than to be completely random?

Assuming that (a) you're following your procedure consistently, and (b) the deck was properly shuffled to begin with, it doesn't make a difference.

What your move-to-back procedure achieves is nothing more or less than what you could achieve by drawing the tiles in the ordinary order but filling up the spots on the game board(?) in a different, but fixed order. You could figure out exactly what the different order would be by starting with a stack of cards numbered 1 to 14 (in that order!) and deal them onto the board using the move-to-back order. The exact sequence you'd find is not important, but it is important that the initial position of a tile in the deck completely determines where on the board it will end up -- no matter which of the ways to deal you use.

Now, the assumption that the deck is properly shuffled to begin means that each of the $14!$ ways the deck could have been arranged is equally likely (for simplicity we're assuming that there's some way to tell "identical" tiles apart, even though this difference may have no gameplay effect). This again means that each of the $14!$ ways the tiles can end up on the board are equally likely.

However, the move-to-back dealing achieves exactly the same thing! Each of the $14!$ possible ways the tiles can end up on the board corresponds to one and only one way the deck must have been arranged before the move-to-back dealing started. And since each of those were assumed to be equally likely, each resulting distribution is also equally likely.

You would need a more involved argument than this if you're accused on choosing on the fly how to continue your deal based on which tiles you have already placed. A more careful analysis would still vindicate you (because each tile you see will reduce the number of possible configurations of the rest of the deck in such a symmetric way that you cannot get any information-theoretic advantage from it), but it would be rather less accessible.

And if the tiles were imperfectly shuffled to begin with, your procedure probably cannot make it worse -- unless the imperfection of the initial shuffling was of a very specific sort, fine-tuned to be made worse by your procedure.

Actually, they're right.

Suppose you have a list:

['a', 'b', 'a', 'b']

And you pull your swapping trick, you'd wind up with:

place 'a',
move 'b' to bottom,
place 'a',
move 'b' to bottom

Which is equivalent to:

['a', 'a', 'b', 'b']

Now, that's not to say that you're cheating. Just drastically affecting the order of the list. This process doesn't help you, just has the potential to completely undo all of your shuffling.


Is this linear functional bounded? Find the norm. $$\ell^2\ni (x_n)\rightarrow2x_{1}+28x_2+35 x_{3}$$

I think it can be bounded:

$$|2x_{1}+28x_2+35 x_{3}| \le |2x_{1}|+|28x_2|+|35 x_{3}| \le 65 (\sum_{n=0}^{\infty}|x_n|^2)^{1/2}$$

But I can't find norm.

What is the norm of the linear functional $L(x)$? It is the smallest constant $M$ such that
$$|L(x)| \leq M ||x||_2$$
holds for every $x \in \ell^2$.

Okay, so now note that your linear functional is of the form $\langle a, x \rangle$, so apply the Cauchy inequality to get

$$|L(x)| = |\langle a, x \rangle| \leq ||a||_2 ||x||_2.$$

This means $M \leq ||a||_2$. Now use the fact that $a_1 x_1 + a_2 x_2 + a_3 x_3 = ||a||_2 ||x||_2 \cos \theta$, provided $x_n = 0$ for $n>3$, to show that if $M < ||a||_2$, you could pick an $\theta$ and hence an $x$, such that $|\langle a, x \rangle| > M ||a||_2 ||x||_2$.

Because the dual space of $l^2$ is $l^2$.


Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that

  $$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$

This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.

Any way, I tried hard on this one but no success yet. here is what I did:

Using the identity

$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$

$$H_{n/2}=H_n-2n\int_0^1 x^{2n-1}\ln(1+x^2)\ dx$$

We can write

$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$

where

\begin{align}
\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\
&=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\
&=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\
&=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)
\end{align}

Therefore

$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$

The sum can be done using the following identity

$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$

Differentiate both sides with respect to $a$ then set $a=1/2$ we get

$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$

and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks

Cornel's way to make it easy. Replace the harmonic number in the numerator by Digamma function, using that $H_{n/2}= \psi(n/2+1)+\gamma$, and then splitting the series using the parity, we have

$$ S=\sum_{n=1}^{\infty} \frac{ \psi(n/2+1)+\gamma}{(2n+1)^3}=\sum_{n=1}^{\infty} \frac{ \psi(n+1)+\gamma}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ \psi(n+1/2)+\gamma}{(4n-1)^3}$$
$$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ 2H_{2n}-H_n-2\log(2)}{(4n-1)^3}$$
$$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}-\sum_{n=1}^{\infty} \frac{H_n}{(4n-1)^3}-2\log(2)\sum_{n=1}^{\infty} \frac{1}{(4n-1)^3}+2\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3},$$
and since the first two series are straightforward using Cornel's Master Theorem of Series from A master theorem of series and an evaluation of a cubic harmonic series, which is also given in the book, (Almost) Impossible Integrals, Sums, and Series, and then noting that

$$\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3}=\frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{H_{n}}{(2n-1)^3}-\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n-1)^3}\right),$$

where for the first series we can use the same mentioned master theorem, and then the second one is already known in the form  $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n+1)^3}$ (it's easy to rearrange the series according to our needs), and you may find its value here together with a solution in comments, we're done.

End of story.


Subadditive Sequence Convergence Given:  A sequence ($a_n$) is called subadditive if $a_{m+n}$ ≤ $a_m$ + $a_n$ for all m, n ∈ N.
Prove that if ($a_n$) is a subadditive sequence of positive real numbers, then $(\frac{a_n}{n})$ converges.

I am unaware of how to start this proof, but I know that it I need to show that $\frac{a_n}{n}$ --> inf{$\frac{a_k}{k}$: k $\geq$ 1}.

Any help would be appreciated.

Let $\epsilon >0$. Choose k such that $\alpha +\epsilon > \frac {a_k} k$ where $\alpha$ is the infimum of $\frac {a_n} n$. This is possible by definition of infimum. Now consider $\{1,k,2k,...\}$. Any positive integer $m$ lies between $nk$ and $(n+1)k$ for some $n$. Now $a_m \leq a_{nk} + a_{m-nk}$. Note that $a_{m-nk} \leq max\{a_j:1\leq j \leq k\}$. Call this maximum as M. Then $\frac {a_m} m \leq \frac {a_{nk}} m +\frac M m$. The last term approaches 0 as $m$ approaches $\infty$ and $ a_{nk} \leq {n a_k} $. Hence $a_{nk}<nk(\alpha +\epsilon)$. Divide by $m$ and note that $\frac {nk} m \leq 1$. Since $\epsilon$ is arbitrary we see that $limsup \frac {a_m} m$ does not exceed $\alpha$. Trivially, $\alpha$ does not exceed $\lim inf \frac {a_m} m$.

From $a_{n+m}\leq a_n+ a_m$, we have $(n+m)b_{n+m} \leq nb_n + mb_m$ for $b_n:=a_n/n$. Now, we show that the sequence $\{b_n\}_{n=1}^{\infty}$ is convergent. Indeed, by setting $m=1$, we have $(n+1) b_{n+1} \leq n b_n + b_1$. This is equivalent to $(n+1) (b_{n+1}-b_1) \leq n(b_n-b_1)$. It follows that $\{n(b_n-b_1)\}_{n=1}^{\infty}$ is a non-increasing function, so $\lim_{n\to \infty} n(b_n-b_1) =\beta$ for some $\beta \in R\cup \{-\infty\}$. Now, if $\beta =-\infty$, we have $b_n<0$ for $n$ sufficiently large, which contradicts to the assumption that $a_n \geq 0$. Hence, we must have $\beta \in R$ or $|\beta|<\infty$, so $b_n \to b_1$.  In summary, we have $a_n/n \to a_1$ as $n\to \infty$.


Reducing Subspace and invariant subspace I know that If $A\in\mathbb{B}(\mathbb{H})$,$\mathbb{M}\subset\mathbb{H}$, and $P=P_{\mathbb{M}}$(Projection) , then T.F.A.E (a)~(b)

(a)$\mathbb{M}$ is invariant for A

(b)PAP=AP

Also, T.F.A.E (c)~(d)

(c) $\mathbb{M}$ reduces A

(d) PA=AP

Let $\mu$=Area measure on $D=\{z\in\mathbb{C}:\vert z\vert <1\}$ and define $A:L^2(\mu)\rightarrow L^2(\mu)$ by $(Af)(z)=zf(z)$ for $\vert z \vert<1$ and f in $L^2(\mu)$.Find a nontrivial reducing subspace for A and an invariant subspace that does not reduce A.

How to find Invariant subspace and reducing subspace?

Let $D_{\frac12}=\left\{z\in\mathbb C:|z|<\frac12\right\}$ and $$L^2\left(D_{\frac12}\right)=\left\{f\in L^2(D):f(z)=0, |z|\geqslant \frac12\right\} $$ the set of $L^2$ functions that vanish outside of $D_{\frac12}$. Then if $f\in L^2\left(D_{\frac12}\right)$, we have $zf(z)=0$ for $|z|\geqslant\frac12$ so $Af\in L^2\left(D_{\frac12}\right)$ and hence $L^2\left(D_{\frac12}\right)$ is invariant under $A$. Similarly, $$L^2\left(D_{\frac12}\right)^\perp = \left\{f\in L^2(D):f(z)=0,|z|<\frac12 \right\} $$ is invariant  under $A$, so $L^2\left(D_{\frac12}\right)$ is a reducing subspace for $A$.

For an invariant subspace that is not reducing, I am still trying to find one.

I want to thank to Math1000 for a great idea finding a nontrivial reducing subspace.
For an invariant but not reducing subspace of A, the answer would be Bergmann space $A^{2}(\mathbb{D})$, where $A^2(\mathbb{D})$:={$f \in L^2(\mathbb{D})~|~ f$ is holomorphic on $\mathbb{D}$}. The reason is followed:

Since a closed linear subspace reduces $A$ if and only if it is invariant under $A$ and its adjoint $A^{*}$, the idea of finding such "invariant but not reducing" subspace becomes to find a closed subspace invariant under $A$ but not under $A^{*}$.

For our given linear operator $A$, its adjoint $A^{*}$ is computed by
$$A^{*}(f)=\bar{z}f$$

Then, we can easily check the following claims:

$A^{2}(\mathbb{D})$ is a closed subspace of $L^2(\mathbb{D})$.
$A^{2}(\mathbb{D})$ is invariant under $A$
$A^{2}(\mathbb{D})$ is not invariant under $A^{*}$

All of above claims can be proved by some simple techniques in Complex Analysis.
(Hint: For 1, use Morera's Theorem and for 2,3 use definition of holomorphic functions)

I think these claims have no problem, but please mail me if it has an logical error.


Is this version of the Hanoi towers problem NP-complete? This was really inspired by Solitaire, but a few people reacted with ``oh, it's like the towers of Hanoi, isn't it?'' so I'll try to pose the problem in terms of discs here.

Let's start. There are n disks on the real line, one of size 1 at position $x_1$, one of size 2 at position $x_2$, ..., and one of size n at position $x_n$. Your goal is to make a tower with all n discs, consuming as little energy as possible in the process. You are allowed to move a tower whose base is a disk of size k only on top of the disk with size k+1 (which may be the top of another mini-tower). The energy you consume to perform such a move is the distance traveled by the moved mini-tower. For example, the energy consumed by the first move is $|x_k-x_{k+1}|$.

Now, you'd like to write a program that tells you whether the energy you have is enough to perform the task. It just needs to say Y or N. (If the answer is Y, then clearly the list of moves is proof enough that the answer is correct, so the problem is NP. If the answer is N, there's no point in even attempting the task---you are too tired.)

What's the fastest such program you can find? Is the problem NP-complete? If it helps, consider a simplifications that restrict $x_k$'s to be rational, integers, integers in a certain range, etc.

Here's an upper bound: $O(n2^n)$. Represent the initial state by the list $x_1$, $x_2$ ,... ,$x_n$. A move affects the state by deleting a number in this list and the energy consumed by the move is the absolute difference between the deleted number and the number coming after. Clearly there are $2^n$ states with at most $n$ moves each. (See my blog post for an example run of this algorithm if it's not clear. The description there is in terms of cards.)

Note: Some of the comments below refer to older versions of the questions. See the history of edits if they seem confusing.

Here's a polynomial time solution given to me by Javi Gomez. Let (i,j) with $i\le j$ represent the situation where disks i, i+1, ..., j are on top of each other in position $x_j$, and let E(i,j) represent the energy needed to obtain that configuration. Clearly E(i,i) is zero for all i. Also, $E(i,j)=\min_k E(i,k)+E(k,j)+|x_k-x_j|$. What we want is E(1,n).

(I made this a community wiki since the answer wasn't really found by me. Feel free to edit.)

You said that "If the answer is Y, then clearly the list of moves is proof enough that the answer is correct, so the problem is NP".
My question is, can you enumerate the list of moves in polynomial time on a nondeterministic TM ? I would say no. Actually, to enumerate all the moves, you require exponential space, regardless of the type of approach you use: deterministic or nondeterministic.
Therefore, your problem is not in NP.


Does a set include its elements or just its subsets? I am a little confused about the basic definition of inclusion.

I understand that, for example, $\{4\}\subset\{4\}$.

I also understand that $4\in\{4\}$, and that it is false to say that $\{4\}\in\{4\}$.

However, is it possible to say that  $4\subset\{4\}$?

First of all, sets can be elements of other sets too. For example if $X$ is a set then $\mathcal P(X)$ is the power set of $X$, and it is a set whose elements are all sets. But now that's out of the way, let us focus on the question whether or not $4\subseteq\{4\}$ makes sense.

It is possible if you interpret $4$ as a set. In naive set theory we often work under assumptions closer to type theory. There are real numbers, and there are vectors, and functions, and there are sets and there are other sort of type of mathematical objects.

In modern set theory we often work under the assumption that everything is a set. We construct surrogate sets to interpret other concepts such as the integers, or the real numbers, as sets.

For example, one of the mainstream ways to interpret the ordered pair $\langle x,y\rangle$ is by considering the set $\{\{x\},\{x,y\}\}$. Even though ordered pairs are "not sets", we can represent them using sets.

Similarly for integers, we can represent them as sets too. Often we choose the following encoding, $0=\varnothing$ and $n+1=n\cup\{n\}=\{0,\ldots,n\}$. In that case $4=\{0,1,2,3\}$. Clearly under this interpretation $4\nsubseteq\{4\}$. But under this interpretation, $0\subseteq\{0\}=1$.

No, because $4$ is not a set, but an element.


Center of circle given 4 points I've been struggling trying to understand how to find the center of a circle given 4 points in the circumference $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$

Please help me. I don't understand when the polygon created is not a rectangle p.ex..

The question states that all the points are on the circumference. Three non-collinear points already determine a unique circle passing through them, so any three of the four given points may be chosen and the fourth will automatically lie on the circle.

The problem of finding the centre of the circle through three points is well-known. Wikipedia itself gives the following solution $(R_x,R_y)$ for the three points $(A_x,A_y), (B_x,B_y), (C_x,C_y)$:

$$R_x = \left[(A_x^2 + A_y^2)(B_y - C_y) + (B_x^2 + B_y^2)(C_y - A_y) + (C_x^2 + C_y^2)(A_y - B_y)\right] / D$$
$$R_y = \left[(A_x^2 + A_y^2)(C_x - B_x) + (B_x^2 + B_y^2)(A_x - C_x) + (C_x^2 + C_y^2)(B_x - A_x)\right]/ D$$
$$D = 2\left[A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)\right]$$

However, there is no circle passing through four or more points in general position. For example, if four points are given as $(0,0),(2,0),(0,2)$ and $(-1,-1)$, the first three points determine a circle with centre $(1,1)$ and radius $\sqrt2$, but the fourth point does not lie on this circle. If this arises, the best you can do is minimise the sum-distance of each point to the circle itself, which becomes a least-squares fitting problem. Many resources for this are available too, like this one from Stony Brook.

Unless you delete/discard one out of the four given points and you do not have a unique circle with a center and radius.This is over-determined but consistent.


Using inverses to solve congruences The example I am looking at is for the question "what are the solutions of the linear congruence 3x ≡ 4 (mod7) ?

Let's say we know -2 is an inverse of 3mod7.

So we multiply by it: -2 * 3x ≡ -2*4 (mod 7)

*The book then says the answer is x ≡ -8 ≡ 6 (mod 7)

Why? I don't understand why 6 is positive.

$-8$ is congruent to $-1$ modulo $7$, and $7-1 = 6$.

Is this what you don't understand?

I don't know if I have the right answer, but I know what you want to ask.

We know $3x\equiv4\pmod7$ from the question. We know $-6\equiv1\pmod7$ from the inverse, and we can get $-24\equiv4\pmod7$ by multiplying 4.
Then we can have $3x=-24$, namely $x=-8$ as a single constant solution.

However, -8 is not the only answer to solve this congruence.

we can take -8 as $x_0$ and say $3(x_0+7k)\equiv4\pmod7$ where k is any integer.
We want to know, -8 is congruent to what modulo m, then we find $6\pmod7$.

I don't know why people don't understand this question. It really troubles me.


Proving that $8^n-2^n$ is a multiple of $6$ for all $n\geq 0$ by induction I have the following induction problem:

  $8^n-2^n$ is a multiple of $6$ for all integers $n\geq 0$.

So far this is what I've done:

Base case: $n = 0$

$8^0-2^0 = 6$

$1 - 1 = 6$

$0 = 6$

This means that it's a multiple of $6$.

Assume $n = k\colon 8^k - 2^k = 6m$

Where I'm getting stuck now is factoring out $8^k - 2^k = 6m$. The example I'm following on YouTube was $6^k + 4 = 5m$, which factored out nicely to $(5m-4) \cdot 6 + 4$. Where do I go from here?

For $n\geq 0$, let $S(n)$ denote the statement
$$
S(n) : 6 \mid (8^n-2^n)\Longleftrightarrow 8^n-2^n=6m, m\in\mathbb{Z}.
$$

Base case ($n=0$): $S(0)$ says that $6\mid (8^0-2^0)$, and this is true.

Inductive step: Fix some $k\geq 0$ and assume that $S(k)$ is true where
$$
S(k) : 6\mid (8^k-2^k)\Longleftrightarrow 8^k-2^k=6\ell, \ell\in\mathbb{Z}.
$$
To be shown is that $S(k+1)$ follows where
$$
S(k+1) : 6\mid (8^{k+1}-2^{k+1})\Longleftrightarrow 8^{k+1}-2^{k+1}=6\eta, \eta\in\mathbb{Z}.
$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
8^{k+1}-2^{k+1} &= 8\cdot 8^k-2\cdot 2^k\tag{by definition}\\[0.5em]
&= 8(8^k-2^k)+6\cdot 2^k\tag{rearrange}\\[0.5em]
&= 8(6\ell)+6\cdot 2^k\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&= 6(8\ell+2^k)\tag{factor out $6$}\\[0.5em]
&= 6\eta,\tag{$\eta=8\ell+2^k; \eta\in\mathbb{Z}$}
\end{align}
we end up at the right-hand side of $S(k+1)$, completing the inductive step.

Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 0$. $\blacksquare$

First of all, your base case should be $0=6\times0$, thus it is a multiple of $6$.

Now as @Brenton said, our inductive step is as follows:

Assume $6$ divides $8^k-2^k$, for some $k$ (we write this $6\mid8^k-2^k$).

We want to show that $6\mid8^{k+1}-2^{k+1}$, using this assumption.

Ok so $$6\mid8^k-2^k$$ so $$6\mid8(8^k-2^k)=8^{k+1}-8\times2^k=8^{k+1}-4\times2^{k+1}$$

Can you finish the proof from here?


Discretization of normal distribution over a finite range If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?

If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $\mu$ and $\sigma$, you need to obey the constraints
$$ \begin{align}\sum p_i&=1\\
 \sum ip_i &= \mu\\
\sum i^2p_i &= \sigma^2+\mu^2\end{align}$$
These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_i\ge 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
$$ \begin{align}\sum i^3p_i &= \mu^3+3\mu\sigma^2\\
\sum i^4p_i &= \mu^4 + 6\mu^2\sigma^2 + 3\sigma^4.\end{align}$$
So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $\ge0$ in your specific case.

If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal).  The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).

I know the thread is very old; I just couldn't resist.


positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root

but the actual answer is simply   $x=\sqrt3$

I am unable to perform the simplification any help would we helpful

Rather than pulling $\sqrt{13+4\sqrt3}=1+2\sqrt3$ magically out of a hat, write $\sqrt{13+4\sqrt3}=a+b\sqrt3,$

as suggested by Parcly Taxel in a comment.  Squaring both sides, $13+4\sqrt3=(a^2+3b^2)+2ab\sqrt3.$

$13=a^2+3b^2$ and $4=2ab$ means $b=2/a$ and $13=a^2+\dfrac{12}{a^2},$ i.e., $a^4-13a^2+12=0$, i.e., $(a^2-1)(a^2-12)=0$, i.e., $a=\pm1$ or $a=\pm2\sqrt3$ and respectively $b=\pm2$ or $\pm\dfrac 1{\sqrt3}.  $  Note that $\sqrt{13+4\sqrt3}>0.$  Can you take it from here?


What is the definition of polynilpotent Lie algebras? I am looking for definition of polynilpotent Lie algebras. Is there any equivalent concept for that?

Usually poly-P means that there is a subnormal series (or normal series; here it doesn't matter) in which all subquotients have Property P.

When P is "abelian" or "solvable", clearly poly-P is the same as solvable. Since nilpotent is between solvable and abelian, poly-nilpotent would just be the same as solvable.

So the definition has no interest. Unless one counts the number of steps. Then being $n$-step poly-nilpotent, i.e. polynilpotent with a subnormal series of length $n$ has a reasonable meaning. For instance for $n=2$ it is called meta-nilpotent.

Over a field, every solvable finite-dimensional Lie algebra is nilpotent-by-abelian, hence 2-step polynilpotent.

In infinite dimension one can construct solvable Lie algebra of arbitrary large minimal step of polynilpotency.

If $L$ is a Lie algebra then by iteration the lower central series is defined $L^1=L$; $L^{i+1}=[L; L^{i}]$; $i = 1,2, \dots$. Now $L$ is called nilpotent of class $s$ iff $L^{s+1}=0$ and $L^{s} \neq 0$.  All Lie algebras of class $s$ form a variety $N_s$. So, $L$ is called polynilpotent with tuple $(s_q, \dots, s_2, s_1)$ iff there is a chain of ideals $0=L_{q+1} \subset L_{q} \subset  \dots L_{2} \subset L_{1}=L $ with $L_{i} / L_{i+1} \in N_{s_{i}}$.


Hartshorne exercise 1.3.8 I'm trying to solve exercise 1.3.8 from Hartshorne's Algebraic Geometry:

  Let $ H_{i} $ and $ H_{j} $ be the hyperplanes in $ \mathbb{P}^n $ defined by $ x_i = 0 $ and $ x_j = 0 $ with $ i \neq j $. I Show that any regular function on $\Bbb{P}^n - (H_i \cap H_j)$ is constant.

I found a solution set online claiming that $ \mathcal{O}(\mathbb{P}^n - H_i) \cong (k[x_0,\ldots,x_n]_{x_i})_{0} $. This is used to show that any regular function is equal to $ f_i/x_i^a = f_j/x_j^b $ and then this forces the function to be constant.

I don't know where the first isomorphism comes from. I can't find such a theorem in the book up to the end of section 3 in chapter 1. The difficulty I'm having is that $ \mathbb{P}^n - H_i $ is not closed, so it is not a projective variety. The theorems up until that point in the book are only applicable to projective varieties as far as I can tell.

I feel that I'm missing something obvious here. Any help would be appreciated.

This isomorphism appears in the first paragraph of the proof of Theorem 3.4.

More generally, if $Y$ is projective, $Y - \{f=0\}$ is isomorphic to the affine variety $\operatorname{Spec} (({S(Y)}_f)_0)$.  We have $K(Y)=S(Y)_0$, and a function is defined on all of $Y-\{f = 0\}$ exactly when its denominator is a power of $f$.


Height of a spherical cap from volume and radius $\hskip 2.2 in$

I'm trying to calculate the height of a spherical cap on a sphere where you know the volume of the cap and the radius of the sphere.

I was able to find the equation for the volume of the cap with radius $r$ and height $h$ which is

$$V=\frac{\pi h^2}{3}(3r-h)$$

I feel really dumb but for the life of me I can't seem to rewrite the equation in such a way to find $h$ when given a known volume and radius.

You have to solve the cubic
$$
\frac{\pi}{3}h^3-\pi r h^2+V=0
$$

that always have a real solution for $h$, and, in general, can be solved with  Cardano's formula.

Given $V$ in order to find $h$ we have to find roots of the cubic equation.
EDIT1:
It is more convenient to express given volume as a fraction$f$ of the full sphere$ \;4\pi r^3/3$
$$\dfrac{\pi h^2(3r-h)}{3}=f \dfrac43 \pi r^3 \;\;\rightarrow 4fr^3+h^3-3r h^2=0$$
The following is given by Mathematica for $h$ value.
  Solve[-3 r h^2 + h^3 + 4 f r^3 == 0, h]

$$ \left(r+\frac{1}{2} \left(1+i \sqrt{3}\right) \sqrt[3]{2 \sqrt{f^2 r^6-f r^6}-2 f r^3+r^3}+\frac{\left(1-i \sqrt{3}\right) r^2}{2 \sqrt[3]{2 \sqrt{f^2 r^6-f r^6}-2 f r^3+r^3}}\right)
$$
in which the imaginary part cancels on evaluation.
Else if you want a graphic solution read off $h$ for given $ (V,r)$  or vice-versa as shown:


Regular polygon inside another? Inspired by this question, I was wondering if one can generalize to the case of an $n$-gon.

For example, when $n=5$ we have this picture:

where $ABCDE$ is a regular pentagon, $AA_1=BB_1=\cdots=EE_1$. Also $A,A_1,B_1$ are colinear in that order, and so on.

Can we conclude that $A_1B_1C_1D_1E_1$ is regular? Or is there a counterexample for that?

Update

Won't include the trig as rather lengthy, but included as visual complement to give genreal idea:

Note that a shutter on a traditional camera works on this same principal:

Note also that Narasimham's observation of the paths taken offer visual proof of the conjecture, and for greater $n$, are quite visually compelling:

Related

Code at Narasimham's request:

Manipulate[
Show[\[Theta] = LambertW[1]//N; length = -10; g = 0.2;
tt = {1.337, 1.236, 1.18, 1.144, 1.121, 1.104, 1.091, 1.081, 1.073,
1.066, 1.061, 1.056, 1.052, 1.049, 1.046, 1.043, 1.041, 1.039};
r = c*(rr =
  If[n < 2*Pi,
   Pi + n*(-1 + 2*Pi) +
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] -
       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] +
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]],
   Pi + n*(-1 + 2*Pi) +
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] +
       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] -
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]]]);
 aa = Table[{Cos[1 + (2*k*Pi)/n]*Cos[\[Theta]] -
  Sin[1 + (2*k*Pi)/n]*Sin[\[Theta]],
 Cos[\[Theta]]*Sin[1 + (2*k*Pi)/n] +
  Cos[1 + (2*k*Pi)/n]*Sin[\[Theta]]}, {k, Join[{n}, Range[n]]}];
 li = Table[{(2*
    Sin[r/n]*(-Sin[1 + ((2*s + 1)*Pi)/n + \[Theta]] +
      Cos[Pi/n]*Sin[(n + (2*s + 1)*Pi + r + n*\[Theta])/n]))/(3 +
    Cos[Pi/n]^2 - 4*Cos[Pi/n]*Cos[r/n] -
    Sin[Pi/n]^2), (2*(Cos[1 + ((2*s + 1)*Pi)/n + \[Theta]] -
      Cos[Pi/n]*Cos[(n + (2*s + 1)*Pi + r + n*\[Theta])/n])*
    Sin[r/n])/(3 + Cos[Pi/n]^2 - 4*Cos[Pi/n]*Cos[r/n] -
    Sin[Pi/n]^2)}, {s, 1, n}];
 list = Join[Drop[li, n - 2], Take[li, n - 2]];
 Graphics[{Opacity[If[TrueQ[JamesBond] == True, 1, 0]], Black,
 Polygon[{{-(1 + g), -(1 + g)}, {1 + g, -(1 + g)}, {1 + g,
   1 + g}, {-(1 + g), 1 + g}}], {Red, Thick,
 Opacity[If[TrueQ[JamesBond] == True, 0,
   If[TrueQ[circles] == True, 1, 0]]], Circle[]}, {Red, Thick,
 Opacity[If[TrueQ[JamesBond] == True, 0,
   If[TrueQ[circles] == True, 1, 0]]], Circle[{0, 0}, Cos[Pi/n]]},
 Opacity[If[TrueQ[JamesBond] == True, 0, 1]], Thick, Blue,
Line[aa], Opacity[If[TrueQ[JamesBond] == True, 1, 0]], White,
Polygon[list], Opacity[1],
If[TrueQ[JamesBond] == True, White, Blue], Thick,
bb = Table[Line[{aa[[u]], list[[u]]}], {u, 1, n}],
Opacity[If[TrueQ[JamesBond] == True, 0, 1]], Red,
PointSize[Large], Point[list],
If[TrueQ[JamesBond] == True, White, Red],
Opacity[If[TrueQ[JamesBond] == True, 1,
  If[TrueQ[lines] == True, 1, 0]]],
Table[Line[{bb[[u, 1,
    1]], {bb[[u, 1, 2,
      1]] + ((bb[[u, 1, 2, 1]] - bb[[u, 1, 1, 1]])*length)/
      Sqrt[Abs[(bb[[u, 1, 1, 1]] -
            bb[[u, 1, 1, 2]])*2 + (bb[[u, 1, 1, 2]] -
            bb[[u, 1, 2, 2]])*2]],
    bb[[u, 1, 2,
      2]] + ((bb[[u, 1, 2, 2]] - bb[[u, 1, 1, 2]])*length)/
      Sqrt[Abs[(bb[[u, 1, 1, 1]] -
            bb[[u, 1, 1, 2]])*2 + (bb[[u, 1, 1, 2]] -
            bb[[u, 1, 2, 2]])*2]]}}], {u, 1, n}]}],
 PlotRange -> {{-(1 + g), 1 + g}, {-(1 + g), 1 + g}}], {{c,
 If[TrueQ[min] == True, 1.18, 1]},
If[TrueQ[min] == True, Evaluate[tt[[n - 2]]], 1],
Dynamic[
1*n*(Pi/If[n < 2*Pi,
   Pi + n*(-1 + 2*Pi) +
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] -

       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] +
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]],
   Pi + n*(-1 + 2*Pi) +
    n*ArcTan[
      Cos[1] + Cos[1 - (4*Pi)/n] + 2*Cos[1 - (2*Pi)/n] +
       2*Sqrt[2]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*Sin[1 - (2*Pi)/n]^2],
       4*Cos[Pi/n]^2*Sin[1 - (2*Pi)/n] -
       2*Sqrt[2]*Cot[1 - (2*Pi)/n]*
        Sqrt[(3 + Cos[(2*Pi)/n])*Sin[Pi/n]^2*
          Sin[1 - (2*Pi)/n]^2]]])]}, {{n, 5}, 3, 20,
 1}, {{circles, False}, {True, False}}, {{lines, False}, {True,
 False}}, {{JamesBond, False}, {True, False}}, {{min, False}, {True,  False}}]

paths calculated separately with

n = 10;
Show[\[Theta] = LambertW[1]//N;
aa = Table[{Cos[1 + (2 k \[Pi])/n] Cos[\[Theta]] -
 Sin[1 + (2 k \[Pi])/n] Sin[\[Theta]],
Cos[\[Theta]] Sin[1 + (2 k \[Pi])/n] +
 Cos[1 + (2 k \[Pi])/n] Sin[\[Theta]]}, {k, Join[{n}, Range[n]]}];

Graphics[{Thick, Blue, Line[aa]}], Show[\[Theta] = LambertW[1];

 Graphics[Thick, Blue, Line[aa]],
  r = c*(Pi + 20*(-1 + 2*Pi) +
  20*ArcTan[(4*Cos[Pi/20]^2*Sin[1 - Pi/10] -
       2*Sqrt[2*(3 + Sqrt[5/8 + Sqrt[5]/8])]*Cos[1 - Pi/10]*
        Sin[Pi/20])/(Cos[1] + Cos[1 - Pi/5] + 2*Cos[1 - Pi/10] +
       2*Sqrt[2*(3 + Sqrt[5/8 + Sqrt[5]/8])]*Sin[1 - Pi/10]*
        Sin[Pi/20])]);

  Table[ParametricPlot[\[Theta] = LambertW[1]//N; {(
  2 Sin[r/n] (-Sin[1 + ((2 s + 1) \[Pi])/n + \[Theta]] +
    Cos[\[Pi]/n] Sin[(n + (2 s + 1) \[Pi] + r + n \[Theta])/n]))/(
  3 + Cos[\[Pi]/n]^2 - 4 Cos[\[Pi]/n] Cos[r/n] - Sin[\[Pi]/n]^2), (
  2 (Cos[1 + ((2 s + 1) \[Pi])/n + \[Theta]] -
    Cos[\[Pi]/n] Cos[(n + (2 s + 1) \[Pi] + r + n \[Theta])/
      n]) Sin[r/n])/(
  3 + Cos[\[Pi]/n]^2 - 4 Cos[\[Pi]/n] Cos[r/n] -
  Sin[\[Pi]/n]^2)}, {c, 0, 0.5}, Axes -> False, PlotRange -> All,
  PlotStyle -> {Red, Thick}], {s, 1, n}]]]

just change n as desired.

Original answer

Idea

Since all regular polygons can be inscribed in the unit circle, a circle can then be inscribed in the polygon (dashed) and a further polygon inside that. If lines then join the vertices of the outer polygon to the corresponding ones of the inner polygon, the outer polygon can be fixed while the inner is rotated. The adjoining lines will then form a third polygon that decreases in size to $0$ when the inner polygon is rotated a half turn. Since the outer is fixed, the relation follows.

Details

A polygon inscribed in a unit circle has vertices at $$e^{i(1 + 2\pi k/n)}$$ for all $k=1$ to $n$, where $n$ is the number of sides of the polygon. This can be described in cartesian coordinates as $$\{\Re\  e^{i(1 + 2\pi k/n)},\Im\  e^{i(1 + 2\pi k/n)}\}.$$ The polygon can then be rotated about the origin so that the uppermost vertex lies on the $y$-axis by multiplying by matrix

\begin{pmatrix}
\Re \ e^{i\Omega} & -\Im \ e^{i\Omega}\\
\Im \ e^{i\Omega}& \Re \ e^{i\Omega}\\
\end{pmatrix}

where $\Omega$ is the Omega constant.

A smaller circle, with radius $\Re \ e^{i\pi/n}$, where $n$ is the number of sides of the polygon, can then be inscribed inside it.

The polygon to be inscribed and rotated within that will then have coordinates

\begin{align}
\Re \ e^{i\pi/n}
\begin{pmatrix}
\Re \ e^{i\Omega_{1}} & -\Im \ e^{i\Omega_{1}}\\
\Im \ e^{i\Omega_{1}}& \Re \ e^{i\Omega_{1}}\\
\end{pmatrix}
\cdot \{\Re\  e^{i(1 + 2\pi k/n)},\Im\  e^{i(1 + 2\pi k/n)}\}
\end{align}

where $\Omega_{1}=\dfrac{r}{n}+\Omega$, with $r$ ranging from $0$ to $n\pi.$

Mathematica code to play with:

Manipulate[ Show[\[Theta] = LambertW[1] // N;
c = Re[E^( I Pi/n)]; \[CapitalTheta] = r/n + \[Theta];
Graphics[{(*Circle[]*){Circle[{0, 0}, c]},
Line[aa = (Table[{{Re[E^( I \[Theta])], -Im[E^( I \[Theta])]},
{Im[E^( I \[Theta])], Re[E^( I \[Theta])]}}.{Re[ E^(I  (1 + 2 \[Pi] k/n))],
Im[E^(I (1 + 2 \[Pi] k/n))]}, {k, Join[{n}, Range[n]]}])],
Line[bb = (c Table[{{Re[E^( I \[CapitalTheta])], -Im[E^( I \[CapitalTheta])]},
{Im[E^( I \[CapitalTheta])],
Re[E^( I \[CapitalTheta])]}}.{ Re[E^(I ( 1 + 2 \[Pi]  k/n))],
Im[E^(I ( 1 + 2 \[Pi]  k/n))]}, {k,Join[{n}, Range[n]]}])],
Line[Table[{aa[[q]], bb[[q]]}, {q, 1, n}]] }], Axes -> False,
PlotRange -> {{-1, 1}, {-1, 1}}], {{r, 0}, 0, n Pi}, {{n, 3}, 3, 20, 1}]

I think it is true.

The construction (and hence the proof) will be easier to start with inclined angle instead of length.

Starting from $A$, the ray $AB_1$ is drawn inclining at angle $x_1$ to the adjacent side $AB$. Then the ray $BC_1$ with $x_2 = x_1$. The process is continued until $EA_1$ is formed.

Note that all x’s are equal. Therefore all y’s are also equal because the polygon is regular.

Then, $⊿ABB_1$ is congruent to $⊿BCC_1$ …… (ASA)

Therefore, $BB_1 = CC_1 = …$.

This further means $A_1B_1 = B_1C_1 = ….$, implying that the constructed polygon is also regular.


Proof that a Kernel of a Linear Mapping is a Subspace In my linear algebra classes, we often just assume that a kernel is a subspace.
However, how do we prove this?

My current idea is to apply subspace theorem (closed under addition, multiplication, and contains the zero vector). However when I try working it out, I can not seem to get it.

Does someone know how to actually prove this?

Suppose $T$ is a linear transformation $T:V \rightarrow W$
To show $Ker(T)$ is a subspace, you need to show three things:

1) Show it is closed under addition.

2) Show it is closed under scalar multiplication.

3) Show that the vector $0_v$ is in the kernel.

To show 1, suppose $x,y\in Ker(T)$. Then $T(x+y)=T(x)+T(y)=0_w+0_w=0_w$ Hence $x+y$ also exists in the kernel and so the kernel is closed on addition.

To show 2, Assume $\lambda \in F,x\in Ker(T)$ so it follows that $T(\lambda x)=\lambda T(x)=\lambda 0_w=0_w$ So again $\lambda v\in Ker(T)$ so it is closed on scalar multiplication.

Finally it is simple to see 3, $\forall v \in V,T(0_v)=T(v+(-v))=T(v)+T(-v)=T(v)-T(v)=0_w$ so indeed $0_v$ is in the kernel.

So you have shown that $Ker(T)$ is a subspace of $V$.

The kernel of F is a subspace of V.
Proof Since F(O) = 0, we see that 0 is in the kernel. Let v, w be in
the kernel. Then F(v + w) = F(v) + F(w) = 0 + 0 = 0, so that v + w is
in the kernel. If c is a number, then F(cv) = cF(v) = 0 so that cv is also
in the kernel. Hence the kernel is a subspace.


$C$ is an uncountable set. Show that $B(x,\epsilon)\cap C$ is uncountable. Suppose $X$ is a separable metric space, and $C$ is an uncountable subset of $X$. Prove that there is a point $x \in C$ such that for each $\epsilon>0$, $B(x;\epsilon)\cap C$ is uncountable.

I missed first couple of classes so I don't really know what is the general strategy to prove something is uncountable (or countable)? The only thing that comes to my mind is that, since $X$ is separable, then for any open cover, it has a countable sub cover (may not be finite though), and I was hoping I could get a contradiction out of here?

I tried contradiction, but that would means for every $x\in C$, there is a corresponding $\epsilon(x)$ such that $B(x;\epsilon(x))\cap C$ is countable. But I cannot deduce that $C$ is then countable because that intersection may be empty.

Any idea?

Even more is true, and not much harder to prove.

HINT: Let $\mathscr{B}$ be a countable base for $X$. Let $$\mathscr{B}_0=\{B\in\mathscr{B}:B\cap C\text{ is countable}\}\;.$$

Let $C_0=C\setminus\bigcup\mathscr{B}_0$. Show that $C_0$ is actually an uncountable set such that $B\cap C_0$ is uncountable for every open nbhd $B$ of every $x\in C_0$.

Initially I was proving existence of $x\in X$ since I misread the problem this way. Now I made the small changes to prove existence of such $x\in C$.

Let $\{x_n\}\subset C$ be dense in $C$. Assume no such $x\in C$ exists. Then there are $r_n>0$, largest, such that $B(x_n,r_n)\cap C$ is countable because, in particular, none of the $x_n$ is a candidate for $x$.

But $\bigcup_n B(x_n,r_n)\supset C$. This implies that $C$ is countable.

Assume $y\in C\setminus\bigcup_n B(x_n,r_n)$ (in particular $y\neq x_n$ for all $n$). Let $r>0$ be such that $B(y,r)\cap C$ is countable, which by the assumption exists. Let $x_n$ be inside $B(y,r/4)$. Then $r_n\leq d(y,x_n)<r/4$. But $B(x_n,3r/4)\subset B(y,r)$ and therefore $B(x_n,3r/4)\cap C$ is countable contradicting the choice of $r_n$ (since $3r/4>r/4>r_n$).

Therefore $\bigcup_n B(x_n,r_n)\supset C$ as claimed.

Existence of the largest $r$:

Under the assumptions that $C$ is uncountable, and that $x$ is such that there is $\epsilon>0$ such that $B(x,\epsilon)\cap C$ is countable. Let $r:=\sup\{s\in\mathbb{R}^+:\ B(x,s)\cap C\text{ is countable}\}$. Since $B(x,r)=\bigcup_{n\in\mathbb{N}}B(x,r-1/n)$ we have that $B(x,r)\cap C=\bigcup_{n\in\mathbb{N}}\left(B(x,r-1/n)\cap C\right)$ is countable.