How to compute $\int_0^1 [\ln(1/x)]^5 dx$ Why is $$\int_0^1 [\ln(\frac{1}{x})]^5  dx=120$$? More generally,  is $$\int_0^1 [\ln(1/x)]^n  dx$$ equal to n!? If so why?

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\begin{align}
\int_{0}^{1}x^{\mu}\,\dd x&={1 \over \mu + 1}\quad\imp\quad
-\int_{0}^{1}x^{\mu}\ln^{5}\pars{x}\,\dd x=-\,\totald[5]{}{\mu}\bracks{1 \over \mu + 1}
={5! \over \pars{\mu + 1}^{6}}
\end{align}

  Set $\ds{\mu = 0}$:
  $$\boxed{\vphantom{\Huge {A \over B}}\color{#00f}{\displaystyle\large%
\quad\int_{0}^{1}\ln^{5}\pars{1 \over x} = 120}\quad}
$$

Prove it inductively!
$$\int_0^1 \ln x dx= (x\ln x -x)\bigg|_0^1=-1$$
Claim:
$$\int_0^1 \ln^nxdx=(-1)^nn!$$
Assume the claim is true for $n=k$. Then for $n=k+1$ we have
$$\int_0^1\ln^{k+1}xdx=x\ln^{k+1}x\bigg|_0^1- \int_0^1 x (k+1) \ln^kx \frac{1}{x}dx$$
$$= -(k+1)\int_0^1\ln^kxdx=-(k+1)\cdot(-1)^k k!=(-1)^{k+1}(k+1)!$$
so the induction is finished.

Then write $(\ln(1/x))^n= (-\ln x)^n=(-1)^n (\ln x)^n$ and you have what you want.


What is the line described by $x + y = y + z = \frac{x+z}{2}$ I'm asked to find the line in $R^3$ representated by $x + y = y + z = \frac{x+z}{2}$

I can see that $$x = z$$ $$y = 0$$

And I know that a line in $\mathbb{R}^3$ can be represented by $$ax+by+c=0$$

How do I proceed?

There may be faster approaches based on intuition, but I find the following approach based on solving a linear system of equation to be quite systematic.

Firstly, write it in matrix-vector form:
\begin{equation}
\left(\begin{array}{rrr}
1&0&-1\\
0&1&0
\end{array}\right)
\left(\begin{array}{c}
x\\
y\\
z
\end{array}\right)
=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) \quad .
\end{equation}

Secondly, derive the extended coefficient matrix from it. Make sure to add a third line, so there is one line for each variable:
\begin{equation}
\left(\begin{array}{rrr|r}
1 & 0 & -1 & 0\\
0 & 1 &  0 & 0\\
0 & 0 &  0 & 0\\
\end{array}\right) \quad .
\end{equation}

Thirdly, solve the underdetermined system of equations by introducing a parameter $\lambda$:
\begin{equation}
\left(\begin{array}{rrr|r}
1 & 0 & -1 & 0\\
0 & 1 &  0 & 0\\
0 & 0 &  1 & \lambda\\
\end{array}\right) \rightarrow
\left(\begin{array}{rrr|r}
1 & 0 &  0 & \lambda\\
0 & 1 &  0 & 0\\
0 & 0 &  1 & \lambda\\
\end{array}\right) \quad .
\end{equation}

Et voilà, you have a parametric representation of your line:
\begin{equation}
\vec{x} = \lambda\left(\begin{array}{c}
1\\
0\\
1
\end{array}\right) \quad .
\end{equation}

From x+y=y+z
It can be deduced that x=z
And from x+y = 0.5(x+z) while x= z
x + y = 0.5(x+x) = x

Thus y = 0

thus the equation can be concluded that
X=Z
Which is the equation of the line


If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following. If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$

I have tried putting

\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}

Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}

Somehow it does not match the statement given.

You correctly computed
$$
(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)=(1+q+s)^2-(p+r)^2
$$
only that this is not what was asked for.

But you were close: $\alpha^2+1 = (\alpha-i)(\alpha+i)$ etc, and therefore
$$
 (1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2) = P(i) P(-i)
$$
which expands to
$$
(1-ip-q+ir+s)(1+ip-q-ir+s) = (1-q+s)^2+(p-r)^2 \, .
$$

Your approach ends up proving $(α^2 - 1)(β^2 - 1)(γ^2 - 1)(δ^2 - 1) = (1 - q + s)^2 - (p - r)^2$. This means we will have to do something with that minus sign and to do that we can use the identity $(x + ιy)(x - ιy) = x^2 + y^2$. Putting $P(ι)$, and $P(-ι)$ you can easily solve the problem with the same approach.


Help explaining divergence theorem example I am looking at an application of the divergence theorem, and I don't understand what's going on. Could anyone explain how to go from the first expression to the second expression (which can then be translated into an integral on $\partial{D}$):

$$\int_{D} f(x,y) \Delta f(x,y) dV = \int_{D} ((\text{div}(f \nabla f) - \nabla{f} \cdot \nabla{f})dV$$

I know that if the expression within the integral was simply $\Delta{f(x,y)}$, then you could make it $\text{div}(\nabla{f(x,y)})$, but I don't know what to do when you're multiplying the Laplacian of $f$ by $f$ itself.

A vector identity is used:

$$\nabla \cdot (f \nabla f) = f\Delta f +\nabla f \cdot \nabla f.$$

($\Delta f, \nabla f, \nabla \cdot f$ are the Laplacian, gradient, and divergence of $f$, respectively.  The Wikipedia example uses two different functions, but there's only one function in your case.)

As far as the mechanics of multiplying $f$ by its Laplacian, they're just two functions of $x$ and $y$.  Given $f$ you can calculate its Laplacian explicitly.


Function with positive Fourier coefficients decaying as $\frac{1}{n}$ Is there an example (I'm not looking for a sufficient or necessary condition but just for an example) of a bounded Rieman-integrable function $f\colon [-\pi,\pi]\rightarrow\mathcal{R}$ with Fourier coefficients
$c_n(f) = \int_{-\pi}^\pi{f(t)e^{-int}\,dt}$ that are positive and decay as $\frac{1}{n}$ to zero (so that the Fourier coefficients are not summable)?

From this discussion a sufficient condition would be to find a function which has Fourier coefficients decaying as $n^{-1/2}$ and I could then take the convolution of this function with itself to obtain what I want.

No, there is no such function.

    Suppose $f(t)\sim\sum_nc_ne^{int}$ where $c_n\ge0$ and $\sum c_n=\infty$. Then $f$ is not bounded (hence not Riemann integrable).

Not-quite proof: Let $t=0$: it follows that $f(0)=+\infty$.

Of course that's not quite a proof, since there's no reason a priori that the series should converge to $f(0)$ for $t=0$. It does make it clear that the answer must be no, and if you have any feeling for real analysis it seems clear that the condition $c_n\ge0$ means that it can't be hard to fix the argument.

Hint for an actual proof: Show that if $f$ is bounded then the Fejer means (or the Abel means) of the Fourier series must be uniformly bounded.

Edit: It appears that the hint was not sufficient. The argument is very simple. First, if $f$ is bounded then the Fejer means are uniformly bounded, as hinted: $\sigma_n=f*K_n$, so $$||\sigma_n||_\infty\le||f||_\infty||K_n||_1=||f||_\infty.$$

On the other hand, it's obvious that if $c_n\ge0$ and $\sum c_n=\infty$ then the Fejer means are not uniformly bounded:
$$\sigma_n(0)=\sum_j(1-|j|/n)^+c_j\ge\frac12\sum_{|j|<n/2}c_j.$$

Now if I additionally assume $c_k = c_{-k}$ (which is the case for my problem, as $c_k$ is the autocovariance sequence of a real-valued time series), then $f$ will be symmetric. In that case Rieman-integrability would imply that the function admits right and left limits at zero, and by symmetry the left and right limits should be the same implying that $f$ can be made continuous at 0 without changing the Fourier coefficients, leading to a contradiction as the Fejer means diverges at zero. So I think that solves my problem in such a case.


Finding extreme value of function Given $f(x,y,z) = 3 \log( x^2+y^2+z^2) -2x^2-2y^3-2z^3$, $(x,y,z) \neq (0,0,0)$, find the extremum value of $f$.

$f_x = \frac{6x}{x^2+y^2+z^2} - 4x$

$f_y = \frac{6y}{x^2+y^2+z^2} - 6y^2$

$f_z = \frac{6z}{x^2+y^2+z^2} - 6z^2$

How do I proceed ?

Hint: Equating $f_x,f_y,f_z$ to $0$ gives (remember $x^2+y^2+z^2\neq 0$)
$$
\left\{
\begin{aligned}
x(3-2(x^2+y^2+z^2)) &= 0\\
y(1-y(x^2+y^2+z^2)) &= 0\\
z(1-z(x^2+y^2+z^2)) &= 0
\end{aligned}\right.
$$
Now solve by considering cases.


Can you provide me historical examples of pure mathematics becoming "useful"? I am trying to think/know about something, but I don't know if my base premise is plausible. Here we go.

Sometimes when I'm talking with people about pure mathematics, they usually dismiss it because it has no practical utility, but I guess that according to the history of mathematics, the math that is useful today was once pure mathematics (I'm not so sure but I guess that when the calculus was invented, it hadn't a practical application).

Also, I guess that the development of pure mathematics is important because it allows us to think about non-intuitive objects before encountering some phenomena that is similar to these mathematical non-intuitive objects, with this in mind can you provide me historical examples of pure mathematics becoming "useful"?

Here are few such examples

Public-key cryptosystems based on elliptic curves, factorization, trapdoor functions, lattices and hyperelliptic curves.
Use of algebraic topology  in distributed computing and sensor networks. Topology has uses in various other branches engineering as well.
Use of differential geometry for computer graphics, computer vision algorithms, robotics and general relativity.
Error-correcting codes based on algebraic geometry. Algebraic geometry
has also applications in robotics.
Lattice theory is used in program analysis and verfication.
Group theory is used in chemistry as well as physics
Tropical geometry, a branch of algebraic geometry, has applications in mathematical biology.
Digital electronics is impossible without Boolean algebra.
Topos theory has been applied to music theory

This may be a little late but I think the most basic and pure proof and yet the most astonishing was Euclid's

Wikipedia Entry on Euclid

There are uses in Digital Media, Cryptography, Physics and Engineering.

A lot of Pure math is knowing how to apply it, most theories have a specific problem set they are known to solve because they are designed that way or were found to solve that problem set that way but when you apply theories in ways which are not typical of the solution you incite innovation and expand your horizons.


Formal Languages - Regular Expression I've been battling the following two questions for more than a day today.

Write a regular expression (comprised of {a, b}) that contain at least two b and do not contain abb.
Write a regular expression (comprised of {a, b}) that contain abba and do not contain baa.

For the first part I wrote a basic one to start off with that doesn't handle the must two-b:
b* OR a* OR (a+ AND b)* OR a*

I ran a few "truth tables" inputs over it and it seems ok. But the moment I add the requirement, things go haywire. Mostly because when pushing in those "b"s, you can suddenly use the OR to break things up.
Same thing on the second one....
Any help would be greatly appreciated!
Norah

For the first language
$$
b^2b^*(1+a) + (ba + aa^*ba)a^*b(1 + a)
$$
For the second language
$$
(aa^*b + bb^*ab)(ab + bbb^*ab)^*ba(bb^*a)^*
$$

You have b*(a+b)*a* that matches words without abb.

Split the string on its first a. On the left it will have b* and on the right it will have (a+b)*a*. The two b can be both on left, one on left and one on right or both on right side. That's three cases. In each case two things of the form x* have to turn into xx*.

bbb*(a+b)*a*
bb*(a+b)(a+b)*a*
b*(a+b)(a+b)(a+b)*a*


A criterion for a group to be abelian I noted a discussion on groups being abelian under a certain restriction on powers of elements, e.g. http://tiny.cc/chs45. Maybe this result (probably not too well-known) concludes it all.

Let $m$ and $n$ be coprime natural numbers. Assume that $G$ is a group such that $m$-th powers commute and $n$-th powers commute (that is for all $g, h$ $\in$ $G$: $g^mh^m=h^mg^m$ and $g^nh^n=h^ng^n$). Then $G$ is abelian.

I am  also confused with  the  question ! Isn`t  the  term  " Abelian "  synonymous  of " "commutative" ? Meaning  that  a  group  G  is  " Abelian " if $ab=ba$ for  all $a,b \in G$


Solving AXB=C equation I would like some help with finding $X$ for the equation $AX B=C$. I understand how to get $X$ by itself ($X=A^{-1}AXBB^{-1}=A^{-1}CB^{-1}$) and I have found the inverse for $B$. The problem is that I can't find the inverse of $A$ which I think is inconsistent. The answer is supposed to be a parameter solution. Here are the matrices:

$$A=\begin{bmatrix}
    6 & -2 \\
    -3 & 1 \\
   \end{bmatrix}$$

$$B=\begin{bmatrix}
    -7 & -3 \\
    -2 & 1 \\
   \end{bmatrix}$$
$$C=\begin{bmatrix}
    28 & 12 \\
    -14 & -6 \\
   \end{bmatrix}$$
$$Answer=\begin{bmatrix}
    -2/3+s/3 & t/3 \\
    s & t \\
   \end{bmatrix}$$

You can check that $A$ is not invertible, because its determinant is $6\times 1 - (-2)\times(-3) = 0$.
This is why you cannot find an inverse for $A$, and also why your answer features parameters $s$ and $t$.

Notice that, as you yourself showed in the opening question, if $A$ and $B$ were both invertible then the answer would be unique $($that is, it would not depend on the value of some parameters$)$.


Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$ Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$

I can not understand how to prove it. Please help me.

HINT :

$(n-1)n(n+1)(n+2)$

is "consective numbers".

The first very obvious thing to note is that $n(n^2 - 1)(n+2)$ can be written as $n(n - 1)(n+1)(n+2)$ which means that they are four consecutive integers. Obviously one of these must be divisible by $4$.
The proof:
Any number $n$ is either of the form $2m$ or $2m+1$.

Case I:  $n = 2m$
$n(n^2 - 1)(n+2)$ = $2m(4m^2 - 1)(2m + 2) = 2m(4m^2 - 1)*2(m+1) = 4x$
Case II: $n = 2m+1$
$n(n^2 - 1)(n+2) = (2m+1)((2m+1)^2 - 1)(2m+1 + 2) = (2m+1)((4m^2 + 4m +1) - 1)(2m+3) = 4x$
Hence proved.

The link provided by @lab bhattacharjee is quite useful.


Showing Bernoulli function is constant on streamlines An incompressible inviscid ﬂuid, under the inﬂuence of gravity, has the velocity ﬁeld
$$\textbf u = (− \cos(x)\sin(y), \, \, \sin(x)\cos(y), \, \, 0)$$
with the $z$-axis vertically upwards, where $g$ is the acceleration due to gravity.

Show that the Bernoulli function $H = p/ρ + \textbf{u}^2/2 + gz$ is constant on the streamlines.

I got the pressure as $$p= - \frac{\rho}4 \bigg( \cos (2x) + \cos (2y) \bigg) - \rho g z$$ and the equations of streamlines as $$ \cos (x) \cos (y) = A $$ where $A$ is a constant.
Then $H$ becomes $$H =- \frac{1}4 \bigg( \cos (2x) + \cos (2y) \bigg) + \frac12 \bigg( \cos ^2 (x) \sin ^2 (y) + \sin^2 (x) \cos ^2 (y) \bigg)$$
but I can't get it in the form of the streamline equations.

When you have ln|cos(x)|=ln|sec(x)+c, try making everything to the power of e. You will get cos(x)+cos(y)=A. Are these your new streamlines?


Find the area and the angles formed by the diagonals of the rhombus given its height and the relationship between the height and the side length Find the area and the angles formed by the diagonals of the rhombus, if its height is two times smaller than the length of its side and is $4.2$.

My attempt

Let's write down the height:
$$h=\frac{a}{2}=4.2=\frac{21}{5}$$
From that we have
$$\frac{a}{2}=\frac{21}{5}\leftrightarrow5a=42\leftrightarrow a=\frac{42}{5}=8.4$$
Therefore, the area is
$$A=\frac{21}{5}\cdot\frac{42}{5}=\frac{882}{25}=35.28$$
I'm quite sure that I've got the area right. I need help with finding the angles. So... can someone help me find them?

hint: Area= Ht*width/2 ; diagonals cut orthogonally!


Relationship between XOR and "AND" I want to XOR the password (0x3d), byte by byte, with 0x42, then 0x51, then 0xF7, then 0x6F. this would give me 0xb6..... But, Is there a shortcut to this operation?

XOR is associative. Therefore, denoting XOR by ^, you can do

>>> 0x3D ^ 0x42 ^ 0x51 ^ 0xF7 ^ 0x6F
182

but also

>>> 0x3D ^ (0x42 ^ 0x51 ^ 0xF7 ^ 0x6F)
182

which is

>>> 0x3D ^ 0x8B
182

So you can XOR with 0x8B instead. In other words, "multiple xor" is equivalent to a single one.

Associativity follows because XOR is associative on each bit.

(p XOR q) = (¬p AND q) OR (p AND ¬q), if that was what you were asking.


What is the result of $5-0\times3+9/3=$? Possible Duplicate:
  Answer of $5 - 0 \times 3 + 9 / 3 =$

What will be the answer of

$$ 5-0\times3+9/3 = ?$$

if we follow BODMAS? Please explain each step.

$$5-0\times 3+9/3=5-0+9/3=5-0+3=5+3=8$$

Assuming, of course, the above is what you meant to write.

The original expression becomes 5-0+3, which is 8.
By the way, this shouldn't have the tag (linear-algebra). Put it in arithmetic.


What is the order of $U$? Is $\{1, 7\}$ an orbit from $U$? Let $G: = S_7$ and $U: = \{(17), (1273)\}$. What is the order of $U$? Is $\{1, 7\}$ an Orbit from $U$?
Attempt:
I know/see $(17)$ has the order $2$ and $(1273)$ the order $4$, but I don't know how to show the order for the subgroup.
Orbit: $1^U = \{1,7\}$ (with $1^{id}= 1, 1^{(1273)}= 2, 1^{(1372)} = 3, 1^{(17)}= 7$), so no orbit?


Integration with Half space Gaussian I have a problem to solve and I have something that i don't know how to do.

The half-space Gaussian integral is given :

$$\int_{0}^\infty \exp(-ax^2)dx = \frac{1}2 \sqrt{\frac{\pi}{a}}$$

I have to calculate

$$\int_{0}^\infty \exp \left(-ax^2 - \frac{b}{x^2} \right)dx$$

a and b are real and positive

Is someone have an idea ? :)

Thanks for your answers

Mathieu

Note that in THIS ANSWER, I presented a solution to a more general version of the integral of interest herein.

Let $I(a,b)$ be the integral given by

for $a>0$ and $b>0$.

$$I(a,b)=\int_0^\infty e^{-\left(ax^2+\frac{b}{x^2}\right)}\,dx \tag 1$$

Enforcing the substitution $x\to \sqrt[4]{b/a}x$ into $(1)$ reveals

$$\begin{align}
I(a,b)&=\sqrt[4]{\frac{b}{a}}\int_0^\infty e^{-\sqrt{ab}\left(x^2+\frac{1}{x^2}\right)}\,dx \tag 2
\end{align}$$

Next, noting that $x^2+\frac{1}{x^2}=\left(x-\frac1x\right)^2+2$, we can write $(2)$ as

$$\begin{align}
I(a,b)&=\sqrt[4]{\frac{b}{a}}e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,dx\tag 3
\end{align}$$

Enforcing the substitution $x\to 1/x$ in $(3)$ yields

$$I(a,b)=\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,\frac{1}{x^2}\,dx\tag4$$

Adding $(3)$ and $(4)$, we obtain

$$\begin{align}
I(a,b)&=\frac12\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,d\left(x-\frac{1}{x}\right)\\\\
&=\frac12\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab}\,x^2}\,dx\\\\
&=\frac1{2\sqrt a} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-x^2}\,dx\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\sqrt{\pi}}{2\sqrt a}e^{-2\sqrt{ab}}}
\end{align}$$

$$\exp \left(-ax^2 - \frac{b}{x^2} \right)$$
$$ = \exp \left( -ax^2\right)*\exp \left( -\frac{b}{x^2} \right)$$
$$=\exp \left( -ax^2\right)*\exp \left( -\frac{b*x^2}{x^4} \right)$$
$$=\exp \left( -ax^2\right)*\exp \left( -bx^2\right)*\exp \left( -\frac{1}{x^4} \right)$$

Now something with partial integration?


Isomorphism among semi-direct product Assume that K is a cylic group, H is an arbitrary group and $\phi_{1}$ and $\phi_2$ are homomorphisms from K into Aut(H) such that $\phi_1(K)$ and $\phi_2(K)$ are conjugate subgroups of Aut(H). If K is infinite assume $\phi_1$ and $\phi_2$ are injective. Prove by constructing an explicit  isomorphism that $H \rtimes_{\phi_1} K \cong H \rtimes_{\phi_2} K$.

I am having problems in the infinite case and for the finite case I have no idea how to do it.

Here is what I did suppose $\sigma$ does this conjugation i.e $\sigma \phi_1(K) \sigma^{-1} = \phi_2(K)$, so we get since K is cyclic we get $\sigma\phi_1(k)\sigma^{-1} = \phi_2(k)^a \ \forall\  k \in K$. I showed that the map defined as $\psi : H \rtimes_{\phi_1} K \rightarrow H \rtimes_{\phi_2} K$, such that $(h,k) \mapsto (\sigma(h),k^a)$ is a homomorphism now I need to construct the 2-sided inverse. I have no idea how to do it in finite or infinite case.

Well, only one such $a$ can describe it all!

Take the generator element $k$ of the cyclic group. Then $\sigma\phi_1(k)\sigma^{-1}=\phi_2(k)^a$, as you said, for some $a\in\Bbb Z$. But then, for any element $k^n$ of $K$, we have
$$\sigma\phi_1(k^n)\sigma^{-1}=\phi_2(k)^{an}\,.$$

Now, for the infinite case,  $\,a$ must be $\pm1$, else we wouldn't get the whole $\phi_2(K)$ by conjugating $\phi_1(K)$ by $\sigma$.

For the finite case, the condition implies $|\ker\phi_1|=|\ker\phi_2|$, and as $K$ is cyclic, we must have $\ker\phi_1=\ker\phi_2$, and thus we can reduce the problem to the injective homomorphisms $\phi_i': K/\ker\phi_i\to {\rm Aut}(H)$.
Say, the order of this cyclic quotient group is $m$.
Finally, the above argument shows that $a$ must be relatively prime to $m$.


what will be the answer either 30 or 31 for this time based question? A red light flashes 3 times per minute and a green light flashes 5 times in two minutes at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?

according to me solution is
Red light flashes every 20 seconds

Green light flashes every 24 seconds

Therefore, they will flash together every 120 seconds

In an hour they will flash 3600/120 = 30

Now since they started with flashing at the same time then 1 more flash will be added isn't it ?
so answer will be 30 + 1 = 31 that is at time 0 , 120 , 240 .....3600 => total = 31
so the answer should be 30 or 31 ?

I see this (31) as correct in the context of a math competition question depending on the wording. However, for realism, the flash of light would not be instantaneous, so 30 flashes occur in a one hour block.

to be more clear, the question wants to know how often they flash together in "each hour," so we want 30 because it flashes every two minutes and there are 60 minutes in an hour. Suppose we thought about an hour that started one second after both flashes occurred. In this case, there would still be 30 flashes in the hour.  However, if we wanted to know how many flashes occurred in the first hour, assuming that they start at the same time, and the flashes take 0 time, then it is 31.

Consider the following. You have 100 feet of fencing. You want to make a fence that is 100 feet long. But you need to put up a fence-post every ten feet. How many fence posts do you need. Well for every ten feet, you need to add one more fence post, but for 100 feet, you need 11 fence-posts because you need to have a post at both the beginning and the end of the fence.

If you have established the convention that the occurrence of the simultaneous flash (when the second-hand is perfectly vertical) is to be associated with the minute which FOLLOWS that event; then, when that event occurs next, you cannot (logically) change your mind to say that the event is to be associated with the minute which PRECEDES it. You must be consistent. Thirty gets my vote!


Most probable value of $x$ I was asked :

  We performed $n$ different trials in which an event $A$ can occur with probability $p$. Let $X$ be the number of occurrences of $A$. Determine the most probable value of $X$.

My try -

$P_X(x)=\binom{n}{x}p^x(1-p)^{n-x}$

By common sense, we know that the middle term $\lfloor \frac n2 \rfloor$ much be the one with largest $nC_x$. Apart from that, I am unable to proceed further to prove that a particular $x$ is most probable.

Or am I reading the question wrong? Are they asking the Expectation?

Thanks in advance.

Note that: $$\frac{P_X(x+1)}{P_X(x)}=\frac{p(n-x)}{(1-p)(x+1)}$$

Thus:

$P_X(x+1)>P_X(x)$ when $\frac{p(n-x)}{(1-p)(x+1)}>1$ or simplified:  $x<pn-\left(1-p\right)$
$P_X(x+1)=P_X(x)$ when $x=pn-\left(1-p\right)$
$P_X(x+1)<P_X(x)$ when $x>pn-\left(1-p\right)$

$P_X$ increases up to a point and then decreases. Find the last $\tilde{x}$ for which $\tilde{x}\leq pn-\left(1-p\right)$. The next value must be (weakly) larger, but none can be larger than that. $\tilde{x}+1$ is a mode. But note there are two modes when this holds with equality.

Some pictures:

Because we have a discrete number of events with two possible outcomes, this is just a binomial distribution. The mean number of successes $X$, given $n$ trials with probability of success $p$, is $X = np$. (This can be proved by summing over all possible numbers of successes from $0$ to $n$; see the proof on Wikipedia here.)


Sequence of functions converges normally to $e^z$ Consider the functions $f_n(z)=(1+\frac{z}{n})^n$ for each $n=1,2,3,...$ Use the fact that $f_n(x)$ $\rightarrow e^x $ for each $x \in \Bbb R $ to prove that $f_n(z) \rightarrow e^z$ normally in $\Bbb C$.

My work: Suppose $f_n(z)=(1+\frac{z}{n})^n$ for each $n=1,2,3,...$ Since $1+x \le e^x$ for $x \ge 0$, it follows that $|f_n(z)| \le e^{|z|}$, for all $z$. So, ${f_n}$ is a normal family. And $f_n(x)$ $\rightarrow e^x $ for each $x \ge 0$. Did I miss anything? Now, How to use Vitali's theorem here to prove $f_n(z) \rightarrow e^z$ uniformly on compact sets. Your kind help will be appreciated. Thank you so much!

I have no idea how you deduced that your family is normal. But in order to that $(f_n)_{n\in\mathbb N}$ converge normally to the exponential function, you use the fact that, for each $z\in\mathbb C$ and each natural $n$, if you expand $\left(1+\frac zn\right)^n$, then you get a polynomial $1+z+a_{n,2}z^2+a_{n,3}z^3+\cdots+a_{n,n}z^n$ with $a_{k,n}\leqslant\frac1{k!}$ and therefore\begin{align}\left|e^z-\left(1+\frac zn\right)^n\right|&=\left|\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)-\left(1+z+a_{n,2}z^2+a_{n,3}z^3+\cdots+a_{n,n}z^n\right)\right|\\&=\left|\left(\frac1{2!}-a_{2,k}\right)z^2+\left(\frac1{3!}-a_{3,k}\right)z^3+\cdots+\left(\frac1{n!}-a_{n,k}\right)z^n+\frac{z^{n+1}}{(n+1)!}+\cdots\right|\\&\leqslant e^{|z|}-\left(1+\frac{|z|}{n!}\right)^n.\end{align}So, if $K$ is compact and $M=\max_{z\in K}|z|$, you can use the fact that$$(\forall z\in K)(\forall n\in\mathbb{N}):\left|e^z-\left(1+\frac zn\right)^n\right|\leqslant e^M-\left(1+\frac Mn\right)^n.$$

Let f and g {/belongs to }R[a,b] .Let p be a partition such that P=[X0,X1],.......[Xn-1,Xn]. Suppose X0=a and Xn=b.let f
and g continuous and f≥g then g(x)-F(x)≤0 so g(x)-f(x) belongs to R[a, b] then {${/int_a^b(g(x)-f(x)}=/Int_a^X1(g(x)-f(x))+Int_X1^X2(g(x)-f(x))...........Int_Xn-1^b(g(x)) -f(x)) ≤0*(b-a) so Int_a^X1(g) +.......-(Int_a^X1(f)) ≤0 so Int _a^b(g) -Int_a^b(f) ≤0$


Show that this fractal contains every odd, positive integer Let $f(x)=2x+\frac{1}{3}$

Let $g(x)=\frac{2x-1}{3}$

$f^n$ represents composition of $f$

Let $G_0=\{1\}$

Let $F_{m}=\{f^n(x):x\in G_{m}, n\in\mathbb{N_{\geq0}}\}$

Let $G_{m+1}=\{g(x):x\in F_m\}$

  Show that for any given odd, positive integer $p$ there is some sufficiently high $m$ for which $p$ is in $F_m$

For what it's worth (and this may well not be helpful at all) but I've got some insight into why it's true but I can't translate that into a solution.

I can see that the denominator of $F_m$'s elements is a power of $3$ increasing with each successive $m$.  I can see that these functions $f$ and $g$ (and restricted to the positive integers) form a Euclidean function where the orbits of $f$ and $g$ are in some sense orthogonal away from the vicinity of $1$.

I think this fact, combined with the fact that $F_m\cap F_{m-1}$ is somehow dense in the odd integers within the range from $G_m$ to $G_{m+1}$, proves the statement.  But I have no idea by what metric $F_m\cap F_{m-1}$ are between $G_m$ and $G_{m+1}$.

I'm pretty sure it's a 2- and 3- adic number problem.

It is also a fact that this is a fractal. Each application of $F$ adds, for every element, a new set of numbers spaced exponentially, just above or below those already included.

Since $n=0$ is allowed, $\bigcup F_n$ is the set of numbers that ca be obtained by an arbitrary finite sequence of $f$ and $g$ from $1$.
Thus the claim is that for every $p\in\Bbb N$, there exists a sequence $x_0,x_1,\ldots, x_N$ with $x_0=1$, $x_N=p$ and $x_{k+1}\in\{2x_k+\frac13, \frac{2x-1}3\}$.

Let $y_k=6x_k+2$. Then $y_0=8$, $y_N=6p+2$, and $y_{k+1}\in\{2y_k,\frac23 (y_k +1) \}$.
We see that $y_k\in\Bbb Z[\frac13]$, and that $y_k\notin\Bbb Z$ implies $y_{k+1}\notin\Bbb Z$. As $y_N\in\Bbb Z$, we must have $y_k\in\Bbb Z$ for all $k$. In particular, $y_{k+1}=\frac23(y_k+1)$ is possible only if $y_k\equiv -1\pmod3$. We also see from this that all $y_k$ are even.

Let $z_k=-\frac12y_{N-k}\in\Bbb Z$. So $z_0=-3p-1$, $z_N=-4$ and $z_{k+1}\in\{\frac12z_k,\frac12(3z_k+1)\}$.
Now the procedure becomes determined because me must have $z_{k+1}\in\Bbb Z$:
If $z_k$ is odd, we must have $z_{k+1}= \frac12(3z_k+1)$, and if $z_k$ is even, we must have $z_{k+1}=\frac12z_k$.
This allows us to extend the sequence beyond $z_N=-4$ as $$\tag1\ldots\to -4\to -2\to-1\to -1\to\ldots.$$
This reminds fatally of the (as of today unsolved) Collatz $3n+1$ problem, which states that for any positive integer $z_0$ we will end at $\ldots \to 1\to 2\to 1\to\ldots$ with the above recursion. (Collatz' original formulation splits the odd case into $z_{k+1}=3z_k+1$ and a necessarily following $z_{k+2}=\frac12z_{k+1}$).
However, we are dealing with negative $z_0\equiv-1\pmod 3$ here, a slightly different problem. As far as I (and Wikipedia) know, this generalization to negative integers is also unsolved, but at least it is known that there are some additional limit cycles possible, namely apart from  the $\ldots \to -1\to\ldots$ we are looking for at least also
$$\tag2\ldots\to -5\to -7\to-10\to-5\to\ldots$$
and
$$\tag3\ldots \to -17\to -25\to -37\to -55\to -82\to-41\to\\\to -61\to -91\to -136\to -68\to-34\to-17\to\ldots.$$
So our task is to show that for every $p\in\Bbb N$, the choice $z_0=-3p-1$ leads to $(1)$ and not to $(2)$ or $(3)$ or any as of now unknown cycle or unbounded behaviour.
Unfortunately, $p=3$ leads to $z_0=-10$, leads to $(2)$, and we find many other such counterexmaples (e.g., $p=8$ leads to $(3)$).
In other words, numbers such as $3$ or $8$ are not in any $F_m$.

This answer is a work in progress...

We need to show that the orbits of $f$ and $g$ are in some sense orthogonal in the positive, odd integers greater than $3$, or that they are in some sense Euclidean under composition in the same range.

The closed form for $f^n(x)$ is given by:

$$f^n(x)=2^nx+\frac{2^n-1}{3}$$

The closed form for $g^m(x)$ is given by:

$$g^m(x)=\left(\frac{2}{3}\right)^mx-\frac{1}{3}\sum_{k=1}^m\left(\frac{2}{3}\right)^{k-1}\\=\left(\frac{2}{3}\right)^m\cdot x+\left(\frac{2}{3}\right)^m-1\\=\left(\frac{2}{3}\right)^m(x+1)-1$$

If $f^n(x)=2^nx+\frac{2^n-1}{3}$ is an action of $\mathbb{N}$

Let $F_x=\{f^n(x)\mid n\in\mathbb{N}\}$ be the orbit of the point $x$ through $\mathbb{Q_{>2}}$ under this action.

And $g^m(x)=\left(\frac{2}{3}\right)^m(x+1)-1$ is also an action of $\mathbb{N}$

Let $G_x=\{g^m(x)\mid n\in\mathbb{N}\}$ be the orbit of the point $x$ through $\mathbb{Q}$ under this action.

  To show that there is some metric space in which $F_{g^{-1}(x)}< F_x<F_{g(x)}$ for all $x\in F_x$ for all $F$ will prove the conjecture.

Update: Ok I've worked out that this metric space is a Euclidean combination of; the inequality:

$\log_2{x}-\lfloor\log_2{x}\rfloor>1-\log_2 3$

And the $2-$adic norm of $x$

I'm sorry if that's a bit vague; it needs more thought.  Truth or otherwise of the inequality determines whether the highest power of $2$ which is less than the odd factors of any successor of any $F_{g^{-1}(x)}$, will be $4$ times that of its predecessor in $F_{g(x)}$, or only $2$ times.

The $2-$adic norm tells us whether the binary string of any $F_{g^{-1}(x)}$'s odd factors shrinks from the right by $1$ digit, or more.


Why is this seemingly valid argument using modus tollens not valid? How is this an example of "begging the question"? I faced a question in the book "Discrete Mathematics" by Rosen. The question is this:

Determine whether the argument is valid:
If $n$ is a real number with $n > 3$, then $n^2 > 9$ : Suppose that $n^2 \le 9$. Then $n \le 3$.

I really think that this argument is valid by using modus tollens, but the answer mentioned at the end of the book is this:
"Fallacy of begging the question".
Can someone explain this for me?
Thanks.

It is not clear from the phrasing in your quote whether the statement "If $n$ is a real number with $n > 3$, then $n^2 > 9$" is meant to be a given premise or the statement that we're trying to prove.

If it's a premise, and we're trying to use it to prove that $n \le 3$, given the additional assumption that $n^2 \le 9$, then the argument is indeed correct: given $A \implies B$, we can deduce $\lnot A$ from $\lnot B$.

If, instead, the first statement in the quoted argument is supposed to be the theorem that we're trying to prove, then the proof is invalid.  While we could indeed deduce $n > 3 \implies n^2 > 9$ from $n^2 \le 9 \implies n \le 3$ (those two statements being logically equivalent), the quoted argument contains nothing that could be used to prove $n^2 \le 9 \implies n \le 3$ in the first place, unless we incorrectly assume the statement that we're trying to prove (or something else that has not been explicitly stated).

Presumably, the book you're reading explains how such arguments given in it are to be parsed.  In particular, if the arguments in the book are written in the format "Theorem to be proved: Proof of theorem," then this particular argument is invalid; if they're instead meant to be parsed as "Established premises: Assumptions. Result," then it's valid.

Unfortunately, while there exist several established conventions like this, none are quite universal enough to be safely assumed without knowing which convention the author follows.  When writing proofs yourself, to avoid such confusion it's usually a good idea to clearly and explicitly indicate which statements are axioms or prior theorems, which are temporary assumptions, and which are the new theorems that you're trying to prove.

The problem is: What if $n \leq 3$ were true even when $n^2 > 9$?
You didn't prove that is impossible!


Continuous function from $(0,1)$ onto $[0,1]$ While revising, I came across this question(s):

A) Is there a continuous function from $(0,1)$ onto $[0,1]$?

B) Is there a continuous one-to-one function from $(0,1)$ onto $[0,1]$?

(clarification: one-to-one is taken as a synonym for injective)

I figured the answer to A is yes, with $\frac{1}{2}\sin(4\pi x)+\frac{1}{2}$ as an example.

The answer to part B is no, but what is the reason?

Sincere thanks for any help.

There isn't such a function. If $f$ is such a function, then $f$ is monotonic, and $f^{-1}$ too. But a monotonic function defined on an interval is continuous iff its range is an interval. So we have that $f$ is an homeomorphism from $(0,1)$ to $[0,1]$, which is impossible since one is compact and the other is non compact.

If such a function exists then (0,1) and [0,1] are homeomorphic. But if you delete 0 from [0,1] you get (0,1] which is connected. Deleting the point x in (0,1) which corresponds to 0 in [0,1] leaves you with (0,x) u (x,1) which is not connected ---> contradiction.


Why is not this function periodic? The function is:

$f(x)= x-[x]+sinx$ where [x] is the greatest integer function.

The time period of $x-[x]$ is $1$ and $sin x$ is $2\pi$

And the time period of $f(x)$ is the LCM of the above given time periods(source)

So here,

$LCM(1,2\pi)=2\pi$

Therefore $f(x)$ must have $2\pi$ as it's time period.But if I plot is graph  $f(x)$ doesnot look like a periodic function.Why?

It isn't periodic because the least common multiple of $1$ and $2\pi$ doesn't exist (multiples here mean integer multiples). What can be done is to change $x$ in $\sin x$ to $\sin(2\pi x)$ (or really any integer multiple of $\pi$) or change $x-\lfloor x\rfloor$ to $\frac{x}{2\pi}-\lfloor\frac{x}{2\pi}\rfloor$.

The key is that after the change, the periods of both functions differ by a rational (hence a natural number) multiple... This guarantees the existence of a least common multiple.

$\pi$ is irrational, so that $1$ and $2\pi$ have no LCM.


Is there a name/study for "connecting" vectors? (not addition) Is there a name/study for this kind of thing?

Edit: I'm not looking for vector addition -- the origins of each arrow should be preserved

The idea is that the vectors are combined so that the end of one vector is the origin of the next vector (all go in the same direction). I'm not that familiar with linear algebra but wanted to learn more about work that's been done on this kind of stuff.

I'm specifically wondering if there's a generalized notion for this kind of thing from abstract algebra.

Yes, there is; it is called addition.

Every vector space by definition has two operations associated with it: Addition of vectors, which is just what you have described, and stretching a vector by a number (i.e. increasing the length by a factor while keeping the same direction).

If you want to learn more about this, you should get familiar with the abstract definitions of groups, fields, and vector spaces (maybe also take a look at inner products). Greetings,

Intergalakti


Conditional Probability I have a question on conditional probability, which goes as follows:

In London, half of the days have some rain. The weather forecaster is correct 2/3 of the time, i.e., the probability that it rains, given that she has predicted rain, and the probability that it does not rain, given that she has predicted that it won’t rain, are both equal to 2/3. When rain is forecast, Mr. Pickwick takes his umbrella. When rain is not forecast, he takes it with probability 1/3.

Find

(b) the probability that he brings his umbrella, given that it doesn’t rain.

I cannot seem to understand why the probability that he brings his umbrella is $\frac{5}{12}$. I seem to get $\frac{5}{9}$ Using the total probability theorem given that we are told it rains.

Please help and thanks in advance.

Let $p_{ab}$ be the probability for $a,b\in\{r,n\}$ (for "rain" and "no rain"), with $a$ the forecast and $b$ the actual outcome. Then $p_{rr}=p_{nn}=\frac13$ and $p_{rn}=p_{nr}=\frac16$. Given that it doesn't rain, we're left with $p_{nn}=\frac13$ and $p_{rn}=\frac16$, so with probability $\frac23$ no rain is forecast and with probability $\frac13$ rain is forecast. In the former case Mr. Pickwick brings his umbrella with probability $\frac13$, and in the latter case with probability $1$. Thus he brings it with probability

$$
\frac23\cdot\frac13+\frac13\cdot1=\frac59\;,
$$

as you correctly calculated.

A simpler way to derive the probabilities $\frac13$ and $\frac23$ for the prediction is to use the symmetry between the prediction and the outcome: We can consider the actual outcome to be a "prediction" of the prediction, which is "correct" precisely if the actual prediction is correct, and thus with the same probabilities.

The answer 5/12 is for another question: the probability P(It does not rains, given the he brings his umbrella). Some books edition puts answers for changed exercises. Then P(not rain|brings umbrella) = (5/9 * 1/2) / (12/18) = 5/12 !!!


Can we find this infinite root in term of elementary function? Let $f(x)=\left(x+f(x+1)\right)^\frac{1}{x}$. What is the value of $f(2)$ ?
More precisely,  how to find the value of $$\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\cdots}}}~?$$

Thank you.

I got the answer as $\sqrt{3+2\sqrt[3]6}/\sqrt[6]{6}$.I took the fact that among $1,\sqrt[2]2,\sqrt[3]3,\sqrt[4]4,…\sqrt[n]n$,…to infinity ∛3  is the largest one. And I have used the best approximation for this.


Proving $ B(x) = 2 \sqrt{x} L(1, \chi) + O(1) $ where $ B(x) = \sum_{n \le x} \frac{1}{\sqrt{n}} \sum_{d \mid n} \chi(d) $ Theorem 6.20 of Introduction to Analytic Number Theory (Tom M. Apostol) states:

Theorem 6.20 For any real-valued nonprincipal character $\chi \bmod k$, let
$$ A(n) = \sum_{d\mid n} \chi(d) \hspace{2em} \text{ and } B(x) = \sum_{n \le x} \frac{A(n)}{\sqrt{n}}. $$
Then we have:
(a) $ B(x) \to \infty$ as $x \to \infty$.
(b) $ B(x) = 2 \sqrt{x} L(1, \chi) + O(1)$ for all $x \ge 1$.
Therefore $L(1, \chi) \ne 0$.

I have understood most of the proof that follows except a single step at the end and that single step seems to involve just addition of terms but I am unable to figure it out. I am presenting that single step at the end below:
\begin{align}
B(x)
& = \sum_{n \le \sqrt{x}} \frac{\chi(n)}{\sqrt{n}} \left\{ 2\sqrt{\frac{x}{n}} + A + O\left( \sqrt{\frac{n}{x}} \right)\right\} \\
& +
\sum_{n \le \sqrt{x}} \frac{1}{\sqrt{n}} \left\{ B + O\left(\sqrt{\frac{n}{x}}\right)\right\} - 2Bx^{1/4} + O(1) \\
&= 2\sqrt{x} \sum_{n \le \sqrt{x}} \frac{\chi(n)}{n} + A\sum_{n \le \sqrt{x}}\frac{\chi(n)}{\sqrt{n}} + O\left(\frac{1}{\sqrt{x}} \sum_{n \le \sqrt{x}} | \chi(n) |\right) \\
& +
B \sum_{n \le \sqrt{x}} \frac{1}{\sqrt{n}} + O\left(\frac{1}{\sqrt{x}} \sum_{n \le \sqrt{x}} 1\right) - 2Bx^{1/4} + O(1) \\
&= 2\sqrt{x}L(1, \chi) + O(1).
\end{align}
How do we get $ 2\sqrt{x}L(1, \chi) + O(1) $ from the preceding step?
$ L(1, \chi) $ is defined as $ L(1, \chi) = \sum_{n=1}^\infty \frac{\chi(n)}{n} $, so it is a sum with infinite terms. But I don't see any sum with infinite terms in the preceding step. So how are we able to go from a step that does not have any infinite series to a step that has an infinite series?

For any $x>1$ we have $$L(1,\chi)=\sum_\limits{n\leq\sqrt{x}}\frac{\chi(n)}{n}+\sum_\limits{n>\sqrt{x}}\frac{\chi(n)}{n}.$$ Now using Theorem 6.18 we conclude that $$\sum_{n>\sqrt{x}}\frac{\chi(n)}{n}=O\Big(\frac{1}{\sqrt{x}}\Big).$$
Thus we have $$L(1,\chi)+O\Big(\frac{1}{\sqrt{x}}\Big)=\sum_\limits{n\leq \sqrt{x}}\frac{\chi(n)}{n}.$$


Convexity of the set $C = \{x \in \Bbb R^n \mid x^T Ax \leq b\}$ where $A \in S^n$, $A \succeq 0$ and $b \geq 0$ I have this question in my homework and neither I nor my colleagues can solve it.

  Show that the set $$C = \{x \in \Bbb R^n \mid x^T Ax \leq b\}$$ where $A$ is symmetric and positive semidefinite and $b \geq 0$, is convex.

I personally find it similar to halfspace and was thinking that maybe somehow proving that this is a halfspace we can prove that it is convex too. Do you have any ideas?

The function $x \mapsto x^TAx$ is convex. Indeed, the Hessian of this function is $A \succeq 0$. $C$ is just a sublevel set of this function.

Let $r := \mbox{rank} (\mathrm A)$. Since $\mathrm A$ is symmetric and positive semidefinite, there is a matrix $\mathrm Q \in \mathbb R^{r \times n}$ such that $\rm A = Q^\top Q$. Hence,
$$b - \mathrm x^\top \mathrm A \,\mathrm x = b - \mathrm x^\top \mathrm Q^\top \mathrm Q \,\mathrm x \geq 0$$
Using the Schur complement, we obtain the following linear matrix inequality (LMI) in $\mathrm x \in \mathbb R^n$
$$\begin{bmatrix} \mathrm I_r & \mathrm Q \,\mathrm x\\ \mathrm x^\top \mathrm Q^\top & b\end{bmatrix} \succeq \mathrm O_{r+1}$$
Hence, the set is a spectrahedron and, thus, convex.

convex-analysis schur-complement linear-matrix-inequality spectrahedra


A square can't be divided into 2, 3 or 5 smaller squares I have just read that a square can't be divided into 2, 3 or 5 smaller squares (not necessarily distinct). I am unable to think of a way to prove it. Can anyone help me by providing a hint?

If $n<4$, at least one of the squares must cover two vertices of the original square. These vertices must also be vertices of the smaller square and it must not protrude; we conclude that it has side length $\ge 1$ and covers the big square by itself.
For $n=5$, there must be exactly one small square that does not cover a vertex of the original square. Only few configurations of neighbourship are possible, and none of the works.

When a square gets divided in to more squares there are (divisions)*(divisions) new squares. If you divide a square two times you get 4 squares.


Prove that $\frac{1}{1+x_1}+\frac{1}{1+x_2}+\cdots+\frac{1}{1+x_n} \geq \frac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}$ If $x_1,x_2,\ldots,x_n$ are real numbers larger than $1$, prove that $$\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} \geq \dfrac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}$$

Attempt

AM-GM doesn't work here since we will get an upper bound. I don't see Cauchy-Schwarz working either. Thus, I think a substitution might work, but I am unsure of which one to use.

Consider the function $f(x)=\frac{1}{1+e^x}$ which is convex for $x>0$ .

Now use Jensen's inequality :

$$f( \ln x_1)+f( \ln x_2)+\ldots+f( \ln x_n) \geq n f \left (\frac{\ln x_1+\ln x_2+\ldots+\ln x_n}{n} \right)$$

This is exactly your inequality :

$$\frac{1}{1+x_1}+\frac{1}{1+x_2}+\ldots+\frac{1}{1+x_n} \geq \frac{n}{1+\sqrt[n]{x_1x_2\ldots x_n}}$$


Do factorials really grow faster than exponential functions? Having trouble understanding this. Is there anyway to prove it?

If you're not quite in the market for a full proof:

$$a^n=a\times a\times a\times a...\times a$$
$$n!=1\times 2\times 3\times 4...\times n$$

Now what happens as $n$ gets much bigger than $a$? In this case, when $n$ is huge, $a$ will have been near some number pretty early in the factorial sequence. The exponential sequence is still being multiplied by that (relatively tiny) number at each step, while $n!$ is being multiplied by $n$. So even if $n!$ starts out small, it'll eventually start being multiplied by gigantic numbers at each step, and quickly outgrow the exponential. If $a=10$ and $n=100$, then $a^n$ has around $100$ digits, while $n!$ has over $150$ digits. Note that near $n=100$, $n!$ is having roughly 2 digits added per step (and that rate will only increase), while $a^n$ is still only ever going to get one more with every step. No contest.

$n!> k^n$ if $n\ge ke$

try it for yourself with various combinations of $n$ and $k$

use $\log(n!)$ and $k \log(n)$ for large $n$


Does 17% have to be equal to 0.17? I have a friend who believes that 17% doesn't have to be equal to 0.17.  Even though he says that 17% is equal to 0.17 on its own, he says that 17% at any other time is not equal to 0.17, referring to the argument that $17\%x \neq 0.17$. No matter how I try to explain it to him, he won't believe me when I say that 17% is always equal to 0.17, no matter what. Does anyone have a good explanation for this?

"17 per cent" on its own is $\frac{17}{100} = 0.17$. That's what it means in English language and I'm pretty sure it's the same in most languages.

However $17\%$ of something, say $x$, will be $\frac{17}{100}x = 0.17x$ which of course isn't $0.17$ except for the special case $x = 1$ but that's not very interesting.

If this still doesn't convince you friend, you could take an example :
Say we have an object with a certain price $x$. Then $1\%$ of the price is like $1$ hundredth of the price which is :$$\frac{x}{100} = \frac{1}{100}x = 0.01x$$

$17\%$ of the price of the object is $17$ times greater than $1\%$ of the price therefore it is :$$17\times\frac{1}{100}x = \frac{17}{100}x = 0.17x$$

While 17% is mathematically .17 of whatever item you have, remember that we don't always work with real numbers. Sometimes we are limited to discrete values, like integers, and so there are times when your friend may be right.

As an example, imagine you have a formula for dividing a certain quantity of items among several people, such that one of those people gets 17%. Now say you have 100 items. Of course that person gets exactly 17, or .17 of the total. But now lets say that instead of 100 items to distribute you have 101 items. You can't break the items apart, but the formula still says this person should get 17%. So what happens? He still gets 17 of them. However, in this case, that 17% did not work out to exactly .17 of the total. Instead, a quick check of the calculator shows the result comes to .16831683168.


Why does the determinant of $(A- \lambda I) $ give its eigenvector? In question 6, why does determinant of $(A- \lambda I)$ gives its eigenvector ?

Let's take $A-1I$ as an example.

We know, from the fact that $\lambda = 1$ is an eigenvalue of $A$, that $A$ is singular, i.e. it has non-trivial nullspace. Here is a useful general fact:

  The nullspace of a matrix $M$ is orthogonal to the column space of $M^T$.

Once we have this fact, it is clear to see what they are doing: They have taken the second and third columns of $(A - I)^T$ (it's not really important which columns we pick, as long as they're linearly independent; the $\lambda = 3$ example picks columns 1 and 2) and they calculate the cross product of those two columns using a standard "determinant" mnemonic. This gives a vector which is orthogonal to both the columns. Since we know the column space of $(A-I)^T$ has dimension $2$ (it is a singular $3\times 3$ matrix, but the columns aren't multiples of one another, so the rank isn't $1$ or $0$), a vector which is orthogonal to those two must be orthogonal to all of the column space of $(A-I)^T$, and it must therefore be contained in the 1-dimensional nullspace of $A-I$.

So, why is this fact true? Well, let's take an arbitrary vector $v$ in the nullspace of $M$, and an arbitrary vector $M^Tw$ in the column space of $M^T$. Their dot product, $v^TM^Tw$, is a $1\times 1$ matrix, which is therefore symmetric. Which is to say, we can transpose it without changing the value:
$$
v^TM^Tw = (v^TM^Tw)^T = w^TMv = w^T(Mv) = w^T0 = 0
$$
so $v$ and $M^Tw$ are orthogonal, which is what we wanted to prove.

The roots of the equation $\det(A-xI)=0$ are the eigenvalues, not the eigenvectors. This is like so because for a root $\lambda$ of this equation, the linear application $v \mapsto A-\lambda I$ is not injective, i.e. its kernel is not reduced to the zero vector.

Then for a given eigenvalue $\alpha$, the non-zero vectors $v$ for which $(A-\alpha I)v=0$ are the eigenvectors associated to the eigenvalue $\alpha$.


Where's the problem with a false "proof": $\;1^0 = 1^2 \overset{?}\implies 0 = 2$ What's wrong with this:

$$\large 1^0=1^2$$

Since bases are same, therefore

$$\large 0=2$$

My thinking:

Since the function $\,f(x)=1^x\,$ is not one to one, therefore whenever $\,f(x)=f(y),\,$ $x\,$ need not be equal to $\,y$.

Question:

Is my reasoning sound?

Yes. That's fine reasoning.

Indeed $f(x) = 1^x = 1 \;\forall x \in \mathbb R$.

Certainly, as you note, $f$ fails to be injective, so it is NOT the case that $\forall x, y \in \mathbb R, \; f(x) = f(y) \implies x = y$.

The best possible explanation could be that  f(x)=1^x is not bijective hence making the inverse non existent and so x = y does not mean any thing.


Example of two dependent random variables that satisfy $E[f(X)f(Y)]=Ef(X)Ef(Y)$ for every $f$ Does anyone have an example of two dependent random variables, that satisfy this relation?

$E[f(X)f(Y)]=E[f(X)]E[f(Y)]$
for every function $f(t)$.

Thanks.

*edit: I still couldn't find an example. I think one should be of two identically distributed variables, since all the "moments" need to be independent: $Ex^iy^i=Ex^iEy^i$. That's plum hard...

Here is a counterexample. Let $V$ be the set $\lbrace 1,2,3 \rbrace$. Consider random variables $X$ and $Y$ with values in $V$, whose joint distribution is defined by the following matrix :

$$
P=\left(
\begin{matrix}
\frac{1}{10} & 0 & \frac{1}{5}  \\
\frac{1}{5} & \frac{1}{10} & 0  \\
 \frac{1}{30} & \frac{7}{30} & \frac{2}{15}
\end{matrix}
\right)=
\left(
\begin{matrix}
\frac{3}{30} & 0 & \frac{6}{30}  \\
\frac{6}{30} & \frac{3}{30} & 0  \\
 \frac{1}{30} & \frac{7}{30} & \frac{4}{30}
\end{matrix}\right)
$$

Thus, for example, $P(X=1,Y=2)=0$ while $P(X=1)P(Y=2)=(\frac{1}{10} + \frac{1}{5})(\frac{1}{10} + \frac{7}{30}) >0$. So $X$ and $Y$ are not independent.

Let $f$ be an  ARBITRARY (I emphasize this point because of a comment below) function defined on $X$ ; put $x=f(1),y=f(2),z=f(3)$. Then

$$
\begin{array}{rcl}
{\mathbf E}(f(X)) &=& \frac{3(x+y)+4z}{10} \\
{\mathbf E}(f(Y)) &=& \frac{x+y+z}{3} \\
{\mathbf E}(f(X)){\mathbf E}(f(Y)) &=& \frac{3(x+y)^2+7(x+y)z+4z^2}{30} \\
{\mathbf E}(f(X)f(Y)) &=&  \frac{3x^2+6xy+3y^2+7xz+7yz+4z^2}{30} \\
\end{array}
$$

The last two are equal, qed.

If $X$ and $Y$ are dependent then this equality can not hold for every function $f$.You may find some $f$ such that this is true but not for all $f$. This is by definition of independent random variables.


proving there is an injection from natural numbers to a set without the axiom of choice Let $\kappa$ be an infinite cardinal number. Prove, without appealing to the Axiom of Choice, that
  $$\aleph_0 \leq 2^{2^\kappa}$$

I am struggling to answer this question. If I had the axiom of choice available, I would have that there is an injection from natural numbers to k, by cardinal comparability. I'm not sure how to approach this question without axiom of choice being available.

If $S$ is an infinite set, we can define an injection $f:\mathbb N\to\mathcal P(\mathcal P(S))$ by setting
$$f(n)=\{X\subset S:X\text{ is an }n\text{-element set}\}.$$
This shows that $\aleph_0\le2^{2^\kappa},$ a result which I believe is due to Bertrand Russell. Of course it's actually a strict inequality; there is no bijection between $\mathbb N$ and $\mathcal P(\mathcal P(S)).$

By assumption, there exists a non-surjective injective map $f\colon\kappa\to\kappa$.
Define $g_0=f$ and  recursively $g_{n+1}=f\circ g_n$. Finally, define $\Bbb \omega\to 2^{2^\kappa}$, $n\mapsto g_n(\kappa)$.


Solving $a \sin 2x = \sin (x + \gamma)$ I am trying to solve the following equation:

$$a \sin 2x = \sin (x + \gamma)$$

or, equivalently:

$$2 a = \frac{\cos \gamma}{\cos x} + \frac{\sin \gamma}{\sin x}$$

where $a$ and $\gamma$ are constants.
I tried for a long time, and searched the web a lot, but can only find solutions to very simple trigonometry equations, or complex ones where the constants happen to be such that the formula can be reduced to another form. I am surprised that I cannot find a solution to such a simple equation.

Any idea how this can be solved?

Hint: after differentiating both sides, you will get
$$\begin{align}
2a\ cos(2x)=\cos(x + \gamma) \tag{1}\\
a\ sin(2x)=\sin(x + \gamma) \tag{2}\\
\end{align}$$

Squaring and adding $(1)$ and $(2)$, you'll end with
$$a^2 = \frac{3\sin(x + \gamma)^2}{4} + \frac{1}{4}.$$


Is a sum of a set of linearly independent functions and their time-reversed versions also linearly independent? Let $\{f_i (t) : [0,T] \mapsto \mathbb{R}\}_{i\in I}$ be a set of linearly independent functions, i.e.:

\begin{equation}
\sum_{i\in I} c_i \, f_i (t) = 0   \Rightarrow   c_i = 0,
\end{equation}

for all $i\in I$ and for all $t$ in the support of $f_i$.

Now let the 'time-reversed' versions of $f_i$, namely $\{f_i (T-t)\}_{i\in I}$ also
be linearly independent, i.e.

\begin{equation}
\sum_{i\in I} c_i \, f_i (T-t) = 0  \Rightarrow c_i = 0,
\end{equation}

for all ${i\in I}$ and for all $(T-t)$ in the support of $f_i$.

Question: do these linear independence assumptions (and any other assumptions that you might need to make $f_i$ 'well-behaved') imply that the set of functions:

\begin{equation}
\{ f_i (t) + f_i (T-t)\}_{i \in I}
\end{equation}

are linearly independent? I.e. does it follow that:

\begin{equation}
\sum_{i\in I} c_i \, \big(f_i (t) + f_i (T-t)\big) = 0   \Rightarrow  c_i = 0,
\end{equation}

for all $i\in I$ and for all $t$ in the support of $f_i (t) + f_i (T-t)$.

(You can make any assumptions you need on the smoothness or
differentiability etc of $f_i$ to make this work!)

Thanks for any help! :)

OK, well, clearly a set of one function is linearly independent and if it is also an odd function (about $T/2$) then $f(t)+f(T-t)=0$ everywhere but this is a very pathological case. How about for functions that are not odd about $T/2$ (and for sets, $I$, which have cardinality greater than one--- i.e. where there is a set of linearly independent functions $\{f_i(t), i \in I\}$)?


Analytic and circle of convergence We know that a function f(z) is analytic iff it can be given by the locally convergent power series. If it is analytic everywhere it must be convergent for all z. What about the analyticity of the function outside the circle of convergence and on the circle of convergence? How to evaluate the singularities from such a power series expansion of the function?

Let $f : U \to \mathbb C$ be analytic on an open $U \subsetneqq \mathbb C$. For $z_0 \in U$ let $r = \inf \{ \lvert v - z_0 \rvert \mid v \in \mathbb C \setminus U \}$. Then $U_r(z_0) = \{ z \in \mathbb C \mid \lvert z - z_0 \rvert < r \}$ is the biggest open disk centered at $z_0$ which is contained in $U$. Note that the boundary $S_r(z_0) = \{ z \in \mathbb C \mid \lvert z - z_0 \rvert = r \}$ contains at least one point not belonging to $U$.

It is well-known that there exists a power series $\sum_{n=0}^\infty a_n(z-z_0)^n$ with radius of convergence $R \ge r$ such that $f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n$ for $z \in U_r(z_0)$.

It is possible that $R > r$ (as a simple example let $f$ be the restriction of an analytic $g : \mathbb C \to \mathbb C$ to an open $U \subsetneqq \mathbb C$). In this case, what can be said about $f(z)$ and $\sum_{n=0}^\infty a_n(z-z_0)^n$ for $z \in (U_R(z_0) \cap U) \setminus U_r(z_0)$? Note that $U_R(z_0) \cap U$ is the set of points where both $f(z)$ and $\sum_{n=0}^\infty a_n(z-z_0)^n$ are defined.

It is very well possible that $f(z) \ne \sum_{n=0}^\infty a_n(z-z_0)^n$. As an example consider the principal branch of the logarithm  $\ln: \mathbb C_{sliced} = \mathbb C \setminus \{ x \in \mathbb R \mid x < 0 \} \to \mathbb C$. For $z_0 = -4 + 3i$ we have $r= 4$ and $\ln(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$ for $z \in U_4(z_0)$. However, the radius of convergence of the power series is $R = 5$, but for $z \in U_5(z_0)$ with $\text{Im}(z) < 0$ we have $\ln(z) \ne \sum_{n=0}^\infty a_n (z-z_0)^n$.

However, if $V$ is any connected component of $U \cap U_R(z_0)$ such that$V \cap U_r(z_0) \ne \emptyset$, then $f(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$ for $z \in V$. This is due to the identity theorem for analytic functions.

Your question is unclear but I have a few advices :

List the power series of $\frac{1}{z}$ at every point $z \ne 0$ and their radius of convergence and how you can find one from the other. Do the same with its primitive $\log z$.

The main theorem is the Cauchy integral formula : if $f$ is analytic on $|z| < r$ (note a priori $f$ is not given by a single power series on that disk) then for $|z|< R< r$, $f(z)  = \frac{1}{2i\pi}\int_{|s|=R} \frac{f(s)}{s-z}ds$, expanding $\frac{1}{1-z/s}$ in geometric series shows $f$ is indeed given by a single power series.

From there we obtain the second main theorem : (the analytic continuation of) a power series always has a singularity of the boundary of its disk of convergence.

The 3rd theorem is that the Cauchy integral formula holds for holomorphic functions : holomorphic functions are analytic (from which we obtain things like the removable singularity theorem).


$X_1,...,X_n$ be independent RVs and $X_i \perp \mathcal F $ for $1\ \leq \forall i \leq n$. Show that $\sigma (X_1,...,X_n) \perp \mathcal F$ Let $X_1,...,X_n$ be mutually independent RVs.
Suppose $X_i \perp \mathcal F $ for $1\ \leq \forall i \leq n$.

How can I show that:

$$\sigma (X_1,...,X_n) \perp \mathcal F$$ ?

What I have tried:

$\sigma (X_1...X_n) = \sigma(\cup_{i=1} ^n \sigma(X_i))$ holds. I think I need some Dynkin's lemma-ish argument but can't see how.

I doubt the conclusion. I remember that there is something like "pairwise independent does not imply independent".
More precisely, let $X_1$, $X_2$, $X_3$ be random variables such that for any $i\neq j$, $X_i$ and $X_j$ are independent. However, $X_1$, $X_2$, $X_3$ need not be independent. If my memory is correct, we put $\mathcal{F}=\sigma(X_3)$, then we will arrive a contradiction. You may consult any textbook or search "pairwise independent does not imply independent".

I remember that there is such a counter-example:

There is a probability space $(\Omega,\mathcal{F},P)$ and $A_{i}\in\mathcal{F}$,
for $i=1,2,3$ such that $P(A_{i}A_{j})=P(A_{i})P(A_{j})$ whenever
$i\neq j$. However, $P(A_{1}A_{2}A_{3})\neq P(A_{1})P(A_{2})P(A_{3})$.

Then we define $X_{i}=1_{A_{i}}$, for $i=1,2,$ and define $\mathcal{G}=\sigma(X_{3})$.
Obviously, $X_{1}$ and $X_{2}$ are independent and $X_{i}\perp\mathcal{G}$.
However, it is false that $\sigma(X_{1},X_{2})\perp\mathcal{G}$.
For, if it is true, then $A_{1}A_{2}\in\sigma(X_{1},X_{2})$ and $A_{3}\in\mathcal{G}$,
then $P(A_{1}A_{2}A_{3})=P(A_{1}A_{2})P(A_{3})=P(A_{1})P(A_{2})P(A_{3})$.

I'll answer for $n=2$. This should be sufficient to show that if it's not true for $n>2$.

  If $\sigma(X),\sigma(Y)$ and $\mathscr {F}$ are independent, then $f(X,Y)$ is independent from $\mathscr{F}$.

In this respect, your question is similar to asking about pairwise independence. If $\sigma(X)$ and $\mathscr {F}$ are independent and $\sigma(Y)$ and $\mathscr {F}$ are independent, then it is not sufficient to conclude $\sigma(X,Y)$ is independent from $\mathscr{F}$. Of course $\sigma(f(X,Y))$ merely $\subseteq \sigma(X,Y)$. That's gotta be some $f$ to achieve independence.

  If $\sigma(X,Y) = \sigma(\sigma(X) \cup \sigma(Y))$ and $\mathscr {F}$ are independent, then $f(X,Y)$ is independent from $\mathscr{F}$.

In this respect, your question is similar to asking if the conclusion holds simply for $\sigma(X) \cup \sigma(Y)$ and $\mathscr {F}$ being independent: $\sigma(X) \cup \sigma(Y)$ isn't even a $\sigma$-algebra!


Intuition behind $\zeta(-1)$ = $\frac{-1}{12}$ When I first watched numberphile's 1+2+3+... = $\frac{-1}{12}$ I thought the sum actually equalled $\frac{-1}{12}$ without really understanding it.

Recently I read some wolframalpha pages and watched some videos and now I understand (I think), that $\frac{-1}{12}$ is just an associative value to the sum of all natural numbers when you analytically continue the riemann-zeta function. 3Blue1Brown's video really helped. What I don't really understand is why it gives the value $\frac{-1}{12}$ specifically. The value $\frac{-1}{12}$ seems arbitrary to me and I don't see any connection to the sum of all natural numbers. Is there any intuition behind why you get $\frac{-1}{12}$ when analytically continue the zeta function at $\zeta(-1)$?

EDIT(just to make my question a little clearer):
I'll use an example here. Suppose you somehow didn't know about radians and never associated trig functions like sine to $\pi$ but you knew about maclaurin expansion. By plugging in x=$\pi$ to the series expansion of sine, you would get sine($\pi$) = 0. You might have understood the process in which you get the value 0, the maclaurin expansion, but you wouldn't really know the intuition behind this connection between $\pi$ and trig functions, namely the unit circle, which is essential in almost every branch of number theory.

Back to this question, I understand the analytic continuation of the zeta function and its continued form for $s < 0$ $$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$ and how when you plug in s = -1, things simplify down to $\frac{-1}{12}$ but I don't see any connection between the fraction and the infinite sum. I'm sure there is a beautiful connection between them, like the one between trig functions and $\pi$, but couldn't find any useful resources on the internet. Hope this clarified things.

There is a really nice answer by master Terence Tao on his blog.

The Euler-Maclaurin formula, Bernoulli numbers, The zeta function and real variable analytic continuation/

It shows that smoothed sums $\eta$ for $\sum\limits_{n\le N}n^s\,\eta(n/N)$ have a divergent part in $N^{s+1}$ and a convergent part $-\frac{B_{s+1}}{s+1}$.

The second part of the paper shows how it is related to analytics continuation in the complex plane.

The values of $\zeta$ for negative integers can be directly calculated from the Bernoulli numbers, from:

$$\zeta(-n)=(-1)^n\frac {B_{n+1}}{n+1}$$

and $B_2=\dfrac 16$.


Is the converse of a false conditional always true as in the Truth Table? Accroding to the Truth Table,

If $p$ is TRUE, and $q$ is FALSE, then $p\implies q$ is FALSE.
And the converse, $q \implies p$, is TRUE.

If the conditional statement is

"If two angles are congruent, they are not equal."
this is a FALSE statement.

Its converse is
"If two angles are not equal, they are congruent."
The converse is also FALSE.

Why does this example contradict the Truth Table??

The source of the confusion is that you are not dealing merely with implications in propositional logic, where, as you said, the truth tables ensure that $(p\implies q)\lor(q\implies p)$ is always true.  Your statements about angles are universally quantified statements, even though the English language lets you hide the quantifiers.  Your first statement really means "For every two angles $x$ and $y$, if $x$ and $y$ are congruent then $x$ and $y$ are not equal."  Similarly for the second statement.

So the logical form of these statements is $(\forall x)(\forall y)\,(P(x,y)\implies Q(x,y))$ and $(\forall x)(\forall y)\,(Q(x,y)\implies P(x,y))$.  Because of the quantifiers, it is entirely possible for both of these to be false.  All that's needed for that to happen is that some particular $x_0$ and $y_0$ satisfy $P$ but not $Q$, while a different pair $x_1$ and $y_1$ satisfy $Q$ but not $P$.

If quantifiers are not involved, for example if you have two particular angles and are making statements about just this pair, not angles in general, then a proposition and its converse will not both be false.

It doesn't. In your example let $q = $"two angles are not equal" be false so that means they are equal. Hence the statement "If two angles are not equal, they are congruent" is vacuously true.


Integer solutions to $x^x=122+231y$ How can I find the integer solutions to the following equation (without a script or trial and error)?

$$x^x=122+231y$$

The function $f_a:\Bbb N\to\Bbb Z_p$ defined as $f_a(x)=a^x$ where $a\in\Bbb Z_p^*$ is periodic and its period is a divisor of $p-1$. Then the modular equation $x^x\equiv 2 \pmod 3$ has only to be checked for $x\in\{1,\ldots,6\}$. And
$$1^1\equiv 1\pmod 3$$
$$2^2\equiv 1\pmod 3$$
$$3^3\equiv 0\pmod 3$$
$$4^4\equiv 1\pmod 3$$
$$5^5\equiv 2\pmod 3$$
$$6^6\equiv 0\pmod 3$$
Therefore, if $x$ is a solution, then $x\equiv 5\pmod 6$.

Now, write $x=5+6k$ and try to find solutions for
$$(5+6k)^{5+6k}\equiv 3\pmod 7$$
but, by LFT, this is equivalent to
$$(5+6k)^{-1}\equiv 3\pmod 7$$
or
$$5+6k\equiv 5\pmod 7$$
so $k$ is a multiple of $7$, that is, $x=5+42j$.

Now we have to deal with the last prime factor of $231$, that is, $11$.
$$(5+42j)^{5+42j}\equiv(5-2j)^{5+2j}\equiv 1\pmod {11}$$
There are some possibilities now:

$5-2j\equiv1\pmod{11}$ which gives $x=462m+89$.
$5-2j\equiv -1\pmod{11}$ and $5+2j$ is even. But this is impossible.
$5-2j\equiv 4,5,9\text{ or }3\pmod{11}$ and $5+2j$ is a multiple of $5$, which gives $x=2310m+5$, $845$, $1895$ or $2105$.
$5+2j$ is multiple of $10$, which is impossible.

To sum up, the solutions are the positive integers of the form
$$x=\left\{
\begin{array}{l}
89+462m\\
5+2310m\\
845+2310m\\
-405+2310m\\
-25+2310m
\end{array}
\right.$$
where $m$ is an integer.

Some solutions for $x$: $5$, $89$, $551$, $845$, $1013$, $1475$, ...

We are attempting to solve:

$$ x^x \equiv 122 \  \mod \ 231  $$

Thus we naturally would want to factor 231 which is

$$ 3*7*11 $$

Thus we can break this into a system of 3 modular equations

$$ x^x \equiv 2 \mod \ 3$$
$$ x^x \equiv 3 \mod \ 7$$
$$ x^x \equiv 1 \mod \ 11$$

upon exploring the equation it appears there are solutions of the form respectively for
$$ x \equiv 5 \mod \ 6$$ (appears to solve the mod 3)

$$ x \equiv 5 \mod \ 84 $$ (appears to solve the mod 7)

And:
$$ x \equiv 1 \mod \ 11 $$ (solves the mod 11 but isn't the only case)

Why these in particular? I don't know...
Its worth noting

$$ x \equiv 0 \mod 5 $$ (the other apparent case)

Appears to imply

$$ x \equiv 1 \ or \ -1 \mod 11 $$

Again: it appears now that:

$$ j*84*10^k + 5 $$ is constant for all k > 0 for fixed j

specifically if j = 0 or j = 1 then for all k > 0 this appears to be solution:

sample solutions include:

5,89, 845, 8405, 84005 etc...


Maximum Word Problem Two positive numbers who have a sum of 18, and the product of the first number and the square of the other is a maximum.

I got one number is 6, and the other is 12. Not sure if correct.

Any insight?

$x+y=18$. Let $P=xy^2$. You have to find $x$ and $y$ which make $P$ a maximum. $P=xy^2=(18-y)y^2=18y^2-y^3$. When $P$ is maximum $\frac{dP}{dy}=0$.

$\frac{dP}{dy}=36y-3y^2=y(36-3y)$. So, $\frac{dP}{dy}=0$ when $y=0$ or $y=12$. But when $y=0, P=0$ and when $y=12, P>0$. Hence $P$ takes its maximum when $y=12$. so $x=6$

$x + y = 18$ , so $y = 18 - x$
Then you can rewrite $xy^{2}$ ... and you just have to find the maximum of a fonction only depending of $x$, with $0<x<18$ an integer.

By the way you can see that your answer is wrong by calculating $17^{2}$ !


Why can't you prove the law of the excluded middle in intuitionistic logic (for layman)? I am learning about the difference between booleans and classical logics in Coq, and why logical propositions are sort of a superset of booleans:

Why are logical connectives and booleans separate in Coq?

The answer there explains the main reason for this distinction is because of the law of excluded middle, and how in intuitionistic logic, you can't prove this:

Definition excluded_middle :=
  forall P : Prop, P \/ ~ P.

I have seen this stated in many different books and articles, but seem to have missed an explanation on why you can't prove it.

In a way that a person with some programming/math experience can understand (not an expert in math), what is the reason why you can't prove this in intuitionistic logic?

If you could prove the law of the excluded middle, then it would be true in all systems satisfying intuitionistic axioms.  So we just need to find some model of intuitionistic logic for which the law of the excluded middle fails.

Let $X$ be a topological space.  A proposition will consist of an open set, where $X$ is "true", and $\emptyset$ is "false".  Union will be "or", intersection will be "and".

We define $U\implies V$ to be the interior of $U^c \cup V$, and $\neg U = (U\implies \emptyset)$ is just the interior of $U^c$.

The system I just described obeys the axioms of intuitionistic logic.  But the law of the excluded middle says that $U\cup \neg U = X$, which fails for any open set whose complement is not open.  So this is false for $X=\mathbb{R}$, for example, or even for the two-point Sierpinski space.

In fact, intuitionistic logic can generally be interpreted in this geometric kind of way, though topological spaces are not general enough and we need things called elementary topoi to model intuitionistic logic.

The law of excluded middle can be proved in the following way:

$$
\cfrac{\cfrac{\cfrac{(\lnot\phi)^{(1)}}{(\phi\lor(\lnot\phi))}(\lor I) \qquad \lnot(\phi\lor(\lnot\phi))^{(3)}}{(1)\cfrac{\bot}{\phi}(RAA)}(\lnot E) \qquad \cfrac{\cfrac{\phi^{(2)}}{(\phi\lor(\lnot\phi))}(\lor I) \qquad \lnot(\phi\lor(\lnot\phi))^{(3)}}{(2)\cfrac{\bot}{(\lnot\phi)}(\lnot I)}(\lnot E)}{(3)\cfrac{\bot}{(\phi\lor(\lnot\phi))}(RAA)}(\lnot E)
$$

which is a derivation of $\vdash(\phi\lor(\lnot\phi))$ using the natural deduction rules $\lor I$, $\lnot E$ (equivalent to $\rightarrow E$, since $(\lnot\phi)$ treated as $(\phi\rightarrow\bot)$), $\lnot I$ and $RAA$. Of these only the latter (reductio ad absurdum) is not part of intuitionistic logic and hence LEM cannot be proved in it.

Furthermore, $RAA$ is equivalent to double negation (in particular, deriving $\phi$ from $\bot$ in $(1)$ is actually a derivation of $(\lnot(\lnot\phi))$, the same holds for $(2)$), so if either was included in intuitionistic logic, the proof would hold.

References:

Chiswell and Hodges - Mathematical Logic p.28-29
Van Dalen - Logic and Structure p.156


Is there any geometric way to characterize $e$? Let me explain it better: after this question, I've been looking for a way to put famous constants in the real line in a geometrical way -- just for fun. Putting $\sqrt2$ is really easy: constructing a $45^\circ$-$90^\circ$-$45^\circ$ triangle with unitary sides will make me have an idea of what $\sqrt2$ is. Extending this to $\sqrt5$, $\sqrt{13}$, and other algebraic numbers is easy using Trigonometry; however, it turned difficult working with some transcendental constants. Constructing $\pi$ is easy using circumferences; but I couldn't figure out how I should work with $e$. Looking at

made me realize that $e$ is the point $\omega$ such that $\displaystyle\int_1^{\omega}\frac{1}{x}dx = 1$. However, I don't have any other ideas. And I keep asking myself:

Is there any way to "see" $e$ geometrically? And more: is it true that one can build any real number geometrically? Any help will be appreciated. Thanks.

For a certain definition of "geometrically," the answer is that this is an open problem. You can construct $\pi$ geometrically in terms of the circumference of the unit circle. This is a certain integral of a "nice" function over a "nice" domain; formalizing this idea leads to the notion of a period in algebraic geometry. $\pi$, as well as any algebraic number, is a period.

It is an open problem whether $e$ is a period. According to Wikipedia, the answer is expected to be no.

In general, for a reasonable definition of "geometrically" you should only be able to construct computable numbers, of which there are countably many. Since the reals are uncountable, most real numbers cannot be constructed "geometrically."

From the hyperbolic identity:

          cosh(x) + sinh(x) = e^x  ...in general,

we have

          cosh(2) + sinh(2) = e^2  ...in particular.

If we take individually, the square of the square root of each of the first two terms of the second equation, we have a Pythagorean identity, where the sides are:

F  =  √(cosh(2)),  G  =  √(sinh(2)), and the hypotenuse of the right-angled triangle,  H  =  e.

I hear you say, not as good as π, the ratio of the circumference of a circle to its diameter, since one can measure the circumference and the diameter, by some means.  One needs the formula or Taylor series for cosh(2) and sinh(2).

(My original edit here was pointless.  It involved the additional identity:

         1  +  sinh^2(x)  =  cosh^2(x)

which is also a Pythagorean identity, but since it can be derived from the previous case, and vice versa, it was pointless in calculating a value for 'e').

N.B.  'e', the base of natural logarithms, was created as a valid idea in calculus, to make differentiation, and therefore integration, easier.  It is not a geometrical constant, but a mathematical constant, and is the limit of:

                      (1 + t)^(1/t), as t tends to 0

which occurs in the differentiation of log(x).


What should I learn next for pure math? I can do basic single variable calculus which is essentially all you do at A Level in the UK. I also just read "What is mathematics?" by Richard Courant which I found very good. I would like to know where to go after I finish my proof writing and basic discrete maths books I am currently reading. Should I study real analysis? What book? Linear algebra?

I would appreciate advice on where to go next and what book would be good. I am currently 14 and my end goal would to become a pure mathematician.

The reason I ask this question is because I was trying to create a structure and I thought about the Gerard t'Hooft physics one or even the Pure mathematician or statiscian plan but they don't seem particularly specific in what you are meant to do where i.e they call a topic calculus and then a later one vector calculus and then analysis but as a pure mathematician I want to go straight into analysis, of reals obviously. In conclusion, I am asking this because I am going to go by steps now and choose the next topic as it comes.

Thank you in advance.

EDIT:

Would this work:

In order...

Elementary Discrete Maths,
Real Analysis,
Linear Algebra,
ODE's,
Probability,
Fourier Analysis,
Complex Analysis,
PDE's,
Graduate stuff which I will get to when needed.

I'm not sure if probability is necessary but I think it'll be interesting.

I think you are correct to lean toward analysis. It is the usual first step to pure math as you describe your inclination. Linear algebra can be done on several levels, but I think is best done in a rigorous context after real analysis.

In real analysis you will become familiar with proofs, begin to accumulate a working, ingrained math vocabulary as well as learn about topology, metric spaces, concepts such as convergence, continuity, etc. There is also usually a component presenting the material of calculus on a rigorous basis. This will give you a feel for the difference between operational mechanics and the pure math behind it.

I would suggest you take a look at this free set of notes of lectures given by Vaughan Jones (Fields Medal winner, equiv. Nobel Prize in math). They are really beautiful, are self-contained, and build nicely from a level that does not  require prior experience.

https://sites.google.com/site/math104sp2011/lecture-notes

Start with calculus and after that do linear algebra. For calculus I can recommend you Calculus Early Transcendentals by Stewart, and for Linear Algebra I recommend Linear Algebra: A modern Introduction by Poole. I'd start with calculus though. That'll be challenging enough at this stage.


Power set of the Universal Set In set theories with a universal set $U$, can we take the powerset of $U$? In particular, can we look at $\mathcal{P}(U)$ and compare with $U$, i.e. can we say something like $U \in \mathcal{P}(U)$, $U \notin \mathcal{P}(U)$, $|U| < |\mathcal{P}(U)|$, or $U \neq \mathcal{P}(U)$?

Yes, you can do this at least some of the time. There are certain things which have to be true - e.g. we must have $\mathcal{P}(U)\in U$ if $\mathcal{P}(U)$ is a set, and we must have $U\in\mathcal{P}(U)$ by definition of $\subseteq$ - but on other points we have some flexibility.

The standard example I think is Quine's New Foundations with urelements, NFU (note that NF alone is not known to be consistent - that said, Holmes has a claimed consistency proof). Since the formula $$\psi(x)\equiv x=x$$ is stratified, we may apply stratified comprehension to it; this gives us the universal set $U$. Similarly, the formula $$\varphi(x; z)\equiv\forall y(y\in x\implies y\in z)$$ (here $z$ is a parameter) is always stratified, so applying stratified comprehension we have that powersets of arbitrary sets exist. In particular, taking $z=U$ we get the powerset of the universal set.

Now things get interesting. If there are no urelements (that is, in NF -
 then every set is a subset of $U$, hence $\mathcal{P}(U)=U$. However, since urelements are not sets of sets, we can have $\mathcal{P}(U)\subsetneq U$ in NFU!

Forster's book has more details on this sort of thing.

Some nonwell-founded set theorizing (§4.4) has mappings $\wp(V) \rightarrow V$ and vice versa. I'm not sure if this is a theorem of the system's axioms or an axiom itself (it appears to be a conjunction or alternation of axioms; the mappings correspond to an initial algebra and a final coalgebra, but I won't pretend to understand that difference!).
In paraconsistent set theory, things are even weirder. It appears to be the case here that a cardinal for ORD is equivalent to its own powerset while sort of being smaller than itself as such. Or at any rate, the situation is so exotic that it's hard to tell how the mapping between $V$ and $\wp(V)$ works except to say that it's not the same as usual!


Convergence of the random harmonic series $\sum_{n=1}^{\infty}\frac{X_{n}}{n}$ Let $(X_{n})_{n \in \mathbb{N}}$ be independent with Rademacher distribution:

\begin{equation}
\mathbb{P}(X_{n} = -1) = \frac{1}{2} = \mathbb{P}(X_{n} = 1).
\end{equation}

I have to investigate

\begin{equation}
\sum_{n=1}^{\infty}\frac{X_{n}}{n}
\end{equation}
for convergence. It was given in a textbook and I'm very interested in the solution. It is something between the harmonic series $\sum_{n=1}^{\infty}\frac{1}{n}$ and the series $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$, but I know the sign changes randomly.

Define for $N \in \mathbb{N}$
\begin{equation}
Y_{N} = \sum_{n=1}^{N}X_{n}
\end{equation}
Now
\begin{equation}
\mathbb{E}(\frac{1}{n}X_{n}) = 0.
\end{equation}
Therefore
\begin{equation}
\mathbb{E}(Y_{N}) = 0.
\end{equation}
Also
\begin{align}
var(Y_{N}) &= var(\sum_{n=1}^{N}X_{n})\\
&= \sum_{n=1}^{N}\frac{1}{n^{2}}var(X_{n})\\
&=\sum_{n=1}^{N}\frac{1}{n^{2}}\\
&\leq\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}.
\end{align}

For $M < N$ we have
\begin{align}
\mathbb{E}[|Y_{N}-Y_{M}|^{2}] &= \mathbb{E}[|\sum_{n=M+1}^{N}\frac{1}{n}X_{n}|^{2}]\\
&=var(\sum_{n=M+1}^{N}\frac{1}{n}X_{n})\\
&=\sum_{n=M+1}^{N}\frac{1}{n^{2}} \xrightarrow{N,M \to \infty} 0,
\end{align}
i.e. $(Y_{N})_{N \in \mathbb{N}}$ is a cauchy-sequence with respect to the $\mathcal{L}^{2}$-norm. Now the Riesz–Fischer theorem says, that $\mathcal{L}^{2}$ is a banach-space. Especially every cauchy sequence in $\mathcal{L}^{2}$ converges. Now $\mathcal{L}^{2}$-convergence implies onvergence in probability and distribution. We also have almost sure convergence, since if $(X_{n})_{n \in \mathbb{N}}$ are independet random variables with $S_{n} = \sum_{i=1}^{n}X_{i} \xrightarrow{p} S$ then $S_{n} \rightarrow S$ almost surely.

Use the following theorem about sums of zero-mean independent variables in $\mathcal L^2$ from David Williams' Probability with Martingales Chapter 12.2.

Let $\{X_n\}_{n \in \mathbb N}$ be a sequence of random variables. We
have $$\sum_n X_n < \infty \ \text{a.s.}.$$ if the following 4
conditions hold:

$$\{X_n\}_{n \in \mathbb N} \text{is independent.}$$

$$E[X_n] = 0 \ \forall n \in \mathbb N$$

$$\sigma_n^2 := Var[X_n] < \infty \ \forall n \in \mathbb N$$

$$\sum_{n \in \mathbb N} \sigma_n^2 < \infty$$

In our case:

$$\{\frac{X_n}{n}\}_{n \in \mathbb N} \ \text{is independent} \ \because \{X_n\}_{n \in \mathbb N} \text{is independent.}$$

$$E[\frac{X_n}{n}]=\frac{1}{n}E[X_n] = \frac{1}{n} 0 = 0 \ \forall n \in \mathbb N$$

$$Var[\frac{X_n}{n}] = \frac{1}{n^2} Var[\frac{X_n}{1}] = \frac{1}{n^2} (1)  = \frac{1}{n^2} < \infty \ \forall n \in \mathbb N$$

$$\sum_{n \in \mathbb N} Var[\frac{X_n}{n}] = \sum_{n \in \mathbb N} \frac{1}{n^2} < \infty \ \text{by integral test}$$

$\therefore,$ by theorem in Chapter 12.2 of David Williams' Probability with Martingales, $$\sum_{n \in \mathbb N} \frac{X_n}{n} < \infty \ \text{a.s.}.$$ QED


solve nonlinear equation with logarithms We were asked to find the inflection points of:

$$f(x) = x^\frac{1}{x}$$

After a lot of derivation I reached:

$$f''(x)=x^{1/x} \left( \frac{\ln^2(x)+2x\ln(x)-2\ln(x)-3x+1}{x^4} \right)$$

I don't know how to separate the $x$'s from the terms containing $\ln(x)$.

Alternatively, we are allowed to use Taylor Series expansions for $\ln(1+x)$ but if there is a way to do it I wasn't able to find it. And I'd love to learn.

And ideas?

Thanks

The question has been changed, but I stick to analysing the numerator of the new $f$ in the question, since as the questioner has rightly pointed out, it is the significant part of the function when it comes to finding roots.

First and foremost, let $f(x) = \ln^2(x) + 2x\ln x - 2\ln x -3x+1$ (Since the question has changed, I apologize for retaining notation). Note that the dominating term in $f(x)$ (the term which,for sufficiently large $|x|$, is the largest contributor) is $x \ln x$. Hence, as $x\to \infty$, $f(x) \to \infty$. Also, as $x\to 0$, this function is dominated by $\ln^2 x$, which is positive, so we know how the graph of this function looks at the end points: it comes crashing down from infinity as we start moving right from zero, and certainly flies off to infinity as $x \to \infty$.

As usual, without techniques of calculus we'd be nowhere on this one.

Find the double derivative, it is just $\frac{2}{x^2}(x-\log(x)+2)$. Note that if $x>0$, the fraction is anyway positive. For the other expression, use calculus again : differentiating gives that the extrema must satisfy $1 - \frac 1x = 0$ so $x=1$ but at $x=1$ this expression is just $3$. Therefore, the function is a product of positive functions and hence is positive.

Therefore, since the double derivative is strictly positive, we see that the first derivative,  which is initially negative and later positive (I leave you to see this by actual differentiation) will increase strictly and therefore attain a zero at precisely one point.

A final observation is that $f(x)$ must have some roots : $f(1) = -2$ , so in fact $f$ must have at least two roots, by the intermediate value theorem.

However, $f$ must have exactly two distinct roots, since if there were more than $3$, then by Rolle's theorem $f'$ would be zero at two points, a contradiction to our earlier statement.

Therefore, $f$ has precisely two roots in the interval $(0,\infty)$.

Nothing more can be said analytically. I am sure that convexity of $f$ will allow the use of excellent approximation algorithms for finding the roots, but other than evaluation you do not expect to find anything further revealing about the roots of $f$.

As it turns out, $x = 0.582$ and $x = 4.368$ would solve this equation, and are the inflection points of $x^{\frac 1x}$ also.

Using the quadratic formula with $a=1$, $b=2x-2$ and $c=1-3x$ you can separate the ln(x) from the x's and get,
$$ln(x) = \frac{2-2x \pm \sqrt{(2x-2)^2 -4(1)(1-3x)}}{2}$$

$$ln(x)=1-x \pm \sqrt{x^2+x}$$


Density of Baire’s Theorem for Complete Metric Spaces Let $\{G_n\}$ be a sequence of dense open subsets of a complete metric space $X$. Show that $ \bigcap_{n \geq 1}G_n$ is nonempty.

Proof:

Consider some point $p_1 \in G_1$. There exist real numbers $0< r_1 <r<1$ such that
$$N_1 = N_{r_1} (p_1) \subset N_r (p_1) \subset G_1.$$

Let $q$ be a limit point of $N_1$. For any $\epsilon > 0$ we can find $x \in X$ such that $x \in N_1$ and $x \in N_\epsilon (q)$. Hence
$$d(p_1, q) \leq d(p_1, x) + d(x, q)<r_1 + \epsilon.$$
Since $\epsilon$ was arbitrary it follows that
$$d(p_1, q) \leq r_1.$$

Thus, if $q$ is any limit point of $N_1$, $q$ is an interior point of $G_1$ (to see this, simply consider the neighbourhood of radius $0<\epsilon<r-r_1$) and so we can write

$\overline{N_1} \subset G_1.$

Since the $G_n$ are dense, either $p_1$ is a point in $G_2$ or a limit point of $G_2$. In either case, arguing as above, we can find a point $p_2 \in G_2 $ and a neighbourhood of $p_2$ with radius $r_2 < \min\big(r_1, \frac{1}{2}\big)$ such that
$$\overline{N_2} \subset \overline{N_1}$$
and, since the $G_n$ are all open, we also choose $r_2$ small enough that
$$\overline{N_2} \subset G_2.$$

Continuing this process, we can construct the sequence of closed subsets $\Big\{\overline{N_n}\Big\}$. Further, by the way we have chosen the $r_n$ (i.e., $0<r_n<\frac{1}{n}$), it follows $r_n \to 0$, so we have actually constructed a nested sequence of closed bounded sets $\Big\{\overline{N_n}\Big\}$ such that $diam \overline{N_n} \to 0.$ Since $X$ is a complete metric space, the intersection of this sequence is nonempty and since each $$\overline{N_n} \subset G_n$$ the result follows. $\qquad \square$

Assuming this proof is correct, I am wondering if it can be extended to show that the intersection of the $G_n$ is actually dense, or if that requires a completely different line of argument?

That the intersection of the $G_n$ is dense, is a small modification of the above argument: let $O$ be any non-empty open set in $X$, and start with an open ball $N_0 \subseteq O$ and stay inside $N_0$ with all subsequent steps. This hardly takes any effort at all, but does show that the $x \in \cap G_n$ is also in $N_0$ hence in $O$. So the intersection of the $G_n$ intersects every non-empty open set, hence is dense.

The construction of the $N_n$ can be a bit simplified:

Start with $p_0 \in O$ and $N_0 := B(p, r_0) \subseteq O$.
$N_0$ is open, so $G_1 \cap N_0$ is non-empty (as $G_1$ is dense) and open (as $G_1$ is open). So pick $p_1 \in G_1 \cap N_0$ and let $0< r_1 < 1$ be small enough that $\overline{B(p_1, r_1)} \subseteq G_1 \cap N_0$. Define $N_1 = B(p_1, r_1)$.
$N_1$ is open and again we have that $p_2 \in N_1 \cap G_0 \cap G_1$ exists by denseness and this set is open so there exists $0<r_2 < \frac12$ such that $\overline{B(p_2, r_2}) \subseteq (N_1 \cap G_0 \cap G_1)$. Define $N_2 = B(p_2, r_2)$.
continue this process recursively.

No distinguishing limit points etc. Just go straight to the goal.

Then the $\overline{N_n}$ $n \ge 1$ form the required nested family that the Cantor intersection theorem can be applied to. The promised $p \in \bigcap_{n \ge 1} \overline{N_n} \subseteq O \cap \bigcap_{n \ge 1}O_n$ witnesses the denseness of $\bigcap_{n \ge 1} G_n$.

The proofs of @Bollo and @Brandsma are great and I try to supplement a situation when the metric space is finite.

Supplement proof:

When the metric space $X$ is a finite set, it is always complete. However, there is no such open set $N_r(p_1) \subset G_1$ (in your proof) because every point is an isolated point. Therefore the only dense open subset of $X$ is itself. So the intersection is itself, which is dense.


Why is a positive angle defect sufficient for the existence of convex regular polytopes? If we want to build a convex regular n-polytope, we can start with a regular (n-1)-polytope, arrange $k$ copies around each (n-3)-dimensional ridge, and fold into n-space. This gives an easy necessary criterion for the existence of the polytope $\{X,k\}$, where $3\le k$, namely
$$
\theta\{X\}<\frac{2\pi}{k}
$$
$\theta\{X\}$ is the dihedral angle of the regular (n-1)-polytope $X$.

What has bothered me for the longest time is this:

  Why is this also a sufficient condition?

I.e., how come the construction always works? I find it conceivable that you could start building a polyhedron around one corner, but then something would go wrong as you went around and connected more polygons to your shape.

You can of course just construct all the polytopes to prove existence. But I always thought there might be some underlying reason for it always working. I would also be happy for an intuitive/heuristic argument. Anything, really.

Okay, when now additionally restricting to convex regular polytopes, then you could refer to any of the classical text books on that subject, e.g. Coxeter's Regular Polytopes.

They all use your argument as a necessary restriction only. But then, for sufficiency of existance, they turn towards an explicite construction. (Which then mostly follows Wythoff's kaleidoscopical construction.)

The only thing on that track to be known additionally in advance is the group thoeretical derivation of the possible (here: finite) Coxeter (i.e. reflection) groups.

--- rk

Well, regularity usually is equated to flag transitivity. And so the local rule clearly is enough. It must not close somewhere up, rather it could wrap around the n-ball over and over again. That would be fully legal.

--- rk


Summation of the given series Is there anyway to find the sum of:
$\cos(A)+\cos^2(2A)+\cos^3(3A)+....$ upto 'n' terms.

Actually original question was to find sum of :
$\cos(A)+\cos(2A)+\cos(3A)+...$ upto 'n' terms and I found it with help of De-Movier's Theorem which also gave me sum of $\sin(A)+\sin(2A)+.....$. But now I am thinking can we also find sum of $\cos(A)+\cos^2(2A)+\cos^3(3A)+....$ upto 'n' terms by some method?

Use the fact $$\cos^k ( nA) =\left(\frac{e^{inA} +e^{-inA}}{2}\right)^k$$


can a ring homomorphism map an integral domain to a non integral domain? i understand that if two rings are isomorphic, and one ring is an integral domain, so must the other be.
however, consider two rings, both commutative rings with unity.  is it possible that one ring contains zero divisors and one does not while there exists an ring homomorphism between the two?
there could not be a isomorphism between the two rings because there would be no one to one or onto mapping between the two rings. but could there be an operation preserving mapping between an integral domain and a commutative ring with unity and with zero divisors?
clearly if such a mapping did exists it would not seem to be one to one or onto, but does this rule out the potential of a homomorphism existing between the two?

Given any unital ring $R$ (with multiplicative identity $1_R$, say), there is a unique ring homomorphism $\Bbb Z\to R$ (take $1\mapsto 1_R$ and "fill in the blanks" from there).

This may be an injective map, even if $R$ has zero divisors. For example, take $$R=\Bbb Z[\epsilon]:=\Bbb Z[x]/\langle x^2\rangle.$$ Surjective examples are easy to come by. You are of course correct that such maps cannot be bijective if $R$ has zero divisors.

Simply define f from Z to Z4 ={0,1,2,3} with  addition and multiplication  modulo 4 as operations, as f(a)  = r, where r is the least positive remainder obtained when a is divided  by 4.Then f is a ring homomorphism & while Z is an integral domain Z4 is only a commutative  ring with unity.


Euler lagrange equation is a constant I'm working through exercises which require me to find the Euler-Lagrange equation for different functionals.
I've just come across a case where the Euler Lagrange equation simplifies to
$$1=0.$$
Please could someone explain what can be concluded about the set of extremals to problems where this is the case.

Notation $$L[u]=\int(f(u)u_{x}+u)dx=\int F(u,u_{x})dx $$

$$\frac{d}{dx}\left( \frac{\partial F}{\partial u_{x}}\right)-\frac{\partial F}{\partial u}=0 $$

As $F(u,u_{x})=f(u)u_{x}+u$  we have

$$\frac{\partial F}{\partial u_{x}}=f(u) $$

$$ \frac{\partial F}{\partial u}=\frac{df}{du}\frac{du}{dx}+1$$

so $$ $$

Hence $$\frac{d}{dx}(f(u)+1)-\frac{df(u)}{du}\frac{du}{dx}=0 $$

$$\frac{df}{du}\frac{du}{dx}-\frac{df}{du}\frac{du}{dx}=0 \rightarrow 0=0$$

This is breeze can no man do calculus of variations no more? Kmt

$$
L[u]=\int(f(u)u_{x}+u)dx
$$

$$
\frac{d}{dx}\left( \frac{\partial F}{\partial u_{x}}\right)-\frac{\partial F}{\partial u}=0
$$

$$
f'(u)u_x +1 -f'(u)u_x=0
$$

$$
1=0
$$

so for any $u$, the E-L equations is 1=0, thus the set of extremals is empty - hence the functional has no stationary value - consider the graph of $e^x$. Infimum, not minimum.
(Do you even math?)


Find the area bounded with the curve $x=1+t-t^3, \ \ y=1-15t^2.$ Find the area bounded with the curve $$x=1+t-t^3, \ \ y=1-15t^2.$$

I drew its graph and looks like this:

It seems to be symmetric about $x=1$. So, the idea that came to me was to cut by half and study any of the pieces. However, I don't know how... Any help is appreciated.

If $1+t-t^3=1+u-u^3,\,1-15t^2=1-15u^2,\,t\ne u$ then$$t\ne0,\,u=-t,\,1+t-t^3=1-t+t^3\implies 2t(1-t^2)=0\implies t=\pm1.$$So the area is from $t=-1$ to $t=1$. As $t$ increases from $-1$ to $0$, we trace out the left-hand part of the area's circumference; from $t=0$ to $t=1$, we follow the right-hand part. The shape vertically stretches from $t=\pm1,\,y=-14$ to $t=0,\,y=1$. From $y$ to $y+dy=y-30tdt$, a strip exists of width $2t(1-t^2)$, so the area is$$\int_0^160t^2(1-t^2)dt=60\int_0^1(t^2-t^4)dt=8.$$As a sanity check, the area fits in a rectangle of height $15$ and width $4/(3\sqrt{3})$ (because $x$ is extremised at $t=\pm1/\sqrt{3},\,x=1\mp2/(3\sqrt{3})$), so of area $20/\sqrt{3}>8$.

Cut the curve into left and right values for $x$ at each $y$ and make use of the symmetry.
$$ {{\int_{-14}^{1} x_r-x_\ell\,\mathrm d y }={2\int_0^1 (x(t)-1)\,\left\lvert\frac{\mathrm d y(t)}{\mathrm d t}\right\rvert\,\mathrm  d t}}$$


the minimum and maximum values of the determinant of order $3\times3$ matrix with entries $\{0,1,2,3\}$ The minimum and maximum values of the determinant of order $3\times3$ matrix with entries $\{0,1,2,3\}$.

Brute force enumeration gives a maximum of $54$ (and hence a minimum of $-54$ since $\det (-A) = - \det A$ for $n$ odd).

A maximising matrix would be $\begin{bmatrix} 0 & 3 & 3 \\
3 & 0 & 3 \\
3 & 3 & 0
\end{bmatrix}$.

In Python:

import numpy
import itertools

N = 3
cols = list(itertools.product(range(0, N+1), repeat=N))
mats = itertools.product(cols, cols, cols)
sup = 0
sup_p = None
for p in mats:
    d = numpy.linalg.det(p)
    if d > sup:
        sup = d
        sup_p = p

print sup
print sup_p

Note:

A slight improvement would notice that the problem can be written
as $\max \{ \det A \,|\,  [A]_{ij} \in \{0,1,2,3\} \}$ and since the cost is
affine in each matrix element, the maximiser must have extreme values
for all entries. Hence the problem is equivalent to
$\max \{ \det A \,|\, [A]_{ij} \in \{0,3\} \}$.

Since the problem is also equivalent to
$\max \{ |\det A| \,|\, [A]_{ij} \in \{0,3\} \}$, the order of rows (or columns)
doesn't matter and so we can choose the first row from
$\{(0,0,0), (3,0,0), (3,3,0), (3,3,3) \}$. This reduces the search space.

Since $(0,0,0)$  (or repeating a row) will result in a zero determinant,
the search space can be reduced further to at most
$3 \cdot \binom{7}{2} = 63$ possibilities, which is a little better
than the original $4^9 = 262,144$.

And finally, although irrelevant to the problem, we note that
this solves the problem $\max \{ |\det A| \,|\, [A]_{ij} \in [0,3] \}$,
the maximum volume of a box whose sides have $l_1$ length no greater than $3$.

I don't know anything about Hadamard's inequality...
But I have a better technique.
Here the smallest element is $0$ and the largest element is $3$
So, the determinant will be max when $0$ is the diagonal element and $3$ is the non-diagonal element, meaning that the answer is $54$.