Confusion on notation in the definition of reflexivity. This is a very silly question i assume to most of you but how would I read this
$\forall x \in X : (x,x) \in R$
I don't quite know what the colon indicates and this definition is used to define reflexivity .
Thanks in advance.

It is a hold over of the dot notation used in Principia Mathematica, which has mainly been replaced by bracketing.   It is basically just a punctuation mark to visually separate the quantifier's restriction from its scope. $$\forall x\in S:P(x)\\\forall x\in S~\big(P(x)\big)\\\forall x\in S~P(x)$$

Normally the colon is read as "such that"
∀ ∈  :(,) ∈
For all $x$ in {the set} X such that $(x,x)$ {$x$ maps to itself} in {the binary relation} R {which is a subset of X}.


Vector Subspace question $X,Y,Z$ are subspaces of vector space $V$ s.t. $X + Z = Y + Z$, then is $X= Y.$ How to prove it?

If $X \oplus Z= Y\oplus Z$ then is $X = Y.$

No. Let $V=\mathbb{R}^2$, $X=\{\alpha(1,0)\colon \alpha \in \mathbb{R}\}$, $Y=\{\alpha(1,1)\colon \alpha \in \mathbb{R}\}$, and $Z=\{\alpha(0,1)\colon \alpha \in \mathbb{R}\}$. Then $X \oplus Z=Y \oplus Z=\mathbb{R}^2$, but $X \ne Y$.

This is not true. $\mathbb{R}^2 \oplus \mathbb{R}^3 \cong \mathbb{R}^3 \oplus \mathbb{R}^2$ but $\mathbb{R}^2 \neq \mathbb{R}^3$. And there's not just the trivial example of order. In infinite dimensions, it can even be the case that $A \oplus B \cong C \oplus D$ and $A \cong C$ but $B \not \cong D$.


Asymptotics of Gaussian integral over the unit sphere I would like to evaluate the integral asymptotically over the unit sphere surface
$$
Z =\int e^{a \cos^2 \theta + b \sin^2\theta\cos2\phi + c\cos\theta} d\Omega = \int\limits_{0}^{\pi}\int\limits_{0}^{2\pi} e^{a \cos^2 \theta + b \sin^2\theta\cos2\phi + c\cos\theta} \sin\theta d\phi d\theta
$$
for $a\rightarrow \pm\infty$ and $b\rightarrow\infty$, if $c$ is set implictly by the constraint $$ L = \frac{\partial \ln Z}{\partial c}$$ for a fixed $0<L<1$.
With the variables change $s=\cos\theta$,
\begin{align}
Z &= 2\pi\int_{-1}^1 e^{a s^2+c s}\, I_0[b (1-s^2)] ds\,, \\
L &= \frac{\int_{-1}^1 s\, e^{a s^2+c s} I_0[b (1-s^2)]\, ds }{\int_{-1}^1 e^{a s^2+c s} I_0[b (1-s^2)]\, ds }\,.
\end{align}

where $I_0(x)$ is the Bessel function of the first kind. (Alternatively, the $\theta$ integral may be evaluated for any $\phi$ using the error function.)

Is there a general method to look for the $a\rightarrow \pm\infty$ asymptotics of this integral?

Note that this is the integral of a Gaussian on the sphere in the sense that
$$Z_1=\int_{\|\mathbf{x}\|=1} e^{-{\mathbf{x}\cdot \mathbf{M x} + \mathbf{v}\cdot\mathbf{x}}} dS$$
where the integration is over the unit sphere's surface, $\mathbf{M}$ is a $3\times 3$ symmetric traceless matrix, and $\mathbf{v}$ is a vector parallel to one of the eigenvectors. The  integral in spherical coordinates  $\mathbf{x}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ and $dS = \sin\theta d\theta d\phi$ aligned with the eigenvectors, reduces to the integral given in the question as $Z_1=e^{-a/3} Z$, where $a=\frac32\lambda_3$, $b=\frac12(\lambda_1-\lambda_2)$, $\mathbf{v}=c\, \mathbf{u}_3$, and $\lambda_{1,2,3}$ and $\mathbf{u}_{1,2,3}$ are the three eigenvalues and normalized eigenvectors of $\mathbf{M}$.

The following is just the sketch of an idea. With some luck, somebody will be able to complete it (or show that it is wrong, of course!)

Differentiate with respect to $c$ to obtain
\begin{align}
\frac{\partial Z}{\partial c} = & \int_0^\pi\int_0^{2\pi}\left(e^{a\cos^2\theta}\cos\theta\sin\theta\right)e^{b\sin^2\theta\cos2\phi+c\cos\theta}d\phi d\theta\\
= & \int_0^\pi\int_0^{2\pi}\left(-\frac{1}{2a}\frac{\partial e^{a\cos^2\theta}}{\partial\theta}\right)e^{b\sin^2\theta\cos2\phi+c\cos\theta}d\phi d\theta.
\end{align}
Integrating by parts, we obtain
\begin{align}
\frac{\partial Z}{\partial c} = &\left[-\frac{1}{2a}\int_0^{2\pi}e^{a\cos^2\theta+b\sin^2\theta\cos2\phi+c\cos\theta}d\phi\right]_0^\pi\\
& +\frac{1}{2a}\int_0^\pi\int_0^{2\pi}e^{a\cos^2\theta+b\sin^2\theta\cos2\phi+c\cos\theta}\left(2b\sin\theta\cos\theta\cos2\phi - c\sin\theta\right)d\phi d\theta.
\end{align}
Now hopefully the term on the second line will go to zero as $a\to\infty$, and the first term can be evaluated using Bessel functions and the error function.

Edit: As remarked by @tired, the second term will not go to zero, as the exponential is increasing too fast as $a\to\infty$. I will leave this here anyway, so if someone wants to try something passing from partial integration as I did above they won't have to do the computation over again.


Stokes theorem to get $\oint \vec{F}d\vec{R}$ I have the vector field

\begin{equation*}
\vec{F}=(ye^x,x^2+e^x,z^2e^z)
\end{equation*}

and the curve $C$ that us given by

\begin{equation*}
\vec{r}(t)=(1+\cos t, 1+\sin t, 1-\cos t-\sin t),~0\leq t\leq 2\pi.
\end{equation*}

What is $\oint_C \vec{F}d\vec{R}$?

I know that my curl is $(0,0,2x)$, but what is my final integral? How do I use Stokes' theorem here?

Surface Integral

You would need to calculate
$$
I = \int_A\limits \mbox{curl} F \cdot dA =
\int_A\limits (0, 0, 2x) \cdot dA =
2 \int_A\limits x \, dA_z
$$
where $A$ is a surface with boundary $\partial A = C$.
The open task to choose some feasible $A$ which allows for simple calculation of the last term.

So what is $C$? We had
$$
r(t) = (1+\cos t, 1+\sin t, 1-\cos t-\sin t)^t
$$
this gives
$$
r(t) \cdot (1,1,1)^t = 3 \iff
r(t) \cdot n = \sqrt{3} \quad n = (1,1,1)^t/\sqrt{3}
$$
So the $r(t)$ endpoints form a curve $C$ which lies in a plane $E$ with normal vector $n$ and origin $\sqrt{3} n = (1,1,1)^t$:
$$
E = \{ x \mid \left(x - \sqrt{3} n\right) \cdot n = 0 \}
$$
Let $A$ be the part of $E$ that is inside $C$. From the observation that the projection of $A$ onto the $x$-$y$-plane is a circle of radius $1$ with center $(1,1)$ we assume the parametrization:

$$
A(r, t) = (1 + r \cos t, 1 + r \sin t, 1 - r \cos t - r \sin t)^t
\quad (r \in [0, 1], t \in [ 0, 2\pi ])
$$
with the partial derivatives along the parameters
$$
\partial_r A = (\cos t, \sin t, - \cos t - \sin t)^t \\
\partial_t A = (-r \sin t, r \cos t, r \sin t - r \cos t)^t
$$
then we get
$$
(\partial_r A \times \partial_t A)_x =
r (\sin t)^2 - r \cos t \sin t + r (\cos t)^2  + r \cos t \sin t =
r \\
(\partial_r A \times \partial_t A)_y =
r \cos t \sin t + r (\sin t)^2 - r \cos t \sin t + r (\cos t)^2 =
r \\
(\partial_r A \times \partial_t A)_z =
r (\cos t)^2 + r (\sin t)^2 = r
$$
this gives
$$
dA
= \partial_r A \times \partial_t A \, dr \, dt
= r \, (1,1,1)^t \, dr \, dt
= n \, \sqrt{3} \, dA^{xy}
$$
So we have
\begin{align}
I
&= 2 \int\limits_A (1 + r \cos t) r dr dt \\
&= 2 \int\limits_0^1 dr \, \int\limits_0^{2\pi} dt \, (r + r^2 \cos t) \\
&= 2 \int\limits_0^1 dr \, 2\pi r \\
&= 2 \pi
\end{align}

Line Integral

With the help of a computer algebra system (link) we get
\begin{align}
I
&= \int\limits_C F \cdot dr \\
&= \int\limits_C (ye^x, x^2+e^x, z^2 e^z) \cdot dr \\
&= \int\limits_0^{2\pi} (ye^x, x^2+e^x, z^2 e^z) \cdot \dot{r} dt \\
&= \int\limits_0^{2\pi} (ye^x, x^2+e^x, z^2 e^z) \cdot (-\sin t, \cos t, \sin t - \cos t) dt \\
&= \int\limits_0^{2\pi} \left(-(1+\sin(t)) e^{1+\cos(t)} \sin(t) + \cos(t) ((1+\cos(t))^2 + e^{1+\cos(t)}) + (\sin(t) - \cos(t)) (1-\cos(t) - \sin( t))^2 e^{1-\cos(t) - \sin(t)}\right) dt \\
&= \left[ t+\frac{7}{4}\sin(t)+\frac{1}{12} \sin(3 t)+\sin(t) \cos(t)+(\sin(t)+1) e^{\cos(t)+1}+(\sin(2 t)+2) e^{-\sin(t)-\cos(t)+1}\right]_0^{2\pi} \\
&= 2 \pi
\end{align}

You know what $F$ is, you know whar $dr$ is. Compute $F \cdot dr$ and then formulate $\int_{0}^{2\pi}F \cdot dr$ and evaluate the integral above.

EDIT
Stokes theorem
$$\iint_{\Sigma} \nabla \times \mathbf{F} \cdot \mathrm{d}\mathbf{\Sigma} = \oint_{\partial\Sigma} \mathbf{F} \cdot \mathrm{d} \mathbf{r}.$$


Word problem of probability In a group of people, 21% own both phones and computers, 49% own just phones and 7% own just computers. One person is randomly chosen, what are the odds that they own a phone if we know that they own a computer?

Let $A$ represent the event that the person owns a cellphone, and let $B$ represent the event that a person owns computers.

Note that it is given that $P(A \cap B) = 0.21, P(A - B) = 0.49$, and $P(B - A) = 0.07$. We then need to find $P(A)$ and $P(B)$. However, one can see that $P(A) = P(A \cap B) + P(A-B)$ and similarly $P(B) = P(A \cap B) + P(B-A)$. We then have $P(A \cap B) = 0.21$, $P(A) = 0.7$, and $P(B) = 0.28$. By Bayes, we know that $P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.21}{0.28} = 3/4$.


How to solve a PDE with in which the coefficient which depends on time but singular at a point? I form a PDE of the following type:

$$0=\frac{\partial V}{\partial \tau}+\frac{\beta -z}{v-\tau}\frac{\partial V}{\partial z}+\frac{1}{2}\frac{\partial^{2} V}{\partial z^{2}}$$

such that $\lim_{\tau\rightarrow v}V(\beta,\tau)=1$. I know that it is difficult to solve, but I would like to know if there exists any literature talking about PDE in which $\tau$ is undefined at one point here. I try to change it to $u=v-\tau$ and solve it, but it can be seen that the new PDE is not defined at $u=0$. Any literature talk about it? What is this type PDE called?

Hint:

$\dfrac{\partial V}{\partial\tau}+\dfrac{\beta-z}{v-\tau}\dfrac{\partial V}{\partial z}+\dfrac{1}{2}\dfrac{\partial^2V}{\partial z^2}=0$

$\dfrac{\partial V}{\partial\tau}+\dfrac{z-\beta}{\tau-v}\dfrac{\partial V}{\partial z}+\dfrac{1}{2}\dfrac{\partial^2V}{\partial z^2}=0$

Let $\begin{cases}\tau_1=\tau-v\\z_1=z-\beta\end{cases}$ ,

Then $\dfrac{\partial V}{\partial\tau_1}+\dfrac{z_1}{\tau_1}\dfrac{\partial V}{\partial z_1}+\dfrac{1}{2}\dfrac{\partial^2V}{\partial z_1^2}=0$ with $\lim\limits_{\tau_1\to0}V(0,\tau_1)=1$

With reference the substitution concept in Change variables into Fokker-Planck PDE,

Let $\begin{cases}z_2=\dfrac{z_1}{\tau_1}\\\tau_2=\tau_1\end{cases}$ ,

Then $\dfrac{\partial V}{\partial z_1}=\dfrac{\partial V}{\partial z_2}\dfrac{\partial z_2}{\partial z_1}+\dfrac{\partial V}{\partial\tau_2}\dfrac{\partial\tau_2}{\partial z_1}=\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}$

$\dfrac{\partial^2V}{\partial z_1^2}=\dfrac{\partial}{\partial z_1}\left(\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}\right)=\dfrac{\partial}{\partial z_2}\left(\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}\right)\dfrac{\partial z_2}{\partial z_1}+\dfrac{\partial}{\partial\tau_2}\left(\dfrac{1}{\tau_1}\dfrac{\partial V}{\partial z_2}\right)\dfrac{\partial\tau_2}{\partial z_1}=\dfrac{1}{\tau_1^2}\dfrac{\partial^2V}{\partial z_2^2}$

$\dfrac{\partial V}{\partial\tau_1}=\dfrac{\partial V}{\partial z_2}\dfrac{\partial z_2}{\partial\tau_1}+\dfrac{\partial V}{\partial\tau_2}\dfrac{\partial\tau_2}{\partial\tau_1}=-\dfrac{z_1}{\tau_1^2}\dfrac{\partial V}{\partial z_2}+\dfrac{\partial V}{\partial\tau_2}$

$\therefore-\dfrac{z_1}{\tau_1^2}\dfrac{\partial V}{\partial z_2}+\dfrac{\partial V}{\partial\tau_2}+\dfrac{z_1}{\tau_1^2}\dfrac{\partial V}{\partial z_2}+\dfrac{1}{2\tau_1^2}\dfrac{\partial^2V}{\partial z_2^2}=0$ with $\lim\limits_{\tau_2\to0}V(0,\tau_2)=1$

$\dfrac{\partial V}{\partial\tau_2}+\dfrac{1}{2\tau_2^2}\dfrac{\partial^2V}{\partial z_2^2}=0$ with $\lim\limits_{\tau_2\to0}V(0,\tau_2)=1$

$2\tau_2^2\dfrac{\partial V}{\partial\tau_2}+\dfrac{\partial^2V}{\partial z_2^2}=0$ with $\lim\limits_{\tau_2\to0}V(0,\tau_2)=1$

Which converts to a separable PDE.

Differential equations like that one are called singular equations, and the points where the coefficients are undefined are called singular points. If you want to figure out how to solve these in general, look at the Frobenius method. It basically tries to look at how the function would have to scale $V \sim x^{\alpha} t^{\beta}$ near the singular points, then looking at an equation for $U$ where $V = x^{\alpha} t^{\beta} U(x,t)$. Sometimes your boundary conditions or other physical requirements (if this comes from modeling some real system) can limit the number of solutions and determine the powers so that you don't have to do the full general analysis.

I tried applying something like the Frobenius method to your problem, using the expansion
$$ V(\tau_1,z_1) = \sum_{k=0}^{\infty} A_k(z_1) {\tau_1}^{r+k} $$
to get a sequence of differential equations for $A_k(z_1)$. Solving those with general $r$, I keep getting $A_k \sim {z_1}^{-(r+k+2)}$ for everything but $A_0$. That is, the required powers of $z_1$ increase faster than for $\tau_1$, so the result would not be well-defined at $(\tau_1,z_1) = (0,0)$ unless the sequence terminates after the first term. All that does is give $r=1$ and the very simple solution $V(\tau_1,z_1) = z_1/\tau_1$. This doesn't have the limit you asked for, but it does have a limit where $z_1$ and $\tau_1$ both go to zero simultaneously. I don't think your problem has a well-defined solution. (This isn't a proof that it doesn't, but it's all I can point you at for now.)


When the characteristic of a finite field is not equal to the order? Can you give me one example where the characteristic of a field is not equal to the number of elements in the field?
I know that the characteristic of a prime field $GF(p)$ is $p$, but even for a non-prime $q$, say $q = 6$, the characteristic is equal to 6.
So when the characteristic and the number of elements in a fields will be unequal?

The field $\Bbb{F}_{q}$ for $q=p^k$ is a field of characteristic $p$, but contains $p^k$ elements. To construct this you take $\Bbb{F}_p[x]/(m(x))$ for $m(x)$ a monic irreducible polynomial of degree $k$. Alternatively, this can be seen as the splitting field of $x^q-x\in \Bbb{F}_p[x]$.

Just take any field $GF(p^n)$ for $p$ prime and $n\gt1$.  The characteristic is $p$.
Also, there's no field of order $6$.


Area of the triangle a,b,c How to prove in any triangle that the area $X$ is given by:

$$X=\frac{1}{4}(a+b+c)^2\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}$$

assume that c is the longest side and that d is perpendicular to c forming 2 right triangles.  The angle between a and d should be called D. The line along c from a to d is c'.

$$Area=\frac 14(a+b+c)^2\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}$$
$$Area=\frac 14(a+c'+d)^2\tan\frac{D}{2}\tan\frac{B}{2}\tan45+\frac 14(b+d+c-c')^2\tan\frac{A}{2}\tan\frac{C-D}{2}\tan45$$
$$\tan 45=1$$
$$d^2=a^2-c'^2=b^2-(c-c')^2$$
IFF we can show the following that works, then the full one should too!
$$\frac 14(a+c'+d)^2\tan\frac{D}{2}\tan\frac{B}{2}=\frac 12c'd$$
$$B=90-D$$
$$\tan \frac B2 = \frac{\sin B}{1+\cos B}$$
$$\tan \frac D2 = \frac{\sin D}{1+\cos D}$$
$$\sin B = \cos D = \frac da$$
$$\sin D = \cos B = \frac {c'}a$$
$$\tan\frac{D}{2}\tan\frac{B}{2} = \frac{c'd}{a^2+ac'+ad+c'd}$$
$$\frac 14(a+c'+d)^2\frac{c'd}{a^2+ac'+ad+c'd}=\frac 12c'd$$
$$(a+c'+d)^2=2(a^2+ac'+ad+c'd)$$
$$a^2+c'^2+d^2+2ac'+2ad+2c'd=2a^2+2ac'+2ad+2c'd$$
$$a^2+c'^2+d^2=2a^2$$
$$c'^2+d^2=a^2$$
which is true.... if we repeat for the other right triangle this works.... Doing it from the Heron is easier but only, I guess, if you know the Heron and can prove it.This uses http://en.wikipedia.org/wiki/Tangent_half-angle_formula so .... yeah....

From this, $$\tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\text{  where  } s=\frac{a+b+c}2$$

This can be easily derived using Tangent half-angle formula and the Law of cosines

Now we know Heron's formula


Motivation Behind Tschirnhaus Transformation and Cardano's Formula I have a general question about the cubic equation:

  Let $a,b,c \in \mathbb{C}$ with $a \neq 0$. The general cubic equation is $t^3+at^{2}+bt+c = 0$. To get Cardano's Formula, we first transform the equation so that $a= 0$ (i.e. $t^3+bt+c = 0$). Then we reduce this to a quadratic that we can solve. What is the motivation behind the transformations that reduce $t^3+at^2+bt+c = 0$ to $t^3+bt+c = 0$?

In particular, if we let $y = t+ \frac{a}{3}$, then $t = y-\frac{a}{3}$ which cancels the $at^2$ term. This is called a Tschirnhaus transformation. We are left with an equation of the form $$y^3+py+q = 0$$

But how do we know what transformations to use to get Cardano's Formula, where $$p = \frac{a^2-2a^3+3b}{3}\quad\mathrm{and}\quad q = \frac{2a^3-9ab+27c}{27}\quad ?$$ Finally if we let $y = \sqrt[3]{u}+ \sqrt[3]{v}$ we eventually get a quadratic which leads to Cardano's Formula.

I don't think it's really all that mysterious. The motivation behind using the Tschirnhaus Transformation seems clear; it's better to have fewer terms and fewer coefficients when solving an equation. It next probably occurred to Cardano to try $y = u + v$ and see what happens. Rewrite the equation as
$$u^3 + v^3 + 3uv(u + v) + p(u + v) + q = 0$$
You have a little leeway in choosing $u$ and $v$, since you only need them to sum to a root of the equation. So you might want to impose a second condition on $u$ and $v$ to simplify things (and two equations in the two variables $u$ and $v$ should specify their values). To reduce the above to a quadratic equation, you can stipulate that $uv = -{p \over 3}$. Namely you have
$$u^3 + v^3 + q = 0$$
which is equivalent to
$$u^3 -{p^3 \over 27}{1 \over u^3} + q = 0$$
Or just
$$u^6 + qu^3 - {p^3 \over 27} = 0$$
This is a quadratic equation in $u^3$, which you can solve. $u$ can be any of the three cube roots of this, and $v$ is determined by the equation $3uv = -p$. Thus you have three possible values of $u + v$ which will be the three roots of your equation.

In Cardano's time they didn't really know much about complex numbers (i.e. that every number has $n$ complex $n$th roots) but he was smart enough to basically understand what was going on.

That bugs me a little too, but I think the great advantage is this one: by making $t=u+v$, $y=t+k$ or whatever, you are giving your problem more degrees of freedom. Since there is more than one tuple $(u,v)$ that satisfies this relation, you will be able to impose another condition on the new coefficients in order to make the problem similar to something you can solve. By annulating one of them, for example.

It is also interesting to think the other way around. At first you don't know how to solve $ax^2+bx+c=0$ (I). However, you can always solve the simpler equation $x^2-p=0$ (II). If you can prove that by substituting $x$ for $[mx+n]$ in (II) you can always arrive at (I) for appropriate $m$, $n$ and $p$, you'll also be able to solve (I). Problem is: in this case we have 3 equations - one for comparing each coefficient - and three variables. As the degree gets bigger*, we keep having three variables but the number of equations increases, yielding the possibility of no solutions to the system.

*$a_0+a_1 x+\dots+a_k x^k=0 = (mx+n)^k-p=0$


Shortest Path Possible Corridors Question You find yourself in a maze, as shown below, and want to escape by the shortest route possible. The lines represent corridors, and you may only travel along corridors. How many different paths can you take? Note that both exits are 21 segments from the start point, so you should only count paths of length 21.

I do not want to know what the answer to this is... just an outline for a strategy. I was thinking about breaking this part up into each square and using permuations... is this logical?

Hint: There are $a+b\choose a$ ways to cross an $a\times b$ rectangle as in the problem statement.

Divide it into cases

Total Ways=number of ways to go to Exit $1$ $+$ number of ways to go to Exit $2$


Determine if there's a linear map such that... Today in the class we were solving a following problem:
Determine if there's a linear map $\phi : R^3 \rightarrow R^3$ such that $\phi(2,1,0)=(3,2,1), \phi(4,2,1) = (4,1,0), \phi(2,2,1) = (2,3,3), \phi(2,2,2) = (0,0,1)$. It was then solved by creating an augmented matrix with the inputs for $\phi$ written in rows in the left hand side of the matrix and outputs of $\phi$ in the right hand side, also in rows. Then the matrix has been row reduced until an identity appeared in the left hand side... it seemed to me like some black magic. I have no idea why or how it works. This whole method doesn't make any sense to me. Is there any better way of doing it or some explanation why something like this works?

You must have studied some verson of the following:
Theorem: Any linear map $\;T:V\to W\;$ , with $\;V,\,W\;$ linear spaces over the same field, is uniquely and completely determined once we know the action of $\;T\;$ on any basis $\;A=\{v_1,...\}\;$ of $\;V\;$ , meaning: $\;\{Tv_1,...\}\;$ completely and uniquely determines $\;T\;$ .
The above theorem holds for both infinite and finite dimesnional spaces, and from here it is not hard to see that
For any basis $\;A=\{v_1,...\}\;$ of $\;V\;$ and any set $\;B=\{w_2,...\}\;$ of elements in $\;W\;$ , with $\;|A|=|B|\;$ , any function $\;t: A\to B\;,\;\;tv_i=w_i\;$ can uniquely be extended by linearity to a linear map $\;T: V\to W\;$ such that $\;Tv_i=w_i\;$ .
Thus, in your case, you only have to check that the set of vectors on which the action of $\;\phi\;$ is given is a basis of $\;\Bbb R^3\;$ ...


How to find perpendicular vector to another vector? How do I find a vector perpendicular to a vector like this: $$3\mathbf{i}+4\mathbf{j}-2\mathbf{k}?$$
Could anyone explain this to me, please?

I have a solution to this when I have $3\mathbf{i}+4\mathbf{j}$, but could not solve if I have $3$ components...

When I googled, I saw the direct solution but did not find a process or method to follow. Kindly let me know the way to do it. Thanks.

There exists an infinite number of vectors in 3 dimension that are perpendicular to a fixed one.
They should only satisfy the following formula:
$$(3\mathbf{i}+4\mathbf{j}-2\mathbf{k}) \cdot v=0$$

For finding all of them, just choose 2 perpendicular vectors, like $v_1=(4\mathbf{i}-3\mathbf{j})$ and $v_2=(2\mathbf{i}+3\mathbf{k})$ and any linear combination of them is also perpendicular to the original vector: $$v=((4a+2b)\mathbf{i}-3a\mathbf{j}+3b\mathbf{k}) \hspace{10 mm} a,b \in \mathbb{R}$$

The dot product of two perpendicular vectors are always $0$ so if you
$(ai+bj+ck)\cdot(di+ej+fk)=0$ you can solve for the different variables.
If you have one vector than the infinite amount of perpendicular vectors will form a plane that is perpendicular to the original vector. If you know one or two of the coordinates of the desired perpendicular line than you can find the corresponding vector(s) on that plane.


Why is the reciprocal used in fraction division? I don't know if this is a basic question or whatever, but I can't seem to find an answer.

As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.

For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.

Any response is appreciated.

I think you're asking why the rule for division of fractions,
$$\frac{p}{q} \div \frac{r}{s} = \frac{p}{q} \cdot \frac{s}{r},$$
works.
And I'm assuming that you're already comfortable with how to multiply fractions.

We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $A\div B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $A\div B$ means the $X$ that solves the equation $$ X\cdot B = A $$

When our $A$ and $B$ are fractions, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $\frac pq \div \frac rs$ we need to solve the equation
$$ X \cdot \frac rs = \frac pq $$
And indeed setting $X=\frac pq\cdot \frac sr = \frac{ps}{qr}$ does this:
$$ \frac{ps}{qr}\cdot\frac rs  = \frac{ps\cdot r}{qr\cdot s} = \frac{p\cdot sr}{q\cdot sr} = \frac pq$$
like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).

This computation hopefully also gives some ides why it works, at least part way. In $\frac{ps}{qr}$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.

Writing the solution $\frac{ps}{qr}$ as $\frac pq\cdot \frac{\vphantom{p}s}{r}$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.

Your question isn't completely clear but what I understood is that you don't get why $$\frac{\frac{a}{b}}{\frac{c}{d}}= \frac{a}{b}*\frac{d}{c}$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^{-1}$$ with $f^{-1}$ the reciprocal of $f$, therefore $$\frac{\frac{a}{b}}{\frac{c}{d}}={\frac{a}{b}}({\frac{c}{d}})^{-1}$$ and since $$\frac{c}{d}*{\frac{d}{c}}=1$$ we have $$\frac{\frac{a}{b}}{\frac{c}{d}}= \frac{a}{b}*\frac{d}{c}$$ our result


If $f(x+1) +f(x-1) =\sqrt{3}\,f(x)$ and $f(2) =2$, what is the value of $f(4)$? My Attempt

$f(2)=2$. So,  $f(1) + f(3)=2\sqrt{3}$ and  $f(2) + f(4)=\sqrt{3}\,f(3)$.  After solving these equations I got the value of $f(3)=2\sqrt{3}$ and $f(4)=4$.  But are there any other methods than this? Any suggestions are welcome.

Update:- @ProfessorVector pointed out that the above solutions are only true if $f(1)=0$. After checking I find that it is true. So, my above attempt is a failure. Is there a way to solve this question?

Update 2:- Is there a way to find the period of this function?

I'll be more general.

$f(n+1)+f(n-1)
= cf(n)
$.

Suppose
$f(n)
= b^n
$.

Then
$b^{n+1}+b^{n-1}
= cb^n
$
or
$b^2-cb+1
= 0
$.

Then
$b
=\dfrac{c\pm\sqrt{c^2-4}}{2}
$.

If $c^2=4$,
$b = c/2 = \pm 1$,
so
$f(n)=1$
or
$(-1)^n$.

If
$c^2 > 4$,
then
$b$ has two possible values,
one with
$|b|>1$
and one with
$|b|<1$.

If
$c^2 < 4$,
then
$b$ has two possible
complex values
$b_1
=\dfrac{c+\sqrt{c^2-4}}{2}
=\dfrac{c+i\sqrt{4-c^2}}{2}
$
and
$b_2
=\dfrac{c-i\sqrt{4-c^2}}{2}
$.
Note that
$b_1b_2
=\dfrac{c+\sqrt{c^2-4}}{2}\dfrac{c-\sqrt{c^2-4}}{2}
=\dfrac{4}{4}
=1
$
(as can also be deduced
from the quadratic
specifying $b$)
and that
$|b_k|^2
=\dfrac{c^2+4-c^2}{4}
=1
$.

Since
$|b| = 1$,
$b
=e^{it}
=\cos(t)+i\sin(t)
$
where
$\cos(t)
=c/2
$.

In your case,
$c = \sqrt{3}$
so
$t
=\arccos(\sqrt{3}/2)
=\pi/6
$.

Therefore
the two possible solutions are
$f(n)
=e^{\pm ni\pi/6}
$
and any linear combination of these.

So any solution
is of the form
$ue^{in\pi/6}+ve^{-in\pi/6}
=(u+v)\cos(n\pi/6)+(u-v)\sin(n\pi/6)
$.

If the solution is real
for an $n$
such that
$\sin(n\pi/6) \ne 0$,
then
$u=v$,
so it is
$f(n)=2u\cos(n\pi/6)
$.

If
$f(2) = 2$,
then
$2=f(2)
=2u\cos(\pi/3)
=u
$
so
$u = 2$
and the solution is
$f(n)
=4\cos(n\pi/6)
$.

Putting $n=4$,
$f(4)
=4\cos(4\pi/6)
=-2
$.


Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ (both arbitrary functions)

To make more clear what I'm talking about I will give a example.

Rewrite $f(x,y) = 1+x-y-xy$ as $g(x)h(y)$

If we choose $g(x) = (1+x)$ and $h(y) = (1-y)$ we have

$f(x,y) = g(x) h(y) \implies (1+x-y-xy) = (1+x)(1-y)$

I'm trying to do the same with $f(x,y) = 1-x^2y^2 = (1-xy)(1+xy)$.

New question:

  Is there also a contradiction for $f(x,y) = \frac{xy}{1-x^2y^2}$ ? Or it's possible to write $f(x,y) $ as $g(x)h(y)$ ?

Suppose we had functions $g,h$ with $g(x) h(y) = 1 - x^2 y^2$.  Then substituting in $x:=1, y:=1$, we would have $g(1) h(1) = 0$, so either $g(1) = 0$ or $h(1) = 0$.  But we also must have $g(1) h(0) = 1$ which implies $g(1) \ne 0$, and similarly $g(0) h(1) = 1$ which implies $h(1) \ne 0$.  Putting these requirements together, we get a contradiction.

Therefore, no such functions $g,h$ can exist.

Without loss of generality let $f(x) = ax^2 + b$ and $g(y) = cy^2 + d$. Now
$$
1-x^2y^2 = f(x)g(y) = acx^2y^2 +adx^2 = cby^2 +bd
$$
By comparing terms we obtain:
\begin{align}
ad&=0\\
cb&=0\\
ac&=-1\\
bd&=1
\end{align}
By $bd=1$ we obtain $a=c=0$. Hence this problem has no solution.


A matrix $A$ such that $Ax = b$ has either 1 or infinitely many solutions, depending on $b$ How can I find a matrix like this:

  Find a matrix $A$ such that $Ax = b$ that has either 1 or infinitely many solutions depending on $b$.

Thank you.

The problem is to identify a matrix $A$ such that $Ax = b$ has infinitely many solutions for some $b$ and exactly one solution for another righthand side $b$.

There is no such square matrix $A$, as either the matrix is singular (and for some $b$ there is no solution while for other $b$ there are infinitely many solutions) or the matrix is nonsingular (and there exists a unique solution for every $b$).

Indeed the logic extends even to nonsquare matrices, that such $A$ is impossible.  Suppose for some particular $b$ there are infinitely many solutions.  Pick two distinct solutions, $x_1 \neq x_2$ such that $Ax_1 = Ax_2 = b$.  Then $A(x_1 - x_2) = 0$, so that there are infinitely many solutions to $Ax = 0$ (taking any multiple of $x_1 - x_2$).

It follows that for any righthand side $b$ the system has either infinitely many or zero solutions.  For if one solution exists, then to that any nonzero multiple of $x_1-x_2$ may be added to obtain another.

I'm assuming you're working over $\mathbb{R}$ (Over a finite field there would only be a finite number of solutions, never infinitely many).  This question can be thought of as

  Think of an $n \times m$ matrix as a function from $\mathbb{R}^m$ to $\mathbb{R}^n$.  Find an $n \times m$ matrix $A$ such the "image" of $A$ is all of $\mathbb{R}^m$, in other words find a matrix that maps (surjectively) onto $\mathbb{R}^m$ (i.e. $A$ is an onto map)

So, now that we have some big words thrown at us, what does that mean?  This means we want to have $m$ "linearly independent columns" in our matrix.  This will mean that $rank(A) = m$, and hence $\dim(Im(A)) = m$, and $Im(A) = \mathbb{R}^m$.

So choose a square nonsingular square matrix or a "fat" matrix of rank $m$, and you'll be all set.  Examples of these are $$I = \begin{pmatrix}1 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \vdots & \vdots \\0 & 0 & \ldots & 1   \end{pmatrix}$$ and for a nonsquare matrix $$I^* = \begin{pmatrix} I &|& * \end{pmatrix}$$
Where the second matrix is a block matrix with one block the identity and one block anything.


Reflexive relation on set of $n$ elements How many reflexive relations are there on a set of $n$ elements? I did the problem and I got the answer $2 ^ {n ^ 2}$. Is it correct? Thanks for the help..!!

Hint: Note that if $X$ is a set with $n$ elements, then there are $n^2-n$ elements in $X\times X$ that are not of the form $\langle x,x\rangle$ for some $x\in X.$ Any, all, or none of these $n^2-n$ elements can be members of a reflexive relation on $X$. (Why?) What about the other $n$ elements of $X\times X$?

$2^{n^2}$ is find for how many number of relations on a set on n elements.
$2^{n(n-1)}$ is find for how many number of reflexive relations on set with $n$ elements.


The Riemann Surface of Logarithmic Function Who can tell me why the compactification of RS of Log is just Riemann Sphere, please?

The Riemann surface is just a copy of $\mathbb C$. Compactifying it (by adding one point) gives therefore the usual sphere.

In general, if $f:\mathbb C\to U$ is an surjective holomorphic function with non-zero derivative with image an open subset of $\mathbb C$, the Riemann surface of the inverse function $f^{-1}:U\to\mathbb C$ (which is in general not-single valued, so I am abusing notation a bit here...) is just the graph of $f$, the set $\Gamma=\{(z,f(z)):z\in\mathbb C\}$. The restriction of the first projection $\mathbb C\times\mathbb C\to\mathbb C$ gives a holomorphic function $\Gamma\to\mathbb C$ which is a biholomorphism. Your question involves the case wher $f(z)=\exp(z)$.

The Riemann surface of $\log z$ is an infinite cylinder since $\log z = \log|z| + \mathrm{arg}(z) + 2 \pi i k$.

To compactify, we're going to add two points at $z = \pm i\infty$, closing the cylinder to make a sphere.


solve for y: $xy'=y^2-2y$ $$xy'=y^2-2y$$ $$y(1) = 1, x >= 0$$

I get $C = 0$ which gives:

$$\left| \frac{y}{y-2}\right|=x^2$$ When solving for $y$ it gives $$\frac{2x^2}{1+x^2}$$, however the correct answer should be $$\frac{2}{1+x^2}$$.

Is $$\left| \frac{y}{y-2}\right|=x^2$$ correct? And if it is, how to continue from there?

(2018-11-25) "Amusing" revenge downvote, four years later.

Instead, try $$\left| \frac{y-2}{y}\right|=x^2.$$


Need help with logical expression involving existential and universal quantifier P is the set of "all exams at university". E(x) denotes the predicate "x is easy". Translate the following sentence to formal logic. "Only one exam is easy." Solution:

I understand the first part, that there exists a test "x" for which the predicate "is easy" holds. But I don't understand the second part. I read it as such that for all y,z element of the set of exams, if y is easy and z is easy and there exist an exam x that is easy it follows that y = z.

The second part says that if you pick any two easy exams, then they will turn out to be one and the same exam. This will rule out that there are two (or more) different easy exams. Indeed, you could equivalently write that second part as:
$$\neg \exists y, z \in P: (E(y) \land E(z) \land y \neq z)$$
Finally, here's an efficient way to say that there is exactly one easy exam:
$$\exists x \in P: \forall y \in P: (E(y) \leftrightarrow x = y)$$

The appropriate translation "Only one exam in the university is easy" is
$$\exists x \in P\Big(E(x) \land \forall y\in P(E(y) \to (x=y))\Big).$$
The translation you posted says there does exist an easy exam, and for all exams, if they are both easy,  they are equal.


How to test if a polynomial GCD is correct I'm studying the Zippel algorithm for computing multivariate polynomial GCDs, which is probabilistic in the sense that it uses some random numbers, and assumes that if a polynomial coefficient is zero, then it is actually zero (as opposed to only zero at the randomly chosen values).
In [Zi79], the author makes the following claim:

Since the results for GCD and factorization can be checked by
division, one is guaranteed to obtain the correct answer, if need be,
by performing the calculation twice.

I can certainly see how we can check if a polynomial is a common divisor, but how can we be sure it is a greatest common division?
I just don't understand how "results for GCD... can be checked by division".
Can anybody help me?
[Zi79] Richard Zippel, PROBABILISTIC ALGORITHMS FOR SPARSE POLYNOMIALS, Laboratory for Computer Science, Massachusetts Institute of Technology

If $d$ divides both $a$ and $b$, then $d=\gcd(a.b)$ iff $\gcd ({a\over d}, {b\over d})=1.$


Derivatives and Integrals are both what of a function? This is more a question on semantics.  I'm writing a document and would like to describe a collection of functions that are both derivative and integral representations of a specific function.  Is there any term or phrase to describe this group?

If you want to use a precise term that is not too common, various derivatives (first, second, etc.) and anti-derivatives could all be called "differintegrals" of the original function.

Depending on whether it is obvious to your audience from the context that anti-derivatives are to be included (they're almost like differentiating a negative number of times), I'd probably just accept the imprecision and use "derivatives" in your situation.

I can't think of anything else short for this, even the vague "results after applying operators" is too long.

Upto a constant the integral and differential coefficients of $ c e ^{\pm x} $ form the same set of functions.

Second differential order invariant function..


Find limit when $\theta$ tends to $0$ of $\tan(\theta) /\theta$ I understand that $\lim_{\theta\to0}(\sin(θ)/θ) = 1$ but what is $x$ when,
$\lim_{\theta\to0}(\tan(θ)/θ) = x$ where $x$ is a real constant value.

Please help me, I will be eternally great full :D

We have that $\dfrac{\tan\theta}{\theta}=\dfrac{1}{\cos\theta}\cdot \dfrac {\sin\theta}{\theta}.$

Then recall that the limit of a product is equal to the product of the limits (when those limits do in fact exist.)

$$\lim_{\theta\to 0} \dfrac{\tan\theta}{\theta}= \lim_{\theta\to 0} \dfrac{\sin\theta}{(\cos\theta)\cdot \theta}=\lim_{\theta \to 0} \dfrac{1}{\cos\theta}\cdot \dfrac {\sin\theta}{\theta} = \lim_{\theta \to 0} \dfrac 1{\cos \theta} \cdot \lim_{\theta\to 0} \dfrac{\sin \theta}{\theta} = 1\cdot 1 = 1.$$

$x=1$

Let $\theta=y$. Then $$\tan y=\frac{\sin y}{\cos y}$$
and $\lim_{y \to 0} \frac{\sin y}{y}=1$ and $\cos 0=1$ so $\lim_{y\to 0}\frac{\tan y}{y}=0$.


Solve the equation $x(\log \log k - \log x) = \log k$ I want to solve this equation by expressing $x$ in function of $k$. Is it possible?

Thanks.

Yes, but you're gonna need the Lambert W function.

\begin{align}
x(\log \log k - \log x) &= \log k\\
\log x - \log \log k &= \frac{-\log k}x\\
\frac x{\log k} &= e^{-(\log k)/x}\\
-1 &= \frac{-\log k}{x}\cdot e^{-(\log k)/x}\\
W(-1) &= \frac{-\log k}{x}\\
x &= \frac{-\log k}{W(-1)}\\
\end{align}

Note that $W(-1)$ is a complex number.

differentiating both sides with respect to x and taking  $$loglogk = p$$
$$d(px)/dx - d(xlogx)/dx = 0$$
$$p - x*d(logx)/dx - logx = 0$$
$$p - 1 - logx = 0$$
$$logx = p - 1$$
$$x = e^p/e$$
$$x = (logk)/e$$


Finite group of order $mn$ with $m,n$ coprime I have a problem. It states that:

  Let $G$ is a group and $|G|=mn$, $(m,n)=1$. Assume that $G$ has exactly one subgroup $M$ with order $m$ and one subgroup $N$ with order $n$. Prove: $G$ is a direct product of $M$ and $N$.

Here is my approach:

Obviously, we have $M\cap N = 1$. By Product Formula we have: $|MN|=|M|\cdot|N|=mn$.

Let $m_1$, $m_2$ be in $M$ and $n_1$, $n_2$ in $N$. If  $m_1n_1=m_2n_2$ then $m_2^{-1}m_1=n_2n_1^{-1}$. By $M\cap N = 1$ we must have $m_1=m_2$, $n_1=n_2$. So those elements in the form $m_in_j$ with $m_i\in M$, $n_j\in N$ are distinct. Because $|G|=mn$ we must have $G=MN$.

I'm kind of confused because it seems the uniqueness of subgroups $M$ and $N$ is useless. Is my proof still correct without this hypothesis or I mislead at certain point? Please explain to me.

In general, $MN=\{mn\mid m\in M, n\in N\}$ is not a group, but in this case $MN$ is a group by considering its cardinality as comment said. But $G=MN$ does not implies $G\simeq M\times N$. To show that $G$ is isomorphic to $M\times N$, we have to show that elements of $M$ and $N$ commute, i.e. $mn=nm$ for all $m\in M, n\in N$. By the uniqueness condition we can show that $M, N$ are normal subgroups of $G$. Then $mnm^{-1}n^{-1}\in M\cap N$ (why?), so $mnm^{-1}n^{-1}=e$ and they commute. Then $(m,n)\to mn$ became an isomorphism between $M\times N$ and $MN=G$.

You say "obviously , we have $M\cap N=1$". This "obvious" is not obvious at all. At this point you dont use the hypothesis.

This is generaly how one detect mistakes in proofs : when the author write "obviously", or "it is easy to see that", there is a serious doubt.

To prove the "obvious", just says that $M\cap N$ hase order dividing the order of $M$ and $N$ (Lagrange), so  the order of this group is 1.

By the way, if $p$ is a prime number, let $G=S_p$ the symmetric group on $\{1,...p\}$ letters, $M=S_{p-1}$ the symmetric group on $\{1,...p-1\}$ letters and $H$ the subgroup of $G$ generated by a cyclic permutation $(1,2,...p)$. Then $G=MN$ but this is not a product group.


Maximum area of a rectangle whose vertices lie on ellipse $x^2+4y^2=1$ Maximum area of a rectangle whose vertices lie on ellipse $x^2+4y^2=1$.

I try to do it by lagrange multiplier as

$F(x,y,t)= xy + t(x^2+4y^2-1=0)=0$. Differentiating w.r.t to x,y and solving i get $x=\frac{1}{\sqrt2}$ and $y=\frac{1}{2\sqrt2}$. So area=0.25. But textbook states answer =1. I like to know where i am wrong?

Thanks

The area of the rectangle is not $xy$. If the rectangle has a vertex at some point $(x,y)$, then the area will be $4xy$. Hopefully the crude drawing below will help you understand why.

Note that, if you consider the rectangle whose bottom-left vertex is at the origin, then the sides have length $x$ and $y$, so the area would indeed be $xy$. This is the problem you solved, which gives $A=1/4$. Since this is a quarter of the total rectangle in question, we can multiply by $4$ to see that the answer is in fact $1$.

By symmetry, the rectangle with the largest area will be one with its sides parallel to the ellipse's axes. Consider any point B(x1,y1)B(x1,y1) on the ellipse located in the first quadrant.

You can easily see that A≡(−x1,y1)A≡(−x1,y1), D≡(x1,−y1)D≡(x1,−y1), and C≡(−x1,−y1)C≡(−x1,−y1).

So, Area=4x1y1Area=4x1y1

We also have the relation:

x21+4y21=1x12+4y12=1

⇒x21=1−4y21⇒x12=1−4y12

⇒x1=1−4y21−−−−−−√⇒x1=1−4y12

We've taken the positive value since we chose this point to be in the first quadrant.

Area=4y11−4y21−−−−−−√Area=4y11−4y12

The possible values of y1y1 for which there lies a point on the ellipse are [0,12][0,12] in the first quadrant. Let's differentiate the area to find its point of maxima.

dAdy=yddy1−4y2−−−−−−√+1−4y2−−−−−−√ddyydAdy=yddy1−4y2+1−4y2ddyy

dAdy=y121−4y2√(−8y)+1−4y2−−−−−−√dAdy=y121−4y2(−8y)+1−4y2

dAdy=−8y2+2−8y221−4y2√dAdy=−8y2+2−8y221−4y2

dAdy=2−16y21−4y2√dAdy=2−16y21−4y2

For maxima,

dAdy=0dAdy=0

⇒2−16y21−4y2√=0⇒2−16y21−4y2=0

⇒y=18√⇒y=18

Corresponding to this,

x=1−4y2−−−−−−√x=1−4y2

⇒x=1−12−−−−−√⇒x=1−12

⇒x=12√⇒x=12

Thus,

Areamax=412√18√Areamax=41218

Areamax=1sq.units


Can the golden ratio accurately be expressed in terms of $e$ and $\pi$ I was playing around with numbers when I noticed that $\sqrt e$ was very somewhat close to $\phi$
 And so, I took it upon myself to try to find a way to express the golden ratio in terms of the infamous values, $\large\pi$ and $\large e$
The closest that I've come so far is:
$$
\varphi \approx \sqrt e - \frac{\pi}{(e+\pi)^e - \sqrt e}
$$

My question is, Is there a better (more precise and accurate) way of expressing $\phi$ in terms of $e$ and $\pi$ ?

The following is exact. :-)

$$\phi=\frac{\frac{\pi}{\pi}+\sqrt{\frac{e+e+e+e+e}{e}}}{\frac{e}{e}+\frac{\pi}{\pi}}$$

An approximation with an accuracy similar to that of $\pi\approx3$ (error<5%) is given by the sixth root of Gelfond's constant,

$$e^{\frac{\pi}{6}}\approx \phi$$

with rational term series

$$e^{\frac{\pi}{6}}=\sum_{k=0}^{\infty}\frac{\left(e^{\frac{\pi}{2}} - (-1)^k e^{-\frac{\pi}{2}}\right)\Gamma\left(\frac{k+i}{2}\right)\Gamma\left(\frac{k-i}{2}\right)}{4\pi k!}$$

Rational series representation of $e^\pi$


Show that the unit circle centered at the origin is an attractor of the system $\dot{x}=-y +(1-x^2-y^2)$, $\dot{y}=x+(1-x^2-y^2)$ Given:

$\dot{x}=-y +(1-x^2-y^2)$

$\dot{y}=x+(1-x^2-y^2)$

Definition given:
A closed invariant set A ⊂ X is an attracting set under the flow if there
exists N(A) such that $\phi(x)_{t} ∈ N(A)$ for all $x ∈ N(A)$ and $t ≥ 0$, and

inf$_{y ∈ A} ∣ \phi_{t}(x) − y∣ → 0$ as $t → ∞, ∀x ∈ N(A)$.

If in addition, for some x ∈ A, the forward orbit O
+
(x) is a dense subset
of A, then we say that A is an attractor.

My Question:
Is there a way to use this definition of an attractor to show that the unit circle centered at the origin is an attractor for the above system? Or is my attempt sufficient?

My Attempt: Transformed to polar co-ordinates and found that:

$\dot{r}=r^2(1-r^2)$

$\dot{\theta}=1$

I then substituted $r=1+\delta$ (with $|\delta| <<1 $) into the equation for $\dot{r}$ to show that $\dot{r}<0$ for $\delta>0$ and $\dot{r}>0$ for $\delta<0$.

You might as well solve ( separation of variables) the diffeq in $r$ and see that $r(t) \to 1$ as $t \to \infty$ for any solution. That will guarantee that the circle is an attracting set.  Moreover, it's easy to see that if you start on the circle you get the solution covering the circle. Thus, all the checks are done.

If one sets

$R(r) = r^2 - r^4, \tag 1$

then

$\dot r = R(r) = r^2 - r^4, \tag 2$

and one sees that

$R(1) = 1^2 - 1^4 = 0 \tag 3$

and

$\displaystyle \dfrac{dR}{dr} = 2r - 4r^3; \tag 4$

thus

$\dfrac{dR(1)}{dr} = -2  < 0; \tag 5$

thus one sees that $r = 1$ is a hypberbolic equilibrium of the differential equation (2) and thus by (5), since $dR(1)/dr < 0$, it is attractive; thus $r(t) \to 1$  as $t \to \infty$ provided $r(t_0) \in (1 - \epsilon, 1 + \epsilon)$ for some $t_0 \in \Bbb R$ and $\epsilon$ sufficiently small; furthermore, since

$R(r) < 0 \; \text{for} \; r > 1, \tag 6$

and

$R(r) > 0 \; \text{for} \; r < 1, \tag 7$

we see that the convergence of $r(t) \to 1$ is strictly monotonic.  One also notes that

$\dot \theta = 1 \tag 8$

implies

$\theta(t) = t - t_0 + \theta(t_0); \tag 9$

then it is easily seen that the point $(r(t), \theta(t)) = (r(t), t - t_0 + \theta(t_0)) \to (1, t  - t_0 + \theta(t_0))$ as $t \to \infty$; indeed the Euclidean distance $d(t)$ 'twixt $ (r(t), t  - t_0 + \theta(t_0))$ and $(1, t - t_0 + \theta(t_0))$ is

$d(t) = \sqrt{(r(t) \cos \theta(t) - 1 \cdot \cos \theta(t))^2 + (r(t) \sin \theta(t) - 1 \cdot \sin \theta(t))^2}$
$= \sqrt{((r(t) - 1) \cos \theta(t))^2 + ((r(t) - 1) \sin \theta(t))^2} = \sqrt{(r(t) - 1)^2 (\cos^2 \theta(t) + \sin^2 \theta(t))}$
$= \sqrt{(r(t) -1)^2} = \vert r(t) - 1 \vert \to 0 \tag{10}$

as $t \to \infty$; it is a reasonably straightforward matter to see that such $d(t)$ is the minimum possible distance twixt $(r(t), \theta(t))$ and the circle $r = 1$, since it is their separation along a radial line.  Thus it may be concluded that the circle is an attracting set for the flow of the system (2), (8); and, as indicated by orangeskid in his answer, any trajectory originating in the circle $r = 1$ passes through every point in said circle; this is very easy to see, geometrically, since the point $(1, t  - t_0 + \theta(t_0))$ returns to the point $(1, \theta(t_0))$ for $t = t_0 + 2\pi$, covering the entire circle $r = 1$ in transit.

Thus one sees that the circle $r = 1$ is an attractor for the system (2), (8).


Expanding a generating function in a series For a given recurrence relation the generating function is A(x)=$\frac{x}{(1-x)(1-2x)}$.

Then the book says that if we want to find an explicit formula for the $a_n$'s we
would have to expand A(x) in a series. The partial fraction of A(x) is $$ x \{\frac{2}{1-2x}-\frac{1}{1-x}\}$$ and this is clear to me. But how did the author expand this
 $$ x \{\frac{2}{1-2x}-\frac{1}{1-x}\} = \{2x+2^2x^2+2^3x^3+...\}-\{x+x^2+x^3+..\} $$

Hope my question is clear. Thanks in advance for your help.

It only took me a few more minutes to get it. From the geometric series evaluation, we know that
$\sum_nx^n=\frac{1}{1-x}$ for $|x|=1$.

And know taking the summation of our partial fraction : $ \frac{1}{1-2x}-\frac{1}{1-x}$ we get

$$ \sum_n \frac{1}{1-2x}-\frac{1}{1-x}=\sum_n\frac{1}{1-2x}-\sum_n\frac{1}{1-x} = (2x)^n - x^n$$ and hence we have $$\{2x+2^2x^2+2^3x^3+...\}-\{x+x^2+x^3+...\}$$


A doubt about lower nil radical while proving 2-primality of ring.( Baer-McCoy Radical) I was proving that a reversible ring is 2-Primal for an exercise in T.Y Lam's book, but I got stuck. Here is where I'm stuck:
let $a$ be a nilpotent element of $R$ with $a^n=0$. Then using reversibility of ring he concluded $(RaR)^n=0$, but then he says this implies that
$RaR\subset \mathrm{Nil}_\ast(R)$ which clearly gives 2-primality as then $\mathrm{Nil}_\ast(R)=$ set of all nilpotent elements.

  But why is $RaR\subset \mathrm{Nil}_{*}(R)$?

$\mathrm{Nil}_\ast(R)$ is the set of elements $x$ s.t. every m-system containing $x$ intersects $0$. It is always contained in the set of all nilpotent elements of $R$. "2-primal" means that the nilpotent elements are all contained in $\mathrm{Nil}_\ast(R)$.

please suggest something..

also tell me if we have to calculate $\mathrm{Nil}_{*}(R)$ for some ring what should we aim at. we can't find all m-systems. we know it is a nil ideal so every element is nilpotent and it is contained in $rad(R)$ and intersection of all prime ideals but we also can't find all prime ideals. so what should we do to find $\mathrm{Nil}_{*}(R)$

I am guessing the intent is to prove that $RaR$ is nilpotent to show that it is contained in all prime ideals, and then use the characterization of $Nil_\ast(R)$ as the intersection of all prime ideals to conclude that $RaR\subseteq Nil_\ast(R)$.

If you haven't proven that equivalent definition of $Nil_\ast(R)$, it is highly recommended, and may even be carried out in the text already.

I'm also not totally sure why $(aRa)^n=\{0\}$ would actually help (is that really what's written?) Knowing that $aRaaRaaRaa\ldots =\{0\}$ does not seem to directly imply that $RaRaRa\ldots aR=\{0\}$.

But it is easy enough to show that $aRaRa\ldots aRa=\{0\}$ whenever there are $n$ or more $a$'s. Let $m\geq n$ and $r_i\in R$. Then

$$a^mr_1=0\implies a^{m-1}r_1a=0\\
a^{m-1}r_1ar_2=0\implies a^{m-2}r_1ar_2a=0\\
\ldots\\
a^2r_1ar_2a\ldots r_{m-1}=0\implies ar_1ar_2a\ldots r_{m-1}a=0$$

So $aRaR\ldots aRa=\{0\}$ whenever there are at least $n$ $a$'s, and that means $(RaR)^n=R(aRaRa\ldots aRa)R=\{0\}$. Thus $RaR$ is contained in every prime ideal, and therefore it is contained in $Nil_\ast(R)$.

Note: an element $a\in R$ is called strongly nilpotent if there exists a nonnegative integer $k$ such that $ar_1ar_2a\ldots ar_{k-1} a=0$ for all choices of $r_i\in R$. Just as the nilradical of a commutative ring is the set of all nilpotent elements, the lower nilradical $Nil_\ast(R)$ is the set of all strongly nilpotent elements.

Edit Poster has since changed $(aRa)^n$ to $(RaR)^n$. In that case, it looks like you might have overlooked the fact that for an ideal $A$ and prime ideal $P$, $A^n\subseteq P\implies A\subseteq P$. This is all you need to prove that if $RaR$ is nilpotent, it's contained in all prime ideals. It's analogous to the idea that nilpotent elements of commutative rings are contained in all prime ideals.

Consider the $n=1$ case with $aRa=0$. Then $(RaR)^2=RaR\cdot RaR = R\cdot a(R)a \cdot R=0$ since multiplication in the ring is closed.


Proving the inverse of a continuous function is also continuous Let $E, E'$ be metric spaces, $f: E\to E'$ a continuous function. Prove that if $E$ is compact and $f$ is bijective then $f^{-1}:E' \to E$ is continuous.

I know one way to prove it is by showing that if $S\subset E$ and $S$ is closed then $f(s)\subset E'$ is also closed where $s\in S$. Since $S$ is closed then $p_n \in S$ and $p_n \to p_0$ in $E$ then $p_o\in S$. Since $E$ is compact there is a convergent subsequence. How can I do this proof?

For proving it we use the if and only if condition that continuous function bring back closed sets to closed sets.

Let $F$ be a closed set in $E$.
Since $E$ is compact, $F$ is compact.
hence $f(F)$ is compact and hence closed.

Now $(f^{-1})^{-1}(F)=f(F)$ is closed.

Since the choice of $F$ was arbitrary, $(f^{-1})$ brings back closed sets to closed sets. hence $f^{-1}$ is continuous.

Theorem 1:  A mapping f of a metric space X into a metric space Y is continuous on X if and only if $f^{-1}(V)$ is open in X for every open set V in Y.

Theorem 2: $f$ is a continuous mapping of a compact metric space X into a metric space Y. Then $f(X)$ is compact.

Based on Theorem 1, it is suffice to show that $f(V)$ is an open set in Y for every open set V in X.

Fix such a set V. $V^c$ is closed in X, hence compact.  Theorem 2 tells us $f(V^{c})$ is a compact subset of Y and so is closed in Y. Since Y is one-to-one and onto, $f(V)$ is the complement of $f(V^{c})$. Hence $f(V)$ is open.


Show it is impossible using Chinese Remainder Theorem $n = pq$, where $p$, $q$ are primes.

Let $m$ be a universal exponent, i.e. $\,a^m \equiv 1 \pmod n\ $ for all $a$ coprime to $n$

Then there are four cases:
\begin{cases} a^{m/2} \equiv\ \ \ 1\ \pmod p\\
a^{m/2} \equiv\ \ \ 1\ \pmod q \end{cases}
\begin{cases} a^{m/2} \equiv\ \ \ 1 \pmod p\\
a^{m/2} \equiv -1 \pmod q \end{cases}
\begin{cases} a^{m/2} \equiv -1 \pmod p\\
a^{m/2} \equiv\ \ \ 1\ \pmod{\! q} \end{cases}
\begin{cases} a^{m/2} \equiv -1 \pmod p\\
a^{m/2} \equiv -1 \pmod q \end{cases}

My question is that, for all $a$ such that $\gcd(a,n)=1$, why can't half of '$a$'s satisfy the first case and half satisfy the last case.( i.2. show it impossible to only have case 1 and 4).

I think I can use Chinese Remainder Theorem to solve it.

It gives exactly half of '$a$'s satisfy $a^{m/2} \equiv -1 \pmod n$ and the rest satisfy $a^{m/2} \equiv 1 \pmod n$. But I still didn't get why it is impossible.

Assume $p$ and $q$ are odd, distinct primes (if one of the primes is $2$, then $1 \equiv -1 \pmod 2$, so the question becomes somewhat meaningless in that case). Your question can then be proven using contradiction.

Assume that only both case $1$ and case $4$ apply, i.e.,

\begin{cases} a^{m/2} \equiv\ \ \ 1\ \pmod p\\
a^{m/2} \equiv\ \ \ 1\ \pmod q \end{cases}
\begin{cases} a^{m/2} \equiv -1 \pmod p\\
a^{m/2} \equiv -1 \pmod q \end{cases}

Consider all $a_b = bp + 1$ for $0 \le b \le q - 1$. First, note $a_b^{m/2} \equiv 1 \pmod p$ for all $a_b$. Thus, by the assumption, $a_b^{m/2} \equiv 1 \pmod q$ for all these $a_b$ which are relatively prime to $q$. However, note that modulo $q$, the set of $a_b$ covers all values from $0$ to $q-1$, inclusive, because, if we have $b_1 p + 1 \equiv b_2 p + 1 \pmod q$, then $(b_1 - b_2)p \equiv 0 \pmod q \implies b_1 - b_2 \equiv 0 \pmod q \implies b_1 = b_2$, since $p$ and $q$ are distinct primes, for any $0 \le b_1, b_2 \le q - 1$. Thus, this means that $c^{m/2} \equiv 1 \pmod q$ for all $1 \le c \le q - 1$. As such, there are no values of any $a$ where $a^{m/2} \equiv -1 \pmod q$, unless $1 \equiv -1 \pmod q$, but this is only true if $q = 2$, which is not allowed as $q$ is an odd prime. This shows only case $1$ would apply and, thus, you can't have just both cases $1$ and $4$ apply.

As for using the Chinese Remainder Theorem to prove this, note this theorem is really only an existence theorem, so I'm not sure how you could use it. If you perhaps meant using something like the Extended Euclidean Algorithm, note that just by inspection you can see the question comes down to trying to prove $a^{m/2} = kn + 1$ for some $k \in \mathbb{Z}$ (i.e., $a^{m/2} \equiv 1 \pmod n$), or $a^{m/2} = kn - 1$ for some $k \in \mathbb{Z}$, (i.e., $a^{m/2} \equiv -1 \pmod n$), with at least one case of each, for all $a$ relatively prime to $p$ and $q$. If this is what you mean, you could use basically the same proof technique I used, although I believe the notation & handling would not be quite as convenient.


Do not understand what is meant by "in terms of n" I took calculus 5 years ago so I am unaware of what is meant by in terms of the variables in this summation example.
specifically these two questions. please help me understand what the steps are to solving these.
I was unable to google my malfunction.

What is the value of the following summation in terms of $n$: $\sum\limits_{=+1}^{3} (2−3)$?

Determine the value of this summation, in terms of $a$, $b$, $c$ and $d$. Assume that $a < b$ and that all four are positive integers. $\sum\limits_{=}^{b} (+)$.

When we say expression in terms of $n$, your final answer should be an expression involving $n$.
For your problem, I will help you with the first one but I will leave the second one to you.
\begin{align}\sum_{i=n+1}^{3n}(2i-3)&=2\sum_{i=n+1}^{3n}i - 3\sum_{i=n+1}^{3n}1 \\
&=2[(n+1) + (n+2) + \ldots + 3n]-3(2n)\\
&= 2 \cdot \frac{2n}{2}\left[(n+1)+3n \right]-6n\end{align}
Try to simplify the expression above.  As you can see, the final answer involves $n$.

To compute the second sum, you can write
$\sum_{i=a}^b (ci + d) = c(\sum_{i=1}^b i - \sum_{i=1}^{a-1} i) + d (b-a+1)$
and apply the formula for the sum of consecutive integers $\sum_{k=1}^n k = n(n+1)/2$.


Straight vs Partial derivative Does it make sense to write $\frac{d}{dx}u(x,t)$ or can one only write $\frac{\partial}{\partial x}u(x,t)$?

For convention's sake, I'll pretend you asked about $\frac{d}{dt}$ and $\frac{\partial}{\partial t}$. It's the same exact question this way, and you'll see the reason for the change in notation afterwards.

Getting to the point, say you have a function of two variables $$u:\Bbb R^2\to\Bbb R^k\atop (x,t)\mapsto u(x,t)$$

For example, let $k=1$ and we have a scalar field (which we can think of as dependent on "position" $x$ and time $t$, if we wish). Now consider a real function $x(t)$ (forgive me for using the same letter $x$ for a completely new term). This determines a path through $\Bbb R^2: \gamma (t) =(x(t),t)$. We can compose the two functions to get a real function of $t$: $u\circ\gamma$. Now, this is a sort of liberal use of notation, but we can "identify" $u$ with $u\circ\gamma$. Now the derivative of $u$ is $$\frac{d}{dt}u = \frac{d}{dt}u(x(t),t) = \frac{\partial u }{\partial x}x'(t) + \frac{\partial u}{\partial t}$$

You can see for yourself that both $\frac{\partial}{\partial t}$ and $\frac{d}{dt}$ are applied to $u$, to mean different things. Again, this is artificious notation (at least, I haven't seen it formalized), but it's usually just comfortable to write this way, so many people do.

Actually, the $\frac{dy}{dx}$ maybe considered as a quotient of two differentials, since $dy= y' \cdot dx$, whereas $\frac{\partial y}{\partial x}$ is an indivisible symbol.

According to this, it's convinient to write such type of equations $\frac{d(f(g(x))}{dx}= \frac{\partial{f}}{\partial{g}} \cdot \frac{\partial{g}}{\partial{x}}$


Simplify the following Boolean expression. Can someone explain to me how this step done? I got a different answer than what the solution said.

Simplify $x(y+z)(\bar{x} + y)(\bar{y} + x + z)$

what the solution got

$x(y+z)(\bar{x} + y)(\bar{y} + x + z)$ = $x(y + z\bar{x})(\bar{y} + x + z)$ (Using distrubitive)

What I got

$x(y+z)(\bar{x} + y)(\bar{y} + x + z)$ = $x(y\bar{x} + y + z\bar{x} + zy)(\bar{y} + x + z)$ (Using distrubitive)

$y\bar{x} + y + z\bar{x} + zy$

=  $y(\bar{x} + 1) + z\bar{x} + zy$

= $y + z\bar{x} + zy$

= $y(1 + z) + z\bar{x}$

= $y + z\bar{x}$

Direct rule -

X + YZ = (X+Y)(X+Z)

So we have -

$(y + z)(y + \bar x)$

= $(y + z \bar x)$

Hint : $ab + ac = a(b+c) = a$ and $a+a=a$.
Do you see the '$b$' and '$c$' (and also the '$a$') in your case?


How to prove that $a<S_n-[S_n]<b$ infinitely often Let $S_n=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$, where $n$ is a positive integer. Prove that for any real numbers $a,b,0\le a\le b\le 1$, there exist infinite many $n\in\mathbb{N}$ such that
  $$a<S_n-[S_n]<b$$
  where $[x]$ represents the largest integer not exceeding $x$.

This problem is from China 2012 China Second Round (High school math competition) competition last problem, I think this problem has more nice methods, maybe using analytic methods.

I think the direct approach works pretty well here.

Let $N > \max(a^{-1},( b - a ) ^ {-1})$, so that $S_{n+1} - S_{n} < ( b - a )$ and $S_{n+1} - S_{n} < a $ when $n > N$.  Now suppose there are only finitely many $n$ with the desired property, and increase $N$ so that it's larger than the greatest index for which the property holds.  Finally, let $n_{0} > N$ be smallest such that $a <S_{n_0 + 1} - \lfloor S_{n_0 + 1} \rfloor $.  Then $\lfloor S_{n_0} \rfloor = \lfloor S_{n_0 + 1} \rfloor $ and $a > S_{n_0} - \lfloor S_{n_0} \rfloor $ since $S_{n_0 + 1} - S_{n_0} = \frac{1}{n_0 + 1} < a$.  Then we also have $S_{n_0+1} - \lfloor S_{n_0 + 1} \rfloor < b$ since $$b > a + \frac{1}{n_0 + 1} > S_{n_0} - \lfloor S_{n_0 } \rfloor  + \frac{1}{n_0+1} = S_{n_0 + 1} - \lfloor S_{n_0 + 1} \rfloor $$  Which means we've found another index where the property holds, contradicting the assumption that $N$ is larger than the largest index for which the property holds.


Osculating circle given parametric coordinates in polar form I have the following parametric equation for a curve in polar form:
\begin{align}
r(t) &= r_0-kt\\
\phi(t)&=h\ln(r_0-k_0t)+a
\end{align}
where $(r_0,h,a,k_0,k)\in\mathbb{R}$
what is the radius of the osculating circle $R(t)$?
Thanks in advance.

Use Frenet- Serret relations with vectors $(N,T)$ in normal plane in the usual differential form after converting  from polar to rectangular coordinates.
rz = 0.75; kz = 0.4; k = 0.25; h = 0.2; a = 1.25;
rho[t_] = (rz - k t); th[t_] = h Log[rz - kz t] + a ;
ParametricPlot[rho[t] {Cos[th[t]], Sin[th[t]]}, {t, -5, 2},
 GridLines -> Automatic, AspectRatio -> 1.25]

In the above  Mathematica used function definition f[t_].

This is the Mathematica output, using the parameter names given.
Let me elaborate it a bit. I just started with $r'$ and $\phi'$, then rotated the vector $(r',r \phi')$ by an angle of $\phi$ to put them in XY coordinates. The norm of the derivative of the normalized resulting vector is the curvature, i.e., the inverse of the radius.


Must the complement of countably many disjoint arcs be connected? Must the complement of the countable union of disjoint arcs (in the plane) be connected?

A (Jordan) arc is the injective image of a continuous function from $[0,1]$ to $\Bbb R^2$. It is known that, if I have finitely many pairwise disjoint arcs, the complement of their union must be connected, and even path-connected. However, this no longer is true when we allow ourselves infinitely many arcs (take the union of line segments of the form $\{x\}\times[0,1]$ for $x\in\Bbb R$).

In fact, the "path-connected" part is not even true for countably many arcs!

  I know three counterexamples (special thanks to Balarka and Alessandro):

  Let $U_r$ be a closed circular arc of $270$ degrees, with the $90^\circ$ opening facing up, of radius $r$. Let $D_r$ be the same but with the opening facing down. Then the complement of:
  $$U_{1/2}\cup D_{2/3}\cup U_{3/4}\cup D_{4/5}\cup\dotsb$$
  is not path-connected.
  The complement of:
  $$\bigcup_{n=1}^\infty(\{1/n\}\times[-n,n])$$
  is not path-connected. This gives an example made of only straight lines, though it's unbounded.
  Let $S_1=[-1,1]\times\{1\}$ and $S_{-1}=[-1,1]\times\{-1\}$ be the top and bottom sides of a square. Let $T_{-1,n}=\{-1-\frac1n\}\times[-1,1]$ and $T_{1,n}=\{1+\frac1n\}\times[-1,1]$ be sequences of segments approaching the left and right sides of the same square. Then the complement of:
  $$S_{-1}\cup S_1\cup\bigcup_{n=1}^\infty T_{-1,n}\cup\bigcup_{n=1}^\infty T_{1,n}$$
  is not path-connected. This gives an example that is bounded, is made of straight lines, and has all of its lines the same length.

However, while these complements are all not path-connected, they all are connected. Hence the question: Must the complement of countably many disjoint arcs be connected?

I've thought about this all day and I haven't been able to solve it. Homology seems useless, since it can only measure path components, not connected components. On the other hand, if we replace $\Bbb R^2$ with other spaces, it can become false (take for example the plane minus open disks of radius $\frac13$ centered at the integers on the $x$-axis), so the proof must use some special property of $\Bbb R^2$.

Connectedness should follow from Sitnikov Duality theorem if you consider subsets $A$ (union of arcs) of $S^2$ rather than $R^2$:
$$
0=H^c_1(A)\cong \check{H}^0(S^2 -A).
$$
It follows that $S^2-A$ is connected. I am still thinking about subsets of $R^2$.

I think if we take $\gamma_{n}:[-1,1]\rightarrow\mathbb{R}^2,\ \gamma_{n}(t)=(1/n,nt)$ we have the desired set.


How many significant figures are needed in base 2? $x \in \mathbb{R}$

$2^{500}<x<2^{501} $

How many significant figures are needed in base 2, to know in high approximation whether $2^x$ is integer?

Basically you want to estimate the $\delta$ such that $2^{x+\delta} - 2^x = 1$. This means $2^\delta = 1+\dfrac{1}{2^x}$, so that $\delta$ might be estimated as $\dfrac{1}{2^x \times \ln2}$, or, taking the upper bound for $x$, $\delta$ might be estimated as $\dfrac{1}{2^{501} \times \ln2}$.

The number of required significant digits (after the decimal point) is about $-\log_{10}\delta = \log_{10}(2^{501}) + \log_{10}\ln2$, which is about 151, plus-minus a digit. Or, if you're working in base 2, the number of required significant digits is about $-\log_{2}\delta = \log_{2}(2^{501}) + \log_{2}\ln2$, which is about 502.

Given such a number of digits after the decimal point, changing the remaining digits won't change $2^x$ by more than $1$, so that, if $2^x$ is an integer, you can say what integer it is.

However, it is impossible to tell for sure whether $2^x$ is an integer, given only its rounded value, independent of the accuracy, as adding a small value beyond the accuracy limits to $x$ will turn $2^x$ from integer to non-integer and vice versa.

For $2^x$ to be an integer where x has a finite decimal representation, x must also be an integer. Am I missing something here?


Combinatorial proof of the identity: $$\sum_{k=1}^n {(k^3)} = \left(\sum_{k=1}^n {k}\right)^2 $$

How could I approach this problem to make a combinatorial proof?

Try induction.  Suppose the statement holds for $n-1$.  To prove it for $n$, write:
$$
\left( \sum_{k=1}^{n} k \right)^2 = \left( n + \sum_{k=1}^{n-1} k \right)^2 = n^2 + 2 n \sum_{k=1}^{n-1} k + \left( \sum_{k=1}^{n-1} k \right)^2
=
n^2 + 2 n \sum_{k=1}^{n-1} k +
\sum_{k=1}^{n-1} k^3.
$$
Now use
$$
\sum_{k=1}^{n-1} k = {(n-1) \, n \over 2}.
$$


Books on Number Theory for Layman Books on Number Theory for anyone who loves Mathematics?(Beginner to Advanced & just for someone who has a basic grasp of math)

A Classical Introduction to Modern Number Theory by Ireland and Rosen  hands down!

In my opinion, "the theory of numbers" by Neal H. Mccoy contains all number theory knowledge that a common person should have.


Compute Gradient from Jacobian I have some trouble understanding a formula from a report :
https://www.samba.org/tridge/UAV/madgwick_internal_report.pdf

It is formula (20) (Page 7). Could you tell me where it comes from?
I can't find anything that resembles in litterature...

Here is an image of the formula:
http://www.les-mathematiques.net/phorum/addon.php?4,module=embed_images,url=http%3A%2F%2Fs21.postimg.org%2Ft89ej68k7%2Fpourlesmaths.png

The author claims that "Equation (20) computes the gradient of the solution
surface defined by the objective function and its Jacobian"and I don't even understand what he means by gradient since f is a function that goes from R^4 into R^3.

Thanks in advance for your answer

I was stuck on this also, but I think I figured this out and I hope I can
explain clearly without being able to type in math (I don't know how).

In equation 15 he defines what he calls his objective function f,
which is a 3-dimensional function of 4 variables; i.e. f is the column vector

f = [fx(q1,q2,q3,q4), fy(q1,q2,q3,q4), fz(q1,q2,q3,q4)]'  (' means transpose)

However, I think this is not really the objective function he's using;
rather he is using the real-valued function

F(q1,q2,q3,q4) = ||f||^2 = fx^2 + fy^2 +fz^2

The gradient of F, grad-F, is

grad-F = [dF/dq1, dF/dq2, dF/dq3, dF/dq4]'

(the "d's" in the above should be script d's to mean partial derivatives,
but I don't know how to type those here...).

So

grad-F = [ 2fx dfx/dq1 + 2fy dfy/dq1 + 2fz dfz/dq1
           2fx dfx/dq2 + 2fy dfy/dq2 + 2fz dfz/dq2
           (similar for /dq3)
           (similar for /dq4) ]

This can be written as

grad-F = [ dfx/dq1  dfy/dq1  dfz/dq1   * [ fx
           dfx/dq2  dfy/dq2  dfz/dq2       fy
           dfx/dq3 ...                     fz ]
           ... ]

The left-hand 4x3 matrix in the above is just the transpose of the
3x4 Jacobian, so can write this as:

grad-F = J' * f

Compare with Madgwick's eq. 20. I think you'll agree this is what
he was getting at.

yeah, got the same question :), this is what i think:

we use gradient descent algorithm.
 A) If your target function is a single function (whatever variables
    number is) it is simple case described in all literature - you compute
    gradient without this multiplication from equation (20).

B) If your target function is composed of more functions - it is a vector of function - like in this algorithm by Madgwick :

d - s = [dx dy dx] - [sx sy sz]  -> [dx-sx   dy-sy   dz-yz]  [f1  f2  f3]

In this case you also compute gradient for each function, wchich forms a vector, which is a Jacobian matrix.

but how you want to optimize all functions at once???  when you optimzie your step for  (dx-sx), you could ruin value of other function e.g. (dz-sz) - (not sure if this is true :)

How can you solve this?   you need to multiply each target function component by its value, so the step for those which are already optimized (d-s ~= 0)(x,y or z) would have weight around 0.

Thats mean this multiplication is necessary for this algorithm to produce some solution, but it is actually synthetic, artifical. I feel like this is synthetic.

I didnt found any serious literature that states that ,,if you use vector target function you need to multiplu the gradient vector (jacobian matrix) by target function values vector'', so i cant provide proof for my elaborate but this is how i understand this. I suppose it is preety much fair.

Maybe this is so obvious nobody mention this. It is popular approach.

regards :)


Using digits {1, 3, 4, 6, 7}, create the largest five-digit even number that is divisible by 3. To clarify, no digits can repeat either. Can someone outline their thought process to solve? Here's mine:

I used a process of elimination to determine what numbers I could use in the ten-thousands, thousands, and hundreds place. This gave me $7 6 4$. Is this thought process safe so far?

Where I think I began to mess up(if not earlier). I was left with $3$ and $1$. Choosing between $7 6 4 1 3$ and $7 6 4 3 1$, I made the unsafe assumption that division by $3$ would not be possible with a five-digit number ending with a $1$. Little did I know that $76431 /3  = 25477$. Aside from getting to this point and manually checking what's possible, can anyone explain a better approach?

The number we seek to obtain must follow 2 conditions:
 a-The number must be divisible by 3
 b-The number must be even

You should not worry about condition (a) since a number is divisible by 3 if and only if the sum of the digits is divisible by 3, which is already obtained (1 + 3 + 4 + 6 + 7 = 21 = 3 * 7). So this condition will always be satisfied, whatever permutation you decide on.
Condition (b) implies that your number must end with an even digit(Either 4 or 6). To obtain the largest number possible, you should place the smallest of these integers in the least significant positions, namely the ones.
Also, intuitively, you must place the largest digits in the most significant places.
So your answer would be 76314.

The sum of the digits, 21,
is divisible by 3,
so any number formed
from these digits is
divisible by 3.

So just take the digits
in descending order,
ensuring that the last digit is even
(which I carelessly overlooked):
76314.

As a check,
76431/3
=25477.


The order of $\sqrt{\epsilon(1-\epsilon)}$ and $4\pi^2\epsilon$ as $\epsilon \rightarrow 0$? I was reading on the big O/little O notation etc. and I understand the definitions, but how exactly would I use it to find the order of an expression/function?

I am asked to determine the order of $\sqrt{\epsilon(1-\epsilon)}$ and $4\pi^2\epsilon$ as $\epsilon \rightarrow 0$. But how would I do that?

For $\sqrt{\epsilon(1-\epsilon)}$,
note that $(1-\epsilon)$ isn't small at all
as $\epsilon \to 0$,
so all that matters is the $\epsilon$.

For $4\pi^2\epsilon$,
the $4 \pi^2$
is just a constant,
so it does not affect the
order of the zeroness.

Big-Oh means that, simply speaking, asymptotically, $ \forall \epsilon <  \epsilon' \ \exists \ C>0$ s.t. the ratio of the functions $f,g$ are upper-bounded by $C$. In such case we write $f=O(g)$. In your case:

$$
\lim_{\epsilon \to 0}\frac{\sqrt{\epsilon(1-\epsilon)}}{4 \pi^2 \epsilon} = \frac{1}{4 \pi^2} \lim_{\epsilon \to 0}\sqrt{\frac{\epsilon(1-\epsilon)}{\epsilon^2}} = \infty
$$
this means that there are no such $ \epsilon', \ C$ that $ \forall \epsilon <\epsilon' \ f(\epsilon) \leq C g(\epsilon)$. Since the limit of the ratio is $\infty$, we write $f=\omega(g)$ or equivalently $g=o(f)$.


The Locus of Q consider point A and B Consider two fixed points A(0,-2) and B(0,4) on a rectangular coordinate plane.

A moving point Q such that QA is always perpendicular to QB.
I want to know how would the graph be like and the equation of the Locus of Q?

Thank you!

the locus of $Q$ is a circle whose diameter is $AB$you derive the equation if you take $Q=(x,y)$
(1) find the slopes of $QA$ and $QB$ by using the formula $rise/run.$

(2) you also need to use that the product of slopes of two perpendicular lines is negative one.

see if you can do these two steps.

your first comment is right. in the second one you forgot to cross multiply the $x^2$  and $-1$

$$-1 = {(y-(-2) \over (x - 0)} {(y - 4) \over( x - 0)} = {y^2 - 2y-8 \over x^2}$$ which simplifies to $$x^2 + y^2 - 2y + 8 = 0 \text { or  } x^2 + (y-1)^2 = 9.$$ which is a circle centered at (0,1) and radius 3 units.

If u will plot the points A and B on the cartesian plane and try to locate Q such that QA is perpendicular to QB, u can see that it's a circle.
 Its obvious because angle in a semicircle is always a right angle and Q will lie on a circle with AB as its one of the diameters. The centre will be the midpoint of A and B that is
C(0, 1) and radius will be AC=BC=3 units.
So concluding
The locus is a circle centered at (0, 1) with radius 3.


On evaluation of $\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$ Evaluate the following limit:$$\lim_{h\to0} \sqrt[h^2]\frac{{1+h}}{e^h}$$

Answer is $\frac{1}{\sqrt{e}}$ as on Wolfram Alpha query and evaluated by sympy (Python). Also here is the math search engine query which returned nothing substantial.

I think the answer is $1$. Why?
If $f(x)\to1$ and $g(x)\to\infty$ then the limit ${f(x)}^{g(x)}=e^{({f(x)-1})\cdot g(x)}$. For those who are curious of the proof:

 $$\lim_{x\to a} [1+(f(x)-1)]^{\frac{1}{f(x)-1} (f(x)-1)g(x)} = \lim_{x\to a} [(1+(f(x)-1))^{\frac{1}{f(x)-1}}]^{\lim_{x\to a} (f(x)-1)g(x)}$$

and then a very popular limit to sum it up $\lim_{x\to0} (1+x)^{\frac{1}{x}}=e$

Applying the limit on the numerator, we get $e^{\frac{1}{h}}$ which is exactly equal to the denominator. So hence, the answer $1$. Only my close friend desmos supports me here:

Whereas Wolfram shows

$$\text{My Questions}$$
$1$. Why does the method I adopted shows a speculatively wrong result?
$2$. Is there any way to correct it?

Source code for sympy evaluation.
#Python 3.8
import math
from sympy import *

x = symbols('x')
expr = ((1+x)**(1/x**2))/(math.e)**(1/x);

print("Expression : {}".format(expr))

# Use sympy.limit() method
limit_expr = limit(expr, x, 0)

print("Limit of the expression tends to 0 : {}".format(limit_expr))

Run the code:
Expression : 2.71828182845905**(-1/x)*(x + 1)**(x**(-2))
Limit of the expression tends to 0 : 0.606530659712633

For your first question see Daniel's comment. To get W|A's answer we just need L'Hopital:$$\lim_{h\to0}\left(\frac{1+h}{e^h}\right)^{1/h^2}=\exp\lim_{h\to0}\frac{\log((1+h)/e^h)}{h^2}=\exp\lim_{h\to0}\frac{\frac{e^h}{1+h}\cdot\frac{-h}{e^h}}{2h}=\frac1{\sqrt e}.$$

I assume that you are familiar with L'Hospital's Rule for evaluating limits.
Applying the formula which you mentioned in the question,
$$lim_{h\to0}\frac{e^{-h}(1+h) - 1}{h^2} = \frac{-1}{2}$$
by the application of L'Hospital's Rule.
So the final limit is $\frac{1}{\sqrt(e)}$


Find a real matrix $B$ such that $B^3 = A$ Given $$A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}$$ find a real, invertible matrix $B$ such that $B^3 = A$

I think I am doing something wrong here, so let me describe my attempt:

1) So I started off with diagonalizing the matrix $A$ with finding the eigenvalues $\lambda_1 = -8$ and $\lambda_2 = 1$ and the corresponding eigenvectors $ \vec v_1 = \begin{bmatrix}1 & 1\\0 & 0\end{bmatrix} = x + y = 0 \Rightarrow -x = y \Rightarrow \begin{bmatrix}1\\-1\end{bmatrix}$ and $ \vec v_2 = \begin{bmatrix}1 & -\frac{1}{2}\\0 & 0\end{bmatrix} = x - \frac{1}{2}y = 0 \Rightarrow 2x = y \Rightarrow \begin{bmatrix}1\\2\end{bmatrix}$

2) With that being done I proceeded with computing $D = \begin{bmatrix}-8 & 0\\0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}$ and check everything with $D = PAP^{-1}$

3) Now I thought I will simple find a diagonal matrix $M = PBP^{-1}$ and $M^3 = D$ and the easiest solution I came up with was $M = \begin{bmatrix}\sqrt[3]{-8} & 0\\0& 1\end{bmatrix}$ so basically $M = D^{\frac{1}{3}}.$ So that $B = PMP^{-1}$. But now come the tricky part, if I compute $B$ it results in a complex matrix not a real. //It is real!

Have I perhaps overlooked something here or miscalculated the solution for $B$?

Edit:
As Cameron pointed out my calculator and I totally failed as it was in complex mode and computed one of the non-real cube roots instead of -2. So $M = \begin{bmatrix}-2 & 0\\0 & 1\end{bmatrix}$ and consequentially $B = \begin{bmatrix}-1 & 1\\2 & 0\end{bmatrix}$

$$A=\begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}=PDP^{-1}$$

Where $$P=\begin{bmatrix}1& 1\\-1 & 2\end{bmatrix}$$is the matrix of eigenvectors

and $$D=\begin{bmatrix}-8& 0\\0 & 1\end{bmatrix}$$ is the matrix of eigenvalues.

Thus $$ B = PD^{1/3}P^{-1} = \begin{bmatrix}-1& 1\\2 & 0\end{bmatrix}$$


What is the simplest proof of the pythagorean theorem you know? Maybe enough so to explain it to children.

I wouldn't call  this a proof, but it's a convincing argument good demonstration of what the theorem means, and I thought it was pretty cool. Hopefully should help with explaining it to kids, as you asked.

What I will say will be the worst in terms of beauty but you need to ask yourself what is a square triangle. How is it defined ?

Well, the answer is that 2 vectors are square when their scalar product is 0, which leads (after 1 or 2 lines) to the Pythagore Theorem

Edit :

Vectors are A and B ; C = A - B to "complete the triangle"
In the general case, ||C||2 = (A-B).(A-B) = ||A||2 + ||B||2 - 2 A.B
We see that ||C||2 = ||A||2 + ||B||2 if and only if A.B == 0 (i.e are square by definition of being square)


Why is according to the law of sines $sinA=\frac{a}{2R}$ question
Why is according to the law of sines $sinA=\frac{a}{2R}$

...
i understand the law of sines but I do not understand how this is possible, i need an explanation of where the radius came from

reminder
R is the radius of a circumcircle

You just need to see the proof of law of sines. See this Or you just take a look at this picture:

The law of sines says that ($d$ - diameter of circumcircle)
$$\frac a {\sin A} =\frac b {\sin B} =\frac c {\sin B} = d $$
$$\Rightarrow \frac {\sin A} {a} = \frac 1 d $$
$$\Leftrightarrow \sin A = \frac a d = \frac a {2r} $$


Computing the maximum of a (Classic) function involving factorial I'm trying to compute the global maximum (or an upper bound that does not depends on x) for the expression
$$1/x!\binom{x+n-1}{x},$$
as a function of x. Where $n$ in a positive integer parameter.
An equivalent problem would be finding the maximum of
$$\frac{\Gamma(x+n-1)}{\Gamma(x)^2}$$
If it helps, we can assume that x is a natural number and bound the above only for such.
The best case scenario is if someone can point out a known closed form for the maximum.
Other then that, a reasonable tight bound that does not not depends on x would also help.
Thanks.

Looking at f(x+1)/f(x) solves it as one of the comments suggests.


If $\lim_{x\to a} f(x) = L$, is the value of $L$ a function of $a$? The $\varepsilon$-$\delta$ definition is: for a function $f: x \to f(x)$, the limit as $x$ tends to $a$ equals to $L$ as long as for every positve number $\varepsilon$, there exists a positive number $\delta$, such that if the distance between $x$ and $a$ (greater than 0) is less than $\delta$, then the distance between $f(x)$ and $L$ is less than $\varepsilon$.
To better illustrate what I mean, I've chosen this image below, where $x_{0}$ represents $a$. As you can see, it seems as if $L$ is a function of $a$

Epsilon-Delta Definition of a Limit. (n.d.). Brilliant. Retrieved April 23, 2021, from https://brilliant.org/wiki/epsilon-delta-definition-of-a-limit/

we're approaching a

This notion of 'approaching' is only an informal way to understand. In the precise definition, it is not there. See wiki

Note that $x$ = {$(a- \delta, a + \delta) - a$}. That is, $x$ $\neq$ $a$ because we get 0 $<$ 0 $<$ $\delta$ , which is false, and $f(x)$ may be equal to $L$, according to the definition.

This part is kinda sketch, it is true that $x \neq a$ because we want $0< |x-a|<\delta$;notice the strict inequality.

Why? As far as I have understood, L is a function of a, we're approaching a and, as a consequence, we approach L.

I'd say the way you are thinking about it is brilliant here and has made me post my own question (see here). However, as far as 'standard' basic concepts go, the limit is informally dependent on the neighbourhood of 'a' rather than of that point 'a' itself.
I wouldn't personally call it a 'function' because it is non standard, a better way to say it would be to say that the limit is dependent of neighbourhood around 'a' rather than 'a' itself.

I'm asking this question because, from what've read, $f(x)$ can be equal to $L$. See Limits: why $f(x)$ can be equal to $L$ and $x$ can't be equal to $c$. Also, the book I'm working on also says $f(x)$ = {$(L - \varepsilon, L + \varepsilon)$}

Here it depends on your definition of $f(x)$ and it's kind of a subtle point. I had a similar doubt a long while back, have a look in this post.

One way of thinking of this is
$$\lim_{x\to a}f(x)=f(a)=L$$
So you can think of $L=f(a)$, indeed a function of $a$.


Distribution of $28$ objects into 4 heaps In how many ways can $28$ different things can be formed into $4$ heaps so that each may contain $7$ things?

Could someone give me slight hint for this particular question.

I'm assuming that the heaps are not labeled.

Object nr.  $28$ can be associated with $6$ other objects in ${27\choose 6}$ ways. The highest ranking left over object can be associated with $6$ other left over objects in ${20\choose6}$ ways, and finally the now highest ranking left over object can be associated with $6$ other left over objects in ${13\choose 6}$ ways. The last heap is what's left over now. The total number of ways therefore comes to
$$N={27\choose 6}\cdot{20\choose6}\cdot{13\choose 6}=19\,688\,264\,481\,600\ .$$

I would say, count of ways = $\binom{28}{7}\cdot\binom{21}{7}\cdot\binom{14}{7}\cdot\binom{7}{7}$


Integral solutions of linear equation How does one find the number of integral solutions of equations like,

$2x + 5y + 4z = 20$

I know how to do this counting if the coefficient of every variable on the LHS is $1$.
But I don't know how to do this counting with arbitrary coefficients.

The linear equation $2x + 5y + 4z = 20$ has a two-parameter family of infinite solutions in integers. Note that $y$ has to be even. Further any even $y$ will give a one parameter family of infinite solutions for the equation $2x + 4z = 20 - 5y$.

Let $y = 2y_1$. Then we need to solve $2x + 4z = 20 - 10y_1 \implies x + 2z = 10 -5y_1$.

Hence, the set of all integer solutions is the two parameter family $$(10-5y_1-2z_1,2y_1,z_1)$$ where $y_1,z_1 \in \mathbb{Z}$

If you are interested in non-negative integral solutions for $(x,y,z)$, then we need $y_1 \geq 0$,$z_1 \geq 0$ and $5y_1 + 2z_1 \leq 10$.

This implies $y_1 \in \{0,1,2\}$.

If $y_1 = 0$, then $z_1 = \{0,1,2,3,4,5\}$.

If $y_1 = 1$, then $z_1 = \{0,1,2\}$.

If $y_1 = 2$, then $z_1 = \{0\}$.

Hence, the set of all non-negative solutions are
$$\{ (10,0,0),(8,0,1),(6,0,2),(4,0,3),(2,0,4),(0,0,5),(5,2,0),(3,2,1),(1,2,2),(0,4,0) \}$$

The set of all positive solutions are
$$\{(3,2,1),(1,2,2)\}$$

Let $x' = 2x$, $y' = 5y$ and $z' = 4z$. Then $2 \leq x' \leq 18$, $5 \leq y' \leq 15$ and $4 \leq z' \leq 16$ for $x',y',z' \in \mathbb{Z}^{+}$. So you can apply the same techniques for counting the number of integral solutions if all the coefficients on the LHS are $1$.

So the number of solutions would be the coefficient of $x^{20}$ in $$\left[(x^2)^{a_1}+(x^2)^{a_1+1}+ \cdots+ (x^a)^{a_2} \right] \cdot \left[(x^5)^{b_1}+(x^5)^{b_1+1}+ \cdots+ (x^5)^{b_2}\right] \cdot \left[(x^4)^{c_1}+(x^4)^{c_1+1}+ \cdots+ (x^4)^{c_2} \right]$$

where $a_1 \leq x \leq a_2$, $b_1 \leq y \leq b_2$ and $c_1 \leq z \leq c_2$.


On the number of integers with an even number of distinct prime factors Let $A(n)$ be the number of integers from 2 to $n$ with an even number of distinct prime factors, and let $B(n)$ be the number of integers  from 2 to $n$ with an odd number of distinct prime factors. (By "distinct" I mean, for example, that $2^{100}$ has one distinct prime factor.)

A little python program shows that as $n$ ranges from 2 to 1000000, there are only 9437 values of $n$ for which $A(n)>B(n)$, whereas there are  990450 values of $n$ for which $A(n)<B(n)$. There are 112 values of $n$ for which $A(n)=B(n)$, the largest of which is 12099.

Is there some precise theorem or conjecture or even plausible heuristic to the effect that $A(n)<B(n)$ "most of the time", or perhaps "eventually"? Or is this just an illusion that dissolves as $n$ grows larger?

The Generalized Prime Number Theorem gives a decent estimate of the number of numbers with k factors less than n. You can use this to see that the number of numbers with 2,4,6,... prime factors should not in the large differ from the number of numbers with 1,3,5,... factors.

The Generalized PNT states:

$$\pi_k(x) \sim \frac{x}{\log x}\frac{(\log\log x)^{k-1}}{(k-1)!} $$

in which $\pi_k(x) $ is the number of numbers with k factors less than or equal to x.

See also problem 168307 (Generalized PNT in limit as numbers get large) which shows that in a sense even numbers have more prime factors than odd ones, and cites a result of Ramanujan that the "normal order" of prime factors of a number n, with or without repetitions, is about $\log\log n.$

For the last see Ramanujan, Collected Works, p. 274. For the Generalized PNT see, for example, G.J.O. Jameson, The Prime Number Theorem, p.145.

Edit in response to comment: This is similar to Chebyshev's bias and it could well be that generalized primes have similar properties. I don't think the proposition in the OP has been proven either way and even the proof of Chebyshev's bias is conditional.

The analysis using Landau's result was interesting but the error in Landau's result is $\sim O(\frac{1}{\log(\log(n)})$ and hence greater than the error derived $\sim (\frac{1}{\ln(n)^2})$, so unless we prove a tighter bound for k-almost primes than Landau did the rest of the argument won't apply.


Show that $\frac{1}{n^{\frac{n+1}{n}}}$ decreases How can one show that $\frac{1}{n^{\frac{n+1}{n}}}$ decreases?  I am using this to show that $\Sigma\frac{1}{n^{\frac{n+1}{n}}} $ diverges using Cauchy's condensation test.

If you know it decreases, then
$$ \varphi(n) \text{ is decreasing } \iff \varphi_{n+1} \leq \varphi_n$$

that is,

$$ (n+1)^{\frac{n+2}{n+1}} \geq n^{\frac{1+n}{n}}$$
can you take it from here? (hint: $\log\varphi_{n+1}\geq \log\varphi_n$)

If we look at the exponent in the expression: $\frac{n+1}{n}$, this simplifies to $1 + \frac{1}{n}$.

We clearly see that $\lim_{n\to\infty} \frac{1}{n} = 0$, which means the exponent is just $1 + 0 = 1$.

Then $\frac{1}{n^{\frac{n+1}{n}}}$ becomes $\frac{1}{n^{1}} = \frac{1}{n}$. But we already know what happens with $\frac{1}{n}$ when $n\to\infty$.

Thus giving you: $$\lim_{n\to\infty} \frac{1}{n^{\frac{n+1}{n}}} = 0$$

As the expression approaches $0$ when $n\to\infty$, it obviously decreases. Then use that fact in Cauchy's condensation test.


Completing the square to decompose quadratic forms in three variables Given the quadratic form $Q(\textbf{x}) = x^2 + 2xy - 4xz +2yz -4z^2$, the question asks to decompose $Q$ into sums of squares, first by eliminating terms in $z$, then terms in $y$, and then terms in $x$.

I know the signature of $Q$ is $(2,1)$ from the solution, but I'm at a loss for how to actually perform what the question is asking. For instance, I recognize that I can take the terms in $x$ and $z$ and form $(x-2z)^2$, but I don't really see what to do next. Any suggestions?

a method. Usually we take the matrix $H$ to be the Hessian matrix of second partial derivatives. In the special case of integer coefficients with all the mixed coefficent terms even, we can get $H$ all integers by taking half the Hessian matrix.

Outcome this time:

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
 - 1 & 1 & 0 \\
 - 1 & 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
1 & 0 & 1 \\
 - 2 & 1 &  - 4 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &  - 1 &  - 1 \\
0 & 1 & 3 \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 &  - 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
1 & 1 & 0 \\
 - 2 &  - 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 &  - 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
0 & 1 &  - 3 \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
1 & 0 & 1 \\
 - 2 & 1 &  - 4 \\
\end{array}
\right)
$$

It is the diagonal elements of $D$ and the rows of $Q$ that give the expression,
$$ \color{red}{ (x+y-2z)^2 - (y-3z)^2 + z^2 } $$

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
 https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
1 & 0 & 1 \\
 - 2 & 1 &  - 4 \\
\end{array}
\right)
$$
$$  D_0 = H  $$
$$ E_j^T D_{j-1} E_j = D_j  $$
$$  P_{j-1} E_j = P_j  $$
$$ E_j^{-1} Q_{j-1}  = Q_j  $$
$$  P_j Q_j = Q_j P_j = I  $$
$$ P_j^T H P_j = D_j  $$
$$ Q_j^T D_j Q_j = H  $$

$$ H = \left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
1 & 0 & 1 \\
 - 2 & 1 &  - 4 \\
\end{array}
\right)
$$

==============================================

$$ E_{1} = \left(
\begin{array}{rrr}
1 &  - 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$  P_{1} = \left(
\begin{array}{rrr}
1 &  - 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rrr}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rrr}
1 & 0 &  - 2 \\
0 &  - 1 & 3 \\
 - 2 & 3 &  - 4 \\
\end{array}
\right)
$$

==============================================

$$ E_{2} = \left(
\begin{array}{rrr}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$  P_{2} = \left(
\begin{array}{rrr}
1 &  - 1 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{2} = \left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{2} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 &  - 1 & 3 \\
0 & 3 &  - 8 \\
\end{array}
\right)
$$

==============================================

$$ E_{3} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 3 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$  P_{3} = \left(
\begin{array}{rrr}
1 &  - 1 &  - 1 \\
0 & 1 & 3 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{3} = \left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
0 & 1 &  - 3 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{3} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 &  - 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$

==============================================

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
 - 1 & 1 & 0 \\
 - 1 & 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
1 & 0 & 1 \\
 - 2 & 1 &  - 4 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &  - 1 &  - 1 \\
0 & 1 & 3 \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 &  - 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
1 & 1 & 0 \\
 - 2 &  - 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 &  - 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
0 & 1 &  - 3 \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
1 & 1 &  - 2 \\
1 & 0 & 1 \\
 - 2 & 1 &  - 4 \\
\end{array}
\right)
$$


Is there an expression for a function that maps integers to one and non-integers to zero? Is there a function that can be built with addition, multiplication, exponentiation, trigonometric functions, integrals, (and all of their inverses i.e subtraction, division, taking logarithms, $\arcsin(x)$, derivatives, etc) that would take an integer and output $1$ and take a non-integer and output $0$?

I looked at the Dirichlet Function but it separated rationals from irrationals, and the only solution that I could come up with was with the Fourier Series of $$x-\lfloor x \rfloor = \frac 12 - \frac 1\pi \sum_{k=1}^\infty \frac 1k \sin (2 \pi k x)$$
But this uses infinite series. Is there a way to build such a function? If there is not, why not?

Maybe this is cheating, but I don't think it's possible otherwise, since all of the other functions are continuous and a composition of continuous functions is continuous, integrals preserve continuity too, and even though differentiation doesn't in general, all of the functions you listed are not just continuous but analytic or infinitely differentiable so their derivatives are continuous as well at all orders.  Anyway, here is my example:

$f(x) = 0^{\sin(\pi \cdot x)^2}$

For this to be correct you have to define $0^0=1$ which is not an unreasonable choice, but usually this expression is left undefined.

Well.. you can always built it as a distribution:

$$
\Upsilon(x) =
\begin{cases}
1 ~~~ \text{for} ~~~ x\in\mathbb{N}
\\
0 ~~~ \text{for} ~~~x\not\in\mathbb{N}
\end{cases}
$$

lol


hermitean structure vs hermitean metric Define a hermitean structure $H$ on a complex linear space $V$ as $H: V\times V \to \mathbb{C}$ s.t.

i. $H(u,v)$ is $\mathbb{C}$-linear in $u$ for every $v \in V,$

ii. $H(u,v)=\overline{H(v,u)}$

iii. $H(u,u)> 0\quad \forall u\neq 0.$

Define a hermitean metric on an almost complex manifold $M$ as $h: \mathbb{C}T_xM \times \mathbb{C}T_xM \to \mathbb{C}$ that depends smoothly on the point $x$ such that

i. $h(\overline{Z},\overline{W})=\overline{h(Z,W)} \quad \forall Z,W\in T_\mathbb{C}M$

ii. $h(Z,\overline{Z})>0 \quad \forall Z>0$

iii. $h(Z,W)=0 \quad \forall Z,W\in T^{1,0}M \quad \text{and} \quad \forall Z,W\in T^{0,1}M$

My question is: how do I prove that
$$h_{i\overline{j}}:=h\left(\frac{\partial}{\partial z^i},\frac{\partial}{\partial \overline{z}^j}\right)=\overline{h_{j\overline{i}}} \ ?$$

Using h.i and the symmetry of $h$ (h must be Riemannian metric itself) we have
the thesis.

Moreover, the correct relation is $h=\operatorname{Re}(H).$


determine if the series converge or diverge How can I go about determining if the following series converge or diverge.
$$\sum_{n=1}^{\infty} (-1)^n a_n$$

where
$$a_n=\begin{cases}
1/n,  & \text{if $n$ is odd} \\
1/n^2, & \text{if $n$ is even}
\end{cases}$$

Since
$$S=\sum_{n=1}^{\infty}(-1)^n a_n=\sum_{k=1}^{\infty}-\frac{1}{2k-1}+\sum_{k=1}^{\infty}\frac{1}{(2k)^2}=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k-1/2}$$
And noting that the first term of the right side is less than 1
$$-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k-1/2}<S<-\frac{1}{2}\sum_{k=2}^{\infty}\frac{1}{k-1/2}$$
And since both of those series diverge, then $S$ is also divergent.

Since both the odd an even subsequences of the sequence $(a_{n})$ converge to zero, we can conclude that the entire sequence converges to zero.

There is a well-known result on real analysis which says that such series converge.

Lemma: If the sequence $(a_{n})_{n=1}^{\infty}$ converges to $0$, then the series
$$
\sum_{n=1}^{\infty}(-1)^{n}a_{n}
$$
converges.

You can find the proof of such fact on any book of Real Analysis. Actually, you can find it here.


The effect of negative eigenvalues for a matrix $A$ in a linear system Let $A$ be a $n$x $n$ matrix and  $x'=Ax$ be a linear system.

Prove if true, give a counter example if it is not.

(i) Suppose that all the eigenvalues of $A$ are distinct and have a negative real part. Then every solution of $x'=Ax$ satisfies $$|x(t)|\leq |x(s)|$$ for all $t>s$.

(ii) Suppose that all the eigenvalues of $A$ have a negative real part. Then every solution of $x'=Ax$ satisfies $$|x(t)|\leq |x(s)|$$ for all $t>s$.

(iii) Suppose $A$ is symmetric and all of its eigenvalue have a negative real part. Then every solution of $x'=Ax$ satisfies $$|x(t)|\leq |x(s)|$$ for all $t>s$.

I would say that I should give a proof for (iii) since the eigenvalue a symmetric matrix is real and hence negative real which implies that $$|x(t)|=\sqrt{(c_1e^{\lambda_1 t}v_{1_{\lambda_1}}+...+c_ne^{\lambda_n t}v_{1_{\lambda_n}})^2+...+(c_1e^{\lambda_1 t}v_{n_{\lambda_1}}+...+c_ne^{\lambda_n t}v_{n_{\lambda_n}})^2} \leq \sqrt{(c_1e^{\lambda_1 s}v_{1_{\lambda_1}}+...+c_ne^{\lambda_n s}v_{1_{\lambda_n}})^2+...+(c_1e^{\lambda_1 s}v_{n_{\lambda_1}}+...+c_ne^{\lambda_n s}v_{n_{\lambda_n}})^2}= |x(s)|$$

for any $t>s$, where $c=(c_1,...,c_n)$ referes to the constant for the general solutions and $v_{{\lambda_i}}=(v_{1_{\lambda_i}},...,v_{n_{\lambda_i}})$ refers to the eigenvectors of $\lambda_i$.

So it holds for (iii), I can not see that is true for i or ii but I can not find an example for that, I would appreciate any help with that. Thank you in advance.

Assume that $A$ is real. For (i), there exists a real invertible matrix $P$ such that $P^{-1}AP$ is block-diagonal,i.e. has diagonal elements and/or diagonal blocks of the form $$\begin{bmatrix}a&-b\\b&a\end{bmatrix}$$ corresponding to complex eigenvalues $a \pm bi$ where $a$ and $b$ are real. Then if we define new variables $y_1,...,y_n$ by $$[x_1,...,x_n]^T=P[y_1,...,y_n]^T$$each $y_i=e^{at}(c \cos(bt)+ d \sin(bt))$. Thus, because each $a$ is negative, each $y_i$ is bounded and therefore each $x_i$ is also bounded. For (ii), here is a brief sketch. We put the matrix into block Jordan form where there are some 2x2 identity matrices along the super-diagonal. The difference from (i) is that when we solve for the $y_i$ we obtain expressions of the form $$ e^{at}f(t)(c \cos(bt)+ d \sin(bt)) $$ where $f$ is a polynomial, which you can show are also bbounded.