Find the limit to $\lim\limits_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$ My problem is evaluating the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$$
The answer should be 0. I tried to convert the limit into polar form, but it didn't help because I couldn't isolate the $r$ and $\theta$-variables of the expression. My "toolbox" for solving problems like these is very limited... If polar form doesn't work, then I usually have no clue on how to continue.

Edit: I think this is the solution.
$$
\lim_{(x,y)\to(0,0)}\left|\frac{x^5+y^2}{x^4+|y|}\right| = \frac{|x^5+y^2|}{|x^4+|y||}
$$
Applying the triangle inequality gives
$$
\frac{|x^5+y^2|}{|x^4+|y||} \leq \left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right|
$$
Inspecting the denominators on the RHS gives:
$$
\left|\frac{x^5}{x^4+|y|}\right| \leq |x|, \quad\left|\frac{y^2}{x^4+|y|}\right| \leq |y|
$$
So
$$
\left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right| \leq |x| + |y|
$$
Since $|x| + |y| \to 0$ when $x,y\to 0$, the sandwich theorem states that $|\frac{x^5+y^2}{x^4+|y|}| \to 0$. And if $\lim |f(x)|=0$ then $\lim f(x)=0$ which solves the original problem.

Hint:)

For paths $x=0$ as $y\to0$ and $y=0$ as $x\to0$ the result of limit clearly is zero, furthermore
$$\Big|\frac{x^5+y^2}{x^4+|y|}\Big|\leq\Big|\frac{x^5}{x^4+|y|}\Big|+\Big|\frac{y^2}{x^4+|y|}\Big|\leq|x|+|y|$$

A common tool is to homogenise the denominator by setting $\displaystyle u=\frac{y}{x^4}$, then discuss the limit in function of $u$.

$\displaystyle f(x,y)=\frac{x^5+y^2}{x^4+|y|}=\frac{x^5+u^2x^8}{x^4+|u|x^4}=\frac{x+u^2x^4}{1+|u|}=\frac{x+uy}{1+|u|}=x\left(\frac 1{1+|u|}\right)+y\left(\frac u{1+|u|}\right)$

Since both $\frac 1{1+|u|}$ and $\frac u{1+|u|}$ are bounded (by $1$ in absolute value)

Then $|f(x,y)|\le |x|+|y|\to 0$


How to calculate the value of $\cos \theta$ if $\cos 3\theta$ is known If $\cos 3\theta = \frac{-11}{\sqrt{125}}$ then how can i calculate the the value $\cos \theta$ from there. I already tried so solve this by using $4\cos^3 \theta-3\cos \theta=\cos 3\theta $ but i was unable to solve the cubic equation.

The tricky part here is to put the cubic equation into a form where factors can be identified.  We could eliminate the radical by squaring the equation or by deriving an equation whose roots are squared, but these methods introduce extraneous roots.
The superior method is to render $\sqrt{125}=(\sqrt5)^3$ and with that, define $x=(\sqrt5)\cos\theta$.  Thereby
$4(x/\sqrt5)^3-3(x/\sqrt5)+11/(\sqrt5)^3=0$
Now the $\sqrt5$ appears to odd powers in all terms allowing it to be cancelled.  Clearing the remaining fractions then gives:
$4x^3-15x+11=0$
which can now be factored if the standard rational root search can give a root for $x$.  Finding the root $x=1$ we obtain the factorization
$(x-1)(4x^2+4x-11)=0$
and the quadratic roots are found by the usual methods.  Remember to render $\cos\theta=x/\sqrt5$ at the end.

Since you know complex numbers:
$$e^{3i\theta} = e^{(i\theta)^3} $$
which leads to:
$$ cos(3\theta) + i \cdot  sin(3\theta) = (cos(\theta) + i \cdot sin(\theta))^3$$
Group the real and imaginary parts to get $cos(3\theta)$ and $sin(3\theta)$ as functions of powers of $sin(\theta)$ and $cos(\theta)$.
The equation you'll get for $sin(3\theta)$ will be a second-degree one. Use the identity $cos(x)^2 + sin(x)^2 = 1$ when necessary.


Show that a vector field is not conservative (example) Let $\Omega=\mathbb{R^2}\smallsetminus\{(0,0)\}$ and $$\vec{F}(x,y)=-\frac{y}{x^2+y^2}\vec{i}+\frac{x}{x^2+y^2}\vec{j}$$ To show that the vector field $F$ is not conservative $$\vec{\nabla}\times \vec{F}=0\,\vec{i}+0\,\vec{j}+\bigg(\frac{y^2-x^2}{(x^2+y^2)^2}-\frac{y^2-x^2}{(x^2+y^2)^2}\bigg)\vec{k}=\vec{0}$$ which indicates that it is conservative(?). Alternatively, the scalar function

$$f(x,y)=-\arctan\bigg(\frac{x}{y}\bigg)$$ is the potential function of the field $F$. If $$\int_C\vec{F}\cdot\mathrm{d}\vec{r}\neq f\big(x(\beta),y(\beta)\big)-f\big(x(\alpha),y(\alpha)\big)$$ $\big($where $r(t)$ is the parametrization of an arbitrary curve $c$ and $\alpha$ and $\beta$ are its starting and end point respectively$\big)$

then the fundumental theorem for line integrals is not satisfied and thus $F$ is not conservative.
I tried using $r(t)=t\vec{i}+t\vec{j}, \space t\in[\alpha,\beta]$, but it didn't work.

What curve would be a better choice for $C$ and what's the deal with $\mathrm{rot}\,F$ being zero?

One of the most important "functions" in classical analysis is the ${\rm arg}$ function, giving the polar angle of a point $(x,y)$, resp., of $z=x+iy$,  in the punctured plane "up to an additive constant $2k\pi$". Locally, i.e., in suitable neighborhoods of points $(x_0,y_0)\in\dot{\mathbb R}^2$ this function has well-defined real representants. E.g., in the half plane $x>0$ we may choose the representant $$\phi(x,y):=\arctan{y\over x}\ .$$
Now this ${\rm arg}$ "function" has a well-defined gradient
$$\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)\qquad\bigl((x,y)\ne(0,0)\bigr)\ .$$It turns out that your ${\bf F}$ is nothing else but $\nabla{\rm arg}$. Since, locally,  ${\rm arg}$ is represented by smooth functions $(x,y)\mapsto\phi(x,y)$ it follows that ${\rm curl}\,{\bf F}\equiv0$.

But it is impossible to concatenate the local polar angles $\phi(x,y)$ to one single real-valued polar angle function $\phi:\>\dot{\mathbb R}\to{\mathbb R}$. This then implies that the given ${\bf F}$ is not conservative. A definitive proof of this fact comes from computing $\int_\gamma {\bf F}\cdot d{\bf z}$ along the unit circle $\gamma$. Along this circle the argument increases continuously by $2\pi$, hence the value of this integral is $2\pi\ne0$.

If a differentiable field $\vec{F}(x,y)=F_1(x,y)\vec{i}+F_2(x,y)\vec{j}$ is conservative then you must have $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$ in every point of the domain. You can easily check that your field doesn't have this property.


How can find domain and range of $f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor}$? How can find $D_{f(x)},R_{f(x)}$ when $$f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor} \ ?$$
I help some problem like this for a question here ... (Domain and range of a floor function)
But I get stuck on this to find Domain and Range of above function .
If You can help me to find out , I will be very thankful .

So, with the standard notation for the  floor function, we are to study
$$ \bbox[lightyellow] {
y = \sqrt {\sqrt 2 \left\lfloor x \right\rfloor  - \left\lfloor {\sqrt 2 x} \right\rfloor }
} \tag {1}$$

Let's first introduce some additional notation concerning the fractional part $\{x\}$
$$
x = \left\lfloor x \right\rfloor  + \left\{ x \right\}\quad 0 \le \left\{ x \right\} < 1
$$
and the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$

Then the radicand can be written as
$$ \bbox[lightyellow] {
\eqalign{
  & r = \sqrt 2 \left\lfloor x \right\rfloor  - \left\lfloor {\sqrt 2 x} \right\rfloor  =   \cr
  &  = \sqrt 2 \left( {x - \left\{ x \right\}} \right) - \left( {\sqrt 2 x - \left\{ {\sqrt 2 x} \right\}} \right) =   \cr
  &  = \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\} \cr}
} \tag {2}$$

and, since the fractional part ranges within $[0,1)$, the radicand is bound to
$$ \bbox[lightyellow] {
 - \,\sqrt 2  < r = \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\} < 1
} \tag {3}$$

If the function is defined in the real field, then its Domain of definition will be given by the values of $x$ such that
$$ \bbox[lightyellow] {
x:\;\;0 \le \left\{ {\sqrt 2 x} \right\} - \sqrt 2 \left\{ x \right\}\quad  \Leftrightarrow \quad \sqrt 2 \left\{ x \right\} \le \left\{ {\sqrt 2 x} \right\}
} \tag {4}$$

It can be easily seen that the fractional part of a sum can be written as
$$ \bbox[lightyellow] {
\eqalign{
  & \left\{ {z + y} \right\} = \left\{ {\left\{ z \right\} + \left\{ y \right\}} \right\} =   \cr
  &  = \left\{ z \right\} + \left\{ y \right\} - \left\lfloor {\left\{ z \right\} + \left\{ y \right\}} \right\rfloor  =   \cr
  &  = \left\{ z \right\} + \left\{ y \right\} - \left[ {1 \le \left\{ z \right\} + \left\{ y \right\}} \right] \cr}
} \tag {5}$$
so that the RHS of the inequality (4) above becomes
$$ \bbox[lightyellow] {
\eqalign{
  & \left\{ {\sqrt 2 x} \right\} = \left\{ {\sqrt 2 \left\lfloor x \right\rfloor  + \sqrt 2 \left\{ x \right\}} \right\} =   \cr
  &  = \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right] \cr}
} \tag {6}$$
and  the inequality transforms into
$$ \bbox[lightyellow] {
\eqalign{
  & \sqrt 2 \left\{ x \right\}\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right]  \cr
  & \quad  \Downarrow   \cr
  & \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor  + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} - \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right]  \cr
  & \quad  \Downarrow   \cr
  & 0 \le \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor  + \left[ {1 \le \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\}} \right]\; \le \;\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} < 1 \cr}
} \tag {7}$$

Now, the Poisson bracket cannot be one (because of the <1), so it shall be null, i.e. its inequality false.
Also, the floor before it (which is either $0$ or $1$) cannot be but $0$.

Thus we can conclude that the Domain of definition (y real) is given by
$$ \bbox[lightyellow] {
\left\{ \matrix{
  \left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \left\{ {\sqrt 2 \left\{ x \right\}} \right\} < 1 \hfill \cr
  \left\lfloor {\sqrt 2 \left\{ x \right\}} \right\rfloor  = 0 \hfill \cr}  \right.\quad  \Rightarrow \quad \left\{ \matrix{
  \left( {0 \le } \right)\left\{ x \right\} < 1/\sqrt 2  \hfill \cr
  \left( {0 \le } \right)\left\{ {\sqrt 2 \left\lfloor x \right\rfloor } \right\} + \sqrt 2 \left\{ x \right\} < 1 \hfill \cr}  \right.
} \tag {7.a}$$
and the Codomain is
$$ \bbox[lightyellow] {
0 \le y < 1
} \tag {7.b}$$

Because of the irrationality of $\sqrt{2}$ inequality (7.a) does not add much to the original (4), and cannot be further simplified.

At first I tried some manipulation but could not complete it. If we take $x=n+\sqrt{p}$ where $0 \leq p<1$ we obtain
$$\sqrt{2}n-[\sqrt{2}n+\sqrt{2p}]$$
we can deduce that $p<0.5$.

Now if we take $x=n+p$ and $\sqrt{2}=1+p_1$ (where $p_1$ is about $0.41$, also this can be considered for $\sqrt{y}[x]-[\sqrt{y}x]$ with $y=m+q$ where $0\leq q<1$) we have
$$(1+p_1)\times[n+p]\geq [(1+p1)\times(n+p)] $$
$$n+np_1\geq[n+np_1+p+pp_1]$$
$$np_1\geq[np_1+p(1+p_1)] \quad or \quad [p_1(n+p)+p]$$
from inequalities above
$$[np_1+p(1+p_1)]=[np_1]$$
if we take $[np_1]=m$ and $r=np_1-m$ we can obtain
$$p(1+p_1)<1-r=1-np_1+[np_1]$$
So that
$$p<\frac{1-np_1+[np_1]}{1+p_1}$$
For example for $n=1$ we have:
$$0\leq p<\sqrt{2}-1$$

For $n=2$, $0\leq p<\frac{3-2\sqrt{2}}{\sqrt{2}}$ which is about $0.1213$. The domain is:
$$\bigcup {[n,n+p]},\quad p<\frac{1-np_1+[np_1]}{1+p_1}$$

I tried both with MATLAB and it is true. The domain depends on the number $n$ and also the decimals of $\sqrt{2}$, $p_1$ which are known. This can be generalized to domain of $\sqrt{\sqrt{y}[x]-[\sqrt{y}x]}$.


Derivatives of $\sinh x$ and $\cosh x$ Can someone give me an intuitive explanation about the derivatives of $\sinh x$ and $\cosh x$? Something similar to:

Intuitive understanding of the derivatives of $\sin x$ and $\cos x$

Thanks!

In my opinion, the correct role of the trig functions in intuition is the other way around -- you shouldn't be trying to understand trig functions in terms of your intuition about other things: you should be trying to develop your intuition so that you understand other things in terms of trig functions.

(and, incidentally, this is in line with the general strategy that a good way to understand a mathematical idea, you need to understand how to use that idea to do things)

But anyways....

The unit circle is to the circular trig functions as the unit rectangular hyperbola is to the hyperbolic trig functions. The points $(\cosh u, \sinh u)$ trace out the points on the rightward-opening hyperbola defined by

$$ x^2 - y^2 = 1 \qquad \qquad x \geq 0 $$

The asymptote to this equation are the lines $y = \pm x$.

The parameter $u$ is the arclength from the point $(1,0)$ to the point $(\cosh u, \sinh u)$ along the hyperbola. Arclength is signed: so going "up" increases arclength, and "down" decreases arclength.

(note that the $\theta$ in $\cos \theta$ can be understood as arclength around the unit circle as well; the circle just has the nice fact that arclength and angle are the same)

If you studied hyperbolas as much as you studied circles, you would immediately know, for example, how to find the tangent line to a point on the hyperbola. But hyperbolas aren't studied so much these days, so it's unlikely that this has much utility.

There is a direct connection between the circular and hyperbolic functions in terms of the complex numbers: e.g.

$\cos x = \cosh ix$
$\sin x = -i \sinh ix$
$\cosh x = \cos ix$
$\sinh x = i \sin ix$

which, IMO, conveys intuition that any fact about the circular functions can be translated into an analogous fact about hyperbolic functions. e.g. I know that there is a double-angle formula for $\cos$. Therefore, there should be a similar double-angle formula for $\cosh$.

Intuitive not really, but why don't you look at the definition of for example $sinhx$? Take the derivative of the e-powers and due to the chain rule of the negative exponent ,it turns out you end up with $coshx$. Other than the fact that $sinhx$ is all increasing and derivative $coshx$ is always positive, there isn't much intuitive about it in my view.


Lost on "Simple Computations" I have come across the follow assertion: $$\text{for } x,y,z >0, xyz = x+y+z+2$$ may be rewritten as $$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1$$

and that proving this is a matter of 'simple computations'. However, I can't seem to see the way about this.

I would say what I've tried so far, but every approach I've tried so far as petered out after a step or two.

I would go reverse engineering for that one $$\frac{1}{x+1} + \frac{1}{1+y} + \frac{1}{1+z} = 1$$

First you add $\frac{1}{x+1} + \frac{1}{1+y}$ to get $$\frac{1+y+1+x}{(x+1)(y+1)}= \frac{2+x+y}{(x+1)(y+1)}$$ now we add this result to $\frac{1}{z+1}$ to get  $$\frac{2+x+y}{(x+1)(y+1)} +  \frac{1}{z+1} = \frac{(2+x+y)(1+z) + (x+1)(y+1)}{(x+1)(y+1)(z+1)} =1$$

now you have $(2+x+y)(1+z) + (x+1)(y+1) = (x+1)(y+1)(z+1) $ Now Subtract $(x+1)(y+1)$ from each side , you will end up with $$(2+x+y)(1+z) = (x+1)(y+1)(z+1)-(x+1)(y+1)$$ now we have $$(2+x+y)(1+z) = (x+1)(y+1)(z+1-1)$$ and so we have

$$(2+x+y)(1+z) = (x+1)(y+1)z$$ Now expand the LHS and the RHS to get $$2 + 2z + x + xz + y + yz = xyz + zx + zy + z$$. Now $xz$ cancels with $xz$ and $yz$ cancels with $yz$ and you end up with $$ 2 + 2z + x + y = xyz + z$$ Now final step , you subtract $z$ from each side to end up with $$2 + z + x + y = xyz$$ and you are done !!!!

If you don't like reverse engineering , you can reverse what I did, and you would be able to go from that to the fraction form. reverse reverse = direct :)


Is there a spiral whose arc-length between two points on it is proportional to the difference in its radius of curvature between these two points? This is a strange spiral, which is from the differential equations as follows
$$
\left\{\begin{aligned}
\dot x-\dot y=\ddot x\sqrt{\dot x^2+\dot y^2} \\
\dot x+\dot y=\ddot y\sqrt{\dot x^2+\dot y^2}
\end{aligned}\right.
$$
I don't know how to solve the equations. However, the strange spiral has an interesting property, which is the difference in its radius of curvature between two points on it is proportional to its arc-length between these two points.

This property is from a problem of physics. We consider a particle with mass $m$, and the velocity of it is $1$ along with the x-axis. There are two forces whose magnitudes are constant on the particle. One is always along with the velocity of the particle, and the other one is always perpendicular to the velocity of the particle. And we assume that the cross product of these two forces is constant.

If we set $m=1$ and $F_1=F_2=1$, we can easily get the differential equations above. I have calculated it by numerical integration as the picture shows.

Actually, I think it is like a spiral, but I don't know what kind of spiral it is. And we can calculate its arc between two points
$$
s(t)-s(t_0)=\int^{t}_{t_0}v(t)\text dt=v(t_0)(t-t_0)+\frac{(t-t_0)^2}2
$$
If we set $t_0=0$ and $v(t_0)=0$, we will get $s(t)\propto t^2$. Actually we cannot do this because $v(0)=0$ isn't what we want, but we can make an approximation that $t\gg t_0$ and $t=F_1t/m\gg v(t_0)$ which can make $s(t)\propto t^2$ approximately correct. Notice that the magnitude of $F_2$ is constant and it is always perpendicular to the velocity, therefore
$$
\frac{F_2}m=a_\perp=\frac{v^2}r=\text{const}
$$
where $r$ is the radius of curvature. Also, we can get that $v^2\propto t^2$, which means
$$
\frac{s(t)}{r(t)}=\text{const}
$$
Is it really a spiral? And what more properties does it have? Or are these differential equations solvable?

As is long for a comment I leave some ideas, not an answer.

Calling $u=\dot x, v=\dot y$ we have the system

$$
\cases{
\dot u = \frac{u-v}{\sqrt{u^2+v^2}}\\
\dot v = \frac{u+v}{\sqrt{u^2+v^2}}
}\ \ \ \ \ \ \ (1)
$$

now making the variables change

$$
\cases{
u = r\cos\theta\\
v = r\sin\theta
}
$$

the system $(1)$ is transformed into

$$
\dot r = 1\\
\dot\theta = \frac 1r
$$

solved this ODE we arrive at the solution curves

$$
r = c_0 e^{\theta}\ \ \ \text{or}\ \ \ \sqrt{\dot x^2+\dot y^2} = c_0 e^{\arctan\left(\frac{\dot y}{\dot x}\right)}
$$

Answer：The spiral is a logarithmic spiral (or a part of a logarithmic spiral)  whose equation is $$  \rho = Ce^{\pm a\theta } $$ in polar coordinates where $$ a=2\frac{a_{//} }{a_{\perp } } $$
The particle has an acceleration a∥ which is always along with the velocity of the particle, and an acceleration a⊥ which is always perpendicular to the velocity of the particle. Let α be the angle between the direction of the velocity and a certain axis (You may assume the axis is x axis or others). Then ω is the angular velocity of the direction of the particle’s velocity i.e. ω=dα/dt. So $$ \omega =\frac{v}{r} =\frac{v^2}{r} \frac{1}{v} =\frac{a_{\perp }}{a_{//}t} \propto \frac{1}{t} $$ For simplicity, we assume that the initial velocity is zero i.e. $$ v\mid _{t=0} \, =0 $$
Use integration we can easily find that for any finite nonzero t0：
$$\begin{align}
\int_{0}^{t_{0} } \omega \mathrm{d}t  = 2\pi \alpha \mid _{0}^{t_{0} }  = +  \infty  &&
 \int_{t_{0} }^{+  \infty} \omega \mathrm{d}t  = 2\pi \alpha \mid _{t_{0} }^{+  \infty }  = +  \infty
\end{align}$$
which means the particle will spiral for infinite cycles both when the particle get close to the origin point and when the particle goes to infinite far away. Thus, we can understand that the spiral has no asymptote when the curve extends to infinite far away, and has infinite curvature in the origin point. Look at the appendix of a textbook of calculus, you can find some common curves, and you may find that the logarithmic spiral is likely to be the answer. You can try to test and verify it.
Because it’s a spiral, it’s much more convenient to use use polar coordinates instead of cartesian coordinate system to study it. We will use polar coordinates to find the solution.
The differential equation of this curve comes from the formula of the curvature radius:$$ \frac{\left ( \rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta }  \right ) ^2\right ) ^{\frac{3}{2} }  }{ \rho ^2+2\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta }  \right ) ^2-\rho \frac{\mathrm{d}^2\rho }{\mathrm{d}\theta ^2}  }=r $$ where ρ and θ are radial coordinate and angular coordinate, respectively. The expression of r is:
$$ r=\frac{v^2}{a_{\perp } } =\frac{\left ( a_{//} t \right ) ^2}{a_{\perp } }=\frac{a_{//}}{a_{\perp }} a_{//}t^2=2\frac{a_{//}}{a_{\perp }}s  $$
where s is the distance traveled from t=0 and v=0.
The expression of s in polar coordinates is:
$$ s=\int\limits_{0}^{t} \mathrm{d}s=\int\limits_{0}^{t} \sqrt{\rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right)^2} \mathrm{d}\theta $$
From the three equations above, we have:
$$　\frac{\left ( \rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta }  \right ) ^2\right ) ^{\frac{3}{2} }  }{ \rho ^2+2\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta }  \right ) ^2-\rho \frac{ \mathrm{d}^2\rho }{ \mathrm{d}\theta ^2}  }=r=2\frac{a_{//}}{a_{\perp }}s=2\frac{a_{//}}{a_{\perp }}\int\limits_{0}^{t} \sqrt{\rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right)^2} \mathrm{d}\theta $$
You can test the solution by simply substitute ρ=Ce^(±aθ) into the integral equation above, hence find the dimensionless parameter a is equal to the ratio between the accelerations i.e. a=2a∥/a⊥.
$$\because \rho = Ce^{\pm a\theta }\therefore \left ( \frac{\mathrm{d} \rho }{\mathrm{d} \theta }  \right ) ^2=a^2\rho ^2;\frac{\mathrm{d} \rho }{\mathrm{d} \theta ^2} =a^2\rho$$
Substitute it into the equation, we have:
$$ \rho =2\frac{a_{//} }{a_{\perp } } \int\limits_{0}^{t} \rho \mathrm{d}\theta \Leftrightarrow Ce^{\pm a\theta } =\frac{2}{\pm a} \frac{a_{//} }{a_{\perp } }\int_{-\infty }^{\pm a\theta } Ce^{\pm a\theta }\mathrm{d}\left ( \pm a\theta \right ) \Leftrightarrow \pm a=2\frac{a_{//} }{a_{\perp } }$$
So the solution fit this equation very well.
Or you can solve the equation directly, it looks difficult, but it’s not such difficult.
$$\frac{\mathrm{d}  } {\mathrm{d} \theta } \left [  \frac{\left ( \rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta }  \right ) ^2\right ) ^{\frac{3}{2} }  }{ \rho ^2+2\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta }  \right ) ^2-\rho \frac{ \mathrm{d}^2\rho }{ \mathrm{d}\theta ^2}  } \right ] =2\frac{a_{//}}{a_{\perp }} \sqrt{\rho ^2+\left (\frac{\mathrm{d}\rho}{\mathrm{d}\theta } \right)^2} $$
By differentiation，we have：
$$\frac{3\left ( \rho {\rho }'+ {\rho }'{\rho }'' \right )\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }''  \right ) -\left ( \rho ^2+{\rho}' ^2 \right ) \left ( 2\rho {\rho }'+4{\rho }'{\rho }''-{\rho }'{\rho }''-\rho {\rho }'''\right )  }{\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }''  \right ) ^2} \sqrt{\rho ^2+{\rho}' ^2}=2\frac{a_{//}}{a_{\perp }}  \sqrt{\rho ^2+{\rho}' ^2}$$
Where ρ’ denotes dρ/dθ; ρ”=dρ’/dθ; ρ”’=dρ”/dθ
Rearranging:
$$(\rho ^3{\rho }'+4\rho {\rho }'^3 +3{\rho }'^3{\rho }''-3\rho^2{\rho }'{\rho }''-3\rho{\rho }'{\rho }''^2+\rho ^3{\rho }'''+\rho {\rho }'^2{\rho }''') \sqrt{\rho ^2+{\rho}' ^2}=2\frac{a_{//}}{a_{\perp }}  \sqrt{\rho ^2+{\rho}' ^2}\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }''  \right ) ^2$$
$$ (\rho ^3{\rho }'+4\rho {\rho }'^3 +3{\rho }'^3{\rho }''-3\rho^2{\rho }'{\rho }''-3\rho{\rho }'{\rho }''^2+\rho ^3{\rho }'''+\rho {\rho }'^2{\rho }''')^2 \left ( \rho ^2+{\rho}' ^2 \right ) =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2  \left ( \rho ^2+{\rho}' ^2 \right )\left ( \rho ^2+2 {\rho}' ^2-\rho{ \rho }''  \right ) ^4$$
This is a high order nonlinear homogeneous differential equation, so its solutions have a form of $$ \rho =Ce^{\lambda \theta } $$
where C can be any constant. Substitute it into the equation, we got its Characteristic equation：
$$ \left (  \lambda +4\lambda ^3 +3\lambda ^5-3\lambda ^3-3\lambda ^5+\lambda  ^3+\lambda ^5  \right )^2\left ( 1+\lambda ^2 \right ) =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2  \left ( 1+\lambda ^2 \right )\left ( 1+2\lambda ^2-\lambda ^2 \right )   ^4 $$
$$ \left (  \lambda +2\lambda  ^3+\lambda ^5  \right )^2\left ( 1+\lambda ^2 \right ) =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2  \left ( 1+\lambda ^2 \right )\left ( 1+\lambda ^2 \right )   ^4 $$
$$\therefore  \lambda ^2\left ( 1+\lambda ^2 \right )^5 =4\left ( \frac{a_{//}}{a_{\perp }} \right )^2  \left ( 1+\lambda ^2 \right )  ^5$$
Then we got the solution:
$$ \therefore \left | \lambda  \right | =2\frac{a_{//}}{a_{\perp }}\quad or \quad \lambda =\pm i $$
$$\therefore \rho =Ce^{\pm a\theta } ,a=2\frac{a_{//} }{a_{\perp } } \quad or \quad \rho =C_{1} \cos\theta +C_{2} \sin\theta $$
The latter solution are not suitable for the integral equation above, one of the reasons is that it doesn’t contain the parameter 2a∥/a⊥.
So we got the solution ρ=Ce^(±aθ). Remember that we got this solution under the condition that the initial velocity is equal to zero, so the geometric center of the spiral is the origin point of the coordinate. If the initial velocity isn’t zero, the trajectory of the particle will still be a part of a logarithmic spiral, you can simply get the trajectory by translation and rotation of the trajectory of the particle which has the same a∥/a⊥ and the same direction  . The shape of a logarithmic spiral is only related to the magnitude of a∥/a⊥. The direction of a logarithmic spiral is only related to the sign of  $$\frac{\overrightarrow{a_{//}}\times \overrightarrow{a_{\perp}}}{\left | \overrightarrow{a_{//}}   \right |\left | \overrightarrow{a_{\perp}}  \right |  } $$
The geometric center of the spiral can be obtained from the momentary position of the particle, the direction of the momentary velocity, the angle between the line connecting the momentary position and the direction of the velocity (which can be easily obtained from the magnitude of a∥/a⊥) and the linear distance between the momentary position and the center (which can be obtained from the magnitude of v²/a∥ where v represents the momentary velocity) . Then we get all of the information about the spiral.
The logarithmic spiral has a lot of interesting properties：
It’s also called equiangular spiral，because the line connecting any of a point on the spiral and the origin point have the same included angle of arccot(a) with the spiral. You can easily prove it by calculating  dρ/ρdθ . That’s why we can know the direction of the geometric center of the spiral.
If two logarithmic spiral are similar to each other, the two spiral are actually congruent to each other.
Also, the particle has a ‘momentary rotation center’ when it spirals. The center is located in the perpendicular line of the tangent line of the spiral, and the distance between them is equal to the square of the velocity divided by the perpendicular acceleration i.e. r=v²/a⊥. Hence find the position of the ‘momentary rotation center’. You can easily prove that r is proportional to ρ for a certain spiral hence find the ratio between them. Then you can find the trajectory of the ‘momentary rotation center’. It is very interesting that the trajectory is also a logarithmic spiral which has the same shape and same direction of the trajectory of the particle. (You can prove it by simply using geometric methods. Then you can find the relation between the two spirals by using The Law of Cosines.)You can call this trajectory ‘adjoint spiral’ or ‘companion spiral’ of the logarithmic spiral. Moreover, another interesting property is that the triangle which the vertexes are the geometric center, the particle and the ‘momentary rotation center’ (i.e. the center of curvature) is always a Rectangular triangle of which the geometric center is the Rectangular vertex. (You can easily prove it by using The Law of Cosines or other methods.) You can calculate the angle difference △θ between the two spirals. Specially, when $$\frac{\ln{a} }{a} =\frac{\pi }{2} +2k\pi ,k\in Z,k< 0$$
(where k can be any negative integer i.e. k∈Z,k<0)  i.e. △θ=2kπ, the ‘adjoint spiral’ and the original spiral coincide.
You can learn more and detailed information about logarithmic spiral from Wikipedia. Further, you can search 'list of spirals' in Wikipedia for other interesting spirals. (If you are in Mainland China and cannot open the pages of Wikipedia, you can Search ‘wanweibaike’ i.e. https://en.wanweibaike.com/ )


Show $A =\{x\in \ell_2 : |x_n|\le\frac{1}{n}\},n=1,2,...$ is a closed set in $\ell_2$ but $B=\{x\in \ell_2 :|x_n|< \frac{1}{n}\}$ is not an open set. Show that the set $A =\{x\in \ell_2 : |x_n|\le\frac{1}{n},n=1,2,...\}$ is a closed set in $\ell_2$ but that $B=\{x\in \ell_2 :|x_n|< \frac{1}{n}\}$ is not an open set.

I think if I show each $A_n=\{x\in\ell_2,|x_n|\le\frac{1}{n}\}$ is closed, I can get $A=\displaystyle\bigcap_{n}A_n$ that is closed. Can I do that? However, I didn't know how to prove $A_n=\{x\in\ell_2,|x_n|\le\frac{1}{n}\}$ is closed.

Are there any other ways to prove this?

As @SangchulLee comment convergence in $\ell_2$ implies coordinate wise converge, therefore $x_n\overset{\ell_2}\rightarrow x$ implies that for all $k:x_n^{(k)}$ converges to $x^k$ in $\mathbb R$.

Now the point is that $|{x}|\leq \frac 1 n$ is a closed set, therefore if $|x_n^{(k)}|\leq \frac 1 k$ and it converges the limit element will also satisfy $|x^{(k)}|\leq\frac 1 k$ and so it effectively lies in $A$.

For the second case you have that $x_n^{(k)}<\frac 1 k$ which is an open condition so that the limit can escape. For example you can take the sequence: $$x_n=\begin{cases} 1-\frac 1 n,\quad k=1\\0\end{cases}$$

this is clearly in $B$ bu the limit is the sequence $(1,0,0,\dotsc)$ which does not lie in $B$.

You could try showing it's the inverse image of a closed set under a continuous map.


Two definitions of simply connected I'm showing that the first definition here implies the second (the other implication is obvious). My thoughts: Let $p,q$ be two paths in the space $X$. Then since $X$ is path connected there are two paths $f,g$ connecting the two endpoints of $p,q$ respectively. The concatenation of $q.g.p.f$ is a map $S^1 \to X$ and by assumption it extends continuously to some $F : D^2 \to X$.

Here is where I'm stuck. May be it's not even the right ideas I have. Can you please help me?

As in the second definition, take two paths $p:[0,1]\rightarrow X$ and $q:[0,1]\rightarrow X$ that have the same start and end point, let $\tilde{q}$ denote the reverse path, ie the one that goes the other way, so $p(1)=\tilde{q}(0)$. The concatenated path $p\cdot \tilde{q} :[0,1]\rightarrow X$ satisfies $p\cdot \tilde{q}(0)=p\cdot \tilde{q}(1)$, so is equivalent to a map $p\cdot \tilde{q}: S^1\rightarrow X$. Intuitively, this map does $p$ on one half of the circle and $\tilde{q}$ on the other half.

The first definition now tells us that $p\cdot \tilde{q}$ extends to a map $F:D^2\rightarrow X$, where we can think of $p$ mapping from one side of the boundary and $q$ mapping from the other. Now define a homotopy from $p$ to $q$ by "moving across" the disk $D^2$ using $F$. To write this down succinctly we can write $F$ as

$F:[0,1]\times[0,1]\rightarrow X$

with $F(0,t)=F(1,t)$ for all $t\in[0,1]$. Then define the homotopy $F_t$ from $p$ to $q$ by $F_t(s)=F(s,t)$. This is definitely continuous, it fixes $\{0,1\}$ (so is a homotopy rel $\{0,1\}$) and it satisfies $F_0=p$ and $F_1=q$ by construction.

Assume WLOG that the paths have  length $1$. Consider $w:[0,1]\rightarrow \mathbb{S}^1$ send $x$ to $(\cos(\frac{x}{2\pi}),\sin(\frac{x}{2\pi}$)). Verify that $w$ is an identification map. Consider $f:\mathbb{S}^1\rightarrow X$ that sends $(m,n)$ to $(-q+p)[arctan(\frac{m}{n})]$. Verify that $fw=-q+p$. Since $w$ is an identification map and $-q+p$ is continuous. Therefore $f$ is a continuous function. Thus, there exists a continuous function $F:D^2\rightarrow X$ such that $F|\mathbb{S}^1=f$. Now remember that $D^2\cong [0,1]\times[0,1]$. Let $h:D^2\rightarrow [0,1]\times [0,1]$ be a homeomorphism. Verfiy that $h^{-1}F$ is a homotopy from  $p$ to $q$.


By induction showing that $m^3\le2^m$ for $m\ge10$ Show that $m^3\le2^m$ for $m\ge10$

My try:

Base case is true for $m=10$

Inductive Hypothesis: Assume $P(k)$ is true $\implies k^3\le2^k$

Now showing that $P(k+1)$ is true

$(k+1)^3\le2^{k+1}$

$\implies (k+1)^3\le k^3+1+3k^2+3k$

$\le 2^{k+1}+3k^2+3k+1($ from inductive hypothesis$)$

From here I could not proceed.

Can anyone explain how to proceed from here.

Hint: Use the fact that $k\geq 10$ so $2^k \geq 3k^2+3k+1$

General solution:

For $m\ge m_0$,
$$m^k\le ab^m\implies m^{k+1}=m^k\left(\frac{m+1}m\right)^k\le m^k\left(\frac{m_0+1}{m_0}\right)^k\le ab^m\left(\frac{m_0+1}{m_0}\right)^k\le ab^{m+1}$$

provided

$$\left(\frac{m_0+1}{m_0}\right)^k\le b.$$

The base case is $$m_0^k\le ab^{m_0}.$$

Here we have a specific instance with

$$k=3,m_0=10,a=1,b=2.$$

We do verify $1.331\le2$ and $1000\le1024$.

Note that for any $b>1$, you can pick

$$m_0=\left\lceil\frac1{\sqrt[k]b-1}\right\rceil$$ and $$a=\frac{m_0^k}{b^{m_0}}$$ which proves that all powers grow slower than all growing exponentials.


If you have $\max(a,b)$ and $\min(a,b)$, how would one find expressions for these using addition, subtraction, and absolute value? Just a quick question. If you have $\max(a,b)$ and $\min(a,b)$, how would one find expressions for these using addition, subtraction, and absolute value?
My thought process was that suppose $\max(a,b) = a$. Then, $a > b \Rightarrow a-b > 0$. This is an idea I had that I do not know how to continue with.

$max(a,b) = \dfrac{a+b+|a-b|}{2}$

Try to get the other one.


$100$-th derivative of the function $f(x)=e^{x}\cos(x)$ I've got this task I'm not able to solve. So i need to find the 100-th derivative of $$f(x)=e^{x}\cos(x)$$ where $x=\pi$.

I've tried using Leibniz's formula but it got me nowhere, induction doesn't seem to help either, so if you could just give me a hint, I'd be very grateful.

Many thanks!

HINT:

$e^x\cos x$ is the real part of $y=e^{(1+i)x}$

As $1+i=\sqrt2e^{i\pi/4}$

$y_n=(1+i)^ne^{(1+i)x}=2^{n/2}e^x\cdot e^{i(n\pi/4+x)}$

Can you take it from here?

There is a pattern, which is easy to see in this problem,
(e^x = @)

@cosx

1st derivative - @cosx -@sinx

2nd derivative - -2@sinx

3rd ..         - -2(@sinx + @cosx)

4th ..         - -4@cosx

100 = 0mod4

So, 100th derivative of @cosx should be of the form a@cosx and a is (-4)^25.

So, at x = π, 100th derivative is 4²⁵e^π.


Classifying Types of Paradoxes:  Liar's Paradox, Et Alia The well-known Liar's Paradox "This statement is false" leads to a recursive contradiction:  If the statement is interpreted to be true then it is actually false, and if it is interpreted to be false then it is actually true.  The statement is a paradox where neither truth value can be assigned to it.

However, "This statement is true" also leads to a paradox where either truth value can be assigned to it with equal validity.  If the statement is perceived to be true then it is actually true, and if the statement is perceived to be false then it is actually false.

These two statements demonstrate two different classes of paradox.

The same paradox states exist in set theory.  Consider "The set of all sets that do not contain themselves" leads to the former paradox (neither solution is valid), and "the set of all sets that do contain themselves" leads to the latter paradox (either solution is valid.)

My question is:  How many classifications of paradox exist? Is there any development in classifying types of paradoxes and applying them to mathematical logic, computer science, and set theory?  What implications would classes of paradoxes have on Gödel's incompleteness theorems--could a system that allows and classifies paradoxes be demonstrably consistent?

Here is a preliminary version of a paper by Noson Yanofsky on paradox and self-reference that may be of interest. There is also a final version in the Bulletin Of Symbolic Logic.

Paradoxes arise any time you try to make deductions from an inconsistent set of axioms. The Liar paradox is just a paradox derived from a pair of axioms (namely, 1="All Cretans are liars." and 2="A Cretan said, "All Cretans are liars.") that should be assumed to be true, a priori, because all of the inference rules of logic are truth-preserving transformations. There is no rule of logic that allows you to assume that any statement is false! If you don't start with a true statment and apply a truth-preserving transformation, you are just breaking the rules of logic! So, you have no warrant to expect the result to be true. The feeling that there is a paradoxical conclusion arises when your intuition tells you one thing and your brain tells you something else. If you break the rules of logic, your intuition tells you that there is something fishy while your head keeps telling you that you did everything right.

The statement, "This statement is false," is an example of a paradox arising from a single axiom.

Arrow's Impossibility Theorem is a case where any two of three given axioms are consistent, but the set of three axioms as a whole leads to a contradiction. We don't ordinarily think of this as a paradox, but it really is a case of what we might call cyclic self-reference, and hence another class of paradox.

So, my answer is that your hunch is right: there is an infinite heirarchy of paradox classes according to the maximum number of axioms in any given set of axioms and inference rules that don't lead to a contradiction. I don't know, but I can imagine that there may be more complicated cases that might be mapped on graphs that are not simple polygons.


Find the domain of the functions of the type $f(x)^{g(x)}$. Find the domain of $(\frac{2+x}{1-x})^{\frac{1}{x}}$

I tried to find the domain of this function but could not find.Then i referred the symbolab.com domain calculator,and it showed me a technique to find the domain of the functions of the type $f(x)^{g(x)}$.It found out the doamin as $[-2,0)\cup(0,1)$
It says for finding the domain of the functions of the type $f(x)^{g(x)}$,The condition is $f(x)\geq0$
The reason it gave was
$\sqrt{f(x)}=f(x)^{\frac{1}{2}}$(or any even root) has real values only when $f(x)\geq 0$
Therefore a function of the form $f(x)^{g(x)}$ is only defined for $f(x)\geq 0$ since $g(x)$ may take values like $\frac{1}{2},\frac{3}{4}...$.
I am confused here because i think $g(x)$ may take values like $\frac{1}{3},\frac{1}{7}...$ also in which $f(x)<0$ is also allowed.
But the domain it gave was correct because i verified from the graphing calculator desmos.com.But i am confused why it took only $f(x)\geq 0$ and not $<0$
Please help me clarify this confusion or please suggest me some other method to find the domain of the functions of the type $f(x)^{g(x)}$.

Consider first, $\frac{2+x}{1-x}$. Then for this quantity to be negative $x<-2$ and $x>1$, so there is no intersection. Hence this is always positive. So $(\frac{2+x}{1-x})^{\frac1x}$ is always positive $\forall x$. Clearly $x\neq 1$.

Now consider $h(x)=(\frac{2+x}{1-x})^{\frac1x}$. Taking log both sides (why is this justified ? ) we get

$ln\ h(x)= \frac1x\ ln(\frac{2+x}{1-x})$. Now for this equation to hold we must have $x\neq 0$ and $(\frac{2+x}{1-x})>0$. So either both $2+x$ and $1-x$ are positive or both of them are negative. Analysinf the corresponding cases gives $x\in [-2,0)\cup(0,1)$.

for non-elementary reasons there is no unique way to define non-integral powers of negative numbers as there are finitely (if the power is rational) and infinitely (if the power is irrational) many choices and they are all complex (non-real) numbers unless the exponent is integral and they all coincide and are real so there is a unique such, while there is a standard way (like the usual plus square root say) to define any real power of a positive number (zero is special and has only positive powers obviously), though there are again more such powers (finite in the rational case, infinite in the non-rational case) if the exponent is not integral, but there is a unique positive one which is taken as standard

so in general when you do f(x)^g(x), you need f(x) positive, g arbitrary but defined, or f(x)=0, g(x) positive, or f(x) negative, g(x) integer

in your case note that 1/x integral makes f(x) positive, so +/- 1/3 and the like work because f(1/3) is positive, not because the exponent is integral


Show that $\left|PGL(2,q)\right| = (q-1) q (q+1)$ I have thought about quite a while, I figured that if $q = 2$ or $q = 3$, then $(q + 1)! = (q + 1)(q)(q − 1)$ because $0! = 1$
and $1! = 1$ respectively. I'm sure that I can use this somehow but I can't figure out how, maybe someone knows and elegant way to show that $\left|PGL(2,q)\right|$ is indeed equal to  $(q-1) q (q+1)$

Help is appreciated :)

Let us count all the regular (=invertible) matrices, i.e. the elements of $\;GL(2,q)\;$ : look at the vector space of ordered pairs $\;V:=(\Bbb F_q)^2\;$ . From this $\;q^2\;$ vectors, there are $\;q^2-1\;$ which can be a regular matrix's first column, and then for the second column we can choose all the vectors in $\;V\;$ but those which are a multiple scalar of the first column, so we're left with $\;q^2-1\;$ choices, and all in all\

$$(q^2-1)(q^2-q)=q(q-1)(q^2-1)=q(q-1)^2(q+1)$$

To get $\;|PGL(2,q)|=|SL(2,q)|\;$ now just observe that this is the kernel of the homomorphism $\;\det: GL(2,q)\to\Bbb F_q^*\;$  , so its image's order is $\;|\Bbb F_q^*|=q-1\;$ .


Isometry Preserves Curvature Let $$f:(0,\pi/2) \longrightarrow S^1 $$ be given by $f(t)=(\cos t,\sin t)$. Let us assume we have the induced metric on $S^1$ from the standard embedding into $\mathbb{R^2}$. Call the inclusion map i. Now we know that with respect to the induced metric, $S^1$ has constant positive curvature.
Let us pull back the metric to $(0,\pi/2)$.

$$ f^*(i^*(dx\otimes dx+dy \otimes dy))=(i\circ f)^*(dx\otimes dx+dy \otimes dy)\\=(i\circ f)^*dx \otimes (i\circ f)^*dx+(i\circ f)^*dy \otimes (i\circ f)^*dy\\=(-\sin t)dt \otimes (-\sin t)dt + (\cos t)dt \otimes (\cos t)dt=dt\otimes dt$$

Now we have the standart metric on $(0,\pi/2)$. Moreover f is a diffeomorphism onto its image and hence an isometry.
Isometries preserve curvature but $S^1$ has positive curvature and $\mathbb{R}$ has 0 curvature. What am i missing?

As I understand it, isometries preserve the metric in a certain sense, which means that intrinsic curvatures will not change, but other notions of curvature certainly can.

Note that isometry preserves the Riemann curvature tensor, which means that it also preserves the Ricci tensor and scalar curvature $R$. When you showed that there is an isometry from the circle to the uncurved line, you essentially showed that the circle has zero (Riemannian) curvature (as does any 1D Riemannian manifold).

However, this "intrinsic" curvature being zero does not mean there is no curvature.
For instance, a space curve under arclength parameterization $\gamma(s)$ has curvature $\kappa(s)=||\ddot{\gamma}(s)||_2$. (Yet still has zero Riemannian curvature; perhaps calling $\kappa$ "curvature" is a disservice to students).

Another example is for 2D Riemannian manifolds in 3D, where the Theorema Egregium of Gauss applies.
It essentially states that Gaussian curvature $K$ is invariant under isometry.
However, the mean curvature $H$ is not. This is not so intuitive, since $H$ is simply the average of the principal curvatures, while $K$ is the product; why should one be invariant and the other not?
However, in this case, $R=2K$, i.e. the Gaussian curvature is just half the Ricci curvature scalar, which we know must be preserved. The mean curvature is "extrinsic", on the other hand. (Indeed, the shape operator and thus the principal curvatures appear to depend on embedding, so one reasonably expects $H$ and $K$ to as well; the fact that $K$ is in fact an intrinsic invariant was actually rather surprising, hence the very remarkable name given to the theorem by Gauss.)

This curvature is defined from the normal vector of the curve. The isometry does not act on the normal of the interval.


Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!

After moving the $x$ to the right side and squaring, it's essentially the quartic equation $$144 x^4 - 840 x^3 + 937 x^2 + 840 x - 1225 = 0.$$
This can be factorized over $\Bbb Q$: $$(12x^2 - 35x - 49)(4x - 5)(3x - 5) = 0$$ and you easily get all four real roots from here.

EDIT:
I should confess that I didn't do the factorisation by hand. Instead, I used Sage, with the following code (which you can also try on the linked page):
R.<x> = ZZ[]
(144*x^4 - 840*x^3 + 937*x^2 + 840*x - 1225).factor()

and it gave the result in no time.

Complete the square.  Then solve to find $x = \dfrac54$ or $x=\dfrac53$.


Find left hand limit of a function $f(x)=\frac{2x^2+|x|}{x}$, show that $\lim_{x\to 0^-} f(x)=-1$

My solution:
$x < 0$
Step 1: Therefore $y=(2x^2-x)/x$
Step 2: $y=2x-1$
Step 3: As $x$ approaches $0$ from left, the difference between $0$ and $x$ diminishes
$\implies$ $y$ approaches $-1$ from left and the difference between $-1$ and $y$ diminishes
Therefore $$|y-(-1)|=1-y=1-(2x-1)=2-2x=2(1-x)$$
Step 4: Let $2(1-x) < ε$
$\implies 1-x < ε/2$
$\implies x>1-ε/2$
I have to prove that $ε/2<x<0$

Where did I go wrong?

There is a little minus sign glitch in the calculation.  In the displayed line, we should have
$$|y-(-1)|=|y+1|=|(2x-1)+1|=|2x|.$$
Now it is easy to see that if $|x|<\epsilon/2$, and $x$ is negative, then $|y-(-1)|<\epsilon$. (We need to have $x$ negative so that the expression $2x-1$ for $y$ will be correct.)

Comment: I assume that you are expected to use "$\epsilon$-$\delta$." But at the informal level, it is clear that $\lim_{x\to 0^-} (2x-1)=-1$.

My solution:

First of all devide numerator and denominator by $x$

$$\lim_{x \to 0^-} \frac{2x+\frac{|x|}{x}}{\frac{x}{x}}$$

Observe that we are interested only in case when $|x|=-x$ so we may write:

$$\lim_{x \to 0^-} \frac{2x+\frac{|x|}{x}}{\frac{x}{x}}=\lim_{x \to 0^-} \frac{o+\frac{-x}{x}}{1}=\lim_{x \to 0^-}\frac{0-1}{1}=-1$$


Why is the support defined as a closure? In the definition of the support of a real function $f$ on $X$, why is it important to consider the closure of the set $S=\{x\in X:f(x)\neq0\}$ and not just $S$ itself?

Why is the closure of $S$ called the "support" of $f$ or how did this name come about?

An equivalent, and possibly more natural definition of the support is the following:  a point $x$ is in the support of $f$ precisely when $f$ does not vanish identically on a neighborhood of $x$. With this definition, it is obvious that the support is closed.

Vanishing on a neighborhood of $x$ is a much more significant property than simply vanishing at $x$, and for many purposes, it is the natural condition to consider.

Well in fact it is common (and sensible) to require the closure only if $X$ is a topological space. But actually the term more generally applies for any set, and there you cannot use the closure. (E.g. when dealing with the so called free vector space, or with sequences, one sometimes talk of functions with finite support.) As for why one takes the closure, closed sets are better behaved than arbitrary sets; if the function is defined on euclidean space (and this is a very important case) then bounded support even means compact support, and compact sets are in general very well behaved. Also one often wants to restrict some integral of a product to the support of some of the functions, and you can only integrate over measurable sets (so arbitary sets won't do), think of the Lebesgue measure. How this name came about I can only guess; the support is the only interesting subset of the domain of the function if we concentrate on the properties of the function. Often one defines functions so as to have some prescribed support, e.g. the unit ball.


Is the following a valid mathematical statement? For all $f:\mathbb N\to\{1,2,3,\ldots,100\}$,
 If $f$ is a one to one correspondence,
 Then $f^{-1}(2)=3$

It seems as though this should not be a valid statement, since the implication fails to remain a statement when the hypothesis becomes false. This is because when $f$ is not one to one and onto, the syntax $f^{-1}(2)$ is undefined. So, the implication on a whole becomes undefined under such a condition. Is this true?

An instructor asked the class whether all cell phones in the classroom are shut off.

If by some fluke, there are no cell phones in the classroom, then the answer is "yes".  "Yes" means there are no turned-on cell phones in the classroom.

Same thing here: there are no one-to-one correspondences from $\mathbb N$ into a set of only $100$ members; therefore, any statement you could make about such a correspondence is true.

Here are two ways to think about your question going back to the philosophy of logical reasoning:

(1) If we trust that classical propositional-logic is an adequate account of truth-preservation in mathematical reasoning then, looking at the truth-table for the material-conditional in classical propositional-logic, we see that the material-conditional is only false when the antecedent is true and the consequent is false. $Every$ of the 3 remaining cases is going to make the material conditional true. In this case, we have a false antecedent so it doesn't matter what the truth-value of the consequent is - that is, it doesn't matter whether $f^{-1}(2) = 3$ or not since the antecedent is already false. This is the idea behind Michael Hardy's comment.

(2) If (1) is counter-intuitive, as it may very well be as it is counter-intuitive for many people, then consider that $p \rightarrow q$ is logically-equivalent classically-speaking to $\sim p \vee q$. Seen this way, either it is false that there is a bijection from $\mathbb{N}$ to $\{1,2,3,...,100\}$ or it is true that $f^{-1}(2) = 3$. In our case, it is false that there is a bijection from $\mathbb{N}$ to $\{1,2,3,...,100\}$, making $\sim p$ true, so the statement is valid.


Module over a simple ring. Let $A$ be a simple ring. Is any $A-$module semi-simple? Why/Why not? (For simplicity, assume that the module is of finite length over $A$.)

EDIT: When is an A-module semi-simple and how?

FFR: "Any" is an ambiguous quantifier and is generally avoided in these statements.

If you mean "Is every $A$-module semisimple?" the answer is no: it is necessary for $A$ to be additionally Artinian for this property to hold for every module.

If you mean "Does there exist a simple $A$ module?" the answer is yes because all rings (with identity) have simple modules.

  The first line of this proof states that $M_1$ and $M_2$ are each just a sum of (different number of) copies of the unique simple $R$−module $M$. How?

Let $S$ be a simple right ideal. Then the sum of all right ideals isomorphic to $S$ is an ideal, hence all of $R$. Thus $R$ is a semisimple module, and all right ideals are sums of copies of $S$ too. In particular if you have a module homomorphism $R/M\cong S'$ where $S'$ is simple, then by decomposing $R=M\oplus T$ you get that $T$ is just a copy of $S$, and $S\cong S'$. This proves every simple $R$ module is isomorphic to $S$.

You can further look up why the modules of simple Artinian rings decompose into simple modules, and then you will have that every module is a sum of copies of the one type of simple submodule.


check uniform convergence of $f_n(x)=\begin{cases} 1-nx &\text{if}\;x \in [0,1/n]\\\\0 &\text{if}\;x \in [1/n,1] \end{cases}$ Let $$f_n(x)=\begin{cases} 1-nx &\text{if}\;x \in [0,1/n]\\\\0 &\text{if}\;x \in [1/n,1] \end{cases}$$  Then

$ \lim_{n \to \infty}f_n(x)$ defines a continuous function on $[0,1]$.
$\{f_n\}$ converges uniformly on $[0,1]$.
$ \lim_{n \to \infty}f_n(x)=0$ for all $x \in [0,1]$.
$ \lim_{n \to \infty}f_n(x)$ exists for  all $x \in [0,1]$.

  In this question I am not able to evaluate the limit function $f(x)=\lim_{n \to \infty}f_n(x)$.
  as I tried to form the sequence
   $f_1(x)=1-x, x\in[0,1]$ ,
  $f_2(x)=1-2x,x \in [0,1/2]$ and $f_2(x)=0 ,x \in [1/2,1]$
  $f_3(x)=1-3x,x\in [0,1/3]$ and $f_3(x)=0,x\in[1/3,1]$......
  now $f_n(1/2)=\{1/2,0,0...0\}$ so $ \lim_{n \to \infty}f_n(1/2)=0$
  $f_n(1/3)=\{2/3,1/3,0,0...0\}$ so $ \lim_{n \to \infty}f_n(1/3)=0$
  and by archimedean property for any $a \in (0,1]$ there exists $n\in \Bbb N$ such that $1/n\lt a$ so $ \lim_{n \to \infty}f_n(a)=0$ $a \in (0,1]$ but $f_n(0)=\{1,1,1...1\}$ so  $\lim_{n \to \infty}f_n(0)=1$.
  so $\lim_{n \to \infty}f_n(x)=1,x=1$ and $\lim_{n \to \infty}f_n(x)=0,x\in(0,1]$.
  Is This limit function right?
  If yes then option $4$ is right?

Yes, the fourth option is right and you computed correctly the limit of $\bigl(f_n(x)\bigr)_{n\in\mathbb{N}}$, for each $x$. One could add that:

The function is not continuous.
The convergence is not uniform, since otherwise the limit would be continuous.

Let $x > 0$. There exists $N$ such that $x > \frac{1}{N}$. So for all $n \geq N$, $f_n(x)=0$. So $\lim_{n \rightarrow +\infty} f_n(x)=0$.

For $x=0$ now, you have $f_n(x)=1$ for all $n$ so $\lim_{n \rightarrow +\infty} f_n(x)=1$.

That means that the simple limit of $(f_n)$ is the function $f$ such that $f(0)=1$ and $f(x)=0$ if $x > 0$.

In particular, $f$ is not continuous meanwhile all the $f_n$ are. This implies that the convergence cannot be uniform, otherwise the simple limit would also be continuous.


Joint density function properties here are the properties of the joint density function, can someone explain me the 4° point?
(it says: for every... as a point of continuity of f)

In a simpler one-dimension case, we have:
$$ f(x)=\frac{dF(x)}{dx} $$
and in the contrary, the integral:
$$ F(x)=\int_{-\infty}^x f(x)dx $$
We call $f(x)$ the probability density function(PDF), and $F(x)$ the cumulative density function(CDF). $f(x)$ denotes the probability density on the point x, $F(x)$ denotes the integral of probability density from $-\infty$ to the point x.


cubic root of negative numbers Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.

But when I checked Wolfram Alpha for $\sqrt[3]{-8}$, real it tells me it doesn't exist.

I came to trust Wolfram Alpha so I thought I'd ask you guys, to explain the sense of that to me.

-8 has three cube roots: $ -2 $, $1 + i \sqrt{ 3 } $ and $1 - i \sqrt{ 3 } $. So you can't answer the question "Is $ \sqrt[3]{-8} $ real" without specifying which of them you're talking about.

For some reason, WolframAlpha is only giving $1 + i \sqrt{ 3 } $ as an answer -- that looks like a bug in WolframAlpha to me.

Although the equation  $x^3+8 = 0$  has actually three roots (real $-2$ and two conjugated complex roots) still the third root of $-8$ does not exist. Nth roots are defined only for nonnegative real values. Please, consider the $$(-8)^{1/3} = (-8)^{2/6}$$  that gives either  $64^{1/6} = 2$  or  $(\sqrt{  -8})^2 $ = nonsense.   Let $(-8)^{1/3} = (-8)^{2/6}$ you never get $(-2)$ as the result. Therefore Nth root exists only of nonnegative real numbers.


Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c >  0$$

I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with differentiation.

Yes, you can do it by differentiation too. In your terms, the LHS becomes
$$
x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}.
$$
By differentiation, you can show that
$$
t+\frac{1}{t}\ge 2,\quad t>0.
$$
It gives the result.

This is a standard example: it's equivalent to
$$(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge9$$
which is AM/HM or Cauchy-Schwarz (provided all $a$, $b$, $c$ are non-negative).


$q^{3}\leq aq^{2}+bq+c$ finite amount of q problem Im a bit confused why this is true it's simple but i seem to have some problems with it $q^{3}\leq aq^{2}+bq+c$ is only true for a finite amount of $\forall q \epsilon \mathbb{N}$

First, you can assume that $a, b, c \ge 0$, for if they're negative, then $q^3 \le aq^2 + bq + c \le |a|q^2 + |b|q | c$, and if there are at most finitely many solutions the the latter, then there are to the former as well. For the same reason, we can assume $c \ge 1$, so that $a+b+c \ge 1$.

Claim: For $q \ge a+b+c$, the left hand side is larger than the right hand side. That shows that there are at most $a+b+c$ solutions.

Why is the claim true? Because

\begin{align}
q^3 &\ge  (a+b+c)^3 \\
& \ge a(a+b+c)^2 + b(a + b + c)^2 + c(a+b+c)^2 \\
& \ge a(a+b+c)^2 + b(a + b + c)^1 + c(a+b+c)^0 \text{, because $a+b+c > 1$} \\
& \ge aq^2 + bq + c \text{, by substitution, $q=a+b+c$} \\
\end{align}

At first,  we assume $a>0$.

we have

$\displaystyle{\lim_{q\to +\infty} \frac{ q^3   }{  aq^2+bq+c }  =+\infty }$

so for enough large $q$ ($q\geq$ than a certain $q_0$),

$$\frac{ q^3}{ aq^2+bq+c  }>1$$

in other terms

$\forall q\geq q_0   \frac{ q^3 }{ aq^2+bq+c }>1$.

this answers your question.

if $a<0$, the denominator goes to $-\infty$ and we use the same reasonning.

if $a=0$, you apply this to $b$ and to $c$ if $b=0$.


Partial Derivative which when evaluated gives $0/0$ My function is $f(x,y)=xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ EDIT defined $f(0,0)=0$ (How does that impact the evaluation of partial derivatives?)

I need to evaluate the function at $f_{xy}(0,0)$ and similarly for $f_{yx}(0,0)$ My problem is that upon evaluating the function at (0,0), I get cases of $0/0$ How can I get around it? Should I try rearranging terms with algebra?

You'll need to use the definition of derivative to find $f'_x$ and $f'_y$ at $(0,0)$; for $(x,y)\neq (0,0)$ you just use differentiate the expression for $f$ as usual. Then you have formulas $f'_x(x,y)$ equal to something at $(0,0)$ and something else elsewhere, and similarly for $f'_y(x,y)$. Then you again use the definition of derivative on those functions to get $f''_{xy}$ and $f''_{yx}$ at $(0,0)$.

Let $y=ax$ and take the limit as $x \rightarrow 0$.

Edit: If the result of taking the above limit is not a constant but a function of $a$, the limit does not exist. You can, however, find the value of $f_{xy}(0,0)$ or $f_{yx}(0,0)$ from a particular direction by picking $a$ such that $y=ax$ defines the line along which you want to approch $(0,0)$.


What are the functions that make composite functions called? What do you call the functions that make a composite function.

Example : $e^{\sin x} $ is made up of $\sin x$ in the argument of $e^x$

So what are $\sin x$ and $e^x$ called here in this context? Basic functions or something?

(Please don't tell me the former is trig and latter is exponential function, I know that, but that's not what I'm asking here, please try to understand)

Hope I made myself clear.

To short circuit the comments back and forth I'll provide a (community wiki) answer.

I know of no established terminology for this. If you need one for something you are writing I think "components" would be appropriate. The "compo" prefix nicely suggests "composition". Be sure to define that term for your readers, and note that the order in which you compose matters.

-- Ethan Bolker

You could also (if you were writing an article discussing this extensively, and had to use the words over and over) invent your own words, in analogy with "summand" or "multiplicand"; because of the asymmetry of the relation, I'd aim for an analogy with division or subtraction: you might, for instance, call the outer function the "composer" and the inner the "composand". Words like this are not (as far as I know) in common use, perhaps because the need for them hasn't arisen as often as those for the elementary arithmetic operations.

-- John Hughes

I'm not sure I understood your question correctly, but maybe the term you want is elementary function?

Or if, as Thomas Andrews suggested, what you want is analogous to the term "factors" for the factors $a,b,c$ in a product $a\cdot b\cdot c$, I think the word you want is "component". The components in your expression have been composed into a larger express, so "component"is just right.

In some contexts, combinatory logic in particular, the function $g(x)$ in $f(g(x))$ is commonly called the applicand of $f$, but the context you're asking about is so far removed from this that "applicand" might just confuse people.  Or maybe not! But if there's a corresponding term for the $f$ part of $f(g(x))$ I can't think what it is.


Proving inequalities with fractions: $(\frac{1}{2^n+1}+\frac{1}{2^n+2}+\frac{1}{2^n+3}+\ldots+\frac{1}{2^{n+1}})>\frac{1}{2}$ Having an immense amount of trouble trying to prove the general case for inequalities involving fractions. For instance the proof is :Prove for all natural numbers,

$(\frac{1}{2^n+1}+\frac{1}{2^n+2}+\frac{1}{2^n+3}+\ldots+\frac{1}{2^{n+1}})>\frac{1}{2}$

I understand the pattern after a little help but I do not know how to finish the proof.

I am having difficulty writing a summation for the general pattern of repeating fractions and therefore using correct notation to prove the result. How should I go about doing this?

For $n=1$

$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

For $n=2$

$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$

I find it often easier
to express a result
that uses ellipses
("...")
in terms of explicit summation
(or product) signs.

In your case,
you want to prove
$\sum_{k=1}^{2^n} \dfrac1{2^n+k}
\gt \dfrac12
$.

Since
$k \le 2^n$
for $1 \le k \le 2^n$
and
$k \lt 2^n$
for $1 \le k \lt 2^n$,
$2^n+k \le 2^{n+1}$
for $1 \le k \le 2^n$
and
$2^n+k \lt 2^{n+1}$
for $1 \le k \lt 2^n$,
so
$\dfrac1{2^n+k} \ge \dfrac1{2^{n+1}}$
for $1 \le k \le 2^n$
and
$\dfrac1{2^n+k} \gt \dfrac1{2^{n+1}}$
for $1 \le k \lt 2^n$.

Therefore
$\sum_{k=1}^{2^n} \dfrac1{2^n+k}
\gt \sum_{k=1}^{2^n} \dfrac1{2^{n+1}}
= \dfrac{2^n}{2^{n+1}}
=\dfrac12
$.


Integrating $\int_{0}^{2 \pi} \cos^{2020}(x)dx$, $\int_{0}^{\pi/2} \frac{1}{\tan^{\sqrt{2020}}(x)+1}dx$, $\int_{0}^{\infty} x^5 e^{-x^4}dx$ I've been working through the 2020 MIT Integration Bee qualifier questions (20 total) for fun, and there are three that I haven't been able to crack yet. (The complete list of problems and answers are all online (PDF link via mit.edu). However, there are no worked solutions with them.)
(9) $\quad\displaystyle\int_{0}^{2 \pi} \cos^{2020}(x) \, dx =  2^{-2019}\pi\binom{2020}{1010}$
(15) $\quad\displaystyle\int_{0}^{\pi/2} \frac{1}{\tan^{\sqrt{2020}}(x)+1} \, dx = \frac{\pi}{4}$
(20) $\quad\displaystyle\int_{0}^{\infty} x^5 e^{-x^4} \, dx = \frac{\sqrt\pi}{8}$
I think the binomial theorem might be needed for (9) since $2020 \choose 1010$ appears in the solution. I've tried substitution and integration by parts for (20) with no luck. Haven't made too much progress with (15), probably need a clever algebra trick. Any ideas would be much appreciated.

Is rather simple with $\cos x=\dfrac{e^{ix}+e^{-ix}}{2}$
$$\cos^n(x)=\frac{1}{2^n}\sum\limits_{k=0}^n
{n\choose k}e^{ikx}\cdot e^{-i(n-k)x}
=\frac{1}{2^n}\sum\limits_{k=0}^n
{n\choose k}e^{i(2k-n)x}$$
Then we group terms: first with last, second with second last, etc to get cosines back
$$2\cos^n(x)=\frac{1}{2^{n-1}}\sum\limits_{k=0}^n
{n\choose k}\frac{e^{i(2k-n)x}+e^{i(n-2k)x}}{2}=
\frac{1}{2^{n-1}}\sum\limits_{k=0}^n
{n\choose k}\cos((n-2k)x)$$
But with integration over $[0;2\pi]$ all the terms cancels except for $n-2k=0$ thus $$\int_0^{2\pi}\cos^n(x)\,\mathrm{d}x=\frac{1}{2^n}{n\choose k}\cdot 2\pi$$
where $k=\frac{n}{2}$ for an even $n$.

hint
For the first
$$\int_0^{2\pi}\cos^{2020}(x)dx=$$
by $t=x-\pi$
$$2\int_0^\pi\cos^{2020}(t)dt=$$
by $ u=t-\frac{\pi}{2}$
$$4\int_0^{\frac{\pi}{2}}\sin^{2020}(u)du$$
This is known as Wallis integral.


help to find SOR optimal parameter $w$ Please let following linear system as $Ax=b$:
$$\begin{array}{l}
6a{x_1} + {x_2} + {x_3} = 1\\
{x_1} - 3a{x_2} + 4{x_3} = 2\\
{x_1} + {x_2} - 2a{x_3} = 3
\end{array}$$
Help me to prove that the optimal value $w$ in SOR method is $\dfrac{6a}{1+6a}$.

Please see https://scicomp.stackexchange.com/questions/851/are-there-any-heuristics-for-optimizing-the-successive-over-relaxation-sor-met. The eigenvalues of $D^{-1} A$ are $1-\frac{1}{3a}$, $1-\frac{1}{2a}$, and $1+\frac{5}{6a}$. Using the formula $\omega_{opt}=\frac{2}{a+b}$ in the answer, you will get the result.

The SOR method is convergent if eigenvalues are positive. Therefore a should be greater than o.5 or smaller than -5/6. The optimal parameter w=2/(1+sqrt(lambda(min)*lambda(max))). See S.M.Grzegorski, On optimal parameter not only for the SOR method, in Applied and Computational Mathematics, November 2019


What does it mean when a function is finite? When someone says a real valued function $f(x)$ on $\mathbb{R}$ is finite, does it mean that $|f(x)| \leq M$ for all $x \in \mathbb{R}$ with some $M$ independent of $x$?

In Elias Stein's Real Analysis,  at the beginning of Chapter 4.1, it reads "We shall say that $f$ is finite-valued if $-\infty<f(x)<\infty$ for all $x$."

No, It means there are only finitely many $y$ such that $f(x)=y$, for example $f(x)=0,1,2,\dots,n$ where $n\in\mathbb{N}$ is finitely valued, although it is also bounded, but not all bounded functions are finitely valued for example $f(x)=x$ on $\mathbb{R}$ or $\mathbb{Q}$ takes uncountably many values within any $|x_i-x_j|<\epsilon$


Fractional calculus in complex analysis According to Fractional calculus, we know that $$(J^\alpha f) ( x ) = { 1 \over \Gamma ( \alpha ) } \int_0^x (x-t)^{\alpha-1} f(t) \; dt$$

It's in real analysis, but what about in complex analysis? As we know, if $\alpha$ is not a integer, then $(x-t)^{\alpha-1}$ may returns more than one values.

So my question is can we find a method to let fractional calculus working on complex field?

There are a lot of definitions of fractional derivatives. If you want an (almost complete) answer for your question, then try the book Samko, Kilbas, Marichev, Fractional integrals and derivatives: theory and applications (1993). Specially, Ch 4, § 22 Fractional Integrals and Derivatives
in the Complex Plane . "We emphasize that any work with definitions  requires precision aimed to single out a branch of the multivalued function. It is usually achieved by means of a cut which goes from the branching point to infinity or by fixing $\arg(t - z)$ in one or another way. Different choices of a cut, which fixes the branch
of the function $(t - z)^{1+\alpha}$ , and of the curve, gives different values of
$f^{(\alpha)}(z)$ in general.

Its easy. Just do the integral and then just replace your real x with complex z.


How find the postive $m,n$,such $a^n\equiv 1\pmod m$ Find all positive integer pairs $(m,n)$ with $m,n \ge 2$ such that for all $a\in\{1,2,\cdots,n\}$, $$a^n\equiv 1\pmod m$$

If $(a,m)=1$, Euler's Theorem tells us that
$$a^{\phi(m)}\equiv 1\pmod m.$$
So one family of solutions is $(n,m) = (\phi (m), m)$. Is it possible to describe all families?

A proof based in the roots of $X^n-1$

  Answer: The only pairs satisfying your statement are $(p,p-1)$ for primes $p$

Given a pair of integers $(m,n)$ such that $n>1$ suppose that for every $k\in\{1,\cdots,n\}$ we have:
$$k^n\equiv 1 \mod m\tag1$$
As noticed by Barry Cipra $n$ must be smaller than $p$ the smallest prime divisor of $m$ because $p^{n}\not\equiv 1\mod m$ hence $n<p$, And as consequence of $(1)$the roots of $x^n-1$ in $\Bbb Z_p$ are exactly $1,\cdots,n$.

Now let's prove that $n\geq p-1$, by contradiction, assume that $n<p-1$:

If $n$ is even then $p-1$ is another root of $x^n-1$ in $\Bbb Z_p$ which is absurd.
If $n$ is odd then $n+1=2a$ with $a,2\leq n$ hence $n+1$ is another root of the polynomial ($2^n\equiv 1\mod p $ and $a^n\equiv 1\mod p$ implies $(n+1)^n\equiv 1\mod n$) which is absurd.

As a conclusion $n=p-1$. Now given another prime $q$ which divides $m$, we can prove using the same method (as we did for $p$) that $n\geq q-1$ and because $n=p-1$ we have $p=q$.
If $m$ is divisible by $p^2$, we know that:
$$(p-1)^{p-1}\equiv -p(p-1)+1 \mod p^2 $$
which is absurd because $(p-1)^{p-1}\equiv 1\mod m\equiv 1\mod p^2$ hence $m=p$

Finally we proved that $n=p-1$ and $m=p$ is aprime

You are looking for something called the "order" of an element belonging to the set of integers modulo m. Now, this term comes from abstract algebra, as the order of an element in some group (don't worry about the axioms for being a group, just know that the set of residues mod m is a group) is the exponent $k$ such that $a^k = e$.

Now, $e$ denotes the identity element of a group under some operation. In our case, this is the identity element of the integers under multiplication. This is pretty clearly seen to be 1. Thus, the order of some element modulo m is the integer $k$ such that $$a^k \equiv 1 (\bmod m)$$ Now, as you probably know, a is restricted to being a member of the reduced residue set modulo m, as in, any integer such that $(a,m)=1$. Thus, for $a$, we have $\phi (m)$ choices for a. Now, the integer $a$ such that $$a^{\phi (m)} \equiv 1 (\bmod m)$$ is known as a primitive root in number theory, or, a generator of the cyclic group of reduced residues modulo m in abstract algebra. So, you have the primitive roots, and then you have the elements of order $k <\phi (m)$. An element cannot have an order greater than $\phi (m)$.

Something to note is that the order of an element must divide $\phi (m)$. This can easily be proven using the division algorithm for $\phi (m)$, rewriting is as $kq+r$, and showing that $r=0$.

Another important theorem asserts that for any $prime$ modulus, the amount of primitive roots mod m is $\phi (p-1)$, where $p$ is prime. This can help you narrow down the rest of the coprime elements. Also, for checking the order of elements, since you know that the order must divide your phi function, it is only necessary to check powers that are divisors of $\phi (n)$, which is significantly faster.

I won't prove them (unless you really want..), but there are two theorems that are pretty helpful for moduli of certain forms. Firstly, for any odd $a$, $$a^{\phi (2^{\alpha})/2} \equiv 1 (\bmod 2^{\alpha})$$

And then we also have a theorem asserting that a primitive root exists if and only if $m = 1, 2, 4, p^{\alpha}$ or $2p^{\alpha}$, where m is you modulus.

Now, I'm not sure exactly what you meant by families, but I can say that the order of any element modulo m will divide $\phi (m)$, so the families would be all elements that have an order of a divisor of $\phi (m)$.


Maximum number of real roots of this equation Question:

What is the maximum number of real roots for this equation? Assume $a,b,c$ are positive real numbers.
$$(ax^2+bx+c)(bx^2+cx+a)(cx^2+ax+b)=0$$

My attempt:
Let us assume the equation has 6 real roots.
So, the discriminants are $b^2-4ac>0,c^2-4ab>0,a^2-4bc>0$.
This implies that $b>2\sqrt{ac},c>2\sqrt{ab},a>2\sqrt{bc}$
Multiply all of them to get $abc>8abc$ which is a contradition for positive real numbers.
Now let us assume the equation has 5 real roots.
So, the discriminants are $b^2-4ac>0,c^2-4ab>0,a^2-4bc=0$. (modifying the last discrimnant here, but we could choose any other discriminant as well since all are equivalent)
This implies that $b>2\sqrt{ac},c>2\sqrt{ab},a=2\sqrt{bc}$
Multiply the first two to get $bc>4a\sqrt{bc}=>\sqrt{bc}>4a$ which is a contradition since $a=2\sqrt{bc}$.
Now let us assume the equation has 4 real roots.
So, the discriminants are $b^2-4ac>0,c^2-4ab>0,a^2-4bc<0$. (modifying the last discrimnant here, but we could choose any other discriminant as well since all are equivalent)
This implies that $b>2\sqrt{ac},c>2\sqrt{ab},a<2\sqrt{bc}$
Multiply the first two to get $bc>4a\sqrt{bc}\implies a<\sqrt{bc}/4$ which is possible for a particular combination of $a,b,c$ like $a=1,b=16,c=27$

I attempted this question on my own (self-study) and wish to know if my method is correct as I usually miss important assumptions and related things.
I hope the question is clear enough. In case of any ambiguity please comment.

If find it fairly good. There is only a little mistake (for copy-pasting, probably) in the case of $4$ real roots, when you say "multiply all of them", but further reading makes clear what you mean.

Another detail. If the equation has $4$ real roots, another possibility is that two discriminants are zero and the other is positive, but this leads to contradiction too. Perhaps you can get rid of zero discriminants at once if you assume at first that all the discriminants are non negative, instead of strictly positive. You get $abc\ge 8abc$ which is also a contradiction.

EDIT: See also DHMO's comment ("OP managed to construct an example. However, the OP has not checked if the equation actually has 4 roots.")

The assumption that a, b and c are 'positive' numbers limits the range of answers. Once you assume that a, b and c are just real (positive or negative) numbers, the answer to your question is 6 real roots. This maximum number of 6 real roots is obtained when for instance b is positive and c is negative or vice versa. You can check it with this example: a equals 1, b equals minus 5 and c equals 6.


Polynomial equation: $P(\sin t) = P(\cos t)$ Let $P(X)$ be a polynomial with real coefficients such that $P(\sin t) = P(\cos t), \, \forall t \in \mathbb R$.

Prove that there exists a unique polynomial $Q(Y)$ with real coefficients, such that $P(X) = Q(X^4-X^2)$.

(The converse is trivially true.)

NOTE:  This problem has almost nothing to do with trigonometry, despite appearances to the contrary. It's really about polynomials.

Edit:  It occurred to me I should add this - for full disclosure: I created this problem many years ago, for the math Olympiad in the East European country I grew up in; I am not looking for a solution for myself. I am offering it as a fun challenge to the math fans on this forum.

From $P(x)=P(\sqrt{1-x^2})$ we conclude $P$ is comprised of even powers, so $P(x)=L(x^2)$.

Now the condition reads $L(u)=L(1-u)$. Divide $L(u)$ by $u(1-u)$ to obtain quotient $q(u)$ and remainder $r(u)$. Substitute $L(u)=q(u)\,u(1-u)+r(u)$ into $L(u)=L(1-u)$ then reduce modulo $u(1-u)$ to obtain $au+b\equiv a(1-u)+b$, hence $a=0$. Then the functional equation descends to $q(u)=q(1-u)$ for the quotient $q$. Induct on degree to get $L(u)=Q(u(1-u))$.

Therefore $P(x)=L(x^2)=Q(x^2(1-x^2))$.

We have $$P(-\sin t)=P(\sin t)$$ because $$P(- \sin t)=P(\sin (-t))=P(\cos (-t))=P(\cos t)=P(\sin t)$$ It follows that the polynomial $P$ is an even function since a polynomial of degree $n$ is completely determined by $n+1$ distinct points. Hence $P(x)$ have even exponents only.

On the other hand, it is easy to verify the following double identity:
$$\sin^4 t-\sin^2 t=\cos^4 t-\cos^2 t=-(\sin^2 t \cos^2 t)$$
Because of $P$ is an even function, one has $P(-(\sin x \cos x)^2)=P((\sin x \cos x)^2)$ so (to work with positive but not necessary)
$$ P((\sin x \cos x)^2)=P(\sin^4 x-\sin^2 x)$$
The function $g(x)=(\sin x\cos x)^2$ is an even function of domain $\mathbb R$ which has a máximum equal to $\frac 14$ at each point $x=\frac{(1+2k)\pi}{4}$ .

We have
$$P(g(X))=Q(X)=P(X^4-X^2)\qquad (*)$$
The equality of polynomials  $(*)$ is valid for all value $X=\sin t$ so is valid for all $X$.

FINAL NOTE.- We have proved that there is a polynomial $Q(X)$ such that $Q(X)=P(X^4-X^2)$ but the proposition ask about  $P(X)= Q(X^4-X^2)$ Are these two equalities equivalent or is there a typo?. Anyway,the relation $(*)$  is true.


For what k values are there more than one solution to the following equation? Suppose we have the equation, where k is some constant:

$0 = (x+k)e^{-(x+k)^2} + (x-k)e^{-(x-k)^2} + xe^{-x^2}$

A trivial solution exists, where x = 0.

How can I figure out what range of values for k enable additional solutions for x to be found in addition to the trivial x = 0 solution?

Context:

If $y =e^{-(x+k)^2} + e^{-(x-k)^2} + e^{-x^2}$

Then y is the sum of three normal distributions, with means 0, k and -k.  If k is small, y will have one maximum at x=0.  If y is larger, there will be 3 maximums.  Therefore if we set y' = 0, we should find what k values enable more than one solution, and thus find the value of k required for three peaks to be distinguishable.

The process to solve this equation is similar to the other one you posted:

$(x+k)e^{-x^2-2kx-k^2}+(x-k)e^{-x^2+2kx-k^2}+xe^{-x^2}=0$

Multyplying both sides for $e^{x^2+2kx+k^2}$ we get:

$(x+k)+(x-k)e^{4kx}+xe^{4kx+k^2}=0$

$x+k+e^{4kx}(x(1+e^{k^2})-k)=0$

Dividing first for $x+k$ and then for $1+e^{k^2}$ leads us to:

$e^{4kx}\frac {(x+ak)} {x+k}=a$ where $a=-\frac 1 {1+e^{k^2}}$

Then the solution(s) is/are (in terms of generalized Lambert's function):

$x=-ak+\frac 1 {4k}W_{-ae^{4ak^2}}(4kae^{4ak^2}(k-ak))$

Now if you want to know how many solutions there are you must check the value of $-a$

1) iff $-a\ge\frac 1 {e^2}$ you have only one solution;

2) iff $0\lt-a\lt\frac 1 {e^2}$ you have three solutions (one positive and two negatives);

3) iff $-a\lt0$ you have two solutions (one positive and the other negative).

Numerically it's pretty simple -- and if there's a computational tool available to you (specifically Matlab) I could advise on finding a solution. But numerically I'm not sure if a sufficiently efficient method exists as not to be a complete waste of your time.


Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:

  Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .

  Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$

  $(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$

  Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$

  Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.

  Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
  So degree of $p(x)$ is atleast $4$.

  But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .

  So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$

But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$

Kindly help with this.

That fact comes from Galois Theory. Under the assumption that $K/F$ is Galois, the (monic) minimal polynomial $f$ of an element $\alpha \in K$ (over $F$) is of the form
$$f(x) = \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma \alpha).$$
This implies that $\sigma \alpha$ are all roots of the minimal polynomial for $\alpha$.

In your case, let $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ and $F = \mathbb{Q}$. Then $K/F$ is Galois and there are 4 automorphisms in $\text{Gal}(K/F)$: namely

identity on $K$
the one that sends $\sqrt{2} \mapsto -\sqrt{2}$ keeping $\sqrt{3}$ fixed: this automorphism sends $\sqrt{2} + \sqrt{3}$ to $-\sqrt{2} + \sqrt{3}$; hence $-\sqrt{2} + \sqrt{3}$ must be a root of $p(x)$.
the one that keeps $\sqrt{2}$ fixed and sends $\sqrt{3} \mapsto -\sqrt{3}$: like above, we deduce $\sqrt{2} - \sqrt{3}$ is root of minimal polynomial $p(x)$
the one that negate both roots: finally $- \sqrt{2} - \sqrt{3}$ must also be a root.

Assume that $m$ and $n$ are square-free coprime integers. Then, if we prove that $Q(\sqrt m +\sqrt n) = Q(\sqrt m, \sqrt n)$, we can conclude that that $f(x) = x^4 -2(m+n)x +(m-n)^2$ must be the minimal polynomial of $\sqrt m +\sqrt n$ since $Q(\sqrt m +\sqrt n)$ is a four-dimensional $Q$-vector space. But if $\xi = (\sqrt m + \sqrt n)$, then ${\frac {{\xi}^2-(m+n)} {2}}={\sqrt m}{\sqrt n}$. Thus, $({\frac {{\xi}^2-(m+n)} {2}})\xi = m{\sqrt n}+n{\sqrt m}$. Hence, $$({\frac {{\xi}^2-(m+n)} {2}})\xi -n{\xi} = (m-n){\sqrt n},$$ and $$({\frac {{\xi}^2-(m+n)} {2}})\xi -m{\xi} = (n-m){\sqrt n}.$$ Thus, $Q(\sqrt m +\sqrt n)$ contains $\{\sqrt m, \sqrt n \}$, and so $Q(\sqrt m +\sqrt n)\supset Q(\sqrt m ,\sqrt n)$. The reverse set inclusion is immediate. The case $m =2$, and $n = 3$ is of course what inspired this train of thought which is hopefully correct.


Is this weird "such that" notation valid? I've seen that the phrase "such that" can be notated as a $|$ or  a $:$, specially when you are using a set builder notation. My Geometry teacher also occasionally uses an $\in$ but "backwards" when stating theorems.

But now my Calculus teacher uses a weird symbol that I've never seen before. It's like an normal uppercase E, but "laying down", "rotated -90°", or like three parallel vertical lines with a horizontal line over them, like this:

Does this symbol even exist?
Have you seen it before? Where?
Is it correct to use it?

Thanks.

I've searched over 900 symbols using this tool and I wasn't able to find such one, nor have I ever seen it used.

My advice would be to just ask your professor if he invented that symbol, and if he did, for what reason.

I've seen it used before, not sure where. Doesn't seem to have caught on. It's $\ni$ = "\ni". See this tex question.


Express polynomial in terms of one coefficient Take the polynomial

$$2 a b n^2 - a^2 b^2 n$$

where $n$ is the variable, and coefficients $a$ and $b$ are integers.

Is it possible to derive an expression for $b$ in terms of $a$ and $n$, or an expression for $a$ in terms of $b$ and $n$?

UPDATE:

It seems I wasn't as clear as I thought. My bad. Here is what I meant to say... Given the expression $2 a b n^2 - a^2 b^2 n$, is it possible to completely factor out either $a$ or $b$ so that, for two functions $f$ and $g$, the expression $2 a b n^2 - a^2 b^2 n$ can be rewritten as

$$2 a b n^2 - a^2 b^2 n=f(b)g(a,n)$$

or

$$2 a b n^2 - a^2 b^2 n=f(a)g(b,n)$$

Hopefully that's clearer. You were right to pull me up. Thank you.

Suppose there exist function $f$ and $g$ such that
$$2 a b n^2 - a^2 b^2 n=f(a)g(b,n).$$
First note that if $n$ is nonzero, then for any nonzero $b$ with $b\neq2n$ we have
$$f(1)g(b,n)=2bn^2-b^2n=bn(2n-b)\neq0,$$
so in particular $g(b,n)\neq0$. Then also $bn$ is nonzero and $b\neq 2bn$, so $g(b,bn)\neq0$ and
$$f(2n)g(b,bn)
=2\cdot(2n)\cdot b\cdot(bn)^2-(2n)^2\cdot b^2\cdot(bn)=0,$$
which shows that $f(2n)=0$ for all nonzero $n$. But for all nonzero $n$ we also have
$$f(n)g(n,n^2)=2\cdot n\cdot n\cdot n^4-n^2\cdot n^2\cdot n^2=n^6\neq0,$$
a contradiction. Hence no such functions $f$ and $g$ exist.

Given only $2abn^2-a^2b^2n$, let's assume standard form $2a^2b^2n-2abn^2+0=0$. Then we can use the quadratic equation to solve for either $a$ or $b$, here solved for $a$.

$$a=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2bn^2\pm\sqrt{(2bn^2)^2-4(b^2n)(0)}}{2(2b^2n)}=\frac{2bn^2\pm 2bn^2}{2(2b^2n)}=\frac{4bn(n)\lor 0}{4bn(b)}=\frac{n}{b}\lor0$$
Since the powers of $a$ and $b$ are the same, $b=\frac{n}{a}\text{ or }0$ and this answers the original question. I'm not sure what you want to do with the part about f(a)g(...


If $a^n-b^n$ is integer for all positive integral value of $n$, then $a$, $b$ must also be integers. If  $a^n-b^n$ is integer for all positive integral value of n with a≠b, then a,b must also be integers.

Source: Number Theory for Mathematical Contests, Problem 201, Page 34.

Let $a=A+c$ and $b=B+d$  where A,B are integers and c,d are non-negative fractions<1.

As a-b is integer, c=d.

$a^2-b^2=(A+c)^2-(B+c)^2=A^2-B^2+2(A-B)c=I_2(say),$  where $I_2$ is an integer

So, $c=\frac{I_2-(A^2-B^2)}{2(A-B)}$ i.e., a rational fraction $=\frac{p}{q}$(say) where (p,q)=1.

When I tried to proceed for the higher values of n, things became too complex for calculation.

Denote $I_n=a^n-b^n$. Note that if $(a,b)$ has the property, then so has $(ma,mb)$ for $m\in\mathbb N$ as this just replaces $I_n$ with $m^nI_n$.
We have $I_1\ne 0$ because $a\ne b$ and therefore find that $a=\frac{I_2+I_1^2}{2I_1}\in \mathbb Q$, say $a=\frac uv$ with $u\in\mathbb Z$, $v\in\mathbb N$. Then $w:=vb=u-vI_1$ is an integer, hence $b=\frac wv\in\mathbb Q$. If $v=1$, we are done.
And if $v>1$, let $p$ be a prime dividing $v$. By the observation above about multiples, we may assume wlog. that $v=p$.
Write $a=A+\frac rp$ with $0<r<p$ and $A\in \mathbb Z$. Wirth $B:=A+I_1$ we obtain $b=B+\frac rp$. From $a\ne b$ we find $A\ne B$ and hence can write $A-B=p^st$ with $s\ge 0$ and $p\not\mid t$.

Pick $m\in\mathbb N$ with $mp\ge s+3$ and $p\not\mid m$.
Then
$$\begin{align}(pa)^{mp}&=(pA+r)^{mp}=r^{mp}+mp^2Ar^{mp-1}+\sum_{k=2}^{mp-1}{mp\choose k}p^kA^kr^{mp-k}+p^{mp}A^{mp}\end{align}$$
and hence
$$\begin{align}p^{mp}I_{mp}
&=mp^2r^{mp-1}(A-B)+\sum_{k=2}^{mp-1}{mp\choose k}p^kr^{mp-k}(A^k-B^k)+p^{mp}(A^{mp}-B^{mp})\\
&=mp^{2+s}r^{mp-1}t+p^{3+s}t\sum_{k=2}^{mp-1}\frac{{mp\choose k}}pp^{k-2}r^{mp-k}\frac{A^k-B^k}{A-B}+p^{mp}(A^{mp}-B^{mp})\\\end{align}$$
where everything that is written as a fraction is in fact an integer by well-known divisibilities.
Now we have a contradiction
because the left hand side and all summands on the right hand side except the first are multiples of $p^{3+s}$.

assuming $a \neq b$

if $a^n - b^n$ is integer for all $n$, then it is also integer for $n = 1$ and $n = 2$.

From there you should be able to prove that $a$ is integer.


Prove that the product of a non-zero rational and irrational number is irrational. Could you please confirm if this proof is correct?

Theorem: If $q \neq 0$ is rational and $y$ is irrational, then $qy$ is irrational.

Proof: Proof by contradiction, we assume that $qy$ is rational. Therefore $qy=\frac{a}{b}$ for integers $a$, $b \neq 0$. Since $q$ is rational, we have $\frac{x}{z}y=\frac{a}{b}$ for integers $x \neq 0$, $z \neq 0$. Therefore, $xy = a$, and $y=\frac{a}{x}$. Since both $a$ and $x$ are integers, $y$ is rational, leading to a contradiction.

As I mention here frequently, this ubiquitous property is simply an instance of complementary view of the subgroup property, i.e.

THEOREM  $\  $ A nonempty subset $\rm\:S\:$ of abelian group $\rm\:G\:$
comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where  $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$

Instances of this are ubiquitous in concrete number systems, e.g.

a is irrational, whereas b is rational.(both > 0)

Q: does the multiplication of a and b result in a rational or irrational number?:

Proof:

because b is rational:
    b = u/j where u and j are integers

Assume ab is rational:
ab = k/n, where k and n are integers.
a = k/bn
a = k/(n(u/j))
a = jk/un

before we declared a as irrational, but now it is rational; a contradiction. Therefore ab must be irrational.


Prove that for every $x>0$, it follows that $\frac{\ln x}{x} \leq\frac{1}{e}$ I need to prove that for every $x>0$ $$\frac{\ln x}{x} \leq\frac{1}{e}$$

I tried to show that when the limit goes to $+\infty$ the function $\frac{\ln x}{x} -\frac{1}{e}$ goes to $-1/e$ but that of course doesn't guarantee it won't suddenly “jump” far and beyond somewhere along the way and then come back.

When I take the derivative I get that for some of the range the function is going up rather than down, so that it being smaller than $0$ cannot be guaranteed.   As a result don't I know how to prove this.

defining $$f(x)=\frac{x}{e}-\ln(x)$$ then we get:
$f(x)$ has a Minimum in $x=e$ and $$\lim_{x\to 0^+}f(x)=+\infty$$ and $$\lim_{x\to \infty}f(x)=+\infty$$ thus we have $$\frac{\ln(x)}{x}\le \frac{1}{e}$$ for $x>0$

$$
\frac{ln(x)}{x} \leq \frac{1}{e} \iff x^e \le e^x. e^x \in C(\mathbb R), x^e \in C(\mathbb R).
$$
$e^x \gt x^e, x \in [0,e)$ and $(e,+\infty)$ because both of them are continious and have only one point of intersaction (e), after that point $e^x \gt x^e$ as before, so $\frac{ln(x)}{x} \leq \frac{1}{e}$ is proofed.


Function Domain of cubic root Does the domain of the function $y=\sqrt[3]{x^3+1}$ include $x<-1$? If yes, why is Mathematica and Wolfram Alpha not plotting that part of the function?

The function:

$y=(x^3+1)^{\frac{1}{3}}$

Should include a domain of all real numbers because negative numbers also can have a cube root. So, yes, it should include $x \lt -1$. I'm not sure why those websites are acting up.


Finding eigenvectors of symmetric matrices when the general solutions of eigenvalues inserted return several vectors Given a symmetric matrix

$ A = \left(\begin{array}{rrr}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2
\end{array}\right). $

By solving the characteristic equation I found out that the eigenvalues are as follows:
$\lambda{_1} = 4$,
$\lambda{_2} = 1$,
$\lambda{_3} = 1$.

Solving $(A-\lambda{_1}I) = 0$ and the general solution to it's RREF returns:

$\vec w_2 = x_3 \left(\begin{array}{rrr}
1 \\
1 \\
1
\end{array}\right). $

Solving $(A-\lambda{_2}I) = 0$ and the general solution to it's RREF returns:

$\vec w_3 = x_2 \left(\begin{array}{rrr}
1 \\
-1 \\
0
\end{array}\right). $

$\vec w_4 = x_3 \left(\begin{array}{rrr}
1 \\
0 \\
-1
\end{array}\right). $

So since this is a symmetric matrix as we can see easily, the eigenvectors are supposed to be orthogonal to each other.
I can immediately see that $\vec w_4$ is orthogonal to $\vec w_2$.
The question now is in order to find the 3rd eigenvector, do you simply take the cross-product of $\vec w_4$ and $\vec w_2$ finding a vector orthogonal to both of them? Doing this I found this vector $\left(\begin{array}{rrr}
1 \\
-2 \\
1
\end{array}\right). $

The last vector happen to be the right answer, but my question is if this really is the correct way to do it, or if that was simply luck (will it work in every case)? If the last is the case what's the correct method?

Correct answer is supposed to be:

Note that the matrix can be written as $I+ee^T$, where $e$ is a vector of ones.

The $I$ just shifts the eigenvalues by one, just concentrate on $A=e e^T$.

Note that $A e = n e$, so this gives one eigenvalue. Let $v$ be any vector
such that $v \bot e$, then $Av=0$. Hence all other eigenvalues are zero.

Hence the eigenvalues of the original matrix are $1,....,1,n+1$.

In the problem above, you can pick any two vectors that are orthogonal to $e$, they are not unique.

In given matrix sum of the elements in all the rows is 2 + 1 + 1 is equal to 4. Since this sum is same in all the rows so one of the eigen values is 4
Now, the trace of the matrix is equal to the sum of the eigenvalues so, sum of all the eigenvalues is equal to 6. Also, the determinant of the matrix is equal to the product to the eigenvalues which is equal to 4.
Now, λ₁+λ₂+λ₃=6 & λ₁*λ₂*λ₃=4
so, λ₁=4, λ₂=1, λ₃=1


Prove that certain elements are not in some ideal I have the following question:

Is there a simple way to prove that

$x+1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$ and

$x-1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$

without using the fact that $\mathbb{Z}[x]/\langle2, x^2+1\rangle$ is an integral domain?

Thanks for the help.

First a comment: it looks like the goal of the problem is to show that $(2,x^2+1)$ isn't a prime ideal (and hence the quotient isn't a domain), because $(x+1)^2=x^2+1+2x$ is in the ideal, and we hope to show that $x+1$ isn't in the ideal.

If $x+1\in (2,x^2+1)$, then we would be able to find two polynomimals in $\Bbb Z[x]$, say $a$ and $b$, such that $x+1=2a+(x^2+1)b$.

Looking at the equation mod 2, you get $x+1=(x^2+1)\overline{b}$, where all the terms are in $\Bbb F_2[x]$. But since $\Bbb F_2[x]$ is a domain, the degrees on both sides have to match. If $\deg(b)>0$, then the degree of the right hand side would be at least 3, and even if the degree of $b$ were 0, the right hand side would have degree 2. It is impossible then, for such an expression to be equal to $x+1$. By this contradiction, we conclude $x+1$ is not in the ideal.

Finally, you can note that $x+1$ is in the ideal iff $x-1$ is, since $x-1+2=x+1$.

Thus we have shown that $(2,x^2+1)$ is not a prime ideal, and it isn't even a semiprime ideal.

$x-1 \in I \implies x-1=2p+q(x^2+1) \implies q=0$ (reasoning on degree) so that you have $x-1=2p$ which is clearly impossible.


What does it mean by piecewise smooth boundary? I will be highly obliged if anyone can give me any reference where i can get the definition of domain (in $\mathbb{R^n}$) with piecewise smooth boundary.

My question is when a domain in $\mathbb{R^n}$ is said to have a piecewise smooth boundary?

I tried a lot to find in Google but i didn't get.Please help me out.

I do not think there is a commonly agreed on definition. Here is one, based on the discussion in

William Thurston, Three-dimensional geometry and topology. Vol. 1. Edited by Silvio Levy. Princeton Mathematical Series, 35. Princeton University Press, Princeton, NJ, 1997.

A  domain $\Omega$ in $R^n$ has piecewise smooth boundary if for each $x\in cl(\Omega)$ there exists a neighborhood $U$ of $x$ in $R^n$ and a diffeomorphism $f: U\to U'\subset R^n$, such that $f(cl(\Omega))\cap U$ equals $\Omega'\cap U'$, where $\Omega'\subset R^n$ is a convex polyhedron.

I got stuck on the definition of piecewise smooth boundary while reviewing material for a calculus course I am taking. I can't intuitively see the what the definition means but it happens to solve your question.

The definition is: if the boundary of a set is a finite union of piecewise, simple closed curves, then the set has a piecewise smooth boundary and moreover, the set itself called regular region.

My own interpretation is that if every collection of curves which forms the boundary of such a set does not self-intersect in any way, then the boundary formed from those curves is said to be piecewise smooth boundary.


Use Linear Approximation to estimate $\Delta f=f(3.02)-f(3)$ I have a homework question that is asking to use the linear approximation to estimate $\Delta f=f(3.02)-f(3)$ for $f(x)=x^2$.

I know that the linear approximation is $L(x) = f'(a)(x-a)+f(a)$.

I however do not quit understand what the problem is asking in the first place. Couldn't I just estimate $\Delta f=f(3.02)-f(3)$, by plugging both 3.02 and 3 into the original function of $f(x) = x^2$ and subtracting them. How do I do this with linear approximation?

You can indeed compute: $$f(3.02)-f(3) = (3.02)^2-3^2 = (6.02)\cdot 0.02 = 0.1204,$$ the exact value. The point of the exercise is avoiding this computations. Using linear approximation gives: $$L(3.02) = f(3) + f'(3)\cdot0.02,$$ so that $\Delta f \approx f'(3)\cdot 0.02 = 6 \cdot 0.02 = 0.12$. Good, no?

here is what you can do to approximate $3.02^2 - 3^2$ using difference of squares:

$$3.02^2 - 3^2 = (3.02 - 3)(3.02 + 3) \simeq 0.02*6 = 0.12 $$


Differential Equation By Method of Integrating factors The differential equation is,

$$\frac{dy}{dx}=x^2-2x^2y$$

I have calculated $μ(x)$ as $μ(x)=e^{2x^3/3}$

thus

$$y=e^{-2x^3/3}\int x^2e^{2x^3/3}\,dx$$

After this I have been having a bit of trouble solving for $y$. I have tried integration by parts but it hasn't helped, probably due to a mistake on my part.

If you are asking how to solve the integral you have, you can use u-substitution.

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

$du = 2x^2$

So

$\int x^2e^{\frac{2x^3}{3}} = \frac{1}{2} \int e^u du = \frac{1}{2} e^{\frac{2x^3}{3}} + C$

If you're going to integrate by factors, couldn't you just write the question as:

$dy/dx = x^2(1-2y)$

This makes it much easier, because you now can integrate x and y separately, as it's own thing, etc.

$(1-2y)^{-1}\ dy=x^2\ dx$

Integrate both sides to get:

$-\frac12 \ln(1-2y)=x^3/3 + C_1$

$\ln(1-2y) = -\frac{2x^3}3 + C_2$

$1-2y=Ae^{-2x^3/3}$

$y=\frac12(1-Ae^{-2x^3/3})$


How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? How can I evaluate
$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$?
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.

In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$

No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Let's find a general formula for the following sum: $$S_{m}=\sum_{n=1}^{m}nr^{n}.$$

Notice that
\begin{align*}
S_{m}-rS_{m} & = -mr^{m+1}+\sum_{n=1}^{m}r^{n}\\
  & = -mr^{m+1}+\frac{r-r^{m+1}}{1-r} \\
& =\frac{mr^{m+2}-(m+1)r^{m+1}+r}{1-r}.
\end{align*}
Hence
$$S_m = \frac{mr^{m+2}-(m+1)r^{m+1}+r}{(1-r)^2}.$$
This equality holds for any $r$, but in your case we have $r=\frac{1}{3}$ and a factor of $\frac{2}{3}$ in front of the sum.    That is
\begin{align*}
\sum_{n=1}^{\infty}\frac{2n}{3^{n+1}}
& = \frac{2}{3}\lim_{m\rightarrow\infty}\frac{m\left(\frac{1}{3}\right)^{m+2}-(m+1)\left(\frac{1}{3}\right)^{m+1}+\left(\frac{1}{3}\right)}{\left(1-\left(\frac{1}{3}\right)\right)^{2}} \\
& =\frac{2}{3}\frac{\left(\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^{2}} \\
& =\frac{1}{2}.
\end{align*}

Added note:

We can define $$S_m^k(r) = \sum_{n=1}^m n^k r^n.$$  Then the sum above considered is $S_m^1(r)$, and the geometric series is $S_m^0(r)$.  We can evaluate $S_m^2(r)$ by using a similar trick, and considering $S_m^2(r) - rS_m^2(r)$.  This will then equal a combination of $S_m^1(r)$ and $S_m^0(r)$ which already have formulas for.

This means that given a $k$, we could work out a formula for $S_m^k(r)$, but can we find $S_m^k(r)$ in general for any $k$?  It turns out we can, and the formula is similar to the formula for $\sum_{n=1}^m n^k$, and involves the Bernoulli numbers. In particular, the denominator is $(1-r)^{k+1}$.

What about the Cauchy product?
$$
\sum_{n=0}^\infty(n+1)x^n=\sum_{n=0}^\infty\sum_{\ell=0}^nx^n=\left(\sum_{n=0}^\infty x^n\right)\left(\sum_{m=0}^\infty x^m\right)=\frac{1}{(1-x)^2}.
$$


$\epsilon - \delta$ proof. Need help I have been knocking my head against this proof for an hour now and I'm no closer to solving it. I simply have no idea where to go. The thing I need to prove is, given
$$f(x)=\frac{2x+3}{x+1},$$
show that $f(x)\to 3$ for $x\to 0$.

I have tried to factor it out but I simply have no luck. Can anybody give me a hint of where to go with this?

Start with $| \frac{2x+3}{x+1}-3|=|\frac{-x}{x+1}|$.

For $|x|< \delta$, you have $|-x|=|x|<\delta$.

For the denominator, use the reverse triangle inequality to get:
$|1+x| \geq 1 - |x| >1-\delta$.

Put everything together to get: $|\frac{-x}{x+1}|<\frac{\delta}{1-\delta}$, and choose $\delta$ to be less than $\frac{\epsilon}{1+\epsilon}$.

if $\lim_\limits {x\to a} f(x) = l$ and $\lim_\limits{x\to a} g(x) = m$ and $m\ne 0$ then $\lim_\limits{x\to a} \frac {f(x)}{g(x)} = \frac {l}{m}$

$\forall \epsilon_1 > 0, \exists \delta >0 : |x-a|<\delta_1  \implies |g(x) - {m}|<\epsilon_1$

$x<\delta_1 \implies |g(x)| > 0$

let $M= \inf \{|g(x)|: |x-a|<\delta_1|\}$

$\forall \epsilon_2 > 0, \epsilon_1 < \frac 1M \implies |\frac {1}{g(x)} - \frac 1{m}|<\epsilon_2$

$\forall \epsilon_3 > 0, \exists \delta_2 >0 : |x-a|<\delta  \implies |f(x) - l|<\epsilon_3$

$\forall \epsilon > 0, \exists \delta >0 : |x-a|<\delta_1  \implies |\frac {f(x)}{g(x)} - \frac {l}{m}|<\epsilon$

$|f(x) \frac {1}{g(x)} - l\frac {1}{m}|\\
|f(x) \frac {1}{g(x)} -f(x)\frac {1}{m} + f(x)\frac {1}{m}- l\frac {1}{m}|\\
|f(x) (\frac {1}{g(x)} -\frac {1}{m}) + (f(x)-l)\frac {1}{m}| < |f(x) |(\frac {1}{g(x)} -\frac {1}{m})| + |(f(x)-l)|\frac {1}{m}| < |f(x)|\epsilon_2 + |\frac {1}{m}|\epsilon_3$