Showing kinetic and potential energy is constant Consider the initial value problem:

  \begin{cases}
u_{tt} &= c^2 u_{xx} \ \ & \text{for} \ -\infty < x < \infty, \  0 \leq t < \infty\\
u(x,0) &= \phi(x)  \ \ & \text{for} \ -\infty < x < \infty\\
u_t(x,0) &= \psi(x) \ \ & \text{for} \ -\infty < x < \infty\\
\end{cases}
  where $\phi$ has compact support (that is, outside some bounded interval, $\phi$ is zero), and $\psi(x) = 0$. Define the kinetic energy $KE = \frac{1}{2}\int_{-\infty}^{\infty} \rho u_t^{2} dx$ and the potential energy $PE = \frac{1}{2} \int_{-\infty}^{\infty} T u_x^{2} dx$. Show not that, for large enough times $t$, each of $KE$ and $PE$ is itself constant, and they are equal to each other. Can you prove the same thing if the inital velocity $\psi$ merely has compact support, instead of being zero?

I am not sure how to start this, how am I to show that $KE$ and $PE$ are constant? I usually post some work but I am not sure how to start this. Any help would be useful.

Define $E_k(t) \equiv \frac{1}{2}\int_{\mathbb{R}}u_t^2(x,t){\rm d}x$ and $E_p(t) \equiv \frac{1}{2}\int_{\mathbb{R}}c^2u_x^2(x,t){\rm d}x$. Then taking the derivative and using integration by parts we get
$$\frac{d}{dt}(E_k(t) + E_p(t)) = \int_{\mathbb{R}}u_t[u_{tt} - c^2u_{xx}]{\rm d}x = 0$$
so the total energy is conserved. To show that $E_p(t) = E_k(t)$ for large $t$ we need an expression for the solution. This is given by d'Alemberts formula

$$u(x,t) = \frac{\phi(x+ct) + \phi(x-ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s){\rm d}s$$

from which you can compute $u_t$, $u_x$ and derive

$$E_k(t) - E_p(t) = \frac{1}{2}\int_{\mathbb{R}}[\psi(x+ct)+c\phi'(x+ct)][\psi(x-ct)-c\phi'(x-ct)]{\rm d}x$$

To make the algebra above simpler note that $u_t^2 - c^2u_x^2 = (u_t - cu_x)(u_t+cu_x)$. Now since $\phi$ and $\psi$ has compact support then there is a $M>0$ such that $\phi'(x) = \psi(x) = 0$ for all $|x| > M$. Now if $ct > \frac{M}{2}$ then either $|x-ct| > M$ or $|x+ct|>M$ for all $x$ so the integrand above is identical to zero and $E_p = E_k$ follows. Finally since we know that $E_k + E_p$ is constant it follows that $E_p$ and $E_k$ has to be constant for large $t$.

We have
\begin{equation}
\begin{cases}
u_{tt} = c^2 u_{xx} \ \ & \text{for} \ -\infty < x < \infty, \  0 \leq t < \infty\\
u(x,0) = \phi(x)  \ \ & \text{for} \ -\infty < x < \infty\\
u_t(x,0) = \psi(x) \ \ & \text{for} \ -\infty < x < \infty
\end{cases}
\end{equation}
Therefore
\begin{equation}
\phi(x) = 0 \ \ \ \text{and} \ \ \ \psi(x) = 0
\end{equation}
and
\begin{equation}
KE = \frac{1}{2}\int_{-\infty}^{\infty} \rho u_t^{2} dx
\end{equation}
\begin{equation}
PE = \frac{1}{2} \int_{-\infty}^{\infty} T u_x^{2} dx
\end{equation}
From equation $(1)$ $u_{tt} = c^2 u_{xx}$ then
\begin{equation}
u_{tt} - c^2 u_{xx} = 0
\end{equation}
Here, $A = -c^2, B = 0$, and $C = 1$. Thus
\begin{equation}
B^2 - 4AC = 4C^2 > 0
\end{equation}
So the equation is hyperbolic. The equation characteristics are
\begin{equation}
\frac{dt}{dx} = \pm\frac{1}{C}
\end{equation}
or $$\xi = x - ct = \ \text{const} \ \ \ \text{and} \ \ \ \eta = x + ct = \ \text{const}$$
Now, in terms of new coordinates $\xi$ and $\eta$ then $$u_{xx} = u_{\xi \xi} + 2u_{\xi \eta} + u_{\eta \eta} \ \ \ u_{tt} = c^2(u_{\xi\xi} - 2 u_{\xi\eta + u_{\eta\eta}}$$
Thus, equation $(1)$ becomes
\begin{equation}
-4c^2 u_{\xi\eta} = 0
\end{equation}
Therefore $c\neq 0$,
\begin{equation}
u_{\xi \eta} = 0
\end{equation}
So, integrating $u_{\xi\eta} = 0$ twice we then get the solution
\begin{equation}
u(\xi \eta) = \phi(\xi) + \psi(\eta)
\end{equation}
where $\phi$ and $\psi$ are arbitrary functions. So, in terms of $x$ and $t$
\begin{equation}
u(x,t) = \phi(x - ct) + \psi(x + ct)
\end{equation}
Now from
\begin{equation}
u(x,0) = \phi(x) \ \ \ \text{and} \ \ \ u_t(x,0) = \psi(x)
\end{equation}
So,
\begin{equation}
\phi(x) + \psi(x) = \phi(x)
\end{equation}
and
\begin{equation}
-c \phi^{\prime}(x) + c\psi^{\prime}(x) = \psi(x)
\end{equation}
Integrating equation $(14)$
\begin{equation}
-c \phi(x) + c\phi(x) = \int_{x_0}^{x}\psi(\tau)d\tau
\end{equation}
where $x_0$ is an arbitrary constant. Now equation $(13)$ and $(15)$ becomes
$$\phi(x) = \frac{1}{2}\phi(x) - \frac{1}{2c}\int_{x_0}^{x}\psi(\tau)d\tau$$
and
$$\psi(x) = \frac{1}{2}\psi(x) + \frac{1}{2c}\int_{x_0}^{x}\psi(\tau)d\tau$$
Thus, equation $(11)$ gives the solution (D'Alembert solution) of the Cauchy problem as
$$u(x,t) = \frac{1}{2}\left[ \phi(x - ct) + \phi(x + ct) \right] + \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(\tau)d\tau$$
So we have
$$u_t(x,t) = \frac{1}{2}\frac{\partial}{\partial t}\left[ \phi(x - ct) + \phi(x + ct) \right] + \frac{1}{2c}\frac{\partial}{\partial t}\left[ \int_{x-ct}^{x+ct} \psi(\tau)d\tau \right]$$
and
$$u_{x}(x,t) = \frac{1}{2}\frac{\partial}{\partial x}\left[ \phi(x-ct) + \phi(x+ct) \right] + \frac{1}{2c}\frac{\partial}{\partial x}\left[ \int_{x - ct}^{x+ct}\psi(\tau)d\tau \right]$$
therefore
$$KE = \frac{1}{2}\int_{-\infty}^{\infty}\rho u_{t}^{2}dx$$
So,
$$KE = \frac{1}{2}\int_{-\infty}^{\infty}\rho\left[ \frac{1}{2}\frac{\partial}{\partial t} \left[ \phi(x-ct) + \phi(x + ct) \right] \right] + \frac{1}{2c}\frac{\partial}{\partial t}\left[\int_{x - ct}^{x+ct}\psi(\tau)d\tau \right]^2dx$$
and $$PE = \frac{1}{2} \int_{-\infty}^{\infty} T u_x^{2} dx$$
So,
$$PE = \frac{1}{2}\int_{-\infty}^{\infty}T\left[ \frac{1}{2}\frac{\partial}{\partial x}\left[ \phi(x - ct) + \phi(x + ct) \right] + \frac{1}{2c}\frac{\partial}{\partial x}\left[ \int_{x-ct}^{x+ct} \psi(\tau)d\tau \right] \right]^2dx$$
Thus from above, we can see that for $t > 0$, $KE$ and $PE$ is constant thus $KE = PE$ at some stage.


How to show that the inverse Gaussian density integrates to 1? How to prove
$\int_{0}^{\infty}\left[\frac{\lambda}{2\pi x^3}\right]^{1/2}\exp\left\{\frac{-\lambda(x-\mu)^2}{2\mu^2 x}\right\}dx=1$?

You can use the closed form formula for the more general integral

  $$\int_{ 0 }^{\infty} (ax^m)^{s} e^{\frac{-b(x-\mu)^2}{x}} dx = 2\,{a}^{s}{\mu}^{ms+1}{{\rm e}^{2b\mu}}
{{\rm K_{m s+1}}\left(2\,b\mu\right)},$$

where $\rm{K}_{\nu}(x)$ is the modified Bessel function  of the second kind.

Remember that google is your friend. Here are some sources where you can find the proof

Youtube:

http://www.youtube.com/watch?v=g_bCDcNWcgU

Math StackExchange:

How to show the normal density integrates to 1?

Wikipedia:

http://en.wikipedia.org/wiki/Gaussian_integral


Definition of quadratic equation? What is a quadratic equation and what is its simplified and cannonic form?

Quadratics are usually written in the following three ways:

Expanded form

This form is written canonically as

$$f(x)=ax^2+bx+c$$

This is a very general form and it is easy to find $f(0)$ if needed.  However, finding the roots are a bit more difficult.

Factored form

This form is written as

$$f(x)=a(x-x_1)(x-x_2)$$

This form is very convenient if you need the roots of $f(x)$, as they are simply $x_1$ and $x_2$.

Vertex Form (complete the square)

This form is written as

$$f(x)=a(x-h)^2+k$$

This form is convenient for finding the vertex $(h, k)$.  This is the form that is used to derive the quadratic formula.

Expanded form is usually considered to be the most basic form, so this is probably what you want.

In Algebra, Equation with the highest degree 2 is a quadratic equation. And its simplest form is
$$
ax^2+bx+cx = 0,
$$
where $x$ represents variable and $a$,$b$, and $c$ represents constant or numbers. Also, $a$ should not be equals to zero because if $a = 0$ then the equation will become a linear one. And its graphical representation is a parabolic shape.


How to compute the $\lim_{(x,y)\to(0,0)}\frac{x^{2}}{x^{2}+y^{2}}$? I have computed the limite of $\lim_{(x,y)\to(0,0)}\frac{x^{2}}{x^{2}+y^{2}}$ as follows:

I want to know if with the first two paths -Path 1 and Path 2-, are sufficient to say that the limit does not exist, because the given paths are different -$1 \neq \frac{1}{2}$.

What about the third one?

Is this procedure correct?

Yes, it is enough to conclude from the first two paths that the limit does not exist.

Yes. This means as you get arbitrsrily close to the origin, you have some function values that do not get arbitrarily close to any potential limit (because they get arbitrarily close to some other potential limit)


Geometric series and Fourier transform In my book solution example, I have this signal
discrete square wave
I don't understand how the right hand side is attained from the expression on the left
$$\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}$$
Where the last step is accomplished by changing variables and simplifying using the fact that
$$\omega_0 = \frac{2\pi}{p}$$
Is there some book can I read to understand things like that?

Starting with the last expression
$$e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}$$
$$=\sum_{m=0}^{M-1}e^{iMk\omega_0}e^{-imk\omega_0}$$
$$=\sum_{m=0}^{M-1}e^{-i(m-M)k\omega_0}$$
Here we can shift the index by $p-M$ and now we have
$$\sum_{m=p-M}^{M-1+p-M}e^{-i(m-M-p+M)k\omega_0}$$
$$=\sum_{m=p-M}^{p-1}e^{-i(m-p)k\omega_0}$$
$$=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}e^{ipk\omega_0}$$
Since $\omega_0=\frac{2\pi}{p}$, we have
$$\sum_{m=p-M}^{p-1}e^{-imk\omega_0}e^{2\pi ik}$$
$$=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}$$
Therefore,
$$\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}$$

Using index shifting property that states:
$$\sum_{s=p}^{N}f(x)=\sum_{s=p+M}^{N+M}f(x-M)$$
Beginning from the R.H.S and using shift of $(M-p)$:
$$R.H.S=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=\sum_{m=p-M+\bf{M-p}}^{p-1+\bf{M-p}}e^{-ik\omega_0(m-(\bf{M-p}))}=\sum_{m=0}^{M-1}e^{-imk\omega_0}e^{iMk\omega_0}e^{-impk\omega_0}$$
$\because e^{iMk\omega_0}$ is a constant and $e^{-impk\omega_0}=e^{-i2\pi mk}=1$
$$\therefore L.H.S=\sum_{m=p-M}^{p-1}e^{-imk\omega_0}=e^{iMk\omega_0}\sum_{m=0}^{M-1}e^{-imk\omega_0}=R.H.S $$


How to factor $x^4-81$? So I was trying to evaluate the following limit:

$$f(x) = \lim \limits_{x\to 3} \frac{x^4 - 3x^3}{x^4-81}$$

To do so, you would need to factor the denominator to $(x^2+9)(x+3)(x-3)$ and cancel out $(x-3)$. If I wouldn't have seen it being factored this way, it would've been pretty difficult for me to factor the denominator in 3 terms. Is there a simple way/method to factor cases like that, that would work in most cases?

Thanks in advance!

Hopefully you know that $a^2-b^2=(a+b)(a-b)$

We can use this on the initial equation to get $$x^4-81=(x^2+9)(x^2-9)$$

We can then use the same technique on the second bracket to get $$x^2-9=(x+3)(x-3)$$

This gives us our final answer $$x^4-81=(x^2+9)(x+3)(x-3)$$


prove ot disprove $\frac{\sin{1}-\sin{x}}{1-x}-\cos{\left( \frac{x+1}{2}\right)}>0.$ let $0<x<1$, prove or dispove
$$arcos{\left(\dfrac{\sin{1}-\sin{x}}{1-x}\right)}<\dfrac{x+1}{2}$$
or
$$\dfrac{\sin{1}-\sin{x}}{1-x}>\cos{\left(
\dfrac{x+1}{2}\right)}.\tag{1}$$

if $(1)$ is right.then I find a method solve  inequality

I try $$\dfrac{\sin{1}-\sin{x}}{1-x}=\cos{\xi},x<\xi<1$$
it suffuce to show
$$\cos{\xi}>\cos{\dfrac{x+1}{2}},x<\xi<1$$

Method 1. Let $a=(1+x)/2$  and $b=(1-x)/2.$ Then $a+b=1$ and $a-b=x . $ So $$\sin 1-\sin x= \sin (a+b)-\sin (a-b)=$$ $$=(\sin a \cos b+\sin b \cos a)-(\sin a \cos b-\sin b \cos a)=$$ $$=2\sin b \cos a.$$ And $(1-x)^{-1} =(2b)^{-1}.$ So the inequality is equivalent to $$(2b)^{-1}(2 \sin b \cos a)>\cos (1+x)/2=\cos a.$$ Since $x\in (0,1)\implies a=(1+x)/2\in (0,1) \implies \cos a>0,$ we may divide out the term $\cos a.$ So the inequality is equivalent $$(2b)^{-1} (2\sin b)>1.$$ Since $b>0,$ this is equivalent to $\sin b>b,$ which is FALSE.

Method 2. With $a=(1+x)/2$ and with $f(y)=\sin y$ for all $y,$ the inequality is equivalent to $f(1)-f(x)-(1-x)f'(a)>0 .$ To show this is false, we have $$(i).\quad f(1)-f(a)=(1-a)f'(a) +\frac {1}{2}(1-a)^2f''(c) \text { for some } c\in (a,1).$$ $$(ii).\quad f(x)-f(a)=(x-a)f'(a)+\frac {1}{2}(x-a)^2f''(d) \text { for some } d\in (x,a).$$  Subtracting (ii) from (i), and using  $(1-a)^2=(x-a)^2$ we have $$(iii).\quad f(1)-f(a)=(1-x)f'(a)+\frac {1}{2}(1-a)^2 (f''(c)-f''(d)).$$ Now $0<x<d<a<c<1$ so $0<d<c<1.$ And $f''(y)=-\sin y$ is strictly decreasing for $y\in  (0,1), $ so $$f''(c)-f''(d)<0.$$  Applying this to  $(iii),$ considering that $(1-a)^2=((1-x)/2)^2>0,$ we have $$(iv).\quad f(1)-f(a)-(1-x)f'(a)=\frac {1}{2}(1-a)^2(f''(c)-f''(d))<0.$$


Find the Eigenvector of a matrix Find the eigenvectors of the matrix

$$\displaystyle\begin{bmatrix}
0 &2  &3 \\
-2 &0  &5 \\
-3 &-5  &0
\end{bmatrix}.$$

So I start with $|A-\lambda I|=0$

$$\begin{bmatrix}
-\lambda &2  &3 \\
-2 &-\lambda  &5 \\
-3 &-5  &-\lambda
\end{bmatrix}=0$$

$$-\lambda(\lambda^2+25)-2(2\lambda+15)+3(10-3\lambda)=0 $$

$$\lambda^3+38\lambda=0$$

The given eigenvalue is -2, which is later used for finding the eigenvector. My equation won't give -2. This is also a part of a multiquestion series, so I cannot think that the question has something wrong in it.

What did I do wrong here ?

Your characteristic polynomial is right, it is $\lambda^3 + 38 \lambda = 0$. But $-2$ is not a root of it, since $(-2)^3 + 38 \cdot (-2) = -84$. If you transform the equation like this $\lambda^3 + 38 \lambda = \lambda ( \lambda^2+38)=0$, then you find that $\lambda_1 = 0$ is a real eigenvalue, and there are two complex $\lambda_{2,3} = \pm i \sqrt{38}$. I let you the rest to solve.

You have not done anything wrong. I have checked your
answer with gp (available from http://pari.math.u-bordeaux.fr/).
I am awful in numerical calculations!
Check whether
you have copied the matrix correctly.


Prove that $|V(G)| \leq \alpha(G)\cdot \omega(G)$ (clique number, independency number) How can you prove that $|V(G)| \leq \alpha(G)\cdot \omega(G)$?

(Number of vertices in a graph is less than or equal to the graph's clique number * its independency number.)

This is false, consider the self complementary graph $C_5$

If you build a random graph by taking e.g. 10'000 vertices and for every pair throwing a (fair) coin to decide whether to put the edge or not, you probably end up having alpha < 3 log |V| (and same for omega by symmetry), so alpha * omega < 9 (log |V|)^2.

There's however a conjecture that if you only look at graphs with no induced copy of some graph H (no matter how you choose H), then alpha * omega > |V|^c where c is some positive constant depending only on H. (Note that the above example isn't a counter-example to that). It's a very hard conjecture, only confirmed for very small H -- but not for H being the cycle on five vertices, for example.


Dedekind Cut additive inverse Let $\alpha$ be Dedekind cut and define $\alpha^* :=\{x\in\mathbb{Q}|\exists r>0\space \text{such that} -x-r\notin\alpha\}$. I need to show that $\alpha^*$ is a Dedkind Cut and the additive inverse of $\alpha$. Note the the additive identity is defined as $0^*:=\{x\in\mathbb{Q}|x<0\}$.  Any help is greatly apprecyed and needed. Thanks in advance

For the 1st part I can show that $\alpha^*$ is downwardly closed as if $y\in\mathbb{Q}$ and $x\in\alpha^*$ with $y<x$ then for some $r>0$, $-x-r>a$,  $\forall a\in \alpha$, but as $-y>-x$, we have $-y-r>-x-r>a$ $\forall a \in\alpha $ i.e. $-y-r\notin \alpha$ so $y\in\alpha^*$.  But I am stuck showing that $\alpha^*$ has no top element and that $\alpha^*\neq\emptyset$ and $\alpha^*\neq\mathbb{Q}$, so any help please.

For the second part I can show that $\alpha+\alpha^*\subset 0^*$ as if $w\in\alpha+\alpha^*$ then $w=u+v$ where $u\in\alpha$  &  $v\in\alpha^*$. Therefore for some (rational) $s>0$

$-v-s>u$ that is $u+v<-s<0$.  But I am stuck with the reverse inclusion.  So please any help will be greatly appreciated and please let me know if what I've done is correct and on the right track.  Thanks

Well, for the first, suppose that $x_0\in\alpha^*.$ Then by definition, there is some rational $r>0$ such that $-x_0-r\notin\alpha.$ But then $\frac r2$ is also a positive rational, and $x_1=x_0+\frac r2$ is rational and greater than $x_0,$ and $-x_1-\frac r2=-x_0-r\notin\alpha.$ Thus, for any $x\in\alpha^*,$ we can find a $y\in\alpha^*$ with $x<y$.

For the second, we will need the following result:

  Lemma:  Given a Dedekind cut $\alpha$ and any $r\in\Bbb Q$ with $r>0,$ there exist $a,b\in\Bbb Q$ with $a\in\alpha$ and $b$ a non-least element of $\Bbb Q\setminus\alpha$ such that $0<b-a<r$.

To prove this we let $z_0$ be the least element of $\Bbb Q\setminus\alpha$--if there is such an element. We define a function $m:\Bbb Q\times\Bbb Q\to\Bbb Q$ by $$m(x,y)=\frac{x+y}2,$$ we fix any $a_0\in\alpha,$ and any non-least $b_0\in\Bbb Q\setminus\alpha.$ Then we define sequences $\langle a_n\rangle_{n=0}^\infty$ and $\langle b_n\rangle_{n=0}^\infty$ recursively as follows:

(1) If $m(a_n,b_n)=z_0$ for some $n,$ then for all integers $k\ge n$ we let $a_{k+1}=m(a_k,z_0)$ and $b_{k+1}=m(z_0,b_k).$

(2) If $m(a_n,b_n)\in\alpha,$ let $a_{n+1}=m(a_n,b_n)$ and let $b_{n+1}=b_n.$

(3) If $m(a_n,b_n)$ is a non-least element of $\Bbb Q\setminus\alpha,$ then let $a_{n+1}=a_n$ and $b_{n+1}=m(a_n,b_n).$

It can be shown that $m$ is well-defined, that $\langle a_n\rangle_{n=0}^\infty$ is a well-defined sequence of elements of $\alpha,$ that $\langle b_n\rangle_{n=0}^\infty$ is a well-defined sequence of non-least elements of $\Bbb Q\setminus\alpha,$ and that for all integers $n\ge 0$ we have $$0<b_0-a_0=2^n\cdot(b_n-a_n).$$ By the Archimedean Property of the rationals, there exists some positive integer $n$ such that $n\cdot r>b_0-a_0,$ so since $2^n>n>0,$ we have $$2^n\cdot r>n\cdot r>b_0-a_0=2^n\cdot(b_n-a_n)>0,$$ whence $$0<b_n-a_n<r,$$ as desired.

I leave the details to you (unless, of course, you already have that result). Now, to show that $0^*\subseteq\alpha+\alpha^*,$ take any $q\in0^*,$ and put $r=-q,$ so $r\in\Bbb Q$ and $r>0.$ By the Lemma, there exist some $a\in\alpha$ and some non-least $b\in\Bbb Q\setminus\alpha$ such that $0<b-a<r.$ Then $q=-r<a-b=a+-b.$ It remains only to show that $-b\in\alpha^*,$ which again I leave to you.

I actually found an easier proof of this without requiring the above lemma.  So, let $M,$ $N,$ and $Z$ be Dedekind cuts.  Let $N$ be defined as containing the elements of $\Bbb Q$ such that the elements are negative and not in $M.$  It is easy to show the $M + N$ consists of rationals that are negative and less than zero.

$\Bbb Q$ and $\Bbb R$ are the respective number systems

Let $m,n ∈\Bbb R$

Define:

$M = \{q_1 ∈\Bbb Q : q_1 < m\}$

$N = \{q_2 ∈\Bbb Q : \text{for some rational }r ∈\Bbb Q\setminus M, q_2 < -r\}$

$M + N = \{q_1 + q_2 : q_1 ∈ M\text{ and }q_2 ∈ N\}$

$Z = \{q_3 ∈\Bbb Q : q_3 < 0\}$

$\Bbb Q\setminus M,$ by the way, is another way of saying $\Bbb Q-M,$ or subtract set $M$ from $\Bbb Q$

It can be shown that $\Bbb Q\setminus M$ is $> M$ (just do a simple proof by contradiction).  So, this implies that for any $r ∈\Bbb Q\setminus M,$ $r > m > q_1,$ for any $q_1$ in $M.$

Thus, $q_2 < -r \implies -q_2 > r > m > q_1.$  So, $-q_2 > q_1$ and $0 > q_2 + q_1.$

So, for any $q_1 + q_2 ∈ M + N,$ $q_1 + q_2 < 0.$  That is, we can redefine $M + N$ as

$M + N = \{q_1 + q_2 ∈\Bbb Q : q_1 + q_2 < 0\}$

Now, the strategy is this, to show that any element in $Z$ (which contains negative rationals less than zero) must also be in $M + N.$  This is now pretty easy.  So, let
$z ∈ Z.$  So, $z < 0.$  If $z \leq q_1 + q_2,$ then we know $Z$ is a proper subset of $M + N$ (this comes from either the definition of Dedekind cuts or a simple lemma to show that $M$ is a subset of $N$ if $m < n,$ and vice versa).  So, for this case, we are done.  Suppose, on the contrary, that $q_1 + q_2 < z.$  Then, since $M + N$ is also a Dedekind cut, there must be a larger element $p = q_3 + q_4$ in $M + N$ such that $z < p.$  $\frac{q_1 + q_2}{2}$ is one such example.  It is less negative, and therefore larger, than $q_1 + q_2.$ It is also another quotient and belongs in $M + N.$  Continuing this train of thought, $\frac{q_1 + q_2}{2^k}$ must be larger than the $\frac{q_1 + q_2}{2^{k-1}}.$  So, there must be a $k$ such that $z \leq \frac{q_1 + q_2}{2^k}.$  So, for any $z,$ $z \leq p,$ for some $p$ in $M + N.$  Thus, $Z$ must be a proper subset of $M + N,$ and we are done.

I personally found this proof to be easier to follow that Cameron's proof.


What is an isosurface? I am trying to understand the marching cubes algorithm. I would like very much an easier definition of an isosurface than what is available online. Could anyone please explain it?

Thanks.

Isosurface is another way to call a surface defined by the implicit equation

$$F(x,y,z)=f$$ where $F$ is a function of space and $f$ a constant, often $0$. The prefix iso- indicates that the function $F$ takes the same value ($f$) all over the surface.

The marching cube algorithm is able to construct the iso-surfaces (decomposed in triangles) for a given $F$, by sampling over a regular grid.

Just knowing the function values at the vertices of many tiny cubes allows you to estimate the intersections of the faces with the surface by interpolation, giving you skewed polygons that approximates the surface inside each cube, in a way that ensures continuity across the cubes.

As there are 8 values to be considered, hence 8 signs/zero, there are 6561 possible configurations per cube. Fortunately, this number can be reduced by means of symmetries.

Just read this: http://www.econym.demon.co.uk/isotut/isobars.htm

THE best isosurface tutorial on the web.

An isosurface is the surface of a 2d or 3d shape where all the values of the graph are the same (ISO=same). The surface can be infinite, like a gyroid, or contained, like a sphere. It does not use a parametric formula. X,Y and Z are all in the same equation and the result of the equation is a gradient, i.e. from - infinity to + infinity, and the surface of the 3d shape can be all points at zero or all points at one for example.

So, the mathematician writes a formula of a sphere of a gyroid or any other generative shape where (X Y Z = a value), and he scans all the points in space, and marks all those that return the value he wants, like 0 is a smaller surface, and if he changes the value to 1 it's a larger surface.

The tutorials online are illustrated and make a heck of a lot more sense than anyone could be expected to do on stack exchange, using 20 formulas, each illustrated, and even online ISOsurface plotting webGL pages.


Prove that $\frac{q(r+1)}{(\beta,\beta)}=\frac{q'(r'+1)}{(\alpha,\alpha)}$ Let  $\alpha $,$ \beta$ $\in$ $\Phi$. Let the $\alpha$-string through $\beta$ be $\beta-r\alpha$,$\ldots$,$\beta+q\alpha$,and let the $\beta$-string through $\alpha$ be
$\alpha-r'\beta$, $\ldots$,$\alpha+q'\beta$.Prove that
$$\frac{q(r+1)}{(\beta,\beta)}=\frac{q'(r'+1)}{(\alpha,\alpha)}.$$

$r-q=\langle \beta,\alpha \rangle$

Note that the reflaction along $\alpha$, which  maps $\alpha\mapsto-\alpha$, $\beta\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha$ leaves the $\alpha$ string invariant, hence exchanges the ends of the string, i.e.
$\beta-r\alpha\mapsto\beta+q\alpha$.
On the other hand, by linearity, $\beta-r\alpha\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha+r\alpha$ so that we conclude
$$r-q = 2\frac{\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}. $$
This shows that the second result $$ r-q=\langle\beta,\alpha\rangle$$ holds if and only if $\langle\alpha,\beta\rangle=0$ or $\langle\alpha,\alpha\rangle=2$.


A finite dimensional algebra over a field has only finitely many prime ideals and all of them are maximal Let $K$ be a field and let $R$ be a $K$-algebra with unity which is finite dimensional as a $K$-vector space. Prove that $R$ has only finitely many prime ideals all of which are maximal. (Hint: Chinese remainder theorem)

My attempt: We have $R \simeq K^n$, for some $n$, as vector spaces. I claim that there are finitely many ideals of $K^n$. Let $I$ be an ideal of $K^n$ and $(a_1, a_2,...,a_n) \in I$. For every entry $a_i$, we have two choices, either $a_i=0$ or not.  So we get $2^n$ many cases, and any case give a particular ideal of $K^n$ which are distinct from others.

Is my proof correct or am I missing some point? If it is true, how can we give a proof by using the Chinese remainder theorem? Thanks!

A finite dimensional (commutative or not) algebra over a field is an artinian ring, because it certainly satisfies the descending chain condition on left or right ideals.

Let $A$ be a commutative artinian ring and let $P$ be a prime ideal. Then $A/P$ is an artinian domain, which is a field: if $D$ is an artinian domain, consider $d\in D$ and the chain of ideal $(d)\supseteq(d^2)\supseteq\dots\supseteq(d^n)\supseteq\dotsb$. This chain stabilizes, so $(d^n)=(d^{n+1})$ for some $n$, which means that $d^n=td^{n+1}$. Since $D$ is a domain, $1=td$ or $d=0$. Therefore all prime ideals of $A$ are maximal.

Consider now a minimal element in the family of products of finitely many maximal ideals, let it be $M_1\cdots M_k$. If $M$ is a maximal ideal, then
$$
MM_1\cdots M_k\subseteq M_1\cdots M_k
$$
so, by minimality, $MM_1\cdots M_k=M_1\cdots M_k$ and
$$
M_1\cdots M_k\subseteq M
$$
Since $M$ is prime, it follows that $M_i\subseteq M$, for some $i=1,\dots,k$; therefore $M_i=M$.

Here we don't use any general fact about artinian rings, just that any nonempty set of ideals has a minimal element (which is general theory of lattices). The Chinese remainder theorem can be used to get an upper bound on the number of maximal ideals; let $M_1,\dots,M_k$ be the set of (pairwise distinct) maximal ideals; then the Jacobson radical $J(A)=M_1\cap\cdots\cap M_k=M_1\cdots M_k$ by coprimality; so
$$
A/J(A)\cong A/M_1\times\cdots\times A/M_k
$$
by the CRT. If $A$ is a finite dimensional $K$-algebra, then $k\le\dim_K A/J(A)\le\dim_K A$. The upper bound can be reached: the product algebra $K^n$ has exactly $n$ maximal ideals.

I have the impression that the underlying additive structure is that of a $K$-vector space. But can one conclude that $R=K^n$ (isn't the ring $\mathbb{F}_p\times \mathbb{F}_{p^2}$ also a $\mathbb{F}_p$-algebra?). In any case, $R$ is a direct sum of fields $\oplus^k_{i=1} F_i $, so every ideal is of the form $\oplus_{i\in S} F_i$ with $S\subseteq\{1,\ldots,k\}$, and the maximal ideals are those where $S=\{1,\ldots,i-1,i+1,\ldots,k\}$.


Prove or Disprove: For all $x \in \mathbb{R}$, there exists a positive real number$\epsilon$ such that either $x \leq (-\epsilon) or x \geq \epsilon$ Can you guys help me prove or disprove this?
Also this is what I got for the negation of this statement: There exists $x \in \mathbb{R}$ for all positive real numbers $\epsilon$ such that either $x > (-\epsilon)$ or $x < \epsilon$

Is my negation correct?
Where should I start with my proof?

Your negation is not quite correct, because the negation of

  either $x\le-\epsilon$ or $x\ge\epsilon$

is not

  either $x>-\epsilon$ or $x<\epsilon$;

it’s

  $x>-\epsilon$ and $x<\epsilon$.

However, the original statement is not true: it has one (and only one) exception. What real number is that exception?

Let $x \in R$.

Let $\epsilon \in R : 0 < \epsilon < x$ or $ \epsilon < -x < 0$.

Since $x + \epsilon \in R$, and $0 < \epsilon$, your statement is true.


For $f(x) = e^x + x^3 - x^2 + x$ find the limit $\lim\limits_{x\to \infty} \frac{f^{-1}(x)}{\ln x}$. I have the function:

$$f : \mathbb{R} \rightarrow \mathbb{R} \hspace{2cm} f(x) = e^x + x^3 -x^2 + x$$

and I have to find the limit:

$$\lim\limits_{x \to \infty} \frac{f^{-1}(x)}{\ln x}$$

(In the first part of the problem, I had to show that the function is strictly increasing and invertible. I don't know if that's relevant to this, since I could show that the function is invertible, but I can't find the inverse.)

So this is what I tried:

I showed

$$\lim_{x \to \infty} f(x) = \infty$$

and so I concluded that $\lim\limits_{x \to \infty} f^{-1}(x) = \infty$. I'm not sure if this is correct, it might be wrong. But if would be right, then we could use l'Hospital, knowing that:

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

but after trying to use all of this on paper, I got nowhere. It just complicated thing a lot more.

So how should I solve this limit?

Hint : $$\forall x\geq 1\quad 2e^x \geq e^x+x^3-x^2+x\geq e^x+1$$

So $$\ln(\frac{x}{2})\leq f^{-1}(x)\leq \ln(x-1)$$

So we deduce that your limit is a constant by the sandwich theorem .

For $x \to \infty, f(x) \sim e^x$ so $f^{-1}(x) \sim \ln(x)$. So the answer is $1$.


Compositions of homotopic maps are homotopic I'm reading some lecture notes on homotopy, and the author has just proved the theorem:

  If $f_{0} \simeq f_{1}$ and $g_{0} \simeq g_{1}$, then $g_{0} \circ f_{0} \simeq g_{1} \circ f_{1}$

Where, for each $i \in \{0,1\}$, $f_{i} : X \to Y$ and $g_{i}: Y \to Z$, and $\simeq$ denotes homotopy equivalence.

He does this by showing $g_{0} \circ f_{0} \simeq g_{0} \circ f_{1} \simeq g_{1} \circ f_{1}$
For the first equivalence he uses the homotopy $g_{0} \circ F$ (where $F$ is a homotopy from $f_{0}$ to $f_{1}$) and for the second he uses $G \circ (f_{1} \times Id_{[0,1]})$. I completely understand that. But I don't see why you couldn't (or wouldn't) use the quicker:

$$H(x,t)=G(F(x,t),t)$$

This is continuous (unless there's something key I've misunderstood), and $H(x,0)=G(F(x,0),0)=g_{0}(F(x,0))=g_{0}(f_{0}(x))$ and similarly for $t=1$.

There's no strong reason not to use $G(F(x,t),t)$.  One reason the author may not have is that they may have had in mind the general principle that you can concatenate homotopies, and that this often lets you build homotopies that "do multiple things" by doing each one separately and concatenating the pieces along the time coordinate.  In this particular example, it is actually possible to do both of them at once as you have pointed out, but it isn't always (see for instance my answer to this question).

Edit: as in the comments, this is clearly false.

I think the difficulty lies in showing rigorously that $G(F(x,t),t)$ is continuous.

While I see no reason that it shouldn't be, I can't see a way to show it is continuous that is as simple as the appeal to the glueing lemma that shows the construction given in the notes is continuous.


Verification: Give a definition of a $G$-Set so that it may be viewed as an algebra A $G$-Set for those who may not be familiar with the terminology is as follows:

Let $G$ be a group and $S$ be a set. $G$ acts on $S$ define by the map $\star : G \times S \to S $ where $e \star s = s$ and $g\star (h \star s) = gh \star s, \forall s \in S, g,h \in G $ and so $\langle  S, \star \rangle$ is a $G$-set

An algebra is a pair $\langle A, F \rangle $ where $A$ is the universe and $F$ is the family of operations on $A$

Now I need to give a definition that presents $G$-sets as an algebra which are equivalent to the above definition of a $G$-set

I thought one such definition would be :

$\langle S, \star \rangle =\langle S, \star_{g},\star_{h} ... \rangle$ (***) where we have $\star = \{\star_{g} : g \in G\}$

$\star_{g} s = g \star s, \forall g \in G, \forall s \in S$

That is for all elements of $G$, each element becomes associated with a unary operation in the signature of (***) .

so we would have $\star_{e}s = e \star s = s, \forall s \in S$
for $\star_{g}(\star_hs) = \star_g(h \star s) = g \star (h \star s)) = hg (\star s) = \star_{gh}s$

This seems valid as we have our universe $S$ with a signature of operations and can axiomatize it by the above equations.

Yes, what you describe is the standard way of describing a $G$-set as an algebra. This is for fixed $G$. You can also give equational axioms for the theory of group actions (where the group is also allowed to vary), but you have to use multi-sorted logic.

You can take a $\Bbb{K}$-vector space $V$ (for some field $\Bbb{K}$), whose dimension is equal to the cardinality of the set $S$. That is, for each $s \in S$, you have a basis element $e_s \in V$. Then your construction is essentially the so-called "group algebra" $\Bbb{K}[G]$, realized as a sub-algebra of the endomorphisms of $V$. So what you call $\star_g$ is realized concretely here as the linear map defined by

$$ \star_g(e_s) = e_{g \star s} $$

So what I'm getting at is that $\mathrm{End}_\Bbb{K}(V)$ is an algebra, and the maps $\star_g$ generate a sub-algebra.


Number of ways in which 6 rings can be worn on the 4 fingers of one hand The way I solved this is -

The 1st finger can have any of the 6 rings, $\therefore 6$ ways

The 2nd finger can have any of the 5 remaining rings, $\therefore 5$ ways

The 3rd finger can have any of the 4 remaining rings, $\therefore 4$ ways

The 4th finger can have any of the 3 remaining rings, $\therefore 3$ ways

$\therefore$ total number of ways = 5*4*3*2 = 120 ways.

But every website I check for the solution, they are different and I'm very confused. Is my approach correct?

You can order the $6$ rings in $6! = 720$ ways. Now you have to split the string of $6$ rings into $4$ groups, one for each finger. One way to see how to do this is stars and bars, you have $6$ stars ($*$, the rings) and $4 - 1 = 3$ bars ($\mid$, separations between groups), this is to select $4 - 1$ positions for the $\mid$ among $6 + 4 - 1$ positions in all. Pulling all together:

$\begin{align}
6! \cdot \binom{6 + 4 - 1}{4 - 1} = 720 \cdot 84 = 60480
\end{align}$

See you can choose the fingers in ${5\choose 4}$ ways . and then normal calculations which you have done $=5.6.5.4.3=1800$ hope its clear.


Find the point on ellipse $4x^2+y^2=4$ that is furthest from $(1,0)$ The equation of an ellipse is $4x^2 + y^2 = 4$. Find the furthest point $(x,y)$ from $(1,0)$.

Additional information from comment: First, I transformed this equation into the general formula and the result I got was: $$\frac {x^2}1 + \frac{y^2}{2^2} = 1.$$

My professor used this formula to find the distance between the 2 points: $$\sqrt{(x-1)^2 + (y-0)^2}$$ but I don't understand where this formula comes from.

The diagram below depicts the ellipse, and the right triangle involved in the distance formula:

In this diagram, the point on the ellipse has coordinates $(x, y)$ in general.  (Here, I've chosen $x = -\frac12$ and $y = \sqrt{3}$, but it could be any point on the ellipse.)  The two legs are represented by the blue line segment, which has length $|x-1|$, and the green line segment, which has length $|y|$.  The hypotenuse—the orange line segment—is the desired distance $r$.

From Pythagoras, we have

$$
r^2 = (|x-1|)^2+(|y|)^2 = (x-1)^2+y^2
$$

where we can get rid of the absolute value because any real value $z$ and its absolute value $|z|$ have the same square.  Take the square root of both sides (since $r \geq 0$), and we get

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

Let $d$ be the distance from the point $(1,0)$ to a point on the ellipse $(x,y)$, then $d^2 = (x-1)^2+y^2= (x-1)^2 + 4-4x^2= x^2-2x+1+4-4x^2 = -3x^2-2x+5 $. This is a quadratic function and you know it attains a max at $x = -\frac{b}{2a} = - \frac{-2}{2(-3)} = -\frac{1}{3}$. Thus $d^2_{\text{max}} = -3\left(\frac{-1}{3}\right)^2-2\left(\frac{-1}{3}\right)+5 = \frac{16}{3}$ . So $d_{\text{max}} = \dfrac{4}{\sqrt{3}}$. Hope this helps !


The most general linear map of $\mathbb{R}^n$ to $\mathbb{R}$ Find the most general linear map of $\mathbb{R}^n$ to $\mathbb{R}$.

Obviously, dot product is one. However, I'm not sure how do I show that it's the most general of them? Or maybe, something else is even more general? Hint appreciated!

As with any two finite dimensional vector spaces, $V$, $W$, you can fix a basis for $V$ and a basis for $W$ and then the linear transformations from $V$ to $W$ become rectangular matrices which you perform matrix multiplication with.

So, one could say that a general linear transformation from $\Bbb R^n$ to $R$ can be identified with a matrix $[a_1,\dots a_n]$ where $T(x_1,\dots x_n):=[a_1,\dots a_n][x_1,\dots, x_n]^\top \in \Bbb R$.

You can see that this means all linear maps arise as the dot product of some fixed vector with other vectors.


The Hurried Duelers brainteaser This question is similar as this other one asked in the forum, but I am trying to give it a different twist. Unfortunately, I am not getting to the same answer, so there might be something wrong in my math.

The problem in this case is the following:

Duels in the town of Discretion are rarely fatal. There, each contestant comes at a random moment between 5 a.m. and 6 a.m. on the appointed day and leaves exactly 5 minutes later, honor served, unless his opponent arrives within the time interval and then they fight. What fraction of duels lead to violence?

My solution is the following. Let X and Y be uniformly distributed random variables on $[0,60]$, each corresponding to the time of arrival of each dueler. The desired probability is
$$
P(X + 5 < Y) + P(Y+5<X)
$$
which, by symmetry, equals $2\,P(X+5<Y)$. I have done the following:
$$
2P(X+5<Y) = 2\iint_{X<Y-5}f_{X,Y}(x,y)\,dx\,dy = 2\int_5^{60}\int_0^{y-5}\left(\frac{1}{60}\right)^2\,dx\,dy
$$

But this does not lead to the desired probability $\frac{1}{6}$. Is perhaps my approach wrong?

Your calculation should be $$1- \displaystyle 2\int_5^{60}\int_0^{y-5}\left(\frac{1}{60}\right)^2\,dx\,dy$$  which would be $1-\frac{55^2}{60^2} \approx 0.1597$ or slightly less than $\frac16$.  I am not sure you have multiplied by $2$ or subtracted from $1$. Perhaps this picture helps

For the violence to occur next man should come in $5$ mins after the first one so the probability of that is $5/60$. but these man can interchange the position so answer will be $2*5/60=1/6$.


Term for "existence" and uniqueness, when you already have solution candidate? For any given property $P(x)$ defined for $x\in X$, we can ask "does there exist an $x\in X$ such that $P(x)$?", and "if so, is this $x$ the unique element of $X$ that satisfies this".

This is called "existence and uniqueness".

However, assume now that the $x$ has already been identified as a candidate for having property $x$, and that we have to prove that $P(x)$ holds, and is unique.

What shorthand term would we use for "this particular $x$ has the property $P(x)$"?

Basically, I am looking for an equivalent version of the statement "prove existence and uniqueness" in the case where the candidate $x$ is already known: "prove ..... and uniqueness".

The reason I'm asking, is in order to shorten the phrase "prove that $x$ satisfies the property $P(x)$ and uniquely so".

EDIT: what I want to do is be able to write a proof like this:

First we prove [existence] (substitute existence for the term im looking for).

.....

Now we prove uniqueness.

....

Let object $x$ and property $P(\check{x})$ be given, where $\check{x}$ is a free variable.

If you want clarity, then you can go with

  "Prove $P(x)$ holds and for each object $y$, if $P(y)$ holds, then $y=x$."

If you want brevity, then you can go with

  "Prove $x$ is the unique object satisfying $P$."

If you have a specific $x$ in mind, then you are able to identify that $x$ because it is the unique element which satisfies some other formula / property $Q$. Then the sentence "For all $y$, $Q(y)$ iff $P(y)$" says that there is only one element which satisfies $P$, and that element is $x$.


Determine whether the region $\text{Im}(\frac{z+1}{z-1}) \leq 3$ is open, closed or neither. Determine whether the region $\text{Im}\left(\frac{z+1}{z-1}\right) \leq 3$ is open, closed or neither.

My attempt and the memo for the question is below.
My answer is somewhat similar to the memo's except that I didn't have the $z \neq 1+0i$ condition.
Is this additional $z \neq 1+0i$ part correct?
While I was working on the solution it didn't seem immediately evident to me, that that condition was necessary.
If the memo is correct, do you have any tips on how I can spot such conditions in similar questions in the future?
Thank you.
Memo answer
My answer

See what happens when you substitute $z= 1+ 0i = 1\;$ into your inequality.
Then $$\text{Im}\left(\frac{z+1}{z-1}\right) = \text{Im}\left(\frac 20\right).$$
You can't do that, because $\frac 20$ is undefined; so we must forbid $z = 1 + 0i = 1$.


Show that Set in $M:=\{x\in \Bbb R^3 : x_1^2\ge2(x_2^3+x_3^3) \}$ is closed I have to show this regarding the Euclidean metric.

I've already shown that it isn't bounded by showing that the $d(x,y)\:\forall x,y \in M$ isn't bounded.

I know that in order to show the closedness i have to show that the complement is open, but how do I show that $$M:=\{x\in \Bbb R^3 : x_1^2\lt 2(x_2^3+x_3^3) \}$$ is open? (I know that a set is open if the $\epsilon$-ball around every element is also in the set but how do I show this in this particular case)

I have to determine whether the set is closed and whether it's bounded for the following set too, but I don't have any ideas how to show either:
$$M:=\{x\in \Bbb R^3 : x_1 \ge -1, \:x_2\lt2, ||x||_2\le\sqrt5\}$$
Thanks in advance

Let $f(x_1,x_2,x_3)= x_1^2-2(x_2^3+x_3^3)$, which is a polynomial, hence continuous. Then the set in question is $f^{-1}([0,\infty))$ which is closed, hence by the definition of continuity, the inverse image of this closed set is closed.

You need to show that for every $x\in M$ you can find an open ball that is still in $M$. The open Ball definition is :
\begin{align}
B_r(x) = \{ y\in \mathbb{R^3} | \sqrt{\sum_{i=1}^3 (y_i-x_i)^2} < r)\}
\end{align}

So you have an inequality for all y in the ball. You have an inequality for all x int the set $M$. Combining the two if you can find a $r$ that satisfy both, you are done !


I have a question about group $G$ which satisfies Inn$(G) $ char Aut$(G)$ and $Z(G)$=$\{1\}$. Let $G$ be a group which satisfies $Z(G)=\{1\}$ and ${\rm Inn(G)} \space \mathbb{char} \space {\rm Aut(G)}$; then every automorphism of $A={\rm Aut(G)}$ is an inner automorphism. ($H \space \mathbb{char} \space G$ means that $H$ is a characteristic subgroup of $G$. Note that we can assume $G \subseteq A$, since $Z(G)=\{1\}$ so $G \cong{\rm  Inn}(G)$.)

I am given a hint, and it says that $C({\rm Inn}(G)) \unlhd {\rm Aut}(A)$, so one derives $C({\rm Inn}(G)) \subseteq C(A)$. I'm stuck only on this point. Why can we say that? ($C(H)$ means the centralizer of $H$ in ${\rm Aut}(A)$.)

Let $I = {\rm Inn}\, G$ and $C = C_{{\rm Aut}(A)}(I)$. Then $C \unlhd {\rm Aut}(A)$ and $A \unlhd {\rm Aut}(A)$. Since $A$ is by definition the group of automorphisms of $G \cong I$, no nontrivial element of $A$ can centralize $I$; i.e. $C \cap A = 1$. Hence $[C,A] \le C \cap A = 1$; i.e. $C \le C(A)$.

I post my solution here:(to prove $C_{AutA}(I)=1$)
http://www.artofproblemsolving.com/community/c7t435f7h622064_automorphism_group_of_automorphism_group_of_simple_group_g


Diagonalization versus s.d. product for non-commuting Hermitian matrices Although the application of the following is in quantum physics, the question per se is mathematical:

I have seen two characterizations of the problem in measuring a discrete variable of a state $\psi$ exactly with each of two non-commuting Hermitian operators $A$ and $B$:

The product of the standard deviations ( $=\sqrt{\langle \psi | A^2 | \psi \rangle - \langle \psi | A | \psi \rangle^2}$ and ditto for $B$) $\geq 1$.
One cannot simultaneously diagonalize the matrix representations of $A$ and $B$
(i.e., if $A = U^\dagger CU$ and $B = V^\dagger DV$, for unitary $U$ and $V$ and diagonal $C$ and $D$, with $\dagger$ denoting the adjoint), then $U \neq V$.

Where is the link between these two?

In quantum mechanics, the mean of an observable $A$ in the state $\psi$ is given by

$$\langle A \rangle_\psi = \langle \psi |A|\psi \rangle,$$

and from this the variance of $A$ in $\psi$ is usually defined by

$$\langle(\Delta A)^2 \rangle_\psi := \langle (A-\langle A \rangle_\psi)^2 \rangle_\psi = \| (A - \langle A \rangle_\psi) \psi \|^2 = \langle A^2 \rangle_\psi - \langle A \rangle_\psi^2.$$

The standard deviation is then given by taking the square root.

  Theorem: (Heisenberg's Uncertainty Principle)

  $$\langle (\Delta A)^2 \rangle_\psi \langle (\Delta B)^2 \rangle_\psi \geq \frac{1}{4} | \langle [A,B] \rangle_\psi |^2$$

Proof: Using Cauchy-Schwarz, we obtain
$$| \langle \psi | [A,B] \psi \rangle | = | \langle A \psi| B \psi \rangle - \langle B \psi| A \psi \rangle| \leq 2 \| A \psi \| \| B \psi \|.$$

Square and substitute $A \to A - \langle A \rangle_\psi$, $B \to B- \langle B \rangle_\psi$ to finish the proof. Q.E.D.

You can observe directly that the product of the standard deviations has lower bound $>0$ if and only if $A$ and $B$ are non-commuting operators. This is statement (1).

From linear algebra, you might remember the following

  Theorem: (without proof) A set of diagonalisable matrices commutes if and only if the set is simultaneously diagonalisable.

You can find the proof for that theorem in most linear algebra textbooks, in case you are interested.

Putting these two theorems together, you can see the connection between statement (1) and statement (2), assuming you have two non-commuting operators $A$ and $B$. I am however a bit confused because you claim the lower bound to be $1$, which is not the case in general.

For example, taking $\mathcal{H} = L^2(\mathbb{R}^3)$ with the two operators

$$x_k: \psi(x) \mapsto x_k \psi(x), \; p_k: \psi(x) \mapsto \frac{\hbar}{i} \frac{\partial \psi}{\partial x_k}$$

which are the components of position and momentum, you can compute

$$i [p_k,x_\ell] = \hbar \delta_{k \ell}$$

and thus

$$\langle (\Delta p_k)^2 \rangle \langle ( \Delta x_\ell )^2 \rangle \geq \frac{\hbar^2}{4} \delta_{k \ell}.$$


To prove that $B\subset A\rightarrow A=A\cup B$ is it necessary to consider the cases $A\cup B=\emptyset$ and $A\cup B\neq \emptyset$? For what I know, to prove $B\subset A\rightarrow A=A\cup B$ it suffices to prove that $A\subset (A\cup B)$ and $(A\cup B)\subset A$.

Prove that $A\subset (A\cup B)$ is trivial, because $A\cup B=\{x:x\in A\vee x\in B\}$.

My teacher in proving that $(A\cup B)\subset A$ separated the proof into two parts: $A\cup B=\emptyset$ and $A\cup B\neq \emptyset$. But I didn't understand why the need to separate this proof in two cases since

$x\in (A\cup B)\to x\in A \vee x\in B$

We know, by hypothesis, that $B\subset A$, so we can conclude that $(x\in A\vee x\in B)\leftrightarrow(x\in A)$ is tautology(using truth table). So saying $x\in A\vee x\in B$ is equivalent to saying $x\in A$ which implies

$x\in (A\cup B)\to x\in A \vee x\in B\equiv x\in A\Rightarrow (A\cup B)\subset A$ (note: the implication $p\to q$ is true whenever $p$ is false).

In my opinion in the proof that $(A\cup B)\subset A$ there was no need to consider two cases, but as my teacher has great knowledge about set theory I feel that I am making a mistake by not considering the two cases. That is why I ask for help in this doubt.

Obs.: If I used some logical symbol wrongly let me know, please, since I still know few things about propositional logic.

You are right, and I see no errors. We can also observe that  $A\subset B\iff \forall x\;(x\in B\implies x\in A).$ Therefore if $B\subset A$ then for all $x$ we have  $$x\in A\cup B)\implies  (x\in A\lor x\in B)\implies$$ $$\implies (x\in A\lor x\in A)\implies(x\in A)\implies$$ $$\implies(x\in A \lor x\in B)\implies(x\in A\cup B).$$

From this we infer that for all $x$ we have $x\in A\cup B\iff x\in A$.

This is in my style, which is almost never brief.

The reason you feel there is no reason to consider the two cases is because you feel that one case is trivial.   Indeed the conclusion from those premises is a vacuuously truth:

$$\begin{array}{lrcl}\because&(B\subseteq A)~,~ (A\cup B=\emptyset) &\vdash& \forall x ~\big((x\in A\cup B) \to (x\in A)\big)\\\hline\therefore&(B\subseteq A)~,~ (A\cup B=\emptyset)&\vdash&(A\cup B\subseteq A)\end{array}$$

However, not all proofs will have a trivial result from the empty case; for instance any proof involving an existance claim may well lead to an exception.   Your teacher just wants you to get into the habit of verifying a proof does not fail in the empty case.


Is free product of groups always bigger that direct product? My question is as follow:

Let $\{G_\alpha\}, \alpha \in A$ be a class of groups. Is it always true that there exists a surjective homomorphism $\phi$
$$\phi:*_\alpha \,G_\alpha\, \to \prod_\alpha G_\alpha.\,$$

For example, there is a surjective homomorphism from $\mathbb{Z}*\mathbb{Z}$ to $\mathbb{Z} \times \mathbb{Z}$.

And how to type the * bigger?

Can you have a surjective homomorphism from
$$
 \underset{{i\in\mathbb{N}}}{{\LARGE{\ast}}}\mathbb{Z}_2 \to \prod_{i\in\mathbb{N}}\mathbb{Z}_2 ?
$$What would be the image of $(1,1,1,1,\ldots)   $?


Induction: Using P(n) vs P(n+1) is bad style? I'm in a proofs class and we were discussing induction. One of the most common ways we (the students) had seen induction was to represent "Statement P hold for $n$" by $P(n)$. Thus, we take the inductive hypothesis of assuming $P(n)$ is true and turning attention to $P(n+1)$. Of this format, my professor said "It's not really the best style and we won't use it that often".

Why would he say that? I've only ever (previously) seen induction in that format and he didn't really explain what he meant.

In real life, you'll be explicit about what $P$ is, and you won't abbreviate it.

  To prove that all naturals are greater than or equal to $0$, we proceed by induction. The base case is trivial: $0 \leq 0$ because $0 = 0$. Now, suppose $0 \leq n$. We have $n < n+1$ because [whatever reason - e.g:] there exists a nonzero natural we can add to $n$ to make $n+1$ - namely $1$. Therefore we also have $n \leq n+1$. Then $$\underbrace{0 \leq n}_{\text{inductive hypothesis}} \leq n+1$$ and so by transitivity $0 \leq n+1$, which completes the proof.

No $P$'s required. In general you want your proofs to look like prose unless you're being really formal; only use symbols if they actually make things clearer.

Hint: The general proof is like this:

Induction base: $P(0)$.

Induction step: $\forall n\geq 0 [ P(0)\wedge P(1)\wedge \ldots \wedge P(n)\Rightarrow P(n+1)]$.

Then we have proved $\forall n\geq 0[P(n)]$. Why? Consider modus ponens.

For most assertions, the induction step is simpler:

$\forall n\geq 0 [P(n)\Rightarrow P(n+1)]$.


Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$ How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$
I have not been sucessful to do this.

This is an infinite series, but it is not geometric because there is no common ratio.

So, let
$$S = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} + \dots + n \cdot \frac{1}{2^n} + \dotsb.$$
Multiplying this equation by 2, we get
$$2S = 1 \cdot 1 + 2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{4} + \dots + n \cdot \frac{1}{2^{n - 1}} + \dotsb.$$
Subtracting these equations, we find
$$S = 1 \cdot 1 + 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + \dots + 1 \cdot \frac{1}{2^n} + \dotsb.$$
So, $S=1+\dfrac12+\dfrac14+\dfrac18+\dotsb.$ Even though it didn't begin as one, we've managed to rewrite $S$ as an infinite geometric series. Thus, we may easily find its sum:
$$S = \frac{1}{1 - 1/2} = \boxed{2}.$$


Is there a non trivial normal subgroup of a group $G$, where $|G|=pm, \ \gcd(p,m)=1$? In fact, the exercise I'm in is this:

Suppose you have an irreducible polynomial $f(x)\in \mathbb{Q}[x]$ of degree $p$, where $p$ is a prime. Also suppose that $K$ is a splitting field of $f(x)$ over $\mathbb{Q}$.

Prove that:

$[K:\mathbb{Q}]=pm$, where $\gcd(p,m)=1$
If $H$ is a normal subgroup of $\mathrm{Gal}(K,\mathbb{Q})$ of order $|H|=m$  then $m=1$.

The first part was easy but the second part has something to do with Sylow theorems in which I'm not very keen on.

Answer to your second question: Since $H$ is normal in $G=Gal(K/\mathbb{Q})$, $H$ corresponds to a field $E$ such that $\mathbb{Q} \subseteq E \subseteq K$ (more preciously $E=K^{H}$). Now $H$ being normal in $G$ we must have that $E$ is a normal extension of $\mathbb{Q}$. (Because $\bigcap_{g\in G} gHg^{-1}$ corresponds to $\prod_{\sigma \in G}\sigma(E)$ and $H$ being normal in $G$ we've $\prod_{\sigma \in G}\sigma(E)$=E, which forces $E/\mathbb{Q}$ is anormal extension). And as $K/\mathbb{Q}$ is itself separable (as it is a Galois extension), $E/\mathbb{Q}$ is also separable and hence Galois of degree $p$.

So now by Primitive Element theorem there exists $b\in E$ such that $E=\mathbb{Q}(b)$. Now if $b$ is root of $f(x)$ then since $E/\mathbb{Q}$ is normal $E$ would become a splitting field of $f(x)$ forcing $m=1$. So WLOG $b$ is not a root of $f(x)$. Now pick a root of $f(x)$ $a\in F$ to and consider $\mathbb{Q}(a,b)/\mathbb{Q}$.

Note that $[\mathbb{Q}(b):\mathbb{Q}]=p$ so now if we can show $[\mathbb{Q}(a,b):\mathbb{Q(b)}]=p$ then we will get $p^2 \mid [K:\mathbb{Q}]$ which is a contradiction from first part. So, now we're just left to prove $[\mathbb{Q}(a,b):\mathbb{Q(b)}]=p$

Suppose not, which means $f(x)$ is reducible over $\mathbb{Q}(b)$. Now note that $f(x)$ is irreducible over $\mathbb{Q}$ and $\mathbb{char}(\mathbb{Q})=0$so $f(x)$ is seperable, so it has no multiple roots in any extension of $\mathbb{Q}$ and so, $f$ must split into linear factors over $\mathbb{Q}(b)[x]$ (if not then its has a factor of form $g(x)^r$ with $r>1$ and $g$ is irreducible over $\mathbb{Q}(b)[x]$, but then any root of $g$ in some extension comes as a root of $f$ with multiplicity more than $1$, contradiction) and so $b$ is a root of $f$ which is a contradiction to our assumption that $b$ is not a root of $f$ and so we're done.

To answer to the question of the title. Consider the group $G=\mathbb{Z}/6\mathbb{Z}$, we have $\mid G \mid = 3\times 2$ and $\gcd(2,3)=1$. Since $G$ is abelian then $H=\langle 2 \rangle \simeq \mathbb{Z}/3\mathbb{Z}$ is a non trivial normal subgroup of $G$.


Proving $x+\sin x-2\ln{(1+x)}\geqslant0$ Question: Let $x>-1$, show that
  $$x+\sin x-2\ln{(1+x)}\geqslant 0.$$

This is true. See http://www.wolframalpha.com/input/?i=x%2Bsinx-2ln%281%2Bx%29

My try: For
$$f(x)=x+\sin x-2\ln{(1+x)},\\
f'(x)=1+\cos{x}-\dfrac{2}{1+x}=\dfrac{x-1}{1+x}+\cos{x}=0\Longrightarrow\cos{x}=\dfrac{1-x}{1+x}.$$
So

  $$\sin x=\pm\sqrt{1-{\cos^2{x}}}=\pm \dfrac{2\sqrt{x}}{1+x}$$

If $\sin x=+\dfrac{2\sqrt{x}}{1+x}$, I can prove it. But if $\sin x=-\dfrac{2\sqrt{x}}{1+x}$, I cannot. See also http://www.wolframalpha.com/input/?i=%28x-1%29%2F%28x%2B1%29%2Bcosx

This inequality seems nice, but it is not easy to prove.

Thank you.

$\def\e{\mathrm{e}}\def\peq{\mathrm{\phantom{=}}{}}$Caution: This proof uses following inequalities of explicit values:$$
2π - 2\ln(1 + 2π) > 1, \quad \frac{2}{π^2} (π - \ln(1 + π)) > \frac{1}{π},\\
\frac{8}{5} - 2\ln 5 > \frac{4}{5} - \frac{3}{5} \left( π + \arccos \frac{3}{5} \right).
$$
(For the sake of completeness, rigorous proofs of these three inequalities are given at the end.)

Case 1: $-1 < x < 0$. This case is easy to prove because $x > \ln(1 + x)$ and $x < \sin x$.

Case 2: $x \geqslant 2π$. Since $x + \sin x - 2\ln(1 + x) \geqslant x - 2\ln(1 + x) - 1$ and $x - 2\ln(1 + x) - 1$ is increasing for $x \geqslant 1$, then$$
x + \sin x - 2\ln(1 + x) \geqslant x - 2\ln(1 + x) - 1 \geqslant 2π - 2\ln(1 + 2π) - 1 > 0.
$$

Case 3: $0 \leqslant x \leqslant π$. First, define $f_1(x) = \dfrac{2}{x^2} (x - \ln(1 + x))$, then$$
f_1'(x) = -\dfrac{2}{x^3} \left( \dfrac{(x + 2)x}{x + 1} - \ln(1 + x) \right).
$$
Define $f_2(x) = \dfrac{(x + 2)x}{x + 1} - \ln(1 + x)$, then $f_2'(x) = \dfrac{x^2 + x + 1}{(x + 1)^2} > 0$. Thus $f_2(x) \geqslant f_2(0) = 0$, which implies $f_1'(x) \leqslant 0$ and $f_1(x) \geqslant f_1(π)$, i.e.$$
x - 2\ln(1 + x) \geqslant \frac{2}{π^2} (π - \ln(1 + π)) x^2 - x. \quad \forall x \in [0, π]
$$

Next, define $f_3(x) = \dfrac{x^2}{π} - x + \sin x$. To prove that $f_3(x) \leqslant 0$ for $0 < x \leqslant π$, note that $f_3(π - x) = f_3(x)$, thus it suffices to prove for $0 < x \leqslant \dfrac{π}{2}$. Because $\cos x$ is concave on $\left[ 0, \dfrac{π}{2} \right]$, then$$
\cos x \geqslant \frac{\cos\dfrac{π}{2} - \cos 0}{\dfrac{π}{2} - 0}(x - 0) + \cos 0 = -\frac{2}{π}x + 1, \quad \forall x \in \left[0, \frac{π}{2} \right]
$$
which implies $f_3'(x) = \dfrac{2}{π}x - 1 + \cos x \geqslant 0$ and $f_3(x) \geqslant f_3(0) = 0$. Thus$$
-\sin x \leqslant \frac{x^2}{π} - x. \quad \forall x \in [0, π]
$$

Therefore for $0 \leqslant x \leqslant π$,$$
x - 2\ln(1 + x) \geqslant \frac{2}{π^2} (π - \ln(1 + π)) x^2 - x \geqslant \frac{x^2}{π} - x \geqslant -\sin x,
$$
i.e. $x + \sin x - 2\ln(1 + x) \geqslant 0$.

Case 4: $π < x < 2π$. Note that $f_4(x) = x - 2\ln(1 + x)$ is convex on $(π, 2π)$ and $f_5(x) = -\sin x$ is concave on $(π, 2π)$, thus for $π < x < 2π$,$$
f_4(x) \geqslant f_4'(4)(x - 4) + f_4(4) = \frac{3}{5} x + \left( \frac{8}{5} - 2 \ln 5 \right),
$$\begin{align*}
f_5(x) &\leqslant f_5'\left( π + \arccos\frac{3}{5} \right) \left( x - \left( π + \arccos\frac{3}{5} \right) \right) + f_5\left( π + \arccos\frac{3}{5} \right)\\
&= \frac{3}{5}x + \left( \frac{4}{5} - \frac{3}{5} \left( π + \arccos\frac{3}{5} \right) \right),
\end{align*}
which implies $f_4(x) \geqslant f_5(x)$, i.e. $x + \sin x - 2\ln(1 + x) \geqslant 0$.

To prove the three inequalities at the beginning, the following three identities are needed:$$
π = 2\sum_{n = 0}^∞ \frac{n!}{(2n + 1)!!}, \quad \e = \sum_{n = 0}^∞ \frac{1}{n!}, \quad \ln\frac{1 + x}{1 - x} = 2\sum_{n = 0}^∞ \frac{x^{2n + 1}}{2n + 1},\\
\arccos(1 - x) = \sqrt{2x} \sum_{n = 0}^∞ \frac{(2n - 1)!!}{(2n)!!}·\frac{x^n}{(2n + 1) 2^n},
$$
where $|x| < 1$. The first one see Wiki. The second one is well-known. The third one can be proved by noting that$$
\ln\frac{1 + x}{1 - x} = \ln(1 + x) - \ln(1 - x),\\
(\ln(1 + x))' = \frac{1}{1 + x} = \sum_{n = 0}^∞ (-x)^n, \quad (\ln(1 - x))' = -\frac{1}{1 - x} = -\sum_{n = 0}^∞ x^n.
$$
The fourth one can be proved by noting that$$
(\arccos(1 - x))' = -\frac{1}{\sqrt{2x}}·\frac{1}{\sqrt{1 - \dfrac{x}{2}}} = -\frac{1}{\sqrt{2x}} \sum_{n = 0}^∞ \binom{-\frac{1}{2}}{n} \left( -\frac{x}{2} \right)^n.
$$

Now, to prove that $2π - 2\ln(1 + 2π) > 1$, note that $2π - 1 > 5$ and$$
2\ln(1 + 2π) < 5 \Longleftrightarrow (1 + 2π)^2 < \e^5.
$$
Since $(1 + 2π)^2 < 9^2 < 2.5^5 < \e^5$, then $2π - 2\ln(1 + 2π) > 1$.

Next, note that$$
\frac{2}{π^2} (π - \ln(1 + π)) > \frac{1}{π} \Longleftrightarrow \frac{\ln(1 + π)}{π} < \frac{1}{2}.
$$
Define $f_6(x) = \dfrac{\ln(1 + x)}{x}$, then $f_6'(x) = \dfrac{1}{x^2} \left( \dfrac{x}{1 + x} - \ln(1 + x) \right).$ Define $f_7(x) = \dfrac{x}{1 + x} - \ln(1 + x)$, then $f_7'(x) = -\dfrac{x}{(1 + x)^2} \leqslant 0$, which implies $f_7(x) \leqslant f_7(0) = 0$ for $x \geqslant 0$. Thus $f_6'(x) \leqslant 0$ for $x \geqslant 0$, which implies$$
\frac{\ln(1 + π)}{π} = f_6(π) \leqslant f_6(3) = \frac{\ln 4}{3}.
$$
Since $\dfrac{\ln 4}{3} < \dfrac{1}{2} \Leftrightarrow \e^3 > 16$ and $\e^3 > \left( \dfrac{8}{3} \right)^3 > 16$, then $\dfrac{2}{π^2} (π - \ln(1 + π)) > \dfrac{1}{π}$.

Finally, note that$$
\frac{8}{5} - 2\ln 5 > \frac{4}{5} - \frac{3}{5} \left( π + \arccos \frac{3}{5} \right) \Longleftrightarrow 3\left( π + \arccos\frac{3}{5} \right) + 4 > 10\ln 5.
$$
Since$$
π = 2\sum_{n = 0}^∞ \frac{n!}{(2n + 1)!!} > 2\sum_{n = 0}^6 \frac{n!}{(2n + 1)!!} = \frac{141088}{45045},
$$\begin{align*}
\arccos\frac{3}{5} &= \arccos\left( 1 - \frac{2}{5} \right) = \frac{2}{\sqrt{5}} \sum_{n = 0}^∞ \frac{(2n - 1)!!}{(2n)!!}·\frac{\left( \dfrac{2}{5} \right)^n}{(2n + 1) 2^n}\\
&> \frac{2}{\sqrt{5}} \sum_{n = 0}^1 \frac{(2n - 1)!!}{(2n)!!}·\frac{1}{(2n + 1) 2^n} \left( \frac{2}{5} \right)^n = \frac{31}{15\sqrt{5}},
\end{align*}\begin{align*}
\ln 5 &= \ln\frac{1 + \dfrac{2}{3}}{1 - \dfrac{2}{3}} = 2\sum_{n = 0}^∞ \frac{1}{2n + 1} \left( \frac{2}{3} \right)^{2n + 1}\\
&< 2\sum_{n = 0}^5 \frac{1}{2n + 1} \left( \frac{2}{3} \right)^{2n + 1} + 2\sum_{n = 6}^∞ \frac{1}{13} \left( \frac{2}{3} \right)^{2n + 1}\\
&= 2\sum_{n = 0}^5 \frac{1}{2n + 1} \left( \frac{2}{3} \right)^{2n + 1} + \frac{2}{13}·\frac{1}{1 - \left( \dfrac{2}{3} \right)^2}·\left( \frac{2}{3} \right)^{13} = \frac{6427830866}{7979586615},
\end{align*}
then\begin{align*}
&\peq 3\left( π + \arccos\frac{3}{5} \right) + 4 > 3\left( \frac{141088}{45045} + \frac{31}{15\sqrt{5}} \right) + 4\\
&> 10·\frac{6427830866}{7979586615} > 10\ln 5.
\end{align*}


Conditional probability question with three independent events I have the following probabilities:
$$P(A_1)= P(A_2) = P(A_3)= .2,$$
where $A_1, A_2,$ and $A_3$ are all independent.
I wish to find
$$P(A_1\mid A_1 \cup A_2 \cup A_3).$$
I know that
$$P(A_1 \cup A_2 \cup A_3) = 3(.2) - 3(.2)(.2) + (.2)(.2)(.2) = .488.$$
I do not know how to proceed after this, however.  I do not understand how to incorporate the conditional part of the problem into my answer.
Any guidance is much appreciated.  Thanks!

Just use the definition of Conditional Probability, and the fact that $A_1\subseteq A_1\cup A_2\cup A_3$.
$$\mathsf P(A_1\mid A_1\cup A_2\cup A_3)=\dfrac{\mathsf P(A_1)}{\mathsf P(A_1\cup A_2\cup A_3)}=\dfrac{0.200}{0.488}$$


How to compute $\lim _{x\to \infty }\left(2\sqrt{x^2-4}-2x+1\right)$? I have a limit that I don't know how to compute: $$ \lim _{x\to \infty }\left(2\sqrt{x^2-4}-2x+1\right) $$
What I've tried : $$ \lim _{x\to \infty }\left(2\sqrt{x^2-4}-2x+1\right) = \lim _{x\to \infty }\left(x\left(2\sqrt{1-\frac{4}{x^2}}-2+\frac{1}{x}\right)\right) = \lim _{x\to \infty }\left(\frac{\left(2\sqrt{1-\frac{4}{x^2}}-2+\frac{1}{x}\right)^{'}}{\left(\frac{1}{x}\right)^{'}}\right) = \lim _{x\to \infty }\left(\frac{\left(\frac{1}{\sqrt{1-\frac{4}{x^2}}}+-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}\right) = \lim _{x\to \infty }\left(\left(-\frac{x^2}{\sqrt{1-\frac{4}{x^2}}}+1\right)\right) = -\infty $$ But the real answer is $1$. Where am I wrong? Thanks for any response.

Hint $2\sqrt{x^2 - 4} \to 2\sqrt{x^2} \to 2x$ as $x \to \infty$, simplifying your expression $\underbrace{2\sqrt{x^2 - 4}}_{\to 2x} - 2x + 1$ to...?


What actually is a polynomial? I can perform operations on polynomials. I can add, multiply, and find their roots. Despite this, I cannot define a polynomial.
I wasn't in the advanced mathematics class in 8th grade, then in 9th grade I skipped the class and joined the more advanced class. This question isn't about something I don't understand; it's something I missed.
My classes have not covered what really a polynomial is. I can generate one, but not define one. The internet has yielded incomplete definitions: "Consisting of multiple terms" or "A mathematical expression containing 2 or more terms and variables."
Take the following expressions for example:
$2x^2-x+12-2x^2+x-12$. Consists of multiple terms, but can also be expressed as $0$. Is zero a polynomial?
What about $x^{-1}$? $x^{-1}$ have -1 zeroes?">I've been told this one isn't a polynomial, but I don't understand why.
Is $x^2+x+1-x^{-1}-x^{-2}-x^{-3}$? a polynomial? It contains both positive and negative exponents?

tl;dr: What actually is the mathematical definition of a polynomial? Is $0$ a polynomial, and why isn't $x^{-1}$ a polynomial under this definition?

A polynomial (in one variable) is an expression of the form $$ p(x) = a_0+a_1x+a_2x^2+\ldots+a_nx^n$$ where the coefficients $a_i$ are some kind of number (or more generally they're elements of a Ring). The exponents $1,2,\ldots n$ must all be integers.

Unless we've been silly and $a_n=0,$ $n$ is called the degree of the polynomial. We can formalize this by defining the largest $n$ such that $a_n\ne0$ as the degree.

Notice that constants are allowed. $p(x) = 3$ is a zero-th degree polynomial.

You asked about zero. Yes, $p(x) =0$ is considered to be a polynomial. However, you'll notice that there is a problem with the definition of degree here since there is no coefficient that is nonzero. The degree of the zero polynomial is thus undefined.

This allows us to say that if we multiply two polynomials $w(x)=p(x)q(x)$ with $p$ of degree $n$ and $q$ of degree $m,$ then $w$ has degree $n+m.$ (Notice how the zero polynomial would mess this up if its degree were defined to be zero like the other constants.)

You're right that simplification is important. The $x$ is just a symbol and we can always "combine like terms" $$ a_lx^l+b_lx^l= (a_l+b_l)x^l.$$ We always combine all the terms together and simplify in order to get an expression into the form above with only one term for each power before we do things like consider the degree.

Notice we can add two polynomials according to the simplification rule and get a polynomial as a result. This is a good reason to consider zero to be a polynomial... it allows the sum of two polynomials to always be a polynomial. Likewise we can multiply two polynomials according to the the distributive property, the rule $$ (a_mx^m)(a_lx^l) = a_ma_l x^{m+l},$$ and the additive simplification rule. The result will be another polynomial.

Yes, the exponents all need to be positive. Of course other expressions are possible but they aren't called polynomials. Terms like $x^{-3}$ are considered part of the family of rational functions (or as a commenter noted, the Laurent polynomials, not to be confused with the (unqualified) polynomials). This is just a definition and thus somewhat arbitrary (though good definitions are important for organization). It's just like saying $-4$ is an integer but not a natural number. It's true by definition, and yes, a bit arbitrary, but nonetheless useful and a nearly universal convention.

EDIT
As Paul Sinclair pointed out in the comments, there are also polynomials in multiple variables. For instance $$p(x,y) = A + Bx + Cy +Dx^2+Exy+Fy^2$$ is the general degree two polynomial in two variables. The degree of a term is just the sum of the degrees with respect to the individual variables. So a term like $3xy$ has degree two and a term like $3x^4y^5z$ would have degree $4+5+1=10.$ The degree of a polynomial is the degree of its highest-degree term with nonzero coefficient.

Polynomial is an object on some particular algebra $\mathbb{A}$ that can be created with addition and multiplication of elements of $\mathbb{A}$.

If the particular algebra is also a field $\mathbb{F}$, then we can have a nice form for the polynomial, e.g. $a_1 x a_2 x a_3=a_1 a_2 a_3 x^2$, where $a_i,x \in \mathbb{F}$.


Finding the value of $\sum k/3^k$ Possible Duplicate:
  How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$

please help me with this
$$\sum\limits_{k=0}^{\infty}\frac{k}{3^k}$$

I need just a hint, not a full answer. Thanks!

There are several ways to do this.  One involves the fact that
$$
\frac{k}{3^k} = kx^k = x\cdot kx^{k-1} = x\cdot \frac{d}{dx} x^k
$$
where $x=1/3$, and $\displaystyle\sum_{k=0}^\infty x\cdot \frac{d}{dx} x^k$ can be found.

Another looks like this:
$$
\sum_{k=0}^\infty \frac{k}{3^k} = \sum_{k=1}^\infty \frac{k}{3^k}
= \left\{\begin{array}{cccccccccccccccc}
& & 1/3 \\
& + & 1/9 & + & 1/9 \\
& + & 1/27 & + & 1/27 & + & 1/27 \\
& + & 1/81 & + & 1/81 & + & 1/81 & + & 1/81 \\
& + & \cdots
\end{array}\right.
$$

Then you can sum the first column, then the second column, etc., and finally find the sum of all of the column sums.


Simple Closed Contour Integral with a simple pole equalling 0? Out of curiosity, can a function with simple poles have their contour integral equalling to 0?
I ask because, say for the function $\ f(z) = \frac{1}{z^2}$, if we integrate it around the closed contour $|z| = 1$, it will be equal to 0 by the fundamental theorem of calculus. However, for $g(z) = \frac{1}{z}$ we cannot say the same because the antiderivative would not be holomorphic along the contour - it has a discontinuity along the negative real axis. Can this be generalised to all functions with simple poles within the contour?

Let $z_0 \in \mathbb C$, r>0, $D=\{z \in \mathbb C: 0<|z-z_0|<r\}$ and suppose that $f:D \to \mathbb C$ is holomorphic and has a simple pole at $z_0$. Then , by Laurent, we have

$$f(z)= \frac{a}{z-z_0}+g(z)$$

with $a \ne 0$ and $g: \{z \in \mathbb C: |z-z_0|<r\} \to \mathbb C$ holomorphic.

Now let $0< \rho <r$. Then we have

$$ \int_{|z-z_0|= \rho} f(z) dz= a\int_{|z-z_0|= \rho} \frac{dz}{z-z_0}+\int_{|z-z_0|= \rho} g(z) dz.$$

Since $\int_{|z-z_0|= \rho} g(z) dz=0$ and $\int_{|z-z_0|= \rho}\frac{dz}{z-z_0}=2 \pi i$, we derive

$$ \int_{|z-z_0|= \rho} f(z) dz= a 2 \pi i \ne 0.$$

To answer your very first question, sure. Take $\frac{\sin z}{z}$ $-$ it has a residue of $0$ at its pole.


Limits and common sense I'm stuck in understanding of limits.
It all makes sense, but at a certain point my answers which seem logical to me are not true.
Please can somebody explain why as a huge number gets divided by a big number plus the relatively small value $1$, the result gets closer and closer to zero:
$$
lim_{x\to \infty} \frac{\infty!}{(\infty + 1)!} =0.
$$
The next one is not equal zero as well.
$$
lim_{x\to \infty} \frac{\infty}{\infty-8}=1
$$
The relatively huge number gets divided by another relatively huge number minus 8, which is so little we could ignore it. Why is it approaching 1 then?
It can make sense from, I don't know, a visual point of view where
$$
lim_{x\to \infty} \frac{\infty}{\infty-8}=\frac{\infty}{\infty}=1
$$
but I could as well do that with first one and also get 1, which is wrong.

This answer is divided in two parts. In the first part I will share some of the theory behind infinity, and in part $2$ we will look at how to use this theory to solve limits.
Part 1 (Theory)
Before we start, I wanna make a comment on something you said. You said in a comment that you think we deal with same-size infinities all around. This is not true. Consider two limits
$$\lim _{x \to \infty} x = \infty$$
$$\lim _{x \to \infty} 2x = \infty$$
While we use the same symbol for the results of both of these limits, the infinity resulting from $2x$ is in some sense "bigger" or  (as I like to think about it) "stronger" than the infinity resulting from $x$. I think this is the root cause of your confusion, and it's the reason why in your question, one limit is equal to $0$, while another is equal to $1$. Anyway, here's the intro to the theory:

The symbol $\infty$ is not a number. Technically it's an extended real number. But no need to get into technicalities. Going by intuition is best.
The symbol $\infty$ behaves like a number in some cases, but not in others.
Here are some of the things you can always do with infinity. These should all be very clear intuitively. If they are not, let me know:
A) $\infty + a = a + \infty = \infty$ for any number $a$
B) $\infty + \infty = \infty$
C) $\frac {a}{\infty} = 0$ for any number $a$
D) $\frac {\infty}{a} = \pm \infty$, depending on the sign of $a$.
E) $\infty \cdot \infty = \infty$
F) $\infty \cdot a = a \cdot \infty = \pm \infty$, depending on the sign of $a$.
G) $\infty^2 = \infty, \infty^3 = \infty,$ $\infty^4 = \infty$, etc.
If you understand these intuitively, there should be no doubt that the above are true. I did not include everything here, but if you understand these intuitively, you should hopefully be able to come up with more of these by yourself (another one: $2^{\infty} = \infty$).
The following operations with $\infty$ are undefined (in the same way that, for example, $\frac 10$ is undefined).

$\infty - \infty$
$\frac {\infty}{\infty}$

Why is this? Consider for example $\infty + \infty$ compared to $\infty - \infty$. In the first expression, there is no doubt that the result should be $\infty$. If you add two huge numbers, you get a huge number. The huge numbers
are working together to make an even bigger number.
But for the second expression, there is doubt. If you subtract two really huge numbers, it totally depends on how big the two numbers are compared to each other. For example, $100000 - 1000$ is really big, but $1000 - 1000000$ is really small, and $100000 - 100000$ is zero. The two huge numbers are working against each other. The first number wants to pull the result up, but the second number wants to pull the result down. It all depends on which huge number is "stronger".
Here are examples with limits which demonstrate that with $\infty - \infty$, anything can happen, depending on how "strong" the specific infinities involved are. In their original form, all of these limits are of the type $\infty - \infty$ which is undefined. But after some easy manipulations, that are not of that type anymore, and we can compute them easily.

$\displaystyle\lim_{x \to \infty} (5x - x) = \lim_{x \to \infty} 4x = 4 \times \infty = \infty$
$\displaystyle\lim_{x \to \infty} [(x+4) - x] = \lim_{x \to \infty} 4 = 4$
$\displaystyle\lim_{x \to \infty} (x - x) = \lim_{x \to \infty} 0 = 0$
$\displaystyle\lim_{x \to \infty} [(x-4) - x] = \lim_{x \to \infty} -4 = -4$
$\displaystyle\lim_{x \to \infty} (x - 5x) = \lim_{x \to \infty} -4x = -4 \cdot \infty = - \infty$

These may seem like silly examples, but they illustrate the point.
Same story with $\frac {\infty}{\infty}.$ The huge number in the numerator wants to pull the result up (because for example $\frac {\infty}{5} = \infty$) but the huge number in the denominator wants to pull the result down towards zero (because for example $\frac {5}{\infty} = \infty$). Again, the result will depend on which huge number is "stronger".
Here are examples which illustrate that again, anything can happen, depending on the "strength" of the specific infinities involved.

$\displaystyle \lim_{x \to \infty} \frac {x^2}{x} = \lim_{x \to \infty} x = \infty$

$\displaystyle \lim_{x \to \infty} \frac {4x}{x} = \lim_{x \to \infty} 4 = 4$

$\displaystyle \lim_{x \to \infty} \frac {x}{x^2} = \lim_{x \to \infty} \frac 1x =  \frac {1}{ \infty} = 0.$

$\\$
Part 2 (Application)
So, how do you actually apply this stuff to solve a limit of the form $\lim_{x \to \infty} f(x)$? You start by plugging in $\infty$, and see what happens. There are two possibilities:
1) Once you plug in, you only have allowable operations to do. In this case, you do those operations, and get the answer. For example
\begin{align*}
 \lim_{x \to \infty} \dfrac {2x^3 + 3x - 4}{5} = \dfrac {2 \cdot \infty^3 + 3\cdot \infty - 4}{5} = \dfrac {\infty + \infty - 4}{5} = \frac{\infty}{5} = \infty
\end{align*}
2) Once you plug in, you get one of the forbidden operations, such as $\frac {\infty}{\infty}.$ If this happens, then you need to go back to the original and try to "fix" the issue. Either you do an algebraic manipulation, or (more advanced) use L'Hopital's Rule, or (even more advanced) Taylor series, etc.
Once you have done the manipulation (or whatever else) then you plug in $\infty$ again, and hope that you end up in case $1$. (notice that this is exactly what I did in the examples above when showing that $\infty - \infty$ or $\frac {\infty}{\infty}$ could be anything).If you end up in case $2$ again, then you repeat.
An example:
$$\lim_{x \to \infty} \frac {x^2 - 4}{x-2}$$
Plugging in gives us $\frac {\infty^2 - 4}{ \infty - 2} = \frac {\infty}{\infty }$ which is undefined, so we go back to the original and manipulate.
$$\lim_{x \to \infty} \frac {x^2 - 4}{x-2} = \lim_{x \to \infty} \frac {(x-2)(x+2)}{x-2} = \lim_{x \to \infty} x + 2 = \infty + 2 = \infty.$$

Now, this is what happens in your two examples in the question. Presumably, the first question started as
$$\lim_{n \to \infty} \dfrac {n!}{(n+1)!}$$
here we think of $n$ as a whole number, because factorials are only defined for whole numbers.
If you plug in $\infty$, you end up with $\frac{\infty}{\infty}$. So we go back and manipulate:
$$\lim_{n \to \infty} \dfrac {n!}{(n+1)!} = \lim_{n \to \infty} \dfrac {1 \cdot 2 \cdot 3 \cdots \cdot n}{1 \cdot 2 \cdot 3 \cdots \cdot n \cdot (n+1)!} = \lim_{n \to \infty} \frac {1}{n+1} = \frac {1}{\infty + 1} = \frac {1}{\infty} = 0.$$
In your second question. Presumably the problem started as
$$\lim_{x \to \infty} \frac {x}{x-8}$$
If you plug in you get $\frac {\infty}{\infty}$. So we go back and manipulate.
$$\lim_{x \to \infty} \frac {x}{x-8} = \lim_{x \to \infty} \frac {x}{x(1-\frac 8x)} = \lim_{x \to \infty} \frac {1}{1-\frac 8x} = \frac {1}{1 - \frac {8}{\infty}} = \frac {1}{1-0} = 1$$
These manipulations (especially for the second problem) might seem like they come out of nowhere. But if you study limits, you will see the same 3-4 types of manipulations over and over, and you will know how to use them. The ones I used are very common.

Well when you have limits as $x$ approaches infinity of a quotient. The limit of the quotient is the limit of the highest power in the numerator over the highest power in the numerator. So
for example in your second case:

$\lim_{x\rightarrow\infty}\frac{x}{x-8}=\lim_{x\rightarrow\infty}\frac{x}{x}=1$

For your first case: $\lim_{x\rightarrow\infty}\frac{x!}{(x+1)!}=\lim_{x\rightarrow\infty}\frac{x!}{(x+1)x!}=\lim_{x\rightarrow\infty}\frac{1}{x+1}=0$

Edit: You shouldn't be treating infinities as numbers because there are different kinds of infinities.


What is the difference between regex operations in math and regex in UNIX / Linux? What is the difference between regular expression operations (union, concatenation, kleene star) and regular expression (implemented in UNIX and can be used together with the grep command)?

Are there such methods that are implemented in UNIX and can be used together with the grep command, that cannot be recreated using only the regular expression operations (union, concatenation, kleene star)

regular expression (implemented in UNIX and can be used together with the grep command) :

(.,*,^,$,\,\<,|>,[],+,?,&,\n))

Regular Expressions in formal languages and automaton theory are the basis for regular expressions in UNIX. Some of the syntax is a bit different in grep than in the formal theory. For example, in grep, you can use $(a+b)^{2}$, while this is not acceptable syntax theoretically (though it would make sense to have it theoretically). Grep also has a larger set of symbols it allows in regular expression syntax.

Grep (at least some versions of it) use finite state automaton to handle the search.

I am not a *Nix expert, nor am I a formal languages expert. Someone may have more enlightening answers than I did. Hopefully this helps some, though.

Regular Expressions is a formal language defined by induction with a finite set of rules. You have some info over here: http://mathworld.wolfram.com/RegularExpression.html


Is there a relation between polynomials $f(a)$ with $f(b)$ and $f(ab)$? Let $f(x) = c_n x^n + \dots + c_1x + c_0$ be a polynomial of degree $n$ with integer coefficients.

Is there a relation between $f(a)$ with $f(b)$ and $f(ab)$ in general?

For example: $f(ab) = f(a)f(b) \big( \cfrac{1}{c_n + \dots + c_0} \big)$.

I'm not sure, but I just want to say that sometimes it's very difficult to work from some generalization backwards to identify some relationship, even though it might look cooler to say it the way you're saying it - ask yourself the first couple cases and check it out

Case n = 0

$f(a) = c_{a0} \\
 f(b) = c_{b0}\\
f(ab) = c_{ab0}$

Case n = 1

$f(a) = c_{a1}a + c_{a0} \\
 f(b) = c_{b1}b + c_{b0}\\
f(ab) = c_{ab1}ab + c_{ab0}$

etc

then you can start messing around with them - maybe look at some morphisms, throw them in some rings, I don't know - this math world is your oyster :)


Uniqueness of solution of differential equation $y'=y$ In school, I have recently been learning about simple differential equations. We know that the solution of $y'=y$  is $y=Ae^x$, where $A$ is a constant. But how can we know that it is the only solution? The only thing I can figure out is that $y$ is continuously differentiable. Help me, please.

Let $z$ be a solution to $y'=y$. Consider $z(t)e^{-t}$. We have
$$
\frac{d}{dt}(z(t)e^{-t})=z'(t)e^{-t}-z(t)e^{-t}=0,
$$
therefore
$$
z(t)e^{-t}=Const\implies z(t)=Ae^{t}.
$$

Suppose there is another solution $y=f(x)$. Then we must have that $y=f(x)-Ae^x$ is a solution. Now working this out gives us:
$$y'=\frac{\partial }{\partial x}(f(x) - A e^x) = f'(x) - A e^x$$
Note that we must have $y=y'$ so $f(x)=f'(x)$ clearly this only holds for $f(x) = A e^x$.


Calculate $ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $ How evaluate the following limit?

$$ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $$

I cannot apply L'Hopital because $ f(x) =  3 -\sqrt{5 -x} \neq 0 $ at $x = 5$

Let $x=5\cos4y$  where $0\le4y\le\pi$

$x\to4\implies\cos4y\to\dfrac45$

But as $\cos4y=2\cos^22y-1, 2y\to\arccos\dfrac3{\sqrt{10}}=\arcsin\dfrac1{\sqrt{10}}$

$$F=\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim_{2y\to\arcsin\frac1{\sqrt{10}}}\dfrac{\dfrac3{\sqrt{10}}-\cos2y}{\dfrac1{\sqrt{10}}-\sin2y}$$

If we set $\arccos\dfrac3{\sqrt{10}}=\arcsin\dfrac1{\sqrt{10}}=2A\implies\cos2A=?,\sin2A=?,\tan2A=\dfrac13$

$$F=\lim_{2y\to2A}\dfrac{\cos2A-\cos2y}{\sin2A-\sin2y}=\lim_{y\to A}\dfrac{-2\sin(A-y)\sin(A+y)}{2\sin(A-y)\cos(A+y)}=-\tan(A+A)=?$$


Find the Generating Function from the sequence (0,1,0,3,0,5,...) Find the Generating Function from the sequence (0,1,0,3,0,5,...)

I can't conceptualize this one and I know it should be easy because it is just odds, but I am having a hard time figuring out how to cancel the even terms.

I was thinking like k$x^k$ but that doesn't cancel the evens.

Any help would be much appreciated.

The answer is simple.
Just iterate for x = 0 to Infinite
And the xth term is

f(x) = (x%2)*x

Where % is modulus operator which returns the remainder of the two operands

Here x%2 cancels out all the even terms leaving all the odd terms behind


Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$? Suppose $-a\sin(s) - b\cos(s)  = 0$ and given that $a^2 + b^2 \le 1$, then $a^2 + b^2 = 1$?

I am having trouble getting the above identity. I vaguely recall that

$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$

where $$\tan(\theta) = \frac{b}{a}$$

But I still don't see why it's obvious?

I don't see how the identity could be true. In the identity that you "vaugely recall", the only way for the right side to be identically zero is for the part in front of $sin$ to be zero. This requires that both $a$ and $b$ are zero, so $a^2+b^2$ would be zero.

The equation is homogeneous in $a$ and $b$, so that if $a,b$ fulfill it, so do $\lambda a,\lambda b$, where $\lambda$ is a free parameter. Then, there is no reason that $\lambda^2(a^2+b^2)=1$. On the opposite, the only way for $\lambda^2(a^2+b^2)$ to remain bounded is to be zero !


$\int_{1}^\infty\frac{\sin(x^2)}{x^p}dx$ for which p values does the integral converge in condition $$\int_1^\infty\frac{\sin(x^2)}{x^p} \, dx$$
for which p values does the integral converge in condition? and for which values does it converge absolutely?

I managed to find the values of $p$ which will make the integral to converge absolutely , they are $p>1$, but i could not manage to solve for the condition values of $p.$

$$\int_1^\infty\frac{\sin(x^2)}{x^p} \, dx $$ assign $t=x^2$ you will get :
$$\frac{1}{2}\int_1^\infty\frac{\sin(t)}{t^\frac{p+1}{2}} \, dt$$ $$\int_1^\infty \left|\frac{\sin(t)}{t^\frac{p+1}{2}}\,dt\right| \le \int_1^\infty\frac{1}{t^\frac{p+1}{2}}\,dt$$
hence if $\frac{p+1}{2}>1 \to p>1$ the integral converges.

In the answers the value for the conditional convergence is for $0<p\le1$ can't understand why

Integrating by parts, (for $n>0$ only (why?))
$$\int^\infty_1 \frac{\sin x}{x^n}dx=\cos 1-(1+n)\int^\infty_1\frac{\cos x}{x^{n+1}}dx$$

By integrating by parts repeatedly, you can always make the power in the denominator as large as possible to obtain absolute convergence, so the only condition is $n>0$.

Correspondingly, $p>-1$ (which is quite counter-intuitive to me).

To show the type of convergence (absolute or conditional), we may consider
$$\begin{align}
&~~~~\int^\infty_1\frac{|\sin x|}{x^n}dx \\
&\ge\int^\infty_1\frac{\sin^2 x}{x^n}dx \\
&=\frac12\int^\infty_1\frac{1-\cos 2x}{x^n}dx \\
&=\frac12\int^\infty_1\frac1{x^n}dx-\frac12\int^\infty_1\frac{\cos 2x}{x^n}dx
\end{align}
$$

We have shown that the second term always converges. Therefore, absolute convergence of the original integral is equivalent to the convergence of the first term, which can be examined by p-test.

The result is the convergence is absolute for $1<p$.

If you graph the integrand and consider the sequence of positive and negative areas, you'll see that you have an alternating series with terms of decreasing absolute value, for any $p>0.$


Evaluating $\int e^{\sin x}\ dx$ in terms of a Taylor series. NOW before anyone gets too hasty in saying "oh, it's impossible," I would like to presume that it is possible despite the fact that it obviously is not, and theoretically search for some way of simply representing the solution. I came up with the idea of switching $e^{\sin x}$ into its taylor series and integrating every term. Now I have a new series, but how do I "de-Taylor" this series into some combination of elementary functions? Is this even possible with $e^{\sin x}$? Even if it isn't, is there a method for being able to de-Taylor a series? My equations:
$$
\int e^{\sin x}\ dx = x+\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{40}-\frac{x^6}{90}-\frac{x^7}{1680}\dots
$$
Surely there is some algorithm that satisfies those denominators, right? There just has to be! And of course, I know how naive this may seem to pursue, but I have taken an interest in this problem and am actively seeking a solution.

To get the power series for
$e^{\sin x}$,
you can do this:

Let
$f(x) = e^{\sin x}$.
Then
$f'(x)
= \cos x\,f(x)
$.

Therefore,
if
$f(x)
=\sum_{n=0}^{\infty}a_n x^n
$,
then
$f'(x)
=\sum_{n=1}^{\infty}n a_n x^{n-1}
=\sum_{n=0}^{\infty}(n+1) a_{n+1} x^{n}
$.

Since
$\cos(x)
=\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}
$,

$\begin{array}\\
f'(x)
&=\cos(x)f(x)\\
&=(\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!})(\sum_{n=0}^{\infty}a_n x^n)\\
&=(\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!})(\sum_{n=0}^{\infty}a_n x^n)\\
&=\sum_{m=0}^{\infty} x^m \sum_{k=0}^{\lfloor m/2 \rfloor} (-1)^k \frac{1}{(2k)!}a_{m-2k}\\
&=\sum_{m=0}^{\infty} x^m \sum_{k=0}^{\lfloor m/2 \rfloor} (-1)^k \frac{a_{m-2k}}{(2k)!}\\
\end{array}
$

Therefore
$(n+1) a_{n+1}
=\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{a_{n-2k}}{(2k)!}
$.

Since
$f(0) = 1$,
$a_0 = 1$.

Then,
with $n=0$,
$a_1
=1
$;
with $n=1$,
$2a_2
=a_1
=1
$
so
$a_2 = \frac12$;
with $n=2$,
$3a_3
=a_2-a_0/2
=\frac12-\frac12
=0
$;
with $n=3$,
$4a_4
=a_3-a_1/2
=0-\frac12
=-\frac12
$
so
$a_4 = -\frac18
$.

So
$f(x)
=1+x+\frac{x^2}{2}-\frac{x^4}{8}+...
$.
Integrating,
$\int f(x)
=x+\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{40}+...
$.

This agrees with Wolfy
(there is an error in
your coefficient of $x^5$ -
you divided by 4 instead of 5).

Note that method
(which is not original by me)
can be used to get
the power series for
$e^{g(x)}$
where $g(0) = 0$:
let
$f(x) = e^{g(x)}$.
Then
$f'(x)
= g'(x)e^{g(x)}
=g'(x)f(x)
$.
Substituting power series
for $f$ and $g$,
we get a recursion
for the coefficients
of $g$
in terms of the
coefficients of
$f$.

Read generatingfunctionology
(it's free!)
at
https://www.math.upenn.edu/~wilf/DownldGF.html
for lots more neat stuff.

There is no such thing as "de-Taylor". This integral can't be combined of elementary functions. Believe me because it's one of the simplest and most famous integrals that can't be combined of elementary functions. In fact only a small minority of integrals and taylor series have a presentation with elementary functions.


Calculating the limit $\lim_{x \to \infty} \frac{(x+2)^b-(x+1)^b }{(x+1)^b-x^b}$ I want to prove/find the following limit:

$$\lim_{x \to \infty} \frac{(x+2)^b-(x+1)^b }{(x+1)^b-x^b}$$

where $1<b\leq1.5$ is fixed.

if trying different values for $b$, wolframalpha suggests that the limit is 1. But if so, I would really like how to prove it, especially because the term has this nice structure.

Please help!

If you set $x=\frac 1t$ and consider $t\to0^+$ you get

$$\frac{(x+2)^b-(x+1)^b }{(x+1)^b-x^b} = \frac{(1+2t)^b-(1+t)^b }{(1+t)^b-1}$$
$$=\frac{(1+2t)^b-1 +1 - (1+t)^b }{t}\cdot \frac{1}{\frac{(1+t)^b-1}{t}}$$

Now, seeing that these are derivatives of $(1+2t)^b$ and $(1+t)^b$ at $t= 0$ you can proceed (taking into consideration that $b>1$)

$$\stackrel{t\to 0^+}{\longrightarrow}(2b-b)\cdot \frac 1{b}=1$$

Some tests using bc seem to indicate that a good approximation for
$$f(x,b) = {{(x+2)^b-(x+1)^b}\over{(x+1)^b-x^b}}$$
is
$$g(x,b) = 1 + {{b-1}\over{x}}$$
Since the next term involves $1\over{x^2}$,  this seems to be close.  Note that the restriction on $b$ does not seem to be needed.

Now I'm not sure how to prove $$\lim_{x \to \infty}{f(x,b)} = \lim_{x\to\infty}g(x,b)$$ but $$\lim_{x\to\infty}g(x,b) \to 1$$ is easy.


Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? In this post we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question:

Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

I know these as $\mathbb{Q}$-Vector spaces, are isomorphic from the  linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it.

More generally: suppose $d$ and $d'$ are both squarefree integers, both different from $1$, and consider $F_1 = \mathbb{Q}(\sqrt{d})$ and $F_2 = \mathbb{Q}(\sqrt{d'})$.

They are both isomorphic as $\mathbb{Q}$-vector spaces, since they are both of dimension $2$; or more explicitly, every element of $F_1$ can be written uniquely as $a+b\sqrt{d}$ with $a,b\in\mathbb{Q}$ (unique because  $\sqrt{d}\notin\mathbb{Q}$), and every element of $F_2$ can be written uniquely as $x+y\sqrt{d'}$ with $x,y\in\mathbb{Q}$. The map $f\colon F_1\to F_2$ given by $f(a+b\sqrt{d}) = a + b\sqrt{d'}$ is additive and $\mathbb{Q}$-homogeneous, clearly bijective, so $F_1$ and $F_2$ are isomorphic as vector spaces over $\mathbb{Q}$.

However, they are never isomorphic as fields; clearly, $d$ is a square in $F_1$. I claim $d$ can only be a square in $F_2$ if $d=d'$. Indeed, if$(x+y\sqrt{d'})^2 = d$. That means that $x^2 + d'y^2 + 2xy\sqrt{d'} = d$, hence $2xy = 0$ and $x^2+d'y^2=d$. If $x=0$, then $d=d'y^2$, so clearing denominators you get $da^2 = d'b^2$ for some $a,b\in\mathbb{Z}$, $\gcd(a,b)=1$; since both $d$ and $d'$ are squarefree, it follows that $|a|=|b|=1$, so $d=d'$. If $y=0$, then $d=x^2$, so $d$ is the square of a rational, contradicting the fact that it is a squarefree integer different from $1$. Thus, of $d$ is a square in $F_2$, then $d=d'$. Hence, if $F_1\cong F_2$, then $d=d'$ (converse is immediate).

Now, since every quadratic extension of $\mathbb{Q}$ is equal to $\mathbb{Q}(\sqrt{d})$ for some squarefree integer $d$ different from $1$, you conclude that any two quadratic extensions are either identical or not isomorphic as fields.

The only possibility for an isomorphism of the extension fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$, fixing $\mathbb{Q}$ (the prime subfield) would be $\varphi:\sqrt{2}\mapsto \sqrt{3}$. But $\varphi$ preserves products and so:

$2 = \varphi(\sqrt{2}\sqrt{2}) = \varphi(\sqrt{2})\varphi(\sqrt{2}) = \sqrt{3}\sqrt{3} = 3$ contradiction. So the aforementioned fields can't be isomorphic.


Is there a real number, that is proven to be normal in every base, whose digits can be enumerated by an algorithm? To clarify, I mean every natural number base $b$ where $b \geq 2$. If so, what is the algorithm to generate the number (and what is the number, if it has a name)?

Wikipedia reveals that the answer is yes, citing the paper "An example of a computable absolutely normal number" by Becher and Figueira (Theoretical Computer Science 270(1-2), 2002).


Dimension of cohomology group is a birational invariant I read the fact somewhere that the dimension of cohomology group $H^i(X,\mathcal{O}_X)$ is a birational invariant.

For surfaces, the reference for this fact is Hartshorne Proposition 3.4 saying $H^i(X, \mathcal{O}_X) \cong H^i(\tilde{X}, \mathcal{O}_{\tilde{X}})$ for all $i \geq 0$ where $\tilde{X}$ is a blow-up of $X$ at $1$ point, combined with the fact that any birational morphism between surfaces factors as a finite number of blow-ups (of a point) and an isomorphism (Beauville-Complex Algebraic Surfaces, Theorem II.11).

What is the source for varieties of higher dimensions?

Unfortunately, the reference you provide is only valid for smooth surfaces. I don't know if this should hold for blow-ups at singular points. At the start of Chapter 5 on Surfaces, Hartshorne assumes that all surfaces in the Chapter are smooth. (This should be a comment, although I don't have the reputation to post this as one.)


Finding a conformal map from this domain into the unit disc Problem: Find a conformal map $f$ from $ A = \left\{ z \in \mathbb{C} \mid \text{Im}(z) > 0, |z| > 1 \right\}$ into the unit disk.

Attempt: I started off with a map $F_1: z \mapsto z + \frac{1}{z}$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.

I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?

Help is appreciated.

Hint: With conformal mapping
$$w=\left(\dfrac{z+1}{z-1}\right)^2$$
you map region $A=\{z\in\mathbb{C}:{\bf Im}\ z>0,\ |z|>1\}$ to lower half plane $\{z\in\mathbb{C}:{\bf Im}\ z<0\}$.

It seems that your region $A$ is the limit as $R \to + \infty$ of
$$
A_R=\{z \in \mathbb C, \Im(z)>0, 1<|z|<R \}=\{r\exp(i\theta), \theta \in (0,\pi), r \in (1,R) \}.
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form


Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?

Proposition 1: $\frac{0}{0} = 0$

Proof: Suppose that $\frac{0}{0}$ is not equal to $0$

$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$  $\Rightarrow$  $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction

Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$

Therefore, $\frac{0}{0} = 0$.

Q.E.D.

Update (2015-12-01) after your answers:

Proposition 2: $\frac{0}{0}$ is not a real number

Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:

Suppose that $\frac{0}{0}= x$, where $x$ is a real number.

Then, either $x = 0$ or $x$ is not equal to $0$.

1) Suppose $x = 0$, that is $\frac{0}{0} = 0$

Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $

Contradiction

Therefore, it is not the case that $x = 0$.

2) Suppose that $x$ is not equal to $0$.

$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction

Therefore, it is not the case that $x$ is a real number that is not equal to $0$.

Therefore, $\frac{0}{0}$ is not a real number.

Q.E.D.

Update (2015-12-02)

If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.

Proposition 3: $\frac{0}{0}$ is not a real number

Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.

$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$

$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$

Q.E.D.

Update (2015-12-07):

How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?

Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$

Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.

$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$

$\therefore \frac{0}{0}=0$

Q.E.D.

Suggested definition of division of real numbers:

If $b \ne 0$, then

$\frac{a}{b}=c$ iff $a=bc$

If $a=0$ and $b=0$, then

$\frac{a}{b}=0$

If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.

A somewhat more minimalistic version:

Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.

Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.

$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$

$\therefore \frac{0}{0}=0$

Q.E.D.

The error is in the very first implication.  If $0/0$ is not equal to zero, there is no reason why $0/0$ must equal some $x$.  There is no reason to believe that we can do this division and get a number.

Therefore you have a proof that $0/0$ cannot equal any nonzero $x$.  Combine this with a proof that $0/0$ cannot equal zero, and you have proved that $0/0$ is not a number.

0/0 is indeterminate. Which means it has infinite number of values.

(0)(1) = 0     =>     0/0 = 1
(0)(2) = 0     =>     0/0 = 2
(0)(4) = 0     =>     0/0 = 3 .....

From above it can be shown that 0/0 has infinite number of values which make it meaningless. Hence the term.


How to equally divide a circle with parallel lines? How can I "draw" $n$ parallel lines in such a way that they will divide a circle (disc) in $n+1$ equal areas ?

Even the case $n=2$ is hard, according to this link.

For any circle surely the distance between the lines drawn is a constant fraction of the diameter of the circle, depending on the divisions required. Is this too simple?