Why does choose in combinatorics work? Say, the question asks for the number of $8$ bit strings with exactly $2$ $1$'s, why is it $8\choose 2$? I don't know how the formula relates to all the ways the two bits can be arranged in the string. Also, I'm not completely sure why it says it's a factorial, when the total number of strings is $2^8$. I'm confused about the relation between all the things. How does it come to the answer?

Visualize: You have a muffin tin with eight holes in it. You can fill up $2$ holes with ones and leave to rest empty with zeros. The question boils down to nothing more or less than: how many ways are there to choose which $2$ of the $8$ holes to fill.

So that is why $ {8\choose 2}=$ number of ways to choose $2$ out of $8$ elements is the correct answer.

From the comments, you seem to understand why "number of ways to choose $k$ out of $n$ elements" $= \frac {n!}{k!(n-k)!}$ so I won't try to explain that.  (Unless you want me to.)

It's a factorial rather than $2^8$ because you do not have any choice in how many holes to fill. If you have a choice for all eight holes there are $2^8$ options. But once you choose the two that will be $1$ you will not have any more choices. If you manage to get to $7$ and only chose one $1$ you will not have any choice for the last bit; it must be $1$. If you get to 6 with all $0$s, you will have no choice for the last two; they must both be $1$.

If you want to think of this in terms of $2^8$ there are $2^8$ ways to filled $8$ bits.  But you must toss out all $11xxxxxx$ except for $11000000$ which you may keep.  SO that that leaves $2^8 - (2^6 - 1)$.

Now you must also toss out all $101xxxxx$ except for $10100000$ which you must keep and you must toss out all but one $1001xxxx$ and all but one $10001xxx$ etc.

So that leaves  $2^8 - (2^6 - 1) - (2^5 - 1) - (2^4 - 1) - (2^3 - 1) - (2^2 - 1) - (2^1 - 1)$

But now we mus toss out all the $011xxxxx$ except $01100000$ and all the $0101xxxx$ but one and so on.  so the answer is:

$2^8 - (2^6 - 1) - (2^5 - 1) - (2^4 - 1) - (2^3 - 1) - (2^2 - 1) - (2^1 - 1)$

$-(2^5-1) -(2^4-1)-(2^3-1) - (2^2-1) -(2^1-1)$

$-(2^4-1)-(2^3-1)-(2^2-1) -(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^5-1) -(2^4-1)-(2^3-1) - (2^2-1) -(2^1-1)$

$-(2^4-1)-(2^3-1)-(2^2-1) -(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^4-1)-(2^3-1)-(2^2-1) -(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^1-1)=K$

And as it turns out  $K = \frac {8*7}{2}$.  (!!) I'll leave it to you to contemplate an algebraic reason that may be so.

But I recommend that you study up on "stars and bars" to get a sense of how these different methods of interpretation tactilely yield the same and correct answers.

Because C(8,2) is simpler notation for 8!/6!2!

This question looks pretty bad on you as it shows you didn't try to properly understand the concept of a combination before you decided to create an account only so you could ask this.


What method can be used for solving this fokker Planck equation and how? Let's have this equation:
\begin{equation}
\frac{\partial p(x,t)}{\partial t} = - a \frac{\partial p(x,t)}{\partial x} + \frac{1}{2} b \frac{\partial^2 p(x,t)}{\partial x^2}
\end{equation}

a and b are constant and $0<x<L$.

boundary conditions
\begin{equation}
p(0,t)= f(t) \qquad
p(L,t)= g(t)
\end{equation}

I need to know which method can work? I have no idea how to deal with this boundary value problem

Many thanks for your help in advance

Your PDE is second order linear parabolic and homogeneous PDE which can be written as:

$$ \partial_t p + a \nabla p = D \, \nabla^2 p, \quad p = p(x,t), \quad 0<x<L, \quad t >0,$$

where $a$ and $D = b/2$ are constants. Since you have a bounded domain on $x$ and your equation is linear, separation of variables can do the trick here. Assume you are given an initial condition such that $p(x,0) = p_0(x)$ and you define the non-dimensional variable $\xi = x/L$ (you can do the same with $t$ and $p$) so you have:

$$\partial_t p + k \partial_\xi p = \alpha \partial_{\xi}^2p, \quad p = p(\xi,t), \quad 0  < \xi < 1, \quad t > 0, \tag{1} $$ with $\alpha = D/L^2$, $p(0,t) = f(t), \ p(1,t) = g(t)$ and $p(\xi,0) = p_0(L \xi) \equiv \tilde{p}_0(\xi)$. We cannot proceed further with separation of variables because of the inhomogenous boundary conditions, so we have to make use of the superposition principle defining:

$$p = u+v,$$

where $u$ satisfies homogenous boundary conditions and $v$ is the simplest function that satisfies $v(0,t) = f(t)$ and $v(1,t) = g(t)$. Set $v(x,t) = A(t) x+ B(t)$ and solve for $A$ and $B$. Write the problem for $u$:

$$\partial_t u + k \partial_\xi u  = \alpha \partial_\xi^2 u \underbrace{ -\partial_tv-k\partial_\xi v + \alpha\partial^2_\xi v}_{W(x,t)}, \tag{2}$$ the known term $W$ makes your equation for $u$ non-homogenous so you must solve the following new problem, which I create (Fredholm's alternative):

$$ \theta_t + k \theta_x = \alpha \theta_{xx}, $$

together with $\theta(0,t) = \theta(1,t) = 0$ and I don't care about the initial conditions (which for $u$ turn to be $u(x,0) = p(x,0)-v(x,0)$).

Set now:

$$\theta(\xi,t) = P(\xi)Q(t), \quad P \neq 0, \ Q \neq 0$$ and substitute back in the PDE for $\theta$ to have:

$$PQ' + k P' Q = \alpha P'' \implies \frac{Q'}{Q} = \frac{\alpha P'' - k P'}{P} = \lambda \in \mathbb{R}^- \cup \{0\} \mathbb{R}^+. \tag{3} $$

Eq. $(3)$ yields to two different problems for $P$ and $Q$. Solve for $P$ and find the so-called eigenfunctions and eigenfunctions associated to the homogenous Dirichlet boundary conditions $P(0) = P(1) = 0$. Expand then the solution $u$ in terms of the eigenfunctions (Sturm-Liouville theory):

$$u = \sum^\infty_{n} Q_n(t) P_n(x),$$ and find the Fourier coefficients $Q_n(t)$ by introducing this in eq. $(2)$ and making use of the property of orthogonality of $P_n(x)$, i.e.,

$$ \int^1_0 P_n(x) P_m(x) r(x) \, \mathrm{d} x = \delta_{nm}, $$ where $\delta$ is the Kronecker delta and $r(x)$ is the weight function of the self-adjoint problem for $P$ (which is $r(x) = 1$ here).

Hope you find this helpful. Cheers!

Is Eq 3 correct?

There is a 'constructive' solution to this where you in-cooperate elements of physics. Use de Broglie's theory and Kac-Moody algebra.


Complex Measures: Decomposition Given a complex measure: $\mu:\Sigma\to\mathbb{C}$.

Consider its decomposition into positive measures:
$$\mu=\Re_+\mu-\Re_-\mu+i\Im_+\mu-i\Im_-\mu=:\sum_{\alpha=0\ldots3}i^\alpha\mu_\alpha$$

  Does it split into disjoint regions: $\mu_\alpha(E)=|\mu|(E\cap A_\alpha)\quad(A_\alpha\cap A_\beta=\varnothing)$

Sure. Set $A$ to be the set on which $\mu$ is positive and $B$ the set on which its negative.

Consider the Radon-Nikodym derivatives:
$$\mu_\alpha(E)=\int_Eu_\alpha\mathrm{d}|\mu|$$
They give rise to the constraints:
$$u_\alpha\geq0:\quad|u|=1\implies u_\alpha=\chi_{A_\alpha}$$

  So the above can only hold for: $\mathrm{im}u\subseteq\{1,i,-1,-i\}$

As an example when this fails consider:
$$\Omega:=[0,1]:\quad\mu(A):=\int_Ae^{it}\mathrm{d}t$$
(This way it becomes clear since it takes other values, too.)


Prove that $B_{0}=\{x\in (0, 1) \mid x=m/2^{n} \;\text{for some} \; m\in\mathbb{N} \;\text{and}\; n\in\mathbb{N}\}$ is countable. Haven't got any clue yet, should we prove there is a bijection between $B_0$ and $\mathbb{N}$ ?

Another attempt:

Let $B:=${$m/2^n| m,n \in \mathbb{N}$}.

$f: \mathbb{N}×\mathbb{N}$ $\rightarrow$ $B$,

$f:(m,n) \mapsto m/2^n.$

$f$ is a surjective map from a countable

set $ \mathbb{N}×\mathbb{N}$ to $B$.

Hence $B$ is countable.

$B_0 \subset B$, i.e

$B_0$ as a subset of a countable set $B$ is countable.

Consider the map $f: \mathbb{N}\times \mathbb{N} \to B_0$ given by: $f(m,n) = \dfrac{m}{2^n}$, and $f$ is a bijection and the latter set is countable since former also is.


An infinite-dimensional subspace of $\mathbb{R}^\infty$ not equal to $\mathbb{R}^\infty$? Let $V=\mathbb{R}^\infty$  be be the vector space of real-valued sequences. Does
  there exist an infinite-dimensional subspace $U \subseteq V$ such that $U$ is not equal to $V$?

I'm not even sure where to start with this one and a push in the right direction would be greatly appreciated!

Given a natural number $n$, we can consider the subspace $U$ of $\mathbb{R}^n$ defined by
$$U=\{(a_1,\ldots,a_n)\in\mathbb{R}^n:a_1=0\}$$
You should know (and be able to prove) that $\dim(U)=n-1$.

However, if $n$ were not a natural number, but instead $\infty$...

There are several sequence spaces which are quite natural and some of them are quite important in functional analysis.

I will mention just two examples:

The spaces $c_0$ consisting of all sequences convergent to zero (i.e., all null sequences).
The space $c_{00}$ consisting of all sequences which have only finitely many non-zero terms. (They are also called sequences with finite support or eventually constant sequences.) You may notice that this is exactly the span of the vectors $e^{i}$, $i\in\mathbb N$, where $e^i$ denotes the sequence which has one on $i$-th coordinate and all other coordinates are zeroes. So, in a sense, this reminds generating $\mathbb R^n$ using standard basis.

It should not be difficult to show that both $c_0$ and $c_{00}$ are infinite-dimensional proper subspaces of what you denote $\mathbb R^\infty$. (For some reasons my personal preference is the notation $\mathbb R^{\mathbb N}$. You can see in some other posts on this site that it is not entirely uncommon notation. This notation has already been mentioned in Peter Franek's comment. Some people also use the notation $\mathbb R^\omega$.)

Basically any sequence space from the the Wikipedia article I linked above should work. (With the caveat that in that article sometimes $\mathbb C$ is used as the base field, so you are interested only in real case.) I have chosen two examples which seemed to be relatively simple and also somewhat natural.


How to prove Schwarz inequality for Hermitian forms? I'm trying to do something like the proof of the Schwarz inequality for inner product.

If $h(y,y)\neq 0$, then we can take $\alpha=-h(x,y)/h(y,y)$ and calculate $h(x+\alpha y,x+\alpha y)$ which is a nonnegative number. The desired conclusion follows of this calculation.

If $h(y,y)=0=h(x,y)$, then the inequality is trivial.

My question is: how do we know that the case $h(y,y)=0$ and $h(x,y)\neq 0$ is not possible? When we are working with an inner product, this case is not possible because, by definition, $\langle y,y\rangle=0\Rightarrow y=0\Rightarrow\langle x,y\rangle=0$. But we don't have this condition for hermitian forms.

Thanks.

If $h(y,y) = 0$ but $h(x,y) \ne 0$, then taking $\alpha = - t h(x,y)$ with $t > 0$
we would have
$h(x + \alpha y,  x + \alpha y) = h(x, x) - 2 t |h(x,y)|^2$,
which would be negative for sufficiently large $t$.

Instead of considering $h(y,y)$, you should consider $h(x,y)\neq 0$, then put $\alpha=\frac{-h(x,x)}{h(y,x)}$ – it always works. For $h(x,y)=0$, since $h(x,x)\ge 0$ we get the inequality.


If the curl of some vector function = 0, Is it a must that this vector function is the gradient of some other scalar function? I know of course that If the curl of a vector function is equal to zero, then the vector function is the gradient of some other scalar function, but is this a must?

if so, please give mathematical proof.

First I would like to note the difference between the following two statements:

If a vector field is the gradient of a scalar function then the curl of that vector field is zero.

If the curl of some vector field is zero then that vector field is a the gradient of some scalar field.

I have seen some trying to prove the first where I think you are asking for the second

I apologize for not giving full details on math here because I'm doing this on my tablet.

Anyway, if the curl of a vector field $F$ is zero $(\nabla\times F=0)$ then the surface integral of the resulting vector field over any arbitrary surface $S$ is also zero.

Stokes theorem (read the Wikipedia article on Kelvin-Stokes theorem) the surface integral of the curl of any vector field is equal to the closed line integral over the boundary curve.

Then since $\nabla\times F=0$ which implies that the surface integral of that vector field is zero then (BY STOKES theorem) the closed line integral of the boundary curve of that (arbitrary) selected surface is also zero.

Since the selection of the surface is arbitrary the. We can say the closed line integral of $F$ over any arbitrary closed curve is zero.

This implies that the line integral of the vector field $F$ is path independent which means the line integral over any curve only depending the initial and final position (not necessarily a closed curve)

To prove this just divide your closed path into two paths from point $P_{1}$ two point $P_{2}$, call those paths $A$ and $B$, the line integral over a closed path $C$ is equal to the summation of the line integral over paths $A$ and $B$ so:

$$
\oint\limits_C F \mathrm{d}\ell = \int\limits_{A_{P_1 \to P_2}} F \mathrm{d}\ell + \int\limits_{B_{P_2 \to P_1}} F \mathrm{d}\ell =0
$$
Then
$$
\begin{split}
\int\limits_{A_{P_1 \to P_2}}\! F \mathrm{d}\ell &=- \int\limits_{B_{P_2 \to P_1}} F \mathrm{d}\ell\\
&\Updownarrow\\
\int\limits_{A_{P_1 \to P_2}}\! F \mathrm{d}\ell &= \int\limits_{B_{P_1 \to P_2}} F \mathrm{d}\ell
\end{split}
$$
( please note that those are alkaline integral and that there should be a for product sign between the two f and dl)

This latter equality implies that it doesn't matter your choice of the path $A$ or $B$ or any path because the result will be the same and it will only depend on the vector field $F$ and the two end points.

And since it only depends on the two points $P_{1}$ and $P_{2}$, then we can DEFINE a scalar field  $\Phi(P)$ (note that the points $P_{1}$ and $P_2$ are position vectors) such that
$$\Phi(P_{2}) - \Phi(P_{1}) = \int\limits_{P_{1}}^{P_{2}}F \mathrm{d}\ell$$

(Note that the integral doesn't depend on the path and that is the only reason we can write it this way).

Now from the gradient theorem ( look for the Wikipedia article on gradient theorem ) $F=\nabla \Phi$.

Also look for the Wikipedia article on conservative fields.

It is rather sufficient to prove that the curl of a vector function $\mathbf{F}$ which is the gradient of a scalar-function $\phi$ is 0.

Let $\phi(x,y,z)$ be a scalar-function. Then its gradient will be $$\nabla\phi(x,y,z) = \frac{\partial \phi(x,y,z)}{\partial x}\hat{\mathbf{x}} + \frac{\partial \phi(x,y,z)}{\partial y}\hat{\mathbf{y}} + \frac{\partial \phi(x,y,z)}{\partial z}\hat{\mathbf{z}} \quad.$$ By assumption, $$\mathbf{F} = \nabla \phi$$. We have to show that $$\nabla \times \nabla \phi(x,y,z) = \mathbf{0}$$.

Since $\nabla \times \mathbf{h} = \text{vector}$, we can write the components of this vector by the rule of cross-product i.e. $$(\nabla \times \mathbf{h})_x = \nabla_y h_z - \nabla_z h_y$$. So, we can write $$\text{curl}\mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{ F_y}{\partial z}\right)\hat{\mathbf{x}} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\hat{\mathbf{y}} +\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{\mathbf{z}} $$. But we know $$F_x = \nabla\phi_x = \frac{\partial \phi(x,y,z)}{\partial x} \\\\\ F_y = \nabla\phi_y =\frac{\partial \phi(x,y,z)}{\partial y} \\\\\\\ F_z = \nabla \phi_z =  \frac{\partial \phi(x,y,z)}{\partial z}$$. Substituting these in the definition of $\text{curl} \mathbf{F}$, we get $$\text{curl} \mathbf{F} = \nabla \times \nabla \phi = \left(\frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y}\right)\hat{\mathbf{x}} + \left(\frac{\partial^2 \phi}{\partial z \partial x} - \frac{\partial^2 \phi}{\partial x \partial z}\right)\hat{\mathbf{y}} + \left(\frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x}\right)\hat{\mathbf{z}} = \mathbf{0} \qquad .$$


Number of $4 $ digit numbers with no repeated digit. Number of $4$ digit numbers with no repeated digit is

$4536$
$3024$
$5040$
$4823$

Well, I am very much weak in combinatorics. Please help.

Ok so lets write down any old $4$ digit number

$abcd$

How many choices do we have for the digit $a$? We have $9$ choices (since the first digit cannot be $0$). Now for each possible choice of $a$ we have $9$ choices for $b$ (since we want $b$ to be a different digit to $a$ and we now allow $0$).

So for choice of the $ab$ part we have $9*9 = 81$ possibilities.

Now for each of these we have $8$ choices for $c$ (to avoid $c$ being the same as either $a$ or $b$). And for each of these we have $7$ choices for $d$ (to avoid $d$ being the same as either $a,b$ or $c$).

So in total there are $9*9*8*7 = 4536$ possible numbers.

When you say "combinations" I also get the sense you mean that order doesn't matter. Sometimes people say "combinations" when they mean "permutations" where the order matters.

So for example, if there are three people April, Bruce and Chandra and I want to know how many different pairs of people there could be there are three: April and Bruce, Bruce and Chandra, or Chandra and April. The orders don't matter because April and Bruce is the same as Bruce and April. This is a "combination" of people. Thus, the number of ways of choosing 2 people from 3 is three.

On the other hand, if those three people competed in a math club, then how many different ways could you choose a winner and a runner-up? It could be April first and Chandra second, or Chandra first and April second, or four other possibilities involving Bruce. Thus, the number of ways of choosing 2 people from 3 where order matters is six. This is working with "permutations".

The two ideas are related. If you take the number of permutations and divide it by the number of ways of arranging two people (2) then you get the number of combinations.

Now to your question:
Since you asked about combinations, I will say that order doesn't matter.

Start by finding the permutations: For the first choice, you have 10 possible digits to choose from. For the second choice, you have 9 digits because you used one for the first choice. The third choice comes from 8 possibilities and the fourth from 7 possibilities. Now we multiply these together: 10 x 9 x 8 x 7 = 90 x 56 = 5040. That's the number of permutations. No digits repeat, but 0123 is different from 0321.

Now to find the number of combinations, I have to know how many different ways there are of arranging four digits. That's the same kind of problem: the first position could be from 4 possibilities, the second from 3 possiblities, the third from 2 choices and the last has to be the 1 left. So there are 4 x 3 x 2 x 1 = 24 possible ways of arranging 4 items.

Therefore I divide 5040 / 24 = 210. So there are 210 different combinations of four digits chosen from 0-9 where the digits don't repeat.


Showing $\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$ $$\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$$

We cannot use the Herschfeld's Convergence Theorem because you can see that it's negative. I have tried some generalization but nothing good. I have looked for some algorithm on the web like Landau's algorithm but we speak about and infinite nested radicals so I was wondering myself if there is not a recursive formula existing.

So if you have some papers or a proof, I will be thankful.

Any helps is greatly appreciated

Thanks a lot.

I suppose the right-hand side is defined (that is, interpreted) as $\lim\limits_{n\to\infty}a_n(1)$, where $$a_1(x)=x,\qquad a_{n+1}(x)=a_n\big(\sqrt{1-2^{-n}x}\big).$$ Now observe that $a_n(1-2^{-n-1})=3/4$ for each $n$; to prove it, use induction and $$a_{n+1}(1-2^{-n-2})=a_n\left(\sqrt{1-2^{-n}(1-2^{-n-2})}\right)=a_n(1-2^{-n-1}).$$ Thus, it remains to prove $\lim\limits_{n\to\infty}\big(a_n(1)-a_n(1-2^{-n-1})\big)=0$. This can be done using the mean value theorem: we have $a_n(1)-a_n(1-2^{-n-1})=2^{-n-1}a_n'(x_n)$ for some $x_n\in(1-2^{-n-1},1)$, and the (very crude) estimate $$x\in(0,1)\implies|a_n'(x)|\leqslant 1\quad\color{gray}{\left[\impliedby a_{n+1}'(x)=-\frac{a_n'\big(\sqrt{1-2^{-n}x}\big)}{2^{n+1}\sqrt{1-2^{-n}x}}\right]}$$ suffices (of course, a much more precise result can be obtained this way).


How to use the inverse function theorem? I have a function $F (x, y) = (x^2+y^2, xy)$ and I need to show that it has an inverse. How do I find the inverse of this function using the inverse function theorem? I have not learned this before in all of my math classes and I just need some brushing up on how to use it.

Thank you for your assistance.

edit:
On the set
{(x, y) : −x < y < x}

For $F(x,\, y) = (x^2+y^2,\; xy)$ denote
$$\begin{cases}u = x^2+y^2, \\
v = xy \end{cases}$$
Then
$$\begin{cases}
u+2v=(x+y)^2, \\
u-2v=(x-y)^2
\end{cases} $$
On the given set $\{(x, y):\;\; −x < y < x \} $ we have $$ x+y>0,\;\;x-y>0, \quad J_F(x,\, y)=2(x^2-y^2)\ne{0},$$ therefore $F$ is invertible and
$$
\begin{cases}
x+y=\sqrt{u+2v}, \\
x-y=\sqrt{u-2v}
\end{cases}$$
Thus
$$\begin{cases}
2x=\sqrt{u+2v}+\sqrt{u-2v},\\
2y=\sqrt{u+2v}-\sqrt{u-2v}.
\end{cases}$$

when you calculate jacobian matrix and found value p,where this matrix's determinant is not zero,there maybe you could you this formula. i would be happy if it helps you.also  you could check  all formula in wikipedia's site
.for single variable .let suppose that

$y=\sqrt{x}$

and   let us replace  places for x and y,or we get
$x=\sqrt{y}$

from which
$y=x^2$


Proving that a field isomorphism exists I have a question that I need some helping solving:

Show if $G$ is a field with exactly three elements, then there is a field isomorphism $f: \Bbb Z_3 \to G$. Note that $\Bbb Z_3$ is defined $\Bbb Z_3=(\{0,1,2\}, +, .)$ where $x+y=(x+y) \mod 3$ and $xy=xy \mod 3$.

Another answer proceeds by showing that the additive groups are isomorphic and the multiplicative groups are isomorphic.  For the convenience of the readers of this question, here is a counterexample to show that this is not in general sufficient for the fields to be isomorphic.  (Thus in my opinion this other answer is totally bogus.)

Let $F_1 = \mathbb{Q}(\sqrt{-7})$ and $F_2 = \mathbb{Q}(\sqrt{-11})$.  These are nonisomorphic fields.  However, the additive groups of each one are two dimensional $\mathbb{Q}$-vector spaces, hence are isomorphic.  Each multiplicative group is isomorphic to the direct product of a cyclic group of order $2$ with a free commutative group of countable rank.  This is especially easy to see here since the ring of integers of each of $F_1$ and $F_2$ is a PID, but in fact for any number field $K$, the group $K^{\times}$ is isomorphic to the product of the finite cyclic group of roots of unity in $K$ with a free commutative group of countable rank.

I will also add that in my opinion it is always better to construct an explicit isomorphism than to argue that two structures are isomorphic, if possible.  In the case of a finite field $F$ of prime order $p$, it is not only possible but easy: as for any ring, there is a unique ring homomorphism $\varphi: \mathbb{Z} \rightarrow F$.  Since $F$ is finite of order $p$, $\varphi$ is surjective and its kernel is $p\mathbb{Z}$, so $\varphi$ induces an isomorphism $\varphi: \mathbb{Z}/p\mathbb{Z} \stackrel{\sim}{\rightarrow} F$.  If now $F_1$ and $F_2$ are two finite fields of order $p$, then $\varphi_2 \circ \varphi_1^{-1}: F_1 \rightarrow F_2$ is an explicit isomorphism.  (Of course this is really just a slightly fancier/cleaner version of the straightforward argument involving $f(1+\ldots+1) = f(1) + \ldots + f(1) = 1 + \ldots + 1$.) It is easy to see that this isomorphism is unique.

A slightly more abstract approach: G and Z3 are both fields of order 3.  A field is a group both under addition and multiplication.  Now both G and Z3 have additive subgroups of order 3, and since there is only one group of order 3 (to within isomorphism), it must be that the additive group in G is isomorphic to the additive group in Z3. Same with the multiplicative group.So the 2 fields are isomorphic.

What we did here was show there must be an isomorphism without actually constructing it.  One nice consequence of this is that the same proof works for fields of any finite prime order.


slant asymptote of $y=x\operatorname{arccot}(x)$ help me with this problem

Find the slant asymptote of

$$y=x\operatorname{arccot}(x)$$

I know that the slant asymptote is $y=ax+b$ if

$$\lim_{x\to \infty} f(x)-(ax+b)=0$$

the answer given to me is:
$$\lim_{x\to +\infty} \frac{y}{x}=\lim_{x\to +\infty}\operatorname{arccot}(x)=0$$
$$\lim_{x\to -\infty} \frac{y}{x}=\lim_{x\to -\infty}\operatorname{arccot}(x)=\pi$$
$\lim_{x\to -\infty}(y-\pi x)=\lim_{x\to -\infty}\frac{\operatorname{arccot}(x)-\pi}{\frac{1}{x}}=\lim_{x\to -\infty}\frac{-\frac{1}{1+x^2}}{-\frac{1}{x^2}}=1$(L'Hospital)

conclusion: $y=\pi x +1$ is the slant asymptote of the function

i don't understand this answer can someone explain the answer for me or have any other ways to do this problem?

note: this is from the last year exam of my college so i don't think the question is wrong, the answer was made by students though so they could be wrong.

Note that,

$$\lim_{x\to \infty} x \cot^{-1}(x) = \lim_{x\to \infty} \frac{ \tan^{-1}\frac1x}{\frac1x}=1$$

Thus, the asymptote $y=b=1$ is horizontal, not slant.

Edit:

The range of the inverse cotangent function is $(-\frac\pi2,0),(0,\frac\pi2)$, which does not permit a slant asymptote as $x\rightarrow -\infty$. Some source may specify its range as $(0,\pi)$, as in your answer. Then, a slant asymptote could be identified, which may be questionable.


Continuous function and extreme value theorem. Suppose $f(x)$ is a continued function defined on $[-1,1]$ with $f(0)>0$ and $f(\pm 1)=0$. Prove that there exist constants $a>0$ and $b$ such that $g(x)=-a\vert x\vert +b\geq f(x) (-1\leq x \leq 1)$, and for which there exists $c\in (-1,1)$ such that $g(c)=f(c)$.

I don't have a clue about how to solve this problem. The book gives a hint that tells me to use the extreme value theorem, and I don't know how to make use of it. And it seems that if I want to use contradiction I would have to construct a very complicated negation.

Pick $a$ s.t. $0 < a < f(0)$ (for example $\dfrac{f(0)}{2}$ will do).

Let $h(x) = a\vert x\vert+ f(x)$. Since $h$ is continuous on $[-1,1]$ and $h(1) = h(-1) = a < h(0) = f(0)$, it has a maximum $c$ strictly inside the interval, so $c$ is in $(-1,1)$.

Then if $b = h(c)$, you get your $g(x) = -a\vert x\vert + b, g(c) = f(c)$ and $g(x) \geq f(x)$ since $f(x) - g(x) = h(x) - h(c)$ and $c$ is the maximum of $h$ by definition.

Because $[-1,1]$ is compact, $f$ has a max, suppose k and note that $k>0$ because exist a point (0) where f is positive. If we put $b=k+1$ and $a=1$ is easy check that works. The second item is not necessary true...


construct entire function Can anyone please explain to me how to use Weierstrass product theorem to construct entire function?
If I want to construct entire function with simple zeros on the positive real axis at the points $\sqrt n , n\ge1$ and double zeros on the imaginary axis at the points $ \pm i \sqrt n , n\ge1$ with no other zeros.

There is a trick to the Weierstrass product. If $a_k$, $k\geq 1$ is a sequence tending to zero then $$\prod_{k\geq 1} (1-a_k z)$$ need not be convergent. You may, however, recover convergence by multiplying by a factor that has no zeros. You start by looking at $$1/(1-a_k z) = \exp( -\ln (1-a_k z))=\exp(\sum_{j\geq 1} \frac{(a_kz)^j}{j})$$ but you truncate the sum in the exponential to a finite order $p_k$.
By choose suitable $p_k$ (which may tend to infinity) the product:$$\prod_{k\geq 1} (1-a_k z) \exp\left(\sum_{j= 1}^{p_k} \frac{(a_kz)^j}{j}\right) $$
is an entire function.

let be the sets of zeros $ x_i $ of a function $ f(x_i) $ then the entire function can be constructed in the form

$$ f(x)= \prod _{n=0}^{\infty}(1-\frac{x}{x_{i}}) $$

then this $ f(x) $ is an entire function, see the product

$$ \prod _{n=0}^{\infty}x_{i}$$

must be zeta regularizable


Is it true that E [ X | E [ X | Y] ] = Ex [ X | Y] ? Does this law have a name? Let $X$ and $Y$ be two random variables (say real numbers, or vectors in some vector space). It seems to me that the following is true:

E [ X | E [ X | Y ] ] = E [ X | Y]

Note that E [ X | Y ] is a random variable in it's own right. Also note that equality here is point-wise, for every point in the sample space of the joint distribution on on $(X,Y)$. My question, assuming I'm not missing something and the above is true, is whether this law has a name, or is written down / proved somewhere.

Let $Z=E[X\ | \ Y]$. Your equation states: $E[X \ | \ Z]=Z$. This follows from the following fact.

Tower Property of Conditional Expectation:

$$E[E[X\ | \ \mathcal{F}]\ | \ \mathcal{G}]=E[X\ | \ \mathcal{G}],\text{ whenever }\mathcal{G}\subset \mathcal{F}.$$

Proof of your equation:

We apply the tower property with $\mathcal{G}=\sigma(Z)$ and $\mathcal{F}=\sigma(Y)$. Note that $\sigma(Z)\subset \sigma(Y)$ follows from the construction of $Z$ as a conditional expectation w.r.t. $Y$.

Plugging in to the tower property,
$$
\begin{align*}
E[E[X\ | \ \sigma(Y)]\ | \ \sigma(Z)]&=E[X\ | \ \sigma(Z)]\\
\implies E[Z\ | \ Z]&=E[X\ | \ Z]\\
\implies Z&=E[X\ | \ Z].
\end{align*}$$

Essentially the law of iterated expectation, perhaps more commonly written like $$\operatorname{E_X} [X] = \operatorname{E}_Y [ \operatorname{E}_{X \mid Y} [ X \mid Y]].$$

For a discrete case, the essence of the proof is
$$\operatorname{E}_Y [ \operatorname{E}_{X \mid Y} [ X \mid Y]] = \sum_y \sum_x x \cdot \operatorname{P}(X=x \mid Y=y) \cdot \operatorname{P}(Y=y) =\sum_x x \cdot \operatorname{P}(X=x) =\operatorname{E_X} [X].$$


Let $A = \{1,2,3,4\}$. Number of functions $f :A \to A$, such that $f(f(x))=x~\forall x \in A$ , is. Let $A = \{1,2,3,4\}$. Number of functions $f :A \to A$, such that $f(f(x))=x,  \forall x \in A$ , is.

For this , what I thought was 2 cases are possible. Either an element maps to itself or 2 elements map to each other.

Eg either $1 \to 1$ or $1 \to 2$ and $2 \to 1.$

So for this I can group the elements in groups of 2 in $^4C_2 $ways. Then say I group $(a,b)$ and$ (c,d)$ . Then $ a$ can either map to $a$ or $b$. The other element from that group will automatically be defined. Similarly for $(c,d)$

So the answer should be $^4C_2× 2×2 = 24$ . But the answer is

  $13$

Please help.

Case 1: There are 4 fixed points. The only such involution is the identity.

Case 2: There are 2 fixed points. In this case, once 2 elements are fixed, the remaining 2 elements must map to each other. Hence, the number of involutions with 2 fixed points is $\binom{4}{2} = 6$.

Case 3: The involution is a derangement (fixed-point-free). In this case, the number 1 must map to 2, 3, or 4, and whatever 1 maps to must be mapped to 1. Then, the remaining 2 elements must map to each other. Hence, the number of involutions without any fixed points is 3.

In total, there are 10 involutions on a 4-element set like $A = \{1, 2, 3, 4\}$.

There are 3 cases:

each element is mapped onto itself $x\to x$
The pair $(a,b)$ is mapped onto $(b,a)$ and the pair $(c,d)$ is mapped on $(d,c)$. There are 6 ways you can choose $(a,b)$
The pair $(a,b)$ is mapped onto $(b,a)$ and $c\to c$ and $d\to d$. There are 6 ways you can choose $(a,b)$

Total 13 cases


Degree of polynomial interpolating the primes The polynomial $p_3(x)$ passes through the points
$(1,2), (2,3), (3,5)$, where $2,3,5$ are the first three primes:
$$
p_3(x) = \frac{x^2}{2}-\frac{x}{2}+2 \;.
$$
Similarly, one can form an interpolating polynomial $p_n(x)$ that
passes through the first $n$ primes.
For example:
$$
p_5(x) = \frac{x^4}{8}-\frac{17
   x^3}{12}+\frac{47
   x^2}{8}-\frac{103 x}{12}+6 \;.
$$
One can check that
\begin{eqnarray}
p_5(1) &=& 2 \\
p_5(2) &=& 3 \\
p_5(3) &=& 5 \\
p_5(4) &=& 7 \\
p_5(5) &=& 11 \;.
\end{eqnarray}

My question is:

  Q. Is the degree of $p_n(x)$ ever strictly less than $n{-}1$, for any $n$?

The answer to Q is positive if a "coincidence" occurs,
such that a smaller degree
polynomial captures those $n$ prime points.
Do such coincidences ever occur?

The degree of $p_n(x)$ is always $n-1$. The proof is by induction.

Note that $p_1(x) = 2$ has degree $0$. Now assume that $p_{n}(x)$ has degree $n-1$. We want to prove that $p_{n+1}(x)$ has degree $n$. Assume otherwise, so $p_{n+1}(x)$ also had degree at most $n-1$. Then since $p_{n+1}(x)$ and $p_n(x)$ agree on the first $n$ values, it must be the case that $p_{n+1}(x) = p_n(x)$. In particular, to obtain a contradiction, it suffices to show that

$$p_n(n+1) \ne^{?} p_{n+1}.$$

In fact, we simply will prove that $p_n(n+1)$ is always even which does the job.

We can write down a formula for $p_n(x)$, namely

$$p_n(x) = \sum_{i=1}^{n} p_i \cdot
\frac{(x-1)(x-2) \ldots \widehat{(x-i)} \ldots (x - n)}{(i-1)(i-2)
\ldots \widehat{(i-i)} \ldots (i - n)},$$

where the hat indicates the term is omitted. This is clearly a polynomial of degree at most $n-1$ and $p_n(i) = p_i$. (This is the general formula for Lagrange interpolation specialized to this case.)

Hence

$$\begin{aligned} p_n(n+1) = & \  \sum_{i=1}^{n} p_i  \cdot\frac{ n!/(n+1-i)}{(i-1)! (n-i)!  (-1)^{n-i}}\\
= & \  (-1)^{n-1} \sum_{i=1}^{n} p_i \cdot \frac{ n!}{(i-1)! (n+1-i)!} (-1)^{i-1} \\
=  & \ (-1)^{n-1} \sum_{i=1}^{n} p_i \cdot \binom{n}{i-1} (-1)^{i-1}\\
= & \  (-1)^{n-1} \sum_{i=0}^{n-1} p_{i+1}  \binom{n}{i} (-1)^i\end{aligned}$$

Now we use the fact that, with the exception of $p_1 = 2$, the primes are all odd. It follows that

$$p_{n}(n+1) \equiv \sum_{i=1}^{n-1} (-1)^i \binom{n}{i} \mod 2.$$

But now

$$\sum_{i=1}^{n-1} (-1)^i \binom{n}{i}
= (1-1)^n - 1 - (-1)^n \equiv 0 \mod 2,$$

is even for $n > 0$, and hence $p_{n}(n+1)$ is even, and thus $\ne p_{n+1}$, as desired.

Yes. Given $n$ unique data points, there is a unique polynomial of degree $n-1$ that interpolates the data. This is one of the first results in the interpolation section of any numerical analysis/methods course. Using data points constructed with primes is a specific case of this. If you write out the interpolation conditions, you'll see that this is equivalent to solving an $n \times n$ linear system.


Filtered Colimit of Subalgebras Let $K/k$ be a finitely generated field extension of characteristic $0$, and let $\mathcal{S}$ be the set of field extensions of $k$ which have finite index in $K$. Is it true that $\text{Hom}_{k \text{-linear}} (K, K) \cong \text{colimit}_{F \in \mathcal{S}} \text{Hom}_{F} (K, K)$ as $k$-algebras under composition?

No.  For instance, if $k=\mathbb{Q}$ and $K=\mathbb{Q}(x)$, then $\mathcal{S}$ is countable and $\operatorname{Hom}_F(K,K)$ is countable for any $F\in S$, so the colimit is countable.  However, $\operatorname{Hom}_k(K,K)$ is uncountable since $K$ has infinite dimension over $k$.

As stated, the statement would certainly not be true, because the only element in the set on the right hand side is the identity. Thus, I assume that there has been a typographical mistake, and $S$ are in fact the finite field extensions of $k$ which are subfields of $K$.

In this case, observe, in the case of field towers $k \le S \le T \le K$ the existence of a forgetful functor
$$
\mathcal C_T \to \mathcal C_S,
$$
where $\mathcal C_T$ is the category whose only object is $K$, together with the $T$-linear morphisms. This then induces the morphisms of algebrae
$$
\operatorname{Aut}_T(K) \to \operatorname{Aut}_S(K)
$$
which then form the inductive system over which the colimit is taken.

Note that in this case, $S = k$ is a minimum (or maximum, whichever way you define inductive systems), and it is quite a general theorem that when a minimum exists, this minimum is precisely the inductive limit.


Verify that $\bigl(p\to(q\to r)\bigr)\to \bigl((p\to q)\to (p\to r)\bigr)$ is a tautology. Verify that $\bigl(p\to(q\to r)\bigr)\to \bigl((p\to q)\to (p\to r)\bigr)$ is a tautology.

I am confused on this whole tautology even after looking at examples both in my book and on-line. I started a truth table and this is what I have so far. Can someone please explain this to me in a way that I can understand.

\begin{array}{ccc}
p & q & r & ? & ? & ? & ?\\
\hline
T & T & T\\
T & T & F\\
T & F & T\\
T & F & F\\
F & T & T\\
F & T & F\\
F & F & T\\
F & F & F
\end{array}

A completely different approach (called a tableau proof, and a bit on the informal side for one). Feel free to ignore.

  1) Assume $p \implies (q \implies r)$.

    2) Assume $p\implies q$.

      3) Assume $p$.

        4) By (2) and (3) and modus ponens, $q$.

        5) By (1) and (3) and modus ponens, $q \implies r$.

        6) By (4) and (5) and modus ponens, $r$.

      7) By (3)–(6) and implication, $p \implies r$.

    8) By (2)–(7) and implication, $(p\implies q)\implies (p\implies r)$.

  9) By (1)–(8) and implication, $\bigl(p\implies (q\implies r)\bigr)\implies \bigl((p\implies q)\implies (p\implies r)\bigr)$.

Your well-formed formula is an instance of axioms schema 3 of the Łukasiewicz axiomatization of propositional logic.   By the completeness of propositional logic, your wf is a tautology.


How many ways are there to pick 18 letters from 12 A's and 12 B's? How many ways are there to pick 12 letters from 12 A's and 12 B's?
How many ways are there to pick 18 letters from 12 A's and 12 B's?

Let's look at the generating function. We have each letter with at most $12$ elements, so we have the generating function:
$$f(x) = (\frac{1-x^{13}}{1-x})^{2}$$

Your first problem can easily be solved using the stars and bars solution: $\binom{12 + 2 - 1}{12}$ is your answer. This comes from the expansion of $\frac{1}{(1-x)^{2}}$.  The expansion of $(1-x^{13})^{2} = \sum_{i=0}^{2} \binom{2}{i} (-1)^{i} x^{13i}$, so yields $1$ as the only viable term for $12$ letters. Since it is multiplied with the expansion of the denominator of the generating function, nothing changes.

With $18$ letters, we have inclusion-exclusion, which brings the expansion of $(1-x^{13})^{12}$ into play.

So we consider $(1 - 2x^{13})$ as the relevant portion of the expansion of the numerator. The $x^{26}$ term counts $26$ letter selections, which we don't care about, so we ignore it.

Now what terms in the expansion of $\frac{1}{(1-x)^{2}}$ do we multiply with terms in $(1 - 2x^{13})$ to get the count. We care about $x^{18}$ and $x^{5}$. So we have $\binom{18 + 2 - 1}{18} - 2 \binom{5 + 2 - 1}{5}$ as our answer for the number of ways to choose $18$ letters.

How many ways are there to pick 18 letters from 12 A's and 12 B's

Count the letters you are throwing away. 12 + 12 - 18 = 6 so the answer is 2 choose 6 with repetition

${2+6-1\choose 6}$


Spectral radius and dense subspace Let $X$ be a Banach space, and let $E$ be a dense subspace of $X$. Let
$A: X \to X$ be a bounded operator on $X$ that maps $E$ to itself. Assume that the spectral radius of $A$ restricted to $E$ is $r>0$. Is it true that the spectral radius of $A$ on $X$ is equal to $r$ (or can it "jump up")?

Edit (precisions)
I am assuming that $E$ is a linear vector space, and that the spectral radius of $A$ on $E$ is equal to a finite $r>0$. I am asking if this implies that the spectral radius of $A$ on $X$ is at most $r$.

Consider the Banach space $X = C[0,1]$ of continuous functions on $[0,1]$, with the operator $A$ of multiplication by $x$ (i.e. $Af(x) = x f(x)$).  This has spectrum $[0,1]$.  Let $E$ be the subspace of $X$ consisting of polynomials.  This is invariant under $A$.  However, $A - \lambda I$ is never surjective as an operator from  $E$ to $E$, e.g. there is no polynomial $g(x)$ such that $(x - \lambda) g(x) = 1$.  Therefore the spectrum of the restriction of $A$ to $E$ is all of $\mathbb C$.

EDIT: In the other direction, the spectrum of $A$ is always contained in the spectrum of the restriction $A|_E$ of $A$ to $E$: if $\lambda$ is not in the spectrum of $A|_E$, i.e. $A|_E - \lambda I|_E$
has a bounded two-sided inverse $R_\lambda: E \to E$, then $R_\lambda$ extends by continuity to a bounded linear operator $\overline{R}_\lambda$ on $X$ which is a two-sided inverse of $A-\lambda I$ there, so $\lambda$ is not in the spectrum of $A$.  Thus the spectral radius can never "jump up".

For an example where $A|_E$ has finite spectral radius, consider the same $A$ and $X$ as in my previous example, but let $E$ be the space of restrictions to $[0,1]$ of functions analytic in the disk $\{z: |z| < 2\}$. If $|\lambda| > 2$, $A|E - \lambda I$ has a bounded inverse on $E$, namely multiplication by $1/(x - \lambda)$.  But if $|\lambda| < 2$, $A|_E - \lambda$ is not surjective.  Thus the spectral radius "jumps down" from $2$ to $1$.

$X=R^2=Vect\{e_1,e_2\}$, $A(e_1)=e_1, A(e_2)=2e_2$, $E=R^2-Re_2$. The spectrum of $A$ restricted to $E$ is $1$ and the spectrum of $A$ is $\{1,2\}$.


Intersection of $\{ [n\sqrt{2}]\mid n \in \mathbb{N}^* \}$ and $\{ [n(2+\sqrt{2})]\mid n \in \mathbb{N}^* \}$ Find the intersection of sets $A$ and $B$ where
$$A = \{ [n\sqrt{2}]\mid n \in \mathbb{N}^* \}$$
$$B = \{ [n(2+\sqrt{2})]\mid n \in \mathbb{N}^* \}.$$

([$x$] is the integer part of $x$)

Using the computer, we found common elements.

Does anyone have an idea to solve?

These are known as Beatty sequences and they've been quite thoroughly studied (you can find one of your two sequences at http://oeis.org/A001951 ).  The most interesting property (and most relevant here) is that if $r$ and $s$ are irrational numbers with $r,s \gt 1$ and $\frac1r+\frac1s=1$, then the sequences $R=\left\{\lfloor (nr)\rfloor\right\}$ and $S=\left\{\lfloor (ns)\rfloor\right\}$ are complementary: $R\cap S=\emptyset$ and $R\cup S=\mathbb{N}$.


Binomial expansion for approximation I've just had to do a homework on binomial expansion for approximation:

$1.07^9$

so:

$(1+0.07)^9$

To do binomial expansion you need a calculator for the combinations button (nCr), so why would use a more complicated method, which only gives an approximation be used over just typing 1.07^9 into a calculator?

(or is this never done in real life, and it's just a homework?)

Expanding the whole thing using Binomial Theorem gives you an exact value. Not an approximation.

To get an approximation you can consider a few terms from the expansion.

For instance, for "small" $x$, $1+nx$ is a "reasonable" approximation for $(1+x)^n$.

Notice that this corresponds to picking the first two terms from the binomial theorem expansion $(1+x)^n = 1 + \binom{n}{1} \ x + \binom{n}{2}\  x^2 + \dots + x^n$.

For example

$1.0007^9 \approx 1 + 9\times 0.0007 = 1.0063$ which agrees with $1.0007^9 = 1.0063176688422737867054812736724$ upto $4$ decimal places.

Depending on how accurate you want it, you could consider more terms from the binomial expansion.

This is based on the fact that for small $x$, as the power $r$ of $x$ gets larger, the term $x^r$ becomes small quite fast.

This looks like a badly thought-out problem to me. The first-order approximation is 1.63, but the true value is greater than 1.83. Can you give us the exact wording of the question?

As for the practicality of this, I would say that being able to estimate such expressions is a useful skill in many branches of mathematics and statistics. Not essential, but useful.


$\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$. I'm starting Spivak's Calculus and finally decided to learn how to write epsilon-delta proofs.

I have been working on chapter 5, number 3(ii). The problem, in essence, asks to prove that

  $$\lim\limits_{x \to 1} \frac{1}{x} = 1.$$

Here's how I started my proof,

  $$\left| f(x)-l \right|=\left| \frac{1}{x} - 1 \right| =\left| \frac{1}{x} \right| \left| x - 1\right| < \epsilon \implies \left| x-1 \right| < \epsilon |x|$$

I haven't made any further progress past this point. Is it possible to salvage this proof? Should I try an alternate approach?

Update 2/19/2018: It appears that this answer has received a lot of attention, which I'm very glad to know about. When you're reading through this answer and you're trying to learn about $\delta$-$\epsilon$ proofs for the first time, I would recommend skipping the sections labeled Addendum. on your first read. Please let me know of any other clarifications that you would like with this answer.

Whenever I am doing a $\delta$-$\epsilon$ proof, I do some scratch work (note, this is NOT part of the proof) to figure out what to choose for $\delta$. I always tell students to think about the following:

  What are you given?
  What do you want to show?

In the definition of the limit, you are given an arbitrary $\epsilon > 0$ and you want to find $\delta$ such that  $$0 < |x - 1| < \delta$$ implies $$\left|\dfrac{1}{x} - 1 \right| < \epsilon\text{.}$$ You have control over what to choose for your $\delta$ in this case. The idea of this $\delta$-$\epsilon$ proof is to work with the expression $|x - 1| < \delta$ and get $\left|\dfrac{1}{x} - 1 \right| < \epsilon$ at the end.

Let's do some scratch work (again, NOT part of the proof).

Scratch Work

Let's start with what we want to show for our scratch work (starting with what you want to show is bad to do $100\%$ of the time when you're doing proofs - again, this is scratch work and not actually part of the proof).

We want to show that $\left|\dfrac{1}{x} - 1 \right| < \epsilon$. Let's work backwards and try to turn the expression $\left|\dfrac{1}{x} - 1 \right|$ into some form of $|x-1|$.

So, note that $$\left|\dfrac{1}{x} - 1 \right| =\left|\dfrac{1-x}{x} \right| = \left|\dfrac{-(x-1)}{x} \right| = \left|\dfrac{x-1}{x} \right|$$ since $|y|=|-y|$ for all $y$ in $\mathbb{R}$.

The last expression can be rewritten as $\dfrac{\left|x-1 \right|}{\left| x \right|}$. Looking at this expression, we do have $|x-1|$ in the numerator, which is good. But we have that pesky $|x|$ in the denominator.

Since we do have control of what $|x-1|$ is less than (this is our $\delta$), let's choose a really convenient, small number to work with that is greater than $0$. Let's say $\delta = \dfrac{1}{2}$.

Well, if $|x - 1| < \dfrac{1}{2}$, then $$-\dfrac{1}{2} < x-1 < \dfrac{1}{2} \implies \dfrac{1}{2} < x < \dfrac{3}{2} \implies \dfrac{1}{2} < |x| < \dfrac{3}{2}\text{.}$$ So if we choose $\delta = \dfrac{1}{2}$, $\dfrac{1}{2} <|x| < \dfrac{3}{2}$.

  Addendum. In many examples, $\delta$ is usually chosen to be $1$. Why did we elect not to do that in this case?

  It's because it wouldn't work.

  Intuitively, here's why it doesn't: when you consider the neighborhood of radius $1$ centered around $x = 1$, you get the interval $(0, 2)$. $f(x) = \dfrac{1}{x}$ doesn't have a finite limit at $x = 0$, so this makes $\delta = 1$ a bad choice.

  This isn't the case if $\delta = 1/2$. The neighborhood of radius $1/2$ around $x = 1$ is $(1/2, 3/2)$. $f$ has limits at every $x$-value in the interval $(1/2, 3/2)$, including the endpoints.

  In terms of the algebra, if we had chosen $\delta = 1$, the algebra wouldn't have worked out. We would've gotten $0 < x < 1$ and would not have been able to obtain a finite upper bound for $\dfrac{1}{x}$. That is,
  $$0 < x < 1 \implies 1 < \dfrac{1}{x} < \infty\text{.}$$
  We do not have a finite upper bound for $\dfrac{1}{x}$ in this case, and hence why $\delta = 1$ will not work for this purpose.

If $\dfrac{1}{2} <|x| < \dfrac{3}{2}$, then $$\dfrac{2}{3} <\dfrac{1}{|x|} < 2$$ and $$\dfrac{1}{|x|} < 2 \implies \dfrac{\left|x-1 \right|}{\left| x \right|} < 2\left| x-1 \right|\text{.}$$ Now we have control over what $|x-1|$ is less than. So to get $\epsilon$, we choose $\delta = \dfrac{\epsilon}{2}$.

But, wait - didn't I say that we chose $\delta = \dfrac{1}{2}$ earlier? A simple solution would be to minimize $\delta$, i.e., make $\delta = \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$.

  Addendum. To see why $\delta = \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$ works, suppose $\dfrac{\epsilon}{2} > \dfrac{1}{2}$, so that $\delta = \dfrac{1}{2}$. Then $\epsilon > 1$. Then $$\dfrac{\left|x-1 \right|}{\left| x \right|} < 2|x-1| < 2 \cdot \dfrac{1}{2} = 1 < \epsilon\text{.}$$

  Now suppose  $\dfrac{\epsilon}{2} \leq \dfrac{1}{2}$, so that $\delta
 = \dfrac{\epsilon}{2}$.

  Then $$\dfrac{\left|x-1 \right|}{\left| x \right|} < 2|x-1| < 2 \cdot \dfrac{\epsilon}{2} = \epsilon\text{.}$$

  In both cases, we have $\dfrac{\left|x-1 \right|}{\left| x \right|}  < \epsilon$, as desired.

  See also Why do we need min to choose $\delta$?.

So now we've found our $\delta$ and can use this to write out the proof.

The Proof

  Proof. Let $\epsilon > 0$ be given. Choose $\delta := \min\left(\dfrac{\epsilon}{2} , \dfrac{1}{2} \right)$. Then
  $$\left|\dfrac{1}{x} - 1 \right| = \left|\dfrac{x-1}{x} \right| =
 \dfrac{\left|x-1 \right|}{\left| x \right|} < 2\left| x-1 \right|$$
  (since if $|x - 1| < \dfrac{1}{2}$, $\dfrac{1}{|x|} < 2$) and $$2\left| x-1 \right| < 2\delta \leq 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{. }\square$$

Success at last.

  Addendum. Note that the end goal above was achieved, namely to show that $$\left|\dfrac{1}{x}-1\right| < \epsilon\text{.}$$

  In the step $$2\left| x-1 \right| < 2\delta \leq
 2\left(\dfrac{\epsilon}{2}\right) = \epsilon\text{,}$$ textbooks
  usually omit the step with the $\delta$ and just write $$2\left| x-1
 \right| < 2\left(\dfrac{\epsilon}{2}\right) =
 \epsilon\text{.}$$

  Addendum. It may seem that the note "(since if $|x - 1| < \dfrac{1}{2}$, $\dfrac{1}{|x|} < 2$)" may be an additional assumption added to the problem - i.e., that $\delta$ has to be $\dfrac{1}{2}$. This is not the case for the following reason: given $|x-1| < \delta$, we have
  $$|x-1| < \min\left(\dfrac{\epsilon}{2}, \dfrac{1}{2}\right)\text{.} $$
  Obviously, if $\epsilon \geq 1$, we end up with $|x - 1| < \dfrac{1}{2}$, as stated above. But let's suppose that $\epsilon < 1$. Then
  $$|x - 1| < \dfrac{\epsilon}{2} < \dfrac{1}{2}$$
  and you end up with $|x - 1| < \dfrac{1}{2}$, so the $\dfrac{1}{|x|} < 2$ implication holds in either case.

$\tag 1 |\frac{1}{x} - 1| < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; -\varepsilon < \frac{1}{x} - 1 < \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; 1 -\varepsilon < \frac{1}{x} < 1 + \varepsilon \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{1}{1 + \varepsilon} < x < \frac{1}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

$ \text{ iff } \; \frac{-\varepsilon}{1 + \varepsilon} < x - 1 < \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

We are trying to find our $\delta \gt 0$ 'setup', $|x -1| \lt \delta$ contrained by

$\tag 2 \frac{-\varepsilon}{1 + \varepsilon} \le -\delta \lt x - 1 \lt +\delta \le  \frac{+\varepsilon}{1 - \varepsilon} \text{ and } 0 \lt \varepsilon \le \frac{1}{2}$

so that $\text{(2)}$ implies $\text{(1)}$ (we can use our developed "$\text{iff }$-logic-chain" in reverse).

Setting

$\quad \delta = \frac{\varepsilon}{1 + \varepsilon}$

sets up the left side of $\text{(2)}$, but also takes care of the rights side since

$\quad \delta = \frac{\varepsilon}{1 + \varepsilon} \le \frac{\varepsilon}{1 - \varepsilon}$

Note: Here we don't use the 'min delta' approach. Using the same mechanical technique found here, we simply 'turn-the-crank'. As we work things out we realize that we want to constrain $\varepsilon$.

Instructional Example 1: You are challenged with an $\varepsilon$ that is greater than or equal to $1/2$.
Then set $\delta = 1/3$.

Note that if $\varepsilon = 1/500$, we take $\delta = 1/501$, a 'looser' number than the $\varepsilon/2 = 1/1000$ found in other (excellent) answers in this thread.


Divisibility Question. Show that if $a$ and $b$ are odd positive integers, and $b$ does not divide $a$, then there are integers $s$ and $t$ such that $ a = bs + t$ where $t$ is odd and $|t| <b$.

Let a = 2k +1 and b = 2j + 1 where k,j are positive integers.

since b does not divide a then there is a non-zero remainder on division by b.
we can write;

$a = bs + t$

now to show that $t$ is odd;

$(2k +1) = (2j + 1)s + t$

rearranging;

$2(k-2js) - ( s-1) = t$

$2(k-2js) + (1-s) = t$

$2[(k-2js) -s/2] + 1  = t$

Hence t is odd. but i have not shown that $|t| <b$.

In fact, i am questioning whether i have shown anything.

The most straightforward method is by using the euclidian algorithm:

So by the euclidian algorithm, we find $q$ and $r$ such that
$$a = bq+r$$
and $0<|r|<|b| = b$. We consider two cases:

$r$ is odd. We are done.
$r$ is even, but in this case, $r-b$ is odd, so $$a = b(q+1)+(r-b)$$ and $-b < r-b < 0$ so $|r-b| < b$.

I think that almost every proof will require a (small) case-study.

Note: The following is more or less a rewrite of JefLaga's answer with a small additional hint in order to answer the detail which is asked by OP in the bounty text.

  It's a good idea to start as JefLaga did and use the Euclidean algorithm, because this way we can already assume the existence of non-negative integers $q$ and $r$ such that
  \begin{align*}
a=bq+r\qquad\qquad 0\leq|r|<b
\end{align*}
  Since $b$ does not divide $a$ we observe
  $$0<|r|<b$$

  So, the only thing we have to show is that $r$ is odd.

Let's recall the parity when adding/subtracting even and odd numbers:
\begin{array}{lcll}
\pm&|&even&odd\tag{1}\\
\hline&|\\
even&|&even&odd\\
odd&|&odd&even
\end{array}

  And as Jef did, we distinguish according to the parity of $r$

  $r$ is odd: Nothing to show
  $r$ is even: Now since $b$ is odd, we observe according to (1) that $r-b$ is odd (even minus odd = odd)

  and again we have a valid representation
  $$a=b(q+1)+(r-b)$$
  with an odd number $r-b$ as remainder and since $-b<r-b<0$ we observe:
  \begin{align*}
|r-b|<b\qquad\qquad&\\
&\Box
\end{align*}


Calculating Rotation from centroid I have a polygon as such:

where the green polygon is the rotated polygon and the purple is the extent of the polygon.

Is there a way to calculate the angle of rotation of the green polygon from the extent?

I am using python to do this as well, so if you have it in python code, that would be helpful too, but not needed.

I created the following code:

def distance(ptA, ptB):
    """ calculates theta """
    diffX = math.pow((ptB[0] - ptA[0]), 2)
    diffY = math.pow((ptB[1] - ptA[1]), 2)
    return math.sqrt((diffX + diffY))
def theta(p1, p2, p3):
    """
    """
    radians = math.atan((distance(ptA=p2, ptB=p3)/distance(ptA=p1, ptB=p3)))
    degree = radians * 360 / math.pi
    return degree

p1,p2, and p3 are [X,Y] pairs.

Now I am using the following point:

The result doesn't seem to be the correct rotation value. Any suggestions?

Thank you

When the angles are 90 degrees.

Make sure that the left bottom co-ordinate is $(0,0)$.

Then you have $(x,0)$ and $(0,y)$ to form a triangle.

Then you have

  $$
\tan(\phi) = \frac{y}{x}
$$

thus

  $$
\phi = \arctan\left( \frac{y}{x} \right).
$$

When you have arbitrary points of the triangle, say

$$
(x_0,y_0), \quad (x_1,x_1), \quad (x_2,y_2)
$$

you have

  $$
\tan(\phi) = \frac
{ \sqrt{ \big( x_1 - x_0 \big)^2 + \big( y_1 - y_0 \big)^2 } }
{ \sqrt{ \big( x_2 - x_0 \big)^2 + \big( y_2 - y_0 \big)^2 } }
$$

thus

  $$
\phi = \arctan\left( \frac
{ \sqrt{ \big( x_1 - x_0 \big)^2 + \big( y_1 - y_0 \big)^2 } }
{ \sqrt{ \big( x_2 - x_0 \big)^2 + \big( y_2 - y_0 \big)^2 } } \right)
$$

For the adjusted image.

Let

  $$
\theta = \angle p_1 p_2 p_3
$$

Then

  $$
\theta = \arctan\left( \frac{|p_1-p_3|}{|p_2-p_3|} \right)
$$

Rotated from what?  Are we to assume that the lower left edge of the rectangle was initially at or parallel to the lower edge of the "extent"?  If so then:

Set up a coordinate system with $(0, 0)$ at the lower left corner of the "extent", the $x$-axis along the lower edge, and the y-axis along the left edge.  Let the $x$-coordinate where the rectangle touches the lower edge be $x_0$ and the $y$-coordinate where the rectangle touches the left edge be $y_0$.  Then angle is given by $\operatorname{arccot}\left(\frac{y_0}{x_0}\right)$.


How To Know in a Application Sequence/Series Problem Which Variable is $a_{0}$ or $a_{1}$? How do you determine what value takes on $a_{0}$ or $a_{1}$ in order to use either the infinite sum formula or $a_{n}$ formula?

For example, consider the following two problems:

A certain culture initially contains 10,000 bacteria and increases by 20% every hour.

(a) Find a formula for the number N(t) of bacteria present after t hours.

In this case, $10,000$ would be taken on by $a_{0}$, so the formula would be $N(t)=10000*(1.2)^{10}$. However, when I initially solved the problem, I thought that $10,000$ was $a_{1}$ and got $N(t)=10000*(1.2)^{10-1}$.

How can I distinguish between these in a word problem? Is $10,000=a_{0}$ because nothing has been "done" to it yet? But $a_{1}$ is the "starting value" in a sequence, so why isn't $10000=a_{1}$?

A certain drug has a half life of 2 hours in the bloodstream. The drug is formulated to be administered in doses of D milligrams every 4 hours, but D is yet to be determined.

Find a formula to get the milligrams of drug in the bloodstream after     the $n$th dose. Show that this sum is $\frac{4}{3}D$.

Going by the logic used in the last problem, $D$ would be the starting value, or $a_{0}$. However, that's not the case because the answer is $a_{n}=D*(1/4)^{n-1},$ so clearly $D$ is $a_{1}$, giving the sequence $D+\frac{1}{4}D+...+\frac{1}{4}D^{n-1}$, where $D=a_{1}$. Why is this the case? Also, the sum is $\frac{a_{1}}{1-r}=\frac{4}{3}D$, again with $D=a_{1}$.

Why is $D$ in this case $a_{1}$ rather than $a_{0}$? I know this may seem rather obvious to some of you but I'm really confused and I need to know this for my test.

Thank you!

The difference between the first and second problems you have listed is the starting point.  Think about each problem logically ... the first problem is asking how much of something there is if it grows at a rate of 20% each hour.  If zero hours have passed, zero growth has happened, so you will start at zero.

The second problem ... again, thinking logically you are starting with 4 and working back down towards zero.  You would get closer and closer to zero without ever reaching zero.  (Poorly written word problem because it is essentially saying that the drug NEVER leaves your system ... which is creepy to imply).

Also, you mention that "a of 1" is the starting value in a sequence and your statement is correct but incomplete.  Yes, the starting value is 1 unless otherwise specified.  Your problems "otherwise specify" the starting points through the context of the word problems.


degree of smooth maps from 2-sphere to 2-torus Why any smooth map from the 2-sphere to the 2-torus has zero degree?
Can we show that there is no surjective smooth map from 2-sphere to 2-torus?

There can't be a surjective map from the two sphere to the torus. There exists a surjective map from $f$ from $T^2$ to $S^2$ obtained by collapsing 1-cells in the canonical CW structure of $T^2$ see:  Is it possible to obtain a sphere from a quotient of a torus?

If you have a surjective map $g:S^2\rightarrow T^2$, then $f\circ g:S^2\rightarrow S^2$ is surjective. This is impossible since $\pi_2(T^2)=1$, $f$ is homotopic to a constant and so is $f\circ g$, so its degree is zero. the answer here

https://mathoverflow.net/questions/20712/do-surjective-degree-zero-maps-exist

implies that $f\circ g$ must miss a point

Here's a proof in the spirit of Schoen & Yau's 2017 proof of the positive-mass theorem in general dimensions. Schoen & Yau's proof crucially uses (and shows) the following fact:

  If $M$ is a compact oriented smooth $n$-manifold (without boundary) and there is
  a smooth map $F: M \rightarrow T^n$ to the $n$-torus with non-zero degree,
  then $M$ cannot possibly carry a metric with strictly positive scalar
  curvature. (In particular, $M$ cannot be a sphere.)

In contrast to the higher-dimensional case, for $n=2$ the proof is easy and answers OP's claim right away:

Suppose $M$ carried a positive scalar curvature metric. Then every connected component of $M$ has total curvature $> 0$ and is thus a topological 2-sphere by Gauß-Bonnet's theorem and the classification of compact 2-surfaces without boundary.

Now consider the form $dx^1 \wedge dx^2$ on $T^2$ where $x^1, x^2$ are canonical coordinates on $T^2 = S^1 × S^1$, normalized such that $\int_{T^2} dx^1 \wedge dx^2 = 1$. Consider the closed 1-forms $\theta^i = F^*(dx^i)$ on $M$ and compute:

$$
  \int_M \theta^1 \wedge \theta^2 = \int_M F^*(dx^1 \wedge dx^2) = \operatorname{deg}(F) \int_{T^2} dx^1 \wedge dx^2 = \operatorname{deg}(F) \neq 0
$$

But this means that $\theta^1 \wedge \theta^2$ has non-zero integral on some connected component $S^2$ of $M$.

We will show that the closed 1-forms $\theta^i$ restricted to $S^2$ lie in linearly independent (in particular non-zero) cohomology classes $[\theta^1], [\theta^2]$ of $S^2$. But then $H^1(S^2)$ is at least 2-dimensional, a contradiction since $H^1(S^2) = 0$.

So suppose their cohomology classes were linearly dependent, i.e. $\lambda [\theta^1] = [\theta^2]$ for some $\lambda \in \mathbb{R}\setminus\{0\}$ (without loss of generality). Then $\lambda \theta^1 - \theta^2 = df$ for some function $f$ on $S^2$. Then:

$$
  \theta^1 \wedge \theta^2 = \theta^1 \wedge (\lambda^{-1} \theta^1 - df) = 0 - \lambda^{-1} \theta^1 \wedge df = d(\lambda^{-1} f \theta^1)
$$

So the 2-form $\theta^1 \wedge \theta^2$ would be exact on $S^2$, so in particular it would have zero integral on $S^2$ by Stokes' theorem – a contradiction.


False proof of $\pi=0$ In an attempt to prove to myself that the equation $e^{i\pi} = -1$ is indeed true, I attempted to 'reverse engineer' it into a state that I already know to be true using Year 12 methods, (i.e. able to verify it in a basic calculator), but ended up with $\pi = 0$ and was wondering where I went wrong. I don't doubt it comes down to me knowing pretty much nothing about imaginary numbers other then '$i$' is defined as the square root of $-1$.

Anyway here was my attempt:

\begin{align}
  e^{i\pi} &= -1 \\
  \ln(-1) &= i\pi \\
  2 \ln(-1) &= 2i \pi, \text{ multiplied all by $2$} \\
  \ln((-1)^2) &= 2 \sqrt{-1} \pi \\
  \ln(1) &= 2 \sqrt{-1} \pi, \text{ it was at this point I realised I'd messed up} \\
  0 &= 4 \cdot -1 \cdot \pi^2, \text{ pointless squaring} \\
  0 &= -4 \cdot \pi^2 \\
  0 &= \pi
\end{align}

Here's a quick way to summarize the essence of your paradox:
$$
e^{2 \pi i} = e^0 \overset{?}\implies 2 \pi i = 0
$$
The first equation is true, the second is not. The problem is that the log is no longer uniquely defined.  If it were defined in that sense, then taking the "log of both sides" would take us from the first equation to the second.

With real numbers, we can only have $e^x = e^y$ when $x = y$. This is not true if $x$ and $y$ are complex.

Since you are learning about complex numbers for the first time, it is worth noting that for real numbers $x$, we have Eulers formula:
$$
e^{xi} = \cos(x) + i \sin (x)
$$
Notably, both cosine and sine are periodic, so this function will be periodic (over the real values of $x$) with period $2 \pi$.  For more on all this, I recommend this video (mathologer) and this video (3blue1brown).

You should be knowing that e^(npii) where n is any integer is equal to one,so that means ln(1) has multiple values ie,npii,not only 0
So you can only take the case in which it is equal,ie ln 1 equals to 2*pi*i


Solve the recurrence relation : $f(n) = 1 + \frac{f(n) + f(n-1) + \cdots + f(1)}{n+1}$ For naturals $n$, $f(n) = 1 + \dfrac{f(n) + f(n-1) + \cdots + f(1)}{n+1}$. What is $f(n)$? This is not a homework problem. Is there a general method to solve these recurrence relations? I will appreciate if someone directs me to a short tutorial/book to learn about solving recurrence relations.

In a matrix-scheme the given equality looks like
$$ \small \begin{array}{}
 \; & &\begin{array} {|r|} 1\\f(1)\\f(2)\\f(3)\\ \vdots \end{array} \\
 & * & \\
  \begin{array} {r|rrrrr|}  &  1 \\
             \phantom{f(1)} &  1 & 1/2 \\
             \phantom{f(1)} &  1 & 1/3 & 1/3\\
             \phantom{f(1)} &  1 & 1/4 & 1/4 & 1/4\\
                            & \vdots & \vdots & \vdots & \vdots & \ddots\\
  \end{array}
 & = &
  \begin{array} {|r|} 1\\f(1)\\f(2)\\f(3)\\ \vdots \end{array} \\
\end{array}$$
The sequence is obviously the normed eigenvector with the eigenvalue 1 of the bottom-left matrix.

The eigenvector (thus the solution for your function at positive integers) begins then with
$[1,2,5/2!,17/3!,74/4!,...]$ which means $f(1)=2$, $f(2)=2.5$ etc.

I just found a good explanation for the general term, I've got with the (unsigned) Stirlingnumbers first kind as sequence $s_{1:r,2}=\{0,1,3,11,50,274,...\}_{r \ge 0}$
 $$f(k) =_{k \ge 0} 1+ s_{1:k,2}/k! \qquad \qquad f(0)=1$$

Looking at this in terms of generating-functions we find that
$$ g(x) = \sum_{k=0}^\infty f(k) x^k \qquad \underset {|x| \lt 1}{=} \qquad {1\over1-x} +  {-\log(1-x) \over 1-x}  $$

Remark: the matrix-ansatz proved useful for me in many instances, so I think it's often useful to approach similar problems with this method (of course only when a simpler one runs into the void...). A very well known example where a recursion is handled (and indepth studied) by the matrix-method is the Fibonacci-sequence, and that matrix-solution leads then also very naturally to a Binet-type formula. If you are interested in more examples you can find some other answers of mine here in MSE. For the matrix-method with the Fibonacci sequence there is very nice and exhaustive material at Rob Johnsons's website.


When does $x^3 - x^2 - 2 x + 1$ split mod $p$? Conjecture: the primes for which the polynomial $x^3 - x^2 - 2 x + 1$ splits mod $p$ are the primes $\equiv 1$ or $6$ mod $7$ (OEIS sequence A045472).  Is this correct?

If $\alpha$ is a root of $x^3-x^2-2x+1$ (in any field) then the latter splits as $$(x-\alpha)(x-\alpha^2+\alpha+1)(x+\alpha^2-2).$$ Also if $\alpha \neq -2$ and $\beta$ is a root of $$x^2+\alpha x + 1$$ then $\beta$ is a primitive seventh root of unity. Combining all this for the field $\mathbb{F}_p$ where $p\neq 7$: If $x^3-x^2-2x+1$ has a root in $\mathbb{F}_p$ then $\mathbb{F}_{p^2}$ contains a primitive seventh root of unity and so $p^2 \equiv 1 \pmod 7$. The other way around: If $p^2\equiv 1 \pmod 7$ then $\mathbb{F}_{p^2}$ contains a primitive seventh root of unity $\beta$ and $\alpha = -\beta-\beta^{-1}$ is a root of $x^3-x^2-2x+1$. Moreover $\beta^p = \beta$ (if $p\equiv 1 \bmod 7$) or $\beta^p = \beta^{-1}$ (if $p \equiv -1 \bmod 7)$. Either way $\alpha^p = \alpha$ so $\alpha \in \mathbb{F}_p$. For $p=7$ $$x^3-x^2-2x+1\equiv (x+2)^3 \pmod 7$$ as pointed out by @lhf in a comment.

Yes, this is because the zeros of the polynomial is $-(\zeta^j+\zeta^{-j})$
for $\zeta=\exp(2\pi i/7)$ and $j\in\{1,2,3\}$. So it splits iff $\zeta+\zeta^{-1}\in\Bbb F_p$ for $\zeta$ now a primitive seventh root of unity
in an extension of $\Bbb F_p$. This is the case iff $(\zeta+\zeta^{-1})^p
=\zeta+\zeta^{-1}$ in characteristic $p$, and that is the case iff
$p\equiv\pm1\pmod7$.


What is "multiplication by juxtaposition"? I was reading http://www.purplemath.com/modules/orderops2.htm it shows

 = 16 ÷ 2[2] + 1   (**)
 ...
 = 5

  The general consensus among math
  people is that "multiplication by
  juxtaposition" (that is, multiplying
  by just putting things next to each
  other, rather than using the "×" sign)
  indicates that the juxtaposed values
  must be multiplied together before
  processing other operations

However when talking to certain people they all have said there is no such thing as this. There is shorthand which uses normal multiplication order and no "multiplication by juxtaposition" and etc.

Is there a "general consensus among math people" or is this simply incorrect?

So, the question is whether $a/bc$ means $(a/b)c$ or $a/(bc)$. And the answer is, DON'T WRITE $a/bc$, because it will only cause confusion. Some people/software/whatever will make one interpretation, some will make the other, neither one has been endorsed by the Dalai Lama or any other great leader. Put in enough parentheses to make your writing foolproof.

This question is more about how we deal with trolling and nuisances.

Read this page: http://knowyourmeme.com/memes/48293

Then maybe we can start a meta article about identifying and dealing with these threads.

Between the math forums that I moderate and otherwise frequent, and dozens of other forums

(a short list is here: http://www.mymathforum.com/viewtopic.php?f=13&t=20148&p=79150#p79150),

I'd guess that thousands of hours have been wasted on this garbage.


Given basis of an ordered square seems to be only open intervals The basis given here What is the Basis of an Ordered Square? seems to be different from the basis given in Topology by James Munkres, if any:

From Chapter 14

From Chapter 16

It seems like the topology that defines the ordered square will be generated by a basis of open intervals. I believe I will not get the answers to the following exercise (also in this question) if my basis consists of only open intervals because of the issues with $0 \times 0$ and $1 \times 1$.

Which of the following explains what's going on?

Munkres implies dictionary order topology that forms the ordered square is generated by a basis of open intervals, and Munkes is right.
Munkres implies dictionary order topology that forms the ordered square is generated by a basis of open intervals, and Munkes is wrong.
Munkres does not imply that the dictionary order topology that forms the ordered square is generated by a basis of open intervals. Instead (insert explanation here).
Munkres actually does not well-define the dictionary order topology that forms the ordered square.
Other

The ordered square is the set $[0,1] \times [0,1]$ in the order topology from the order that the square inherits from $\mathbb{R} \times \mathbb{R}$ with the lexicographic order (the lexicographically ordered plane).

Munkres already spent some time warning the reader that this is not the same topology that $[0,1] \times [0,1]$ gets as a subspace of the ordered plane.

In general, if $(X,<)$ is an ordered set with order topology $\mathcal{T}_<$ and $A \subseteq X$ we can consider two topologies on $A$: the subspace topology $\mathcal{T}_<|_A = \{O \cap A: O \in \mathcal{T}_<\}$ that $A$ inherits from $(X,\mathcal{T}_X)$ and the order topology that $A$ gets when we restrict the order first to $A$ and then generate the topology from that restricted order.

And these topologies concide when $A$ is order-convex: $\forall a_1 \in A, a_2 \in A: \forall b \in X: (a_1 < b < a_2 ) \to b \in A$.

Even in the reals you can see this phenomenon: Take $A = [0,1) \cup \{2\}$. As a subspace of the reals (ints order topology), $2$ is an isolated point, i.e. $\{2\}$ is open, while in the restricted order topology $2$ lies "just above" $[0,1)$ and in the restricted order topology the set is homeomorphic to $[0,1]$. Note that $A$ is not order convex.

Also $A = [0,1]\times [0,1]$ is not order convex in the ordered plane: e.g. $(0,1), (1, 0)$ are both in $A$, while $(0,2)$ which lies between them lexicographically is not.

Munkres does take the square in its (restricted) order topology, not the subspace topology, so the subbase for its topology (as in any order topology) is the set of all open segments both up and down, and the derived base has the two special cases involving $(0,0) = \min(A)$ and $(1,1) = \max(A)$. So not only open intervals, also sets like $[(0,0), (0,t))$ and $((1,t), (1,1)]$ are basic open.

Number Three (3).

We know

that standard topology on $\mathbb R$ is the order topology on $\mathbb R$ and
that the standard topology on $\mathbb R^2$ is not the order topology on $\mathbb R^2$.

But actually the dictionary topology on $\mathbb R^2$ is the order topology on $\mathbb R^2$:

Observe that Munkres says the dictionary topology on $\mathbb R^2$ aka order topology on $\mathbb R^2$ can be generated by a basis of open intervals because $\mathbb R^2$ has no $\max$ or $\min$. Thus, there is no reason to conclude that a topology that is on a subset of $\mathbb R^2$ with a $\max$ or $\min$ and that is inherited from the dictionary topology of $\mathbb R^2$, again, aka order topology on $\mathbb R^2$ (and that is not inherited as a subspace) can be generated by a basis of open intervals. Instead, the subset's dictionary topology's basis goes back to the definition of order topology:

The basis does not consist of only open intervals.


integer solution of $(x-y)(x+y)xy=z^2$ By wolfram alpha, integer solution of  $(x-y)(x+y)xy=z^2$ is $x=y=z=0$.
How to show that there are not another solutions with $z \neq 0$.
Thanks.

If $\gcd(a,b) = d$, and $(x,y,z) = (a,b,c)$ is a nontrivial solution, then $(\frac{a}{d},  \frac{b}{d},\frac{c}{d^2})$ is also a solution, where $\gcd(\frac{a}{d},\frac{b}{d}) = 1$.  So if there is no solution $(a,b,c)$ having $\gcd(a,b) = 1$, then there are no nontrivial solutions.

Assume $\gcd(a,b) = 1$.  We want to find $a$ and $b$ where the expression $(a+b)(a-b)ab$ would be a perfect square.  But $(a+b)$, $(a-b)$, and their product $(a^2-b^2)$ are all coprime to both $a$ and $b$.  So it is necessary that $a$ and $b$ are both squares, because any unsquared factors they may have will not be able to have their squares completed by the other terms.

$a = a'^2$ and $b = b'^2$.  $(a+b)(a-b) = a^2 - b^2 = a'^4 - b'^4$.  If there is a solution, then there must be some $c$ where $c^2 = a'^4 - b'^4$, or $c^2 + b'^4 = a'^4$.  Fermat proved this was impossible, so there cannot be any nontrivial solutions.

If $(x,y)=k$ then $k^4(x_1-y_1)(x_1+y_1)x_1y_1= z^2$ which implies an equation (we use the same notation) $(x-y)(x+y)xy=z^2$ with $(x,y)=1$; it follows that $(x+y,x-y)=1$  ($x+y=mX$ and $x-y=mY$ give clearly a contradiction). Hence $x-y,  x+y, x, y$ are four coprime integers and each of them must be a square, $x, y$ being of different parity (if not then $2|(x-y,x+y)=1$).

We have three pythagorean triples to which we apply the well known parametrization
$(a,b,c) =(p^2-q^2, 2pq, p^2+q^2)$:

(1)………………$x^2=y^2+u^2$

(2)……………..$x^2+y^2=v^2$

(3)…………….$x^4+y^4=(uv)^2$

If $y$ were odd then $x$ would be odd in (1) and even in (2), absurde, so $y$ must be even.

Now from (3)
$x^2=t^2+s^2, y^2=2ts$ implies $ {x^4-y^4=(t^2-s^2)^2}$ , $t$ and $s$ of distinct parity.
Since $t^2+s^2=x^2$, we write $t=m^2-n^2, s=2mn$ and $x=m^2+n^2$  with $(m,n)=1$.

It follows $y^2=2(m^2-n^2)(2mn)=4(m^2-n^2)mn$
Hence $m=p^2, n=q^2$ and $y^2=4(p^2-q^2)p^2q^2$ and since y is even,$(p^2-q^2)(p^2+q^2)p^2q^2= w^2$.

Thus, from $(x^2-y^2)(x^2+y^2)x^2y^2=z^2$ we have got another equation
$(p^2-q^2)(p^2+q^2)p^2q^2= w^2$ similar to the first one with the values of $p,q$ less than the values of $x, y$. This process can be iterated indefinitely therefore, by infinite descent,  we finish the proof.


Right Triangle Trig: Why are the Angles the Same? On Khan Academy, the problem involves this picture of a triangle:

In the problem, we must solve for the theta angle, so in the explanations, the triangle was broken up into a square so that there is the same angle theta on another triangle.

Here, how do you explain the same angles in this triangle and the one in triangle KGJ? Perhaps an idea being a certain rule of right triangle trig to be followed.

It must be noted that they are corresponding angles. Under the "Corresponding Angles Axiom", we state that
"When a transversal (here, the line $t$) cuts two parallel lines, then the corresponding angles ($\alpha$ and $\alpha_1$) are equal."
This is an axiom, and can be visually shown by superposition.

In your case, the line through $K$ at $HF$ is parallel to the line $GF$, and line $GH$ is the transversal. Hence, the angles under consideration are equal.

Call the vertex with $90^{\circ}$ between side lengths $(5,7)$ as $L$. The angle $ HGL$ should not be labelled as $\theta$ already.
The angles $KLF,LFG$ add up to $180^{0},\;. $ So lines $FG,LK$ are parallel.
$LK$ should make the same angle to the transversal cutting line $HKG$ by virtue of parallelism property of corresponding angles.


Sine of angle between direct common tangents of two circles? This is a question I found: Given two circles intersecting orthogonally having the length of common chord 24/5 units the radius of one of the circles is 3 units then what is the sine of the angle between the direct common tangents? The answer given is 4√6/25. How do I approach this problem? I know that direct common tangents meet at a point that externally divides line joining centres of circles in ratio of their radii, but I am getting negative values when I try to solve this. How do I do this?

Using pythagoras theorem and similarity of triangles, you can easily find the other radius which is $4$ and distance between centres is $5$.

Now let centre of circle with radius $r_1=3$ be $C_1$ and with radius $r_2=4$ be $C_2$. Let intersection of direct tangent and line through $C_1C_2$ be $P$.

Using similarity,

$$\frac{PC_1}{3} = \frac{PC_1+C_1C_2}{4}= \frac{PC_1+5}{4}$$
This gives $PC_1 = 15$. So now let half of the required angle be $\alpha$, this is the angle between direct tangent and line through $C_1C_2$.

$$\sin \alpha = \frac{r_1}{PC_1} = \frac{1}{5},\,\, \cos\alpha = \frac{2\sqrt{6}}{5}$$

So required answer is $\sin 2\alpha = \frac{4\sqrt{6}}{25}$

To make things simpler. There's this formula you can memorise or derive as you may like by similarity and Pythagoras.

Alpha{angle between DCT}= 2sin^-1[|r2-r1|/ d]
Where R1 and R2 are radii and d is distance between centres.

Here, sin a/2 comes out to be 1/5.
Thus sin a = 2 sin a/2 cos a/2

= 4✓6/25


Triangles in the Gauss Plane What is the image of $z$ ($z=a+ib$) in $C$ ($z$ is complex number) under reflection in the line which passes through $O$ at angle $\theta$ to the real axis?
Since one rotation can be factored into a product of two reflections, which means the question contains 3 reflections? Anyone has hint? What should I do?

The slope of the specified line is $\tan\theta$ or $\frac{\sin\theta}{\cos\theta}$. The slope of any normal to the specified line is the negative reciprocal of that (or $- \frac{\cos\theta}{\sin\theta}$). Plug $a$, $b$ and that value into the point-slope equation of a line. Solve that equation and the equation of the original line $\left(\sin\theta\right)x=\left(\cos\theta\right)y$ simultaneously to find the intersection of the two (say, $(c|d)$. The coördinates of the desired reflection point are $(2c-a|2d-b)$ or: $- a + 2 a \cos^2 θ + 2 b \cos θ \sin θ + (- b + 2 a \cos θ \sinθ + 2 b \sin^2 θ) i$.


Showing that $(f_1,f_2,\dots,f_m)$ is a measurable function from $(\mathbb R^m,\mathcal B(\mathbb R^m))$ into itself If $\{f_i, 1 \le i \le m\}$ is a set of real valued Borel functions on $\mathbb R$, how to show a vector of functions, $(f_1, f_2,..., f_m)$ is a measurable mapping from $(\mathbb R^m, \mathcal{B}(\mathbb R^m))$ to $(\mathbb R^m, \mathcal{B}(\mathbb R^m))$ itself?

It seems to be an obvious result. However, I don't know how to prove it.

I think it is equivalent to show i-th entry of projection of a open set in $\mathbb R^n$ is still open in $\mathbb R$.

Since $\{f_i, 1 \le i \le m\}$ is a set of Borel measurable functions
$$
\{x:f_i(x)<c_i\}\subset \mathcal{B}(\mathbb R)
$$
for all possible $c_i\in\Bbb{R}$.
So
$$
\{x:f_1(x)<c_1\}\times \{x:f_2(x)<c_2\}\times\cdots\times\{x:f_m(x)<c_m\}\subset \mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)
$$
for all possible $c_1\cdots c_m\in\Bbb{R}$. This means $(f_1, f_2,..., f_m)$ is Borel measurable.

Edit: $\quad$ We prove $\mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)$. The idea is inspired by PhoemueX's comment.

Let $\overrightarrow{A_i}=\mathbb {R^i}×A_i×\mathbb {R^{m−i−1}}$, where $A_i\:(1\leqslant i \leqslant m)$ are Borel set. Then clearly $\overrightarrow{A_i}\subset \mathcal{B}(\mathbb R^m)$.

Now for any set $B\in\mathcal{B}(\mathbb R)^m$
$$
B=A_1\times\cdots \times A_m=\bigcap_{i=1}^m \overrightarrow{A_i}\in\mathcal{B}(\mathbb R^m)
$$
for $\overrightarrow{A_i}\subset \mathcal{B}(\mathbb R^m)$ and Borel $\sigma$-algebra is closed under countable intersection. So we have
$$
\mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)
$$

You're right. you have that the image of an open set under projection is open, then the preimage of that is measurable in $\mathbb{R}^1$. Then, the product of measurable sets is measurable, and you are done.


Bijection between real and natural numbers. I know, I know, this question has been asked several times, but I feel mine is a little bit different.
Imagine a correspondence between $[0,1]$ and natural numbers in the following sense:
$$ 0.12 \longleftrightarrow 21 $$
$$ 0.443 \longleftrightarrow 344 $$
$$0.12345 \longleftrightarrow 54321 $$

Now, before you argue that in fact I'm leaving rationals as $\frac{1}{3}$, let me explain a bit more.

What this correspondence does is: given a real numer $ r \in [0,1] $, which can be expressed as (or can not?) $ r= x_1 \cdot 10^{-1} + x_2 \cdot 10^{-2} + x_3 \cdot 10^{-3} +$ ... (For example: $\frac{1}{3} = 0.33333...=0.3+0.03+0.003+...$). I now define $n_r$ as $n_r=x_1 \cdot 10^0 + x_2 \cdot 10^1 + x_3 \cdot 10^2 + ...$ in a recursive way. So I will have $ n_{\frac{1}{3}} = 3 + 30+300+... $ as a limit of that recursion.

What's the problem with my reasoning? If it is about that last series, isn't the nature of $\mathbb{N}$ precisely to be defined in an inductive way? So I suppose $ n_r$ defined like that is still a natural number.

PD: just an undergrad student who doesn't like infinity

Welcome to Mathematics Stack Exchange.

A problem with your proposed correspondence $r\leftrightarrow  n_r$

between elements of $[0,1]$ and elements of $\mathbb N$

is that $n_{1/3}$ would be $3+30+300+\cdots\not\in\mathbb N$.

Even if you define it like that, there is no guarantee that two of these series would sum to the same number. The entire point of creating a bijection is to show that the mappings are unique. If you can demonstrate uniqueness of the series sum for each rational, then and only then would it be considered a bijection. I'm assuming that's the difficult bit


can you disprove this rule PEDMSA?-(division before multiplication, subtraction before addition) I know the proper teaching of PEMDAS/PEDMAS/BODMAS/BOMDAS..  is that multiplication and division have equal priority and are evaluated going through the arithmetic expression left to right (So, doing division first if division is left of multiplication in the arithmetic expression, and multiplication first if it is to the left of a division). Ditto for addition and subtraction.

And I know PEMDAS if taken as multiplication always before division, gives wrong answers. And addition always before subtraction, also gives wrong answers. e.g. 3-2+1 addition before subtraction would be wrong result.  And 6/2*3 multiplication before division would be wrong result.

So one can break(get a wrong result from), a literal reading of PEMDAS, i.e. doing multiplication always before division, and ditto with additionan and subtraction.

But how about a literal reading of   PEDMSA?   so, taking it literally..  division always before multiplication.  Subtraction always before addition.

Can you break it(get a wrong result from it)?

This question brings to light an excellent example of how we can introduce ambiguity into a language by removing one interpreter rule from a language/interpreter system.

A language and the corresponding interpreter go hand in hand. When we construct a language, and the corresponding interpretation rules, the onus is on the writer to phrase statements in a way that will always give the desired result by the corresponding interpreter. While there may be many speakers of one language, there is only one interpreter for languages of this sort.

Polish notation would be an example of an alternative language for arithmetic operations.

As you know, the "PEMDAS" rule is nothing more than a mnemonic device to assist in interpreting what is currently our most common language for operation order. The left to right scanner priority rule is however an integral part of this device, and PEMDAS would be best written as $P-E-(M \vee D)-(A\vee S)$ "with repeated left to right scan priority after each atomic operation execution".

We can make a new language and interpreter at will. We can also modify a language, but once you decide to modify an extant language/interpreter, like say "remove the left to right rule", then all bets are off. You now have a new language, it is not the old language, and things may need to either be written differently or interpreted differently for this new language, otherwise there will be problems, ambiguity being one such problem.

If you modify the language, you probably need to modify the interpreter. If you modify the interpreter, you probably need to modify the language.

Most of this is above my head, but I take issue with the OP that 3-2+1 cannot be done with addition before subtraction.

Subtraction is just adding a negative number, so that calculation is actually (+)3+(-2)+1

No matter what order you do that on your going to get the same result.

3+(-2)+1 = 2

3+1+(-2) = 2

(-2)+3+1 = 2

(-2)+1+3 = 2

1+(-2)+3 = 2

1+3+(-2) = 2

Multiplication is just repetitive addition,

8×3 = 3+3+3+3+3+3+3+3

Division is inverse multiplication so

8÷3 = 8×⅓ = ⅓+⅓+⅓+⅓+⅓+⅓+⅓+⅓

Now division before multiplication is more difficult.

6÷2×3

This is either

(6÷2)×3 =

(6×½)×3 =

(½+½+½+½+½+½) + (½+½+½+½+½+½) + (½+½+½+½+½+½) =

3+3+3 = 9

Or it is

6÷(2×3) =

6÷6 = 6×⅙ =

⅙+⅙+⅙+⅙+⅙+⅙ = 1

In that example, the division it written first, so reading left to right it would be done first, and the answer should be 9

But what if the equation was written the other way around?

3×6÷2 ???

LTR

(3×6)÷2 =

18÷2 =

18 × ½ =

½+½+½+½+½+½+½+½+½+½+½+½+½+½+½+½+½+½ = 9

DBM

3×(6÷2) =

3×(6×½) =

(½+½+½+½+½+½) + (½+½+½+½+½+½) + (½+½+½+½+½+½) =

3+3+3 = 9

So interestingly, in the latter calculation the order made no difference, but in the former, it made a huge difference.

I can't actually say, based on above that DBM will always yield the incorrect result!


What is the difference between the three types of logarithms? In complex analysis I came across three types of logarithms
 namely $\ln$, $\log$ and $\text{Log}$. What is the difference between the three?

ln is always a natural logarithm (of base $e$).
log if it has a base like $\log_5x$ then the base is the specified one, otherwise it is either base $e$ or base $10$. It depends on people. Some people (mostly high school people) use logarithms without a base as base $10$ and other people as base $e$. It is most often used as a natural logarithm though but you have to watch out. If you see a ln button on a calculator, then log without base is guaranteed to be base 10 on the same calculator.

(Edit: As @ClementC. pointed out, in computer science, $log$ without a base most often means a binary logarithm.)

Log stands for complex logarithm in its principal branch. The principal branch has its imaginary part in the interval $(−π,π]$. (Basically it's the inverse of the complex exponential function with its imaginary part in that interval.)

I thing if $z=re^{i\theta}$ then
$\log z=\ln r+i\theta$. Now if in the arguement $\theta ,n=1$ then "$\log$" becomes "Log". Also "$\ln$" is generally used with real part of complex number as per the notations used by C.V.Chirchill


Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian Trying to get my head around the commutator subgroup. This is an excercise from Artin's Algebra:

Let $C$ be the commutator subgroup of $G$. Prove that $G/C$ is abelian.

Here is what I've done:

Let $xC,yC \in G/C$ then $xyx^{-1}y^{-1}C = C$ since $xyx^{-1}y^{-1}$ is a commutator hence belongs to $C$. But then $xyC = yxC$ so $xC$ and $yC$ commute in $G/C$. This can be done for any elements, so $G/C$ is abelian.

This seems somewhat surprisingly short. Is that all there is to it?

Regards

For some reason the OP won't post, or can't post, an answer, so summarizing the comments:

$(1)\,\,\forall\,g,x,y\in G\,$ , and putting $\,a^b:=b^{-1}ab\,\,,\,a,b\in G\,$:
$$[x,y]^g:=g^{-1}[x,y]g:=g^{-1}x^{-1}y^{-1}xy g=\left(x^{-1}\right)^g\left(y^{-1}\right)^gx^gy^g=[x^g,y^g]\in G'\Longrightarrow G'\triangleleft G$$and thus the quotient $\,G/G'\,$ is a group.

$(2)\,\,$ Let now $\,N\,$ be any normal subgroup of $\,G\,$  s.t. $\,G/N\,$ is abelian, then:
$$\forall\,x,y\in G,\,\,xNyN=yNxN\Longleftrightarrow xyN=yxN \Longleftrightarrow (yx)^{-1}xy\in N \Longleftrightarrow [x,y] \in N$$
and since $\,G':=\langle\,[x,y]\;:\;x,y\in G\,\rangle\,$ , then $\,G'\leq N\,\Longrightarrow \,G'$ is the minimal (normal) subgroup of

$\,G\,$ s.t. its quotient is abelian -- "minimal" wrt set inclusion --.

Exercise:  Explain the parentheses around "normal" above, i.e. show that any subgroup of G containing the commutator subgroup is normal.

It is easy to show that the commutator subgroup is a characteristic subgroup , hence it is a normal subgroup. Proving G/C abelian is straightforward recalling that C is the subgroup of G generated by all commutators.


Bounded sets do have a least upper bound I think this is very fundamental property of real numbers, as it allows us to discuss about limits. I do not know, how the limits are playing an important role to complete the following one. Please explain.

" Any non-empty set of real numbers which is bounded above has a supremum"

This is available in standard books. But, how to complete the proof of the cited above statment by introducing LIMITS, I don't know. Advanced thanks to MSE members, who can help me.

The fact that a nonempty set of real numbers bounded above has a supremum is often taken as an axiom in analysis courses. One can construct the set of real numbers from the set of rational numbers via Dedekind cuts, and then prove this as a theorem instead. This is done for example in Rudin's Principles of Mathematical Analysis, and the proof does not have anything to do with limits at all!

Let $U\subset\mathbb{R}$ be bounded above. Then there exists $x'\in\mathbb{R}$ such that $u\leq x$, for all $u\in U$. Therefore, $x'$ is the upper bound of $U$. Let, $X=\{x\in \mathbb{R}:u\leq x\,for\,u\in U\}$ be the set of all upper bounds of $U$. Then obviously $x'\in X$ and $x^*=\min X$ is the least upper bound of $U$. As $u\leq x^*$ for all $u\in U$ and $x^*\leq x$, for all upper bounds $x$ of $U$.


$\mathrm{Hom}_R(N,M)$ is an essential extension of $\mathrm{Hom}_R(N,L)$ $M$ is an $R$-module and an essential extension of $L$. $N$ is a finitely generated submodule of $M$. Then $\mathrm{Hom}_R(N,M)$ is an essential extension of $\mathrm{Hom}_R(N,L)$.

I tried to use the definition of essential extension: intersection of every submodule of $\mathrm{Hom}_R(N,M)$ with $\mathrm{Hom}_R (N,L)$ is not trivial.
But I can not. Can you help me?

$L\subset M$ is essential, and $N\subset M$ is finitely generated. Then $\operatorname{Hom}(N,L)\to\operatorname{Hom}(N,M)$ is essential.

Let $f\in\operatorname{Hom}(N,M)$, $f\ne0$.

Since $N$ is finitely generated and $L$ is essential in $M$ there is $a\in R$ such that $af(N)\subseteq L$ and $af(N)\ne0$. (This can be shown as follows: set $N=Rx_1+\cdots+Rx_n$, and suppose that $f(x_i)\ne 0$ for all $i$. Then there is $a_1\in R$ such that $a_1f(x_1)=y_1\in L\setminus\{0\}$. If $a_1f(x_i)=0$ for all $i\ge2$ then set $a=a_1$. Suppose $a_1f(x_2)\ne0$. Then there is $a_2\in R$ such that $a_2a_1f(x_2)=y_2\in L\setminus\{0\}$. If $a_1a_2f(x_i)=0$ for all $i\ge3$ then set $a=a_1a_2$, and so on.)

Thus $af\ne0$, $af\in\operatorname{Hom}(N,L)$, and we are done.

this is false SINCE for example hom(M, N) has Z module structure and not R module structure.


Calculate the optimal angle for the roof of this dog house. I'm new to optimization and I'm stuck with the following problem:

Given four square planks 10 cm long each one, without cutting or changing any shape, we use two of them to make the walls and the other two to build the roof (by forming an isosceles triangle). What should be the optimal size of the angle $\alpha$ between the walls and the roof for the house to have the biggest possible volume?

I've spent to much time with this, even though I'm pretty sure the answer is quite simple.

After many trials, the only thing I imagined was that the total volume of the house could be expressed as $$V=2a^2b+abh=200b+10b(\sqrt{100-b^2})$$
whose derivative find a positive local maximum at $b = 5\sqrt{2}\sqrt[4]{3}$, leading to an optimal angle around $167º$, which is definitely absurd.

Thank you in advance.

EDIT

I'll explain my $a$, $b$ and $h$. If we divide in two de base of the prism that forms the roof, we get two right triangles. $a$ is the the side of each plank, $b$ is the base of both right triangles and $h$ is the hight of those.

My algebra gives b=√(√7500) or 9.30604 cm, (same as your answer) leading to an (internal) angle of 111.47°--which agrees with @Bence's result.


4 Spheres all touching each other?? If there are 4 spheres all touching each other and 3 of them have diameters 4, 6 and 12 what is the diameter of the fourth one?
 I imagine it like 3 balls on a flat table touching each other and then we are supposed to put another one on top of them but in my imagination the top sphere could be any size basically right?

Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres.

I’m guessing this problem is asking for the minimum diameter of the fourth sphere which guarantees contact with all of the other three.

Imagine all three balls touch each other. The fourth one must exactly fit in the middle of these three balls. Only then each ball touches every other ball. (of course the assumption here is, that it's all in 2D).

If we say, that speres are allowed inside other spheres or these are more than 2D spheres, then there are many other possibilities, too...


Integral by parts of this function I have this integral to solve:

$\int t \sqrt{1+t}\,dt$

The book says to solve it by part but i have no clue about how to do so. Can someone help me?

Apply the integration by parts formula, $$\int u\,\frac{dv}{dt}dt = uv - \int v\,\frac{du}{dt}dt$$
Start by letting $u = t$. Then $du = dt$, and $$dv = \sqrt{1+t} \;dt = (1+t)^{1/2} \;dt\implies v = \dfrac 23 (1+ t)^{3/2}$$

In this case, we have $$\begin{align}\int t\,\sqrt{1+t}\;dt & = \frac 23 t\,(1+t)^{3/2} - \frac 23 \int (1+t)^{3/2}\, dt\\ \\ & = \frac 23 t\,(1+t)^{3/2} - \frac 4{15}(1+t)^{5/2} + C\end{align}$$


The lifetime of special light-bulbs is normally distributed. The lifetime of special light-bulbs is normally distributed. A sample of 81 bulbs has produced a mean of 738 hours and a standard deviation of 38.2 hours.

Test a hypothesis $_0 ∶  = 747.5$ versus $-1 ∶  ≠ 747.5,$ at the significance levels of $_1 = 0.05$ and $_2 = 0.01.$ Find the p-value of the test, and use the p-value to verify your answers.

This is how I approached it but not sure if this was right:

$z=\frac{738-747.5}{38.2/√81} = -2.23.$

p-value = $P(z=-2.23)$ --> (1-0.9871) = 0.01

Reject at 0.05 because it is higher than p-value.
Accept at 0.01 because == p-value.

I am extremely unsure about this so any help/verification would be great. Thank you for reading!

You calculated z correctly, however, I noticed when I calculated, I got z=-2.238, so I'd go round to z=-2.24 (rather than z=-2.23), though it doesn't change any results.

You also calculated the p-value correctly. I don't know how many decimals you're required to give, but it may be useful to give one more (as then you can see $p>\alpha_2$).

When it comes to testing hypotheses, if $p<\alpha$ then you reject the null hypothesis, $H_o$, and accept the alternative hypothesis, $H_1$.

So, your answers are all correct. Good job.


For an uncountable subset $E$ in $[0,1]\times[0,1]$, show there is a condensation point of $E$. I'm not sure how to approach this question:
Given $E\subseteq[0,1]\times[0,1]$ and $E$ is uncountable, show that there is some $p\in [0,1]\times[0,1]$ such that $\forall \epsilon>0, E\cap B(p,\epsilon)$ is uncountable.
I tried using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. I defined a set $A$ to be the collection of all open balls centered at rational points with some $\epsilon$ such that the unit square is completely covered. Surely at least one of the balls (centered at say $x$) contains uncountably many points in $E$. But, I don't know how to make this work (or even if it could work). I'm concerned about the overlap of the balls and how to find some $x$ that doesn't move around as $\epsilon$ changes. This is my introduction to analysis, so I don't have a good intuition for much of these concepts yet. Any help would be greatly appreciated!

Hint:I would divide the square into four smaller $1/2\times 1/2$ (closed) squares: one of them must contain uncountably many elements of $E$. Then, I would divide that square further into four (closed) $1/4\times 1/4$ squares, etc.
We get a chain of ever smaller squares, which (due to compactness of the original square) must have nonempty intersection. Let $p$ be a point in all those squares: it should now be an easy check that $p$ satisfied the requirement from your question.

By contradiction, assume that for any $p \in [0,1] \times [0,1]$ there exists $\epsilon_p >  0$ such that $E \cap B(p, \epsilon_p)$ is countable.
Consider the union $B = \bigcup B(p,\epsilon_p)$ over all rational points $p$ in $[0,1] \times [0,1]$ (that is, points where both components are rational). Then $E \subseteq B$ since $\mathbb{Q}$ is dense in $\mathbb{R}$. But $B$ is the countable union of countable sets, so $B$ is countable and thus $E$ is countable, a contradiction.


A formula that yields a particular graph shape I would like a formula for a function whose graph has the following properties:

$f(0) = 0$.

$\lim\limits_{x\to\infty}f(x) = y$.

The shape of the function is approximately the following:

It should have an exponential or a logarithm in the formula.

Any function like this?

The graph of $y=5-5e^{-x}$ has the desired characteristics. More generally, if you want $\lim\limits_{x\to\infty}f(x)=a>0$, the function $f(x)=a-ae^{-x}$ works.

Also your graph doesn't have anything shown for x<0, so presumably its equal to zero there?

i.e. Use a piecewise defined function:

f(x) = a-a*exp(-x) : for positive x and f(x)=0 for non-positive x.