A point can be viewed as a circle? In analytic geometry, can a point be viewed as a circle? In analytic geometry, can the point $(0, 0)$ be view as the circle of zero radius with center $(0, 0)$?

The equation $x^2+y^2=0^2$ has a single point as its solution. Whether that that means that a point "can be viewed as a circle" depends more on the exact meaning you assign to "can be viewed as" than on the nature of points and circles in themselves.

It certainly doesn't mean that all properties of circles are also properties of points. For example, a circle has the property that for every point on the circle there is exactly one line that passes through that point and not through other points on the circle. This is not true about a singleton set, of course.

Every circle of radius $r$ and circumference $C$ satisfies $C/r = 2\pi$, but your "circle"  has $r=0$.


Finding an angle in a figure with a regular octagon and regular pentagon I've recreated the diagram using Adobe Illustrator.

The pink polygon is a regular octagon and the grey is a regular pentagon.

I've tried to solve this for a while now but get nowhere. It seems like all I do is create more random angles that don't contribute to the solution. I've tried extending lines but that doesn't seem to help me at all.
Most other problems I've been given are relatively easy.

Assuming the common sides to be unit and setting the origin at the bottom left vertex, the angle $x$ is determined by its apex at $(0,1+\sqrt2)$ and the points $(1/\sqrt2,1+1/\sqrt2)$ and $(\cos2\pi/5,\sin2\pi/5)$.

The tangent of $x$ is given by the ratio of the cross-product of the two legs over their dot product.

$$x=\arctan\frac{\frac1{\sqrt2}\left(\frac{\sqrt{10+2\sqrt5}}4-1-\sqrt2\right)+\frac1{\sqrt2}\frac{\sqrt5-1}4}{\frac1{\sqrt2}\frac{\sqrt5-1}4-\frac1{\sqrt2}\left(\frac{\sqrt{10+2\sqrt5}}4-1-\sqrt2\right)}.$$

Hint:

The sum of the internal angles of a convex polygon with $n$ sides (the angles between two consecutive sides) is equal to $\; (n-2)\pi$ (in radians).


Find all $C^1$ functions from the reals to the reals satisfying $f'(a-x)=f(x)$ for some $a>0$. My try : it is easy to see that $f$ is twice differentiable so by differentiations we get :
$f''(x)=-f'(a-x)=f(x)$ and so $f''(x)+f(x)=0$ and the solution must be of the form $a\cos(x)+b
\sin(x)$ for some constants $a,b$.
Now plugging this in the original equation i get some ugly equations (well, actually not that ugly, but it seems kinda strange) on $a$ that i have to solve and i am unnsure of whether this is the right thing to do or not.
Is it the right approach ? If not not, could you show me what i've done wrong ?
By the way, if anyone could post good resources to learn basic strategies for solving differential equations of order 1 and 2 at beginner's level i would be extremely grateful to them.


If I have two random variables $X,Y$ , $X>0$ when $Y0$ then $\operatorname{cov}(X,Y)<0$? So as I said in the title- I'm talking about the covariance of $X$ and $Y$.

Also, is it correct to say that if I know that $Y>0$ iff $X>0$, and also $Y<0$ iff $X<0$ then $\operatorname{cov}(X,Y)>0$?

It makes sense to me but I just can't see it by the definition of covariance.

This is certainly true when the expected values of $X$ and $Y$ are both $0$, as is easily verified.

Now consider $\{(-1,-1)\} \cup \{ (j,1000-j) : j=1,\ldots,1000\}.$ Software is telling me the correlation is $-0.9939881.$ Look at the graph.


Calculate $\sin^{-1} (\sin (45^\circ))$ without a calculator We know that in this triangle, $h^2=a^2+a^2$. Therefore,

$$
h=\sqrt{a^2+a^2}=\sqrt{2a^2}=\sqrt{2}\:a\:.
$$

As $h$ can be written as $\sqrt{2}\:a$, we now know how to write an expression for $\sin(45^\circ)$,

$$\sin(45^\circ)=\frac{a}{\sqrt{2}\:a}=\frac{1}{\sqrt{2}}\:.$$

Substitute $\sin(45^\circ)=\frac{1}{\sqrt{2}}$ into the expression $\sin^{-1} (\sin (45^\circ))$. Then

$$\sin^{-1}  \left ( \frac{1}{\sqrt{2}} \right ) \:.$$

This means that

$$\alpha=\sin^{-1}  \left ( \frac 1 {\sqrt 2} \right ) \tag 1$$

But how do I calculate $(1)$ without a calculator? It should be very simple, I think I'm just missing a concept here. I may have chosen the wrong method to calculate this expression, though.

What is the function $\sin^{-1} $? $\sin^{-1}(x) $ is asking for the angle whose sine is $x $.

What is $\sin^{-1}(\sin(x)) $? That is asking for the angle whose sine is the sine of $x $. Can you see the wordplay? What is the angle whose sine is the sine of $x $? Well, breaking it into smaller parts, let us assume the sine of $x$ is $y $. We have $\sin^{-1}(y) $: what is the angle whose sine is  $y $? Well, from definition, $y = \sin(x) $ so $x $ is the angle whose sine is $y $. Thus the answer to "what is the angle whose sine is $y $?" is $x $. But that is the same as asking "What is the angle whose sine is the sine of $x $?" which has $x $ as an answer.

If I understood your question correctly you're asking how to calculate  $arcsin \left(\frac{1}{\sqrt(2)}\right)$ without a calculator (since you are referring to formula (1)). I would suggest the unit circle.

It's not that hard to memorize and a very important tool. But if you want to do it without memorizing you'll probably need to look at the origin of the sinus and cosines functions.


Show that a locally compact subspace of a locally compact Hausdorff space is open in its closure Let $Y$ be a locally compact Hausdorff space and let $X$ be a locally compact subspace of $Y$. Show that $X$ is open in $\overline{X}$.

It seems I need to show that there exists an open set $U$ of $Y$ such that $X=U\cap\overline{X}$. Alternatively, since $Y$ is Hausdorff, I could show that $\overline{X}\setminus{X}$ is compact or closed. However, I can't think of a way to do either of these things.

By definition of locally compact, given $x\in Y$, there exists an open set $U$ and a compact subspace $C$ such that $x\in U\subset C$. Since $Y$ is Hausdorff, this definition is equivalent to saying given $x\in Y$ and a neighborhood $U$ around $x$, there exists a neighborhood $V$ around $x$ such that $\overline{V}\subset U$ and $\overline{V}$ is compact. I believe this equivalent definition could be useful, but I am not sure how it will come into play.

We prove the following more general result:

Each locally compact subspace $X$ of a Hausdorff space $Y$ is open in $\overline{X}$.

It suffices to consider the case $\overline{X} = Y$ and to show that $X$ is open in $Y$.

Each $x \in X$ has an open neighborhood $U$ in $X$ such that the set $\overline{U}^X = \overline{U} \cap X$ is compact and in particular closed in $Y$. But $U \subset \overline{U} \cap X$, hence $\overline{U} \subset \overline{U} \cap X \subset X$. Let $W$ be open in $Y$ such that $W \cap X = U$. Then

$$x \in W \subset \overline{W} = \overline{W \cap X} = \overline{U} \subset X .$$

Here we used a well-known fact: If $\overline{X} = Y$ and $W$ is open in $Y$, then $\overline{W} = \overline{W \cap X}$. This is true because if we have an open $V \subset Y$ such that $V' = V \cap W \ne \emptyset$, then $V \cap (W \cap X) = V' \cap X \ne \emptyset$.

Let $x\in X$, since $Y$ is locally compact and $X$ is a locally compact subset of $Y$. There exists an open neighborhood $V$ of $x$ whose adherence in $X$ is compact. We can write $V=U\cap X$ where $U$ is an open subset of $Y$. Let $y\in \bar X\cap U$. Since $Y$ is Hausdorff, there exists a sequence $x_n\in X$ such that $lim_nx_n=y$, since $x_n\in V$, we deduce that $y=lim_nx_n\in X$, since $V$ is a compact subset of $X$. It results that  $\bar X\cap U=X\cap U$.


Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:

$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
$f(0)=0$ and $f(1)=0$;
the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.

Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.

$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$

Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?

$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$

Find the Shape of the Graph

We wish to minimize
$$
\int_0^1\sqrt{f'(x)^2+1}\,\mathrm{d}x\tag{1}
$$
while keeping
$$
\int_0^1f(x)\,\mathrm{d}x=1\tag{2}
$$
This means that we wish to find an $f$ so that the variation of length is $0$
$$
\int_0^1\frac{f'(x)\,\delta f'(x)}{\sqrt{f'(x)^2+1}}\,\mathrm{d}x=0\tag{3}
$$
which, after integration by parts, noting that $\delta f(0)=\delta f(1)=0$, becomes
$$
\int_0^1\frac{f''(x)\,\delta f(x)}{\sqrt{f'(x)^2+1}^{\,3}}\,\mathrm{d}x=0\tag{4}
$$
for all variations of $f$, $\delta f$, so that the variation of area is $0$
$$
\int_0^11\,\delta f(x)\,\mathrm{d}x=0\tag{5}
$$
This means that $\frac{f''(x)}{\sqrt{f'(x)^2+1}^{\,3}}$ is perpendicular to all $\delta f$ that $1$ is. This is so only when there is a $\lambda$ so that
$$
\frac{f''(x)}{\sqrt{f'(x)^2+1}^{\,3}}=\lambda\tag{6}
$$
However, $(6)$ just says that the curvature of the graph of $f$ is $\lambda$. That is, the graph of $f$ is an arc of a circle.

Find the Length of the Arc

Since the length of the chord of the circle we want is $1$, we have
$$
2r\sin\left(\frac\theta2\right)=1\tag{7}
$$
Since the area cut off by this chord is $1$, we have
$$
r^2\left[\frac\theta2-\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)\right]=1\tag{8}
$$
Square $(7)$ to get
$$
2r^2(1-\cos(\theta))=1\tag{9}
$$
and rewrite $(8)$ to get
$$
\frac12r^2(\theta-\sin(\theta))=1\tag{10}
$$
Solve  $4(1-\cos(\theta))=\theta-\sin(\theta)$ to get
$$
\theta=4.3760724130128873845\tag{11}
$$
and then $(7)$ gives
$$
r=0.61313651252231835636\tag{12}
$$
This would lead to a minimum length of
$$
L=r\theta=2.6831297778598481320\tag{13}
$$

Problem

Unfortunately, since $\theta\gt\pi$, the minimizing curve is an arc that cannot be represented by a function.

The minimizing curve that is closest to the graph of a function is the curve that joins $(0,0)$ and $(1,0)$ to the endpoints of
$$
y=1-\frac\pi8+\sqrt{x-x^2}\tag{14}
$$

which has a length of
$$
2+\frac\pi4=2.7853981633974483096\tag{15}
$$
However, this curve is not the graph of a function.

A Sequence of Approximations

$$
f_n(x)=\frac1{c_n}\left(1-\frac\pi8+\sqrt{x-x^2}\right)\left(x-x^2\right)^{1/n}\tag{16}
$$
where
$$
c_n=\left(1-\frac\pi8\right)\frac{\Gamma\left(1+\frac1n\right)^2}{\Gamma\left(2+\frac2n\right)}+\frac{\Gamma\left(\frac32+\frac1n\right)^2}{\Gamma\left(3+\frac2n\right)}\tag{17}
$$
As $n\to\infty$, the length of $f_n$ approaches $2+\frac\pi4$.

At $n=100$, we get a length of $L=2.7857313936$, less than $\frac1{3000}$ above the minimum:

At $n=1000$, we get a length of $L=2.7854017568$, less than $\frac1{250000}$ above the minimum.

The most recent overkill contribution - which is at least somewhat interesting, though not the optimal - is the following behemoth: $f(x)=-0.00010156(b^{x-1/2}+b^{-(x-1/2)})+1.120357$ where $b=121716670.4$ You can think of this as a hyperbolic cosine with a different base. This has arc length $2.8788364$. The above answer does beat polynomials of the form $p(x)=a(x-1/2)^n+b$. Such polynomials can do no better than 2.8940115.


Sara has a white, a pink, a blue, a yellow and an orange flower. Sara has a white, a pink, a blue, a yellow and an orange flower.
She wants to plant these $5$ flowers in a row, but she does not want the orange flower to be the first
or the last flower in the row.
How many different arrangements are possible?

There are three positions in which to place the orange flower (the middle three): that gives $3$ ways to plant that flower. The remaining four spots can be filled in $4!$ ways: That gives us, in total: $$3\times 4! = 3 \cdot (4 \cdot 3\cdot 2\cdot 1) = 72$$ arrangements in which to plant the flowers.

My answer is $3\times 2\times 1$ which is $6$.


Proof Errors: 4=5 Using sin Many fake proofs are somewhat obvious involving division by 0 for example. I've come across this one which I have not seen discussed anywhere and I'm not too sure where the error lies, if there even is one.

I'm not so good with MathJax but the opening line given is:

"4 = 5 iff 4/5 = 1" : Not sure if use of double arrow is correct.

The 'proof' is as follows:

$$ 4=5 \iff \frac{4}{5} = 1$$
$$ \Rightarrow \frac{4}{5} = \sin\left(\frac{\pi}{2}\right)$$
$$Let \space \theta = \frac{\pi}{2} $$
$$\Rightarrow \sin \left( \theta\right) = \frac{4}{5}$$
$$\Rightarrow \sin\left(\pi - \theta\right) = \frac{4}{5}$$
$$\Rightarrow \pi - \theta = \arcsin\left(\frac{4}{5}\right) $$
$$\Rightarrow\pi - \arcsin\left(\frac{4}{5}\right) = \arcsin\left(\frac{4}{5}\right) $$
$$\Rightarrow\arcsin\left(\frac{4}{5}\right) = \frac{\pi}{2} $$
$$\Rightarrow\frac{4}{5} = \sin\left(\frac{\pi}{2}\right) = 1$$
$$\therefore 4 = 5$$

Is there an error with restricted domain of $\arcsin?$ Is it that there are a family of solutions when $\sin$ is introduced? Or is it just a fancy manipulation which ultimately does nothing and leaves the original statement? My gut feeling is that it's just a string of fancy ways to re-write $1$, but I'm really not sure and am therefore asking here.

What's the way to argue against this?

As I pointed out in the comments, the main flaw of the solution is poor exposition, which makes it difficult to find the finer, intended error. A better version of this puzzle would be:

Let $\theta = \arcsin \frac{4}{5}$ so that $\sin \theta = \frac{4}{5}$. Then also $\sin( \pi - \theta ) = \frac{4}{5}$, thus $\pi - \theta = \arcsin \frac{4}{5}$. Matching it together, we get $\theta = \pi - \theta$ so $\theta = \frac{\pi}{2}$, therefore $\frac{4}{5} = \sin \frac{\pi}{2} = 1$.

The obvious fallacy is drawing the conclusion that $\pi - \theta = \arcsin \frac{4}{5}$ from the equation $\sin( \pi - \theta ) = \frac{4}{5}$, since $\arcsin$ is only a partial inverse of $\sin$.

One could even argue I just made up the puzzle above myself, because there is so little similarity to the original puzzle in the structure of the proof. Well, I wouldn't deny.

But maybe the mess in the puzzle was intended to make it more obscure...

Many times in mathematics your "assumption" can be false, but it can lead to a satisfying and believing proof.

All the fake proofs start off with something false, and so whatever the result is from then on, its all gibberish, and can be true, or it can be false. We don't know.

The proper proofs assume something that is known to be true.

This is just another case where it just "works", but since your assumption that $4=5$ is already false from the start, the proof from then on has no meaning.


Name of Formula $x^3+y^3=z^3+1$ I encountered the formula
$$x^3+y^3=z^3+1$$
with the condition, that
$$x<y<z$$
and wonder, whether it has got a specific name or whether it can be easily transformed into another well-known (family of) formula(s).

Have a look at http://www.mathpages.com/home/kmath071.htm

There you will find $$(1\pm9m^3)^3+(9m^4)^3+(-9m^4\pm3m)^3=1$$ and another similar-but-more-complicated formula, also it says it is known that there are infinitely many such formulas, and it is not known whether every solution is part of such an infinite family.

$$X^3+ Y^3+ Z^3=1$$ is the formula which is known as harder factor and yours is a distorted and conditional form of harder factor

If $X+Y+Z=0$ then $X^3+ Y^3+ Z^3=1$.

In your question $X$ is less than $Y$ and $Y$ is less than $Z$ means the minimum possible difference between $X$ and $Y$, $Y$ and $Z$ is $1$.
At the same time the minimum possible difference between $X$ and $Z$ will be $2$.

So there will in all the cases except $X=-2$, $Y=-1$, $Z=3$ where $X+Y+Z$ is not equal to zero then it must be that $X^3+ Y^3+ Z^3$ is not equal to $1$. So $X^3+ Y^3+ Z^3$ must be greater/less than $1$. As it is given that $Z>Y>X$ then $X^3+ Y^3$ must be unequal to $Z^3$. It means
$X^3+ Y^3$ may be equal to $Z^3+1$.

In this way $X^3+ Y^3+ Z^3=1$ is related to the question asked by the poster


It it possible to "compress" a list of large numbers using their prime factors? On a computer I can have integers on arbitrary size thanks to GMP, so it's represented in base 2 in memory.

I'm wondering if it's possible in theory to use less memory if I store only prime factors and their exponent, I think it's worthless for most numbers, but I wonder if it works for a very small minority of numbers with large prime factors.

How can I prove me wrong ?

No, it is not possible. In general, there is no algorithm that compresses every $n$ bit string, or reduces the length of a random string of bits (in expectation).

Your algorithm may compress some numbers but then it will use more memory to store others. If your data is random, there is no benefit in using your compression scheme.

Actually the fact that no such compression scheme exists can be used to prove that there are infinitely many prime numbers — otherwise, if there were only finitely many prime numbers your compression scheme would be very efficient!

There are excellent lecture notes on Kolmogorov Complexity by Alexander Shen that discuss “incompressability”; in particular, Shen writes about your compression algorithm in Section 18.

Something which sort of implies that the answer might be yes in some circumstances (I specialize in precision!) is that solving linear equations with large integer coefficients can be done by solving the system modulo a number of primes (or relatively prime values such as $2^n-1$ for appropriate $n$) and combining the results using the Chinese remainder theorem. You have to have enough moduli so the their product is larger than any of the results.


Variables defined as floor and fraction part from exponentially distributed random variable Random variable $X$ has exponential distribution with parameter $\lambda>0$. Let $T=\lfloor X\rfloor $, $R=\{X\}$ where $\lfloor x\rfloor$ is floor from number $x\in\mathbb{R}$ and $\{x\}$ is it's fraction part. What is the distribution and expected value of $T$ and $R$?

I tried doing that the following way:

$$\mathbb{P}(T=k)=\mathbb{P}(X\in[k,k+1))=\int_k^{k+1}\lambda e^{-\lambda x}=-e^{-\lambda x}\Big|_k^{k+1}=\frac{e^{\lambda}-1}{e^{\lambda +k}}$$ so $$\mathbb{E}T=\sum_{k=0}^{\infty}\frac{k(e^{\lambda}-1)}{e^{\lambda +k}}=\frac{e^{\lambda}-1}{e^{\lambda}}\sum_{k=0}^{\infty}\frac{k}{e^{\lambda k}}=\frac{e^{\lambda}-1}{e^{\lambda}}\frac{e^{\lambda}}{(e^{\lambda}-1)^2}$$

I took the last equality from Wolfram, yet I'm unsure of it's origins. Where does it come from? Secondly, I'm clueless about $R$ distribution. Could you help?

Since the pdf of $X$ is $\lambda e^{-\lambda x}\mathbb{1}_{x\geq 0}$ we have:
$$\forall m\in\mathbb{N},\quad\mathbb{P}[\lfloor X\rfloor\leq m]=\int_{0}^{m+1}\lambda e^{-\lambda x}\,dx = 1-e^{-\lambda(m+1)}$$
hence the pdf of $\lfloor X \rfloor$ is $(e^{\lambda}-1)e^{-\lambda(m+1)}\mathbb{1}_{m\in\mathbb{N}}$ and:
$$\mathbb{E}[\lfloor X\rfloor]=(e^{\lambda}-1)\sum_{m=0}^{+\infty} m e^{-\lambda(m+1)}=\frac{1}{e^\lambda-1}$$
since for any $-1<y<1$:
$$\sum_{m=0}^{+\infty} m y^m = y\cdot\frac{d}{dy}\sum_{m=1}^{+\infty}y^m=\frac{y}{(1-y)^2}.$$
For the second part, we have:
$$\forall t\in(0,1),\qquad \mathbb{P}[\{X\}\leq t]=\sum_{n\in\mathbb{N}}\int_{n}^{n+t}\lambda e^{-\lambda x}dx = (1-e^{-\lambda t})\sum_{n\in\mathbb{N}}e^{-\lambda n}=\frac{1-e^{-\lambda t}}{1-e^{-\lambda}}$$
hence the pdf of $\{X\}$ is $\frac{\lambda}{1-e^{-\lambda}}e^{-\lambda x}\mathbb{1}_{x\in(0,1)}$. By the linearity of $\mathbb{E}$,
$$\mathbb{E}[\{X\}]=\mathbb{E}[X]-\mathbb{E}[\lfloor X\rfloor] = \frac{1}{\lambda}-\frac{1}{e^{\lambda}-1}.$$

To calculate the expected value, you can also notice that the floor function follows a geometric distribution (number of failures before the first success) with parameter p=1-exp{-lambda}.


Matrix derivative formula using the matrix chain rule Let $X \in \mathbb{C}^{m \times n}$ be a matrix. Let $F(X) \in \mathbb{C}^{m \times m}$ be a matrix, function of $X$, e.g. $F(X) = I_m + X X^{\dagger}$, where $^\dagger$ means conjugate-transpose and $I_m$ is the identity matrix of dimension $m$. Finally, let $\mathbf{g}(X)$ be a (column-)vector-valued function of $X$, e.g. $\mathbf{g}(X) = u - Xv$, with $u,v$ column-vectors of appropriate dimensions. Then,
$$
Q(X) = \mathbf{g}(X)^\dagger F(X) \mathbf{g}(X)
$$
is clearly a scalar. What I want to find is a formula for
$$
\frac{\partial \mathbf{g}(X)^\dagger F(X) \mathbf{g}(X)}{\partial X} = \ ?
$$

Edit: By Leibniz's rule,
$$
\frac{\partial Q(X)}{\partial X} = \frac{\partial \mathbf{g}^{\dagger}(X)}{\partial X} F(X) \mathbf{g}(X) + \mathbf{g}^{\dagger}(X) \frac{\partial F(X)}{\partial X} \mathbf{g}(X) + \mathbf{g}^{\dagger}(X) F(X) \frac{\partial \mathbf{g}(X)}{\partial X}
$$

To begin with, as discussed in the comments, one should understand what $dX$ in the denominator means. The space of matrices is a vector space, and so, all maps in question are multi-variable maps. Hence, every map from the space of matrices to another space has a differential which can be thought of as a bunch of partial derivatives. In other words, describing the differential of such a map is equivalent to specifying all the partial derivatives.

So, let $e_i$ be a basis of the space of matrices, and let $\frac{\partial}{\partial x^i}$ denote the directional derivative in the $e_i$ direction. By the Leibniz rule,$$\frac{\partial}{\partial x^i}(f_1\cdot\ldots\cdot f_k)=\frac{\partial f_1}{\partial x^i}f_2\ldots f_k+\ldots+f_1\ldots f_{k-1}\frac{\partial f_k}{\partial x^i}.$$ Note that if the $f$'s are matrix-valued (and they are in your example), then you can't change the order in the above equation, as $AB\neq BA$ for general matrices. Taking transpose and/or conjugation commutes with differentiating, and so, transpose and $\dagger$ simply carry through.

There is a nice answer in Seber(08), page 360, result 17.25, for differentiation by vector.

  Suppose y=w'Az (scalar) where: w(mX1), A(mxn), z (nx1) and all are functions of vector x. consider: "kp"=kroneker product, ' the transpose.
  ∂y/∂x'. Note that y=vec(y)=(z'"kp"w')vec(A)=(z"kp"w)'vec(A)=[vec(wz')']'vec(A).
  Since y=y' so w'Az=z'A'w, Hence
  ∂y/∂x'=z'A'∂w/∂x'+ [vec(wz')]'∂vec(A)/∂x'+ w'A∂z/∂x'"

SEBER, G. A. F. A Matrix Handbook for Statisticians. Wiley, 2008.


Prerequisits for Gauss-Green theorem Consider the following theorem from the appendix C from Evans PDE book:

I know about integration in $\mathbb{R}^n$ but not about how to make sense of the integrals on the right-hand side. As my time is very limited I can't read 50+ pages on the general theory integration on manifolds, I am in desperate need of a text that in a minimum amount of space manages to get my aquainted with those integral and the proof of (i) from above.
It's a plus if the text has some worked examples (the proofs from those statements of the text, that are intuitively plausible, I want to skip anyway because I don't have sufficient time).

(Note:This question is a total re-edit. For further details please consult the edit history.)

Perhaps this not-too-long notes will help
http://www.owlnet.rice.edu/~fjones/chap14.pdf

It gives some rather elementary explanations and proof of the Gauss theorem in $\mathbb{R}^n$. For the complete set of notes (used for an honors calculus class at Rice University) covering other related topics, refer to http://www.owlnet.rice.edu/~fjones/

What you're really asking for a textbook giving a modern and careful presentation of vector calculus/calculus of functions of several variables.Since you're reading Evan's text,I'll assume you've had a least a good undergraduate course in real analysis of one variable on metric spaces a la "baby Rudin" or Pugh. I'll also assume you've got a decent background in linear algebra, which is really important in understanding a careful proof of the variants of Stokes' Theorem, which is really what the Guass-Green theorem is a special case of.

The standard books for learning this material are Calculus On Manifolds by the legendary Micheal Spivak and Analysis on Manifolds by James Munkres. Spivak's book is basically a problem course with quite a few pictures. It's quite rough going,but it's worth the effort if you've got the patience. Munkres is more of a standard textbook and covers the same material with much more detail.Since you're looking for a rapid introduction that'll cover the material in minimum space, Spivak really will fill your needs best if you're willing to work through it.Indeed, the entire point of the book is to build to careful proofs of the variants of Stokes' theorem. Munkres sacrifices brevity for clarity, so if you need more detail, that's a better choice.Be warned,though-Munkres has a bit more in the way of prerequisites then Spivak. Spivak basically assumes only good courses in linear algebra  and rigorous $\epsilon-\delta$ single variable calculus while Munkres assumes in addition, a good working knowledge of basic topology in metric spaces.Both books work mostly in $R^n$ and discuss abstract manifolds at the end in passing. Both are still really good choices-again, it depends on how much time you have to invest.

A third book you might want to check out for an excellent rigorous discussion  of Green and Stokes' theorems is Loring Tu's An Introduction To Manifolds. The purpose of Tu's book is to give a clear and relatively short introduction to the theory of differential forms and manifolds with only a background in undergraduate algebra and real analysis. Specifically, the purpose of the book is to provide all the essential background in manifolds that's needed to read the classic text on algebraic topology he co-wrote with Raoul Bott, Differential Forms In Algebraic Topology. That being said, the book contains all the basics with many pictures and is extremely readable. Chapters 21-24 contain an in-depth presentation of statements and proofs of both the Green-Gauss theorem and the general Stokes' Theorem, which the GG theorem is a special case of. I think this book will also serve your needs very well.

I think any of these books will be good choices for you. But again,if brevity is paramount here and you're willing to work through it with a pen and paper, Spivak is probably your best choice.

Good luck!


Tangent lines to ellipse from external point I'm given an ellipse with the equation
 $$\frac{x^2}{5^2} +\frac{y^2}{4^2} -1 = 0 $$

and I have to find the tangents to this ellipse  passing through a point $P(10, -8)$.

Use auxiliary coordinates $u:=\dfrac x5,v:=\dfrac y4$ to turn the ellipse to the unit circle. The external point is $(2,-2)$, which is at distance $\sqrt8$ of the origin, in direction $-45°$.

The point, the circle center and the tangency points form right triangles of small side $1$ and hypothenuses $\sqrt8$. Thus the small angle is given by $\tan\theta=\dfrac1{\sqrt{8-1}}$.

Hence, the slope of the tangents are given by

$$\tan(\pm\theta-45°)=\frac{\pm\sqrt7-1}{1\mp\sqrt7}.$$
After converting back to the $(x,y)$ coordinates,

$$m=\frac45\frac{\pm\sqrt7-1}{1\mp\sqrt7}.$$

The straight line has equation

$$y+8=m(x-10).$$ We plug that in the equation of the ellipse and get

$$\frac{x^2}{25}+\frac{(m(x-10)-8)^2}{16}-1=0.$$

This quadratic equation in $x$ has a double root (which ensures tangency) when its discriminant is zero, i.e.

$$75m^2+160m+48=0.$$

There are two solutions.


if a bounded sequence xn in [a,b] converges to x0 then x0 belongs to [a,b] If a bounded sequence xn in [a,b] converges to x0 then prove that x0 belongs to [a,b]. I don't understand here how can x0 belong to [a,b], it may be slightly greater than b too. Please help.

The sequence is convergent and thus bounded, so there exists $M$ and $g$ such that

$$
|a_{n}|<M\qquad\text{and}\qquad\lim_{n\rightarrow\infty}a_{n}=g.
$$

This holds because we can find $M$ such that $[a,b]\in[-M,M].$ Then

$$
|g|=|\lim_{n\rightarrow\infty}a_{n}|=\lim_{n\rightarrow\infty}|a_{n}|<\lim_{n\rightarrow\infty}M=M
$$
So
$$
g\in[-M,M]\qquad\text{and}\qquad{a_{n}}\in[-M,M],
$$
as desired.

A convergent sequence is surely bounded. Since $a\le x_n\le b$, permanence of sign tells you that
$$
a\le\lim_{n\to\infty}x_n\le b
$$

If you want a detailed proof, assume by contradiction that $x_0<a$. Then there exists $N$ such that, for $n>N$,
$$
|x_n-x_0|<\frac{a-x_0}{2}
$$
which implies
$$
x_n<x_0+\frac{a-x_0}{2}=\frac{a+x_0}{2}<a
$$
Contradiction.

Similarly for proving that $x_0\le b$.


How can $\frac12 \log(x) = \log(\sqrt{x})$? How can $\frac{\log (x)}{2}= \log \left(\sqrt{x}\right) \\$?

How would I come to this conclusion?

I assume that you know $\log(ab)=\log a+\log b$. Then
$$
\log x=\log(\sqrt{x}\cdot\sqrt{x})=\log\sqrt{x}+\log\sqrt{x}=2\log\sqrt{x}
$$
and so
$$
\log\sqrt{x}=\frac{1}{2}\log x.
$$
Anyway, it's just a special case of the general law
$$
\log(x^k)=k\log x
$$
(for $x>0$ and any $k$).

Firstly you should know that this is a direct result of the Logarithms of Powers Rule:
$$\Large\log(\sqrt{x}) = \log(x^{1/2}) = \frac{1}{2} \log(x)$$

Here's my attempt at a proof nonetheless,

To Prove: $ \frac{1}{2} \log(x) = \log(\sqrt{x})$

Proof: Let us assume that the base of the logarithm is $b$.

We know that $x = b^{\log x} \implies \sqrt{x} =  b^{\log \sqrt{x}}$
$$
\Rightarrow x = b^{2\log \sqrt{x}} \\
\Rightarrow b^{\log x} = b^{2\log \sqrt{x}} \\
\Rightarrow \log(x) = 2\log(\sqrt{x}) \\
\Rightarrow \frac{1}{2} \log(x) = \log(\sqrt{x})
$$

Aliter:
$$\large\frac{1}{2} \log(x) = \log(\sqrt{x}) \Leftrightarrow \log(x) = 2\log(\sqrt{x}) \Leftrightarrow x = b^{2\log(\sqrt{x})}$$

$$
\large\text{RHS} = b^{\log(\sqrt{x}) + \log(\sqrt{x})}
=  b^{\log(\sqrt{x})} \times b^{\log(\sqrt{x})}
= \sqrt{x} \times \sqrt{x} = x = \text{LHS}$$


converting the sum of numbers to logarithms If I have three number, $[0.2, 0.3, 0.4]$ then I can take the sum simply by adding the term: $0.2+0.3+0.4=0.9$. The proportion of the first element is then $0.2/0.9 = 2/9$.

Now I don't know what the three numbers are but I do know the logarithms (which are so small I cannot take the exponentiation of it). How can I get the proportion of the elements?

So given: $[-1.6094,   -1.2040,   -0.9163]$, how do I find out that the first component has a proportion of $2/9$?

If this is a practical problem then note you can use an arbitrary precision floating point library ( like MFPR , which is free ) to write a short application to do the exponentiation and addition and division to any precision you care to.


What's wrong with this solution of Tarski's circle-squaring problem? Tarski's circle-squaring problem asks whether it is possible to cut up a circle into a finite number of pieces and reassemble it into a square of the same area. Note that this is different from the problem of squaring the circle (which is about compass and straightedge constructions) and also from the Banach-Tarski paradox (which involves a change of volume). Laczkovich showed in 1990 that the answer was yes, and proved the existence of a solution with about $10^{50}$ pieces. The proof was nonconstructive.

What is the error in the following proof that the answer to the problem is no, contrary to Laczkovich? At a given stage in the process of cutting and reassembly, let $u$ be the total length of all convex edges that are circular arcs whose radius equals the radius $r$ of the original circle, and similarly let $v$ be the total length of all such concave edges. Initially, $u=2\pi r$ and $v=0$. At the end of the construction, we would have to have $u=v=0$. But $u-v$ is conserved by both cutting and reassembly, so this is impossible.

[EDIT] Edited to remove mistake about Banach-Tarski.

According to the Wikipedia article you link to, the pieces in Laczkovich's decomposition are non-measurable and do not have Jordan curve boundary, so it is impossible to define the length of their edges.

Like you I have a simple (and correct) proof that the answer is no.  The error in Laczkovich's decomposition is that he's not actually cutting up a circle, he's performing an imaginary operation similar to cutting but not constrained by the need for cuts to obey the "Jordan curve boundary" and therefore, basically, anything is possible.


Does every group homomorphism from $(0,\infty),\times)$ to $(\mathbb{R},+)$ send $1$ to $1$? I just have some true/false questions I am revising with and I'm not sure about this.

Let $f:((0,\infty),\times)\to(\mathbb{R},+)$ be a group homomorphism, then $f(1)=1\tag{1}$

I know that a group homomorphism is when you have 2 groups $(G,*)$ and $(H,\odot)$ such that a function $f:G \to H$ gives $f(a*b)=f(a) \odot f(b)$ for all $a,b \in G$

If someone could explain that (or give me hint), it'd be awesome!

Okay actually as I was typing this I may have had an epiphony!

I think that (1)is  false because $1\times 1 \neq 1+1$

I also have,

Let $f:((0,\infty),\times)\to(\mathbb{R},+)$ be a group homomorphism, then $f(\frac{1}{2})=-f(2)\tag{2}$

But please don't tell me the answer to this one! I'll try figure it out based on feedback from the first one and go from there

Clearly $(1)$ is not true. In general, given $f:G\rightarrow H$ group homomorphism, we know that the identity element $e_G\in G$ maps to the identity element $e_H \in H$.

You just need to identify these elements in your groups and substitute in
$$f(e_G)=e_H. \tag{1}$$

Then it should be easy to figure out $(2)$, since
$$f(1)=f\left(\frac{1}{2}\cdot 2\right) = f\left(\frac{1}{2}\right) + f(2).$$

Use the same idea to view $\frac 1 2$ and $2$.
Try to view this
$$f(\frac  1 2)=f(\frac 1 4.2)=2\frac 1 4\tag 1$$
but $$ -f(2)=-f(\frac 1 2.4)=4\frac 1 2\tag 2$$ which is not same.


Solution for $x$ and $y$. If $x^6-y^6 = 665$, and $x^3y+xy^3=78$, find $x$ and $y$.

I have tried this question in the following way:
We know that $x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$. By substituting the values we can yield $665=(x-y)(x^5+78x+78y+y^5)$. Further simplification made it complicated.

These are some initial thoughts. I will add more if I have more ideas.

Let $u = xy$, $v=x+y$, $w=x-y$. Note that $$x^6-y^6=vw\cdot(v^2-u)\cdot(w^2+u)$$And $$x^3y+xy^3=u\cdot(v^2-2u)=u\cdot(w^2+2u)$$This gives us $3$ equations with $3$ variables, and all of the equations are at most cubic.

Let us first limit our search for solutions to the reals. Note that the second equation forces $x,y$ to have the same sign, and by @Arthur's comment, $(x,y)$ being a solution implies $(-x,-y)$ is a solution. So, we can limit our search in the reals to positive $x,y$.

However, note that $x^3y+y^3x$ is strictly increasing on both $x,y$, so if $x_0,y_0$ and $x_1,y_1$ are both solutions to that equation, $x_0,x_1,y_0,y_1>0$, then WLOG $x_1>x_0$, $y_0>y_1$. This is a problem, however, as if $x_0,y_0$ and $x_1,y_1$ are solutions to the first equation, then $x_1>x_0$ implies $y_1>y_0$. But this is a contradiction, implying that there can be at most one set of positive $x,y$ that solves the system. @DavidG.Stork already found this solution to be $(2,3)$.

So, we've shown that the only real solutions are $\color{red}{(2,3)}$ and $\color{red}{(-2,-3)}$. Solving for complex roots is likely to be tedious.


How prove this $\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\frac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$ Prove or disprove:

if $f(x)\ge 0,\forall x\in [-\pi,\pi]$,show that
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\dfrac{\pi}{2}\int_{-\pi}^{+\pi}f^2(x)dx$$

I can prove this if $2\pi$ takes the place of $\dfrac{\pi}{2}$

because  use Cauchy-schwarz inequality  we have
$$\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2\le\int_{-\pi}^{\pi}\sin^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$

$$\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2\le\int_{-\pi}^{\pi}\cos^2{x}dx\int_{-\pi}^{\pi}f^2(x)dx$$
add this two  inequality, we have
$$\begin{align*}\left(\int_{-\pi}^{\pi}f(x)\sin{x}dx\right)^2+\left(\int_{-\pi}^{\pi}f(x)\cos{x}dx\right)^2 &\le \int_{-\pi}^{\pi}f^2(x)dx\int_{-\pi}^{\pi}(\sin^2{x}+\cos^2{x})dx\\
&=2\pi\int_{-\pi}^{+\pi}f^2(x)dx\end{align*}$$

see this Discrete form of inequality：Prove this inequality with Cauchy-Schwarz inequality

So far, I haven't found any counterexamples，such  $f(x)=1,\sin{x}+1$ it such this inequality

Note that
\begin{align*}
\int_{-\pi}^\pi f(x) \sin x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) + f(\pi - x) - f(-x)) \sin x \,dx \\
&= \int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx
\end{align*}
and similarly
\begin{align*}
\int_{-\pi}^\pi f(x) \cos x \,dx
&= \int_0^{\pi/2} (f(x) - f(x - \pi) - f(\pi - x) + f(-x)) \cos x \,dx \\
&= \int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx
\end{align*}
where we define $g, h : [0, \pi/2] \to \mathbb{R}$ by $g(x) = f(x) - f(x - \pi)$ and $h(x) = f(\pi - x) - f(-x)$.

Then by Cauchy-Schwarz,
\begin{align*}
\left(\int_0^{\pi/2} (g(x) + h(x)) \sin x \,dx\right)^2
&\leq \int_0^{\pi/2} \sin^2 x \,dx \int_0^{\pi/2} (g(x) + h(x))^2 \,dx \\
&= \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx
\end{align*}
and
\begin{align*}
\left(\int_0^{\pi/2} (g(x) - h(x)) \cos x \,dx\right)^2
&\leq \int_0^{\pi/2} \cos^2 x \,dx \int_0^{\pi/2} (g(x) - h(x))^2 \,dx \\
&= \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx
\end{align*}
hence
\begin{align*}
\left(\int_{-\pi}^\pi f(x) \sin x \,dx\right)^2 &+ \left(\int_{-\pi}^\pi f(x) \cos x \,dx\right)^2 \\
&\leq \frac{\pi}{4} \int_0^{\pi/2} (g(x) + h(x))^2 \,dx + \frac{\pi}{4} \int_0^{\pi/2} (g(x) - h(x))^2 \,dx\\
&= \frac{\pi}{2} \int_0^{\pi/2} (g(x))^2 + (h(x))^2 \,dx \\
&\leq \frac{\pi}{2} \int_0^{\pi/2} (f(x))^2 + (f(x - \pi))^2 + (f(\pi - x))^2 + (f(-x))^2 \,dx \\
&= \frac{\pi}{2} \int_{-\pi}^\pi (f(x))^2 \,dx
\end{align*}
as desired, where the last inequality holds because $f$ is nonnegative.

This is a consequence of parseval identity. Let $\mathcal{H}$ a Hilbert space (dont worry all you need to know is that you have an inner product and the concept of orthogonal vectors, i.e. when $\langle f, g\rangle = 0$ and that the norm is given by $\langle f, f \rangle = \lVert f \rVert^2$) and let $\{e_n\}$ be a numerable orthonormal set then if $f \in \mathcal{H}$
$$
\sum_n \lvert \langle e_n, f\rangle \rvert^2 \leq \lVert f \rVert^2.
$$
with equality if $\{e_n\}$ is a basis. In this case
$$
\lVert f \rVert^2 = \int_{-\pi}^{\pi} \lvert f(x) \rvert^2 \,dx
$$
and
$$
e_1(x) = A\cos(x), \quad e_2(x) = A\sin(x)
$$
where $A$ is chosen so they are normal.

Parseval identity is deduced from the fact that for orthonormal (it also works with orthogonal, but the last equality is not true) vectors $u_n$ one has
$$
\lVert \sum a_n u_n \rVert^2 = \sum \lVert  a_n u_n \rVert^2 = \sum \lvert  a_n \rvert^2
$$
and you can write
$$
f = \langle e_1,f \rangle e_1 + \langle e_2, f\rangle e_2 + (f - \langle e_1,f \rangle e_1 + \langle e_2, f\rangle e_2)
$$
where the 3 parts of the sum are orthogonal and you have that the norm of $f$ is bigger than the norm of the first 2 terms.

EDIT: it seems the constants are off. It does not coincide with the one in parseval's.


How to integrate $\frac{x^{2}\log {\sin x}}{1+x^{6}}$ I recently stumbled upon a question
$$\int_0^{\infty}\frac{x^{m-1}\log^{a}x}{1+x^n}dx$$
I was able to evaluate it,but I am curious if there exists a closed form for,
$$\int_0^{\pi/2}\frac{x^{2}\log{\sin x}}{1+x^6}dx$$
It numerically evaluates to -0.1392432458.

My attempt-

$$\int \frac {x^2}{1+x^6}dx=\frac13 \int \frac {d(x^3)}{1+x^6}=\frac13 \arctan {x^3}$$
Then,by applying integration by parts,
$$\int_0^{\pi/2}\frac{x^2\log\sin x}{1+x^6}dx=-\int_0^{\pi/2}\frac13\arctan {x^3} \cot x dx$$.But now I'm stuck.

The integration you want to solve doesn't have an analytical solution, so using integration by parts won't help you much. However, you can use a number of numerical methods out there for approximating integrals.


Rotated functions and taylor series If one rotates a function such as the sine function about the origin, is there a general method to find the taylor series for the rotated function? Assuming of course that the rotated function is still a single valued function.

I have limited time to post the answer, but I think this is it.  The individual coefficients get multiplied such that $b_n = a_n\exp(2n\pi i k)$, where k is the rotation angle, some rational or irrational rotational coefficient scaled to be between zero and one, such that $\theta=\exp(2\pi i k)$


Convergence in Norm to Zero If a sequence, $\{f_n\}$, of nonnegative functions with $\lim_{n\to\infty}\int_{X}\sqrt{f_n}dm=0$, must it be the case that $\lim_{n\to\infty}f_n=0$ $m$-a.e. on $X$?

No, though  $\liminf_{n\to\infty}f_n=0$, $m$-a.e., by Fatou's Lemma; and also, $f_n\to 0$ in measure, by Markov's inequality. For a counterexample, use the "typewriter sequence" of functions (Example 4 in https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/)

Yes. The definition of the $L^2$ norm is $\|f\| = (\int_X f^2\,dm)^{1/2}$, so your hypothesis says that $\|f_n^{1/4}\| \to 0$. Since pointwise multiplication of functions is continuous in the 2-norm, it follows that $\|f_n\| \to 0$ as well. Since the $0$ element of $L^2$ is the class of functions that are $0$ $m$-almost everywhere, this gives you your answer.


Describe all ring homomorphisms from $\mathbb{R}[T] \rightarrow \mathbb{R}[T]$ One of the problems in a problem set I was given as homework in my Algebra course proposes the next problem:

  Describe all ring homomorphisms $\mathbb{R}[T] \rightarrow \mathbb{R}[T]$. Which of them are isomorphisms?

I would like some suggestions towards the right direction, not the answer to the problem.

This is what I've got so far:

Given a ring homomorphism $f:\mathbb{R}[T] \rightarrow \mathbb{R}[T]$:

Given an arbitrary polinomial $p(T) = a_{0} + a_{1}T + \ldots + a_nT^n$ we have that $f(p) = f(a_0) + f(a_1)f(T) + \ldots + f(a_n)f(T)^n$. So we get that $f$ is completely determined by the values it assumes on $\mathbb{R}$ and $f(T)$.

So this problem may now be separated in two:

Classifying all homomorphisms of the form $f:\mathbb{R} \rightarrow \mathbb{R}[T]$
Classifying all possible values $f(T)$

With respect to (1):

Conjecture
The only ring homomorphism is $f(x)=x$ (we put as a condition that f(1) = 1, on the definition the professor gave us, so that discards $f(x)=0$)

I've shown by induction that $f(n) = n, ~\forall n\in\mathbb{N}$, then $f(m) = m, ~\forall m\in \mathbb{Z}$, then $f(q) = q, ~\forall q\in \mathbb{Q}$ but I
am having problems showing that $f(\alpha) = \alpha, ~\forall \alpha \in \mathbb{Q}'$ because I don't really know if $f(\alpha) \in \mathbb{R}$.

If I knew that $f(\mathbb{R}) \subset \mathbb{R}$ then I could do something like $f(\alpha) = f(\sqrt{\alpha})^2 > 0$ if $\alpha > 0$. And this would help me prove that $f(\beta) = \beta$ for all irrationals too. The only problem is that, what happens if say, $f(\alpha) = T$. Then $>$ would make no sense.

I think I solved this problem but I am not sure, maybe here is where you guys can help me a little.

If we suppose that $f(\alpha)$ is a polynomial with degree $n$, we can then compute $f(\alpha^\frac{1}{n+1}) = f(\alpha)^\frac{1}{n+1}$ and that would be in $\mathbb{R}[T]$ only if the degree, $n$, of $f(\alpha)$ were $0$ thus proving that indeed $f(\alpha) \in \mathbb{R}$.

Is there any mistake or an easier way? Or is there any usefull comment anyone wants to make that could help me out. Thanks in advance :)

The only Homomorphism on $\mathbb R$ that fixes $1$, is the identity map. For the proof you can see Ring homomorphisms $\mathbb R\to\mathbb R$ .
Since $f$ is a homomorphism that fixes $1$, image of every invertible elements in $\mathbb R[T]$ should be invertible, and we should have $f_{|\mathbb R}=Id_{\mathbb R}$ .

$f$ completely determined by the image of $T$, and so

$f(a_0+a_1T+...+a_nT)=a_0+a_1(p(T))+...+a_n(p(T))^n$ where $p(T)=f(T)$.

At last $f$ is isomorphism iff $f(T)=aT+b$

(Big) hint: Show that a (unital) ring homomorphism from $\mathbb{R}[T]$ to any other unital ring $A$ is completely determined by the image of $T$.


Proving the real numbers are complete In Rudin's book, the following proof is published:

Let $A$ be the set of all positive rationals $p : p^2 < 2$. Let $B$ be the set of all positive rationals $p : p^2 > 2$. $A$ contains no largest number and $B$ contains no smallest.
Let q be a rational. More explicitly, $$\forall p \in A, \exists q \in A : p < q$$
and $$\forall p \in B, \exists q \in B : q < p$$
This needs proof.

Now he introduces a couple of equations which are not clear where they are derived from. As follows,

To do this, we associate with each rational $p > 0$ the number $$q = p - {\frac{p^2-2}{p+2}} = {\frac{2p+2}{p+2}}$$
then $$q^2 - 2 = {\frac {2(p^2-2)}{(p+2)^2}}$$

He then goes on to prove how $p \in A \rightarrow q \in A$; similarly for the set B.
My question is: How did he arrive at those equations - particularly, the first one (since the second is clear)?

If $p\in\mathbb{Q}$ is smaller than $\sqrt{2}$, we want to find another rational $p<q<\sqrt{2}$. Now, this can easily be handled with decimal fractions, but if you want something neater that relies solely on $p$, then what we basically want to do is find a rational $r\in\mathbb{Q}\cap(0,\sqrt{2}-p)$ and define $q=p+r$.

Note that $\sqrt{2}-p\not\in\mathbb{Q}$, but $2-p^2 = (\sqrt{2}-p)(\sqrt{2}+p)\in\mathbb{Q}$. If we want it to be small enough, it suffices that $\frac{\sqrt{2}+p}{2+p} < 1$, so that
$$\frac{2-p^2}{2+p} = (\sqrt{2}-p)\frac{\sqrt{2}+p}{2+p}<\sqrt{2}-p.$$

if $p^2<2$
let's add 2p to both sides of the inequality
then $p^2+2p<2+2p$ then $p(p+2)<2(p+1)$ then $p<\frac{2(p+1)}{p+2}$ so let's set $q=\frac{2(p+1)}{p+2}$ then we have if $p^2<2$ then $p<q$. what is left to show is that $q^2<2$
and that would show that $0<p<q$ and $q \in A$


Ramanujan's method for solving a Magic Square Ramanujan came up with a formula for solving a magic square.

  Theorem 1: Let $m_1,m_2$ denote the sums of the middle row and middle column respectively for a $3\times 3$ square array of numbers. Let $c_1,c_2$ denote the sums of the main diagonal and secondary diagonal respectively. And lastly, let $S$ denote the sum of all nine elements of the square. If $x$ denotes the center element, then$$x=\frac 13\left(m_1+m_2+c_1+c_2-S\right)\tag1$$
  Suppose that the sum of each row and column is equal to $r$. Then, we have$$x=\frac 13\left(c_1+c_2-r\right)\tag2$$
  And note that that implies that $x=\frac r3$, so $r$ is a multiple of $3$.
  And lastly, in a $3\times 3$ magic square, the elements in the middle row, middle column and diagonals are in arithmetic progression. More clearly, let $a,b$ be the first and third elements respectively. Then$$a+r/3+b=r\tag3$$Hence$$b-r/3=r/3-a\tag4\\\vdots$$

Although the formula is clear, I'm just wondering how you would go about actually solving a magic square.

  Question: Using Theorem 1, how would you solve a magic square? (Example provided below)

For example:

Construct a magic square with $r=15$ and all numbers are odd!

My Attempt:  Since $r=15$, we have $x=15/3=5$. Thus, the middle digit$^{[1]}$ is $5$. From $(3)$, we get a Diophantine equation, and solving we get $$(a,b)=(1,9)\\(a,b)=(3,7)\\(a,b)=(5,5)\tag5$$
After that, I'm not sure what to do. I have $3$ possible $a,b$ values and I'm not sure which two I should use. Any help would be accepted!

$\scriptsize{[1]:\text{Maybe..?}}$

Why it's impossible:

There are actually 4 more arithmetic sequences that will always be present in a magic square.
Let the square:
$$a,b,c$$
$$d,e,f$$
$$g,h,i$$
So from $(3)$ and $e=\frac{r}{3}$ we get the equations:
$$a-e=e-i$$
$$b-e=e-h$$
$$c-e=e-g$$
$$d-e=e-f$$
$(6)$ From these we get:
$$2e=a+i=b+h=c+g=d+f$$
Next take:
$$r=a+b+c=a+d+g\implies b+c=d+g$$
Then from $(6)$, substitute in $g=b+h-c$ to get:
$$b+c=d+(b+h-c)$$
Rearrange:
$$d-c=c-h$$
The same "trick" can be done too for $(b,g,f),(b,i,d),(h,a,f)$ either using the same method or the symmetries of the magic square.

$(7)$ We have:
$$d-i=i-b$$
$$b-g=g-f$$
$$f-a=a-h$$
$$h-c=c-d$$
Note the middle elements are the corners, which is helpful for remembering these.

Back to your question now.

For $r=15$ we have 4 sequences across the center: $(1,9),(2,8),(3,7),(4,6)$

From these we can make 5 more arithmetic sequences: $(1,2,3),(2,4,6),(1,4,7),(3,6,9),(7,8,9)$

All 5 progressions have at least one even number so no such square is possible!

Finding the smallest odd magic square:

Since the square with $5$ in the center (and a total of $15$) is impossible. Let's try the square with $7$ in the center.

The sequences about 7 are $(1,13),(3,11),(5,9)$

From these we can pull the sequences $(1,3,5),(1,5,9),(9,11,13),(5,9,13)$

The middle elements will become the centers and the left/right elements will become the middles of the edges of the square. This implies $5$ and $9$ will be duplicated. So no distinct square exists with center element $e=7$

However for $e=9$ we can take the pairs about $9$: $(1,17),(3,15),(5,13),(7,11)$

Now note for the equations in $(7)$, each equation's last variable is the first variable of the equation following it. That is to say, we must pull 4 sequences from the elements of the pairs above such that they form a sort of loop with their first and last elements. The right sequence can be found with a small bit of work:
$$\rightarrow(1,7,13)\rightarrow(13,15,17)\rightarrow(17,11,5)\rightarrow(5,3,1)\rightarrow$$

Now we construct the square, starting with $(1,7,13)$:
$$a,[1],c$$
$$[13],[9],f$$
$$g,h,[7]$$
Then $(13,15,17)$:
$$a,1,[15]$$
$$[13],9,f$$
$$g,[17],7$$
Then $(17,11,5)$:
$$[11],1,15$$
$$13,9,[5]$$
$$g,[17],7$$
Finally $(5,3,1)$:
$$11,[1],15$$
$$13,9,[5]$$
$$[3],17,7$$
And we're done. The smallest all odd magic square has $r=27$
$$11,1,15$$
$$13,9,5$$
$$3,17,7$$

You can always arrange (rotate, reflect, etc.) the numbers in a magic square so that the smallest is top-middle and the next is bottom-right. Then there are two possible arrays of ranks of the $9$ numbers.
\begin{array}{cc}
   \text{TYPE 1} & \text{TYPE 0} \\
      \begin{pmatrix}
         8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2
      \end{pmatrix}
   &
      \begin{pmatrix}
         8 & 1 & 7 \\ 4 & 5 & 6 \\ 3 & 9 & 2
      \end{pmatrix}
\end{array}
The first is itself a magic square, the second is not. The nicest examples of a TYPE$1$ and a TYPE $0$ magic square are
$$
   M_1 = \begin{pmatrix}
         8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2
         \end{pmatrix}
  \qquad
   M_0 = \begin{pmatrix}
         8 & 0 & 7 \\ 4 & 5 & 6 \\ 3 & 10 & 2
         \end{pmatrix}
$$

The first $9$ odd numbers are
$$1,3,5,7,\color{red}9,11,13,15,17$$
with $9$ in the middle. So the smallest magic sum has to be $3\times 9 = 27$.
That means that $r=15$ is not possible.
The smallest TYPE$1$ magic square would be
$$2M_1 - 1 = \begin{pmatrix}
         15 & 1 & 11 \\ 5 & 9 & 13 \\ 7 & 17 & 3
         \end{pmatrix}$$
The smallest TYPE$0$ magic square would be
$$2M_0 + 1 = \begin{pmatrix}
                17 & 1 & 15 \\ 9 & 11 & 13 \\ 7 & 21 & 5
             \end{pmatrix}$$


prove by induction that $(a^n - b^n)$ is a multiple of $(a - b)$ for $n \geq 1$ Okay so I just finished a final in discrete mathematics and I could not figure out how to finish this proof:

"Prove by mathematical induction, that $(a^n - b^n)$ is a multiple of $(a - b)$ when $a$ and $b$ are integers and $n \geq 1$.

$\underline{Base case: n = 1}$

$(a^1 - b^1) = (a - b)x$

$x = 1$

$a - b = a - b$

$\underline{Inductive Hypothesis:}$

$n = k$ for some $k$

$(a^k - b^k) = (a - b)x$

$\underline{Inductive Step:}$

$n = k + 1$

$(a^{k + 1} - b^{k + 1}) = (a - b)x_1$

$(a \cdot a^{k}) - (b \cdot b^{k}) = (a - b) x_1$

$a \cdot [(a - b)x + b^k] + b \cdot [(a - b)x - a^k]= (a-b)x_1$

$(a -b)xa + ab^k + (a - b)xb - ba^k = (a - b)x_1$

$(a -b)xa + (a - b)xb - ab^k + ba^k = (a - b)x_1$

$(a -b)xa + (a - b)xb + ba^k - ab^k = (a - b)x_1$

.....

I don't know if I did it right but I couldn't get any further than this.


Find the sum of the series of $\frac{ 1}{n(n-2)!}$ from $n=2$ to infinity I am trying to find the sum of the series from $n=2$ to infinity of $\frac{ 1}{n(n-2)!}$

I was originally given: 1/2 + 1/3 + 1/8 + 1/30 + 1/144 ... ? and I had to figure out the next term (which is 1/840) and then figure out the sum of the series. I deduced what the formula is to make the series but I have no clue how to find the sum. (Brute forcing the sum on a calculator gave me a number very very close to 1 so I'm guessing that's the sum, but I don't know how to find that sum using calculus) Please help!


Prove that $\dim(V)$ is even Let $V$ be a finite dimensional vector space. Let $A_1,A_2: V\rightarrow V$ be commuting linear operators such that $A_1+A_2=-I$ where $I$ is the identity operator. Also $A_1,A_2$ have no negative eigenvalues. Prove that $\dim(V)$ is even.

How to approach this problem?

(Updated)

The assumption that $A_1$ and $A_2$ commute is superfluous. This already follows from $A_1+A_2=-I$.

It is tacitly assumed that the ground field is ${\mathbb R}$; for otherwise $A_1=[i]$, $A_2=[-1-i]$ would be a counterexample.

Now for the hint: Show that $A_1$ and $A_2$ can have no real eigenvalues at all, and think about the degree of the characteristic polynomial: it has to be even.

$A_1 + A_2 = -I$ => $A_2 = -I - A_1$

$ A_1A_2 - A_2A_1 = A_1(-I-A_1) - (-I-A_1)A_1 = 0 $

So for every $A_1$ there is an $A_2$ to commute with.

If $\lambda > 0$ is an eigenvalue of $A_1$ and s is eigenvector:

$A_2s = (-I-A_1)s = -(1+\lambda)s $
So if $A_1$ has only positive eignevalues, $A_2$ has only negative ones.


Reverse Triangle Inequality Proof I've seen the full proof of the Triangle Inequality
\begin{equation*}
|x+y|\le|x|+|y|.
\end{equation*}
However, I haven't seen the proof of the reverse triangle inequality:
\begin{equation*}
||x|-|y||\le|x-y|.
\end{equation*}
Would you please prove this using only the Triangle Inequality above?

Thank you very much.

$$|x| + |y -x| \ge |x + y -x| = |y|$$

$$|y| + |x -y| \ge |y + x -y| = |x|$$

Move $|x|$ to the right hand side in the first inequality and $|y|$ to the right hand side in the second inequality. We get

$$|y -x| \ge |y| - |x|$$

$$|x -y| \ge |x| -|y|.$$

From absolute value properties, we know that $|y-x| = |x-y|,$ and if $t \ge a$ and $t \ge −a$ then $t \ge |a|$.

Combining these two facts together, we get the reverse triangle inequality:

$$|x-y| \ge \bigl||x|-|y|\bigr|.$$

For all $x,y\in \mathbb{R}$, the triangle inequality gives
\begin{equation}
|x|=|x-y+y| \leq |x-y|+|y|,
\end{equation}

\begin{equation}
|x|-|y|\leq |x-y| \tag{1}.
\end{equation}
Interchaning $x\leftrightarrow y$ gives
\begin{equation}
|y|-|x| \leq |y-x|
\end{equation}
which when rearranged gives
\begin{equation}
-\left(|x|-|y|\right)\leq |x-y|. \tag{2}
\end{equation}
Now combining $(2)$ with $(1)$, gives
\begin{equation}
-|x-y| \leq |x|-|y| \leq |x-y|.
\end{equation}
This gives the desired result
\begin{equation}
\left||x|-|y|\right| \leq |x-y|. \blacksquare
\end{equation}


homeomorphic topology of quotient space of $S^1$ I got stuck on the problem about quotient space from General Topology of Stephen Williard. Here is the problem:

    Let $\sim$ be the equivalence relation $x \sim y$ iff $x$ and $y$ are diametrically opposite, on $S^1$. Which topology is the quotient space $S^1/\sim$ homeomorphic to?

I tried to build a continuous function $S^1$ such that 2 points which are diametrically have the same images, but I couldn't find it. For each point in $S^1$, we can write it as $(\cos(\phi), \sin(\phi))$, then what is the function which satisfies the previous requirement. Can anyone help me with this? I really appreciate.

Take a rubber band; that’s your $S^1$. Now fold it into a figure eight: 8. Finally, fold the $8$ about its horizontal midline to get a double circle. Thus, you’ve gone from O to 8 to doubled o. Check that this brings diametrically opposite points of the original O together: they’re simply on different copies of the o.

The space you will get is $\mathbb{RP}^1$, the one dimensional real projective space which is by definition of all lines through the origin in $\mathbb{R}^2$.


Example for algebraic homomorphism between topological groups which is not continuous I am quite sure there should be an easy example of:

(algebraic) homomorphism between topological groups which is not continuous.

However, I do not see one immediately.

The identity map from the real numbers with the usual topology to the real numbers with the discrete topology.

Consider
$$\begin{align}f\colon\quad\mathbb Q[\sqrt 2]&\to \mathbb Q[\sqrt 2],\\ a+b\sqrt 2&\mapsto a-b\sqrt 2\end{align}$$
where the topology is that inherited from $\Bbb R$.

   Define two sequences of integers as follows: Let $a_1=-1,b_1=1$ and  recursively $a_{n+1}=a_n^2+2b_n^2, b_{n+1}=2a_nb_n$. Then with $x_n:=a_n+b_n\sqrt 2$ we have $x_{n+1}=x_n^2$ and from $|x_1|=\sqrt 2-1<1$ we see that $x_n\to 0$. On the other hand $x_nf(x_n)=a_n^2-2b_n^2\ne 0$ because $\sqrt 2$ is irrational and $a_n=b_n=0$ (i.e., $x_n=0$) does not occur. We conclude $|f(x_n)|=\frac{|a_n^2-2b_n^2|}{|x_n|}\ge \frac1{|x_n|}\to \infty$. If $f$ were continuous, we'd expect $f(x_n)\to f(0)=0$ instead.


Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime) I want to show without using Sylow theorem that

  Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)

My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$ Then $H=\{e,a\}\leq G.$ Let $S$ be the set of all left cosets of $H$ in $G.$ Now $S$ forms a group w.r.t. the composition $g_1Hg_2H=(g_1g_2)H.$ The mapping $\sigma:S\to G:gH\mapsto g$ is an $1-1$ homomorphism. Then $S\simeq\sigma(S)\leq G.$ Thus $G$ has a subgroup of order $p.$

Please tell me whether I'm right!

  I'm skecptical as I didn't use that $p$ is prime.

We only need to consider the case that $p$ is odd.If there is no element of order $p$ in the group, then every non-identity element of the group has order $2$. But a group in which every non-identity element as order $2$ is Abelian:
($1 = abab,$ so $aba = b$ and $ba = ab).$ The group has more than one subgroup of order $2,$ so since it is Abelian, it has a subgroup of order $4,$ contrary to Lagrange's theorem.

Why the set of left cosets of $H$ forms a group? The question is a special case of Every group has a subgroup of prime order?


Showing a sequence defined recursively is convergent Given the recursively defined sequence

$$ a_1 = 0, a_{n+1} = \frac 1{2+a^n} $$

Show it converges. I'm working with Cauchy sequences, and proved in a previous question that any sequence of real numbers $(a_n)$ satisfying:

$$|a_n - a_{n+1}| \leq \frac1{2^n} $$

is convergent (by showing $(a_n)$ was Cauchy), and was told that it may be helpful to use induction to prove that the above recursive sequence satisfies the inequality and is therefore convergent.

Work:

Base Case: When $n = 1 , a_1 = 0, a_2 = \frac12,$ so $$ \left|0-\frac12\right| \leq \frac1{2^{1}}$$ and the base case checks out

Inductive Step: Want to show $$|a_{n+1} - a_{n+2}| \leq \frac1{2^{n+1}} $$

So $$\left|\frac1{2+a_n} - \frac1{2+a_{n+1}}\right| \leq \frac1{2^{n+1}}$$

And I'm not sure where to proceed from here. Any help would be greatly appreciated, thanks.

To prove the sequence is convergent using the inequality

$$ |a_{n+1}-a_n| < \frac{1}{2^n} $$

you need the fact which states that "if $\sum_{n} |a_{n+1}-a_n|< \infty $ then the sequence $a_n$ is Cauchy" which implies the sequence is convergent.


Deriving $\frac{\partial{v}}{\partial{z}}$ using chain rule A problem from one of past exams at my uni reads:

  Suppose $xy^2 + yz^2 = 2$ and $v = xy^3$ where $x, y, z$ and $v$ are all variables. Using a chain rule, express $\partial{v}/\partial{z}$ as a function of $x, y, z$.

I just don't think I really get multivariable calculus... I know (?) I can write $$\frac{\partial v}{\partial z}= \frac{\partial v}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial z}$$, and clearly $\frac{\partial v}{\partial x}=y^3$ and $\frac{\partial v}{\partial y}=3xy^2$, so that $$\frac{\partial v}{\partial z}= y^3\frac{\partial x}{\partial z}+3xy^2\frac{\partial y}{\partial z}.$$ Now, $xy^2+yz^2=2$, so $x=\frac{2}{y^2}-\frac{z^2}{y}$, and I probably can write $$\frac{\partial x}{\partial z} = -2\frac{z}{y}, $$ but how do I go about deriving $\frac{\partial y}{\partial z}$? Or does all of it make no sense and I should proceed in a completely different way?

We have that $xy^2+yz^2 = 2$. Then,

$$\dfrac{\partial}{\partial z} (xy^2+yz^2) = \dfrac{\partial}{\partial z}2 =0$$

Taking the partial derivative will yield

$$\begin{align}
\dfrac{\partial}{\partial z} (xy^2+yz^2) &=y^2\dfrac{\partial x}{\partial z}+x\dfrac{\partial}{\partial z}y^2 + z^2\dfrac{\partial y}{\partial z}+y\dfrac{\partial}{\partial z}z^2
\\
&=-2zy+2xy\dfrac{\partial y}{\partial z} + z^2\dfrac{\partial y}{\partial z}+2zy
\\
&=0
\end{align}$$

Thus, we have that $$\dfrac{\partial y}{\partial z}=0$$


checking if a function is positive using Fourier coefficients Given a function $$f(x) = \sum_{k=0}^N a_k\ \sin(k\pi x)$$ defined over the region $S = [0, 1]$, is there some way to check if $f(x) \geq 0$ for all $x \in S$ using the coefficients $\{ a_k;\ k \leq N \}$? In particular, I was hoping there'd some inequality involving the coefficients that'll enable us to check this quickly.

I do know a few cases in which positivity can be proved. For instance if for each $k$, $a_k \gt 2 a_{k+1}$. A more general situation in which positivity holds is if for some $s \in (0,1)$ one can write $a_k=c_k s^k$ where $c_k-(1+s)c_{k+1}+s c_{k+2} \gt 0$.


Relation between roots and coefficients - manipulation of identities The polynomial $x^3+3x^2-2x+1$ has roots $\alpha, \beta, \gamma$ . Find $$\alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)$$

I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.

Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta + \beta \gamma + \gamma \alpha)x - \alpha\beta \gamma = x^3+3x^2-2x + 1
$$
that
$$
\alpha+\beta+\gamma = -3, \quad \alpha\beta + \beta \gamma + \gamma \alpha = -2, \quad
\alpha\beta \gamma = -1.
$$
Note that
$$
-2\alpha = \alpha(\alpha\beta + \beta \gamma + \gamma \alpha) = \alpha^2(\beta+\gamma) +\alpha\beta\gamma = \alpha^2(\beta+\gamma) - 1.
$$
Thus $\alpha^2(\beta + \gamma)=1-2\alpha$.
Similarly, $\beta^2(\alpha + \gamma) = 1-2\beta$ and $\gamma^2(\alpha + \beta) = 1-2\gamma$.
Therefore,
\begin{align}
\alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)
&= (1-2\alpha) + (1-2\beta) + (1-2\gamma) \\
&= 3 -2(\alpha+\beta+\gamma) = 3 +6 =9.
\end{align}


How do you derive a normal vector from the equation of a line? I've seen that to find a normal vector to a line such as $3x+4y-1=0$ people take the coefficients and say $(3,4)$ is a normal vector?

Why does this work? How are the coefficients related to the normal?

Two vectors $a,b$ are normal iff $a\cdot b=a_1b_1+...+a_2b_2=0$.

If you have a vector, $(x,y)$ and you want to find vectors that are normal to it you want to find a vector $(a,b)$ such that $ax+by=0$. So a normal vector to the line $3x+4y=0$ is simply $(3,4)$.

If you have a constant on the right side, it just moves the line up or down. It doesn't change anything else. So a normal vector will still be $(3,4)$.

You want to understand this stuff? There's an infinite number of lines (in 2 dims) that rise 7 whenever they run 3. Draw any one of them. Call it L1.

Now, there's an infinite number of lines that are perpendicular to L1. Draw any one. Now use geometry (basic triangle laws) to answer this about this second line: how much does it rise when it runs 7?

--

Added: It's not clear what level you are at, but you will need proficiency with both the $y = mx + b$ and $Ax + By + C = 0$ forms of equations of a line (and converting between them) in order to benefit from my suggestion.


Show that $(a,b)+(c,d) = (a+c,b+d)$ Let $a<b$ and $c<d$ be real numbers. Show that $(a,b)+(c,d) = (a+c,b+d)$.

I don't understand the question. Since $(a,b)$ and $(c,d)$ are intervals, what does it mean to add them?

Hint: The set inclusion $(a,b) + (c,d)\subset (a+c, b+d)$ is easy to prove. The reverse inclusion is trickier. For the reverse inclusion, first prove the simpler case $a=c=0$, then apply this result to the general case, using the identity
$$(a,b)+(c,d)=\{a\}+(0,b-a)+\{c\} + (0,d-c).$$

Let function $f : [a,b] \times [c,d] \to \mathbb R$ be defined by

$$f (x,y) := x + y$$

We can find the minimum and maximum of $f$ by solving two linear programs, namely,

$$\begin{array}{ll} \text{minimize} & x+y\\ \text{subject to} & a \leq x \leq b\\ & c \leq y \leq d\end{array}$$

and

$$\begin{array}{ll} \text{maximize} & x+y\\ \text{subject to} & a \leq x \leq b\\ & c \leq y \leq d\end{array}$$

Since $\nabla f (x,y) = 1_2$, we conclude that:

the minimum of $f$ is $a+c$, which is attained at $(x,y) = (a,c)$.
the maximum of $f$ is $b+d$, which is attained at $(x,y) = (b,d)$.

Thus, interval addition is defined as follows

$$[a,b] + [c,d] := [a+c,b+d]$$


time and work problem A man can begin a work at his maximum rate; but afterwards the rate at which he works follows a cyclic pattern. Every 2 hr,it reduces by half but after 8 hrs,it comes back to its maximum level.He can complete a job in 151 Hrs at his maximum rate.How many hours would he take to complete the job if his rate follows the cyclic pattern?

Options are:

A. 301
B. 601
C. 641
D. 321

i am unable to understand this question.

Please help

The pattern is the following :

0h $\rightarrow$ 2h : Work rate = 100% efficiency
2h $\rightarrow$ 4h : Work rate = 50% efficiency
4h $\rightarrow$ 6h : Work rate = 25% efficiency
6h $\rightarrow$ 8h : Work rate = 12.5% efficiency

And every time he reaches 8 hours of work, he gets back to 0h and works at 100% efficiency for 2 hours, etc.

The question is : if the worker was working at 100% every time, he would need 151 hours to get the job done. However he doesn't work at 100% every time, so you have to use the cycle to determine how long it would take.

$a= \text{maximum efficiency}$
$$X\times \frac{(a+\frac{a}{2}+\frac{a}{4}+\frac{a}{8})}{4} = a\times 151$$ solving for $x$ we get $$x=151\times \frac{32}{15} =321$$
$X$ gives us the effective working time


Fundamental period of piecewise defined function. Fundamental period of the function $f(x) = \left\{\begin{matrix}
 2\;\;\;\;,x\in \mathbb{Q} \\\\
-2\;\;\;\;,x\notin \mathbb{Q} &
\end{matrix}\right.$ is

As we know that function $f(x)$ is periodic function, If it satisfy the relation $f(x+T)=f(x)$

and smallest positive value of $T$ is called period of that function $f(x)$.

But here i did not understand How can i calculate period of that constant function

Help me,Thanks

This function has no fundamental period. Suppose that $T$ were such a period. Then it would have to be rational, for $f(0) = F(T) = 2$. Consider $T' = T/2$. This, too, will be a period, for if $x$ is rational, so is $x + T'$, and if $x$ is irrational, so is $x + T'$.

In fact, the numbers $T = 1, 1/2, 1/4, \ldots$ are all periods of this function, so there's no smallest positive period.

As per definition $T$ for which $f\left(x\right)=f\left(x+T\right)$ Other than $T=0$ is called period of $f\left(x\right)$
For constant Function $f\left(x\right)=c$ ,$f\left(x\right)=f\left(x+T\right)$
For all T thus we can say that constant function is periodic but choice of T is ours


check the convergence of the series So I have the following series:
$$\sum_{n=2}^{\infty}\frac{e^n}{e^{n\sqrt[n]{n}}(ln(n))^2}$$
I thought that this series does not converge. $\sqrt[n]{n}\rightarrow 1$ so I thought that for large $n$ it will act like $\sum\frac{1}{ln(n)^2}$.
But I checked this with wolframalpha and it told me that this series converges.

Where did I make a mistake?

Note that $\sqrt[n]{n}=\exp((\ln n)/n)\geqslant1+(\ln n)/n$ hence $n\sqrt[n]{n}\geqslant n+\ln n$ and the $n$th term of the series is at most $$\frac{\mathrm e^n}{\mathrm e^{n+\ln n}(\ln n)^2}=\frac1{n(\ln n)^2},$$ from which the convergence of the series follows.

  Where did I make a mistake?

When you "deduced" from $\sqrt[n]{n}\to1$ (true) that $\mathrm e^{n\sqrt[n]{n}}/\mathrm e^n\to1$ (false).


Find an example of a set $A$ and functions $f :B\to B$ and $g:B\to B$ such that $f\circ g=id_B$ but $g\circ f\ne id_B$ Find an example of a set $A$ and functions $f :B\to B$ and $g:B\to B$ such that $f\circ g=id_B$ but $g\circ f\ne id_B$.

I am not sure what $id_B$ means, so I don't understand how to find a set of functions for this. What is $id_B$ and how do you find functions for it?

I know what it means to find functions such that $f\circ g=f$ and similar to this, but now idea how to do this for $id_B$.

An example for $f\circ g=f$ is
$$
f(x)=x+3; \qquad g(x)=x
$$
then $f\circ g=f(g(x))=f(x)=x+3$.

Is there an easier way to find these than trying random equation?

(I'm reading $A$ as $B$.)

For any set $X$ the map ${\rm id}_X$ is the identity map of $X$, and is defined as
$${\rm id}_X:\quad X\to X,\qquad x\mapsto{\rm id}(x):=x\ .$$
Here is an example of the required kind:

Let $B:={\mathbb N}_{\geq0}=\{0,1,2,3,4,\ldots\}$, and define
$$g:\ B\to B,\quad k\mapsto g(k):=2k;\qquad f:\>B\to B,\quad n\mapsto f(n):=\left\lfloor{n\over2}\right\rfloor\ .$$
Then $f\circ g={\rm id}_B$. On the other hand $g\circ f\ne{\rm id}(B)$ since $g\circ f(1)=0\ne1$.

Hint: If $f \circ g = \textsf{id}_B$, then it follows that $f$ is surjective (onto) and $g$ is injective (one-to-one). Thus, try finding a set $B$ and any function that is surjective but not injective (or injective but not surjective). Work out a pseudo-inverse for the other function, and you're done!


For every sufficiently large $m$ there exists $k$ such that $m = k + \tau(k)$ Let $\tau(k)$ , be the number of positive divisors of natural number $k$. Is it true, that there exists $n_0$ , such that for every $m\geq n_0$ there exists $k  \in\mathbb{N}$ such that:
$$
m = k + \tau(k)
$$

I have tried to use the following formula for $\tau$:
$$
\tau(p_1^{k_1}\ldots p_{s}^{k_s}) = (k_1 +1)\cdot\ldots \cdot(k_s +1),
$$
Where $p_1, \ldots, p_s$ are different prime numbers.

Intuitively, I think that the answer will be no. So, we can assume the contrary (that such $n_0$ exists) and try some infinite series of numbers (primes, factorials, primorials, etc.), which can't be written in the form $k + \tau(k)$ for every $k$. But my attempts weren't successful.

So, I will be grateful for hints and ideas.

contest problem related to divisor function
Look at Loppukilpailija's answer. I think he solved your problem negatively.

We are looking at, a specific composition. Included, is a multiplicative partition( number of divisors, in this case). Highly composite numbers are likely to have the highest number of multiplicative partitions due to highest divisor number up to a point. These may not be so dense on the naturals however. 12 divisors comes about in one of 4 ways:$$p^{11}\\p^5q\\p^3q^2\\p^2qr$$


choose from implication and logical and in write assertions in first-order logic I am a student and I get confused in translating some sentence to logic assertion.

For example: Joe does not have a lawyer, i.e. is not a customer of any lawyer.

The right way to translate is:

"For all $p$, if $p$ is lawyer then Joe is not a customer of $p$"

or

"For all $p$, if $p$ is lawyer ∧ Joe is not a customer of $p$"

What the difference between these two clause?

If you are having problems with this sort of very very elementary point, that suggests you need to do some general homework (a one-off answer to this particular question won't get to the roots of your difficulty). Literally dozens of people have written excellent introductory logic books covering exactly this kind of thing, with detailed chapters on question of translation to and from ordinary language to regimented first-order formal languages. You need to be spending some time carefully reading the relevant chapters of such a book. If you are a student, get yourself to the library. It will be time well-spent.

I'd recommend, for example, the coverage in P-t-r Sm-th's Introduction to Formal Logic. Though having no access to a library is no excuse either, as various texts are available online. For example, Paul Teller's Formal Logic Primer is freely available from his website and is also very good on such questions of translation and the principles behind them.


Under what conditions does $\lim (a_n)^{1/n}$ converge? Under what conditions does $\lim (a_n)^{1/n}$ converge?  I was motivated by trying to solve limits and it would be quite useful to know if this converges and when $\lim(a_n)^{1/n}=\lim(a_n)$, or $\lim(a_n)^{1/n}=c$ for some constant $c$.

I have narrowed this down a little bit. This is what I have figured out so far:

If $a_n$ is constant and $a_n>0$ then this converges to $1$, if $a_n=0$, this converges to $0$,
If $a_n$ grows very large it clearly does not hold, for example if $a_n=e^{e^{e^n}}$ for all $n$ then the sequence diverges,
If $a_n=n$ for all $n$, or $a_n=cn$ for some constant $c$, then $\lim (a_n)^{1/n}=1$,
If $a_n$ is of the form $(b_n)^n$ then this will obviously converge iff $(b_n)$ does, and will share its limit; this way, for any $x\in\mathbb{R}_+$, I could construct sequences $(a_n)$ such that $\lim(a_n)^{1/n}=x$,
If $a_n$ is bounded and positive then this will converge to $1$.

I wonder whether there is a general rule for this?

(All limits in this question are taken as $n$ tends to $+\infty$.)

Thank you!

Assume that $a_n^{1/n}$ converges to some number $A$. As pointed out in the other answer, we must have $A \geq 0$, since otherwise the exponent $\frac1n$ doesn't make sense.

Now let $e > 0$, and look at $b_n = (A+e)^n$. I claim that $b_n > a_n$ for all sufficiently large $n$. Proof: By definition of limit, we have that $|a_n^{1/n} - A| < \frac e2$ for all $n$ larger than some natural number $N$. But for any such $n$ we have
$$
a_n^{1/n} < A+\frac{e}2 < A+e < b_n^{1/n}\\
a_n < b_n
$$
This means that if $a_n^{1/n}$ converges, then there must be some number $B$ such that $a_n < B^n$, for all sufficiently large $n$, which again implies that there is such a $B$ that works for all $n$.

Well, first $a_n$ has to be positive if $n$ is even, otherwise $a_n^{1/n}$ wouldn't make sense. Write down $a_n=e^{log(a_n)}$, then, since the exponential is continuous and has continuous inverse, it follows that $(a_n^{1/n})_{n\in\mathbb N}$ converges if and only if $\left(\frac{log(a_n)}{n}\right)_{n\in\mathbb N}$ does.


Computing $\bigcup_{n=1}^\infty C_n$ and $\bigcap_{n=1}^\infty C_n$ for $C_{n}=\{(x,y) \in \mathbb{R^2} : y^{n} \le x \}$ I am looking for some help to understand the following because I am feeling really lost.

Say $$C_{n}=\{(x,y) \in \mathbb{R^2} : y^{n} \le x \}$$

then how can we find

$$\bigcup_{n=1}^{\infty}C_{n}$$

and

$$\bigcap_{n=1}^{\infty}C_{n}$$

My thoughts were to start off by listing some such as

$$C_{1}=\{(x,y) \in \mathbb{R^2} : y \le x \}$$
$$C_{2}=\{(x,y) \in \mathbb{R^2} : y^{2} \le x \}$$
$$C_{3}=\{(x,y) \in \mathbb{R^2} : y^{3} \le x \}$$

for example, and drawing out the graphs. I also solved for x in those cases to see if I could use that.

I am just quite lost, I am not sure the way to best approach this.

Overview

Here is a visualization using the GeoGebra software:

The purple curve and area is $y^{101} \le x$.
The blue curve and area is $y \le x$.

Union

First candidate for the union seems to be the union of the coloured areas here:

To get the two wanted sets precise one seems to need a good case distinction for each.

First we have $C_1 = \{ (x,y) \in \mathbb{R}^2 \mid y \le x \}$, thus the line $y = x$ and all points below
For $x \in I_1 = (-\infty, -1)$ we note that the sets $C_k$ with odd $k$ (shown is $C_3 = \{ (x,y) \in \mathbb{R}^2 \mid y^3 \le x \}$ in green) approach the half line $L_1 = \{ (x,-1) \mid x \in I_1 \}$ from below but do not reach $L_1$. That boundary $L_1$ is open. I used red dotted lines in the image above for such.
For $x \in I_2 = (0,1)$ we note that the sets $C_k$ (all $k$) cover more and more of the area bounded by $L_2 = \{ (0,y) \mid y \in (0,1] \}$ and $L_3 = \{ (x,1) \mid x \in [0,1) \}$ and $y=x$ but do not reach $L_2$ or $L_3$.

Intersection

First candidate for the intersection seems to be the area bounded by the purple curves.

It is based on this overview:

Shown are $C_2 = \{ (x,y) \in \mathbb{R}^2 \mid y^2 \le x \}$ in cyan and
$C_4 = \{ (x,y) \in \mathbb{R}^2 \mid y^4 \le x \}$ in green.

On $I_2 = (0,1)$ for positive $y$ the line $y=x$ is the boundary and included in the intersection. For negative $y$ on $I_2$ the line $y^2 = x$ is the boundary.

On $I_3 = (1, \infty)$ all points from $I_3 \times [-1,1]$ are part of the intersection. This area is included in all $C_k$, especially the $C_k$ with even $k$ approach it from outside.

The way you are approaching it is correct.

Identify the graphs that matter
Sketch in the regions of the graphs, which correspond to points in the set
Take the union/intersection of those regions

The function in question here is $y^n$, for all $n\ge1$. This takes on three major different forms, which you'll discover if you graph $y$ to a few powers.

Another helpful tip about these functions is that they always pass through (1,1) and (0,0), no matter how fast they grow.


Bicycle Route Optimization Puzzle I tried to make some expressions about where each person stops the bike, but I couldn't solve it :(
There are three people who would like to cross the road.

It takes $a$ minutes for the first person to cross the road by walking.

It takes $b$ minutes for the second person to do so.

It takes $c$ minutes for the third person to do so.

They also have one bike - all three of them can cross the road in $k$ minutes - where $k<a,b,c$.

However, only one person at a time can ride the bike.

Now they want to cross the road as fast as possible. What is the shortest time possible for all three of them to arrive at the end of the road?

Please note, that these people cannot go outside the road.

The below solutions have some cases where people go outside the road - also possibly implying negative time.

First, it will be useful to notice two principles

All walkers, no matter how many they are, have to get to the end of the road at the same time, always. If some walker got before another, it could have given some of her "bicycle time" to the walker that arrived the last to decrease the total time.
It doesn't make sense for any walker at any time to go backwards to give the bicycle to other walker. Going back is less efficient than just letting the bicycle in the road and keep walking.

The total crossing time is determined by the proportion of walking to riding time, that should be adjusted so that all walkers get to the end of the road at the same moment.

If you let me, I will use speeds rather than times. We'll call the speeds of the walkers $v_1,v_2,v_3$ and the one of the bicycle $V$. Also, lets assume, without loss of generality, that
$$v_1<v_2<v_3<V$$
The simplest procedure the walkers could choose to adjust their riding to walking ratio is the following. The slowest walker takes the bicycle up to a distance $d_{c1}$ (the $c$ is for "cut") then leaves it and walks through the end $d$. The middle walker walks until $d_{c1}$ then takes the bicycle up to $d_{c2}$ and walks to the end. Finally, the fastest walker walks to $d_{c2}$ and takes the bicycle to the end.
We'll call the total crossing times for the three walkers $t_1,t_2,t_3$. They have the following expressions
$$t_1=\frac{d_{c1}}{V}+\frac{d-d_{c1}}{v_1}$$
$$t_2=\frac{d_{c1}}{v_1}+\frac{d_{c2}-d_{c1}}{V}+\frac{d-d_{c2}}{v_2}$$
$$t_3=\frac{d_{c2}}{v_3}+\frac{d-d_{c2}}{v_2}$$
Now, our first principle sets the constraints $t_1=t_2=t_3=t_T$ (total time). With this we have three linear equations to determine three unknowns, $t_T,d_{c1}$ and $d_{c2}$. Actually I just put the equations in Mathematica and obtained

$$t=\frac{d}{V}\frac{2V^3+v_1v_2v_3-V^2(v_1+v_2+v_3)}{3 v_1v_2v_3+V^2(v_1+v_2+v_3)-2V(v_1v_2+v_1v_3+v_2v_3)}$$
$$d_{c1}=d\frac{-2Vv_1v_3+v_1v_2v_3+V^2(-v_1+v_2+v_3)}{3 v_1v_2v_3+V^2(v_1+v_2+v_3)-2V(v_1v_2+v_1v_3+v_2v_3)}$$
$$d_{c2}=2d\frac{(V-v_1)(V-v2)v_3}{3 v_1v_2v_3+V^2(v_1+v_2+v_3)-2V(v_1v_2+v_1v_3+v_2v_3)}$$

The most interesting point is that the total time is symmetric to permutations of $v_1,v_2,v_3$, i.e. we could have made the fastest or the middle walker to take the bicycle first, it doesn't matter, as long as the bicycle times proportions are kept. Besides the grueling algebra, that could be perhaps simplified with some tricks, I think this problem would be easy to generalize for more walkers.

WARNING

The formula given for Case$1$ doesn't necessarily give the optimal time, trial and error can throw up better answers (see comments below the answer), but the formula for Case $2$ should be ok, as all reach the end simultaneously.

There are two assumptions I make at start:

(i) the bike is faster than each of the walkers (reasonable !)

(ii) zero time is wasted in switching from bike to walk or vice versa (unreasonable but customary !)

WLOG, I take time to bike entire distance as $1$, and that of the walkers as $a,b,c,\;\; a < b < c$

[ The cases where two or three have the same walking speed are progressively simpler cases ]

First consider the two slowest walkers.

Let (say) the faster one start with the bike, leave it at fraction $f$ of the distance, and walk the rest of the way, whereas the other walker starts by walking, and rides the remaining distance on reaching the bike.

Let $t$ be the time needed for both of them to simultaneously reach the end, then

$f + (1-f)b = (1-f) + f*c = t\;\; \Rightarrow f = \dfrac{b-1}{b+c-2}$

and $\;t = \dfrac{bc-1}{b+c-2}$

Two cases can now arise:

Case $1:\;t > a$

Example:$\; a = 1.5, b = 2, c= 20 \to t= 1.95$

Decision rule: Let $a$ walk the whole distance. Optimal group reach time $= 1.95$

Case $2:\;t \le a$

Example:$\; a = 1.5, b = 1.6, c= 1.7 \to t= 1.323$ (considering only $b$ and $c$)

Decision rule: All three should share the bike for appropriate fractions of the distance and arrive simultaneously.

Appropriate fractions

Let $f_1,f_2,f_3$ be the fractions of full distance the bike is used by $a,b,c$
and the time taken for them to reach simultaneously be $t$

$f_1 + (1-f_1)a = t \to f_1 = \dfrac{a-t}{a-1}$
$f_2 + (1-f_2)b = t \to f_2 = \dfrac{b-t}{b-1}$
$f_3 + (1-f_3)c = t \to f_3 = \dfrac{c-t}{c-1}$

Now $f_1+f_2+f_3 = 1$, which yields:

$t = \dfrac{(2 a b c - a b  - a c
 - b c  + 1)}{(a b + a c + b c - 2 a  - 2 b  - 2 c  + 3 )}$

Using the above decision rule brings down the optimal time from $1.5$ (due to laggard $a$ to $\approx 1.3925$


Propositional Logic - List of N integers Today in our propositional-calculus class, we were asked this question:

Suppose you have a list of N integers: you write a computer program to
analyze the list. If every integer in the list is greater than 10,
then the returned output is the value TRUE. Else, the output is the
value FALSE. If N = 0, and therefore, the list is empty, what is the
output?

While explaining how to interpret the problem, my teacher said that the returned value is FALSE, because, since the list is empty, there aren't any counterexamples against that statement.
I didn't seem to grasp that. First of all, isn't the else statement a condition in itself? Namely, the condition:

If there's at least one number less than (or equal to) 10 [...]

There aren't any counterexamples to that either, so why favouring a condition over the other?
Honestly, my solution (as wrong as it may be) would be to claim that the expressed conditions simply do not cover the "empty list" case, so the output value cannot be determined with the information we possess.
Otherwise, is the "else" to be interpreted as "in every other case"? In this eventuality, the output would be FALSE, but to be fair, I'm not sure of this answer.
So, what is the solution?

I suspect your teacher just got muddled with which output represents what. If there are is a counterexample to the statement "every integer in the list is greater than 10", it should output FALSE. Since there can't be a counterexample, it should output TRUE. So the reasoning is correct, but your teacher has linked it to the other output, presumably by mistake.
FWIW if I was to program this in python, it would look like
def all_greater(mylist):
    for x in mylist:
        if x<10:
            return False
    return True

This will return True if the list is empty, because the loop will execute 0 times.

It depends on how the code is written. I'm pretty sure it can be False or True depending on the code itself.


Probability in Minesweeper Suppose I click a random tile during a Minesweeper game. It is a 1. During each time I click an adjacent square, what are the chances of hitting a mine? How would this change if it were a 2 or another number?

I think this is a fantastic question, because you actually arrive at a pretty good strategy for how to begin a game. Sorry it's such a long answer, I try to be thorough so you can see this is the correct answer - there is a summary at the end. I'm going to rephrase your question in the way I interpret what you mean.

At the start of a game of minesweeper, if you click adjacent cells what is the chance of there being a mine in a given cell around these clicked cells? Is there any way to use this for a good opening strategy?

Firstly I'd like to define the following parameters:

$n$ is the number of mines in a grid;
$N$ is the number of cells in a grid;
$\rho=n/N$ is the density of mines in a grid, and the probability of hitting a mine in a random cell.
$$\begin{array}{c c c c}
\mathrm{Difficulty}   & n  &  N  & \rho\\
\hline
\mathrm{Beginner}     & 10 &  64 & 0.156\\
\mathrm{Intermediate} & 40 & 256 & 0.156\\
\mathrm{Expert}       & 99 & 480 & 0.206
\end{array}$$

Secondly I'd like to define the following random variables and events:

$C=\bigcup_i C_i$ is the event of the grid configuration, where $C_i$ is the event of a particular mine configuration (see below). Note that the $C_i$ are disjoint events, so $\mathbb{P}(C)=\sum_i\mathbb{P}(C_i)$.
$X_i=\{\mathrm{mine\ in\ cell\ }i|C\}$ is the event there is a mine in cell $i$, given the situation of the board (these are the probabilities we want to find).
$Y_i$ is a random variable representing the number of mines in a group of cells as specified later on. $\mathbf{Y}$ is the corresponding vector.

Let's assume we neither hit a mine or an opening, as these events are trivial. Instead, let's start by supposing our first click, placed somewhere in the middle of the grid, reveals a 1, which is the most likely case. The probability of any of the 8 surrounding cells containing a mine is $\frac{1}{8}=0.125$, which is lower than $\rho$ on any normal game, so the safest move is to choose an adjacent cell. If the first click were to reveal a 2, you should not click an adjacent cell as $\frac{2}{8}=0.25>\rho$. One possible resulting configuration is displayed below (I use '$(i)$' to label a cell as 'cell $i$').
$$\begin{array}{c c c c c c c}
_{(1)} &|& _{(4)}     &|& _{(5)}     &|& _{(8)}\\
\hline
_{(2)} &|& \mathbf{1} &|& \mathbf{1} &|& _{(9)}\\
\hline
_{(3)} &|& _{(6)}     &|& _{(7)}     &|& _{(10)}
\end{array}$$

The way to calculate the probabilities $\mathbb{P}(X_i)$ for $1\leq i\leq 10$, is to consider all possible configurations of mines in these 10 cells. Cells 1, 2 and 3 are effectively equivalent, as they have the same probabilities and are independent to the other cells in each configuration. In this case there are 3 such groups of cells, $\{1, 2, 3\}, \{4, 5, 6, 7\}, \{8, 9, 10\}$, which $Y_1, Y_2, Y_3$ refer to respectively.

There are two configurations: $\mathbf{Y}= (1,0,1)$ or $(0,1,0)$, which I refer to as $C_1$ and $C_2$ respectively, so $\mathbb{P}(C_1)=\mathbb{P}(Y_1=1,Y_2=0,Y_3=1)$. These are illustrated below.
$\begin{array}{*{7}c}
\star &|&   &|&   &|& \star\\ \hline
\star &|& 1 &|& 1 &|& \star\\ \hline
\star &|&   &|&   &|& \star
\end{array}\qquad$
$\begin{array}{*{7}c}
 &|& \star &|& \star &|& \\ \hline
 &|& 1     &|& 1     &|& \\ \hline
 &|& \star &|& \star &|&
\end{array}$

Now we use conditional probability, and the fact that if $Y_1=1$ then we are in configuration 1.
$$\begin{align}
\mathbb{P}(Y_1=1|C) &= \frac{\mathbb{P}(\{Y_1=1\}\cap C)}{\mathbb{P}(C)}\\
  &= \frac{\mathbb{P}(C_1)}{\mathbb{P}(C_1)+\mathbb{P}(C_2)}
\end{align}$$

To calculate $\mathbb{P}(C_1)$, think about what is required for this to occur - we need exactly 1 mine in a given 3 cells, none in a given 4 cells, and 1 in a different 3 cells. Now to simplify the problem, we will assume that the mine density doesn't change, which is a good approximation at least for large grids like expert. We get
$$\begin{align}
\mathbb{P}(C_1) &= \left[\binom{1}{3}\rho(1-\rho)^2\right] \times (1-\rho)^4 \times \left[\binom{1}{3}\rho(1-\rho)^2\right]\\
  &= 9\rho^2(1-\rho)^8,
\end{align}$$
which corresponds to two situations of choosing 1 mine in 3 cells, 8 cells with no mine, and 2 cells with mines. Thinking of it this way makes it much easier to just write down probabilities, and we can see $\mathbb{P}(C_2)=4\rho(1-\rho)^9$.

This gives
$$\begin{align}
\mathbb{P}(Y_1=1|C) &= \frac{9\rho^2(1-\rho)^8}{9\rho^2(1-\rho)^8+4\rho(1-\rho)^9}\\
  &= \frac{9\rho}{9\rho+4(1-\rho)}\\
  &= \frac{9\rho}{4+5\rho}.
\end{align}$$

Finally, divide this by $3$ to get $\mathbb{P}(X_1),\mathbb{P}(X_2),\mathbb{P}(X_3)$, and use this to deduce the other cells' probabilities. So we get the following probabilities (for expert grid):
$$\begin{array}{*{7}c}
0.123 &|& 0.158      &|& 0.158      &|& 0.123\\
\hline
0.123 &|& \mathbf{1} &|& \mathbf{1} &|& 0.123\\
\hline
0.123 &|& 0.158      &|& 0.158      &|& 0.123
\end{array}$$
So we conclude that the best next move would be to click on one of the side cells on the left or right, next to the clicked cells, as $0.123<\rho$, so there is a lower chance of hitting a mine.

To speed up these calculations for other situations, notice it is easier to think of the ratio of the configuration probabilities, only having to consider the number of combinations and the relative number of mines. For example, if our second click was a 2 instead of a 1, to give
$\begin{array}{*{5}} \hline
|& 1 &|& 2 &| \\ \hline
\end{array}$
 the two combinations $C_1$ and $C_2$ are $\mathbf{Y}=(1,0,2)$ or $(0,1,1)$.
$$\frac{\mathbb{P}(C_1)}{\mathbb{P}(C_2)}=\frac{\binom{1}{3}\binom{2}{3}\rho}{\binom{1}{4}\binom{1}{3}(1-\rho)}$$

However, we can approximate $\rho\approx 0.2$ so that the relative chance of hitting a safe cell to hitting a mine is $\frac{1-\rho}{\rho}=4$. The calculations then become very easy:
$$\begin{align}
4\times12\mathbb{P}(C_1) &= 9\mathbb{P}(C_2)\\
\mathbb{P}(X_1) &= \frac{1}{3}\mathbb{P}(Y_1=1|C)\\
  &= \frac{\mathbb{P}(C_1)}{3(\mathbb{P}(C_1)+\mathbb{P}(C_2))}\\
  &= \frac{9}{3\times(4\times12+9)} = \frac{1}{19}\\
\mathbb{P}(X_4) &= \frac{1}{4}\left(\frac{4\times12}{4\times12+9}\right) = \frac{4}{19}\\
\mathbb{P}(X_7) &= \frac{2}{3}\left(\frac{9}{4\times12+9}\right) + \frac{1}{3}\left(\frac{4\times12}{4\times12+9}\right) = \frac{22}{57}.
\end{align}$$

So we see the cells that are only adjacent to the 1 are very safe.
$$\begin{array}{*{7}c}
0.05 &|& 0.21       &|& 0.21       &|& 0.39\\
\hline
0.05 &|& \mathbf{1} &|& \mathbf{2} &|& 0.39\\
\hline
0.05 &|& 0.21       &|& 0.21       &|& 0.39
\end{array}$$

We would see a similar result if it were a 3 on the right next to the 1. In any of these cases, our best option is to click one of $X_1, X_2, X_3$. Let's consider the case that we click $X_2$. We're fairly likely to hit an opening at this point, but lastly, I will analyse the other likely situation, of hitting another number 1, giving us
$$\begin{array}{*{9}c}
_{(1)} &|& _{(4)}     &|& _{(6)}     &|& _{(8)} &|& _{(10)}\\
\hline
_{(2)} &|& \mathbf{1} &|& \mathbf{1} &|& \mathbf{1}\ \mathrm{or}\ \mathbf{2} &|& _{(11)}\\
\hline
_{(3)} &|& _{(5)}     &|& _{(7)}     &|& _{(9)} &|& _{(12)}
\end{array}$$

The equivalence groups are $\{1,2,3\}, \{4,5\}, \{6,7\}, \{8,9\}, \{10,11,12\}$. For the 1 1 1 case... the configurations are
$\mathbf{Y}\in\{(1,0,0,1,0), (0,1,0,0,1), (0,0,1,0,0)\}$, and
$\mathbb{P}(C_1):\mathbb{P}(C_2):\mathbb{P}(C_3) = 3\cdot2:2\cdot3:4\cdot2 = 3:3:4$

For the 1 1 2 case... the configurations are
$\mathbf{Y}\in\{(1,0,0,1,1), (0,1,0,0,2), (0,0,1,0,1)\}$, and
$\mathbb{P}(C_1):\mathbb{P}(C_2):\mathbb{P}(C_3) = 3\cdot2\cdot3:2\cdot3:4\cdot2\cdot3 = 3:1:4$

Working out how many mines result in each section with these configurations, we arrive at the following probabilities.
$$\begin{array}{*{9}c}
0.1 &|& 0.15       &|& 0.2        &|& 0.15       &|& 0.1\\
\hline
0.1 &|& \mathbf{1} &|& \mathbf{1} &|& \mathbf{1} &|& 0.1\\
\hline
0.1 &|& 0.15       &|& 0.2        &|& 0.15       &|& 0.1
\end{array}$$

$$\begin{array}{*{9}c}
0.13 &|& 0.06        &|& 0.25        &|& 0.19     &|& 0.38\\
\hline
0.13 &|& \mathbf{1} &|& \mathbf{1} &|& \mathbf{2} &|& 0.38\\
\hline
0.13 &|& 0.06        &|& 0.25        &|& 0.19     &|& 0.38
\end{array}$$

In summary:
A good opening strategy is as follows:

Click randomly in the middle of the grid (20% chance of losing).
If it was a 1, click an adjacent cell, say the one to the right (13% chance of losing).

If we now have a 1 1 configuration, click one of the six cells adjacent to only one of the revealed 1s, say the one directly to the right (12% chance of losing).
If we have a 1 2 configuration, click one of the 3 cells only next to the 1, say the one directly to the left (5% chance of losing)

If we now have a 1 1 1 configuration, click one of the 6 cells on the ends, next to only one of the 1s (10% chance of losing)
If we have a 1 1 2 configuration, click one of the two cells shared by only the two 1s (6% chance of losing), or maybe click one of the 3 cells on the far left for a higher chance of getting an opening?? (13% chance of losing)

For any tile in a MineSweeper game, the number revealed when clicked (if not a mine) tells the player the number of mines adjacent to that tile.

For example, assuming your tile is not on the edge of the playing field, it will have 8 adjacent tiles- three above, one on the left, one on the right, and three below.

So, if our tile $T$ reveals some $x \in {1, 2, 3, 4, 5, 6, 7, 8}$ , then the probability a randomly-selected adjacent tile to $T$ is a mine would be $\frac{x}{8}$ ..

The probability calculations for the number of edge cases (on the playing field) is similar.

EDIT

The key strategy of the game then, is to deduce locations of mines from the already-revealed values of neighboring tiles to T.  Then you can 'flag' correct tiles as mines.


Lower bound for the norm of the resolvent I need to prove next statement (I want to do it for general case)
$\|R_A(z)\| = \lVert \frac{1}{A-zI} \rVert \ge \text{dist}(z,\sigma(A))^{-1}$
I think it could be like this
let $a\in \sigma(A) z \notin \sigma (A)$
then $(A-a)^{-1}$ is unbounded.
$(A-z-(a-z))=(A-z)(I-(a-z)(A-z)^{-1})
||(a-z)(A-z)^{-1}||>1$
then we get statement.
but my proof doesn't cover eigenvalue.

See Corollary 6.9(1), p242 in Invitation To Operator Theory (here).


Is this equality true Is the following equality true?

$$\int_{0}^{+\infty} \frac{x+1}{3x+4} = \int_{0}^{+\infty} \frac{x}{3x+4}+\int_{0}^{+\infty} \frac{1}{3x+4}$$

$$ \int ( f(x) + g(x))dx = \int f(x) dx + \int g(x)dx$$

So, your equality is correct.

[Wiki article]

http://en.m.wikipedia.org/wiki/Sum_rule_in_integration


How to prove Linear Independence of piecewise functions? Suppose we have two functions: $f_1(x)=x^2,x\geq 0$ and $f_1(x)=0,x\leq 0$ and $f_2(x)=0,x\geq 0$ and $f_2(x)=x^2, x\leq0$. Show that these two functions $f_1(x)$ and $f_2(x)$ are linearly independent.

My attempt: Consider the interval $[0,\infty]$. Then $f_1(x)=x^2$ and $f_2(x)=0$. The Wronskian $W(f_1,f_2)=0$. Since the Wronskian is zero for every $x\in [0,\infty]$, we conclude that over this interval the functions are linearly dependent.  Similar reasoning shows that for $x\in[-\infty,0]$ the functions are linearly dependent. Thus, one might conclude for all $x$ the functions are linearly dependent. However, this is not true. What is wrong with my argument?

I know that the Wronskian being equal to zero is not a sufficient condition of linear dependency, but then several solutions make use of this condition directly to establish linear independency.

remember that a zero vector can never be a vector in a free system; what you have to demonstrate is that the restrictions of $f_1 $ and of $ f_2$ to every  interval are linearly dependent and not the functions themselves

to be clear in my answer,  let $I=]-\infty,0]$ and $J=[0;\infty[$ for $i=1,2$ let $f_{i,I}$
and $f_{i,J}$ the restriction of $f_i$ respectively to $I$ and
$J$. I said in my answer above that for showing that the system
$\{f_{1,I},f_{2,I}\}$ (resp. $\{f_{1,J},f_{2,J}\}$)  is linearly
dependant, you    need not in this case to use the Wronksian of
$(f_{1,I},f_{2,I})$ (resp.$(f_{1,J},f_{2,J})$ because we know that
a zero vector can not be in a linearly independent system.

in the
seconde affirmation that if two maps $f$ and $g$ him restriction
$f_{A},g_{A}$ to A (resp. $f_{B},g_{B}$ to B) are linearly
dependent, where $A\cup B=D$ is the domain on which $f$ and $g$
are defined, the system $\{f,g\}$ can be not linearly dependant,
as shown your example, note in your example you can not us the
Wronksian for $f$ and $g$ because $f$ and $g$ are not derivable
functions on $\Bbb{R}$, but you can use the definition directly to show
that are linearly independents.