问:
Morita theory and non basic path algebra
I'm learning about representation theory of associative algebras. In my studies arrised less technical questions involving Morita theory and quivers that will exposed.

Morita theory give us a way to find a more "economic" algebra for a given algebra such that they have same modular theory. For Artin algebras we have a complete set o idempotents which allow us to construct this basic algebra.

1) There examples of non basic algebras?  There are explicit examples of algebras without this complet set of idempotents? That is, an algebra with not only the trivial idempotents, but in wich we can't show that there are only those (could has a infinite number).

2) What is the quiver of a non basic algebra?
The book (Assem-Simson-Skowronski) always talk about quiver involving basic and connected algebras, but don't show the issues arrised when consider an algebra without this properties. There exist quiver representation for non connected non basic algebras?

I know that is a lot of questions, but any help/commentarie/references will be really apreciated. Thanks in advance!

答:
There are several questions here, and I will try to answer them in order.

First, an example of an algebra which is not basic would be the $k$-algebra $M_2(k)$ of $2\times 2$ matrices over a field $k$.  To see that it is not basic, note that as a right module over itself, it is the direct sum of the two submodules $R_1$ and $R_2$ of matrices with zero entries in row $2$ and row $1$, respectively.  Since $R_1$ and $R_2$ are isomorphic as right $M_2(k)$-modules, the algebra is not basic.

Second, on the existence of certain sets of idempotents.  I am assuming here that you are interested in complete sets of pairwise orthogonal primitive idempotents, that is, collections of elements $e_1, \ldots, e_n$ such that $e_i\cdot e_j$ is $e_i$ if $i=j$ (idempotence) and $0$ otherwise (orthogonality); $\sum_{i=1}^n e_i = 1$ (completeness); and none of the $e_i$ can be written as a sum of two orthogonal non-zero idempotents (primitiveness).  If an algebra is artinian, then such a set of idempotents always exists.  An algebra which does not admit such a set is $\prod_{\mathbb{N}} k$, the product of a countably infinite number of copies of a field $k$.  This algebra admits complete sets of pairwise orthogonal idempotents, but we cannot require that these idempotents are all primitive.  Another example would be the $\mathbb{R}$-algebra of continuous functions from $\mathbb{R}$ to $\mathbb{R}$.

Next, you ask about the quiver of a non-basic algebra.  This is somewhat ill-defined.  Let's review the theory over an algebraically closed field $k$.  In this setting, for any basic, finite-dimensional $k$-algebra $A$, there exists a unique quiver $Q$ and a (non-unique) admissible ideal $I$ of $kQ$ such that $A$ is isomorphic to $kQ/I$.  Moreover, any algebra of the form $kQ/I$ as above is basic.  It follows from these results that if $A$ is not basic, then it cannot be of the form $kQ/I$, with $I$ an admissible ideal of $kQ$.

In the above discussion, the operative word is admissible.  For any finite-dimensional $k$-algebra, you can find lots of quivers $Q$ and (not necessarily admissible) ideals $I$ such that $A$ is isomorphic to $kQ/I$.  For matrix algebras $M_n(k)$, some such descriptions are given in this mathoverflow question and its answers.  Still, for non-basic algebras, it will never be possible to find an admissible ideal, and there will be no good notion of the uniqueness of the quiver.

Finally, if an algebra is basic but not connected, then the description by a quiver with an admissible ideal of relations works fine; all that happens is that the quiver is not connected.


问:
How do I turn this sequence into a sigma function?
$$3\times2+1$$
$$ 3\times3+1$$ $$3\times3+2$$
$$3\times4+1$$ $$3\times4+2$$ $$3\times4+3$$
$$3\times5+1$$$$3\times5+2$$$$3\times5+3$$$$3\times5+4$$

And so it goes on for a number of terms dependent on the n-value, though exactly how it is dependent I'm not sure. You can leave the number of terms blank, as I'll figure that out myself. However, how I make this sequence into a sigma function I have no idea. Note that something like $(3\times k)(k-1)+\frac{(k-1)k}{2}$ will not work, because the terms are the individual operations, and not the bigger groupings: i.e., the last term doesn't need to be $3\times k + (k-1)$. The last term can be anything from $3\times k + (k-(k-1))$ all the way to the aforementioned one.

As such, the sequence becomes: 7, 10, 11, 13, 14, 15, 16, 17, 18, 19, 19, 20, 21, 22, 23, 22, 23, 24, 25, 26, 27, and so on.
So, any thoughts?

EDIT:

$$\sum_k^? \lceil \frac{n-x}{2} \rceil$$

"?" is something related to the n-value, and "x" is related to the k-value, and is somehow supposed to give the sequence. However, it appears that the current answers to this post give the answer for the objective of subtracting from "n" the entire sequence-parts, instead of the individual terms. As in, those answers make it so that "x" does this: $n - ((3\times4+1) + (3\times4+2) + (3\times4+3))$. However, "x" doesn't need to be all of that. The last "x" can be $3\times4+2$, the number added doesn't need to be k-1.

答:
This is a double summation since you have two sets of values (columns) which are increasing with conditions on each.

$$ \sum \sum 3\times a + b$$ should be the required form. Now, note that if you fix $a$, then the values $b$ go from $b=1$ upto $a-1$. For instance, if $a=4$ your terms are
$$ 3\times 4+1$$
$$ 3\times 4+2$$
$$ 3\times 4+3$$
You can update your sigma formulation as

$$ \sum \sum_{b=1}^{a-1} 3\times a + b$$

Finally, your $a$ values range from $2$ to $5$. Thus, the sigma formulation is

$$ \sum_{a=2}^{5} \sum_{b=1}^{a-1} 3\times a + b.$$

If you are looking for a general $k$-th term, it is possible but that requires triangular numbers $T(n) = n(n+1)/2$. The idea is that if $k$ is an integer with the property that
$$ T_n < k \leq T_{n+1}$$ then your general term is of the form
$$ a_k = 3\times(n+2) + (T_n-k).$$

ContentLicense: CC BY-SA 4.0


问:
A question on the center of a group
Given a finite nonabelian group $G$, let $p$ be the smallest divisor of $|G|$. Then I want to show that

$p^2 |Z(G)| \le |G|$

If $N \le G$ is normal and $|N|=p$ then $N \le Z(G)$.

For the first one I think that since $G$ is non-abelian, $|G/Z(G)|>1$ and since $|G/Z(G)| $ divides $|G|$ then $|G/Z(G)|$ is at least $p$. However, in this case $G$ would be abelian. Then we must have $|G/Z(G)| \ge p^2$ thus $|G|\ge p^2 |Z(G)|$.

How can I show the second one?

答:
$G/Z(G)$ is not cyclic, and so $|G/Z(G)|$ is not a prime. Thus, $|G/Z(G)|\ge p^2$.

For the second question, it suffices to show that $C_G(N)=G$. Note that $N_G(N)/C_G(N)\lesssim\operatorname{Aut}(N)\cong \mathbb{Z}_{p-1}$ (see this question). As $p$ is the smallest prime divisor, it implies that $N_G(N)/C_G(N)=1$.

Title: A question on the center of a group


问:
Why do I get a math error for this when trying to compute $\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx$?
I tried to evaluate
$$
\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx
$$
in the following way:
$$
\left[\frac{x^3}{3}+\frac{1}{x}+3x\right]_{-1}^1=\frac{26}{3} \, .
$$
But when I typed the integral into a calculator I got a math error. Why did this happen?

答:
Say $F$ is an antiderivative of $f$—that is, if you differentiate $F$, then you get $f$. What you're doing to compute the integral is the following:
$$
\int_{a}^{b} f(x) \, dx = [F(x)]_a^b=F(b)-F(a) \, .
$$
This method of evaluating integrals is an application of the fundamental theorem of calculus—which connects antiderivatives with integrals. To be clear,
$$
\int_{a}^{b} f(x) \, dx
$$
represents the area under the graph of $f(x)$ between $x=a$ and $x=b$. Meanwhile,
$$
\int f(x) \, dx
$$
represents the family of functions which, when differentiated, gives you $f(x)$. It should surprise you that these two ideas should should be so intimately connected. (This also explains why the integral and antiderivative notations are so similar.) However, the fundamental theorem of calculus only applies in certain cases. The example you gave,
$$
\int_{-1}^1 x^2-\frac{1}{x^2}+3 \, dx \, ,
$$
is not one of those cases, since the function is undefined at $x=0$. This is a problem many students have: they are so used to applying the fundamental theorem of calculus that they don't realise when it doesn't apply. In fact,
$$
\int_{-1}^1 x^2-\frac{1}{x^2}+3 \, dx
$$
is not even an integral at all in the strict sense of the word. Just take a look at the graph of $y=x^2-\frac{1}{x^2}+3$:

What would 'the area under the graph' mean to you in this case? What we are dealing with is called an 'improper integral'. The example you gave
$$
\int_{-1}^1 x^2-\frac{1}{x^2}+3 \, dx \, ,
$$
is merely a shorthand for
$$
\lim_{a\to0^-}\int_{-1}^{a} x^2-\frac{1}{x^2}+3 \, dx + \lim_{b \to 0^+}\int_{b}^{1} x^2-\frac{1}{x^2}+3 \, dx \, .
$$
Since neither of these limits exist, the 'integral' is said to be divergent. Here are some useful links:

The Wikipedia articles on the fundamental theorem of calculus and improper integrals.

I also gave an introduction to improper integrals here.


问:
Finding the differential equation of a circle touching three curves $f(x,y)=0$, $g(x,y)=0$, $h(x,y)=0$

Find the differential equation of a circle tangent to three given curves:
$$ f(x,y)=0,\; g(x,y)=0,\; h(x,y)=0$$

Found a simple locus recently using analytical geometry.

Circle Touching 2 lines & a Circle

My thoughts relating to generalization are the following:

In a particular case let $ f(x,y)= y+ \log(x) = a$

Implicit differentiation gives

$$ y'_{tgt} + \frac{1}{x}=0\; y'_{norm}= x$$

Equation of normal

$$\dfrac{y-y_f}{x-x_f} =x \tag1$$

Two more such equations of the normal would be available and their intersection points can be found. To be concurrent at circle center  their determinant should vanish.

That provides one condition.

Next, from the polar form of straight line of each tangent

$$  \tan \alpha = y'_{tgt},\quad x \cos \alpha  + y \sin \alpha = p_f $$

we can find  $ p_f,p_g,p_h$ which are equated, second condition.

The procedure is computation intensive; wondering if any aspect of geometric invariant theory of curves etc. could be relevant. Seeking to set up DE from viewpoint of differential geometric generality.

A simple situation comes to mind is the Clairaut's DE where all tangents to a family of separate roted curves has an associated invariant:

$ r \sin \psi = r_{min\; circle}$

Your help and pointers here would be much appreciated. Regards.

答:
This problem can be easily equated as a minimization problem. It's unique solution depends on convexity. Giving three functions $f(x,y)=0,g(x,y)=0,h(x,y)=0$ and calling

$$
\cases{
\vec n_f=\frac{\nabla f}{||\nabla f||}\\
\vec n_g=\frac{\nabla g}{||\nabla g||}\\
\vec n_h=\frac{\nabla h}{||\nabla h||}\\
p_f = \{x_f,y_f\}\ |\  f(x_f,y_f)=0\\
p_g = \{x_g,y_g\}\ |\ g(x_g,y_g)=0\\
p_h = \{x_h,y_h\}\ |\ h(x_h,y_h)=0\\
o_f = p_f+\lambda_f\vec n_f\\
o_g = p_g+\lambda_g\vec n_g\\
o_h = p_h+\lambda_h\vec n_h\\
}
$$

the minimization problem can be formulated as

$$
\min_{p_f,p_g,p_h,\lambda_f,\lambda_g,\lambda_h}||o_f-o_g||^2+||o_g-o_h||^2+||o_f-o_h||^2\ \ \text{s. t.}\ \ \ |\lambda_f|=|\lambda_g|=|\lambda_h|
$$

Follows a MATHEMATICA script for

$$
\cases{
f(x,y) = x^x-y=0\\
g(x,y) = \ln x^{\ln x}-y=0\\
h(x,y) = \left(\ln x\right)^2-y=0
}
$$

```<code>f[x_] := x^x
g[x_] := Log[x]^Log[x]
h[x_] := Log[x]^2
pf = {x1, f[x1]};
pg = {x2, g[x2]};
ph = {x3, h[x3]};
nf = Grad[y1 - f[x1], {x1, y1}];
ng = Grad[y2 - g[x2], {x2, y2}];
nh = Grad[y3 -   h[x3], {x3, y3}];
of = pf + lambda1 nf/Norm[nf];
og = pg + lambda2 ng/Norm[ng];
oh = ph + lambda3 nh/Norm[nh];
obj = Norm[of - og] + Norm[of - oh] + Norm[og - oh];
sol = NMinimize[{obj, 1.5 > x1 > 0.5, 1.5 > x2 > 1, 1.5 > x3 > 0.5, lambda1^2 == lambda2^2 == lambda3^2}, {lambda1, lambda2, lambda3, x1, x2, x3}, Method -> "DifferentialEvolution"]
plot = Plot[{f[x], g[x], h[x]}, {x, 0, E}, PlotRange -> {0, 1}];
{x0, y0} = of /. sol[[2]]
r = lambda1 /. sol[[2]]
circ = (x - x0)^2 + (y - y0)^2 - r^2;
grc = ContourPlot[circ == 0, {x, 0, 2}, {y, 0, 1}];
Show[plot, grc, AspectRatio -> 1/3]
</code>```

NOTE

Additional restrictions were included to avoid calculations involving complex numbers.

Another example (a little more general) with

$$
\cases{
f(x,y) = y-x-1.25=0\\
g(x,y) = x^2+y^2-9=0\\
h(h,y) = x^2+y = 0
}
$$

Follows the new script

```<code>Clear[f, g, h]
f[x_, y_] := y - x - 1.25
g[x_, y_] := x^2 + y^2 - 9
h[x_, y_] := x^2 + y
pf = {x1, y1};
pg = {x2, y2};
ph = {x3, y3};
nf = Grad[f[x1, y1], {x1, y1}];
ng = Grad[g[x2, y2], {x2, y2}];
nh = Grad[h[x3, y3], {x3, y3}];
of = pf + lambda1 nf/Norm[nf];
og = pg + lambda2 ng/Norm[ng];
oh = ph + lambda3 nh/Norm[nh];
obj = Norm[of - og]^2 + Norm[of - oh]^2 + Norm[og - oh]^2;
sol1 = NMinimize[{obj, f[x1, y1] == g[x2, y2] == h[x3, y3] == 0, Abs[lambda1] == Abs[lambda2] == Abs[lambda3], -3 < x1 < 0, -3 < x2 < 0, -3 < x3 < 0, -3 < y1 < 0, -3 < y2 < 0, -3 < y3 < 0}, {lambda1, lambda2, lambda3, x1, x2, x3, y1, y2, y3}]
sol2 = NMinimize[{obj, f[x1, y1] == g[x2, y2] == h[x3, y3] == 0, Abs[lambda1] == Abs[lambda2] == Abs[lambda3], 0 < x1 < 3, 0 < x2 < 3, 0 < x3 < 3, -0.3 < y1 < 2, -0.3 < y2 < 2, -0.3 < y3 < 2}, {lambda1, lambda2, lambda3, x1, x2, x3, y1, y2, y3}, Method -> "DifferentialEvolution"]

plot = ContourPlot[{f[x, y] == 0, g[x, y ] == 0, h[x, y] == 0}, {x, -3, 3}, {y, -3, 3}];
{x0, y0} = of /. sol1[[2]]
r = Norm[of - pf] /. sol1[[2]]
circ = (x - x0)^2 + (y - y0)^2 - r^2;
grc1 = ContourPlot[circ == 0, {x, -3, 3}, {y, -3, 3}];
{x0, y0} = of /. sol2[[2]]
r = Norm[of - pf] /. sol2[[2]]
circ = (x - x0)^2 + (y - y0)^2 - r^2;
grc2 = ContourPlot[circ == 0, {x, -3, 3}, {y, -3, 3}];
Show[plot, grc1, grc2]
</code>```

Body:
This problem can be easily equated as a minimization problem. It's unique solution depends on convexity. Giving three functions $f(x,y)=0,g(x,y)=0,h(x,y)=0$ and calling

$$
\cases{
\vec n_f=\frac{\nabla f}{||\nabla f||}\\
\vec n_g=\frac{\nabla g}{||\nabla g||}\\
\vec n_h=\frac{\nabla h}{||\nabla h||}\\
p_f = \{x_f,y_f\}\ |\  f(x_f,y_f)=0\\
p_g = \{x_g,y_g\}\ |\ g(x_g,y_g)=0\\
p_h = \{x_h,y_h\}\ |\ h(x_h,y_h)=0\\
o_f = p_f+\lambda_f\vec n_f\\
o_g = p_g+\lambda_g\vec n_g\\
o_h = p_h+\lambda_h\vec n_h\\
}
$$

the minimization problem can be formulated as

$$
\min_{p_f,p_g,p_h,\lambda_f,\lambda_g,\lambda_h}||o_f-o_g||^2+||o_g-o_h||^2+||o_f-o_h||^2\ \ \text{s. t.}\ \ \ |\lambda_f|=|\lambda_g|=|\lambda_h|
$$

Follows a MATHEMATICA script for

$$
\cases{
f(x,y) = x^x-y=0\\
g(x,y) = \ln x^{\ln x}-y=0\\
h(x,y) = \left(\ln x\right)^2-y=0
}
$$

f[x_] := x^x
g[x_] := Log[x]^Log[x]
h[x_] := Log[x]^2
pf = {x1, f[x1]};
pg = {x2, g[x2]};
ph = {x3, h[x3]};
nf = Grad[y1 - f[x1], {x1, y1}];
ng = Grad[y2 - g[x2], {x2, y2}];
nh = Grad[y3 -   h[x3], {x3, y3}];
of = pf + lambda1 nf/Norm[nf];
og = pg + lambda2 ng/Norm[ng];
oh = ph + lambda3 nh/Norm[nh];
obj = Norm[of - og] + Norm[of - oh] + Norm[og - oh];
sol = NMinimize[{obj, 1.5 > x1 > 0.5, 1.5 > x2 > 1, 1.5 > x3 > 0.5, lambda1^2 == lambda2^2 == lambda3^2}, {lambda1, lambda2, lambda3, x1, x2, x3}, Method -> "DifferentialEvolution"]
plot = Plot[{f[x], g[x], h[x]}, {x, 0, E}, PlotRange -> {0, 1}];
{x0, y0} = of /. sol[[2]]
r = lambda1 /. sol[[2]]
circ = (x - x0)^2 + (y - y0)^2 - r^2;
grc = ContourPlot[circ == 0, {x, 0, 2}, {y, 0, 1}];
Show[plot, grc, AspectRatio -> 1/3]

NOTE

Additional restrictions were included to avoid calculations involving complex numbers.

Another example (a little more general) with

$$
\cases{
f(x,y) = y-x-1.25=0\\
g(x,y) = x^2+y^2-9=0\\
h(h,y) = x^2+y = 0
}
$$

Follows the new script

Clear[f, g, h]
f[x_, y_] := y - x - 1.25
g[x_, y_] := x^2 + y^2 - 9
h[x_, y_] := x^2 + y
pf = {x1, y1};
pg = {x2, y2};
ph = {x3, y3};
nf = Grad[f[x1, y1], {x1, y1}];
ng = Grad[g[x2, y2], {x2, y2}];
nh = Grad[h[x3, y3], {x3, y3}];
of = pf + lambda1 nf/Norm[nf];
og = pg + lambda2 ng/Norm[ng];
oh = ph + lambda3 nh/Norm[nh];
obj = Norm[of - og]^2 + Norm[of - oh]^2 + Norm[og - oh]^2;
sol1 = NMinimize[{obj, f[x1, y1] == g[x2, y2] == h[x3, y3] == 0, Abs[lambda1] == Abs[lambda2] == Abs[lambda3], -3 < x1 < 0, -3 < x2 < 0, -3 < x3 < 0, -3 < y1 < 0, -3 < y2 < 0, -3 < y3 < 0}, {lambda1, lambda2, lambda3, x1, x2, x3, y1, y2, y3}]
sol2 = NMinimize[{obj, f[x1, y1] == g[x2, y2] == h[x3, y3] == 0, Abs[lambda1] == Abs[lambda2] == Abs[lambda3], 0 < x1 < 3, 0 < x2 < 3, 0 < x3 < 3, -0.3 < y1 < 2, -0.3 < y2 < 2, -0.3 < y3 < 2}, {lambda1, lambda2, lambda3, x1, x2, x3, y1, y2, y3}, Method -> "DifferentialEvolution"]

plot = ContourPlot[{f[x, y] == 0, g[x, y ] == 0, h[x, y] == 0}, {x, -3, 3}, {y, -3, 3}];
{x0, y0} = of /. sol1[[2]]
r = Norm[of - pf] /. sol1[[2]]
circ = (x - x0)^2 + (y - y0)^2 - r^2;
grc1 = ContourPlot[circ == 0, {x, -3, 3}, {y, -3, 3}];
{x0, y0} = of /. sol2[[2]]
r = Norm[of - pf] /. sol2[[2]]
circ = (x - x0)^2 + (y - y0)^2 - r^2;
grc2 = ContourPlot[circ == 0, {x, -3, 3}, {y, -3, 3}];
Show[plot, grc1, grc2]


答:
Recall when a function is Riemann integrable:

A bounded function on a compact interval $[a, b]$ is Riemann integrable if and only if it is continuous almost everywhere.

So, is $x^2-\dfrac{1}{x^2}+3$ bounded on $[-1,1]$?

Title: Why do I get a math error for this when trying to compute $\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx$?

Body:
I tried to evaluate
$$
\int_{-1}^1(x^2-\frac{1}{x^2}+3) \, dx
$$
in the following way:
$$
\left[\frac{x^3}{3}+\frac{1}{x}+3x\right]_{-1}^1=\frac{26}{3} \, .
$$
But when I typed the integral into a calculator I got a math error. Why did this happen?


问:
Find Aut$(G)$, Inn$(G)$ and $\dfrac{\text{Aut}(G)}{\text{Inn}(G)}$ for $G = \mathbb{Z}_2 \times \mathbb{Z}_2$
Find Aut$(G)$, Inn$(G)$ and $\dfrac{\text{Aut}(G)}{\text{Inn}(G)}$ for $G = \mathbb{Z}_2 \times \mathbb{Z}_2$.

Here is what I have here:

Aut$(G)$ consists of 6 bijective functions, which maps $G$ to itself, since Aut$(\mathbb{Z}_2 \times \mathbb{Z}_2) \approx S_3$.

I think the next part goes wrong.  For Inn$(G)$, letting $\kappa_x : G \rightarrow G$ to be the conjugate function, I found $\{e, \kappa_{(0,1)}, \kappa_{(1,0)}, \kappa_{(1,1)}\}$.

I can't determine $\dfrac{\text{Aut}(G)}{\text{Inn}(G)}$ yet since I need to correctly determine Inn$(G)$.

Any advices or comments?

答: Hint: to expand on Tyler's comment, can you prove that, more generally, $\operatorname{Inn}(G) = \{0\}$ iff $G$ is abelian?

ContentLicense: CC BY-SA 3.0


问:
Find all triplets (x, y, z) of positive integers such that 1/x + 1/y + 1/z = 4/5
Here's what i did :-

i wrote Find all triplet $$1/x + 1/y + 1/z = 4/5$$ as $$1/x + 1/y = 4/5 - 1/z$$

$(4z-5)/5z = 1/x + 1/y$ which can be written as $1/x + 1/y = 1/(5z)/(4z-5)$

Let $(5z)/(4z-5) = n$ therefore $1/x + 1/y = 1/n$ which on factorization gives $(n-x)(n-y) = n^2$ and $n$ should be an integer , i am not sure about the proof but i got $(5z)/(4z-5)$ is never an integer except $z = <o,1>$ but z cannot be 0 .

Here's my problem :-
if we take $n = 1$ we get 0 solutions of $<x,y,-1>$
is what i did correct ?

答:
Just to give another way (I like that). It is known the only $n$ such that its decimal expression is finite are $n=2^a\cdot5^b$ Since we look for
$$\frac 1x+\frac 1y+\frac 1z=0.8$$ we begin with $x=2$ which gives
$$\frac 1y+\frac 1z=\frac45-\frac12=\frac{3}{10}=\frac{6}{20}$$ from which the solutions
$$(x,y,z)=(2,5,10),(2,4,20)$$ and obviously all the permutations so the smallest integer to be considered is $x=8$.

It is easily verified that all other such as $x=8,16,25,....$ are not solutions.


问:
How to solve inequality equation with ceil operator product
I am trying to solve this inequality for $x$:

$$ \left \lceil{\frac{c_1}{x}}\right \rceil \cdot \left \lceil{\frac{c_2}{x}}\right \rceil \leq N $$

With: $c_1, c_2, x, N \in \mathbb{N}$ and $c_1, c_2, N$ being just constants.

I could make my system work with a solution assuming that $ x \leq c_1, c_2 $ but I would like ot have a general solution if someone can help me. The context behind this equation is how to split an image of $[H,W]=[c_1, c_2]$ in square patches of size $[H_{patch}, W_{patch}]=[x,x]$.

What I have tried yet:

Solve the equation for the lower bound of each inequality: I does not work.

$$\frac{c_1c_2}{x^2} \leq N \rightarrow \sqrt{\frac{c_1c_2}{N}} \leq x$$

Solve the equation for the upper bound: I am stuck getting x out of this expression.

$$ (\frac{c_1}{x^2}+1) \cdot (\frac{c_1}{x^2}+1) =\frac{c_1c_2}{x^2} + \frac{c_1}{x} + \frac{c_2}{x} + 1\leq N$$

Forgive my possible mistakes, I am not a mathematician. Thank you for your help in advance.

答:
As discussed in the comments, you need a minimum value for $x$, let's call it $x_{min}$. This is what you mean by lower bound for $x$. I think I also understand what did you mean by upper bound: a "safe" value of $x$ for which the relation is definitely true but could be improved. Let's call it $x_{safe}$.

Your attempt by having $x \leq c_1, c_2$  is just the other way round: $x \geq c_1, c_2$, or as I prefer to denote $x_{safe} = max(c_1, c_2)$. But this is far from $x_{min}$ in cases $N \neq 1$, and you just need to do trial-and-error by brute-force or something to get $x_{min}$ from $x_{safe}$.

First of all, let's make it clear, that if $N = 1$, then $x_{min} = max(c_1, c_2)$.

Secondly, when $N \neq 1$, I can show you how to get $x_{safe}$ closer to $x_{min}$. I can not tell you how to get $x_{min}$ with a formula. You still need to get $x_{min}$ from $x_{safe}$ if you really need the minimum value. To see how it is improving, let $c_1 = 10, c_2 = 4$.

For this, let's look at your attempt on lower bound:

$$\sqrt{\frac{c_1c_2}{N}} \leq x$$

If $N = 8$, then $x \geq 2.23... \Rightarrow x = 3 \Rightarrow  \left \lceil{\frac{c_1}{x}}\right \rceil \cdot \left \lceil{\frac{c_2}{x}}\right \rceil = 8$, correct.

If $N = 7$, then $x \geq 2.39... \Rightarrow x = 3 \Rightarrow  \left \lceil{\frac{c_1}{x}}\right \rceil \cdot \left \lceil{\frac{c_2}{x}}\right \rceil = 8$, not correct. Using $x = 4 \Rightarrow  \left \lceil{\frac{c_1}{x}}\right \rceil \cdot \left \lceil{\frac{c_2}{x}}\right \rceil = 6$, it is now correct.

This attempt does not give you $x_{safe}$. You need to increase $x$ until it is OK. Or go with $x_{safe} = max(c_1, c_2) = 10$, which is far.

This can be improved.

Let $\alpha \in \mathbb{R}$. We can have a value for $\alpha$ so that if we only had $\lceil{\frac{c_1}{x}} \rceil \leq N$, $\lceil{\frac{c_1}{\alpha}} \rceil = \frac{c_1}{\alpha} = N \Rightarrow \alpha = \frac{c_1}{N}$, and as we need an integer for $x$, we have $x \geq \alpha \Rightarrow x = \lceil{\alpha} \rceil \Rightarrow  \lceil{\frac{c_1}{x}} \rceil \leq \lceil{\frac{c_1}{\alpha}} \rceil = N$.

But we have 2 ceiled expressions. So let's examine if $c_1 = c_2 \Rightarrow {\lceil{\frac{c_1}{x}} \rceil}^2 \leq N$:

let:

$$\frac{c_1}{\alpha} = \sqrt{N} \Rightarrow \alpha = \frac{c_1}{\sqrt{N}}$$

From this:

$$x \geq \alpha \Rightarrow x = \lceil{\alpha} \rceil \Rightarrow x = \lceil{\frac{c_1}{\sqrt{N}}} \rceil \Rightarrow \frac{c_1}{x} = \frac{c_1}{\lceil{\frac{c_1}{\sqrt{N}}} \rceil} \leq {\frac{c_1}{\frac{c_1}{\sqrt{N}}}} = \sqrt{N}$$

So we have $\frac{c_1}{x} \leq \sqrt{N}$, but this does not guarantee $\lceil{\frac{c_1}{x}} \rceil \leq \sqrt{N}$, so still need to adjust to have $x_{safe}$.

$$x = \lceil{\frac{c_1}{\sqrt{N} - 1}} \rceil \Rightarrow \frac{c_1}{x} = \frac{c_1}{\lceil{\frac{c_1}{\sqrt{N} - 1}} \rceil} \leq {\frac{c_1}{\frac{c_1}{\sqrt{N} - 1}}} = \sqrt{N} - 1$$

So we have $\frac{c_1}{x} \leq \sqrt{N} - 1 \Rightarrow \lceil{\frac{c_1}{x}} \rceil \leq \lceil{\sqrt{N} - 1} \rceil < \sqrt{N}$. Now this is OK if $c_1 = c_2$, so one more adjustment, and we have:

$$x_{safe} = \lceil{\frac{max(c_1, c_2)}{\sqrt{N} - 1}} \rceil$$

Because if $c_i = max(c_1, c_2) \Rightarrow \lceil{\frac{c_i}{x_{safe}}} \rceil < \sqrt{N}$ from the above and therefore:

$$c_j \leq c_i \Rightarrow \lceil{\frac{c_j}{x_{safe}}} \rceil \leq \lceil{\frac{c_i}{x_{safe}}} \rceil \Rightarrow \left \lceil{\frac{c_1}{x_{safe}}}\right \rceil \cdot \left \lceil{\frac{c_2}{x_{safe}}}\right \rceil < N$$

If $N = 7$, then $x \geq 6.07... \Rightarrow x = 7 \Rightarrow  \left \lceil{\frac{c_1}{x}}\right \rceil \cdot \left \lceil{\frac{c_2}{x}}\right \rceil = 2$, correct. Closer to 4 than 10, but in general I can not show how close it would be. But it is for sure:

$$\lceil{\frac{max(c_1, c_2)}{\sqrt{N} - 1}} \rceil \leq max(c_1, c_2), \sqrt{N} - 1 > 1 \Rightarrow N > 4$$

I leave you to investigate $1 < N \leq 4$

So if you need $x_{min}$ you either use your lower bound formula and keep increasing until its sufficient, or use this upper bound and keep decreasing.


问:
Solve the linear congruences
To solve $\,26x\equiv 6\pmod{\!110},\,$ note $\,\gcd(26,110)=2\,$ so there are $\,2\,$ solutions.

Next, $\ 104x\equiv 24\pmod{\!110}\ $ follows by scaling the above by $\,4.$

i.e. $\,\ \ \ \ \ {-}6x\equiv 24\pmod{\!110}$

Hence $\ \ \ \ \,x\equiv -4\pmod{\!55}\ $ by dividing prior by $\,6$.

What is the final answer?

答: It is very simple. After rescaling it to $$13x \equiv 3 \pmod{55}$$, you can simply find the modular inverse of $13$ mod $55$, which in this case is $17$(Denoted by $13^{-1}$, and also note that it exists as 13 and 55 are coprime). Multiplying both sides by $17$ you get $$13 \times 13^{-1} x \equiv 3 \times 13^{-1} \pmod {55} $$  $$13 \times 17x \equiv x \equiv 3 \times 17 \pmod {55} $$  $$x \equiv 51 \pmod {55} $$

Title: Solve the linear congruences


问:
Mistake in number of ways in which we can assign $7$ projects to $4$ people such that each person get at least one project
Number of ways in which we can assign $7$ projects to $4$ people such that each person get at least one project:

My approach to this question was:

First assign $1$ project to everyone.

For this I first choose $4$ jobs out of $7$ and count every way i.e.$\binom{7}{4}4!$.

Next thing I did was applying the combinations with replacement as

J # J # J #

J J J # # #

J # # # J J

.

.

.

where 2 # are kind of boundaries and 3rd is also a boundary which also covers one position.

(Sorry but I don't remember the formula and always use it like this)

I got the solution $\binom{6}{3}$.

Now I thought I got all the cases and by product rule
$$\binom{7}{4}4!\binom{6}{3}$$

would give me answer. But the answer that I got was $16800$ while the answer in my book is $8400$. So where did I go wrong and why do I have to divide with $2$?

答:
By designating particular projects as the one received by each person and then distributing the remaining projects, you count the same distribution multiple times.  However, you made a second mistake.  You forgot to choose which which additional project is received by which person.  Had you done that, you would have had a much larger answer:

Since $7$ can be partitioned into four parts in three ways:
\begin{align*}
7 & = 4 + 1 + 1 + 1\\
  & = 3 + 2 + 1 + 1\\
  & = 2 + 2 + 2 + 1
\end{align*}
either one person receives four projects and each of the other people receives a single project; one person receives three projects, a second person receives two projects, and each of the other two people receives a single project; or three people each receive two projects and the other person receives one.

One person receives four projects and each of the others receives a single project:  Choose which of the four people receives four projects, choose which four of the seven projects that person receives, then distribute the remaining three projects to the remaining three people so that each person receives one.  This can be done in
$$\binom{4}{1}\binom{7}{4}3!$$
ways.

One person receives three projects, a second person receives two projects, and each of the others receives a single project:  Choose which person receives three projects, choose which three of the seven projects that person receives, choose which of the remaining three people receives two projects, choose which two of the remaining four projects that person receives, then distribute the remaining two projects to the remaining two people so that each person receives one.  This can be done in
$$\binom{4}{1}\binom{7}{3}\binom{3}{1}\binom{4}{2}2!$$
ways.

Three people each receive two projects and the remaining person receives one:  Choose which three of the four people each receive two projects, choose which two of the seven projects the most senior of these three people receives, choose which
two of the remaining five projects the next most senior of these three people receives, choose which two of the remaining three projects the least senior of these three people receives, then give the remaining project to the remaining person.  This can be done in
$$\binom{4}{3}\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}$$
ways.

Total:  The number of ways the seven projects can be distributed to four people so that each person receives at least one project is
$$\binom{4}{1}\binom{7}{4}3! + \binom{4}{1}\binom{7}{3}\binom{3}{1}\binom{4}{2}2! + \binom{4}{3}\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1} = 8400$$

Had you remembered to choose which person receives which additional project, you would have found that there were
$$\binom{4}{1}\binom{3}{3} + \binom{4}{1}\binom{3}{2}\binom{3}{1}\binom{1}{1} + \binom{4}{3}3! = 64$$
such distributions, which would have given you the incorrect total
$$\color{red}{\binom{7}{4}4!\left[\binom{4}{1}\binom{3}{3} + \binom{4}{1}\binom{3}{2}\binom{3}{1}\binom{1}{1} + \binom{4}{3}3!\right] = 53,760}$$

Your method counts each case in which a person receives four projects and each of the others receives one four times, once for each way of designating one of the four projects the person who receives four projects as the project that person receives.

Your method counts each case in which a person receives three projects, a second person receives two projects, and each of the others receives a single project six times, once for each way of designating one of three projects the person who receives three projects as the project that person receives and once for each way of designating one of the two projects the person who receives two projects receives as the project that person receives.

Your method counts each case in which three people each receive two projects and the other person receives one eight times, once for each of the two ways you could designate one of the two projects each of three people who receives two projects as the project that person receives.

Notice that
$$\color{red}{4} \cdot \binom{4}{1}\binom{7}{4}3! + \color{red}{3 \cdot 2} \cdot \binom{4}{1}\binom{7}{3}\binom{3}{1}\binom{4}{2}2! + \color{red}{2 \cdot 2 \cdot 2} \cdot \binom{4}{3}\binom{7}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1} = \color{red}{53,760}$$


问:
Frobenius norm of the difference of two symmetric matrices.
Let A and B be the $n\times n$ symmetric matrices. Let $\lambda_A, \lambda_B$ be the largest eigen values of A and B. Show that $$|\lambda_A-\lambda_B|\leq \|A-B\|_F.$$
Here $\|A\|_F$ is the Frobenius norm.
I want to use the fact that $$\lambda_A=\max_{\|v\|=1, v\in\mathbb{R}^n}v^TAv.$$ Please help how to do this.

答: Hint. Prove that $\lambda_\max(A)\le\lambda_\max(A-B)+\lambda_\max(B)\le\|A-B\|_F+\lambda_\max(B)$. It follows that $\lambda_\max(A)-\lambda_\max(B)\le\|A-B\|_F$ and by symmetry, also that $\lambda_\max(B)-\lambda_\max(A)\le\|B-A\|_F$. Hence the conclusion follows.


问:
Are there real numbers that have no finite description?
Here's my argument:

Descriptions of finite length are countably infinite (we can enumerate through all descriptions of max length 1 with 1 distinct symbols, then max length 2 with 2 distinct symbols, etc.)

Some finite length descriptions specify a single real number including the axioms/definitions needed to define the real number.

Therefore the real numbers that have a finite length description are a subset of all finite length descriptions.

Since there are uncountable reals, then some reals have no finite length description.

Someone may argue, we cannot define which of those descriptions correspond to real numbers and which do not.

Agreed - due to Godel's incompleteness theorem, some of those descriptions will correspond to real numbers that cannot be proven to be real numbers from the axioms and definitions in the description.

Therefore, given a finite description of axioms and definitions and a candidate real number, we cannot decide whether or not it's a real number from the axioms and definitions in the description.

The enumeration of defineable real numbers, then is not defineable.

However, every defineable real number is an element of descriptions of finite length.

A similar question is here: Are there real numbers that cannot be uniquely expressed with a finite number of symbols?

I don't understand this part of this answer:

Externally, it is possible (but not necessary) that every real number can be uniquely expressed in the same metalanguage you use to describe the set theory you're using. Here is a related question asking whether, not just each individual real number, but every set can be defined.

Apparently there's a model in ZFC that can specify each set - and thus real number. How does that work with my argument? Does that model require an infinite length description? If so, then to fully describe those real numbers, you need to fully describe the model of ZFC, which requires an infinite description. That's compatible with my argument that says some real numbers require an infinite description.

答:
I read a book by Y. Manin where he presents a very similar argument for why there exist uncomputable functions. Then he gives a critique of that argument, which I will try to apply here:

We don't know what a $\bf description$ formally is. This may not seem to be a problem, since for any reasonable way to formalize the notion it is going to be a finite sequence of letters under a finite alphabet. But can we assume that we can work with a single such formalization? Maybe for any real number, we can find an intuitively valid description of it, but there is no formalization that can capture them all.

Externally, it is possible (but not necessary) that every real number can be uniquely expressed in the same metalanguage you use to describe the set theory you're using.

I believe that here $\textit{to be uniquely expressed in a metalanguage}$ means $\textit{to have an intuitively valid description}$.

Apparently there's a model in ZFC that can specify each set - and thus real number.

I am not very knowledgeable in this area and can be totally wrong, but here is what I think:

Models do not specify anything. Formal theories specify elements of their models. The answer to the question suggests that there is a model of ZFC, each element of which can be specified in ZFC. So we are not talking about specifying all sets in general, but all sets that are elements of that particular model (and there must be countably many of them, I believe).


问:
Proving quasiconvexity of $x^3 + y^3$
I was studying quasiconvex functions, and suddenly I have stumbled upon the function $f(x,y)= x^3+y^3$.

Is it quasiconvex? From the graph it seems like.

答:
It is not quasiconvex nor is it quasiconcave.

In general, for $a,b,c>0$ and $a>b$, consider local extrema of $x^3+y^3$ along the line $ax+by=c$. substituting the constraint gives us a cubic in $x$ and cubics of a real variable always have range $\mathbb{R}$. The cubic you get will always have two distinct critical points, a local max, and a local min, and so it is not quasiconvex or quasiconcave. Therefore, picking two points with appropriate $x$ on such a line will violate the definitions of both quasiconvexity and quasiconcavity for $f$.

Title: Proving quasiconvexity of $x^3 + y^3$


问:
Proof of Segner's lemma for the number of sign alternations in polynomials
In the context of Descartes' Rule of Signs, Segner's lemma is often stated as follows:

The polynomial $(x-1)p(x)$ has more alternations of signs than $p(x)$, for any real polynomial $p(x)$.

This is discussed e.g. in arXiv:1309.6664, but I was looking for a more focused treatment about this particular aspect, possibly including images to better visualise what's going on with the argument.

答:
Write the polynomial as $P(x)=\sum_{k=0}^n a_k x^k$ with real coefficients $a_k\in\mathbb R$. We can always represent this polynomial via its sequence of coefficients $\boldsymbol a\equiv (a_0,...,a_n)$.

The number of alternations of signs, denoted with $v(P)$, is the number of times in the sequence $\boldsymbol a$ that we go from a positive to a negative sign, or vice versa. Equivalently, it's the number of indices $0\le k\le n-1$ such that $a_k a_{k+1}<0$.
This definition has to be somewhat adapted when $c_k=0$ for some $k$. Here we can simply ignore these so-called lacunary coefficients. This means that we count as an alternation of sign only a pair of successive nonzero coefficients with different signs. For example, $\boldsymbol a=(1,0,0,-1,1)$ has two alternations, as it is equivalent to $\boldsymbol a=(1,-1,1)$ for our purposes.

Multiplying by $(x-1)$ (or $(x-\alpha)$ for any $\alpha>0$ for that matter), gives the polynomial
$$(x-1)P(x) = -a_0 + (a_0-a_1)x + + \cdots + (a_{n-1}-a_n)x^n + a_n x^{n+1}.$$
Our question is thus how the number of alternations of signs in $(x-1)P(x)$ relates to $v(P)$.
Denote with $\boldsymbol a'$ the coefficients of $(x-1)P(x)$. Note that $a_k'=-a_k+a_{k-1}$, with the understanding that in this formula $a_{-1}=a_{n+1}=0$.

TL;DR

We consider how the signs of the coefficients of $P(x)$ relate to the signs of the coefficients of $(x-1)P(x)$.

We observe that trailing and leading zeros in the sequence of coefficients of $P$ can be neglected for the purpose of counting sign alternations. Similarly, we observe that we can replace any subset of repeated signs with a single one.

We are left considering sequences made of $+1,-1,$ and $0$s, without repetitions. We observe that the $0$s can also be neglected as far as sign alternations are concerned.

We are left with simple sequences of alternating $\pm1$ signs. For these we can easily count the number of sign alternations and reach the conclusion.

Full argument

Because we only really interested in the signs of the coefficients, let us denote with $\boldsymbol s$ and $\boldsymbol s'$ the signs corresponding to $\boldsymbol a$ and $\boldsymbol a'$. We chose $s_k,s_k'=\pm1$ if the corresponding coefficients $a_k,a_k'\lessgtr0$, or $s_k,s_k'=0$ if $a_k,a_k'=0$.
There are a few main observations we can make here:

$s'_0=-s_0$

If $s_k s_{k+1}<0$ (a sign alternation in the original sequence) then $s'_{k+1}=s_k$

If $s_k s_{k+1}>0$ (a sign permanence in the original sequence) then $s'_{k+1}$ cannot be determined from $\boldsymbol s$ alone.

If $s_k=0$, then $s'_{k}=s_{k-1}$ and $s'_{k+1}=-s_{k+1}$.

These rules convert the problem into a puzzle in which given a few ground rules we want to figure out how many sign changes are in the second sequence given those in the first one. Here is a bunch of examples of these rules applied to random $6$-elements sequences:

Here, each second row gives $\boldsymbol s'$ corresponding to the $\boldsymbol s$ reported in the first row. I'm denoting with a question mark the signs we cannot infer from $\boldsymbol s$.

A few things that are clear from observing these examples (or equivalently, from analysing the abovementioned rules) are:

Leading and trailing zeros can be safely ignored, as they don't change the number of sign alternations.

Similarly, repeated adjacent zeros can be replaced with a single zero

Let us then focus, without loss of generality, on the sequences without these features. Here's a few examples:

Observe now that, whenever we have repeated adjacent identical signs in the first row, we correspondingly get question marks in the second row. Question marks represent our inability to assess sign changes with the given knowledge. Still, these question marks can never decrease the number of sign changes in the second row, only possibly increase it. If we then forego trying to get an exact estimate, but limit ourselves to get a lower bound on the number of sign changes on the second row, we can ignore these question marks. This corresponds to ignoring sign permanences in the first rows. Let us then do this and consider sequences that do not have sign permanences. Here is a few examples:

We are almost there.

Focus now on the $0$ elements. Observe that we cannot simply remove these. For example, a $(+,0,+)$ pattern results in $(\square,+,-)$ (square denoting something that depends on the elements on the left of the pattern), while without the zero we would get a question mark due to the repeated $+$.

Instead, we can have two different types of patterns around a zero. Either $(s,0,s)$ with $s=\pm$, or $(s,0,-s)$ with $s=\pm$. In the former case, the second row has the pattern $(-s,s,-s)$, while in the latter case the second row has the pattern $(-s,s,s)$.
So when we have the pattern $(s,0,s)$, we get two additional sign changes in the second row, while with the pattern $(s,0,-s)$ there is no additional sign change.

All this is telling us that we can neglect all zeros from our sequences, and in the process only neglect zero or two additional sign changes.

Once the zeros have also been removed safely, we are left with simple sequences of alternating signs. More precisely, we have sequences of the form $(s,-s,s,-s,...,\pm s)$ of length $n$. Any such sequence has exactly $n-1$ sign alternations. Any such sequence becomes another sequence of the form $(-s,s,-s,s,...,\pm s)$ of length $n+1$, which has exactly $n$ sign alternations.
We thus proved that $(x-1)P(x)$ has at least one sign alternation more than $P(x)$.

To generate the images I used the following Mathematica snippet:

```<code>signsAfterProd[listOfSigns_] := ReplacePart[
    listOfSigns,
    {
     1 -> -listOfSigns[[1]],
     j_ :> Which[
        listOfSigns[[j]] == 0 && listOfSigns[[j - 1]] == 0, 0,
        listOfSigns[[j]] == 0, listOfSigns[[j - 1]],
        listOfSigns[[j - 1]] == 0, -listOfSigns[[j]],
        listOfSigns[[j]] == 1 && listOfSigns[[j - 1]] == -1, -1,
        listOfSigns[[j]] == -1 && listOfSigns[[j - 1]] == 1, +1,
        True, 5
        ] /; 1 < j <= Length@listOfSigns
     }
    ] // Append[#, Last@listOfSigns] &;

prettyDoubleRowGrid[listOfLists_] := listOfLists /. {
     1 -> Style["+", Red, Background -> LightRed],
     -1 -> Style["-", Blue, Background -> LightBlue],
     0 -> Style["0", Background -> LightGray],
     5 -> Style["?", Background -> LightPurple]
     } //
   Grid[#, Frame -> All, ItemStyle -> Directive[Larger, Bold]] &;

makeNewSequence[sequence_] := {#, signsAfterProd@#} &@sequence;

Table[makeNewSequence@RandomChoice[{+1, 0, -1}, 6] //
   prettyDoubleRowGrid, 5, 5] // Grid

Table[RandomChoice[{+1, 0, -1}, 6], 1000] //
    Select[#[[1]] != 0 && #[[-1]] != 0 &&
       Not[Or @@
         Thread[Greater[Length /@ SequenceCases[#, {0 ..}], 1]]] &] //
    prettyDoubleRowGrid@*makeNewSequence /@ #[[;; 25]] & //
  Partition[#, 5] & // Grid

Table[RandomChoice[{+1, 0, -1}, 6], 1000] //
    Select[#[[1]] != 0 && #[[-1]] != 0 &&
       Not[Or @@
         Thread[Greater[
           Length /@ SequenceCases[#, {(0 .. | 1 .. | -1 ..)}],
           1]]] &] //
   prettyDoubleRowGrid@*makeNewSequence /@ #[[;; 25]] & //
  Partition[#, 5] & // Grid
</code>```

Body:
Write the polynomial as $P(x)=\sum_{k=0}^n a_k x^k$ with real coefficients $a_k\in\mathbb R$. We can always represent this polynomial via its sequence of coefficients $\boldsymbol a\equiv (a_0,...,a_n)$.

The number of alternations of signs, denoted with $v(P)$, is the number of times in the sequence $\boldsymbol a$ that we go from a positive to a negative sign, or vice versa. Equivalently, it's the number of indices $0\le k\le n-1$ such that $a_k a_{k+1}<0$.
This definition has to be somewhat adapted when $c_k=0$ for some $k$. Here we can simply ignore these so-called lacunary coefficients. This means that we count as an alternation of sign only a pair of successive nonzero coefficients with different signs. For example, $\boldsymbol a=(1,0,0,-1,1)$ has two alternations, as it is equivalent to $\boldsymbol a=(1,-1,1)$ for our purposes.

Multiplying by $(x-1)$ (or $(x-\alpha)$ for any $\alpha>0$ for that matter), gives the polynomial
$$(x-1)P(x) = -a_0 + (a_0-a_1)x + + \cdots + (a_{n-1}-a_n)x^n + a_n x^{n+1}.$$
Our question is thus how the number of alternations of signs in $(x-1)P(x)$ relates to $v(P)$.
Denote with $\boldsymbol a'$ the coefficients of $(x-1)P(x)$. Note that $a_k'=-a_k+a_{k-1}$, with the understanding that in this formula $a_{-1}=a_{n+1}=0$.

TL;DR

We consider how the signs of the coefficients of $P(x)$ relate to the signs of the coefficients of $(x-1)P(x)$.

We observe that trailing and leading zeros in the sequence of coefficients of $P$ can be neglected for the purpose of counting sign alternations. Similarly, we observe that we can replace any subset of repeated signs with a single one.

We are left considering sequences made of $+1,-1,$ and $0$s, without repetitions. We observe that the $0$s can also be neglected as far as sign alternations are concerned.

We are left with simple sequences of alternating $\pm1$ signs. For these we can easily count the number of sign alternations and reach the conclusion.

Full argument

Because we only really interested in the signs of the coefficients, let us denote with $\boldsymbol s$ and $\boldsymbol s'$ the signs corresponding to $\boldsymbol a$ and $\boldsymbol a'$. We chose $s_k,s_k'=\pm1$ if the corresponding coefficients $a_k,a_k'\lessgtr0$, or $s_k,s_k'=0$ if $a_k,a_k'=0$.
There are a few main observations we can make here:

$s'_0=-s_0$

If $s_k s_{k+1}<0$ (a sign alternation in the original sequence) then $s'_{k+1}=s_k$

If $s_k s_{k+1}>0$ (a sign permanence in the original sequence) then $s'_{k+1}$ cannot be determined from $\boldsymbol s$ alone.

If $s_k=0$, then $s'_{k}=s_{k-1}$ and $s'_{k+1}=-s_{k+1}$.

These rules convert the problem into a puzzle in which given a few ground rules we want to figure out how many sign changes are in the second sequence given those in the first one. Here is a bunch of examples of these rules applied to random $6$-elements sequences:

Here, each second row gives $\boldsymbol s'$ corresponding to the $\boldsymbol s$ reported in the first row. I'm denoting with a question mark the signs we cannot infer from $\boldsymbol s$.

A few things that are clear from observing these examples (or equivalently, from analysing the abovementioned rules) are:

Leading and trailing zeros can be safely ignored, as they don't change the number of sign alternations.

Similarly, repeated adjacent zeros can be replaced with a single zero

Let us then focus, without loss of generality, on the sequences without these features. Here's a few examples:

Observe now that, whenever we have repeated adjacent identical signs in the first row, we correspondingly get question marks in the second row. Question marks represent our inability to assess sign changes with the given knowledge. Still, these question marks can never decrease the number of sign changes in the second row, only possibly increase it. If we then forego trying to get an exact estimate, but limit ourselves to get a lower bound on the number of sign changes on the second row, we can ignore these question marks. This corresponds to ignoring sign permanences in the first rows. Let us then do this and consider sequences that do not have sign permanences. Here is a few examples:

We are almost there.

Focus now on the $0$ elements. Observe that we cannot simply remove these. For example, a $(+,0,+)$ pattern results in $(\square,+,-)$ (square denoting something that depends on the elements on the left of the pattern), while without the zero we would get a question mark due to the repeated $+$.

Instead, we can have two different types of patterns around a zero. Either $(s,0,s)$ with $s=\pm$, or $(s,0,-s)$ with $s=\pm$. In the former case, the second row has the pattern $(-s,s,-s)$, while in the latter case the second row has the pattern $(-s,s,s)$.
So when we have the pattern $(s,0,s)$, we get two additional sign changes in the second row, while with the pattern $(s,0,-s)$ there is no additional sign change.

All this is telling us that we can neglect all zeros from our sequences, and in the process only neglect zero or two additional sign changes.

Once the zeros have also been removed safely, we are left with simple sequences of alternating signs. More precisely, we have sequences of the form $(s,-s,s,-s,...,\pm s)$ of length $n$. Any such sequence has exactly $n-1$ sign alternations. Any such sequence becomes another sequence of the form $(-s,s,-s,s,...,\pm s)$ of length $n+1$, which has exactly $n$ sign alternations.
We thus proved that $(x-1)P(x)$ has at least one sign alternation more than $P(x)$.

To generate the images I used the following Mathematica snippet:

signsAfterProd[listOfSigns_] := ReplacePart[
    listOfSigns,
    {
     1 -> -listOfSigns[[1]],
     j_ :> Which[
        listOfSigns[[j]] == 0 && listOfSigns[[j - 1]] == 0, 0,
        listOfSigns[[j]] == 0, listOfSigns[[j - 1]],
        listOfSigns[[j - 1]] == 0, -listOfSigns[[j]],
        listOfSigns[[j]] == 1 && listOfSigns[[j - 1]] == -1, -1,
        listOfSigns[[j]] == -1 && listOfSigns[[j - 1]] == 1, +1,
        True, 5
        ] /; 1 < j <= Length@listOfSigns
     }
    ] // Append[#, Last@listOfSigns] &;

prettyDoubleRowGrid[listOfLists_] := listOfLists /. {
     1 -> Style["+", Red, Background -> LightRed],
     -1 -> Style["-", Blue, Background -> LightBlue],
     0 -> Style["0", Background -> LightGray],
     5 -> Style["?", Background -> LightPurple]
     } //
   Grid[#, Frame -> All, ItemStyle -> Directive[Larger, Bold]] &;

makeNewSequence[sequence_] := {#, signsAfterProd@#} &@sequence;

Table[makeNewSequence@RandomChoice[{+1, 0, -1}, 6] //
   prettyDoubleRowGrid, 5, 5] // Grid

Table[RandomChoice[{+1, 0, -1}, 6], 1000] //
    Select[#[[1]] != 0 && #[[-1]] != 0 &&
       Not[Or @@
         Thread[Greater[Length /@ SequenceCases[#, {0 ..}], 1]]] &] //
    prettyDoubleRowGrid@*makeNewSequence /@ #[[;; 25]] & //
  Partition[#, 5] & // Grid

Table[RandomChoice[{+1, 0, -1}, 6], 1000] //
    Select[#[[1]] != 0 && #[[-1]] != 0 &&
       Not[Or @@
         Thread[Greater[
           Length /@ SequenceCases[#, {(0 .. | 1 .. | -1 ..)}],
           1]]] &] //
   prettyDoubleRowGrid@*makeNewSequence /@ #[[;; 25]] & //
  Partition[#, 5] & // Grid


问:
Show that there is an $m\in\mathbb{N}_0$ So that $\ker(A)\subsetneq ...\subsetneq \ker(A^m)=\ker(A^{m+1})=...$
Let $A:V\to V$ be an endomorphism of the finite dimensional vector space $V$. For $n\in \mathbb{N}$ set $U_n := \ker(A^n)$ and $W_n := \text{Im}(A^n) $. Show that there is an $m\in \mathbb{N}_0$ so that:
\begin{equation}
W_1\supsetneq ...\supsetneq W_m=W_{m+1}=W_{m+2}=...
\end{equation}
and
\begin{equation}
U_1\subsetneq ...\subsetneq U_m=U_{m+1}=U_{m+2}=...
\end{equation}

I don't know how could I start solving this. Could somebody give me a hand?

答:
Observe that $$v\in \ker(A^n)\Rightarrow A^nv=0\Rightarrow AA^nv=0\Rightarrow A^{n+1}v=0 \Rightarrow v\in\ker(A^{n+1})$$
and thus $U_n\subseteq U_{n+1}$. Similarly, if $w\in Im(A^n)$, then there exists $v\in V$ such that $A^nv=w$. Setting $Av=u$ gives us $A^{n-1}(Av) = A^{n-1}u = w$ which shows that $w\in Im(A^{n-1})$ and thus $W_{n+1}\subseteq W_n$. We have proved the subset inclusions.

To show these sequences of sets have a maximal element, you can observe that each of your spaces in the sequences $\{U_n\}_{n=1}^{\infty}$ and $\{W_n\}_{n=1}^{\infty}$ are subspaces of the finite vector space $V$. Note that the dimensions of $U_n$ are increasing(non-decreasing) and the dimensions of $W_n$ are decreasing(non-increasing). Let $n$ be the dimension of your vector space $V$. If there is no $m$ such that $U_{m+i} = U_m$ for all positive integers $i$, then after finite steps, $U_M$ for some $M>m$ will equal the entire vector space $V$. I hope you can formulate this rigorously. Similar, argument holds for the sequence of $W_n$s (which will tend to the trivial vector space $\{0\}$ if it is not equal to some finite space $W_{M'}$.

Once it is established for $U_n$s you can apply the rank nullity theorem to get the result for $W_n$s.

(If you vector space is nilpotent, then $U_M = V$ and $W_{M'}=\{0\}$.)


问:
What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

What do we mean when we say "Let $x$ be an element of the set $\mathbb{R}$"?

Does $x$ represents only a single  element of set $\mathbb{R}$? or does $x$ represent all the element of set $\mathbb{R}$ simultaneously at the same time?

People say that if $x\in\mathbb{R}$ then $x$ is any real number; that means $x$ represents all real numbers. But if $x$ is any real number then let's say $x=1$; so $x$ is one, then how it can represent all real numbers?

Please help me I am very confused.

答:
I know an answer has already been accepted, but I'll try to answer the heart of the question anyway. All of the following statements mean the same thing:

Let $ x \in \mathbb{R}$. Then $x^2 \in$ {$u \in \mathbb{R}: u \geq 0$}.

Let $ x \in \mathbb{R}$. Then $x^2 \in \mathbb{R}$ and $x^2 \geq 0$.

$ x \in \mathbb{R} \implies x^2 \in$ {$u \in \mathbb{R}: u \geq 0$}.

$ x \in \mathbb{R} \implies \left(x^2 \in \mathbb{R} \text{ and } x^2 \geq 0 \right)$.

If $ x \in \mathbb{R}$, then $x^2 \in$ {$u \in \mathbb{R}: u \geq 0$}

If $ x \in \mathbb{R}$, then $\left(x^2 \in \mathbb{R} \text{ and } x^2 \geq 0 \right)$

If $x$ is any real number then its square is a non-negative real number.

$$$$

The statements are true. However, note that their truth value is independent to whether or not the statements mean the same thing, In other words, false statements can also be equivalent.

Personally, I avoid using the statement with "any" in it, because it can be confusing and I'd rather just get on with the maths than spend time being confused. If someone else uses the statement with "any" in it, then I translate it into one of the other statements.


问:
Is $[0,5]$ open in $[-5,5]$?
I'm writing examples about relatively open sets. Can an open subset $U\subset\Bbb{R}$ be found such that $[0,5]=U \cap[-5,5]$, or is this maybe a nonsensical question because of some theorem?

答: Nope, since every open set in $\mathbb{R}$ is made by open intervals (basis for the euclidean topology) and there is no way to intersect an open set of $\mathbb{R}$ with $[-5,5]$ such that $[0,5]$ is open in the subspace topology. $[0,5]$ is a closed set (easy to verify) and indeed it cannot be open otherwise you would reach the contradiction of having $[-5,5]$ a disconnected space.

Title: Is $[0,5]$ open in $[-5,5]$?


答:
First note that $G$ is abelian, therefore $Z(G) = G$

($Z$ is the Center of $G$).

Then you can use a little proposition (Humphreys pg.73-74) that tells you $\text{Inn}(G) \approx \frac{G}{\text{Z}(G)}$

The proof of this prop in a few words:

define an automorphism (prove it) $\varphi_x(g)=xgx^{-1}$, then define $\phi : G \to \text{Aut}(G)$ by $x \mapsto \varphi_x$  and find its image and its kernel, then apply first theorem of homomorphism and you conclude.

Body:
Find Aut$(G)$, Inn$(G)$ and $\dfrac{\text{Aut}(G)}{\text{Inn}(G)}$ for $G = \mathbb{Z}_2 \times \mathbb{Z}_2$.

Here is what I have here:

Aut$(G)$ consists of 6 bijective functions, which maps $G$ to itself, since Aut$(\mathbb{Z}_2 \times \mathbb{Z}_2) \approx S_3$.

I think the next part goes wrong.  For Inn$(G)$, letting $\kappa_x : G \rightarrow G$ to be the conjugate function, I found $\{e, \kappa_{(0,1)}, \kappa_{(1,0)}, \kappa_{(1,1)}\}$.

I can't determine $\dfrac{\text{Aut}(G)}{\text{Inn}(G)}$ yet since I need to correctly determine Inn$(G)$.

Any advices or comments?


问:
Product of $x$ with $x$ and $x+1$ being coprime to $n$

For an odd positive integer $n \gt 1$, let $S$ be the set of integers $x$, $1 \le x \le n$, such that both $x$ and $x+1$ are relatively prime to $n$. Show that $$\prod_{x\in S} x \equiv 1 \pmod n$$

I know there is already a solution on this site here but I don't understand the solution fully and the thread was created over 3 years ago so that is why I decided to create a new thread. Can anyone retranslate the solution presented in the linked thread. On my approach I thought of using $CRT$ but I couldn' t solve the problem using it, so if you know solution that involves $CRT$ please post it.

Any help appreciated.

答: Since $x$ is relatively prime with $n$ there exists some $y$ s.t. $xy  \equiv 1 \mod {n}$. Now assume that $x \ne y \mod{n}$. Then observe that $y+1 \equiv \frac {1}{x} +1 \equiv \frac {1+x}{x}\mod {n}$. Since both $x+1$ and $x$ are relatively prime with $n$, they have muliplicative inverses respective to modulo $n$. Thus, from the previous equality, $y+1$ necessarily has a multiplicative inverse. But this implies that $ y+1 $ and $n$ are relatively prime. Moreover $ y $ and $n$ are relatively prime(since $x$ is the multiplicative inverse of $y$).Hence $y \in S$. If $x = y$, then $x^2 \equiv 1 \mod{n}$ which implies $(x-1)(x+1) \equiv 0 \mod{n}$. But since $x+1$ is relatively prime with $n$, it has an inverse and so this implies that $x\equiv 1 \mod {n}$. Hence, if $x \in S $ and x is not equal to its multiplicative inverse $y$ in modulo $n$, $y \in S$,else $x\equiv 1 \mod {n}$ .


问:
Suppose $A\subseteq{C}\And{B}\subseteq{D}$ Prove that: $A\times{B} = \left(A\times{D}\right)\cap\left(C\times{B}\right) $
I'm not sure how to approach it. So far I've come to this:
\begin{array}{|c|}
            \hline
            a\in{A}\rightarrow{}a\in{C} \\
            b\in{B}\rightarrow{}b\in{D} \\
            \hline
\end{array}
$$\left(x,y\right)\in \left(A\times{D}\right)\cap\left(C\times{B}\right)$$
$$\Downarrow$$
$$\left(x,y\right)\in\left(A\times{D}\right)\land\left(x,y\right)\in\left(C\times{B}\right)$$
$$\Downarrow $$
\begin{array}{c|c}
x\in{A} & x\in{C} \\
y\in{D} & y\in{B}
\end{array}
True
$$\Downarrow$$
$$A\times{B} = \left(A\times{D}\right)\cap\left(C\times{B}\right)$$
But I don't think it's correct at all.

答: If $(a,\,b)\in A\times D\cap C\times B$ then $a\in A\cap C=A$ and $b\in B\cap D=B$, so $(a,\,b)\in A\times B$. On the other had, if $(a,\,b)\in A\times B$ then $(a,\,b)\in A\times D$ and $(a,\,b)\in C\times B$.


问:
Proof that we can find rational numbers arbitrarily close to $\sqrt{2}$: direct approach.
The statement of the proposition:

Proposition. For every rational number $\epsilon > 0$, there exists a non-negative rational number $x$ such that $x^{2} < 2 < (x+\epsilon)^2$.

The most common approach to proving the proposition is by using contradiction (1,2).

My question is: is it possible to prove the proposition directly? More concretely, is it possible to find a function $f: \mathbb Q^+\rightarrow \mathbb Q^+$ such that for arbitrary positive rational $\epsilon$, we have

$$f(\epsilon)^2 < 2 < (f(\epsilon) + \epsilon)^2 $$

?

答:
Define $f(\varepsilon)$ to be the truncation of $\sqrt{2}$ to $n$ decimal places, where $10^{-n} \leq \varepsilon$ is the nearest power of $10$ from below.

This ensures that $$f(\varepsilon) < \sqrt{2} < f(\varepsilon) + 10^{-n} \leq f(\varepsilon) + \varepsilon$$

To illustrate, if $\varepsilon=0.2$ then $n=1$ and the inequality reads
$$1.4 < \sqrt{2} < 1.5 \leq 1.6$$


问:
Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$.
Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$

While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it through other means, say by AM-GM? Any hints would be much appreciated.

答:
A bit late but I thought worth mentioning it:

Consider the equivalent inequality

$$4x^2+12x+ \frac 4x = (2x)^2 + 6(2x)+\frac 8{2x}\geq 15$$

Setting $y=2x$, you get

\begin{eqnarray*} y^2+6y+\frac 8y
& = & y^2 - 2y +8\left(y+\frac 1y\right)\\
& = & (y-1)^2-1+8\left(y+\frac 1y\right) \\
& \geq & 0-1+8\cdot 2 = 15
\end{eqnarray*}
with equality if and only if $y=1 \Leftrightarrow x=\frac 12$.


问:
Defined Integral of a function defined by parts
The function is the following:

The objetive of the exercise is to calculate
.

For that I used the following steps:

My resolution is wrong because the result has to be 2.

答:
So, you have two calculations wrong.  The biggest problem is that, even though at x = 1, y = 2, this is an infinitesimal point.  Therefore, it has no bearing whatsoever on the integral.  If y was equal to 2 for some period of time (like 0.8 > x > 1.2), then it would have an effect.  But, for the infinitely small slice of time where x = 1, y isn't 2 long enough to make any difference.  Therefore, you can treat it as just an ordinary integral of $x\,dx$ from 0 to 2.

The other problem you did is that, when you split up the integral into two pieces, you did the right hand one wrong.

$$\int_1^2 x\,dx = \frac{x^2}{2} \biggr\rvert_1^2 = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$$

When added to your first one, you would get 2.


问:
Line integral - Parametric representation
the question is: C is the following area:

$$C = \left\lbrace (x,y) \; \middle| \;
         x^2 + y^2=\frac{\pi^2}{4}, \;  x \geq 0  \right\rbrace$$

Begins in: $\displaystyle \left( 0,\frac{\pi}{2} \right)$ end in $ \displaystyle \left( 0, -\frac{\pi}{2} \right)$

we want to calculate the integral : $$I = \int_{C} \left( e^x\sin{y} \; + 24 y \right) \, dx + e^x \cos{y} \,dy$$

First I'm required to use green's theorem, - I completed C to a complete circle and calculated :
$$ \displaystyle
  J   =  \int_{C\cup C_1 }(e^x\sin{y} \; + 24 y) \,dx + e^x\cos{y} \, dy
      =  \iint 24 dxdy = {3\pi ^3} $$

Now I'm told to subtract the following line integral with $y(t) = t$:
$$
\int_{C_1} ( e^x \sin{y} \; + 24 y ) \, dx + e^x \cos{y} \, dy = {} \rule{60pt}{0pt}
$$

Where $C_1$ is the "rest of the circle"
So its:
$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} *something* dt$

I Can't figure out what that something should be.. help would be greatly appreciated.

答:
First I'm required to use green's theorem, on a complete circle

Here is an alt approach, consider the half disc on right side and do greens on that,

$$ \int_{\partial C} \vec{F} \cdot ds = \int_{C} \nabla \times \vec{F} dA$$

Or,

$$ \int_C (e^x \sin y + 24 y) dx + (e^x \cos(y) ) dy = \int \nabla \times < e^x \sin y + 24 y, e^x cos y> dA = \int (e^x \cos(y) -e^x \cos(y) +24) dx dy = 24 \frac{\pi r^2}{2}$$

Now that you have this substract the vertical line integral from which goes from $(0 , \frac{\pi}{2} ) \to ( 0 , - \frac{\pi}{2})$ i.e: over the line $ x=0$ from y=$  \frac{\pi}{2} \to \frac{-\pi}{2}$. It's parameteric equaton is:$$ \vec{r(t)} =  \pi t \vec{j}$$

From $ t= \frac{1}{2}$ $\to$$ t= - \frac{1}{2}$


答:
$\color{blue}{g(x)=2 \text{ when } x = 1}$ doesn't mean $\color{red}{g(x)=2 \text{ when } x\in[1,2](\text{or some other intervals with length 1})}$.

And recall the definition of the integral.

Title: Defined Integral of a function defined by parts

Body:
The function is the following:

The objetive of the exercise is to calculate
.

For that I used the following steps:

My resolution is wrong because the result has to be 2.


答:
An integral from $a$ to $b$ isn’t simply the sum of $f(x)$ for each $x\in[a,b]$, but instead the sum of $f(x) dx$ (times an infinitesimal) for each $x\in[a,b]$. So, when you’re dividing the integral into parts, you should get:

$$\int_0^2 g(x) dx = \int_0^1 x dx + g(1) dx + \int_1^2 x dx $$ and $g(1) dx \to 0$ so it can be neglected.


问:
An unbounded sequence s.t. $\forall t>0 |[a_nt]-a_nt|\rightarrow 0$
While studying probability I accidentally proved the result below by what I think is an overly complicated proof (a sketch of this proof below). Is there an easy way to prove it?

The result is the following.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals such that for all $t>0$ we have $\lim_{n} |[a_nt]-a_nt|=0$, where we denote with $[x]$ the nearest integer to x.

Then $\{a_n\}_{n\in\mathbb{N}}$ converges to $0$.

This is equivalent to another result.

Let $\{a_n\}_{n\in\mathbb{N}}$ be a sequence over the reals and $a$ a real number such that for all $t>0$ we have $\lim_{n} \text{e}^{\text{i}a_nt}=e^{\text{i}at}$

Then $\{a_n\}_{n\in\mathbb{N}}$ converges.

Proof. Let us suppose that $\{a_n\}_{n\in\mathbb{N}}$ doesn't converge. We will study two possible scenarios, both leading to contradiction.

First case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is bounded.
Then, choosing a small enough $t$, we get that $\forall n |a_nt|<\pi$, and we conclude, using the continuity of the complex logarithm, that the sequence converges to $a+2k\pi$ for some $k\in\mathbb{Z}$.

Second case. The sequence $\{a_n\}_{n\in\mathbb{N}}$ is unbounded. Then, we consider a sequence $\{X_n\}_{n\in\mathbb{N}}$ of random variables such that $\forall n X_n\sim N(a_n,1)$ and their respective characteristic functions $\{\text{exp}(\text{i}a_nt-t^2/2)\}_{n\in\mathbb{N}}$. Our initial assumptions imply that the sequence of characteristic functions is pointwise convergent to $\text{exp}(\text{i}at-t^2/2)$, so, by Paul-Levy's theorem, the sequence of random variables $\{X_n\}_{n\in\mathbb{N}}$ converges in distribution to a random variable $X$ such that $X\sim N(a,1)$. However, it can be proved via Chebishev inequality that if the sequence of numbers $\{a_n\}_{n\in\mathbb{N}}$ is unbounded then the sequence of random variables doesn't converge in distribution to any random variable.

答:
In order to redeem myself from the stupid comments the extremely late other night, I thought that I'd share a different proof of the second statement. Primarily, we notice that

$$ 2 \left(1- \frac{\sin(a_n - a)}{(a_n - a)} \right) = \int_0^1 \lvert e^{ia_nt}-e^{iat} \rvert^2 dt \rightarrow 0 \qquad,\, n\rightarrow \infty.
$$
This follows from the fact that $e^{ia_n t} \rightarrow e^{iat}$ pointwise, for all $t>0$ and the dominated convergence theorem. There might be a simpler way to prove this particular step, but for the moment we could content ourselves with this. It readily follows that
$$
\lim_{n \rightarrow \infty} \frac{\sin(a_n -a)}{(a_n - a)} = 1.
$$

For the sake of obtaining a contradiction, suppose that $a_n \nrightarrow a$, as $n \to \infty$. Then there exists a $\delta >0$ and sequence of integers $\{n_j \}$, such that $\lvert a_{n_j}-a \rvert \geq \delta$, for all $j\geq 1$. Since $t \mapsto \frac{\sin(t)}{t}$ is (uniformly) continuous with global maximum at $t=0$, there consequently exists a number $0<\eta <1$, such that
$$
\frac{\sin(a_{n_j}-a)}{(a_{n_j}-a)} \leq (1-\eta), \qquad, \, \forall j\geq 1.
$$
This implies that
$$ \bigl| 1- \frac{\sin(a_{n_j}-a)}{(a_{n_j}-a)}\bigr| = \left( 1- \frac{\sin(a_{n_j}-a)}{(a_{n_j}-a)}\right) \geq \eta \qquad, \forall j\geq 1, $$
which contradicts the previously established fact that $\frac{\sin(a_n -a)}{(a_n -a)} \to 1$, as $n\to \infty$.


问:
Number of possible triplets of form $(\alpha,\beta,\alpha\beta)$
Consider the $r\times 2^r-1$ matrix

$$\mathbf{G} =  \left(\begin{matrix}1 & 0 & 1 & 0 & 1 & 0 & 1\\ 0 & 1 & 1 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 & 1 & 1 & 1 \end{matrix}\right)\text{,}$$

created by considering all possible combinations $\left(c_i + c_j \mod 2 \right)$ of the three independent columns:

$$  \left(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\right)  \left(\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\right) \left(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right)\text{.} $$

Now I want to find all possible triplets of the form $(c_1,c_2,c_3)$ for $c_i \in \mathbf{G}$ so that $\left(c_3 = c_1 + c_2 \mod 2\right)$. I know that for $\mathbf{G}$, there are 7 such triplets:

$c_1,c_2,c_3$

$c_1,c_4,c_5$

$c_2,c_3,c_6$

$c_1,c_6,c_7$

$c_2,c_5,c_7$

$c_3,c_4,c_7$

$c_3,c_5,c_6$

However I would like to know if there is a formula to know the number of such triplets for a given value of $r$, i.e. for $r=3$ $n_{triplets}=7$, for $r=4$ $n_{triplets}=35$, etc...

答: There are $\frac 16(2^r-1)(2^r-2)$ triplets.  As you say, there are $2^r-1$ columns in $G$.  You can take any two of them in $(2^r-1)(2^r-2)$ ordered ways.  You find the third by doing the bitwise XOR of the first two columns.  Then you put the indices in order, which picks one order out of $6$ for the three columns.


问:
Finding the intersecting points on two circles
Given 2 circles on a plane, how do you calculate the intersecting points?

In this example I can do the calculation using the equilateral triangles that are described by the intersection and centres of the 2 circles, however, I need a more general formula that will provide the coordinates of the 2 intersection points C and D, on circles that are not placed so conveniently, or of the same radius.

答: Each circle can be described by an equation of the form $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is its radius.  Given the equations of the two circles, expanding the squared terms and subtracting the equations should yield an equation of the line through the two points of intersection.  Solving that for $x$ or $y$ and substituting into one of the original circle equations should yield the coordinates of your two points of intersection.</p>


问:
Find minimal polynomial of given matrix
I want to find the minimal polynomial of this matrix
$$\left[\begin{array}{ccccc}
2 & 0 & 0 & 0 & 0\\
4 & 2 & 0 & 0 & 0\\
1 & 4 & 2 & 0 & 0\\
0 & 0 & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 3
\end{array}\right]$$

I found it charastic polynomial $\chi_{A}=\left(2-x\right)^{3}\left(3-x\right)^{2}$

I also know that the minimal polynomial irreducible factors have the same irreducible factors as charastic polynomial so
$$m_{A}\left(x\right)\in\left\{ \left(2-x\right)^{3}\left(3-x\right)^{2},\left(2-x\right)^{3}\left(3-x\right),\left(2-x\right)^{2}\left(3-x\right),\left(2-x\right)\left(3-x\right),\left(2-x\right)\left(3-x\right)^{2}\right\}  $$
but I don't know how to continue from here

答: The minimal polynomial is $(X-2)^3(X-3)$. This a triangular matrix with a $2\times 2$ diagonal block whose eigenvalue is $3$ thus whose minimal polynomial is $X-3$. There exists a block which is a $3\times 3$ matrix which has $2$ on the diagonal. The minimal polynomial of this block is $(X-2)^3$.

Title: Find minimal polynomial of given matrix


问:
Difficulty in proving that closure of an open ball of radius $r$ is a closed ball.
Let $X$ be a metric space and $A$ be a subset of $X$ such that for some $p\in X$ and $\delta \gt 0$, we have $A=\{a\in X: d(a,p)\lt \delta\}$. Let's define $B=\{b\in X: d(b,p)\le\delta\}$.

Let $\bar A=$ Closure of $A=A\cup A'$, where $A'$ is the set of limit points of $A$.

I am trying to prove that $\bar A= B$.

For any $x\in B$, we have $d(p,x)\le \delta$ so either $d(p,x)\lt \delta $ or $d(p,x)=\delta$.

If $d(p,x)\lt \delta $, then $x\in A\subseteq A\cup A'$

However, if $d(p,x)=\delta$, then I am having difficulty in showing $x\in \bar A$. I tried as follows:

Let $0\lt\epsilon \lt \delta$, choose $z\in A$ such that $\delta \gt d(p,z)\gt \delta -\epsilon$, then we must have

$d(x,p)+d(p,z)\ge d(x,z)\implies d(x,z)\le 2\delta$. I am stuck here.

I want to show that $d(x,z)\lt \epsilon$ which would prove that $x$ is a limit point of $A$ i.e., $x\in A'\subseteq A\cup A'$

Please help. Thanks.

答: That is generally not true unless it is a normed vector space. For example, consider the metric space $(-\infty, -1] \cup [0, \infty)$. The closure of the unit ball is $[0,1]$, which does not include $-1$.


问:
What general techniques are there to show that two finitely-generated groups are bi-Lipschitz equivalent?
We define bi-Lipschitz equivalency as follows.

We say two metric spaces $(X,d)$, $(Y,e)$ are bi-Lipschitz
equivalent if there exists a bijection $f : X \to Y$ and some $ L
 \geq 1$ such that  $$ \frac 1 L \ d(x,y) \leq e(f(x),f(y)) \leq L \
 d(x,y), $$ for all $x,y \in X$.

Viewing finitely-generated (f.g.) groups as metric spaces under their word metric, we have that any two word metrics on a f.g. group are bi-Lipschitz equivalent. Thus it makes sense to discuss whether two f.g. groups $G$, $H$ are bi-Lipschitz equivalent, without fixing any generating sets.

I understand that to show two groups are not bi-Lipschitz,  we may compare their growth functions, for example. However I am not aware of any general tools for showing two groups are bi-Lipschitz equivalent, outside of just exhibiting such a map $f$ as above (and even that seems incredibly difficult).

Any references to literature which discusses this topic or contains examples would be appreciated. Thanks!

答: Whyte, Kevin, Amenability, bi-Lipschitz equivalence, and the von Neumann conjecture, Duke Math. J. 99, No. 1, 93-112 (1999). ZBL1017.54017.


问:
IMO Shortlist $2009$ - Problem C$1$

Consider $2009$ cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two player, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of $50$ consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins.
(a) Does the game necessarily end?
(b) Does there exist a winning strategy for the starting player?

Sol - a) Take the standard binary interpretation with gold=$1$ and black=$0$. Note that the total must decrease with each step, and thus must terminate.

b) No. In fact, the starting player loses no matter what moves are made. To see this, let $a_i=1$ if the $i$th card from the left is gold, and $0$ otherwise, and let $S = a_1+a_{51}+\dots+a_{2001}$, and remark that the game can only terminate if $S=0$. But on each move, $S$ changes by $1$, so during the second player's turn, $S$ will always be odd and thus never $0$.

Why game can only terminate if $S=0$ ?

If $2001$th card is $1$ and all previous are $0$'s, game can still end ?

thankyou

答:
The given solution is incorrect, for the reason you mentioned; if $a_{2001}=1$ and all other cards are zero, then $S=1$, yet the game is over. The correct solution is to let $T=a_{1960}+a_{1910}+a_{1860}+\dots+a_{10}$, and apply the same logic to $T$. Indeed, if $T>0$, then any nonzero term of $T$ gives a legal move.

They made two mistakes. First of all, they confused left and right, and meant to say that $a_i$ is the $i^{th}$ card from the right end. Second, they seemed to be thinking that the rules were as follows: you can select any card which is gold, and then flip it and the $49$ cards to its right to the other side. If there are fewer then $49$ such cards, then you just flip all that you can. With these two changes, their value of $S$ works.

Title: IMO Shortlist $2009$ - Problem C$1$


问:
Symmetric difference and equivalence relation
Let $P(\mathbb{N})$ denote the set of all subsets of $\mathbb{N}$. Define a relation $D$ on $P(\mathbb{N})$ as follows: $(A, B) \in D$ if the symmetric difference $A\Delta B := (A \cup B) \setminus (A \cap B)$
is finite. (That is: $A$ and $B$ only differ by finitely many elements).

I have already proved that $D$ is an equivalence relation and that all finite subsets of $\mathbb{N}$ are in the same equivalence class. Then the problem asks to prove that every equivalence class is a countable collection of sets and that there are uncountably many equivalence classes. These two questions come with the following hint: fix $A_{0} \in P(\mathbb{N})$, then for each $n \in \mathbb{N}$ define $A_{0}^{(n)} = \{A \subseteq \mathbb{N} : A \text{ differ from } A_{0} \text{ by } n \text{ numbers}\}$. The issue is that I still do not know how to prove $A_{0}^{(n)}$ is countable. Could you give me some hints or the solution, please?

答:
If we fix $n$ digits that are the different ones (let's say the first $n$), then the difference can only happen in finitely many ways, since each of these digits can only vary in finitely many ways. Clearly, there countably many ways of choosing these $n$ digits, so you get a countable union of finite elements, which is countable.

We then union these over all $n$, which gives a countable set (countable union of countable sets).


答:
Lemma $ $ If $\,S\,$ is a subset of $\,\Bbb Z_n\,$ closed under inverses then the product of its elements equals the product of the set of roots of $\,x^2-1\,$ lying in $\,S$.

Proof $\ S$ partitions into pairs $\,(x,x^{-1})$ with $\,x\not\equiv x^{-1}$ & singletons $(x)$ with $\,x^{-1}\equiv x.\,$ Pairs contribute $\,1\,$ to the product, so it reduces to the product of the singletons = roots of $\,x^2\equiv 1$.

Your $S\,$ is closed under inverses, i.e. $\,(n,x(x\!+\!1))=1\Rightarrow\,(n,x^{-1}(x^{-1}\!+\!1))=1,\,$ since $\,x^2\, x^{-1}(x^{-1}\!+\!1) = 1+x\,$ is coprime to $\,n\,$ hence so too are its factors $\,x^{-1},\ x^{-1}\!+\!1.$

If $\,x\in S\:\!$ and $\,x^2\!-1\equiv (x\!-\!1)(x\!+\!1)\equiv 0\, $ then $\,x\!-1\equiv 0,\,$ by $\,x\!+\!1\,$ is cancellable, being coprime to $\,n.\,$ So the set of roots of $\,x^2\!-1\,$ in $\,S\,$ is $\{1\},\,$ with product $1 = \prod S\,$ via Lemma.

Remark $ $ This is closely related to this elementary proof of a generalization of Wilson's theorerm to $\,\Bbb Z_n,\,$ based on similar reflection (involution) symmetry afforeded by inversion.

Title: Product of $x$ with $x$ and $x+1$ being coprime to $n$

Body:
For an odd positive integer $n \gt 1$, let $S$ be the set of integers $x$, $1 \le x \le n$, such that both $x$ and $x+1$ are relatively prime to $n$. Show that $$\prod_{x\in S} x \equiv 1 \pmod n$$

I know there is already a solution on this site here but I don't understand the solution fully and the thread was created over 3 years ago so that is why I decided to create a new thread. Can anyone retranslate the solution presented in the linked thread. On my approach I thought of using $CRT$ but I couldn' t solve the problem using it, so if you know solution that involves $CRT$ please post it.

Any help appreciated.


问:
Compact operator $T_{A,B}$
Let $A,B \in \mathcal{B}(X)$ where $X$ is a Banach space. Define $T_{A,B}: \mathcal{B}(X) \rightarrow \mathcal{B}(X)$ by $T_{A,B}(S)=ASB$.

Prove that if $A$ and $B$ are compact then $T_{A,B}$ is compact.

I consider a bounded sequence $\{S_n\}$ in $\mathcal{B}(X)$ but I can't find a subsequence of $T_{A,B}(S_n)$ such that is convergent. Any hint?

答:
Since $B$ is compact the image of the unit ball under $B$ is pre-compact, in  particular separable. Let $x_m$ be a countable dense subset of this image. Now consider for $m$ fixed the sequence $(S_n x_m)_{n\in\Bbb N}$, since $A$ is compact you have a sub-sequence so that $A S_{n_{k}}x_m$ converges. By the diagonal argument you may assume that $n_k$ is so that $AS_{n_k}x_m$ converges for every $m$.

Denote with $A\!S$ the map $\{x_m\mid m\in\Bbb N\}\to X$ given by $x_m\mapsto \lim_k A S_{n_k}x_m$. Since $S_n$ is bounded you have that
$$\|A\!S(x_m) -A\!S(x_l)\| ≤ \sup_n\|AS_n\|\,\|x_m-x_l\|$$
and the map is Lipschitz. You may then extend it to have domain $\overline{\{x_m\mid m\in \Bbb N\}}$, which is the closure of the image of the unit ball under $B$ (denote this map by $\overline{A\!S}$). You can again use the boundedness of $S_n$ to get that $AS_{n_k} x\to \overline{A\!S}x$ for all $x$ in the image of the unit ball under $B$:
$$\|\overline{A\!S}x - AS_{n_k}x\| ≤ \|\overline{A\!S}(x-x_m)\|+\|\overline{A\!S}(x_m)- AS_{n_k}(x_m)\|+\|AS_{n_k}(x_m-x)\|$$
you may choose $x_m$ to make the first and the second terms as small as you like, the middle term converges to zero by the arguments before - hence for any $m$ you can find $K$ after which it is as small as you like.

Continue with arguments of this kind to get that $\overline{A\!S}$ is linear and then also itself Lipschitz - you may thus extend it to a continuous linear map $\overline{A\!S}:\overline{B(X)}\to X$. Whats left is to check that $AS_{n_k}B$ converges uniformly on bounded sets to $\overline{A\!S}B$ (ie in operator norm).

Supposing this were not the case you would get a sequence $y_k$ in the unit ball of $X$ with $\|(AS_{n_k}B-\overline{A\!S}B)y_k\|>\epsilon$ for some $\epsilon$. Well you know what to do: $B(y_k)$ has a convergent sub-sequence so just assume that $B(y_k)$ converges with limit $x$. But we have already done this calculation
$$\|(\overline{A\!S}-AS_{n_k})(By_k)\| ≤ \|\overline{A\!S}(B(y_k)-x)\|+\|(\overline{A\!S}-AS_{n_k})(x)\| +\|AS_{n_k} (B(y_k)-x)\|$$
Each term converges by the same arguments as before.

Title: Compact operator $T_{A,B}$


问:
Proof verification: approximate identity for function $f(x)= (1-|x|) \chi_{[-1,1]}(x) $
I want to construct an approximate identity for the function $f(x)= (1-|x|) \chi_{[-1,1]}(x)  $ in following way:

$\phi _t (x) = \frac{1}{t} f(\frac{x}{t}) = \frac{1}{t} (1- \frac{|x|}{t}) \chi_{[-t,t]}(x) $ ,  when
$t\rightarrow 0$ from right.

I verified that  $\phi _t (x) \geq 0$ and $\int_{\mathbb{R}}  \phi _t (x) dx =1$ and I tried to prove the third property i.e. $\int_{|x| \geq \delta}  \phi _t (x) dx  \rightarrow 0$ when $t  \rightarrow 0$, for every $\delta > 0$. Can someone just check if this is correct:

We take arbitrary $\delta > 0$. Since  $t  \rightarrow 0$, we can find $t_0 < \delta $ such that for all $t \leq t_0$ holds:  $\chi_{[-t,t]}(x) =0$. Then it follows that $(1-\frac{|x|}{t}) \chi_{[-t,t]}(x) \rightarrow 0 $ when $t  \rightarrow 0$. This function is dominated by a integrable function and according to Lebesgue's dominated convergence theorem, it follows that $\int_{|x| \geq \delta}  \phi _t (x) dx  \rightarrow 0$ when $t  \rightarrow 0$, for every $\delta > 0$.

Could someone please check this solution? I can't find any similar solved tasks on this topic, so I would also be very thankful if someone recommends a book or link. Thanks a lot in advance.

答:
It's not necessary to use dominated convergence theorem.

Fix $\delta > 0$. Suppose that $|t| < \delta$ and  $|x| \ge \delta$. Hence $\chi_{[-t,t]}(x) = 0$ and it follows that $\int_{|x| \ge \delta} \phi_t(x)dx = 0$.


问:
Does the Bunyakovsky conjecture apply for every cyclotomic polynomial?
The Bunyakovsky conjecture states that for every irreducible polynomial $\ f\in \mathbb Z[x]\ $ with positive leading coefficient, there are infinite many primes of the form $\ f(m)\ $ , where $\ m\ $ is a positive integer , if $\ \gcd(f(1),f(2),f(3),\cdots)=1\ $

The cyclotomic polynomials are well known to be irreducible and the leading coefficient is $\ 1\ $.

But what about the condition $\ \gcd(f(1),f(2),f(3),\cdots)=1\ $ ? Does it hold for all cyclotomic polynomials ? If yes, how can we prove it ?

答:
Let $n\geq2$ be a positive integer, and let $\Phi_n$ denote the $n$-th cyclotomic polynomial. Then
$$\Phi_n(1)=\begin{cases}p&\text{ if }n=p^k\\1&\text{ otherwise}\end{cases},$$
where in the first case $p$ is a prime number and $k$ is a positive integer, i.e. $n$ is a prime power. For a proof, see this question. It immediately follows that if $n$ is not a prime power, then
$$\gcd(f(1),f(2),f(3),\ldots)=1.$$
Finally, if $n=p^k$ then $\Phi_n(p)\equiv1\pmod{p}$ and so $\gcd(\Phi_n(1),\Phi_n(p))=1$.


问:
Proof for 1/k! using n choose k as n approaches infinity and its relation to the gamma function
Prove that
$\lim_{ n \to \infty }\binom{n}{k}(1/n)^k =\frac{1}{k!}$

How is this related to the gamma function?

答:
$\displaystyle lim_{n \rightarrow \infty}{n \choose k}\frac{1}{n^k} = \frac{1}{k!} lim_{n\rightarrow \infty}\frac{\frac{n!}{(n-k)!}}{n^k} = \frac{1}{k!}$

It turns out that this result can be used to derive a Taylor series, as I show in this post Why does a power series with binomial coefficients reduce to a Taylor series?

$f(t)= \displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^{n}{n \choose k} \left( \frac{t}{n}\right)^kf^{(k)}(0)= \sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}\lim_{n\rightarrow \infty}\frac{\frac{n!}{(n-k)!}}{n^k}=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}$


问:
prove that $\lim_{x \to 1^{-}}\frac{(x+2)}{2x^{2}-3x+1} = -\infty$
Prove that $\lim_{x \to 1^{-}}\frac{(x+2)}{2x^{2}-3x+1} = -\infty$

Without loss of generality, assume tha $M<0$ and $\delta>0$,

choose $0<\delta<\frac{1}{2}$ such that $1-\delta<x<1$ implies   $\frac{3}{M}<{2x^{2}-3x+1}<0$ for  that is $\frac{M}{3}>\frac{1}{2x^{2}-3x+1}$

since $\frac{1}{2}<x<1$ also implies $\frac{5}{2}<x+2<3$ it follows that $f(x)<M$ for all $1-\delta<x<1$

I don't know whether this is correct or not if there is any easy method to prove it using definition show it. Also tell me is this correct?

Body:
$$\frac{x+2}{2x^2-3x+1} = \frac{x+2}{(x-1)(2x-1)} = \frac{1}{2(x-1)} \frac{2x+4}{2x-1}= \frac{1}{2(x-1)} \cdot \left(1 + \frac{5}{2x-1}\right)$$

If $x>\frac 12$ then $1 + \frac{5}{2x-1}> 1$.

Therefore if $|x-1| < \delta: = \min\left(\frac 12,\frac{1}{2|M|}\right)$ then $\frac{x+2}{2x^2-3x+1}<-|M|$


问:
What type of logical statement is this statement and how do you prove it?
I'm learning proof techniques. I understand "if, then" statements and how they correspond to "if p, then q" and "hypothesis, and conclusion". But, I don't understand what kind of statement the following is considered to be:

"The sum of two even integers is always even"

How do I look at this statement logically, using p and q? Is there a hypothesis and conclusion here? Is this considered a non-conditional statement? And lastly, how I do I prove such a statement?

This was one of the first statements I came across when I started learning how to do proofs, so I have seen the proof for it. I just don't understand it logically.

Thank you for the help.

UPDATE:

Bram28 asked, "Do you have to use formal logic for this? Or would a mathematical proof suffice?"

What would formal logic look like for this proof? Or, am I misunderstanding something?

I've been reading more about ZFC Theory and http://us.metamath.org/mpegif/opoeALTV.html for instance. And, I can see how these proofs are being constructed.

That's why I am so curious as to how Bram28's idea of formal logic would apply to my original post.

答: Here is a formal proof ... with the comments making up a mathematical proof:


问:
Using the chain rule to relate inverse function's derivative to function's derivatives
We know the inverse functon identity:

$$ f(g(x) ) = x$$

then,

$$ \frac{df(g)}{dg} \frac{dg(x)}{dx} = 1$$

However, if we have

$$ q(x) = \frac{df(x)}{dx}$$

Is,

$$ q(g)  = \frac{df(g) }{dx} = \frac{df(g)}{dg} \frac{ dg}{dx} \tag{1}$$

What's wrong with the above equation?

Consider a counter example for (1):

$$ f = 2x \tag{2}$$

And,

$$ g = \frac{x}{2} \tag{3}$$

Combining (2) and (3),

$$ f= 4 \frac{x}{2} = 4g$$

Notice that

$$ \frac{df}{dg} \frac{dg}{dx} = 2 \neq 1$$

What went wrong?

Tl;dr Is plugging inverse then derivative equal to  derivative then plugging in inverse? Why/  why not?

答:
Well your counter example doesn't make sense .There is a diffrence between $\frac{df}{dg}\frac{dg}{dx}$ and $\frac{df(g)}{dg}\frac{dg}{dx}$  In case of your example $$f(x)=2x,g(x)=x/2$$ $$\Rightarrow f(g)=2(x/2)=x$$ $$\frac{df(g)}{dg}\frac{dg}{dx}=1$$ as you obtained initially.

I mean to say $$\frac{df}{dg}\frac{dg}{dx}=2$$ is not a counterexample to yourclaim as they are simply different


问:
Proof that we can find rational numbers arbitrarily close to $\sqrt{2}$: direct approach.
The statement of the proposition:

Proposition. For every rational number $\epsilon > 0$, there exists a non-negative rational number $x$ such that $x^{2} < 2 < (x+\epsilon)^2$.

The most common approach to proving the proposition is by using contradiction (1,2).

My question is: is it possible to prove the proposition directly? More concretely, is it possible to find a function $f: \mathbb Q^+\rightarrow \mathbb Q^+$ such that for arbitrary positive rational $\epsilon$, we have

$$f(\epsilon)^2 < 2 < (f(\epsilon) + \epsilon)^2 $$

?

答: Take $$\epsilon\left\lfloor\frac{\sqrt2}\epsilon\right\rfloor.$$ This rational is less than $\epsilon$ away from $\sqrt2$.


问:
Why do we assume the set of all sets that do not contain themselves exists?
It seems to me that the set of all sets that do not contain themselves is very similar to the set
\begin{equation}
X = \{x | \ x \not\in X\}
\end{equation}
which is again very similar to the number $x$ such that
\begin{equation}
x\neq x
\end{equation}
For this number $x$ we simply state,
\begin{equation}
\mathrel{\nexists} x \in \mathbb{R} : x\neq x
\end{equation}
Why isn't the same done in the case of sets?
\begin{equation}
\mathrel{\nexists} X : X = \{x | \ x \not\in X\}
\end{equation}
and
\begin{equation}
\mathrel{\nexists} X : X \text{ contains all the sets that do not contain themselves}
\end{equation}

答:
We do!  Assuming the existence of such a set, we can reach a contradiction.  This proves that no such set exists.  So, it is a theorem of set theory that there does not exist a set of all sets that do not contain themselves.

The problem (in set theory with unrestricted comprehension) is, it is also a theorem that such a set does exist.  Namely, the unrestricted comprehension axiom says that for any property $P$, there exists a set whose elements are exactly the sets that satisfy $P$.  In particular, if $P(x)$ is $x\not\in x$, this means that there is a set $\{x:x\not\in x\}$.

So, it is both a theorem that $\{x:x\not\in x\}$ does not exist and a theorem that it does exist.  This is a contradiction which makes set theory with unrestricted comprehension inconsistent.

In modern set theory, though, the axiom of unrestricted comprehension is not accepted (since it leads to a contradiction!).  So, Russell's paradox becomes simply a theorem that $\{x:x\not\in x\}$ does not exist, not an actual paradox.


问:
How can I find the solutions to this trigonometric equation in a specific interval?
since days I'm wondering how I can solve this equation:

$0 = \sin( \pi * \sqrt{x})-0.5$

The solution should be in the interval $[196,256]$.

My solution so far:

$\sin( \pi * \sqrt{X})= 0.5$

Substitution: $u = \pi * \sqrt{x}$

$\sin(u) = 0.5$

$u = \pi / 6$ +2∗∗,∈ℤ

Resubstitution: $u = \pi * \sqrt{x}$

$2*k*\pi+\pi/6 = \pi * \sqrt{x}.  $

$(12*k+1)^2/36 = x$

$-(12*k+1)^2/36 = x$

Now I've tried to set up an inequality with x >=196 and x<= 256 to see in which interval the variable k has to be. But now I don't know what I should do next...

As I've used Photomath, I know that there are more solutions. But I just can't find them. I'm also not able to solve the equation in the interval $[196,256]$.

I'd really appreciate it if someone could help me.

Have a nice day!

答:
Your point 4. is incomplete

$\pi\sqrt x=\pi/6+2k\pi;\;\pi\sqrt x=\pi-\pi/6+2k\pi$

which leads to two possibilities

$x_1=\left(\frac{1}{6}+2k\right)^2;\;x_2=\left(\frac{5}{6}+2k\right)^2,\;k\in\mathbb{Z}$

Both of these set of values must be in the interval $[196,256]$

$196\le \left(\frac{1}{6}+2k\right)^2\le 256\to 14\le\frac16+2k\le16\to k=7\to x=\frac{7225}{36}\approx 200.694$

In a similar way we find

$14\le \frac{5}{6}+2k\le 16\to x=\frac{7921}{36}\approx 220.028$

Confirmed by the graph below


问:
Is there any specific terminology to refer to an initial sequence of a sequence?
Lets say you have a sequence  $S = (0, 1, 2, 3, 4, 5, 6, 7, 8)$

And another sequence $T = (0, 1, 2, 3)$

Is there any specific mathematical term that defines the relationship between $S$ and $T$, that specifically says that $S$ starts with $T$?

I thought $T$ would be called an initial sub-sequence, but this is incorrect because a sub-sequence seems to be a any subset of the elements of the sequence in the same order of the sequence (so even $(2, 4, 6, 8)$ would be a sub-sequence, while I want the prefix sub-sequence part, i.e. a sub-sequence that the sequence starts with)

答:
In computer science (which is, at least at the beginning, maths), you call them words or strings over an alphabet $X$ which contains your letters.

For this alphabet $X$, you define $X^n$ by the usual Cartesian product and $X^*=\bigcup\limits_{n\in \Bbb N}X^n$. $X^*$ is the set of all words written with letters of $X$.

Then you can define a product $\cdot$ on $X^*$: $\left(u_1,\dots,u_p\right)\cdot \left(v_1,\dots,v_q\right)=\left(u_1,\dots,u_p,v_1,\dots,v_q\right)$

And so saying a word $u$ start with $v$ is saying : $\exists w \in X^*, u=v\cdot w$. And $v$ is therefore called a left factor of $u$, or a prefix.


问:
With some theory prove that A⊂Cl(Int(ClA))
A⊂Cl(Int(ClA))

Yes or No?
Counter-Example: A=Q, IntQ = ∅, ClQ = R,
Cl(Int(Cl(Q))) = Cl(Int(R)) = ClR = R ⊃ Q

But what about complete proof??

And second one: A⊃Int(Cl(IntA))

答:
This is false.

Let $A$ be the Cantor set.

Hence, $Cl(A)=A$, and $Int(Cl(A))=\varnothing$, so $Cl(Int(Cl(A)))=\varnothing$.

Let $B$ be the complement of the Cantor set in $(0,1)$.

Thus, $B$ is open, and so, $Cl(Int(B))=(0,1)$ since it is dense, and so $Int(Cl(Int(B)))=(0,1)$, which is a contradiction to the second claim.


问:
Conceptual question of Linear Transformations and its eigen values
Let's think about a linear transformation $T:\mathbb{R}^n \to \mathbb{R}^n$. And say $U$ is a subspace of $\mathbb{R}^n$ of dimension $k \lt n$. Then I want to see if each eigen values of restriction of $T$, $T_u$ to $U$ are eigen values of $T$ and how ? When I asked some fellow students they said that it's true because all the elements of $U$ are also the elements of $\mathbb{R}^n$ but I don't see how...can someone help me with a formal proof?

Note: This is just something conceptual that I want to have a clear understanding about.

答:
The linear transformation $T$ need not even fix $U$. Indeed if you consider $T:\mathbb{R}^2\to\mathbb{R}^2$ given by

$$
T\left(\begin{pmatrix}x\\y\end{pmatrix}\right) = \begin{pmatrix}1&1\\-1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}x+y\\-x+y\end{pmatrix},
$$
then $T$ does not fix any non-trivial subspace of $\mathbb{R}^2$. Therefore the question of eigenvalues does not make sense.

If, for the sake of argument, we assume that $T(U)\subset U$, then by the definition of eigenvector and eigenvalue, we can see that eigenvalues of $T_u$ is a subset of eigen values of $T$.


问:
If p and q are two distinct primes then $\sqrt{pq}$ is irrational.
There are a lot of good proofs already available for this question here, but I am trying to write the proof on my own. It'll be a great help if I can get an idea whether my proof is right or not.

$Proof.$ The given statement is equivalent to its contrapositive.
$$If\;\sqrt{pq}\;is\;rational\;then\;p\;or\;q\;is\;composite\; or\; p = q$$
$\sqrt{pq}$ is rational only when $pq$ is a perfect square.

Every natural number can be written as a product of its prime factors.

So, $p = p_1^{r_1} p_2^{r_2} p_3^{r_3}...$, where $p_i$ represent a prime and $r_i$ represent its power.

Similarly, $q = p_1^{s_1} p_2^{s_2} p_3^{s_3}...$, where again $p_i$ represent a prime and $s_i$ represent its power.

Each $r_i \geq 0$ and $s_i \geq 0$. Also, the above series will be finite as $p$ and $q$ are finite.

So, $pq = p_1^{r_1 + s_1} p_2^{r_2 + s_2} p_3^{r_3 + s_3}...$

Now, $pq$ can be a perfect square only when for each $p_i$, $r_i + s_i$ is even.

Now, let us take $p\; or\; q\; is\; composite\; or \; p = q$ to be false, i.e $p\; and\; q\; both\; must\; be\; distinct\;prime$ is the true statement.

Now, if $p$ and $q$ are prime then $$p = p_k^{r_j}$$where $p_k = p$ and $r_j = 1$.

Similarly, $$q = p_l^{s_t}$$where $p_l = q$ and $s_t = 1$.
Hence, $pq = p_k^{r_j}p_l^{s_t}$, where both $r_j$ and $s_t$ are odd and $gcd(p_k, p_l) = 1$, which means $pq$ is not a perfect square and hence irrational.(a contradiction).

So, $p\; or\; q\; is\; composite$ is true, which implies
 $$If\; \sqrt{pq}\;is\;rational \;then \;p \;or\; q \;is\; composite.$$

Hence, proved.

答: If $\sqrt{pq}=a/b$ for naturals $a, \, b$ then $a^2=b^2pq$. But $p$ divides the left (right) side an even (odd) number of times.


问:
Differentiation problem of power to infinity by using log property
Problem:

Find $\frac{dy}{dx}$ if $y =\left(\sqrt{x}\right)^{x^{x^{x^{\dots}}}}$

Let ${x^{x^{x^{\dots}}}} =t. (i)$ Taking $\log$ on both sides $ \implies {x^{x^{x^{\dots}}}}\log x = \log t$

This can further be written as $ t
\cdot\log x = \log t$

Differentiating w.r.t. $t$ we get :

$\frac{dx}{dt}=\left(\frac{1-\log t}{t^2}\right)x$. (ii)

Now taking $\log$ on both sides of original equation :

$\log y = {x^{x^{x^{\dots}}}} \log\sqrt{x} \implies \log y = t \log \sqrt{x}$

Now differentiating both sides w.r.t. $x$ we get :

$\frac{1}{y} \frac{dy}{dx} = t\log\sqrt{x} = t \frac{1}{2\sqrt{x}} + \log\sqrt{x}\frac{dt}{dx}$. (iii)

Now putting the values of $y$, $\frac{dt}{dx} $, and $t$  from (i), (ii) and (iii) we get:

$$\frac{dy}{dx} = \left(\sqrt{x}\right)^{x^{x^{x^{\dots}}}}\left[\frac{{x^{x^{x^{\dots}}}}}{2\sqrt{x}}+\frac{\log\sqrt{x} \cdot{x^{x^{x^{\dots}}}}}{(1-\log{x^{x^{x^{\dots}}}})x}\right]$$

Please confirm whether it is the correct answer or not. Thanks.

答:
HINT:

If $y =\left(\sqrt{x}\right)^{x^{x^{x^{\dots}}}}=\left(\left(x\right)^{x^{x^{x^{\dots}}}}\right)^\frac12$

$$\implies y^2=\left(x\right)^{x^{x^{x^{\dots}}}}=(x)^{y^2}$$

Now, taking log wrt $e,$  $$2\ln y=y^2\ln x$$


问:
Does there exist a form of Pythagorean quadruples gives that they are relatively prime?
I am working on how to find the Frobenius number of Pythagorean quadruples. In this problem we have
$$a=m^2+n^2-p^2-q^2\\
b=2(mq+np)\\
c=2(nq-mp)\\
d=m^2+n^2+p^2+q^2\\
a^2+b^2+c^2=d^2$$
Where $m, n, p, q, a, b, c, d$ are all non-negative integers.
And I would like to find conditions on the integers $m,n,p,q$ that would ensure $\gcd (a,b,c,d)=1$.

Ps: Obviously there exist the situation that the four numbers are relatively prime but they are not fit in the form of $a, b, c, d$.

答: The additional condition is that $m+n+p+q \equiv 1 \pmod 2,$  which means $m+n+p+q$ must be odd. That is all


问:
Projection of a hypercube along a long diagonal. What is this polytope called?
I was wondering what people call a certain type of shape. It is the shape formed by an orthogonal projection of a hypercube along one of its longest diagonals.

In other words, fill in the missing entries:

```<code>n-simplex : Triangle ->      Tetrahedon        ->  5-cell    ...
 n-cube   :  Square  ->         Cube           -> Tesseract  ...
   ?      :  Hexagon -> Rhomibic Dodecahedron  ->     ?      ...
</code>```

IMPORTANT: I'm looking for the name of the class of shapes, not just the 4D analogue.

答:
The polytopes you describe are the duals of a family of uniform polytopes.
So, they are higher-dimensional analogues of the Catalan solids.

I have currently no better idea for how to describe these uniform polytope to you other than by their Coxeter diagrams:

$\qquad\qquad$

They belong to the $A_d$-family of uniform polytopes, that is, they have (an extended version of) the symmetry group of the $d$-simplex.

According to a link posted by A. Schulz, these polytopes are also called "cats-eye polytopes", which I think is pretty cool.