答:
This image is intended to show that $\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}$.
The width of this cuboid can be computed in two ways, one way by simple addition.
$$\frac{a}{b}+\frac{c}{d}$$
Another way is to compute the volume of the cuboid first. The volume of the green cuboid is the area $a$ times the depth $d$, the volume of the blue cuboid is the area $c$ times the height $b$. Adding these volumes gives $ad+bc$. Therefore the width of this cuboid is the volume of the cuboid divided by the height $b$ and depth $d$.
$$\frac{ad+bc}{bd}$$

Title: Visual proof for $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$

Body:
I've found visualizations of simple mathematical concepts to be a really useful tool in building intuition for more complex mathematical concepts. For example, this visualization of $(a + b)(c + d) = ac + ad + bc + bd$ can be used to visualize the calculus product rule.

That being said, does anyone know of a visual way to show the following equation?
$$
\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}
$$


问:
How to prove $4\times{_2F_1}(-1/4,3/4;7/4;(2-\sqrt3)/4)-{_2F_1}(3/4,3/4;7/4;(2-\sqrt3)/4)\stackrel?=\frac{3\sqrt[4]{2+\sqrt3}}{\sqrt2}$
I have the following conjecture, which is supported by numerical calculations up to at least $10^5$ decimal digits:
$$4\times{_2F_1}\left(-\frac{1}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)-{_2F_1}\left(\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)\,\stackrel?=\,\frac{3\sqrt[4]{2+\sqrt{3}}}{\sqrt{2}},$$
where $_2F_1$ denotes the hypergeometric function.

Can you suggest any ideas how to prove it?

The conjectural closed form was obtained using WolframAlpha query

```<code>ToRadicals[RootApproximant[2.94844576626425580599908814238570067699233]]
</code>```

答:
Edited on request of moderator team

Gauss hypergeometric function satisfies a linear transformation formula
$$ (c-a) _2F_1(a-1,b,c,z)+(2a-c-az+bz){}_2F_1(a,b,c,z)+a(z-1){}_2F_1(a+1,b,c,z)=0.$$
Setting $a=b$ makes $z$ disappear from the 2nd prefactor, and one obtains
$$(c-a) _2F_1(a-1,a,c,z)+(2a-c){}_2F_1(a,a,c,z)+a(z-1){}_2F_1(a+1,a,c,z)=0.\tag{1}$$

Also, sometimes hypergeometric function simplifies to elementary functions. For example,
$_2F_1(\alpha,\beta,\alpha,z)=(1-z)^{-\beta}$. Setting in this formula $\alpha=a+1$, $\beta=a$ and combining it with (1) with $c=a+1$, one obtains
$$ _2F_1(a-1,a,a+1,z)+(a-1){}_2F_1(a,a,a+1,z)=a(1-z)^{1-a}.\tag{2}$$

Setting in (2) $\displaystyle a=\frac34$ and $\displaystyle z=\frac{2-\sqrt{3}}{4}$
gives the quoted relation.

ContentLicense: CC BY-SA 3.0


问:
What is the derivative of $\mathbf{a}^TX^2\mathbf{a}$ with respect to the symmetric matrix $X$?
Given a constant vector $\mathbf{a}\in{\rm I\!R}^n$ and a real symmetric matrix $X\in{\rm I\!R}^{n\times n}$, what is the derivative of $\mathbf{a}^T X^2 \mathbf{a}$ with respect to $X$?

I tried a simple example using $n=2$, with the following vector and matrix :
$$
\mathbf{a} = \begin{bmatrix}a\\b\end{bmatrix} \qquad X = \begin{bmatrix} x & z \\ z & y \end{bmatrix}
$$

What we want to differentiate is
$$
\mathbf{a}^T X^2 \mathbf{a} = a^2 (a^2 + z^2) + 2 ab (xz+yz) + b^2 (z^2 + y^2)
$$

Which gives the following result
$$
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial X} =
\begin{bmatrix}
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial x} & \frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial z} \\
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial z} & \frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial y}
\end{bmatrix}
=
2
\begin{bmatrix}
a^2 x + abz &  a^2z + abx + aby + b^2z \\
a^2z + abx + aby + b^2z & b^2 y + abz
\end{bmatrix}
$$
I tried to re-write this to end up with something meaningful, but I could only write

$$
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial X} = 2X\mathbf{a}\mathbf{a}^T + 2
\begin{bmatrix}
0 & a^2 z + ab y \\
abx + b^2z & 0
\end{bmatrix}
$$
and I do not know what to do with the matrix on the right...

What I would like is a valid expression for any $n > 1$ involving only $X$ and $\mathbf{a}$.

答:
$\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}\def\p#1#2{\frac{\partial #1}{\partial #2}}$Let $U$ be an unconstrained matrix
and use a colon denote the trace function in product form, i.e.
$$A:B = {\rm Tr}(A^TB) = B:A$$
Write the function using the colon product and calculate the unconstrained derivative.
$$\eqalign{
\phi &= aa^T:U^2 \\
d\phi &= aa^T:(U\,dU+dU\,U) \\
 &= (Uaa^T+aa^TU):dU \\
\p{\phi}{U} &= Uaa^T+aa^TU \;\doteq\; G \qquad({\rm gradient}) \\
}$$
Here is recipe for converting an unconstrained gradient $G$ into the desired form
$$\eqalign{
G_S &\doteq G+G^T - {\rm Diag}(G) \\
 &= (Xaa^T+aa^TX) + (Xaa^T+aa^TX)^T -{\rm Diag}(Xaa^T+aa^TX) \\
 &= 2(Xaa^T+aa^TX) - {\rm Diag}(Xaa^T+aa^TX) \\
\\
}$$
Apply this general result to your $2\times 2$ example.
$$\eqalign{
A = Xaa^T &= \m{a^2x+abz & abx+b^2z \\ a^2z+aby & abz+b^2y} \\
B = A+A^T &= \m{2(a^2x+abz) & (abx+b^2z+a^2z+aby) \\ (a^2z+aby+abx+b^2z) & 2(abz+b^2y)} \\
G_S = 2B-{\rm Diag}(B)
  &= \m{2(a^2x+abz)&2(abx+b^2z+a^2z+aby)\\2(a^2z+aby+abx+b^2z)&2(abz+b^2y)} \\
}$$
which is the same result that you obtained.

Having provided you with the formula that you were searching for, I must warn you that it is nonsense.

What you should do is extract a vector of fully independent parameters from the $X$ matrix using the half-vec operation
$$\eqalign{
p &= {\rm vech}(X) = \m{x\\z\\y} \\
}$$
and solve whatever problem you have in mind in terms of this vector.

Everyone agrees that the following vector gradient is valid and unambiguous
$$\eqalign{
g\doteq \p{\phi}{p} &=
\m{2(a^2x+abz) \\ 2(abx+b^2z+a^2z+aby) \\ 2(abz+b^2y)} \\
}$$
However, casting this vector into matrix form using the reverse of
the ${\rm vech}()$ function creates a thing which is difficult to interpret as a gradient, and hard to use (properly) in algorithms such as gradient descent.

Instead, you should leave the gradient in vector form and use it to
optimize/solve for the $p$ vector. Then, as a post-processing step, you can cast the solution back into matrix form $$X = {\rm unvech}(p)$$

ContentLicense: CC BY-SA 4.0


问:
Is there an algorithm for constructing a Triangle Hierarchy for maximal planar graphs?
In the research I am currently involved in, I am given a planar, maximal graph $G$ of $n$ vertices, and its embedding. I require to find a structure, which I call a "triangle hierarchy" of $G$, in deterministic time $O(n)$. Informally, this structure is a tree rooted at the outer face of $G$ (which is a triangle, and by triangle I mean any $K_3$ in $G$), which we can traverse down, to the "inner triangles" of $G$. Two triangles $t$ and $s$ are connected in that tree ($t$ is a parent of $s$), if $s$ is contained in $t$ (in the given embedding) and there is no triangle $r$ such that $r$ is contained in $t$ but not in $s$. This looks like a useful structure for many algorithms, but I failed to find any reference about it. I am hoping someone from this community can point me to some paper.

Here is the formal definition with the example:

[inner trinagles] A triangle in the planar graph embedding that is not a face we call an inner triangle.

[pivotal triangles]
A triangle in a maximal, planar graph embedding of $G$ is called pivotal, if it is an inner triangle in $G$ or an outer face of $G$.

[triangle inside]
Let $t$ be a triangle in a maximal, planar graph $G$. The inside of $t$ is a plane fragment bounded by the edges
of $t$ in the embedding of $G$. Edges and vertices of the triangle are not part of its inside.

[contains relation] A triangle $t$ contains a triangle $s$ iff the inside of $s$ is a proper subset of the inside of $t$. We say that a triangle $t$
contains a vertex $v$, if $v$ belongs to the inside of $t$. Let $G$ be a maximal, planar graph, embedded on a plane, and let $A$ be the set of its pivotal triangles.
We define a relation $\sqsubset$ on $A$, such that for any $t,s \in A$, $s \sqsubset t$ if and only if $t$ contains $s$.

By the abuse of notation we will write $v \sqsubset t$, where $v$ is a vertex, to say that a triangle $t$ contains $v$.
Let $A$ be the set of triangles of the embedding of some maximal, planar graph. Observe that $(\sqsubset, A)$ is a partially ordered set.

[triangle hierarchy]
Let $G$ be a maximal, planar graph, embedded on a plane, and let $A$ be the set of its pivotal triangles. Let $D=(A,F)$ be a directed graph,
with vertices corresponding to the set of pivotal triangles $A$, and the set of edges $F$ defined as follows:
for each $t,s \in A$, $\langle t,s \rangle \in F$ iff $s \sqsubset t$ and there is no $r \in A \setminus \{t,s\}$ such that $s \sqsubset r \sqsubset t$.
We call graph $D$ a triangle hierarchy of $G$.

In any planar graph embedding, edges do not intersect (except at endpoints) and every triangle consists of three vertices, therefore any two triangle insides are either
disjoint or one contains the other. Therefore a triangle hierarchy of $G$ is a tree rooted at the outer face of $G$.

Let $D=(A,F)$ be a triangle hierarchy of some maximal, planar graph $G=(V,E)$. For each triangle $t \in A$ we define $\lambda(t)$ to be the set of vertices
forming $t$, and $\sigma(t)=\{ v \in V \, : \, v \sqsubset t \wedge \forall_{s\in A}(s \sqsubset t \Rightarrow v \not \sqsubset s) \}$. Informally, $\sigma(t)$
is the set of vertices, such that $t$ contains them, but no triangle lower in the hierarchy contains them.

The example:

Sample graph

Its triangle hierarchy

答:
As a starting point, here is an $O(n^2)$ algorithm. I have no strong reason to think that this is optimal.

First, we can find all the triangles in $O(n^2)$ time. If $xy$ is an edge, there are at most $n-2$ possibilities for a third vertex $z$ such that $(x,y,z)$ form a triangle, so we can check them all in $O(n)$ time (by checking if $xz$ and $yz$ are also edges). There are $3n-6$ edges, which is also $O(n)$.

It took us $O(n^2)$ time to find the triangles, but in reality there are only $O(n)$ triangles. We can prove a bound of $10n$ by induction: let $G$ be any planar graph, and let $v$ be a vertex of degree less than $6$, which is known to exist. Then there are at most $\binom{\deg v}{2} \le 10$ triangles containing $v$, and by the inductive hypothesis at most $10(n-1)$ triangles in $G-v$.

To find the triangle hierarchy, start with a partial hierarchy that initially only contains the exterior triangle. To add a triangle $t$ into the hierarchy:

Start by letting $r$ be the top of the hierarchy; it will be an invariant that $t \sqsubset r$.

If $s_1, \dots, s_k$ are the children of $r$, check if $s_i \sqsubset t$ and if $t \sqsubset s_i$ for each of them.

If $t \sqsubset s_i$ for some $i$, set $r = s_i$ and recurse on the subtree rooted at the new $r$.

Otherwise, make $t$ another child of $r$. For every $s_i$ such that $s_i \sqsubset t$, make $s_i$ a child of $t$ instead of a child of $r$.

Add the triangles into the hierarchy one at a time as you find them. This will also include the triangles which are faces, and you didn't want those, but they will all be leaves of the tree, and we can pluck them at the end if we need to.

This will certainly take at most $O(n^2)$ time, because we compare any two triangles at most once in each direction. It may take $O(n^2)$ time if there is a chain of $O(n)$ nested triangles and we add them to the hierarchy in a bad order. Regardless, even if there is a better way to do this, there is no point improving it unless we can also find the triangles more quickly.


问:
How can I "see" that calculus works for multidimensional problems?
Let's say I have some function f(x) = x^2 + b. I can see what's going on, I can count the slope geometrically even without knowing the rules of derivatives.

When I need to minimize some cost function for linear problem (linear regression with gradient descent), I just need to picture 3d, and I can "see", and be quite confident why and how it works.

How can I "see" or get intuition that calculus works for multidimensional problems? Let's say I have a problem with many variables like:

f(area_km, bedrooms, ...) = theta + theta1 * area_km + theta2 * bedrooms etc

If I want to apply gradient descent, I know I need to calculate partial derivatives and multiply it with a learning rate etc. It works. But it's kinda magical that it works.

I am sorry this is a silly question, I am just beginning.

答:
$\def\p{\partial}
\def\vr{{\bf r}}
\def\th{\theta}
\newcommand\pder[2][]{\frac{\partial #1}{\partial #2}}
\newcommand\der[2][]{\frac{d #1}{d #2}}$Nontrivial geometrical intuition can be obtained by considering temperature as a function of position in three dimensions.
Suppose temperature is given by $T(\vr)$, where $\vr = \langle x,y,z\rangle$.
Surfaces of constant temperature ("equipotentials") are given by $T(\vr) = \mathrm{const}$.
We wish to minimize the temperature starting from $\vr=\vr_0$ where $T(\vr_0)=T_0$, that is, we wish to move from this equipotential to another with lower temperature following the shortest possible path.
We will argue that we must move in a direction orthogonal to the equipotential.
In fact, we must move in the direction opposite that of the gradient of temperature.

Consider a path through $\vr_0$ parametrized by $t$.
By multivariable chain rule, the rate of change of $T$ along the path is given by
\begin{align*}
\frac{d T(\vr(t))}{dt} = \pder[T]{x}\der[x]{t}
+\pder[T]{y}\der[y]{t}
+\pder[T]{z}\der[z]{t}
= \nabla T\cdot\vr'(t),\tag{1}
\end{align*}
where $\nabla T = \langle \p T/\p x,\p T/\p y,\p T/\p z\rangle$ is the gradient of temperature.
Assume that $|\vr'(t)|=1$, independent of path (that is, the paths are parametrized by arclength).
The derivative $dT/dt$ is then a measure that can be used to compare the rate of change of $T$ for different paths through $\vr_0$.
Note that
$$\left|\pder[T(\vr(t))]{t}\right|
= |\nabla T|\cos\th,$$
where $\th$ is the angle between $\nabla T$ and $\vr'(t)$.
The magnitude of the rate of change is clearly maximized for $\th=0,\pi$.
The value of $T$ does not change for paths with $\th=\pi/2$.
These paths are tangential to the equipotential at $\vr_0$.
Thus, $dT/dt$ is maximally negative if $\vr'$ is antiparallel to $\nabla T$ (and thus perpendicular to the equipotential).
Since $\vr'$ is tangential to the path, we expect $T$ to decrease maximally in the direction given by $-\nabla T$.
Thus, we let $\Delta\vr = -\varepsilon\nabla T$, where $\varepsilon>0$ is sufficiently small.

Example

If $T(\vr) = x^2+y^2+z^2$, the equipotentials are spheres and $\nabla T = 2\langle x,y,z\rangle = 2\vr$.
Thus,
$\Delta r = -2\varepsilon \vr_0$.
This displacement is directly towards the origin, where the temperature has a minimum value of $T({\bf 0}) = 0$.
See the figure below.

Figure 1. Equipotentials and displacement vector.

One dimension

Equation (1) and its interpretation generalize nicely to higher and lower dimensions.
In one dimension we have
$$\der[T(x(t))]{t} = \der[T]{x} \der[x]{t}.$$
Clearly this value is maximally negative if $dT/dx$ and $dx/dt$ are opposite in sign.
Thus, we let
$\Delta x = -\varepsilon dT/dx$.

Example

If $T(x) = x^2$, we find
$\Delta x = -2\varepsilon x_0$.
This displacement is again directly towards the minimum at $x=0$, independent of the sign of $x_0$.


问:
How to solve a system of 4 coupled non-linear differential equations?
The system of equations I have is:
$$\frac{dx_1}{dt} = k_2 \left( \frac{y_2 - y_1}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$
$$\frac{dy_1}{dt} = k_2 \left( \frac{x_1 - x_2}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$

$$\frac{dx_2}{dt} = k_1 \left( \frac{y_1 - y_2}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$
$$\frac{dy_2}{dt} = k_1 \left( \frac{x_2 - x_1}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$

where $k_1,k_2$ are constants. How would I go about solving this for $x_1(t),y_1(t),x_2(t),y_2(t)$?

For some context, these differentials equations determine how 2 vortex velocity fields of arbitrary circulation ($k_1, k_2$) interact with each other. The x-y time derivatives represent the component-wise speed of each vortex due to the other vortex.

My attempt:

So far, I multiplied the first and second equation by $(x_1−x_2)$ and $(y_1−y_2)$ and added them to get $(x_1−x_2)(dx_1/dt)+(y_1−y_2)(dy_1/dt)=0$. Doing the same thing for the third and fourth equations, I got $(x_1−x_2)(dx_2/dt)+(y_1−y_2)(dy_2/dt)=0$. Don't know where to go from here.

答:
Try to substitute $\Delta=x_1-x_2$ and $\Psi=y_1-y_2$. Then substract equation $1$ and equation $3$. And also equation $2$ and equation $4$. With two differential equations it will be easier to solve.
$$\Delta'= \dfrac {A\Psi}{\Psi^2+\Delta^2}$$
$$\Psi'= \dfrac {B\Delta}{\Psi^2+\Delta^2}$$
Where $A=-(k_1+k_2),B=k_1+k_2$.

Then try to solve:
$$\dfrac {d \Delta}{d\Psi}=-\dfrac { \Psi}{ \Delta}$$


问:
Visual proof for $\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$
I've found visualizations of simple mathematical concepts to be a really useful tool in building intuition for more complex mathematical concepts. For example, this visualization of $(a + b)(c + d) = ac + ad + bc + bd$ can be used to visualize the calculus product rule.

That being said, does anyone know of a visual way to show the following equation?
$$
\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}
$$

答: Here is my version which depends on similar triangles.


问:
Can we always "locate" a transitive model inside some larger transitive model?
We work in $\mathsf{ZFC}$ + "Every set is an element of a transitive model of $\mathsf{ZFC}$."

Suppose $M$ is a transitive set model of $\mathsf{ZFC}$. Say that $M$ is locatable iff there is some transitive set $N$ such that:

$N\models\mathsf{ZFC}$,

$M\in N$, and

$M$ is parameter-freely-definable in $N$.

The requirement that $N\models\mathsf{ZF}$ prevents coding $M$ into $N$ in a "silly" way. To see how this can be an issue, let $\alpha=\min\{\gamma: L_\gamma\models\mathsf{ZFC}\}$. If $N$ is a transitive model of $\mathsf{ZFC}$ with $L_\alpha\in N$, then $N$ sees that $L_\alpha$ is countable and so contains lots of (say) reals which are Cohen-generic over $L_\alpha$. Suppose $r,s$ are such reals; how is $N$ to distinguish between $L_\alpha[r]$ and $L_\alpha[s]$ in a parameter-free way? This seems to be an obstacle in general to the locatability of $L_\alpha[c]$ for $c$ Cohen-generic over $L_\alpha$. However, such an $L_\alpha[c]$ is nonetheless locatable: force over $L_\beta$, where $\beta=\min\{\gamma: \gamma>\alpha, L_\gamma\models\mathsf{ZFC}\}$, to code $c$ into the continuum pattern appropriately.

Indeed, such coding tricks give:

Every countable $M$ is locatable.

Note that the locating model is very simple - it's an element of $L(M)$. For uncountable $M$s this seems like a nontrivial further condition, and in particular none of the absoluteness theorems I'm familiar with seem to directly apply.

Similarly, we have:

If $L_\theta\models\mathsf{ZFC}$ then $L_\theta$ is locatable, even if $\theta$ is uncountable.

(For example, every ordinal $\theta$ is parameter-freely-definable in the $(\vert\theta\vert^++\theta+1)$th level of $L$ satisfying $\mathsf{ZFC}$ since $(i)$ that level of $L$ can "count" the $\mathsf{ZFC}$-satisfying levels below it, and $(ii)$ we can recover $\theta$ from $\vert\theta\vert^++\theta$ as there is a unique way to write the latter in the form $\kappa+\gamma$ for $\kappa$ a cardinal and $\gamma<\kappa$.)

However, the general situation is not clear to me:

Question: is every transitive set model of $\mathsf{ZFC}$ locatable?

Of course it is quite plausible that this will depend on additional set-theoretic hypotheses; for example, I think an affirmative answer is more likely assuming $\mathsf{V=L}$ than assuming the existence of large cardinals. But I don't immediately see how to show any of this.

答:
A partial positive answer belatedly occurs to me, motivated by Jason Zesheng Chen's comments above (although I'm not sure this is what they had in mind):

Suppose $\mathsf{V=L}$ and $M\models\mathsf{ZFC}$. Let $M$ be the $\theta$th set in the $L$-ordering, and let $\delta$ be the $(\vert\theta\vert^++\theta+1)$th index of a level of $L$ satisfying $\mathsf{ZFC}$. Then $M\in L_\delta$ and $\theta$ is parameter-freely definable in $L_\delta$ (by the argument that every level of $L$ is locatable); since the $L$-ordering is absolute to $L_\delta$, this gives the parameter-free-definability of $M$ in $L_\delta$.

So assuming $\mathsf{V=L}$, then every transitive set model of $\mathsf{ZFC}$ is locatable. (And I suspect the same trick will work for similar fine-structural models.)

Interestingly, at a glance this doesn't seem to generalize to $\mathsf{V=HOD}$. The $L$-ordering is playing two roles in the above argument, both getting sets from ordinals and building arbitrarily "big" models which recognize a given ordinal (the map $\theta\mapsto L_{\vert\theta\vert^++\theta+1}$), and while the former is handled by $\mathsf{HOD}$ the latter seems genuinely problematic to me. But I could be missing something.


问:
Surjective bounded linear map bound
Let $X,Y$ be Banach spaces.

Let $T: X \rightarrow Y$ be a surjective bounded linear map.

Show that there is a constant $M>0$ such that for each $y\in Y $ there is a solution to $Tx=y$ with $\| x\| \leq M \| y\|$.

Let $B_Y(0,r)= \{ y \in Y : \| y \| < r\}$.

I have shown that there exists $r>0$ such that $B_Y(0,r) \subset T(B_X(0,1))$.

Not sure how to conclude from here.

Thank you in advance!

答: The open mappin theorem implies that there exits $r>0$ such that $B_Y(0,r)\subset T(B_X(0,1))$. Let $y\in Y, y\neq 0$, ${r\over {\|y\|}}y\in B_Y(0,r)$ implies that there exists $x'\in B_X(0,1)$ such that $T(x')={r\over{\|y\|}}y$. We have $T({{\|y\|}\over r}x')=y$ and $\|{{\|y\|}\over r}x'\|\leq {{\|y\|}\over r}$. Take $M={1\over r}$.

Title: Surjective bounded linear map bound


答:
For the most general case, think about a mixing board.

Each input argument to the function is represented by a slider with an associated piece of a real number line along one side, just like in the picture. If you are thinking of a function which can accept arbitrary real number inputs, the slider will have to be infinitely long, of course, which of course is not possible in real life, but is in the imaginary, ideal world of mathematics. This mixing board also has a dial on it, which displays the number corresponding to the function's output.

The partial derivative of the function with respect to one of its input arguments corresponds to how sensitive the readout on the dial is if you wiggle the slider representing that argument just a little bit around wherever it's currently set - that is, how much more or less dramatic the changes in what is shown are compared to the size of your wiggle. If you wiggle a slider by, say, 0.0001, and the value changes by a factor 0.0002, the partial derivative with respect to that variable at the given setting is (likely only approximately) 2. If the value changes in an opposite sense, i.e. goes down when you move the slider up, the derivative is negative.

The gradient, then, is the ordered list of signed proportions by which you have to "wiggle" all the sliders so as to achieve the strongest possible, but still small, positive wiggle in the value on the dial. This is a vector, because you can think of vectors as ordered lists of quantities for which we can subject to elementwise addition and elementwise multiplication by a single number.

And of course, when I say "small" here I mean "ideally small" - i.e. "just on the cusp of being zero" which, of course, you can make formally rigorous in a number of ways, such as by using limits.

Title: How can I "see" that calculus works for multidimensional problems?

Body:
Let's say I have some function f(x) = x^2 + b. I can see what's going on, I can count the slope geometrically even without knowing the rules of derivatives.

When I need to minimize some cost function for linear problem (linear regression with gradient descent), I just need to picture 3d, and I can "see", and be quite confident why and how it works.

How can I "see" or get intuition that calculus works for multidimensional problems? Let's say I have a problem with many variables like:

f(area_km, bedrooms, ...) = theta + theta1 * area_km + theta2 * bedrooms etc

If I want to apply gradient descent, I know I need to calculate partial derivatives and multiply it with a learning rate etc. It works. But it's kinda magical that it works.

I am sorry this is a silly question, I am just beginning.


答: Those two cosets do indeed form $\Bbb Z_8/H$ (under the operation $(a+H)+(b+H):=(a+b)+H$). Well done! You could improve your answer by noting that $0+H=H$.

Title: Constructing factor-/quotient groups

Body:
Given:$$\Bbb Z_8 =\{0,1,2,3,4,5,6,7\}$$ and $$H=\{0,2,4,6\} = \left<[2]\right>$$, construct the factor group $\frac{\Bbb Z_8}{H}.$

Definition:

For a normal subgroup $H$ of $G$, the group $\frac{G}{H}$ is called the factor-/ or quotient group of $G$ modulo $H$. The set $\frac{G}{H}$ = {$aH | a \in G$} (left-coset) is a group under the operation $(aH)(bH)=abh$.

\begin{align}\\
\frac{\Bbb Z_8}{H}
& =\{gH|g\in \Bbb Z_8\} \\
& =\{g+H|g\in \Bbb Z_8\}
&\text{with }
& g+H=\{g+h|h\in H\}\\
\frac{\Bbb Z_8}{H}
& =\{H, 1+H, 2+H, ..., 7+H\}\\
\end{align}

So we have:
\begin{align}\\
& 0+H=\{0,2,4,6\}&& 4+H=\{4,6,8,10 \}\\
& 1+H=\{1,3,5,7\}&& 5+H=\{5,7,9,11 \}\\
& 2+H=\{2,4,6,8\}&& 6+H=\{6,8,10,12\}\\
& 3+H=\{3,5,7,9\}&& 7+H=\{7,9,11,13\}\\
\end{align}

Taking modulo 8, we get:
\begin{align}\\
&0+H\mod 8=\{0,2,4,6\}&& 4+H\mod 8=\{4,6,0,2\}\\
&1+H\mod 8=\{1,3,5,7\}&& 5+H\mod 8=\{5,7,1,3\}\\
&2+H\mod 8=\{2,4,6,0\}&& 6+H\mod 8=\{6,0,3,4\}\\
&3+H\mod 8=\{3,5,7,1\}&& 7+H\mod 8=\{7,1,3,5\}\\
\end{align}

We see that:
\begin{align}\\
(1+H)\mod 8 =(3+H)\mod 8 =(5+H)\mod 8 =(7+H)\mod 8 &&=\{1,3,5,7\}
\\
(2+H)\mod 8 =(4+H)\mod 8 =(6+H)\mod 8 &&=\{0,2,4,6\}
\end{align}

I calculated the cosets of subgroup $H=\{0,2,4,6\}=\left<[2]\right>$ of $\Bbb Z_8$:
\begin{align}\\
& 0+\left<[2]\right> & =\{0,2,4,6\} \\
& 1+\left<[2]\right> & =\{1,3,5,7\} \\
\end{align}

I don't know how to proceed from here, or even if I'm on the right track. I thougt I would end up with a set of pairs, but I keep ending up calculating the same two sets.


问:
Riemann Integrability of the indicator function-- small clarification

Suppose $a\le s<t\le b$. Define $f:[a, b]\mapsto \mathbb{R}$ by
$$f(x) = \begin{cases}
1 \; \; s<x<t \\
0 \; \; \text{otherwise}
\end{cases}$$
Prove that $f$ is Riemann integrable on $[a, b]$ and that $\int_a^b f = t-s$.

I am trying to construct a proof of this statement, but I don't see how the assertion $\int_a^b f = t-s$ is true. If we think of $\int_a^b f$ in terms of area under a curve, I would say that $\int_a^b f = t-s$ would hold iff $f(x) = 1$ for when $s\le x\le t$ (replacing strict inequalities with non-strict ones) and $0$ otherwise. Can someone please explain how $\int_a^b f = t-s$ holds given how $f$ is defined originally?

Note: Please don't give a proof of the statement since that is my HW assignment. I'd appreciate just some intuitive clarifications/ explanations.

答: Notice that $$\int_a^bfdx=\int_{a}^sfdx+\int_s^tfdx+\int_t^bfdx=\int_s^t1dx$$ When the function it is the value $1$, what the integral really means, it's the lenght of the interval. Or in terms of area, it means the area of a rectangle of length $1$ and width $t-s$.


问:
Proving every Cauchy sequence in $\mathbb{R}$ is convergent
I've seen a lot of proofs of how to do this and I don't quite like any of them so I tried proving this myself. Let me know if I am making any logical errors here.

Proof

Let $\varepsilon > 0$ be given.

Since $\{x_n\}_{n=1}^\infty$ is Cauchy, $\forall \varepsilon >0, \exists N \in \mathbb{N} : |x_n-x_m|<\varepsilon$ when $n,m \geq N$

Since $\{x_n\}$ is Cauchy, $\{x_n\}$ is bounded. We know by Bolzano-Weierstrass this means that $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ that converges to $l$.

Put another way: $\forall \varepsilon > 0, \exists N' \in \mathbb{N}: |x_{n_k}-l|<\varepsilon$ when $k>N'$

Suppose $n >$ max$\{N,N'\}$

Then $|x_n-l|=|x_n-x_{n_k}+x_{n_k}-l| \leq |x_n-x_{n_k}|+|x_{n_k}-l| < \varepsilon + \varepsilon = 2\varepsilon.$ $\square$

(Note: I'm not concerned with the fact that my proof ends with $2\varepsilon$. I just wanna know if my statements are true and if my proof logic is good.)

答: That's fine as long as you have a definition of the real numbers that leads to Bolzano Weierstrass without using Cauchy sequences.</p>


问:
How are these two equations equal?
$$\dfrac{1}{1+e^{-x}} = \dfrac{e^x}{1+e^x}$$

I was told to sketch a curve but couldn't figure out the first step. The solution manual rewrote the left hand side of the equation above as the right hand side. I cannot figure out what they did to get this. Could someone explain this to me?

答: They just multiplied the fraction by $1$, in the form $\dfrac{e^x}{e^x}$.</p>

Title: How are these two equations equal?


问:
Given three customers arrive in the first hour, find probability third arrives in second half
Assume customers arrive according to a Poisson process with rate $\lambda = 3$.

Given three customers arrive in the first hour, find the probability that the third customer arrives in the second half of the hour.

So that is if $X_3$ is the third arrival time: $P(X_3 \in [0.5,1] \mid N(1) = 3)$

and we have $X_3 \sim$ Gamma$(3,3)$.

To solve this I have cases : $\bigg \{X_1,X_2 \in [0,.5]$ and $X_3 \in [.5,1]\bigg\}$ or

$\bigg \{X_1 \in [0,.5]$ and $X_2,X_3 \in [.5,1]\bigg\}$ or $\bigg \{X_1,X_2, X_3 \in [.5,1] \bigg \}$

However, isn't there a way to solve this using the Uniform distribution?

答: I think the key to solving this problem is to identify the following relationship: $$\{N(1)=3\}=\{X_3\leq 1,X_4>1\}$$ This way we can say $$P\bigg(\frac{1}{2}\leq X_3 \leq 1\bigg|N(1)=3\bigg)=\frac{P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)}{P(N(1)=3)}$$ Using the total law of probability, $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)=\int_{1/2}^1P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg|X_3=t\bigg)f_{X_3}(t)dt$$ Since inter$-$arrival times are independent and $\sim \exp(3)$ we have $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg|X_3=t\bigg)=\int_{1-t}^{\infty}3e^{-3t}dt=e^{3t-3}$$ Using the fact that $X_3 \sim \text{Erlang}(3,3)$ and $N(1)\sim \text{Poisson}(3)$ we get $$P\bigg(\frac{1}{2}\leq X_3 \leq 1,X_4>1\bigg)=\int_{1/2}^1e^{3t-3}f_{X_3}(t)dt=\frac{63}{16e^3}$$ $$P(N(1)=3)=\frac{27}{6e^3}$$ Dividing these two gives us our desired probability: $$P\bigg(\frac{1}{2}\leq X_3 \leq 1\bigg|N(1)=3\bigg)=\frac{7}{8}$$


问:
Showing integral $\int_0^k \text{sinh}^{-2/3}(x)\mathrm{d}x$, has a hypergeometric solution.
For my undergraduate physics research I encountered the integral,
\begin{align}
I=\int_0^k\frac{1}{\text{sinh}^{2/3}x}\mathrm dx, \ \ \ k=\text{arsinh}(\alpha),\ \ \  \alpha\in(0,\infty)\tag{1}
\end{align}
This has the solution, as is given by WA,
\begin{align}
I=a^{1/3}\ _2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right).\tag{2}
\end{align}
I've convinced myself it is correct, as I've plotted the result and evaluated it numerically via SymPy, but I'm stuck trying to derive the answer myself.

My plan of attack has been to transform Eqn. [1] into the standard integral representations of $_2F_1$ I can find online, those being
\begin{align}
_2F_1(a,b;c;z)=&\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1\frac{t^{b-1}(1-t)^{c-b+1}}{(1-zt)^a}\mathrm dt,\tag{3}\\
\\
_2F_1(a,b;c;z)=&\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^\infty\frac{t^{c-b+1}(1+t)^{a-c}}{(1-z+t)^a}\mathrm dt.\tag{4}\\ \\
_2F_1(a,b;c;z)=&\frac{1}{\Gamma(b)}\int_0^\infty e^{-t}t^{b-1}\ _1F_1(a;c;zt)\mathrm dt\tag{5}
\end{align}
Substituting my values for $a, b, c,$ and $z$ into Eqns. [3] & [4] I get
\begin{align}
_2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right)&=\frac{\Gamma(7/6)}{\pi^{1/2}}\int_0^1\frac{t^{1/2}}{(1-t)^{1/3}(1+\alpha^2t)^{1/6}}\mathrm dt,\ \ \ \ \ \ \text{or}\\ \\
_2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right)&=\frac{\Gamma(7/6)}{\pi^{1/2}}\int_0^\infty\frac{t^{5/4}}{(1+t)(1+\alpha^2+t)^{1/6}}\mathrm dt.
\end{align}

For $I$, Using the definition of sinh$(x)$, substituting $e^x=t$, and factoring I arrive at
\begin{align}
I=2^{2/3}\int_1^{e^k}\frac{\mathrm dt}{t^{1/3}(t-1)^{2/3}(t+1)^{2/3}}.
\end{align}
I've really just got a hundred questions on what exactly to do.

答:
Define the function $\mathcal{I}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{I}{\left(a\right)}:=\int_{0}^{\operatorname{arsinh}{\left(a\right)}}\mathrm{d}\eta\,\frac{1}{\left[\sinh{\left(\eta\right)}\right]^{2/3}}.$$

You're probably going to kick yourself when realize what the first substitution should be, but sometimes the most obvious paths are most overlooked.

Given $a\in\mathbb{R}_{>0}$, we obtain

$$\begin{align}
\mathcal{I}{\left(a\right)}
&=\int_{0}^{\operatorname{arsinh}{\left(a\right)}}\mathrm{d}\eta\,\frac{1}{\left[\sinh{\left(\eta\right)}\right]^{2/3}}\\
&=\int_{0}^{a}\mathrm{d}x\,\frac{1}{x^{2/3}\sqrt{1+x^{2}}};~~~\small{\left[\eta=\operatorname{arsinh}{\left(x\right)}\right]}\\
&=\int_{0}^{a^{2}}\mathrm{d}y\,\frac{1}{2y^{5/6}\sqrt{1+y}};~~~\small{\left[x=\sqrt{y}\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{a^{2}}{2a^{5/3}t^{5/6}\sqrt{1+a^{2}t}};~~~\small{\left[y=a^{2}t\right]}\\
&=\frac{a^{1/3}}{2}\int_{0}^{1}\mathrm{d}t\,\frac{1}{t^{5/6}\sqrt{1+a^{2}t}}.\\
\end{align}$$

You should now have no problem translating this into the hypergeometric form using the first integral representation you listed, so I leave you to take it from here.


问:
Proof of a theorem in Washington's book on Cyclotomic fields.
I am reading chapter 13 - 'Iwasawa theory of $\mathbb Z _p$-extensions' in Washington's book 'Cyclotomic fields'.

On page 267 it is claimed that for extension $F$ (which is infinite as it is maximal unramified extension outside $p$), there is a closed subgroup $H \subset J$(the idèle group) containing $K^*$ such that $J/H \simeq Gal(F/K)$.

Milne states the surjectivity of the reciprocity map for number fields (once the map is subjective the existence of closed subgroup is fine as the reciprocity map is continuous and the subgroup $\{1\}$ is closed in Krull topology). So I am ok with this.

Later on the same page the author claims that for $$U''=\prod _{\mathfrak p \nmid p} U_{K_\mathfrak p } $$ $\overline {K^* U''} \subset H$ and because $F$ is maximal, and this is the part which I do not follow, $H=\overline {K^* U''}$.

Can I prove that under the reciprocity map if $I_K/\overline {K^*U''} \simeq Gal(L/K)$ then $L$is unramified outside $p$ ?

Any help is appreciated.

答: In Milne's notes, V, 5.7, it is explained what happens for infinite extensions. In the number field case, the map from the idele group is always surjective with, as you observe, closed kernel. The kernel is always the closure of the obvious stuff that it contains. I don't think that's proved in Milne's notes --- perhaps in Tate's article in Cassels-Frohlich or Artin-Tate. (Obvious stuff: $K^{\times}$ times the product of whatever is in the kernels of the local Artin maps.)


答:
My version of Mathematica gives a slightly different result. I got
$$I=\frac{(-1)^{1/6}\Gamma(1/6)\sqrt{\pi}}{\Gamma(2/3)}+3\alpha^{1/3}{}_{2}F_1\left(\left[\frac{1}{6},\frac{1}{2}\right];\frac{7}{6},-\alpha^2\right)$$
With a whole bunch of conditions for convergence after that.

Title: Showing integral $\int_0^k \text{sinh}^{-2/3}(x)\mathrm{d}x$, has a hypergeometric solution.

Body:
For my undergraduate physics research I encountered the integral,
\begin{align}
I=\int_0^k\frac{1}{\text{sinh}^{2/3}x}\mathrm dx, \ \ \ k=\text{arsinh}(\alpha),\ \ \  \alpha\in(0,\infty)\tag{1}
\end{align}
This has the solution, as is given by WA,
\begin{align}
I=a^{1/3}\ _2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right).\tag{2}
\end{align}
I've convinced myself it is correct, as I've plotted the result and evaluated it numerically via SymPy, but I'm stuck trying to derive the answer myself.

My plan of attack has been to transform Eqn. [1] into the standard integral representations of $_2F_1$ I can find online, those being
\begin{align}
_2F_1(a,b;c;z)=&\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1\frac{t^{b-1}(1-t)^{c-b+1}}{(1-zt)^a}\mathrm dt,\tag{3}\\
\\
_2F_1(a,b;c;z)=&\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^\infty\frac{t^{c-b+1}(1+t)^{a-c}}{(1-z+t)^a}\mathrm dt.\tag{4}\\ \\
_2F_1(a,b;c;z)=&\frac{1}{\Gamma(b)}\int_0^\infty e^{-t}t^{b-1}\ _1F_1(a;c;zt)\mathrm dt\tag{5}
\end{align}
Substituting my values for $a, b, c,$ and $z$ into Eqns. [3] & [4] I get
\begin{align}
_2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right)&=\frac{\Gamma(7/6)}{\pi^{1/2}}\int_0^1\frac{t^{1/2}}{(1-t)^{1/3}(1+\alpha^2t)^{1/6}}\mathrm dt,\ \ \ \ \ \ \text{or}\\ \\
_2F_1\left(\frac{1}{6},\frac{1}{2};\frac{7}{6};-\alpha^2\right)&=\frac{\Gamma(7/6)}{\pi^{1/2}}\int_0^\infty\frac{t^{5/4}}{(1+t)(1+\alpha^2+t)^{1/6}}\mathrm dt.
\end{align}

For $I$, Using the definition of sinh$(x)$, substituting $e^x=t$, and factoring I arrive at
\begin{align}
I=2^{2/3}\int_1^{e^k}\frac{\mathrm dt}{t^{1/3}(t-1)^{2/3}(t+1)^{2/3}}.
\end{align}
I've really just got a hundred questions on what exactly to do.


问:
Prove $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$ is convex
I came across a question which contained this $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$

I needed to prove $x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5\geq0$ which I've done by just showing that the minimum is 3 and it's convex function.

The second part asked to show that the whole problem is convex but I didn't know how to prove $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$  is convex which is same as asking $\sqrt{x^TAx+b^Tx+c}$ is convex where $x^TAx+b^Tx+c\geq0$. any help?

答:
$f(x)=\sqrt{x^TAx+b^Tx+c}$ is convex where $x^TAx+b^Tx+c\geq0$

$A$ should be assumed positive definite (e.g., $\sqrt{25-x^2}$ is not convex on $[0,5]$).

Proof

Some nonnegative polynomials can be written as a sum of squares. By Hilbert's Theorem, that is the case for every second degree polynomial. This allows you to write the function as a norm.

Failed attempt below

For convenience I assume that $A$ is symmetric. The gradient of $f$ is
$$Df=\frac{2Ax + b}{2\sqrt{x^TAx+b^Tx+c}}$$
and the Hessian is:
$$Hf =\frac{4\sqrt{x^TAx+b^Tx+c}A - (2Ax+b)(2Ax+b)^T/\sqrt{x^TAx+b^Tx+c}}{4(x^TAx+b^Tx+c)}$$
Multiplying the numerator and denominator with $\sqrt{x^TAx+b^Tx+c}$ yields:
$$Hf =\frac{4(x^TAx+b^Tx+c)A - (2Ax+b)(2Ax+b)^T}{4(x^TAx+b^Tx+c)^{3/2}}$$
The denominator is positive on the interior of the feasible region, so all we need to check is if the numerator is positive semidefinite. In the one dimensional case, the numerator simplifies to $4ac-b^2$, which is indeed nonnegative because $ax^2+bx+c=0$ has at most one solution. Looking at derivations of the abc-formula, completing the square might be the key to a solution, but I do not yet see how it can be done. Another issue in higher dimensions is that $x^TAxA$ does not cancel against $Axx^TA$. Perhaps the statement is false after all.


Title: How are these two equations equal?

Body:
$$\dfrac{1}{1+e^{-x}} = \dfrac{e^x}{1+e^x}$$

I was told to sketch a curve but couldn't figure out the first step. The solution manual rewrote the left hand side of the equation above as the right hand side. I cannot figure out what they did to get this. Could someone explain this to me?

Body: $$\frac{1}{1+e^{-x}}=\frac{e^x}{e^x}\frac{1}{1+e^{-x}}=\frac{e^x}{e^x+e^x\cdot e^{-x}}=\frac{e^x}{e^x+1}$$


问:
Proof that given $y\geq x$, $\frac{y}{\sqrt{y^2+4}} \geq \frac{x}{\sqrt{x^2+4}}$
Given $x,y\in \mathbb{R}$ such that $x\leq y$, then $$\frac{y}{\sqrt{y^2+4}} \geq \frac{x}{\sqrt{x^2+4}}$$
Any hint?

答:
Let $x = 2\tan a$ and $y = 2\tan b$ such that $a,b \in (-\frac\pi2,\frac\pi2)$. ( tan obtains all possible values in this interval)

In this interval $\tan$ is increasing. So, $\boxed{x\le y \iff a\le b}$

Now, $\frac{x}{\sqrt{x^2+4}} = \frac{2\tan a}{2\sec a} = \sin a$. Similarly $\frac{y}{\sqrt{y^2+4}} = \sin b$

In the interval $(-\frac\pi2,\frac\pi2)$, $\sin $ is also increasing. So $\boxed{a\le b \iff \sin a\le \sin b \text{ or } \frac{x}{\sqrt{x^2+4}} \le\frac{y}{\sqrt{y^2+4}}}$


问:
Minimum of difference is difference of minimum
Suppose $g: \mathbb R^n \to [0,\infty]$ and $h: \mathbb R^n \to [0,\infty)$. Let $f = g-h$. For any function $j: \mathbb R^n \to [0,\infty]$, let $j_M(x) = \min\{j(x),M\}$ for $M \in \mathbb N$.

Is it true that $f_M = g_M - h_M$?

Attempt: Let $x \in \mathbb R^n$. If $f_M(x) = M$, then $g(x) \geq M + h(x) \geq M$, so $g_M(x) = M$. But there's no reason why $h(x)=0$, so it doesn't look like the result will hold. Is this the right idea?

Body:
No. Consider $g(x)=2x, h(x)=x$. Then $f(x)=x$. If $x=\frac{3M}{4}$, then $g_M(x)=\min(\frac{3M}{2},M)=M, f_M(x)=h_M(x)= \frac{3M}{4}$. So
$$ f_M(x)=\frac{3M}{4} > \frac{M}{4}= g_M(x)-h_M(x)$$


问:
Properties of Chebyshev’s $\psi$ Function
Given the definition $\psi(x)=\sum_{n\le x}\Lambda(n)$

How can one arrive to the conclusion below:

$$\psi(x)=\sum_{p\le x}\left\lfloor{\frac{\log x}{\log p}}\right\rfloor\log p $$

答:
$$
\begin{aligned}
\left\lfloor\log x\over\log p\right\rfloor
&=\sum_{k\le\log x/\log p}1=\sum_{k\log p\le\log x}1 \\
&=\sum_{p^k\le x}1
\end{aligned}
$$

Thus we have

$$
\begin{aligned}
\psi(x)
&=\sum_{n\le x}\Lambda(n)=\sum_k\sum_{p^k\le x}\log p \\
&=\sum_{p\le x}\left\lfloor\log x\over\log p\right\rfloor\log p
\end{aligned}
$$


答:
Extended hint: a non-negative quadratic function with positive-definite $A$ can be written as
$$
x^TAx+b^Tx+c=\|M(x-x_0)\|^2+d^2.
$$
It can be done by completing the square in $x$ or simply identifying $M$, $x_0$ and $d$ from the equation: $A=M^TM$, $d^2$ is the minimum value ($3$ in your case) and $x_0$ is the minimizer.

Now the function you are dealing with is
$$
f(x)=\left\|\begin{matrix}M(x-x_0)\cr d\end{matrix}\right\|.
$$
It could be proven to be convex directly by definition using triangle inequality (or as a composition of the convex norm-function and the affine one).

Title: Prove $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$ is convex

Body:
I came across a question which contained this $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$

I needed to prove $x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5\geq0$ which I've done by just showing that the minimum is 3 and it's convex function.

The second part asked to show that the whole problem is convex but I didn't know how to prove $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$  is convex which is same as asking $\sqrt{x^TAx+b^Tx+c}$ is convex where $x^TAx+b^Tx+c\geq0$. any help?


问:
How to prove $4\times{_2F_1}(-1/4,3/4;7/4;(2-\sqrt3)/4)-{_2F_1}(3/4,3/4;7/4;(2-\sqrt3)/4)\stackrel?=\frac{3\sqrt[4]{2+\sqrt3}}{\sqrt2}$
I have the following conjecture, which is supported by numerical calculations up to at least $10^5$ decimal digits:
$$4\times{_2F_1}\left(-\frac{1}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)-{_2F_1}\left(\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)\,\stackrel?=\,\frac{3\sqrt[4]{2+\sqrt{3}}}{\sqrt{2}},$$
where $_2F_1$ denotes the hypergeometric function.

Can you suggest any ideas how to prove it?

The conjectural closed form was obtained using WolframAlpha query

```<code>ToRadicals[RootApproximant[2.94844576626425580599908814238570067699233]]
</code>```

答:
This is the identity 15.5.12 from DLMF, with $a=-1/4$, $b=3/4$, $c=7/4$ and the special form
$$ F(b,a,a,x)=(1-x)^{-b}. $$
Is this how you got your identity?</p>


问:
Evaluating $\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt$
I need help evaluating any of:

$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt$$

$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi (2n - 1) t)}{t^z} dt$$

Where $z \in \Bbb{C}$

Also I would like to ask if this is true:

$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt = \int_{1}^{\infty} \frac {\sum_{n=1}^{\infty} \cos(2 \pi n t)}{t^z} dt$$

If yes then how to in clever way calculate:
$$\sum_{n=1}^{\infty} \cos(2 \pi n t)$$

答:
I will solve the first sum only, since the second one may be tackled in a similar way. Assume that $\operatorname{Re}(z) > 0$. Then

\begin{align*}
\sum_{n=1}^{N} \int_{1}^{\infty} \frac{\cos(2\pi n t)}{t^z} \, \mathrm{d}t
&= \int_{1}^{\infty} \left( \sum_{n=1}^{N} \cos(2\pi n t) \right) \frac{\mathrm{d}t}{t^z} \\
&= z \int_{1}^{\infty} \left( \sum_{n=1}^{N} \frac{\sin(2\pi n t)}{2\pi n} \right) \frac{\mathrm{d}t}{t^{z+1}}
\end{align*}

Now, from the knowledge on Fourier series, we know that

$$ S_N(t) := \sum_{n=1}^{N} \frac{\sin(2\pi n t)}{2\pi n} $$

is bounded uniformly in $N$ and $t$ and that

$$ \lim_{N\to\infty} S_N(t) = \frac{1}{4} - \frac{t - \lfloor t \rfloor}{2} $$

for each $ t \in \mathbb{R}\setminus\mathbb{Z}$. So by the dominated convergence theorem,

\begin{align*}
\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac{\cos(2\pi n t)}{t^z} \, \mathrm{d}t
&= z \int_{1}^{\infty} \left( \frac{1}{4} - \frac{t - \lfloor t \rfloor}{2} \right) \frac{\mathrm{d}t}{t^{z+1}}.
\end{align*}

The last integral can be computed by decomposing it over the subintervals of the form $[k, k+1]$ for $k = 1,2, \dots$, and the result is

$$ \frac{\zeta(z)}{2} - \frac{1}{4}\left(\frac{z+1}{z-1}\right). $$

This function is analytic on all of $\operatorname{Re}(z) > 0$.


问:
How to get the matrix of a computation (MIT 18.06SC)
I have this matrix A:

```<code>| 1 0 0 0 |
| 1 1 0 0 |
| 1 2 1 0 |
| 2 3 2 1 |
</code>```

If I multiply the first row with (-1), and add it up to the second row, we have the matrix B:

```<code>| 1 0 0 0 |
| 0 1 0 0 |
| 1 2 1 0 |
| 2 3 2 1 |
</code>```

And E is an unknown matrix where:

E*A = B

What is the value of the matrix E?

Basically, I'm trying to figure out the matrix which is equivalent to a computation in a matrix. I know that an option is to compute the inverse of A. However, I'm looking for a solution without inverse (as it's in the original problem I provided in the link below).

Originally this is the problem 2.2 from Algebra MIT OCW 18.06SC but simplified, as I'm struggling with this precise step.

答:
Try to convince yourself of the following by looking at the matrix multiplication
formaula:

Let $E, A$ and $B$ be $n \times n$ matrices such that $ E A =B$.   Let $\{E_{ik}\}_{k=1}^n$ the $i$th row elements of $E$ and let $A_k$ denote the rows of $A.$ (Note that $A_k$ is a $1\times n$ matrix). Then the $k$th row of B is given by
$$ \sum_{k=1}^n E_{ik}A_k.$$
In your example:

$\begin{pmatrix}
1&&&\\
-1&1&&\\
&&1\\
&&&1\\
\end{pmatrix}
\begin{pmatrix}
1& 0&0&0 \\
1 &1 &0& 0 \\
1 &2 &1& 0\\
2 &3&2&1\
\end{pmatrix}  =\begin{pmatrix}
1 & 0&0&0 \\
0 &1 &0& 0 \\
1 &2 &1& 0\\
2 &3&2&1
\end{pmatrix}$

The second row of $B$ is given by  minus one times first row of $E$ plus one times the  second row of $E$ plus $0$ times the third row etc.


问:
Compound interest vs cumulative growth rate
My friend gave me the following problem. Imagine you made \$10 in
January. You ended up making \$100 by the end of December. What would
the growth rate have to be on the income if you were to make \$100 cumulative
in 12 months? His answer was:
$$
10+10(1+r)+10(1+r)^{2}+...=100
$$
and then solve for $r.$ Mine was to use the compound interest formula:
$$
10\left(1+r\right)^{12}=100
$$
What is the conceptual difference between the two formulas? In both
cases, the 10 grows to 100. I undersand what is going on mathematically here, but aren't the two conceptually super related? If not, where is the key difference?

答: In his formula if the ... goes on forever, it will converge only for negative r.  Yours is the formula for monthly compounding, which is not specified.


问:
Chebyshev’s $\theta_1(x) = \sum_{n \leqslant x } (x-n) \Lambda_1 (n)$
Let $\theta_1 = \int_{1}^{x}\theta (t)dt$, for $x \gt 1$ where $\theta(x)$ is the Chebyshev’s function.

Letting $\Lambda_1(n) = \log n,\;$ if $n$ is prime, then $\Lambda_1(n)=0.$ Otherwise prove that:

$$\theta_1(x) = \sum_{n \leqslant x } (x-n) \Lambda_1 (n).$$

答:
Method 1: Geometric interpretation

From single-variable calculus, integrals can be interpreted as area under the curve. To calculate the area under the step-function's "curve," we consider calculating them vertically. Since each rectangle has area $(x-n)\Lambda_1(n)$, we conclude

$$
\vartheta_1(x)=\sum_{n\le x}(x-n)\Lambda_1(n)
$$

Method 2: Abel's summation (aka Riemann-Stieltjes integration)

Consider reversing Abel's summation:

$$
\begin{aligned}
\vartheta_1(x)
&=\int_1^x\vartheta(t)\mathrm dt
=t\vartheta(t)|_1^x-\int_1^x t\mathrm d\vartheta(t) \\
&=x\vartheta(x)-\sum_{1<n\le x}n\Lambda_1(n)
\end{aligned}
$$

Since $\Lambda_1(1)=0$, it does not make any difference if we change the summation interval to $1\le n\le x$. As a result, we get

$$
\vartheta_1(x)=x\sum_{n\le x}\Lambda_1(n)-\sum_{n\le x}n\Lambda_1(n)=\sum_{n\le x}(x-n)\Lambda_1(n)
$$


问:
Derive Cauchy-Bunyakovsky by taking expected value
In my notes, it is said that taking expectation on both sides of this inequality $$|\frac{XY}{\sqrt{\mathbb{E}X^2\mathbb{E}Y^2}}|\le\frac{1}{2}\left(\frac{X^2}{\mathbb{E}X^2}+\frac{Y^2}{\mathbb{E}Y^2}\right)$$ can lead to the Cauchy-Bunyakovxky (Schwarz) inequality $$\mathbb{E}|XY|\le\sqrt{\mathbb{E}X^2\mathbb{E}Y^2}$$ I am not really good at taking expected values, may anyone guide me how to go about it?

Note: I am familiar with the linearity and monotonicity of expected values, what I am unsure about is the derivation that leads to the inequality, especially when dealing with double expectation.

Thanks.

答:
You can simplify your inequality as follows, for the left side:

$|\frac {XY}{\sqrt{EX^{2}EY^{2}}}|=\frac {|XY|}{\sqrt{EX^{2}EY^{2}}}$

for the right side, take the expectation:

$\frac{1}{2}E\left( \frac{X^2}{EX^2}+\frac{Y^2}{EY^2}\right)= \frac{1}{2} E \left( \frac{X^2 EY^2+X^2 EY^2}{EY^2 EX^2} \right)$

Now, $E(X^2 EY^2+X^2 EY^2)=2*EX^2EY^2 $ using the fact that $E(EY^2)=E(Y^2)$

Plug in and you get the result.


问:
Distributive modulo?
I would like to know if the modulo operation has distributivity like this:

$$A+B+C \pmod{M} = (A+B)\pmod{M}  +C \pmod{M}$$?

Does the equality hold true?

答:
In fact, Distribution Properties for module M holds for addition, subtraction and multiplication for integers.

Check this section for examples:

https://en.wikipedia.org/wiki/Modular_arithmetic#Integers_modulo_n

Title: Distributive modulo?


问:
Is it even possible to compare two smples with the data I have?
I'm pretty new to statistics. and recently I learned how to check if the samples are equal. I want to learn how to find out which of the samples is greater. I have some friends in college, that are good at statistics, and they said that's impossible to find it out with the data I have.
For example, I have two drugs ($x_1$, $x_2$), and I have to find out which of them is more effective. Here is the data:

Sample mean:$\bar x_1=5.9$, $\bar x_2=4.1$

Sample standard deviation: $\sigma_1=4.1$, $\sigma_2=3.7$

Sample size: $n_1=42$, $n_2=47$

And I want to test:
$$H_0:\mu_1-\mu_2>0$$
$$H_a:\mu_1-\mu_2<0$$

答:
I have some quibbles with your notation:

(a) The sample standard deviations should be
denoted $S_1, S_2.$ Notations $\sigma_1$ and
$\sigma_2$ are for population standard deviations.

(b) It is pointless to test $H_0: \mu_1 - \mu_2 \ge 0$
against $H_1: \mu_1 - \mu_2 < 0,$ because you say
$\bar X_1 = 5.9 > \bar X_2 = 4.1.$ So you have no
evidence that $\mu_1 - \mu_2 < 0$ might be true.

I will test $H_1: \mu_1 - \mu_2 \le 0$ against
$H_1: \mu_1 - \mu_2 > 0.$  You know that the first
sample mean is greater than the second. The
question whether it is enough greater that the
difference would be called 'statistically significant'.

(3) Also, notice that the null hypothesis $H_0$ must always
contain an equal sign (maybe as $=, \le$ or $\ge).$

Finally, you have given summary data rather than the
actual observations. So I have no way to check whether
the data are approximately normally distributed, as
is required for the Welch two-sample t test to be valid.
So I will simply assume your data are nearly normal.

I am using a Welch two-sample t test instead of a 'pooled' two-sample t test because the two sample standard deviations you provide are different. Thus the population standard deviations may also be different, which would
violate an assumption of the 'pooled' test.

Here is printout from a recent release of Minitab
statistical software:

```<code>Two-Sample T-Test and CI

Sample   N  Mean  StDev  SE Mean
1       42  5.90   4.10     0.63
2       47  4.10   3.70     0.54

Difference = μ (1) - μ (2)
Estimate for difference:  1.800
95% lower bound for difference:  0.417
T-Test of difference = 0 (vs >):
  T-Value = 2.16  P-Value = 0.017  DF = 83
</code>```

Because the P-value $0.017$ is smaller than $0.05 = 5\%,$
we reject the null hypothesis and conclude that $\bar X_1 = 5.9$ is significantly larger than
$\bar 4.1$ at the 5% level of significance. (You could also
declare significance at the 2% level.)

"Proving" that one drug is better than another
is a practical situation requires evidence from
thousands of randomly selected subjects and
significance levels much smaller than 5%.

If required to show computations by hand using
a calculator, you can look for the Welch t statistic in a statistics text (or online). Also notice that there is a special formula for
the degrees of freedom DF in a Welch t test;
it uses sample sizes and sample standard deviations.

Body:
I have some quibbles with your notation:

(a) The sample standard deviations should be
denoted $S_1, S_2.$ Notations $\sigma_1$ and
$\sigma_2$ are for population standard deviations.

(b) It is pointless to test $H_0: \mu_1 - \mu_2 \ge 0$
against $H_1: \mu_1 - \mu_2 < 0,$ because you say
$\bar X_1 = 5.9 > \bar X_2 = 4.1.$ So you have no
evidence that $\mu_1 - \mu_2 < 0$ might be true.

I will test $H_1: \mu_1 - \mu_2 \le 0$ against
$H_1: \mu_1 - \mu_2 > 0.$  You know that the first
sample mean is greater than the second. The
question whether it is enough greater that the
difference would be called 'statistically significant'.

(3) Also, notice that the null hypothesis $H_0$ must always
contain an equal sign (maybe as $=, \le$ or $\ge).$

Finally, you have given summary data rather than the
actual observations. So I have no way to check whether
the data are approximately normally distributed, as
is required for the Welch two-sample t test to be valid.
So I will simply assume your data are nearly normal.

I am using a Welch two-sample t test instead of a 'pooled' two-sample t test because the two sample standard deviations you provide are different. Thus the population standard deviations may also be different, which would
violate an assumption of the 'pooled' test.

Here is printout from a recent release of Minitab
statistical software:

Two-Sample T-Test and CI

Sample   N  Mean  StDev  SE Mean
1       42  5.90   4.10     0.63
2       47  4.10   3.70     0.54

Difference = μ (1) - μ (2)
Estimate for difference:  1.800
95% lower bound for difference:  0.417
T-Test of difference = 0 (vs >):
  T-Value = 2.16  P-Value = 0.017  DF = 83

Because the P-value $0.017$ is smaller than $0.05 = 5\%,$
we reject the null hypothesis and conclude that $\bar X_1 = 5.9$ is significantly larger than
$\bar 4.1$ at the 5% level of significance. (You could also
declare significance at the 2% level.)

"Proving" that one drug is better than another
is a practical situation requires evidence from
thousands of randomly selected subjects and
significance levels much smaller than 5%.

If required to show computations by hand using
a calculator, you can look for the Welch t statistic in a statistics text (or online). Also notice that there is a special formula for
the degrees of freedom DF in a Welch t test;
it uses sample sizes and sample standard deviations.


问:
Integer solutions of $xy+9(x+y)=2006$ [Completing a product / rectangle]
How many integer solutions does $xy+9(x+y)=2006$ have? Here $x$ and $y$ are both integers.

My trying:
      I have tried to solve this problem. But I have no idea to solve this. Please help

答:
Just to give another way to solve.

$$p+9s=2006\Rightarrow p\equiv8\pmod9\Rightarrow p=8+9n$$Then $s=t-n$ for some $t$ which is deduced from
$$(8+9n)+9(t-n)=2006\Rightarrow t=222$$ so we have the general solution for the sum and the product $$\begin{cases}p=8+9n\\s=222-n\end{cases}$$
Now  we have for values of $x$ and $y$
$$X^2-(222-n)X+(8+9n)=0$$ where for confort we put $2n$ instead of $n$ so we have the equation $$X^2-2(111-n)X+(8+18n)=0\Rightarrow X=111-n\pm\sqrt{n^2-240n+12313}$$ We need $$n^2-240n+12313=Y^2\\(n-120)^2+2087=Y^2\\2087=(Y+n-120)(Y-n+120)$$ Since $2087$ is prime we have an easy factorization giving setting values of $n$.

(By the way, $2087$ is the same integer reached in the concise and beautiful proof by @heropup).


问:
How to evaluate a function with a quadratic argument or higher? --how to proceed?
example
if I have a function of the form
$$f(2x+5) = 4x-3$$ and I am asked to evaluate $$f(8)$$ ,I make $$t=2x+5$$ from where
$$x= (t-5)/2$$
then $$f(t) = 4( (t-5)/2 ) -3$$ , I take $$t=8$$ and that's it.**

Doubts

but how do I do it here, if I do the above I get a quadratic equation or a cubic one

a)$$f( x^2 +1/x^2) =x+1/x$$ , find $$f(4)=?$$ , x>0
b)$$f( x^3 +1/x^3) =x+1/x$$ , find $$f(4)=?$$ , x>0

thank you in advance
what is in bold is edited

答:
Sometimes it helps to see things a different way. Instead of solving for $x$, I would have looked at the equation as

$$f(2x+5) = 4x-3 = 4x+10-10-3 = 2(2x+5)-13$$

which means $f(x) = 2x-13$. Taking the second equation we have

$$f\left(x^2+\frac{1}{x^2}\right) = x+\frac{1}{x} = \sqrt{\left(x+\frac{1}{x}\right)^2} = \sqrt{x^2+2+\frac{1}{x^2}} = \sqrt{\left(x^2+\frac{1}{x^2}\right)+2}$$

which means $f(x) = \sqrt{x+2}$ if $x>0$ (you will need to clarify whether this is the case). Can you try the last one?


问:
How are these two equations equal?
$$\dfrac{1}{1+e^{-x}} = \dfrac{e^x}{1+e^x}$$

I was told to sketch a curve but couldn't figure out the first step. The solution manual rewrote the left hand side of the equation above as the right hand side. I cannot figure out what they did to get this. Could someone explain this to me?

Title: How are these two equations equal?

Body:
Multiply the left hand side by $1 = e^x/e^x$.  So, you have

$$\frac{1}{1+e^{-x}}\cdot\frac{e^x}{e^x} = \frac{e^x}{1+e^{x}}$$
$$\frac{e^x}{e^x + e^{x-x}} = \frac{e^x}{1+e^{x}}$$
$$\frac{e^x}{e^x + 1} =  \frac{e^x}{1+e^{x}}$$

($e^0 = 1$)


问:
Given three customers arrive in the first hour, find probability third arrives in second half
Assume customers arrive according to a Poisson process with rate $\lambda = 3$.

Given three customers arrive in the first hour, find the probability that the third customer arrives in the second half of the hour.

So that is if $X_3$ is the third arrival time: $P(X_3 \in [0.5,1] \mid N(1) = 3)$

and we have $X_3 \sim$ Gamma$(3,3)$.

To solve this I have cases : $\bigg \{X_1,X_2 \in [0,.5]$ and $X_3 \in [.5,1]\bigg\}$ or

$\bigg \{X_1 \in [0,.5]$ and $X_2,X_3 \in [.5,1]\bigg\}$ or $\bigg \{X_1,X_2, X_3 \in [.5,1] \bigg \}$

However, isn't there a way to solve this using the Uniform distribution?

答: Given that exactly three events were observed in the first hour, the distribution of their event times are independent and uniform over that hour.  So the probability the third event occurs in the second half hour is equal to the complement of the probability that all three events occur in the first half hour.  But the probability of all three events occurring in the first half hour is simply $(1/2)^3 = 1/8$; therefore, the desired answer is $1 - 1/8 = 7/8$.


问:
Need help on simplifying an analytical solution to quadratic programming problem
I'm trying to find the analytical solution to the following problem

\begin{align}
\min_c & \; c^T Wc \\
\text{s.t. } & \; Xc = x_0
\end{align}
where $c,x_0 \in\mathbb R^n$, $W$ is an invertible $n\times n$ diagonal matrix, $X$ is a $m\times n$ matrix.

It's easy to show with Lagrange multiplier that the solution $c^*$ satisfies
$$
\begin{pmatrix}
W & -X^T\\
X & 0
\end{pmatrix}
\begin{pmatrix}
c^*\\
\lambda
\end{pmatrix} =
\begin{pmatrix}
0\\
x_0
\end{pmatrix}
$$

With the method in this answer, we can find that
$$
\begin{pmatrix}
W & -X^T\\
X & 0
\end{pmatrix}^{-1}
 = \begin{pmatrix}
\left[W^TW+X^TX + W^T(-X^T)(XX^T)^{-1}(-X)W \right]^{-1}  &  **\\
** & **
\end{pmatrix}
\begin{pmatrix}
W & X^T\\
-X & 0
\end{pmatrix} \\= \begin{pmatrix}
** & \left[W^TW+X^TX + W^T(-X^T)(XX^T)^{-1}(-X)W \right]^{-1}X^T\\
** & **
\end{pmatrix}
$$

(** are irrelevant elements in calculation.)

Thus
$$
c^* = \left[W^TW+X^TX + W^T(-X^T)(XX^T)^{-1}(-X)W \right]^{-1}X^Tx_0
$$

It seems complicated. I read in a note that
$$
c^* = WX^T (XWX^T)^{-1} {x}_0
$$

Are they equivalent and how can I simplify the expression? Any reference is appreciated.

答: The linear system boils down to $Wc^* = X^T\lambda$ and $Xc = x_0$. The first equation gives you $c^*=W^{-1}X^T\lambda$, and plugging that into the second equation you get $XW^{-1}X^T\lambda=x_0$. That means $\lambda=(XW^{-1}X^T)^{-1}x_0$, so $c^*=W^{-1}X^T(XW^{-1}X^T)^{-1}x_0$.


问:
Proof by Induction for $\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)}$
$$\sum_{i = 1}^{n - 1} \frac{1}{i(i+1)} = \frac{n- 1}{n}$$

How would you go about proving this by induction? the base case $n=1$ gives you $1/2 = 0$

The only thing I can think of using is $\sqrt{2}$, but that is not a natural number.

答:
Your understanding of the base case is incorrect.  When $n = 1$, the sum on the left-hand side is empty, because the upper index is $n - 1 = 0$, meaning there are no terms to add.  This gives $0 = 0$ as desired.

The induction step is straightforward.  If we let $$S_n = \sum_{i=1}^{n-1} \frac{1}{i(i+1)},$$ then $$S_{n+1} = S_n + \frac{1}{n(n+1)}.$$  So if the induction hypothesis $$S_n = \frac{n - 1}{n}$$ is true, what does that say about $S_{n+1}$?


问:
Can we always "locate" a transitive model inside some larger transitive model?
We work in $\mathsf{ZFC}$ + "Every set is an element of a transitive model of $\mathsf{ZFC}$."

Suppose $M$ is a transitive set model of $\mathsf{ZFC}$. Say that $M$ is locatable iff there is some transitive set $N$ such that:

$N\models\mathsf{ZFC}$,

$M\in N$, and

$M$ is parameter-freely-definable in $N$.

The requirement that $N\models\mathsf{ZF}$ prevents coding $M$ into $N$ in a "silly" way. To see how this can be an issue, let $\alpha=\min\{\gamma: L_\gamma\models\mathsf{ZFC}\}$. If $N$ is a transitive model of $\mathsf{ZFC}$ with $L_\alpha\in N$, then $N$ sees that $L_\alpha$ is countable and so contains lots of (say) reals which are Cohen-generic over $L_\alpha$. Suppose $r,s$ are such reals; how is $N$ to distinguish between $L_\alpha[r]$ and $L_\alpha[s]$ in a parameter-free way? This seems to be an obstacle in general to the locatability of $L_\alpha[c]$ for $c$ Cohen-generic over $L_\alpha$. However, such an $L_\alpha[c]$ is nonetheless locatable: force over $L_\beta$, where $\beta=\min\{\gamma: \gamma>\alpha, L_\gamma\models\mathsf{ZFC}\}$, to code $c$ into the continuum pattern appropriately.

Indeed, such coding tricks give:

Every countable $M$ is locatable.

Note that the locating model is very simple - it's an element of $L(M)$. For uncountable $M$s this seems like a nontrivial further condition, and in particular none of the absoluteness theorems I'm familiar with seem to directly apply.

Similarly, we have:

If $L_\theta\models\mathsf{ZFC}$ then $L_\theta$ is locatable, even if $\theta$ is uncountable.

(For example, every ordinal $\theta$ is parameter-freely-definable in the $(\vert\theta\vert^++\theta+1)$th level of $L$ satisfying $\mathsf{ZFC}$ since $(i)$ that level of $L$ can "count" the $\mathsf{ZFC}$-satisfying levels below it, and $(ii)$ we can recover $\theta$ from $\vert\theta\vert^++\theta$ as there is a unique way to write the latter in the form $\kappa+\gamma$ for $\kappa$ a cardinal and $\gamma<\kappa$.)

However, the general situation is not clear to me:

Question: is every transitive set model of $\mathsf{ZFC}$ locatable?

Of course it is quite plausible that this will depend on additional set-theoretic hypotheses; for example, I think an affirmative answer is more likely assuming $\mathsf{V=L}$ than assuming the existence of large cardinals. But I don't immediately see how to show any of this.

答:
Here's another case where you get a positive answer, one which is quite far from $V=L$. The idea is, with the right hypothesis the strategy Noah outlined for coding countable $M$ into the continuum pattern of some larger model works even if $M$ is uncountable. There's some annoying details to make it work, but thats the core of it.

Work over ZFC + the Inner Model Hypothesis. For the sake of the reader who is unfamiliar, let me say a bit about the IMH, introduced by Sy Friedman. (The reader who already knows can skip to the next paragraph.) The IMH is a sort of width-reflection principle, saying that if something happens in an outer model you can locate it in an inner model. Specifically, it states that if a parameter-free assertion $\varphi$ is true in an inner model of a width-extension of $V$, then it is true in an inner model of $V$. The IMH implies there are inner models with measurable cardinals of arbitrarily large Mitchell order, so in particular it implies $V=L$ badly fails. On the other hand, it implies there are no large cardinals in $V$. For consistency, you can get a model of the IMH from a Woodin cardinal with an inaccessible above.

You might worry about how you actually formulate the IMH, given that it involves quantifying over arbitrary outer models. But for the purposes of this answer, it's enough to just look at extensions from a (tame) class forcing. This is formalizable as a single assertion in second-order set theory. Call this principle the CIMH.

There is a class-forcing extension $V[H]$ of $V$ where we add the following: for each transitive model $M$ of ZFC and each $N \in V$ with $M \in N$, an $N$-generic $G_{N,M}$ for the forcing which codes $M$ into the continuum pattern of an extension of $N$, starting at, say, $\aleph_{13}^N$.
Let me note that the point of doing all the coding at once is that to use the CIMH we need the $\varphi$ we are reflecting to not use parameters, and so we cannot just do the coding for one pair $(M,N)$.

By the CIMH, we have an inner model $W$ which thinks that for all appropriate pairs $(M,N)$ that $G_{N,M}$ exists. This is because that is a parameter-free assertion satisfied by $V[H]$. It's now immediate that if a pair $(M,N)$ as above is in $W$ then $M$ is locatable in $V$. This is because $W$ must have $N[G_{N,M}]$ as an element, and this model can define $M$ by looking at the continuum pattern. But of course then $N[G_{N,M}]$ must be in $V$, so $V$ thinks $M$ is locatable.

We are not done yet, however, since we have only seen that every $M \in W$ is locatable, not every $M \in V$. This takes a small bit more.

Suppose otherwise. Then we have a least-rank counterexample, and the rank $\alpha$ of the least-rank counterexample is definable without parameters. So consider the following forcing extension. First ensure that some appropriate $M$ of rank $\alpha$ is coded into the continuum pattern in $V[k]$ at a definable location, then force over $V[k]$ to get $V[k][H]$ like above, in such a way that we don't destroy the coding of $M$. We now reflect this property of $V[k][H]$ down to an inner model $W'$, by the CIMH. That is, some transitive model $M$ of ZFC of rank $\alpha$ must be in $W'$ and so $M$ is locatable in $V$, just like the $W$ argument. This is the desired contradiction.


答: $$\theta_1(x) = \int_1^x \theta(t) dt= \int_1^x \left(\sum_{n\le t} \Lambda_1(n) \right)  dt =\sum_{n\le x} \int_{n}^x \Lambda_1(n) dt = \sum_{n \le x}(x-n)\Lambda_1(n).$$

Title: Chebyshev’s $\theta_1(x) = \sum_{n \leqslant x } (x-n) \Lambda_1 (n)$

Body:
Let $\theta_1 = \int_{1}^{x}\theta (t)dt$, for $x \gt 1$ where $\theta(x)$ is the Chebyshev’s function.

Letting $\Lambda_1(n) = \log n,\;$ if $n$ is prime, then $\Lambda_1(n)=0.$ Otherwise prove that:

$$\theta_1(x) = \sum_{n \leqslant x } (x-n) \Lambda_1 (n).$$


问:
Equation of a line passing through a point and forming a triangle with the axes
How can I find the equation of a line that;

is passing through the point (8, 6) and

is forming a triangle of area 12 with the axes

?

So I tried to start using $A = |{\frac{mn}{2}}|$ and got that $m\cdot n$ is either -24 or 24. What now?

答: You might want to convince yourself that such a line is determined uniquely by it's intersection point with the x-axis, which allows you to calculate it's intersection point with the y-axis and the area of the resulting triangle.</p>


问:
Prove $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$ is convex
I came across a question which contained this $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$

I needed to prove $x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5\geq0$ which I've done by just showing that the minimum is 3 and it's convex function.

The second part asked to show that the whole problem is convex but I didn't know how to prove $\sqrt{x_1^2+4x_1x_2+5x_2^2+2x_1+6x_2+5}$  is convex which is same as asking $\sqrt{x^TAx+b^Tx+c}$ is convex where $x^TAx+b^Tx+c\geq0$. any help?

答:
Note that
$$
x^\top A x + b^\top x + c = \begin{bmatrix}x^\top&1\end{bmatrix}\begin{bmatrix}A&b/2\\b^\top/2 & c\end{bmatrix}\begin{bmatrix}x\\1\end{bmatrix} = \tilde{x}^\top\tilde{A}\tilde{x}.
$$

Now, we test $\tilde{A}$ for positive definiteness by calculating its principal minors (I'll skip most of the calculations). We have
$$
A = \begin{bmatrix}1&2\\2&5\end{bmatrix},\quad b=\begin{bmatrix}2\\6\end{bmatrix},\quad c = 5,
$$
thus, the first principal minor is $1$, and we have $\det(A)>0$, which is the second principal minor of $\tilde{A}$. All that remains is to test $\det{\tilde{A}}$, which is (using Schur):
$$
c\det(A - \frac{bb^\top}{4c}) = 5\begin{vmatrix}1 - 1/5& 2-3/5\\2-3/5 &5-9/5\end{vmatrix} = 3 > 0.
$$

Since $\tilde{A}$ is positive definite, then $\lVert \tilde{x} \rVert_{\tilde{A}} = \sqrt{\tilde{x}^\top \tilde{A} \tilde{x}}$ is a norm, and thus convex.


问:
A question on Inversion of a Permutation
There is a defination of inversion their which states that the number of "inversions" of the word: the number of pairs of positions for which the leftmost position of the pair holds a letter 1 and the rightmost position holds a letter 0 in the word. For example, there is one word with 0 inversions, 0011. There is 1 with only a single inversion, 0101. There are two words with 2 inversions, 0110, and 1001. There is one with 3, 1010, and finally one word with 4 inversions, 1100.Could anyone what does this mean and i know what the inversion of a Permutation is but i did'nt get this defination.</p>

答: For example, in $1010$ the inversions are the pairs $\{1,2\}$ (because the $1^{\text{st}}$ letter is $1$ and the $2^{\text{nd}}$ is $0$), $\{1,4\}$ (because the $1^{\text{st}}$ letter is $1$ and the $4^{\text{th}}$ is $0$), and $\{3,4\}$ (because the $3^{\text{rd}}$ letter is $1$ and the $4^{\text{th}}$ is $0$).


答:
Just take that as:

Find $x+\dfrac1x$ given $x^2+\dfrac{1}{x^2}=4$.

And this is easily solvable by algebraic methods.

EDIT

Find $x+\dfrac1x$ given $x^3+\dfrac{1}{x^3}=4$.

Let $x+\dfrac1x=k$. Then,

We know,
$$(a+b)^3=a^3+b^3+3ab(a+b)$$
Putting $a=x$ and $b=\dfrac1x$, we get
$$k^3=x^3+\dfrac{1}{x^3}+3\times x\times\dfrac1x\times k$$
$$k^3-3k-4=0$$
Solving gives
$$k=\boxed{\sqrt[3]{2-\sqrt3}+\sqrt[3]{2+\sqrt3}}$$

Hope this helps. Ask anything if not clear :)

Title: How to evaluate a function with a quadratic argument or higher? --how to proceed?

Body:
example
if I have a function of the form
$$f(2x+5) = 4x-3$$ and I am asked to evaluate $$f(8)$$ ,I make $$t=2x+5$$ from where
$$x= (t-5)/2$$
then $$f(t) = 4( (t-5)/2 ) -3$$ , I take $$t=8$$ and that's it.**

Doubts

but how do I do it here, if I do the above I get a quadratic equation or a cubic one

a)$$f( x^2 +1/x^2) =x+1/x$$ , find $$f(4)=?$$ , x>0
b)$$f( x^3 +1/x^3) =x+1/x$$ , find $$f(4)=?$$ , x>0

thank you in advance
what is in bold is edited


问:
Is there a way to simplify $2^{2^{n-1}}$?
A question from a math noob...

Is there a way to simplify the following?
$$2^{2^{n-1}}$$

Thanks in advance!

答:
Not really. $2^{2^{n-1}}$.

If it’s part of an equation you can take the log $2^{n-1} log 2$. Another log and you get $(n-1)log 2 + log log 2$.


问:
Are A, B and C mutually independent?
Chose a number uniformly at random on $[0,1]$. Consider the events

$A= [0,1/2] \quad B= [1/3,2/3], \quad C= [11/24,17/24]$

Are A, B, C independent (i.e. mutually independent)?

I realize that I need to show that $\mathbb{P}(A\cap B\cap C)=\mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)$. But I don't understand what the events are in the first place. How are we choosing those two numbers on $[0,1]$?

答:
To show that two events are independent, you need only show that $P(A\cap B)=P(A)P(B)$, but to show that three events are independent, you need to not only show $P(A\cap B\cap C)=P(A)P(B)P(C)$ but also pairwise independence $P(A\cap B)=P(A)P(B),P(A\cap C)=P(A)P(C), \text { and }P(B\cap C)=P(B)P(C)$.

For this problem, A is the event that the random number falls into the interval given, and similarly for B and C. It is easiest to draw the intervals on a number line to see where they overlap.

$P(A)=\frac 1 2, P(B)=\frac 1 3, P(C)=\frac 1 4, P(A\cap B\cap C)=\frac 1 2 - \frac {11} {24}=P(A)P(B)P(C)$, so this condition holds.

Check the conditions for pairwise independence:

$P(A\cap B)=\frac 1 2 - \frac 1 3=P(A)P(B)$

$P(A\cap C)=\frac 1 2 - \frac {11} {24} \approx .042 \ne P(A)P(C)=\frac  1 8 =.125$. Hence the three events are not independent.

In general, to show that a set of events $A_1,...,A_k$ is independent, you must show that for every subset $A_{1_{j_l}},...,A_{i_{j_l}};j=2,...,k;i=1,...,j;l=1,...,{k \choose j}$, where j is the size of the subset, i runs from 1 to the subset size, and l runs from 1 to the number of subsets of that size, that $P(A_{1_{j_l}}\cap...\cap A_{i_{j_l}})=P(A_{1_{j_l}})...P(A_{i_{j_l}})$.

Title: Are A, B and C mutually independent?


问:
Consider the experiment of rolling two dice. If X =(second die−first die), then find the probability mass function of X.
What should be the probability mass function for this question??</p>

答:
Just write down all possible values of $X$ and find the probabilities for each of those events. First find the range of $X$. In this case the range will be $1-6 = -5$ to $6-1 = 5$. So PMF can be found as follows:
$$P(X<-5) = P(X>-5) = 0$$
$$P(X = -5) = P(die2 = 1)P(die1 = 6) = \frac{1}{6}\frac{1}{6} = \frac{1}{36}$$
$$P(X = -4) = P(die2 = 2)P(die1 = 6)+ P(die2 = 1)P(die1 = 5) = \frac{2}{36}$$
$$P(X = -3) = \frac{3}{36}$$ and so on


答:
Left-hand side of the equation is

$$\dfrac{1}{1+e^{-x}} = \dfrac{1}{1 +\dfrac{1}{e^{x}}}=\dfrac{1}{\dfrac{e^{x} +1}{e^x}}=\dfrac{e^x}{e^x+1},$$

which is equal to the right-hand side $\dfrac{e^x}{1+e^x}$.

Title: How are these two equations equal?


答: Quick answer: same distribution as sum, but shifted so that it is centered at zero.  $P(X=0)=6/36$, $P(X=\pm 1)=5/36$ each, ..., $P(X=\pm 5)=1/36$ each.

Title: Consider the experiment of rolling two dice. If X =(second die−first die), then find the probability mass function of X.

Body: What should be the probability mass function for this question??</p>


问:
Solve an equation with mod operator
Consider the following equation
$$mod(x+100/x,5)=1.$$
How can we solve it with $x>0$?

First of all, the function $x+100/x$ is a continuous function with range $[20, \infty)$, therefore, it must has solutions. Then, it can be checked that the solutions are discrete, but how can we express them?

答:
Using that the RHS must be an integer, show that $x$ must be an integer such that $1 \le x \le 100$ and that it must divide $100,$ then use that $100 = 1 \cdot 2^2 \cdot 5^2$ to find all the divisor of $100,$ then test which ones work.

Spoiler alert: Only $1$ and $100$ work.

Title: Solve an equation with mod operator


问:
When equality holds in C-S integral inequality?
In general C-S inequality :
$(a^2+b^2)(x^2+y^2)\geq (ax+by)^2$

We know that if $\frac{a}{x}=\frac{b}{y}$ then equality holds.

Then, when equality holds in C-S integral inequality? :
$$ (\int_a^bf(x)g(x)dx)^2\leq \int_a^b f(x)^2dx\int_a^bg(x)^2dx$$

Is there any condition (in $f$ and $g$ for equality holds?

答:
The general form of the Cauchy-Schwarz inequality is not $(a^2 + b^2)(x^2 + y^2)\geq(ax+by)^2$—that's just another specific example of it (here, for the real vector space $\mathbb{R}^2$ with the dot product as your inner product). The general inequality holds for all inner product spaces and says that $|\langle u, v\rangle|^2 \leq \langle u,u\rangle\cdot\langle v,v\rangle$, with equality precisely when $u$ and $v$ are linearly dependent, i.e. when there exists some scalar $c$ such that $u=cv$. (If you're looking for a proof of this general version, you'll find one in basically any linear algebra book, as well as on Wikipedia.) If you consider the vector space $C([a,b])$ over $\mathbb{R}$ of continuous functions on $[a,b]$, with an inner product defined by

$$\langle f, g\rangle =\int_a^b f(x)g(x)dx$$

then you get the version of the Cauchy-Schwarz inequality you're discussing:

$$\left( \int_a^b f(x)g(x)dx\right)^2 \leq \left(\int_a^b f(x)^2 dx\right) \left(\int_a^b g(x)^2 dx\right)$$

As with the general statement of the inequality, equality holds precisely when $f$ and $g$ are linearly dependent.


问:
How to argue that variance integral actually converges?
I'm just starting out with a course where were being taught about expectation and variance of continuous distributions via probability density functions. I was given a definition of how the variance of a random variable $X$ is calculated given its probability density function $f(x)$, but the definition seems all to suspicious to me. In particular, if $X$ is a random variable and its PDF is $f(x)$, then it's variance is computed via
$$
\text{Var}(X) = \int _{\mathbb{R}}x^2 f(x) \, dx
$$
where I've assumed the expected value of $X$ is $0$ just to make things a bit simpler.

Now, since $f(x)$ is assumed to be a PDF, then you'd expect
$$
\int _{\mathbb{R}}f(x) \,dx = 1
$$
which seems to hint that the function is (sort of) integrable. But, how can you even make sure that integration of $x^2 f(x)$ actually yields a finite value? I've tried a back-of-the-envelope calculation using integration by parts, and the situation seems to hinge on whether or not $x^2 f(x) \rightarrow 0$ as $|x| \rightarrow 0$, yet I can't find how this is justified. Any and all help will be appreciated :)

答:
The integral does not always converge, even if $\lim_{|x|\to\infty} x^2f(x)=0$. There are valid random variables which have an infinite variance: for instance, consider  a random variable with density
$$
f(x) = \frac{3\sqrt{3}}{4\pi}\cdot\frac{1}{1+|x|^3}, \qquad x\in\mathbb{R}.
$$

Worse, there even are valid random variables for which the expectation is not defined. A standard example is those following the Cauchy distribution, e.g., one with density function
$$
f(x) = \frac{1}{\pi}\cdot \frac{1}{1+x^2}, \qquad x \in\mathbb{R}.
$$
(You can check easily that $\int_{\mathbb{R}} f(x)dx = 1$, yet $\int_{\mathbb{R}} xf(x)dx$ does not exist.) A fortiori, the random variables without a well-defined expectation do not have a finite variance.

Some vocabulary: a random variable is in $L^1$ if its expectation exists. It is in $L^2$ if its variance exists. By the Cauchy—Schwarz inequality, $L^2\subseteq L^1$.

Finally, you can even find density functions  $f$ (hence, random variables) such that
$$
\lim_{|x|\to\infty} x^2f(x)=0
$$
yet which don't have a well-defined expectation (not in $L^1$). An example is
$$
f(x) = \frac{1}{C}\cdot\frac{1}{1+x^2\log(1+x^2)}, \qquad x\in\mathbb{R}
$$
where $C = \int_{\mathbb{R}}\frac{1}{1+x^2\log(1+x^2)} \approx 1.36$.


问:
Find the area of an equilateral triangle
Point $M$ lies on the side $AB$ of an equilateral triangle $ABC$ such that $AM:BM=1:2$. If $CM=m$, find the area of the triangle $ABC$.

 The answer given in my book is $\dfrac{9m^2\sqrt3}{28}$.

So we have $$S_{ABC}=S_{AMC}+S_{BMC}\iff\\\iff\dfrac{AB\cdot CH}{2}=\dfrac{AM\cdot CH}{2}+\dfrac{BM\cdot CH}{2}\\AB\cdot CH=AM\cdot CH+BM\cdot CH.$$
This doesn't seem to help very much. Any help would be appreciated.

答:
I like your approach of looking at the areas of smaller triangles. You nearly showed that $S_{BMC}=\frac23 S_{ABC}$ , which means if we find $S_{BMC}$ , we can simply multiply it by $\frac32$ to get the desired $S_{ABC}$ . Let's see how we can do that!

If we choose point $N$ on $BC$ such that $BN : CN = 1 : 2$ then $S_{MNC} = \frac23 S_{BMC}$. Besides, $\triangle MNB$ is a right triangle (why?). Now, this helps us calculate side length $a$ of $\triangle ABC$ in terms of $m$. Can you take it from here?


问:
How to compute the elements of the field $\mathbb{R}[x] / \langle x^2+1\rangle $?
Let $I$ be the ideal generated by the polynomial $f(x) = x^2 +1 $. Hence, $I = \langle x^2 + 1 \rangle $.

Generally, how Can I compute the elements of the field $\mathbb{R}[x] / \langle x^2+1\rangle $ ?

I see how to compute things like $\,x^2 = -1\,\overset{\large \times\ x}\Longrightarrow\, x^3 = -x,\ $ but it is not immediately clear how to do general computations.

答:
Question asked in Two exercises about Euclidean algorithm in Euclidean ring $\mathbb{Z}_2[x]$. , was stated as duplicate of above. and I don't have enough points to comment or reopen it. so, providing answer here.

if someone reopen and post below answer there, it would be helpful to @Vizik.

@Vizik
you have a field with elements $${\{1,x,x^2,1+x,x+x^2,1+x+x^2,1+x^2\}}$$  generated by irreducible polynomial ${1+x+x^3}$ where ${1+x=x^3}$

Now it is easy to see that ${(1+x) \cdot (x+x^2) = 1 (mod(1+x+x^3))}$

Hence the inverse of ${(1+x)}$ is ${(x+x^2)}$


问:
Alternate form of quotient rule
Given a function $f(x)=h(x)g(x)$, the product rule gives us
\begin{equation}
f'(x)=g(x)h'(x)+h(x)g'(x)
\tag{1}
\end{equation}

We can rearrange this to get the quotient rule, as follows.

Given a function $g(x)=\frac{f(x)}{h(x)}, \; h(x) \neq 0$, equation 1 gives us
\begin{align}
g'(x)&=\frac{f'(x)-g(x)h'(x)}{h(x)}\\
     &=\frac{f'(x)h(x)-g(x)h(x)h'(x)}{[h(x)]^2}\\
     &=\frac{f'(x)h(x)-f(x)h'(x)}{[h(x)]^2}
\tag{2}
\end{align}

I can see the standard version of the quotient rule in line 3 of equation 2 has the advantage(?) of not mentioning $g$ on the RHS, but I find line 1 easier to understand geometrically. Is there any reason not to use line 1 as a version of the quotient rule?

答:
Well, think about it.

$g(x)$ is not a function you know explicitly, if $f(x)$ and $h(x)$ are different 'classes' of functions you will have a hard time figuring out $h(x)$, if both of them are polynomials then you still have to divide them, which takes up a considerable amount of time depending on your division skills and the length of the polynomials.

Easy case:
$$ \left(g_1(x)=\dfrac{f(x)}{h(x)}=\dfrac{\sin(x)}{\cos(x)}\right)'\Rightarrow g_1(x)=\tan(x)$$

Harder case:
$$ \left(g_2(x)=\dfrac{f(x)}{h(x)}=\dfrac{x^7+2x^6+2x^4-x^2}{x^3-5x+7}\right)'\Rightarrow g_2(x)=x4+x3+x2+x+1 \text{ with a remainder of } -3x^2-2x-7$$
Of course you could just calculate with $\frac{f(x)}{h(x)}$ but at that point you are using the classical quotient rule.

Non-solvable case (by simple algebraic manipulations):
$$ \left(g_1(x)=\dfrac{f(x)}{h(x)}=\dfrac{\sin(x)}{\sinh(x)}=\dfrac{e^{ix}-e^{-ix}}{e^x-e^{-x}} \right) $$

I cannot comment on the visual appeal of your geometric understanding of it as i don't really have that, but that's also subjective.

Title: Alternate form of quotient rule


问:
Can the sum of two non-degenerative random variables ever be uniform?
Background

A rather famous problem is regarding if the sum of loaded dice can be uniformly distributed over $\{2, \cdots, 12\}$; the answer is no. We can see that the top answer in the above link can be generalized to an $n$-sided dice for odd $n$. But what about even $n$? This isn't so hard to prove through a different technique; denote $\mathbb{P}(X = k) = p_k$ and $\mathbb{P}(Y=k) = q_k$ for $k \in \{0, \cdots, n-1\}$ (it's more convenient to work with zero indexing). In order for $X+Y$ to be uniform over $\{0, \cdots, 2n-2\}$, we must have
\begin{align*}
\mathbb{P}(X+Y = 0) = p_0q_0 = \frac{1}{2n-1} \qquad \text{and} \qquad \mathbb{P}(X+Y = 2n-2) = p_{n-1}q_{n-1} = \frac{1}{2n-1}
\end{align*}
But
\begin{align*}
\mathbb{P}(X+Y = n-1) &= \sum_{k=0}^{n-1}\mathbb{P}(X = k)\mathbb{P}(Y = n-1-k) \\
&\ge p_{0}q_{n-1} + p_{n-1}q_0 \\
&\ge 2\sqrt{p_{0}q_{n-1}p_{n-1}q_0} \\
&= \frac{2}{2n-1} \\
&> \frac{1}{2n-1}
\end{align*}
We see that the above strategy would've also worked for the odd case as well.

Generalizations
I have the following few problems. Is it possible to construct two (real) independent random variables $X, Y$ such that $X+Y$ is

uniformly distributed over a finite set $S \subseteq \mathbb{R}$ (e.g. $S = \{1, 1.6162, 2.71828, 3.14159\}$, it doesn't even have to be integers for that matter). In other words, the measure associated with $X+Y$ is a normalized counting measure over $S$.

uniformly distributed over finite-Lebesgue measure set $S \subseteq \mathbb{R}$ (e.g. Cantor set, it doesn't have to be an interval). In other words, the measure associated with $X+Y$ is a normalized Lebesgue measure over $S$.

If the above settings are two limited, then we can extend $X$ and $Y$ to be $p$-dimensional random vectors and ask the associated existence question for $S \subseteq \mathbb{R}^p$.

If still not possible, perhaps ask if it's possible to construct a sequence $\{X_k\}$ and $\{Y_k\}$ with $X_k, Y_k$ independent and converging to non-degenerate $X, Y$ respectively such that the limit of the measure associated with $X_k + Y_k$ converges to either a normalized counting or normalized Lebesgue measure.

I conjecture the answer to the first 3 problems is not possible because, intuitively in the continuous case, the density of $X+Y$ must be ``smooth'' because convolutions generally smooth the functions they are working with, but a uniform density is discontinuous; a similar principle applies to the discrete case. Problem 4 would be interesting to think further.

答: If $X$ or $Y$ (or both) is discrete then it is possible to find examples that sum to a uniform distribution. For example, let $X$ assign a weight of $1/2$ to each of $0$ and $1$, and let $S$ be any subset of $\mathbb R$ such that $S$ and $S+1$ are disjoint. Then if the measure of $Y$ is the uniform (or counting, if $S$ is finite) measure over $S$ then $X+Y$ is uniformly distributed over $S\cup (S+1)$.


问:
Approximate (continuous) functions by step functions - Convergence Rate
I want to approximate a measurable and bounded function $S:[0,1]^2 \rightarrow \mathbb{R}$ by step functions. So, assume we have a uniform partition $$
\{I_j\}_{j=1}^n,  \textit{   }  j=1,...,n
$$ of $[0,1]$. Then a step function $S_n$ with $n$ steps is defined as
$$
S_n(u,v) = \sum_{i, j \leq n} a_{i,j}\chi_{I_i}(u) \chi_{I_j}(v).
$$
What (continuity-)conditions do we have to assume for $S$ and how are we supposed to choose the coefficients $a_{i,j}$ to achieve "good" convergence rates for the approximation, for instance
$$
\|S_n - S\|_{L^{2}} \in \mathcal{O}(f(1/n))
$$
for some $f$ (ideally the identity or better) for $n \rightarrow \infty$.

答:
Since
$$ ||S - S_n||_2^2 = \sum_{i,j} \int_{I_i \times I_j} |f - a_{ij}|^2 $$
it's fairly easy to see that the best choice is
$$ a_{ij} = \frac{1}{N^2} \int_{I_i \times I_j} f $$
you can prove this by differentiating
$$ \alpha \in \mathbb{R} : \varphi(\alpha) = \int_{I_i \times I_j} |f - \alpha|^2$$
and noticing it is convex.


问:
$P^2=P, Q^2=Q, (P+Q)^2=P+Q$, show $PQ=QP=0$.
Let $P,Q$ be linear operator from $n$ dimensional vector space $V$ to $V$. If $P^2=P, Q^2=Q, (P+Q)^2=P+Q$, show $PQ=QP=0$.

It is easy to see $PQ+QP=0$. What to do next?

答:
Multiply by $I-P$ on the right, to get $PQ(I-P)=0$, or $PQ=PQP$.

Multiply by $I-P$ on the left, to get $(I-P)QP=0$, or $QP=PQP$.


问:
How to make a cos function with decreasing wavelength
How could I decrease the wavelength, as x increases in a cos function?</p>

答:
There are many ways to achieve that, but the simplest is to use
$$\cos(x^k)$$
where the real constant $k \gt 1$ determines how rapidly the wavelength decreases as $x$ increses.


问:
Parametrization of Contour $C$
For the function $f(z)=1$ $(z\in \mathbb{C})$ and C is an arbitrary contour from any fixed point ${z}_{1}$ to any fixed point ${z}_{2}$ in the $z$ plane. Use parametric representations for $C$ to evaluate $\int _{C}^{}f(z)dz$.

Here is my solution: Let ${z}_{1}$ and ${z}_{2}$  are two fixed point and $C$ is straight line. Then $z=(1-t){z}_{1}+t{z}_{2}$ where $t\in \left[0,1\right]$, from here $dz=-{z}_{1}dt+{z}_{2}dt$. Then we get

$\int _{{z}_{1}}^{{z}_{2}}f(z)dz=\int _{0}^{1}1({z}_{2}dt-{z}_{1}dt)=\int _{0}^{1}{z}_{2}dt-\int _{0}^{1}{z}_{1}dt={z}_{2}-{z}_{1}$

Hence,

$\int _{C}^{}f(z)dz=\int _{{z}_{1}}^{{z}_{2}}f(z)dz=\int _{{z}_{1}}^{{z}_{2}}1dz={z}_{2}-{z}_{1}$

Is that my solution right ? I feel that I'm missing something. Any help will be appreciated.

答:
Your proof is fine but restricted to the case where $C$ is a staight line. You are asked to prove the result for an arbitrary contour.

According to the definition given in the comments, a contour $C$ is the concatenation of smooth curves $C_1,\dots,C_K$. The curve $C_k$ has endpoints $c_{k-1},c_{k}$ and is parametrized by the $C^1$ function $\gamma_i:[0,1] \mapsto \mathbb{C}$. In particular, $c_0 = z_1$ and $c_K = z_2$.

By the formula for countour integration, for each $C_k$:

$$\int_{C_k} f(z) dz = \int_0^1 f(\gamma_k(t)) \gamma_k'(t) dt.$$

For the specific case $f(z) =1$, it yields:

$$\int_{C_k} f(z) dz =\int_0^1 \gamma_k'(t) dt=\left[ \gamma_k (t)\right]_0^1=c_k-c_{k-1}.$$

The concatenation of the smooth curves $C_k$ to create $C$ leads to:

$$\int_C f(z) dz = \sum_{k=1}^K \int_{C_k}f(z) dz= \sum_{k=1}^Kc_k-c_{k-1}$$

This sum is telescopic, and we are left with, as wanted:

$$\int_C f(z) dz = c_K-c_0=z_2-z_1.$$

Further note : this shows that the countour integration of the constant function $1$ is path-independent, it only depends on the endpoints. This property is in fact a characterization of holomorphic functions, as shown here. Holomorphic function is the complex equivalent of being derivable for real ones, so it can be seen as an extension of the fundamental theorem of calculus to the complex domain.

Title: Parametrization of Contour $C$


问:
How to show $E^0\subseteq E?$
Prove that  $E$ is open  if and only if $E^{0}=E$ ?

My attempt : to prove $E^{0}=E$ we must show that $E^0\subseteq E$ and $E^0\supseteq E$

To show  :$E^0\supseteq E$

proof : Suppose $E$ is open and $p \in E$ , then there exist $r >0$ such that $N(p,r) \subseteq E $ $\implies$ that $p$ is an interior point

therefore $p \in E^0$ .Hence $E^0\supseteq E$

Here im confused that how to show  $E^0\subseteq E?$

答:
Here it is a proof in the context of metric spaces.

Let us prove the implication $(\Rightarrow)$ first.

Then we are assuming that $E$ is open. This means that $E\cap\partial E = \varnothing$.

Consequently, we have that
\begin{align*}
E = E\cap X & = E\cap(\text{int}(E)\cup\partial E\cup\text{ext}(E))\\\\
& = (E\cap\text{int}(E))\cup(E\cap\partial E)\cup(E\cap\text{ext}(E))\\\\
&= E\cap\text{int}(E)
\end{align*}

Hence we conclude that $E\subseteq\text{int}(E)$.

Once we always have that $\text{int}(E)\subseteq E$, we deduce that $E = \text{int}(E)$.

Indeed, if $p\in\text{int}(E)$, there is an open ball such that $B(p,r)\subseteq E$, whence $p\in E$.

We may now prove the implication $(\Leftarrow)$.

Therefore we have that $E = \text{int}(E)$.

Hence we have that $E\cap\partial E = \text{int}(E)\cap\partial E = \varnothing$

Thence we conclude that $E$ is open.

Hopefully this helps!

Title: How to show $E^0\subseteq E?$


答: Second inclusion is done by definition. Interior of set $A$ is the largest open set contained in $A.$

Title: How to show $E^0\subseteq E?$

Body:
Prove that  $E$ is open  if and only if $E^{0}=E$ ?

My attempt : to prove $E^{0}=E$ we must show that $E^0\subseteq E$ and $E^0\supseteq E$

To show  :$E^0\supseteq E$

proof : Suppose $E$ is open and $p \in E$ , then there exist $r >0$ such that $N(p,r) \subseteq E $ $\implies$ that $p$ is an interior point

therefore $p \in E^0$ .Hence $E^0\supseteq E$

Here im confused that how to show  $E^0\subseteq E?$


问:
Show that $ 3^n$ is not $\mathcal{O}(2^n)$
To show that $2^n$ is $\mathcal{O}(3^n)$ is straightforward. On the other hand, if we want to show the opposite that $3^n$ were $\mathcal{O}(2^n)$, then we would have $3^n\le C·2^n$ for all sufficiently large $n$. This is equivalent to $C\ge (\frac{3}{2})^n$, which is clearly impossible, since $(\frac{3}{2})^n$ grows without bound as $n$ increases.

Question:is it clearly impossible based on the question because $C$ is supposed to be constant in the definition of Big-$\mathcal{O}$?

答:
This is too long for a comment, but worth pointing out. Rosen's approach for analyzing asymptotics is nice, because it doesn't rely on Calculus techniques. So in that sense, it is more accessible. On the other hand, the Calculus techniques are quite nice for showing things like $f(n) \not \in \mathcal{O}(g(n))$.

Fix $k \geq 0$, and suppose that $f, g : \mathbb{Z}_{\geq k} \to \mathbb{R}$ satisfy that the following limit exists:
\begin{align}
L := \lim_{n \to \infty} \left| \frac{f(n)}{g(n)} \right|.
\end{align}

We have the following:

If $L = 0$, then $f(n) \in o(g(n))$. That is, $f(n) \in \mathcal{O}(g(n))$ and $f(n) \neq \Theta(g(n))$. Put another way, $f(n) \in \mathcal{O}(g(n))$ and $g(n) \not \in \mathcal{O}(f(n))$.

If $0 < L < \infty$, then $f(n) = \Theta(g(n))$. Put another way, $f(n) \in \mathcal{O}(g(n))$ and $g(n) \in \mathcal{O}(f(n))$.

If $L = \infty$, then $f(n) \in \omega(g(n))$. That is, $f(n) \not \in \mathcal{O}(g(n))$, but $g(n) \in \mathcal{O}(f(n))$.

Computing limits is sometimes much easier than trudging through these tedious inequalities.


问:
Question About Maximal Smooth Atlas
Definition of Maximal Smooth Atlas: In the text GTM $218$ by John Lee, the author defines a smooth altas $\mathcal{A}$ to be a maximal smooth atlas if any chart that is smoothly compatible with every chart in $\mathcal{A}$ is already in $\mathcal{A}$.

The author makes a comment that, if a topological manifold $M$ can be covered by a single chart, the smooth compatibility condition is trivially satisfied, so any such chart automatically determines a smooth structure on $M$, which means a maximal smooth altas on $M$.

Now let's consider an example.

Take the manifold $M=\mathbb{R}^n.$ Let $\mathcal{A}_1=\{ (\mathbb{R}^n,Id_{\mathbb{R}^n})\}$ and  $\mathcal{A}_2=\{ (B_1(x),Id_{B_1(x)}):x\in \mathbb{R}^n\}$.

By author's comment, $\mathcal{A}_1$ should be a maximal smooth atlas on $M$. But it is obvious that every chart in $\mathcal{A}_2$ is compatible with the only chart in $\mathcal{A}_1$. And by the definition of maximal smooth atlas, $\mathcal{A}_1$ is not maximal because at least it doesn't contain those charts in $\mathcal{A}_2$.

I'm quite confused now. Thanks for help.

答:
No, I didn't say that $\mathscr A_1$ is a maximal smooth atlas, I said it determines a maximal smooth atlas. This is a reference to Proposition 1.17(a), which says "every smooth atlas $\mathscr A$ for $M$ is contained in a unique maximal smooth atlas,
called the smooth structure determined by ${\mathscr A}$."

Title: Question About Maximal Smooth Atlas


问:
Fixed point of curvature operator
Let $y=f(x)$ the graph of a real-valued function. We define its curvature by : $$curv(f) = \frac{|f''|}{(1+(f')^2)^{3/2}}$$

I would like to know if there is any function (apart from the trivial anwser $f(x)=0$) whose curvature is itself. So what is the fixed point of $curv$ ?

答:
If, as @Gae. S. did, we try to solve the equation

$$y'' - y (1 + [y']^2)^{\frac 32}=0$$ I suppose that there is no explicit solution.

However, switching variables, the equation
$$y \left(1+\frac{1}{[x']^2}\right)^{\frac 32}+\frac{x''}{[x']^3}=0$$ shows explicit solutions in terms of elliptic integrals.

Title: Fixed point of curvature operator


问:
problem with Multinomial Theorem
I am trying to find the constant term in this expansion:

$$\bigg(1 + x + 2y^{2} - \frac{1}{x^{2}y}\bigg)^{15}$$

I have been trying for hours but I hit a wall when working out the values $(r_1, r_2, r_3, r_4)$. I get $2$ equations $r_3 = r_2/4$ and $r_4 = r_2/2$ and can't get any further. maybe I'm just being stupid but any help is greatly appreciated

Title: problem with Multinomial Theorem

Body:
$$\bigg(1 + x + 2y^{2} - \frac{1}{x^{2}y}\bigg)^{15}=\sum\frac{15!}{(n_1)!(n_2)!(n_3)!(n_4)!}(1)^{n_1}(x)^{n_2}(2y^2)^{n_3}\Bigl(\frac{-1}{x^2y}\Bigr)^{n_4}$$
$$=\sum\frac{15!}{(n_1)!(n_2)!(n_3)!(n_4)!}(2)^{n_3}(-1)^{n_4}(x)^{n_2-2n_4}(y)^{2n_3-n_4}$$
$$2n_3-n_4=0$$
$$n_2-2n_4=0$$
$$n_2=4n_3=2n_4$$
$$n_1+n_2+n_3+n_4=15$$
$$n_1+4n_3+n_3+2n_3=15$$
$$n_1+7n_3=15$$
$$(n_1,n_3)=(15,0),(8,1),(1,2)$$
Thus,
$$(n_1,n_2,n_3,n_4)=(15,0,0,0) , (8,4,1,2), (1,8,2,4)$$

$$C=\sum\frac{15!}{(15)!(0)!(0)!(0)!}(2)^{0}(-1)^{0}(x)^{0-0}(y)^{0-0}+\sum\frac{15!}{(8)!(4)!(1)!(2)!}(2)^{1}(-1)^{2}(x)^{4-2*2}(y)^{2*1-2}+\sum\frac{15!}{(1)!(8)!(2)!(4)!}(2)^{2}(-1)^{4}(x)^{8-2*4}(y)^{2*2-4}$$
$$C=1+\frac{2\times 15! }{(8!)(2!)(4!)}+\frac{4\times 15! }{(8!)(2!)(4!)}=1+\frac{6\times 15! }{(8!)(2!)(4!)}=4054051$$


问:
Can I take limsup and liminf on both sides of the inequality?
I have $g(n)<f(n)$ but I only know that $g(n)$ converges as $n\to \infty$. The problem is that I don't know if the limit of $f(n)$ exists, so instead of proving that one, I want to take liminf on both sides to get $\lim_{n\to\infty}g(n)\leq \liminf_{n\to\infty}f(n)$. Would there be any leap of logical step in my argument? I'd appreciate your help.

答:
There is an increasing sequence of integers $a_n$ (in particular, $a_n\to+\infty$) such that $\lim_{n\to\infty} f(a_n) = \liminf_{n\to\infty} f(n)$.
Since $g( a_n ) \leq f( a_n )$ for all $n$
we have $\lim_{n\to\infty} g(a_n) \leq \lim_{n\to\infty} f( a_n )$,
i.e., $\lim_{n\to\infty} g \leq \liminf_{n\to\infty} f$.


问:
Consider the experiment of rolling two dice. If X =(second die−first die), then find the probability mass function of X.
What should be the probability mass function for this question??</p>

答:
Let A, B be the rolls of the first and second die, respectively, and $X=A-B$.

$\begin{matrix}
Rolls &A+B &Probability& A-B&Rolls\\
1,\textbf 1&2&\frac 1{36}&\textbf{-5}&1,6\\
1,\textbf2;2,\textbf1&3&\frac 2{36}&\textbf{-4}&1,5;2,6\\
1,\textbf3;2,\textbf2;3,\textbf1&4&\frac 3{36}&\textbf{-3}&1,4;2,5;3,6\\
...&...&...&...&...\\
6,\textbf4;5,\textbf5;4,\textbf6&10&\frac 3 {36}&\textbf{3}&4,1;5,2;6,3\\
6,\textbf5;5,\textbf6&11&\frac 2{36}&\textbf{4}&6,2;5,1\\
6,\textbf6&12&\frac 1{36}&\textbf{5}&6,1
\end{matrix}$

Notice that the pmf of $A-B$ follows a similar pattern as the pmf for $A+B$. In particular, the center is now at 0 (instead of 7), and the second dice now runs in the "opposite direction". The pmf is then, as stated in the quick answer, $P(A-B=\pm5)=\frac 1 {36} \text { each},...,P(A-B=\pm1)=\frac 5 {36} \text{ each},P(X=0)=\frac 6 {36}$.