问:
For $\psi : R \to \prod^n_{i=1}R/I_i$, do we get $\ker(\psi)=\cap^n_{i=1}\ker(\pi_i)$?
Let $R$ be a ring and $I_i$ be ideals. Let also $\pi_i: R \to R/I_i$ be the projection and  $\psi : R \to \prod^n_{i=1}R/I_i$ be a homomorphism. Is it always true, that $\ker(\psi)=\cap^n_{j=1}\ker(\pi_j)$?

答:
Clearly $\ker(\pi_i)=I_i$. By definition of $\ker$:

$$ \ker(\psi)=\{x\in R\mid \pi_i(x)=0\  \forall\,i\} = \{x\in R\mid x\in I_i\  \forall\,i\} = \bigcap_{i=1}^n I_i $$


问:
Compactness of $K(S)$, if $K$ is an infinite-dimensional compact operator in Hilbert space

$H$ is infinite-dimensional Hilbert space. $K: H \rightarrow H$ is infinite-dimensional compact operator. Let $S$ be a unit sphere in H. The task is to proof that $K(S)$ couldn't be compact.

I know that:

Really we need to proof that $K[S]$ is not closed. If B is unit ball, $[B]$ - closure, then $K([B])$ is also closed(for compact operator in Hilbert space).

If $\mathrm{Ker} B$ is empty, i.e. $0 \notin K(S)$ then there is a sequence in $K(S)$ converging to $0$, so $K(S)$ isn't compact. But I don't understant how to proof noncompactness of $K(S)$ even if one of the basis vectors is mapped to $0$.

On the other side we can try to proof that if $K(S)$ is compact, then $\mathrm{Im}(K)=K(H)$ shuold have finite dimension. It is easy if $K(B)$ contains some ball(like a subspace in $K(H)$), but it's far from being true.

答:
In fact, $K(S)$ can be compact. Let $H = \ell^2(\mathbb{N})$, i.e.
$$H = \Biggl\{ f \colon \mathbb{N} \to \mathbb{C} \biggm\vert \sum_{k = 0}^{\infty} \lvert f(k)\rvert^2 < +\infty\Biggr\}$$
(using function notation to make indexing of sequences in $H$ clearer) and $K = C \circ D$, where
$$(Df)(k) = f(k+1)$$
and
$$(Cf)(k) = 2^{-k}f(k)\,.$$
Then $K(S)$ is closed, hence compact.

Let $g \in \overline{K(S)}$, and $(f_n)_{n \in \mathbb{N}}$ a sequence in $S$ with $K(f_n) \to g$. If $g = 0$, then we have $K(e_0) = g$ where $e_m(k) = \delta_{m,k}$, hence we can assume $g \neq 0$. For every fixed $k$ we have
$$f_n(k+1) \to 2^kg(k)$$
and therefore
$$\sum_{k = 0}^{m} 2^{2k}\lvert g(k)\rvert^2 = \lim_{n \to \infty} \sum_{k = 0}^{m} \lvert f_n(k+1)\rvert^2 \leqslant 1$$
for all $m$. It follows that $h \in H$, where
$$h(k) = 2^kg(k)\,.$$
Furthermore, we have $\lVert h\rVert \leqslant 1$, hence there is $h_1 \in S$  with $h = Dh_1$. By construction, $Ch = g$, thus
$$g = Kh_1 \in K(S)\,.$$

ContentLicense: CC BY-SA 4.0


问:
How can I calculate $\log(e^{e^i})$?
I'm studying complex analysis and I'm wondering how to calculate the following multivalued function (using the expression $\log(z)=\ln|z| + i\operatorname{Arg(z)}$):

$$\Large \log(e^{e^i})$$

Thank you!

答:
The exponential function $e^w$, with $w=a+ib$,  is defined as $e^{w}=e^{a}\cos b+ie^{a}\sin b$. So, for $w=i$, $a=b=1$, we have the particular value of the Euler's formula
$e^{i}=\cos \left( 1 \right) +i\sin \left( 1 \right)$.

Now let $z=e^{e^{i}}$. Since $e^{w\pm i2k\pi}=e^{w}e^{\pm i2k\pi}=e^{w}\times 1=e^{w}$, we then have

\begin{equation*}
z=e^{e^{i}}=e^{\cos \left( 1\pm 2k\pi \right) }e^{i\sin \left( 1\pm 2k\pi \right) }
\end{equation*}

and

$$
\left\vert z\right\vert =\left\vert e^{\cos \left( 1\pm 2k\pi \right)
}\right\vert \left\vert e^{i\sin \left( 1\pm 2k\pi \right) }\right\vert
=\left\vert e^{\cos \left( 1\pm 2k\pi \right) }\right\vert\times 1 =e^{\cos \left(
1\right) }.
$$

The principal argument of $z$, $\operatorname{Arg}(z)$, is such that $-\pi <\operatorname{Arg}(z)\leq \pi $. So $\operatorname{Arg}(z)=\sin \left( 1\right) $ occurs for $k=0$, and, by definition, the principal value of $\log (z)$, $\operatorname{Log}(z)$, is equal to

\begin{equation*}
\operatorname{Log}(z)=\log(\left\vert z\right\vert) +i\operatorname{Arg}(z)=e^{\cos \left(
1\right) }+i\sin \left( 1\right) ,
\end{equation*}

where $\log(|z|)$ denotes the natural logarithm of $|z|$.

Title: How can I calculate $\log(e^{e^i})$?


问:
Smallest Possible Value

Among $100$ students, $x_1$ have birthdays in January, $x_2$ have birthdays in February, and so on. If $x_0 = \max(x_1, x_2, \dots, x_{12})$, then the smallest possible value of $x_0$ is...

I came across this question and have been trying to get my head around it and the way to solve this.
I am not able to think of the way this question should be approached. A little help or hint would be very helpful.

答:
Note that $x_0 \ge x_1$, $x_0 \ge x_2$, ... , $x_0 \ge x_{12}$. Thus

$$100=x_1 + x_2 + \cdots + x_{12} \le x_0 + x_0 + \cdots + x_0 = 12 x_0$$
This shows that
$$100 \le 12 x_0$$
This gives us a lower bound: $x_0$ is at least $100/12 \approx 8.6$, i.e. $x_0$ is at least $9$.

Now you have to show that $x_0$ could be $9$. Can you conclude from here?

Title: Smallest Possible Value


问:
Does $\,x>0\,$ hint that $\,x\in\mathbb R\,$?

Does $x>0$ suggest that $x\in\mathbb R$?

For numbers not in $\,\mathbb R\,$ (e.g. in $\mathbb C\setminus \mathbb R$), their sizes can't be compared.

So can I omit "$\,x\in\mathbb R\,$" and just write $\,x>0\,$?

Thank you.

答:
Easiest solution is to just say
$$
x\in\mathbb{R}^+
$$
Expresses both conditions in one hit.</p>

ContentLicense: CC BY-SA 3.0


问:
Which equilateral triangles does the P-hexiamond (the "sphinx") tile?
There's been lots of work investigating the polyominoes which can tile a square (equivalently, a rectangle).

However, as far as I can tell there's been less investigation into the polyiamonds which can tile an equilateral triangle.

Two obvious restrictions come to mind. First, the polyiamond has to actually fit into a corner - i.e., there is some 60-degree angle the tile covers the tip of without exceeding its bounds. Second, we have a simple coloring argument: the number of right-side-up and upside-down triangles in the tile can't be the same, because in an equilateral triangle there are more of one than the other.

Applying these two conditions to all the polyiamonds on at most 4 triangles as listed e.g. here, every case is either immediately ruled out or can be easily seen to work for every triangle of a plausible side length (i.e., $s$ such that $s^2$ is a multiple of the number of triangles in the tile). For the pentiamonds, we have:

C$5$: Can't fit in the corner, ruled out.

L$5$: If placed in the corner, it leaves a $60$ degree angle void which must be filled by a piece in the same orientation, which repeats infinitely and never fits into the other corner of an equilateral triangle:

P$5$: Can be ruled out by a small amount of casework: there's only one way to place the first piece in the corner, and then a second triangle in the "concave part" of the piece can only be filled in one way, but placing those two pieces leaves a void that can't be filled.

I$5$: Very difficult; it turns out the answer is yes for all triangles with side length $5k\ge 30$, as discussed in this MathOverflow thread.

I'm interested in extending this to the hexiamond case. The two considerations outlined above rule out every piece but F$6$ and P$6$.

For F$6$, I've managed to show that the tiles must join into parallelograms in a way that prevents them from ever lining up with a third side of an equilateral triangle.

For P$6$, however, things are more complicated. (Since it is a rep-tile, it can certainly tile an infinite triangular "quadrant" (hexant?).)

Via an exhaustive search with a computer program, I have verified that P$6$ does not tile an equilateral triangle of side length $6$. However, it does tile an equilateral triangle of side length $12$:

A search for side length $18$ did not reveal any solutions after running for a while, but it was nowhere near completion when I halted the search - the tools I'm using don't seem very tractable for this or higher cases.

Which equilateral triangles can be tiled by the sphinx polyiamond?

Edit: I think that computing the tile homotopy group of the sphinx, as described in this paper, may be promising; as Theorem $7.16$ in the paper shows, it can be used to answer questions of this type for very similar kinds of tiles. Tile homology techniques may work as well, although not if it turns out that there is a signed tiling of the size-$6$ equilateral triangle with the sphinx.

答:
The triangle of side $18$ is not tilable by P6. I ran a backtracking program on it for a few hours. The fact that a side $12$ triangle is tilable means that all triangles of side 12n are tilable trivially with $n^2$ copies of the side $12$ triangle.

Side $12n-6$ triangles are not tilable for $n=1$ and $n=2$, I conjecture that a proof could be found that no $12n-6$ triangle is tilable.

I'm running my solver on the side $30$ triangle now but I don't anticipate it will finish in realistic time. Update: After $500$ hours it has placed the second last piece (out of $150$) about $12.2 billion$ times, and the third last piece $25 billion$ times. This is no sort of proof, but it's the sort of thing you get when some constraint like parity is blocking a tiling. And I'm stopping it now as I estimate it's zero percent complete (to around $30$ decimal places).

I counted the tilings for the side $12$ case, I got $830$. My solver doesn't detect symmetric tilings for triangular grids so it found asymmetric tilings 6 times each and symmetric tilings fewer times depending on their symmetry class.


问:
Distribution of point-dependent function
I want to find the distribution function of $$(XY)^{-Z},$$ where $(X, Y, Z)$ are the coordinates of a point randomly selected from the cube $[0;1]^3$. I don't know how to solve it and how to start. Can you give me a hint, please?

答:
You need to compute the volume of the part of the cube where $(XY)^{-Z} < C$.

It will end up being a triple integral.  You integrate the value 1 and manipulate the bounds so that the equation is true.

Starting with $Z$:

$$(XY)^{-Z} < C$$
$$-Z\ln(XY) < \ln(C)$$
Direction reverses because $\ln(XY)$ is always negative.
$$-Z > \frac{\ln(C)}{\ln(XY)}$$

$$Z < \frac{-\ln(C)}{\ln(XY)}$$

So you have the bounds of Z in terms of C,X,Y.   Now you need to integrate for the $X,Y$ directions.


问:
Area between paraboloid cut out by cylinder

Find area between surface $z^2=x^2+y^2$ and $x^2+y^2=2x$.

So, after using polar coordinates $x=r\cos \phi,y=r\sin \phi$ i get $0\le r \le 2\cos\phi$ and $0\le \phi \le \frac{\pi}{2}$. Than i plugged that into formula $\iint \sqrt{1+p^2+q^2}\,dx\,dy$ where $p=\frac{\partial z}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$ and $q=\frac{\partial z}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}$ and got $\frac{\sqrt{2}\pi}{2}$ but correct answer is $2\sqrt{2}\pi$. I probably need to multiply area with $4$ but honestly I'm not sure when to do that, on what it depends etc. I sketched this in plane(paraboloid and cylinder) and I can only see two symmetric areas (instead of four). If someone could clear this for me i would be very thankful.

答:
It's a basic algebra exercise to see that the horizontal cross-section of your cyclinder is a circle centered at $(1,0,0)$ on the $xy$-plane.

$$\Huge{\lvert \hspace{-15px} \bigodot}$$

object
code
$\text{equation}$

$\lvert$
$\lvert$
$x = 0$

$\bigodot$
$\bigodot$
$(x-1)^2 + y^2 = 1$

It's clear that in your cylindrical coordinates $\begin{cases}x = r\cos \phi& \\ y = r\sin \phi,& \end{cases}$, $\phi$ represents the angle between the segment from $(0,0)$ to $(x,y)$ and the $x$-axis.  By setting $0\le \phi \le \dfrac{\pi}{2}$, you've only swept the first quadrant.  However, by observing the symmetry along the $xz$-plane (i.e. $y = 0$), it suffices you have to multiply your calculated value by $2$ to get the correct answer area for one slice.

As @MathLover has pointed out, there's another slide due to the symmetry along the $xy$-plane, so you have $$\frac{\sqrt2 \pi}{2} \cdot 4 = 2\sqrt2 \pi,$$ which is your desired answer.


问:
Is $ \arctan(f(x))\leq f(x)$ whenever $|f(x)|<1\ \ \forall x\in\mathbb{R} $?
Is the following always true?
$$
\arctan(f(x))\leq f(x),\ \ \text{where}\ \ |f(x)|<1\ \ \forall x\in\mathbb{R}
$$

For example, is it correct to say that
$$
\arctan\left(\frac{\sin x}{\sqrt{x+1}}\right)\le\frac{\sin x}{\sqrt{x+1}}\ \ \forall x\in\mathbb{R},
$$
since $\arctan t=t-\frac{t^3}{3}+\frac{t^5}{5}-\dots$, where each term is not greater than the one before?

I'm confused because in the example above $\arctan$'s argument can be negative (i.e., $\sin x$ can be negative).

答:
If you derive the function $x\mapsto \text{Arctan} (x) -x$, you get that:

if $x>0$ then $\text{Arctan} (x) < x$

but if $x<0$ then $\text{Arctan} (x) > x$.


问:
Finding side of rectangle using given information
Really simple question but I am stuck. The following information is given:
$$BD=8,\quad AB = 6,\quad  ED  =5,\quad EF = EC$$
and we want to find $AF$.

If we have three $90^\circ$, what does that really mean, and how I can find $AF$?

答:
Alternative way to prove this:

Since $EF+EC=6$ and $EF=EC$, this implies that $EC=3$

Notice that triangles ABD and ECD are similar in sharing a right angle, the angle at D. There is a scaling factor of 2 as $AB=6$ and $EC=3$.

Since $BC=CD$ and $BD=8$ this implies $BC=4$ as this is half the overall distance.

$AF$ is the same length as the rectangle has opposite sides of the same length.


问:
Partial Derivative of a Relation?
I always thought that it only made sense to take the partial derivative of a function $z=f(x_{1},x_{2},x_{3},...,x_{n})$ with respect to one of its input variables, like ${\partial{f}}/{\partial{x_1}}$. But then I encountered this question:

Compute all first and second partial derivatives, including mixed derivatives, of the following function:$$x^{2}+y^{2}=\sin(xy)$$
Which leads me to think there might be some notion of "implicit partial differentiation" or something thereabouts, but I am quite confused how I should understand this, or what to do. Thanks.

答:
Think of the top most surface of a saddle on a horse.
If you think of x and y in the horizontal plane and the elevation of the surface as z and then describe the elevation in terms of the x and y location, then you have a relation.

At some point $(x, y)$ you have defined the z, elevation, and the partials describe how rapidly z changes as you move along x or y from that point.

(This doesn't work very well on a phone.)

Title: Partial Derivative of a Relation?


问:
Clarify about the connecting homomorphisms and the snake lemma
In class we said (without proof) that given a an exact sequence of chain complexes $$0\to \mathcal A\xrightarrow{f} \mathcal B\xrightarrow g \mathcal C\to 0$$ one can construct the long exact sequence $$\dots \xrightarrow{\delta_{q+1}} H(A_q)\xrightarrow{f_q} H(B_q) \xrightarrow{g_q} H(C_q)\xrightarrow{\delta_q} H(A_{q-1})\xrightarrow{f_{q-1}}\dots  \ ,$$ where the $\delta _q$ are the connecting homomorphisms. Now, I read on wikipedia how these homomorphisms are obtained, but I don't understand well why the snake lemma works in this context. In fact the vertical maps of the lemma here are the boundary homomorphisms of the corresponding complex (let's call them $\alpha_q$, $\beta_q$ and $\gamma_q$ for respectively $A_q$, $B_q$ and $C_q$); and the lemma builds a homomorphism from $\text{ker}(C_q)$ to $\text {coker }(\alpha_q)$. However, shouldn't we need an homomorphism that goes in $H(A_{q-1})\subseteq\text {coker }(\alpha_q)$? My idea is that when the vertical maps are the boundary homomorphisms one can prove that $\text {Im} (\delta_q)\subseteq \text{ker} (\alpha_{q-1})$, but I first want to know if I'm approaching things from the right point of view. Thanks for any clarify

答:
Apply the snake lemma not to
$$\begin{array}{ccccccccc}
& & A_q & \rightarrow & B_q & \rightarrow & C_q & \rightarrow & 0 \\
& & \downarrow & & \downarrow & & \downarrow & & \\
0 & \rightarrow & A_{q-1} & \rightarrow & B_{q-1} & \rightarrow & C_{q-1}, & &
\end{array}$$
but to
$$\DeclareMathOperator{\coker}{coker}
\begin{array}{ccccccccc}
& & \coker(\alpha_{q+1}) & \rightarrow & \coker(\beta_{q+1}) & \rightarrow & \coker(\gamma_{q+1}) & \rightarrow & 0 \\
& & \downarrow & & \downarrow & & \downarrow & & \\
0 & \rightarrow & \ker(\alpha_{q-1}) & \rightarrow & \ker(\beta_{q-1}) & \rightarrow & \ker(\gamma_{q-1}). & &
\end{array}$$

Make sure you see where the maps come from, that the diagram commutes, and that the rows are exact.


问:
A pigeonhole riddle

John offers Dean the following challenge:

Dean would choose 8 natural numbers so that $10\leq n \leq 36$.
If John manages to create two equal sums using only numbers that Dean chose, at most once each, he wins.

Assuming John has adequate time to check all options, will he always win? and why?

My thoughts were counting the number of options that Dean has for choosing 8 numbers, which is $27 \choose 8$, and also counting the number of options John has of creating sums, which is $\binom{27}{8}\cdot\sum^8_{i=2} \binom{8}{i}$. But that doesn't bring me anywhere.

I feel like it's true that John would always win, and I know I need to use the pigeonhole principle somehow, which means showing that for sure there are more possible ways to arrange sums than the number of sums possible. So, there are at least two arrangements for the same sum. (By arrangement for the same sum I mean two different combinations of numbers that summed give the same answer).
But I'm kind of stuck there.

Would appreciate assistance!

答:
It is being asked if John can always find from any $8$-set, two non-empty, proper, disjoint subsets having same sum.

There are $2^8-2=254$ non-empty, proper subsets. Hence $254$ possible sums for any $8$-selection.

The minimum and maximum sum for any non-empty, proper subset is $10$ and $30+31+\ldots+36=231$ respectively.

Hence due to Pigeonhole principle, John always wins.

Note that either John obtains two disjoints subsets right away or two non-disjoint subsets. In the latter case, he removes the common terms to obtain disjoint ones.

Title: A pigeonhole riddle


问:
Determinant of $n\times n$ Matrix Linear Algebra
So, I have a matrix

$$
  A = \begin{pmatrix}
  0 & 1 & 1 & ... & 1 \\
  1 & 0 & x & ... & x \\
  1 & x & 0 & ... & x \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & x & x & ... & 0
  \end{pmatrix}
$$
I need to evaluate it's determinant. At first I calculated $\det A$ for $n=2,3,4$. And I got a pattern $\det A=(-1)^{n-1}(n-1)x^{n-2}$ for every $n\ge2$. But I need to solve it differently. I added all rows to the first, then multiplied every row (except 1) by $(1+x(n-2))$ and subtracted first row multiplied by $x$. This is what I got:
$$
  \begin{pmatrix}
  n-1 & 1+x(n-2) & 1+x(n-2) & ... & 1+x(n-2) \\
  1-x & -x(1+x(n-2)) & 0 & ... & 0 \\
  1-x & 0 & -x(1+x(n-2)) & ... & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  1-x & 0 & 0 & ... & -x(1+x(n-2))
  \end{pmatrix}
$$
Now I can multiply diagonal elements, but I don't know what can I do with the rest of it. Any hints will be helful.

答:
First factor out $1+x(n-2)$ from all columns except the first to get
$$\det A(x)=\begin{vmatrix}
  n-1 & 1 & 1 & \cdots & 1 \\
  1-x & -x & 0 & \cdots & 0 \\
  1-x & 0 & -x & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  1-x & 0 & 0 & \cdots & -x
  \end{vmatrix}$$
and note that this cancels the multplication by $(1+x(n-2))^{n-1}$ from before. Now multiply the first column by $x$ and the rest of them by $(1-x)$ and then add them all to the first one:
$$\det A(x)=\frac1{x(1-x)^{n-1}}\begin{vmatrix}
  n-1 & 1-x & 1-x & \cdots & 1-x \\
  0 & -x(1-x) & 0 & \cdots & 0 \\
  0 & 0 & -x(1-x) & \cdots & 0 \\
  \vdots & \vdots & \vdots & \ddots & \vdots \\
  0 & 0 & 0 & \cdots & -x(1-x)
  \end{vmatrix}$$

which yields

$$\det A = \frac{(n-1) (-x(1-x))^{n-1}}{x(1-x)^{n-1}} = (-1)^n(n-1)x^{n-2}.$$

Note that this calculation was assuming that $x \notin \left\{0,1,-\frac1{n-2}\right\}$ so these cases should be considered separately. Alternatively, $\det A(x)$ is a continuous function in $x$ equal to $(-1)^n(n-1)x^{n-2}$ on the dense set $\Bbb{R}\setminus \left\{0,1,-\frac1{n-2}\right\}$ so it must be $\det A(x) = (-1)^n(n-1)x^{n-2}$ everywhere.


问:
Is is true that for sets $A,B,C$: $(A \cap B \cap \overline {C}) \cup (A \cap C \cap \overline {B} )=A$
Is is true that for sets $A,B,C$:

$$(A \cap B \cap  \overline {C}) \cup (A  \cap C \cap \overline {B} )=A$$

I tried:

$$(A \cap B \cap  \overline {C}) \cup (A  \cap C \cap \overline {B} )$$
$$=A \cap ((B \cap  \overline {C})\cup (C \cap \overline {B}))$$
$$=A \cap ((B \cup C) \cap (\overline {C} \cup \overline {B} ))$$

But don't know how to continue.

Body:
No, consider $A = \{1,2,3\}$, $B=\{2\}$, $C = \{3\}$

$$A\cap B\cap C^c = \{2\}$$
$$A \cap C \cap B^c = \{3\}$$

But clearly $A \ne \{2,3\}$


问:
On integral of a function over a simplex
Help w/the following general calculation and references would be appreciated.

Let $ABC$ be a triangle in the plane.
 Then for any linear function of two variables $u$.
$$
\int_{\triangle}|\nabla u|^2=\gamma_{AB}(u(A)-u(B))^2+
\gamma_{AC}(u(A)-u(C))^2+\gamma_{BC}(u(B)-u(C))^2,
$$
 where
$$
\gamma_{AB}=\frac{1}{2}\cot(\angle C),
\gamma_{AC}=\frac{1}{2}\cot(\angle B),
\gamma_{BC}=\frac{1}{2}\cot(\angle A).
$$

What is a good reference for the formula? Is it due to R. Duffin?

Is there generalization to linear functions of three variables? The number of parameters fits (in any $nD$), and it seems like one needs to calculate 6x6 Cayley-Menger like determinants, but it's difficult and the geometric interpretation is not clear.
Thank you.

答:
It is always easier to work on an affine coordinate system, Barycentric coordinate system, when the integration is performed locally over a piecewise linear structure, than to work on the global Cartesian coordinates.

First let's replicate the formula you gave in a triangle (2-simplex). For the following triangle $T = \triangle ABC$:

denote the top vertex as $A$, lower left vertex as $B$, and lower right vertex as $C$.
With slightly abuse of notation, $A$ also denotes the Cartesian coorinates: $A = (x_A,y_A)$, etc.
The unit outward normal vector $\nu_{A}$ opposite to the vertex $A$, etc. The edge vector $e_A = C- B$, $e_B = A - C$, $e_C = B- A$, rotating counterclockwisely.

Then for any point $P$ inside or on the boundary can be written as the weight sum :
$$
P = A\lambda_A  +  B \lambda_B +  C\lambda_C
$$
where $\lambda_A(A)= 1$, and $\lambda_A(B)= \lambda_A(C) = 0$. This is, for any vertex $V$, $\lambda_V$ is a linear function valued 1 on the vertex $V$, 0 on two other vertices, also we have:
$$
\lambda_A(P) + \lambda_B(P) +\lambda_C(P) = 1.
$$
Any linear function $u(x,y) = ax + by +c$ restricted on this triangle is a plane over this triangle:

Now We can rewrite $u(P)$ using the barycentric coordinate system (for you to check):
$$
u(P) =u\big(A\lambda_A(P)  +  B \lambda_B(P) +  C\lambda_C(P) \big) =  u(A)\lambda_A(P)  +  u(B) \lambda_B(P) +  u(C)\lambda_C(P) .
$$
in short we just write
$$
u = u(A)\lambda_A  +  u(B) \lambda_B +  u(C)\lambda_C
$$

The gradient of $u$ over this triangle is:
$$
\nabla u = u(A)\nabla\lambda_A  +  u(B) \nabla\lambda_B +  u(C)\nabla\lambda_C .
$$
For vertex $A$, the gradient lies in the direction of descent of a plane 1 at $A$, 0 at $B$ and $C$, and is inversely proportional to the height $h_A$ on $e_A$:
$$
\nabla \lambda_A = -\frac{1}{h_A} \nu_A = -\frac{|e_A|}{2|T|}\nu_A. \tag{1}
$$
Then
$$\begin{aligned}
&|\nabla u |^2 = \nabla u \cdot \nabla u \\
=& (u(A)\nabla\lambda_A  +  u(B) \nabla\lambda_B +  u(C)\nabla\lambda_C)\cdot (u(A)\nabla\lambda_A  +  u(B) \nabla\lambda_B +  u(C)\nabla\lambda_C)
\\
=&  u(A)^2 |\nabla\lambda_A|^2  +  u(B)^2 |\nabla\lambda_B|^2 +  u(C)^2 |\nabla\lambda_C|^2
\\
&+ 2u(A) u(B)\nabla\lambda_A\cdot \nabla\lambda_B + 2u(B) u(C)\nabla\lambda_B\cdot \nabla\lambda_C  + 2u(C) u(A)\nabla\lambda_C\cdot \nabla\lambda_A
\end{aligned}$$
By (1):
$$
\nabla\lambda_A\cdot \nabla\lambda_B = \frac{|e_A|}{2|T|}\nu_A\cdot \frac{|e_B|}  {2|T|}\nu_B = -\frac{|e_A||e_B|}{4|T|^2} \cos(\angle C)
$$
we have:
$$
\begin{eqnarray}
|\nabla u |^2 &= u(A)^2\frac{|e_A|^2}{4|T|^2} + u(B)^2\frac{|e_B|^2}{4|T|^2} + u(C)^2\frac{|e_C|^2}{4|T|^2}
\\
&\quad - 2u(A) u(B)  \frac{|e_A||e_B|}{4|T|^2} \cos\angle C - 2u(B) u(C)  \frac{|e_B||e_C|}{4|T|^2} \cos\angle A
\\
&\quad - 2u(C) u(A)  \frac{|e_C||e_A|}{4|T|^2} \cos\angle B.\tag{2}
\end{eqnarray}
$$
To get the formula, denote the height $h_A$ touches the base $e_A$ at $A'$, similar for height $h_B$ and $h_C$. Notice $u(A)^2$'s coefficient times area $|T|$ is:
$$
\frac{|e_A|^2}{4|T|} = \frac{|e_A|}{2 h_A}  = \frac{|BA'|+|A'C|}{2 h_A} = \frac{1}{2} (\cot\angle B + \cot\angle C) .
$$
similar for other two. The $u(A)u(B)$'s coefficient times area $|T|$ is:
$$
\frac{|e_A||e_B|}{4|T|} \cos\angle C = \frac{|e_A||e_B|}{2 |e_A| h_A} \frac{|A'C|}{|e_B|} = \frac{|A'C|}{2h_A} =\frac{1}{2} \cot \angle C
$$
Therefore, (2) becomes:
$$
\begin{eqnarray}
|\nabla u |^2 &= u(A)^2\frac{1}{2|T|} (\cot\angle B + \cot\angle C)\\
&\quad + u(B)^2\frac{1}{2|T|} (\cot\angle C + \cot\angle A)
\\
&\quad + u(C)^2\frac{1}{2|T|} (\cot\angle A + \cot\angle B)
\\
&\quad - 2u(A) u(B)  \frac{1}{2|T|} \cot \angle C
\\
&\quad - 2u(B) u(C) \frac{1}{2|T|} \cot \angle A
\\
&\quad - 2u(C) u(A)  \frac{1}{2|T|} \cot \angle B.\tag{3}
\end{eqnarray}
$$
Finally for we are integrating a constant number on this triangle $T$:
$$
\int_{T} |\nabla u|^2 = |\nabla u|^2 |T|,
$$
and (3) multiplying with the area $|T|$ is the formula you gave.

For a $u(x,y,z) = ax + by +cz +d $, the integration of $\nabla u$ on a 3-simplex $T = V_1 V_2 V_3 V_4$ (a tetrahedron), denote the barycentric coordinate of vertex $V_i$ as $\lambda_i$, $\lambda_i(V_j) = \delta_{ij}$. Also let $u(V_i) = u_i$, then:
$$
\nabla u = \sum_{i=1}^4 u_i \nabla \lambda_i.
$$
And $\nabla \lambda_i$ is pointed in the opposite direction of the unit outer normal vector $\nu_i$ to the face $F_i$(a triangle) opposite to vertex $V_i$, also inversely proportional to the height:
$$
\nabla \lambda_i = -\frac{1}{h_i}\nu_i = -\frac{|F_i|}{3|T|} \nu_i.
$$
In the cross terms of $|\nabla u|^2$:
$$
\nu_i\cdot \nu_j = \cos \angle F_i F_j
$$
where $\angle F_i F_j$ is the dihedral angle (angle between two planes, notice the angle is defined using normal vectors' inner product, for higher dimensions, I am afraid we have to use normals instead of trigonometric functions), working the same routine as in 2 dimensional case, we will get a similar formula:
$$
\int_T |\nabla u|^2 = \sum^4_{i=1}u_i^2\frac{|F_i|^2}{9|T|} - 2\sum_{i,j} u_i u_j\frac{|F_i||F_j|}{9|T|}\cos\angle F_i F_j, \tag{$\dagger$}
$$
Consider $|F_1|^2/(9|T|)$:
$$
\frac{|F_1|^2}{9|T|} = \frac{|F_1|}{3 h_1}.
$$
We draw the height of base $F_1$ from vertex $V_1$, denote $P_1$ as the point where this height $h_1$ touches the base $F_1$. $F_1$ has three sides: $e_{12}$, $e_{13}$, and $e_{14}$, where $e_{ij}$ is the common edge of $F_i$ and $F_j$.

From $P_1$ draw three perpendicular line segment $m_{12}$, $m_{13}$, and $m_{14}$ to $e_{12}$, $e_{13}$, and $e_{14}$ (this happens on the triangle $F_1$), then we can decompose the area $|F_1|$ into three parts:
$$\begin{aligned}
\frac{|F_1|}{3 h_1} &= \frac{|m_{12}| |e_{12}| + |m_{13}| |e_{13}| + |m_{14}| |e_{14}|}{6 h_1}
\\
&= \frac{1}{6}( \cot\angle F_1 F_2 |e_{12}| + \cot\angle F_1 F_3 |e_{13}| + \cot\angle F_1 F_4 |e_{14}|)
\end{aligned}\tag{5}
$$
we can decompose the term for other three vertices also in this fashion.

Now for coefficient $|F_1||F_2|\cos\angle F_1 F_2 /(9|T|)$ (we have six this types of coefficient corresponding to six edges of a tetrahedron):

$$
\frac{|F_1||F_2|}{9|T|}\cos\angle F_1 F_2 = \frac{|F_1||F_2|}{3|F_2| h_2}\cos\angle F_1 F_2  = \frac{|m_{12}||e_{12}|}{6 h_1} = \frac{1}{6} \cot\angle F_1 F_2 |e_{12}| \tag{6}
$$
doing this for other 5 edges also. Now plugging (5) and (6) into $(\dagger)$ we have:
$$
\int_T |\nabla u|^2 = \sum_{i<j} \big(u_i - u_j \big)^2 \frac{1}{6}  |e_{ij}|\cot\angle F_i F_j, \tag{$\ddagger$}
$$
which is a formula in three dimension. Just a reminder for notation: $u_i$ is the value of $u$ at vertex $V_i$, $F_i$ is the face opposite to $F_i$, $\angle F_i F_j$ is the dihedral angle between face $F_i$ and $F_j$, $e_{ij}$ is the common edge of $F_i$ and $F_j$, not the edge from vertex $i$ to $j$.

The result can be generalized to the integration the gradient square of a linear $u:\mathbb{R}^n\to \mathbb{R}$ on an $n$-simplex, $\triangle^n$, which is the convex hull of $(n+1)$ points.

Things are similar to the case in 3-simplex, $F_i$ is the $(n-1)$-face (codimension is 1) "opposite" to $V_i$, $\nu_i$ is unit outward normal to this $(n-1)$-face. We should be able to get a formula like $(\dagger)$, but I am afraid I could not see a trigonometrical interpretation like $(\ddagger)$.


问:
Proving two sigma algebras coincide
Let $(S, \mathcal{S})$ be a measurable space and $\Omega= \prod_{i=0}^{\infty} S$, let $X_i:\Omega \to S$ be the "coordinates" and $\theta_n$ the shift operators on $\Omega$: $w=(x_0,x_1,...) \mapsto (x_n,x_{n+1},...)$. Finally let $\mathcal{A}=\sigma(X_n,  n\geq 0)$ and $ \mathcal{G}_n=\sigma(X_n,X_{n+1},...)$. How would you prove that $\mathcal{G}_n=\theta_n^{-1}(\mathcal{A})$? I am puzzled because it seems to me that the antiimage of an element of $\mathcal{A}$ could very well be in $\mathcal{A}$ and not in $\sigma(X_n,X_{n+1},...)$. This claim is in the context of canonical Markov Chains, Ionescu-Tulcea theorem.

答:
It's known (see, e.g. math.stackexchange.com/questions/7881) that
$$f^{-1}(\sigma(C))=\sigma(f^{-1}(C))$$
and that $A$ is generated by cylinder sets $\mathcal{B}$.

Hence $\theta_n^{-1}(\mathcal{A}) = \theta_n^{-1} (\sigma(\mathcal{B})) = \sigma(\theta_n^{-1} (\mathcal{B})) = \sigma( \mathcal{\tilde{B}}) = \mathcal{G}_n $,

where $\tilde{B}$ are cylinder sets, that generate $\mathcal{G}_n$.

Addition:
elements of $\mathcal{B}$ have the next form: $F_{t_1, \ldots, t_N, D_1, \ldots, D_N} = \{(y_0, y_1, \ldots): y_{t_1} \in D_1, y_{t_2} \in D_2, \ldots, y_{t_N} \in D_N \}$, where $D_i$ are Borel sets of real line and $t_i \ge 0$.

Elements of $\tilde{\mathcal{B}}$ have the next form: $\tilde{F}_{s_1, \ldots, s_N, D_1, \ldots, D_N} = \{(y_0, y_1, \ldots): y_{s_1} \in D_1, y_{s_2} \in D_2, \ldots, y_{s_N} \in D_N \}$, where $D_i$ are Borel sets of real line and $s_i \ge n$.

Thus $\theta_n^{-1} (\mathcal{B}) = \mathcal{\tilde{B}}$ because $$\theta_n^{-1} (\mathcal{B}) =  \theta_n^{-1} ( \{ F_{t_1, \ldots, t_n, D_1, \ldots, D_N}\}) = \{ \theta_n^{-1} (F_{t_1, \ldots, t_N, D_1, \ldots, D_N}) \} = \{ \tilde{F}_{t_1+n, \ldots, t_N+n, D_1, \ldots, D_N} \} = \mathcal{\tilde{B}}.$$

Addition2: $$\theta_n^{-1} ( {F}_{t_1, \ldots, t_N, D_1, \ldots, D_N} ) = \{ (y_0, y_1, \ldots) : \theta_n (y_0, y_1, \ldots) \in {F}_{t_1, \ldots, t_N, D_1, \ldots, D_N}  \}$$
$$= \{ (y_0, y_1, \ldots) :  (y_n, y_{n+1}, \ldots)  \in {F}_{t_1, \ldots, t_N, D_1, \ldots, D_N}  \} = $$
$$ = \{ (y_0, y_1, \ldots) :  y_{n+t_1} \in D_1, y_{n+t_2} \in D_2, \ldots  \}  = \tilde{F}_{t_1+n, \ldots, t_N+n, D_1, \ldots, D_N}.$$

Title: Proving two sigma algebras coincide


答:
The claim is false, since whenever $\xi<0$, $\arctan(\xi)>\xi$. On the other hand, if we include absolute values, the inequality
$$(\forall \xi\in[-1,1])\quad|\arctan(\xi)|\leq|\xi|$$
is true.

Title: Is $ \arctan(f(x))\leq f(x)$ whenever $|f(x)|<1\ \ \forall x\in\mathbb{R} $?

Body:
Is the following always true?
$$
\arctan(f(x))\leq f(x),\ \ \text{where}\ \ |f(x)|<1\ \ \forall x\in\mathbb{R}
$$

For example, is it correct to say that
$$
\arctan\left(\frac{\sin x}{\sqrt{x+1}}\right)\le\frac{\sin x}{\sqrt{x+1}}\ \ \forall x\in\mathbb{R},
$$
since $\arctan t=t-\frac{t^3}{3}+\frac{t^5}{5}-\dots$, where each term is not greater than the one before?

I'm confused because in the example above $\arctan$'s argument can be negative (i.e., $\sin x$ can be negative).


答:
It is not true; assuming you have a universal set $X$ and $\overline{C}$ refers to the complement of $C$ with respect to $X$, the equality is actually
$$A = (A \cap B \cap C) \cup (A \cap \overline{B} \cap C) \cup (A \cap B \cap \overline{C}) \cup (A \cap \overline{B} \cap \overline{C})\text{.}$$
To understand this intuitively, obviously each part of the union must include $A$.

Then, with respect to $B$ and $C$, one considers the cases with complements on none of them, one of them, and then with both of them. Union these to get $A$.

Title: Is is true that for sets $A,B,C$: $(A \cap B \cap \overline {C}) \cup (A \cap C \cap \overline {B} )=A$

Body:
Is is true that for sets $A,B,C$:

$$(A \cap B \cap  \overline {C}) \cup (A  \cap C \cap \overline {B} )=A$$

I tried:

$$(A \cap B \cap  \overline {C}) \cup (A  \cap C \cap \overline {B} )$$
$$=A \cap ((B \cap  \overline {C})\cup (C \cap \overline {B}))$$
$$=A \cap ((B \cup C) \cap (\overline {C} \cup \overline {B} ))$$

But don't know how to continue.


问:
Prove that $\frac{d}{dx}e^x=e^x$
I would like to know how to prove that
$$\frac{d}{dx}e^x=e^x$$
An argument I've seen used is that this is the defnition of the number $e$; we can say that for some number, let's call it $a$, the derivative of $a^x$ is $a^x$ and we've just called this number $e$. But a problem I have with this approach is that we are assuming that there exists a number $a$ that satisfies this property; how do we know it exists in the first place?

I've also seen an approach used that I think is very circular in its reasoning; they use the Maclaurin expansion of $e^x$ to prove by first principles that $\frac{d}{dx}e^x=e^x$. But the derivation of this series makes use of the very fact that $\frac{d}{dx}e^x=e^x$.

Any other proof will be very appreciated; an explanation of the $2$ above proofs would also be very useful.

Thank you for your help.

答:
One nice approach is to define the function $\ln x$ (for $x\gt0$) as the integral

$$\ln x=\int_1^x{dt\over t}$$

and then show that this function has the property of a logarithm, namely $\ln(ab)=\ln a+\ln b$, after which one can define the function $\exp(x)$ to be the inverse of $\ln x$ and show it has the properties of an exponential, namely $\exp(x+y)=\exp(x)\cdot\exp(y)$. The number $e$ is defined as the (unique) value for which $\ln x=1$.

Title: Prove that $\frac{d}{dx}e^x=e^x$


问:
Define a $4 \times 4$ operator matrix so that pre-multiplying any matrix $X$ with this matrix will result in...
addition of $c_i$ times the $i$th row of $X$ to the $2$nd row of $X$ for each $i \in \{1, 2, 3, 4\}$ in one shot. Additionaly, Show that this matrix can be expressed as the product of three elementary addition matrices and a single elementary multiplication matrix.

My attempt:

$\begin{bmatrix}1 & 0 & 0 & 0\\ c & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 1 & c & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 0 & c\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ c & 1 & c & c\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$

I am unsure about the part for $i \in 2$ (would it just be a $1+c$ in the $(2,2)$ position of the matrix?) or how to express it as a product of three elementary addition matrices and a multiplication matrix.

SOURCE: Linear Algebra and Optimization for Machine Learning (page 23, problem 1.3.1)

答:
You can imagine that rows of $X$ are simply numbers: $$\begin{bmatrix} 1 & 0 & 0 & 0 \\
c_1 & 1+c_2 & c_3 & c_4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \\ X_4\end{bmatrix} = \begin{bmatrix} X_1 \\ X_2 + (c_1X_1+c_2X_2+c_3X_3+c_4X_4) \\ X_3 \\ X_4 \end{bmatrix}$$


答:
One way to do this without making any assumptions is by starting with the definition of the natural logarithm, and working from there. This approach is outlined by Barry Cipra (+1). I posted an answer to How can we come up with the definition of natural logarithm?, which I will reproduce here. Although it seems unnatural to start with the logarithm and use it to define exponents, this approach is easy to make rigorous and can also be made fairly intuitive. As I mention in the post, the reason we might do choose to do this is because it is fairly difficult to define something like $2^\pi$. It's clear what $2^x$ would mean for positive integers, even for rational numbers, but not for irrational numbers. Starting with the logarithm, we can avoid this issue.

The study of the exponential function and the logarithm is motivated by the desire to find a way of modelling a particular kind of growth, where the rate of change of a quantity is proportional to the value of that quantity at any given time. Consider how the population of a colony of bacteria might change over time—perhaps it doubles every hour. Or how your wealth compounds when you put it into a bank. In both cases, we are dealing with exponential growth. Later, when Calculus makes rigorous the notion of 'instantaneous change', we may define an exponential function as a solution to the differential equation
$$
\frac{dy}{dx}=ky \, .
$$
Solving this equation, we find that
$$
kx+C = \int \frac{1}{y} \, dy \, . \tag{*}\label{*}
$$
It seems like we have hit a roadblock, since we cannot integrate $1/y$ using the rule
$$
\int y^n \, dy = \frac{y^{n+1}}{n+1}+C \, ,
$$
since that would lead to division by zero. However, the fundamental theorem of calculus tells us that the set of antiderivatives of $1/y$ must be
$$
\int_{a}^{y} \frac{1}{t} \, dt \, .
$$
This is a set of functions that differ by a constant as we vary the value of $a$. However, since there is already a constant on the LHS of $\eqref{*}$, we can assume that $a=1$ without loss of generality. This yields
$$
kx+C = \int_{1}^{y} \frac{1}{t} \, dt \, .
$$
Let $f(y)=\int_{1}^{y} \frac{1}{t} \, dt$. This function represents the area of the region bounded by the hyperbola $1/t$, the horizontal axis, and the vertical lines $t=1$ and $t=y$:

As the value of $y$ gets larger, the area of this region also gets larger. We can then verify that $f$ is strictly increasing, meaning that $f^{-1}$ exists. Hence,
$$
kx+C = f(y) \implies y = f^{-1}(kx+C) \, .
$$
Since growth of this kind is called exponential growth, it only seems right to bestow upon the function $f^{-1}$ the new name $\exp$. The statement reads more nicely as
$$
y=\exp(kx+C) \, .
$$
Of all of the types of exponential growth, the most pleasing is that in which the rate of change of a quantity is equal to the value of that quantity (not just proportional to it). Since the set of solutions to $dy/dx = ky$ is $y=\exp(kx+C)$, the set of solutions to $dy/dx = y$ is
$$
y=\exp(x+C) \, .
$$
And of these solutions, you will no doubt agree that the simplest and most aesthetic is
$$
y=\exp(x) \, ,
$$
where $\exp(x)$ is defined as the inverse of $f(x)=\int_{1}^{x} \frac{1}{t} \, dt$. For obvious reasons, this function is called the natural exponentional function. There seems to be clear parallels between $\exp$ and functions of the form $g(x)=a^x$. If we return to our examples of exponential modelling—the colony of bacteria that doubles in size every hour—it certainly seems that $2^x$ is an exponential function. After all, the growth of a bacterial colony is proportional to its population at any given point in time. This suggests that
$$
2^x = \exp(kx+C)
$$
for some values of $k$ and $C$. (So far, we haven't actually defined what $2^x$ means when $x$ equals $\sqrt{2}$, say, but we'll leave to the side for the moment, and use an 'intuitive' definition of irrational exponents, where $2^{\sqrt{2}} \approx 2^{1.41421}$.) Suspecting that $a^x$ and $\exp(x)$ are somehow linked, we try to recall the essential properties of $a^x$, hoping that $\exp(x)$ shares these properties. The most important of these properties is that
$$
a^x \cdot a^y = a^{x+y} \, .
$$
And it turns out that $\exp(x)$ also has this property! In other words,
$$
\exp(p) \cdot \exp(q) = \exp(p+q) \, .
$$
However, proving this takes a little work. Since $\exp(x)$ is defined as the inverse of $f(x)=\int_{1}^{x}\frac{1}{t} \, dt$, it seems sensible to use this integral in our proof. Showing that $\exp(p) \cdot \exp(q) = \exp(p+q)$ is actually equivalent to showing that $f(r) + f(s) = f(rs)$, since
\begin{align}
&f(r) + f(s) = f(rs) \\
\iff &f^{-1}(f(r)+f(s)) = rs \\
\iff &f^{-1}(f(r)+f(s)) = f^{-1}(f(r)) \cdot f^{-1}(f(s)) \\
\iff &\exp(p+q) = \exp(p) \cdot \exp(q)
\end{align}
(And this should make intuitive sense, too. Notice how $f$ changes multiplication into addition, and so it is only natural that $f^{-1}$ changes addition into multiplication.) There is also a slick way to prove that $f(r)+f(s)=f(rs)$:
$$
\int_{1}^{rs}\frac{1}{t} \, dt = \int_{1}^{r}\frac{1}{t} \, dt + \int_{r}^{rs}\frac{1}{t} \, dt
$$
Let $z=t/r$, meaning that $dz=\frac{1}{r}dt=\frac{z}{t}dt$. Then we have
\begin{align}
\int_{1}^{rs}\frac{1}{t} \, dt &= \int_{1}^{r}\frac{1}{t} \, dt + \int_{1}^{s}\frac{1}{z} \, dz \\
f(rs) &= f(r) + f(s) \, .
\end{align}
Now that we have established that $\exp(p) \cdot \exp(q)=\exp(p+q)$, it seems reasonable to conjecture that $\exp(x)$ is actually equal to $a^x$ for some base $a$. To find that base, we plug in $x=1$, yielding $\exp(1)$, which we will abbreviate as $e$. This means that
$$
e^x = \exp(x) \, .
$$
This notation, apart from being very convenient, is also entirely valid. To reiterate, $\exp(p) \cdot \exp(q)=\exp(p+q)$, and so
$$
e^p \cdot e^q = e^{p+q}
$$
as expected. Geometrically, $e$ represents the number $a$ for which $\int_{1}^{a}\frac{1}{t} \, dt = 1$. This property follows directly if one recalls the definition of $\exp$. And since $\log_a(x)$ is defined as the inverse of $a^x$, it is evident that $\log_e(x)$ is the inverse of $e^x$. But we know that the inverse of $\exp(x)$ is $f(x)=\int_{1}^{x}\frac{1}{t} \, dt$, and putting these statements together, we find that
$$
\log_{e}(x) = \int_{1}^{x}\frac{1}{t} \, dt \, .
$$
In fact, this logarithm is so important, so natural, that we often dispense with the base $e$ entirely and write
$$
\log(x) = \int_{1}^{x}\frac{1}{t} \, dt \, .
$$
Earlier, we mentioned the difficulty of defining $a^x$ when $x$ is an irrational number. The logarithm solves this problem. For rational $x$, we know that
$$
a^x = e^{x\log(a)} \, .
$$
But the RHS makes sense for all $x$, and so we may define
$$
a^x = e^{x\log(a)} \, ,
$$
for $a>0$. And defining exponents in this way means that
$$
a^x \cdot a^y = a^{x+y} \, ,
$$
still holds even for irrational exponents! The proof of this fact follows smoothly if we unravel the definition of $a^x$:
$$
a^x = e^{x\log(a)} = \exp(x\log(a)) \, ,
$$
where $\exp(x)=\log^{-1}(x)$ and $\log(x) = \int_{1}^{x}\frac{1}{t} \, dt$. Armed with this knowledge, we return to the differential equation we were trying to solve earlier, and finally obtain the solution we were expecting.
\begin{align}
\frac{dy}{dx} &= ky \\
\int k \, dx &= \int\frac{1}{y} \, dy \\
kx + C &= \log(y) \\
y&=e^{kx+C} = e^{kx} \cdot e^{C} = Ae^{kx} \text{ where $A=e^C$}
\end{align}
Since $a^x$ simply represents $e$ being raised to another base, functions such as $2^x$ are also included in the solution set. In particular,
$$
\frac{d}{dx}(a^x) = \frac{d}{dx}(e^{x\log(a)}) = e^{x\log(a)} \cdot \log(a) = a^x \cdot \log(a) \, ,
$$
and so $k=\log(a)$. Note that the derivative of $e^x$ is simply a special case of this where $\log(e)=1$. Hopefully, this should provide the motivation behind defining the logarithm as $\int_{1}^{x}\frac{1}{t} \, dt$.

There is another way of understanding why functions of the form $f(x)=a^x$ grow in the way they do. If we differentiate $a^x$ with respect to $x$ from first principles, we get
\begin{align}
\lim_{\Delta x \to 0}\frac{a^{x+\Delta x}-a^x}{\Delta x} &= a^x \cdot \lim_{\Delta x \to 0}\frac{a^{\Delta x} - 1}{\Delta x} \\
&= a^x \cdot f'(0) \, .
\end{align}
In other words, the growth rate is proportional to the value of $a^x$ at any given point, with the gradient at $x=0$ being the proportionality constant. The natural logarithm comes in handy here:
$$
f'(0) = \lim_{\Delta x \to 0}\frac{a^{\Delta x} - 1}{\Delta x} \, .
$$
Exercise: prove that this limit is equal to $\log(a)$. This is tricky if you want to avoid using L'Hôpital's rule.

Title: Prove that $\frac{d}{dx}e^x=e^x$

Body:
I would like to know how to prove that
$$\frac{d}{dx}e^x=e^x$$
An argument I've seen used is that this is the defnition of the number $e$; we can say that for some number, let's call it $a$, the derivative of $a^x$ is $a^x$ and we've just called this number $e$. But a problem I have with this approach is that we are assuming that there exists a number $a$ that satisfies this property; how do we know it exists in the first place?

I've also seen an approach used that I think is very circular in its reasoning; they use the Maclaurin expansion of $e^x$ to prove by first principles that $\frac{d}{dx}e^x=e^x$. But the derivation of this series makes use of the very fact that $\frac{d}{dx}e^x=e^x$.

Any other proof will be very appreciated; an explanation of the $2$ above proofs would also be very useful.

Thank you for your help.


答:
Define $e$ as
$$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$
Thus $$e^x=\lim_{n\to\infty}\left(1+\frac1n\right)^{xn}.$$
Set $m=nx$ and note that $n\to\infty$ implies that $m\to\infty$, assuming $x$ is real. Then
$$e^x=\lim_{m\to\infty}\left(1+\frac1{m/x}\right)^m=\lim_{m\to\infty}\left(1+\frac{x}{m}\right)^m.$$
Then $$\frac{d}{dx}e^x=\lim_{m\to\infty}\frac{d}{dx}\left(1+\frac{x}{m}\right)^m=\lim_{m\to\infty}\left(1+\frac{x}{m}\right)^{m-1}.$$
This is $$\frac{d}{dx}e^x=\lim_{m\to\infty}\frac{\left(1+\frac{x}{m}\right)^m}{1+x/m}=\frac{\lim_{m\to\infty}\left(1+\frac{x}{m}\right)^m}{\lim_{m\to\infty}(1+x/m)}=\frac{e^x}{1}=e^x.$$

A verification that $\frac{d}{dx}\lim=\lim\frac{d}{dx}$.

Let $C^1=C^1(\Bbb R)$ be the space of continuously differentiable functions $\Bbb R\to\Bbb R$. Similarly, let $C^0=C^0(\Bbb R)$ be the space of continuous functions $\Bbb R\to\Bbb R$. Then the differential operator map $D:C^1\to C^0$ given by $\phi(x)\mapsto \phi'(x)$ is a linear transformation.

To show this, note that for all $f,g\in C^1$ and $\alpha\in\Bbb R$,
$$\begin{align}
D(f+g)&=D(f)+D(g)\\
D(\alpha f)&=\alpha D(f).
\end{align}$$
These are easy to check.

Since $D$ is a linear transformation it is continuous. That is, every convergent sequence of functions in $C^1$ is mapped by $D$ to a convergent sequence of functions in $C^0$: $$D\left(\lim_{n\to\infty}f_n\right)=\lim_{n\to\infty}D(f_n),$$
for a given convergent sequence of functions $(f_n)\subset C^1$. Clearly, $f_n(x)=(1+x/n)^n$ converges to $e^x$, so the above applies.

Title: Prove that $\frac{d}{dx}e^x=e^x$


答:
If $0\leq f(x) <1$, yes, of course. Since $\frac{d}{dy}\arctan y=\frac{1}{1+y^2}\leq1$ (and $\arctan(0)=0$), you'll have
$$ \arctan y \leq y, \quad\text{for }\ y\geq0. $$
Clearly, the situation is reversed if $y<0$, because both functions are negative. You can say that $|\arctan f(x)|\leq|f(x)|$ with no restrictions on $f(x)$, though.


问:
Tighter tail bounds for subgaussian random variables

Let $X$ be a random variable on $\mathbb{R}$ satisfying $\mathbb{E}\left[e^{tX}\right] \leq e^{t^2/2}$ for all $t \in \mathbb{R}$. What is the best explicit upper bound we can give on $\mathbb{P}[X \geq x]$ for $x > 0$?

A well-known upper bound can be obtained by applying Markov's inequality to the moment generating function of X: For $t>0$, $$\mathbb{P}[X \geq x] = \mathbb{P}\left[e^{tX}\geq e^{tx}\right] \leq \frac{\mathbb{E}\left[e^{tX}\right]}{e^{tx}} \leq e^{t^2/2-tx}.$$ Setting $t=x$, yields $$\mathbb{P}[X \geq x] \leq e^{-x^2/2}.$$
How tight is this?

Examples of such $X$ include the standard Gaussian and Rademacher distributions. Both satisfy stronger tail bounds, as illustrated by the following plot. This makes me wonder whether stronger bounds hold for all such $X$.

Some more precise questions:

Define $$f(x) = \sup \left\{ \mathbb{P}[X\geq x] : X \text{ a random variable satisfying } \mathbb{E}\left[e^{tX}\right] \leq e^{t^2/2} \text{ for all } t \in \mathbb{R} \right\}.$$

What is $f(1)$?

(We know $f(1) \in \left[\frac12,e^{-1/2}\right]$, by the Rademacher example and the above tail bound.)

 What is $\limsup_{x \to \infty} f(x) \cdot e^{x^2/2}$?
(We know it is $\leq 1$, by the above tail bound.)

 What is $\limsup_{x \to \infty} f(x) \cdot e^{x^2/2} \cdot {x}$?
(We know it is $\geq 1/\sqrt{2\pi}$, by the Gaussian example [link].)

Any improved bounds on $f$ would interest me. Is there a closed form expression for $f$? or some way to numerically approximate $f$?

A follow-up question would be: Are there are any natural assumptions that in addition to subgaussianity would imply $\mathbb{P}[X \ge x] \le O\left(\frac{e^{-x^2/2}}{x}\right)$.

答: Consider a r.v. $X$ with distribution function $F(x) = (1 - e^{-\frac{x^2}2})I_{x \ge 0}$. Then $E e^{tX} \le \sqrt{2 \pi} e^{\frac{t^2}2} $ and condition $P(X \ge x) = e^{-\frac{x^2}2}$ holds for all $x \ge 0$. It's not a solution of the problem, but it shows that the estimate $ e^{-\frac{x^2}2}$ looks like a rather sharp  estimate.


答: If you can see triangles $\widehat { AFE } $ and $\widehat { DCE } $ are the same, then everything becomes easy. Because $\widehat { AFE } $=$\widehat { AFE } $, $AF$=$CD$, because of the rectangle $AF$=$BC$, then $AF$=$BC$=$CD$. Finally $AF$=$BC$=$4$ because $BC$ is half of the $BD$.  Without math you can solve the question.</p>

Title: Finding side of rectangle using given information

Body:
Really simple question but I am stuck. The following information is given:
$$BD=8,\quad AB = 6,\quad  ED  =5,\quad EF = EC$$
and we want to find $AF$.

If we have three $90^\circ$, what does that really mean, and how I can find $AF$?


问:
$f:X \to Y$ is a continuous map. Suppose it induces an isomorphism $f_*:H_*(X) \to H_*(Y)$. Does that imply $f:X \to Y$ is a homotopy equivalence?
Suppose that $f:X \to Y$ is a continuous map. Suppose it induces an isomorphism $f_*:H_*(X) \to H_*(Y)$. Does that imply $f:X \to Y$ is a homotopy equivalence?

I think that this is not the case. This is because the following is not True.

Let $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence,

Can anyone give me a simple example.

答:
A simple example in chain complexes is the following: consider the infinite complex
$$\cdots  \to \mathbb Z/4\to \mathbb Z/4 \to  \mathbb Z/4\to\cdots$$
where every map is multiplication by $2$. This is acyclic but not contractible over $\mathbb Z/4$. A similar example is afforded by $\mathbb k[t]/(t^2)$ and multiplication by $t$ over this ring.


答:
This is based on the definition

$$e = \lim_{n \to \infty} \left(1+\frac1n\right)^n$$

and the definition of $\ln$ as inverse of $e^x$.

Let $y=e^x$, then

$$\begin{align*}
x &= \ln y\\
\frac{dx}{dy} &= \lim_{h\to0}\frac{\ln(y+h)-\ln y}{h}\\
&=\lim_{h\to0}\ln\left(1+\frac{h}{y}\right)^\frac1h\\
&=\ln\lim_{h\to0}\left(1+\frac{h}{y}\right)^{\frac yh\cdot \frac1y}\\
&= \ln \left(e^\frac1y\right)\\
&= \frac1y\\
&= \frac1{e^x}\\
\frac{dy}{dx} &= e^x
\end{align*}$$

Title: Prove that $\frac{d}{dx}e^x=e^x$


答:
Suppose we've already proved that $\frac{d}{dx} a^x = g(a) a^x$ for some function $g : \mathbb{R}_{> 0} \to \mathbb{R}$ (see this post for example). The question is then "Does there exist $a \in \mathbb{R}_{>0}$ such that $g(a) = 1$ ?"

Observe that, by the chain rule, for any $k \in \mathbb{R}$:
$$
\frac{d}{dx} a^{kx} = k g(a) a^{kx}
$$
On the other hand
$$
\frac{d}{dx} a^{kx} = \frac{d}{dx} (a^k)^x = g(a^k) (a^k)^x = g(a^k) a^{kx}
$$
Thus $kg(a) = g(a^k)$ for all $a \in \mathbb{R}_{>0}$ and $k \in \mathbb{R}$.

In particular, we may set $k = \frac{1}{g(a)}$ for some $a \in \mathbb{R}$ such that $g(a) \neq 0$ (if $g = 0$, then all exponential functions are constant, which we know is not true). Then
$$
\frac{d}{dx} (a^k)^x = g(a^k) (a^k)^x = kg(a) (a^k)^x = (a^k)^x
$$
Therefore, if $g(a) \neq 0$, $e \equiv a^{\frac{1}{g(a)}}$ is a number such that
$$
\frac{d}{dx} e^x = e^x
$$
By carrying the analysis further, we could convince ourselves that $a^{\frac{1}{g(a)}}$  is a unique number, i.e. does not depend on $a$.

Title: Prove that $\frac{d}{dx}e^x=e^x$


问:
Treating Differentials like Fractions
Why does treating differential elements like fractions work so well? I studied calculus from a pure math perspective first, and have only basic knowledge of it, so can someone explain with a rigorous proof or a simple explanation why this hack works? Some people told me that treating differentials like fractions is a sin. Can anyone give me an example where treating differentials like fractions fails to work?</p>

答: An example where treating differentials as fractions fail to work is given by user @asmeurer here. Also read other answers to this question.</p>

Title: Treating Differentials like Fractions


问:
Show that specific polynomial with $p(a)=p(b)=0$ must be constantly $0$.

Say polynomial $p\in \mathbb{R}[x]$ is such that for all
reals $x,y,z$ such that $xy+yz+zx = 0$, we have
$$ p(x-y) + p(y-z) + p(z-x) = 2p(x+y+z). $$
Prove that $p(x)=0$ for all $x$ if $p(a) = p(b)=0$ for some
different positive $a$ and $b$, without explicitly solving the
equation.

To make it easier, you can choose $a,b$ on your own explicitly.

All I can get is
\begin{align}
 x=y=z=0: &\;\; 3p(0)=2p(0) \implies \boxed{p(0) =0}\\
 x=y=0 : &\;\; p(-z)+p(z) =2p(z) \implies \boxed{p(-z) =p(z),\;\; \forall z}\\
\end{align}
so $p$ is an even polynomial with constant term $0$. From $p(a)=p(b)=0$ we can say that $\deg (p)\geq 6$.

I can't find anything usefull from here.

答:
Set $x=6t$, $y=3t$, and $z=-2t$. Then

$$xy+yz+zx=18t^2-6t^2-12t^2=0$$

and

$$p(x-y)+p(y-z)+p(z-x)=2p(x+y+z)$$

$$p(3t)+p(5t)+p(-8t)=2p(7t)$$

Then with your observation that $p(t)=p(-t)$, this simplifies to

$$p(3t)+p(5t)+p(8t)=2p(7t)$$

Now, we know that if the polynomial is not identically zero, then for very large $t$ (past all the zeros)

$$1=\frac{2p(7t)}{p(3t)+p(5t)+p(8t)}$$

If we denote the leading coefficient of $p(t)$ by $a$, then

$$1=\frac{2p(7t)}{p(3t)+p(5t)+p(8t)}=\lim_{t\to\infty}\frac{2p(7t)}{p(3t)+p(5t)+p(8t)}=\lim_{t\to\infty}\frac{2a7^nt^n}{a3^nt^n+a5^nt^n+a8^nt^n}$$

(where $n$ is the degree of $p(t)$). This simplifies to

$$=\lim_{t\to\infty}\frac{2a7^nt^n}{a3^nt^n+a5^nt^n+a8^nt^n}=\lim_{t\to\infty}\frac{2\cdot 7^n}{3^n+5^n+8^n}=\frac{2\cdot 7^n}{3^n+5^n+8^n}$$

Now, is this equation ever satisfied? Yes, at $n=2$ and $n=4$. However, you showed in your question that $n\geq 6$ (this is where we use the fact that $p(a)=p(b)=0$ for some $a,b>0$). But for these $n$ we have

$$\frac{2\cdot 7^n}{3^n+5^n+8^n}<2\left(\frac{7}{8}\right)^n\leq 2\left(\frac{7}{8}\right)^6=\frac{117649}{131072}<1$$

Thus, the polynomial must be identically zero.


问:
Categorization of Group Scheme of rank 2
I got stuck on this Exercise 2.11 in the book Introduction to Affine Group Scheme of Waterhouse. I really appreciate if anyone can give me a hint.

Let $A$ be a Hopf algebra over $k$ (a base ring) which is free of rank 2. I proved that $I = Ker(\epsilon) = kx$ for some $x$, and $\Delta(x) = x \otimes 1 + 1 \otimes x + b x\otimes x$ for some $b$ in k. Show that $x^2 + ax = 0$ for some $a$, so $A = k[X]/(X^2 + aX)$.

I guess we can obtain the equation $x^2 + ax = 0$ from $\Delta(x) = x \otimes 1 + 1 \otimes x + b x\otimes x$. We can apply multiplication to both sides of the equation, but I don't know how $\Delta$ behaves under multiplication. Thanks in advance.

答: Since $\{1, x\}$ is a basis for $A$, you can write $x^2 = c_1 + c_2 x$ for some $c_1, c_2 \in k$.  Apply $\epsilon$ to this equation to show that $c_1 = 0$, so that $x^2 = c_2 x$.


问:
For which $2\times 3$ matrices $A,B$ does $\operatorname{rref}(A) + \operatorname{rref}(B) = \operatorname{rref}(A+B)$ hold?
Let $A$ and $B$ be $2\times 3$ matrices. For which $A$ and $B$ does the following equation hold?

$$\operatorname{rref}(A)+ \operatorname{rref}(B) = \operatorname{rref}(A+B)$$

($\operatorname{rref}$ is the operation that makes a matrix in reduced row echelon form.)

This problem is from Strang's Introduction to Linear Algebra book, and he says about this problem in the book that, it is silly. I don't know why he said that but I couldn't figure this out for a while. I'd appreciate your help.
Thanks from now.

答:
In equational terms, there are elementary matrices $E,F,G$ such that each of the following terms are in rref, $$EA+FB=G(A+B)\tag{1}.$$ The key is that the sum of two rrefs is not usually an rref.

If $A=0$ or $B=0$ then the problem is trivial. So suppose $A,B\ne0$ and suppose wolog that $B$'s first non-zero column is at or to the right of $A$'s.

Facts

The first row operation on $A$ reduces its first non-zero column to $(1,0)^\top$. The same column of $B$ cannot be non-zero as well, for otherwise its first row operation would also reduce it to $(1,0)^\top$ and the sum of both rrefs would have a column of $(2,0)^\top$. Hence $B$'s first non-zero column is after $A$'s; so $A$ can have at most one leading zero column, and $B$ must have a leading zero column.

The first non-zero column of $A+B$ is identical to that of $A$.

The first $(1,0)^\top$ in rref$(A+B)$ occurs at the same column as $A$'s and must be produced by the same row operations as those for $A$.

Any column $v$ of $A$ is reduced to $Ev=(r,0)^\top$ iff $v$ is a multiple $r$ of the first non-zero column of $A$.

Case 1. A second row operation is needed on $A$, by multiplying the second row by a constant to reduce some column to $(0,1)^\top$ and then reduce the component above it to $0$. For the same reason as above, this same column of $B$ must be $(0,0)^\top$. Thus, two columns of $A+B$ agree with those of $A$, and so exactly the same row operations have to be used to reduce $A+B$ to an rref. Thus $E=G$, so $FB=GB=EB$ by (1), which means that the same row operations $E$ can be used on $B$ to reduce it to rref.

Case 2. It can happen that after this one row operation, the result is a rref, i.e., $\begin{pmatrix}1&*&*\\0&0&0\end{pmatrix}$. So all three columns of $A$ are multiples of the first one. This operation also reduces the first column of $A+B$ to $(1,0)^\top$, and it does not affect $B$'s first column. More operations may be needed on $B$ to reduce it to rref, and these are precisely the ones needed to reduce $A+B$ to rref. Thus $G=F$, so $EA=GA=FA$, and once again we can assume $E=F=G$.

Thus in both cases, $E=F=G$. This forces some columns of $B$ to be the same as $A$'s. For example, since $B$ has rref with a $(1,0)^\top$, then that column must be the same as $A$'s first non-zero column (by fact 4.).

Using the facts above, one can see that $B$ must have one or two zero columns, one column that agrees with the first non-zero column of $A$, and possibly a third non-zero column. Here are the five possibilities for $A+B$ and their reduced forms:

\begin{align}
\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{pmatrix}
+\begin{pmatrix}0&0&a_{11}\\0&0&a_{21}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&0&*\\0&1&*\end{pmatrix}\\
\begin{pmatrix}a_{11}&ca_{11}&a_{13}\\a_{21}&ca_{21}&a_{23}\end{pmatrix}
+\begin{pmatrix}0&a_{11}&0\\0&a_{21}&0\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&*&0\\0&0&1\end{pmatrix}\\
\begin{pmatrix}a_{11}&ca_{11}&0\\a_{21}&ca_{21}&0\end{pmatrix}
+\begin{pmatrix}0&a_{11}&a_{13}\\0&a_{21}&a_{23}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&*&0\\0&0&1\end{pmatrix}\\
\begin{pmatrix}a_{11}&ca_{11}&da_{11}\\a_{21}&ca_{21}&da_{21}\end{pmatrix}
+\begin{pmatrix}0&a_{11}&ba_{11}\\0&a_{21}&ba_{21}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&*&*\\0&0&0\end{pmatrix}\\
\begin{pmatrix}0&a_{12}&a_{13}\\0&a_{22}&a_{23}\end{pmatrix}
+\begin{pmatrix}0&0&a_{12}\\0&0&a_{22}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}0&1&*\\0&0&0\end{pmatrix}\\
\end{align} (The third possibility needs a bit more argumentation, which I've skipped.)

Silly? Well, almost an hour writing an answer for what? a silly (+1)...maybe :)


问:
Normal Order Statistic for Terry-Hoeffding Test
Related to this question:

I am trying to replicate the numbers found in the third column of the table on page 189 from this article (E(V^ri)) in Python by way of the Terry-Hoeffding Normal Scores test. Further explanation is provided on page 184. I am trying to locate Harter, 1961 to include it if needed.

答: Perhaps without realizing it, you've basically asked the same question twice. Eq. [1] in the paper is $E(V^k)=-E(V^{N+1-k})$. In Table 2 rank $1$ scores $-1$ times what rank $8$ scores, so $N=8$. The code I provided in your other question finds $E(1,\,8)$ to be $1.4236$, the value given to rank $7$, or $-1$ times the value given to rank $1$. In other words, $V(r)=-E(r,\,N)$. (My linked code indents with two spaces at a time, instead of the four used in my other answer.)


答:
You can approach this from the limit definition of the derivative

$$ f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$$

With $f(x) = e^x$ this becomes

$$f'(x) = e^x \cdot \lim_{h \to 0} \dfrac{e^h - 1}{h}$$

and you can use your method of choice to show that

$$\lim_{h \to 0} \dfrac{e^h - 1}{h} = 1$$

To do this we'll use the limit definition of $e$:
$$\displaystyle e = \lim_{n \to \infty} (1 + 1/n)^n$$

or equivalently

$$e = \lim_{x \to 0} (1 + x)^{1/x}$$

Introduce a change of variables, $t = e^h - 1$. Rewrite this as $\ln(t + 1) = h$.

$$\begin{align}
\lim_{h \to 0} \dfrac{e^h - 1}{h} &= \lim_{t \to 0} \dfrac{t}{\ln(t+1)} \\
&=\lim_{t \to 0} \dfrac{1}{\frac{1}{t}\ln(t+1)} \\
&= \lim_{t \to 0} \dfrac{1}{\ln(t+1)^{1/t}} \\
&= \dfrac{1}{\ln(e)} \\
&= 1
\end{align}$$

Title: Prove that $\frac{d}{dx}e^x=e^x$


答:
Comment:

$y=e^x$

Due to definition of derivative we can write:

$$y'=\lim_{\Delta x\rightarrow 0}\frac{e^{x+\Delta x}-e^x}{\Delta x}=\lim_{\Delta x\rightarrow 0}e^x(\frac {e^{\Delta x}-1}{\Delta x})=e^x$$

Provided we prove that:

$\lim_{\Delta x\rightarrow 0} \frac {e^{\Delta x}-1}{\Delta x}=1$

Title: Prove that $\frac{d}{dx}e^x=e^x$


答:
Let $y=e^x$.

Taking $\text{ln}$ both sides, we obtain

$\text{ln}y=x\text{ln}e=x$.

Differentiating w.r.t. $x$, we have

$$\frac{1}{y}\frac{dy}{dx}=1$$
$$\frac{dy}{dx}=y$$
$$\frac{d}{dx}e^x=e^x.$$

Title: Prove that $\frac{d}{dx}e^x=e^x$


问:
$x,y,z>0$ and $x+y+z=1$.
$x,y,z>0$ and $x+y+z=1$. Prove:
$$\sum_{cyc}\frac{x^{5}+y^{4}z}{y+z}\geq \frac{1}{27}$$
My tried was to add $3$ and to homogenize with $x + y + z$. Another tried was to write LHS as $\sum_{cyc} \frac{x^{5}}{y+z}+\sum _{cyc}\frac{y^{4}z}{y+z}$.

答:
Hint: Split the terms as

$$ \sum \frac{ x^5 - y^5 } { y+z } + \sum \frac{ y^5 + y^4 z } { y+z} \geq \frac{1}{27}.$$

 Show that

 1. the first summation is $ \geq 0$, (If stuck, see comments)

 2. the second summation is $ \sum y^4 \geq \frac{1}{27}$.

I'm not quite satisfied with this solution because the splitting seems "magical", though I know what steps / similar problems motivated the approach.

I would like to see another distinct-enough (non brute-force) approach.

The issue with $ \sum \frac{ y^4 z } { y+ z }$ is that because of the cross terms, it could get too low (even close to 0, like with $ x =y = \epsilon \rightarrow 0 , z = 1 - 2 \epsilon$).

In the event this happens, the $ \sum \frac{ x^5 } { y+z }$ term would have gotten large, and so we need to "share" it across the split.
Adding a $y^5$ term in the numerator of the second term helps us to factor it, leaving a no-cross-term $y^4$ that makes the split ultimately work.

We then have to subtract a $y^5$ term in the numerator of the first term, and check if it truly satisfies the desired bound.

Title: $x,y,z>0$ and $x+y+z=1$.


问:
To derive quantity $k=\frac{2-p^2}{p+2}$ for positive rational $p$ such that $p^2>2$ and $(p+k)^2>2$
I am reading Baby Rudin and stuck at one small point.
$A$ is set of all positive rationals $p$ such that $p^2<2$ and $B$ is set of all positive rationals $p$ such that $p^2>2$. We wanted to show $A$ contains no largest element and $B$ contains no smallest element. So for any such $p$ in either set, author associated number $q$ as,

$q=p-\frac{p^2-2}{p+2}$, which works fine.

I was trying to figure out how to obtain the quantity $\frac{p^2-2}{p+2}$.

I did it for set $A$ as follows.

So, we have positive rational $p$ such that $p^2<2$. We want some $k>0$ such that $(p+k)^2 <2$

i.e. $p^2+2pk+k^2<2$

As all three quantities are positive, we can say

$p^2+2pk<2$, which gives

$k<\frac{2-p^2}{2p}$.

But as $p^2<2$, we have $p<2 \Rightarrow p+p=2p<p+2$

Therefore, we can choose $k=\frac{2-p^2}{p+2}$.

But I am stuck while deriving it for set $B$.

I did following.

So, we have positive rational $p$ such that $p^2>2$. We want some $k<0$ such that $(p+k)^2 >2$

i.e. $p^2+2pk+k^2>2$.

From this point, I can say $p^2+k^2>2$, but not sure how this will take me to the desired result.

Any hint or help. Thank you.

答:
Let $p\in B.$ I'll change the definition of $k$ slightly to $k=\dfrac{2-p^2}{2p}.$ (So $k<0,$ hence $p+k<p.$) Because $k^2>0,$ we can say

$$(p+k)^2 = p^2 +2pk +k^2 > p^2 +2pk$$ $$ =p^2 +2p\frac{2-p^2}{2p} = p^2 +2-p^2 =2.$$

We're done.


问:
Question About the Group Axioms and Symmetry
Let $f$ be a function $f:X^2\to Y$. Let $G_f$ be a subset of the symmetric group on $X$, such that $G$ consists of all of the permutations $\sigma$ satisfying $f(\sigma(a),\sigma(b))=f(a,b)$ for the chosen $f$. $X$ need not be finite, and also $Y$ need not coincide with $X$, or even have equal cardinality to it; it is just some arbitrary co-domain.

Question:

Is every group just isomorphic to some $G_f$, if we make the correct choice of $f$? Alternatively, for any group does there exist some function $f$ such that $G_f$ if isomorphic to the group in question?

Purpose:

I'm looking for ways to motivate the group axioms by showing that they describe something where the level of abstraction is no more and no less than is required to describe this particular object.

What I already know/have figured out/have tried:

Any $G_f$ satisfies the group axioms. Closure) if $f(\sigma_1(a),\sigma_1(b))=f(a,b)$ and $f(\sigma_2(a),\sigma_2(b))=f(a,b)$, then the composition $\sigma_3=\sigma_1\circ\sigma_2$ or $\sigma_2\circ\sigma_1$ will also satisfy $f(\sigma_3(a),\sigma_3(b))=f(a,b)$ once decomposed, implying closure. Identity) the identity permutation $e$ trivially satisfies $f(e(a),e(b))=f(a,b)$. Inverses) if $f(\sigma(a),\sigma(b))=f(a,b)$, then $\sigma^{-1}$ will also satisfy $f(\sigma^{-1}(a),\sigma^{-1}(b))=f(a,b)$, because $f(a,b)=f(e(a),e(b))=f(\sigma(\sigma^{-1}(a)),\sigma(\sigma^{-1}(b)))=f(\sigma^{-1}(a),\sigma^{-1}(b))$. Associativity) permutations are functions, and function composition is always associative.

As per this question I've asked, I know that every group is the automorphism group of some magma. In the language of my question here, I think that's basically saying that if $f:X^2\to X$ then any group is isomorphic to the set of all $\sigma$ under composition such that $\sigma(f(x,y))=f(\sigma(x),\sigma(y))$, but this doesn't seem exactly the same as what I have above. I think that because my condition here is even less restrictive the automorphism group of a magma, I strongly suspect that the answer to my question is yes, but I absolutely lack the algebraic prowess to justify this hunch beyond intuition. I don't know enough about lattices to follow what's going on in the answer to this question.

Note:

If the answer to this question is affirmative, then I am planning to also - at some point in the future - use this to ask a question about how this approach to compares to the magma-automorphism approach and binary-relations approach of defining the group axioms, and which way and how these concepts actually relate to the applications of group theory, such as symmetry in physics and geometry, the abstract concept of in-variance in math, how to interpret this in the case of, say, a rubrics cube, conjugation and cosets, the isomorphism theorems, & more.

答:
This follows from the beginning of Hagen von Eitzen's answer that I linked to previously. Every group occurs in this way, and we can take $X = Y = G$ and $f : G \times G \to G$ to be the function

$$f(x_1, x_2) = x_1^{-1} x_2.$$

Then we can check that if $g \in G$ and $L_g : X \to X$ denotes the left-multiplication-by-$g$ map $x \mapsto gx$ that we have

$$f(L_g x_1, L_g x_2) = (gx_1)^{-1} gx_2 = x_1^{-1} g^{-1} g x_2 = x_1^{-1} x_2$$

so $L_g$ is an automorphism of $f$ (which defines a kind of graph structure on $G$ where the edges are directed and colored by the elements of $G$ itself; $f$ above is describing the "difference" between two elements of $G$).

Conversely, suppose $\varphi : X \to X$ is an automorphism of $f$ and set $g = \varphi(e)$. Then

$$f(\varphi(e), \varphi(x)) = f(g, \varphi(x)) = g^{-1} \varphi(x) = f(e, x) = x$$

gives that $\varphi(x) = gx$ for all $x \in X$. So $\varphi = L_g$. Hence the automorphisms of $f$ are exactly the left multiplications $L_g$.

The function $f$ above can be thought of as describing the structure of a right $G$-torsor on $G$, and it's classical that the automorphism group of a $G$-torsor is $G$ again; this is a more precise version of Cayley's theorem.


问:
What we get from comparing $q(x)$ , $P_3(x)$ , $e^x$ numerically?

For $f(x)=e^x$, construct a cubic polynomial $q(x)$ for which

$$q(0)=f(0),\quad q(1)=f(1)$$ $$q'(0)=f'(0),\quad q'(1)=f'(1)$$
Numerically compare it to $e^x$ and the Taylor polynomial $P_3(x)$ of
the formula
$P_n(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\cdots+\frac{(x-a)^n}{n!}f^{(n)}(a)$ for $0\le x\le 1
$.

I get $q(x)=(3-e)x^3+(2e-5)x^2+x+1$ and third degree Taylor polynomial of $f(x)=e^x$ at $x_0=0$ is $P_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}$. But my question is what is the point of this problem and what we conclude of we compare $P_3(x) $ and $q(x)$ and $e^x$?

For example for $x=\frac12$ I got these results in calculator (I write it up to fifth decimal):

$e^{0.5}=1.64872$

$q(0.5)=1.64435$

$P_3(0.5)=1.64583$

EDIT:
Why $q(x)$ approximate the function precisely like Taylor polynomial do?

答:
If you have the opportunity to choose four numerical parameters to write a function that approximates $e^x$ on the unit interval how might you proceed?

One way is to find a cubic with the same value and slope at each end of the interval. Another is to use the first four terms of the power series for the exponential.

The problem asks you to compare these strategies. The picture in the comment from @lulu is a good place to start. The Taylor polynomial will match better at the origin but not as well at $x=1$.


问:
Find the number of positive integers $n$ such that $\big(n + 2n^2 + 3n^3 + ... + 2005n^{2005}\big)$ is divisible by $(n-1)$.

Find the number of positive integers $n$ such that $\big(n + 2n^2 + 3n^3 + ... + 2005n^{2005}\big)$ is divisible by $(n-1)$.

What I Tried: I can write the expression as :-
$$\rightarrow n\big(1 + 2n + 3n^2 + ... + 2005n^{2004}\big)$$

We know $(n-1)$ cannot divide $n$, so it must divide that expression. That was only useful I was able to conclude. I am not finding any other ideas on how to tackle this problem using mod, I might put some small values for $n$ but I cannot continue like that, it will take ages.

Can anyone help me?

答:
(Fill in the gaps as needed.)

Hint:

$$n + 2n^2 + 3n^3 + ... + 2005n^{2005} \equiv 1 + 2 + 3 + \ldots + 2005 \pmod{n-1} . $$

Corollary: We need $ n-1 \mid \frac{2005 \times 2006 }{ 2 } $. How many factors does this number have?


问:
Finding marginal pmf given a joint mass function
I want to find the marginal mass function of $\ x $ given

$$\ f_{x,y} (x,y) = \begin{cases} \frac{1}{x} & 0< y<x < 1 \\ \\0 & else \end{cases} $$

if I draw the area of the function

I was thinking I could just integrate $\ y $ to find the pmf of $\ x $. so $\ \int_0^x \frac{1}{x} \ \ dy = 1 $ but it seems wrong to me.

The value of $\ f_{x,y} $ is very high when both $\ x,y $ close to $\ 0 $ (approaching $\ \infty $ ) so when I cancel y over $\ x$, the $\ f_x $ function should have very high values when $\ x $ also closer to infinity and approach to $\ 1 $ as $\ x $ approach $\ 1 $

答:
Note that you know the joint density function $$f_{x,y}(x,y)=\frac{1}{x}\mathbf{1}_{\{0<y<x<1\}}=\left\{\begin{aligned}\frac{1}{x}, \quad 0<y<x<1\\
0, \quad \text{otherwise} \end{aligned} \right.$$
Now, we need to calcule the marginal function of $x$,by definition we have
$$\color{blue}{f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)dy}$$
since that $S_{X,Y}=\{(x,y): 0<y<x<1\}$, where $S_{X,Y}$ denote the support of $(X,Y)$.
So, we have $$f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)dy=\int_{0}^{x}\frac{1}{x}dy=\left.\frac{1}{x}y\right|_{0}^{x}=\frac{x}{x}=1$$
So, we can conclude that $$\boxed{f_{X}(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)dy=1\cdot \mathbf{1}_{\{0<x<1\}}}$$


问:
A simple question about Leibniz Criterion and the limit comparison test for series
In a series we can take a factor that doesn't alternate and proof that it is convergent which by the definition of Leibniz criteria means the whole series is also convergent. I get that.

In lecture we shown that the non alternating factor is convergent by using the limit comparison test $\lim_{n\rightarrow \infty} \frac{c_n}{b_n} = L \in (0, \infty) \text{ and } c_n,b_n > 0 $
But what if one of the $c_n$ or $b_n$ are less than zero.

The example we have done was:

$$\sum^\infty_{n = 1}(-1)^{n-1}\tan(\frac{1}{n\sqrt{n}}) $$
where we take out $\tan(x)$ which doesn't alternate for Leibnitz criteria.
We then showed:
$$ \tan(\frac{1}{n\sqrt{n}}) = c_n \text{ is almost the same as } \frac{1}{n^{3/2}} = b_n$$

Then we solved the proof of convergence by using the limit comparison test as mentioned above.

What should we do if the other factor is negative ? Why did we use limit comparison test here instead of any other (do we always have to?) ?

答:
$0<\tan\left(\frac{1}{n\sqrt n}\right)<\tan 1$ for any $n$ (positivity) and $\underset{n\to \infty }{\text{lim}} \tan\left(\frac{1}{n\sqrt n}\right)=0$ are the conditions of Leibniz test.

The comparison test is not necessary. Even $\sum _{n=2}^{\infty } (-1)^n \log n$ converges.

An  alternating series must have all terms positive or all terms negative, otherwise the Leibniz test doesn't apply. If they are all negative you can collect the minus sign like the following example:
$$\sum _{n=2}^{\infty } \frac{(-1)^n}{1-n^2}=-\sum _{n=2}^{\infty } \frac{(-1)^n}{n^2-1}$$


问:
Find the extrema of $f(x)=(x^2+2x-7)e^{|x+2|}$
I need to find the extrema of $f(x)=(x^2+2x-7)e^{|x+2|}$. The first derivative is $e^{|x+2|}(2x+2+(x^2+2x-7)(|x+2|)')$. Looking at two cases where $x<-2$ or $x>-2$ gives two extrema - at the points -3 and 1. And now comes the real question: Is there an extremum at the point -2 and why? If we replace x with -2, the first derivative equals -2. But the fact that the derivative there isn't $0$ doesn't mean there isn't an extremum. In fact, if I take a look at the graphic of the function, it is obvious that in the region of -2, the value of the function for $x=-2$ is the highest. So, according to the definition, there should be a local maximum. Which one is true? Is $x=-2$ a local maximum and why?

答:
The derivative of $|x+2|$ is $\dfrac {x+2}{|x+2|} = sgn(x+2) := \begin{cases} 1 &\text{ for } x+2 > 0\\-1 &\text{ for } x+2 < 0 \end{cases}$ and is not defined for $x=-2$. Therefore the derivative of $f(x)$ is also not defined for $x=-2$. (Some textbooks include these points where $f'(x)$ is not defined as critical points; regardless one must check whether an extremum exists at these points.)

The first derivative test can still be applied by taking left/right limits. Note that:

$$\lim_{x \to -2^-} f'(x) = \lim_{x \to -2^-} e^{|x+2|}\left(2x+2+(x^2+2x-7)\dfrac {x+2}{|x+2|}\right)=e^0(-4+2+(4-4-7)(-1))=5$$

$$\lim_{x \to -2^+} f'(x) = \lim_{x \to -2^+} e^{|x+2|}\left(2x+2+(x^2+2x-7)\dfrac {x+2}{|x+2|}\right)=e^0(-4+2+(4-4-7)(1))=-5$$

This shows that the function is increasing before $-2$ and decreasing after $-2$, and thus there is a local maximum at $x=-2$.

A similar analysis can be carried out on a simpler function: $|x|$. It is not differentiable at $x=0$, but has a local minimum there.


问:
A simple problem about adjoint operators
Consider the operator $T:L^2(0,1)\rightarrow C([0,1])$ defined by

$$Tx(t) := \int_0^tx(s)ds.$$

Pick any $a \in [0,1] $ and let be $\delta_a$ be the Dirac delta at $a$. If $T'$ is the operator adjoint of $T$, then I am supposed to find some $g \in L^\infty (0, 1)$ which defines $T'\delta_a$. I do not really understand the question, how could some $g \in L^\infty (0, 1)$ define $T'\delta_a$? Please do not give me the function $g$ since this is an exercise that I will be graded on, I just want to understand the question. How should I think here to come up with $g$?

答:
First note that

$$<\delta_a, Tx> = \int_{0}^{a} x(s)ds$$

So is there a $g := T'\delta_a$ such that

$$<g, x> = \int_{0}^{1} g(t)x(t) dt = \int_{0}^{a} x(t)dt$$

??

Answer

 $$ g(t) = 1 $$ for $t < a$ and
 $$ g(t) = 0 $$ for $t > a$


问:
Fundamental group of mapping torus?
Let $f\colon X\to X$ be a homeomorphism between a CW-complex $X$ and iteself.

Let $M_f=X\times [0,1]/(x,0)\sim (f(x),1)$, mapping torus of $X$ from $f$.

I want to calculate the fundamental group $\pi_1(M_f)$ of $M_f$ in terms of $\pi_1(X)$ and $f_*\colon \pi_1(X)\to \pi_1(X)$.

Are there any hint to do this?

Note: This is not a homework problem.

答:
You can use van Kampen's Theorem.  The upshot is that you get a semi-direct product:

$\pi_1M_f\cong\pi_1X\rtimes_{f_*}\mathbb{Z}$.

Title: Fundamental group of mapping torus?


问:
Simplifying $\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$
How do I simplify:

$$\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$$

Should I use modulos or should I factor them?  Or any I suppose to use combinatorics?  Any one have a clue?

Body:
Observe that for integer $m\ge 0,$
   $$\underbrace{11\dots1}_{m\ 1's}=\frac{\underbrace{99\dots99}_{m\ 9's}}9=\frac{10^m-1}9 $$

$$\text{So, }\underbrace{11\dots11}_{2n\ 1's}=\frac{10^{2n}-1}9\text{ and }\underbrace{22\dots22}_{n\ 2's}=2\cdot \underbrace{11\dots11}_{n\ 1's}=2 \cdot\frac{10^n-1}9$$

$$\implies \underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}=\frac{10^{2n}-1-2(10^n-1)}9=\frac{10^{2n}-2\cdot10^n+1}9=\left(\frac{10^n-1}3\right)^2$$

$$\text{Now as }\underbrace{11\dots1}_{m\ 1's}=\frac{10^m-1}9,$$
$$ \frac{10^n-1}3=3\cdot \frac{10^n-1}9=3\cdot(\underbrace{11\dots11}_{n\ 1's})=\underbrace{33\dots33}_{n\ 3's}$$


答: HINT: just find such a $g$ in $L^2(0,1)$. Once you get it, you can check that it is also an element of $L^{\infty}(0,1)$

Title: A simple problem about adjoint operators

Body:
Consider the operator $T:L^2(0,1)\rightarrow C([0,1])$ defined by

$$Tx(t) := \int_0^tx(s)ds.$$

Pick any $a \in [0,1] $ and let be $\delta_a$ be the Dirac delta at $a$. If $T'$ is the operator adjoint of $T$, then I am supposed to find some $g \in L^\infty (0, 1)$ which defines $T'\delta_a$. I do not really understand the question, how could some $g \in L^\infty (0, 1)$ define $T'\delta_a$? Please do not give me the function $g$ since this is an exercise that I will be graded on, I just want to understand the question. How should I think here to come up with $g$?


问:
Separable topological space
Show that if $(X,\tau)$ is a separable topological space, then $\{x \in X : \{x\} \in \tau\}$ is an enumerable set.

I thought this way, if a topological space is separable, this implies that it has a dense and numerable set, that is, if this set is dense, then its closure is the topological space X, which has numberable points, finite or that has bijeção with the natural numbers.
Does this have anything to do with the elements of the topology being numberable? in particular discrete elements, formed by only one element ${x}$. In fact, how can I state that $\{x\}$ is in the topology?

答:
Let $I$ be the set of isolated points, that is
$I=\{x \in X : \{x\} \in \tau\}$ (all singletons that are open).

Let $D$ be any countable dense set. Then $I$ must be a subset of $D$, hence $I$ must be countable too.

Indeed, suppose the contrary, that $I$ is not a subset of $D$, and pick any $x\in I\setminus D$. Then the set $U=\{x\}$ is open and non-empty, but $U\cap D=\varnothing$, contradiction.

Edit/addition. Just to clarify the definition of separable, since in your question it is not clear whether you think that $D$ is numerable or that $X$ is numerable. A space $X$ (with topology $\tau$) is separable if there is a countable (also called numerable) dense subset $D$ of $X$. That is, $D$ is numerable, and the closure of $D$ equals $X$ (but $X$ need not be numerable). To say that $D$ is dense is the same as to say the $D$ intersects every non-empty open subset of $X$.

Take the real line with the usual topology $\sigma$ and isolate all rational numbers. That is, consider the topology $\tau$ with base $\mathcal B=\sigma\cup\{\{q\}:q \mathrm{\ is\ rational\ }\}$. This provides an example of a separable space $X$ with countably many isolated points, and such that $X$ is uncountable and $\tau$ is uncountable too.

Title: Separable topological space


问:
Given a compact space $T^*$ and Hausdorff space $T$ on set $X$ such that $T^* \subseteq T$. Show that $T^*=T$
Let $(X,T)$ be a compact topological space. Let $T^*$ be a topology on $X$ such that $T^*$ is contained in $T$ and suppose that $(X,T^*)$ is  Hausdorff space. Show that $T^*=T$

To prove $T^*=T$, I have to show that $T^*$ is contained in $T$ and vice versa. It is already given that $T^*$ is contained in $T$. I just have to show that "$T$ is also contained in $T^*$". How do I do it? Can anyone help?

答:
You can also do it directly with open covers.

Suppose that $\mathscr{T}^*\subsetneqq\mathscr{T}$, and let $U\in\mathscr{T}\setminus\mathscr{T}^*$. Let $K=X\setminus U$; $K$ is not closed in $\langle X,\mathscr{T}^*\rangle$, but $K$ is closed in $\langle X,\mathscr{T}\rangle$, which is a compact space, so $K$ is compact in $\langle X,\mathscr{T}\rangle$.

Let $\mathscr{V}$ be a cover of $K$ by $\mathscr{T}^*$-open sets. $\mathscr{T}^*\subseteq\mathscr{T}$, so $\mathscr{V}$ is automatically also a $\mathscr{T}$-open cover of $K$, and as such it has a finite subcover $\mathscr{V}_0$. Thus, every cover of $K$ by $\mathscr{T}^*$-open sets has a finite subcover, and $K$ is therefore compact in $\langle X,\mathscr{T}^*\rangle$. But every compact subset of a Hausdorff space is closed, and $\langle X,\mathscr{T}^*\rangle$ is Hausdorff, so $K$ is closed in $\langle X,\mathscr{T}^*\rangle$. This contradiction shows that $\mathscr{T}$ cannot be a proper subset of $\mathscr{U}$.


问:
$T:\mathbb R^2\to\mathbb R^3$; if $(1,1),(2,3)\in\ker(T)$, is $(\pi,3)\in\ker(T)$?
I am not so familiar with linear algebra, so I would really appreciate your help. I think this is quite a simple question, but I really can't find a way through it.

The proposition I want to prove is as the title says:

$T:\mathbb R^2\to\mathbb R^3$; if $(1,1),(2,3)\in\ker(T)$, then $(\pi,3)\in\ker(T)$.

Is this proposition right or wrong?

答:
Notice that $(1,1),(2,3)$ are linearly independent and therefore form a basis of $\mathbb R^2$ as a real vector space (why?).

Also, recall that if $T$ is a linear map, $\ker(T)$ is a linear subspace of the domain of $T$. In this case, what happens if both the vector of a basis of $\mathbb R^2$ are in $\ker(T)$?

Alternatively, you can just consider the second observation ($\ker(T)$ is a vector space); can you write $(\pi,3)$ as a linear combination (with real coefficients) of $(1,1)$ and $(2,3)$, in order to prove that also $(\pi,3)$ lives in $\ker(T)$?


问:
Jacobson radical of a finitely generated algebra (A-M 5.22)
This is regarding exercise 5.22 in Atiyah-MacDonald :

$A$ is a subring of an integral domain $B$ such that $B$ is finitely generated over $A$. Prove that if the Jacobson radical of $A$ is zero, then so is the Jacobson radical of $B$.

From a previous exercise (5.20) we have the following result:

There exist $y_1, \dots, y_r\in B$ algebraically independent over $A$ and an $s\in A\setminus 0$ such that $B_s$ is integral over $B_s'$ where $B' = A[y_1, \dots, y_r]$.

Here's my solution. With $y_1, \dots, y_r, s$ as above, I claim that any non constant $f\in B'$ is not a unit in $B$. Suppose not, say $f\in B'$ has an inverse $g\in B$, then $g/1$ satisfies a monic polynomial over $B_s'$ of some degree $m$. Multiplying such a relation by $f^{m-1}$ gives a relation of the form
$$g-\frac{p}{s^k} = 0, p\in B', k\geq 0.$$
Multiplying again by $f$ and rearranging,
$$s^k = pf.$$
Since $f$ is a non constant polynomial, and $p\neq 0$ we have an algebraic relation in $B'$ which is impossible.

Next, let $b\in B$ be a nonzero non unit and let $b/1$ satisfy a monic polynomial over $B_s'$. Clearing the denominators and rearranging we get an equation of the form
$$bx+f = 0, x\in B, f\in B'.$$
Note that $f$ is a nonzero non unit in $B$. First, suppose $r\geq 1$, then multiplying the above by $y_1$ and adding $1$ we get $bxy_1+(fy_1+1) = 1$ and $fy_1+1$ is a non constant polynomial in $B'$, hence not a unit in $B$. This means that $b$ is not in a maximal ideal containing $fy_1+1$.

So we are left with the case when $r = 0$. This means that there is an $s\in A\setminus 0$ such that $B_s$ is integral over $A_s$. From the arguments above, it suffices to prove that for every $0\neq f\in A$ there is a maximal ideal in $B$ not containing $f$. If $f$ is a unit in $A$ this is obvious, so assume $f$ is not a unit, in which case there's a maximal ideal $\mathfrak{m}$ of $A$ not containing $sf$, hence not containing $s$ and $f$.

Push the maximal ideal to $A_s$, use integrality to obtain a maximal ideal $\mathfrak{m}'$ lying over $\mathfrak{m}$ in $B_s$ and then pull it back to $B$. This is a maximal ideal not containing $f$ as required.

My answer seems correct, am I missing something? Anyway, this is the hint in the book

To take any $v\neq o$ in $B$, localize at $v$, then $B_v$ is finitely generated over $A$. Obtain $s$ as above and a maximal ideal $\mathfrak{m}\not\ni s$ in $A$, and consider the map $A\to A/\mathfrak{m} = k$. Extend this homomorphism to $g\colon B_v\to\Omega = \bar{k}$. Then $g(v) \neq 0$ and $ker(g)\cap B$ is maximal in $B$ not containing $v$.

I don't see how the kernel is maximal in $B$, could someone explain that?

答:
It is clear that given the homomorphism $ g \colon B_v \rightarrow \Omega$ we have $g(v) \neq 0$, since if $g(v)=0$ then $g(1)=g(v v^{-1})=g(v)g(v^{-1})=0$, which leads to a contradiction. Furthermore, it is clear that $ker(g) \cap B$ is a maximal ideal in $B$; note that $ker(g) \cap B \cap A = \mathfrak{m}$ and since $B$ is integral over $A$, this shows that $ker(g) \cap B$ is maximal in $B$ (by Corollary 5.8 in Atiyah-Macdonald). Be careful that when applying (5.8), one must check that $ker(g) \cap B$ is prime in $B$ (this is clear since $B/ker(g)$ is isomorphic to a subring of $\Omega$.)

Hope this helps!


答:
Hint: You can simplify a lot the problem taking the $\log$ of your function and studying the extrema of the new function.

Indeed $f(x)= (x^2 +2x -7)e^{|x+2|}$ has the same extrema of $g(x)= \log[(x^2 +2x -7)e^{|x+2|}]$ and you can easily see that $$g(x)= \log[(x^2 +2x -7)e^{|x+2|}] = \log(x^2 +2x -7) + |x+2|$$

The extrema are the same since the log is a strictly increasing function and the composition of any function with a strictly increasing function does not change the extrema.

Title: Find the extrema of $f(x)=(x^2+2x-7)e^{|x+2|}$

Body: I need to find the extrema of $f(x)=(x^2+2x-7)e^{|x+2|}$. The first derivative is $e^{|x+2|}(2x+2+(x^2+2x-7)(|x+2|)')$. Looking at two cases where $x<-2$ or $x>-2$ gives two extrema - at the points -3 and 1. And now comes the real question: Is there an extremum at the point -2 and why? If we replace x with -2, the first derivative equals -2. But the fact that the derivative there isn't $0$ doesn't mean there isn't an extremum. In fact, if I take a look at the graphic of the function, it is obvious that in the region of -2, the value of the function for $x=-2$ is the highest. So, according to the definition, there should be a local maximum. Which one is true? Is $x=-2$ a local maximum and why?


问:
Finding $y$ value of canonical ellipse.
I have an ellipse:
$$
 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
This may be a simple question, but my mind plays tricks on me at the moment;

Which is the most efficient way if I have $x$, $a$ and $b$ and want to find the value of $y$?

Hope someone can help me - thanks in advance :)!

答:
Given:

$\boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}$

We can subtract $\dfrac{x^2}{a^2}$ from both sides:

$\dfrac{x^2}{a^2}-\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1-\dfrac{x^2}{a^2}$

Next multiplying both sides by $b^2$:

$\left(\dfrac{y^2}{b^2}\right)b^2=\left(1-\dfrac{x^2}{a^2}\right)b^2$

Which becomes:

$\dfrac{b^2y^2}{b^2}=b^2-\dfrac{b^2x^2}{a^2}$

The b's on the left hand side cancel leaving us with:

$y^2=b^2-\dfrac{b^2x^2}{a^2}$

We can take the square root of both sides leaving us with our final answer:

$\boxed{y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}}$

Title: Finding $y$ value of canonical ellipse.


问:
Test the series for conditional convergence: $ \sum_{n=1}^{\infty}a_n,\ \ a_n=\arctan\frac{\sin n}{\sqrt{n+1}} $
Test the series for conditional convergence:
$$
\sum_{n=1}^{\infty}a_n,\ \ a_n=\arctan\frac{\sin n}{\sqrt{n+1}}
$$

I tried to apply the fact that $|\arctan(t)|\le|t|\ \ \forall |t|<1$ and then prove that the initial series converges absolutely (and therefore conditionally).
However, this idea didn't work:
$$
|a_n|<\left|\frac{\sin n}{\sqrt{n+1}}\right|=\frac{|\sin n|}{\sqrt{n+1}}=b_n
$$
And the series $\sum_{n=1}^\infty b_n$ is divergent.

Every other method I know of doesn't help here, because $a_n$ can take both positive and negative values. Dirichlet test doesn't work either, since I have one unsplittable function here, which is $\arctan$.

So, what should I do?

答:
$$\sum_{n=1}^\infty \arctan \frac{\sin n}{\sqrt{n+1}} = \sum_{n=1}^\infty  \frac{\sin n}{\sqrt{n+1}} + \sum_{n=1}^\infty \left(\arctan \frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}}\right)$$
Dirichlet test for the first, and the second converges absolutely.


问:
How to calculate the correlation between the number of heads of 100 toss of coin and the number of heads of the first 10 toss of those 100 tosses?
It is a bit cumbersome to explain:

Toss a coin is a Bernoulli distribution, with the probability of seeing a head is p

if we toss this coin 100 times, we should expect $X_{1}$ times of head. Within that 100 toss (this is important, we are NOT tossing another 10 times), we should see $X_{2}$ heads from the first 10 toss.

How to calculate $corr(X_{1}, X_{2})$ ?

The only thing I can think of is, $X_{1} > X_{2}$, practially, we are doing two sets:

toss a coin 10 times, we see $X_{2}$ heads

independently toss a coin 90 times, we see $X_{3}$ heads

we want to calculate $corr(X_{2}, X_{2} + X_{3})$

答:
Your splitting $X_1$ into $X_2$ and an independent $Y_3$ is a correct approach and you get $$\textrm{Cov}(X_{2}, X_{1})=\textrm{Cov}(X_{2}, X_{2} + X_{3})=\textrm{Var}(X_2)+0=10p(1-p)$$

So $$\textrm{Corr}(X_{2}, X_{1})=\frac{\textrm{Cov}(X_{2}, X_{1})}{\sqrt{\textrm{Var}(X_2)\textrm{Var}(X_1)}}=\frac{10p(1-p)}{\sqrt{10p(1-p)100p(1-p)}} = \frac{1}{\sqrt{10}}$$ as drhab found


问:
Solution verification(double integrals)
I need to find surface area between $z^2=x^2+y^2$ and $(x^2+y^2)^2=2xy$.Using polar coordinates i found bounds for r:$0\le r \le \sqrt{\sin 2\phi}$ and $\phi$ $0\le \phi \le \frac{\pi}{4}$. Then using the formula for surface area i get $\frac{\sqrt{2}}{4}$, but in solution it says $2\sqrt{2}$. Where am I mistaking? Btw for $\phi$ i used this bounds because in $[0,\frac{\pi}{4}]$ is already included every positive value for sine function(because $\sin 2\phi$ needs to be $\ge 0$).

答:
You need to find the surface area of the cone $z^2 = (x^2+y^2) = r^2$ bound by the cylinder $\rho^2 = |\sin2\theta|$.

The cylinder has two petals as shown in the diagram $(0 \leq \theta \leq \frac{\pi}{2} \,$ for the petal in the first octant). The radius of $1$ is maximum at $\theta = \frac{\pi}{4}$ and radius is $0$ at $\theta = 0, \frac{\pi}{2}$.

Please note you need to find surface area in $4$ octants, as the cone $z^2 = x^2 + y^2$ forms both below $z = 0$ and above. So we will find surface area in first octant and then multiply by $4$.

As the cone is a right circular cone with equation $z = \pm r$, we have $\phi = \frac{\pi}{4}$.

Now the cone and cylinder intersect at two points at any given $z$ in the first octant (see the part of the cross section of the cone bound by cylinder, marked in black).

$r^2 = z^2 = \sin 2\theta, \sin (\pi - 2\theta) \implies \theta = \frac{1}{2} \arcsin (r^2), \frac{\pi}{2} - \frac{1}{2} \arcsin (r^2)$.

At $z = 1$, cone touches the cylinder at its maximum radius $1$ and there is no intersection for $z \gt 1$. So $0 \leq r \leq 1$.

$\sqrt {1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y)^2})^2} = \sqrt 2$. In cylindrical coordinates, $dx \, dy = r \, d\theta \, dr$.

So surface area $S = \displaystyle 4 \int_0^1 \int_{\arcsin r^2 / 2}^{(\frac{\pi}{2} - \arcsin r^2)/2} \sqrt 2 \, r \, d\theta \, dr = 2 \sqrt2$

Title: Solution verification(double integrals)


答:
Hint (almost solution):

Statement: $\sum_{n=1}^{\infty}$ converges conditionally.

It's know that $|\arctan(x) - x| \le 100 |x|^3$ for $-\frac{1}{100} \le x \le  \frac{1}{100}$.

A series $100\sum_{n=1}^{\infty} (\frac{\sin n}{\sqrt{n+1}})^3$ converges absolutely, because $|(\frac{\sin n}{\sqrt{n+1}})^3 | \le \frac{1}{n^{\frac32}}$. Hence $\sum_{n=1}^{\infty} (\arctan\frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}})$ converges absolutely.

As

$$\sum_{n=1}^{\infty} \arctan\frac{\sin n}{\sqrt{n+1}} = \sum_{n=1}^{\infty} (\arctan\frac{\sin n}{\sqrt{n+1}} - \frac{\sin n}{\sqrt{n+1}}) + \sum_{n=1}^{\infty} \frac{\sin n}{\sqrt{n+1}},$$

it's sufficient to show that $\sum_{n=1}^{\infty} \frac{\sin n}{\sqrt{n+1}}$ converges conditionally. We may get convergence from Dirichlet's test for convergence.

It's sufficient to show that $\sum_{n=1}^{\infty} \big|\frac{\sin n}{\sqrt{n+1}} \bigr| = +\infty$.

Title: Test the series for conditional convergence: $ \sum_{n=1}^{\infty}a_n,\ \ a_n=\arctan\frac{\sin n}{\sqrt{n+1}} $

Body:
Test the series for conditional convergence:
$$
\sum_{n=1}^{\infty}a_n,\ \ a_n=\arctan\frac{\sin n}{\sqrt{n+1}}
$$

I tried to apply the fact that $|\arctan(t)|\le|t|\ \ \forall |t|<1$ and then prove that the initial series converges absolutely (and therefore conditionally).
However, this idea didn't work:
$$
|a_n|<\left|\frac{\sin n}{\sqrt{n+1}}\right|=\frac{|\sin n|}{\sqrt{n+1}}=b_n
$$
And the series $\sum_{n=1}^\infty b_n$ is divergent.

Every other method I know of doesn't help here, because $a_n$ can take both positive and negative values. Dirichlet test doesn't work either, since I have one unsplittable function here, which is $\arctan$.

So, what should I do?


问:
Composition of systems of equations
Suppose $$2x + 3y = u$$ $$x - 4y = v$$

and further that
$$3u - 5v = c$$ $$2u + 3v = d$$

Express c and d in terms of $x$ and $y$ by matrix multiplication.

It's quite easy by direct substitution but I can;t work out how to use matrix multiplication. Any ideas? Thanks in advance!

答:
$$\begin{align*}
2x + 3y& = u\\
x - 4y &= v
\end{align*}\quad \underset{\substack{\text{convert to}\\ \text{matrix language}}}{\leadsto}\quad \begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}u\\ v\end{bmatrix}$$
$$\begin{align*}
3u - 5v &= c\\
2u + 3v &= d
\end{align*}\quad \underset{\substack{\text{convert to}\\ \text{matrix language}}}{\leadsto}\quad \begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\begin{bmatrix} u\\ v\end{bmatrix}=\begin{bmatrix}c\\ d\end{bmatrix}$$

$$\begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\begin{bmatrix} u\\ v\end{bmatrix}=\begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\Bigg(\begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}\Bigg)=\begin{bmatrix}c\\ d\end{bmatrix}\implies\begin{bmatrix} p & q\\ r &s\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}c\\ d\end{bmatrix}\\[0.4in]
\quad\text{ where }\quad \begin{bmatrix} p & q\\ r &s\end{bmatrix}=\begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}.\\[0.4in]
\begin{bmatrix} p & q\\ r &s\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}c\\ d\end{bmatrix}\quad\underset{\substack{\text{convert to}\\ \text{equation language}}}{\leadsto}\quad \begin{align*}
px + qy& = c\\
rx + sy &= d
\end{align*}$$

Title: Composition of systems of equations


问:
Suppose Division ring D is infinite, Prove D* is not finitely generated.
Suppose Division ring D is infinite, Prove D* is not finitely generated.
Either prove or disprove
Proven for the field.</p>

答:
You say that you've proven the result when $D$ is commutative, which shows that your question in full generality is equivalent to the following open conjecture:

If $D$ is a division ring and $D^\times$ is a finitely generated group, then $D$ is commutative.

By a result of Akbari and Mahdavi-Hezavehi, the conjecture is known to be true in the case that $D$ is finite-dimensional over its center.


问:
Find the domain and the correspondence rule of $F$
$F=f ∘ f∘ f$, $f(x)=\frac{1}{1-x} ⇒ Dom f(x)=\mathbb{R}-\{1\} \land Ran f(x)=\mathbb{R}-\{0\}$, then:
\begin{align*}
 Dom(f∘ f)&=\{ x ∈ Dom(f) \land f(x) ∈ Dom(f)\} \\
           &=\{ x ∈ \mathbb{R}-\{1\} \land \frac{1}{1-x} ∈ \mathbb{R}-\{1\} \} \\
           &=\{ x ∈ \mathbb{R}-\{1\} \land x ∈ \mathbb{R}-\{0\}-\{1\}  \} \\
                &=\{ x ∈ \mathbb{R}-\{1\}  \} \\
  Dom(f∘ f∘ f)&=\{ x ∈ Dom(f∘ f) \land f(f(x)) ∈ Dom(f∘ f)\} \\
          &=\{ x ∈  x ∈ \mathbb{R}-\{1\} \land \frac{1}{1-\dfrac{1}{1-x}} x ∈ \mathbb{R}-\{1\}\} \\
            &=\{ x ∈ \mathbb{R}-\{1\} \land 1-x ∈
              \mathbb{R}-\{1\}\} \\
  &=\{ x ∈ \mathbb{R}-\{1\} \} \\
F&=x
\end{align*}

It's wrong?

From: Lumbreras Editors

答:
Since $f(\cdot)$ only takes values $\neq 1$, $f(f(x))$ is defined for $x \neq 1$ (because we need that the inner $f(x)$ is defined) and all the values of $x$ for which $f(x) \neq 1$, that is, $f(f(x))$ is defined for $x \neq 0,1$. Similarly, $f(f(f(x)))$ is defined for $x \neq 0,1$ (because we need that $f(f(x))$ is defined) and all the values of $x$ for which $f(f(x)) \neq 1$, that is, $f(f(f(x)))$ is defined for $x \neq 0,1$ since $f(f(x))$ is always $\neq 1$.

Thus $\operatorname{dom}(F) = \mathbb R \setminus \{0,1\}$.