答:
$\operatorname{ker}A$ is the set of vectors $\vec x$ that satisfy $A\vec x=0.$ And you are given that $x-2y+z = 0.$ So, if we let $\alpha=z,\ \beta =y$ then $x=2\beta-\alpha$ and so $\vec x=(2\beta-\alpha,\beta,\alpha)^T.$ Collecting this information in a slightly different way, $\begin{pmatrix}
x\\y
\\ z
\end{pmatrix}=\begin{pmatrix}
-1\\0
\\ 1
\end{pmatrix}\alpha+\begin{pmatrix}
2\\1
\\ 0
\end{pmatrix}\beta.$ Set $\vec v_1 = (-1,0,1)^T$ and $\vec v_2=(2,1,0)^T$ and extend to a basis for $\mathbb R^3:$ we may take $\vec v_3=(0,0,1)^T.$

So, in the basis $\{\vec v_1,\ \vec v_2,\ \vec v_3\}$, we have $A\vec v_1=A\vec v_2=0$ and $A\vec v_3=(a,b,c,d)^T$ for some numbers $a,b,c,d.$ The matrix of $A$ in this basis relative to the standard basis of $\mathbb R^4$ is therefore

$A=\begin{pmatrix}
0 &0  &a\\
0 &0  &b \\
 0&0  &c \\
0 &0  &d
\end{pmatrix}.$

Now we compute the change of basis matrix $[\vec v_1,\vec v_2,\vec v_3]\to [\vec i,\vec j,\vec k]:$

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

The matrix of $A$ relative to the standard bases for $\mathbb R^3$ and $\mathbb R^4$ is therefore $AB$.

Title: how to write this linear transformation as a matrix?

Body:
I was trying to write this linear transformation as a matrix,
the exercise is the following:
let "f: $R^3 --> R^4$" be a linear transformation and Kerf = (x, y , z) in $R^3$: $x-2y+z = 0$  is its null space. Determine the canonical matrix.

I know how to solve this type of exercises (generally), as follow:

I use the basis of the first vector space (R3 in this case),

I plug x, y, z, in the equations of the linear transformation,

then, I'm ready to use each component of the vectors (of R4, in this case) to create the matrix. (the canonical matrix has basis of the entire space, for example, 1 0 0, 0 1 0, 0 0 1).

But, when I'm about to solve these exercises, I simply cannot solve them. I don't know why, because I know how to do them theoretically.

Can you help me to solve this exercise?


问:
Exchange property of infinite minimal structures
Here's an exercise from the really nice Hinman's book "Foundamentals of Mathematical Logic". The minimal structures are those whose definable sets are finite or cofinite. The algebraic closure  (acl) of $X$ is a set with elements from some finite set that is definable with parameters from $X$. I've stuck with the last sentence from the hint. I've proved the union contains at most $nm$ elements but can't see how to get a contradiction from it. Will be greatful for all your tips.

P.S. $\exists^{=k}xF$ means there are exactly $k$ values of $x$ s.t. $F$ is true.

答: Since $\mathfrak{A}$ is infinite, because $\bigcup_{i\leqslant m}\neg\psi_{a'_i,c}^\mathfrak{A}$ is finite its complement in $\mathfrak{A}$ must be infinite. Let $b'\in\mathfrak{A}\setminus\bigcup_{i\leqslant m}\neg\psi_{a'_i,c}^\mathfrak{A}$ be arbitrary; then $b'$ satisfies $\psi(a'_i,y,c)$ for each $i\leqslant m$. Recall that $$\psi(a'_i,y,c)\equiv\phi(a'_i,y,c)\wedge\exists^{=m}x\phi(x,y,c).$$ In particular, $\phi(a'_i,b',c)$ holds for each $i\in\{0,\dots,m\}$, and so $\phi_{b',c}^\mathfrak{A}$ has size at least $m+1$. On the other hand, $\exists^{=m}x\phi(x,b',c)$ holds as well, so this gives the desired contradiction.

ContentLicense: CC BY-SA 4.0


问:
Use divergence theorem to prove
Let r = (x, y, z) and $r_ε = \sqrt {x^2+y^2+z^2+ε} $ for any number $ε \ge0$  fε : R
3 → R is
the scalar field fε(r) = $1\over r_ε $

Use the Divergence Theorem to prove that

$$lim_{ε→0} \int \int \int_{v(R)} ∇^2 f_ε dx dy dz = −4π$$

where V (R) is the ball of radius R > 0 centred at the origin.

so far I have proved that ∇fε(r) = $-r\over r_ε^3$

and I tried to do  ∇^2 fε and got $3r^2 (r_ε)^{-5}-3r_ε^{-3} $ but I don't think that is right could someone help me out. If it is right I am not sure where to go from here

答:
$\vec{r} = (x, y, z)$ and $r_\epsilon = \sqrt{x^2+y^2+z^2+ε}, \, \epsilon \geq 0$

$\displaystyle f_\epsilon (r) = \frac{1}{r_\epsilon} = \frac{1}{\sqrt{r^2 + \epsilon}}$

We need to prove $\lim_{\epsilon→0} \iiint_{v(R)} \nabla^2 f_\epsilon \, dx \, dy \, dz = −4π$

Please note $\nabla^2 f_\epsilon$ is the divergence of $\nabla f_\epsilon \, $ (the gradient of the scalar field $f_\epsilon)$ and the surface is a ball of radius $R$ centered at $(0, 0, 0)$.

As per divergence theorem,

$\iiint_S div (\vec{F}) \, dV = \iint_S \vec {F} \cdot \hat{n} \, dS$

So, $\iiint_S \nabla \cdot \nabla f_\epsilon \, dx \, dy \, dz = \iint_S (\nabla {f_\epsilon}) \cdot \hat{n} \, dS$

$\displaystyle \nabla f_\epsilon = - \frac{\vec{r}}{(r^2 + \epsilon)^{3/2}}$

The unit normal vector is $\displaystyle \frac{\vec{r}}{r}$

So the surface integral over a ball of radius $R (\gt 0)$ centered at origin is given by,

$I = \displaystyle \int_0^{2\pi} \int_0^{\pi} - \frac{\vec{r}}{(R^2 + \epsilon)^{3/2}} \cdot \frac{\vec{r}}{R} \, (R^2 \sin \phi) \, d\phi \, d\theta$

$\displaystyle = - \frac{R^3}{(R^2 + \epsilon)^{3/2}} \int_0^{2\pi} \int_0^{\pi} \sin \phi \, d\phi \, d\theta = - \frac{4R^3\pi}{(R^2 + \epsilon)^{3/2}}$

Now as $\epsilon$ tends to $0$, you can see what happens to your surface integral.

Title: Use divergence theorem to prove


问:
Formal power series with coefficient in $\{0,1 \}$ which is not rational
Recall that $f \in \mathbb{C}[[x]]$ is rational if $f = ab^{-1}$ where $a, b \in \mathbb{Q}[x]$ (the inclusion  $ \mathbb{Q}[x] \to \mathbb{C}[[x]]$ is obvious).

How do I obtain an example of  $f \in \mathbb{C}[[x]]$ with coefficients $0,1$ which is not rational?  I have an experience of proving that some series is not rational (say, $exp(x)$).I could't come up with an example myself, though.

答:
One necessary condition for a power series to be rational is for it's image at the rational points to be rational. This is clear since for every $r\in\mathbb{Q}$ with $b(r)\neq 0$, we need to have $f(r)=a(r)/b(r)$. Since $a(x),b(x)\in\mathbb{Q}[x]$, we have $a(r),b(r)\in\mathbb{Q}$.

The transcendece of the Liouville constant (see here) (we choose $r = 10^{-1}$) thus implies that the example provided by @Arthur in the comments is an example of a non-rational power series which has only $0$'s and $1$'s


问:
Show map is not a isomorphism
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$ ($K$ is just a field)

Show that $f:U \rightarrow K^2$ is not a isomorphism

My try

I have only been told what a isomorphism is in a general sense; that it has to be "the same" i.e some structure is still the same. For vectors I guess it could be addition, multiplication and the null vector? I am thinking $f$ must be bijective as well. Other than that I have not been taught much more than "it is the same" – which really does not help me in the slightest.

答:
Obviously $V$ is the null space of the rank-1 matrix $[1 \hspace{0.3cm} -1 \hspace{0.3cm} 2]$. So $V$ is 2 dimensional vector space and any linearly independent two vectors from $V$ can be a basis. Select $v=(1,1,0)$ and $w=(0,2,1)$ in $V$. They are lin independent so $\{v,w\}$ can be a basis of $V$.

$\textbf{Fact:}$ $\{fv,fw\}$ can be a basis for $K^2$ iff $f$ is isomorphism.

But $fv=f(1,1,0)=(1,1)$ and $fw=f(0,2,1)=(-1,-1)$.So they are linearly $\textbf{dependent}$ and they cannot be a basis. By the fact above we showed $f$ is not be an isomorphism.

Title: Show map is not a isomorphism


答:
Hint:

By the rank-nullity theorem, $V$ has dimension $2$, like $K^2$. Hence it is enough to prove that $\ker f$ is non- trivial to prove $f$ is non-injective, and a fortiori non-bijective. This means you have to find a non-zero vector $v\in K^3$ such that
\begin{cases}
x_1-x_2+2x_3=0,\\
x_1-x_3=0,\\
x_2-3x_3=0.
\end{cases}
You can check that  the kernel of $\varphi:K^3\longrightarrow K^2, \enspace\begin{pmatrix}x_1\cr x_2\cr x_3\end{pmatrix}\longmapsto\begin{pmatrix}x_1-x_3\cr x_2-3x_3\end{pmatrix}$ ($\,f$ is its restriction to $V$) is actually contained in $V$, so you only have to determine a basis of $\ker \varphi$.

Title: Show map is not a isomorphism

Body:
Consider $V= \left \{ v= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$ and the linear map $f:V \rightarrow K^2$ defined by

$f\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=\begin{pmatrix}
x_1-x_3\\
x_2-3x_3
\end{pmatrix}$ ($K$ is just a field)

Show that $f:U \rightarrow K^2$ is not a isomorphism

My try

I have only been told what a isomorphism is in a general sense; that it has to be "the same" i.e some structure is still the same. For vectors I guess it could be addition, multiplication and the null vector? I am thinking $f$ must be bijective as well. Other than that I have not been taught much more than "it is the same" – which really does not help me in the slightest.


问:
Equidistant points by modular arithmetic
If $\frac{a}{b}$ is an irreducible rational number, how does one show that the collection of points $\{ \frac{na}{b}\ \mbox{mod}\, 1\}$, $1\leqslant n \leqslant b$ are equidistant along the real line with distance $\frac{1}{b}$?

I’ve tried numerous examples to see how it works but I’m sure how to give a rigorous argument.

答:
$\{\frac {na}b \pmod 1\}, 1⩽n⩽b$ are equidistant along the real line with distance $\frac 1b \iff$

$\{\frac {an}b \pmod 1\} = \{\frac jb \pmod 1|0\le j < b;j\in \mathbb Z\}\iff$

$\{an \pmod b\} = \{j \pmod b\}= \mathbb Z_b\iff$

For all $j$ the equivalence $an \equiv j\pmod b$ is solvable.

Note:  If $an \equiv j \pmod b$ is solvable for all $j$ then $an \equiv 1 \pmod b$ is solvable and $a$ is invertable.

And if $a$ is invertible then $an \equiv j\pmod b$ is solvible by $n = ja^{-1}$.

So back to our argument:

For all $j$ the equivalence $an \equiv j\pmod b$ is solvable.$\iff$

$a$ is invertible $\mod b$

Now you should have a theorem that $a$ is invertible $\mod b\iff \gcd(a,b)=1$.

And $\gcd(a,b) = 1 \iff \frac ab$ is an irreducible fraction.

=== old answer below =====

Key thing is that $\gcd(a,b)=1$ so $a$ is invertible $\mod b$ so there is a an integer $k$ so that $ka \equiv 1 \pmod b$, or in other words $ka = 1+ mb$ for some integer $m$, or in other words that $k\frac ab = m + \frac 1b$, or in other words $k\cdot a \equiv \frac 1b \pmod 1$.

And from there it's immediate that for any $0 < j \le b$ we can find that that $(jk)\frac ab \equiv \frac jb \pmod 1$ for all possible $\frac jb$ values.

Remains to show that there are no other possible values.

Well.... if $m\frac ab = \frac {ma}b$ and $ma$ is an integer, so it must be equivalent $\mod b$ to some integer $j: 0\le j < b$.  That is if $ma \equiv j \pmod b$ then $ma = j + kb$ for some integer $k$ and $\frac {ma}b = k + \frac jb\equiv \frac jb \pmod 1$.

So as $\frac jb\mod 1$ values will be reached and only $\frac jb\mod 1$ values will be reached.

====

I suppose another way of putting it is

$k \frac ab \equiv \frac jb \pmod 1 \iff ka \equiv j \pmod b$.

And the question becomes a matter of proving if $\gcd(a,b) = 1$ prove that for all residue classes $j$ there will exist a $k$ where $ka \equiv j\pmod b$.

And that is nothing more or less than the basic result: $a$ is invertible $\mod b$ if and only if $\gcd(a,b) =1$.


问:
Determining distance with laser and camera
I'm designing a device that measures distance using a laser and a camera. The device has the camera and laser separated by 100mm and they both point in the exact same direction. This is done so the camera can determine how far an object is depending on the distance between the center of the camera and the laser light is. So if the device is pointing towards a far object the light from the laser is closer to the center of the camera but on the contrary, if the object is very close, the distance between the laser light and the center is about 100mm. In my head it seems possible but how would I do this?

I was thinking of posting this to StackOverflow but I decided this was more of a math question rather than a code question.

Edit: I found a video of this method being used here.

答:
See this Wikipedia page about the pinhole camera model.

Assuming that the central target point of the camera and the point $P$ projected by the laser beam lay on a plane parallel to the image plane, you have a situation like in the picture:

You want to compute $x_3$ based on $y_1$, the distance from the center on the image, $f$ the focal length, and $x_1$, the distance between the laser light and the camera aperture $O$:

$$x_3=\frac{f}{y_1}x_1$$

In practice, you don't have $y_1$ on the image sensor but a distance in pixels, so you need to compute a value of $f$ adapted to your image file and you can compute it through a calibration at a known distance $x_{3c}$:

$$f=x_{3c}\frac{y_{1c}}{x_1}$$

You will have errors due to the target not on a plane perfectly parallel to the image plane, $X3$ axis not perfectly parallel to the laser beam, and radial distortion.


问:
Solving Equation Using Algebraic Method
How to solve these equations using an algebraic method?
I need to show my working, don't you do something in reverse, like 7 multiplies by something. I haven't done it in class.

$$\dfrac{5(3y-4)}{2y}=7$$

答:
$$\begin{equation*}
\frac{5(3y-4)}{2y}=7.\tag{0}
\end{equation*}
$$

like 7 multiplies by something

The given equation is only defined when the denominator of the left-hand side is different from $0$. So assume that $y\ne 0$. Then  you can multiply both sides of equation $(0)$ by $2y$ to obtain an equivalent one, i.e. the new equation has the same solution as the original.  $$
\begin{eqnarray*}
\frac{5(3y-4)}{2y}\times 2y &=&7\times 2y \tag{1$\mathrm{a}$} \\
\Leftrightarrow5(3y-4) &=&14y \tag{1$\mathrm{b}$} \\
\Leftrightarrow15y-20 &=&14y,\qquad\text{after expanding the LHS}.\tag{1$\mathrm{c}$}
\end{eqnarray*}$$

You can subtract $14y$ from both sides. The new equation is equivalent to the previous  one:
$$
\begin{eqnarray*}
15y-20-14y &=&14y-14y \tag{2$\mathrm{a}$}\\
\Leftrightarrow y-20 &=&0.\tag{2$\mathrm{b}$}
\end{eqnarray*}
$$

You can add $20$ to both sides. The new equation is equivalent to the previous  one:
$$
\begin{eqnarray*}
y-20+20 &=&0+20 \tag{3$\mathrm{a}$}\\
\Leftrightarrow y &=&20.\tag{3$\mathrm{b}$}
\end{eqnarray*}
$$

Since the solution $y=20\neq 0$, the multiplication in 1 is valid.

Comment. In general to get an equivalent equation one can:

multiply or divide both sides of a given equation by the same value, provided that it is different from $0$.

add or subtract the same value to and from both sides.

Simplify either side according to the algebraic rules as in $(1\mathrm{b})$ to $(1\mathrm{c})$.

ContentLicense: CC BY-SA 3.0

Title: Solving Equation Using Algebraic Method


问:
How do I graph the following curve without using a calculator $y^2=x\sin x$?
I have to graph the following curve by hand without taking the derivative of it.
If you could help in any way that would be lovely.
Thanks in advance!</p>

答:
Let's first understand $y=x\sin x$. It's bound between the lines $y=\pm x$, which form two "cones" touching at the origin. The curve touches each line infinitely often, where $\pi/2|x$. So it looks a bit like $\sin x$, but with larger oscillations as we move away from $x=0$ in either direction. It's an even function rather than an odd one, but the roots - multiples of $\pi$ are the same. You should be able to sketch it.

Now, what about $y^2=x\sin x$? This time, there are three key changes. Firstly, the bounds are now $\pm\sqrt{x}$ instead, so the oscillations don't grow as much. Secondly, the regions where $x\sin x<0$ are deleted, so there are infinitely many width-$\pi$ gaps. Finally, $y=\pm\sqrt{x\sin x}$, so if you work out what happens with $y\ge0$, it's mirrored below the $x$-axis.


问:
Mahalanobis distance connection with Euclidean distance
My question is more about the meaning of Mahalanobis distance when compared to L2 distance.
As a reminder:

Mahalanobis distance is the distance between a point $x$ and a distribution with mean $\mu$ and covariance $\Sigma$ can be calculated as: $d_{\text{Mahalanobis }} = \sqrt{(x-\mu)^T\Sigma^{-1} (x-\mu)}$

Back to the question:
if the Euclidean distance between a point $p_a$ and $p_b$ is for example 0.5 and the Mahalanobis distance between the point $p_a$ and some other distribution $\beta$ is 0.75, could we state that point $p_a$ is closer to $p_b$ than distribution $\beta$?

If not, then under what conditions we could have this comparison valid?

答:
Comparing a Euclidean distance directly to a Mahalanobis distance does not make sense, generally.

However, define distribution $\alpha$, such that $\vec{\mu_{\alpha}} = \vec{p_b}$ and $\Sigma_\alpha = I$.  So $\alpha$ is centered at $\vec{p_b}$ and spherical.

The Mahalanobis distance between $\alpha$ and $\vec{p_a}$ is

$$ d_M{(\alpha,p_a)} = \sqrt{(\vec{p_a}-\vec{\mu_\alpha})^T\Sigma_\alpha^{-1}(\vec{p_a}-\vec{\mu_\alpha})} = \sqrt{\sum_{i=1}^n(p_{a_i}-p_{b_i})^2}$$

Thus the Euclidean distance between $\vec{p_a}$ and $\vec{p_b}$ is the Mahalanobis distance between $\vec{p_a}$ and $\alpha$.

Now you can compare Mahalanobis distances and make a statement that $\vec{p_a}$ is closer to $\alpha$ (which has center $\vec{p_b})$ than to $\beta$.


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

"The sum of two even integers is always even"

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

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

Thank you for the help.

UPDATE:

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

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

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

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

答:
It's almost always helpful to write out definitions in these cases! What's an even integer? Well we say an integer $x$ is even if (and only if) $\exists k \in \mathbb{Z}\ \big(x = 2k\big)$.

So the statement is saying given any two integers $x$ and $y$ that satisfy the property above (that is the property that defines them as being even), their sum is also even. Let's denote the sum $z = x + y$. This means $\exists k \in \mathbb{Z}\ \big(z = 2k\big)$. In other words, that the statement $\exists k \in \mathbb{Z}\ \big(x+y = 2k\big)$ is true.

Since the statement is really saying that for any two integers which are even, their sum is even, we are effectively saying

$$\forall x \in \mathbb{Z}\  \forall y \in \mathbb{Z}\ \Big(\exists k \in \mathbb{Z}\ \big(x = 2k\big) \land \exists k \in \mathbb{Z}\ \big(y = 2k\big) \rightarrow \exists k \in \mathbb{Z}\ \big(x+y = 2k\big)\Big)$$

I know when I first started writing proofs, I would often get confused about why things were universally quantified, when the word 'all' wasn't explicitly mentioned. But over time, you learn to understand the implicit quantification, and in this statement, it is exactly meant to be true for any two even integers.

Also the proof of the statement is easy: Supposing $x$ and $y$ are even, then we can write them as $x = 2k_1$ and $y = 2k_2$ for some integers $k_1$ and $k_2$. So their sum is of course $x+y = 2k_1 + 2k_2$ $= 2(k_1 + k_2)$. But we know any integer plus itself is again an integer, so $x+y = 2k_3$ for an integer $k_3 = k_2 + k_1$. Of course since we assumed nothing about $x$ and $y$ other than the fact that they were even, we have proven the statement for all pairs of even integers.


答:
Strictly speaking, your statement is not an implication, but it can be expressed as a quantified formula in which the universal quantifier governs a logical implication.

Let $P(x)$ mean "$x$ is even". Then your statement can be written as

$$
\forall n, k \in \mathbb{Z} \quad \left(\, P(n) \land P(k) \implies P(n+k) \,\right).
$$

The clause governed by the universal quantifier has the structure you want. The hypothesis is $P(n) \land P(k)$ and the conclusion is $P(n+k)$.

This is an example of a formula in a first-order logic. In simple terms, first-order logic extends propositional calculus which you are familiar with by adding quantifiers (such as $\forall$ meaning "for all"), predicates (such as $P(x)$ which we defined here to mean "$x$ is even"), functions (such as $+$) and constants (such as $0$ which we did not use here).

The statement can be proven for example in Peano's arithmetic.

Title: What type of logical statement is this statement and how do you prove it?

Body:
I'm learning proof techniques. I understand "if, then" statements and how they correspond to "if p, then q" and "hypothesis, and conclusion". But, I don't understand what kind of statement the following is considered to be:

"The sum of two even integers is always even"

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

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

Thank you for the help.

UPDATE:

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

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

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

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


问:
Iterated function?
$$f(n) = \frac{n}{\lg n}$$
$$g(n) = \min (i \ge 0: f^i(n)\le 2)$$

In other words, $g(n)$ is the number of times $f(n)$ needs to be iterated to reduce $n$ to 2 or less.

What's a tight bound on $g(n)$? $\lg$ is the binary logarithm ($\log_2$)

Thanks

答:
Note that $g$ is nondecreasing and $g(x)= 1+g( f(x))$ for $x>2$.
Also not that $f(x)\le \frac{x}{k}$ if $x\ge 2^k$.
As a consequence, for any $k>1$ and $x>2^k$,
$$ \tag1\begin{align}g(x)&\le g\left(2^k\right)+\min\left\{i\ge 0\biggm| \frac{x}{k^i}\le 2^k\right\}\\&=g(2^k)+\left\lceil\frac{\ln(2^{-k}x)}{\ln k}\right\rceil\\&\le g(2^k)+1-\frac{k\ln 2}{\ln k}+\frac{\ln x}{\ln k}.\end{align}$$
Since $\frac1{\ln k}\to 0$ as $k\to\infty$, we immediately obtain
$$\tag2 g(x)=o(\ln x).$$

Assume $g(t)\le c\frac{\ln t}{\ln \ln t}$ for $T_0\le t\le T$ with $T\ge T_0^4$ and $T_0>2$.
For $x$ with $T<x\le 2T$, we can find $k$ with
$T_0\le 2^k\le T$ and $\sqrt[4] x\le 2^k\le\sqrt x$, i.e. $ \frac 14\ln x\le k\ln 2\le \frac 12\ln x$ and thus $k=\theta\ln x$ with $1>\frac1{2\ln2}\ge\theta\ge \frac1{4\ln 2}>0$. With this we obtain
$$\begin{align} g(x)&\le g(2^k)+1-\frac{k\ln 2}{\ln k}+\frac{\ln x}{\ln k}\\
&\le c\frac{k\ln 2}{\ln k+\ln\ln 2}+1-\frac{k\ln 2}{\ln k}+\frac{\ln x}{\ln k}\\
&\le (c-1)\frac{k\ln 2}{\ln k}+1+\frac{\ln x}{\ln k}\\
&=1+\left((c-1)\theta\ln 2+1\right)\frac{\ln x}{\ln \theta+\ln \ln x}\\
&=1+\frac{c+1}2\cdot\frac{\ln x}{\ln \theta+\ln \ln x}.\end{align}$$
In order to conclude $g(x)\le c\frac{\ln x}{\ln\ln x}$ also for $T\le T\le 2T$ from this (and hence by induction for all $x\ge T_0$), it is sufficient to start with big enough $T_0$ (so that $\ln\theta\ll \ln\ln x$) and $c$ (so that both $\frac{c+1}2<c$ and the $1$ can be absorbed).

We conclude $$ g(x)=O\left(\frac{\ln x}{\ln\ln x}\right).$$

Remark: From plotting/numerical experimentation, it looks like
$$ g(x)\sim\frac{\ln x}{\ln\ln x}.$$
For example, with $x=e^{1.000.000.000}$, we have
$$\frac{g(x)\ln\ln x}{\ln x}=1.03442069\ldots$$

Title: Iterated function?


问:
Given $f(x) = \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Find $f^{(80)}(0)$
I want to find $f^{(80)}(0)$, where $f(x) = \displaystyle \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$.

My attempt:

Let $g(x) = \displaystyle \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Now $f(x)$ can be rewritten as
$$ f(x) = g{\left(x^{20}\right)} - g(0). $$

Taking the derivative $80$ times,
$$ f^{(80)}(x) = \sum_{i = 1}^{80}{a_i \times g^{(i)}{\left(x^{20}\right)}}, $$
where $\{a\}$ is a sequence of integers. I tried to find the sum, but little progress was made.

I noticed that if I plugged $x = 0$ into the equation, all of the "$x$-linked" terms would be canceled. hence, only the first few terms would be significant, which is $\displaystyle \sum_{i = 1}^{4}{a_i \times g^{(i)}{\left(x^{20}\right)}}$.

Another observation is that $a_i$ will become $0$, i.e. $g^{(i)}$ will vanish after 20 times the derivative was taken. Thus, the final answer would be $a_4 \times g^{(4)}{\left(x^{20}\right)}$.

I only managed to find $a_1 = 20!$ for the first 20 derivatives and failed to derive explicit formulas for the other three.

I would like to know how to solve this problem appropriately, thanks in advance.

答:
Let $$F(t) = \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$$
Then, we have $f(x) = F(x^{20})-F(0).$

Differentiating once you get
$$
f'(x) = 20x^{19}F'(x^{20}) = 20x^{19}(\sqrt{x^{20}}\sin\sqrt{x^{20}}) \\
f'(x) = 20x^{29}\sin(x^{10})$$
I hope continuing is not difficult.


答:
It could be the case that you've only studied propositional (also called zeroth-order) logic so far. If so, then we may reword your statement slightly and write something like "if $p$ is even and $q$ is even, then $p+q$ is even too".

Nevertheless, such a formalution may feel lacking in some sense. I believe the culprit here is the word "always". To me this suggests something else is at play. The statement as you give it can't really be fully captured using propositional logic, because it contains an implicit quantifier. This means we have to use predicate logic (also called first-order logic) here. Note that what your statement really says is that for any even integers $p$ and $q$, $p+q$ is also even. Notationally, "for all $x$" is written as $\forall x$. Let's write $E(x)$ to express the property of being even. Hence, in formal notation, your statement becomes: $$\forall p \forall q \big(E(p)\land E(q)\to E(p+q)\big).$$


答:
I think the Mahalanobis metric is perhaps best understood as a weighted Euclidean metric. For example, consider distances in the plane. Then, the Euclidean metric coincides with one's geometric intuition of distance, and the Mahalanobis metric coincides with costliness of traveling along that distance, say, treating distance along one axis as "more expensive" than distance along the other axis.

So, if the distance between two points if 0.5 according to the Euclidean metric but the distance between them is 0.75 according to the Mahalanobis metric, then one interpretation is perhaps that travelling between those two points is more costly than indicated by (Euclidean) distance alone.

Does this answer?

P.S. This interpretation is meaningful in statistics where the general multivariate normal distribution (MVN) can be understood as being derived from the MVN with unit variance by means of a scale transformation: effectively replacing the Euclidean metric in the exponential function with a Mahalanobis metric turns a standard Gaussian distribution into a general one (there is also the mean to consider, but that is secondary).

P.P.S. I'm sure that, in machine learning, there are cases in which us a Mahalanobis metric instead of Euclidean corresponds to more/less heavily penalizing incorrect guesses of one feature than of another. As the other answerer pointed out, Mahalanobis metric absorbs Euclidean as a special case.

Title: Mahalanobis distance connection with Euclidean distance

Body:
My question is more about the meaning of Mahalanobis distance when compared to L2 distance.
As a reminder:

Mahalanobis distance is the distance between a point $x$ and a distribution with mean $\mu$ and covariance $\Sigma$ can be calculated as: $d_{\text{Mahalanobis }} = \sqrt{(x-\mu)^T\Sigma^{-1} (x-\mu)}$

Back to the question:
if the Euclidean distance between a point $p_a$ and $p_b$ is for example 0.5 and the Mahalanobis distance between the point $p_a$ and some other distribution $\beta$ is 0.75, could we state that point $p_a$ is closer to $p_b$ than distribution $\beta$?

If not, then under what conditions we could have this comparison valid?


问:
How to prove that an ordered set is a lattice.
I am given an interval $a,b$ with $a \le x$ and $x \le b$ . How can I prove that this let s call it $( L ,\le)$ is  a lattice ? I know it is probably stupidly simple but I would appreciate some help .

答:
According to @amrsa, your question is incomplete. I've answered this by assuming $(L,\le)$ is a chain.

Use the definition. Let $A=\{x_1,x_2\}\subseteq L$. Then you need to show two things.

$A$ has an infimum in $L$

$A$ has a supremum in $L$

For infimum, you need to show that $\exists i \in L$ such that

$i \le x_1$ and $i \le x_2$

$\forall x\in L$ if $x \le x_1$ and $x \le x_2$, then $x\le i$

For supremum, show that $\exists s \in L$ such that

$x_1\le s$ and $x_2\le s$

$\forall x\in L$ if $x_1\le x$ and $x_2\le x$, then $s\le x$

Hint:

Choose $i=\min(x_1,x_2)$ and $s=\max(x_1,x_2)$ and show that they satisfied the above criteria.


问:
Restriction of two variable function on curve
Definition of restriction

Let $f:\mbox{Dom}(f)\subseteq\mathbb{R}^{2}\to\mathbb{R}$ be a function, $A\subseteq\mbox{Dom}(f)$ be a non empty set. The restriction of $f$ on $A$ is the two variable function $f_{|A}:A\to\mathbb{R}$ such that

$$f_{|A}(x,y)=f(x,y) \ \ \ \forall (x,y)\in A$$

moreover $\mbox{Gr}(f_{|A})=\left\{(x,y,z)\in\mathbb{R}^{3}\ \mbox{t.c.}\ (x,y)\in A, z=f(x,y) \right\}$

Restriction of a function on a graph of one variable function

This definition is not from a book, it is my reworking of theory  (just for fun), so it could be pointless.

Let:

$\bullet \ \ \ f:\mbox{Dom}(f)\subseteq\mathbb{R}^2\to\mathbb{R}$ be a two variable function;

$\bullet \ \ \ g:\mbox{Dom}(g)\subseteq\mathbb{R}\to\mathbb{R}$ be a one variable function;

$\bullet \ \ \ A=\{(x,y)\in\mbox{Dom}(f) \ \mbox{s.t.} \ y=g(x)\}$

The restriction of $f$ on $A$ is the function

$$f_{|A}(x,y)=f(x,g(x))=h(x)$$

with domain

$$\mbox{Dom}(h)=\left\{x\in\mathbb{R} \ \mbox{s.t.} \ x\in\mbox{Dom}(g)\wedge (x,g(x))\in\mbox{Dom}(f)\right\}$$

My questions

From the definition of restriction, $h(x)$ is formally a two variable function, isn't it? Moreover is that true that the graph of $h$ is

$$\mbox{Gr}(h)=\left\{(x,y,z)\in\mathbb{R}^{3} \ \mbox{t.c.} \ x\in\mbox{Dom}(h),\, y=g(x),\, z=h(x)\right\}?$$

Can I say that $h(x)$ is a one variable function? If so, I think I can define the graph of $h$ on $Oxz$ as follow: $$\mbox{Gr}(h)=\left\{(x,z)\in\mathbb{R}^{2} \ \mbox{t.c.} \ x\in\mbox{Dom}(h), z=h(x)\right\}$$ right?

I'm really confused.

答:
Ok, I'll try to answer my own questions. Please, check that it is correct.

In strict terms, $f_{|A}(x,y)$ is function of two variables $(x,y)=(x,g(x))$, however $y$  is not an indipendent variable: its value depends on $x$.

In addition, the graph of $f_{|A}$ is a subset of $\mathbb{R}^{3}$.

The function $h:\mbox{Dom}(h)\to\mathbb{R}$ is not the restriction of $f$ to $A$, but the function of one variable which associates $x\in\mbox{Dom}(h)$ the real number $h(x)=f(x,g(x))$, and it's not equivalent to $f_{|A}(x,y)$: their graphs aren't even comparable.


问:
Find the general solution of $x^3y^{\prime \prime \prime} + y = 0$
I tried to use the Frobenius technique, but I am unable to find the solution by infinite series.</p>

答:
From the format the more general ODE $x^ny^{(n)}+y=0$ it is clear that there are numbers $r$ such that $y=x^r$ is a solution.

$$y^{(n)}=r(r-1)\cdots(r-n+1)x^{r-n}$$

Thus the values of $r$ that work obey $r(r-1)\cdots(r-n+1)+1=0$. This is an equation of degree $n$. It only needs to be checked that the $n$ roots are all simple (not always the case, e.g. $n=4$) since it is pretty straight forward to use the Wronskian to show that the $n$ power functions we get are the basis of the general solution for $\mathbb{C}\mapsto\mathbb{C}$ functions. If any pair of roots are $r_{\pm}=a\pm bi\notin\mathbb{R}$ and we want to solve for $\mathbb{R}\mapsto\mathbb{R}$ functions then observe $x^{r_\pm}=x^a\cos(b\ln x)\pm ix^a\sin(b\ln x)$ so that instead of the two basis elements $x^{r_\pm}$ you will use $x^a\cos(b\ln x)$ and $x^a\sin(b\ln x)$.

Therefore if $r_1,\cdots,r_n$ are the solutions to the polynomial equation then the general solution to the ODE is $y=\sum_{k=1}^nC_kx^{r_k}$.

$n=1$: $y=C_1\frac{1}{x}$

$n=2$: $y=\frac{1}{\sqrt{x}}\left(C_1\cos\left(\frac{\sqrt{3}}{2}\ln x\right)+C_2\sin\left(\frac{\sqrt{3}}{2}\ln x\right)\right)$

$n=3$: $y=C_1x^{-r}+C_2x^{r/2}(C_2\cos(b\ln x)+C_3\sin (b\ln x))$ where

$$r=\frac{\sqrt[3]{108+12\sqrt{69}}+\sqrt[3]{108-12\sqrt{69}}}{6}$$
$$b=\frac{\left(\sqrt[3]{108+12\sqrt{69}}-\sqrt[3]{108-12\sqrt{69}}\right)\sqrt{3}}{12}$$

etc.


问:
Calculate integral $\int_{}^{} \frac{1}{\sqrt{1 - \cos x}} \,dx$
The integral is

$$\int_{}^{} \frac{1}{\sqrt{1 - \cos x}} \,dx$$

I've made a bit of progress below, but I'm stuck as to how to move forward.

Using some help, I've used $1 - \cos x = 2\sin^{2} \frac{x}{2}$ to get that this integral becomes
$\frac{1}{\sqrt{2}} \int_{}^{} \frac{1}{\sin \frac{x}{2}} dx$. Because $\sin \frac{x}{2} = \frac{2\tan \frac{x}{4}}{1 + \tan^{2} \frac{x}{4}}$, we get it to be $\frac {1}{\sqrt{2}} \int_{}^{} \frac{1 + \tan^{2} \frac{x}{4}}{2\tan \frac{x}{4}} dx$. I would like some help as to how to proceed from here.

答: The evaluation of the antiderivatives of $\sec x$ and $\csc x$ are frequently taught as follows:  $$\sec x = \frac{\sec x (\tan x + \sec x)}{\tan x + \sec x} = \frac{\sec x \tan x + \sec^2 x}{\tan x + \sec x},$$ and since $$\frac{d}{dx}[\tan x] = \sec^2 x, \quad \frac{d}{dx}[\sec x] = \sec x \tan x,$$  we get with the substitution $$u = \tan x + \sec x, \quad du = (\sec x \tan x + \sec^2 x) \, dx$$  $$\int \sec x \, dx = \int \frac{1}{u} \, du = \log | \tan x + \sec x | + C.$$  A similar approach applies with $\csc x$:  $$\int \csc x \, dx = - \int \frac{-\csc x \cot x - \csc^2 x}{\cot x + \csc x} \, dx = - \log | \csc x + \cot x | + C.$$


问:
Heat equation initial and terminal conditions
I have a question that is bugging me and for which i would appreciate some help or to be pointed out to the litterature on the subject.

Is it possible to solve the heat equation with setting an initial and terminal condition?

that is

$$
\frac{\partial f}{\partial t} = \alpha \frac{\partial^2 f}{\partial x^2}\\
f(0,x)=g(x)\\
f(T,x)=h(x)\\
for \: t \in [0,T], \: x \in \mathbb{R}
$$

Thank you for your help!

答: No. Solutions of the heat equation (if say to limit oneself with bounded functions $u$) is totally defined by the inicial function $g$, so the final function $h$ is determined uniquely by $g$.


问:
Interior of Self-Intersecting Curve
Is there a natural analogue of the notion of the interior (and thus exterior) for a closed curve that is self-intersecting? For some curves I’ve come up with I can visualise it, as shown in the diagram below, and might have a definition. Red depicts the interior and blue the exterior, arrows indicate the nature of the curve as I think multiple curves with the same image could have different interiors.

I think for a closed curve $f:S^1\to\mathbb{R}^2$ you could define it as $\bigcap\left\{\textrm{range}\left(f’\right)\ \middle|\ f’:D^2\to\mathbb{R}^2\land f’|_{S^1}=f\right\}\setminus\textrm{range}\left(f\right)$. However, I don’t know if this set corresponds to the interior for non-intersecting curve, or if it is even non-empty for non-degenerate curves.

答: Here is a 14-sided polygon whose interior has been defined in several different ways (the list is not exhaustive). Note not only the variety of fillings or densities which this leads to, but also the different calculations of its area which arise in consequence. There is no "right" answer, there is only the one which best serves your current purpose.

Title: Interior of Self-Intersecting Curve


问:
Sum of symbol values in wff written in polish notation.
So I am currently dealing with a problem like this:

If we count $\rightarrow , \land , \lor$ and $\leftrightarrow$ each as $+1$ , each statement letter as $-1$ and $\lnot$ as $0$ , prove that an expression $\mathscr B$ in this parenthesis-free notation is a statement form if and only if $(i)$ the sum of the symbols of $\mathscr B$ is $-1$ and $(ii)$ the sum of symbols in any proper initial segment of $\mathscr B$ is nonnegative.

So , if A = "an expression $\mathscr B$ in this parenthesis-free notation is a statement form"

We have to show the following wff to be true,
$$A \leftrightarrow ((i) \land (ii))$$

Here , I assume that "parenthesis free notation" is just polish notation

I was able to prove that $A \rightarrow (i)$ the same way wff are defined with a recursive algorithm (Not sure if mine was valid though because it is polish notation). But couldn't figure out How to prove $A \leftarrow (i)$ .

I have no idea How to prove something like $A \leftrightarrow (ii)$ with a recursive approch because there is also proper initial segments imvolving . At this point , proving $A \leftrightarrow ((i) \land (ii))$ seems to be far for me.

Is there something beyond recursion that I need?

答:
The forward implication is indeed proved by induction. It’s clear that $(i)$ and $(ii)$ hold for a single statement letter. If they hold for a statement form $\varphi$, it’s easy to see that they hold for $\neg\varphi$:

The sum of the symbols is unchanged by prefixing $\neg$.

Every proper initial segment of $\neg\varphi$ is of the form $\neg\sigma$, where $\sigma$ is empty or a proper initial segment of $\varphi$. If $\sigma$ is empty, the sum is $0$, and if not, the sum is the same as that of $\sigma$ and so is non-negative.

Now suppose that $(i)$ and $(ii)$ hold for statement forms $\varphi$ and $\psi$, and let $\diamondsuit$ be one of the binary connectives $\to$, $\land$, $\lor$, and $\leftrightarrow$; we want to show that $\diamondsuit\varphi\psi$ satisfies $(i)$ and $(ii)$. The sum of the symbols in $\diamondsuit\varphi\psi$ is $1+(-1)+(-1)=-1$, so $\diamondsuit\varphi\psi$ satisfies $(ii)$. Now consider a proper initial segment $\diamondsuit\sigma$ of $\diamondsuit\varphi\psi$.

If $\sigma$ is empty, the sum of the segment is $1$.

If $\sigma$ is a proper initial segment of $\varphi$, its sum is non-negative, so the sum of $\diamondsuit\sigma$ is positive.

If $\sigma=\varphi$, the sum of $\diamondsuit\sigma$ is $1+(-1)=0$.

If $\sigma=\varphi\tau$, where $\tau$ is a proper initial segment of $\psi$, the sum of $\diamondsuit\varphi$ is $0$, and the sum of $\tau$ is non-negative, so the sum of $\diamondsuit\sigma$ is also non-negative.

Thus, $\diamondsuit\varphi\psi$ satisfies $(i)$. This shows that every statement form satisfies $(i)$ and $(ii)$.

For the other direction you cannot hope to prove that $(i)$ implies $A$: it’s the conjunction of $(i)$ and $(ii)$ that implies $A$. This can be proved by induction on the length of the string of symbols. I’ll sketch the argument, leaving many of the details to you.

Show that $\sigma$ satisfies $(i)$ and $(ii)$ and has length $1$, it must be a single statement letter and hence a statement form.

Now assume that $\sigma$ has at least two symbols and that every shorter string that satisfies $(i)$ and $(ii)$ is a statement form. There are three possibilities for the first symbol of $\sigma$.

Show that if $\sigma$ satisfies $(i)$ and $(ii)$, its first symbol cannot be a statement letter.

Show that if $\sigma=\neg\tau$, then $\tau$ satisfies $(i)$ and $(ii)$. The induction hypothesis then implies that $\tau$ is a statement form, and that in turn implies that $\sigma$ is a statement form.

The third possibility is that $\sigma=\diamondsuit\tau$; this is the hardest case.

Show that $\sigma$ must have a proper initial segment whose symbols sum to $0$; you’ll use the fact that $\sigma$ satisfies $(i)$.

Let $\diamondsuit\varphi$ be the shortest such segment.

Verify that $\varphi$ satisfies $(i)$ and $(ii)$ and conclude that it must be a statement form. (Why?)

Let $\psi$ be the remainder of $\sigma$, so that $\sigma=\diamondsuit\varphi\psi$. Verify that $\psi$ also satisfies $(i)$ and $(ii)$ and so must be a statement form, and conclude that $\sigma$ is a statement form.

By induction, therefore, every string of symbols satisfying $(i)$ and $(ii)$ is a statement form.


问:
A sum with binomial coefficients
Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$</p>

答:
The only thing I have to add on @robjohn's answer is the derivation of the identities used:

We have that

$\displaystyle\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^j=\frac{d^{j}}{d x^j}\left[(e^x-1)^n\right]\Big|_{x=0}=\frac{d^{j}}{d x^j}\left[x^n+\frac{n}{2}x^{n+1}+\frac{n(3n+1)}{24}x^{n+2}+\mathcal O(x^{n+3})\right]\Bigg|_{x=0}$,

so (for $0\leq j<n$ it is $\displaystyle\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^j=0$,)

for $j=n$ it is $\displaystyle\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^{n}=n!$,

for $j=n+1$ it is $\displaystyle\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^{n+1}=\frac{(n+1)!n}{2}$ and

for $j=n+2$ it is $\displaystyle\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^{n+2}=\frac{(n+2)!n(3n+1)}{24}$,

and that the sums on these identities are related to the Stirling numbers of the second kind (http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind), since they satisfy
$\displaystyle{\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k^j=n!S(j,n)}$.

One can also check out here (http://www.mathematica.gr/forum/viewtopic.php?f=59&t=36932) for a derivation using counting arguments.

Title: A sum with binomial coefficients


问:
How to rewrite $\frac{dx}{dt}=a\cdot x$ into the form $\frac{dx}{x}=a\cdot dt$ using the chain rule
I'm reading about ODE's in a calculus text. The goal is to solve the differential equation: $\frac{dx}{dt}=ax$ for $x(t)$. The text says using the chain rule and substitution rewrite the equation as $\frac{dx}{x}=a\:dt$.

I'm not sure how to use the chain rule to do the rewriting. However after this step I was able to follow to the solution.

答:
To see how this works in a general context, observe that the first order differential equation is separable and is of the form

$$ \frac{dx}{dt}=F(t)G(x). $$
To solve this equation we can divide by $G(x(t))$ to get
$$\frac{1}{G(x(t))}\frac{dx}{dt}=F(t).\tag{1}$$
Then we can find a function $H(x)$ whose derivative with respect to $x$ is
$$H'(x)=\frac{1}{G(x)}\quad\left(\text{solution: $H(x)=\int \frac{dx}{G(x)}$}\right).\tag{2}$$
The chain rule then implies that the left hand side in $(1)$ can be written as
$$\frac{1}{G(x(t))}\frac{dx}{dt}=H'(x(t))\frac{dx}{dt}=\frac{dH(x(t))}{dt}.$$
Therefore $(1)$ is equivalent to
$$\frac{dH(x(t))}{dt}=F(t).$$
In your example
$$F(t)=a,\quad G(x)=x.$$
Therefore
$$H'(x)=\frac{1}{x},\quad a=F(t)=\frac{d}{dt}\left(H(x(t))\right)=\frac{d}{dt}\left(\ln(x(t))\right)=\frac{1}{x}\frac{dx}{dt},$$
which implies
$$\frac{dx}{x}=a\,dt.$$
It is more intuitive to observe that the first order differential equation
$$\frac{dx}{dt}=ax\tag{3}$$
is separable. Thus, it can be rewritten as
$$\frac{dx}{x}=a\,dt.\tag{4}$$


问:
Partial fraction decomposition algorithm
So I was faced by this problem: If $p,q \in K[x]$, with $q = q_{0}^{\alpha_{0}} \dots  q_{m}^{\alpha_{m}}$, all of $q_{i}$ irreducible and $\alpha_{i}$ natural numbers. So, there is some $p_{0} \dots p_{m} \in K[x]$ where
$$
\frac{p}{q}=\sum_{j = 0}^{m}\frac{p_{j}}{q_{j}^{\alpha_{j}}}.
$$

So, at first sight, it looks a lot like the partial fractions decomposition, which we use a lot in Calculus, but I can't prove this. Any help, I will be appreciated. Thanks in advance!

答: After some research, I've found a text that explain this. You may click here. And look for propositions $3.5$ and $3.6$.


问:
On the existence of sets
It is an important question whether mathematical objects in fact exist. Certainly, we can't locate them in the physical world. I want to know whether sets exist. If they in fact do not exist, then all the axioms of ZFC are either false or vacuous. So, how do we establish that sets exist in the first place?</p>

答: All formulations of ZFC imply that at least one set exists, either by implicit semantic rules of first order logic or by stating it directly.</p>

Title: On the existence of sets


问:
Does an integrable function vanish at infinity in a suitable sense?
I'm searching for a generalization of the following observation: If $f\in L^2(\mathbb R)$, then \begin{equation}\begin{split}\int_{[-n,\:n+1]}f(x)\:{\rm d}x&=\left\langle1_{[-n,\:n+1]},1_{\mathbb R\setminus[-n,\:n)}g\right\rangle_{L^2(\mathbb R)}\\&\le\left\|f-1_{[-n,\:n)}f\right\|_{L^2(\mathbb R)}\xrightarrow{n\to\infty}0\end{split}\tag1\end{equation} by the Cauchy-Schwarz inequality and Lebesgue's dominated convergence theorem.

$(1)$ captures the idea that an integrable function should "vanish at infinity" in a suitable sense.

Can we generalize this result to $f\in L^p(\mu)$, where $(E,\mathcal E,\mu)$ is an arbitrary measure space and $p\ge1$? Maybe assuming that $\mu(E)=\infty$ and/or that $E$ is a metric/normed space with $\mathcal E=\mathcal B(E)$?

In the case of a metric/normed space, the intuition tells me that an integrable function should "essentially be supported in a compact set".

What's obviously true is that $$\forall f\in L^p(\mu):\not\exists B\in\mathcal E:\mu(B)=\infty\text{ and }\inf_{x\in B}|f(x)|>0\tag2.$$

答:
A (continuous, say) function $f:\Bbb R^d\to\Bbb R$ is said to vanish at infinity provided for each $\epsilon>0$ there is a compact set $K$ such that $|f(x)|\le\epsilon$ for all $x\in K^c$.

A measure theoretic form of this, for a function $f\in L^1(\mu)$ (with $(X,\mathcal X,\mu)$ an arbitrary measure space) follows from Markov's inequality: Given $\epsilon>0$ choose $\delta :=\|f\|_1/\epsilon$ and define $K:=\{x\in X: |f(x)|> \delta\}$. Then (Markov)
$$
\mu(K)\le {\|f\|_1\over\delta}=\delta<\infty.
$$
Here "set of finite $\mu$-measure" plays the role of compact set, and $|f(x)|\le\epsilon$ for $x$ in the complement of $K$.


问:
Show that $A \subseteq B \Leftrightarrow \overline{A} \cup B = U$
I'd like to show that if $A \subseteq B \Leftrightarrow \overline{A} \cup B = U$

I'd like to understand the way I am suppose to approach set theory challenges. Since this is not something intuitive I don't know where to start when approaching such challenges.

Apart from not being sure how to start, since there is a $\Leftrightarrow$ I suppose that I need to provide in both directions, am I correct? Please help with the right reasoning for this proof.

Updated:
U here is the universe of discourse

答:
Yes, you have to prove both implications. For the forward implication, suppose that $A\subseteq B$; we want to show that $\overline A\cup B=U$. Clearly $\overline A\cup B\subseteq U$, so we really only have to show that $U\subseteq\overline A\cup B$. To do that, let $x\in U$. If $x\in B$, then certainly $x\in\overline A\cup B$, so suppose that $x\notin B$. Then $x\notin A$, since $A\subseteq B$, so $x\in\overline A$, and again we have $x\in\overline A\cup B$. Thus, $U\subseteq\overline A\cup B$, and we conclude that $\overline A\cup B=U$.

For the other direction I suggest proving the contrapositive instead: show that if $A\nsubseteq B$, then $\overline A\cup B\ne U$. Use the fact that if $A\nsubseteq B$, then there is some $a\in A\setminus B$.


问:
How can I prove this question about unbounded?
Let $K= \{x: Ax\leq b, x\geq0\}, b\geq 0$. Show that if there is a nonzero solution $x\geq 0$  of $Ax\leq 0$, then $K$ is unbounded. ($A$ is $m \times n$, $x$ is $n \times 1$ matrix, $b$ is $m \times 1$ matrix)?

答: Let $u$ be a non-zero solution to $x\geq0, Ax\leq0$. Note that $\alpha u \in K$ for any scalar $\alpha > 0$. Thus, the half-line $L = \{\alpha u | \alpha > 0\}$ is an unbounded subset of $K$. Therefore, $K$ is unbounded.


答:
Note
$$\int\frac1{\sin\frac x2}dx=\int\frac1{2\cos\frac x4 \sin\frac x4}dx
=\frac12 \int\frac{\sec^2\frac x4}{\tan\frac x4 }dx=2\ln(\tan\frac x4)+C
$$
Thus
$$\int_{}^{} \frac{1}{\sqrt{1 - \cos x}} \,dx=
\frac{1}{\sqrt{2}} \int_{}^{} \frac{1}{\sin \frac{x}{2}} dx
=\sqrt2 \ln(\tan\frac x4)+C
$$

Title: Calculate integral $\int_{}^{} \frac{1}{\sqrt{1 - \cos x}} \,dx$

Body:
The integral is

$$\int_{}^{} \frac{1}{\sqrt{1 - \cos x}} \,dx$$

I've made a bit of progress below, but I'm stuck as to how to move forward.

Using some help, I've used $1 - \cos x = 2\sin^{2} \frac{x}{2}$ to get that this integral becomes
$\frac{1}{\sqrt{2}} \int_{}^{} \frac{1}{\sin \frac{x}{2}} dx$. Because $\sin \frac{x}{2} = \frac{2\tan \frac{x}{4}}{1 + \tan^{2} \frac{x}{4}}$, we get it to be $\frac {1}{\sqrt{2}} \int_{}^{} \frac{1 + \tan^{2} \frac{x}{4}}{2\tan \frac{x}{4}} dx$. I would like some help as to how to proceed from here.


问:
Seeking references on the geometry of regular four-dimensional polytopes
I've got intrested in the geometry of higher dimensional spaces. Now I'm studying four dimensional geometry. I'm especially interested in regular four dimensional polytopes and their visalization. I've found some intresting information here and here.

I am trying to find some literature focused on the geometry of the polytopes itself but not on the visalization only. Has someone been interested in the topic? What literature can you advice me to read?

答:
The standard reference work is still Coxeter's Regular Polytopes (e.g. Dover edition).

He discusses the two-, three- and four-dimensional examples in some detail, as well as the few which have analogues in higher dimensions still.

He also includes useful background remarks, especially historical detail, and photos of some nice wireframe models made by another investigator.

However, if your interest arises more from their fourth dimension than their regularity, then this is not the best resource. There are innumerable uniform polytopes of four dimensions (A project existed in recent years to enumerate them all, I am not sure if it has succeeded yet) and even more with lower symmetries.


问:
Fixed points of an action of a group $G$ on a set $A$, where $|G|=pq$ and $|A|=p^2-2$, with $p$ and $q$ twin primes.
Let $p$, $q$ be two twin primes, that is such that $q=p+2$.

Let $G$ be a group of order $pq$. Let $A$ be set of order $p^2-2$.

Suppose there is an action

$G \times A \to A$

$(g,a) \mapsto g \cdot a $

Prove that $Fix(G)$ is non-empty, where $Fix(G)=\{ a\in A| g \cdot a=a \,,\forall g \in G \}$

Since there are primes involved, I tried something with Sylow theorems:

Let $n_p(G)$ and $n_q(G)$ be, respectively, the number of $p$-Sylow and $q$-Sylow of $G$.

We must have, because $|G|=pq$, that

$n_p(G) \equiv 1\bmod q\;$,$\;n_q(G) \equiv 1\bmod p$

and

$n_p(G)|q\;$,$\;n_q(G)|p$.

This means that $n_p(G)=n_q(G)=1$. Let $Syl_p(G)=\{P\}$ and $Syl_q(G)=\{Q\}$ the set of $p$-Sylow and $q$-Sylow of G. Let's consider $P$. Since all of $p$-Sylow are conjugated, it means that $\forall g \in G$, we must have $P^g=P$, because $P$ is the only $p$-Sylow in $G$. It follows that $P$ is a normal subgroup of $G$. The same applies to $Q$.

Let's now consider the set $PQ=\{xy|x \in P, y \in Q\} \subseteq G$.

Since both $P$ and $Q$ are normal in $G$, we have that $PQ \leqslant G$.

Because $|P|=p$ and $|Q|=q$, with $p$, $q$ primes, we have that $P \cap Q = \{1\}$.

We know that $|PQ|= \frac{|P||Q|}{|P \cap Q|}$, since everything is finite. By substituting what we know, we get

$\;$$|PQ| = \frac{|P||Q|}{|\{1\}|} = \frac{pq}{1} = pq = |G|$

Thus $PQ \leqslant G$ and $|PQ|=|G|$, it must be then $PQ=G$.

Putting together these facts, we can conclude that $G \simeq P \times Q$.

Now, we know that $|P|=p$ and $|Q|=q$, which means that $P \simeq C_p$ and $Q \simeq C_q$, where $C_p$ and $C_q$ are the cyclic groups of order $p$ and $q$, respectively. But $(p,q)=1$, because of primality, it follows then that $C_p \times C_q \simeq C_{pq}$. We can write then $G \simeq C_{pq}$. Since $C_{pq}$ is cyclic, it must be abelian, which implies that $G$ must be abelian.

This is what I could do for now.

I think I should use this formula in some way, but I don't really know how

$$|A|=|Fix(G)|+\sum_{\delta \in \Delta} |G:Stab_G(\delta)|$$

Where $\Delta$ is the set containing the representatives of the non-trivial orbits of $G$, for its action on $A$, $Stab_G(\delta)=\{g \in G|g \cdot \delta = \delta \}$ and $|G:Stab_G(\delta)|$ is the index of $Stab_G(\delta)$ in $G$ which, thanks to the Orbit-Stabilizer Theorem, is equal to $|\mathcal O(\delta)|$, the order of the orbit of $\delta$, $\forall \delta \in \Delta$.

答:
You are almost there.
$$|A|=|\text{Fix}(G)|+\sum_{\delta \in \Delta} |G:\text{Stab}_G(\delta)|$$

Here, suppose for a contradiction that $|\text{Fix(G)}| = 0$. Then, we have
$$|A| = p^2-2 =\sum_{\delta \in \Delta} |G:\text{Stab}_G(\delta)| $$
But as you suggested, by Orbit Stabilizer Theorem, $|\mathcal O(\delta)|$ must divide $|G| = p(p+2)$. So, we have $|\mathcal O(\delta)| =p,p+2\text{ or }p(p+2), \forall \delta \in \Delta$. Notice that $|\mathcal O(\delta_0)| = p(p+2)$ for some $\delta_0 \in \Delta$ reaches to a contradiction since $p(p+2) > p^2-2 = |A|$. So, there are only two possibilities left for $|\mathcal O(\delta)|$, which are $p$ and $p+2$.

Now, suppose there are $a$ many orbits of size $p$, $b$ many orbits of size $p+2$. Then, we must have

$$ap+b(p+2) = p^2-2 \implies ap+bp+2b+2 = p^2 \implies p \mid (2b+2)$$

Here, since $p$ and $q$ are twin primes, $p \ne 2$. So, by Euclidean Lemma,
$$p \mid 2(b+1) \implies p \mid (b+1)$$

Now, $b < p$ since otherwise, we would have $b(p+2) \ge p^2+2p > p^2-2 = |A|$. Thus, only possibility is $b = p-1$. Then we have
$$ap +b(p+2) = ap + p^2+p-2 =p^2-2 \implies p(a+1) = 0$$

This is a contradiction.


问:
gre division with remainder problem
A certain number of toys were packed into $x$ boxes so that each box contained the same number of toys, with no toys left unpacked. If $3$ fewer boxes had been used instead, then $12$ toys would have been packed in each box, with $5$ toys left unpacked. What's the value of $x$?

$A. 11$

$B. 14$

$C. 28$

$D. 31$

$E. 34$

I'm not sure If I interpreted the problem correctly, but my approach as I will assume(guess) $x = 10$, then with $3$ fewer boxes, $x = 7$, so we packed $12$ toys in each box, each box has $1$ toy, left $5$ toys left unpacked. But none of these answer choices is $10$, so pretty sure I'm wrong.

I feel like this is just remainder problem, any help is highly appreciated.

答:
Hi first time answering a question so bare with me

folowing from lulu's reponse you need

$$y=12(x-3)+5=12x-31$$

where $y$ denotes the total number of toys. To be divisible by $x$ one way to look at this is that

$$\frac{12x-31}{x}=12-\frac{31}{x}$$

Must be an integer.

The only number on that list which satisfies this is 31. So the answer is D.

Title: gre division with remainder problem


答:
Here it is an alternative proof for the sake of curiosity.

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

If $A\subseteq B$, then $A - B = A\cap B^{c} = \varnothing$. Consequently, one has that
\begin{align*}
U & = (A - B)\cup(A - B)^{c}\\\\
& = (A\cap B^{c})\cup(A\cap B^{c})^{c}\\\\
& = \varnothing\cup(A^{c}\cup B)\\\\
& = A^{c}\cup B
\end{align*}
and we are done.

Now we shall prove the converse implication $(\Leftarrow)$.

Indeed, one has that
\begin{align*}
U = A^{c}\cup B & \Rightarrow A\cap U = A\cap(A^{c}\cup B)\\\\
& \Rightarrow A = (A\cap A^{c})\cup(A\cap B)\\\\
& \Rightarrow A = \varnothing\cup(A\cap B)\\\\
& \Rightarrow A = A\cap B\\\\
& \Rightarrow A\subseteq B
\end{align*}

and we are done.

Hopefully this helps!

Title: Show that $A \subseteq B \Leftrightarrow \overline{A} \cup B = U$

Body:
I'd like to show that if $A \subseteq B \Leftrightarrow \overline{A} \cup B = U$

I'd like to understand the way I am suppose to approach set theory challenges. Since this is not something intuitive I don't know where to start when approaching such challenges.

Apart from not being sure how to start, since there is a $\Leftrightarrow$ I suppose that I need to provide in both directions, am I correct? Please help with the right reasoning for this proof.

Updated:
U here is the universe of discourse


问:
Inequality for infimum over intersection of sets
Let $f$ be a function $f: \mathbb{R}^n \to \mathbb{R}$, and let $A,B \subset \mathbb{R}^n$.

Is the following in general true? $$\inf_{x \in A} f(x) \leq \inf_{x \in A\cap B} f(x)$$

答:
I suppose that for "taking values" you mean the domain of the function, otherwise the notion is not well defined in general (you may solve the problem defining orders on vectors in the sets $f(A),f(B)$ you are considering in the codomain.)

It comes straight from the definition of infimum. As in one comment is mentioned, the infimum of a (weakly) bigger set is for sure (weakly) smaller than the infimum of the other set. Since $A \cap B \subseteq A$, the claim follows.

Recall: The infimum of a set $S$ is defined as the maximum of the lower bounds of the set $A$.


问:
Cauchy problem $xu_x+yu_y=u+1$ if $u(x,x^2)=x^2$ (concern on the restrictions)

Solve  the Cauchy problem:
$$xu_x+yu_y=u+1,$$
if $u(x,x^2)=x^2$ for all $x\in \mathbb{R}.$

Attempt. Using the method of characteristics (as in What is the geometric interpretation of the solution to PDE $xu_x+yu_y=0$)
we reach $u(x,y)=x\,f(y/x)-1$ for some $C^1$ function $f$. Since $u(x,x^2)=x^2$, we have $\displaystyle xf(x)-1=x^2$, so $f(x)=x+\frac{1}{x}$ and:
$$u(x,y)=y+\frac{x^2}{y}-1.$$

which is the valid solution, since it satisfies the PDE (just plug it in) and $u(x,x^2)=x^2$.

My main concern is the following: I have seen answers that finish here. On the other hand, we have the restriction that $y$ should be non-zero. So

how may we speak about the solution to the initial problem since we exclude the $y$-axis?

Why the initial problem was not given as: $xu_x+yu_y=u+1$ for $y\neq 0$,
if $u(x,x^2)=x^2$ for all $x\in \mathbb{R}$, in order to avoid the above? (the arguments, as given, would then be enough).

is it a general practice in such problems to be a bit "loose" at this point, regarding the
restrictions? (since i have seen solutions in such problems, that neither inserts restrictions on the initial problem, neither comment on the restrictions on the "solution", when they appear).

All the above hold, unless the initial problem is equivalent to the following one:

Solve, wherever it can be solved, the Cauchy problem:
$$xu_x+yu_y=u+1,$$
if $u(x,x^2)=x^2$ for all $x\in \mathbb{R}.$

Thanks in advance for the help.

答:
This is not an "initial value problem" in the sense that your data are not given on the $x$-axis. It is a Cauchy problem with data on the parabola $y=x^2$. All points on this parabola except for $(0,0)$ are non-characteristic points and then by Kovalevskaya's Theorem a solution exists in a neighborhood of each of them, and it is given by your formula. Observe that the neighborhood shrinks as you approach the origin. You cannot issue a characteristic from $(0,0)$. As for your second point, I would say that would be more appropriate since only under such assumption the method of characteristics can be applied.

Definitely, statements for this kind of problems are usually loose. Probably because it would be annoying to specify all the details,


问:
Does a basis $\mathcal{H} \subseteq \mathbb{R}^3$ exist?
I'm having some trouble with the following question:

Let $f:\Bbb{R}^3 \to \mathbb{R}^3$, $f(x,y,z)=(x-y,y-z,x-z)$ be a linear map. Let $\mathcal{H}$ be a basis for $\mathbb R^3$, such that the matrix of $f$ with respect to the basis $\cal H$ is:
$$M(f,\mathcal{H},\mathcal{H})=\left( \begin{matrix}
2 & -2 & -2 \\
-2 & 2 & 2\\
2 & -2 & -2
\end{matrix}\right)$$
Is it possible for such basis $\mathcal H$ to exist?

I confess I'm a little stuck in this problem. I see no reason for it not to exist but I can't prove that it exists. How can I do this?

答: Hint: The trace of a linear map with respect to a basis does not depend upon the basis.


问:
Given $f(x) = \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Find $f^{(80)}(0)$
I want to find $f^{(80)}(0)$, where $f(x) = \displaystyle \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$.

My attempt:

Let $g(x) = \displaystyle \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Now $f(x)$ can be rewritten as
$$ f(x) = g{\left(x^{20}\right)} - g(0). $$

Taking the derivative $80$ times,
$$ f^{(80)}(x) = \sum_{i = 1}^{80}{a_i \times g^{(i)}{\left(x^{20}\right)}}, $$
where $\{a\}$ is a sequence of integers. I tried to find the sum, but little progress was made.

I noticed that if I plugged $x = 0$ into the equation, all of the "$x$-linked" terms would be canceled. hence, only the first few terms would be significant, which is $\displaystyle \sum_{i = 1}^{4}{a_i \times g^{(i)}{\left(x^{20}\right)}}$.

Another observation is that $a_i$ will become $0$, i.e. $g^{(i)}$ will vanish after 20 times the derivative was taken. Thus, the final answer would be $a_4 \times g^{(4)}{\left(x^{20}\right)}$.

I only managed to find $a_1 = 20!$ for the first 20 derivatives and failed to derive explicit formulas for the other three.

I would like to know how to solve this problem appropriately, thanks in advance.

答: We have $f'(x) = 20x^{19}x^{10}\sin(x^{10}) = 20x^{29}\sin(x^{10})$. What do the first few terms of the Taylor series of $\sin$ tell you?


问:
Fourier transform of complex exponential
I have been assigned to make the Fourier transform of the following equation:
$$\frac{d g(t)}{d t}+A g(t)=B f(t)$$
where:
$$f(t)=\left\{\begin{aligned}
0, & \quad t<0 \\
e^{i \omega t}, & \quad t \geq 0
\end{aligned}\right.$$

Making the Fourier transform of the left-hand side gives:
$$\begin{array}{l}
g(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \hat{g}(\omega) e^{i \omega t} d \omega \Rightarrow \frac{d g}{d t}=\frac{1}{2 \pi} \int_{-\infty}^{\infty} i \omega \hat{g}(\omega) e^{i \omega t} d \omega \\
\Rightarrow F[g(t)]=\hat{g}(\omega), \quad F\left[\frac{d g}{d t}\right]=i \omega \hat{g}(\omega)
\end{array}$$
However, the Fourier transform of $f(t)$ does not seem to converge. $f(t)$ is supposed to represent an oscillating field and I am doing the Fourier transform to find an expression for $g(t)$ "after the transient term".

答:
$$f(t)= e^{i \omega t}1_{t >0}=\lim_{a\to 0^+}f_a(t), \qquad f_a(t)=e^{(i \omega-a) t}1_{t >0}$$
The Fourier transform of $f_a$ converges without trouble $$F[f_a](w) = \frac1{a-i(w-\omega)}$$
But $f$ is not $L^1$ nor $L^2$ and its Fourier transform integral diverges.

The problem is to understand $\lim_{a\to 0^+} F[f_a]$ which converges in the sense of distributions but not in $L^2$ nor in $L^1$.

For a Schwartz function $\phi$ then $$\lim_{a\to 0^+} \int_{-\infty}^\infty \frac1{a-i(w-\omega)}\phi(w)dw=\lim_{h\to 0^+} \int_{-\infty}^{-h}+\int_h^\infty \frac{\phi(w)}{-i(w-\omega)}dw+\pi\phi(\omega)$$
Thus $$F[f](w) = PV(\frac1{-i(w-\omega)})+\pi \delta(w-\omega)$$
($PV$ for principal value)


答:
First, let $\cal B$ denote the standard basis. With respect to this basis, we note that we have

$$A := M(f, \cal B, \cal B) = \begin{bmatrix}1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & -1\end{bmatrix}.$$
Note that $A$ has trace $1$. However, the matrix in your question has $2$. Since trace is invariant under similarity and change of basis gives similar matrices, we see that the given matrix is not possible.

Title: Does a basis $\mathcal{H} \subseteq \mathbb{R}^3$ exist?

Body:
I'm having some trouble with the following question:

Let $f:\Bbb{R}^3 \to \mathbb{R}^3$, $f(x,y,z)=(x-y,y-z,x-z)$ be a linear map. Let $\mathcal{H}$ be a basis for $\mathbb R^3$, such that the matrix of $f$ with respect to the basis $\cal H$ is:
$$M(f,\mathcal{H},\mathcal{H})=\left( \begin{matrix}
2 & -2 & -2 \\
-2 & 2 & 2\\
2 & -2 & -2
\end{matrix}\right)$$
Is it possible for such basis $\mathcal H$ to exist?

I confess I'm a little stuck in this problem. I see no reason for it not to exist but I can't prove that it exists. How can I do this?


答:
The matrix of $f$ in the standard basis $\mathcal{B} = \{[1, 0, 0]^T, [0, 1, 0]^T, [0, 0, 1]^T\}$ is

$$
M(f,\mathcal{B},\mathcal{B})=\left( \begin{matrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
1 & 0 & -1 \\
\end{matrix}\right).
$$

Compute the trace of $\mathrm{tr}M(f,\mathcal{B},\mathcal{B}) = 1 + 1 - 1 = 1$. However,  trace is basis-independent, so your matrix cannot be a matrix of $f$ in any basis.


问:
how to find $\mathbb{P}[Y=0]$ given $\left( Y| X=k\right)\sim \mathsf{Poi}\left(k \ln 3\right)$
Given random variable $X$ uniform distributed on the set $\{1,...,{70}\}$ Let $Y$ be random variable such that $$\left( Y| X=k\right)\sim \mathsf{Poi}\left(k \ln 3\right)$$  calculate $\mathbb{P}[Y=0]$

My attempt:

I thought on finding general exprerssion for $P_Y(y)$

$$\mathbb{P}(Y|X=k)=\frac{k(\ln3)^n}{n!3^k}$$
$$\mathbb{P}(Y=y)=\sum_{k=1}^{70} \mathbb{P}(Y|X=k) \mathbb{P}(X=k)=\sum_{k=1}^{70}\frac{k(\ln3)^n}{n!3^k}\cdot k$$

I dont understand if this is the correct method because I have 2 variables and eventualy the sum will be a function of $n$ so I guess I made a mistake but I have no idea how to solve it

答: You're on the right track but you need to simplify things a bit.  If the only thing you want is $\Pr[Y = 0]$, then you should compute the conditional probability $$\Pr[Y = 0 \mid X = k] = e^{-k \ln 3} \frac{(k \ln 3)^0}{0!} = 3^{-k}.$$  So $$\Pr[Y = 0] = \sum_{k=1}^{70} \Pr[Y = 0 \mid X = k]\Pr[X = k] = \sum_{k=1}^{70} \frac{3^{-k}}{70}.$$  Factor out $1/70$ and then recognize the sum is a finite geometric series with common ratio $1/3$.


问:
Saddle point of function
Let $f:\mathbb{R}^{k}\rightarrow\mathbb{R}$ differentiable $n$ times.
Assume that $\left(Df\right)_{a}=0$ and $\left(D^2f\right)_{a}=0$ and $\left(D^{3}f\right)\neq0$.
Prove that $a$ is a saddle point.

答:
Since as I explained in the comment $f(x)=x^4$ is a counterexample otherwise, I assume you mean $(D^3f)_a\neq 0$.

If $k=1$ we have by Taylor formula $f(x)=f(a)+f'''(a)(a-x)^3+o(|x-a|^3)$. Suppose $f'''(a)>0$; then for $x$ sufficiently close to $a$ we have $f(x)>f(a)$ if $x>a$ and $f(x)<f(a)$ if $x<a$, so $a$ is not a local minimum or nor a local maximum. If $f'''(a)<0$; then again for $x$ sufficiently close to $a$ we have $f(x)>f(a)$ if $x<a$ and $f(x)<f(a)$ if $x>a$, so again $a$ is not a local minimum or nor a local maximum.

Now let $k$ be arbitrary. Since $(D^3f)_a\neq 0$, there exists $v\in \mathbb{R}^k$ such that for the line $\gamma(t)=a+tv$ through $a$ the function $g=f\cdot \gamma$ has $g'''(0)\neq 0$ (otherwise $(D^3f)_a$ is a symmetric trilinear form which vanishes on all triples $(v,v,v)$ for any $v \in \mathbb{R}^k$ and so $(D^3f)_a$ must be zero).

Observe that  $g'(0)=(Df)_a \gamma'(0)=0$ and $g''(0)=\gamma'(0)^T (D^2f)_a \gamma(0)=0$. Thus by the case $k=1$ applied to $g$ we have both points with $g(t)>g(0)$ and with $g(t)<g(0)$ arbitrarily near $t=0$. Their images $x=\gamma(t)$ will attain both values $f(x)>f(a)$ and values $f(x)<f(a)$ arbitrarily close to $x$. This means $f$ does not have either a local minimum nor a local maximum at $a$, as wanted.

Title: Saddle point of function


答:
Depending on the problem, a strategy for multiple-choice questions can be just to check the potential answers. (You commented that you guessed $x=10$, but since it doesn't appear as one of the answers, you know that it's wrong. It's much more reasonable to guess one of the provided answers.)

If the answer were A ($x=11$), then we know that with $8$ boxes, we'd have $12$ toys in each box plus an extra $5$, for a total of $8\cdot12+5=101$. But this number must be a multiple of $x$; it isn't, so eliminate A. Do the same calculation for the others.

Title: gre division with remainder problem

Body:
A certain number of toys were packed into $x$ boxes so that each box contained the same number of toys, with no toys left unpacked. If $3$ fewer boxes had been used instead, then $12$ toys would have been packed in each box, with $5$ toys left unpacked. What's the value of $x$?

$A. 11$

$B. 14$

$C. 28$

$D. 31$

$E. 34$

I'm not sure If I interpreted the problem correctly, but my approach as I will assume(guess) $x = 10$, then with $3$ fewer boxes, $x = 7$, so we packed $12$ toys in each box, each box has $1$ toy, left $5$ toys left unpacked. But none of these answer choices is $10$, so pretty sure I'm wrong.

I feel like this is just remainder problem, any help is highly appreciated.


问:
Is every nowhere dense closed set intersection of closures of non-intersecting open sets?
This is a double-cross-post: I first asked a question on cs.se, then posted a particularly mathematical part of it to mathoverflow; after about month and a half I realised that I don't know answer to a simpler question which looks like an exercise from a general topology textbook. So I decided to ask it here.

Given two non-intersecting open sets, $U\cap V=\varnothing$, it is easy to show that the intersection $F=\bar U\cap\bar V$ of their closures is nowhere dense (i. e. has empty interior, that is, does not contain any open set).

Is the converse true? That is, is any nowhere dense closed set $F$ intersection of closures of two non-intersecting open sets?

答: Let $X=\Bbb N$, and let $\tau=\{\varnothing\}\cup\{U\subseteq\Bbb N:0\in U\}$; $\tau$ is a topology on $X$. $\{1\}$ is closed and nowhere dense but it cannot be the intersection of the closures of two disjoint, non-empty open sets, because the space doesn’t have two disjoint, non-empty open sets.


答:
Some simple rules allow us to find the first derivative

 Namely the fundamental theorem of calculus and the chain rule

Which gives

 $$ \frac{df(x)}{dx} = \frac{df(x)}{d(x^{20})}\cdot \frac{d(x^{20})}{dx} = x^{10} \cdot \sin x^{10}\cdot 20x^{19}= 20x^{29}\sin x^{10}$$

We know that deriving this another 79 times gives the 80th derivative of $f$. To find the 79th derivative of this derivative we use

 The taylor expansion of $\sin u$ at $u=0$ and substitute $u=x^{10}$

Which gives

 $$20x^{29}\sin x^{10} = 20x^{29}\sum_{k=0}^{\infty} \frac{(-1)^k (x^{10})^{1 + 2 k}}{(1 + 2 k)!}$$

Which allows us to use the following fact

 the 79th derivative of this function at zero will be $79!$ multiplied by the coefficient of $x^{79}$ - this is valid because all other terms have either vanished or are stil multiplied by $x=0$

We can find the corresponding $k$ as follows

 $$10(1+2k) = 79-29 \iff k = 2$$

Now we apply arithmatic to find the solution

 $$f^{(80)}(0) = \frac{20\cdot 79!}{5!} =  \frac{79!}{6}$$

Title: Given $f(x) = \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Find $f^{(80)}(0)$

Body:
I want to find $f^{(80)}(0)$, where $f(x) = \displaystyle \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$.

My attempt:

Let $g(x) = \displaystyle \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Now $f(x)$ can be rewritten as
$$ f(x) = g{\left(x^{20}\right)} - g(0). $$

Taking the derivative $80$ times,
$$ f^{(80)}(x) = \sum_{i = 1}^{80}{a_i \times g^{(i)}{\left(x^{20}\right)}}, $$
where $\{a\}$ is a sequence of integers. I tried to find the sum, but little progress was made.

I noticed that if I plugged $x = 0$ into the equation, all of the "$x$-linked" terms would be canceled. hence, only the first few terms would be significant, which is $\displaystyle \sum_{i = 1}^{4}{a_i \times g^{(i)}{\left(x^{20}\right)}}$.

Another observation is that $a_i$ will become $0$, i.e. $g^{(i)}$ will vanish after 20 times the derivative was taken. Thus, the final answer would be $a_4 \times g^{(4)}{\left(x^{20}\right)}$.

I only managed to find $a_1 = 20!$ for the first 20 derivatives and failed to derive explicit formulas for the other three.

I would like to know how to solve this problem appropriately, thanks in advance.


问:
Linear Algebra Transformations Notation
Let A be a basis of vector space V and B be a basis of vector space W.  Let there also be a linear transformation T:V→W.

What does the following notation signify?
$$[T]_B,_A$$

The notation appeared very briefly in the chapter about similarity in my textbook but I am not quite understanding what is being asked of me when I see this notation.

答: $[T]_{B,A}$ denotes the matrix of $T$ with respect to the bases $A$ of $V$ and $B$ of $W$.

Title: Linear Algebra Transformations Notation


问:
Limit of a recursive sequence with $u_n$
It is given that $u_{n+1} =1+\dfrac{1}{u_n}$ and $u_1 =1$. Find the limit of $u_n$ as $n\to\infty$.

The limit is $\frac{\sqrt{5}+1}{2}$ from a calculator. Is there an algebraic way to determine this? You can also determine that sequence is bounded but not monotonic.

答:
Writing $u_n$ as $\frac{p_n}{q_n}$ for two sequences $(p_n)$, $(q_n)$ to be determined. We have:

$$u_{n+1} = 1 + \frac{1}{u_n} \iff \frac{p_{n+1}}{q_{n+1}} = 1 + \frac{q_n}{p_n} = \frac{p_n+q_n}{p_n}$$
Normalize $(p_n)$ and $(q_n)$ appropriately, we can turn the non-linear recurrence equation into a linear one:

$$\begin{pmatrix}p_{n+1}\\q_{n+1}\end{pmatrix}
= \begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}p_n\\q_n\end{pmatrix}$$
The charateristic polynomial of the matrix $\begin{pmatrix}1&1\\1&0\end{pmatrix}$ is
equal to:

$$\lambda (\lambda - 1 ) - 1^2 = \lambda^2 - \lambda - 1 = (\lambda - \varphi)(\lambda + \varphi^{-1})$$
where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. The matrix
has eigenvalues $\varphi$ and $-\varphi^{-1}$ with corresponding eigenvectors:

$$\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}\varphi\\1\end{pmatrix} = \varphi \begin{pmatrix}\varphi\\1\end{pmatrix}
\quad\text{ and }\quad
\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix} = -\varphi^{-1}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix}
$$
Since $u_1 = 1$, we will choose $p_1 = q_1 = 1$. Expressing $\begin{pmatrix}1\\1\end{pmatrix}$ in terms of the eigenvectors:

$$\begin{pmatrix}1\\1\end{pmatrix} = \alpha \begin{pmatrix}\varphi\\1\end{pmatrix} + \beta
\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix}
\implies \alpha = \frac{\varphi+1}{\varphi+2} \;\text{ and }\; \beta = \frac{1}{\varphi+2}
$$
We get
$$ \begin{pmatrix}p_n\\q_n\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1} \begin{pmatrix}p_1\\q_1\end{pmatrix} = \alpha \varphi^{n-1} \begin{pmatrix}\varphi\\1\end{pmatrix} + \beta (-\varphi^{-1})^{n-1}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix}$$
This allow us to derive a closed form expression for $u_n$:
$$u_n = \frac{p_n}{q_n} = \frac{\alpha \varphi^n + \beta (-\varphi^{-1})^n}{\alpha \varphi^{n-1} + \beta (-\varphi^{-1})^{n-1}} = \varphi \frac{\alpha + \beta (-\varphi^{-2})^n}{\alpha + \beta (-\varphi^{-2})^{n-1}}$$
Since $| -\varphi^{-2} | < 1$, this implies $\lim_{n\to\infty} u_n = \varphi \frac{\alpha + \beta*0}{\alpha + \beta*0} = \varphi$.


问:
Spacing of plus-or-minus sign (±) when used with percentage?
I have a question regarding the correct spacing of plus-or-minus sign (±) in a specific situation.

Normally there should be a space on both sides of the sign, e.g., A ± B. If I want to express “20 percent plus or minus 3 percent” it could be 20% ± 3%. But what if I want to omit the first percentage sign for conciseness? Should it be:

20±3%, or

20 ± 3%?

I personally prefer the first one, because the second one could also mean 20(an integer) fluctuate within 3%, i.e., 19.4-20.6?

I couldn’t find the answer anywhere. Thank you in advance for answering the question.

答: This is a typesetting, rather than a mathematical issue, but I'd prefer the latter, personally.</p>


答:
$x=31$ and the toys are $341$ because they say $$k x=12 (x-3)+5$$ where $k$ is the number of toys in the first boxing. Therefore
$$x=\frac{31}{12-k}$$
and the only integer value is $x=31$ for $k=11$.

Title: gre division with remainder problem


问:
Is $\frac{ab}{dc}<1$ if $0<a<c<d<b$ and $a+b<c+d$
I'm trying to prove $\frac{ab}{dc}<1$ if $0<a<c<d<b$ and $a+b<c+d$. I've played around with this a bit and I'm pretty confident its true.

My approach is to use AMGM

Want to show:
$$\frac{ab}{cd}<1$$
$$\iff \sqrt{ab}<\sqrt{cd}$$
but $$\sqrt{ab}\le\frac{a+b}{2}<\frac{c+d}{2}$$
If I could show $\frac{a+b}{2}<\sqrt{cd}$ I would be sorted, but I'm not sure if this stronger condition is true. At this point I haven't use the fact $a<c<d<b$

Body:
First note that

$$
a + b < c + d
$$
$$
(a + b)^2 < (c + d)^2 \tag1.
$$

Next, from $0 < a < c < d < b$ we have

$$
b - a > d - c
$$
$$
(b - a)^2 > (d - c)^2\tag2.
$$

Subtract $(2)$ from $(1)$ to get

$$
(a + b)^2 - (b - a)^2 < (c + d)^2 - (d - c)^2\\
4ab < 4cd \\
\frac{ab}{cd} < 1.
$$


问:
(Vol. 1 Shafarevich) Two questions about finite maps
I am currently reading Basic Algebraic Geometry by Shafarevich. I have two questions about finite maps on affine and quasi-projective varieites.

For a finite map $f:X\rightarrow Y$ between affine varieties $X$ and $Y$, we say that $f$ is finite if $f(X)$ is dense in $Y$ and $k[X]$ is integral over $k[Y]$ (where $k[Y]$ has been identified with a subring of $k[X]$).

Question 1

The first question is about the proof of Theorem 1.13. The statement of the theorem is as follows.

If $f:X\rightarrow Y$ is a regular map of affine varieties, and every $x\in Y$ has an affine neighbourhood $U\ni x$ such that $V = f^{-1}(U)$ is affine and $f:V\rightarrow U$ is finite, then $f$ itself is finite.

The proof itself is not so hard, but I find myself unable to understand the first step, which is to assume that each neighbourhood $U$ in the statement of the theorem above is a principal open set.

As far as I understand it, the neighbourhood $U$ of $x$ may be decomposed into a finite union of principal open sets $U_1,\ldots, U_r$ and for each $U_i$, the set $V_i := f^{-1}(U_i)$ is also a principal open set in $X$. However, I fail to understand why $f$ must be finite when restricted to any of these sets $V_i$.

Question 2

The definition of finite maps on quasi-projective varieties follows the "local definition" above. More precisely,

A map $f: X \rightarrow Y$ of quasi-projective varieties is finite if any point $y\in Y$ has an affine neighbourhood $U$ such that $V = f^{-1}(U)$ is affine and $f:V \rightarrow U$ is a finite map between affine varieties.

This definition makes sense given the theorem above, but it got me wondering regarding the following fairly easy fact about finite maps between affine varieties.

Fact: If $X,Y$ are affine varieties and $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are finite maps, then so is their composition $g\circ f$.

It's not clear to me if the above holds for finite maps between quasiprojective varieties as defined above. The "obvious" proof again runs into a similar obstacle where one needs to show that a restriction of a finite map to some open affine subset is still finite.

Possibly I am missing some simple clean fact about finite maps. If so, I would like to know it.

答:
For question (1), what you are trying to show is that if you have a finite ring map $A \to B$ and some element $f \in A$ then $A_f \to B_f$ is finite (meaning invert the multiplicative subset of powers of $f$ in both rings). That is, we have some $A$-module surjection $A^n \twoheadrightarrow B$. Since localization is exact, and it commutes with direct sums, we get a surjection $A_f^n \twoheadrightarrow B_f$, as desired.

In general you're coming upon the idea of properties of morphisms that are affine-local on the target, where their truth can be verified either by checking every affine open on the target, or by checking it on a cover of the target by affine opens. This idea is discussed very cleanly in Vakil's Rising Sea notes section 5.3.


问:
How do I represent recursion in my filtration set?

Let's take a super simple stochastic process consisting of one coin toss.

Normally I would have $\mathcal{F}=\{\{\},\{T\},\{H\},\{T,H\}\}$

Now let's say that if I get heads, I toss again.

How would my new $\mathcal{F}'$ set look like?

Update: the process is meant to continue indefinitely if I keep getting $H$, so imagine the filtration symbol $\mathcal{F}'$ would have to be re-used in the set itself

答:
You could start with sample space $\Omega=\{HH,HT,TH,TT\}$ and the associated power set (with 16 elements) as $\sigma$-field. But as you won't perform a second toss if the first toss is a head, you don't care to distinguish between $TT$ and $TH$. So an adequate $\sigma$ field would be
$$
\left\{\emptyset,\Omega,\{HH\},\{HT\},\{TH,TT\},\{HH,HT\},\{TH,TT,HT\},\{TH,TT,HH\}\right\},
$$
in which any event containing one of $TH$ or $TT$ also contains the other.

Alternatively, you could take your sample space to be $\{HT, HH,T\}$, acknowledging that you stop with one toss if it's a $T$. With this choice the power set would be the appropriate $\sigma$-field.


问:
The Convergence of Jacobi and Gauss-Seidel Iteration
I have a question, is there any example that the convergence of Jacobi is faster than Gauss-Seidel ?</p>

答:
The answer to that would according to what I know would be No. To see this, I will explain what happens in both iterative methods.

The Jacobi method tells us if there is a matrix $A=M-N,$ where in this method we denote $M=D,$ and $N=L+U,$ then the iteration matrix is :
$$B_{J}=M^{-1} N=
D^{-1}(L+U) $$ For this method to converge, the spectral radius of $B_{J}$ should be less than $1$ i.e.
$$
\rho\left(B_{J}\right)=\max _{\lambda \in \Lambda\left(B_{J}\right)}|\lambda|<1
$$
The Gauss-Siedal method tells us that if there is a matrix $A=M-N,$ where in this method we denote $M=D-L,$ and $N=U,$ then the iteration matrix $$B_{G}=
M^{-1} N=(D-L)^{-1} U $$ For the Gauss-Seidel method to converge, the spectral radius of $B_{G}$ should be less than $1,$ i.e.
$$
\rho\left(B_{G}\right)=\max _{\lambda \in \Lambda\left(B_{G}\right)}|\lambda|<1
$$
These are the necessary conditions to establish a convergence criteria. Indeed, they don't differ much except on how we choose our initial guess point and process the iteration. Now to look at what happens during the corresponding iterations of the two methods, we note that the Gauss-Siedal method is indeed faster since it operates on as much available and updated data as possible unlike the Jacobi method that relies on the updated data after the second iteration and thus takes more iterations so that the relative residual or relative error becomes less than the tolerance set. In fact the convergence rate for Gauss Siedal is twice the speed of convergence of the Jacobi method and this is true especially for strictly diagonally dominant matrices i.e.
$$|A(i, i)|>\sum_{j=1 \atop j \neq i}^{n}|A(i, j)|, \quad \forall i=1,2, \cdots, n$$
To further enhance the rate of convergence of Gauss-Siedal method, one can pick a good choice of $\omega$ in the relaxation method But the choice of the optimal $\omega$ is often difficult to make. I will not enter into further details about convergence of relaxation method but I will end my answer by attaching two interesting links on the speed of convergence of these two methods.

$(1)$ https://www.researchgate.net/publication/234783725_The_Ray_Engine/figures?lo=1

$(2)$ https://mathoverflow.net/questions/48122/the-convergence-of-jacobi-and-gauss-seidel-iteration


问:
Find the minimum of $q$.

Given $\displaystyle p,q\in\mathbb N, \frac p q=0.123456789...$(i.e. the first 9 digits after decimal point are $123456789$). Find the minimum of $q$.

I guessed the answer to be $111111111$ as $\displaystyle\frac {123456789} {999999999}=0.\dot{1}2345678\dot{9}$, but I'm not sure.

Thank you.

Title: Find the minimum of $q$.

Body:
The smallest $q$ is $q=10989019$ corresponding to $$\frac pq=\frac{1356669}{10989019}=0.12345678900000081899940294943524986\ldots$$
To see this, note that
$$ \frac ab=\frac{1356659}{10988938}=0.1234567889999925379504370668029977\ldots$$
is too small and
$$\frac cd=\frac{10}{81}=0.12345679012345679012345679012345679\ldots$$
is too big.
For any fraction with $\frac ab<\frac pq<\frac cd$, we have
$$\frac pq-\frac ab=\frac{bp-aq}{bq}>0\qquad\frac cd-\frac pq=\frac{cq-dp}{dq}>0,$$
hence $$ bp-aq\ge 1\qquad cq-dp\ge 1$$
and finally (because $bc-ad=1$)
$$ q=(bc-ad)q=d(bp-aq)+b(cq-dp)\ge b+d=10989019.$$


问:
Explicit isomorphism between ${\rm SL}(2,{\Bbb R})$ and ${\rm SU}(1,1)$
I have heard that there exists an isomorphism of real algebraic groups as in the title. I am asking for an explicit isomorphism.

Motivation: I need such an isomorphism for a calculation of Galois cocycles (with values in some other group).

答: The elements of $SU(1,1)$ are the matrices of the form$$\begin{bmatrix}\alpha+\beta i&\gamma+\delta i\\\gamma-\delta i&\alpha-\beta i\end{bmatrix}$$with $\alpha,\beta,\gamma,\delta\in\Bbb R$ such that $\alpha^2+\beta^2-\gamma^2-\delta^2=1$. Map each such matrix into$$\begin{bmatrix}\alpha-\delta&-\beta+\gamma\\\beta+\gamma&\alpha+\delta\end{bmatrix}\in SL(2,\Bbb R).$$That will give you your isomorphism.


问:
Is there any known lower bound for the number of polynomia functions over a finite field with fixed number of variables?
Say we have a finite field $K$ of size $q$, is there an easy lower bound for the number of distinct polynomial functions in $K[x_1,x_2,\ldots,x_n]$? What about if we specifically fix the degree $d$, so that we only count those polynomial functions of degree $d$ on $n$ variables?

答:
The total number of distinct polynomial functions is $q^{q^n}$, also equal to the total number of functions $f:K^n\to K$ (proof at end). Every such polynomial is equivalent to one where all variables have degree less than $q$, because of the identity $x^q=x$. As long as we restrict our attention to polynomials with this degree bound, then two polynomials represent the same function if and only if they have the same coefficients; indeed, there are $q^{q^n}$ polynomials where every variable has degree at most $q$, and these represent all functions, so they must all be distinct functions. Therefore, you can count polynomials of degree $d$ by counting the number of representations with degree $d$.

Let $m(n,q,d)$ be the number of monomials with degree at most $d$ where all powers of all variables are at most $q$. Then the number of polynomials of degree exactly $d$ is
$$
q^{m(n,q,d)}-q^{m(n,q,d-1)}
$$
Furthermore, $m(n,q,d)$ is equal to the number of nonnegative integers solutions to the equation $a_1+\dots+a_n\le d$. Using a standard stars-and-bars/inclusion-exclusion argument, we can show
$$
m(n,q,d)=\sum_{k\ge 0}(-1)^k\binom{n}k\binom{n+d-qk}{d-qk}
$$

Proof: To prove there are $q^{q^n}$ polynomial functions, you need to show every function can be represented as a polynomial, which we do now. First, given any vector $v=(v_1,\dots,v_n)\in K^n$, we will make a polynomial $p_v(x_1,\dots,x_n)$ whose value is $1$ when $(x_1,\dots,x_n)=(v_1,\dots,v_n)$, and zero otherwise. Then, any arbitrary function can be made using a linear combination of such polynomials. To do this, for each $i\in \{1,\dots,n\}$, let
$$
g_i(x_1,\dots,x_n)=\prod_{a\in K\setminus \{v_i\}}\frac{x_i-a}{v_i-a}
$$
Note that $g_i(x_1,\dots,x_n)$ is $1$ when $x_i=v_i$, and zero otherwise. Therefore, the desired polynomial is
$$
p_v(x_1,\dots,x_n)=\prod_{i=1}^ng_i(x_1,\dots,x_n)
$$


问:
Find the equation of the equilateral hyperbola given one of its foci points $(1;1)$ and an asymptote $x+y=0$
If the asymptote is $y=-x$, then doesn't it mean that the other one is $y=x$?

But does't that also mean that the foci points are $(c_1,0)$, $(c_2,0)$?

But then how is it $(1,1)$?

答:
HINT

The picture shows a rectangular (equilateral) hyperbola, its asymptotes and one of the foci.

In this particular case we see a square with vertices: the focus, its orthogonal projection on one asymptote, the center of the hyperbola and projection of the focus on other asymptote.

EDIT

This new picture shows the given point $(1,1),$ the asymptote $y+x=0$ and one of convenient hyperbolas.

The second hyperbola corresponds to the green square. Its center is $(1,-1)$ and the second focus is the small red point bellow.


问:
Proof of a property of conditional expectation
I'm new here and this is my first question.

I'm struggling with a proof for my statistics course.

I need to show the following proposition:

$E[Y^2|\bar{X}=\bar{x}]\geq[E(Y|\bar{X}=\bar{x})]^2$

I tried to show it using Jensen's inequality but my teacher told me that is not correct.

Thank you :)

答:
Not to subvert your teacher, but there is a conditional form of Jensen's inquality, which in your situation would be: For a function $g$ defined and convex on the range of $Y$,
$$
E[g(Y)\mid \bar X =\bar x]\ge g\left(E[Y\mid \bar X = \bar x]\right).
$$
(Of course, $Y$ needs to have a finite expectation.)


问:
Does Hilbert's $\varepsilon$-calculus have any real problem when using for incomplete systems?
In Mathias' paper Hilbert, Bourbaki and the scorning of logic (see https://www.dpmms.cam.ac.uk/~ardm/hbslmag2.pdf), he mentions Hilbert's $\varepsilon$-calculus (and thus Bourbaki's $\tau$-calculus) is not suitable for incomplete systems. But he did not elaborate if there is any real problem (like inconsistency or incapability) with it.

Obviously Hilbert hoped to use $\varepsilon$-calculus to prove the completeness of systems like ZFC, and failed as we know. But this does not mean $\varepsilon$-calculus itself has any problems.

So, is  Mathias' criticism out of his personal flavor of logic and educational reasons, or there is any real problem for $\varepsilon$-calculus in incomplete systems?

答:
I am no expert on this subject, and it has been many years since I've read Bourbaki's Theory of Sets and do not have a copy available to me, but I'll point out a couple of things I see in this paper.

First, at least some of his argument appears to be valid. Bourbaki's formalism does lead to some very misleading statements. As an informal conception, in Bourbaki's theory somebody has listed out every possible relation $R_x$, and chosen, once-for-all-time, an object for each - if possible one that if substituted for $x$ would make $R_x$ true. This selected item is written as $\tau_x(R_x)$. In the actual theory, they are just formal symbols and the interpretation is supported by the chosen axioms. So $\tau_x(x = 1) = 1$ while $\tau_x(x \ne x)$ could be anything at all. But the key thing is, it is still some object that can be considered. In particular, for any $R_x$, you can prove
$$(\exists b)(b = \tau_x(R_x))$$
even when $(\exists x)R_x$ is false. ($(\exists x)R_x$ is short-hand for $R_x(\tau_x(R_x(x))$.)

This becomes a problem when combined with the set builder notation: $$\{x\mid R_x\} := \tau_y((\forall x)(x \in y \iff R_x))$$
Therefore, while the statement $$(\exists y)(\forall x)(x \in y \iff x \notin x)$$ is provably false in Bourbaki's formalism, the statement
$$(\exists b)(b = \{x\mid x\notin x\})$$
is provably true. It just means something different that we expect it to mean. The natural, even intended, interpretation of $(\exists b)(b = \{ x\mid x\notin x\})$ is that there is a set consisting of every set that is not an element of itself. But in Bourbaki's theory, $\{ x\mid x \notin x\}$ exists but is an object with no connection to the relation $x \notin x$.

Thus Bourbaki is consistent here (as far as I can see), but does it by subtly changing the meaning of the notation from what we expect.

On the other hand, some of his criticism appears to me to be unfair. For example from page 22:

B-13 Thus there is an acute difference between the normal use, in $\bf ZF$ and many other set theories, of the class-forming operator $\{\ \mid\ \}$ with the Church conversion schema $x \in \{x\mid R\}\iff R$ holding for all classes whether sets or not, and the Bourbaki treatment whereby, magically, conversion holds for a class if and only if that class is a set.

where he apparently fails to understand that in Bourbaki's theory, there is no such thing as a "proper class". The only objects are sets, so "conversion holds for a class if and only if that class is a set" makes no sense, as the only interpretation possible for the word "class" is "set".

Now of course one may prefer to have a theory that allows classes. But the failure to include classes in this theory is not a problem with the theory itself, only in our expectations. To pretend that the theory is mishandling a concept when in fact that concept does not appear in it is disingenuous.


答:
That's a telescoping product so

$\begin{array}\\
\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\dfrac{i}{i+1}
&=\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\dfrac{n}{2k+1}\\
\end{array}
$

but then there is a problem at $n=2$
when there is a division by zero.

I'll change that to $n+2$
and we get

$\begin{array}\\
\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n+2{)}^{2}}\prod _{i=n}^{2k}\dfrac{i}{i+1}
&=\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n+2{)}^{2}}\dfrac{n}{2k+1}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}(n-2)}{n^2}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}n}{n^2}-\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}2}{n^2}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}}{n}-\dfrac{2}{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}}{n^2}\\
\end{array}
$

and these can be handeled
by the usual methods.

Title: Evaluate $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$ for $k=3$

Body:
$\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$

How can I think this for k=3 ?

Here is my idea: $\prod _{i=n}^{6}\frac{i}{i+1}=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}$

then

$\frac{3}{14}\sum _{n=1}^{3}\frac{{\left(-4\right)}^{n+2}}{{\left(n-2\right)}^{2}}=\frac{3}{14}.\left(\frac{{\left(-4\right)}^{3}}{{\left(1-2\right)}^{2}}+\frac{{\left(-4\right)}^{5}}{{\left(3-2\right)}^{2}}\right)=-233,1428571$

Is my idea correct?


问:
Does an integrable function vanish at infinity in a suitable sense?
I'm searching for a generalization of the following observation: If $f\in L^2(\mathbb R)$, then \begin{equation}\begin{split}\int_{[-n,\:n+1]}f(x)\:{\rm d}x&=\left\langle1_{[-n,\:n+1]},1_{\mathbb R\setminus[-n,\:n)}g\right\rangle_{L^2(\mathbb R)}\\&\le\left\|f-1_{[-n,\:n)}f\right\|_{L^2(\mathbb R)}\xrightarrow{n\to\infty}0\end{split}\tag1\end{equation} by the Cauchy-Schwarz inequality and Lebesgue's dominated convergence theorem.

$(1)$ captures the idea that an integrable function should "vanish at infinity" in a suitable sense.

Can we generalize this result to $f\in L^p(\mu)$, where $(E,\mathcal E,\mu)$ is an arbitrary measure space and $p\ge1$? Maybe assuming that $\mu(E)=\infty$ and/or that $E$ is a metric/normed space with $\mathcal E=\mathcal B(E)$?

In the case of a metric/normed space, the intuition tells me that an integrable function should "essentially be supported in a compact set".

What's obviously true is that $$\forall f\in L^p(\mu):\not\exists B\in\mathcal E:\mu(B)=\infty\text{ and }\inf_{x\in B}|f(x)|>0\tag2.$$

答:
I think of two statements.

Let $r_n = f-1_{[-n,n]}f$. Then,

$\int |r_n(x)|\,dx \to 0$ as $n\to\infty.$

$r_n \to 0$ in measure as $n\to\infty$.


问:
Symbolic definition for $i$?
Please help!

I am trying to find a definition for $i$ that doesn't work for $-i$.

let $j$ be either $i$ or $-i$.

saying $j^2=-1$ doesn't help since $(i)^2=-1$ and $(-i)^2 = -1$

saying $j=\sqrt{-1}$ doesn't halp since $(-1)^{\frac{1}{2}}$ is multivalued $i$ and $-i$

saying $j=e^{i\pi/2}$ doesnt help since $e^{i\pi/2}=i$ but $e^{(-i)\pi/2}=-i$ too (yeah I know I said j=$e^{i\pi/2}$ and not $j=e^{j\pi/2}$ I did that to hopefully make it more intuitive)

saying $j=ln(i)/(\pi/2)$ doesn't work since $i=ln(i)/(\pi/2)$ and $-i=ln(-i)/(pi/2)$ (check with a phase plotter, ln is multivalued but includes i and -i in their respective functions shown above http://davidbau.com/conformal/#log(z)%2F(pi%2F2)-z)

saying $Im(j) > 0$ doesn't work because $Im(z)=Re(z/i)$ and $Re(i/i)>0$ but $Re((-i)/(-i))>0$

Is this parity some sort of law? And yet $i\neq-i$ since $i=(-1)\cdot(-i)$. Unless...

Post Scriptum: Irionically, it is very easy to symbolically represent $1$ vs $-1$ even though $(1)^2=1$ and $(-1)^2=1$, we can just say $x$ such that $x=x^2$

EDIT: can quaternions help? Is $i$ a set of 2 numbers? Are they always equal?

答:
You could construct the field of complex numbers $\mathbb{C}$ as the real vector space $\mathbb{R}^2$ with an operation
$$ (a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2, a_1b_2 + a_2b_1). $$
It can "easily" be seen that $(\mathbb{R}^2,\cdot)$ is a field and has the same properties as the $\mathbb{C}$ you know and love. Now you could define the imaginary unit as
$$ i := (0,1). $$
Then $i$ and $(1,0)$ form a basis of $\mathbb{R}^2$ and we can write each element
$$ (a,b) = a + bi $$
by abuse of notation (by identifying $a=(1,0)a$).

Title: Symbolic definition for $i$?


问:
Find conditional probability (′|′)
I'm trying to solve this task but I'm obviously stuck.  Can anybody help me how to solve this ?

If ()=1/6, ()=1/3 , and (∪)=5/12, what is (′|′)?

答:
Based on @YJT's comment, we can proceed as follows:
\begin{align*}
\mathbb{P}(A^{c}|B^{c}) = \frac{\mathbb{P}(A^{c}\cap B^{c})}{\mathbb{P}(B^{c})} = \frac{\mathbb{P}((A\cup B)^{c})}{\mathbb{P}(B^{c})} = \frac{1 - \mathbb{P}(A\cup B)}{1 - \mathbb{P}(B)} = \frac{1-5/12}{1 - 1/3} = \frac{7}{8}
\end{align*}

Title: Find conditional probability (′|′)