问:
SVD in scipy and numpy for tensors
Can someone explain to me the difference between SVD of numpy and scipy for Multidimensional arrays (Tensors)?

```<code>X = np.random.randn(3,3,3)
S1 = numpy.linalg.svd(X)
S2 = scipy.linalg.svd(X)
</code>```

The S1 here is a tuple containing U 3x3x3, Sigma 3x3 and Vh 3x3x3. But the S2 line throws an error saying 'expected Matrix'. Thus, I use the reshape option to unfold the tensor and compute the svd using scipy.linalg as follows:

```<code>Xreshape = np.reshape(X, (9,3))
S2 = scipy.linalg.svd(X)
</code>```

Now, S2 here is a tuple containing U 9x9, Sigma 3x1 and Vh 3x3.
The elements of S1 and S2 are not the same. Could someone explain to me the theory behind it?

答:
It's easy to explain what's going on for the scipy svd method: the scipy method does not permit arrays of dimension 3 or higher as an input. If you feed in a matrix, it yields the SVD of that matrix.

The real question, then, is what does the numpy svd method do with the input? The answer is explained in the documentation, but the explanation is a bit terse. In general (for $k \geq 3$), numpy interprets a shape $n_1 \times n_2 \times \cdots \times n_k$ array as an $n_1 \times \cdots \times n_{k-2}$ array of $n_{k-1} \times n_k$ matrices and it computes the svd of each of these matrices separately.

For your particular case, numpy interprets $X$ as a list of $3$ matrices. Let $X_i$ denote the slice X[i,:,:]. If U,s,Vh denote the components the output of numpy.linalg.svd(X), then these contain and svd for each matrix $X_0$, $X_1$, and $X_2$. In particular, the matrix product

```<code>U[i,:,:] @ numpy.diag(s[i,:]) @ Vh[i,:,:]
</code>```

Should be (close to being) equal to $X_i$ for $i = 0,1,2$. You should be able to verify this with

```<code>numpy.allclose(X[i,:,:],U[i,:,:] @ numpy.diag(s[i,:]) @ Vh[i,:,:])
</code>```

which should return True. As the documentation notes, the product U[i,:,:] @ numpy.diag(s[i,:]) can be more efficiently computed as U[i,:,:] * s[i,None,:], but this is fairly confusing unless you're comfortable with numpy's broadcasting rules for array multiplication.

Title: SVD in scipy and numpy for tensors

Body:
It's easy to explain what's going on for the scipy svd method: the scipy method does not permit arrays of dimension 3 or higher as an input. If you feed in a matrix, it yields the SVD of that matrix.

The real question, then, is what does the numpy svd method do with the input? The answer is explained in the documentation, but the explanation is a bit terse. In general (for $k \geq 3$), numpy interprets a shape $n_1 \times n_2 \times \cdots \times n_k$ array as an $n_1 \times \cdots \times n_{k-2}$ array of $n_{k-1} \times n_k$ matrices and it computes the svd of each of these matrices separately.

For your particular case, numpy interprets $X$ as a list of $3$ matrices. Let $X_i$ denote the slice X[i,:,:]. If U,s,Vh denote the components the output of numpy.linalg.svd(X), then these contain and svd for each matrix $X_0$, $X_1$, and $X_2$. In particular, the matrix product

U[i,:,:] @ numpy.diag(s[i,:]) @ Vh[i,:,:]

Should be (close to being) equal to $X_i$ for $i = 0,1,2$. You should be able to verify this with

numpy.allclose(X[i,:,:],U[i,:,:] @ numpy.diag(s[i,:]) @ Vh[i,:,:])

which should return True. As the documentation notes, the product U[i,:,:] @ numpy.diag(s[i,:]) can be more efficiently computed as U[i,:,:] * s[i,None,:], but this is fairly confusing unless you're comfortable with numpy's broadcasting rules for array multiplication.


问:
Can this extension of $\sf ZFC^-$ intepret Mac Lane set theory?
Let $\sf ZFC^-$ be the theory $``\sf ZF + R \text{ well orders } V- Power"$ (where $\sf R$ is a primitive binay relation symbol).

Can $\sf ``ZFC^- + \exists \aleph_\omega"$ interpret Mac Lane set theory?

答:
If Mac Lane set theory is a theory obtained from Zermelo set theory $\mathsf{Z}$ by restricting Separation to $\Delta_0$-formulas, then yes.

You can see that the usual proof of $V=L\to\mathsf{GCH}$, which uses condensation lemma, still carries over $\mathsf{ZFC^-}$, and the same proof shows $\aleph_\omega$ in $L$ is a strong limit, that is, $L_{\omega_\omega}$ is closed under a power set. Hence $L_{\omega_\omega}$ satisfies $\mathsf{Z}$ with Powerset.

ContentLicense: CC BY-SA 4.0


问:
Why is the power of maximal ideals comaximal in an Artinian ring?
Let $R$ be an Artinian ring, commutative with 1.

We know :

there are only finitely many maximal ideals of $R$.

$Jac(R)^m = 0$ for some natural number $m$, using D.C.C.

Every prime ideal is maximal, using 2.

The backdrop of this question is we want to show $R$ is isomorphic to the product of some Artinian local rings. In the process of doing this, Dummit & Foote uses Chinese Remainder Theorem, which requires that the collection of powers of maximal ideals $\{M_1^m, ..., M_n^m\}$ is pairwise comaximal, where $m$ is the number from 2 above. I do not know why this comaximality is true.

Reference, Dummit & Foote 3ed, 16.1, page 753.

答: If $\mathfrak p + \mathfrak q = (1)$, then $r(\mathfrak p^n + \mathfrak q^m) = r(r(\mathfrak p^n) + r(\mathfrak q^m)) = r(\mathfrak p + \mathfrak q) = r((1)) = (1)$, so $\mathfrak p^n + \mathfrak q^m = 1$.


问:
Why is the difference between two odd squares multiples of $3$ is divisible by $72$?
Why is the difference between two odd squares multiples of $3$ is divisible by $72$?

Here is my solution and I am not sure what should be next.

Note that any odd number is in the form of  $2m+1$

Here we take  $m$,  $n$  so that  $m>n$

$(2m+1)^2−(2n+1)^2$

$=4(m^2−n^2+m−n)$

$=4(m-n)(m+n+1)$

Since $m$ and $n$ are odd, the difference will be even.
Hence, $m-n$ is divisible by $2$, which makes the difference between two odd squares divisible by $8$.
Now, since the given difference between two odd squares must be multiples of $3$, $(m+n+1)$ must be divisible by $3$. This makes the given difference between two odd squares divisible by $24$.

From here, I am not sure how to show that it is divisible by $72$.
Please feel free to share your ideas on how to solve this problem.
Thank you in advance.

答:
Just use the form $6n+3$ the way you used the form $2n+1$.

$$(6n+3)^2 - (6k+3)^2 = 36(n^2 - k^2 +n-k) = 36(n-k)(n+k+1).$$

If $n$ and $k$ have the same parity, then $n-k$ is even.  If they
have opposite parity, then $n+k+1$ is even.


问:
show that for an arbitrary continuous r.v. $X, E(X) = - \int_{-\infty}^0F_x(x)dx+\int_0^\infty(1-F_x(x))dx$
The solution defines $X^+ = max(0,X),X^-=min(X,0)$ so $X = X^++X^-,
E(X)= E(X^+)+E(X^-)$

then it states that $X^+ = \int_0^\infty \mathbb{I}_{\{X^+>x\}}$. This part confuses me because my understanding is that X takes individual x from the event space. As $x$ changes between $0$ and $\infty, \mathbb{I}_{\{X^+>x\}}$ decreases. However, after integrating, we can clearly see that $X^+$ is monotonic increasing. Hence, how can $X^+$ be any arbitrary r.v?

from the previous line it is derived $E(X^+) = \int_0^\infty\mathbb{P}(X>x)dx$ (which I can understand in the discrete case)

答:
Would you agree that the equation
$$
a = \int_0^\infty \mathbb I_{\{a>x\}}\,dx
$$
holds for all nonnegative real numbers, $a$? Once you believe that, you only need the following fact; if a (deterministic) equation $f(a)=g(a)$ holds for all nonnegative values of $a$, then it also holds for a randomly chosen nonnegative value of $a$ with probability one. Therefore, it is true when you substitute $X^+$ for $a$.

To conclude $E[X^+]=\int_0^\infty P(X^+>\infty)\,dx$, you need to use Fubini's theorem to switch the order of integration and $E$. It is allowed in this case since $\mathbb I_{\{X^+>x\}}$ is always nonnegative (in fact, it is always $0$ or $1$).


问:
Continuous semigroup
Let $X$ Banach, suppose that $A: D(A) \subset X \to X$ is the infinitesimal generator of a strongly continuous semigroup. $e^{At}, t \geq 0$. Then is the following number finite?

$$M=\sup_{0 \leq t \leq T} |{e^{At}}|_{L(X)}$$

答:
Yes, it is.  First note that there exists $\delta>0$ such that $\sup_{t\leq \delta}\|e^{tA}\|<\infty$. For if not, there would be a sequence null sequence $(\delta_n)$ such that $\|e^{\delta_n A}\|\to\infty$. By the uniform boundedness principle, this implies the existence of $x\in X$ such that $\|e^{\delta_n A}x\|$ is unbounded, contradicting the strong continuity of $(e^{tA})$.

Now if $T>0$ is arbitrary, there exists $N\in \mathbb{N}$ such that $T/N\leq \delta$, and for $t\in [0,T]$ we have
$$
\sup_{t\leq T}\|e^{tA}\|=\sup_{t\leq T}\|(e^{t/N A})^N\|\leq \sup_{s\leq \delta}\|e^{sA}\|^N<\infty.
$$

Title: Continuous semigroup


问:
Prove a space is totally bounded
Consider the set of $\{0, 1\}$ sequences,

$X = \{(a_1, a_2, . . .) : a_i ∈ \{0, 1\}\}$

with the following metric
$d((a_n),(b_n)) := 1/k$, where $a_k \neq b_k $ and $a_n = b_n$ for all $n < k$.

Prove the space $X$ is totally bounded; that is, for every $ε > 0$ there exists a finite set

$F_\epsilon = \{x^1 , . . . , x^N \} ⊂ X$ such that $ min_i\ d(x^i, x) < ε $ for any $x ∈ X$

答:
Let $\epsilon >0$,  consider an integer $n$ such that ${1\over n}<\epsilon$.

Let $P(n)$ the set whose elements are maps $\{0,..,,n\}\rightarrow \{0,1\}$ for every $y\in P(n)$, consider $x_y$ such that $(x_y)_i=f(i), i\leq n, (x_y)_i=0, i>n$. For every $x\in X$, there exists $x_y$ such that the first $n$-coordinates of $x_y$ and $x$ are equal, it implies that $d(x,x_y)<{1\over n}<\epsilon$.

Title: Prove a space is totally bounded


问:
An example in Murphy's $C^*$- algebras

I was told to check that in this example the Gelfand representation is non-isometry nor surjective. So I'm trying to write a proof for it.

My attempt:

Our Gelfand representation is  $\phi: A \rightarrow C_0(\Omega(A))$, defined by $a \rightarrow \hat{a}$, where $\Omega(A)$ is the set of characters on $A$.

Show it's non-isometry: we want to show $\|\phi a\|= \| \hat{a}\| \ne \| a\|$. This is equivalent to showing that the map is not injective, i.e $Ker(\phi)\ne \{0 \}$. I feel like somehow I need to use that if $f' \in A$, then $f' \in C([0,1])$ to show the kernel is not zero. But I'm not sure what to do from here.

Show it's not surjective: We want to find $\hat{a} \in C_0(\Omega(A)) $ such that nothing in $A$ will be mapped to $\hat{a}$. I'm not sure how to proceed, Again I feel like I'm missing something from that  $f' \in C([0,1])$.

Any help will be appreciated.

答:
It is not hard to see that the arguments that show that $\Omega(C[0,1])$ are the pointwise evaluations still work for $\Omega(A)$.

Now you can take the $x$ from the example (the identity). You have $\|x\|=2$, $\|\hat x\|=1$. So the Gelfand transform is not an isometry. It is injective, though.

If $f\in C[0,1]$ is not differentiable, it cannot be the image of some $g\in A$. This is because the characters are the point evaluations: if $f=\phi(g)$ we would have, for any $a\in[0,1]$,
$$
g(a)=a(g)=\hat g(a)=f(a).
$$
So $\phi$ is not surjective.

Title: An example in Murphy's $C^*$- algebras


问:
Stochastic differential equation problem and applying ito formula
I am given that for $b,a,\sigma >0$ and $x \in (-a,b)$ and $\nu \in \mathbb{R}$, I have the following stochastic differential equation:

$$ dZ_t = \nu \,dt + \sigma\, dW_t$$
$$ Z(0) = x$$

and $\tau = \inf\{t>0:Z(t)=b \text{   or   } Z(t) = -a\}$

and $u(x) = b\mathbb{P}[Z(\tau)=b\mid Z_0 = x] -a\mathbb{P}[Z(\tau)=-a\mid Z_0 = x]$, I want to find

$\mathbb{E}[u(Z(\tau))\mid Z_0 = x] - u(x)$ and the differential eq that $u(x)$ satisfies.

Attempt at solution:
We know that for a starting point $x$, $u(x) = b\mathbb{P}[Z(\tau)=b\mid Z_0 = x] -a\mathbb{P}[Z(\tau)=-a\mid Z_0 = x]$ so then $u(Z(\tau)) = b\mathbb{P}[Z(\tau)=b\mid Z_0 = Z(\tau)] -a\mathbb{P}[Z(\tau)=-a\mid Z_0 = Z(\tau)]$  but $\mathbb{P}[Z(\tau)=b\mid Z_0 = Z(\tau)] = 1/2 = \mathbb{P}[Z(\tau)=-a\mid Z_0 = Z(\tau)]$ So then $u(Z(\tau)) = \frac{1}{2}b + \frac{1}{2}a$ But this seems wrong since it doesnt depend on time? Also how would I apply Ito to $u(x)$? It is a function of a probability of $Z(\tau)$. How would I write it as an explicit function since I cant take partials of probability?

Any help would be greatly appreciated! Thanks!

答:
After "so then", you seem to use the identity that if $v(x)=P[U=b\mid V=x]$ for every $x$ then $v(U)=P[U=b\mid V=U]$. This is wrong, and you could have been suspicious about this, due to the fact that in the setting of the exercise, the event $U=V$ is impossible.

Instead, note that if $U$ takes values in $\{u_k\}_k$, then $\sum\limits_{k}u_kP[U=u_k\mid V=x]=E[U\mid V=x]$, by definition.

ContentLicense: CC BY-SA 3.0


问:
Does analytic imply mixed paritials exist?
Let $f(z,w)$ be defined on $\Bbb{C}\times\Bbb{C}$. Assume for each fixed $z$, $f(z,w)$ is analytic for all $w$. Likewise, assume for each fixed $w$, $f(z,w)$ is analytic for all $z$.

Is it the case that the mixed partial derivatives must exist? That is, WLOG, for a fixed $w_0$, will $$\left.\frac{\partial}{\partial w}f(z,w)\right|_{w=w_0}$$ be analytic for all $z$? If not, what is the counterexample?

答: Yes, this is a famous theorem of Hartogs: If $f$ is separately holomorphic in $z,w,$ then $f$ is locally representable by power series in $z,w.$


问:
Show a random variable is independent of a sigma algebra
I am stuck with the following doubt (for which I cannot find a proof, or a counter example). Let us have $(\mathcal{F}_t)_{t \geq0}$ as a filtration on $(\Omega,\mathcal{F}, \mathbb{P})$. Let us have a process $X = (X_t)_{t \geq0}$

Is it true that:
$$\text{if} \ \ \ \ \ \mathbb{E}\left[f( X_t) | \mathcal{F}_s \right] = a$$
$$\text{and} \ \ \ \ \ \mathbb{E}\left[f( X_t) \right] = a$$
$$\text{then} \ \ \ \ \ \ X_t \ \ \text{ is independent of} \ \ \ \ \mathcal{F}_s ?$$
where $f()$ is a deterministic, continuous function and $a \in \mathbb{R}$.

答:
If this is true for every deterministic continuous function $f$, then you do have $X_t$ is independent of $\mathcal F_s$.  The reason is that you can approximate the indicator function of a Borel set $B$ by continuous functions to show that it holds for indicator functions, then using the definition of conditional expectation we have for every $A \in \mathcal F_s$ that
\begin{align*}
\mathbb{E}[1_A 1_{X_t \in B}] = \mathbb{E}[1_A \mathbb{E}[1_{X_t \in B}|\mathcal F_s]] = \mathbb{E}[1_A]\mathbb{E}[1_{X_t \in B}] = P(A)P(X_t \in B).
\end{align*}
Since $\sigma(X)$ is generated by sets of the form $\{X_t \in B\}$, this shows $X_t$ is independent of $\mathcal F_s$.

If you just know it is true for a particular deterministic continuous function $f$, you can't say anything.  For example, the constant function $f(x)=a$ for all $x$ satisfies those conditions for any $X_t$ and $\mathcal F_s$.


问:
How to show this integral inequality?
For $t \in \mathbb{R}$, let
$$
  I_1(t) = \int_{0}^1 \frac{x e^x}{1 + \cosh(t x)} dx
$$
and
$$
  I_2(t) = \int_{0}^\infty \frac{x e^{-x}}{1 + \cosh(t x)} dx.
$$

I want to show $I_1(t) \geq I_2(t)$ with equality only if $t = 0$. Numerical integration suggests this is true. Any suggestions?

答:
This is only a partial answer:

We have $I_1 (0) = I_2 (0) = \frac{1}{2}$ and since $I_1$ and $I_2$ are symmetric functions we can assume $t > 0$ from now on. Then
\begin{align}
I_1 (t) &= \int \limits_0^1 \frac{x \mathrm{e}^{x}}{1 + \cosh(tx)} \, \mathrm{d} x \stackrel{t x = 2 u}{=} \frac{2}{t^2} \int \limits_0^{t/2} \frac{u \mathrm{e}^{2u/t}}{\cosh^2(u)} \, \mathrm{d} u \, , \\
I_2 (t) &= \int \limits_0^\infty \frac{x \mathrm{e}^{-x}}{1 + \cosh(tx)} \, \mathrm{d} x \stackrel{t x = 2 u}{=} \frac{2}{t^2} \int \limits_0^\infty \frac{u \mathrm{e}^{-2u/t}}{\cosh^2(u)} \, \mathrm{d} u \, .
\end{align}
Using Taylor series and a few known integrals we can derive the asymptotic behaviour of the integrals (working with the $x$-integrals for small and the $u$-integrals for large values of $t$, of course):
\begin{align}
I_1(t) &\sim \begin{cases}\frac{1}{2} \left[1 - \frac{3 - \mathrm{e}}{2} t^2 + \mathcal{O} (t^4)\right] , & t \to 0^+ \\ \frac{2 \log(2)}{t^2} \left[1 + \frac{\pi^2}{6 \log(2) t} + \mathcal{O} \left(\frac{1}{t^2}\right)\right], & t \to \infty\end{cases}, \\
I_2(t) &\sim \begin{cases}\frac{1}{2} \left[1 - \frac{3}{2} t^2 + \mathcal{O} (t^4)\right] , & t \to 0^+ \\ \frac{2 \log(2)}{t^2} \left[1 - \frac{\pi^2}{6 \log(2) t} + \mathcal{O} \left(\frac{1}{t^2}\right)\right], & t \to \infty\end{cases}.
\end{align}
These results show that $I_1(t) > I_2(t)$ holds for sufficiently small or large $t$.

In order to obtain the inequality on a larger interval we write
\begin{align}
I_2 (t) &= 2 \int \limits_0^\infty x \mathrm{e}^{-x} \frac{\mathrm{e}^{-t x}}{(1 + \mathrm{e}^{-t x})^2} \, \mathrm{d} x \stackrel{\text{IBP}}{=} \frac{2}{t} \int \limits_0^\infty \frac{(x-1) \mathrm{e}^{-x}}{1 + \mathrm{e}^{-tx}} \, \mathrm{d}x \\
&= \frac{2}{t} \sum \limits_{k=0}^\infty (-1)^k \int \limits_0^\infty (x-1) \mathrm{e}^{-(1+kt)x} \, \mathrm{d} x = 2 \sum \limits_{k=1}^\infty \frac{(-1)^{k-1} k}{(1+kt)^2}
\end{align}
(interchanging integration and summation is justified by the dominated convergence theorem). Doing the same for $I_1$ we find the series representation
$$ I_1(t) - I_2(t) = \frac{2}{t} \sum \limits_{k=1}^\infty (-1)^{k-1} f(kt) \, , \tag{1}$$
where $f \colon \mathbb{R}^+ \to \mathbb{R}^+$ is defined by
$$ f(r) = r \left(\frac{1 - r \mathrm{e}^{1-r}}{(1-r)^2} - \frac{1}{(1+r)^2}\right) \, . $$
The apparent singularity at $r=1$ is removed by setting $f(1) = \frac{1}{4}$. $f$ is indeed a positive function, since letting $r = \frac{1 - s}{1 + s}$ and rearranging the inequality shows that
$$ f(r) > 0 ~ \forall ~ r \in (0,\infty) \setminus \{1\} ~ \Leftrightarrow ~ \log(1+s) > \frac{s}{1+s} ~ \forall ~ s \in (-1,1) \setminus \{0\}$$
and this elementary inequality is true. Moreover, taking the derivative we see that $f$ increases from $f(0^+) = 0$ to a maximum at $r_0 \simeq 1.8143$ before monotonically approaching zero again. In particular, the sequence $(f(kt))_{k \in \mathbb{N}}$ is strictly decreasing to zero if $t \geq r_0$ (and, in fact, for some smaller $t$ as well). But this implies that the alternating series in equation $(1)$ is positive, so $I_1(t) > I_2(t)$ holds for $t \in [r_0,\infty)$.

This method does not work on the remaining intervall $(0,r_0)$, however (ironically, this is where $I_1(t) - I_2(t)$ is largest numerically). At the moment, I can only think of piecing together a few Taylor expansions and their error estimates in this region. We have, for example ($\operatorname{Li}_2$ is the dilogarithm),
\begin{align}
I_1 (1) &= \frac{\pi^2}{6} - 2 \log(2) + 2 \left[2 \log(1 + \mathrm{e}) + \operatorname{Li}_2 \left(-\mathrm{e}\right) - \frac{\mathrm{e}}{1 + \mathrm{e}}\right] \, , \\
I_2 (1) &= \frac{\pi^2}{6} - 2 \log(2)
\end{align}
and the expression in the square brackets is positive, so the inequality holds in some neighbourhood $(\varepsilon, 2 -\varepsilon)$ of $1$. It should be possible (but probably tedious) to make $\varepsilon > 0$ small using Taylor polynomials. Combining this with the previous results would then complete the proof, but this is clearly not a very elegant method.

Title: How to show this integral inequality?


问:
Showing singularity when eigenvalue $\lambda = 1$
Lets say I have matrix $M$ with all its entries $\geq 0$. Then how can I show that given $M$ has an eigenvalue $\lambda = 1$, then $(I-M)$ is not invertible. So far my attempt considers the system $Mx=x$ where you find that $x = [1  1...1]^T$. However, I am not sure where to go from here to show whether $(I-M)$ is not invertible.

答: Hint: $Mx = x \iff Mx = Ix \iff Mx-Ix = 0 \iff (M-I)x = 0$


问:
commutant of the corner of a von Neumann algebra
Let $M$ be a von Neumann algebra and $p$ is a projection in $M$. Does there exist relationship between the commutant $(pMp)'$ of $pMp$ and $pM'p$.

I know the fact that if $p\in M'$, we have $(pMp)'=pM'p.$

答:
Let $N=M'$. Then $p\in N'$. So, by the fact you know,
$$
(pM'p)'=(pNp)'=pN'p=pMp.
$$
Taking commutant (and here you need to be careful when taking double commutants of degenerate things),
$$\tag1
pM'p+(1-p)B(H)(1-p)=(pMp)'.
$$
If you want to consider $pMp\subset B(pH)$, then
$$
pM'p=(pMp)'.
$$


问:
Do there exist any numbers such that any in-order combination of digits is prime (including the original number)?
I was inspired by this question, in particular, where removing any digit of a prime yielded another prime. My question is, are there any numbers that this holds true for, and will continue holding true for if digits continue to be removed?

I have a strong feeling that this is false. However, I'm very interested in seeing a proof for the falsehood of this, or maybe even a proof that one does exist but is not computable within reason.

答: Found this when I went looking for numbers where any combination of their digits result in a prime, since earlier I happened to stumble upon this property of 3 digit numbers comprised of two 1's and one 3, those being 113, 131, and 311, which are all prime, and all satisfy the "take any 1 digit and a prime will remain" condition, though they all start with a crutch since they're comprised of only 2 unique digits. I also happened to find something interesting while testing the combinations of the digits in the number provided in the other post you linked, there are two of the six combinations which were not prime, 791 and 917, both of which had a 7 and a 3 digit number comprised of two 1's and one 3 as factors, 791 = 7 * 113 and 917 = 7 * 131, it's oddly coincidental how the example managed to connect to my number at all, but I thought it was really interesting. Also I'm 9 years late lmao</p>


问:
distance point to affine subspace
I have $n$ points $x_i\in\mathbb{R}^N$, $n \le N$. Is there a general formula for the (squared) distance of another point $b$ to the subspace defined by the $x_i$? (Note that I'm not only looking for hyperplanes; the codimension can be larger than 1.)

答:
Move all points such that the plane of the $x_i$ passes through the origin, e.g.,
$\tilde{x}_i = x_i - x_0$. Every point $y$ in the new plane can be written as
$$
y = \sum_i \alpha_i \tilde{x}_i
$$
with any $\alpha_i$. This leaves us with the least-squares problem
$$
\|b - A\alpha\|^2 \to \min_\alpha
$$
where $A$ is the column-matrix of the $\tilde{x}_i$. This can be solved in multiple ways (SVD, QR, normal equation etc.). One possibility is to create an orthogonal basis in the space, e.g., by (modified) Gram-Schmidt, and project $b - x_0$ onto it.

Title: distance point to affine subspace


问:
CGA Representation of an Ellipse
In conformal geometric algebra (CGA), I know the representation of a circle $s$ with center $c$ is:

$$
S = H(c) - \frac{1}{2}r^2n_{\infty}
$$

That $H(c)$ is the representation of $c$ (i.e., $n_0 + c + \frac{1}{2}c^2n_\infty$). Notice that $n_\infty$ is also denoted by $e_\infty$. Now my question is what is the representation of an ellipse with centers of $c_1$ and $c_2$ under these notations in CGA?

答: You might be interested by Double Conformal Geometric Algebra, Cl(8,2) https://link.springer.com/article/10.1007/s00006-017-0784-0 "DCGA is an extension of CGA and has entities representing points and general (quartic) Darboux cyclide surfaces in Euclidean 3D space, including circular tori and all quadrics"</p>

Title: CGA Representation of an Ellipse


问:
Isomorphism of quotient rings with Cartesian product
If $I$ and $J$ are ideals in $R$ and $S$ respectively, how do I show that $(R\times S)/(I\times J)\cong(R/I)\times(S/J)$? I started by showing that $I\times J$ is an ideal in $R\times S$ but am unsure how to proceed from here. Is the First Isomorphism Theorem perhaps useful here?

答: Hint: Show that $I \times J$ is the kernel of the surjection $R \times S \to (R/I) \times (S/J)$. Then use the first isomorphism theorem.


问:
Existence of a function satisfying given conditions
I was going through the topic of $Function$, its boundedness, continuity etc. I got a problem.

Does there exist a function defined on the closed interval $[a,b]$ which is....

1. bounded;

2. takes its maximum and minimum values;

3. takes all its values between the maximum and minimum values;

Then can we conclude that then this function is continuous at some points or subintervals on
$[a,b]$.

答: No. Take $f:[0,1]\to[0,1]$ such that $f(x) = x$ if $x$ is rational and $f(x)=1-x$ if $x$ is irrational. Well, okay, that's continuous at $x=\frac{1}{2}$, so just set $f(0) = \frac{1}{2}$ and $f(1/2) = 0$ and use the rule I gave above for all other values of $x$. :)</p>


问:
Why is the power of maximal ideals comaximal in an Artinian ring?
Let $R$ be an Artinian ring, commutative with 1.

We know :

there are only finitely many maximal ideals of $R$.

$Jac(R)^m = 0$ for some natural number $m$, using D.C.C.

Every prime ideal is maximal, using 2.

The backdrop of this question is we want to show $R$ is isomorphic to the product of some Artinian local rings. In the process of doing this, Dummit & Foote uses Chinese Remainder Theorem, which requires that the collection of powers of maximal ideals $\{M_1^m, ..., M_n^m\}$ is pairwise comaximal, where $m$ is the number from 2 above. I do not know why this comaximality is true.

Reference, Dummit & Foote 3ed, 16.1, page 753.

答: A more general fact. If $A, B$ are comaximal, write $1=a+b, a\in A, b\in B$. Given $A^n, B^m$, write $1 = (a+b)^{n+m-1} \in A^n + B^m$ because in the binomial expansion, one of the powers is always greater than or equal to n or m. Then $A^n + B^m = R$.


问:
Partial derivative of matrix w.r.t. its eigenvalue and eigenvector
Standard algorithms have been proposed to compute the partial derivative of eigenvalue and eigenvector w.r.t. the matrix, e.g. http://www.win.tue.nl/casa/meetings/seminar/previous/_abstract051019_files/Presentation.pdf.

However, as far as I know, no solution has derived for the partial derivative of the matrix w.r.t. its eigenvalue and eigenvector. Suppose the matrix $A\in \mathbb{R}^{n \times n}$ with eigenvalue $\lambda\in \mathbb{R}$ and eigenvector $X\in \mathbb{R}^{n}$, i.e. $ AX = \lambda X$,the problem is to compute $\frac{\partial A}{\partial \lambda}$ and $\frac{\partial A}{\partial X}$. Does anyone have some ideas? Thanks for your help!

答:
$\def\L{\left}\def\R{\right}$Since no one has yet addressed it,
here is the second of the requested gradients.

Using the same assumptions as Victor, calculate the differential wrt $X$
$$\eqalign{
 A &= X\Lambda X^{-1} \\
dA &= dX\Lambda X^{-1} + X\Lambda dX^{-1} \\
   &= dX\Lambda X^{-1} - X\Lambda X^{-1}\,dX\,X^{-1} \\
   &= I_n\,dX\Lambda X^{-1} - A\,dX\,X^{-1} \\
}$$
Vectorizing both sides of this expression yields the desired gradient
$$\eqalign{
da &=  \Big((\Lambda X^{-1})^T\otimes I_n - X^{-T}\otimes A\Big)\,dx \\
 &= \L(X^{-1}\otimes I_n\R)^T\L(\Lambda\otimes I_n-I_n\otimes A\R)dx \\
\frac{\partial a}{\partial x}
 &=  \L(X^{-1}\otimes I_n\R)^T \L(\Lambda\otimes I_n - I_n\otimes A\R) \\
}$$


答:
Consider the canonical surjective homomorphism (epimorphism): $R\rightarrow R/I:r\mapsto r+I$ with kernel $I$.

This gives rise to the epimorphism $\phi:R\times S\rightarrow R/I\times S/J: (r,s)\mapsto (r+I,s+J)$.

The kernel is $I\times J$ and so by the Homomophism theorem:

$(R\times S)/\ker(\phi)$ is isomorphic to $im(\phi)= R/I\times S/J$.

Title: Isomorphism of quotient rings with Cartesian product

Body: If $I$ and $J$ are ideals in $R$ and $S$ respectively, how do I show that $(R\times S)/(I\times J)\cong(R/I)\times(S/J)$? I started by showing that $I\times J$ is an ideal in $R\times S$ but am unsure how to proceed from here. Is the First Isomorphism Theorem perhaps useful here?


问:
The cycle structure of the permutation $a \mapsto ma \bmod{n}$
Given an odd $n$, and an $m$ such that $(n,m)=1$, i would like to know what is the cycle structure of the permutation $\pi_{n,m} (a)=ma\bmod{n}$.

Specifically, how do i know if $\pi_{n,m}$ and $\pi_{n,k}$ have the same structure.

Even more specifically, do $\pi_{n,m}$ and $\pi_{n,m^{-1}}$ have the same structure, when $m\cdot{m^{-1}}=1\bmod{n}$.

Thanks!

答: Those last two permutations are inverse to each other, no? What do you know about the cycle structure of a permutation and its inverse?</p>


答:
Function below satisfies all three conditions above but it is
discontinuous at every point on $[-1,1]$

$f(x)=\begin{cases}
1,&\text{if $x = 0$}\\
x,&\text{if x is rational, $x \neq 0$ , $x\neq 1$}\\
-x,&\text{if x is rational, $x \neq 0$ , $x\neq 1$,  $x\neq -1$} \\
0,&\text{if x = 1}
\end{cases}$

It is impossible to draw the graph of the function $y = f(x)$ but the sketch below gives an idea of its behavior.

Hence answer is no.

You can find this example in counter examples in calculus.

Title: Existence of a function satisfying given conditions

Body:
I was going through the topic of $Function$, its boundedness, continuity etc. I got a problem.

Does there exist a function defined on the closed interval $[a,b]$ which is....

1. bounded;

2. takes its maximum and minimum values;

3. takes all its values between the maximum and minimum values;

Then can we conclude that then this function is continuous at some points or subintervals on
$[a,b]$.


问:
Homotop a homotopy-equivalence of $S_g$ to make the inverse image of an embedded circle to be a single circle
Let $\varphi:S_g\to S_g$ be a smooth homotopy equivalence on genus-$g$ surface, where $g\geq 2$. Let $C$ be an embedded circle in $S_g$ such that $\varphi$ is transversal to $C$. Therefore, $\varphi^{-1}(C)$ is an embedded $1$-dimensional submanifold of $S_g$. Using Stack-Record theorem one can show $\varphi\big|\varphi^{-1}(C)\to C$ is a finite covering map. Actually, $\varphi^{-1}(C)$ is disjoint union of finitely many embedded circles in $S_g$.

Is it possible to homotop $\varphi$ so that $\varphi$ is again transverse to $C$ with $\varphi^{-1}(C)$ is a single embedded circle?

Note that $\varphi$ is a degree $\pm1$ map. So, after homotopy, we can not expect $\varphi^{-1}(C)=\varnothing$ as homotopy preserves degree and degree $\pm 1$ maps are surjective.

答:
Yes, this can be done, and in fact this is an important step in the proof of the theorem that every self-homotopy equivalence of $S_g$ is homotopic to a homeomorphism. I think you'll find at least a sketch of this proof (and that step) in the book of Farb and Margalit, and you should also find full references there. You might also look at the paper by Epstein, "Curves on 2-manifolds and isotopies", Acta Math. 115 (1966), 83–107.

Here's an outline.

Step 1: Prove that if $\varphi^{-1}(C)$ is not connected then one of two things occurs:

Some component of $\varphi^{-1}(C)$ bounds an embedded disc in $S_g$; or

Some pair of components of $\varphi^{-1}(C)$ bound an embedded annulus in $S_g$.

The reason this is true is that if two disjoint circles $\gamma_1,\gamma_2 \subset S_g$ are homotopic to each other then either they each bound discs or the pair of them bounds an annulus. I would say that the best modern proof (best both in intuition and rigor) uses hyperbolic geometry to prove that if a simple closed curve does not bound a disc then it is isotopic to a unique simple closed geodesic.

Step 2: Improve Step 1 so that:

Some component of $\varphi^{-1}(C)$ bounds an embedded disc $D \subset S_g$ such that the interior of $D$ is disjoint from $\varphi^{-1}(C)$; or

Some pair of components of $\varphi^{-1}(C)$ bound an embedded annulus $A \subset S_g$ such that the interior of $A$ is disjoint from $\varphi^{-1}(C)$.

To obtain a disc or annulus satisfying 3 or 4, simply take an innermost one (with respect to inclusion) of the collection of discs and annuli satisfying 1 or 2.

Step 3: If one obtains a disc $D$ as in 3, one may then homotope $\varphi$ on a small open neighborhood of $D$ so as to remove $\partial D$ from the set of components of $\varphi^{-1}(C)$. To do this, one uses transversality to find a disc $D'$ whose interior contains $D$, with $\gamma = \partial D'$, such that $\varphi(\gamma)$ lies on a circle $C'$ which is disjoint from but close to $C$ and is isotopic to $C$. The restricted map $\gamma \to C'$ is null-homotopic, so we may replace the restriction $\varphi \mid \hat D$ by a map whose image is contained in $C'$.

Step 4: Assuming that there are no such discs as in 3, take an annulus as in 4. Note in this case that $C$ itself does not bound a disc.

One may then homotope $\varphi$ to remove the two component of $\partial A$ from the list of component of $\varphi^{-1}(C)$. To do this, one first uses transversality to obtain an annulus $A' \subset S$ whose interior contains $A$, with $\partial A' = \gamma_1 \cup \gamma_2$, such that $\varphi(\gamma) \cup \varphi(\gamma_2)$ lies on a circle $C'$ which is disjoint from but close to $C$ and is isotopic to $C$. The circle $C'$, like $C$, is $\pi_1$-injective. Using this one then checks that the restrictions of $\varphi$ to the two boundary circles $\gamma_1,\gamma_2$ represent the same element of $\pi_1 C' = \mathbb Z$, and therefore the restricted map $\varphi : \partial A' \to C'$ may be extended to a map $A' \to C'$.


问:
Confused about why the conormal exact sequence is what it is on a scheme
Consider a composition of morphisms of schemes,
$$Z \stackrel{j}{\longrightarrow} X \stackrel{f}{\longrightarrow} Y,$$
where $j: Z \rightarrow X$ is a closed immersion with sheaf of ideals $\mathscr{I}$. The conormal exact sequence for the sheaves of differentials should read,
$$
j^{*}(\mathscr{I} / \mathscr{I}^{2}) \longrightarrow j^{*} \Omega_{X / Y} \longrightarrow \Omega_{Z/Y} \longrightarrow 0.
$$
I am trying to prove this via the corresponding conormal sequence for modules. So suppose the composition of morphisms of schemes above are of affine schemes. Say $X = \text{spec}B$, $Y = \text{spec}A$, and $Z = \text{spec}C$, with $C \simeq B / I$ for some ideal $I$ of $B$. Then the conormal exact sequence of modules reads,
$$
I / I^{2} \longrightarrow \Omega_{B/A} \otimes_{B} C \longrightarrow \Omega_{C/A} \longrightarrow 0,
$$
where this is an exact sequence of $C$-modules. Now we just have to take the corresponding sequence of quasicoherent $\mathcal{O}_{Z}$-modules. We are viewing $I/I^{2}$ as a $C$-module via the isomorphism,
$$
I \otimes_{B} C \simeq I \otimes_{B} B / I \simeq I / I^{2}.
$$
But then $I/I^{2}$ is just the $B$-module $I$ with the change of ring $- \otimes_{B} C$ applied. In that case, the exact sequence of $\mathcal{O}_{Z}$-modules should just be
$$
j^{*}\mathscr{I}  \longrightarrow j^{*} \Omega_{X / Y} \longrightarrow \Omega_{Z/Y} \longrightarrow 0.
$$
So where is my flaw in reasoning? Are $j^{*}\mathscr{I}$ and $j^{*}(\mathscr{I} / \mathscr{I}^{2})$ somehow canonically isomorphic or what?

答:
Yes, $j^* \mathscr{I}$ and $j^* \mathscr{I}/\mathscr{I}^2$ are canonically isomorphic in this case. Working affine locally we see $j^* \mathscr{I} = I \otimes_B B/I \cong I/I^2$. The last isomorphism is an iso of $B$-modules, but obviously $I \otimes_B B/I$ has the structure of a $C = B/I$-module, so we can give $I/I^2$ the induced $C$-module structure, and this is still an isomorphism of $C$-modules.

Similarly, locally we see $j^* I/I^2 = I/I^2 \otimes_B B/I \cong (I/I^2)/(I(I/I^2)) = (I/I^2)/0 = I/I^2$. I think this should be thought of as saying that $I/I^2$ has a canonical $C$-module structure.

More generally, if $j : Z \hookrightarrow X$ is a closed embedding with ideal sheaf $\mathscr{I}$, then quasicoherent sheaves on $Z$ are equivalent to qcoh $j_* \mathcal{O}_Z$-modules. I think the explicit correspondence is given by $j_*$ and $j^*$. Looking at the closed embedding SES we see $j_* \mathcal{O}_Z = \mathcal{O}_X/\mathscr{I}$. As $\mathscr{I}/\mathscr{I}^2$ is an $\mathcal{O}_X$-module, but also naturally an $\mathcal{O}_X/\mathscr{I}$-module, it can be thought of as a qcoh $\mathcal{O}_Z$-module. This is why the $j^*$ is sometimes ommited when defining the cornormal sheaf. So while $j^* \mathscr{I}$ is isomorphic to $j^* \mathscr{I}/\mathscr{I}^2$, it makes sense to think of $\mathscr{I}/\mathscr{I}^2$ as a sheaf on both $Z$ and $X$, but $\mathscr{I}$ has no natural intepretation as a sheaf on $Z$.

Some other possibly helpful answers are Pullback of ideal sheaf defining closed immersion and Conormal Sheaf (Morphisms of Schemes, Stacks Project).


问:
Problem in Double Integral by Change of Order

$$\text{Evaluate by changing the order} \int^1_0\int^y_{4y}e^{x^{2}}dx\ dy.$$

I am unable to solve the following question. I have tried using the following approach. First I formed the equation of lines using the limits of the inner integral which gave equations of $2$ lines.

$$ y = x $$

$$ y = x/4 $$

and the limits of the outer integral gave me the total bounded region as below

From what I understood, I have to calculate the volume of function $e^{x^2}$ within the bounded area.
Since, from the current order, calculating the integral is difficult, I tried to change the order of integral (as given in question) so this is what I did :

First I thought the current integral divides the current region into small $dA$ and first integrating by $dx$ means that we are taking a strip parallel to $X$-axis with length between $y = x$ and $y = x/4$ and then integrated it all the way above from 0 to 1.

Then I tried to change this order and thought of integrating first w.r.t. $dy$ as this will mean I will divide the region into strips parallel to $Y$-axis but the equation will be divided into two with the inner limits of first being $x/4$ to $x$ and second from $x/4$ to $1$. and the outer limit will change to 0 to 1 for first and 1 to 4 for second. I was solving the integral but then I encountered a problem.

I am unable to integrate the highlighted term any further. Please tell me where I went wrong.

Note: I forgot the $x$ in the first term, it will be $\frac{3xe^{x^2}}{4}$

Body:
$$
\Big[ y e^{x^2} \Big]_{y\,:=\,x/4}^{y\,:=\,x} = xe^{x^2} - \frac x4e^{x^2} = \frac3 4 xe^{x^2}.
$$


问:
How to compute an integral with respect to $K(t, dk)$
Let $X$ be a random variable with distribution $F$ and $K: \mathbb{R}^+\times\mathbb{N}\to[0,1]$ defined by $K(t, \{k\}):=F(t)^{k-1}(1-F(t))$. If $u:\mathbb{R}^+\times\mathbb{N}\to\mathbb{R}^+$, how does one compute something like
\begin{align*}
\int e^{-u(t,k)}\,K(t, dk)\,?
\end{align*}
I am just very confused on the notation here, any help would be greatly appreciated.

答: Your integral is simply $$\sum_{k=1}^\infty e^{-u(t,k)}F(t)^{k-1}(1-F(t)).$$


问:
Prove by Hilbert deduction: $\vdash _{HFOL} \forall x (\neg(A \to \neg B))\to \neg(\forall xA \to \neg(\forall xB))$
I'd really like your help proving:

$\vdash_{HFOL} \forall x (\neg(A \to \neg B))\to \neg(\forall xA \to \neg(\forall xB))$

Where $HFOL$ is the proof system which contains the Hilbert relevant axioms:

$A\rightarrow(B \rightarrow A)$

$(A\rightarrow(B\rightarrow C))\to((A\rightarrow B)\rightarrow(A\rightarrow C))$

$(A\rightarrow B)\rightarrow ((A\rightarrow\bar{B})\rightarrow \bar{A})$

$\bar{\bar{A}} \rightarrow A$

Plus the following two axioms:

$\forall x(B\to A)\to (\exists xB \to A)$

$\forall x(A \to B) \to (A \to \forall x B)$

Restricted to use only while $x$ is not a free variable in $A$.

And other two axioms: $\forall xA \to A \{ t/x \}$, $A\{t/x\} \to \exists xA$ where $t$ is free for assignment instead of $x$ in $A$.

Finally theres the $MP$ rule: From $A, A\to B$ we conclude $B$ and the Gen rule which from $A$ we conclude $\forall xA$.

I tried couple of different ways but didn't come with any smart result.

Any suggestions?

答:
Well, you can use the deduction theorem because ∀x(¬(A→¬B)) has no free variables.
After using it you get: ∀x(¬(A→¬B))⊢HFOL¬(∀xA→¬(∀xB))

Now we will build a proof in HFOL:

∀x(¬(A→¬B)) - by assumption.

∀x(¬(A→¬B))→¬(A→¬B) - first axiom.

¬(A→¬B) - MP on 1,2.

¬(A→¬B)→A - because you can prove it on hpc.

A - MP on 3,4.

∀xA - Gen on 5.

¬(A→¬B)→B - because you can prove it on hpc.

B - MP on 3,7.

∀xB - Gen on 8.

¬(∀xA→¬(∀xB)) - because you can prove it on hpc.


问:
limit of f(x)/x, where f'(x)->0
Say $f:[0,\infty) \to \Bbb R$ is differentiable.

Given $\lim_{x \to \infty}f'(x) = 0$, I need to prove that $\lim_{x \to \infty}f(x)/x = 0$.

I tried looking here and using the Lagrange's mean value theorm, without any success (wrong use of the theorm I suppose)
Any ideas?

thanks.

答:
Since we are given that $\lim_{x\to 0}f'(x)=0$, we have by L'Hopital's rule that

$$
\lim_{x \to \infty}\frac{f(x)}{x}=\lim_{x \to \infty}\frac{f'(x)}{1}=0\,.
$$

Title: limit of f(x)/x, where f'(x)->0


问:
how to prove the kernel of a matrix is 0 if a linear transformation of this matrix is surjective?
Good evening everyone,

I have a question of linear algebra, if $M \in Mat_{nxn}(k)$ and $\phi_M:K^n \to K^n$ a linear transformation which sends $v \in K^n$ to $Mv$. Prove that if $\phi_M$ is subjective then $ker(M)=0$.

I have no idea where shall I begin with, can someone give me some help? thanks in advance.

答:
This follows directly from the rank-nullity theorem. For a more direct proof, consider a basis $e_1, \ldots, e_n$ of $K^n$. As $\phi_M$ is surjective, there exist $v_1, \ldots, v_n$ such that $\phi_M(v_i) = e_i$ for $1 \leq i \leq n$.
These $v_i$ are linearly independent (why?) and as a consequence the kernel must be $0$.


问:
Prove the following relation in vector space
Let $(V,+, \cdot)$ be a $\mathbb{K}$ -vector space. Prove the following relations.

$\mathrm{o} \cdot x=\mathrm{o},$ for all $x \in V$

$\lambda \cdot \mathrm{o}=\mathrm{o},$ for all $\lambda \in \mathbb{K}$.

$(-1) \cdot x=-x$ for all $x \in V$.

Here is my idea, if it has some problems or you have to add for me please kindly share an idea for me.

Since $(V,+,\cdot)$ be a $\mathbb{K}$ -vector space.

$\exists \mathrm{o}=0 \in\mathbb{K}$ for all $x\in V$
$$\mathrm{o}x=0x=0=\mathrm{o}$$

$\exists \lambda\in\mathbb{K}, \lambda\cdot\mathrm{o}=\lambda \cdot 0=0=\mathrm{o}$

Since $-x\in V$ such that $(-1)\cdot x=1\cdot(-x)$

Please kindly share or recommend me. Thank in advance!

答:
For the first one, Note that $0.x = (0 + 0).x = 0.x + 0.x$. So $0.x = 0$ for all $x$. About the second equality, use $\lambda.0 = \lambda.(0 + 0) = \lambda.0 + \lambda.0$ to show that $\lambda.0 = 0$ for all $\lambda$. And in order to prove the last one, Note that:
$$(-1).x + x = ((-1) + 1).x = 0.x = 0 $$. Since the additive inverse of every vector is unique, It follows that $(-1).x = -x$ for all $x$.


问:
Can every rational number can be written as an irreducible fraction?
I just watched a proof which showed that the square root of is irrational. It seemed to me that the reason it was asserted irrational was because it was reducible. Therefore, I thought, that this must mean that all rational numbers can be represented by an irreducible fraction. Is this just a part of the definition of rationality or is there a proof that such is true?

Thanks in advance.

答:
It's not part of the definition, but the proof is easy.

Suppose one has a rational number $\frac{m}{n}$ that is reducible. We can assume $n \ge 1$ (divide the numerator and denominator by $-1$ otherwise).

Take an integer $k \ge 2$ which divides both of $m$ and $n$, and so $m' = m/k$ and $n'=n/k$ are smaller integers satisfying $\frac{m'}{n'} = \frac{m}{n}$.

Repeat this process as long as the fraction remains reducible.

One cannot repeat forever, because the sequence of denominators is a sequence of positive integers that gets smaller and smaller, and no such sequence can continue infinitely (essentially one is using mathematical induction in this step).

When the process stops, one has an irreducible fraction representing the same rational number


问:
(Simple) Examples on Non-Commutative Rings
Looks like it is easier to find example of commutative rings rather than
non-commutative rings. Prabably the easiest examples of the former are
$\mathbb{Z}$ and $\mathbb{Z}_n$. We can find elaborations on these two commutative rings in various
literatures including here and here. These are quite simple and easy to comprehend.

However, the examples on simple (non-commutative kind) are not that easy.
One example is found here and it has been mentioned as "one of the simplest examples of a non-commutative ring". This is also on the easier side.

EDIT: The following two are also commutative rings.

Other examples are given in this enumeration. But as you can see, examples like Gaussian integers or Eisenstein integers are difficult for starters to comprehend.

Do you think you can give one or two simple examples on non-commutative rings, based on every day numbers?

If it is that difficult, perhaps some insight comments why this is difficult would be welcome.

答: How about the ring of quaternions? This is an example of a non-commutative ring and I don't think it's super difficult to wrap your head around. Understanding the ring of quaternions is also in fact very useful in understanding representations of rotations and groups of isometries for polyhedra. </p>


答:
The definition of $x$ being rational is that it can be written as a fraction, i.e. as $x=\frac{m}{n}$ where $m,n$ are integers and $n\neq 0$. Of course, there are many different ways to do this for any given rational $x$.

Suppose $x$ is rational and we choose a way to write $x=\frac mn$ for which $n>0$ but is as small as possible. (Note that we can always make $n>0$ by multiplying top and bottom by $-1$, if necessary, so such a choice always exists.) Now we claim $\frac mn$ must be irreducible. Suppose not, so there is some common factor $h>1$ which divides both $m$ and $n$. Then $\frac{m/h}{n/h}=x$, but the denominator is smaller, contradicting our choice of representation.

Title: Can every rational number can be written as an irreducible fraction?

Body:
I just watched a proof which showed that the square root of is irrational. It seemed to me that the reason it was asserted irrational was because it was reducible. Therefore, I thought, that this must mean that all rational numbers can be represented by an irreducible fraction. Is this just a part of the definition of rationality or is there a proof that such is true?

Thanks in advance.


问:
Addition of 2 subspaces.
The questions goes like this:

Write down the subspace $U + V$ explicitly if

$U = {(t, 2t, 3t)}$ and $V = {(t, -2t, 0)}$ where $t$ is a member of the real number

My answer for the question is:

$U + V$ $=$ {$(x + y, 2x - 2y, 3x)$ | $x$, $y$ are member of the real number}

However, I am confused as other have mentioned that the answer should be instead:

$U + V\{(2t, 0, 3t)\mid t \text{ is a member of the real number}\}$

In my opinion, the I feel that the second answer is wrong as the two sub-spaces are not a subset of each other. Is my thinking wrong?

答:
When we say $l$ is in the set $S=\{f(x)\mid x\in A\}$, we mean by $l$ takes the form $f(x)$. More precisely, $l=f(x)$ for some $x\in A$.

Next we define the set $U\oplus V$ by $$x\in U\oplus V \Longleftrightarrow x=y+z \mbox{ for some } y\in U,z\in V.$$ But then the $y,z$ from $U,V$ have their own form, namely $(t, 2t, 3t)$ and $(t, -2t, 0)$ for some real number $t$. (To not confuse ourselves we use different dummy variable, say $r,s$.) $$\Longleftrightarrow x=(r, 2r, 3r)+(s, -2s, 0) \mbox{ for some }r,s\in\mathbb{R}$$
$$\Longleftrightarrow x=(r+s,2r-2s,3r) \mbox{ for some }r,s\in\mathbb{R}$$
So the set $U\oplus V$ is $\{(r+s,2r-2s,3r)\mid r,s\in\mathbb{R}\}$.

The point that you might have misunderstood is the duplicate misuse of dummy variables $r,s$.

Note that we may use $+$ instead of $\oplus$.

Title: Addition of 2 subspaces.


问:
Find $g(x,y,z);p(x,y,z)\ge 0$ so that $f(x,y,z):=x\cdot g(x,y,z)+p(x,y,z)$
We have the following fact: (I don't remember where I read it, but there is.)

If $f(x)$ is which is non-negative for $x\ge 0,$ then $f(x)=g(x)+x\cdot h(x),$
where $g(x)$ and $h(x)$ are SOS.

So I wanna know how to find $f(x,y,z)$ as $$f(x,y,z)=x\cdot g(x,y,z)+p(x,y,z),$$
where $g(x,y,z),p(x,y,z)$ are all non-negative polynomials.

That also means that my problem is:

Given $f(x,y,z),$ which is a non-negative polynomial where $x,y,z\ge 0.$ Find $g(x,y,z)$ and $p(x,y,z)\ge 0$ (for same condition) such as $f(x,y,z)=x\cdot g(x,y,z)+p(x,y,z).$

There would be something like @Haidangel's question for make

$$H:= m\left ( a, b, c \right )+ kn\left ( a, b, c \right )ca;$$
and $$H:=m\!\left(a,b,c\right)\!-n\!\left(a,b,c\right)\!\left(a-b\right)\!\left(b-c\right)$$
where $m(a,b,c),n(a,b,c)$ are non-negative.

But here, I wanna make it as a sum of polynomials.

I tried $$f(x,y,z)=f(0,y,z)+\left[f(x,y,z)-f(0,y,z)\right]\equiv f(0,y,z)+x\cdot h(x,y,z),$$

But it's not the form I like since $h(x,y,z)$ is not certainly as SOS.

Also, AoPS/@dragonheart6 had told me to use The Unknown coefficients but I don't know how to do it here because there are so many coefficients for the higher degree problems.

Finally, my last purpose is to make $f(x,y,z)$ as SOS where the condition of variables is $x,y,z$ are all non-negative.

$\textbf{Note.}$ We know that if replace $x,y,z$ by $a^2,b^2,c^2$ then we have to prove the inequality for real numbers, but now, the degree is very high.

I wish that someone could have a solution and an example about this.

PS. SOS means Sum of Squares.

答:
We have the following fact: (I don't remember where I read it, but there is.)

If $f(x)$ is which is non-negative for $x\ge 0,$ then $f(x)=g(x)+x\cdot h(x),$
where $g(x)$ and $h(x)$ are SOS.

This result can be found in the book by Pólya and Szegö (1st ed, p. 78):

For higher dimensions it is not so easy. You are trying to use Putinar's Positivstellensatz, meaning you are looking for SOS polynomials $g_1$, $g_2$, $g_3$ and $p$ such that:
$$f(x,y,z)=x\cdot g_1(x,y,z)+y\cdot g_2(x,y,z)+z\cdot g_3(x,y,z)+p(x,y,z).$$

The conditions for that theorem are not satisfied for nonnegativity over $\mathbb{R}^n_+$, so not every nonnegative $f$ will admit such a representation with $g_i$ and $p$.

The Krivine–Stengle Positivstellensatz states that every polynomial $f$ that satisfies $f(x)\geq 0$ for all $x\in\mathbb{R}^3_+$ can be decomposed using SOS polynomials $g_1,\ldots,g_6$ as:
$$f(x)\left(x_1\cdot g_4(x)+x_2\cdot g_5(x)+x_3\cdot g_6(x)\right) = f(x)^{2k} + x_1\cdot g_1(x)+x_2\cdot g_2(x)+x_3\cdot g_3(x)$$
for some $k\in\mathbb{N}$. The classic method to find $g_1,\ldots,g_6$ is using semidefinite optimization (which was already mentioned in the comments). YALMIP is another excellent tool for that. This method requires you to have an upper bound on the degree of the polynomials, which you do not have. You also do not know $k$. What you could do is vary $k$ and gradually increase the upper bound and hope you get 'lucky'. I am not aware of any other general method.

A nice background read is Section 3 (3.3 and 3.6 in particular) of Sums of squares, moment matrices and optimization over polynomials by M. Laurent.


问:
Is it meaningful to define the Dirac delta function as infinity at zero?
I am in discussion with someone online on the subject of the Dirac delta function. This other person wants to say:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

and wants to justify it by saying:

We have that:

$$\delta (x) = \lim_{\epsilon \mathop \to 0} F_\epsilon (x)$$

where:

$$F_\epsilon (x) = \begin {cases} 0 & : x < -\epsilon \\ \dfrac 1 {2 \epsilon} & : -\epsilon \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$$

Therefore:

$$\delta (0) = \lim_{\epsilon \mathop \to 0} F_\epsilon (0) = \dfrac 1 {2 \times 0} = \infty$$

and:

$$ \delta {x \ne 0} = \lim_{\epsilon \mathop \to 0} F_\epsilon (x \ne 0) = 0$$

Therefore:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \infty & : x = 0 \end{cases}$$

This comes across as iffy to me. I don't trust $\delta (0) = \infty$ as it is not well-defined exactly what $\infty$ actually means in this context, whereas in fact $\delta$ is precisely defined by means of the limit definition as given above.

I am also seriously unsure about that $\dfrac 1 {2 \times 0}$ in the middle of the exposition, which is at best meaningless and at worst a blasphemous lie.

However, when I consult a number of mathematical works on my shelves and online, there are many of them which give that above definition quite casually, along the lines "As an obvious consequence of the definition: $\delta (0) = \infty$" or some such. Other works heavily italicise the warning that $\delta$ is not a function, and $\delta (0)$ is undefined.

What is the current mode of thought here? I am trying to craft a set of webpages which are mathematically rigorous, but coworkers on the same site are of the mind "It doesn't really matter, we all sort of know what we mean, and hey, $\delta (0) = \infty$ looks really cool!"

When questioning the matter, he is prepared to compromise and say: $\delta (0) \to \infty$ as $x \to 0$, or even:

$$\delta (x) = \begin{cases} 0 & : x \ne 0 \\ \to \infty & : x = 0 \end{cases}$$

I am insufficiently mathematically sophisticated as to be able to explain why I hate this approach, but I seriously dislike bandying the $\infty$ sign around, when (once we get past the obvious intuitive meaning we learn in elementary school) we really don't understand what it means.

EDIT:

As requested, I have actually dug up one of my texts which defines the delta function as the limit of the rectangle function as its width goes to zero, as follows. It comes from I.N. Sneddon's "Special Functions of Mathematical Physics and Chemistry", appendix.

If we consider the function
$$\delta_a (x) = \begin {cases} 0 & : |x| > a \\ \dfrac 1 {2 a} & : |x| < a \end {cases}$$
then it is readily shown that
$$\int_{-\infty}^\infty \delta_a(x) d x = 1.$$

...

We now define
$$\delta(x) = \lim_{a \to 0} \delta_a (x).$$
Letting $a$ tend to zero in equations [above] we see that $\delta(x)$ satisfies the relations
$\delta(x) = 0$, if $x = 0$
$$\int_{-\infty}^\infty \delta (x) d x = 1.$$

答:
I actually disagree with other answers; yes, it is natural to argue that the $\delta$-function does have an infinite value at the origin. If you think of a function as a density on the real line (for example, mass density), then the "physical" definition of density would be
$$
\lim_{\epsilon\to 0} \frac{m([-\epsilon,\epsilon])}{2\epsilon},
$$
the mass of the interval divided by the length of the interval. The $\delta$-function would be a density of a point mass at the origin; that is, the mass will always be $1$ and the density will be infinite.

The critiques raised in the other answers boil down to the remark that this does not tell the whole story, or that you cannot define the $\delta$-function this way. This is absolutely correct, but it is a common feature of infinity; by "replacing" something with infinity you forget some information that may or may not be relevant.

For example, if you have a sequence $a_n\to +\infty$, you may sometimes wish to "forget" the sequence, retaining only the information that $\lim a_n=+\infty$. This is very useful for solving some problems (for example, calculating $\lim (a_nb_n)$ when $b_n$ is bounded from below by a positive constant) and not useful for others (for example, calculating $\lim (a_nb_n)$ when $b_n\to 0$). The scope of usefulness basically boils down to "extending" $\mathbb{R}$ with $\pm\infty$ and partially extending the operations and relations (for example, $x+\infty=\infty$ for any $x\neq -\infty$) while leaving them undefined in some other cases (as in $\infty\cdot 0$).

So, the main problem with defining the $\delta$-function (or, rather, $\delta$-density) is the following: the main use of the density is that you can recover the mass from it by integration. If you try to do it with the $\delta$-density, you naturally run into the classical $+\infty\cdot 0$ problem (infinite density times zero length), which shows that you must have retained more information than just the infinite value an the origin. But to me, it is not per se an argument against the identity $\delta(0)=+\infty$, similarly to the fact that the $+\infty\cdot 0$ indeterminancy does not invalidate the use of infinity altogether.


问:
How do I graph (-2,30) on a polar grid?
I tried to graph the point $(-2,30)$ on a polar graph and I think I got it wrong, how should it be graphed?

答:
Your diagram is a bit confusing, but your answer may be correct.

If your co-ordinates represent $(r, \theta)$, and if $30$ should be $30^\circ$, and if the red dot is the point in question, and if it is two units away from the origin (there is no indication of units), then you are correct.

This is under the usual convention that $r<0$ is valid and represents a point in the opposite direction from $\theta$ at a distance of $|r|$ from the origin.

The green dot in your diagram beside the label $(-2, 30)$ is in fact $(2, 30^\circ)$. (Again, assuming that it's two units away from the origin.) If the final dot is the red dot, it would be better to place the label beside it.

Title: How do I graph (-2,30) on a polar grid?


答:
Hint: for a solution that doesn't use L'Hopital's rule, note that for all $\varepsilon>0$, you can find $M\in \mathbb{R}$ such that $$f'(x)<\varepsilon \quad \forall x>M.$$

You can use this with the mean value theorem to give a bound to $|f(x)|$ in terms of $x$ for all $x$ large enough. $\lim_{x\to \infty}\frac{f(x)}{x}=0$ will follow from this.

Title: limit of f(x)/x, where f'(x)->0

Body:
Say $f:[0,\infty) \to \Bbb R$ is differentiable.

Given $\lim_{x \to \infty}f'(x) = 0$, I need to prove that $\lim_{x \to \infty}f(x)/x = 0$.

I tried looking here and using the Lagrange's mean value theorm, without any success (wrong use of the theorm I suppose)
Any ideas?

thanks.


问:
Differentiation of Rotation Matrix
I have some troubles on differentiating a time-varying rotational matrix $R(t) \in SO(3)$. If I use a axis-angle representation with the Rodrigues formula we have:

$R(t) = \exp(\,A(t) \theta(t)\,) = I + A(t)\sin(\theta(t)) + A^2(t)(1-\cos(\theta(t))) $

where $\vec{a}(t)$ is a time-varying unit vector ($A(t) \in so(3)$ is its matrix representation), and $\theta(t)$ is the time-varying angle.
I would like to compute $ \frac{d}{dt}( R(t) )$. Any hints? Thanks a lot!

答:
Use the usual rules for derivative, keeping in mind that multiplication of matrices is not commutative.
$$
\begin{split}
\frac{d}{dt} I &=0
\\
\frac{d}{dt}  A\sin \theta &= A'\sin \theta +A( \cos \theta)\theta'
\\
\frac{d}{dt} A^2(1-\cos \theta ) &= (AA'+A'A)(1-\cos \theta )+A^2 (\sin \theta )\theta'
\end{split}
$$  </p>

Title: Differentiation of Rotation Matrix


答:
EDIT: As pointed out in the comments, the following example is not associative.

What about the ring of matrices? It is non-commutative and perhaps as easy to understand as the real numbers. You may also consider the subset of invertible matrices.

You can consider the abelian group of matrices and define the multiplication as

$ X*Y := XY - YX $.

Then you have a non-commutative ring. This is in fact the Lie algebra of all matrices.

Title: (Simple) Examples on Non-Commutative Rings

Body:
Looks like it is easier to find example of commutative rings rather than
non-commutative rings. Prabably the easiest examples of the former are
$\mathbb{Z}$ and $\mathbb{Z}_n$. We can find elaborations on these two commutative rings in various
literatures including here and here. These are quite simple and easy to comprehend.

However, the examples on simple (non-commutative kind) are not that easy.
One example is found here and it has been mentioned as "one of the simplest examples of a non-commutative ring". This is also on the easier side.

EDIT: The following two are also commutative rings.

Other examples are given in this enumeration. But as you can see, examples like Gaussian integers or Eisenstein integers are difficult for starters to comprehend.

Do you think you can give one or two simple examples on non-commutative rings, based on every day numbers?

If it is that difficult, perhaps some insight comments why this is difficult would be welcome.


答:
The mean value theorem is certainly a valid approach.

Let $f$ be such that $\lim_{x \to \infty}f'(x) = 0$. Let $\epsilon > 0$. Let $x_0$ be such that for all $x > x_0$, we have $|f'(x)| < \epsilon/2$. Let $x_1$ be such that $x_1 > \max\{|2f(x_0)|/\epsilon,x_0\}$. Consider any $x$ with $x > x_1$. By the mean value theorem, there exists a value $c \in (x_0,x)$ such that $f'(c) = (f(x) - f(x_0))/(x - x_0)$.
$$
\begin{align}
\left|\frac{f(x)}{x} \right| &\leq \frac{|f(x_0)|}{x} + \frac{|f(x) - f(x_0)|}{x}
\\ & \leq \frac{|f(x_0)|}{x_1} + \frac{|f'(c)| \cdot (x - x_0)}{x}
\\ & < \frac{|f(x_0)|}{2|f(x_0)|/\epsilon} + \frac{|f'(c)| \cdot (x - x_0)}{x- x_0}
 < \epsilon/2 + \epsilon/2 = \epsilon.
\end{align}
$$
That is, for any $\epsilon > 0$, there exists an $x_1 > 0$ such that $|f(x)/x| < \epsilon$ whenever $x > x_1$. The conclusion follows.

Title: limit of f(x)/x, where f'(x)->0


问:
What is the expansion of the operation $\nabla(r^2\phi(r))$ , where $r$ is $ |r|$
What is the expansion of the operation  $\nabla(r^2\phi(r))$ , where $r$ is $|\mathbf{r}|$?

Is it $(r(\mathrm{d}\phi/\mathrm{d}r)+2\phi)\mathbf{r}$ ?

答:
Indeed the identity holds:

$\mathbf{\text{grad}} (r^2\phi(r))=(2\phi(r)+r\phi'(r))\mathbf{r}$ [1]

where $\mathbf{\text{grad}}$ is the gradient operator, $\phi(r)$ a scalar function, $\mathbf{r}$ a three dimensional vector and $r=|\mathbf{r}|$. I just modified a bit the notation of the OP using bold notation for vector properties.

To prove [1], just perform the calculation using:

The i component of the gradient $(\mathbf{\text{grad}}f(\mathbf{r}))_i$ is by definition $\partial_{r_i} f(\mathbf{r})$ for every function $f(\mathbf{r})$

$\partial_{r_i} r = r_i/r$ (in case this is not known, try to prove it :) )


问:
How to solve this limit: $\lim_{x\to 0}\frac{\sin(x\sin(2x))}{x^2}$
How to solve this limit?

$$\lim_{x\to 0} \frac{\sin(x\sin(2x))}{x^2}$$

I tried to solve with trigonometric formulas but no result yet. It seems that need to use this $$ \lim_{x\to 0} \frac{\sin(x)}{x} $$

答: $$\frac{\sin(x\sin 2x)}{x^2} = \frac{\sin (x\sin 2x)}{x\sin 2x} \times \frac{\sin 2x}{x} \to 1\times \frac{\sin 2x}{2x} \times 2 \to 1\times 1\times 2 =2 $$


问:
What is the notation for equivalent cardinalities?
In NBG class theory, for any set $a$ we can define the class $V_a$ of all sets $b$ such that there is a bijection $f: a \to b.$

$a$ and $b$ have equivalent cardinalities if $b \in V_a$ (or $a \in V_b$). One way to write this is
$$|a| = |b|.$$
Using the von Neumann cardinal assignment, what $|a| = |b|$ means is that the least ordinal number in bijection with $a$ equals the least ordinal number in bijection with $b$. The ordinal $|a|$ is a canonical representative in $V_a$, and when we say $|a| = |b|$ we're saying the canonical representative of $V_a$ equals the canonical representative of $V_b$, so $V_a = V_b$.

My question is: is there an established equivalence symbol for $|a| = |b|$ without using $| \cdot |$? I should be able to say two sets are equivalent (equinumerous) without referring to any particular representatives.

For example, "$a \overset{\#}{=} b$", with the meaning "there is a bijection from $a$ to $b$."

答:
I have seen

$$a\approx b\qquad \text{or}\qquad a\approx_f b$$

where $f$ is the witness (i.e. the bijection) of the equinumerous relation.

I prefer to use

$$a\sim b$$

because I also like to use
$$a\lesssim b$$
to indicate that there exists an injection from $a$ to $b$ which is a partial ordering due  to Schröder–Bernstein theorem:$$a\lesssim b\lesssim a\implies a\sim b$$
To answer your question, there is no established relation symbol. Although $|a|=|b|$ and $\#a=\#b$ are quite standard in my opinion.


问:
On the asymptotic behavior of Elliptic integral near $k=1$
I was trying to prove that as $k\to1^-$,

$$
\int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}\sim\frac12\log{1\over1-k}
$$

Since the original integral is difficult to work with, I split it:

$$
\int_0^1{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}=\int_0^k+\int_k^1
$$

in which it is evident that

$$
\begin{aligned}
\int_0^k{\mathrm dx\over\sqrt{1-x^2}\sqrt{1-k^2x^2}}
&=\int_0^k{\mathrm dx\over1-x^2}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}} \\
&=\frac12\int_0^k{\mathrm dx\over1-x}\cdot{\sqrt{1-x^2}\over\sqrt{1-k^2x^2}}+\mathcal O(1) \\
&=\frac12\int_0^k{\mathrm dx\over1-x}+\mathcal O\left(\int_0^k{1-k\over1-x}\mathrm dx\right)+\mathcal O(1) \\
&=\frac12\log{1\over1-k}+\mathcal O(1)
\end{aligned}
$$

I wonder how to show that $\int_k^1$ converges to zero as $k\to1^-$ so that I can complete the proof.

P.S. Alternative methods are welcome since I have been stuck on this for so much time :D

答:
If $0<k<x<1$, then
\begin{align*}
\sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 }  = \sqrt {1 - x} \sqrt {1 + x} \sqrt {1 - kx} \sqrt {1 + kx}   \ge \sqrt {1 - x} \sqrt {1 - k} .
\end{align*}
Hence,
$$
0 \leq \int_k^1 {\frac{{dx}}{{\sqrt {1 - x^2 } \sqrt {1 - k^2 x^2 } }}}  \le \frac{1}{{\sqrt {1 - k} }}\int_k^1 {\frac{{dx}}{{\sqrt {1 - x} }}}  = 2.
$$
This bound together with your estimate for the integral between $0$ and $k$ imply the desired asymptotics.


问:
Finite $\alpha \approx \beta$ implies $\alpha = \beta$
Assume that $\alpha,\beta \in \mathbb{N}$, and $\alpha \approx \beta$ (there exists a bijection $f: \alpha \to \beta$), how should we use the bijection $f: \alpha \to \beta$ to prove that $\alpha = \beta$? This conclusion seems pretty straightforward, but I didn't find a good way to establish $=$ from $\approx$ directly.

答:
If any of $\alpha,\beta$ is $0$, the assertion is true by the uniqueness of the empty set.

Let this assertion be true for $\alpha < n$ for some $n \in \Bbb N$.

Now, for $\alpha=n$, let $g : \beta \to \beta$ be a permutation such that $g \circ f(n-1)=\beta-1$ (identifying $\alpha$ with $\{0,1,\dots,\alpha-1\} \, \forall \alpha \in \Bbb N$).

Then, $(g \circ f)|_{n-1}$ is a bijection from $n-1$ to $\beta-1$. So, by induction hypothesis, $n-1=\beta-1 \implies n=\beta$.


问:
How would I prove $10^x> 1$ for $x>0$ without using logarithms?
I was asked this question and got completely stumped. The person who asked this has instructed me specifically not to use logarithms. I know there is a more sophisticated way to explain exponentiation using limits of sequences (which would work here), but unfortunately limits are outside the reach in this particular context. So I was wondering, is there a simple way to explain why $0<x<1$ can only be expressed as negative powers of $10$ (or any other number greater than $1$, for that matter) using only high-school algebra? I suppose the proof doesn't have to be entirely rigorous, as long as it's sufficiently convincing.

答:
If $x$ is a positive rational number $p/q$ (where $p$ and $q$ have no common factors), then we can define $10^x$ as
$$
\left(\sqrt[q]{10}\right)^p=\underbrace{\sqrt[q]{10} \cdot \sqrt[q]{10} \cdot \sqrt[q]{10} \cdot \ldots \cdot \sqrt[q]{10}}_{\text{$p$ times}} \, .
$$
For any positive integer $q$, the $q$-th root of $10$ is greater than $1$. This means that $10^x$ is greater than $1$ because the product of $p$ numbers that are greater than $1$ is greater than $1$.

There are no hiccups caused when $x$ is irrational, because in that case $10^x$ is defined in such a way that it is strictly increasing.


问:
Given a parabola with equation $y=ax^2+bx+c$ which has as a tip the point $(3, 1)$ and passes through $(2, 0)$. Find out the product $abc$
Given a parabola with equation $y=ax^2+bx+c$ which has as a tip the point $(3, 1)$ and passes through $(2, 0)$. Find out the product $abc$.

I can't solve the question. I am trying to solve it with the use of analytic geometry on parabolas, but it isn't working out for me. I have that $x=-\frac{b}{2a}$ is a tip. Hence $y=\frac{b^2}{4a}-\frac{b^2}{2a}+c$. I don't know how to finish it off from here. I assume we need to get an expression of the form of $abc=...$ but I haven't managed to get that. Could you please explain to me how to solve it?

答:
Hint:

Prove that if $y=ax^2+bx+c$, then it's "tip point" is at $x=-\dfrac{b}{2a}$.

Prove that if a curve passes through a point, then it satisfies the equation of the curve.


问:
A question about ideal of a polynomial and its vanishing set
I'm trying to show the following, but not sure how to proceed:

Let $f(x,y) = x^2 + y^2 - 1 \in \mathbb{C}[x,y]$. Suppose that $g,h \in \mathbb{C}[x,y]$ s.t. for all $(a,b) \in V(f)$ we have $g(a,b)=h(a,b)$. Then the polynomial $g(x,y)-h(x,y) \in \langle x^2 + y^2 -1 \rangle$.

Now I know that on the set of points in $V(f)$ $g(a,b)-h(a,b) =f(a,b)$. Then intuitively I would think that $g(x,y)-h(x,y) = f(x,y)$ but this doesn't need to be the case I guess right? Otherwise the problem would be trivial.

答:
Clearly, $g(x,y)-h(x,y)=0$ in $V(f)$.

Then, $g-h \in I(V(f))=\sqrt{\langle f \rangle}=\langle f\rangle$, since $f$ is prime in $\Bbb C[x,y]$.


答:
Since there is no $xy$ term in the equation, the parabola has axes parallel to coordinate axes. Specifically, it is just the parabola $y=Ax^2$ translated (no rotation).

The new vertex is $(3,1)$ instead of $(0,0)$. Hence the equation of parabola is
$$(y-1)=A(x-3)^2$$

$A$ can be obtained by using that $(2,0)$ lies on parabola whence $a,b,c$ can be found.

Body:
Given a parabola with equation $y=ax^2+bx+c$ which has as a tip the point $(3, 1)$ and passes through $(2, 0)$. Find out the product $abc$.

I can't solve the question. I am trying to solve it with the use of analytic geometry on parabolas, but it isn't working out for me. I have that $x=-\frac{b}{2a}$ is a tip. Hence $y=\frac{b^2}{4a}-\frac{b^2}{2a}+c$. I don't know how to finish it off from here. I assume we need to get an expression of the form of $abc=...$ but I haven't managed to get that. Could you please explain to me how to solve it?


问:
Regular Submanifold and diffeomorphism
For any $x,y\geq 0$, there is a regular submanifold of $\mathbb{R}^{x+y+1}$ which is a diffeomorphism to $S^x\times S^y.$

My idea is what if we let $S$ to be a regular sub manifold of $\mathbb{R}^2$ and start from there. By the definition of regular submanifold, there exists a chart $(U,\phi)$ of $\mathbb{R}^2$ such that $S\cap \mathbb{R}^2\{x\in\mathbb{R}^2\vert x_{\dim(S)+1}=\cdots=m_{\dim(\mathbb{R}^2)}=0\}$. But unsure how to proceed. To show there is a regular sub manifold which is a diffeomorphism to $S^m\times S^n$, maybe it is not a good place to start with $S$. Instead we should start with arbitrary regular sub manifold of $\mathbb{R}^{x+y+1}$ and somehow show a diffeomorphism relation. Any help is greatly appreciated.

答:
You can embedd $\mathbb{S}^{x}\times\mathbb{R}$ into $\mathbb{R}^{x+1}$ as follows. First notice that $f:\mathbb{S}^{x}\times\mathbb{R}^+\rightarrow \mathbb{R}^{x+1}$ given by $f(x,t)=tx$ is an embedding. Then, since $\mathbb{R}^+\cong\mathbb{R}$ diffeomorphic via the natural logarithm, you are done.

Next, as composition of embeddings are still embeddings, and $\mathbb{S}^y$ naturally sits inside $\mathbb{R}^{y+1}$, you get a chain of embeddings $$\mathbb{S}^{x}\times\mathbb{S}^y\hookrightarrow\mathbb{S}^{x}\times\mathbb{R}^{y+1}\cong(\mathbb{S}^{x}\times\mathbb{R})\times\mathbb{R}^y\hookrightarrow\mathbb{R}^{x+y+1}.$$
The same arguments proves that there is an embedding $\mathbb{S}^{n_1}\times\ldots\times\mathbb{S}^{n_k}\hookrightarrow\mathbb{R}^{n_1+\ldots+n_k+1}$.

Title: Regular Submanifold and diffeomorphism


问:
Good non-mathematician book on Game Theory
I'm looking for a good book on Game Theory. I run a software company and from the little I've heard about Game Theory, it seems interesting and potentially useful.

I've looked on Amazon.com but wanted to ask the Mathematicians here.

One that looks good (despite the title:) is The Complete Idiots Guide to Game Theory. But please post that as an answer if you think it's good.

Please post one book per answer so folks can vote on each book.

答:
Ken Binmore's Playing for Real: A Text on Game Theory is suitable for undergraduates and doesn't get heavily mathematical at all.

What is game theory about?

How do I apply game theory?

Why is game theory right?

There's even a marker in the margin in places where you're allowed to skip some of the math and move on to the next bit.

It isn't easy to write in a light-hearted style when presenting mathematical material, but I did my best.

I think he did an excellent job.


问:
Prove that a given recursively defined sequence converges
Given:

Suppose that {$x_n$} is a bounded sequence such that
$$\begin{align*}
x_n&=x_{n-1}+\frac{f(n)}{10^n}\\\
\end{align*}$$
Where $f: \mathbb{N} \longrightarrow \mathbb{R^+}$ and
$ x_1=1 $

How do I go about to formally prove that the sequence converges and show it?

Thank you in advance.

答:
The condition gives
$$x_n -x_{n-1}=\frac{f(n)}{10^n} \ge 0$$
which means the sequence is increasing. Also, $x_n$ is said to be bounded. Therefore, it converges.


问:
Probability additivity axiom on countably infinite set
I just have a simple question about the additivity axiom on countably infinite set, which I can't get my head around it. If any outcome in the sample space has greater or equal 0 probability (a positive number), how can we make sure that the overall probability of this infinite set is less than 1?</p>

答:
Looks like there is some formal discussion about convergent series.  Here’s a more intuitive idea:

Your entire sample space has probability one.
Now pick some event.  The event is made up of disjoint outcomes in the sample space.  In other words , the event is a subset of the sample space.

By monotonicity , the probability of the event has to be less than or equal to one.


答:
Theorem: any increasing bounded sequence converges. Proof: the increasing bounded sequence $\{x_n\}_{n \in \mathbb{N}}$ converges to $\sup \{x_n | n \in \mathbb{N}\}$ (easy to verify with $\delta-\epsilon$).

This is an increasing bounded sequence. Therefore, it converges.

Title: Prove that a given recursively defined sequence converges

Body:
Given:

Suppose that {$x_n$} is a bounded sequence such that
$$\begin{align*}
x_n&=x_{n-1}+\frac{f(n)}{10^n}\\\
\end{align*}$$
Where $f: \mathbb{N} \longrightarrow \mathbb{R^+}$ and
$ x_1=1 $

How do I go about to formally prove that the sequence converges and show it?

Thank you in advance.


问:
Is there the empty topological space?

May the empty set be a topological space?
Is the set $\{\emptyset\}$ its topology?

What about the empty metric space?

答:
Definitely yes and yes. In point of fact, it's the initial object of the category of topological spaces. It is compact, but as far as I know it is currently a matter of debate whether it should be considered connected or not.

I would say yes (like wikipedia does). It's homeomorphic to the empty topological space and it still is the initial object in all the categories I can think of putting on metric spaces (namely: the category with continuous maps, or the category of uniformly continuous maps, or the category of Lipschitz maps, or the category of non-expansive maps).

Title: Is there the empty topological space?


问:
Prove the solution set of a Linear programming problem is a polyhedron
Problem: Prove the solution set of a Linear programming problem is a polyhedron.

I have proved the feasible set of an LPP is a polyhedron (as the constraints are inequations). Now I want to show the solution set is also a polyhedron but I don't know where to start. I'd like to have some hints for the problem.

Thank you.

答: Hint: If $x^*$ minimizes $c^T x$ subject to $Ax \le b$, then every optimal solution satisfies $c^T x \ge c^T x^*$ (in fact, $c^T x = c^T x^*$).


答: By the chain rule$$\frac{\partial}{\partial r_j}f(r)=\frac{1}{2r}\frac{\partial r^2}{\partial r_j}\frac{\partial}{\partial r}f(r)=\frac{r_j}{r}f^\prime(r)\implies\nabla f=\frac{f^\prime}{r}\mathbf{r}.$$You asked about the case $f=r^2\phi$,$$\nabla(r^2\phi)=(r\phi^\prime+2\phi)\mathbf{r}.$$

Title: What is the expansion of the operation $\nabla(r^2\phi(r))$ , where $r$ is $ |r|$

Body:
What is the expansion of the operation  $\nabla(r^2\phi(r))$ , where $r$ is $|\mathbf{r}|$?

Is it $(r(\mathrm{d}\phi/\mathrm{d}r)+2\phi)\mathbf{r}$ ?


问:
Find all subspace vectors
There are two questions but I can do one and the second I don’t know how to do please kindly help me. Thankfully!!

Let $S=\{(x,0)\in\mathbb{R}^2, x\in\mathbb{R}\}$Prove that $S$ is a subspace of $\mathbb{R}^2$

Find all vector subspaces of $\mathbb{R}^2$ containing $S$

For the Question (1) I can prove it. But Q2 I cannot,  please kindly show me how to find in a question (2).

Q1) Since $0=(0,0)\in S$

Let $ u=(a,0), v=(b,0)\in S, \alpha \in \mathbb{R}$
$$u+\alpha v=(a,0)+\alpha(b,0)$$
$$u+\alpha v=( a+\alpha b, 0)\in S$$
Thus, $S$ is a subspace in $\mathbb{R}^2$

Additional
If $\mathbb{R}=\mathbb{K}$, Does the solution of this Q2 change ?

Remark $\mathbb{K}=\mathbb{R}\or\mathbb{C}$

答: Clearly, $S$ and $\Bbb R^2$ are subspaces of $\Bbb R^2$ which contain $S$. There are no others. If $T$ is a subspace of $\Bbb R^2$ containing $S$, then $(1,0)\in T$. If $S\varsubsetneq T$, take $t\in T\setminus S$. Then $t=(x,y)$, with $x,y\in\Bbb R$ and $y\neq 0$. But then $(0,1)\in T$, since$$(0,1)=\frac1y\bigl((x,y)-(x,0)\bigr)\in T.$$Since $(1,0),(0,1)\in T$, $T=\Bbb R^2$.

Title: Find all subspace vectors


问:
Multiple Origin/Destination connection problem Formulation
The multiple origin/destination connection Problem is as follows

Let $G=(V,E)$ be an undirected graph, and let $c_e$ be a given cost for each $e\in E$. Let $(s_k, t_k)$
for $k=1,...,p\ $ be a given list of origin/destination pairs with $s_k, t_k\in V$ for each k. We seek to choose a set of edges Q ⊆ E of minimum cost such that for each O-D pair k, $s_k$ and $t_k$ will remain connected after removal of any set of m-1 edges of Q.

Binary variables $x_e$ are introduced for $e\in E$, where $x_e=1$ if and only if $e\in Q$. The objective is hence to minimize $\sum c_ex_e$.

Formulations by adding additional variables is possible. I just wonder if it is possible to form the problem as an integer problem without introducing any extra variable, i.e., using only the variables $x_e$? If yes, then how is it formed?

答: Yes, it suffices to impose constraints of the form $\sum_{E \setminus S} x_e \ge 1$, where $S \subset E$, $|S|=m-1$, and removal of $S$ separates $s_k$ from $t_k$ for some $k$.  This constraint forces at least one edge outside of $S$ to appear in $Q$.  There are typically exponentially many such constraints, but you can generate them dynamically if they are violated.


问:
$ \int _ {- 1} ^ {3} [| x ^ 2-6x | + \text{sgn} (x-2)]\, \text {d} x $

Calculate the integral $$\int _ {- 1} ^ {3} \bigl[| x ^ 2-6x | + \text{sgn} (x-2)\bigr]\, \text {d} x .$$

We know that the two functions are defined as follows,
\begin{align*}
f (x) &= | x ^ 2-6x |\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x^2-6x>0; \\
6x-x^2, & \text{ if } x^2-6x<0.
\end{cases}\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x\in(-\infty, 0)\cup (6,+\infty); \\
6x-x^2, & \text{ if } x\in(0,6).
\end{cases}.
\end{align*}
And, $$\text{sgn}(x-2)=\begin{cases}
1, & \text{ if } x>2; \\
-1, & \text{ if } x<2.
\end{cases}$$

I know that the integral gives $ 58 / $ 3. But someone can explain to me the step by step, how to find that result.

I considered doing it with the graph of the function which is the following,

Calculating, the area of the triangle on the left from $ -1 $ to $ 0 $, and then calculating the two on the right and adding it, but I don't know how to calculate the area on the right.

答:
You have all the necessary details in your explanation. It's just a matter of cleverly splitting the integral.
$$\int_{-1}^3 [|x^2-6x| + \operatorname{sgn}(x-2)]dx = \int_{-1}^0 [x^2 - 6x -1] dx + \int_0^2 [6x - x^2 -1]dx + \int_2^3[6x - x^2 +1] dx$$
Now you can integrate the three integrals separately

General strategy Say you are trying to integrate a piecewise function on $[a,b]$
$$f = \begin{cases} f_1 & x \in (x_1, x_2) \\
f_2 & x \in (x_2, x_3) \\
\vdots \\
f_n & x \in (x_n, x_{n+1})\end{cases}$$
It helps to split the integral in the following fashion
$$\int_a^b f = \int_{x_1}^{x_2} f_1 + \dots + \int_{x_n}^{x_{n+1}}f_n$$

In your case, the function you are trying to integrate was over $[-1,3]$ and the pieces, like you figured out were
$$f = |x^2 - 6x|+\operatorname{sgn}(x-2) = \begin{cases} x^2-6-1 & x \in (-1,0) \\ 6x-x^2-1 & x\in (0,2) \\ 6x-x^2 +1 & x\in (2,3) \end{cases}$$


问:
Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
This site
states:

Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
  “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”

In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$

Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?

答:
Consider the infinite "multiplication table" array $M$ defined by $M_{i,j}=ij$

$M =
\left[\begin{array}{r}
 1 &  2 &  3 &  4  & 5 \\
 2 &  4 &  6 &  8 & 10 \\
 3 &  6 &  9 & 12 & 15 & \cdots\\
 4 &  8 & 12 & 16 & 20 \\
 5 & 10 & 15 & 20 & 25 \\
&& \vdots &&& \ddots
\end{array}\right]$

Let $L(n) =
\{M_{i,j} : 1 \le i \le n \wedge j=n \} \cup
\{M_{i,j} : i=n \wedge 1 \le j \le n \}$

be sum of the the _| shaped region consisting of all elements of $M$ whose first or second index is $n$ and whose other index is less than or equal to $n$. For example, the elements of $L(4)$ are shown below.

$M =
\left[\begin{array}{r}
 1 &  2 &  3 &  \color{red}4  & 5 \\
 2 &  4 &  6 &  \color{red}8 & 10 \\
 3 &  6 &  9 & \color{red}{12} & 15 & \cdots\\
 \color{red}4 &  \color{red}8 & \color{red}{12} & \color{red}{16} & 20 \\
 5 & 10 & 15 & 20 & 25 \\
&& \vdots &&& \ddots
\end{array}\right]$

It turns out that $L(1)=1,\; L(2)=8,\; L(3)=27,\; L(4)=64,$ and $L(5)=125$.

We show that $L(n) = n^3$.

\begin{align}
   L(n)
   &= \sum_{i=1}^n M_{i,n} + \sum_{j=1}^n M_{n,j} - M_{n,n} \\
   &= n\sum_{i=1}^n i + n\sum_{j=1}^n j - n^2 \\
   &= n^2(n+1) - n^2 \\
   &= n^3
\end{align}

Let $S(n)$ be the sum of the $n^2$ elements in the $n \times n$ subarray of elements in the upper-left corner of $M$.

For example, $S(3) = (1+2+3) + (2 + 4 + 6) + (3 + 6 + 9) = 36$

We will compute $ S(n)$ two ways.

\begin{align}
   \sum_{i=1}^n \sum_{j=1}^n M_{i,j}
   &= \sum_{i=1}^n \sum_{j=1}^n ij \\
   &= \left(\sum_{i=1}^n i \right) \left(\sum_{j=1}^n j \right) \\
   &=\left[\frac{n(n+1)}{2}\right]^2
\end{align}

Also,
\begin{align}
   S(n) &= L(1) + L(2) + \cdots + L(n) \\
   S(n) &= \sum_{i=1}^n i^3
\end{align}

So $\displaystyle \sum_{i=1}^n i^3 = \left[\frac{n(n+1)}{2}\right]^2$


问:
Subspaces of locally compact spaces
A topological space $X$ is called locally compact if every point $x \in X$ has a compact neighborhood. (It seems that this notion of "locally compact" is sometimes called weakly locally compact.)

I know that in a locally compact Hausdorff space all open, closed and locally closed subspaces are again locally compact. What is an example for an arbitrary subspace which is not locally compact?

Also, I am quite lost what happens if one drops the Hausdorff assumption. Do the above results still hold? If not, what are counterexamples?

答: In $\Bbb R$, endowed with its usual topology, $\Bbb Q$ is not locally compact. Actually, in $\Bbb Q$, if $V$ is any neighborhood of any point, then $V$ is not compact.

Title: Subspaces of locally compact spaces


问:
$\sum_{n=1}^{\infty}\frac{\cos(\frac{\pi}{2}n)}{n}$ convergence
I am once again asking about convergence of sums.

Does $\sum_{n=1}^{\infty}\frac{\ln(n^{(-1)^n})}{n^{p}}, p>0$ converge? As an inexperienced math student, I have not seen this type of function before. It looks to be a changing series, except that it is inside of a ln function. So can I still use the changing series test?

Does $\sum_{n=1}^{\infty}\frac{\cos(\frac{\pi}{2}n)}{n}$ converge? Since $-1 \leq \cos(\frac{\pi}{2}n) \leq 1$, it looks like I can compare it with $\frac{1}{n}$ and conclude that it diverges too.


问:
What books do you recommend I should use as an introduction to convex analysis?
A few years ago, I was in undergrad taking an introductory real analysis course. There was a problem I could not solve. I went to the TA. He was able to solve it using convexity. He recommended that I study convexity. What books do you recommend I should use as an introduction to convex analysis?</p>

答: A general reference for working with convexity (i.e. convex functions + sets) is chapters 1 and 2 of Boyd and Vandenberghe. It's available for free on Boyd's website.</p>


问:
How to solve this limit: $\lim_{x\to 0}\frac{\sin(x\sin(2x))}{x^2}$
How to solve this limit?

$$\lim_{x\to 0} \frac{\sin(x\sin(2x))}{x^2}$$

I tried to solve with trigonometric formulas but no result yet. It seems that need to use this $$ \lim_{x\to 0} \frac{\sin(x)}{x} $$

答:
Since $\sin t=t+o(t)$, you have
$$
x\sin2x=2x^2+o(x^2)
$$
and therefore
$$
\sin(x\sin2x)=2x^2+o(x^2)
$$
so your limit becomes
$$
\lim_{x\to0}\frac{2x^2+o(x^2)}{x^2}=2
$$


问:
How does the radius of a 30-60-90 triangle correspond to the radius of the unit circle?
This may have an obvious answer but my brain is riddled with this. Maybe some of my assumptions below are wrong...

If cosine and sine is the x and y coordinate on a circle with radius one, why is the 30-60-90 triangle defined as having a radius of 2? Wouldn't the cosine and sine values derived from it represent the coordinates on a circle with radius two?
Thanks.

Edit: Is this why you divide by the hypotenuse? That would scale it down by a radius of one.

Edit 2: The radius of the triangle is the hypotenuse.

答:
One of the features of a triangle is the circle that inscribes it. for instance this triangle.

If by radius of a triangle you mean the radius of the circle that inscribes it, it is clear that its center is not on any edges and therefore the trigonometry functions you mention have no direct correlation. If you wanted to calculate that radius you could do it in some simple steps.

When referring to the unit circle both cosine and sine are marked assuming a rectangle. That is: cosine is defined as the x coordinate and sine as the y coordinate. That is logical as $sin=\frac{opposite}{hypotenuse}$ and $cos=\frac{adjacent}{hypotenuse}$ and in the unit circle the hypotenuse is by definition 1 unit unit circle


问:
Game theory - self study
I want to self study game theory. Which math-related qualifications should I have? And can you recommend any books? Where do I have to begin?</p>

答:
Secondary-school math is enough to get started. And since game theory is essentially math, along the way you'll learn more math.

Ken Binmore's Playing for Real: A Text on Game Theory is suitable for undergraduates and doesn't get heavily mathematical at all.

What is game theory about?

How do I apply game theory?

Why is game theory right?

There's even a marker in the margin in places where you're allowed to skip some of the math and move on to the next bit.

Title: Game theory - self study


问:
Find all the numbers that are equal to one quarter of the sum of their own digits

The number $1.5$ is special because it is equal to one quarter of the sum of its digits, as $1+5=6$ and $\frac{6}{4}=1.5$   .Find all the numbers that are equal to one quarter of the sum of their own digits.

I was puzzling over this question for a while, but I couldn't find a formula without using the $\sum$ , but I can't really solve generalizations, only come up with them. The only thing I could come up with was to brute-force it, but I can't really come up with any 'special' numbers. Any help?

答:
This is not much better than a brute force method:

The sum of the digits $S$ is a non-negative integer, so a quarter of it $\frac{S}4$ is non-negative, of the form $x$ or $x.25$ or $x.5$ or $x.75$ for some non-negative integer $x$.

We must have $x < 10$ since if $10 \le x\lt 100$ then $40 \le 4x \le S \lt 412$ so the sum of the digits of $x$ must be at least $40-12=28$ but two-digit integers have sums of digits no more than $18$.  Similarly with larger $x$.

So a satisfactory $x$ has sum of digits $4x=S=x$ with solution $x=0$

and a satisfactory $x.25$ has sum of digits $4(x+\frac14)=S=x+7$ with solution $x=2$

and a satisfactory $x.5$ has sum of digits $4(x+\frac12)=S=x+5$ with solution $x=1$

and a satisfactory $x.75$ has sum of digits $4(x+\frac34)=S=x+12$ with solution $x=3$

making the numbers which work $$0, \quad2.25,\quad 1.5,\quad 3.75$$