问:
What is $\frac{\det(\hat{\Sigma}_0)}{\det(\hat{\Sigma})}$ in terms of $\hat{\mu}_1$, $\hat{\mu}_2$ and $\hat{\Sigma}$
Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu_1,\Sigma)$ and let $Y_1,...,Y_{n_2}$ be an independent sample from $N_p(\mu_2,\Sigma)$, for some $\mu_1,\mu_2 \in \mathbb{R}^p$ and some invertible, $p\times p$ positive definite matrix $\Sigma$.

Let $\hat{\mu}_0 := \frac{\sum_{i=1}^{n_1}x_i + \sum_{i=1}^{n_2}y_i}{n_1 + n_2}$, $\hat{\mu}_1 := \frac{1}{n_1}\sum_{i=1}^{n_1}x_i$ and $\hat{\mu}_2 := \frac{1}{n_2}\sum_{i=1}^{n_2}y_i$

Suppose $\hat{\Sigma}_0=\frac{1}{n_1+n_2}\biggl(\sum^{n_1}_{i=1}(x_i-\hat{\mu}_0)(x_i-\hat{\mu}_0)^T+\sum^{n_2}_{i=1}(y_i-\hat{\mu}_0)(y_i-\hat{\mu}_0)^T\biggr)$

and $\hat{\Sigma}=\frac{1}{n_1+n_2}\biggl(\sum^{n_1}_{i=1}(x_i-\hat{\mu}_1)(x_i-\hat{\mu}_1)^T+\sum^{n_2}_{i=1}(y_i-\hat{\mu}_2)(y_i-\hat{\mu}_2)^T\biggr)$

I would like to show that:

$$\frac{\det(\hat{\Sigma}_0)}{\det(\hat{\Sigma})} = 1 + \frac{n_1n_2}{(n_1+n_2)^2}(\hat{\mu}_1 -\hat{\mu}_2)^T\hat\Sigma^{-1}(\hat{\mu}_1 -\hat{\mu}_2) $$

I know that $\frac{\det(\hat{\Sigma}_0)}{\det(\hat{\Sigma})}=\det(\hat{\Sigma}^{-1/2}\hat{\Sigma}_0\hat{\Sigma}^{-1/2})$, but I'm not sure how to continue.

答:
Python simulation showing that the right-hand side must contain $\hat\Sigma^{-1}$ (equation in comments), not $\Sigma^{-1}$ (your original equation)

```<code>import numpy as np
from numpy.linalg import det, inv
from scipy.stats import invwishart, multivariate_normal as normal

n1, n2, p = 10, 12, 5
μ1, μ2 = normal.rvs(size=(2,p))
Σ  = invwishart(df=p+1, scale=np.eye(p)).rvs()
X  = normal(mean=μ1, cov=Σ).rvs(n1)
Y  = normal(mean=μ2, cov=Σ).rvs(n2)

μ0hat = (X.sum(axis=0) + Y.sum(axis=0))/(n1+n2)
μ1hat = X.mean(axis=0)
μ2hat = Y.mean(axis=0)

Σ0hat  = ((X-μ0hat).T@(X-μ0hat) + (Y-μ0hat).T@(Y-μ0hat))/(n1+n2)
Σhat= ((X-μ1hat).T@(X-μ1hat) + (Y-μ2hat).T@(Y-μ2hat))/(n1+n2)
LHS = det(Σ0hat)/det(Σhat)
RHS = 1 +  ((n1*n2)/(n1+n2)**2) * (μ1hat-μ2hat)@inv(Σ)@(μ1hat -μ2hat)
print(LHS-RHS)
corrected_RHS = 1 +  ((n1*n2)/(n1+n2)**2) * (μ1hat-μ2hat)@inv(Σhat)@(μ1hat -μ2hat)
print(LHS-corrected_RHS)
</code>```

Body:
Python simulation showing that the right-hand side must contain $\hat\Sigma^{-1}$ (equation in comments), not $\Sigma^{-1}$ (your original equation)

import numpy as np
from numpy.linalg import det, inv
from scipy.stats import invwishart, multivariate_normal as normal

n1, n2, p = 10, 12, 5
μ1, μ2 = normal.rvs(size=(2,p))
Σ  = invwishart(df=p+1, scale=np.eye(p)).rvs()
X  = normal(mean=μ1, cov=Σ).rvs(n1)
Y  = normal(mean=μ2, cov=Σ).rvs(n2)

μ0hat = (X.sum(axis=0) + Y.sum(axis=0))/(n1+n2)
μ1hat = X.mean(axis=0)
μ2hat = Y.mean(axis=0)

Σ0hat  = ((X-μ0hat).T@(X-μ0hat) + (Y-μ0hat).T@(Y-μ0hat))/(n1+n2)
Σhat= ((X-μ1hat).T@(X-μ1hat) + (Y-μ2hat).T@(Y-μ2hat))/(n1+n2)
LHS = det(Σ0hat)/det(Σhat)
RHS = 1 +  ((n1*n2)/(n1+n2)**2) * (μ1hat-μ2hat)@inv(Σ)@(μ1hat -μ2hat)
print(LHS-RHS)
corrected_RHS = 1 +  ((n1*n2)/(n1+n2)**2) * (μ1hat-μ2hat)@inv(Σhat)@(μ1hat -μ2hat)
print(LHS-corrected_RHS)


问:
$\sum_{k=1}^n m(k)$, where $m(k)$ is defined by $2^{m(k)} || k$.
I'm looking at the sum:
$$f(n) = \sum_{k=1}^n m(k),$$
where $m(k)$ is defined by $2^{m(k)} || k$, i.e. $2^{m(k)}$ is the largest power of $2$ that divides $k$. For example, we have $f(8) = 0+1+0+2+0+1+0+3 = 7$. Here's a table of $n$, $m(n)$, and $f(n)$ for a few small $n$:

```<code>n     m(n)  f(n)
1     0     0
2     1     1
3     0     1
4     2     3
5     0     3
6     1     4
7     0     4
8     3     7
</code>```

The values $m(n)$ in the table are the last $m(k)$-value for the given $n$, i.e. the value of $m(k)$ when $k=n$. I haven't spent much time trying to figure this out, mostly because it's a funny little problem and I thought I'd share it. It seems to me that we have $f(2^j) = 2^j-1 \forall j$ and $f(n)\sim n$.

答:
As a bit of interesting trivia concerning $f(n)$ we see that the Mellin-Perron summation formula applies, giving
$$ f(n) = \frac{1}{2} v_2(n) +
\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} L(s) n^s \frac{ds}{s}
\quad\text{where}\quad
L(s) = \sum_{n\ge 1} \frac{v_2(n)}{n^s}.$$
But we have
$$ L(s) = \sum_{m\ge 1} \frac{v_2(2m)}{(2m)^s} =
\frac{1}{2^s} \sum_{m\ge 1} \frac{1+v_2(m)}{m^s} =
\frac{1}{2^s} (\zeta(s) + L(s)),$$
so that
$$ L(s) = \frac{\zeta(s)}{2^s-1},$$
giving
$$  f(n) = \frac{1}{2} v_2(n) +
\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \frac{\zeta(s)}{2^s-1} n^s \frac{ds}{s}$$

We can apply the Cauchy Residue Theorem to this integral,
treating first the poles at $s=1$ and $s=0$, which yields
$$f(n) \sim \frac{1}{2} v_2(n)
+ n - \frac{1}{2}\log_2 n - \frac{1}{4} - \frac{1}{2}\log_2 \pi.$$
The above approximation is excellent, giving for example $f(200) \sim 196.6023238$ when the correct value is $197$, or $f(411) \sim 405.5827546$ when the correct value is $405.$
Finally, $f(1000)$ is $994$ when the approximation gives $995.4413598.$

The complete asymptotic expansion is obtained by including all the poles on the imaginary axis at $\frac{2\pi i k}{\log 2}$, with $k$ a nonzero integer. These poles are all simple. We have
$$\operatorname{Res}\left(\frac{1}{2^s-1}; s = \frac{2\pi i k}{\log 2}\right) =
\frac{1}{\log 2}.$$
Now introducing $$\rho_k = \frac{2\pi i k}{\log 2}$$ we finally obtain
$$f(n) \sim \frac{1}{2} v_2(n)
+ n - \frac{1}{2}\log_2 n - \frac{1}{4} - \frac{1}{2}\log_2 \pi +
\frac{1}{\log 2}
\sum_{k\in \mathbb{Z}\setminus 0} \frac{\zeta(\rho_k)}{\rho_k} e^{2\pi i k\log_2 n},$$
where the sum is a Fourier series in $\log_2 n$ that describes the fluctuations.

Taking the terms up to $|k|=7$, we obtain $f(200) \sim  197.0498970.$ With $34$ terms, we obtain $f(411)\sim 405.3130370$ and $f(1000) \sim 993.9869087.$

ContentLicense: CC BY-SA 3.0


问:
Degree of minimal polynomial when number of distinct eigenvalues are given
If a square matrix of order $ 10 $ has exactly $ 4 $ distinct eigenvalues, then degree of it's minimal polynomial is

At least $ 4 $

At most $ 4 $

At least $ 6 $

At most $ 6 $

$ 4 $ distinct eigenvalues would mean $ 4 $ factors in the minimal polynomial: $ ( x - a ) ( x - b ) ( x - c ) ( x - d ) $, where $ a $, $ b $, $ c $ and $ d $ are the distinct eigenvalues, but I can't see how the minimal polynomial can have degree more than or less than $ 4 $.

答:
The roots of the minimal polynomial are precisely the eigenvalues, and the degree of a nonzero polynomial is at least the number of its roots, so 1. is true.

With your conditions, it is possible for the minimal polynomial to have any degree between 4 (in the diagonalisable case) and 10, so all the other answers are false.

To prove this, let's call $J_k = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 \end{pmatrix} \in M_k(\mathbb C)$ the usual nilpotent Jordan block.

You can easily check that the minimal polynomial of $J_k$ is $X^k$.

For $1 \leq k \leq 7$, the block-diagonal matrix $\mathrm{diag}(1,2,3) \oplus J_k \oplus \mathrm{diag}(\underbrace{0,\ldots,0}_{\text{$7-k$ zeroes}}) \in M_{10}(\mathbb C)$ has four eigenvalues, and its minimal polynomial is $X^k(X-1)(X-2)(X-3)$.

ContentLicense: CC BY-SA 4.0


问:
If each slice of $f : X \times Y\rightarrow Z$ is continuous then $f$ is continuous?
I'm having trouble to conclude my proof of the next statement.

If $f:X \times Y \rightarrow Z$ is a function such that, $f_x : Y \rightarrow Z$ given by $f_x(y) = f(x,y)$ is continuous for each $x \in X$ and $f_y : X \rightarrow Z \ $ given by $\ f_y(x) = f(x,y)$ is continuous for each $y \in Y\ $ then $f$ is continuous

This is what I've tried, given $(x,y) \in X \times Y$ and $\ U \in \tau_Z$ such that $f(x,y) \in U$ we take $\ f_x, \ f_y$ continuous functions, so $f_x^{-1}(U) \in \tau_Y \text{ and } f_y^{-1}(U) \in \tau_X$ so we take $V = f_y^{-1}(U) \times f_x^{-1}(U)$ wich is open in $X \times Y$ and $(x,y) \in V$.

My problem is when I have to show that $f(V) \subseteq U$. I need some help

Edit: I've saw something similar here and uses strongly compactness, so this is likely false, But I don't know how can be a function wich each are continuous but not the whole function.

答:
At least in the context of $X=Y=[0,1]$ and $Z=\mathbb{R}$, this is essentially the same as

If $(g_n)_{n\geq 1}$ is a sequence of continuous functions $[0,1] \to \mathbb{R}$ that converges pointwise to the continuous function $g$, then $g_n$ converges to $g$ uniformly.

which is false.  For instance, we may consider the sequence $$g_n(x) = \begin{cases}nx(1-nx) & 0 \leq x < \frac{1}{n} \\ 0 & \text{ otherwise}\end{cases}$$
which converges to $g(x) = 0$ pointwise, but $g_n(1/(2n)) = \frac{1}{4} \not\to 0,$ so $g_n$ doesn't converge to $g$ uniformly.

For this particular example, we may create our $f$ by formally replacing $n$ with $1/y$ to get
$$f(x,y) = \begin{cases}\frac{x}{y}\left(1-\frac{x}{y}\right) & 0 \leq x < y \\ 0 & \text{ otherwise}\end{cases}.$$

In general, if we have a sequence $g_n \to g$ pointwise but not uniformly, then we define $f : [0,1]^2 \to \mathbb{R}$ by $$f(x,y) = \begin{cases}\left(n+1-\frac{1}{y}\right)g_n(x) + \left(\frac{1}{y} - n\right)g_{n+1}(x) & \text{ if } y > 0, n=\lfloor 1/y\rfloor \\ g(x) & \text{ if } y=0\end{cases}$$

Since the functions $g$ and $(g_n)_{n\geq 1}$ are continuous, each $f(\cdot, y)$ is continuous.  Since $g_n \to g$ pointwise, each $f(x,\cdot)$ is continuous.

However, one may use the compactness of $X=[0,1]$ along with the assumption $g_n$ doesn't converge to $g$ uniformly to show that $f$ is not continuous at some point $(x,0)$.

Conversely, if $f : [0,1]^2 \to \mathbb{R}$ were a function with continuous slices but not continuous at $(a,b)$, then there is a sequence of points $(a_n,b_n) \in [0,1]^2$ such that $(a_n,b_n) \to (a,b)$ but $f(a_n,b_n) \not\to f(a,b)$.  Then we may define $$g_n(x) = f(x,b_n) \\ g(x) = f(x,b).$$ Once again, since $f$ had continous slices, these functions are continuous and $g_n \to g$ pointwise, but $g_n(a_n) \not\to g(a)$, so the convergence is not uniform.


问:
Sigma Sum Manipulation: How does the second equality hold?
I'm only familiar with multiplying sigma sums by integers and adding them together, and so the second equality is quite confusing for me.

$f_i$ is a polynomial of degree $i$.

How does the solution below isolate $x^j$ like that?

I got up to here

$$\sum_{i=0}^{n}c_i(\sum_{j=0}^{i}a_{ij}x^j)=\sum_{i=0}^{n}(\sum_{j=0}^{i}c_ia_{ij}x^j) = ?$$

答:
Here we use an intermediate representation which might help to better see the  index region. We obtain

\begin{align*}
\sum_{i=0}^nc_if_i&=\sum_{i=0}^n c_i\left(\sum_{j=0}^i a_{ij}x^j\right)\\
&=\sum_{\color{blue}{0\leq j\leq i\leq n}}c_ia_{ij}x^j\\
&=\sum_{j=0}^n\left(\sum_{i=j}^nc_ia_{ij}\right)x^j
\end{align*}


问:
How EXACTLY does Tarski's undefinability of truth interfere with defining $\vDash$ for proper classes, and why doesn't it do so for sets?
My mind is getting a bit twisted with notions relating to formalizing $\vDash$ for sets and classes in the $ZFC$. The end result I believe is:

For sets $A$, we may formalize $\vDash$ in $ZFC$ (and a fragment of it), and we may make statements such as $ZFC \vdash \forall \varphi \in T(A \vDash \ulcorner \varphi \urcorner)$ (where $T$ indicates some godelized set of sentences such as $ZFC$)

Moreover, in the metatheory it is fairly easy to prove that for every formula $\varphi$, $ZFC \vdash (A\vDash \ulcorner \varphi \urcorner) \iff \varphi^A$ (has to be in the metatheory since we're quantifying over actual formulas).

But the same cannot be done for proper classes. For proper classes, the best we can do is stick to relativizations $\varphi^M$ for a proper class $M$. The reason given is "due to Tarski's indefinability of truth". I have multiple questions here:

So if I naively replicated Tarski's recursive definition of truth anyway and tried to define $M \vDash \varphi$ in $ZFC$, using basically the exact same technique as for sets, what goes wrong? I'm guessing it has to do with the fact that we're quantifying over a proper class. But when we have a single proper class $M$, what's the problem? I realize we can't do $\forall M$ (where $M$ proper classes), but what's the problem with saying $\forall a \in M$ (that is, $\forall a M(a)$ where $M(x)$ is the defining formula for $M$) to help define $\vDash \forall x \varphi(x)$? Does it have to do with the fact that recursions have to happen on set like and well founded relations?

Given formulas $\varphi$, the relativization $\varphi^M$ is defined recursively. So again, fixing $M$ proper class, what's stopping us from talking about $\ulcorner \varphi^M \urcorner$, and for instance, saying $ZFC \vdash \forall \varphi (ZFC(\varphi) \rightarrow \ulcorner \varphi^M \urcorner)$. I have the feeling this is a dumb question but I would appreciate it nonetheless. Edit: Nevermind, $\ulcorner \varphi^M \urcorner$ is just a number and just $ZFC \vdash \ulcorner \varphi^M \urcorner$ doesn't make sense without an accompanying $\vDash$ somewhere. Yep that was indeed stupid.

Finally, so Tarski's indefinability of truth, at a high level, is why we can't define the relation $\vDash$ on proper classes. Shouldn't the same apply for sets? Meaning, why doesn't the fact that we can define $\vDash$ for sets contradict Tarski's indefinability of truth.

答:
I think you have things pictured a bit off. (In order to talk freely about classes, the following analysis takes place in $\mathsf{NBG}$ for simplicity; as usual, this can be "$\mathsf{ZFC}$-ified" straightforwardly at the cost of making all the statements annoyingly circuitous and only applicable to definable classes.)

Below, unless stated otherwise "structure" will mean "transitive set or class containing $\mathsf{HF}$ as a subset thought of as an $\{\in\}$-structure." The requirement that $\mathsf{HF}\subseteq M$ is really just an overkill way of guaranteeing that Godel coding works properly; if you prefer, you can instead demand that $M$ satisfy some very weak set theory like $\mathsf{KP}$ (which will in fact imply containment of $\mathsf{HF}$ but may feel more natural).

The set/class distinction really doesn't interact with Tarski at all. Tarski's undefinability theorem applies to every structure, class- or set-sized:

Suppose $M$ is a transitive set or class with $\mathsf{HF}\subseteq M$, thought of as an $\{\in\}$-structure. Then $Th(M)$, thought of as a subset of $\omega$ in the obvious way, is not parameter-freely-definable in $M$.

The size of $M$ plays no role; what Tarski does, regardless of the size of $M$, is limit where a parameter-free definition of $Th(M)$ can take place. For example, we have:

Suppose $M,N$ are set-or-class-sized structures with $M\subsetneq N$. Then Tarski's theorem does not immediately preclude $Th(M)$ from being parameter-freely-definable in $N$.

The only special thing about $V$ is that there is no "outside" we can go to to get a parameter-free definition of truth. But that's the only way $V$ differs from a set-sized structure here. Indeed, with set-sized structures there are very simple ways of "going outside to get the theory" without even going to the level of proper-class-sized structures! For example, suppose $M\subseteq N$ are set-sized structures. Let $M_N$ be the expansion of $N$ by a new unary predicate $U$ with $U^N=M$. Note that $M_M$, while not literally the same structure as $M$, is biinterpretable with $M$ so this doesn't always give us more power. However:

For every set-sized structure $M$, we have $Th(M)$ is parameter-freely-definable in $N_M$ where $N=V_{height(M)+1}$.

This is massive overshooting actually, but it makes the necessary point. And the reason this doesn't work for proper-class-sized structures is simple and has nothing to do with Tarski: if $M$ is a proper class, there is no "level of the cumulative hierarchy above $M$".

(Incidentally, it's not immediately clear to me whether we actually need to expand the language!)

This leaves us with one major interesting question remaining: when is the theory of a class-sized structure parameter-freely-definable in $V$? As noted above, Tarski says nothing directly here for classes other than $V$ (or other than minor modifications of $V$). In fact, there is a simple example of a proper-class-sized structure whose theory is parameter-freely definable in $V$: the class of ordinals, $\mathsf{Ord}$. This is vastly less complicated than one might expect, and its theory is in fact parameter-freely-definable in $(\mathbb{N};+,\times)$ alone!

OK, $\mathsf{Ord}$ doesn't fit our picture since $\mathsf{HF}\not\subseteq\mathsf{Ord}$. Perhaps a better thing to look at is what I'll call "$\mathsf{HF(Ord)}$," the class-sized structure gotten by closing $\mathsf{Ord}$ under pairing. This is now significantly more complicated - it interprets true first-order arithmetic for example - but still extremely simple compared to other structures in the context of set theory, and in particular its theory is still parameter-freely definable in $V$.


问:
How can incremental fees be infinity for the majority of fund managers who don't beat the market?
Kindly see the emboldened sentence below. Is it hyperbole? How can fund managers charge infinity as a fee?!!?!

      But even this recalculation substantially understates the
real cost of active beat-the-market investment manage-ment. Here's why: Broad-based index funds and exchange-
traded funds that reliably produce the market rate of
return—with no more than market risk—are now avail-
able with even lower fees than when the first edition of
Elements was published. Today, market-matching returns

Page 141 below.

are available to all investors at low "commodity" prices,
such as 5 basis points (5/100 of 1 percent). Therefore,
investors should consider the fees charged by active man-
agers not as a percentage of total returns, but as the incre-mental fee for active management as a percentage of the
incremental risk-adjusted returns above the market index.

       Thus correctly stated, management fees are quite high.
Leaving aside several interesting quibbles—such as that,
for individual investors, taxes on short-term gains with
100 percent portfolio turnover can be significant—fees are
remarkably high. Incremental fees are somewhere between
50 percent of incremental returns and, amazingly, infinity
for the majority of fund managers who do not beat the
market. And fees are generally closer to the high end than
the low end of that range. Are fees for any other services
of any kind at such a high proportion of value?

The Elements of Investing: Easy Lessons for Every Investor 2013 Updated Edition.
By Burton Malkiel, Charles Ellis. p 141.

答:
Fund managers set their fees as a percentage of total investment, at most a few percent. The writer is asking what value they produce for you in return for the fee, and says that is what you should compare the fee to. Moreover the only value they produce is the amount by which they can do better than the index funds. If they don't beat the index funds, the value they have delivered to you is zero, so the author claims that any fee they charge is infinitely larger (in proportion) than what you got in return.

That's not a mathematical statement; mathematically, you can cause a lot of trouble by trying to say that a positive number divided by zero is infinity.
But the question (as framed by this author) is much the same as this question: if you try to buy beef from a butcher, and the butcher charges you ten dollars but does not give you any beef,
what is the butcher's price per pound for beef?


Title: Sigma Sum Manipulation: How does the second equality hold?

Body:
I'm only familiar with multiplying sigma sums by integers and adding them together, and so the second equality is quite confusing for me.

$f_i$ is a polynomial of degree $i$.

How does the solution below isolate $x^j$ like that?

I got up to here

$$\sum_{i=0}^{n}c_i(\sum_{j=0}^{i}a_{ij}x^j)=\sum_{i=0}^{n}(\sum_{j=0}^{i}c_ia_{ij}x^j) = ?$$

Body:
Compare coefficients: $$c_0(a_{00}x^0)+c_1(a_{10}x^0+a_{11}x^1)+\dots+c_n(a_{n0}x^0+a_{n1}x^1+\dots a_{nn}x^n)$$

can be rewritten as,

$$(c_0a_{00}+c_1a_{10}+\dots+c_na_{n0})x^0+(c_1a_{11}+c_2a_{21}+\dots+c_na_{n1})x^1+\dots+c_na_{nn}x^n$$


答:
If you examine carefully just which combinations of values of $i$ and $j$ occur in it, you’ll see that the first double sum can be rewritten like this:

$$\sum_{i=0}^nc_i\left(\sum_{j=0}^ia_{ij}x^j\right)=\sum_{0\le j\le i\le n}c_ia_{ij}x^j\,.$$

There is actually one term in the double sum for each pair $\langle i,j\rangle$ with $0\le j\le i\le n$. It may help to look at a small example, say with $n=3$; the following table shows exactly which combinations occur, and each of them occurs only once. If you move the $c_i$ inside the inner sum, the terms are:

$$\begin{array}{c|cc}
i\backslash j&0&1&2&3\\\hline
0&c_0a_{00}x^0\\
1&c_1a_{10}x^0&c_1a_{11}x^1\\
2&c_2a_{20}x^0&c_2a_{21}x^1&c_2a_{22}x^2\\
3&c_3a_{30}x^0&c_3a_{31}x^1&c_3a_{32}x^2&c_3a_{33}x^3
\end{array}$$

Once the $c_i$ multipliers have been brought inside the inner summation, it is adding across the rows of this table; the outer one is then adding up these sums.

Now change the order of summation: add up the columns, and then add the column sums. The sum of column $j$ is

$$\sum_{i=j}^nc_ia_{ij}x^j\,,$$

so the sum of the whole array is

$$\sum_{j=0}^n\sum_{i=j}^nc_ia_{ij}x^j\,.$$

Now the index $j$ is constant within the inner summation, so we can factor out $x^j$, just as $c_i$ was outside the original inner summation:

$$\sum_{j=0}^n\sum_{i=j}^nc_ia_{ij}x^j=\sum_{j=0}^nx^j\sum_{i=j}^nc_ia_{ij}\,.$$

Clearly the inner summation is just the coefficient of $x^j$, so we could just as well write this as

$$\sum_{j=0}^n\left(\sum_{i=j}^nc_ia_{ij}\right)x^j\,.$$


答:
Notice that as we proceed through the second equality, the order of the sums is reversed (from "for each $i$ sum over $j$" to "for each $j$ sum over $i$").  It can be helpful to draw a diagram of which pairs of indices, $(i,j)$ actually appear in the sum.

On the left, we have $i \in [0, n]$ and for each of those $j \in [0,i]$  Let's graph those pairs.

To reverse the order of summation, we need to study the range of $i$s for each $j$.  When $j = 0$, we have terms for $i \geq 0$.  When $j = 1$, for $i \geq 1$.  In fact, for each $j \geq 0$, we have $i \geq j$.  So our sum should be of the shape
$$  \sum_{j=0}^n \sum_{i=j}^n \dots $$

So lets' show a few more steps in the derivation of the equality you ask about:
\begin{align*}
\sum_{i=0}^n c_i \left( \sum_{j=0}^i a_{ij} x^j \right) &= \sum_{i=0}^n \left( \sum_{j=0}^i \left( a_{ij} c_i x^j \right) \right)  &  &\text{distributivity of multiplication over addition}  \\
    &= \sum_{j=0}^n \left( \sum_{i=j}^n \left( a_{ij} c_i x^j \right) \right)  &  &\text{exchange order of sums}  \\
    &= \sum_{j=0}^n \left( \left( \sum_{i=j}^n a_{ij} c_i\right) x^j \right)  &  &\text{(un-)distribute}  \\
\end{align*}

We are able to undistribute $x^j$ out of the inner summand because "$x^j$" does not depend on the inner index, $i$.  We cannot un-distribute $c_i$ or $a_{ij}$ from the final inner sum because these depend on $i$, so are not (necessarily) the same for each term in the sum over $i$.


问:
What is significance of the row space?
I became to study linear algebra a bit on my own. It is already  more than twenty years since my graduation. I ended up to my question due to practical problem that I have already solved. Having said that, my practical problem lead me to trying understand a concept of a column space, row space, null space and left null space.

I have developed a intuition or concept of a column space. In my thinking matrix columns are vectors in given space. Further on, in my thinking a column space is a space spanned by matrix columns. All that makes sense to me. I just rely on interpretation of a matrix as collection of vectors.
Also, somehow I have become to understand a null space as a special space that will lead to matrix equation of Ax=0. In my thinking, importance of a null space is that there can't be inverse for a such transformation.

Now, my difficulties arise in understanding a row space. It is useless to repeat definition of a row space as Col(A'). The definition doesn't really open this for me. I'd like to understand significance of a row space. In my thinking, columns of a matrix are vectors. This understanding doesn't really fit to a row or a row space. Rows are not vectors, hence a row space is not similar space as a vector space  spanned by vectors. Yes, technically it is so or may be, as I have understood. But I am still missing intuition or understanding of concept of a row space.
What is significance of a row space, how does it manifest itself?

And very similarly, what is importance of a left null space, how does it manifest itself?

You see, I am not a mathematician nor student of math. I am an engineer. And due to my personal way of learning, I am always trying to visualize or develop some kind of intuition before diving into the details and rigorous definitions.

Thanks in advance

Poincaré look-alike -

答:
I personally view it from the following perspective: if you have a matrix, you can associate a set of linear equations with it (one row gives you one equation $\sum_{j}a_{ij}x_j=0$). And if you have a bunch of relations, you may want to ask what other linear relations $\sum_j c_jx_j=0$ follow from it? (that is, I write you a bunch of equalities like $x+y+z+t=0, x-y+z=0, y-z+t=0$, etc. but what you really care about in all that jungle is whether you must necessarily have $x=y$). The answer is that the new linear equation follows from your system if and only if the row $c$ defining it is in the row space of your matrix $A$, i.e., there are no non-trivial linear consequences of a system linear equations. In the latter form, the statement can be generalized to linear inequalities and, once you understand what it means there, it tells you pretty much all you need to know about the duality in linear programming though it is not, strictly speaking, about the row space anymore.

That is not the only possible interpretation of it, but I find it quite useful to know and always tell it to the students when teaching a course in linear algebra. See if it clarifies things for you a bit or not :-)

Title: What is significance of the row space?


问:
If for every $x_n$ such that $x_n \rightarrow x$, there exists a $x_{n_k}$ such that $Tx_{n_k} = Tx$, is $T$ continuous?
Let $X$ and $Y$ be Banach spaces and $T$ be the (possibly nonlinear) map $T\colon X \rightarrow Y$. $T$ is continuous if for every $x_n \in X$ such that $x_n \rightarrow x$, then $Tx_n \rightarrow Tx$. Is $T$ also continuous if instead, for every $x_n \in X$ such that $x_n \rightarrow x$, then there exists a subsequence $x_{n_k}$ such that $Tx_{n_k} = Tx$?</p>

答: I misunderstood the question first. Indeed the answer is yes. If $Tx_n$ does not go to $Tx$ one can choose subsequence $x_{n_k}$ that is bounded away from $Tx,$ e.g. $\|Tx_{n_k}-Tx\|\ge\varepsilon.$ This leads to a contradiction.    </p>


问:
Find the almost certain convergence of $(Y_n), \, n \geq 1$
Hi I'm having trouble understanding this problem

Consider a succession of tosses of a balanced die. Let $X_k$ be the result of the $k$-th face and $Y_n = \max(X_k), k ≤ n.$
Find the almost certain convergence of $(Y_n)_{n \geq 1}$.

I'm just don't know how should I find the convergence of this random variable.

Can someone explain/help me a method to find the convergence?

Thank you!

答:
Consider the probability of $Y_n=6$:

$$\begin{align*}
\Pr\{Y_n=6\} & = \Pr\{\text{at least one $X_k = 6$, for $0\leq k\leq n$}\} \\
 & = 1 - \Pr\{\text{no $X_k = 6$, for $0\leq k\leq n$}\} \\
 & = 1-\left(1-\frac{1}{6}\right)^n \rightarrow 1, \quad \text{ as } n\rightarrow\infty
\end{align*}
$$
which proves the convergence in probability. To prove almost surely convergence you need to prove that the probability of the set of all sequences $\{X^n: X^n=\{X_k\}_{k=1}^n, \nexists X_k=6\}$ tends to zero. The proof, aside from some details, follows the same idea as in the convergence in probability.


问:
Reduction of Order On Functions

Suppose $y_1 (x)$ is a solution to $y'' + p(x)y' + q(x)y = 0$.  Use
reduction of order to show that a second solution is $y_2 (x) = y_1
(x) \int \frac{e^{-\int p(x)dx}}{(y_1 (x))^2} dx$.

I plugged the guess $v(x)y_1 (x)$ into the ODE and obtained $v''y_1 + 2v'y_1' + pv'y_1 = 0$.

The substitution $w(x) = v'$ yields the first-order ODE $w' = -\frac{2y_1' + py_1}{y_1}w$, but this is hard to solve because $w$, $y_1$ and $p$ are all functions of $x$.  Additionally, this exercise appears in the section on The Method of Frobenius, so I may be doing something different than the intended approach.

Is this work correct, and if so, how can I obtain the stated result from this point?

Title: Reduction of Order On Functions

Body:
$$y_2=vy_1$$
$$y'_2=v'y_1+vy_1'$$
$$y''_2=v''y_1+2v'y'_1+vy''_1$$
So that:
$$y''_2 + p(x)y'_2 + q(x)y_2 = 0$$
Becomes:
$$ v''y_1+2v'y'_1+vy''_1+ p(x)(v'y_1+vy_1') + q(x)vy_1 = 0$$
$$ v''y_1+2v'y'_1+ p(x)v'y_1  = 0$$

$$ v''y_1+v'(2y'_1+ p(x)y_1)  = 0$$
This is separable.
$$ (\ln v')'=-\dfrac {2y'_1+ p(x)y_1 }{y_1}$$
$$ \ln v'=-\int \dfrac {2y'_1+ p(x) y_1}{y_1}dx$$
$$ \ln v'=-2 \ln (y_1)-\int  { p(x) }dx$$
$$ v'=\dfrac {e^{-\int  { p(x) }dx}}{ y_1^2}$$
$$ v=\int \dfrac {e^{-\int { p(x) }dx}}{ y_1^2}dx$$
Finally:
$$ y_2= {y_1}\int \dfrac {e^{-\int { p(x) }dx}}{ y_1^2}dx$$


问:
Proving inequality that (problably) using AM-GM
For $a, b$ and $c$ positive real number, prove following inequality $$\dfrac{a+b}{c^2}+\dfrac{b+c}{a^2}+\dfrac{a+c}{b^2}\geq 2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$$

答: Hint: prove $a/c^2+c/a^2-1/a-1/c=(a+c)(1/a-1/c)^2$. From this, get an inequality; sum three versions of it viz. the cyclic symmetry.


问:
Generating Euler Bricks
In a problem set assigned to my class, we were asked to show that if $(u,v,w)$ is a Pythagorean triple, then a cuboid with side lengths $u|4v^2-w^2|$, $v|4u^2-w^2|$, and $4uvw$ generated an Euler brick. I succeeded by replacing $w$ with $\sqrt{u^2+v^2}$, and then replacing $u$ and $v$ with $m^2-n^2$ and $2mn$ respectively, the well-known generator for Pythagorean triples. However, I am unsure how this would be derived. It seems like something easy to check but hard to get. I wasn't able to find any sort of derivation on the web, so does anyone have one?

答: The formula regarding the Euler Bricks is from Nicholas Saunderson. You can dig up the original derivation here.</p>

Title: Generating Euler Bricks


问:
Let $f:[0,\infty) \to [0,\infty)$ be increasing and satisfy $f(0)=0$ and $f(x)>0 \forall x>0$.
Let $f:[0,\infty) \to [0,\infty)$ be increasing and satisfy $f(0)=0$ and $f(x)>0
\forall x>0$. If $f$ also satisfies $f(x+y) \leq f(x)+f(y) \forall x,y \geq 0$, then $f \circ d$ is a metric whenever $d$ is metric. Show each of the following conditions is sufficient to ensure that $f(x+y) \leq f(x)+f(y) \forall x,y \geq 0$:

a) $f$ has a second derivative satisfying $f'' \leq 0$;

b) $f$ has a decreasing first derivative.

c) $f(x)/x$ is decreasing for $x>0$.

My Work Thus Far:

Let $g(x)=f(x)x$, for $x>0$. According to the question we have that $f''<0$ then $f$ has a decreasing first derivative, meaning $f'$ is decreasing. Now, taking the derivative of $g$ yields: $g'(x)=xf'(x)-f(x)x^2=f'(x)x-f(x)x^2$, where $x>0$.

I know that $-f(x)x^2$ is decreasing, but how do I know for sure that $f'(x)x$ is decreasing so I can deduce that $g'(x)<0, \forall x>0$ and that $g(x)$ is a decreasing function?

答:
By $c$ you have $f(x+y)\cdot(x+y)\leq f(x)\cdot x$ and $f(x+y)\cdot(x+y)\leq f(y)\cdot y$. Equivalently: $$f(x+y)\cdot\dfrac{x+y}x\leq f(x), \  f(x+y)\cdot\dfrac{x+y}{y}\leq f(y)$$

Summing both inequalities gives: $$f(x+y)\left(\dfrac{x+y}{x}+\dfrac{x+y}{y}\right)\leq f(x)+f(y)$$
From here you can conclude expanding and aplying AM-GM inequality.


问:
Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$
We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.

$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
 = 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$

Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.

答:
A monomial giving a contribution to degree 16 must be of the form

$$x^{a_1}\cdot x^{a_2}\cdot x^{a_3}\cdot x^{a_4}$$

with $a_i <6$ and $a_1+a_2+a_3+a_4=16$.

Suppose  $a_1 \geq a_2 \geq a_3 \geq a_4$, then you have only five possibilities: $(5,5,5,1)$, $(5,5,4,2)$, $(5,5,3,3)$, $(5,4,4,3)$ and $(4,4,4,4)$.

The coefficient of $x^{16}$ equals the cardinality of the set composed by these vectors and the ones obtained permuting their coordinates.

It is easy to prove that this cardinality is equal to:

$$\binom{4}{3}+2\binom{4}{2}+\binom{4}{2}+2\binom{4}{2}+1=35$$


问:
Can cardinality be used to express exactly one?
If I wanted to express the statement in predicate logic: 'There is exactly one apple that is green.'
Would it be correct to say that if $x$ is an apple that is an element of $A$: {all apples} and that $G(x)$: $x$ is a green apple, that: 'There exists an apple $x$ that is an element of the set $A$, $|A|=1$, $G(x)$'?

答: Introducing sets and their sizes is not necessary. If, for $x$ in our universe of apples, $G(x)$ means $x$ is green, then $\exists x\forall y(G(y) \leftrightarrow x = y)$ is true iff there is exactly one green apple.


问:
Condition for a cubic equation to have a single root
If a cubic equation

$$ f(x) = ax^3+bx^2+cx+d$$

Is given, what is the condition for the equation to only have a single root (counting multiple roots as one)

答: Consider $f(x)=ax^3+bx^2+cx+d,$ with $a>0$ WLOG. Then there is a unique solution provided that either $$f'(x)=3ax^2+2bx+c$$ has almost positive sign -- that is, when $b^2-3ac\le 0$ -- or if $b^2-3ac>0,$ then we must have that $$f(r_1)f(r_2)\ge 0,$$ where $r_1,r_2$ are the roots of the quadratic equation $f'(x)=0.$


问:
Prove the convergence of the series.
Let r > 1 be a real number. Prove that the following series is convergent.
$$\sum_{n = 1}^{\infty}\frac{1}{n^r}$$</p>

答:
Back to basics. The partial sums are clearly increasing. So we only need to show they are bounded above. Consider the partial sum up to $n=2^{k}-1$.

Look at the first term, the sum of the next $2$ terms, the sum of the $4$ terms after that, the sum of the $8$ terms after that, and so on.

The first term is $1$.

The sum of the next $2$ terms is $\lt \frac{2}{2^r}=\frac{1}{2^{r-1}}$.

The sum of the next $4$ terms is $\lt \frac{4}{2^{2r}}=\frac{1}{2^{2(r-1)}}$.

The sum of the next $8$ terms, by a similar argument, is $\lt \frac{1}{2^{3(r-1}}$.

And so on. So the sum of the terms up to $2^k-1$ is less than
$$1+\frac{1}{2^{r-1}}+\frac{1}{2^{2(r-1)}}+\cdots+\frac{1}{2^{(k-1)(r-1)}}.$$
This is a finite geometric series, with sum equal to
$$\frac{1-2^{-k(r-1)}}{1-2^{-(r-1)}}.$$
In particular, it has the upper bound $\dfrac{1}{1-2^{-(r-1)}}$.

Remark: Without mentioning it explicitly, we have used the idea behind the Cauchy Condensation Test.

Title: Prove the convergence of the series.


问:
Orthogonal projection over $Z=\overline{\text{span}\{e_k;k\in I\}}$
My colegues and me was trying to solve the following exercise:

Let $\mathcal{B}=\{e_k;k\in I\}$ an orthonormal subset of the Hilbert space $X$. Let consider $Z=\overline{\text{span}(\mathcal{B})}$. Show that the orthogonal projection $P\colon X\to Z$ is given by $$Px=\sum_{k\in I}\langle x,e_k\rangle e_k.$$

Some ideas: Since $Z$ is a closed subspace of a Hilbert space, we have that $Z$ is complete. Then, for each $x\in X$, $Px$ is the unique element of $Z$ which satisfies $$x-Px\perp Z\qquad\Longleftrightarrow\qquad\langle x-Px,z\rangle=0,\ \forall z\in Z.$$ In particular, we have that $$\langle x-Px,e_k\rangle=0,\ \forall k\in I.$$ The next step is try to write $Px$ as some combination of the $e_k$'s and make the computations. But we can't finish this, because if $Px\in Z$, then there exists $(y_n)\subseteq\text{span}(\mathcal{B})$ such that $y_n\to Px$ in the norm induced by the inner product.

Maybe a important fact that we could use here is that:

$\{k\in I;\langle x,e_k\rangle\ne0\}$ is countable.

I have some computations about the $y_n$'s, but since I don't know if it's fine, I will not write here right now.

Thanks for any help!

答:
Let $y = \sum\limits_{k=1}^{\infty} \langle x, e_k \rangle e_k$. Then $y \in Z$. We have to show that $\langle (x- y), z \rangle =0$ for all $z \in Z$. By continuity of the inner product in the second variable it is enough to show that this holds when $z$ is  a finite linear combination of $e_k$'s. By linearity of inner product in the second variable it is enough to show that this holds when $z=e_j$ for some $j$. But in this case $\langle (x- y), z \rangle=\langle x, e_j \rangle-\langle y, z \rangle=\langle x, e_j \rangle-\langle x, e_j \rangle=0$.

[$\langle  y, e_j\rangle =\lim_{N \to \infty} \langle \sum\limits_{k=1}^{N} \langle x, e_k \rangle e_k ,e_j \rangle$ and $\langle \sum\limits_{k=1}^{N} \langle x, e_k \rangle e_k ,e_j \rangle=0$ if $N < j$ and $\langle x, e_j \rangle$ if $N\geq j$].


问:
Given any digraph $D$ is the flip distance to an acyclic reorientation of $D$ equal to the minimum cardinality of an acyclic feedback-set of $D$?
Given an arbitrary digraph $D=(V,A)$ we call any set $F\subseteq A$ a feedback-set of $D$ iff the digraph: $H=(V,A-F)$ is acyclic, while in particular we call $F\subseteq A$ an acyclic feedback-set of $D$ iff $F$ is a feedback-set of $D$ such that $H'=(V,F)$ is acyclic, likewise we let $\phi(D)$ be the minimum number of arcs in $D$ that if flipped will result in an acyclic digraph, or equivalently in symbols we let $\phi$ be the graph invariant such that $\phi(D)=\min\{|R\setminus A|:(V,R)\text{ is a reorientation of }D\}$ while moreover we write $\beta(D)=\min\{|F|:F\text{ is an acyclic feedback-set of }D\}$, now with all of that terminology established is it true that for every digraph $D$, we must have: $\beta(D)=\phi(D)$?

Clearly by definition $\beta(D)\leq\phi(D)$ because if flipping any set of arcs $A'\subseteq A$ in $D$ results in an acyclic digraph $H$ then we have that $(V,A-F)$ and $(V,F)$ are both subgraphs of $H$ thus these are both acyclic digraphs, thus $A'$ is an acyclic feedback-set of $D$ and thus $\beta(D)\leq |A'|$ where by definition we can choose the set of arcs we are flipping $A'\subseteq A$ in $D$ so that we have $|A'|=\phi(D)$ i.e. this means $\beta(D)\leq \phi(D)$. But is it true that $\phi(D)\leq \beta(D)$? I know this is true if $D$ is planar (i.e. this is true when $D$ is an orientation of a planar graph) but is this identity true in general?

答:
Suppose that $F$ is a minimal feedback-set of $D$: the digraph $D-F$ is acyclic, but for every proper subset $F' \subset F$, the digraph $D - F'$ is not acyclic. Then I claim that the graph $D^{-F}$ produced by merely reversing all the arcs in $F$ is also acyclic.

This implies your second inequality and also implies that all minimal feedback-sets are acyclic, which was not obvious to me at first.

To see this, take any arc $(u,v) \in F$. Because $F' = F - \{(u,v)\}$ is not a feedback-set, there must be a cycle in $D - F'$; since this cycle is not in $D$, it must contain arc $(u,v)$. Deleting arc $(u,v)$ from this cycle still leaves a path from $v$ to $u$, which is a path in $D-F$.

Essentially, the existence of a $v$-to-$u$ path for every arc $(u,v) \in F$ means that the reverse arc $(v,u)$ would be redundant: it would not help create any cycles.

More formally, suppose that there is a cycle in $D^{-F}$. Every time that cycle uses an arc $(v,u)$ which is the reverse of an arc $(u,v) \in F$, replace that arc by the $v$-to-$u$ path in $D-F$. After we do this, we have obtained a closed walk in $D-F$, so there is a cycle in $D-F$: this contradicts our assumption that $F$ is a feedback-set.

Therefore $D^{-F}$ must be acyclic.


问:
Why notate the solution to steady-state heat equation as $2\Phi$?
In literature I read the following:

Based on what is said, I think that to obtain Eqn(3) one substitutes Eqn(1.17(1)) into the heat equantion, Eqn(B), and then consider the steady-state condition, $\partial v/\partial t=0$, leading to Eqn(C):
$$\tag{1.17(1)} f_i=-K_{ij} \frac{\partial v}{\partial x_j}$$
$$\tag{B} \rho c \frac{\partial v}{\partial t}+\left(\frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}\right)=0$$
$$\tag{C} K_{ij}\frac{\partial ^2 v}{\partial x_i \partial x_j}=0$$
In deriving Eqn(C) the $K_{ij}$ are constants, and thus are pulled from the derivative. So my question is this: if I correctly derived Eqn(C), and I am correct that it is equivalent to Eqn(3), why then would the authors equate Eqn(3) to $2\Phi$ instead of $0$?

I note that I have not seen the variable $\Phi$ used/defined in this text leading up to this section (Sec. 1.20), nor have I seen it used after.

Some thoughts to consider:

The 2nd-rank tensor $K_{ij}$ is usually taken as symmetric,
$K_{ij}=K_{ji}$. Under this assumption the right-hand side of Eqn(3) may be written as $$K_{11}\partial_{xx}^2v+K_{22}\partial_{yy}^2v+K_{33}\partial_{zz}^2v+2K_{23}\partial_{yz}^2v+2K_{13}\partial_{xz}^2v+2K_{12}\partial_{xy}^2v \tag{D}$$

The scalar magnitude $K$ of the symmetric positive-definite 2nd-rank tensor $K_{ij}$ is $$K=K_{ij}l_il_j=K_{11}l_1^2+K_{22}l_2^2+K_{33}l_3^2+2K_{23}l_2l_3+2K_{13}l_1l_3+2K_{12}l_1l_2 \tag{E}$$ where $l_i$ is a unit vector in the direction in which $K$ is evaluated. So perhaps in some way $2\Phi$ is related to the scalar magnitude of $K_{ij}$?

答:
The notation is a bit unusual, using $v$ for temperature, but they probably use $T$ for something else.  If you start with the steady-state heat equation:
$$-\nabla \cdot (K \nabla v)=Q$$

with a heat coefficient tensor $K$ (or you can choose to see it as a matrix) and source term $Q$. So the heat equation is not necessarily equal to zero since you can have a source term.

For anisotropic materials, I think some authors write the left-hand side of the heat equation as:
$$-\frac{1}{2}\nabla \cdot (K \nabla v)-\frac{1}{2}\nabla \cdot (K^T \nabla v)=$$
$$=-\nabla \cdot (\frac{1}{2}(K+K^T) \nabla v)$$
If the diffusion tensor/matrix $K$ is symmetric, which it usually is, then you get:
$$\frac{1}{2}(K+K^T)=K$$

Maybe the author has written:
$$-\frac{1}{2}\nabla \cdot (K \nabla v)-\frac{1}{2}\nabla \cdot (K^T \nabla v)=Q$$

If we now define: $$\Phi=\frac{1}{2}\nabla \cdot (K \nabla v)$$
we get the heat equation as:
$$-2\Phi=Q$$

This is a highly unusual way of writing things but it is my best guess.


问:
Series evaluation: 2x^2 + 4x^4 + ...
Is there any closed form solution of the following series (assuming $0 \leq x \leq 1$)?
$$2x^2 + 4x^4 + 6x^6 + 8x^8 \ldots$$

答:
some hints:

 geometric-like series

 substitute $w=x^2$

 $\sum_{n=1}^\infty 2n x^{2n} = 2\sum_{n=1}^\infty n w^n = \ldots$

Title: Series evaluation: 2x^2 + 4x^4 + ...


Title: Series evaluation: 2x^2 + 4x^4 + ...

Body:
Is there any closed form solution of the following series (assuming $0 \leq x \leq 1$)?
$$2x^2 + 4x^4 + 6x^6 + 8x^8 \ldots$$


答:
Hint: put $y = x^2$, then
$$
2x^2 + 4x^4 + 6x^6 + 8x^8 + \ldots = 2y(1 + 2y + 3y^2 + 4y^3 + \ldots)
$$
You can get a closed formula for $s(y) = 1 + 2y + 3y^2 + 4y^3 + \ldots$ using the formula for the sum of a geometric series either by considering it as the derivative of the geometric series $y + y^2 + y^3 + y^4 + \ldots$ or by expanding $s(y) - ys(y)$.

Title: Series evaluation: 2x^2 + 4x^4 + ...


问:
Second moment of Silverman kernel
I am learning kernel smoothing method, in the Wiki page , I find that second moment of  Silverman kernel is equal to zero (i.e $\int u^{2} K(u) d u$), I am not sure how to derive this result, that integral look difficult to compute,  it suffices to show that $$
\int_{0}^{\infty} x^{2} e^{-x / \sqrt{2}} \sin \left(\frac{x}{\sqrt{2}}+\frac{\pi}{4}\right) d x=0
$$
(Actually this equal to zero when I use WolframAlpha to verify), can any body help to solve this integral, original integral is

$$
\int_{-\infty}^{\infty} x^{2} \frac{1}{2} e^{-\frac{|x|}{\sqrt{2}}} \cdot \sin \left(\frac{|x|}{\sqrt{2}}+\frac{\pi}{4}\right)dx=0
$$

答:
First let us substitute $y=x/\sqrt 2$. Then we have to show
$$
J:=
\int_0^\infty
y^2\; e^{-y}\; (\sin y + \cos y)\; dy =0\ .
$$
We use partial integration to reduce the degree of the polynomial $y^2$.
$$
\begin{aligned}
J
&=
\int_0^\infty
y^2\; e^{-y}\; (\sin y + \cos y)\; dy
\\
&=
\int_0^\infty
y^2\; (-e^{-y}\; \cos y)'\; dy
\\
&=
-\Big[
y^2\; e^{-y}\; \cos y
\Big]_0^\infty
+
\int_0^\infty
(y^2)'\; e^{-y}\; \cos y\; dy
\\
&=
\int_0^\infty
2y\; e^{-y}\; \cos y\; dy
\\
&=
\int_0^\infty
y\; (e^{-y}\; (\sin y-\cos y))'\; dy
\\
&=
\Big[
y\; e^{-y}\; (\sin y-\cos y)\Big]_0^\infty
-
\int_0^\infty
y'\; e^{-y}\; (\sin y-\cos y)\; dy
\\
&=
-
\int_0^\infty
e^{-y}\; (\sin y-\cos y)\; dy
\\
&=
\int_0^\infty
(e^{-y}\; \sin y)'\; dy
\\
&=
\Big[
e^{-y}\; \sin y\Big]_0^\infty
\\
&=0\ .
\end{aligned}
$$

Title: Second moment of Silverman kernel


问:
Dividing Two Infinities
I am Curious if the following is mathematically correct:

Let $a$ be the infinite set of all nonnegative integers $0,1,2,3...$.

Let $b$ be the infinite set of all nonnegative EVEN integers $0,2,4,6...$.

If I take the sum of $a$ and divide it by the sum of $b$ do I get 2?

Is this correct? Can one divide infinities like this?

If yes, does this mean that SUM $a$ > SUM $b$ (despite that both are infinite)?

(if this is not possible can one do some kind of equivalent division using mapping of sets?)

答:
As pointed out in the comments, your series are divergent, so their sums are not defined.

However, if you have a convergent infinite series, say $b_1 + b_2 + b_3 \cdots$ converges to $S$, and $c$ is a constant, then $cb_1 + cb_2 + cb_3 + \cdots$ converges to $cS$.

Moreover, given a second series $a_1 + a_2 +a_3 \cdots$, if $0 \le a_n \le b_n$ for all $n$, then the second series $a_1 + a_2 + a_3 + \cdots$ converges to a value less than or equal to $S$.  This is known as the comparison test for infinite series.

Title: Dividing Two Infinities


问:
Stability, critical points and similar properties of solutions of nonlinear Volterra integral equations
I've posted also on MathOverflow, but repost here once I think it's an important question and would like to get more attention. Thank you!

I have a system of nonlinear Volterra integral equations of form

$$x(t)=x_0+\int_0^t K(t,s)F(x(s))ds$$

and I am interested on the critical points of $x(t)$, I mean maximum, minimum, increasing and decreasing intervals, nonnegativity etc.

I imagine it's impossible to get complete informations about that, but here I am asking for theorems and general results to help me to study these aspects, once is impossible know the true solution.

Thank you.

EDIT
I am mainly interested in nonnegativity, since we need it for physically coherent solutions. $K$ is nonnegative and the signal of $F$ depends on signal of $x$. Also interested on asymptotic equilibrium when $\int_0^\infty K(t)dt=\infty$ (but $K(t)\to 0$).

答:
Assuming the hypotheses of the Leibniz integral rule apply,
$$  x'(t) = K(t,t)F(x(t)) + \int_0^t \partial_1(K(t,s)) F(x(s)) \,\mathrm{d}s  \text{.}  $$
So, can you find the zeroes of $K$ and $F$?  Can you find the signs of $K$ and $F$ on various intervals?  You don't constrain $K$ or $F$ at all in your statement, so how could anyone possibly make more specific statements?


答:
For convergence you don't even require any result from probability theory. $(Y_n)$ is an increasing sequence of real numbers bounded by $6$ (at each sample point) so it converges at every sample point $\omega$.
To show that the limiting random variable $Z$ has to be $6$ w.p. $1$ consider $P(Z <6)=P(X_k <6 \forall k)=\prod_k P(X_k <6)=\prod_k P(X_1 <6)=0$ since $P(X_1 <6)=\frac  5 6 <1$.

Title: Find the almost certain convergence of $(Y_n), \, n \geq 1$

Body:
Hi I'm having trouble understanding this problem

Consider a succession of tosses of a balanced die. Let $X_k$ be the result of the $k$-th face and $Y_n = \max(X_k), k ≤ n.$
Find the almost certain convergence of $(Y_n)_{n \geq 1}$.

I'm just don't know how should I find the convergence of this random variable.

Can someone explain/help me a method to find the convergence?

Thank you!


问:
How find this maximum $x+y$
let $x,y\in [0,2\pi]$, and $2\sin{x}\cos{y}+\sin{x}+\sin{y}+\dfrac{1}{2}=0$,find the maximum $x+y$

my idea
$$\sin{x}+\sin{y}=2\sin{\dfrac{x+y}{2}}\cos{\dfrac{x-y}{2}}$$

答:
Let us change $t=x-y,\, s= x+y.$ Then your problem is written in terms of $s$ and $t$ as follows .

Maximize $s$
under the constraints $$ \left\{\sin(s) +\sin(t) +2\sin\left(\frac s 2 \right) \cos\left(\frac t 2\right)= \frac 1 2 , 0 \le s, s \le 4\pi, -2\pi \le t,\,t \le2\pi \right\}. $$
It remains to put $s=4\pi$ and to solve the equation $\sin(t) = \frac 1 2$ and  to return to $x$ and $y$. I leave it on your own, stating the answer $y=2\pi,\, x= 2\pi-  \pi /6 .$ The numerical calculations confirm that result.

Title: How find this maximum $x+y$


问:
For given mean $\mu$ of random variable X in [0,1], what is the probability distribution function $p(X)$ that makes $VAR(X)$ maximum?
Given the conditions

$\int_{0}^{1} p(x)dx=1$, $\int_{0}^{1} xp(x)dx=\mu$ and $p(x)\ge0$ for $\forall x \in [0,1]$,

What probability distribution function $p(x)$ makes $Var(X)$=$\int_{0}^{1} x^{2}p(x)dx - \mu^{2}$ maximum?

Is this can be solved using Lagrange multipliers?

答:
Let us prove that the variance is maximized for the corresponding Bernoulli ("might as well put as much probability mass as possible on the extreme points").

Let $X$ be any random variable supported on $[0,1]$, with expectation $\mathbb{E}[X]=\mu\in[0,1]$. Consider the Bernoulli random variable $Y$ such that
$$
Y = \begin{cases}
1 &\text{ with probability } \mu\\
0&\text{ with probability } 1-\mu\\
\end{cases}
$$
Clearly, $\mathbb{E}[Y]=\mu\in[0,1]$, and further $\mathbb{E}[Y^2]=\mu$ as well. Now,
$$
\mathbb{E}[X^2] = \int_0^1 x^2 p(x) dx
\leq \int_0^1 1\cdot x p(x) dx = \mathbb{E}[X] = \mu
$$
so $\mathbb{E}[X^2]\leq \mathbb{E}[Y^2]$. As $\operatorname{Var} X = \mathbb{E}[X^2]-\mu^2$, this means
$$
\operatorname{Var} X \leq \operatorname{Var} Y\,.
$$


问:
Show that $ \operatorname{Hom}(K[s]/I, A)$ is isomorphic to $Z(I;A)$
We have a field $K$, and an ideal $I \subseteq K[s],$ $s=(s_1,..., s_n)$ . The exercise requires me to prove that for any $A$ algebra over $K$,
$Z(I; A) = \operatorname{Hom}(K[s]/I, A),$ and here $Z(I; A)$ is the zero set of $I$ ideal in $A^n$.
I understand what zeros are for $I$ in a ring but what does it mean for an ideal to have zeros in an algebra?

答:
Recall the universal property of the polynomial ring $K[X_1,...,X_n]$. It says, given a $K$-algebra $A$ and a function $\phi:\{1,...,n\} \rightarrow A$ there is a unique $K$-algebra homomorphism $\Phi: K[X_1,...,X_n] \rightarrow A$ satisfying $\Phi(X_i) = \phi(i)$. Hence we can evaluate a polynomial $f\in K[X_1,...,X_n]$ at a point $(a_1,...,a_n)\in A^n$ by considering the image of $f$ under the map $\Phi$ induced by the map $\phi(i)=a_i$. So in other words every point $a\in A^n$ induces a unique homomorphism
$$\begin{array}{rcl}
\operatorname{ev}_a: K[X_1,...,X_n] & \longrightarrow & A\\
f & \longmapsto & f(a_1,...,a_n)
\end{array}$$
and every homomorphism $\Phi: K[X_1,...,X_n] \rightarrow A$ yields a point $(\Phi(X_1),...,\Phi(X_n))\in A^n$. Thus we have mutually inverse maps
$$\begin{array}{rcl}
\operatorname{Hom}(K[X_1,...,X_n],A) & \longleftrightarrow & A^n\\
\Phi & \longmapsto & (\Phi(X_1),...,\Phi(X_n))\\
\operatorname{ev}_a & \leftmapsto & a=(a_1,...,a_n)
\end{array}$$

Now if we restrict this bijection to
$$Z(I;A)=\{(a_1,...,a_n) \in A^n \mid \forall f\in I: f(a_1,...,a_n)=0\}$$ on the right side, which set of homomorphisms do we get on the left? For this we will need another universal mapping property, namely that of the quotient ring. Let $I$ be an ideal in some ring $R$. Given any $R$-module $M$ we have that any morphism $\psi: R \rightarrow M$ satisfying $\psi\vert_I = 0$ induces a unique morphism $\Psi: R/I \rightarrow M$ satisfying $\Psi \circ \pi = \psi$ (where $\pi: R \rightarrow R/I$ denotes the quotient map). In other words we have mutually inverse maps
$$\begin{array}{rcl}
\operatorname{Hom}(R/I, M) & \longleftrightarrow & \{\psi \in \operatorname{Hom}(R,M)\mid \psi\vert_I =0\}\\
\Psi & \longmapsto & \Psi\circ \pi\\
\Psi & \mapsfrom & \psi
\end{array}$$

Can you see how this helps you with the exercise?


问:
Let $y = \sqrt{4 + \sqrt{64 + \sqrt{5096+ ...}}}$ and $\sqrt{4x^2 + \sqrt{16x^2 + \sqrt{64x^2 + ...}}} = 11 .$ What is the value of $x + y$
Let

$ y = \sqrt{4 + \sqrt{64 + \sqrt{5096 + \dots }}} $

and

$ \sqrt{4x^2 + \sqrt{16x^2 + \sqrt{64x^2 + \dots }}} = 11 . $

What is the value of $x + y$ ?

edit(1)

I've tried to squared both side of the first equations :

$ y^2 = 4 + \sqrt{64 + \sqrt{5096 + \dots }} $

from here, I am stuck.

答:
I assume here that it should be 4096.

If you factor out some fours, it's quite easy:
$$y=4\cdot \sqrt{1+4\sqrt{1+4\sqrt{1+4\sqrt{\ldots}}}}=4\cdot \sqrt{1+y} $$
and $y=8+4\sqrt{5}$.

With the same method I found that $x=\pm \frac{3\sqrt{11}}{2}$ because
$$11=2\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+\ldots}}}=2q \text{ with } q=\sqrt{x^2+q}.
$$


问:
A non-cyclic Artinian module has at least two distinct minimal (simple) submodule.
My Question: A non-cyclic Artinian module has at least two distinct minimal (simple) submodule.

My attempt: Let $M$ be a non-cyclic Artinian $R$-module. Then there exists two nonzero element $x,y\in M$ such that $Rx\nsubseteq Ry$ and $Ry\nsubseteq Rx$. Hence, by Zorns Lemma, $M$ has two distinct minimal submodules as $M$ is Artinian.

What is wrong hare?

答:
Firstly, you are taking a lot of liberties with your assertion.

Why do such $x,y$ exist?

Then why do you think that means there are distinct minimal submodules? It isn’t clear.

Why are you using Zorn’s lemma? You have both the DCC, already: you don’t need Zorn.  And it’s not clear how you were applying it anyway.

—-

Let $R=\mathbb R[x,y,z]/(x^2,y^2,xz,yz,z^2-xy)$.  It is known that $R$ is Artinian and has a simple socle, and hence a unique minimal ideal contained in all other nonzero ideals. It is also not a principal ideal ring. So if we take $M$ to be any nonprincipal ideal, we have a counterexample.

—-
Eric Wofsey points out in my comments:

$\mathbb R[x,y]/(x^2,y^2)$ works for the same reasons!


问:
An exercise about zerodivisors
If $A$ is a commutative ring with unity, $f\in A$ and $x\in SpecA$,  with the notation $f(x)$ I mean the coset $x+f\in A/x$. Now look at this exercise:

Prove that a nonzero element $f\in A$ is a zerodivisor if and only if there exists a decomposition $SpecA=X\cup X'$ where $X\subseteq$ $SpecA$ and $X'\subsetneq Spec A$ such that $f(x)=0$ for all $x\in X$.

One direction is easy: ($\Rightarrow$) If $f$ is a zerodivisor, it's not invertible, so $f$ is contained in a maximal ideal $\mathfrak m$. We have that $SpecA= V(f)\cup D(f)$ where $D(f)\neq Spec A$ because $\mathfrak m\notin D(f)$.

I have problems to show that the other direction. Any hint?

答:
Nice for you that you have problems showing the other direction, since it is completely false!

Take for example $A=\mathbb Z$ and $f=2$:

Clearly $f=2$ is not a zero divisor.  However  $\text {Spec}(A)$ can be written as $\text {Spec}(A)=X\cup X'$, with $X=V(2)=\{(2)\}$ , $X'=\text {Spec}(A)\setminus V(2)\subsetneq \text {Spec}(A)$ and $f(x)=0$ for all $x\in X$ .

Title: An exercise about zerodivisors


问:
A free algebra that is not generic
In a previous question, I was told that free algebras on sufficiently many generators are generic algebras. That raises the question, what is an example of a free algebra that is not generic? An algebra $A$ is generic for a variety $V$ iff $A$ generates $V$.

答: For example the free group [or free monoid] on one generator is surely not generic among all the groups [monoids], because it's commutative.</p>

Title: A free algebra that is not generic


答:
It is mathematically incorrect, but the idea is intuitively good in certain contexts. The right way to say things is that it all depends on how you take the limits. Since both series sum to infinity, you need to make your result "$2$" precise. For instance,
$$
\lim_{n \to \infty} \frac{\sum_{i=0}^{n} 2i }{\sum_{i=0}^n i}
$$
is probably what you had in mind when you thought of the result being "two". Of course, this limit is worth two, but we cannot say
$$
\lim_{n \to \infty} \frac{\sum_{i=0}^{n} 2i }{\sum_{i=0}^n i} = \frac{\lim_{n \to \infty} \sum_{i=0}^n 2i}{\lim_{n \to \infty} \sum_{i=0}^n i}
$$
because the two "things" on the numerator and denominator are not numbers, so their ratio is not a priori defined. To understand what I mean better, notice that if we tweak things a little bit :P , it is not hard to show that
$$
\lim_{n \to \infty} \frac{\sum_{i=0}^{n} 2i }{\sum_{i=0}^{n^2} i}= 0, \qquad \lim_{n \to \infty} \frac{\sum_{i=0}^{n^2} 2i }{\sum_{i=0}^{n} i}= \infty.
$$
Hope that helps,

Title: Dividing Two Infinities

Body:
I am Curious if the following is mathematically correct:

Let $a$ be the infinite set of all nonnegative integers $0,1,2,3...$.

Let $b$ be the infinite set of all nonnegative EVEN integers $0,2,4,6...$.

If I take the sum of $a$ and divide it by the sum of $b$ do I get 2?

Is this correct? Can one divide infinities like this?

If yes, does this mean that SUM $a$ > SUM $b$ (despite that both are infinite)?

(if this is not possible can one do some kind of equivalent division using mapping of sets?)


问:
Need help on the general formula for the Taylor series of $\frac{2x}{e^{2x}-1}$
I need help on deriving the general formula for the Taylor series of $\frac{2x}{e^{2x}-1}$

It is known that the series for $\dfrac{x}{e^{x}-1}$ is: $$\dfrac{x}{e^{x}-1}=\sum_{n=0}^{+\infty}\dfrac{B_nx^{n}}{n!}$$

So the question is whether the series for $\frac{2x}{e^{2x}-1}$ is:

$$\frac{2x}{e^{2x}-1}=\sum_{n=0}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\tag{1}$$

or

$$\frac{2x}{e^{2x}-1}=2\left(\dfrac{x}{e^{2x}-1}\right)=2\left(\sum_{n=0}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)\tag{2}$$

So which series is the correct representation for $\frac{2x}{e^{2x}-1}$? $(1)$ or $(2)$

I suspect that it must be $(2)$, because if it is $(1)$, how does one represent the series for $\dfrac{3x}{e^{2x}-1}$

答:
If you make the substitution $u=2x$, it’s clear that

$$\frac{2x}{e^{2x}-1}=\frac{u}{e^u-1}=\sum_{n\ge 0}\frac{B_nu^n}{n!}=\sum_{n\ge 0}\frac{B_n(2x)^n}{n!}\,.$$

And from that we get

$$\frac{3x}{e^{2x}-1}=\frac32\cdot\frac{2x}{e^{2x}-1}=\frac32\sum_{n\ge 0}\frac{B_n(2x)^n}{n!}\,.$$


问:
Help to understand a simple statement in elementary proof of the euclidean theorem.
Well, I found an elementary proof for the Euclidean theorem from here :

It was okay until I got stuck in $x*b=qx_0b+rb$ :

"Indeed, let $x\in E$. We have $x=qx_0+r$ for a unique pair $(q,r)\enspace (0\le r<x_0)$. As $p$ divides $xb$ and $x_ob$, $p$ divides $xb-qx_0b=rb$"

I need to be more rigorously, but I didn't understand why $p$ divides $xb-qx_0b=rb$

答:
Since $p$ divides $xb$ and $x_0b$, there are integers $m$ and $n$ such that $xb=pm$ and $x_0b=pn$. Then

$$xb-qx_0b=pm-qpn=p(m-qn)\,,$$

where $m-qn$ is an integer, so $p$ divides $xb-qx_0b$.


问:
How do I show $\int_0^1 f^2(x)dx=0$ then $f(x)=0$ where $f$ is continuous real-valued function?
This seems very obvious, but feel so stupid that I cannot prove this. How can I do?

I was thinking of using the mean value theorem for integration or uniqueness of such $f$ but they cannot work. I was thinking of using the definition of integration but I'm not sure how I can argue with that.

答: Assume $f(c)\neq 0$ $c\in[0,1]$. Then $f^2(c)>0$. By definition of continuity there exists $\delta\in\mathbb{R}^+$ such that $f^2(x)>0$ for all $x\in B(c,\delta)\cap[0,1]$. From here you can conclude using that $f^2\geq 0$ and properties of integral.


问:
Prove that $\sum_{-\infty}^{\infty}\frac{(a;q)_nt^n}{(b;q)_n}$ is an analytic funciton in $b$.
In the proof of the theorem $$\sum_{-\infty}^{\infty}\frac{(a;q)_nt^n}{(b;q)_n}=\frac{(at,\frac{q}{at},q,\frac{b}{a};q)_\infty}{(t,\frac{b}{at},b,\frac{q}{t};q)_\infty}$$ It is used both sides are analytic in $b$. When $|b/a|<|t|<1$ and $|q|<1$. I can see the right hand side is analytic, but i do not see how the left hand side is. How would i prove this? I assume I need to use the Wierstrass M-test, but do not see a nice bound here. Any help would be appreciated.

答: Each term is analytic and it converges locally uniformly (Morera's theorem).</p>


问:
How to find a polynomial that passes through a set of points (not just $1$)?
I am learning that polynomials come in this form after they are fully factored:

$f(x) = a(x+\text{root $1$})(x+\text{root $2$})(x+\text{root $n$}).$

At which point, problems say, "find a polynomial that passes through, say, $(1, 2)$. Easy, just plug in $1$ for $x$, and $2$ for $f(x)$, and solve for $a$.

But what if I have a set of points I need it to cross through?

Using an example, let's say I have $f(x) = a(x)(x-2)(x-4)$. I want this to be able to pass through the points $(1,2)$ AND $(3,-1)$?

It seems like every time I solve for a using $1$ point, I modify it to where it no longer satisfies the other point I want. It's almost like I have to solve $2$ equations at the same time? Is this possible?

答:
Remember that it if you have $n+1$ points $(x_0,y_0),$ $(x_1,y_1), \ldots,$
$(x_{n+1},y_{n+1}),$ where the $x_i$ values are all different,
there is one and only one polynomial of degree $n$ or less that passes through those $n+1$ points.

Now, obviously once you have found the unique polynomial $p$ of degree $n$ or less that passes through those given $n+1$ points, the same polynomial passes through a lot of other points as well. For any given $x,$ just compute $p(x)$, and the polynomial passes through $(x,p(x))$ as well.

So you might be able to put a polynomial of degree $n$ or less through $n+2$ points -- if the $n+2$nd point just happens to agree with the unique polynomial that goes through the first $n+1$ points.
That is, if $p$ is the unique polynomial of degree $n$ or less that goes through the first $n+1$ points, and someone says it also needs to go through another point that happens to be at $(x_{n+1}, p(x_{n+1}))$, then OK -- you already have a polynomial going through that point.
But if they demand that the polynomial go through $(x_{n+1}, p(x_{n+1}) + 1),$
it simply is not possible to do with only an $n$th-degree polynomial.
You will need at least an $(n+1)$-degree polynomial to satisfy that demand.

In the case where you are given $f(x)=ax(x−2)(x−4),$
you know that the polynomial goes through the points $(0,0),$ $(2,0),$ and $(4,0),$
because $x=0,$ $x=2,$ and $x=4$ are all places at which $f(x) = 0.$
So you've implicitly been told $3$ points your polynomial must go through.
If you are given a $4$th point at a different $x$ coordinate (not $0,$ $2,$ or $4$), that's OK -- we know we can always find a polynomial of degree $3$ through any $4$ points with all different $x$ coordinates.
We also know from the three points $(0,0),$ $(2,0),$ and $(4,0)$
that the formula must be of the form
$f(x)=ax(x−2)(x−4).$
But we can choose $a$ in order to make the polynomial go through that $4$th point.

If we're now given another point the polynomial has to pass through,
that's $5$ points in total for a polynomial that is only of degree $3.$
If you're lucky, the new point happens to be on the unique degree-$3$ polynomial through the first $4$ given points. Otherwise there is no solution.

If the problem is just to find "a polynomial" through all the given points
(the three zeros at $(0,0),$ $(2,0),$ and $(4,0)$ and two other points with different $x$ coordinates), then you can solve the problem with a polynomial of degree $4,$ since you have only $5$ points you need to fit.


问:
Prove $(U_1 + U_2) \cap (U_2 + U_3) \cap (U_3 + U_1) = (U_1+U_2) \cap U_3 + (U_1 + U_3) \cap U_2$
Prove that for any three subspaces $U_1, U_2, U_3$. $(U_1 + U_2) \cap (U_2 + U_3) \cap (U_3 + U_1) = (U_1+U_2) \cap U_3 + (U_1 + U_3) \cap U_2$ is true.

I tried to prove it by definition as I did in the begging of my calculus course with different equalities with sets. However, nothing became more clear after these attempts.

Maybe some other ways to prove it are possible? Of course, we can try to find basis somehow and see that one is the same as another. But it seems even worse as we don’t know anything about their structure.

答:
Use two sided inclusion:

Let $x$ be element of the left hand side, so there are $a_i,b_i,c_i\in U_i$ such that
$$x\ =\ a_1+a_2\ =\ b_2+b_3\ =\ c_1+c_3\,.$$
Then $b_3\ =\ x-b_2\ =\ a_1+a_2-b_2\ \in (U_1+U_2)\cap U_3$, and similarly
$b_2=c_1+c_3-b_3\ \in (U_1+U_3)\cap U_2$.

For the other direction, let $x$ be an element of the right hand side, so that $x=b_2+b_3$ with $b_2\in (U_1+U_3)\cap U_2$ and $b_3\in (U_1+U_2)\cap U_3$, then we already see that $x\in (U_2+U_3)$.

But, as $b_2\in U_1+U_3$ and clearly $b_3\in U_3\subseteq U_1+U_3$, we also get $x=b_2+b_3\in (U_1+U_3)$ and similarly $x\in (U_1+U_2)$.


问:
Complexity of a satisfaction relation
Given $x, y \in \omega^\omega$, we may define an expanded structure $(\mathbb{N}, x, y)$ where $x,y$ are treated as interpretations of function symbols. I want to show that there is a computable set $P \subseteq \omega^{<\omega} \times \omega^{<\omega} \times \omega^{<\omega}$ such that

$\forall x \in \omega^{\omega} \exists y \in \omega^{\omega}  (\mathbb{N}, x, y) \vDash \sigma \iff \forall x \in \omega^{\omega} \exists y \in \omega^{\omega} \forall n \in \omega P(x_{|n}, y_{|n})$

This is an important intermediary step in proving (a version of) the Schoenfield Absoluteness theorem. To be clear, I don't want a proof of the Schoenfield absoluteness theorem. I'm curious about this in particular.

I imagine $P$ has to do with the recursive definition of the $\vDash$ relation, but I'm not sure I can completely see it. I would appreciate the help.

答:
To clarify, the actual structure in question - per the comments below - is $\mathcal{N}_{x,y}:=(\mathbb{N};+,\times,<,x,y)$ or something "morally equivalent" to that (e.g. it's fine if we drop "$<$"). Weakening the language doesn't make the left-to-right direction any harder of course, but it may make the right-to-left direction false.

The right-to-left direction is vastly easier: it basically amounts to the fact that computable relations are definable in $(\mathbb{N};+,\times,<)$. So I'll ignore it here and focus on the interesting part, the left-to-right direction.

The left-to-right direction is rather tedious to write out in full, but the key ideas aren't too hard. This answer is really just a hint for the left-to-right direction but I think it does extract those key ideas, and will make the full solution much simpler to understand. Specifically, I'll indicate how to solve a "toy version" of the actual problem which, while substantially simpler, still in my opinion will clarify the idea.

Below I've written out my rationale for the choice of problem; if you just want to dive in, skip to the subsequent section.

First, I'm going to make a seemingly-trivial change to the desired property of $P$ in question: $$\forall x \in \omega^{\omega} \exists y \in \omega^{\omega}  \mathcal{N}_{x,y} \vDash \sigma \iff \forall x \in \omega^{\omega} \exists \color{red}{z} \in \omega^{\omega} \forall n \in \omega P(x_{|n}, z_{|n}).$$ I want to change the variable name because this $z$ should not be thought of as serving the role of $y$ in the left hand side. Instead, $z$ contains a lot more information. Specifically, and this is the absolutely key idea, we'll think of the desired $z$ as being the "join" of two reals $a,b\in\omega^\omega$ where $a$ is a real such that $\mathcal{N}_{x,a}\models \sigma$ and $b$ is a real witnessing that fact in a precise sense to be developed below.

There are a few different ways to define the join of two reals; it doesn't matter which we use, but for concreteness I'll set $a\oplus b:=(\langle a_i,b_i\rangle)_{i\in\omega}$, where $\langle\cdot,\cdot\rangle$ is the Cantor pairing function and for $u\in \omega^\omega$ we write "$u_i$" for the $i$th bit of $u$. The point is that we can clearly "extract" $a$ and $b$ from $a\oplus b$.

Second, I want to replace computability with topology. This is often a useful simplifying step, under the slogan "computable = effectively tame" for tame $\in\{$ continuous, closed, ...$\}$ - in this case, tame=closed. (The converse is also often useful, e.g. in the proof of the Harrington-Kechris-Louveau theorem, but not relevant right now.)

Finally, I want to fix a specific $\sigma$ which will be easier to work with - I'll use the sentence $$\sigma\equiv\forall m\exists n>n(x_n<y_n),$$ often abbreviated as "$y$ escapes $x$." This is just a very simple sentence to think about, while not being totally trivial.

So with all that, here's the "toy version" of the original problem:

Show that there is a closed set $\mathfrak{Q}\subseteq(\omega^\omega)^2$ such that $$\forall x \in \omega^{\omega} \exists y \in \omega^{\omega}  \mathcal{N}_{x,y} \models \forall m\exists n>m(x_m<y_m) \quad\iff\quad \forall x \in \omega^{\omega} \exists z \in \omega^{\omega} (\langle x,z\rangle\in\mathfrak{Q}).$$

The point of looking at this toy version first is that it lets us focus on the actually "deep" part of the problem: witnessing satisfaction by reals. Fixing reals $x$ and $y$, what does it mean for one real $w$ to witness the fact that $\mathcal{N}_{x,y}\models \forall m\exists n>m(x_n<y_n)$?

Well, think of a witness as a way of convincing me that the given statement holds. The way you'd do this is provide me with a rule giving an appropriate $n$ for each $m$. Now it's too much to hope for an actual rule-rule, but we can "bundle up" all the individual $m$-cases into a single real: for appropriate $x$ and $y$, we'll say $w$ witnesses $\mathcal{N}_{x,y}\models\sigma$ iff for each $i\in\omega$ we have $w_i>i$ and $x_{w_i}<y_{w_i}$.

Taking the idea from the start of this answer, this suggests looking at the following set. First, let $p_0,p_1$ be the projection maps coming from the Cantor pairing function, so that for all $l\in\omega$ we have $l=\langle p_{left}(l), p_{right}(l)\rangle$. Next, let $\mathbb{p}_{left},\mathbb{p}_{right}:\omega^\omega\rightarrow\omega^\omega$ be their "bitwise" analogues for reals, so that for each $c\in\omega^\omega$ we have $c_i=\langle \mathbb{p}_{left}(c)_i, \mathbb{p}_{right}(c)_i \rangle$. We then want to look at: $$\mathfrak{Q}=\{\langle x,z\rangle: (\mathbb{p}_{right}(z)\mbox{ witnesses }\mathcal{N}_{x,\mathbb{p}_{left}(z)}\models\forall m\exists n>m(x_n<\mathbb{p}_{left}(z)_n))\}.$$

Keep in mind that $\mathbb{p}_{left}(z)$ is an element of $\omega^\omega$, so $\mathbb{p}_{left}(z)_n$ is that element's $n$th "bit."

Now on the face of things that doesn't look like a closed set, but you can rewrite it:

 The pair $\langle x,z\rangle$ is in $\mathfrak{Q}$ iff $\forall k\in\omega$ we have $$x_{\mathbb{p}_{right}(z)_k}<(\mathbb{p}_{left}(z))_{\mathbb{p}_{right}(z)_k}.$$ That is, $\mathbb{p}_{right}(z)_k$ tells us a term which is sufficiently large and at which $x$ is less than $\mathbb{p}_{left}(z)$.

Using this rewriting, it's not hard to show that $\mathfrak{Q}$ is closed. (HINT: supposing $\langle x,z\rangle\not\in\mathfrak{Q}$, find some $j$ such that there is no pair of reals $x', z'$ extending $x_{\vert j}, z_{\vert j}$ respectively with $\langle x',z'\rangle\in\mathfrak{Q}$; this will say that $\mathfrak{Q}$'s complement is open.)

We now have to switch from topology back to computability, and generalize to arbitrary $\sigma$.

The computability/topology shift is pretty easy to undo, and goes the same for every $\sigma$: you just have to show that there is some computable relation $Q\subseteq(\omega^{<\omega})^2$ corresponding to our closed set $\mathfrak{Q}$, that is, some computable relation $Q$ such that for all $u,v$ we have $$\langle u,v\rangle\in\mathfrak{Q}\quad\iff\quad\forall n(Q(u_{\vert n}, v_{\vert n})).$$

Generalizing to arbitrary sentences $\sigma$ by contrast is meaningfully difficult, so I'd focus on that first. Think about coding up families of Skolem functions for an arbitrary $\sigma$ by individual reals. For example, suppose $\sigma$ is of the form $\forall q_1\exists q_2\forall q_3\exists q_4\tau$ with $\tau$ quantifier-free and think about $(i)$ what a family of Skolem functions for such a sentence is and how it must behave and $(ii)$ how to "bundle up" such a family of functions into a single real; this case has all the complexity of the general situation already.

Title: Complexity of a satisfaction relation


答:
You cannot divide the sum of the elements of $a$ by the sum of the elements of $b$. The reason is that what we expect out of our "normal" rules of addition, subtraction, multiplication and division become very tricky when infinity becomes involved.

This is seen very commonly in a number of misleading but seemingly correct manipulations. Consider, for instance, the infinite sum $$S = 1 + 2+ 4 + 8 + 16 + \cdots$$

Now multiply $S$ by 2 to find $$2S = 2 + 4 + 8 + 16 + \cdots = S - 1$$

This would imply that $S = -1$, which is an absurd conclusion indeed!

Math is rife with examples like this where infinities are tricky to manage and must be handled with exceptional care. For more examples like these, the excellent notes of Dr. Tao are quite informative:

http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/week1.pdf

Title: Dividing Two Infinities


问:
local property of tangent vector
Given a smooth manifold $M$ and a point $p \in U$, $U \subset M$ open, my book says there is a linear isomorphism between vector spaces $T_pU$ and $T_pM$ by:

$v \rightarrow \tilde{v}$ where for any $f \in C^\infty(M)$, $\tilde{v}(f) = v(f|_U)$. Now I can see the linear part, but I cannot see the bijection.

Edit: Background lemma that might help:
Let $v \in T_pM$. If $f,g \in C^\infty(M)$ are equal on a neighborhood of $p$, then $v(f) =  v(g)$. If $h \in C^\infty(M)$ is constant on a neighborhood of $p$, then $v(h) = 0$

答:
The "background lemma" answers your question.

A tangent vector in $T_pM$ is a derivation $v : C^\infty(M) \to \mathbb R$ at the point $p$. The lemma says that that we may regard $v$ as a function $C^\infty(M)/\sim_p \phantom{.} \to \mathbb R$, where $f \sim_p g$ if $f$ and $g$ are equal on a neighborhood of $p$.

You have already defined a linear map $\phi: T_pU \to T_pM, \phi(v)(f) = v(f|_U)$. Let us define $\psi : T_pM \to T_pU$ as follows:

Let $v \in T_pM$ and $f \in C^\infty(U)$. Choose an open neigborhood $V$ of $p$ in $M$ such that $\bar V \subset U$. You know that $f \mid_{\bar V}$ has a smooth extension $\tilde f : M \to \mathbb R$. There are many choices of $V$ and many choices of extensions $\tilde f$ of $f \mid_{\bar V}$, but clearly all these $\tilde f$ are equivalent with respect to $\sim_p$. Thus $v(\tilde f)$  is well-defined. Now define  $\psi(v)(f) = v(\tilde f)$. It is clear that $\psi \circ \phi =id$ and $\phi \circ \psi =id$ which proves that $\phi$ is an isomorphism.

Title: local property of tangent vector


问:
Prove or disprove: if $f(x)$ has a removable discontinuous point at 0 then $g(x)$ = $x*f(x)$ is differentiable at 0
I need to determine whether it is true that if $f(x)$ has a removable discontinuous point at 0 then $g(x)$ = $x*f(x)$ is differentiable at 0.

I thought to prove it by proving that $g(x)$ is continuous from both positive and negative directions by saying that the limit of $f(x)$ from both positive and negative directions is equal and, because the function $h(x) = x$ is continuous. which will lead me to say that $g(x)$ is continuous and then find the derivative.

But I'm not sure if it's possible to claim that.

答:
Here, it is worth just going back to the definition of a derivative as a limit.

By definition of a removable discontinuity, we have that $f$ is defined at $0$ and that $\lim_{x\to 0} f(x)$ exists, but $f(0) \neq \lim_{x\to 0} f(x)$.

Now, for $g$ as in the question, we have $g(0)=0\cdot f(0)=0$ by definition.

For $x\neq 0$, we can write
$$
\frac{g(x)-g(0)}{x-0} = \frac{g(x)}{x} = \frac{xf(x)}{x} = f(x)
$$
so $\lim_{x\to 0} \frac{g(x)-g(0)}{x-0}$ exists since $\lim_{x\to 0} f(x)$ does. It follows that $g$ is differentiable at $0$, with
$$
g'(0) = \lim_{x\to 0} \frac{g(x)-g(0)}{x-0} = \lim_{x\to 0} f(x)
$$


问:
Any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$. My solution
Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:

If $i=0$, then
\begin{align*}
        n = 5k \Rightarrow n = 5k + 4(0).
\end{align*}

If $ i = 1 $, then
\begin{align*}
       n & = 5k + 1 \\
         & = 5k-5(3) +5(3) +1 \\
         & = 5(k-3) + 15 + 1 \\
         & = 5(k-3) +16 \Rightarrow n = 5(k-3) +4(4).
\end{align*}

If $ i = 2 $, then
\begin{align*}
   n & = 5k + 2 \\
     & = 5k-5(2) +5(2) +2 \\
     & = 5(k-2) + 10 + 2 \\
     & = 5(k-2) +12 \Rightarrow n = 5(k-2) +4(3).
\end{align*}

If $i=3$, then
\begin{align*}
   n & = 5k + 3 \\
     & = 5k-5 + 5 + 3 \\
     & = 5(k-1) +8 \Rightarrow n = 5(k-1) +4(2).
\end{align*}

If $i=4$, then
\begin{align*}
    n = 5k + 4 \Rightarrow n = 5k + 4(1).
\end{align*}

Thus, every positive number can be expressed as a linear combination of $5$ and $4$. Now using that $n>11$, so we have:
\begin{align*}
n       &> 11 \\
5k + i  &> 5(2) +1 \\
5k-5(2) &> 1-i \\
5 (k-2) &> 1-i \\
k-2     &> \frac{1-i}{5} \\
k       &> 2+\frac{1-i}{5}.
\end{align*}
So by increasing over the values that $ i $ takes, we have:
\begin{align*}
k &> 2+ \frac{1-i}{5} \geq 2+ \frac{1-0}{5}\\
k &> 2 + 0.2 = 2.2
\end{align*}
But $k\in\mathbb{Z}_0^+ \Rightarrow k \geq 3 \Rightarrow n \geq 15 $. Thus we have that every positive integer greater than or equal to $15$ is a non-negative linear combination of $5$ and $4$.

Finally, let's look at the cases that are still unverified, which are $12$, $13$ and $14$.
\begin{align*}
12 &= 5(0) +4(3) \\
13 &= 5(1) +4(2) \\
14 &= 5(2) +4(1).
\end{align*}
Therefore, any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$.

I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

答: We proceed by induction on $n.$ Observe that $12 = 3 \cdot 4 + 0 \cdot 5$ is a non-negative linear combination of $4$ and $5.$ We will assume inductively that any integer $n \geq 12$ can be written as a non-negative linear combination $n = 4x + 5y$ of $4$ and $5.$ Given that $x \geq 1,$ we have that $$4(x - 1) + 5(y + 1) = 4x - 4 + 5y + 5 = 4x + 5y + 1 = n + 1$$ is a non-negative linear combination of $4$ and $5,$ as desired. On the other hand, if we have that $x = 0,$ then $n = 5y.$ By hypothesis that $n \geq 12,$ we must have that $n \geq 15$ (because $n$ is a multiple of $5$) so that $y \geq 3.$ Consequently, we find that $$4 \cdot 4 + 5(y - 3) = 16 + 5y - 15 = 5y + 1 = n + 1$$ is a non-negative linear combination of $4$ and $5.$ QED.


答:
To see that $g$ is differentiabile at $0$, we calculate the limit $$\lim_{h\to 0}\frac{g\left(0+h\right)-g\left(0\right)}{h}=\lim_{h\to 0}\frac{\left(0+h\right)\cdot f\left(0+h\right) - 0\cdot f\left(0\right)}{h}=\lim_{h\to 0}\frac{h\cdot f\left(h\right)}{h}=\lim_{h\to 0} f\left(h\right)$$
So, $f'(0)$ exists.

Title: Prove or disprove: if $f(x)$ has a removable discontinuous point at 0 then $g(x)$ = $x*f(x)$ is differentiable at 0

Body:
I need to determine whether it is true that if $f(x)$ has a removable discontinuous point at 0 then $g(x)$ = $x*f(x)$ is differentiable at 0.

I thought to prove it by proving that $g(x)$ is continuous from both positive and negative directions by saying that the limit of $f(x)$ from both positive and negative directions is equal and, because the function $h(x) = x$ is continuous. which will lead me to say that $g(x)$ is continuous and then find the derivative.

But I'm not sure if it's possible to claim that.


问:
Inner product at the boundary
Let $K\subset \mathbb{R}^n$ be a convex set with nonempty interior, i.e. $K^{\circ}\neq \emptyset$. Suppose there exist a nonzero vector $v$ such that

$$\langle x, v\rangle\geq 0$$

for all $x\in K^{\circ}$. I want to show that in this case the property holds for all  $x\in K$.

It is supposed to be a continuity argument. Am having trouble because $x\in K\setminus K^{\circ}$ doesn't imply that there exist a sequence in $K^{\circ}$ converging to $x$.

Any help is greatly appreciated.

答:
But $x \in K \setminus K^\circ$ does imply that there exists a sequence in $K^\circ$ converging to $x$.

Choose any open ball $B \subset K^\circ$ centered at $p$.

Let $\overline{ab}$ be the diameter of $B$ perpendicular to the segment $\overline{xp}$.

The interior of the entire triangle $\triangle xab$ is therefore contained in $K^\circ$, and in particular the half-open segment $[p,x)$ is contained in $K^\circ$.

Title: Inner product at the boundary


答:
You also can prove it by induction on $n>11$.

Here is how the inductive step goes:

Suppose that for some $n>11$, one has $\;n=5x+4y$, $x,y \ge 0$.

If $y >0$, as $1=5-4$, you can write $n+1=5(x+1)+4(y-1)$, and $y_1\ge 0 $;

If $y=0$, $n=5x$, and since $n>11$, you have $x\ge 3$. Now you also can write $1=16-15$, so
$$n+1=5(x-3)+4\cdot 4\qquad(\text{note that } x-3\ge 0)$$

Title: Any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$. My solution

Body:
Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:

If $i=0$, then
\begin{align*}
        n = 5k \Rightarrow n = 5k + 4(0).
\end{align*}

If $ i = 1 $, then
\begin{align*}
       n & = 5k + 1 \\
         & = 5k-5(3) +5(3) +1 \\
         & = 5(k-3) + 15 + 1 \\
         & = 5(k-3) +16 \Rightarrow n = 5(k-3) +4(4).
\end{align*}

If $ i = 2 $, then
\begin{align*}
   n & = 5k + 2 \\
     & = 5k-5(2) +5(2) +2 \\
     & = 5(k-2) + 10 + 2 \\
     & = 5(k-2) +12 \Rightarrow n = 5(k-2) +4(3).
\end{align*}

If $i=3$, then
\begin{align*}
   n & = 5k + 3 \\
     & = 5k-5 + 5 + 3 \\
     & = 5(k-1) +8 \Rightarrow n = 5(k-1) +4(2).
\end{align*}

If $i=4$, then
\begin{align*}
    n = 5k + 4 \Rightarrow n = 5k + 4(1).
\end{align*}

Thus, every positive number can be expressed as a linear combination of $5$ and $4$. Now using that $n>11$, so we have:
\begin{align*}
n       &> 11 \\
5k + i  &> 5(2) +1 \\
5k-5(2) &> 1-i \\
5 (k-2) &> 1-i \\
k-2     &> \frac{1-i}{5} \\
k       &> 2+\frac{1-i}{5}.
\end{align*}
So by increasing over the values that $ i $ takes, we have:
\begin{align*}
k &> 2+ \frac{1-i}{5} \geq 2+ \frac{1-0}{5}\\
k &> 2 + 0.2 = 2.2
\end{align*}
But $k\in\mathbb{Z}_0^+ \Rightarrow k \geq 3 \Rightarrow n \geq 15 $. Thus we have that every positive integer greater than or equal to $15$ is a non-negative linear combination of $5$ and $4$.

Finally, let's look at the cases that are still unverified, which are $12$, $13$ and $14$.
\begin{align*}
12 &= 5(0) +4(3) \\
13 &= 5(1) +4(2) \\
14 &= 5(2) +4(1).
\end{align*}
Therefore, any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$.

I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.


问:
Average of middle number
I was doing a question on average and the question was that - If the average of 1st 12 number is 10.8, average of 1st 6 number is 10.4 and the last 6 number is 11.5. Find the 6th number.
Now according to me i thought the number which is the middle one is counted twice but the actual solution subtracts the sum of last and 1st 6 number sum and subtracts with the sum of 12 numbers.
Can anyone help me why is it correct and does it not double counts the middle number twice?</p>

答: If you have a list of $12$ numbers, the first six numbers are in positions $1-6$ and the last six numbers are in positions $7-12$.  As stated, the data you are given is inconsistent with the list being $12$ numbers.  If the first six average $10.4$ and the last six average $11.5$, the whole list would have to average $10.95$.  Please check the problem and state it clearly.

Title: Average of middle number


问:
Quadratic integers proof verification
Defining $\mathcal{O}$ to be the set of all $x \in \mathbb{Q}(\sqrt{D})$, for $D$ a square-free integer, such that the minimal polynomial has coefficients in $\mathbb{Z}$.

I have to show that $\mathcal{O}$ = $\mathbb{Z}[\omega]$ for $$\omega = \begin{cases}
\sqrt{D},  & \text{if $D \equiv 2,3 \pmod 4$} \\
\frac{1+\sqrt{D}}{2}, & \text{if $D \equiv 1 \pmod 4$}
\end{cases}$$

I was wondering if there were any holes in my proof and any feedback would be appreciated.

Given the minimal polynomial $x^2 -2ax+(a^2-b^2 D)$ of an arbitrary element $a+b\sqrt{D} \in \mathbb{Q}(\sqrt{D})$, the definition requires $2a \in \mathbb{Z}$ and $a^2 -b^2 D \in \mathbb{Z}$. Therefore $a=\frac{i}{2}$ for some $i \in \mathbb{Z}$. We can replace the $a$'s equivalent in the second equation to get $\frac{i^2}{4} -b^2D= j$ for some $j \in \mathbb{Z}$, which after rearranging terms and multiplying both sides by $4$

$$i^2 -4j=4b^2D =k \in \mathbb{Z}$$

because $i,j \in \mathbb{Z}$. $D|k$. $b^2 = \frac{k'}{4}$ and $b=\frac{i'}{2}$ for some $i' \in \mathbb{Z}$, and $\mathcal{O} \subseteq \mathbb{Q}(\sqrt{D})$ must be elements of the form $\frac{i}{2}+\frac{i'}{2}\sqrt{D}$ for $i,i' \in \mathbb{Z}$.

For simplicity I switched $i$ and $i'$ to $a$ and $b\in \mathbb{Z}$, and then I calculated the minimal polynomial of $\frac{a}{2}+\frac{b}{2}\sqrt{D}$
$$x^2 -ax +\frac{a^2-b^2D}{4}$$
where the last term being in $\mathbb{Z}$ implies $a^2 \equiv Db^2 \pmod4$, and with the added knowledge that $z^2 \equiv 0,1 \pmod 4$ for $z \in \mathbb{Z}$:

$D \equiv 1 \pmod 4$ then $a^2 \equiv b^2 \pmod 4$ $a,b$ are both even or odd.

$D \equiv 2,3 \pmod 4$ then $a^2 \equiv (2,3) b^2 \equiv 0 \pmod 4$ since $a^2 \equiv 0,1 \pmod 4$ thus $a,b$ must both be even.

Thus if $D \equiv 2,3 \pmod 4$ then we must have elements of the form $a' + b'\sqrt{D}$ for $a',b' \in \mathbb{Z}$.

If $D\equiv 1 \pmod 4$ then we also have elements of that form $a' + b'\sqrt{D}$ but also elements of the form

$$\frac{2i+1}{2}+\frac{2i'+1}{2}\sqrt{D} = (i -i') +(2i'+1)\frac{1+\sqrt{D}}{2}$$

and noting that $$a'+b'\frac{1+\sqrt{D}}{2} = a'+\frac{b'}{2}+\frac{b'}{2}\sqrt{D} \equiv (c\in\mathbb{Z})+(d\in \mathbb{Z})\sqrt{D}$$ for $b$ even, every form is included in rewriting elements for $D \equiv 1 \pmod 4$ in the form $a'+b'\frac{1+\sqrt{D}}{2}$ for $a', b' \in \mathbb{Z}$.

答: As mentioned in the comments already: this looks good (and I will definitely bookmark this post for further reference!). You might be interested in this post which goes over the proof (maybe) more efficiently.</p>

Title: Quadratic integers proof verification


答: This question is somewhat related to this MO question. The MO question asks for examples of locally finite varieties that are not finitely generated. If a variety is locally finite, then its finitely generated free algebras are finite. If the variety is not finitely generated, then no finite algebra in the variety is generic. So the examples given at the link are of varieties where no free algebra of finite rank is generic.</p>

Title: A free algebra that is not generic

Body: In a previous question, I was told that free algebras on sufficiently many generators are generic algebras. That raises the question, what is an example of a free algebra that is not generic? An algebra $A$ is generic for a variety $V$ iff $A$ generates $V$.


问:
Why is $\frac{1}{x^{-1}}=x$?
I believe I get why $x^{-1}=\frac{1}{x}$ . Because if we would like our exponent rule ($\frac{x^n}{x^m}= x^{m-n}$) to be true for all values of $m$ and $n$ then $x^{-1}$ should be another way to write $\frac{1}{x}$ as [$\frac{x^1}{x^2}=x^{1-2}=x^{-1}=\frac{1}{x}$] even if multiplying a certain number by itself negative number of times makes no sense.

So I can see why $x^{-1}$ is just another way to write $\frac{1}{x}$ but I can't get how we've arrived that $\frac{1}{x^{-1}}$ is equal to $x$; I can't think of any way that will give us a negative power in the doninator in order to proof that $\frac{1}{x^{-1}}$=$x$, we can't subtract from the dominator's power even when we divide. We always end up subtracting from the numerator's power instead, example:

$\frac{1}{x}÷\frac{1}{x^3}=\frac{1}{x}×x^3= \frac{x^3}{x}=x^{3-1}=x^2$

答:
For a nonzero real number $x$, the reciprocal $\frac1x$ is the real number such that when multiplying with $x$, you get $1$. In other words, when $x\ne 0$, $\frac1x$ is the multiplicative inverse of $x$.

For $x\ne 0$, the negative exponential $x^{-1}$ is by definition $\frac1x$, and thus
$$
x\cdot x^{-1}=x^{-1}\cdot x=1
$$
Therefore,
$$
\frac{1}{x^{-1}}=x.
$$

Title: Why is $\frac{1}{x^{-1}}=x$?


问:
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$
Find the limit without the use of L'Hôpital's rule or Taylor series

$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$

Body:
Let $I = \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)$, we have:

$$\begin{align}
I = & \lim_{x\to 0}\left(\frac{1}{(2x)^2} - \frac{1}{\sin^2(2x)}\right)\\
  = & \lim_{x\to 0} \left(\frac{1}{4 x^2} - \frac{1}{4 \sin^2 x\cos^2 x}\right)\\
\implies  4I = & \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)\\
\implies  3I = & \lim_{x\to 0}\left\{\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)
                             -\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)\right\}\\
=&-\lim_{x\to 0} \frac{1-\cos^2 x}{\sin^2 x\cos^2 x} = -\lim_{x\to 0}\frac{1}{\cos^2 x} = -1\\
\implies I = & -\frac{1}{3}
\end{align}$$


问:
Finding all Covering Spaces of a given index.
I am reviewing for an upcoming topology qualifying exam, and I have a question regarding a specific type of question:

Find all connected two-sheeted covering spaces of $S^1\vee \mathbb{R}P^2$ up to equivalence.

The covering space that I was able to think of is picture below.

$p$ sends $a_1$ and $a_2$ to $a$ and $p$ is the anti-podal map on $S^2$ (the universal cover).
I have that $\pi_1(\tilde X,\tilde x_0)=\langle a_1,b_1a_2\overline{b_1}|\rangle$ where $\tilde x_0$ is the point where $a_1$ intersects $S^2$, $b_1$ is the half of the equator pictured and $x_0$ is the point at which $a$ intersects $\mathbb{R}P^2$. Then $p_*(\pi_1(\tilde X,\tilde x_0))=\langle a, \gamma a \gamma^{-1}|\gamma^2=1\rangle =\langle a, \gamma a \gamma|\gamma^2=1\rangle$ where $\gamma$ is the generator of $\pi_1(\mathbb{R}P^2)=\mathbb{Z}_2$. Are there other subgroups of index 2 of $\pi_1(S^1\vee \mathbb{R}P^2)=\langle a,\gamma|\gamma^2=1\rangle$? I could try to argue that the only nontrivial covering space of $\mathbb{R}P^2$ is $S^2$, and thus the covering above is all, but this isn't very rigorous. In general, given a space, how might one go about finding all covering spaces (up to equivalence) of a certain index? I know that given a space $X$ (p-conn., locally p-conn., semi-locally simply connected), there is a bijection between the path-connected basepoint preserving covering spaces $p:(\tilde X,\tilde x_0)\rightarrow (X,x_0)$ (up to equivalence) and the subgroups of $\pi_1(X,x_0)$ where the correspondence is given by $p_* \pi_1(\tilde X,\tilde x_0)$. I also know that if $p:(\tilde X,\tilde x_0)\rightarrow (X,x_0)$ is an $n$-sheeted covering, then $p_* \pi_1(\tilde X,\tilde x_0)$ has index $n$ in $\pi_1(X,x_0)$. So, if I can find all subgroups of a given index, then I at least know how many covering spaces I need to find. However, finding all subgroups of a given index on a free group isn't always easy, and it still doesn't tell me how to actually construct these spaces. For example, given my covering space above,  How can I prove that I have found all $2$ sheeted covering spaces without calculating all subgroups of a given index? In general, can I find all $n$ sheeted covering spaces without calculating the number of subgroups of index $n$ and just guessing as to the actual construction of the covering space ? Also, as an aside question, what would the universal cover of $S^1\vee \mathbb{R}P^2$ be?

答:
You've found an example which contains a 2-sheeted connected covering of $\mathbb RP^2$. You can stretch that idea to try to imagine other examples: maybe one which contains a 2-sheeted connected covering of $S^1$; maybe one which contains 2-sheeted covering spaces of each of $\mathbb R P^2$ and $S^1$.

Also, finding all index 2 subgroups of a finitely presented group $G$ is easier than you think: every index 2 subgroup of $G$ is normal and has quotient is isomorphic to $\mathbb Z / 2 \mathbb Z$. From this, with a bit more thought, you can deduce that the "kernel" operation induces a bijection between the set of surjective homomorphisms $G \mapsto \mathbb Z / 2 \mathbb Z$ and the set of index 2 subgroups of $G$. So, if you use Van Kampen's theorem to write down a presentation of the fundamental group of your $S^1 \vee \mathbb R P^2$ then you should be able to pretty easily write down a list of all possible surjective homomorphisms to $\mathbb Z / 2 \mathbb Z$.


Title: Why is $\frac{1}{x^{-1}}=x$?

Body:
I believe I get why $x^{-1}=\frac{1}{x}$ . Because if we would like our exponent rule ($\frac{x^n}{x^m}= x^{m-n}$) to be true for all values of $m$ and $n$ then $x^{-1}$ should be another way to write $\frac{1}{x}$ as [$\frac{x^1}{x^2}=x^{1-2}=x^{-1}=\frac{1}{x}$] even if multiplying a certain number by itself negative number of times makes no sense.

So I can see why $x^{-1}$ is just another way to write $\frac{1}{x}$ but I can't get how we've arrived that $\frac{1}{x^{-1}}$ is equal to $x$; I can't think of any way that will give us a negative power in the doninator in order to proof that $\frac{1}{x^{-1}}$=$x$, we can't subtract from the dominator's power even when we divide. We always end up subtracting from the numerator's power instead, example:

$\frac{1}{x}÷\frac{1}{x^3}=\frac{1}{x}×x^3= \frac{x^3}{x}=x^{3-1}=x^2$

Body: $\frac{1}{x^{-1}}=\frac{x}{xx^{-1}}=\frac{x}{1}=x$


问:
What is $\frac{\det(\hat{\Sigma}_0)}{\det(\hat{\Sigma})}$ in terms of $\hat{\mu}_1$, $\hat{\mu}_2$ and $\hat{\Sigma}$
Let $X_1,...,X_{n_1}$ be an i.i.d. sample from $N_p(\mu_1,\Sigma)$ and let $Y_1,...,Y_{n_2}$ be an independent sample from $N_p(\mu_2,\Sigma)$, for some $\mu_1,\mu_2 \in \mathbb{R}^p$ and some invertible, $p\times p$ positive definite matrix $\Sigma$.

Let $\hat{\mu}_0 := \frac{\sum_{i=1}^{n_1}x_i + \sum_{i=1}^{n_2}y_i}{n_1 + n_2}$, $\hat{\mu}_1 := \frac{1}{n_1}\sum_{i=1}^{n_1}x_i$ and $\hat{\mu}_2 := \frac{1}{n_2}\sum_{i=1}^{n_2}y_i$

Suppose $\hat{\Sigma}_0=\frac{1}{n_1+n_2}\biggl(\sum^{n_1}_{i=1}(x_i-\hat{\mu}_0)(x_i-\hat{\mu}_0)^T+\sum^{n_2}_{i=1}(y_i-\hat{\mu}_0)(y_i-\hat{\mu}_0)^T\biggr)$

and $\hat{\Sigma}=\frac{1}{n_1+n_2}\biggl(\sum^{n_1}_{i=1}(x_i-\hat{\mu}_1)(x_i-\hat{\mu}_1)^T+\sum^{n_2}_{i=1}(y_i-\hat{\mu}_2)(y_i-\hat{\mu}_2)^T\biggr)$

I would like to show that:

$$\frac{\det(\hat{\Sigma}_0)}{\det(\hat{\Sigma})} = 1 + \frac{n_1n_2}{(n_1+n_2)^2}(\hat{\mu}_1 -\hat{\mu}_2)^T\hat\Sigma^{-1}(\hat{\mu}_1 -\hat{\mu}_2) $$

I know that $\frac{\det(\hat{\Sigma}_0)}{\det(\hat{\Sigma})}=\det(\hat{\Sigma}^{-1/2}\hat{\Sigma}_0\hat{\Sigma}^{-1/2})$, but I'm not sure how to continue.

答:
After fixing the missing $\hat{}$ over the $\Sigma$ on the right-hand side, the equation is trivial to prove via the Matrix Determinant Lemma. It tells us that $\det(A+uv^T) = (1+v^T A^{-1}u)\det(A)$, and we simply need to match the terms: With $A=\hat\Sigma$, $u=v=\frac{\sqrt{n_1 n_2}}{n_1 + n_2}(\hat \mu_1 -\hat\mu_2)$ it follows that

$$\begin{aligned}
\Big(1 + u^T\hat\Sigma^{-1}v\Big)\det(\hat\Sigma)
&= \det(\hat\Sigma + uv^T)
\end{aligned}$$

Now, all we need to do is simplify until we arrive at $\hat \Sigma_0$. Using $\hat\mu_1 = \frac{1}{n_1} X^T 1$, $\mu_2=\frac{1}{n_2}Y^T1$ we have

$$\begin{aligned}
\sum_{i=1}^{n_1} (x_i-\hat\mu_1)(x_i-\hat\mu_1)^T
&= (X-1\hat\mu_1^T)^T(X-1\hat\mu_1^T)
\\&= X^TX-\hat\mu_11^TX -X^T 1 \hat\mu_1 + \hat\mu_1 1^T1 \hat\mu_1^T
\\&=\boxed{X^TX-n_1\hat\mu_1\hat\mu_1^T}
\end{aligned}$$

And similarly $Y^TY-n_2\hat\mu_2\hat\mu_2^T$ for the $y$-sum. On the other hand

$$\begin{aligned}
\sum_{i=1}^{n_1} (x_i-\hat\mu_0)(x_i-\hat\mu_0)^T
&= (X-1\hat\mu_0^T)^T(X-1\hat\mu_0^T)
\\&= X^TX-\hat\mu_01^TX -X^T 1 \hat\mu_0 + \hat\mu_0 1^T1 \hat\mu_0^T
\\&=\boxed{X^TX - n_1\hat\mu_0\hat\mu_1^T - n_1\hat\mu_1\hat\mu_0 + n_1\hat\mu_0\hat\mu_0^T}
\end{aligned}$$

And similarly $Y^TY - n_2\hat\mu_0\hat\mu_2^T - n_2\hat\mu_2\hat\mu_0 + n_2\hat\mu_0\hat\mu_0^T$ for the $y$-sum. Adding both together, and using  $\hat \mu_0 = \frac{n_1 \hat\mu_1 + n_2\hat\mu_2}{n_1 + n_2}$ we find

$$\begin{aligned}
&X^TX - n_1\hat\mu_0\hat\mu_1^T - n_1\hat\mu_1\hat\mu_0 + n_1\hat\mu_0\hat\mu_0^T \\&+Y^TY - n_2\hat\mu_0\hat\mu_2^T - n_2\hat\mu_2\hat\mu_0 + n_2\hat\mu_0\hat\mu_0^T
\\&=X^TX+Y^TY-\hat\mu_0(n_1\mu_1^T + n_2\mu_2^T) - (n_1\mu_1 + n_2\mu_2)\hat\mu_0^T +(n_1+n_2)\hat\mu_0\hat\mu_0^T
\\&=X^TX+Y^TY-(n_1+n_2)\hat\mu_0\hat\mu_0^T - (n_1+n_2)\hat\mu_0\hat\mu_0^T +(n_1+n_2)\hat\mu_0\hat\mu_0^T
\\&=X^TX+Y^TY-(n_1+n_2)\hat\mu_0\hat\mu_0^T
\\&=\boxed{X^TX+Y^TY-\frac{(n_1\hat\mu_1+n_2\hat\mu_2)(n_1\hat\mu_1+n_2\hat\mu_2)^T}{n_1+n_2}}
\end{aligned}$$

Thus, our goal of showing $\hat\Sigma_0 = \hat\Sigma +uv^T$ is equivalent to (after cancelling $X^TX$ and $Y^TY$ terms)

$$
-\frac{(n_1\hat\mu_1+n_2\hat\mu_2)(n_1\hat\mu_1+n_2\hat\mu_2)^T}{(n_1+n_2)^2} =\frac{-n_1\hat\mu_1\hat\mu_1^T-n_2\hat\mu_2\hat\mu_2^T}{n_1+n_2} + \frac{n_1n_2(\hat \mu_1 -\hat\mu_2)(\hat \mu_1 -\hat\mu_2)^T}{(n_1+n_2)^2}
$$

Which is equivalent to

$$
(n_1\hat\mu_1+n_2\hat\mu_2)(n_1\hat\mu_1+n_2\hat\mu_2)^T + n_1n_2(\hat \mu_1 -\hat\mu_2)(\hat \mu_1 -\hat\mu_2)^T =(n_1+n_2)(n_1\hat\mu_1\hat\mu_1^T+n_2\hat\mu_2\hat\mu_2^T)
$$

which is a true statement. q.e.d

Final Remark: Of course, the Matrix Determinant Lemma would also offer a strategy of figuring out the equation in the first place: If we wanted to find the relationship bewtween $\det(\hat\Sigma_0)$ and $\det(\hat\Sigma)$, it would suggest to express $\hat\Sigma_0 = \hat\Sigma + UV^T$. Since both  $\hat\Sigma_0$ and $\hat\Sigma$ are symmetric, we are in fact guaranteed  $\hat\Sigma_0 = \hat\Sigma + UU^T$ for some $(p\times k)$ matrix $U$ with $k\le p$ (generally we would choose the $U$ matrix with $k$ minimal)


答:
I'm not really good at explaining stuff, but this guy definitely is.

Answering you question, $x^{-1}$ is equal to $\frac{1}{x}$.

And $\frac{1}{x^{-1}}$ is equal to $\frac{1}{1}/\frac{1}{x}=\frac{1}{1}*\frac{x}{1}=x$

Title: Why is $\frac{1}{x^{-1}}=x$?