问:
Powers of bounded Sobolev functions
I felt that the claim

$$u\in L^\infty(\Omega)\cap W^{1,p}_{loc}(\Omega)\implies u^k\in W^{1,p}_{loc}(\Omega),$$

for $1<p<\infty$ and $k\geq 1$, should be true but I find it hard to believe. The proof I had in mind went as follows. Since, $u\in L^\infty$, all its powers are also bounded and hence in $L^p$. For the derivative, we find a sequence of smooth functions $u_n$ converging to $u$ in $W^{1,p}(\Omega')$ for $\Omega'\subset\subset\Omega$. Then, it holds true that for any $\phi\in \mathcal{D}(\Omega)$ compactly supported in $\Omega'$, we have
$$\int_{\Omega'}u_n^k \phi_{x_i}dx=-k\int_{\Omega'}u_n^{k-1}(u_n)_{x_i}\phi\,dx.$$

Now, since both integrands converge pointwise almost everywhere for a subsequence, we can apply the Dominated convergence theorem to conclude the expected limits.

Where am I wrong? If it is true, does anyone know of a reference?

答:
You can easily prove that is true by just computing the weak derivative of $u^k$. However, a more general result holds. In fact, $L^\infty(\Omega)\cap W^{1,p}(\Omega)$, $1\leq p\leq \infty$, is an algebra:

If $u,v\in L^\infty(\Omega)\cap W^{1,p}(\Omega)$ with $1\leq p\leq
\infty$, then $u v \in L^\infty(\Omega)\cap W^{1,p}(\Omega)$.

What you ask is the particular case in which $u=v$. The proof is easy and can be found in Brezis's book (Prop.9.4, p.269).

Title: Powers of bounded Sobolev functions


问:
Asymptotic behaviour of quantity related to an ODE
Consider the differential equation

$$u''(t) + u(t) + u^3(t) = f(t) u(t) + g(t),\,\, u(0) = u_0, \,u'(0) = u_1$$

where $u :\mathbb{R}^+ \to \mathbb{R}$ is a bounded solution  and $f, g: \mathbb{R}^+ \to \mathbb{R}$ are given continuous functions such that $$\lim_{t\to \infty} f(t) = \lim_{t\to \infty} g(t) = 0.$$
Define the quantity $E(t) := \frac 12 (u(t))^2 +\frac 12 (u'(t))^2 + \frac 14 (u(t))^4.$
I want to study the asymptotic of $E$ when $t \to \infty.$  First, I considered many examples for $f, g$ using mathematica (See the attached images where I plotted $t\mapsto E(t)$). I'm interested in the case where
$\liminf_{t\to \infty}E(t) >0$ or roughly $\lim_{t\to \infty} E(t) >0.$ I wonder if this is the general case or it depends on the initial values $u_0, u_1$ and $f,g.$ In all the examples that I tested numerically, I found that $\lim_{t\to \infty} E(t) >0.$
My attempt started by differentiating $E$ with respect to $t.$ It seems that in the case where
$u'$ is bounded, we have $$\lim_{t\to \infty} E'(t) =0,$$
since $E' = (f u + g) u'.$ But this does not imply necessarily that $\lim_{t\to \infty}E(t) >0.$
The limit might be zero.

Thank you for any hint.

答:
In order for the system to be well-behaved a sufficient condition is that $f$ and $g$ are integrable at infinity.

Consider some uniform bounds of the RHS in terms of the energy functional (a standard trick): The $f$ term behaves the best. By Cauchy-Schwarz you have $|uu'|\leq E$ whereas for the $g$ term you only have $|u'|\leq \sqrt{2 E}$. We thus get the crude bound:
$$ |E'| \leq |f| E + |g| \sqrt{2E}$$.

A special case is $g\equiv 0$ for which we get (use e.g. a Gronwall Lemma):
$$ |\log(E(\infty)) - \log(E(0)) | \leq \int_0^\infty |f(t)|dt.$$
In this case it suffices that $f$ is integrable in order for
$\lim_{t \rightarrow \infty}  E(t)$ to exist and be strictly positive.

When $g$ does not vanish you may put e.g. $E=\frac12 A^2$  which yields (as long as $A>0$):
$$ |A'| \leq \frac12 |f| A + |g| $$

When e.g. $f$ and $g$ are integrable then also in this case the  system never explodes (use again a Gronwall Lemma) and will have a finite limit  which, however, may be zero. You may also from this get conditions on initial values of $A$ which when large enough compared to the $L^1$ norm of $f,g$ give lower bounds that assures that the limit is strictly positive. For example, a sufficient condition is that
$$ A(0) \ (1 - \frac12 \|f\|_1) > \|g\|_1 $$

ContentLicense: CC BY-SA 4.0


问:
how det(A)=0 implies that the solution isn't unique?
Solution of matrix equation Ax=b, where
$$ A=\left(\begin{matrix}
a_1&a_2&\dots&a_n
\end{matrix}\right), \ a_i \in \mathbb{R}^n,$$

is not unique, if vectors $$ a_1, \ a_2, \dots, \ a_n $$
are linearly dependent. Then by properties of determinant, $$ \det A=0. $$
However, does it always follow, that if det A = 0, column vectors of A are linearly dependent? Can someone present a proof?

答: If the first column is all $0$'s, clear. Otherwise, consider a row with first element $\ne 0$. Permute it so it becomes the first row. The determinant is still $0$, the  system is equivalent to the previous. Now reduce all elements in first column, lower than the first row. Determinant still $0$, system still equivalent. Now, look at the matrix formed by removing first row and column. Determinant is $0$. Apply induction, find a non-zero solution $(x_2, \ldots, x_n)$. Now use the original first equation to get $x_1$. Now we have a non-zero solution to the whole system.


问:
Optimal planning for mail box servicing and processing (for a message queue online consumer scheduling)
Consider I have $n$ mail boxes $\{m_1, m_2, \dots, m_n \}$. Messages come to each box at different rate/sec e.g., mail box $n$ has an arrival rate of messages equal say 5 per second. In general, let the set $a = \{a_1, a_2, a_3, \dots, a_n\}$ denotes the arrival rate of messages into each of the mail boxes respectively.

Now suppose that I have mail workers that can pick up and process messages at uniform rate of say $x$ message/sec (suppose  for example $x=10 $ message/sec)?

How many (minimum) mail workers do I need, so that no message waiting time in the system is more than say 2 seconds (generally $s$ seconds)?

Constraint: every mail box should be assigned to only one mail worker, but a mail worker can be responsible of many mail boxes?

I believe the problem looks similar to bin packing (online)? if that is true what is the best known approximation algorithm for the online and offline  bin packing cases?

Thank you.

答:
This is indeed the offline bin packing problem. The "bins" here are the mail workers with size $x$, and the items are the mail boxes, with size $a_i$. The observation here is that if the total message rate of the mail boxes assigned to a given worker exceeded $x$, then the letters would start to pile up and sooner or later the waiting time would exceed $s$ seconds for some letter. In this way the condition that letters can wait some amount of time is immaterial. (Unless there is a point $t$ in time when letters stop arriving. But this does not change the problem in a substantial way: we only have to increase the bin sizes from $x$ to $x + s/t$.)

The wikipedia page has some nice tables to summarize various algorithms. It's difficult to say what the best algorithm is, because it depends on what your criterium is – "worst-case" or "asymptotic" approximation ratio? Additive gap? The "first-fit decreasing" algorithm is really simple and performs well.


问:
Volume of surface using triple integrals
Using triple integrals I need to find the volume of surface bounded with $$(x^2+y^2+z^2)^2=a^3x$$.
First of all I'm not sure what surface is this. I even used wolfram and geogebra but they did not sketched it as well. From equation it is obvious that if $a>0$ that $x$ can't be negative so four octants are of.
Also what i did is converting to spherical coordinates and got that $r=a\sqrt[3]{\cos\phi\sin\theta}$. Then, since $\phi$ is the angle in $xOy$ is is easy to see that $-\frac{\pi}{2}\le\phi\le\frac{\pi}{2}$. But for $\theta$ not sure, looking at solution $\frac{a^3\pi}{3}$ it should have same bounds as $\phi$ but since i don't know to graph this surface I am not sure how to get these bounds.

答:
Your surface is actually this: in the range of $x$ such that $x^4\le a^3x$ (whatever that range is), you have $\sqrt{y^2+z^2}=\sqrt{\sqrt{a^3x}-x^2}$. Thus, the surface can be obtained by rotating the curve $y=\sqrt{\sqrt{a^3x}-x^2}$ around $x$-axis.

Now this implies that you can use cylindrical coordinates ($x=x, y=\rho\cos\theta, z=\rho\sin\theta$), where $x$ is in the above range, $\rho$ is in the range from $0$ to $\sqrt{\sqrt{a^3x}-x^2}$ and $\theta\in[0,2\pi]$.

The range for $x$ depends on $a$. If $a\ge 0$, it is $[0,a]$. If $a\le 0$, it is $[a,0]$; this is easily checked and so I will leave the details to you.


问:
How can I prove mathematically that $(u+v+w)/3$ represents the centroid of the triangle formed by these vectors?
Given $u ,v, w$ vector in different directions.

I can do it using a cad software.

答: Let $C=\frac{u+v+w}3$. The position vector of the mid-point of side $vw$ is $m=\frac{v+w}2$. It is easy to see that $u,m,C$ are colinear: $C-u\|m-u$. Thus $C$ lies on the median from $u$ to side $vw$. In a similar fashion you can show $C$ lies on the other medians.


问:
Elementary example for the indeterminate form $1^\infty$
I am talking math with a bright middle schooler, who has not even seen logarithms in class yet. (We have successfully introduced logs as the inverse of exponentiation.) She is intrigued by this video and indeterminate forms. We have discussed how "$1^\infty$" is really the same as "$0/0$".

I would now like to discuss an example of "$1^\infty$". Unfortunately, every example I can come up with, and everything I find on the internet, uses that $\frac{\ln(1+t)}{t}\to 1$ as $t\to 0$, which is either "a well known fact", or an application of L'Hospital's rule - both of which I find unsatisfactory.

Is there a non-trivial example for the "$1^\infty$" indeterminate form (so, not just $1^t$ for $t\to\infty$) that can be analyzed using only the definition of the logarithm as the inverse function to exponentiation, without calculus or facts I would need to pull out of a hat?

答:
We seek $f,\,g$ with $f\to1,\,g\to\infty$, say as $x\to0$, so that $f^g$ can have just any limit $L\in[0,\,\infty]$ or none. Examples:

$f=e^{x^2},\,g=x^{-2}\ln L$ for $L>1$

$f=e^{-x^2},\,g=-x^{-2}\ln L$ for $L\in(0,\,1)$

$f=e^{x^4},\,g=x^{-2}$ for $L=1$

$f=e^{-x^2},\,g=x^{-4}$ for $L=0$

$f=e^{-x^4},\,g=x^{-2}$ for $L=\infty$

$f=e^{x^2\sin(1/x)},\,g=x^{-2}$ for $\lim_{x\to0}f^g$ to be undefined.

The replacement $(f,\,g)\mapsto(1/f,\,-g)$ shows $1^{-\infty}$ works the same way, but nobody every lists that separately.


答:
Why not just fix $k>0$ (e.g. $k=2$) and look at $(k^{1/n})^n$?

It is pretty clear intuitively that $k^{1/n}=\sqrt[n]{k}\to 1$ as $n\to\infty$; on the other hand, clearly $n\to\infty$ when $n\to\infty$. Thus, you have the case $1^\infty$ which actually converges to $k$ (and not just converges to $k$ but is constant), which you chose arbitrarily to start with.

Now this is easy to expand with $(k^{1/n})^{n^2}=k^n$ or $(k^{1/{n^2}})^n=k^{1/n}$, which converge to $0$ and $\infty$ (in some order, as long as $k\ne 1$).

Title: Elementary example for the indeterminate form $1^\infty$

Body:
I am talking math with a bright middle schooler, who has not even seen logarithms in class yet. (We have successfully introduced logs as the inverse of exponentiation.) She is intrigued by this video and indeterminate forms. We have discussed how "$1^\infty$" is really the same as "$0/0$".

I would now like to discuss an example of "$1^\infty$". Unfortunately, every example I can come up with, and everything I find on the internet, uses that $\frac{\ln(1+t)}{t}\to 1$ as $t\to 0$, which is either "a well known fact", or an application of L'Hospital's rule - both of which I find unsatisfactory.

Is there a non-trivial example for the "$1^\infty$" indeterminate form (so, not just $1^t$ for $t\to\infty$) that can be analyzed using only the definition of the logarithm as the inverse function to exponentiation, without calculus or facts I would need to pull out of a hat?


问:
Using the pigeonhole principle to show that there are sequences with the same sum
We have $a_1,a_2,...,a_n$, and $b_1,b_2,...,b_m$, all positive integers, with $a_i < m+1$ for all i, and $ b_j < n+1$ for all j. It is known that $m>n$, and that the sum of $b_1,..,b_m$ is strictly larger than the sum of $a_1, a_2,...,a_n$. Show that there is a subset of $a_1,..,a_n$ whose sum is equal to the sum of a subset of $b_1,...,b_m$.

I know this should be solvable using the pigeonhole principle on several sequences, but I just can't seem to find the sequence that works. I tried using sequences that excluded one of the values, but I think that since there are so many possible sequences, this just won't work, and using all possible sums seems pretty hard to do, since there can be multiple occurrences of the same number.

I would greatly appreciate any hints, thanks!

答:
Arrange the numbers so that $a_1 \le a_2 \le  \dotsb \le a_n$ and $ b_1 \le b_2 \le \dotsb \le b_m$, now consider the sums $P_k = a_1+ a_2+ ... + a_k$, $Q_l= b_1+b_2+ \dotsb b_l$, then call $t_k$ the number such that $Q_{t_k} \le P_k < Q_{t_k +1}$. Then, since $b_{t_k+1} \le n$, $P_k -Q_{t_k} < n$, and we have $n$ such differences, so by Pigeonhole theorem we have two equal differences.

WLOG, assume that $P_k -Q_{t_k} = P_m -Q_{t_m}$, then we have $P_k-P_m=Q_{t_k} - Q_{t_m}$ which means $ a_{m+1}+a_{m+2}+ \dotsb + a_k = b_{t_k +1} +b_{t_k +2} +\dotsb + b_{t_m}$.

Q.E.D $\square$


问:
Finding the Degree of a Polynomial, as a function of the Polynomial
Recently, I was wondering about the following question:

Given a polynomial $P(x)$ with real coefficients, express its degree $d$ as a function of only $P(x)$, i.e. $d(P(x))$. Only elementary functions/operations and values of $P$ (or its derivatives/anti-derivatives) at any value may be used.

There must be several creative ways to do this. After some experimentation, I came up with the following:

$$d(P(x)) = \sum_{i=1}^{\infty} \text{sgn}\left(\sum_{j=0}^{\infty} \left| \frac{d^{i+j} P}{dx^{i+j}} (0) \right| \right)$$

Are there simpler functions that work? Any inputs are welcome.

答:
We have
$$ \deg P=\lim_{n\to\infty}\frac{\ln (P(n)^2+n^{-n})}{2\ln n}.$$
The strange addition in the numerator is to ensure that all terms are defined and finite (for $n>1$) and also to correctly lead to $\deg 0=-\infty$.

This may be considered "simpler" than your expression because it does not use any derivatives and uses only a single limit instead of nested series (though your series turn out to be sums after all). Since it is clear that infinitely many evaluations are needed, it seems that this is about as elementary as you can get.


问:
Under what conditions is a product of dominant morphisms dominant?
Let $f:X\to S$ and $g:Y\to T$ be dominant morphisms. Then under what conditions does it hold that $f\times g$ is dominant? Is it enough to suppose that $X,Y$ are irreducible? I need it only in fairly restrictive conditions (e.g. smooth projective over an algebraically closed field), but still I wonder in what generality it holds, and can't seem to find any references for it.

答:
you need some flatness condition. just look at the affine case: for integral affine schemes dominance is equivalent to the condition that $A\to B$ is injective. now if $A\to B$ is injective and $C\to D$ is injective there is no reason for $A\otimes C\to B\otimes D$ to be injective.

but if you are working over an algebraically closed field then all is fine: $f:X\to Y$ is dominant between integral schemes $X,Y$ if and only if there are some open affine subschemes of $X,Y$ like $U=Spec\, A,V=Spec\, B,f(U)\subset V$ such that the induced map between $A\to B$ is injective. by this and the fact that everything is flat over a field(and product of two integral scheme over an algebraically closed field remain Integral) you can deduce the dominance of product.

the problem is topological so you can deduce the same for irreducible schemes over a field form the integral case.


问:
Infinite Series Problem Using Residues
Show that $$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth\pi a+\frac{1}{2a^2}, a>0$$

I know I must use summation theorem and I calculated the residue which is:

$$Res\left(\frac{1}{z^2+a^2}, \pm ai\right)=-\frac{\pi}{2a}\coth\pi a$$

Now my question is: how do I get the last term $+\frac{1}{2a^2}$ after using the summation theorem?

答:
The method of residues applies to sums of the form

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{res}_{z=z_k} \pi \cot{\pi z}\, f(z)$$

where $z_k$ are poles of $f$ that are not integers.  So when $f$ is even in $n$, you may express as follows:

$$2 \sum_{n=1}^{\infty} f(n) + f(0)$$

For this case, $f(z)=1/(z^2+a^2)$ and the poles $z_{\pm}=\pm i a$ and using the fact that $\sin{i a} = i \sinh{a}$, we get

$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2+a^2} = \frac{\pi}{a} \text{coth}{\pi a}$$

The rest is just a little more algebra.

ContentLicense: CC BY-SA 3.0

Title: Infinite Series Problem Using Residues


答:
It would be much easier to use spherical coordinates centered on the $x$ axis rather than the $z$ axis, giving

$r^3 = a^3\cos\theta$

Then from here set up the bounds with the angular integral first

$$\int_0^{2\pi}\int_0^a\int_0^{\cos^{-1}\left(\frac{r^3}{a^3}\right)}r^2\sin\theta\:d\theta dr d\phi = 2\pi\int_0^ar^2-\frac{r^5}{a^3}\:dr=\frac{a^3\pi}{3}$$

Title: Volume of surface using triple integrals

Body:
Using triple integrals I need to find the volume of surface bounded with $$(x^2+y^2+z^2)^2=a^3x$$.
First of all I'm not sure what surface is this. I even used wolfram and geogebra but they did not sketched it as well. From equation it is obvious that if $a>0$ that $x$ can't be negative so four octants are of.
Also what i did is converting to spherical coordinates and got that $r=a\sqrt[3]{\cos\phi\sin\theta}$. Then, since $\phi$ is the angle in $xOy$ is is easy to see that $-\frac{\pi}{2}\le\phi\le\frac{\pi}{2}$. But for $\theta$ not sure, looking at solution $\frac{a^3\pi}{3}$ it should have same bounds as $\phi$ but since i don't know to graph this surface I am not sure how to get these bounds.


问:
Why convex? In Theorem 10.38 Rudin
In Theorem 10.38 Principles of Mathematical Analysis by W Rudin it states without an explanation that $E \subset \mathbb{R^n} $ the open domain of $f$ to be convex is necessary and sufficient for the following claim to hold :

If $1\le p\le n$ and $D_jf(x) =0$ for all $p <j \le n$ then $f$ does not depend on $x_{p+1}, \dots x_n$.

The book and comments in here gives examples of why $D_jf(x) =0$ is not enough to make $f$ independent of $x_j$ but how imposing E to be $convex$ is enough to make $f$ independent of $x_j$? If the answer is "
that allows you to have a set of directions to take partial derivatives along lines" how does this hint applies to my question, rigorously?

答:
Suppose $D_1f = 0$. If $(a, x_2, ..., x_n)$ and $(b, x_2, ..., x_n)\in E$, then the segment from $(a, x_2, ..., x_n)$ to $(b, x_2, ..., x_n)$ is contained in $E$. Now, by the mean-value theorem,
$$f(b, x_2, ..., x_n) - f(a, x_2, ..., x_n) = D_1f(\theta, x_2, ..., x_n)(b - a) = 0,$$
hence $f$ does not depend on $x_1$. (Here, after fixing $x_2, ..., x_n$ we thought of $f$ as a single-variable function, and we needed $E$ to be convex, so the function would be defined on the segment, so we could use the mean-value theorem).

Title: Why convex? In Theorem 10.38 Rudin


问:
Inverse of a product of $n$-tuple in abstract algebra
I am trying to prove $x = (a_1 \cdot ... \cdot a_n)^{- 1} = a_n^{- 1} \cdot ... \cdot a_2^{- 1} \cdot a_1^{- 1}$ (1)

If we have,
$a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_{n - 1} \cdot a_n$
is this the same as
$(a_1 \cdot a_2 \cdot ... \cdot a_{n - 1}) \cdot a_n$ and why?

(If so using $(ab)^{- 1} = b^{- 1}a^{- 1}$ I can easly prove (1).) Thanks.

答:
This is trivial when $n=1$. Let $n>1$ and suppose it is true for $n-1$.

Then $(a_1\cdots a_n)^{-1}=((a_1\cdots a_{n-1})a_n)^{-1}=a_n^{-1}(a_1\cdots a_{n-1})^{-1}$. Now, by the inductive hypothesis we have $(a_1\cdots a_{n-1})^{-1}=a_{n-1}^{-1}\cdots a_1^{-1}$, so we are done.


问:
The boundary of a manifold has empty boundary
It is a known fact, for example well explained here, that the boundary of an $n$ manifold is an $n-1$ manifold, and the proof shows essentialy that properly restricted charts which parametrize $M$ parametrize also $\partial M$, with open set of a lower dimension, $n-1$.

My question is : I have to show that $\partial M$ has no boundary, or is implicit somehow having shown that $\partial M$ is parametrizable with open sets of $\mathbb{R}^{n-1} ?$ If so, why ?

Because for my definition of boundary I don't see how this should be implicit.

My definition is $p \in \partial M \iff p \in \phi(U \cap \left\lbrace x_n = 0\right\rbrace)$ i.e is the image of $\partial \mathbb{H}^n \cap U$, (where $U$ is the open set of $\mathbb{R}^n$ such that the intersection with closed half space $U \cap \mathbb{H}^n$ is a diffeomorphism with an open set of the manifold $V$).

How should I prove it ? It could be useful using this fact ?

I'd like to understand the answer given in this previous question, which seems the way I should prove, it but I don't understand how the fact follows easily from the definitions.

Any help would be appreciated.

答: By definition, you $\partial M$ is endowed with charts that "look like" an open subset $U\cap \Bbb R^{n-1}$.


问:
Question about repeating decimal?
For simple fraction, we can easily convert it to repeating decimal by calculator. Ex. $\frac 1 3 = 0.33333\ldots$, $\frac 1 7=0.(142857),\ldots$ But some fraction fraction like $10/29, 1/97,...$ The repeating part of them are too long, so it can't fully show on the calculator. So is there any algorithm  to find the repeating part for that fraction?</p>

答:
For, say,  $\large \frac{1}{97}$, start by running the Python program

```<code>R = []

numer = 1
denom = 97

for i in range(0,denom):
    r = numer * 10**i % denom
    try:
        ndx = R.index(r)
    except:
        ndx = -1
    if ndx >= 0:
        print(i,ndx)
        break
    R.append(r)
</code>```

The output of the program is

```<code>96 0
</code>```

The period length is

```<code>i - ndx
</code>```

and is therefore equal to $96 - 0 = 96$.

The offset (to the right of the decimal) is $0$ so the period begins immediately after the decimal.

So we need to put a bar over the first $96$ digits.

To get those decimal digits you can conveniently use wolfram, and compute

```<code>integerPart((1 + 1/97) * 10^96)
</code>```

The output is

```<code>1010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567
</code>```

and you have to drop the leading $1$ (used to 'light up' any zeroes right after the decimal point) to get the final answer (with period broken down into blocks of $25$),

$\quad \large \frac{1}{97} =$

$0.\overline{0103092783505154639175257}$

$\; \; \, \overline{7319587628865979381443298}$

$\; \; \, \overline{9690721649484536082474226}$

$\; \; \, \overline{804123711340206185567}$

Title: Question about repeating decimal?

Body:
For, say,  $\large \frac{1}{97}$, start by running the Python program

R = []

numer = 1
denom = 97

for i in range(0,denom):
    r = numer * 10**i % denom
    try:
        ndx = R.index(r)
    except:
        ndx = -1
    if ndx >= 0:
        print(i,ndx)
        break
    R.append(r)

The output of the program is

96 0

The period length is

i - ndx

and is therefore equal to $96 - 0 = 96$.

The offset (to the right of the decimal) is $0$ so the period begins immediately after the decimal.

So we need to put a bar over the first $96$ digits.

To get those decimal digits you can conveniently use wolfram, and compute

integerPart((1 + 1/97) * 10^96)

The output is

1010309278350515463917525773195876288659793814432989690721649484536082474226804123711340206185567

and you have to drop the leading $1$ (used to 'light up' any zeroes right after the decimal point) to get the final answer (with period broken down into blocks of $25$),

$\quad \large \frac{1}{97} =$

$0.\overline{0103092783505154639175257}$

$\; \; \, \overline{7319587628865979381443298}$

$\; \; \, \overline{9690721649484536082474226}$

$\; \; \, \overline{804123711340206185567}$


问:
Show that the following localization ring has one maximal ideal
I was given the following question:

$R=\mathbb{Z}$ and $S$ contains all elements not divisible by $p$ (which is a given prime)

Prove that $S^{-1}R$ has only one maximal ideal

I tried assuming it has two and looking at one element in each that does not exist in the other but I didn't manage to come to any meaningful result so I'm pretty much stuck now, any thoughts on what to try next?

This subject is pretty new to me and I think I lack some intuition as to what the ring $S^{-1}R$ even is so any kind of explanation would be very much appreciated

Thank you

答:
$S^{-1}R$ is the localization of $R$ at the prime ideal $p\mathbb{Z}$. Since $\mathbb{Z}$ is an integral domain, we can describe $S^{-1}R$ as a subring of its field of fractions, i.e. $\mathbb{Q}$. In this case $S^{-1}R$ consists of all those rational numbers which can be written as $\frac{a}{s}$ where $a\in R,s\in S$ that is $a$ and $s$ are integers with $s$ not divisible by $p$.

Now for the actual question regarding the maximal ideals:

Show that the units in $S^{-1}R$ are precisely those elements in $S^{-1}R\setminus pS^{-1}R$. This then shows that $pS^{-1}R$ is the unique maximal ideal of $S^{-1}R$ (why?).


答:
We want to show that $(a_1\dots a_n)^{-1}=a_n^{-1}\dots a_1^{-1}$.

We'll show this property using the induction on the number of elements of the product of the elements $a_i, i\in I$.

(i) For $n=1$ we have nothing to prove.

(ii) Let's prove that $\mathcal P(n-1)\implies \mathcal P(n)$:

$(a_1\dots a_n)^{-1}=((a_1\dots a_{n-1})a_n)^{-1}\underset{Hp}{=}a_n^{-1}(a_1\dots a_{n-1})^{-1}=a_n^{-1}a_{n-1}^{-1} \dots a_1^{-1}$.

Title: Inverse of a product of $n$-tuple in abstract algebra

Body:
I am trying to prove $x = (a_1 \cdot ... \cdot a_n)^{- 1} = a_n^{- 1} \cdot ... \cdot a_2^{- 1} \cdot a_1^{- 1}$ (1)

If we have,
$a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_{n - 1} \cdot a_n$
is this the same as
$(a_1 \cdot a_2 \cdot ... \cdot a_{n - 1}) \cdot a_n$ and why?

(If so using $(ab)^{- 1} = b^{- 1}a^{- 1}$ I can easly prove (1).) Thanks.


答:
$S^{-1}R:=\{a/s:a\in R,p\not\mid s\}$ is a local ring with a unique maximal ideal $(p)$, since one may check that $S^{-1}R\setminus(p)$ is the group of units of $S^{-1}R$.

Suppose there is a prime ideal $0\ne\mathfrak p\subseteq (p)$. Let $a/s\in\mathfrak p$ be nonzero. Then $a/1\in\mathfrak p$. Let $a=up^k$ for $p\not\mid u$. Since $u/1\in S^{-1}A$ is a unit, we must have $p^k\in\mathfrak p$, so that since $\mathfrak p$ is prime, $p\in\mathfrak p$. Thus, in fact $\mathfrak p=(p)$.

Title: Show that the following localization ring has one maximal ideal

Body:
I was given the following question:

$R=\mathbb{Z}$ and $S$ contains all elements not divisible by $p$ (which is a given prime)

Prove that $S^{-1}R$ has only one maximal ideal

I tried assuming it has two and looking at one element in each that does not exist in the other but I didn't manage to come to any meaningful result so I'm pretty much stuck now, any thoughts on what to try next?

This subject is pretty new to me and I think I lack some intuition as to what the ring $S^{-1}R$ even is so any kind of explanation would be very much appreciated

Thank you


问:
Generators of free groups are really generators
The free group $F_X$ generated by the set $X$ is characterized by a map $i:X\longrightarrow F_X$ which has the universal property that for any group $G$ and any map $f:X\longrightarrow G$, there is exactly one homomorphism $\tilde f:F_X\longrightarrow G$ such that $f=\tilde f\circ g$. I want to show that $X$ can indeed be considered the generating set of $F_X$ by showing that $i$ is automatically injective and that the image of $i$ is a generating set of $F_X$.

Injectivity was easy: just let $f$ be injective, then $i$ must also be injective for $f=\tilde f\circ i$ to be injective.

But I struggle with the image of $i$ (let's call it $E$) being a generating set. My intuition is that if the subgroup generated by $E$ isn't $F_X$ itself, we can choose an element that isn't generated by $E$ and map it to some arbitrary element of $G$ so $\tilde f$ wouldn't be unique. But it can't actually be arbitrary. For instance, the homomorphisms $S_3\to A_3$ are fully determined by where the $3$-cycles are mapped, and the transpositions, which aren't generated by $3$-cycles, must be mapped to the identity without any arbitrary choice. How do I get around this?

Edit: I just noticed that $S_3\to A_3$ is a bad example since there are no non-trivial homomorphisms here. But the point still stands that I'm not sure how to show that the homomorphism $f$ isn't unique if $E$ isn't a generating set.

答:
Say that a free group on a set $X$ is a pair $(F, \iota)$ with the universal property you have stated.

Now prove that $(\langle \iota(X) \rangle, \iota')$ is also a free group on $X$ (here $\iota' : X \to \langle \iota(X) \rangle$ is given by $\iota'(x) = \iota(x)$ for $x \in X$).

Then use the universal property for these two free groups, using the other one for $G$, and $\iota$ resp. $\iota'$ for your $f$.


问:
Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $
The limit can be rewritten as

$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$

Recall the inequality,

$$ \cos x < \frac{\sin x}{x} < 1$$

holds for $ x\in (-\pi/2, \pi/2) $.

This provides the new inequalities

$$ 0 < \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] < \frac{1-\cos x}{x},  \text{ with } x>0 $$

$$ \frac{1-\cos x}{x} < \frac{1}{x} \left[1- \frac{\sin x}{x} \right]< 0,  \text{ with } x<0 $$

Applying the Squeeze Theorem, we get

$$ \lim_{x \rightarrow 0^{\pm}}  \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] = 0$$

Note:

$ \lim_{ x \rightarrow 0} \frac{1-\cos x}{x} = 0 $

答:
Here is another way to proceed it you don't wanna use L'Hopital, or series expansions.

Let $ x\in\mathbb{R}^{*} $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{2}}=\frac{x}{2}\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y}\end{array}$} $$

Using the fact that $ \left(\forall t\in\mathbb{R}\right),\ \left|\cos{t}\right|\leq 1 $, we have : \begin{aligned} \left|\frac{x-\sin{x}}{x^{2}}\right|=\frac{\left|x\right|}{2}\left|\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y}\right|&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-y\right)^{2}\left|\cos{\left(xy\right)}\right|\mathrm{d}y}\\&\leq\frac{\left|x\right|}{2}\int_{0}^{1}{\left(1-y\right)^{2}\,\mathrm{d}y}=\frac{\left|x\right|}{6}\underset{x\to 0}{\longrightarrow}0 \end{aligned}

Thus : $$ \lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 $$


问:
Use limit theorems to show that the following function is continuous on $[0,1]$

Use limit theorems to show that the following function is continuous on $[0,1]$.
$$f(x)=\begin{cases}
e^{-\frac{1}{x}}   & :x \neq0 \\
           0   & :  \text{$x= 0$}
\end{cases}$$

Here is the answer which is given to me

$*$ We know that $\space\dfrac{1}{x}$ is continuous for $x \neq 0$

$*$ Hence $\space-\dfrac{1}{x}$ is continuous on $(0,1]$ and $\space e^{-\frac{1}{x}}$ continuous on $(0,1]$

$*$ For $x=0$ we must show that,
$$\lim_{x \to 0}f(x) = f(0)=0$$

$*$ Since $f(x)=e^{-\frac{1}{x}}$ for $x\neq 0$

$$\lim_{x \to 0}f(x) = \lim_{x \to 0}e^{-\frac{1}{x}}$$

$*$ Consider,
$$\lim_{x \to 0^{+}}f(x) = \lim_{x \to 0^{+}}e^{-\frac{1}{x}}=0$$

$\therefore$ The limit of the function as $x \rightarrow 0^+$ is equal to the function value at $x \neq 0$

That is
$$\lim_{x \to 0}f(x) = f(0)=0$$

$\therefore$ $f$ is continuous at $0$

I feel some point are wrong here because $\lim_{x \to 0^{-}}f(x)=\infty$ so we must show for $x=0$
$$\lim_{x \to 0^{+}}f(x) = f(0)=0$$ not $$\lim_{x \to 0}f(x) = f(0)=0$$ this

Is there anything wrong what I am saying please tell me?

答: You are wrong when you claim that $\lim_{x\to0^-}f(x)=\infty$. The domain of $f$ is $[0,1]$, and therefore it doesn't make sense to mention $\lim_{x\to0^-}f(x)$.


问:
Determining conditions on two circles inscribed in square so that a tangent rectangle has integer area
In the accompanying figure, all seven contact points are tangential. It can be verified that if the areas of the major and minor circles, noted $A$ and $a$, are respectively equal to $8\pi$ and $2\pi$ then the area of the blue rectangle is equal to $7$.

Can you find without much work necessary and sufficient conditions between the areas $A$ and $a$ so that the area of the blue rectangle is an integer? Or would it be too difficult?

答:
If the radii of the circles are $r$ and $R$, then the side of the big square is $s=(r+R)(1+\frac{\sqrt{2}}{2})$ and so the sides of the rectangle are $s-2r=R(1+\frac{\sqrt{2}}{2})+r(-1+\frac{\sqrt{2}}{2})$ and $s-2R=R(-1+\frac{\sqrt{2}}{2})+r(1+\frac{\sqrt{2}}{2})$, so (after the calculation) the area of the rectangle is:

$$-\frac{1}{2}R^2-\frac{1}{2}r^2+3Rr=\frac{-A-a+6\sqrt{Aa}}{2\pi}$$

Now, I presume you want $A=u\pi, a=v\pi$, so this area then becomes:

$$-\frac{u+v}{2}+3\sqrt{uv}$$

which gives you necessary and sufficient conditions: (a) $u,v$ are of the same parity; (b) $uv$ is a square.


问:
Logic problem - what kind of logic is it?
I would be most gratefull, if someone could verify my solution to this problem.

```<code> 1) Of Aaron, Brian and Colin, only one man is smart.

 Aaron says truthfully:
    1. If I am not smart, I will not pass Physics.
    2. If I am smart, I will pass Chemistry.

 Brian says truthfully:
    3. If I am not smart, I will not pass Chemistry.
    4. If I am smart, I will pass Physics.

 Colin says truthfully:
    5. If I am not smart, I will not pass Physics.
    6. If I am smart, I will pass Physics.

 While
   I. The smart man is the only man to pass one particular subject.
   II. The smart man is also the only man to fail the other particular subject.

 Which one of the three men is smart? Why?
</code>```

I would say that it could have been any one of them, as the implications in every statement are not strong enough to disprove the statements I,II. But I'm not sure if my solution is enough, as I'm not sure what kind of logic it is.

答:
Here is an example:

Suppose Aaron is the smart one. This means Brian and Colin are not smart. Thus Aaron passes Chemistry (by 2) and fails Physics (by II), but Colin also failed Physics (by 5), contradicting II. Thus Aaron is not the smart one.

So, see if you can finish the problem using this kind of reasoning.


问:
If $0 \to A \to B \to C \to 0$ is split, we say that $B \cong A \oplus C$. Is there a reason we take the direct sum rather than the product?
I'm studying some commutative algebra and a solution to an exercise(concerning commutative $R$-modules) relies on the isomorphisms $$\text{Hom}(C, A \oplus C) \cong \text{Hom}(C, A \times C) \cong \text{Hom}(C, A) \times \text{Hom}(C, C)$$
where the first isomorphism takes the canonical $A \oplus C \cong A \times C$ and the second uses the fact that Hom preserves products in the second variable. I figured I could do this because finite direct sums and finite products are naturally isomorphic, right? This kind of operation with Hom seemed too good to be true. Does this actually work?

The reason I ask is that this solution would be greatly simplified if I took the sequence being split to mean $B \cong A \times C$ instead. That way, I wouldn't have to worry about the first isomorphism.

Thank you!

答:
I guess you know that the categorical concepts of product and sum (or coproduct) are defined by universal properties which are dual to each other.

Consider a three term sequence $A \stackrel{f}{\to} B \stackrel{g}{\to} C$ in a category $\mathcal C$.

Given a morphism $r : B \to A$, we get a unique morphism $\phi : B \to A \times C$ such that $\pi_A \circ \phi =r$ and $\pi_C \circ \phi = g$. Here $\pi_A : A \times C \to A$ and $\pi_C : A \times C \to C$ are the "projection" morphisms belonging to the categorical product of $A$ and $C$. If $\mathcal C$ is the category of $R$-modules, the sequence $0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$ is exact and $r \circ f = id_A$, then one can show that $\phi$ is an isomorphism.

Given a morphism $i : C \to B$, we get a unique morphism $\psi :  A \oplus C \to C$ such that $\psi \circ \iota_A  = f$ and $\psi \circ \iota_C  = i$. Here $\iota_A : A \to A \oplus C$ and $\iota_C : C \to A \oplus C$ are the "embedding" morphisms belonging to the categorical sum of $A$ and $C$. If $\mathcal C$ is the category of $R$-modules, the sequence $0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$ is exact and $g \circ i = d_C$, then one can show that $\psi$ is an isomorphism.

As Qi Zhu mentions in his comment, in additive categories finite products are canonically isomorphic to finite sums. Thus you may interpret $A \oplus C$ and $A \times C$ as the same object. In fact, both can be taken as the set of all pairs $(a,c)$ with $a \in A, c \in C$ with the usual algebraic operations making it an $R$-module. But be aware that the categorical status is different which does not become transparent if we work on the "naive" level of sets.


问:
What is the probability that both of the scores is a $3$?

Question

Someone rolls two identical fair 6-sided dice but does not know the result. Someone was told that one of the numbers on the die was a $3$. Given this information, what is the probability that both of the numbers on Someone's die was a $3$?

Answer

$$ \frac{1}{11} $$

Given that the first number was a $3$, isn't the probability of the $2$nd die being $3$ equal to $\frac{1}{6}$? How is $\frac{1}{11}$ obtained?

Body:
Given that the first number was a $3$, isn't the probability of the $2$nd die being $3$ equal to $\frac{1}{6}$?

No.

Using a Probability Diagram, draw all of the possible results:

$1$
$2$
$3$
$4$
$5$
$6$

$1$
1,1
2,1
3,1
4,1
5,1
6,1

$2$
1,2
2,2
3,2
4,2
5,2
6,2

$3$
1,3
2,3
3,3
4,3
5,3
6,3

$4$
1,4
2,4
3,4
4,4
5,4
6,4

$5$
1,5
2,5
3,5
4,5
5,5
6,5

$6$
1,6
2,6
3,6
4,6
5,6
6,6

Now highlight all of the results with at least one $3$:

$1$
$2$
$3$
$4$
$5$
$6$

$1$
1,1
2,1
3,1
4,1
5,1
6,1

$2$
1,2
2,2
3,2
4,2
5,2
6,2

$3$
1,3
2,3
3,3
4,3
5,3
6,3

$4$
1,4
2,4
3,4
4,4
5,4
6,4

$5$
1,5
2,5
3,5
4,5
5,5
6,5

$6$
1,6
2,6
3,6
4,6
5,6
6,6

If you count them, there are $11$ possible results with a $3$, and there is only one result which has two $3$s (3,3).

\begin{align}
\therefore P(\text{both of the numbers on the die was a 3}) &= \frac{n(\text{number of results with two 3s})}{n(\text{number of results with a 3})} \\
&= \frac{1}{11}
\end{align}


问:
How many rectangles can be found in this shape?

Question: How many rectangles can be found in the following shape?

My first solution:

Let $a_n$ be the number of rectangles in such a shape with side length $n$. Then, the number of the newly added $1 \times k$ rectangles while expanding the shape with side length $n-1$ to the shape with side length $n$ is, $n^2$. Similarly, the number of the newly added $2 \times k$ rectangles while expanding is, $(n-2)^2$, the number of the $3 \times k$ ones is $(n-4)^2$, and so on. So, we have, $$a_n = a_{n-1} + 1^2+3^2+\ldots +n^2$$ or, $$a_n = a_{n-1} + 2^2+4^2+\ldots +n^2$$ depending on the parity of $n$. In either case, we have $$a_n=a_{n-2}+1^2+2^2+ \ldots n^2$$ So, $a_{n+1} = a_{n-1} + 1^2+2^2+ \ldots (n+1)^2$. Subtracting both sides, we have, $a_{n+1}-a_n = a_{n-1} - a_{n-2} + (n+1)^2$. Repeating the same method, we obtain $$a_{n+1}-4a_n+5a_{n-1} = 5a_{n-3} - 4a_{n-4}+a_{n-5}$$ This recurrence relation has characteristic equation $r^6-4r^5+5r^4-5r^2+4r-1=0 \iff (r-1)^5 \cdot (r+1) = 0$. So, using the usual ways to solve recurrence relations, we can obtain $a_n=\frac{n^4}{24} + \frac{n^3}{4} + \frac{11n^2}{24} + \frac{n}{4} = \binom{n+3}{4}$.

However, since the answer is so simple, I tried to solve it with a method which can give the result in a more 'direct' way:

Let us find these rectangles by choosing 2 rows and 2 columns and looking at their intersection. If we choose the right most column, we have $n$ choices for the other column and only $1$ choice for the rows. If we choose column which is the second from right, we have $n-1$ choices for the other column (in order not to over count), and, $\binom{2+2-1}{2}$ (number of ways to choose 2 rows out of 2 rows including repetitions) choices for the row. Using this argument, we can obtain, $$a_n = \sum_{k=0}^{n-1} (n-k) \cdot \binom{(k+1)+2-1}{2} = \sum_{k=0}^{n-1} (n-k) \cdot \binom{k+2}{2}=\sum_{k=0}^{n-1} \frac{(n-k) \cdot (k+2) \cdot (k+1)}{2}$$ After some computation, we can see from here that $a_n=\binom{n+3}{4}$.

However, I think this solution is also overcomplicated and I think that there is a really elegant and/or 'direct' solution for this problem. So, my question is, how can this problem be solved with a 'direct' method (for example, a combinatorial method which immediately gives the number as $\binom{n+3}{4}$)? It would be nice to see your 'nice' solutions too, thank you.

答:
Label the lines starting from $1$ on the left to $n+1$ on right/above. Extend the lines by $1$ unit to create tetherings. There are $n+3$ of them, label these from $1$ to $n+3$. The diagram is for $n=5$.

There is a bijection between the coordinates of four vertices of any rectangle and the coordinates at end of tetherings which determine the four edges of the given rectangle.

For example, the rectangle with vertices at $(1,1),(5,1),(5,3),(1,3)$ corresponds to the unique quadruplet $(1,3,4,8)$.


答: Searching ${n+3 \choose 2}$ at OEIS you can find in A000332 the reference to A004320 and there the document Counting the lattice rectangles inside Aztec diamonds and square biscuits by Teofil Bogdan and Mircea Dan Rus and the solution to problem 3, on page 3. Basically, for each rectangle you have to choose four coordinates for the vertices $(x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2,y_2)$ with $x_1 \lt x_2 \lt y_1 \lt y_2$ among $n+3$ values.

Title: How many rectangles can be found in this shape?

Body:
Question: How many rectangles can be found in the following shape?

My first solution:

Let $a_n$ be the number of rectangles in such a shape with side length $n$. Then, the number of the newly added $1 \times k$ rectangles while expanding the shape with side length $n-1$ to the shape with side length $n$ is, $n^2$. Similarly, the number of the newly added $2 \times k$ rectangles while expanding is, $(n-2)^2$, the number of the $3 \times k$ ones is $(n-4)^2$, and so on. So, we have, $$a_n = a_{n-1} + 1^2+3^2+\ldots +n^2$$ or, $$a_n = a_{n-1} + 2^2+4^2+\ldots +n^2$$ depending on the parity of $n$. In either case, we have $$a_n=a_{n-2}+1^2+2^2+ \ldots n^2$$ So, $a_{n+1} = a_{n-1} + 1^2+2^2+ \ldots (n+1)^2$. Subtracting both sides, we have, $a_{n+1}-a_n = a_{n-1} - a_{n-2} + (n+1)^2$. Repeating the same method, we obtain $$a_{n+1}-4a_n+5a_{n-1} = 5a_{n-3} - 4a_{n-4}+a_{n-5}$$ This recurrence relation has characteristic equation $r^6-4r^5+5r^4-5r^2+4r-1=0 \iff (r-1)^5 \cdot (r+1) = 0$. So, using the usual ways to solve recurrence relations, we can obtain $a_n=\frac{n^4}{24} + \frac{n^3}{4} + \frac{11n^2}{24} + \frac{n}{4} = \binom{n+3}{4}$.

However, since the answer is so simple, I tried to solve it with a method which can give the result in a more 'direct' way:

Let us find these rectangles by choosing 2 rows and 2 columns and looking at their intersection. If we choose the right most column, we have $n$ choices for the other column and only $1$ choice for the rows. If we choose column which is the second from right, we have $n-1$ choices for the other column (in order not to over count), and, $\binom{2+2-1}{2}$ (number of ways to choose 2 rows out of 2 rows including repetitions) choices for the row. Using this argument, we can obtain, $$a_n = \sum_{k=0}^{n-1} (n-k) \cdot \binom{(k+1)+2-1}{2} = \sum_{k=0}^{n-1} (n-k) \cdot \binom{k+2}{2}=\sum_{k=0}^{n-1} \frac{(n-k) \cdot (k+2) \cdot (k+1)}{2}$$ After some computation, we can see from here that $a_n=\binom{n+3}{4}$.

However, I think this solution is also overcomplicated and I think that there is a really elegant and/or 'direct' solution for this problem. So, my question is, how can this problem be solved with a 'direct' method (for example, a combinatorial method which immediately gives the number as $\binom{n+3}{4}$)? It would be nice to see your 'nice' solutions too, thank you.


问:
How to write join of multiple graphs
Let $\overline{K}_{a_i}$ denote a null graph on $a_i$ vertices where $1\le i\le n$.
I want to do join of   $\overline{K}_{a_i}$.

I am not able to  understand how to write join of the graphs $\overline{K}_{a_i}$ mathematically.

Can someone please help me to write it down formally?

答:
In general, you might denote graph join with $+$ (in which case you'd write $\overline{K}_{a_1} + \overline{K}_{a_2} + \dots + \overline{K}_{a_n}$) or with $\vee$ (in which case we have an additional option: $\overline{K}_{a_1} \vee \overline{K}_{a_2} \vee \dots \vee \overline{K}_{a_n}$ can also be $\bigvee_{i=1}^n \overline{K}_{a_i}$).

In this particular case, the join of empty graphs with $a_1, a_2, \dots, a_n$ vertices happens to be the complete multipartite graph $K_{a_1, a_2, \dots, a_n}$ which has its own notation. (Provided that $n>1$.)

Title: How to write join of multiple graphs


问:
Elementary example for the indeterminate form $1^\infty$
I am talking math with a bright middle schooler, who has not even seen logarithms in class yet. (We have successfully introduced logs as the inverse of exponentiation.) She is intrigued by this video and indeterminate forms. We have discussed how "$1^\infty$" is really the same as "$0/0$".

I would now like to discuss an example of "$1^\infty$". Unfortunately, every example I can come up with, and everything I find on the internet, uses that $\frac{\ln(1+t)}{t}\to 1$ as $t\to 0$, which is either "a well known fact", or an application of L'Hospital's rule - both of which I find unsatisfactory.

Is there a non-trivial example for the "$1^\infty$" indeterminate form (so, not just $1^t$ for $t\to\infty$) that can be analyzed using only the definition of the logarithm as the inverse function to exponentiation, without calculus or facts I would need to pull out of a hat?

答:
Who can forget the classic example:

$\underset{n\to\infty}{\lim}\left(1+\dfrac{1}{n}\right)^{n}$?

If we expand $(1+\dfrac{1}{n})^{n}$ with the Binomial Theorem and compare terms with corresponding powers of $1/n$ for different values of $n$, we find that this function increases as $n$ increases without bound, but the function is bounded by the convergent series

$1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...$

So the limit is guaranteed to exis and is thus definable as $e$, from which the rule $[\ln(1+x)]/x\to1$ as $x\to 0$ follows.


问:
How many elements of order $2$ are in the alternating group $A_8$
So for this question I thought the number of elements of order $2$ in the alternating group $A_8$ would be calculated by doing $\frac{8!}{2^44!}+\frac{8!}{2^22!}=5145$ as for order $2$ it must be from $4$ $2$-cycles and 2 2-cycles however the answer I'm getting is incorrect. Can anyone advise on how to get the answer?

答:
Your second summand should be
$$
\frac{8 \cdot 7 \cdot 6 \cdot 5}{2^{2} \cdot 2!} = 210.
$$

And please do not forget to speak of four disjoint $2$-cycles and two disjoint $2$-cycles.


问:
Can someone demonstrate this integral with a cartesian product?
From this question, we can apparently get an integral:

$$\int_{a \times c}^{b \times d}\!{\left(1+e^{i(x+1/2y)}+e^{i(y)}\right)\,d(x \times y)}$$

...I'm not exactly sure that this integral is posed correctly.

I'm wondering how we can solve this integral.  Essentially, I'm wondering how we can solve an integral involving the caresian product.  I'm trying to avoid a nested integral in this instance, so I'm hoping that this will work.

Can someone show how to integrate this integral in extreme detail?

答: It doesn't make much sense to integrate in this case because there is no well-defined set upon which you are integrating. All you are stating is that the integral has to start at a certain point in $\mathbb{R}^2$ and end at a certain point in $\mathbb{R}^2$. Now with a certain class of functions, it turns out it doesn't matter which way you go from the start to end point, but in general it does matter which path is taken.</p>


问:
Double series convergence (without integrals)
Can somebody help me with double summation problem? I don't know at all how to prove that $\sum_{n=2}^{\infty} \sum_{k=2}^{\infty}\frac{1}{k^n}$ converges absolutely.
I would be really glad if somebody helped.

Body:
$$
\sum_{k=2}^\infty \sum_{n=2}^\infty\frac{1}{k^n}=\sum_{k=2}^\infty \frac{\frac{1}{k^2}}{1-\frac{1}{k}}= \sum_{k=2}^\infty\frac{1}{k(k-1)}= 1$$


问:
What's the geometric intuition of the determinant of a covariance matrix?
As explained in intuition of determinant : we know that determinant is the how the volume is scaled when the matrix is regarded as a projection.

Now, if we narrow down the definition of the matrix here: assuming that it is a covariance matrix, i.e., positive semi definitive  and symmetric, will there be any new characteristics/intuition of determinant.

As per Ben's comments:

In a 2D space, covariance matrix represents an ellipse, which captures the distribution of a vector of 2 random variables. In such a context, what does determinant say about these 2 variables?

答: If $\rho_{ij}=\operatorname{Cov}(X_i,\,X_j)$ and $Y_i:=R_{ij}X_j$ for a rotation matrix $R$ then $\operatorname{Cov}(Y_i,\,Y_j)=R_{ik}\rho_{kl}R^T_{lj}$, which for a suitable choice of $R$ is diagonal. So $\det\rho$ is the determinant of this diagonal matrix, i.e. the product of variances of the $Y_i$. This rotation places Cartesian axes along the axes of the ellipse associated with $\rho$, so the rotation-invariant determinant, being a product of eigenvalues, is proportional to the ellipse's area. (Needless to say, in higher dimensions I'd be talking about a hyperellipsoid's measure, hypervolume, volume, or whatever term you prefer.)


问:
Help understanding this question on the Gamma Distribution

If someone could please explain how I go about this question as indicated, that would be amazing!  Revising for exams and this one has stumped me. From a little bit of research, i think its the Erlang distribution in disguise.
Thanks!

答:
Observe that if

$$X_1\sim Gamma(1;\beta)= exp(\beta)$$

it is easy to prove that $E(X_1)=1/\beta$ thus

$$X\sim Gamma(\alpha;\beta)$$

can always be viewed as a sum of independent exponentials an thus its mean is

$$\frac{\alpha}{\beta}$$

Note that in this parametrization $\beta$ is the rate parameter; if you consider it as the scale parameter the result will be modified in $E(X)=\alpha\cdot\beta$


问:
Calculate coordinates of certain function using the transformation matrix of a certain basis.

Let $V$ be a vector space spanned by the functions $f,g:\mathbb{R}\to \mathbb{R}$ where $f(x) =\sin(x)$ & $g(x) =\cos(x)$.

Find $[l]_C$ using the transformation matrix, where:

$B = \{f, g\}$ the standard basis.

$C = \{2\sin(x) +\cos(x), 3\cos(x)\}$

and $l = 5\sin(x) -2\cos(x)$.

I've found that the matrix is:
$M_{B\rightarrow C} = \begin{bmatrix}2&0\\1&3\end{bmatrix}$

But when I multiply the matrix by the coordinates of $l$ by the standard basis, I don't get a correct result.

$\begin{bmatrix}2&0\\1&3\end{bmatrix}· \begin{bmatrix}5\\-2\end{bmatrix} = \begin{bmatrix}10\\-1\end{bmatrix}$

This doesn't make sense, as $10(2\sin(x) +\cos(x)) -1(3\cos(x)) \ne 5\sin(x) -2\cos(x)$

Any help would be appreciated!

答:
The simple answer to your question is that $M_{B\to C}$ is incorrect.

Note that $[f]_C=\begin{bmatrix}1/2\\-1/6\end{bmatrix}$ and $[g]_C=\begin{bmatrix}0\\1/3\end{bmatrix}$, giving$$M_{B\to C}=\begin{bmatrix}1/2&0\\-1/6&1/3\end{bmatrix}$$which gives $[l]_C=\begin{bmatrix}1/2&0\\-1/6&1/3\end{bmatrix}\begin{bmatrix}5\\-2\end{bmatrix}=\begin{bmatrix}5/2\\-3/2\end{bmatrix}$, which is correct since $\frac52(2f+g)-\frac32(3g)=5f-2g$ as desired.


问:
Lipschitz continuity of $\sqrt{A}$
Let $U \subset\mathbb{R}^n$ be an open set, $\mathbb{S}^n$ be the set of all $n\times n$ symmetric real matrices, $A:U\to \mathbb{S}^n$ be a uniformly Lipschitz continuous function.

Suppose $\exists \lambda>0$ s.t. $\forall x\in U, \xi\in\mathbb{R}^n$ $\ ^t\xi A(x)\xi\geq\lambda|\xi|^2$. Then we can define $A^{1/2}:U\to \mathbb{S}^n$, which is the nonnegative sqrt of A at each point.

Then is $A^{1/2}$ also uniformly Lipschitz continuous? If it holds, how to prove it?

答:
If $a,b$ were scalars, $\sqrt a - \sqrt b = (a-b)/(\sqrt a + \sqrt b)$ is the common strategy to prove Lipschitzness of the square root.

Here similarly, if vector $x$ with $\|x\|=1$ is an eigenvector of $\sqrt A - \sqrt B$ with eigenvalue $\mu$ then
\begin{align*}
x^T(A-B)x &=
x^T(\sqrt A - \sqrt B)\sqrt A x
+ x^T\sqrt B (\sqrt A - \sqrt B) x
\\&=
\mu x^T(\sqrt A + \sqrt B)x.
\end{align*}
As $\mu$ can be chosen as $\pm\|\sqrt A - \sqrt B\|_{op}$, this implies
$$\|\sqrt A - \sqrt B\|_{op} \le \|A-B\|_{op} \lambda_{\min}(\sqrt A + \sqrt B)^{-1}
\le
\|A - B\|_{op} /(2\sqrt \lambda)
$$
and Lipschitzness with respect to the operator norm if the eigenvalues of $A$ and $B$ belong to $[\lambda,+\infty)$.

Title: Lipschitz continuity of $\sqrt{A}$


答:
Using the decomposition of a permutation as a product of disjoint cycles, you have that these cycles have to be disjoint transpositions, and as we're in $A_8$, they have to be the product of an even number of transpositions. Explicitly, you should have

either the product of two disjoint transpositions, which are $\frac12\dbinom 82\dbinom 62$ in number,

or the product of four disjoint transpositions, which are $\frac1{24}\dbinom 82\dbinom 62\dbinom 42$ in number.

Title: How many elements of order $2$ are in the alternating group $A_8$

Body: So for this question I thought the number of elements of order $2$ in the alternating group $A_8$ would be calculated by doing $\frac{8!}{2^44!}+\frac{8!}{2^22!}=5145$ as for order $2$ it must be from $4$ $2$-cycles and 2 2-cycles however the answer I'm getting is incorrect. Can anyone advise on how to get the answer?


答:
They're using that $$1+2+3+\dots+n=\frac{n(n+1)}2$$

For example $1+2+3=6=\dfrac{3\cdot 4}{2}$

On the other hand $$\underbrace{1+1+1+\cdots+1}_{n \;\;\rm times}=n$$

Thus $$\frac{1+1+1+\cdots}{1+2+3+\cdots}=\frac{n}{\dfrac{n(n+1)}2}=\frac{2}{n+1}$$

Of course, we're interpreting $$\frac{1+1+1+\cdots}{1+2+3+\cdots}$$ as $$\lim\limits_{n\to\infty}\frac{\sum_{i=1}^n 1}{\sum_{i=1}^n i}$$

If we had, for example $$\lim\limits_{n\to\infty}\frac{\sum_{i=1}^{n^2} 1}{\sum_{i=1}^n i}$$

then we would say that $$\frac{1+1+1+\cdots}{1+2+3+\cdots}=2$$

We can write this more suggestively as $$\frac{{1 + (1 + 1 + 1) + (1 + 1 + 1 + 1 + 1) +  \cdots }}{{1 + 2 + 3 +  \cdots }} = 2$$ to make clear how many terms we add in each step.

The moral of the story is that you cannot have "voids" in your notation. Writing something like $$\frac{1+1+1+\dots}{1+2+3+\dots}=\text{something}$$

should be better replaced by something informative and clear that really says what is going on.

Title: Explain the 1 + 2 + 3 in $ \frac{1 + 1 + 1 + \cdots}{1 + 2 + 3 + \cdots} = \lim_{n \to \infty} \frac{1}{(n+1)/2} $

Body:
$$ \frac{1 + 1 + 1 + \cdots}{1 + 2 + 3 + \cdots} = \lim_{n \to \infty} \frac{1}{(n+1)/2} = 0 $$

If $n$ goes to infinity, we can image that a bit by taking a very big number.

Like $1.000.000.000$

$1.000.000.000+1=1.000.000.001$

$1.000.000.001/2=500.000.000,5$

Now how does the $1 + 2 + 3 + ...$ make sense?


问:
Stochastic integral of bounded variation martingale
I am having trouble wrapping my head around a provided answer regarding a stochastic integration question, and have had no luck searching the forums for identical questions so I will post this here. There were some pre-requisites to the exercise so I'll add the items below respectively.

a) Assume that M = $(M_t)_{t \geq 0}$ is a bounded cts. martingale of finite variation $\implies M$ is a constant.

b) The question then proceeded to defining a new process N = $(N_t)_{t \geq 0}$ by $N_t = M_t^4$ and asked to express N in terms of stochastic integrals w.r.t. M and [M] which I did:
$$
N_t - N_0 = 4\int_{0}^t M_s^3 dMs+ 6 \int_0^t M_s^2d[M]s
$$

The question then asked to show that: if N is also a martingale, then M must be a constant.

Now I understand that in order for N to be a martingale, I must have that the bounded variation integral:

$\int_0^t M_s^2d[M]s \equiv 0$ (i.e. a process without the drift term - which I get)

The solution then states that either $M \equiv 0$ or $[M]_t = c \in \mathbb{R}$ $\forall t$, which brings me to my first question: why is the quadratic variation of the process M constant? I get the feeling I am missing something incredibly basic here but can't quite figure it out.

The solution then continues as follows:

but since $\mathbb{E}[(M_t^2 - M_s^2)] = \mathbb{E}[M_t^2 - M_s^2] = \mathbb{E}[[M]_t -[M]_s] =0$ $\forall s\leq t$ we must have $M_t = M_s$

Which brings me to my other question- why are we now considering $M_t^2 - M_s^2$ and how exactly does this show that M is constant?

Any help/pointers on the above would be greatly appreciated, as I've been at this for a few hours now to no avail!

答:
After some given thought I managed to deduce the solution for anybody who is interested:

The reason why we take $[M]_t = c$ is because it will have differential $= 0$ i.e. $d[M]_t = 0$

If we show that for all possible pairs of s,t that the value of $M_t = M_s$ everywhere, this concludes that the process itself must be constant in value and we can simply take $s=0$ and deduce that $M_t = M_0$ $\forall t$


问:
Let $A,B$ be self-adjoint operators and $A\leq B$. Is $e^A \leq e^B$ true?
If $A,B$ commute, that $e^A\leq e^B$ follows from functional calculus.

Is this still true when $A,B$ do not commute?

Also I wonder if $A^{2n+1}\leq B^{2n+1}$ is true for natural number $n$.

答:
A continuous function $f:\mathbb R\to \mathbb R$ is said to be operator monotone if, for every self-adjoint operators
$A$ and $B$ such that $A\leq B$, one has that $f(A)\leq f(B)$.

Operator monotone functions were characterized by Lowner as being
the functions  which admit a holomorphic extension $\tilde f$
to the upper half plane
$$
  \mathbb C_+ = \{z\in \mathbb C: \Im (z) > 0\},
  $$
such that $\tilde f(\mathbb C_+)\subset \mathbb C_+$.  Since the exponential does not satisfy Lowner's
condition, it is not operator monotone.

The same goes for $f(x)=x^{2n+1}$, for $n\geq 1$.


问:
Proposition 4.3 from "Mixed Hodge structures"
In the book "Mixed Hodge structures", Proposition 4.3 reads ($U$ a smooth variety, $X$ a simple normal crossings compactification):

In inclusion of complexes
$$\Omega_X^\bullet(\log E)\to j_*\Omega_U^\bullet$$
is a quasi-isomorphism and induces a natural identification
$$H^k(U;\mathbb{C})\cong \mathbb{H}^k(X,\Omega_X^\bullet(\log D)).$$
Furthermore, the natural map $j_*\Omega_U^\bullet\to Rj_*\Omega_U^\bullet$ is a quasi-isomorphism.

I do not really understand the logic in this formulation. There are three statements here. The first statement is a local computation (which can be proven in Stein opens), and gives us that for all $k$,
$$\mathbb{H}^k(X,\Omega_X^\bullet(\log E)) = \mathbb{H}^k(X,j_*\Omega_U^\bullet).$$
The formulation suggests that this already gives that $\mathbb{H}^k(X,\Omega_X^\bullet(\log E)) \cong H^k(U;\mathbb{C})$, without using the third statement (this is suggested by the word "furthermore"). So it seems that without the third statement it should be clear that
$$H^k(U;\mathbb{C})\cong \mathbb{H}^k(X,j_*\Omega_U^\bullet).$$
I do not quite see how to deduce this without using point $3$ though. The right hand side is equal to $\mathbb{H}^k(X,R^0j_*\Omega_U^\bullet)$. By the Poincare lemma, this is equal to $\mathbb{H}^k(X,R^0j_*\underline{\mathbb{C}})$. But these terms form only a single column on the $E_2$ page of the Leray spectral sequence for $j$, which converges to $\mathbb{H}^k(U,\underline{\mathbb{C}})=H^k(U;\mathbb{C})$. So it seems that to get the isomorphism we are looking for we need that the Leray spectral sequence degenerates, e.g. by showing that that $R^ij_*\underline{\mathbb{C}}=0$ for $i>0$, but this is essentially point (3).

So what am I missing here? How do we deduce $H^k(U;\mathbb{C})\cong \mathbb{H}^k(X,\Omega_X^\bullet(\log D))$ from the first claim directly without using the "furthermore" claim?

答: The inclusion $\Omega_X^\bullet(\log E) \hookrightarrow j_*\Omega_U^\bullet$ induces an isomorphism $$\mathbb H^k(X, j_*\Omega_U^\bullet) \cong \mathbb H^k(X, \Omega_X^\bullet(\log E)).$$  The fact that $H^k(U, \mathbf C) \cong \mathbb H^k(X, j_*\Omega_U^\bullet)$ follows from the (abstract) de Rham theorem (theorem B.18 in "Mixed Hodge Structures").  See also the first paragraph of the proof of Prop. 4.3.


问:
$f$ is continuous iff $G(f)$ is a closed set in metric spaces
The graph of $f$ is $G(f) = \{(x,f(x)) : x\in X\} \subseteq X\times Y$

$X$ and $Y$ are metric spaces. $Y$ is compact.

$f$ is continuous iff $G(f)$ is a closed set.

I got the closest answer here but I tried it by myself first and got stuck at one point and I need help on that particular situation which I didn't get anywhere else/

$\Rightarrow$ part: Let $(z_n)=(x_n,f(x_n))\in G_f$ be a convergent sequence of $G(f)$. If $(x,y)$ is its limit. We have to show that $y=f(x)$ in other words $(x,y)\in G_f$.

$x_n \to x$ $\Rightarrow$ $f(x_n)\to f(x)$[By continuity of $f$.] $\Rightarrow f(x)=y$ by uniqueness of the limit. Hence $G_f$ is closed.

$\Leftarrow$ part: Let $x\in X$ and $(x_n)$ a convergent sequence with limit $x$. You have to prove that $(f(x_n))$ is convergent in $Y$ with limit $f(x)$. I have used the sequence $z_n=(x_n,f(x_n))$ and $G_f$ is closed in the compact space $Y$ and hence $G_f$ is compact. Then there is subsequence $(x_{n_k},f(x_{n_k})) \to (x,y)\in G_f$. Then we will have $y=f(x)$ but how do I prove that $f(x_n) \to f(x)$? It is true that every subsequence of $f(x_n)$ has a subsequence converging to $f(x)$.

答:
From the comment I got my answer which comes from this lemma:

Lemma Let $Y$ be a compact metric space and $(y_n)$ a sequence whose terms belong to $Y$. If every convergent subsequence of $(y_n)$ converges to the same limit $\ell\in Y$, then $(y_n)$ converges to $\ell$.

Proof Suppose the contrary. Then, there exists $\epsilon>0$, such that :

$$\forall N\in\mathbb{N},\,\exists n\ge N;\,d(y_n,\ell)>\epsilon$$

This allows us to construct a subsequence $(y_{n_k})$ such that :

$$\forall k\in\mathbb{N},d(y_{n_k},\ell)>\epsilon$$

Now extract from $(y_{n_k})$ a convergent subsequence : its limit $\ell$ from the hypothesis and hence we get $0=d(\ell,\ell)\ge\epsilon$ ...

A contradiction!

Now someone can close this answer but I can keep it under my record and if someone will proceed in this way. They will get help from it. I asked the question because I was checking one of the obvious ways that can come to our mind. Thanks a lot!


问:
an example to show separability of a Banach space does not imply separability of the dual space
$X$ is a Banach space and it is separable, is there any simple counterexample to show the dual space $X^\ast$ is not separable?</p>

答: The dual of $\ell^1$ (clearly separable by finite sequences of rational numbers) is $\ell^\infty$ (clearly not separable, as you have $\mathfrak c$-many disjoint balls of radius $1/2$ around indicator functions).</p>


问:
The derivative of $y=x^2$ and other values of x that equal zero?
Put simply, if the derivative of $y=x^2$ is $0$ when $x$ is $0$, how does that explain all the other values being greater than zero? From what I know, a derivative is the slope of a line tangent to a function. It can also be explained as the slope from a point on a graph to another point that is very, very close to it. So if the slope is $0$ when $x$ is $0$, then wouldn't there be two values (one right before and one right after) that are equal to zero as well? And it could be said that this value approaches zero, so it pretty much is zero. In that case, wouldn't the next very, very small point also be zero, and so on forever? How would the function ever increase in value?

答:
For every $x\neq 0$ you have $x^2 > 0$. Values with small distances to $0$ yield small squares, but they are non-zero. The secant through $(0,0)$ and $(x,x^2)$ with $x\neq 0$ has slope $\frac{\Delta y}{\Delta x}=\frac{x^2}{x}=x$ which is non-zero but approaches $0$ when $x$ approaches $0$. Only the tangent through $(0,0)$ has slope equal to $0$, and it doesn't intersect the graph at any other point.

If this doesn't answer your question, please try to be more specific.


答:
Let $ n\geq 2 $, $ \sum\limits_{k\geq 2}{\frac{1}{k^{n}}} $ converges. And we have : \begin{aligned} \sum_{k=2}^{+\infty}{\frac{1}{k^{n}}}=\frac{1}{2^{n}}+\sum_{k=3}^{+\infty}{\int_{k-1}^{k}{\frac{\mathrm{d}x}{k^{n}}}}\leq\frac{1}{2^{n}}+\sum_{k=3}^{+\infty}{\int_{k-1}^{k}{\frac{\mathrm{d}x}{x^{n}}}}=\frac{1}{2^{n}}+\int_{2}^{+\infty}{\frac{\mathrm{d}x}{x^{n}}}&=\frac{1}{2^{n}}\left(\frac{n+1}{n-1}\right)\\&\leq\frac{3}{2^{n}} \end{aligned}

Since $ \sum\limits_{n\geq 2}{\frac{3}{2^{n}}} $ converges, we get that $ \sum\limits_{n\geq 2}{\sum\limits_{k=2}^{+\infty}{\frac{1}{k^{n}}}} $ does also converge (Absolutly since we're dealing with positive terms only).

From the absolute convergence, we can, using Fubini's theorem, switch the order of the summations, to get the value of the sum. As seen in a previous answer, $ \sum\limits_{n=2}^{+\infty}{\sum\limits_{k=2}^{+\infty}{\frac{1}{k^{n}}}}=1 $.

Title: Double series convergence (without integrals)

Body:
Can somebody help me with double summation problem? I don't know at all how to prove that $\sum_{n=2}^{\infty} \sum_{k=2}^{\infty}\frac{1}{k^n}$ converges absolutely.
I would be really glad if somebody helped.


答:
There is no number "right after $0$" to connect to $0$ to form a slope of $0$, or at which the slope will still be $0$.

You need the idea of a limit to build a logically sound way to deal with the fact that there is no smallest positive number. That's what calculus is all about. When you start learning it you rely on developing intuition about limits - look carefully at the algebra that tells you  the derivative of $f(x) = x^2$ is $f'(x) = 2x$.

In later studies you will see a more formal definition of limits.

Title: The derivative of $y=x^2$ and other values of x that equal zero?

Body: Put simply, if the derivative of $y=x^2$ is $0$ when $x$ is $0$, how does that explain all the other values being greater than zero? From what I know, a derivative is the slope of a line tangent to a function. It can also be explained as the slope from a point on a graph to another point that is very, very close to it. So if the slope is $0$ when $x$ is $0$, then wouldn't there be two values (one right before and one right after) that are equal to zero as well? And it could be said that this value approaches zero, so it pretty much is zero. In that case, wouldn't the next very, very small point also be zero, and so on forever? How would the function ever increase in value?


答:
a derivative is the slope of a line tangent to a function

Actually, it's the slope of a line tangent to the graph of a function.

It can also be explained as the slope from a point on a graph to another point that is very, very close to it.

Not really. The derivative of $f$ at $x_0$ is the limit of the slopes of the lines joining $\bigl(x_0,f(x_0)\bigr)$ to points of the form $\bigl(x,f(x)\bigr)$, as $x$ gets closer to $x_0$. And a limit of things which are all different from $0$ may well be $0$.


问:
Does $\sum_{r=0}^n 1=n+1$?
Here's my question:

Is the following true? $$\sum_{r=0}^n 1=n+1$$

Here is my area of confusion. Looking at this superficially it would indeed seem to be the case that the result is true, but the result contradicts my understanding of what the sum from $r=a$ to $b$ of a constant means.

In the case of
$$\sum_{r=1}^n 1$$ I would make sense of the summand by thinking of it as

$$\sum_{r=1}^n r^0$$
so the summand would still be a function of $r$. But in the case of
$$\sum_{r=0}^n 1$$
I can't think of it as
$$\sum_{r=0}^n r^0$$
as $0^0$ is undefined. So, is the result correct, or is meaningless?

Note: I know that $\lim_{x\to0} x^x=1$, but it doesn't seem right to apply this to our case of $0^0$; I seem to remember a different value that can found for the 'value' of $0^0$ by taking a different limit.

Thank you for your help.

答:
Hint: You can approach the serie as

$$\sum_{r=0}^{n} 1 = 1 + \sum_{r=1}^{n} 1 = 1 + n$$

Title: Does $\sum_{r=0}^n 1=n+1$?


问:
How do I use Mean Value Theorem to proof that the speed equals the average speed?
I was watching a lecture and the example given: a car travelled 240 miles in 3 hours and during the interval of [0, 3], there is at least once that the speed equals the average speed.

How do I use the Mean Value Theorem to proof this example?

Please pardon me for asking the answer as I am self-learning Calculus by watching videos.

答:
Think about the following:

Can the speed be always strictly less than the average speed?


问:
Use limit theorems to show that the following function is continuous on $[0,1]$

Use limit theorems to show that the following function is continuous on $[0,1]$.
$$f(x)=\begin{cases}
e^{-\frac{1}{x}}   & :x \neq0 \\
           0   & :  \text{$x= 0$}
\end{cases}$$

Here is the answer which is given to me

$*$ We know that $\space\dfrac{1}{x}$ is continuous for $x \neq 0$

$*$ Hence $\space-\dfrac{1}{x}$ is continuous on $(0,1]$ and $\space e^{-\frac{1}{x}}$ continuous on $(0,1]$

$*$ For $x=0$ we must show that,
$$\lim_{x \to 0}f(x) = f(0)=0$$

$*$ Since $f(x)=e^{-\frac{1}{x}}$ for $x\neq 0$

$$\lim_{x \to 0}f(x) = \lim_{x \to 0}e^{-\frac{1}{x}}$$

$*$ Consider,
$$\lim_{x \to 0^{+}}f(x) = \lim_{x \to 0^{+}}e^{-\frac{1}{x}}=0$$

$\therefore$ The limit of the function as $x \rightarrow 0^+$ is equal to the function value at $x \neq 0$

That is
$$\lim_{x \to 0}f(x) = f(0)=0$$

$\therefore$ $f$ is continuous at $0$

I feel some point are wrong here because $\lim_{x \to 0^{-}}f(x)=\infty$ so we must show for $x=0$
$$\lim_{x \to 0^{+}}f(x) = f(0)=0$$ not $$\lim_{x \to 0}f(x) = f(0)=0$$ this

Is there anything wrong what I am saying please tell me?

答:
To verifying the continuity of $f$ on $[0,1]$, it is enough to check the continuity of $f$ on $(0,1]$, which you have done in the first and second $*$, and to see the right continuity of f at $0$, which you have done in the fifth $*$. The rest is superfluous.

The evaluation of $\lim_{x \to 0^{-}}f(x)=\infty$ is correct, and $f$ is not continuous on $\mathbb{R}$ and in particular at $0$. Nevertheless, it is continuous on $[0,1]$. See its graph!


问:
I have decided to study math on my own and I came across these questions. Would really appreciate if anyone could help and explain how to solve these.

I could not come up with a solution for both of the problems, so help is appreciated

答:
For your first question, notice the following:

$$\text{a}_\text{n}:=\frac{3\text{n}-72}{\text{n}}=\frac{3\text{n}}{\text{n}}-\frac{72}{\text{n}}=3-\frac{72}{\text{n}}\tag1$$

Finding the divisors of $72$ gives:

$$\left\{1,2,3,4,6,8,9,12,18,24,36,72\right\}\tag2$$


答:
we must have $n\mid 72$, which can be listed by brute force.

Do division algorithm and find that $n+2\mid 3$, and continue from here similarly.

Title: I have decided to study math on my own and I came across these questions. Would really appreciate if anyone could help and explain how to solve these.

Body: I could not come up with a solution for both of the problems, so help is appreciated


答:
For the first question, you can rewrite it like this:

$$\frac{3n-72}{n}=\frac{3n}{n}-\frac{72}{n}=3-\frac{72}{n}$$
This means that $\frac{3n-72}{n}$ will only be an integer if $n$ is a factor of $72$ (it may be a positive or negative factor of $72$).

For the second question, try using algebraic long division and use a similar method to the one I've employed in the first question.

If you need any more help, please ask :)


问:
What's the term for elements in (image of codifferential/adjoint exterior differential aka 3rd part of hodge decomposition theorem )?
Part of the Hodge Decomposition Theorem is that smooth $k$-forms on compact oriented Riemannian (smooth) $m-$manifold $(M,g)$ (I think $M$ need not be connected, but assume connected if need be or you want) are decomposed as

$$\Omega^k(M) = B^kM \bigoplus \mathscr H^k(M,g) \bigoplus image(\delta_{k+1}),$$ where

$\Omega^k(M):=$ smooth $k$-forms on $M$ (I think $g$ is not needed here.)

$\mathscr H^k(M,g):=ker(\Delta_k):=$ harmonic $k$-forms on $(M,g)$, where $\Delta_k$ is Laplace operator (or Hodge-Laplace operator or Laplacian or Hodge-Laplacian or whatever)

$B^kM:=image(d_{k-1}):=$ exact $k$-forms on $M$, aka image of exterior derivative/differential $d_{k-1}: \Omega^{k-1}M \to \Omega^{k}M$ (I think $g$ is not needed here.)

$image(\delta_{k+1}):=$ image of codifferential/coderivative $\delta_{k+1}: \Omega^{k+1}M \to \Omega^{k}M$ of exterior derivative/differential $d_{k-1}: \Omega^{k-1}M \to \Omega^{k}M$ (I guess $\delta_{k+1}$ is like 'exterior codifferential/coderivative' or 'co-exterior differential/derivative' or whatever; I believe $g$ is needed here.)

Question 1: What is/are the term/s for elements in $image(\delta_{k+1})$?

Guess: After typing up all this I weirdly came up with the simple answer,

'$\delta$-exact $k$-forms',

in that elements of $B^kM$ are actually 'exact' in the sense of being '$d$-exact'. It's like how for forms on Riemann surfaces we have $d$-closed vs $\partial$-closed vs $\overline \partial$-closed.

Question 2: Any other term besides $\delta$-exact?

答: I think that co-closed (for lying in the kernel of $\delta$) and co-exact (for lying in the image of $\delta$) is quite common terminology.


问:
how to prove thar $|| T\circ R|| =||T|||R|||$
Let $T\in L(\mathbb{R}^{n},\mathbb{R}^{m})$ with $n\leq m$. If $R\in L(\mathbb{R}^{n})$,  how to prove that $|| T\circ R|| =||T|||R|||$. Here i am using two facts: Polar decomposition

When i have $V,W$ Hilbert spaces and $L:V\to W$ linear application then

i) if $dim V\leq dim W$ then exist $S\in Sym (n)$ and $O\in O(V,W)$ such that $L=O\circ S$

ii) If $dim W\leq dim V$  then exist $S\in Sym (n)$ and $O\in O(V,W)$ such that $L=S\circ O^{*}$

In this case, when $L=O\circ S$ or $L=S\circ O^{*}$ then  $||L||:=|Det S|$.

My attemp , using polar decomposition for $T=O\circ S$ imply

$||T\circ R||=||O\circ S\circ R||=|det ( S\circ R)|$ because $R$ is inyective, and using that $ O\circ O^{*}=I$, and (Im not sure if) the las equality if take $ |det ( S\circ R)|=|det ( S\circ O\circ O^{*}\circ R)|$ can to help in something, or maybe take $L_1:=S\circ O$ and $L_2=O^{*}\circ R$, Can somebody help me to continued please, or give one hint, other way to prove that, tahnk you!!!

答:
Note (or verify using the polar decomposition) that $\|M\| = \sqrt{\det(M^*\circ M)}$ for $M \in L(\Bbb R^n, \Bbb R^m)$ (with $n \leq m$) and note that $\det(A\circ B) = \det(A) \det(B)$ for $A,B \in L(\Bbb R^n)$. With that, we have
$$
\begin{align}
\|T \circ R\|^2 &= \det([T \circ R]^* \circ [T \circ R])
\\ & = \det(R^* \circ T^* \circ T \circ R)
\\ & = \det(R^* \circ [T^* \circ T] \circ R)
\\ & = \det(R^*) \det(T^* \circ T) \det(R)
\\ & =  \det(T^* \circ T)[\det(R^*)\det(R)]
\\ & = \det(T^* \circ T) \det(R^* \circ R)  = \|T\|^2\|R\|^2.
\end{align}
$$
Thus, $\|T\circ R\| = \|T\|\cdot \|R\|$.

Proof of the statement: let $M = U \circ S$ be a polar decomposition (so $U$ is orthogonal injective and $S$ is symmetric). It follows that
$$
M^* \circ M = (U \circ S)^* \circ (U \circ S) = S^* \circ U^* \circ U \circ S
= S \circ (U^* \circ U) \circ S = S^2.
$$
It follows that $\det(M^* \circ M) = \det(S^2) = \det(S)^2$. Thus, we have
$$
\sqrt{\det(M^* \circ M)} = \sqrt{\det(S)^2} = |\det(S)| = \|M\|,
$$
which was what we wanted.


答:
It doesn't in general make sense to take ratios of infinite series because convergence may not be assured. In this case, the numerator and denominator diverge so you have to be very careful about what you mean by

$$\frac{1+1+1+\cdots}{1+2+3+\cdots}.$$

The best way to interpret this ratio is as the following:

$$\frac{1+1+1+\cdots}{1+2+3+\cdots} = \lim_{n\rightarrow\infty}\frac{\sum_{i=1}^n1}{\sum_{i=1}^ni} = \lim_{n\rightarrow\infty} \frac{n}{\frac{n(n+1)}{2}} = \lim_{n\rightarrow\infty}\frac{2}{n+1} = 0.$$


问:
Can the plane be covered by open disjoint one dimensional intervals?
I remember I heard this question a while ago but have never figured it out. Let an open interval in $\mathbb{R}$ of the form $(a,b)$. Now, imaging placing it in $\mathbb{R}^2$, anywhere we want while keeping its length but perhaps rotating it if we want. Essentially this creates a line segment without its start/endpoints parametrized by

$P(1-t)+Qt$ for $0<t<1$ where $P$ and $Q$ are its start and end points and with length $||P-Q||=b-a$.

As an example, if we take the unit interval $(0,1)$, we can place it as it is in $\mathbb{R}^2$ as $(0,1)\times \{0\}$ or we can rotate it as $\{1\}\times (0,1)$ etc. I hope it is clear what I mean. My question is the following.

Can we cover the entire $\mathbb{R}^2$ with disjoint such "open intervals" or if you prefer, line segments of finite length without their start/endpoints?

I don't remember if the initial question I read was using only unit intervals, or of any length $(a,b)$ and I don't know if that plays a role in the answer but as a gut feeling, I doubt it does. My impression is that the answer is negative and we can't cover the plane in such a manner, or if it is possible it would use some weird abstract construction perhaps involving AC. Any help is appreciated!

Edit: It seems that I could have phrased my wording a bit better although that was my last paragraph in the post above that I wasn't certain if it mattered whether we need to have fixed length intervals or not. Apparently it does, it is impossible with using only fixed length intervals but it is possible if we allow intervals of different lengths. Proof of impossibility in one case given in a paper given in the answers, and a construction in the other case is given by some answers and also found in that paper itself. Thanks everyone!

答:
My guess is that you can do it.

$\qquad\qquad\qquad$

Notice first that you can tile the plane with open (rectangular) triangles and open squares. In  fact, grid the plane with rotated/translated copies of the segment $(0,1)$, obtaining a regular lattice of open squares (notice you have still to cover the endpoints of the segments).
Now add diagonal segments of length $2\sqrt 2$ in order to cut one square every 2 into 4 rectangular triangles. You can do it by shifting every parallel diagonal by $\sqrt 2$.

But now you can cover any open triangle (that is, without its perimeter) and open square with segments parallel to one of the sides.


答:
Think of the trajectory of the car as a differentiable function $d(x)$, with $x\in[0,3]$. Then, by the mean value theorem there is an $x_0\in(0,3)$ that

$$\frac{d(3)-d(0)}{3-0}=d'(x_0),$$

where $d(3)=240$ and $d(0)=0$. So in the left hand you get the average speed, and in the right hand is the speed in some specific time. Sorry for my english.

Title: How do I use Mean Value Theorem to proof that the speed equals the average speed?

Body:
I was watching a lecture and the example given: a car travelled 240 miles in 3 hours and during the interval of [0, 3], there is at least once that the speed equals the average speed.

How do I use the Mean Value Theorem to proof this example?

Please pardon me for asking the answer as I am self-learning Calculus by watching videos.


问:
About Classification of compact 2-manifolds
I have question about classification of compact 2-manifolds.
I have read in some sources that every compact 2-manifold is diffeomorphic with sphere with n-holes or sphere with m-mobius strips for some natural n or m (e. g. torus is a sphere with 1 hole and klein bottle is a sphere with 2 mobius strips).
But shouldn't there also be a possibility to be a compact 2-manifold with some non-zero number of holes and some non-zero number mobius strips? I don't see a reason why there should be allowed only holes or only mobius strips.</p>

答:
If $\Sigma$ is a closed connected two-dimensional manifold, then

$\Sigma$ is diffeomorphic to the connected sum of $n \geq 0$ tori if $\Sigma$ is orientable, or

$\Sigma$ is diffeomorphic to the connected sum of $m \geq 1$ real projective planes if $\Sigma$ is non-orientable.

What about surfaces which are connected sums of tori and real projective planes? The answer is that such a surface is diffeomorphic to the connected sum of real projective planes only. This follows from the fact that $T^2\#\mathbb{RP}^2$ is diffeomorphic to $\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2$; see here for a proof, and here for a visual illustration. So for $m > 0$, the manifold $nT^2\#m\mathbb{RP}^2$ is diffeomorphic to $(2n + m)\mathbb{RP}^2$.


答:
In fact it is impossible with open intervals, although you can do it with closed intervals:
https://mathoverflow.net/questions/43611/decomposing-the-plane-into-intervals</p>

Title: Can the plane be covered by open disjoint one dimensional intervals?

Body:
I remember I heard this question a while ago but have never figured it out. Let an open interval in $\mathbb{R}$ of the form $(a,b)$. Now, imaging placing it in $\mathbb{R}^2$, anywhere we want while keeping its length but perhaps rotating it if we want. Essentially this creates a line segment without its start/endpoints parametrized by

$P(1-t)+Qt$ for $0<t<1$ where $P$ and $Q$ are its start and end points and with length $||P-Q||=b-a$.

As an example, if we take the unit interval $(0,1)$, we can place it as it is in $\mathbb{R}^2$ as $(0,1)\times \{0\}$ or we can rotate it as $\{1\}\times (0,1)$ etc. I hope it is clear what I mean. My question is the following.

Can we cover the entire $\mathbb{R}^2$ with disjoint such "open intervals" or if you prefer, line segments of finite length without their start/endpoints?

I don't remember if the initial question I read was using only unit intervals, or of any length $(a,b)$ and I don't know if that plays a role in the answer but as a gut feeling, I doubt it does. My impression is that the answer is negative and we can't cover the plane in such a manner, or if it is possible it would use some weird abstract construction perhaps involving AC. Any help is appreciated!

Edit: It seems that I could have phrased my wording a bit better although that was my last paragraph in the post above that I wasn't certain if it mattered whether we need to have fixed length intervals or not. Apparently it does, it is impossible with using only fixed length intervals but it is possible if we allow intervals of different lengths. Proof of impossibility in one case given in a paper given in the answers, and a construction in the other case is given by some answers and also found in that paper itself. Thanks everyone!


问:
Applying Multivariate Ito's Lemma
I am having trouble applying Ito's Formula to the following:

Let $Z_t = W_{1t}^2  e^{W_{1t}+ \int_0^t W_{3s}dW_{2s}}$. Find $dZ_t$. $W_1,W_2,W_3$ are independent Brownian motions.

I know the formula but I am having trouble differentiating the integral with respect to $W_2$ and $W_3$.

Title: Applying Multivariate Ito's Lemma

Body:
Let $A_t := \int_0^t W_{3,s} dW_{2,s}$.  As pointed out by @fesman in the comments, we have $dA_t = W_{3,t}dW_{2,t}$.  Note that this is just notation for how we defined $A_t$, we are not actually differentiating anything.  Then we have $Z_t = W_{1,t}^2 e^{W_{1,t} + A_t}$, and Ito's formula gives

\begin{align*}
dZ_t &= (W_{1,t}^2 e^{W_{1,t}+A_t} + 2 W_{1,t} e^{W_{1,t}+A_t}) dW_{1,t} + W_{1,t}^2 e^{W_{1,t}+A_t} dA_t \\
& \qquad + (W_{1,t}^2 e^{W_{1,t}+A_t} + 4 W_{1,t} e^{W_{1,t}+A_t}  +2e^{W_{1,t}+A_t}) dW_{1,t}dW_{1,t} + W_{1,t}^2 e^{W_{1,t}+A_t} dA_t dA_t \\
&= (W_{1,t}^2 e^{W_{1,t}+A_t} + 2 W_{1,t} e^{W_{1,t}+A_t}) dW_{1,t} + W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}dW_{2,t} \\
& \qquad+ (W_{1,t}^2 e^{W_{1,t}+A_t} + 4 W_{1,t} e^{W_{1,t}+A_t}  +2e^{W_{1,t}+A_t}) dt + W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}^2dt.
\end{align*}

Again, to emphasize, we have not differentiated $Z_t$.  This is just notation for $$Z_T = \int_0^T (W_{1,t}^2 e^{W_{1,t}+A_t} + 2 W_{1,t} e^{W_{1,t}+A_t}) dW_{1,t} +\int_0^T W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}dW_{2,t} + \int_0^T (W_{1,t}^2 e^{W_{1,t}+A_t} + 4 W_{1,t} e^{W_{1,t}+A_t}  +2e^{W_{1,t}+A_t} + W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}^2) dt.$$


问:
Should the unbiased estimator of the variance of the sample proportion have (n-1) in the denominator?
I know that the variance of the sample mean is:

$$\frac{\sigma^2}{n}$$

And that the unbiased estimator for that expression is:

$$\frac{\sigma^2}{n-1}$$

The variance of the sample proportion is:

$$\frac{p(1-p)}{n}$$

Does an unbiased estimator of variance of the sample proportion also need to have (n-1) in the denominator, like the unbiased estimator of the variance of the sample mean?

I feel the resoning for the denominator should be applicable for both situations, but I haven't seen it mentioned anywhere...

答:
If $\hat{p}$ is the sample proportion, then $n\hat{p} \sim \text{Binomial}(n, p)$ so

$$E[\hat{p}(1-\hat{p})] = E[\hat{p}] - E[\hat{p}^2] = E[\hat{p}] - \text{Var}(\hat{p}) - E[\hat{p}]^2 =  p(1-p) (1 - \frac{1}{n})$$
so it seems that the unbiased estimator of $\frac{p(1-p)}{n}$ would be
$$\frac{\hat{p}(1-\hat{p})}{n-1}.$$

I might have made a miscalculation somewhere though.


问:
Proof collecting: A short proof of chage of variable formula
I've been searching for a  proof of the change-of-variable formula for multi-variable Riemann integration, but most of them seem to be too long and cumbersome. I'd like to see one proof that uses clever approximation skills or some advanced techniques, and of course,the shorter, the better.  Thanks in advance :).

By the way, any proof that involves  interesting ideas are also welcomed.

答:
Since, you asked for ideas, one interesting derivation of change of variables is from Radon-Nikodyn theorem about derivative of measures. First, let us state a precise formulation.
Let $U \subset \mathbb{R}^n$ be an open set. If $f:U \to \mathbb{R}^n$ is continuously differentiable, 1-to-1, and $\det Df(x) \neq 0$ for any $x \in U$, then for any integrable $w:f(U) \to \mathbb{R}^n$,
$$
\int_{f(U)} w(y) \, d\mathcal{L}^n(y) = \int_U w(f(x)) \, |\det Df(x)| d\mathcal{L}^n(x)\, . \qquad \text{(C.of.V. 1)}
$$

Proof Sketch:

It suffices to prove C.of.V for characteristic functions of measurable sets, i.e. for any measurable $E\subset U$,
$$
\mathcal{L}^n({f(E)}) = \int_E |\det Df(x)| \, d\mathcal{L}^n(x)\, . \qquad \text{(C.of.V. 2)}
$$
Then the general case c.of.V. 1 follows from approximation by characteristic functions.

I will assume $U=\mathbb{R}^n$ for simplicity. Define a new measure on $\mathbb{R}^n$ by
$$
\mu(E):=\mathcal{L}^n({f(E)}) \, .
$$

Show stuff about this measure. Like it is well-defined, Borel, etc. You'll probably will also need to show $f(E)$ is Lebesgue measurable iff $E$ is.

3.5) Prove that $\mu$ is absolutely continuous w.r.t $\mathcal{L}^n$, i.e. if $\mathcal{L}^n({E})=0$ then $\mu(E)=0$.

Next compute the Radon-Nikodyn derivative:
$$
\frac{d\mu}{d\mathcal{L}^n}(x) = \lim_{r \to 0} \frac{\mu(f(B(x,r))}{\mathcal{L}^n(B(x,r))} = |\det Df(x)| \, .
$$
This follows from approximating $f(B(x,r))$ by $Df(x)(B(x,r))$. The approximation comes from differentiability of $f$, and gets arbitrarily close as $r\to 0$. since $Df(x)$ is a linear map, even before having $r\to 0$, we have $\mathcal{L}^n(Df(x)(B(x,r))) = |\det Df(x)|\mathcal{L}^n(B(x,r))$.

By Radon-Nikodyn, for every $\mathcal{L}^n$-measurable $E$,
$$
\mu(E) = \int_E \frac{d\mu}{d\mathcal{L}^n}(x) \, d\mathcal{L}^n(x) = \int_E |\det Df(x)| \, d\mathcal{L}^n(x),
$$
Which is just C.of.V 2.

Apparently this is done in Rudin's book, but I do not have access to them to verify this! Will hopefully update once I find out.

Note: As one matures in math, what is elementary, easy, natural, elegant, etc keeps changing. So, feel free to criticize this proof! :)


问:
Expansion of $1-\frac{1}{p}$
My professor wrote the following identity in our analysis class:
$$
1-1/p=\exp(-1/p+O(1/p^2))
$$
where $O(h(x))$ means a function $f(x)$ that satisfies $|f(x)|\leq C h(x)$ for some constant $C$.
With that being said, can somebody explain the intuition behind this expansion of $1-1/p$?

答:
This is derived from the expansion of

$$\log(1-x) = -\sum_{n=1}^\infty {x^n\over n}$$

by setting $x = {1\over p}$ and applying the exponential function to both sides.

Title: Expansion of $1-\frac{1}{p}$


答:
The derivative of $y=x^2$ with respect to $x$ is
$$
\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x} \, .
$$
Informally, people regard the derivative as the change in $y$ divided by a small change in $x$. In other words, they regard $(x^2)'$ as
$$
\frac{(x+\Delta x)^2-x^2}{\Delta x}
$$
for some very small value of $\Delta x$. While it is fine to use this notion as an informal conception of what the derivative is, you must remember that this is not quite correct. Instead, the derivative is the limit of the above quotient as $\Delta x \to 0$. When $x=0$, the quotient simplifies to
$$
\frac{(\Delta x)^2}{\Delta x}=\Delta x
$$
Therefore,
$$
(x^2)'|_{x=0}= \lim_{\Delta x \to 0}\Delta x=0 \, .
$$
Even though the slope of the secant line is never zero, the limit of the slope of the secant lines is zero.


答: Look user, you are brining up a great issue here! The numerical expression itself is pretty meaningless if we do not know the context, i.e. the formula where this expression came from. With regrouping symbols I can get virtually anything out of the given numerical expression, because, as stated, it is an infinity / infinity situation, which is indeterminate. The word "indeterminate" is really appropriate here. Sure, the answer could be 1/2 if the problem arises from n/2n, but with grouping symbols, the problem also could arive from 2n/2n in which the answer would be 1. If I group enough 1's together, I can also get 4 as an answer. As long as we do not know from which expression your numerical problem comes from, there is no point arguing what answer it has to be.</p>


问:
Volume of the solid bounded by the paraboloid $z = r^2$
To find the volume of the solid bounded by the paraboloid $z = r^2$ and the plane $z = 9$ using cylindrical coordinates.

So here the limits are $0 \le r \le 3$ and $0 \le \theta \le 2\pi$. Hence the volume is
$$V = \int_0^{2\pi} \int_0^3 r^3 dr d\theta = \frac{81\pi}{2} .$$

Is the solution correct?

答: Since we are measuring the height between $z=9$ and $z=r^2$ and not $z=r^2$ and the $xy$ plane, the integral is$$\int_0^{2\pi}\int_0^3(9-r^2)r~dr~d\theta = \frac{81\pi}{2} .$$


问:
Is the matrix square root uniformly continuous?
Let $\operatorname{Psym}_n$ be the cone of symmetric positive-definite matrices of size $n \times n$.

How to prove the positive square root function $\sqrt{\cdot}:\operatorname{Psym}_n \to \operatorname{Psym}_n$ is uniformly continuous?

I am quite sure this is true, since on any compact ball this clearly holds, and far enough from the origin, I think the rate of change should decrease (analogous to the one-dimensional case where $(\sqrt{x})'=\frac{1}{2\sqrt{x}}$ tends to zero when $x \to \infty$).

A naive approach is to try to use the mean value inequality:

For that we need to show the norm $\|d(\sqrt{\cdot})_A\|$ is bounded for $\|A\|$ large enough. We know the derivative satisfies:

$$d(\sqrt{\cdot})_A(B) \cdot \sqrt{A} + \sqrt{A} \cdot (\sqrt{\cdot})_A(B)=B$$ for every $B \in \operatorname{sym}_n$.

Thus,

$$ \|B\| \le \| d(\sqrt{\cdot})_A(B) \cdot \sqrt{A}\| +\| \sqrt{A} \cdot (\sqrt{\cdot})_A(B) \|\le 2 \| \sqrt{A}\| \| d(\sqrt{\cdot})_A(B)\|,$$

so we only get a bound from below: $$\|d(\sqrt{\cdot})_A\|_{op} \ge \frac{1}{2\|\sqrt{A}\|}$$

答:
If vector $x$ with $\|x\|=1$ is an eigenvector of $\sqrt A - \sqrt B$ with eigenvalue $\mu$ then
\begin{align*}
x^T(A-B)x &=
x^T(\sqrt A - \sqrt B)\sqrt A x
+ x^T\sqrt B (\sqrt A - \sqrt B) x
\\&=
\mu x^T(\sqrt A + \sqrt B)x.
\end{align*}
Now for any $\epsilon>0$, if $\|A-B\|_{op}\le \delta=\epsilon^2$, choose $\mu=\pm\|\sqrt A - \sqrt B\|_{op}$ and as in the scalar case,
\begin{align*}
\|\sqrt A - \sqrt B\|_{op}^2
&
= (x^T(\sqrt A - \sqrt B)x)^2
\\&
\le
|x^T(\sqrt A - \sqrt B)x|
~x^T(\sqrt A + \sqrt B)x
\\&=|x^T(A-B)x|
\\&\le \delta = \epsilon^2.
\end{align*}


问:
Indefinite integral $\displaystyle \int \frac 1 {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\mathrm{d}x$
Working on $\displaystyle \int \frac 1 {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\mathrm{d}x$.

According to Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968) item $14.422$ it should work out as:

$$\frac {-1} {a \sqrt {p^2 + q^2} } \tan \left({\frac \pi 4 - \frac {ax + \arctan (q/p)} 2}\right)$$

First thing I do is a Weierstrass substitution: $u = \tan \dfrac {a x} 2$ which leads (after algebra) to the expression:

$$\dfrac 2 {a (\sqrt {p^2 + q^2} - q) } \int \frac {\mathrm d u} {\left({u + \dfrac p {\sqrt {p^2 + q^2} - q} }\right)^2}$$

This is a standard integral, giving (again after algebra):

$$\frac {-2} {(\sqrt {p^2 + q^2} - q) u + p}$$

Replacing $u = \tan \dfrac {a x} 2$ gives us:

$$\frac {-2} {(\sqrt {p^2 + q^2} - q) \tan \dfrac {a x} 2 + p}$$

which is worlds away from what the book gives.

I can't reconcile the two expressions. I'm fairly sure of the exactness of the square on the bottom, because it's the result of a quadratic in $u$ with a discriminant of zero. Fiddly and messy, but ultimately tractable.

I have tried deriving the expression in the book w.r.t. $x$ but all I get is a complicated and daunting squared-secant expression which may work out in the end to what I started with, but eugh.

How on earth does one arrive at Spiegel's result?

(Sidenote: It's a special case of this integral:

$$\displaystyle \int \frac {\mathrm d x} {p \sin a x + q \cos a x + r} = \begin{cases}
\dfrac 2 {a \sqrt {r^2 - p^2 - q^2} } \arctan \left({\dfrac {p + (r - q) \tan \dfrac {a x} 2} {\sqrt {r^2 - p^2 - q^2} } }\right) + C & : p^2 + q^2 < r^2 \\
\dfrac 1 {a \sqrt {p^2 + q^2 - r^2} } \ln \left| {\dfrac {p - \sqrt {p^2 + q^2 - r^2} + (r - q) \tan \dfrac {a x} 2} {p + \sqrt {p^2 + q^2 - r^2} + (r - q) \tan \dfrac {a x} 2} }\right| + C & : p^2 + q^2 > r^2
\end{cases}$$

but where the discriminant equals zero and so cannot be used.)

答:
Integrate as follows

\begin{align}
& \int \frac {dx} {p \sin ax + q \cos ax + \sqrt {p^2 + q^2} }\\
=&\frac1{\sqrt {p^2 + q^2}  } \int \frac {dx} {\cos(\tan^{-1}\frac pq -ax)+1 }\\
 =&\frac1{\sqrt {p^2 + q^2}  } \int \frac {dx} {\cos(\frac\pi2-ax -\tan^{-1}\frac qp  )+1 }\\
 =&\frac1{2\sqrt {p^2 + q^2}  } \int  \sec^2\left(\frac\pi4-\frac {ax+\tan^{-1}\frac qp}2\right)dx\\
 =&-\frac1{a\sqrt {p^2 + q^2}  }   \tan\left(\frac\pi4- \frac {ax+\tan^{-1}\frac qp}2\right)+C\\
\end{align}