问:
How does the Cooley-Tukey-Algorithm work?
I have tried to understand the Cooley Tukey Algorithm for some time. I read many articles and watched some videos like these:

https://www.youtube.com/watch?v=iTMn0Kt18tg

https://www.youtube.com/watch?v=h7apO7q16V0

https://www.youtube.com/watch?v=E8HeD-MUrjY

https://www.youtube.com/watch?v=mPVtovydY1k

(Source: YouTube)

First my background:
I know how the normal Fourier Transform, the DFT, Laplace and so on work.
I would be able to explain them to other people and have no problems with complex numbers.
I programmed the DFT many times and was able to program a Spectrum-Analyzer where you put in an audio file and it outputs a spectrogram. (Just by using DFT but very limited Frequency range) I use Python. Here is an example DFT in Python:

```<code>import numpy
import matplotlib.pyplot

frequency = 4
samples_power = 6
f_min = 0
f_max = 10

samples = 2**samples_power
signal = numpy.sin(frequency*2*numpy.pi*numpy.linspace(0,1,samples))

f, t = numpy.meshgrid(numpy.linspace(f_min,f_max,samples) ,numpy.linspace(0,1,samples))
n = f*t

dft_matrix = 1/numpy.sqrt(numpy.shape(n)[0])*numpy.power(numpy.e,-2*numpy.pi*1j*n)

dft = numpy.sum(signal*dft_matrix,axis=1)
amplitude = numpy.abs(dft)
phase = numpy.arctan(dft.imag/dft.real)

matplotlib.pyplot.plot(numpy.linspace(f_min,f_max,samples),amplitude)
matplotlib.pyplot.xlabel("Frequency (Hz)")
matplotlib.pyplot.ylabel("Energy / Amplitude")
matplotlib.pyplot.show()
</code>```

But this gets very slow very soon. I want to learn how the FFT works, so that I can use it.
What I know about it, that is has something to do with the "nth root of unity". And even and odd indexes are separated into two separate matrices. There are some redundancies in the DFT, if you analyze just integer frequencies. They are all at 0 in the middle. Even and odd frequencies also cross zero on different fractions. And the Cooley-Tukey-Algorithm/FFT makes use of it.

What I dont understand:

-the connection between all these things (zero points on different fractions, even and odd, nth root of unity ...)

-what seperating even and odd indexes does

-what axis is split into even and odd and what happens with the other because everyone seems only to be talking about one axis.

-what points are not used in the smaller matrices

-if the even and odd matrices are multiplied with the even and odd indexes of the signal

-what operations have to be done with the two resulting vectors to get the output

-what frequencies are used/allowed and if they have to be in a specific order

How does the Cooley Tukey Algorithm Work?

I need it explained in right order with logical connections and as much detail as possible (especially in Math).
I would appreciate it, if someone could do that. Thanks for the answers in advance!

答:
I found a great Video on Youtube from a German University:

https://www.youtube.com/watch?v=HFlIIIi5cEM

(don't really need audio because the drawings are self-explaining, but you can use subtitles if you want)

And this lecture from Rensselaer Polytechnic Institute:

https://www.youtube.com/watch?v=1mVbZLHLaf0

(Source Youtube)

Now I know what the missing puzzle piece was. It is the "Twiddle Factors" or how you want to call them.
I didn't know three things:

Why you need them at all

Why they are power of $k$

What happens with the other side of the Matrix

The German guy explains it very good by drawing the DFT-Matrix.

Why do you need them?

-> Because you can just calculate the DFT of the even rows and by multiplying it with $W_N^k$ you get the odd rows for "free"

Why are they a Power of $k$?

-> The further right you look at the DFT-Matrix, the bigger the $W_N^{kn}$ gets, and the twiddle factors have to get bigger too

What about the other side?

-> In even row's the other side is the same, and in odd row's it's the opposite.
If you would calculate an $N=8$ Matrix this would mean that for as an example the first row (even, in the 8x8) you add samples 0 to 3 and 4 to 7 and then multiply the length-4 vector by the first row of the 4x4 Matrix and you have just calculated the first frequency. For the odd rows (in the 8x8), you would subtract samples 0 to 3 - 4 to 7 and multiply the resulting length-4 vector by the twiddle factors. Then multiply the vector by the first row of the 4x4 and suprise, you just got the the second frequency.

It is so easy, you just need the right teacher. And in the Internet it is really hard to find someone who's explaining something the way you understand it. Thankfully I did.


问:
Differentiability of $f(x,y) = | |x| - |y| | - |x| - |y|$ at $(0,0) \in \mathbb{R}$
I have an exercise in a multivariable calculus book that asks me to find out whether
$f(x,y) = | |x| - |y| | - |x| - |y|$ is differentiable at $(0,0) \in \mathbb{R}^{2}$. I have tried to do this without success.
Can someone help provide a hint or an answer for this problem ? I have found that the partial
derivatives $f_{x}(0,0)$ and $f_{y}(0,0)$ exist. I have also found that the directional derivatives (denoted $\boldsymbol{D_{u}}f(0,0)$) at $(0,0)$ do not exist exist. Note here
$\boldsymbol{u} \in \mathbb{R}^{2}$ is such that $\lVert u \rVert = 1$.

答: Suppose that $f$ is differentiable at $(0,0)$ and consider the map $g\colon\Bbb R\longrightarrow\Bbb R^2$ defined by $g(x)=(x,x)$, which is differentiable. So, if $f$ was differentiable at $(0,0)$, then $f\circ g$ would be differentiable at $0$. But$$(\forall x\in\Bbb R):f\bigl(g(x)\bigr)=-2|x|.$$

ContentLicense: CC BY-SA 4.0


问:
Two strange integrals
I was asked to solve the two integrals

$$
\int\frac{1}{x^2e^{1/x}-x}dx
$$
$$
\int\frac{x^2+1}{x^3-x^2+x+1}dx
$$
I think the first one is not soluble with the known methods of calculus and the second one is involved with complex roots of cubic equation.

Any improvements appreciated.

Thanks.

答:
For the integral

$$
\int\frac{x^2+1}{x^3-x^2+x+1}dx
$$

note that the cubic polynomial in the denominator has one real root $r= -0.5437$ and a pair of complex roots, which allows the factorization

$$x^3-x^2+x+1= (x-r)(x^2+(r-1)x-1/r)
$$

Then, decompose the integrand as
\begin{align}
\frac{x^2+1}{x^3-x^2+x+1}=\frac{A}{x-r}+\frac{Bx+C}{x^2+(r-1)x-1/r}
\end{align}

and integrate accordingly.

For the integral
$$
\int\frac{1}{x^2e^{1/x}-x}dx
\overset{x\to\frac1x}=-\int \frac1{e^x-x}dx
=-\sum_{k\ge0}  \int x^ke^{-(k+1)x}dx
$$

where each term can be integrated separately.

Title: Two strange integrals


问:
Linear map which sends all but finitely many basis vectors to zero is continuous
I was thinking over the following problem for some time, but didn't get much idea:

Let $X,Y$ be normed linear spaces, and let $ f : X \rightarrow Y$  be a linear map. Further assume that there exists a Hamel basis $\mathfrak{B}$ of X having the property that only finitely many elements of $\mathfrak{B}$ are mapped to non zero elements of $Y$. Then is $f$ necessarily continuous?

Any hint will be appreciated. Also, what happens if we assume that $X$ is complete?

答: No, this is not the case, even assuming $X$ is complete. Take your favourite discontinuous linear functional $f$ on your favourite infinite-dimensional normed linear space $X$. The kernels of discontinuous linear functionals on $X$ are codimension-$1$ subspaces of $X$, and hence any Hamel basis of $\operatorname{ker} f$ can be extended to a Hamel basis of $X$ by adding a single vector. Thus, $f$ sends every vector in this basis for $X$ to $0$, except that single added vector, but $f$ is not continuous.


问:
Example of the norm not acting as a euclidean function in $\mathbb{Z}[\sqrt{-3}]$
It is clear that $\mathbb{Z}[\sqrt{-3}]$ is not a UFD and so cannot be a euclidean domain. However, after some consideration of elements of small norm I have not managed to find the norm failing to act in a euclidean manner. That is, I am looking for an example  $a,b \in \mathbb{Z}[\sqrt{-3}] $ such that for all $q,r \in \mathbb{Z}[\sqrt{-3}]$, with  $a=bq+r$, have that $N(r)\geq N(b)$ with $N$ the norm.

答: $R=\mathbb{Z}[\sqrt{-3}]$ is "barely" non-Euclidean (with respect to the usual norm $\sqrt{z\bar{z}}$).  Open unit disks centered at elements of $R$ cover $\mathbb{Q}[\sqrt{-3}]$ except for $R+\frac{1+\sqrt{-3}}{2}$.  So take any $a$, $b$ with $a/b\in R+\frac{1+\sqrt{-3}}{2}$ to get an example.


问:
$f(n)=O(g(n))$ implies $f(n)^2=O(g(n)^2)$
I have trouble proving that $f(n)=O(g(n))$ implies $f(n)^2=O(g(n)^2)$.
I was thinking about using the definition
$$f(n) \leq c \cdot g(n)$$ and then just
$$f(n)^2 \leq c^2 \cdot g(n)^2$$
But that doesn't seem to be correct.

答:
Your intuition is right, you just need to verify it rigorously.

The definition gives
$$\begin{align}
|f(n)| \le Cg(n) \implies |f(n)|^2 &\le Cg(n)|f(n)| \\&\le Cg(n)\cdot Cg(n) \\&= C^2g(n)^2\end{align}$$
Hence, $f(n)^2 = O(g(n)^2)$.


问:
When $X_t = {W_t}^n - k\int^{t}_{0} {W_s}^{n-2} \, ds$ is a martingale?
I know that  $X_t = {W_t}^3 - 3\int^{t}_{0} W_s \, ds$ is a martingale, but my general question is:
for what values of $k$, $X_t = {W_t}^n - k\int^{t}_{0} {W_s}^{n-2} \, ds$ is a martingale?

答:
Hint

You wish to find a function $f\in \mathcal C^2(\mathbb R)$ s.t. $$X_t=\int_0^t f(W_s)\,\mathrm d W_s=W_t^n-k\int_0^t W_s^{n-2}\,\mathrm d s.$$

According to Itô formula
$$\int_0^tf(W_s)\,\mathrm d W_s=f(W_t)-\frac{1}{2}\int_0^t f''(W_s)\,\mathrm d s.$$
I let you conclude.


问:
Doubt in chapter $9$ of Rudin( functions of several variables)
My doubt over here is in page no $212$ of Rudin chapter 9-section (Differentiation)

We are defining the differentiation as :

$f(x+h)-f(x)=f'(x)h+r(h)$ where $r(h)$ is the remainder so that $lim_{h \to o}r(h)/h=0$.

Then it states that :

we are defining the derivative of $f$ at $x$ as a linear function that takes $h$ to $f'(x)h$.

I am not sure of why this statement is true as we are defining $f(x+h)-f(x)$ as the linear function that takes $h$ to $f'(x)h$ and what does this statement even mean(Why do we even have to do this?).I also have a problem in the way he is trying to link things with linear transformation.Can someone just sum it up.

Edit 1:2. Derivative of a linear transformation. If $f:S \to \mathbb{R}$ is linear transformation, then $f (a + hy) =f(a) + hf(y)$ and the difference quotient $\frac{f(a + h) -f(a)}{h}$ is equal to f(y) for every $ h \ne 0$. In this case,
$f’(a; y)$ always exists and is given by
$f’(a; y) =f(y)$
for every $a$ in S and every $y$ in $R”$. In other words, the derivative of a linear transformation
with respect to y is equal to the value of the function at y.

This a paragraph from Apostol.I was just wondering whether this has a link with the way Rudin has defined things?What does the derrivative of a linear transformation mean?

答:
For $f:\mathbb{R}^n\to\mathbb{R}^m$, the derivative of $f$ at $x$ (if it exists) is the unique linear function $L$ such that
$$
\lim_{|h|\to0}\frac{|f(x+h)-f(x)-L(h)|}{|h|}=0.
$$
If $f:\mathbb{R}\to\mathbb{R}$, this reduces to $L$ being multiplication by the number $f'(x)$.

Some examples:

In single variable calculus, to estimate the change in a function near a fixed $x$, we multiply the change in $x$ by $f'(x)$, i.e.
$$
f(x+h)-f(x)\approx f'(x)h=L(h),
$$
and think of this as depending on $h$.

In in multivariable calculus, we have the Jacobian matrix $L=\left(\partial f_i/\partial x_j\right)_{ij}$.

If $f:\mathbb{R}^n\to\mathbb{R}$ we have the gradient $L(h)=\nabla f\cdot h$ for $h\in\mathbb{R}^n$.


问:
False logic argument
I have to proof the following theorem $$(w ⇒ p ∨ s) ∧ (t ∧ p ⇒ r) ⇒ (t ∧ ¬ r ⇒ ¬ q)$$

Using the assumptions, I would like to proof that $(t∧p⇒r)⇒(t∧¬r⇒¬q)$, but I got only $(p⇒r)⇒(t∧¬r⇒¬q)$. I thought I could use $(t∧p⇒r) ⇒ (p⇒r)$, but it seems to be false. How can I modify/finish this proof?

The answer of @DonAntonio is great, but I can't work with this. I would need a complete, linear proof.

答:
I will gently disagree with DonAntonio: a linear proof does not have to be long and tiring, but it may very well be the way you do it (no offense, Leo!).

Your proof

First, let me address your proof. You need to make coarser steps. I don't know what class you're in (it looks like you are using perhaps a Gries and Schneider text), but please don't use contrapositive to turn $p \Rightarrow \neg q$ into $\neg\neg q \Rightarrow \neg p$ and then a separate step to replace $\neg\neg q$ with $q$. It is fine to have one form of contrapositive be simply: $p \Rightarrow \neg q \ \equiv \ q \Rightarrow \neg p$.

A similar comment applies to the steps of your calculation where you replace $s \Rightarrow \neg t$ with $\bf true$, then commute $\bf true$ to the other side of the formula for some reason, then eliminate it using the identity property of $\wedge$. This level of detail is not appropriate for the complexity of proof we are dealing with here.

If we clean up that mess, here is your calculation:

$\quad t \wedge \neg r \;\Rightarrow\; \neg q$

$\Leftarrow \quad \{$ using $q \Rightarrow s \vee p$ and monotonicity $\}$

$\quad t \wedge \neg r \ \Rightarrow \ \neg(s \vee p)$

$\equiv \quad \{$ de Morgan $\}$

$\quad t \wedge \neg r \ \Rightarrow \ \neg s \wedge \neg p$

$\Leftarrow \quad \{$ monotonicity $\}$

$\quad (t \Rightarrow \neg s) \;\wedge\; (\neg r \Rightarrow \neg p)$

$\equiv \quad \{$ contrapositive, twice $\}$

$\quad (s \Rightarrow \neg t) \;\wedge\; (p \Rightarrow r)$

$\equiv \quad \{$ using $s \Rightarrow \neg t$ $\}$

$\quad p \Rightarrow r$

The problem is, we only have $t \wedge p \Rightarrow r$, which is weaker than $p \Rightarrow r$, so this proof is simply not viable. The reason it is not is because of the third step, where you strengthened the calculandum into something too strong to prove.

My general suggestion to you is to be more careful with strengthening steps, and only use equivalence where possible.

A different approach

If you are calculating with booleans or anything else, there is a certain trade-off you have to be aware of. Either you do some work to "clean up" the given formulae, or you will have to do that work while calculating.

To be more specific, let me label the formulae in question:

$(0) \quad q \,\Rightarrow\, s \vee p$

$(1) \quad s \Rightarrow \neg t$

$(2) \quad t \wedge p \,\Rightarrow\, r$

$(3) \quad t \wedge \neg r \,\Rightarrow\, \neg q \quad.$

Now, if I don't wish to do any work up front, I might try to prove $(3)$ by strengthening $\neg q$ into $t \wedge \neg r$:

$\quad \neg q$

$\Leftarrow \quad \{$ using $(0)$ and antimonotonicity $\}$

$\quad \neg (s \vee p)$

$\equiv \quad \{$ de Morgan $\}$

$\quad \neg s \wedge \neg p$

$\Leftarrow \quad \{$ using $(1)$ and antimonotonicity $\}$

$\quad t \wedge \neg p$

$\Leftarrow \quad \{$ using $(2)$ in the form $t \wedge \neg r \,\Rightarrow\, \neg p$ and monotonicity $\}$

$\quad t \wedge t \wedge \neg r$

$\equiv \quad \{$ idempotence $\}$

$\quad t \wedge \neg r$

Personally, I think this is a bad proof: too much work is crammed into each step.

Cleaning up the givens

My taste is to instead do some work up front to clean up the given formulae. First, I use the contrapositive to rewrite $(3)$ as the equivalent:

$(3') \quad t \wedge q \,\Rightarrow\, r \quad.$

Given the shape of $(2)$:

$(2) \hphantom{'} \quad t \wedge p \,\Rightarrow\, r \quad,$

this suggests that we try to prove $(3')$ by strengthening $r$ into $t \wedge q$. We start as follows:

$\quad r$

$\Leftarrow \quad \{ \ (2) \ \}$

$\quad t \wedge p \quad.$

Now we need to strengthen $p$, so we rearrange $(0)$ as the equivalent:

$(0') \quad q \wedge \neg s \,\Rightarrow\, p \quad.$

Thus we can continue our calculation:

$\quad r$

$\Leftarrow \quad \{ \ (2) \ \}$

$\quad t \wedge p$

$\Leftarrow \quad \{ \ (0') \ \}$

$\quad t \wedge q \wedge \neg s \quad.$

Finally, we need to strengthen $\neg s$ so we rewrite $(1)$ as the equivalent:

$(1') \quad t \Rightarrow \neg s \quad,$

and finish the calculation as follows:

$\quad r$

$\Leftarrow \quad \{ \ (2) \ \}$

$\quad t \wedge p$

$\Leftarrow \quad \{ \ (0') \ \}$

$\quad t \wedge q \wedge \neg s$

$\Leftarrow \quad \{ \ (1') \ \}$

$\quad t \wedge q \wedge t$

$\equiv \quad \{$ idempotence $\}$

$\quad t \wedge q \quad.$

The cleanliness and clarity of each step suggests that this was a nice rewriting of the givens. We simply used shunting and contraposition to rewrite everything compatibly. For reference, we used the equations in the following forms:

$(0') \quad q \wedge \neg s \,\Rightarrow\, p$

$(1') \quad t \Rightarrow \neg s$

$(2) \hphantom{'} \quad t \wedge p \,\Rightarrow\, r$

$(3') \quad t \wedge q \,\Rightarrow\, r \quad.$

Note that because of the shape of $(1)$, it was inevitable that one of our symbols would need to be negated. In our case, we used $\neg s$. However, note that we never used $s$ without the negation symbol. Thus, letting $S = \neg s$ we could write everything as:

$(0') \quad q \wedge S \,\Rightarrow\, p$

$(1') \quad t \Rightarrow S$

$(2) \hphantom{'} \quad t \wedge p \,\Rightarrow\, r$

$(3') \quad t \wedge q \,\Rightarrow\, r \quad,$

and see that, by doing the work up front, negation no longer has anything to do with the structure of our argument.

The moral

The moral of the story here is that in any field of mathematical inquiry, it is important to massage our symbols and concepts so they are compatible with each other. Perhaps this is a simple problem from logic, but with a mess of symbols in front of you, even that simple problem confused you. How can we hope to prove complicated theorems in algebra, analysis, etc, if we can't even neatly tackle simple problems?

Be clean, be organized. It's no guarantee of a proof but it's the best we can do.

Appendix

If you are working in a rigid framework that doesn't allow you to do this sort of "cleaning" work up front, here is an alternate, "wide" calculation using the original formulae:

$(0) \quad q \,\Rightarrow\, s \vee p$

$(1) \quad s \Rightarrow \neg t$

$(2) \quad t \wedge p \,\Rightarrow\, r$

$(3) \quad t \wedge \neg r \,\Rightarrow\, \neg q \quad.$

Here is the calculation of $(3)$:

$\quad t \wedge \neg r \,\Rightarrow\, \neg q$

$\Leftarrow \quad \{$ using $(2)$ and $(0)$ and monotonicity $\}$

$\quad t \wedge \neg(t \wedge p) \,\Rightarrow\, \neg(s \vee p)$

$\equiv \quad \{$ de Morgan, twice $\}$

$\quad t \wedge (\neg t \vee \neg p) \,\Rightarrow\, \neg s \wedge \neg p$

$\equiv \quad \{$ absorption on the left side $\}$

$\quad t \wedge \neg p \,\Rightarrow\, \neg s \wedge \neg p$

$\Leftarrow \quad \{$ monotonicity of $\wedge$ $\}$

$\quad t \Rightarrow \neg s$

$\equiv \quad \{$ contrapositive $\}$

$\quad s \Rightarrow \neg t$

$\equiv \quad \{ \ (1) \ \}$

$\quad {\bf true} \quad.$

To clarify, I don't endorse this sort of calculation, generally speaking!

Title: False logic argument


问:
Is the "derivative limit theorem" true  in the higher dimension?
<s>I am wondering if the following statement is a canonical theorem in real analysis. Does anybody here know the exact reference for it? It maybe a corollary of the mean value theorem, I think. Motivated by the answer to this question, I curious about that if this statement is still true in the higher dimension case.</s>

The following statement is a corollary of the Mean Value Theorem.

Let $x_0\in{\bf R}$ and $U(x_0)$ be
  the neighborhood of $x_0$.

$$f:U(x_0)\to{\bf R}$$

is continuous on $U(x_0)$ and
  differentiable on $U(x_0)\setminus\{x_0\}$.
   If the limit

$$\lim_{x\to x_0;x\in
U(x_0)\setminus\{x_0\}}f'(x)$$

exists, then $f$ is differentiable at
  $x_0$ and

$$f'(x_0)=\lim_{x\to x_0;x\in
U(x_0)\setminus\{x_0\}}f'(x)$$

Here is my question:

Is this statement still true for the higher dimension if
$$f:U(x_0)\subset{\mathbb R}^n\to{\mathbb R}^m ?$$

Since I don't know any generalization of the MVT in the higher dimension, I think one may need to approach it directly by definition. On the other hand, if one can deal with it "component-wise", one may be able to reduce it to the one dimension case.

答:
It is an essential feature of modern multivariate calculus that it can and should be done denominator-free.

We may assume that $x_0=f(x_0)=\lim_{x\to 0}f'(x)=0$ and have to prove that $f'(0)$ exists and is $=0$, that is to say: For any $\epsilon>0$ there is a $\delta>0$ with $|f(x)|<\epsilon|x|$ as soon as $|x|<\delta$.

For the proof we have to resort to the mean-value theorem for functions $\phi: [a,b]\to{\mathbb R}$, because in this case it is enough that $\phi$ is differentiable in the interior $]a,b[$ and continuous on all of $[a,b]$.

So let an $\epsilon>0$ be given. By assumption there is a $\delta>0$ such that $\|f'(x)\|<\epsilon$ for all $x$ with $0<|x|<\delta$. ${\it Fix}$ such an $x$. If $f(x)=0$ there is nothing to prove. Otherwise put $u:=f(x)/|f(x)|$ and consider the auxiliary function
$$\phi(t):=u\cdot f(t x)\quad (0\leq t\leq 1)\ .$$
Then by the chain rule we get
$$|f(x)|=|\phi(1)-\phi(0)|=|\phi'(\tau)|=|u\cdot (f'(\tau x).x)|\leq \|f'(\tau x)\| |x|< \epsilon|x|\ ,$$
as required.

ContentLicense: CC BY-SA 3.0


问:
Comparing $2^n$ to $n!$
Comparing $\;2^n\;$ and $\;n!\;$

I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.

Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?

答:
Normally, I would take the logarithm when trying to re-express a power like 2n,
  but I don't think this will work, since I don't know how to take the logarithm of a
  factorial, if at all.

$\log(n!) = \log(1) + \log(2) + \log(3) + ... + \log(n)$

You can approximate the sum by integrating the log function from 1 to $n$.  Recall that the anti-derivative of the natural log is $x \ln{x} - x + C$.  From this, you get log(n!) = O(n log n).  Which is larger than $\log(2^n) = O(n)$.

If you're studying sorting algorithms in computer science, this is why comparison-based sorting algorithms (e.g., quicksort or mergesort) require O(n log n) time: These algorithm can be analyzed as a binary decision tree with $n!$ leaves (the possible permutations of the array).  Taking the (base-2) logarithm gives you the depth of the tree (if it's balanced), which equals the number of comparisons required.

Title: Comparing $2^n$ to $n!$


问:
Introduction to the theory of distributions, Friedlander, Exercise 1.7
Here is the question:

(Sorry that I don't know how to type Latex, so I include the question here as an image.)

I have proved the first part but I am stuck on the second. The question seems to ask me to prove that (b) holds for all epsilon between 0 and 1.

The fact that the integral of (b) tends to 1 is obvious as epsilon goes to 0 but I don't know how to prove (b) for all epsilon between 0 and 1.

Anyone have an idea?

Thanks a lot.

答:
That implication is obvious nonsense.

Say $(f_\epsilon)$ satisfies (a) and $f_\epsilon\to\delta$. Define $$g_\epsilon=\begin{cases}2f_\epsilon&(\epsilon=1/2),\\f_\epsilon,&(\epsilon\ne1/2).\end{cases}$$Then $(g_\epsilon)$ also satisfes (a) and $g_\epsilon\to\delta$, but $(f_\epsilon)$ and $(g_\epsilon)$ cannot both satisfy (b).


问:
Why are pull backs always defined for 1 forms but not push forwards?
So I know that given a differentiable function $\mathcal{F}: \mathcal{M} \rightarrow \mathcal{N}$, we can define the push forward map:
$$
\mathcal{F}_*: T_p(\mathcal{M}) \rightarrow T_{\mathcal{F}(p)}(\mathcal{N})
$$

However, the pull back map for vectors would require the inverse function $\mathcal{F}^{-1}$ which may not necessarily exist and this is why we cannot define a pull back for vector fields.

Now coming to forms, I do not understand why the pull back is always defined while the push forward is not? That is, given that I know $\langle \omega, v \rangle$, shouldn't I be able to define $\langle \mathcal{F}_* (\omega), \mathcal{F}_*(v)\rangle$? Why is this not possible?

答:
Your question really comes down to sets, and functions, and compositions of functions.

Let's just do 1-forms.

A differential 1-form $\omega$ on $\mathcal M$ is a certain kind of function whose domain is the set $T \mathcal M$: it inputs a tangent vector on $M$ and outputs a number, i.e. it outputs an element of the set $\mathbb R$:
$$(1) \qquad \omega : T \mathcal M \to \mathbb R
$$
A differential 1-form $\rho$ on $\mathcal N$ similarly inputs a tangent vector on $N$ and outputs a number:
$$(2) \qquad \rho : T \mathcal N \to \mathbb R
$$

Now consider a smooth function $\mathcal F : \mathcal M \to \mathcal N$ with corresponding pushforward map
$$(3) \qquad \mathcal F_* : T \mathcal M \to T \mathcal N
$$
How can we chain these function diagrams together? What choices to we have for composing these functions?

Well, first of all, we can easily compose (2) and (3) to get
$$T \mathcal M \xrightarrow{ \mathcal F_*} T \mathcal N \xrightarrow{\rho} \mathbb R
$$
and that's the definition of the pullback $\mathcal F^*(\rho) : T \mathcal M \to \mathbb R$.

But how do you propose composing (1) and (3)? There's simply no composition to be made. And that's why the "pushforward" of forms is not defined. Whether or not $\mathcal F$ has an inverse is irrelevant.


答:
It seems you misunderstood the definition of the differential  of a function: it is the best linear approximation $l(h)$  of $f(x+h)-f(x)$ in a very precise sense, based on asymptotic analysis:
$$f(x+h)=f(x)+l(h)+o(\Vert h\Vert).$$

Title: Doubt in chapter $9$ of Rudin( functions of several variables)

Body:
My doubt over here is in page no $212$ of Rudin chapter 9-section (Differentiation)

We are defining the differentiation as :

$f(x+h)-f(x)=f'(x)h+r(h)$ where $r(h)$ is the remainder so that $lim_{h \to o}r(h)/h=0$.

Then it states that :

we are defining the derivative of $f$ at $x$ as a linear function that takes $h$ to $f'(x)h$.

I am not sure of why this statement is true as we are defining $f(x+h)-f(x)$ as the linear function that takes $h$ to $f'(x)h$ and what does this statement even mean(Why do we even have to do this?).I also have a problem in the way he is trying to link things with linear transformation.Can someone just sum it up.

Edit 1:2. Derivative of a linear transformation. If $f:S \to \mathbb{R}$ is linear transformation, then $f (a + hy) =f(a) + hf(y)$ and the difference quotient $\frac{f(a + h) -f(a)}{h}$ is equal to f(y) for every $ h \ne 0$. In this case,
$f’(a; y)$ always exists and is given by
$f’(a; y) =f(y)$
for every $a$ in S and every $y$ in $R”$. In other words, the derivative of a linear transformation
with respect to y is equal to the value of the function at y.

This a paragraph from Apostol.I was just wondering whether this has a link with the way Rudin has defined things?What does the derrivative of a linear transformation mean?


问:
Use the quantifiers to give the logical form of the following statement. Also negate it and prove it True or False.
For almost all great enough natural numbers $n$ we have:
$$\frac{n-2}{n^2-4n+2}<0.07$$

I am not sure the "for almost all" here. Does it mean there exists great enough n for which the statement is not satisfied:
$$\exists N\in \mathbb{N}: \exists n≥N:\frac{n-2}{n^2-4n+2}≥0.07$$
Or just simply there exists some great enough n for which the statement is satisfied:
$$\exists N\in \mathbb{N}: \exists n≥N:\frac{n-2}{n^2-4n+2}<0.07$$
Or something else?

答:
What's meant is

$$\exists N\in \mathbb{N}: \forall n\ge N:\frac{n-2}{n^2-4n+2}<0.07$$

so that the statement holds for all $n$, except at most finitely many (the ones that are $<N$ for which we have no info)

The negation is

$$\forall N\in \mathbb{N}: \exists n\ge N:\frac{n-2}{n^2-4n+2}\ge 0.07$$

which implies there are infinitely many $n$ for which the inequality would fail.

But the first one is true of course and follows from the well-known fact that $$\lim_{n \to \infty} \frac{n-2}{n^2 + 4n +2} = \lim_{n \to \infty}\frac{\frac{1}{n} - \frac{2}{n^2}}{1 + \frac4n + \frac{2}{n^2}} = \frac{0+0}{1+0+0} = 0$$ using that $\frac1n \to 0$ and continuity of arithmetic operations.


问:
The cohomology ring for disjoint union and wedge sum
I have a question concerning example 3.14 of Hatcher's Algebraic Topology. There are a few details that confuse me, so I suppose the goal of my questions is more or less clarification - I apologize if they seem trivial. The first (ring)-isomorphism written is
$$H^*(\coprod_\alpha X_\alpha;R)\stackrel{\cong}{\longrightarrow}\prod_\alpha H^*(X_\alpha;R)\quad(*)$$
Have I understood it correctly when I say that the argument is based on the fact that the inclusions $i:X_{\alpha_0}\hookrightarrow\coprod_\alpha X_\alpha$ induce, on the level of groups, the isomorphism $(*)$ and since continuous maps induce morphisms that respect the $\cup$-product, then $(*)$ is in addition in fact a ring-isomorphism?

If so, then I can not quite understand why we need he additional assumption on the pair $(X_\alpha,x_\alpha)$ when arguing that we have an isomorphism similar to $(*)$ for the wedge sum. Consider the following reasoning: On the level of groups we have
$$H^*(\bigvee_\alpha X_\alpha;R):= \bigoplus_{i=0}^\infty H^i(\bigvee_\alpha X_\alpha;R)\stackrel{\prod i^*}{\to}\prod_\alpha\bigoplus_{i=0}^\infty H^i(X_\alpha;R)=\prod_\alpha H^*(X_\alpha;R)$$
where $\prod i^*$ is induced by the inclusion forming an isomorphism by the Eilenberg-Steenrod axiom for the wedge sum. Again, since continuous maps induce ring-morphisms, the isomorphism is in fact a ring-isomorphism and the claim follows. Of course something is wrong because I do not apply the additional assumption Hatcher makes. If anyone would care to point out where my mistake is and explain how one implement the assumption correctly, I would appreciate that very much. Thank you in advance!

答:
"is induced by the inclusion forming an isomorphism by the Eilenberg-Steenrod axiom for the wedge sum"

No it is not. The Eilenberg-Steenrod axioms in Hatcher are stated for CW-pairs, for which the "well-pointed axiom", that there is a neighborhood of $x_\alpha$ deformation retracting to it, is true for an arbitrary point in the CW complex.

The statement you want to prove relies on an argument precisely dual to the one for homology. In homology Hatcher's book states this assumption: see Corollary 2.25. It is the assumption that $(X, x_\alpha)$ is a "good pair".

If you want to state Eilenberg-Steenrod axioms outside the category of CW-pairs you need "good-pair" type assumptions.


问:
Limit of a function $f(x,y)$ within $\mathbb{R}^{2}$ that has a different defintion when $x$ and $y$ are both rational
I have come to a problem in a multivariable calculus text book that I'm having trouble solving.

The problem goes :

"Let $f : \mathbb{R}^{2} \rightarrow \mathbb{R}$ be defined by $f(x,y) := x^{2} + y^{2}$ if
both $x$ and $y$ are rational, and $f(x,y) := 0$ otherwise. Determine the points of
$\mathbb{R}^{2}$ at which (i.) $f_{x}$ exists, (ii.) $f_{y}$ exists."

Here $f_{x}$ is the partial derivative with respect to $x$ and $f_{y}$ is the partial
derivative with respect to $y$. I know $f_{x}$ is given by the following limit :
\begin{equation}
f_{x}(x,y) = \lim_{h \rightarrow 0} \frac{ f(x+h,y) - f(x,y) }{h}
\end{equation}
Now let the irrational numbers be denoted $\mathbb{F} = \mathbb{R} \setminus \mathbb{Q}$.
I know from the internet :
\begin{align}
a,b \in \mathbb{Q} & \Rightarrow a + b \in \mathbb{Q}\\
a \in \mathbb{Q} \text{ and } b \in \mathbb{F} & \Rightarrow a + b \in \mathbb{F} \\
a,b \in \mathbb{F} & \Rightarrow a + b \in \mathbb{F}
\end{align}
Now let's assume for example $(x_{r},y_{r}) \in \mathbb{Q}^{2} \subset \mathbb{R}^{2}$.
So we have :
\begin{align}
f_{x}(x_{r},y_{r})
  & = \lim_{h \rightarrow 0} \frac{f(x_{r}+h,y_{r}) - f(x_{r},y_{r})}{h}\\
  & = \lim_{h \rightarrow 0} \frac{f(x_{r}+h,y_{r}) - (x_{r}^{2} + y_{r}^{2})}{h}
\tag{1}
\end{align}
We know
\begin{align}
h \in \mathbb{Q} & \Rightarrow f(x_{r}+h,y_{r}) = (x_{r}+h)^{2} + y_{r}^{2} \\
h \not \in \mathbb{Q} & \Rightarrow f(x_{r}+h,y_{r}) = 0
\end{align}
This is where I am stuck. I am not sure how to evaluate the limit in equation (1) to
determine if it exists.

(and can someone show me how to properly use the tag and ref commands so that I can tag
equation (1) and reference it correctly).

答:
After arriving at
$$\begin{align}
f_{x}(x_{r},y_{r})
  & = \lim_{h \rightarrow 0} \frac{f(x_{r}+h,y_{r}) - f(x_{r},y_{r})}{h}\\
  & = \lim_{h \rightarrow 0} \frac{f(x_{r}+h,y_{r}) - (x_{r}^{2} + y_{r}^{2})}{h}
\end{align}$$
note that if this limit exists, then it exists for all $\{h_n\}$ such that $h_n \to 0$ and converges to the same value.

First, let $h_n \in \mathbb Q$. Then,
$$\begin{align}
\lim_{n \rightarrow \infty} \frac{f(x_{r}+h_n,y_{r}) - f(x_{r},y_{r})}{h_n}
&= \lim_{n\to\infty} \frac{(x_r+h_n)^2 + y_r^2 - x_r^2-y_r^2}{h_n} \\
&= \lim_{n\to\infty}\frac{h_n(2x_r+h_n)}{h_n} \\
&= 2x_r
\end{align}$$
Now, let $h_n \in \mathbb R \setminus \mathbb Q$. Then, $x_r + h_n \in \mathbb R \setminus \mathbb Q$ and therefore
$$\lim_{n \rightarrow \infty} \frac{f(x_{r}+h_n,y_{r}) - f(x_{r},y_{r})}{h_n} = \lim_{n\to\infty} \frac{ - x_r^2-y_r^2}{h_n} = \infty$$
Hence, the limit does not exist at $(x_r, y_r) \in \mathbb Q^2$.

Otherwise, i.e. if $(x,y) \notin \mathbb Q^2$, the limit exists because
$$\lim_{h\to0}\frac{f(x+h, y) - f(x,y)}{h} = \lim_{h\to 0}\frac{0}{h} = 0$$


问:
Matrix operator $p$-norm and eigenvalues
Denote by $\|\cdot\|_p$ the operator $p$-norm for matrices in $\mathcal M_{d\times d}(\mathbb R)$.

Let $a_n \to 0$ be a sequence of positive real numbers, and $\mathbf A$ a positive definite matrix.

We know that when $p=2$, $\|\mathbf I-a_n \mathbf A \|_2 = 1-a_n \lambda_{\min}(\mathbf A)$ when $n$ is large.

Question: When $1<p<2$ or $p>2$, is there a constant $K$, such that

$$  \|\mathbf I-a_n\mathbf A \|_p \leqslant 1-Ka_n $$

holds for large $n$?

答:
I finally solved this problem in my paper.

Behold the preprint of it (see Section 3.1 onwards): https://arxiv.org/abs/2102.10346


问:
Why do we prove that $x_1$ = $x_2$ when looking if a function is injective?
For a function to be considered injective, we consider that $\forall x_1, x_2 \in X, x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$.

Now, when it comes to proving this, we prove that $\forall x_1, x_2 \in X, x_1 = x_2 \Rightarrow f(x_1) = f(x_2)$, and if it turns out that $x_1 = x_2$, that means that the function is injective, but injectivity means that every $x$ has maps to only one, unique $y$, and we proved that it is not. (?)

So, why do we try to prove the opposite, and if it proves correct we consider it injective?

答: To prove "Any two different inputs give two different outputs" it is enough to prove the equivalent statement: "if two inputs give the same output, then actually they are the same input".</p>


问:
Does it exist a function for which the derivative changes sign more than countably many times?
Does there exist any function $f \in C^2[0,1]$; $f: [0,1] \mapsto [0,1]$, for which the derivative changes sign more than countably many times?</p>

答: How about a function $f'$ whose zero set is the Cantor set.  And on complementary intervals with length $3^{-k}$, $k$ odd, the function is positive, while on complementary intervals with $3^{-k}$, $k$ even, the function is negative.  Then could we say $f'$ "changes sign" at each point of the Cantor set (since there are positive values and negative values clustering at that point)?


问:
Velocity as a function of position, given velocity as a function of time
Knowing the expression for the acceleration as a function of time:
$$ \frac{dv}{dt} = - c v^n$$
(for some constant c >0 and n >1), one needs to find the velocity as a function of time and as a function of position. Solving for the velocity as a function of time is pretty straightforward and has lead me to the following:
$$ v(t) = [ (n-1)ct + v_{O}^{(1-n)}]^{\frac{1}{1-n}} $$
Given the initial condition that $ v(0) = v_0$. Now, to find $v(x)$, I tried to use

$$ \frac{dv}{dx} = \frac{dv}{dt} \frac{dt}{dx} = a(t) \frac{1}{v(t)} $$
This leads to
$$ \int \frac{dv}{v^n} = \int \frac{-c}{ [ (n-1)ct + v_{O}^{(1-n)}]^{\frac{1}{1-n}}} dx = -ct $$
Where the last equality is obtained from the first DE after separating variables.  I don't see how this could possibly give me $v(x)$. How could one change variables in the integral without being in the possession of a relationship between $t$ and $x$ (Or am I? Am I overseeing  something?), which if I had I could simply insert into the RHS of the equation, without having to worry about the integral at all.
Any push into the right direction is appreciated.

答:
I want to emphasize that the '$t$' in the fourth equation should be treated as a function of $x$(displacement).

Let me just rewrite your third formula with a input $\tilde{x}$ (You probably have applied the inverse function theorem.) :
$$
\frac{dv}{dx} (\tilde{x})= \frac{dv}{dt}(\,t(\tilde{x})\,)\,\, \frac{dt}{dx} (\tilde{x}) = a( t(\tilde{x}) ) \frac{1}{\frac{dx}{dt} (t (\tilde{x}))} =  \frac{a( t(\tilde{x}) )}{v (t (\tilde{x}))}
$$

, where $t(\tilde{x})$ is the inverse function of the time-displacement function.

This slightly different notation lets us understand the integral more precisely. Indeed, your integral is precisely
$$
\int_{v_0}^{v(\tilde{x})} \frac{dv}{v^n} = \int_{0}^{\tilde{x}} \frac{-c}{ [\, (n-1)c\,\,t(x) + v_{O}^{(1-n)}\,]^{\frac{1}{1-n}}} dx \quad \text{}.
$$
i.e.
$$
\frac{1}{1-n}  \left( v(\tilde{x})\,^{1-n} \,-\,v_0\,^{1-n} \right) = \mathrm{D}(\tilde{x}).
$$

( D is the result of our integral. (i.e our Dream function) )

Then our $v(x)$ can be derived as follows.
$$
v(x) = \left[ {v_0}^{1-n} - (n - 1)\, D(\tilde{x}) \right]^{1 / {1-n}}\, .
$$


答:
I'm putting my second reply to @DonAntonio into its own answer. If we take a substitution-of-values approach, the ideal candidate to me seems to be to assume $t$ — that is, to let $t = \bf true$.

Then our givens reduce to:

$q \,\Rightarrow\, s \vee p$

$\neg s \quad$ (that is, $s = \bf false$)

$p \Rightarrow r$

and our demonstrandum reduces to:

$\neg r \Rightarrow \neg q \quad.$

But from $s = \bf false$, the first given reduces to $q \Rightarrow p$, which in combination with $p \Rightarrow r$ yields $q \Rightarrow r$, which is just the contrapositive of what we're trying to show.

Of course, if $\neg t$ then our demonstrandum is vacuously true. (This was the reason we chose to perform a case analysis on $t$.) So the theorem is proved in any case.

Title: False logic argument

Body:
I have to proof the following theorem $$(w ⇒ p ∨ s) ∧ (t ∧ p ⇒ r) ⇒ (t ∧ ¬ r ⇒ ¬ q)$$

Using the assumptions, I would like to proof that $(t∧p⇒r)⇒(t∧¬r⇒¬q)$, but I got only $(p⇒r)⇒(t∧¬r⇒¬q)$. I thought I could use $(t∧p⇒r) ⇒ (p⇒r)$, but it seems to be false. How can I modify/finish this proof?

The answer of @DonAntonio is great, but I can't work with this. I would need a complete, linear proof.


问:
If two vectors are orthogonal then $\|a + b\|^2=\|a\|^2 + \|b\|^2$
Can we say that if $a$ and $b$ are orthogonal (meaning $\langle a,b \rangle = 0$) then

$\|a + b\|^2=\|a\|^2 + \|b\|^2$

must be correct?

Reasoning:

$\|a+b\|^2 = \langle{a+b,a+b}\rangle \underset{Orthogonality}{=} \langle{a,a}\rangle + \langle{b,b}\rangle = \|a\|^2 + \|b\|^2$

答: Take @GEdgar's advice. In particular, $\langle a+b,\,a+b\rangle=\langle a,\,a\rangle+\langle a,\,b\rangle+\langle b,\,a\rangle+\langle b,\,b\rangle$ by bilinearity, but two of the terms are $0$.


问:
Composing words from a limited letter supply
A typesetter working with an $n$-letter alphabet has $k$ copies of each letter in his type case; how many $n$-letter words can he compose?

What is the number of $n$-letter words composed from an $n$-letter alphabet such that no letter appears more than $k$ times?

Denoting this number by $T_k(n)$, we have $T_1(n)=n!$ in a trivial way.

For $k=2$ we have, say, $T_2(1)=1$, $T_2(2)=4$, and $T_2(3)=24$; subsequent terms can be computed from
$$ T_2(n) = \sum_{k=0}^{n/2} \frac{(n!)^2}{2^k(k!)^2\,(n-2k)!}. $$
The sequence $T_2(n)$ is OEIS A012244. Asymptotically,
$$ T_2(n) = C\,(1+\sqrt 2)^n\, (n/e)^n (1+o(1)) $$
where $C=(1+1/\sqrt 2)^{1/2}$.

What happens for $k\ge 3$? It would be great to have an asymptotic formula with an explicit remainder term; that is, without assuming that $k$ is fixed.

答:
This problem is discussed in overwhelming detail in the following paper on arXiv: Coalescence under preimage constraints by Benjamin Otto. The relevant result is Corollary $7.8$, which implies that for fixed $k\ge 2$,

$$
T_k(n)=n!(\rho_k)^{-n}n^{-1/2}(2\pi \tau_k\rho_k \cdot e_{k-2}(\tau_k))^{-1/2}(1+o(1))\qquad \text{as }n\to\infty
$$
where

$
e_k(x)=1+x+x^2/2!+\dots+x^k/k!,$

$\tau_k$  is the unique $x\in \mathbb R_{+}$ for which $e_k(x)-te_{k-1}(x)=0,$

$\rho_k=\tau_k/e(\tau_k)$.

I have not checked the details, but this agrees with your $k=2$ result.

I am paraphrasing their notation. They handle the more general case where the number of times each letter is allowed to appear is a set $\mathcal P$, when you are only concerned with $\mathcal P=\{0,1,\dots,k\}$. Whereas they have $e_{\mathcal P}(x)=\sum_{i\in \mathcal P}x^i/i!$, I replaced this with $e_k(x)$.


问:
$E$ is open iff union of open balls
Let $(A, d)$ be a metric space and $\emptyset \neq E \subset A$ a subset. Define $B(a) = \{b \in A : d(a, b) < \epsilon\}$. Show that $E$ is open $\iff$ it can be written as the union of a family of open balls of the form $B(a)$ above.

I have a general proof for an open set involving the union of balls:

Theorem: If $(A, d)$ is a metric, then $E \subset A$ is open $\iff$ it is the union of open balls.

Suppose $E$ is open then by definition; for every $x \in E$, there is some $r>0$ such that $B(x,r) \subset E$. A claim: $E = \bigcup_{x\in E} B(x,r)$. Choose $x \in E$ then can find $r>0$ such that $x \in B(x,r) \subseteq \bigcup B(x,r)$. Suppose, by a converse argument, $y \in \bigcup B(x,r) \implies y \in B(x,r) $ for some $x$. But $B(x,r) \subseteq E$ for some $x$ so $y \in E$. So $E$ is the union of open balls.
Going the other way, suppose $E = \bigcup_{\phi} O_{\phi} $ where $O_{\phi}$ is open ball. Take $x \in E$, then $x \in O_{\phi} $ for some $\phi$. But $O_{\phi} \subseteq \bigcup O_{\phi} = E$. So, we have found an open ball inside $E$, and $E$ must be an open set.

QED.

Unfortunately I am not sure how to do the problem with the specific form $B(a)$. Can somebody show me how to do this problem? Any hints or assistance is appreciated.

答:
Assuming that in your original problem (stated at the top), $a$ is a fixed point (if it is not fixed then I don't understand what $B(a)$ is supposed to denote):

The theorem is clearly wrong.

Counterexample:

Let the metric space be $\mathbb{R}$ and let $E = (1,2) \cup (3,4)$. $E$ is clearly open however, it cannot be union of family of open balls of form $B(a)$ since such a union must be a open segment centered at $a$.

This is probably not what you meant, so maybe you can edit the question to make it clear.

Title: $E$ is open iff union of open balls


问:
Does every non-archimedean absolute value on field take value in $\mathbb{Q}$
Let $K$ be a field, a non-archimedean absolute value is defined to be a map $K\to \mathbb{R}$ satisfying $|x|=0\Rightarrow x=0$, $|x|\cdot|y|=|xy|$ and $|x+y|\leq\max(|x|,|y|)$.

Is there an example that the valuation of a non-archimedean field not take value in $a^\mathbb{Q}$ for any fixed
real number $a$?

答: Think to $x^{\sqrt{2}}$ as a function $(0,\infty)\to \Bbb{R}$, no problem to look at the ring $\Bbb{R}[x,x^{\sqrt2}]$ and its fraction field $K$ with the natural valuation $$v(\sum_{n,m}c_{n,m}x^{n+m\sqrt2})=\inf \{ n+m\sqrt2, c_{n,m}\ne 0\}, \qquad |f/g|=2^{v(f)-v(g)}$$


问:
How to show that $1 + p$ has order $p^{n-1}$ in $(\mathbb{Z}/p^n \mathbb{Z})^{\times}$ for $p$ an odd prime?
$(\mathbb{Z}/p^n \mathbb{Z})^{\times}$ is the group of invertible elements of $\mathbb{Z}/p^n \mathbb{Z}$.

To find the order of $1+p$, we start by expanding $(1 + p)^{p}$ using the binomial theorem:

$(1 + p)^p = 1 + p \cdot p + {p \choose 2}\cdot p^2 + \dots \equiv 1 + p^2 \pmod {p^3}$. Similarly,

$(1 + p)^{p^2} \equiv 1 + p^3 \pmod {p^4}$

$\cdots$

$(1 + p)^{p^{n-1}} \equiv 1 + p^{n} \pmod{p^{n+1}}$.

I'm thrown off by $\pmod{p^{n+1}}$. I need $p^{n}$ where there is $p^{n+1}$.

答: If you accept those identities, then $(1+p)^{p^{n-1}} \equiv 1+p^n \pmod{p^{n+1}}$ implies $(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n}$, so the order divides $p^{n-1}$, so the order is a power of $p$, but the lower powers of $p$ cannot give $1$ by the lower identities.


问:
Properties of Entropy
When someone writes $H(X_1, X_2, X_3) = H(X_1) + H(X_2\mid X_1) + H(X_3\mid X_2, X_1)$, how should that last term be interpreted/read? As the joint entropy between 2 variables where variable 1 is $X_3\mid X_2$ and variable 2 is $X_1$? Or As the entropy of $X_3$ conditioned on both $X_2$ and $X_1$? In other words is it:

$$H[ (X_3\mid X_2) , (X_1) ]\text{ or }H[ (X_3) \mid (X_2, X_1) ]$$

Are they the same? If so could someone show me how? If not, could someone tell me the correct way to read/interpret/write both possible interpretations?

答: The intended interpretation in this formula is $H(X_3\mid (X_2,X_1))$.  Your other option won't work because $X_3\mid X_2$ is not a random variable.</p>

Title: Properties of Entropy


问:
How to find the slope of certain lines matching random points
I have a large set of random points, $(x_i,y_i)$, which, for unknown reasons, seem to align along certain lines:

I want to calculate the slope of those lines (also their other parameters, but I'm asking about the slope here)

Separating the lines algorithmically, seems to be too complex. I guess that I would need a classification algorithm, and those never give stable results.

That's why I am only talking about the slope.

What I did:

I made a synthetic sample generating random lines:

The average point $\color{blue}{(c_x,c_y)}$ is drawn in blue, and on this chart, the lines have $-45$° angle.

I have the intuition that the sum of all points distances to a line passing by the blue point, should be maximal/minimal when the line matches the slope I'm looking for (or the perpendicular slope).

I calculated the distance summed of all points to a line passing for the center, with angle $\alpha$, for different angles (from $0$ to $360$), and I got this chart:

I drawn the point lines at $-45$°, and the distance seems to be maximal just around the normal angle $+45$, so it suggest that I could use a maximization solver to find the slope normal to the lines, by maximizing the distance of all points to a line passing by the center (blue point).

It seems to work for different angles I tried, but I'm not sure if that's a correct procedure.

Also, the distance is not minimal when the slope $\tan(\alpha)$ is parallel to the lines.

EDIT: I just noticed that if the points are clustered in two distant groups, independent of the orientation of the lines, the maximum distance would be the one separating the groups.
Maybe Fourier transforms would be able to detect the orientation of the lines?

答:
The method employed by us:

Convert the picture issued by the OP to Windows Bitmap format

Load the BMP file into a computer program, developed for the purpose

Map the color content onto real (double precision) numbers

Define a zero level of this mapping; in our case white = - 0.2

Define empirically limitations for the area that looks interesting

Use a contouring module for making isolines at level zero

Use all sort of limitations for distinguishing relevant data from the rest

Calculate second order moments (variances) for the relevant isolines

Take a look at the theory in the webpage Two-dimensional Moments

Especially take notice of the formula $\;\tan(2 \theta) = 2 \sigma_{xy}/(\sigma_{xx} - \sigma_{yy})$

Calculate mean value $\pm$ spread of relevant angles $\alpha=-\theta$ and Output

The numerical end result obtained is:
```
Alpha = 38 +/- 2 degrees
```
Picture produced:

Note.

A more robust method may be to consider instead the angles of the minor axes of the ellipses of inertia $\;\sigma_{yy}(x-\mu_x)^2-2\sigma_{xy}(x-\mu_x)(y-\mu_y)+\sigma_{xx}(y-\mu_y)^2=\sigma_{xx}\sigma_{yy}-\sigma_{xy}^2\;$ with the x-axis.

Free accompanying source-only software (Delphi Pascal) has been made available at the site:
MSE publications / references 2021

Disclaimer. Anything free comes without guarantee :-(and without referee)

Body:
The method employed by us:

Convert the picture issued by the OP to Windows Bitmap format

Load the BMP file into a computer program, developed for the purpose

Map the color content onto real (double precision) numbers

Define a zero level of this mapping; in our case white = - 0.2

Define empirically limitations for the area that looks interesting

Use a contouring module for making isolines at level zero

Use all sort of limitations for distinguishing relevant data from the rest

Calculate second order moments (variances) for the relevant isolines

Take a look at the theory in the webpage Two-dimensional Moments

Especially take notice of the formula $\;\tan(2 \theta) = 2 \sigma_{xy}/(\sigma_{xx} - \sigma_{yy})$

Calculate mean value $\pm$ spread of relevant angles $\alpha=-\theta$ and Output

The numerical end result obtained is:

Alpha = 38 +/- 2 degrees

Picture produced:

Note.

A more robust method may be to consider instead the angles of the minor axes of the ellipses of inertia $\;\sigma_{yy}(x-\mu_x)^2-2\sigma_{xy}(x-\mu_x)(y-\mu_y)+\sigma_{xx}(y-\mu_y)^2=\sigma_{xx}\sigma_{yy}-\sigma_{xy}^2\;$ with the x-axis.

Free accompanying source-only software (Delphi Pascal) has been made available at the site:
MSE publications / references 2021

Disclaimer. Anything free comes without guarantee :-(and without referee)


问:
What is it that determines the degree of a tile, and to what extent does the shape of a tile determine its degree?
I'd be interested to learn if/what the geometric or algebraic approach to acquiring a tile of degree $2, 3, \ldots, n$ (i.e. a tile of an arbitrary degree) would be? Asked another way- what is it that determines the degree of a tile, and to what extent does the shape of a tile determine its degree?

By degree of a tile, I mean the number of other tiles which are adjacent to it, and by adjacency, I mean tiles which share a common boundary.

And I am referring to any tiling of the plane initially - though maybe it would be simpler to look at a regular tesselation of the plane, first. Then, would there be a finite or infinite number of ways of dividing the plane to acquire tiles of degree $2, 3, \ldots, n$? A related question could be, to what extent does the type of tiling affect my question? Presumably the degree of a tile and the symmetry group must be interconnected?

答:
The degree of a tile is not determined by its shape. Consider the following two  tilings of $1\times 2$ rectangles:

One of these has degree $4$ at every tile, and one has degree $6$ at every tile.

If all tiles meet along some portion of their border, and not just at a corner, then (so long as there are finitely many shapes of tile used) the average degree will always be $6$; this is a consequence of Euler's formula for planar graphs.


问:
Are there densely defined symmetric operators from a Banach space into its algebraic dual which do not map into the continuous dual?
The notion of symmetric operator on Hilbert space can be generalized to locally convex spaces as follows: A map $T \colon D \to X^a$ between a subspace $D$ of a locally convex topological vector space $X$ and its algebraic dual $X^a$ is called symmetric, if

$$(Tx,y)_{X^a,X} = (x,Ty)_{X,X^a}$$,

where the brackets denote duality pairings. This condition automatically ensures that $T$ is linear.

For the special case where $X$ is a Banach space $B$ and $T$ is defined a) on a dense linear subspace or b) on all of $B$, does this symmetry condition ensure that $T$ must map into the continuous dual $B^{\ast}$?

As the trivial example of an operator defined on $\{0\}$ shows, this is not true in general.

Here are some related results:

If $T \colon B \to B^{\ast}$ is symmetric, then it is continuous (this is a generalization of the Helling-Toeplitz theorem on Hilbert spaces and can be found as Proposition 1.1 on p. 145 in the book "Probability on Banach spaces" by Vakhania, Tarieladze and Chobanyan). From this

If  $T \colon B^{\ast} \to B^{\ast a}$ is the covariance operator of some Borel measure on $B$ (so in particular positive symmetric), then $T$ maps into $B^{\ast\ast}$. This is claimed (referring to an article in Russian and not easily available) on page 12, before stating Theorem 6 of Covariance operators of probability measures in locally convex spaces by Vakhania and Tarieladze.

答:
Let $H$ be any infinite-dimensional Hilbert space, and let $\psi:H\to\mathbb C$ be an unbounded linear functional (these exist in ZFC). Define $T:H\to H^a$ by
$$
Tx=\psi(x)\,\psi.
$$
Then $T$ is symmetric, defined everywhere, and $TH\subset H^a\setminus H^*$.

If you don't want Choice, then you lose a lot of functional analysis; but in any case the above can be done on a dense subspace.


问:
Intermediate Value Theorem Question to show another function:
How would I go about showing this question:

Suppose that the function $f:[0,1] \rightarrow [0,1]$ is continuous.  Use the Intermediate Value Theorem to prove that there exists $c \in  [0,1]$ such that $$f(c) = c(2-c^2)$$

My thoughts are we can define a function say $g(x) := f(x) - x(2-x^2)$ and then $g(0) = f(0) $and then $g(1) = f(1) - 1 $ and then I'm stuck on how to complete it to get to the point where

$$g(0) < 0 < g(1)$$ so there would exist a c s.t. $g(c) = 0$ [by IVT] and thus $f(c) = c(2-c^2)$

If anyone could help, would be much appreciated thank you!

答: Hint: Try to prove instead that $g(0) \ge 0 \ge g(1)$.


问:
Bounds on triple integral (Cartesian)
I want to setup a triple integral for the volume of the surface in the ordering $dy \hspace{1mm} dx \hspace{1mm} dz$:

So far I have that for $0\leq z \leq 1, 0 \leq y \leq x$ and for $1 \leq z \leq 2, 0 \leq y \leq \sqrt{2-z}$. I'm having trouble setting up bounds for $x$. It looks from the projection like $0 \leq x \leq 1$ for both integrals, but it doesn't give me the right value for volume (should be $\frac{11}{12}$ based on the other differential orderings.)

答:
For $0 \leq z \leq 1$, the volume is simply bound by $y = 0$, $y = x$ and $x=1$.

For any given value of $z \in (1,2)$, there are two bounds of $y$: below $x = \sqrt{2-z}$, it is simply bound by plane $y = x$ and above $x = \sqrt{2-z}$, by the parabolic cylinder.

So the integral should be

$\displaystyle \int_0^1 \int_0^1 \int_0^x \ dy \ dx \ dz$ +

$\displaystyle \int_1^2 \int_0^{\sqrt{2-z}} \int_0^x \ dy \ dx \ dz$ +

$\displaystyle \int_1^2 \int_{\sqrt{2-z}}^1 \int_0^{\sqrt{2-z}} \ dy \ dx \ dz$

Title: Bounds on triple integral (Cartesian)


问:
About smooth approximation in a Sobolev space
I want to prove the following fact :

Consider $\Omega \subset R^n$ a bounded and open set. Let $v \in H^{1}_{0}(\Omega)$ a nonnegative function. Then exists a sequence $v_m$ in $C^{\infty}_{0}(\Omega)$ of nongative functions converging to $v$ in $H^{1}( \Omega)$.

I know how to prove this fact (I think this fact can help): If $u_m$ is a sequence converging to $u$ in $H^1(\Omega)$ , then ${u^{+}_m} \rightarrow  u^+$.

Someone can give me a hint ?

Thanks in advance

答:
By your known fact, if the definition of $u^+$ is
$$
u^+ = \begin{cases} u \quad \text{ when }u\geq 0,
\\
0 \quad \text{ when }u< 0.\end{cases}
$$
Then for any non-negative $v\in H^1_0(\Omega)$, we can find a sequence of $u_m \in C^{\infty}_0(\Omega)$ converging to $v$. Hence by the proposition in your question:
$$
u_m^+\to v^+ = v \quad \text{ in }\;H^1(\Omega).
$$
Now we can construct a smoothing sequence $w_{m,k} \to u_m^+$ for any $m$ using mollifier $\phi_k$ compactedly supported near 0: $$w_{m,k} = \int_{\Omega} \phi_k(x-y)u_m^+(y)dy \geq 0.$$
Now let $v_m =w_{m,m}$ which is non-negative and in $C^{\infty}_0(\Omega)$, then we have:
$$
\|v - v_m\|_{H^1}\leq \|v - u_m^+\|_{H^1}+ \|u_m^+ - w_{m,m}\|_{H^1} \to 0.
$$</p>


问:
Hamilton equations-Symplectic Euler method
We know that $\dot{q} = \frac{\partial H}{\partial p}$ and $\dot{p} = -\frac{\partial H}{\partial q}$, and we also know the values $Q$ and $P$ respectively of $q$ and $p$ at a later time step $\Delta t$. How could we prove that the quantities
$$
\begin{align}
Q &= q + {\Delta}t\frac{\partial H}{\partial p}(q,p),\\
P &= p - {\Delta}t\frac{\partial H}{\partial q}(q,p)
\end{align}
$$
are not symplectic, while
$$
\begin{align}
Q &= q - {\Delta}t\frac{\partial H}{\partial p}(q,p),\\
P &= p + {\Delta}t\frac{\partial H}{\partial Q}(Q,p)
\end{align}
$$
are symplectic?

Clarification:
The sets of equations define different numerical integrators: in the first case (qi+1,pi+1) directly in terms of (qi,pi), and in the second case qi+1 in terms of (qi,pi), and pi+1 in terms of (qi+1,pi).

答:
See canonical transformations via generating functions.

You might also directly compute the preservation of the symplectic form,
\begin{align}
\sum _i dP_i∧dQ_i  &=\sum_i \left(dp_i-Δt \sum_j[H_{q_iq_j}(Q,p)dQ_j+H_{q_ip_j}(Q,p)dp_j]\right)∧ dQ_i
\\
&=\sum_i dp_i ∧ dQ_i-Δt \sum_i\sum_j[H_{q_iq_j}(Q,p)dQ_j+H_{q_ip_j}(Q,p)dp_j]∧ dQ_i
\\
&=\sum_i dp_i ∧ dQ_i-Δt \sum_jdp_j∧ \sum_i  H_{q_ip_j}(Q,p) dQ_i
\\
&=\sum_i dp_i ∧ \left[dQ_i-Δt   \sum_j  H_{p_iq_j}(Q,p) dQ_j\right]
\\&=\sum_i dp_i ∧  dq_i
\end{align}
Due to the symmetric coefficient matrix $\sum_{i,j}\sum_i\sum_j H_{q_iq_j}(Q,p)dQ_j∧ dQ_i=0$ and the same for the $p$ coordinates.

Note that except for a separable Hamiltionian like $H(q,p)=T(p)+V(q)$, the first equation
$$
Q=q+ΔtH_p(Q,p)
$$
would be implicit.

For the Hamiltonian compare
\begin{align}
H(q,p)&=H(Q-ΔtH_p,p)=H-ΔtH_q·H_p+O(Δt^2)
\\
\text{ and }
H(Q,P)&=H(Q,p-ΔtH_q)=H-ΔtH_p·H_q+O(Δt^2),
\end{align}
the arguments on the right all $(Q,p)$, to see that it stays largely constant. One can modify this to $\tilde H=H+\tfrac12ΔtH_q·H_p$ where the Taylor expansion to one order higher gives
\begin{align}
\tilde H(q,p)&=\tilde H(Q-ΔtH_p,p)=H-\tfrac12ΔtH_q·H_p-\tfrac12H_{qp}[H_p,H_q]+O(Δt^3)
\\
\text{ and }
\tilde H(Q,P)&=\tilde H(Q,p-ΔtH_q)=H-\tfrac12ΔtH_p·H_q-\tfrac12H_{pq}[H_q,H_p]+O(Δt^3).
\end{align}
This means that this modified energy functional has a global error of $O(Δt^2)$, the global $O(Δt)$ first-order error will mainly manifest as a time dilation.


问:
Finding the eigenvalues and an orthornormal basis of eigenvectors of an orthogonal mapping
Let $\alpha = \{\vec{a_1},\vec{a_2},\vec{a_3},\vec{a_4}\}$ be an orthonormal basis for $\mathbf{R}^4$. Let  $\mathbf{A}:\mathbf{R}^4 \rightarrow \mathbf{R}^4$ be a linear orthogonal map with the following properties: $\mathbf{A}^2=\mathbf{I}$, $\mathbf{A}\vec{a_1}=\vec{a_3}$, and $\mathbf{A}\vec{a_2}=\vec{a_4}$. Find the eigenvalues and an orthonormal basis of eigenvectors of $\mathbf{A}$.
I know that the eigenvalues are either 1, -1 or both because the length is invariant in an orthogonal mapping, which means the real roots of the characteristic polynomial are either 1 or -1. But I don't know how I know which eigenvalue is the correct one, and how I would go about finding the eigenvectors.

答: You know that $A.\vec{a_3}$ has norm $1$ and it is orthogonal to both $\vec{a_3}\left(=A.\vec{a_1}\right)$ and to $\vec{a_4}\left(=A.\vec{a_2}\right)$. Therefore, it is of the form $\gamma\vec{a_1}+\beta\vec{a_2}$, with $\gamma^2+\beta^2=1$. And you know that $A.\vec{a_4}$ has norm $1$ and it is orthogonal to $\vec{a_3}\left(=A.\vec{a_1}\right)$, to $\vec{a_4}\left(=A.\vec{a_2}\right)$, and to $\alpha\vec{a_1}+\beta\vec{a_2}\left(=A.\vec{a_3}\right)$. Therefore, it is equal to $\mp\beta\vec{a_1}\pm\gamma\vec{a_2}$. So, the matrix of $A$ with respect to $\alpha$ is$$\begin{bmatrix}0&0&\gamma&\mp\beta\\0&0&\beta&\pm\gamma\\1&0&0&0\\0&1&0&0\end{bmatrix}.$$But$$\operatorname{Id}=\begin{bmatrix}0&0&\gamma&\mp\beta\\0&0&\beta&\pm\gamma\\1&0&0&0\\0&1&0&0\end{bmatrix}^2=\begin{bmatrix}\gamma&\mp\beta&0&0\\\beta&\pm\gamma&0&0\\0&0&\gamma&\mp\beta\\0&0&\beta&\pm\gamma\end{bmatrix},$$amd therefore $\gamma=1$, $\beta=0$, and the matrix of $A$ with respect to $\alpha$ is, in fact$$\begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}.$$Can you take it from here?


问:
A doubt on metric space.
Let $c$ be the set of all convergent sequences $x=(x_n)$, then $c$ is a metric space with metric

$$d(x,y)=\text{sup}_{1\leq n<\infty}|x_n-y_n|.$$

I have to check all the conditions of metric space.

$0=|x_n-x_n|\geq \text{sup}_{1\leq n<\infty}|x_n-x_n|$, if $\text{sup}_{1\leq n<\infty}|x_n-x_n|<0$, then the distance function $d$ will be negative, which is not possible thus $d(x,x)=\text{sup}_{1\leq n<\infty}|x_n-x_n|=0.$

If this is correct, then I can proceed the rest of  conditions.

Thanks!

答:
We will check the conditions one by one.

(Distance zero iff the convergent sequences are exactly same) If the sequences are exactly same for all $n$, $| x_n-y_n| = 0$, hence the supremum is also $0$. On the other hand, if the supremum is zero, all points in the set must be $0$ as the set contains only points which are non-negative (due to the modulus). Hence, it follows that the sequences are same.

(Distance between two distinct points is positive) If the sequences are different, at least one term will be different, for that particular $n$, $| x_n-y_n| > 0$, so the $sup>0$.

(Symmetry) Follows from the symmetry of the modulus function.

(Triangle Inequality) Hint: Use the triangle inequality for modulus for every $n$ seperately.

Title: A doubt on metric space.


问:
Why aren't these two integrals equivalent when using the substitution $x=\frac{1}{t}$?
Why aren't these two integrals $$\int_{-1}^{1}\frac{1}{\left(1+x^2\right)^2}\,\mathrm{d}x$$
and $$\int_{-1}^{1}\frac{-t^2}{\left(1+t^2\right)^2}\,\mathrm{d}t$$ equal to each other, despite using the substitution $x=\frac{1}{t}$, which yields the second integral when the substitution is used on the first one?

Could it be that $t$ is undefined at $x=0$ since the limits are from $-1$ to $1$?

Body: For even $f$,$$\int_{-1}^1f(x)\mathrm{d}x=2\int_0^1f(x)\mathrm{d}x=2\int_1^\infty\tfrac{f(1/t)}{t^2}\mathrm{d}t.$$


问:
How can I prove $2\sup(S) = \sup (2S)$?

Let $S$ be a nonempty bounded subset of $\mathbb{R}$ and $T = \{2s : s \in S \}$. Show $\sup T = 2\sup S$

Proof

Consider $2s  = s + s \leq \sup S + \sup S = 2\sup S $. $T \subset S$ where T is also bounded, so applying the lub property, we must have $\sup T \leq 2 \sup S$.

On the other hand $2s + s - s \leq \sup T + \sup S - 3\sup S \implies 2\sup S \leq 2s + 2\sup S \leq \sup T $. Which gives the desired result.

Okay I am really worried about my other direction. Especially  $2\sup S \leq 2s + 2\sup S$, do I know that $2s$ is positive?

Also in the beginning, how do I know that $\sup S \leq 2 \sup S$? How do I know that the supremum is positive?

答:
You can't assume $s$ is positive, nor can you assume $\sup S$ is positive.
Your proof also assumes a couple of other weird things:

$T \subset S$ is usually not true.

$2s + s - s \le \sup T + \sup S - 3\sup S$ is not necessarily true.  Why would $-s \le -3\sup S$?

The first part of your proof is actually correct, ignoring the $T \subset S$ statement.
What you are saying is that any element of $T$, say the element $2s$, is bounded above by $2 \sup S$; thus $2 \sup S$ is an upper bound on $T$; thus $2 \sup S \ge \sup T$ by the least upper bound property.

For the second part of the proof, you need to show that $\sup T \ge 2 \sup S$.
To do this, you need to show that $\frac{\sup T}{2}$ is an upper bound on $S$.
This will imply $\frac{\sup T}{2} \ge \sup S$ by least upper bound property.

Title: How can I prove $2\sup(S) = \sup (2S)$?


答: If $x=\frac1t$, then, since $x\in[-1,1]$, you should get$$\int_{-\infty}^{-1}\frac{t^2}{(1+t^2)^2}\,\mathrm dt+\int_1^\infty\frac{t^2}{(1+t^2)^2}\,\mathrm dt.$$

Title: Why aren't these two integrals equivalent when using the substitution $x=\frac{1}{t}$?

Body:
Why aren't these two integrals $$\int_{-1}^{1}\frac{1}{\left(1+x^2\right)^2}\,\mathrm{d}x$$
and $$\int_{-1}^{1}\frac{-t^2}{\left(1+t^2\right)^2}\,\mathrm{d}t$$ equal to each other, despite using the substitution $x=\frac{1}{t}$, which yields the second integral when the substitution is used on the first one?

Could it be that $t$ is undefined at $x=0$ since the limits are from $-1$ to $1$?


问:
Proving that the matrix representation of a nilpotent linear operator is upper-triangular with diagonal entries $ = 0$
Let T be a nilpotent operator on an $n$-dimensional vector space $V$, and suppose that $p$ is the smallest positive integer for which $T^P=T_0$ (zero-transformation).

I have so far proved the following subquestions:

$N(T^i) \subseteq N(T^{i+1})$ for every positive integer $i$

There is a sequence of ordered bases $\beta_1, \beta_2, ..., \beta_p$ such that $\beta_i$ is a basis for $N(T^i)$ and $\beta_{i+1}$ contains $\beta_i$ for $1 \leq i \leq p-1$

Now, I must show that if $\beta = \beta_p$ is the ordered basis for $N(T^p) = V$ constructed so that it contains all the previous $\beta_i$, then $[T]_\beta$ is upper-triagular with diagonal entries being zero

I'm having a hard time showing this, and even seeing why this is true.

The two statements I was asked to prove above implies that I can have a scenario where $T^1$ has an empty nullspace. This doesn't violate $N(T) \subseteq N(T^2)$ nor $\{\emptyset\} = \beta_1 \subseteq \beta_2$.

But in that case, the first element of $\beta$ (denoted $v_1$) is not in the kernel of $T$. So $T(v_1) \neq 0$, and this contradicts what I am supposed to prove.

Could someone help me understand where my logic is breaking down, and how one could show the above?

答:
Your logic is breaks down when you try to show $T$ is nilpotent. If $T$ has trivial nullspace (we can never have an empty nullspace), then $T$ has full rank and is invertible. Invertible matrices can never be nilpotent; we can just multiply $T^n = 0$ both sides by $T^{-1}$ $n$ times to get $I = 0$, a contradiction.

To prove this result, note that $T$ maps $N(T^{i+1})$ into $N(T^i)$. So, if $v \in \beta_{i+1}$, then $Tv \in \operatorname{span}(\beta_i)$. So, when you form your basis $\beta$, the image of each basis vector lies in the span of the basis vectors before it in the basis (not including itself). That is, if $\beta = (v_1, \ldots, v_n)$, then
$$T(v_i) = a_{1i} v_1 + a_{2i} v_2 + \ldots + a_{(i-1)i}v_{i-1} + \color{red}0v_i + 0v_{i+1} + \ldots + 0v_n.$$
When forming the matrix for $T$ with respect to basis $\beta$, this turns into a column vector, placed in the matrix so that the red $0$ lies on the diagonal. This makes the matrix upper triangular, with a $0$ diagonal.


答:
The substitution is not defined for $x=0$ or at $t=0$. So in fact if you make the substitution, what you have is

$$\begin{align}
\int_{-1}^1{1\over(1+x^2)^2}dx&=\int_{-1}^0{1\over(1+x^2)^2}dx+\int_0^1{1\over(1+x^2)^2}dx\\
&=\int_{-1}^{-\infty}{-t^2\over(1+t^2)^2}dt+\int_\infty^1{-t^2\over(1+t^2)^2}dt\\
&=\int_{-\infty}^{-1}{t^2\over(1+t^2)^2}dt+\int_1^{\infty}{t^2\over(1+t^2)^2}dt
\end{align}$$

(And as J.G. points out, the symmetry of the integrand allows you to reduce to just one integral, from $1$ to $\infty$.)


问:
find $p$ that maximizes $P[X = 10]$ where $X$ is the total number of attempts.
Suppose we keep tossing a coin until we see $4$ heads, and suppose that the probability of
seeing a head on each toss is an unknown value $p$. If we actually perform this experiment and get the fourth head on the $10$th toss, what is the value of $p$ that maximizes the chance
of what we obtained? (that is, find $p$ that maximizes $P[X = 10]$ where $X$ is the total
number of attempts.)

My attempt:

We can write $P[X=10] = \sum \binom{X}{k} p^k (1-p)^{10-k}$

Buy how to continue from here to find $p$?

答:
If you get the fourth head on the $10$th toss this means that there were exactly $3$ heads in the first $9$ tosses. So
$$
P(X=10)=\binom{9}{3}p^4(1-p)^6 \quad 0\leq p\le1
$$
The goal is to maximize this probability (note that is a function of $p$ and can be maximized using calculus for example).


问:
How to solve $e^{ix} = i$
I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.

Thanks.

答:
A completely different approach:

$$ e^{i\pi} = -1 \implies \sqrt {e^{i\pi}} = \sqrt {-1} \implies e^{\frac{i\pi}{2}} = i $$

Title: How to solve $e^{ix} = i$


问:
Periodic points of $\sin(x)$ on $(-1,1)$
Let $f:(-1,1)\rightarrow (-1,1):$
$$f(x)=\sin(x)$$
Find all the periodic points of $f(x)$.

Note: a point $p$ is called periodic if $\exists n\in\mathbb N: f^n(p)=\underbrace{f\circ f\circ f\circ \dots \circ f}_{ \text{$n$ times }}(p).=p.$

Obviously, a fixed point of $f$ is a periodic point of $f$ (with period $1$), so

$f(0)=0$ is the only fixed point of $f(x) \implies 0$  is periodic,

I'm not sure how to prove (or disprove) that is unique.

Is it enough to show that
$$\big\{ x\in(-1,1): \lim_n f^n(x) = 0  \big\} = (-1,1)?$$
Thank you.

答: Suppose by contradiction that there exists a non-zero periodic point $x$ of $f$, i.e., $f^p(x)=x$ for some $p\in \mathbb{N}$. Assuming Gary's hint as a fact, then $|\sin^p(x)|=|\sin(\sin^{p-1}(x))|<|\sin^{p-1}(x)|<\ldots <|\sin(x)|<|x|$. Then $|x|<|x|$ contradiction.


答:
The probability you need $n$ independent trials to get exactly 4 successes is a Negative Binomial

$$\mathbb{P}[X=n]=\binom{n-1}{4-1}p^4(1-p)^{n-4}$$

Where $n=4,5,6,\dots$

In order to maximize it w.r.t. $p$ you can consider that the function to be maximized is

$$L \propto p^4(1-p)^6$$

... take its log

$$l=4\log p+ 6\log(1-p)$$

... take the derivative w.r.t. $p$

$$l^*=\frac{4}{p}-\frac{6}{1-p}$$

set it $=0$ and solve w.r.t. $p$ obtaining

$$\hat{p}=\frac{4}{10}$$

Body:
Suppose we keep tossing a coin until we see $4$ heads, and suppose that the probability of
seeing a head on each toss is an unknown value $p$. If we actually perform this experiment and get the fourth head on the $10$th toss, what is the value of $p$ that maximizes the chance
of what we obtained? (that is, find $p$ that maximizes $P[X = 10]$ where $X$ is the total
number of attempts.)

My attempt:

We can write $P[X=10] = \sum \binom{X}{k} p^k (1-p)^{10-k}$

Buy how to continue from here to find $p$?


问:
Equivalent definitions of Borel $\sigma$-algebra
I am reading Christopher Heil's Introduction to Real Analysis and there I found the following definition of the Borel $\sigma$-algebra. First denote $\mathcal{E}$ as a collection of subsets of $X$. Then the $\sigma$-algebra generated by $\mathcal{E}$, $\Sigma(\mathcal{E})$, is the smallest one containing $\mathcal{E}$. That is
$$
\Sigma(\mathcal{E}) = \cap \left\{{\Sigma : \textrm{$\Sigma$ is a $\sigma$-algebra on $X$ and $\mathcal{E} \subseteq \Sigma$}}\right\}
$$
Now let $\mathcal{U} = \left\{{U \subseteq X : \textrm{$U$ is open}}\right\}$ be the collection of all open subsets of $X$. Then the Borel $\sigma$-algebra on $X$ is $\Sigma(\mathcal{U})$. Moreover, the Borel sets are the members of $\Sigma(\mathcal{U})$.

On the other hand, the Wikipedia page of Borel sets says that Borel sets are the ones that can be written as countable combination of unions, intersections and complements of open sets, and that the Borel $\sigma$-algebra is the $\sigma$-algebra containing the Borel sets.

My question is the following: How can we conclude that the first definition implies that any member of $\Sigma(\mathcal{U})$ can be written as countable combination of unions, intersections and complements of open sets?

EDIT: From my understanding we can state something like this. Define the following four functions acting on $\mathcal{P}(X) \times \mathcal{P}(X) \rightarrow \mathcal{P}(X)$:
$$
\begin{align}
s(A,B) &= A \cup B \\
d(A,B) &= A \cap B \\
c(A,B) &= X - A \\
i(A,B) &= A
\end{align}
$$
Then for any Borel set $B$ there exists a sequence of open sets $\left\{{U_k}\right\}_{k \in \mathbb{N}}, U_k \in \mathcal{U}$ and a sequence of functions $\left\{{f_k}\right\}_{k \in \mathbb{N}}, f_k \in \left\{{s,d,c,i}\right\}$ such that (if that even makes sense)
$$
B = \cdots f_3(f_2(f_1(U_1,U_2),U_3),U_4) \cdots
$$
My question is how the intersection definition of the Borel $\sigma$-algebra implies the aforementioned representation of Borel sets.

答: Consider $W = \{S \in \Sigma(U) : S $ can be formed from countable unions/intersections/complements of open sets$\}$. It can immediately be verified that since $\Sigma(U)$ is a $\sigma$-algebra, so too is $W$. And clearly $W$ contains all open sets. Then $\Sigma(U) \subseteq W$.


答:
As $x$ goes from $0$ to $+1,$ $t$ goes from $+\infty$ to $1.$

As $x$ goes from $-1$ to $0,$ $t$ goes from $-1$ to $-\infty.$


问:
Show that $(\tau^n(x))$ is eventually in every neighborhood of $A$ iff $\lim_{n\to\infty}d(\tau^n(x),A)=0$
Let $(\Omega,d)$ be a metric space, $\tau:\Omega\to\Omega$, $A\subseteq\Omega$ be compact and $$B:=\{x\in\Omega:(\tau^n(x))_{n\in\mathbb N}\text{ is eventually in every neighborhood of }A\}.$$

How can we show that $B=\tilde B:=\{x\in\Omega:d(\tau^n(x),A)\xrightarrow{n\to\infty}0\}$?

In general, I know that $d(x,C)=0$ iff $x\in\overline C$. Now the crucial point must be the compactness of $A$; but how can we utilize it?

If $x\in B$ and $\varepsilon>0$, then $N:=\{x\in\Omega:d(x,A)<\varepsilon\}$ is clearly a neighborhood of $N$. So, it holds $B\subseteq\tilde B$; even when $A$ is not compact.

But how can we show the other inclusion? My guess is that we need the compactness of $A$ to ensure (can we?) that if $N$ is a neighborhood of $A$, there is a $\varepsilon>0$ with $A\subseteq O:=\{x\in\Omega:d(x,A)<\varepsilon\}\subseteq N$. This would be impossible if $A$ is not bounded (but I'm unsure why it would fail if it is not closed as well).

答:
Lemma

If $A$ is compact and $A \subseteq O$ where $O$ is open, then there is some $r>0$ such that $\{y \in X\mid \exists a \in A : d(a,y) < r\} \subseteq O$.

This is quite straightforward from the open cover definition of compactness: for each $a \in A$, there is some $r(a)>0$ such that $B(a,2r(a)) \subseteq O$.
Then $\{ B(a, r(a))\mid a \in A\}$ is an open cover of $A$ so finitely many, say $B(a_1, r(a_1)), \ldots, B(a_n, r(a_n))$ also cover $A$. If now $a \in A$ and $d(y,a) < r= \min_{i=1}^n r(a_i)$, then $a \in B(a_j, r(a_j))$ for some $1 \le j \le n$ and then $d(y,a(j)) \le d(y, a) + d(a(j),a) < r(a(j)) + r \le 2r(a(j))$ so that $y \in O$ by how the balls were chosen.

I think this little lemma will allow you to show the required equivalence.


问:
Is there a chance to prove this sequence to be right?
Can anyone please, set me on right path to prove this ?

We have Real numbers $a_1,a_2,a_3,\ldots,a_n$. Then, $M = \frac{(a_1+2a_2+3a_3+\ldots+na_n)}{n}$.
Prove there exists $i \in \{1,2,\ldots,n\}$ such that $a_i \leq \frac{M}{i}$.

I would be really grateful if anyone could at least help me how to solve this, thanks in advance.

答:
Suppose instead that $a_i > \frac{M}{i}$ for all $i \in \{1,...,n\}$. We will show this is not possible (this strategy is called proof by contradiction).

If $a_i > \frac{M}{i}$, then $ia_i > M$ for all $i$. Then $$M = \frac{\sum_{i=1}^n ia_i}{n} > \frac{\sum_{i=1}^n M}{n} = \frac{nM}{n} = M,$$ which is clearly not possible. Therefore, there must exist some $i$ such that $a_i \leq \frac{M}{i}$.


答:
Since $v+1/n\to v$, it suffices to approximate $v+1/n$. As in your previous question Sobolev spaces - about smooth aproximation you can use a smooth threshhold function $\phi:\mathbb R\to\mathbb R$ such that

$\phi(x)=x$ when $x\ge 1/n$

$\phi\ge 0$ everywhere

$0\le \phi'\le 1$ everywhere

The proof stays the same: pick a smooth approximation $u_m\to v+1/n$, compose with $\phi$, argue that a subsequence of $\phi\circ u_m$ converges to $v+1/n$.

Title: About smooth approximation in a Sobolev space

Body:
I want to prove the following fact :

Consider $\Omega \subset R^n$ a bounded and open set. Let $v \in H^{1}_{0}(\Omega)$ a nonnegative function. Then exists a sequence $v_m$ in $C^{\infty}_{0}(\Omega)$ of nongative functions converging to $v$ in $H^{1}( \Omega)$.

I know how to prove this fact (I think this fact can help): If $u_m$ is a sequence converging to $u$ in $H^1(\Omega)$ , then ${u^{+}_m} \rightarrow  u^+$.

Someone can give me a hint ?

Thanks in advance


答:
By Faulhaber,$$1^3+2^3+3^3+\cdots n^3=\frac{n^2(n+1)^2}4$$

and your sum can be written as

$$\frac14+\frac44+\frac94+\cdots\frac {n^2}4.$$

By Faulhaber again, this is

$$\frac{n(n+1)(2n+1)}{24}.$$

Title: Is there a formula for (0.5)²+(1)²+(1.5)²+(2)²+...+(25)²

Body:
(1/4)(1³) + (1/9)(1³+2³) + (16)(1³+2³+3³) + ... + (1/2704)(1³+2³+3³+...+51³) = ?

Initially, I factored it as 1³ (1/2² + 1/3² + ... + 1/52²) + 2³ (1/3² + ... + 1/52²) + 3³ (1/4² +...+ 1/52²) + ... + 51³ (1/52²), but got stuck to split them as separate series.

Alternatively, I observed this pattern

1³/4 + (1³/9 + 2³/9) + (1+8+27)/16 + 100/25 + ... + (1+2+3+...+51)/52

= (1/2)² + (3/3)² + (6/4)² + (10/5)² + ... + ((1+...+51)/52)²

= (1/2)² + (2/2)² + (3/2)² + (4/2)² + (5/2)² ... + (51/2)²

= (1/2)² (1+2²+3²+...+51²)

= (1/2)² (1/6)(51)(2(51)+1)(51+1)

= (1/2)² (1/6)(51)(103)(52)

= (1/2)² (17)(103)(26)

= 45526/4

= 11381.25

Does this seem correct?


问:
Real Analysis - Does my counter example disprove this lemma?
I was reading a real analysis text that gave the following lemma after introducing the axiom of completeness and least upper bounds, it reads:

Assume $s \in \mathbb{R}$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s = sup A$ if and only if, for every choice of $\epsilon \gt 0$, there exists an element $a \in A$ satisfying $s - \epsilon \lt a$.

When thinking about this lemma I was testing it with some example sets, and I stumbled across one that seemed to disprove it. The set is simply the set of real numbers on the interval $[0,1)$, our supremum is $s = 1$, and I set $\epsilon = 1 - a$. Putting it all together we get

$$1 - (1 - a) \lt a$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a \lt a$$
As far as I can tell this set fulfills all of the requirements for A, 1 is the supremum, and $1-a$ is a valid choice for $\epsilon$, as no choice of $a \in [0,1)$ will make it less than or equal to zero. But it also feels very unlikely this lemma would be disproved so easily, so I'm sure I went wrong somewhere, but I'm not sure where. If anyone has some insight into this it would be greatly appreciated, I can provide further context if necessary.

答: For every $\epsilon > 0$, there exists some $b \in A$ such that $b > \sup A - \epsilon$. In your example, you showed that for $\epsilon = 1 - a$, $b = a$ does not work. But there are other $b \in A$ that will work! For example, $b = \frac{1+a}{2}$ will work in your example.


问:
Combinatorics Question with vowels
How many 8-letter words contain exactly 5 vowels (a,e,i,o,u)? What if repeated letters were not allowed?

This question has two parts to be answered.

The first part is,"How many 8-letter words contain exactly $5$ vowels (a,e,i,o,u)?" so what I did did $8$ boxes, it can have all $5$ vowels and I am left with $3$ boxes left. So the answer is $(8C5)\cdot 5^5 \cdot 21^3$

now for the last part of the question,"What if repeated letters were not allowed?" so again I used the 8 boxes, and I can't use the same letters are vowels, so it's 5! because I can't repeat the letters. Each vowel is 5 boxes of the 8; $(8C5)$. Now I have 3 boxes left, this is where I am stuck, the answer has $(8C5)\cdot 5! \cdot (21P3)$. Why are they using permutation for the last three boxes?

Is it correctly done thus far?

答:
It says no letters are repeated (which means none of the $8$ letters).

Once we find all combinations of $5$ places for vowels in the word, which is $\displaystyle {8 \choose 5}$, we only need to now permute vowels and consonants within themselves. Hence the answer.

Another way to look at it is that we choose $3$ consonants in $\displaystyle {21 \choose 3}$ ways and then permute all $8$ letters of the word, which is $8!$

So the answer is $\displaystyle {21 \choose 3} \cdot 8! \ $ which is same as the answer in your book.

Title: Combinatorics Question with vowels


问:
What is a in the series $1 + 1/a + 1/a^2 + 1/a^3 ... = 3$? And how to solve it?
I have this summation series $1/a^i$, where a is a constant, and i starts at zero and goes to infinity. But the thing is, I want the sum to be 3, then solve for the value of a.

In other words, $1 + 1/a + 1/a^2 + 1/a^3$ ... = 3. What is the value of a?

And hopefully someone can explain the rules how to solve for a. I don't think normal algebra rules alone will do the job, since this is an infinite series.

Optional Background:

In case you're wondering where I got this question... First of all, I'm designing an RPG game where your meters upgrade based on your level. I also know the classic assymptotic increase towards 2:

1 + 1/2 + 1/4 + 1/8 ... = 2

I.e., $1/a^i$, i goes from zero to infinity, and the sum should equal 2. In this case, a equals 2. I forget what this series is called but I know it's a classical one, dating to zeno's arrow paradox I believe.

But then I thought, hmm, wouldn't it be interesting if I let your meters approach 3x their original value instead of 2? What would the series be then?

I actually found two ways to do it based on the classical one.

In the classical sum to 2, if we truncate the series and look at the values, we see 1 , 3/2 , 7/4 , 15/8, etc, all of which are closer and closer to 2 as it should be. The pattern here is $(2^{i+1} - 1)/2^i$. It was easy to extrapolate this to a sum to 3: 1 , 8/3 , 26/9 , 80/27, etc, all of which are closer and closer to 3 as they should be. The pattern there is $(3^{i+1} - 1)/3^i$.

However, to my eye, that pattern jumps a little too quickly towards 3. The jump from 1 to 8/3 = 2.667 is pretty big, so I rejected it.

The second way I found was to just go closer to 3 by 50% every time. So 1, 2, 2.5, 2.75, 2.875, etc. This doesn't "jump" like the last time, is still assymptotically approaching the value I want, and is an infinite series. However, it appears to use the exact same scheme as the classical sum to 2, just adding 1 to every term. So I consider it the same series as 1 + 1/2 + 1/4 ... but just offset by a constant. I guess you could say I aesthetically don't accept it.

(In fact, I think you can choose any value greater than or equal to 50% by which you go closer to your goal. E.g., 75% would be 1, 2.5, 2.875, 2.96875 etc. I think if it's less than 50%, the series won't converge, but this still leaves a lot of choices for this scheme.)

Then I thought of the third way, $1 + 1/a + 1/a^2 + 1/a^3$ ... = 3. But I don't know what a should equal.

Body: $\left(1+\dfrac1a+\dfrac1{a^2}+\cdots\right)a=a+\left(1+\dfrac1a+\dfrac1{a^2}+\cdots\right)$ and $3a=a+3.$


答:
for every choice of $\epsilon>0$, there exists an element $a \in A$ ...

means that we choose the number $a$ depending on the value of $\epsilon$. Since $a$ must be allowed to depend on $\epsilon$, you can't have $\epsilon$ depending on $a$.

In other words, when you say $\epsilon = 1-a$, what is this number $a$ you're using? No such variable has been introduced yet.

In formal logic, we call $a$ "bound" by the phrase "there exists an element $a \in A$ satisfying $s - \epsilon < a$" - the variable exists for the purpose of that quantified phrase, and has no meaning outside that phrase.

Title: Real Analysis - Does my counter example disprove this lemma?

Body:
I was reading a real analysis text that gave the following lemma after introducing the axiom of completeness and least upper bounds, it reads:

Assume $s \in \mathbb{R}$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s = sup A$ if and only if, for every choice of $\epsilon \gt 0$, there exists an element $a \in A$ satisfying $s - \epsilon \lt a$.

When thinking about this lemma I was testing it with some example sets, and I stumbled across one that seemed to disprove it. The set is simply the set of real numbers on the interval $[0,1)$, our supremum is $s = 1$, and I set $\epsilon = 1 - a$. Putting it all together we get

$$1 - (1 - a) \lt a$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a \lt a$$
As far as I can tell this set fulfills all of the requirements for A, 1 is the supremum, and $1-a$ is a valid choice for $\epsilon$, as no choice of $a \in [0,1)$ will make it less than or equal to zero. But it also feels very unlikely this lemma would be disproved so easily, so I'm sure I went wrong somewhere, but I'm not sure where. If anyone has some insight into this it would be greatly appreciated, I can provide further context if necessary.


问:
Fixed Point Property of $(0,1]$
Does $(0,1]$ have Fixed-point property? I can't wrap my head around any examples of continous map $f:(0,1]\rightarrow(0,1]$ that doesn't have fixed point.

答: $f:(0,1] \to (0,1]$ defined by $f(x) = \frac{1}{2}x$ has no fixed point.

Title: Fixed Point Property of $(0,1]$


问:
How can we find the determinant of $2A^TA$ while only knowing the determinant of A and the order of the matrix?
If $A$ is a square matrix of order 3 and $det(A)=5$, then how much is $det(2A^TA)$?

Assuming the product of a matrix and its transpose is nothing special how do we solve this question?

This was a previous year test question so I don't think it's a long process, there must be some rule I am missing.

答:
Hint 1: if $B$ is an $n\times n$ matrix and $a$ is a scalar, then $\det(aB)=a^n\det B$.

Hint 2: $\det(A^T)=\det A$

Do you need a third hint? I hope not.


答: No, there are many more Borel sets than those. The Borel sets contain (by definition ) all countable intersections of open sets, all complements of open sets ( ie all closed sets), so all countable unions of closed sets too. These are just the first steps in the Borel hierarchy which has $\omega_1$ levels for separable complete metric spaces like $\Bbb R$. It’s not an exhaustive list of the Borel sets and how they can be built up from open sets.

Title: Equivalent definitions of Borel $\sigma$-algebra

Body:
I am reading Christopher Heil's Introduction to Real Analysis and there I found the following definition of the Borel $\sigma$-algebra. First denote $\mathcal{E}$ as a collection of subsets of $X$. Then the $\sigma$-algebra generated by $\mathcal{E}$, $\Sigma(\mathcal{E})$, is the smallest one containing $\mathcal{E}$. That is
$$
\Sigma(\mathcal{E}) = \cap \left\{{\Sigma : \textrm{$\Sigma$ is a $\sigma$-algebra on $X$ and $\mathcal{E} \subseteq \Sigma$}}\right\}
$$
Now let $\mathcal{U} = \left\{{U \subseteq X : \textrm{$U$ is open}}\right\}$ be the collection of all open subsets of $X$. Then the Borel $\sigma$-algebra on $X$ is $\Sigma(\mathcal{U})$. Moreover, the Borel sets are the members of $\Sigma(\mathcal{U})$.

On the other hand, the Wikipedia page of Borel sets says that Borel sets are the ones that can be written as countable combination of unions, intersections and complements of open sets, and that the Borel $\sigma$-algebra is the $\sigma$-algebra containing the Borel sets.

My question is the following: How can we conclude that the first definition implies that any member of $\Sigma(\mathcal{U})$ can be written as countable combination of unions, intersections and complements of open sets?

EDIT: From my understanding we can state something like this. Define the following four functions acting on $\mathcal{P}(X) \times \mathcal{P}(X) \rightarrow \mathcal{P}(X)$:
$$
\begin{align}
s(A,B) &= A \cup B \\
d(A,B) &= A \cap B \\
c(A,B) &= X - A \\
i(A,B) &= A
\end{align}
$$
Then for any Borel set $B$ there exists a sequence of open sets $\left\{{U_k}\right\}_{k \in \mathbb{N}}, U_k \in \mathcal{U}$ and a sequence of functions $\left\{{f_k}\right\}_{k \in \mathbb{N}}, f_k \in \left\{{s,d,c,i}\right\}$ such that (if that even makes sense)
$$
B = \cdots f_3(f_2(f_1(U_1,U_2),U_3),U_4) \cdots
$$
My question is how the intersection definition of the Borel $\sigma$-algebra implies the aforementioned representation of Borel sets.


问:
Prove $\text{Hom}(K,K)=\{0,id_K\}$ for $K=\mathbb{Q}$
I've been solving some problems from my Galois Theory course, and I need help with the final step of this one:

Prove $\text{Hom}(K,K)=\{0,id_k\}$ for $K=\mathbb{Q}$.

The work I did so far:

Given $K$ is a field, it's only ideals are the trivial ones, to be said, $0$ and $K$. So, noticing that ring homomorphisms $\varphi$ verify that $ker\varphi$ is ideal of the origin ring, so from that I get that, being
$$\varphi: K\to K$$
a ring homomorphism, then $ker\varphi\unlhd K$, so $ker\varphi\in\{0,K\}$. Now, we study both cases separatedly:

If $ker\varphi = K$, then $\varphi=0$.

If $ker\varphi = 0$, then $\varphi$ is ring monomorphism, and from the fact that $\varphi$ goes from $K$ to $K$ we get also that it's bijection. My problem is that I don't know how to deduce $\varphi = id_K$ from this.

Is the work I did correct? How can I finish this proof? Any help will be appreciated, thanks in advance.

答:
It's not true for general $K$. E.g. for $K = \Bbb C$, you have a lot of homomorphisms from $\Bbb C$ to itself.

It is true in the case $K = \Bbb Q$. Let $\phi:\Bbb Q \rightarrow \Bbb Q$ be a ring homomorphism, i.e. $\phi(x + y) = \phi(x) + \phi(y)$ and $\phi(xy) = \phi(x)\phi(y)$ for all $x, y \in \Bbb Q$.

We consider $\phi(1)$. Since $\phi(1) = \phi(1^2) = \phi(1)^2$, we see that $\phi(1)$ is either $0$ or $1$.

If $\phi(1) = 0$, then we have $\phi(x) = \phi(x \cdot 1) = \phi(x)\phi(1) = 0$ for any $x\in \Bbb Q$.

Now assume $\phi(1) = 1$. Using $\phi(x + 1) = \phi(x) + \phi(1) = \phi(x) + 1$, we may show by induction that $\phi(n) = n$ for all integers $n \geq 0$. Using $\phi(-x) = -\phi(x)$ (as $\phi$ is a homomorphism of the additive group), we get $\phi(n) = n$ for all $n\in \Bbb Z$.

Therefore for any rational number $q = a/b$ with $a, b \in \Bbb Z$ and $b \neq 0$, we have $$b\phi(q) = \phi(b) \phi(q) = \phi(bq) = \phi(a) = a,$$ which gives $\phi(q) = a/b = q$.


问:
For which rational coefficients is $a+b\alpha+c\alpha^2+d\alpha^3+e\alpha^4$ constructible, where $\alpha=3^{1/5}$?

Let be $\alpha=3^{1/5}$, and be $\gamma$ the number
$$
\gamma=a+b\alpha+c\alpha^2+d\alpha^3+e\alpha^4
$$
with $a,b,c,d,e \in \mathbb{Q}$ . For which $a,b,c,d,e \in \mathbb{Q}$ is $\gamma$ constructible?

I know the field of constructible numbers is closed by square roots and the Wantzel theorem for constructibility, but I didn't get anywhere.

Any hint?

答:
The number $\gamma$ lives in the field $\Bbb Q(\alpha)$, which is a degree $5$ extension of $\Bbb Q$.

If $\gamma$ is constructible, then the extension $\Bbb Q(\gamma)/\Bbb Q$ must have power-of-two degree.

But this degree should also be a factor of $5$, as $\Bbb Q(\gamma)/\Bbb Q$ is a subextension of $\Bbb Q(\alpha)/\Bbb Q$.

Therefore the only possibility is that the degree is equal to $1$, i.e. $\gamma \in \Bbb Q$. This again is equivalent to $b = c = d = e = 0$, as $1, \alpha,\dots, \alpha^4$ form a $\Bbb Q$-basis of $\Bbb Q(\alpha)/\Bbb Q$.


答:
Let me give a hint, using numbers having the form $a+b\sqrt[3]2+c\sqrt[3]4$, with $a,b,c\in\mathbb{Q}$.  Whatever irreducible polynomial with rational coefficients has this root, will have a set of three conjugate roots including the given one ($\omega=$ a non-real cube root of unity):

$r_1=a+b\sqrt[3]2+c\sqrt[3]4$

$r_2=a+b\omega\sqrt[3]2+c\omega^2\sqrt[3]4$

$r_3=a+b\omega^2\sqrt[3]2+c\omega\sqrt[3]4$

and these would also be constructible.  But the combination

$r_1+\omega^2r_2+\omega r_3=3b\sqrt[3]2$

would also be constructible, which might pose a minor problem unless $b=0$.  A similar contradiction using the combination $r_1+\omega r_2+\omega^2r_3$ similarly forces $c=0$, and so the only constructible blnumvers contained in my set are just the rational numbers.

Now try a similar method with your numbers.  All complex fifth roots of unity, like the cube roots of unity described above, are constructible.

Title: For which rational coefficients is $a+b\alpha+c\alpha^2+d\alpha^3+e\alpha^4$ constructible, where $\alpha=3^{1/5}$?

Body:
Let be $\alpha=3^{1/5}$, and be $\gamma$ the number
$$
\gamma=a+b\alpha+c\alpha^2+d\alpha^3+e\alpha^4
$$
with $a,b,c,d,e \in \mathbb{Q}$ . For which $a,b,c,d,e \in \mathbb{Q}$ is $\gamma$ constructible?

I know the field of constructible numbers is closed by square roots and the Wantzel theorem for constructibility, but I didn't get anywhere.

Any hint?


问:
Why the initial space of $U$ is $\mathcal M\ $?

$\textbf {Polar decomposition}$ $:$ Let $\mathcal H$ and $\mathcal K$ be Hilbert spaces. Then any operator $T \in B(\mathcal H, \mathcal K)$ admits a decomposition $T = UA,$ where $U \in B(\mathcal H, \mathcal K)$ is a partial isometry and $A \in B (\mathcal H)$ is a positive operator.

The proof of the above theorem goes as follows $:$

For any $x \in \mathcal H$ we have

$$\left \langle Tx,Tx \right \rangle = \left \langle T^*Tx,x \right \rangle = \left \langle |T|^2 x, x \right \rangle = \left \langle |T|x, |T|x \right \rangle$$ where $|T|^2 = T^*T.$ Thus the map $V_0 : \text {ran} (|T|) \longrightarrow \text {ran} (T)$ defined by $V_0 (|T|x) = Tx, x \in \mathcal H$ is a well-defined isometry and hence it extends to an isometry $U_0 : \overline {\text {ran} (|T|)} \longrightarrow \overline {\text {ran} (T)}$ such that $U_0 |T| = T.$ Let $\mathcal M =  \overline {\text {ran} (|T|)}.$ Let $P$ be the projection onto $\mathcal M.$ Consider the operator $U = U_0 P \in B(\mathcal H,K).$ Then it is easy to see that $UU^*U = U$ and hence $U$ is a partial isometry, where the initial space of $U$ is $\mathcal M$ and the final space of $U$ is $\mathcal N = \overline {\text {ran} (T)}$ and it is easy to see that $U|T| = T.$

Can anybody please make it clear to me as to why the initial space of $U$ is $\mathcal M\ $? The initial space of $U$ has to be ${(\text {ker}\ U)}^{\perp}.$ Since $\mathcal M$ is the projection onto $\mathcal M$ it follows that $\mathcal M^{\perp} \subseteq  \text {ker}\ U.$ Hence $(\text {ker}\ U)^{\perp} \subseteq \mathcal M.$ How to prove the equality here?

Any help in this regard will be warmly appreciated. Thanks for your time.

答: Let us see that $\ker U\subset\mathcal M^\perp$. Let $x\in\ker U$, $x = u+v$ with $u\in\mathcal M$ and $v\in\mathcal M^\perp$. Then $0 = \|Ux\| = \|Uu\| = \|u\|$, so $u=0$ and thus $x = v\in\mathcal M^\perp$.


问:
Determine the linear production function given relationship between labor and capital productivity
Determine the linear production function in the form $Q=aK+bL$ if it takes $3.5$ labor ($L$) hours to produce what a machine ($K$) can do in one hour?

I was given the answer is: $Q=3.5 \times K+L$

But I don't understand why the function is not $Q=K+3.5 \times L$

答:
Great question, it trips up intro econ students all the time. Think about it in terms of marginal products, and it's a lot easier.

Given $Q = aK + bL$, the marginal amount that $L$ can produce per unit of labor is the marginal product of labor, $\text{MPL}$, which is

$$ \text{MPL} = \frac{\partial Q}{\partial L} = b . $$

Similarly, the marginal amount that capital can produce per unit of capital is the marginal product of capital, $\text{MPK}$, which is

$$ \text{MPK} = \frac{\partial Q}{\partial K} = a .$$

Then, if capital can produce $3.5$ times what labor can produce in one hour, we have

$$ \text{MPK} = 3.5 \text{MPL} $$
$$ \Rightarrow a = 3.5 b . $$

Finally, substitute that back into $Q$, and you have

$$ Q = 3.5bK + bL . $$

The answer you provided is the case when $b = 1$, but it holds for any value of $b$.


问:
Sheaf associated to global sections of a quasicoherent sheaf is the pushforward by the structure morphism
Let $A$ be a commutative unital ring and $X$ a separated, quasi-compact scheme over $A$ where the structure map is denoted $f:X\to \text{Spec }A$. Suppose $\mathcal{F}$ is a quasicoherent $\mathcal{O}_X$-module. Are these hypotheses strong enough to ensure that $f_*\mathcal{F}=\widetilde{\mathcal{F}(X)}$?

The notation on the right hand side refers to the sheaf associated to the module $\mathcal{F}(X)$. I know  that $f_*\mathcal{F}$ is quasicoherent, but I am not sure if this is enough to conclude. Certainly I hope this is true. If not, what assumptions could one make on the scheme $X$ to ensure it is true? Any and all help is appreciated.

答:
There's an equivalence of categories between quasicoherent sheaves on $\operatorname{Spec} A$ and $A-$Modules. We can send $\mathcal{F}\mapsto \Gamma(\operatorname{Spec}A,\mathcal{F})$ in one direction and in the other we send $M\mapsto \widetilde{M}$. In particular, since you know that $f_*\mathcal{F}$ is quasicoherent (assuming $X$ Noetherian will do), the correspondence for affine schemes between quasicoherent sheaves and modules shows us
$$f_*\mathcal{F}\cong \widetilde{\Gamma(\operatorname{spec} A,f_*\mathcal{F})}=\widetilde{\Gamma(f^{-1}(\operatorname{spec} A),\mathcal{F})}=\widetilde{\Gamma(X,\mathcal{F})}=\widetilde{\mathcal{F}(X)}.$$