问:
Why the fundamental group of a base acts trivially on homology of the fiber - configuration spaces
My question is connected with the answer to this question:
https://mathoverflow.net/questions/344491/zero-differential-in-serre-spectral-sequence-for-configuration-spaces

I cannot see why from the fact that $\pi_1(C_n)\to\pi_1(C_{n-1})$ is a split epimorphism we have that the action of $\pi_1(C_{n})$ on the cohomology of a fiber is trivial.
This seems like a sort of a general statement:
If in a fibration $F\to E\to B$ there exists a section $B\to E$, then the action of the fundamental group of $B$ on the cohomology of $F$ is trivial. Is it true? And if so, why?

答:
It is not true in general that if a bundle has a section then the monodromy on the cohomology of the fiber is trivial. For example, a projectivized vector bundle always has two sections ($0$ and $\infty$) but the monodromy on the top cohomology (for odd rank) is trivial iff the bundle is orientable.

In the case of the bundle $C_n \to C_{n-1}$, it is not hard to explicitly describe the monodromy. The action on $\pi_1(F)$ is by conjugation (by what?) and hence trivial on $H^1$. Since the fiber is a wedge of circles this is all that needs to be checked.


问:
Critical region for one-sided test of binomial proportion at 5% level

For this question's part C), the answer key says you must test the upper tail, but it isn't the alternative $p < 0.3$? So shouldn't you be testing the lower tail and get $x\le2$ for the critical region? Instead the answer key says it's $x\ge 10$.

答:
I will show some computations from R.
You do not say whether you are expected to use the binomial PDF formula to compute probabilities. If so, I'll leave that part to you.
You are correct that you need to reject in the lower tail.

$H_0: p = .3$ vs. $H_a: p < .3,$ where $p$ is the probability of a defect. Using $n = 20,$ the null distribution is that the number $X$ of observed defects has $X \sim \mathsf{Binom}(n=20, p=.3).$

You need a critical value for a test at the 5% level (or perhaps a little lower). $P(X \le 2 \,|\, p=0.3) = 0.03548.$ So the critical value is $c = 2.$
[In R pbinom is a binomial CDF.]

```<code>pbinom(3, 20, .3)
[1] 0.1070868
pbinom(2, 20, .3)
[1] 0.03548313
</code>```

If you reject for $X \le c = 2,$ you will have a test at level
$\alpha = 0.355.$

Extra: Power computation.

If the new version has $p = 0.1,$ then what is the power (probability of rejection)? Answer: About 68%.

```<code> pbinom(2, 20, .1)
 [1] 0.6769268
</code>```

Here is output for Minitab's power and sample size procedure.

```<code>Power and Sample Size

Test for One Proportion

Testing p = 0.3 (versus < 0.3)
α = 0.05

              Sample
Comparison p    Size     Power
        0.20      20  0.221724
        0.20      30  0.303238

        0.10      20  0.680418  <- Compare R above
        0.10      30  0.872633

        0.05      20  0.952676
        0.05      30  0.997631
</code>```

ContentLicense: CC BY-SA 4.0

Body:
I will show some computations from R.
You do not say whether you are expected to use the binomial PDF formula to compute probabilities. If so, I'll leave that part to you.
You are correct that you need to reject in the lower tail.

$H_0: p = .3$ vs. $H_a: p < .3,$ where $p$ is the probability of a defect. Using $n = 20,$ the null distribution is that the number $X$ of observed defects has $X \sim \mathsf{Binom}(n=20, p=.3).$

You need a critical value for a test at the 5% level (or perhaps a little lower). $P(X \le 2 \,|\, p=0.3) = 0.03548.$ So the critical value is $c = 2.$
[In R pbinom is a binomial CDF.]

pbinom(3, 20, .3)
[1] 0.1070868
pbinom(2, 20, .3)
[1] 0.03548313

If you reject for $X \le c = 2,$ you will have a test at level
$\alpha = 0.355.$

Extra: Power computation.

If the new version has $p = 0.1,$ then what is the power (probability of rejection)? Answer: About 68%.

 pbinom(2, 20, .1)
 [1] 0.6769268

Here is output for Minitab's power and sample size procedure.

Power and Sample Size

Test for One Proportion

Testing p = 0.3 (versus < 0.3)
α = 0.05

              Sample
Comparison p    Size     Power
        0.20      20  0.221724
        0.20      30  0.303238

        0.10      20  0.680418  <- Compare R above
        0.10      30  0.872633

        0.05      20  0.952676
        0.05      30  0.997631


问:
How to find set of points to satisfy a vector paralel to a known vector?
I am given two points:
$C(4,-2,3)$ and $U(1,0,-2)$.

How do I find the set of points $P(x,y,z)$, knowing that $\vec{UC} \cdot \vec UP=0$?

I know that $\vec{UC}$ and $\vec{UP}$ should be parallel, but how do I develop this idea?

答: You said " I know $\vec{UC}$ and $\vec{UP}$ should be parallel ". The fact is they need not be. What made you think so? back to your question, @Don Thousand already addressed it. I just brush it up a bit. Let $P = (x,y,z) \implies \vec{UP} = <x-1,y,z+2>, \vec{UC} = <3,-2,5>$. Use the definition of dot product for $2$ perpendicular vectors, namely: if $ \vec{X} = <x_1,y_1,z_1>, \vec{Y} = <x_2,y_2,z_2>$ then $\vec{X}\cdot \vec{Y} = 0 \iff x_1x_2+y_1y_2+z_1z_2 = 0$. Can you take it from here...?


问:
Do two equal spans have the same basis?
So I'm trying to solve the following question:

Suppose that $\{A, B\}$ is a basis for a vector space $W$. Prove that $\{X, Y\}$ is also a basis for $W$, where $X = 2A + 3B$ and $Y = 3A − 5B$.

I just proved that $\operatorname{Span}(X,Y) = \operatorname{Span}(A,B)$, but I'm not really sure how to go from here. Does having equal spans mean that their basis are equal? Any help is appreciated.

答:
If the two spans are the same, then they span the same vector space.  A vector space can have any number of bases, but the number of elements in any basis is always the same, called the dimension of the vector space.  Any basis for the first span will be a basis for the second span, since they are the same vector space.

Note that once you know the dimension, say $n$, any set of $n$ linearly independent vectors, or $n$ vectors that span, will form a basis.

Title: Do two equal spans have the same basis?


问:
Prove that if $G$ is a finite group with an even class number, then $G$ is of even order. Give an example that the converse fails.

Prove that if $G$ is a finite group with an even class number (the number of conjugacy classes of a group $G$ is called its class number) then $G$ is of even order. Give an example that the converse fails.

My attempt: If $n$ is even $$|G|=\sum_{i=1}^{n} |{\rm orb}(g_i)|$$
and for all $i$, $|{\rm orb}(g_i)|$ divides $|G|$. Let $a,b_i$ be integers. Does

$$a=\sum_{i=1}^{n \text{ even } }b_i$$ with $b_i\mid a$ $\forall i$ imply that $a$ is even?

Actualization: By contradiction, if $a$ is odd, $b_i\mid a$ implies $b_i$ is odd. Because odd+odd=even then $a$ is even a contradiction.

What counterexample can be work?

答:
This can be solved with character theory. Let $\chi \in Irr(G)$, with $\chi \neq 1_G$ (the principal character). Apply the orthogonality relation to $\chi$ and $1_G$, then
$$\sum_{x \in G}\chi(x)=0.$$
Now assume that $\chi$ is real-valued. Then $\chi(x)=\overline{\chi(x)}$ (complex conjugate) for all $x \in G$. From the orthogonality formula above one has
$$\chi(1)=-\sum_{x \in G-\{1\}}\chi(x).$$
But $\chi(1) \mid |G|$, so the degree of $\chi$ is odd. The oddness of $|G|$ also implies that for all $x \in G-\{1\}$, $x\neq x^{-1}$, so
$$\sum_{x \in G-\{1\}}\chi(x)=\sum \chi(x)+\chi(x^{-1}),$$
where the last sum is taken over a set of disjoint pairs $\{x,x^{-1}\}$. Now $\chi(x)$ is the sum of the eigenvalues of a matrix representing $x$, while $\chi(x^{-1})$ is the sum of the eigenvalues of the inverse of that same matrix. The eigenvalues of the matrix are roots of unity and for a root of unity $\zeta$, one has $\zeta^{-1}=\overline{\zeta}$. Since $\chi$ is real-valued one must have $\chi(x)=\chi(x^{-1})$. It follows that
$$\frac{1}{2}\chi(1)=-\sum_{x}\chi(x).$$
We now have a contradiction: the left hand is half an odd positive integer, while the right hand side is an algebraic integer.

Hence we have shown that if $|G|$ is odd, and $\chi \in Irr(G)$ is not principal, then $\overline{\chi} \neq \chi$. It follows that the number of irreducible characters, which is the same as the number of conjugacy classes is odd.

Counterexample for the reverse $G=Q$, the quaternion group of order $8$, it has $5$ conjugacy classes.


问:
$\forall\varepsilon > 0\exists\delta > 0:(|f(x) - \ell| < \varepsilon)\implies(0<|x-a| < \delta )$. $f(x)=x$ satisfies this for all $a$ and $\ell$?
This is to be found in the answer to problem 26 (b) of chapter 5 of Michael Spivak's Calculus 3rd edition. The problem reads as follows:

26 Give examples to show that the following definitions of $\lim_{x\to a}f(x)=\ell$ are not correct.

(b) For all $\varepsilon > 0$ there is a $\delta > 0$ such that if $|f(x) - \ell| < \varepsilon$, then $0<|x-a| < \delta$.

I thought of constant functions because they do not satisfy this condition, but approach a limit "everywhere". This means, let $f(x)=c, c\in \mathbb R$ and let $a\in \mathbb R$ and let $\varepsilon > 0$, then $|f(x) - c|=0 < \varepsilon$ for all $x$ satisfying $0<|x-a| < \delta$ for any $\delta > 0$ and not just some $\delta > 0$. (It is already shown in the text of this chapter that $\lim_{x\to a}c=c$.)

In his answer Spivak also mentions constant functions but then adds the following:

Moreover, the function $f(x)=x$, for example, satisfies this condition no matter what $a$ and $\ell$ are.

What does he mean by that? It seems to me that if $|f(x) - \ell| < \varepsilon$ holds, then $|x - \ell| < \varepsilon$ so $a=\ell$ does hold, which means it does matter what $a$ and $\ell$ are. Unfortunately, I really don't understand what he is getting at.

答:
Edit
JEdwards and fleablood already explained much of what I added. Sorry! Not trying to poach answers. I'll leave mine up in case it's useful but please don't give me points that should go elsewhere!

I've missed the main thrust of what Spivak's getting at. The bad stuff I pointed out is true, however there's a more general issue.

With $f(x)= x$, no matter what we choose for $\ell$, $a$, or $\varepsilon$, we can find some $x$ such that $0<|x-a|<\delta$ and $|f(x) -\ell| < \varepsilon$. We just have to make $\delta$ big enough.

For example, take $a = 0$, $\ell = 10$.

Now, For any $\varepsilon > 0$ there is a $\delta$ ($\delta \geq 11$ works) such that $|f(x) - 10| < \varepsilon$ and $0 <|x-a|< \delta$.

But this suggests that $\lim_{x\to 0}x = 10.$

In other words, this $f$ is such that it approaches every number somewhere so using the criteria of our bad limit definition, if we make $\delta$ big enough we can show that $f(x)$ approaches any value, anywhere.

Original Post

What does he mean by that? It seems to me that if $|()−\ell|<\varepsilon$ holds, then $|−\ell|<\varepsilon$ so $=\ell$ does hold, which means it does matter what $$ and $\ell$ are.

This is true, however, the bad limit condition also says that for all $x$ satisfying $|()−\ell|<\varepsilon$, that $0 < |x-a|$.

For $f(x) = x$, $\lim_{x\to a} x = a$.

For $x = a$, we have $|f(x) - \ell| = |a - a| = 0 < \varepsilon$, but, $0 \not<|x-a|< \delta$

The same thing happens for constant functions. They satisfy $|f(x) - \ell| < \varepsilon$ at $x = \ell$, which is not allowed.

As an aside, this bad limit definition also runs into problems with functions that have the same limit at multiple different points. For example, $f(x) = x^2$ approaches $1$ at both $x = 1$ and $x = -1$.

Here we're forced to make $\delta$ big enough to cover both of them. Consequently, the $\delta$ neighborhood includes many $x$ that do not satisfy the original $\varepsilon$ condition. For cases like this, the $\delta$ neighborhoods stop being useful (even setting aside the earlier $x=a$ problem.)

Edit: Another point: using this definition of limits and considering constant functions, if $f(x) = c$ then then for every $\varepsilon > 0$ and all $x$, $|f(x) - c| < \varepsilon$. The fact that every $x$ satisfies this $\varepsilon$ requirement means that for the limit $c$ to exist at any point $a$, we require some $\delta > 0$ such that $|x-a|< \delta$ for all $x$. No such $\delta$ exists. Constant functions cannot approach limits.


问:
Wiener-Khinchin theorem and fluctuations in flicker noise
Suppose that we have a flicker noise process, that takes values $d(t)\in\{-1,+1\}$. We discretize time into $t\in[0,dt,2dt,\ldots, (N-1)dt]$, and $dt = T/N$ with $T$ being the total length of the 'cycle'. At every time step we flip the state of $d$ with probability $W dt/2$. Ultimately we end up with a list of outcomes composed of $\pm1$ of length $N$, e.g. $d = [-1,-1,-1,\ldots, 1,1,\ldots,-1]$.

The process defined in this way, has correlation $\langle d(t) d(t')\rangle = \exp(- W |t-t'|)$ or $Q(k dt) = \langle d(k dt) d(0)\rangle = \exp(- W k |dt|)$ for discrete times and $k=0,\ldots, N-1$. According to Wiener-Khinchin theorem, I could compute fast Fourier transform (FFT) of my list of outcomes $d$ or of a list of $Q = [1, e^{-W dt}, e^{-2 W dt}, \ldots, e^{-(N-1)W dt}]$ and should get the same results for $N\to \infty$. Taking FFT of $Q$ is easy, since we have a geometric series
$$
\tilde{Q}(k)=\sum_{n=0}^{N-1}e^{-2\pi i \frac{kn}{N}} e^{-W n dt} = \frac{1-e^{-W N dt}}{1-e^{-2\pi i \frac{k}{N}-W dt}}\approx \frac{N}{T} \frac{W}{W^2 + (2\pi \frac{k}{N})^2},
$$
where we Taylor expanded the denominator and kept the first order correction, and numerator was approximated by 1, which is valid for large $N$ (note that $dt = T/N$).

If I calculate absolute value square of FFT of $d$, i.e.
$$
\tilde{V}(k) = \frac{1}{N}\left|\sum_{n=0}^{N-1} e^{-2\pi i \frac{kn}{N}}d(n)\right|^2,
$$
where $d(n)$ is $n$-th outcome of my random process, I should (in the limit of $N\to \infty$) obtain the same result as for $\tilde{Q}(k)$ (isn't this the operational meaning of Wiener-Khinchin theorem?). However, what I got is strongly fluctuating picture, with general trend behaving like $\tilde{Q}$ see the plot, where $\omega_k = 2\pi \frac{k}{NT}$.

The questions that I've got:

shouldn't we observe much smoother behavior of sampled process (sampling procedure explained below), similar to 'analytical' results?

If I repeat the sampling procedure $M$ times, and average all repetitions, I get smooth behavior, which is exactly what I'd expect. Why it doesn't work for a single (large $N$) realization, where 'ergodicity' should smoothen the fluctuations?

Sampling code snippet (in python):

```<code>import numpy as np #standard numerical tools
import matplotlib.pyplot as plt #for plotting
from scipy.fft import fft, ifft #fast fourier transform and it's inverse

'''
Used parameters:
N = 10**7
Trep = 10**5
Wmin = 0.1
cut_off = 10**5  - how many frequencies we're depicting
'''
#drawing random number for flipping rate
W = np.random.uniform(low = Wmin, high = Wmax )

#expressing total cycle time related to flipping rate (i.e. related to correlation length)
T = Trep/W

t_list = np.linspace(0,T, N+1)[:-1] #list from 0, to (N-1) dt

dt = t_list[1] #defining dt = T/N

d = [-1] #wlog setting initial value to -1

for idx in range(N-1):

    I = np.random.rand() < W * dt/2 #defining successful flip

    if I :
        dn.append(d[idx] * -1) #flipping the new outcome
    else:
        dn.append(d[idx] * 1) #leaving the new outcome as the previous one

omega_k = np.linspace(0, N , N+1)[:-1]* 2 * np.pi/T #defining frequencies

Q_WK = N/T * np.real(( W  ) /(W**2 +    omega_k**2)) #approximated geometric series

FT = (abs(fft(d))**2)/(N) #calculating FFT for the outcome dn

plt.rcParams.update({'font.size':30})
plt.figure(100,figsize=(24,16))
plt.grid(True)
plt.plot(omega_k[:cut_off],  0.5 * FT[:cut_off], label = 'sampled FT')
plt.plot(omega_k[:cut_off],  Q_WK[:cut_off], label = r'<span class="math-container">$\frac{W}{W^2 + \omega_k^2}$</span>', linewidth = 4.0)
plt.legend()
plt.show()
</code>```

答:
I think you have run into the classic problem of trying to estimate the spectrum from a single time series observation. It is well known that the amplitude spectrum, i.e., the modulus of the periodogram (Finite Fourier transform) of the signal itself is not a consistent estimator (see, Davenport & Root, An Introduction to the theory of random signals and noise, chapter 6). In other words, as the observation interval increases to infinity the variance of the modulus (or the square of the modulus) does not decrease to zero. This is in contrast to what you would normally expect from the law of large numbers; of course, the samples in the Fourier Sum are not independent, so the LLN cannot just simply be applied.

This is the reason for real-life spectrum analyzers always use a so-called "video filter". (The term "video" here is a historical relic referring to a variable bandwidth and variable shape baseband filter.)

To estimate the spectrum one makes several independent observations of the same time series length whose reciprocal is roughly equal to the effective frequency resolution. The   output of that filter, in your case the Fourier components themselves, are squared and passed through the video filter forming essentially a weighted average of these periodograms. Once properly averaged the estimator is consistent, it may be a biased estimator but it is now consistent so that longer and longer observations will lead to less and less fluctuations but you must align the IF with the video filter; a good commercial spectrum analyzer will have a large set of calibrated IF/video filter pairs built-in and will automatically suggest their use.

(Recall the so-called Gibbs phenomenon where increasing the number of Fourier terms does not help reducing the error fluctuations at discontinuities. If you want to avoid those nasty ripples you must window/taper the signal appropriately.)


问:
Showing a property holds in arbitrary measure space

A measure space $\left(X,\mathcal{M},\mu\right)$ is said to have Property A if for every $A\in\mathcal{M}$ with $\mu(A)>0$ there is a set $B\subset A$, $B\in\mathcal{M}$ such that $0<\mu(B)<\mu(A)$.

Let $\left(X,\mathcal{M},\mu\right)$ be a measure space with Property A. Show that for every $A\in\mathcal{M}$ with $0<\mu(A)<\infty$ and for every $\epsilon>0$ there is a set $B\subset A$, $B\in\mathcal{M}$ such that $0<\mu(B)<\varepsilon$.

I am not really sure how to go about doing this. I'd expect some argument relying on countable additivity to be made, e.g. $$\mu(B)<\sum_n \frac{\varepsilon}{2^n} = \varepsilon.$$ But I am just not sure overall how to do such problem.

答:
Let $\mu(A)=r$ for some $r\in\Bbb R$. By property $A$ we can find a $B\subseteq A$ with $0<\mu(B)<\mu(A)$.

Since $\mu(B)+\mu(A\setminus B)=r$ one of $B$ and $A\setminus B$ must have measure $\leq r/2$. Pick this one and apply property $A$ to it, do you see how to conclude inductively?


问:
Can you add quasiconcave functions with scaling to get another quasiconcave function?
Give two quasiconcave functions $f(t),g(t)$, can we always find some $c>0$ so that $f(t)+cg(t)$ is quasiconcave?

I know that $c=1$ doesn't work in general, see Is the sum of quasi concave functions quasi concave But can you find some $c$ so that it does work? What if you assume that $f,g$ are strictly quasiconcave?

This came up in a research problem and would help a lot.

答:
The same counterexample as in the link you gave: $f(x) = x^3$ and $g(x) = 1 - x^2$ are quasiconcave, but $f(x) + c g(x)$ is not quasiconcave for any $c > 0$ (note that $x=0$ is
a strict local maximum but not a global maximum).


问:
$A\cong A^{op}$ means $A$ is commutative
Sorry for my bad English.

Let $A$ be (not necessarily commutative) ring.

By here, if $A\cong A^{op}$ as ring,  $A$ is commutative, but I cannot proof.

Is this trivial? Please help me.

答:
[This just expands upon my comment, since the answer to the question I asked provides another solution]

Given any ring $A$, the map $$(A\times A^{op})^{op} \to A\times A^{op} : (x,y) \mapsto (y,x)$$
will be a ring isomorphism.  If $A$ were non-commutative, then $B = A \times A^{op}$ is non-commutative but $B^{op} \cong B$.

In particular, if your definition of a ring doesn't require a multiplicative identity, then this is succinctly stated as "Every ring $A$ naturally injects into a ring $B$ such that $B^{op} \cong B$"


问:
show a subset generates a symmetric group
How would I show that the subset $V = \{(1,2), (2,3), (3,4)\}$ generates the symmetric group $S_4$?

I am just being introduced to group theory this is all new. I have unsuccessfully attempted to solve this question. . How do you prove this?

答:
Welcome to MSE!

Here is a hint:

We know that every permutation can be written as a product of transpositions
(see here, for instance). So if you can get every transposition, then you can get every permutation (do you see why?).

Of course, you already have $3$ transpositions available! Can you combine these to get the others?

I hope this helps ^_^


问:
Is the image of a curve under the projection morphism of a fiber product again a curve?
If $X$ is a variety defined over the rationals and $f:X_\mathbb{C}\to X$ is the projection map of the fibered product, then is $f(C)$ is a curve in $X$? (Where $C$ is a curve in $X_\mathbb{C}$.)

答:
No. Let $X=\Bbb A^2_{\Bbb Q}$ and let $C=V(x-\pi)\subset X_{\Bbb C}$. The preimage of the ideal $(x-\pi)\subset \Bbb C[x,y]$ in $\Bbb Q[x,y]$ is the zero ideal since $\pi$ is transcendental over $\Bbb Q$. Thus $f(C)$ contains a point with residue field of transcendence degree $2$ over $\Bbb Q$, so no matter what subscheme structure you put on it, it cannot be a curve over $\Bbb Q$.

Let me point out that what you're doing is a bit strange - in general, the question "what subscheme structure do I put on $f(Z)$ for a subscheme $Z\subset X$ and a map $f:X\to Y$" doesn't really have a good answer and may not even be answerable at all.


问:
Intuitively, why are there 4 classical Lie groups/algebras?
I would like to understand the big picture in mathematics.  Lie groups and Lie algebras seem to play a central role in bridging analysis and algebra.  I'm curious to understand, intuitively, why there are 4 classical Lie groups/algebras $A_{n},B_{n},C_{n},D_{n}$ and what are their various natures.

$A_{n}$ special linear Lie algebra $\mathfrak {sl}_{n+1}$

$B_{n}$ odd-dimensional special orthogonal Lie algebra $\mathfrak {so}_{2n+1}$

$C_{n}$ symplectic Lie algebra $\mathfrak {sp}_{2n}$

$D_{n}$ even-dimensional special orthogonal Lie algebra $\mathfrak {so}_{2n}$

These writings raise my curiosity:

Each of the four types $W, S, H, K$ of simple primitive Lie algebras $(L, L_{0})$ correspond to the four most important types of geometries of manifolds: all manifolds, oriented manifolds, symplectic and contact manifolds. Victor Kac, page 20.

Whenever we pick a Dynkin diagram and a field we get a geometry: $A_{n}$ projective, $B_{n}$ $D_{n}$ conformal, $C_{n}$ symplectic. John Baez, in his blog.

John Baez, in writing about the octonions, combines the orthogonal Lie algebras/groups so as to have three families. These arise naturally as symmetry groups of projective spaces over $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$, respectively. @JakobH and @arctic-tern discuss that at this post.

I imagine that I could study the associated Weyl groups and the infinite families of polytopes (simplexes, cross-polytopes, hypercubes, demicubes) which they are symmetry groups for.  I write my thoughts about that here.  I imagine that I should study the bilinear and sesquilinear forms as described in this Wikipedia article on the classical groups.

Concretely, I've been looking at the Dynkin diagrams and learning how they relate to the Cartan matrices and the orthogonal bases used to describe the root systems.  The Dynkin diagrams are, in each case, chains that differ only at one end.  The links in the chain show that two fundamental roots are not orthogonal but differ by 120 degrees. I think the associated Cartan matrix can be thought of as describing the looseness in the root system, that is, the number of times that a root $\alpha$ can be added to $-\beta$ and stay within the root system. The Cartan matrix needs to describe an independent basis and thus needs to be nondegenerate with determinant nonzero, and in fact, we can write the Cartan matrix as $D\cdot S$ where $S$ is positive-definite. We can thereby show that if the Dynkin diagram grows in certain ways, for example, if it forms a cycle, then the determinant will become zero, the system will collapse. The determinant of the $n\times n$ Cartan matrix for a chain grows by the recursion formula $d_{n+1}=2d_{n}-d_{n-1}$, which is to say, $d_{n}=(d_{n-1}+d_{n+1})/2$. In the case of $A_{n}$ the initial conditions are $d_{1}=2$ and $d_{2}=3$ so it grows further $4,5,6...$.  But in the case of $B_{n}$ and $C_{n}$ we have $d_{1}=2$ and $d_{2}=2$ and so the determinants stay constant $2,2,2...$ and likewise for $D_{n}$ we have $4,4,4...$  The point is that the one end of the chain constrains what is possible at the other end of the chain.  The constraint is given by the fact that the wrong link at the end will make the whole system collapse.  It seems like a system for propagating a signal or for describing entanglement.

But what I'm trying to understand is how to explain the four possible options for the Dynkin diagrams and what they imply about the characteristics of the Lie algebras/groups.  I am trying to explore the relationship between the Cartan matrices and the fundamental roots $\left \{  e_{i}-e_{i+1} | 1\leq i < n\right \}$ with each diagram distinguished by an additional root as follows:

$A_{n}: e_{n}-e_{n+1}$

$B_{n}: e_{n}$

$C_{n}: 2e_{n}$

$D_{n}: e_{n-1}+e_{n}$

Each of these new roots maintains an independent system of roots but the cases $B_{n}, C_{n}, D_{n}$ "economize" and thereby compromise the options for expanding the system.

Finally, I'm noticing based on what I've read at Wikipedia about positive-definite matrices that the formula $z^{\mathrm{T}}Sz=0$, with $D\cdot S$ equalling the Cartan matrix as discussed above, seems to work like a formula in logic. For example, for $A_{2}$ we get $z_{1}^{2}+(z_{1}-z_{2})^{2}+z_{2}^{2}=0$ forcing $z_{1}=0,z_{1}=z_{2},z_{2}=0$, which means that the vector $z=0$. Whereas for a cycle we could get
$(z_{1}-z_{2})^{2}+(z_{2}-z_{3})^{2}+(z_{3}-z_{1})^{2}=0$ allowing a degenerate, nonzero solution $z_{1}=z_{2}=z_{3}$. Setting

$S = \begin{pmatrix}
k & \frac{-1}{k}\\
\frac{-1}{k} & 1
\end{pmatrix}$

to handle the cases $A_{2} (k=1), B_{2} (k=2), G_{2} (k=3)$, we get the equation

$(1-\frac{k}{4})z_{2}^2+k(z_{1}-\frac{z_{2}}{2})^2=0$

which forces us to have solutions $z_{2}=0$ and $z_{1}=z_{2}/2$ and also shows that we we will have degenerate solutions for $k\geq 4$.

I have yet to find an intuitive explanation why there are 4 classical Lie algebras/groups.  I would very much appreciate one.  Otherwise, I suspect an answer might be found by working from the most concrete vantage point rather than the most abstract one. Thank you for helping me.

答:
You may find insight in Coxeter's work on the subject, which leans heavily on classical geometry.  In Regular Polytopes, $\S$11-5, Coxeter achieves the result in question by studying the fundamental region of reflection groups (I also recommend reviewing the historical context in $\S$11-x).  Intuition is subjective, but I find Coxeter's style as a classical geometer highly intuitive.

In the reference above, Coxeter relies heavily on Cartan's paper La géométrie des groupes simples.  This paper is extremely lucid, and has also been a significant source of intuition for me on this subject.

ContentLicense: CC BY-SA 3.0


问:
How to calculate this angle from 2 points in 3d space?
How do I find the following angle $a$ given $2$ points $(x, y, z)$  in $3$-dimensional space?

I've drawn $2$ points, one in green, one in red. The curved black line being the earth, and the normal vector is perpendicular to the earth. Green point is on the ground, and red point will always be above ground. Can assume earth is centered at $(0, 0, 0)$

Edit: can assume red vector is pointing in opposite direction than drawn.

答:
The angle between two vectors can be given by the dot product. More specifically, if $a$ and $b$ are vectors, then
$$a\cdot b = \|a\|\|b\|\cos\theta,$$
where $\|a\|$ is the length of $a$, $\|b\|$ is the length of $b$, and $\theta$ is the angle between the two (in radians!). So here you have $a= (1,1,1)$, $b  =(1,1,1.3)$, then you can compute their lengths and solve for $\theta$.


问:
Taylor series of $\ln(x)$ not uniformly convergent

How do I show that the Taylor series of $\ln(x)$ at point $x=1$ doesn't converge uniformly on any interval $(0,\delta)\subseteq(0,2)$?

My edited approach:

Let be $S_n(x):= \sum\limits_{k=1}^n \frac{(-1)^{k+1}}k(x-1)^k$ the $n$-th Taylor polynomial of $\ln(x)$ at point $x=1$.

The ieda is to prove the statement by showing that the Cauchy-criterion can't be satisfied, namely:

$\exists \epsilon>0$ such that for all $n_0\in\mathbb{N}$ there are two $n>m>n_0$ such that $\Vert S_m-S_n\Vert_{\infty}\geq\epsilon$.

First we assume $0<\delta\leq1$ and define a sequence by $x_n:=1-(\frac{1-\delta}{2})^{\frac{1}{n}}$, clearly $x_n\in(0,\delta)$.

We observe that: $$\Vert S_m-S_n\Vert_{\infty}\geq|S_n(x_n)-S_m(x_n)|=|\sum\limits_{k=m+1}^n\frac{(-1)^{k+1}}k(x_n-1)^k|=|\sum\limits_{k=m+1}^n\frac{(-1)^{2k+1}}k(1-x_n)^k|\\=|-\sum\limits_{k=m+1}^n\frac{(1-x_n)^k}k|\geq \sum\limits_{k=m+1}^n\frac{(1-1+(\frac{1-\delta}{2})^{\frac{1}{n}})^n}k=\frac{1-\delta}{2}\sum\limits_{k=m+1}^n\frac{1}k.$$ As the harmonic series doesn't converge at all the Cauchy criterion can't be satisfied.

However, the case when $1<\delta<2$ doesn't seem so easy. At first glance it looks like that the Taylor series could converge uniformly because there appears an alternating series and $(x-1)^k$ resembles a geometric series!?

$|S_n(x)-S_m(x)|=|\sum\limits_{k=m+1}^n\frac{(-1)^{k+1}}k(x-1)^k|=...$?

Is the first part o.k.? How do I proceed in the second part?

答:
First consider uniform convergence on any interval $(0,\delta)$ where $\delta \leqslant1$.

We have

$$\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|=  \sum_{k=n+1}^{\infty}\frac{(1-x)^k}k \geqslant\sum_{k=n+1}^{2n}\frac{(1-x)^k}k ,$$

and

$$\sup_{x \in (0,\delta)}\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|\geqslant  \sup_{x \in (0,\delta)}\sum_{k=n+1}^{2n}\frac{(1-x)^k}{k} \geqslant \sum_{k=n+1}^{2n}\frac{(1-1/n)^k}{k}\\\geqslant n \cdot \frac{(1-1/n)^{2n}}{2n} =  \frac{(1-1/n)^{2n}}{2}\underset{n \to \infty}\longrightarrow \frac{e^{-2}}{2}\neq 0$$

Hence, the convergence is not uniform on $(0,\delta)$ with $\delta \leqslant 1$.  Of course, that means the series fails to converge uniformly on any interval $(0,\delta)$ with $\delta <2$, since the source of the non-uniform convergence is the behavior near the   endpoint $x = 0$.

On the interval, $[\delta, 1]$ where $\delta < 1$, the series takes the form $-\sum_{k=1}^{\infty}\frac{(1-x)^k}{k}$. Since, $\frac{(1-x)^k}{k} \leqslant (1-\delta)^k$ we have uniform convergence here by the Weierstrass M-test.

You should be able to show that convergence is also uniform on the interval $[1,2]$.  Hint: Dirichlet test.


问:
Finding the PDF
Consider a region $A = \{(x, y) \in \mathbb{R}^2|y - 2 \leq x \leq y + 1, 0\leq y \leq 1\}$. If $(X, Y)$ is jointly uniform on this region, find the joint pdf and marginal pdfs of $X$ and $Y$. Determine if $X$ and $Y$ are independent.

My attempt:

Jointly PDF: In the region $A$, we know that $f(x, y)$ must be of the form:

$$f(x, y)= \left\{
  \begin{array}{lr}
      (\operatorname{Area}(A))^{-1} & \mbox{if } (x, y) \in A \\
      0 & \mbox{otherwise},
      \end{array}
\right.$$
where
$$\operatorname{Area}(A)=\int_{0}^{1}\int_{y-2}^{y+1} \,dx \,dy = 3.$$

We have

$$f(x, y)= \left\{
  \begin{array}{lr}
      \frac{1}{3} & \mbox{if } (x, y) \in A \\
      0 & \mbox{otherwise.}
      \end{array}
\right.$$

Marginal pdfs: Within the region we have
$$f_x(x)= \left\{
  \begin{array}{lr}
      \int_{x+2}^{-1} \frac{1}{3} \,dy = \frac{-1}{3} - \frac{1}{3}(x+2)  & \mbox{if } x \in [-2, 1] \\
      \int_{-1}^{1} \frac{1}{3} \,dy = \frac{2}{3}  & \mbox{if } x \in [-1, 1] \\
      \int_{1}^{x-1} \frac{1}{3} \,dy = \frac{1}{3} (x-2) & \mbox{if } x \in [1, 2], \\
      \end{array}
\right.$$
so
$$f_y(y) = \int_{0}^{1} \frac{1}{3} \,dx = \frac{1}{3}.$$

Is this correct or is it $f_y(y) = \int_{y-2}^{y+1} \frac{1}{3} \,dx = 1$? Are they independent? Clearly, $f(x, y)$ is not equal to $f_x(x) \cdot f_y(y)$. They are not independent.

Is this correct? Did I make a mistake? I'm fairly new to joint distributions and so any assistance is appreciated.

答:
Almost.

The domain is the trapezium: $$\def\<{\langle}\def\>{\rangle}\begin{align}A &= \{\< x, y\> \in \mathbb{R}^2: 0\leq y \leq 1\land y {-} 2 \leq x \leq y {+} 1\}\\[1ex] &= \{\< x,y\>\in\Bbb R^2: -2\leq x\leq 2 \land \max(0,x{-}1)\leq y\leq\min(1,x{+}2)\}\\[1ex] &=\{\<x,y\>\in\Bbb R^2:{{(-2\le x\lt {-1}\land 0\le y\le x{+}2)}\lor\\{(-1\le x\lt 1\land 0\le y\le 1)}\lor\\{(1\le x\le 2\land x{-}1\le y\le 1)}}\lower{6.5ex}\}\end{align}$$

Notice where we will partition $x$ and what is the corresponding domain for $y$ at each part.

Thus as the joint pdf is $f(x,y)=\tfrac 13\mathbf 1_{\< x,y\>\in A}$ (as you calculated), we therefore have::

$$\begin{align}f_{\small X}(x)&=\begin{cases}\int_0^{x+2}\tfrac 13\mathrm d y&:& x\in[-2..{-1})\\\int_{0}^{1}\tfrac 13\mathrm d y&:& x\in[-1..1)\\\int_{x-1}^{1}\tfrac 13\mathrm d y&:& x\in[1..2]\\0&:&\text{elsewhere}\end{cases}\\[2ex]&=\begin{cases}(x+2)/3&:& x\in[-2..{-}1)\\1/3&:&x\in[-1..1)\\(2-x)/3&:& x\in[1..2]\\0&:&\text{elsewhere}\end{cases}\end{align}$$

Likewise, by integrating over $x\in[y{-}2..y{+}1]$ in the support $y\in[0..1]$ we evaluate: $$\begin{align}f_{\small Y}(y)&=\int_{y-2}^{y+1}\tfrac 13\mathbf 1_{y\in[0..1]}\mathrm d x \\[1ex]&= \mathbf 1_{y\in[0..1]}\\[2ex]&=\begin{cases}1&:& 0\leq y\leq 1\\0&:&\text{elsewhere}\end{cases}\end{align}$$

Title: Finding the PDF


问:
Is there $z \in \mathbb{C}$ with $\mathrm{Re}(z) > 0$ such that $e^{2z}+\frac{z+1}{z-1} = 0$?
Let $z$ be a complex number with positive real part. Is it possible that $e^{2z}+\frac{z+1}{z-1} = 0$?

答:
Define:
$$f:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto(z-1)e^{2z}+(z+1),$$
$$g_1:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto|f(x+iy)|^2,$$
$$g_2:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto g(x,y)e^{-2x},$$
$$h:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto(x-1)e^{2x}+1.$$
Now prove this:
$$(\forall x>0)\frac{\partial g_1}{\partial x}>0\Rightarrow(\forall x>0)g_1(x,y)>0,$$
$$(\forall x>0)\frac{\partial g_2}{\partial x}>0\Rightarrow(\forall x>0)\frac{\partial g_1}{\partial x}>0,$$
$$(\forall x>0)\frac{\partial g_2}{\partial x}\geqslant 2x(h(x)+h(-x)),$$
$$(\forall x\neq 0)h(x)+h(-x)>0.$$


问:
Show that $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$

Show that $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$

Letting $y=e^x-1\implies e^x=y+1\implies x=\log(y+1)$ the evaluation is easy.

But I can't understand how to express given function as a composition of two functions so that the following rule can be used limit operation can be done.

Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$

答:
Another way to do it is to define $f: \mathbb{R} \to \mathbb{R}$ by setting $f(x)=e^x$, then note that $f(0) = 1$, in that case the derivative of $f$ at the point $0$ can be given by the limit:

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

In that case your limit is just $f'(0)$ which is $1$ since $f'(0) = e^0$. Of course this only works when you know that $f'(x) = e^x$. This helps you somehow?


问:
Does direct substitution in multivariable limits save us from checking other possible directions?
When dealing with a multivariable limit, if we find that direct substitution gives us a valid answer, then can we assume that the limit does exist and not check the other directions where it might not exist?

For example, given the limit $$\lim_{(x,y)\to(1,1)}\frac{2xy}{x^2+2y^2},$$
we find through direct substitution that it approaches $2/3$. Would we care about approaching it from the $x$-axis, $y$-axis, etc.?

答:
1）
it is necessary to check.other possible approach, if you plug in (0,0), the value2/3 is just the y value at the point (0,0),it doesn't represent the limit of the f(x,y) when（x，y）approaches（0,0）.

2）
in analysis， the existence of limit  at a point has nothing to do with the existence of the function's value of this point.
take some simple function for example
let
f(x)=1/x (x≠0),
=1     (x=0)
when x approaches 0, f(x) approaches ±∞. but if you plug in x=0, the function's value is 1

3)
for your example,let y=kx, y approaches 0 along the line kx,
so you substitute, and you find out that
f(x,kx)(x⇒0)=k/1+2k², this value varies in terms of different k, so the limit dose not exist


问:
How to properly evaluate this expression of a Wiener integral $E\left[ \exp \left( \int_0^t f(s)dB_s \right)^2 \right]$?
Given $f \in L^2$, I am trying to simplify the following expression

$$E\left[  \exp \left( \int_0^t f(s)dB_s \right)^2 \right]$$

What I know is that $\int_0^t f(s)dB_s$ is a Wiener integral, so it has a Gaussian distribution of zero mean and variance $\int_0^t f(s)^2ds$.  I am getting out of my comfort zone but here it goes.

$\int_0^t f(s)dB_s$ is Gaussian, so $\exp \left( \int_0^t f(s)dB_s \right)$ is log-normal. Let $\exp \left( \int_0^t f(s)dB_s \right)=Y_t$, so then $E[Y_t^2]$ is the second moment of the log-normal random variable. Using the moment generating function $E[X^n]=\exp(n\mu + n^2 \sigma^2/2)$ we have
$$E[Y_t^2]=\exp\left(2\int_0^t f(s)^2ds \right)$$

The importance of this is that I would like it to be finite, which it is (if this derivation is correct) since $f \in L^2$. How does this look?

答:
We have $ \int_0^t f(s)dB_s \sim N(0;\int_0^t f^2(s)ds)$.

Let's denote $a =\int_0^t f^2(s)ds$, then it suffices to calculate $E(e^{aX^2})$ with $X \sim N(0,1)$.

And $E(e^{aX^2})$ exists if and only if $a<\frac{1}{2}$. If this condition is satisfied, then $E(e^{aX^2}) = \frac{1}{\sqrt{1-2a}}$

Conclusion: if  $\int_0^t f^2(s)ds < \frac{1}{2}$ then
$$E\left[  \exp \left( \int_0^t f(s)dB_s \right)^2 \right] = \frac{1}{\sqrt{1-2\int_0^t f^2(s)ds}}$$

PS: the expectation $E\left[ \left( \exp \left( \int_0^t f(s)dB_s \right)\right)^2 \right]$ can be calculated as follows

\begin{align}
E\left[ \left( \exp \left( \int_0^t f(s)dB_s \right)\right)^2 \right] &= E\left[  \exp \left( 2\int_0^t f(s)dB_s \right) \right] \\
&= E\left[  \exp \left( 2\sqrt{\int_0^t f^2(s)ds} *X \right) \right] \\
\end{align}
We know that $E[e^{aX}] = e^{\frac{a^2}{2}}$, then
$$E\left[ \left( \exp \left( \int_0^t f(s)dB_s \right)\right)^2 \right] = \exp \left( 2\int_0^t f^2(s)ds \right)$$


问:
Finite Element Method and implementation
I'm studying this problem:

$$
\begin{equation}
\left (\mathcal P\right ) \quad \begin{cases}\label{P}
-u''(x) = f(x), \quad \text{sur} \quad I = ]0,1[\\
u'(0) = 1 \\
u(1) = 0
\end{cases}
\end{equation}
$$

where

$$
f(x) = H\left(x-\frac{1}{2} \right), \quad \text{H = Heaviside function}
$$

$$
\begin{align*}
-u'(x) & = (x-1/2)H\left(x-\frac{1}{2}\right) + c_1, \quad u'(0) = 1 \implies c_1 = 1 \\
& = \frac{1}{2}(2x-1)H\left(x-\frac{1}{2}\right) + 1
\end{align*}
$$

$$
\int (2x-1)H(x-{1}{2}) \, dx   = \left[ \frac{1}{2}(2x-1)^2H\left(x-\frac{1}{2}\right)  \right] - \int (2x-1)H\left(x-\frac{1}{2}\right) \, dx
$$

$$
\implies \int (2x-1)H\left(x-\frac{1}{2}\right) \, dx = \frac{1}{4}(2x-1)^2H\left(x-\frac{1}{2}\right)
$$

$$
u(x) =  -\frac{1}{8}(2x-1)^2H\left(x-\frac{1}{2}\right) + x + c_2, \quad u(1) = 0 \implies c_2 = -\frac{7}{8}
$$

the exact solution is:

$$
u(x) =  -\frac{1}{8}(2x-1)^2H\left(x-\frac{1}{2}\right) + x -\frac{7}{8}, \quad  u\left(\frac{1}{2}\right) =  \frac{-3}{8}
$$

We know that we need to search the solution in H^1(I) with the boundary conditions. How can i show that it's also in H^2(I)?

Also, at the end with the P1 finite element aproximation,  i have this matrix:

$$
A = \begin{pmatrix}
    \dfrac{2}{h}  & -\dfrac{1}{h}  &  0 & \dots & \dots & 0\\
    -\dfrac{1}{h}  & \dfrac{2}{h} & -\dfrac{1}{h}  & 0 & \dots & 0 \\
    0  &   \ddots     & \ddots & \ddots & \vdots & \vdots\\
    0  &   \ddots     & \ddots & \ddots & \vdots &  0\\
    0 & \dots & \ddots & -\dfrac{1}{h}  & \dfrac{2}{h} & -\dfrac{1}{h}  \\
    0  & \dots & \dots & 0  & -\dfrac{1}{h}  & \dfrac{2}{h}
    \end{pmatrix}
$$

which i easily implemented with in Fortran. I'm wondering how can i implement the source term?

The variational formulation of this problem is $$
\begin {cases}
u\in V, \forall v\in V\\
\int_{0}^{1} u'(x)v'(x)  \, dx  = \int_{0}^{1} f(x)v(x) \, dx  -  v(0)
\end {cases}
$$

答:
For the $H^2$-regularity, since the space $V=\{v\in H^1(0,1):v(1)=0\}$, and $C^\infty_0(0,1)\subset V$, from the weak formulation, we immediately get that

$$\int_0^1u'(x)v'(x)=-\int_0^1(-f)v\quad\forall v\in C^\infty_0(0,1),$$

and so $-f$ is the second weak derivative of $u$. Since $-f\in L^2(0,1)$, it follows that $u\in H^2(0,1)$.

For the right-hand side implementation, I assume you have a uniform finite element grid, and that the middle gridpoint lies on the point $x=1/2$, and that the finite element space is piecewise linear. I'm also assuming that $H$ is zero on $[0,1/2)$ and $1$ otherwise.

You then just have to calculate $L(v_i):=\int_0^1 fv_i\,dx-v_i(0)=\int_{1/2}^1v_i\,dx-v_i(0)$ for each linear basis function $v_i$. Interestingly for $v_0$ (the basis function that is $1$ at $x=0$ and $0$ at the first internal gridpoint $x_1=h$, we have $L(v_0) = -v_0(0)=-1$. For all the $v_i$'s that are supported only in $[h,1/2]$, we have $L(v_i)=0$. For the $v_i$ that is $1$ at $x=1/2$, we have $L(v_i) = \int_{1/2}^{1/2+h}v_i = \sqrt{h^2+1}$, then for all the rest of the $v_i$'s $L(v_i)=\int_{1/2}^{1}v_i = 2\sqrt{h^2+1}$ (since we integrate over the full support of $v_i$).


答:
You can get all the transpositions from those three.  There are ${4\choose2}=6$ transpositions total, so there are sort of $3$ things to check.

For instance, $(12)(23)(12)=(13)$.

(One more hint:  the conjugation action is transitive;  that is all the transpositions are conjugate)

So it remains to find $(14)$ and $(24)$.

Title: show a subset generates a symmetric group

Body:
How would I show that the subset $V = \{(1,2), (2,3), (3,4)\}$ generates the symmetric group $S_4$?

I am just being introduced to group theory this is all new. I have unsuccessfully attempted to solve this question. . How do you prove this?


答:
Oh, okay:

Consider this.  Using the weird wrong definition.

$\lim_{x\to 3} x = 591$.

Proof: For any $\epsilon > 0$ let $\delta = 588 + \epsilon$

If $|f(x) - 591| < \epsilon$ then $|x-591| < \epsilon$ so $591 -\epsilon < x < 591 + \epsilon$.  So $588 -\epsilon < x -3 < 588+ \epsilon$.

So $-\delta = -588 - \epsilon < 588-\epsilon < x-3< 588 + \epsilon=\delta$

So $|x-3| < \delta$.

So $\lim_{x\to 3} x = 591$.

Title: $\forall\varepsilon > 0\exists\delta > 0:(|f(x) - \ell| < \varepsilon)\implies(0<|x-a| < \delta )$. $f(x)=x$ satisfies this for all $a$ and $\ell$?

Body:
This is to be found in the answer to problem 26 (b) of chapter 5 of Michael Spivak's Calculus 3rd edition. The problem reads as follows:

26 Give examples to show that the following definitions of $\lim_{x\to a}f(x)=\ell$ are not correct.

(b) For all $\varepsilon > 0$ there is a $\delta > 0$ such that if $|f(x) - \ell| < \varepsilon$, then $0<|x-a| < \delta$.

I thought of constant functions because they do not satisfy this condition, but approach a limit "everywhere". This means, let $f(x)=c, c\in \mathbb R$ and let $a\in \mathbb R$ and let $\varepsilon > 0$, then $|f(x) - c|=0 < \varepsilon$ for all $x$ satisfying $0<|x-a| < \delta$ for any $\delta > 0$ and not just some $\delta > 0$. (It is already shown in the text of this chapter that $\lim_{x\to a}c=c$.)

In his answer Spivak also mentions constant functions but then adds the following:

Moreover, the function $f(x)=x$, for example, satisfies this condition no matter what $a$ and $\ell$ are.

What does he mean by that? It seems to me that if $|f(x) - \ell| < \varepsilon$ holds, then $|x - \ell| < \varepsilon$ so $a=\ell$ does hold, which means it does matter what $a$ and $\ell$ are. Unfortunately, I really don't understand what he is getting at.


问:
we need to know at $z=0$ ,$f(z)={e^z+1\over e^z-1}$ has a
we need to know at $z=0$ ,$f(z)={e^z+1\over e^z-1}$ has a

removable singularity

a pole

essential singularity

residue is $2$

for removable singularity we need $\lim_{z\to 0}zf(z)=0$ but here this is not the case so $1$ is false, clearly $2$ is true, also $3$ is false, I have calculated that $\lim_{z\to 0}zf(z)=2$ so residue is $2$ at simple pole $z=0$ am I right in all cases?

答:
Yes, you are right.

Yes, you are right.

Yes, you are right.

Yes, the residue is 2.

I don't have too much to add unfortunately.


问:
Help with S&P probability

According to this model, what is the probability that the change in the S&P $500$ on a particular day is positive. We have $P(X>0)$ and then converting to standard normal table I get $P(Z>(0-0.75)/25=0.03$ because $\mu=0.75$ and the SD is $25$. Then using standard normal table we get $0.512$

What is the third quartile quartile of daily changes in the S&P $500$? I know that $\mu$ and SD is the same, but would I just use $0.75$ as $X$? that would not make any sense because then you would just be dividing $0/25$? Also is there a way to do 1. without the standard table?

答:
Let me address your last question first: "Also is there a way to do (a) without the standard table?"

Well, one needs to know enough about the density of $X$ to determine the probability $\mathbb{P}(X > 0)$.  Where we get this information, is secondary.  A table is a convenient device for getting an approximation close enough.  Another way is to compute the appropriate integral of the density function, a task no less pleasant than looking into the table.

The third quartile of $X$ is such a value $q$ that $\mathbb{P}(X \leq q) \geq 0.75$.  Your normalized variable is
$$
Z = (X - \mu) / \mbox{SD},
$$
so the condition $X \leq q$, rewritten in terms of $Z$, would be
$$
Z \leq (q - \mu)/\mbox{SD}.
$$
Thus, you would be using the table to find the $q' = (q - \mu)/\mbox{SD}$ such that
$$
\mathbb{P}(Z \leq q') \geq 0.75.
$$
Once you've found the $q'$, solve for $q$.

Title: Help with S&P probability


问:
Is broken Sobolev space a Sobolev space?
The definition of a broken Sobolev space is as follows.
Given infinite-dimensional (but mesh-dependent)
spaces on an open bounded domain $\Omega \in R^3$ with Lipschitz boundary. The mesh, denoted by $\Omega_h$, is a disjoint partitioning of $\Omega$ into open elements $K$ such that the union of their closures
is the closure of $\Omega$. The collection of element boundaries $\partial K$ for all $K \in \Omega_h$, is denoted by
$\partial \Omega_h$. We assume that each element boundary $\partial K$ is Lipschitz. The shape of the elements is
otherwise arbitrary for now. The broken Sobolev space is defined as
$$\hat{H}^{1}\left(\Omega_{h}\right)=\left\{u \in L^{2}(\Omega):\left.u\right|_{K} \in H^{1}(K), K \in \Omega_{h}\right\},$$
where $H^1$ is the standard Sobolev space. According to my understanding, if $\Omega_h$ can be partitioned into a finite number of subdomains, $K$'s, and because the union of the intersections of $K$'s is a set of measure zero, then $\forall u \in \hat{H}^1, \int_{\Omega_h} \sum_{|\alpha|\leq 1} (D^\alpha u)^2 dx = \sum_{K=1}^n \int_K  \sum_{|\alpha|\leq 1} (D^\alpha u)^2 dx$. This seems mean that $\hat{H}^{1}$ is also a Sobolev space.

If the partion of $\Omega_h$ consists of an infinite number of $K$'s, then the equality of the integration above would not hold. In this case, $\hat{H}^{1}$ is not a Sobolev space.

Am I right about the above two statements? Thanks a lot.

答:
For the first question, $\hat{H}^1$ is not a Sobolev space, since the functions in $\hat{H}^1$ do not have a weak derivative belonging to any Lebesgue space, e.g. $H^1$ is the space of functions in $L^2$ with weak derivatives in $L^2$. A broken Sobolev space simply means that the functions belongs to Sobolev spaces on each individual element.

The second question is very interesting, and the paper: https://arxiv.org/abs/2006.07215 considers the notion of limiting broken Sobolev spaces (see the Definition 4.1). Here, the infinite partition $\Omega_\infty$ is obtained by iteratively refining the mesh of the domain, so that one does obtain an infinite collection of elements. The corresponding limit space is not a classical Sobolev space (unless the refinement of the domain eventually refines everywhere), but it does have similar characteristics.


问:
$f$ continuous $\Rightarrow$ $f$ $\mu$ measurable?
Let $f: X\to [0,\infty]$ be a continuous function and $\mu$ an outer measure on X.
For a continuous function I can split it in a sum of step functions, so $$ f(x) = \sum\limits_{i=1}^\infty s_i \chi_{A_i} $$ Is $f$ also $\mu$ measurable and therefore Lebesgue integrable?

If not can you give me an example? I am not even sure if the splitting is possible for every continuous function.

Edit:

I see that splitting isn't possible for every continuous function. After reading a bit in my notes of my lecture, I cant see why I can argument that a function $f$ is $\mu$ measurable if it is continuous.
Has someone an explaining or a hint so I can try the prove myself?

答:
Welcome to MSE!

We need a little bit of bonus information -- $X$ must be a topological space to make sense of continuous functions, and $X$ must also be equipped with a $\sigma$-algebra to make sense of measurable functions.

In order for continuity to imply measurability, these two structures must be compatible (in which case $\mu$ is called a borel measure). Without this constraint, things can go wrong. For instance if $X$ is equipped with the discrete topology, then every function is continuous. As an exercise, you might try to use this idea to find an explicit example of a space $X$ equipped with a topology and a $\sigma$-algebra so that some continuous function is not measurable.

Ok, so with the pedantic caveat out of the way, let's try to see why (for borel measures) a continuous function is automatically measurable. Here's a hint:

Let's look at the definitions of measurable and continuous functions.

$f$ is continuous whenever $f^{-1}((a,b)) \subseteq X$ is open for every $(a,b) \subseteq \mathbb{R}$.

$f$ is measurable whenever $f^{-1}((a,b)) \subseteq X$ is measurable for every $(a,b) \subseteq \mathbb{R}$.

So to show continuous $\Rightarrow$ measurable, it suffices to show that every open subset of $X$ is measurable (do you see why?). Can you use our assumption that $\mu$ is a borel measure to finish up?

I hope this helps ^_^


问:
Confusing dot product and inner product in a weak formulation
I have been struggling with this for a while. Here, as you can see, they define the weak formulation of the Poisson equation as:

$-\int_{\Omega}\nabla u\cdot\nabla v\,ds = \int_{\Omega}fv\,ds \equiv -\phi(u,v)$ (not sure what the $\equiv$ symbol means here(?)), where they specify that the ''." denotes the dot product, $v$ is a test function, $u$ the unknown and $\phi$ an operator (or map). Few lines later; they explain: "Such functions are (weakly) once differentiable and it turns out that the symmetric bilinear map $\phi$ then defines an inner product which turns $H_0^1(0,1)$ into a Hilbert space"

Does this mean that the weak formulation in this case is by default an inner product that uses the dot product?

EDIT: I checked this pdf file that studies the weak formulation of Poisson's equation, apparently the weak formulation defines an inner product.

I just need a confirmation if what I am doing is correct...

答:
The reason it's an inner product is that the inner product axioms are satisfied for $\phi(u,v)=\int_\Omega\nabla u\cdot\nabla v.$

The easy ones to show are symmetry and linearity, i.e., $\phi(u,v)=\phi(v,u)$ and $\phi(au,v)=a\phi(u,v)$ ($a$ constant). One has to also show that $\phi(u,u)\ge0$ for all $u$, and $\phi(u,u)=0$ if and only if $u=0$.

One has that $\phi(u,u)=\int_\Omega|\nabla u|^2\ge0$. Then, if $u=0$, then $\phi(u,u)=\int_\Omega0=0$, and if $\phi(u,u)=0$, then $\int_\Omega|\nabla u|^2=0$, and so $u$ must be constant, but also $u\in H^1_0(\Omega)$, so the constant must in fact be zero, so $u=0$.


问:
Birthday $pmf$ and joint $pmf$ for $n$ people

Let a discrete r.v be denoted $X_1,..,X_n$ denote the birthdays of $n$ people in a room. Assume that $X_1,..,X_n$ are mutually independent and that $X_i$ is a distribution such that $X_i \sim U(\left \{ 1,2,...,365 \right \})$ for all $i \in {(\left \{ 1,...,n\right \}}$.

Find and write out the pmf $p_i: \mathbb{R} \rightarrow \mathbb{R}$ for the r.v $X_i$.

Find the joint pmf $p:\mathbb{R}^n \rightarrow \mathbb{R}$ for the vector $(X_1,..,X_n)$ and show that it is a valid joint pmf.

My try

First: The pmf would just be the pmf of a uniformal distribution. That would mean that $p_i(x)={x_i}^{-1}$ for the $x=1,...,n$. (Now the notation is weird for me as it is for the $i$'th $X_i$). (I know that $p_i: \mathbb{R} \rightarrow \mathbb{R}$)

Second: I know that for it to be a valid pmf the total probability has to sum up to $1$. The joint pmf should be denoted as $p_{X_1,...,X_n}(x_1,...,X_n):=P(X_1=x_1,...,X_n=x_n)$ but no idea on how to write this out from my results. Wouldnt I just have to replace $x_i$ with $1/365$? (I know that $p_i: \mathbb{R}^n \rightarrow \mathbb{R}$)

答:
It was claimed that $X_i\sim\mathcal U\{1..365\}$ for all $i\in\{1..n\}$.

The probability mass function $p(u)$ for a uniform discrete distribution of $\mathcal U\{a{.}{.}b\}$, is $$p(u)=(b-a+1)^{-1}\mathbf 1_{u\in\{a..b\}}$$

So for any $i\in\{1..n\}$, the marginal probability mass function $p_i(x)$ is...

For the second, the keywords are "mutually independent".    There's something about the joint probability mass function of mutually independent random variables ...


答:
The first question is easier than you think: each day is equally likely and there are $365$ possibilities so $p_i(x)=\frac1{365}$ for $x \in \{1,2,3,\ldots,365\}$ and $0$ otherwise

Since the birthdays are independent you have $p(\mathbf{x}) $ $=p(x_1,x_2,\ldots,x_n) $ $= \mathbb P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n) $ $= \mathbb P(X_1=x_1)\mathbb P(X_2=x_2)\cdots\mathbb P(X_n=x_n) $ $= p_1(x_1)p_1(x_2)\cdots p_n(x_n)$.  This will be $\frac{1}{365^n}$ when all of the $x_i \in \{1,2,3,\ldots,365\}$ i.e. when $\mathbf{x}\in \{1,2,3,\ldots,365\}^n$, and $0$ otherwise.  You can easily check that it adds up to $1$.

Title: Birthday $pmf$ and joint $pmf$ for $n$ people

Body:
Let a discrete r.v be denoted $X_1,..,X_n$ denote the birthdays of $n$ people in a room. Assume that $X_1,..,X_n$ are mutually independent and that $X_i$ is a distribution such that $X_i \sim U(\left \{ 1,2,...,365 \right \})$ for all $i \in {(\left \{ 1,...,n\right \}}$.

Find and write out the pmf $p_i: \mathbb{R} \rightarrow \mathbb{R}$ for the r.v $X_i$.

Find the joint pmf $p:\mathbb{R}^n \rightarrow \mathbb{R}$ for the vector $(X_1,..,X_n)$ and show that it is a valid joint pmf.

My try

First: The pmf would just be the pmf of a uniformal distribution. That would mean that $p_i(x)={x_i}^{-1}$ for the $x=1,...,n$. (Now the notation is weird for me as it is for the $i$'th $X_i$). (I know that $p_i: \mathbb{R} \rightarrow \mathbb{R}$)

Second: I know that for it to be a valid pmf the total probability has to sum up to $1$. The joint pmf should be denoted as $p_{X_1,...,X_n}(x_1,...,X_n):=P(X_1=x_1,...,X_n=x_n)$ but no idea on how to write this out from my results. Wouldnt I just have to replace $x_i$ with $1/365$? (I know that $p_i: \mathbb{R}^n \rightarrow \mathbb{R}$)


答:
Start with $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\ldots$$ This means that $$e^x - 1= x+\frac {x^2}{2!}+\frac {x^3}{3!}+\ldots$$
and that $$\frac{e^x - 1}x= 1+\frac {x}{2!}+\frac {x^2}{3!}+\ldots$$

I believe you can take it from here.

Title: Show that $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$

Body:
Show that $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$

Letting $y=e^x-1\implies e^x=y+1\implies x=\log(y+1)$ the evaluation is easy.

But I can't understand how to express given function as a composition of two functions so that the following rule can be used limit operation can be done.

Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$


问:
Proper Way to Solve Mathematical Induction?
Say I have $2^n +1 < n! -n$ for all $n \ge 4$, and $n$ is an integer.

My inductive steps says, consider a $k$ that is a arbitrary integer, assuming $P(k)$.

Thus,
$$\begin{align} 2^{k+1} +1 = 2^k \cdot 2 +1 \\ 2^k\cdot 2+1\overset{\mathrm{IH}}{<} 2(k!-k) < ((k+1)!-k) \\\text{Therefore, we have proven this my mathematical induction.}
 \end{align}$$

I am not too sure on how to use < in mathematical induction. I based this off an example I saw, and wondering it is valid, and maybe some calcification on what this works (if it does)...

答:
Assuming $P(k)$ you have to show $P(k+1)$. In other words, assuming that $$2^k + 1 < k! - k,\tag{$P(k)$}$$ you have to show that
$$2^{k+1} + 1 < (k+1)! - (k+1). \tag{$P(k+1)$}$$
Your inductive hypothesis $P(k)$ can be rewritten as
$$2^k < k! - k - 1\tag{*}\label{*}.$$
Thus
\begin{align}
2^{k+1}+1
&= 2\cdot 2^k + 1\\
&\stackrel{\eqref{*}}{<}2(k!-k-1)+1\\
&=2k! - 2k -1\\
&<2k!-(k+1)\\
&<(k+1)!-(k+1).
\end{align}


答:
In general, mathematical induction on the statement $P(k)$ provides a proof strategy requiring the following steps:

Prove the base case: $P(k=0)$ is true

Assume the inductive hypothesis: $P(k)$ is true for some $k \in \mathbb{N}.$

Show that as a result of the above: $P(k+1)$ is true.

The proof is complete for all $k$.

In this case, your $P(k)$ is the statement $2^k + 1 < k! - k \ \forall k \geq 4 $.
Let us show $P(k=4)$ holds (the smallest value for the inequality): $2^4 + 1 = 17 < 4! - 4 = 24-4=20$ is clearly satisfied.

Now let us assume $2^k + 1 < k! - k$ for some $k \geq 4$ (since we've already proven the case $k=4$).
We want to use this to prove $2^{k+1}+ 1 < (k+1)! - (k+1)$

As you've mentioned, $2^{k+1}=2*2^{k}$. Recalling the inductive step above,
$2*2^{k} < 2*(k!-k - 1).$ Certainly $2k! - 2k - 2 = 2k! - 2*(k+1) < (k+1)! - (k+1)$.

This is because $2k! < (k+1)! = (k+1)k!\  \forall k > 1$.

Now, chaining together the last few inequalities, we arrive at $2*2^k=2^{k+1} < (k+1)! - (k+1)$

Title: Proper Way to Solve Mathematical Induction?

Body:
Say I have $2^n +1 < n! -n$ for all $n \ge 4$, and $n$ is an integer.

My inductive steps says, consider a $k$ that is a arbitrary integer, assuming $P(k)$.

Thus,
$$\begin{align} 2^{k+1} +1 = 2^k \cdot 2 +1 \\ 2^k\cdot 2+1\overset{\mathrm{IH}}{<} 2(k!-k) < ((k+1)!-k) \\\text{Therefore, we have proven this my mathematical induction.}
 \end{align}$$

I am not too sure on how to use < in mathematical induction. I based this off an example I saw, and wondering it is valid, and maybe some calcification on what this works (if it does)...


问:
Questions about the Gamma function
tl;dr What is the import of the Gamma function?

So, I've read everything I can about the Gamma function, including a whole book ("Gamma") on it. I've scoured the internet on the subject. I've watched videos in Hindi, which I don't speak, and no kidding every video on Youtube when you search "Gamma beta function" comes up with a video meant for Indian engineers. Thanks to this site, I found some links to academic articles that discuss the history of the gamma function, too. All cool.

But still, I don't get it.

I do understand the various expressions of it, and even how to solve a problem that involves the Gamma function. What I don't get is the importance of this.

For example, I have learned that the gamma function was developed to deal with the interpolation problem that the Eulers, Bernoullis, Wallis, Stirling and gang were all interested in. I do love the math history. So is it right to say that the main use of the gamma function is to calculate non-integer factorials, which might have use in probability? (It does appear, and I'm not well-tutored in this, that the gamma function is useful to statisticians and coders.)

And why the Gauss/Euler/Weierstrass variations? So, I can express the gamma function in several ways. Gauss uses the product operator; Euler uses the integral; etc. What is the use of all these different forms?

And Weierstrass, it is rarely stated (in my reading) but I found one source who talked about its derivation as coming from using complex numbers in the Euler formula. Is that what makes the Weierstrass version so significant?

In the end, is gamma just an easy way of calculating an integral? That is, most problems I've solved involve rewriting integrals to meet the canonical gamma form. Once done, the integral's value is just gamma(n). Easy. Is that what this is all about? I need a hack to evaluate nasty integrals?

And I could rewrite all of this about the beta function. Hmmmm....... I'd welcome the hivemind to show me what I'm not getting about this.

答:
So this huge post just breaks down to:

What is the use of the Gamma function and it's different representations.

Phew... I don't even know where to start.

First of all - we can extend the definition of the factorial, which definition is purely based on positive integers, to the real and the complex world. How cool is that??

In a mathematic-theoretical aspect one would argue to solve many many problems and bring them, if you see the Gamma function as a closed form, to a closed form. The nice thing about the Gamma function is also, as you said, it's different representations, which are helpful when evaluating different kind of problems. If I may add a personal statement concerning your Weierstrass comments, I am currently working on a proof that depends on the Weiertrass representation of the Gamma function.

Regarding applications in physics there are also a lot, even in the String theory. Physics is not exactly my branch but this paper gives a few insights.

Title: Questions about the Gamma function


答:
Let $\vec{p}_r = (x_r, y_r, z_r)$ be the red point, and $\vec{p}_g = (x_g, y_g, z_g)$ be the green point.  The direction unit vectors are then
$$\hat{r} = \frac{\vec{p}_r - \vec{p}_g}{\left\lVert \vec{p}_r - \vec{p}_g \right\rVert}, \quad
\hat{g} = \frac{\vec{p}_g}{\left\lVert \vec{p}_g \right\rVert}$$
where $\lVert\vec{p}\rVert = \sqrt{\vec{p} \cdot \vec{p}} = \sqrt{x^2 + y^2 + z^2}$.

The angle $\theta$ between the two unit vectors fulfills
$$\begin{aligned}
\cos\theta &= \hat{r} \cdot \hat{g} \\
\sin\theta &= \left\lVert \hat{r} \times \hat{g} \right\rVert \\
\end{aligned}$$
and usually you use $\theta = \arccos(\hat{r} \cdot \hat{g})$.

Title: How to calculate this angle from 2 points in 3d space?

Body:
How do I find the following angle $a$ given $2$ points $(x, y, z)$  in $3$-dimensional space?

I've drawn $2$ points, one in green, one in red. The curved black line being the earth, and the normal vector is perpendicular to the earth. Green point is on the ground, and red point will always be above ground. Can assume earth is centered at $(0, 0, 0)$

Edit: can assume red vector is pointing in opposite direction than drawn.


问:
$L_\infty$-stability for forward heat equation $u_t=\frac12u_{xx}$
I am considering the forward heat equation

$$\frac{\partial u}{\partial t}=\frac12\frac{\partial^2u}{\partial x^2}$$

with $0<x<1$, $t>0$ under the explicit Euler scheme such that

$$U_j^{(n+1)}=U_j^{(n)}+\frac{\Delta t}{2\Delta x^2}\left(U_{j+1}^{(n)}-2U_j^{(n)}+U_{j-1}^{(n)}\right).$$

I have already shown conditional $L_2$-stability using von Neumann analysis, and now would like to prove $L_\infty$-stability conditioned on $\Delta t\leq \Delta x^2$. My attempt at a proof is as follows: denoting $\lambda=\frac{\Delta t}{2\Delta x^2}$, we have

\begin{align*}
|U^{(n+1)}_j|&\leq|U^{(n}_j|+\lambda\left(|U^{(n)}_{j+1}|-2|U^{(n)}_j|+|U^{(n)}_{j-1}|\right)\\
&\leq\max_j|U^{(n)}_j|+\lambda\left(\max_j|U^{(n)}_{j+1}|-2\max_j|U^{(n)}_j|+\max_j|U^{(n)}_{j-1}|\right)\\
&=\max_j|U^{(n)}_j|,
\end{align*}

where I have used $|x+y|\leq|x|+|y|$. However, my attempt therefore leads to unconditional $L_\infty$-stability. Where have I gone wrong?

In general, how should I go about deriving the conditions for $L_\infty$-stability? I have just taken a closer look at the hints provided and it mentions that since

$$U_j^{(n+1)}=pU_{j+1}^{(n)}+(1-2p)U_j^{(n)}+pU_{j-1}^{(n)},$$

where $p=\lambda/2$, I must justify that for $2p\leq1$, $\min_j|U^{(n+1)}_j|\leq U^{(n+1)}_j\leq\max_j|U^{(n+1)}_j|$ $\forall n$, and therefore $\max_j|U^{(n+1)}_j|\leq\max_j|U^{(n)}_j|$. However, where this justification comes from (I suspect some form of discrete maximum principle) and it contradicts what was provided in the comment by @John_Krampf. What has happened here?

Body:
Since $2p\le 1$ implies that $1-2p\ge0$, we have that $|1-2p| = 1-2p$, and $p\ge 0$ implies that $|p|=p$, and so

\begin{align*}
|U^{n+1}_j| & = |pU^{n}_{j+1}+(1-2p)U^{n}_{j}+pU^{n}_{j-1}|\\
&\le|pU^{n}_{j+1}|+|(1-2p)U^{n}_{j}|+|pU^{n}_{j-1}|\\
&=|p||U^{n}_{j+1}|+|(1-2p)||U^{n}_{j}|+|p||U^{n}_{j-1}|\\
&=p|U^{n}_{j+1}|+(1-2p)|U^{n}_{j}|+p|U^{n}_{j-1}|\\
&\le p\max_k|U^{n}_{k}|+(1-2p)\max_k|U^{n}_{k}|+p\max_k|U^{n}_{k}|\\
&=(p+1-2p+p)\max_k|U^{n}_{k}|\\
&=\max_k|U^{n}_{k}|.
\end{align*}
Thus, $|U^{n+1}_j|\le \max_k|U^{n}_{k}|$ for all $j$, and therefore $\max_k|U^{n+1}_k|\le \max_k|U^{n}_{k}|$.


问:
In how many ways can a salesperson visit $15$ different offices during the week?
Consider the following question:

I am able to get the answer matched in the book for the part a and b but not part c.

I tried part a as:

```<code>Factorial[15]
</code>```

1,307,674,368,000

Part b as:

```<code>FactorialPower[15,4]
</code>```

32,760

But can't get the book's answer for part c. Please help.

The book is saying the answer to part c should be:

13,650

答: The book says that you do $\text{FactorialPower}[15,3]*5$. It is a little strange of an answer. It seems like the question is asking how many ways can you visit three different offices on a single day if the day can be any weekday.


答:
So, just to recap:
A) The ways in which to visit the 15 offices is to choose 1 of the 15, then 1 of the 14, then.... 15!

B) To only visit 4 on Monday, choose 1 of the 15, then 1 of the 14, then 1 of the 13, then 1 of the 12.... 15!/11!

C) Given what you reported the answer to be I'm assuming you don't have to visit all 15 offices, just any 3 different ones on a given day. By that logic, on 1 day it would take you 15!/12! ways to visit 3 offices on Monday. With 5 days in the week, it would have 5*(15!/12!) ways in which to visit 3 offices each day.

Edit I think Stacker beat me to it.

Title: In how many ways can a salesperson visit $15$ different offices during the week?

Body:
Consider the following question:

I am able to get the answer matched in the book for the part a and b but not part c.

I tried part a as:

```<code>Factorial[15]
</code>```

1,307,674,368,000

Part b as:

```<code>FactorialPower[15,4]
</code>```

32,760

But can't get the book's answer for part c. Please help.

The book is saying the answer to part c should be:

13,650


答:
Comment. Graph of $\mathsf{NORM}(\mu = 0.75, \sigma=25)$ showing 75th percentile $17.61224.$ (See initial comment moved just below from elsewhere on this page.)

```<code>1 - pnorm(0, .75, 25)
[1] 0.5119665    # area to right of vertical green line

qnorm(.75, .75, 25)
[1] 17.61224     # 75% of area is left of vertical red
</code>```

May be confusing to see two .75s. Suppose you wanted the median (50th percentile). Should get mean $\mu = .75.$

```<code>qnorm(.5, .75, 23)
[1] 0.75
</code>```

```<code>curve(dnorm(x,.75, 25), -75, 75, ylab="PDF", col="blue", lwd=2,
      main="NORM(0.75, 25")
 abline(v=17.61, col="red", lty="dotted", lwd=2)
 abline(h=0, col="green2");  abline(v=0, col="green2")
</code>```

Title: Help with S&P probability

Body:
According to this model, what is the probability that the change in the S&P $500$ on a particular day is positive. We have $P(X>0)$ and then converting to standard normal table I get $P(Z>(0-0.75)/25=0.03$ because $\mu=0.75$ and the SD is $25$. Then using standard normal table we get $0.512$

What is the third quartile quartile of daily changes in the S&P $500$? I know that $\mu$ and SD is the same, but would I just use $0.75$ as $X$? that would not make any sense because then you would just be dividing $0/25$? Also is there a way to do 1. without the standard table?

Body:
Comment. Graph of $\mathsf{NORM}(\mu = 0.75, \sigma=25)$ showing 75th percentile $17.61224.$ (See initial comment moved just below from elsewhere on this page.)

1 - pnorm(0, .75, 25)
[1] 0.5119665    # area to right of vertical green line

qnorm(.75, .75, 25)
[1] 17.61224     # 75% of area is left of vertical red

May be confusing to see two .75s. Suppose you wanted the median (50th percentile). Should get mean $\mu = .75.$

qnorm(.5, .75, 23)
[1] 0.75

curve(dnorm(x,.75, 25), -75, 75, ylab="PDF", col="blue", lwd=2,
      main="NORM(0.75, 25")
 abline(v=17.61, col="red", lty="dotted", lwd=2)
 abline(h=0, col="green2");  abline(v=0, col="green2")


问:
Non-commutativity of differential operators multiplied like a quadratic
I just read that if $v$ and $w$ are smooth functions of $x$ you can define operators $\partial _x + v$ and $\partial _x + w$. From there we find,
$$
(\partial _x + v)(\partial _x + w) = \partial _x ^2 + w' + (v+w)\partial_x + vw \\
(\partial _x + v)(\partial _x + w) = \partial _x ^2 + v' + (v+w)\partial_x + vw
$$
Where do these $w'$ and $v'$ terms come from? I do not follow the logic of how they appear.

答:
A really good way to make sure we don't miss terms like this is to use the method of a "test function." We have for some differentiable function $f(x)$ that

$$ (\partial_x+v)(\partial_x+w)f = (\partial_x+v)(\partial_xf + wf) = \partial^2_xf + v\partial_xf + \partial_x(wf) + vwf. $$

It's from the third term that your $w'$ comes out:

$$ (\partial_x+v)(\partial_x+w)f = (\partial_x+v)(\partial_xf + wf) = \partial^2_xf + v\partial_xf + w'f + w\partial_xf + vwf. $$

After evaluating fully, you can now say that since the final expression is true for "any" differentiable function $f$, the operator $(\partial_x+v)(\partial_x+w)$ is equivalent to the operator

$$ \partial^2_x + v\partial_x + w' + w\partial_x + vw. $$

The logic is the same for the other operator.


问:
Proving $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$
It is straightforward to prove using a connectedness argument that $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^n$, for $n>1$.

How do you prove that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$?

Note: I'm looking for a proof which does not use any algebraic topology,not even Brouwer's fixed point theorem. [I know that there are proofs of Brouwer's theorem using purely analytic methods, but I still do not want to include it - I'm looking for something even simpler]

Me and a few friends have been at this for a couple of weeks, but kept running in circles. Unfortunately, this was a long time ago, and I have completely forgotten what we tried. Any ideas?

Edit: Thanks for all the answers, but I still didn't get what I was looking for. I should have been more explicit : I don't want to use the Jordan curve theorem either. (The simplest proof of that which I've seen involves Brouwer's fixed point theorem). I'm looking for a proof which does not use anything apart from ideas of connectedness and compactness.

Looking at the answers, a second question came to my mind: Given that $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$, can you deduce the Jordan curve theorem?

答:
Hint:

In $\mathbb{R}^2$ any closed curve has an outside and an inside (Jordan curve theorem), but in $\mathbb{R}^3$, what happens then?


问:
For parabolic or hyperbolic PDEs, does the BC ever contain a time derivative?
I am curious if the boundary condition of a parabolic or hyperbolic PDE ever contains the time derivative of a function. I could see this happening if one of the terms in the PDE had both a spatial and time derivative on it. Then when looking at the weak form of the PDE you might get a time derivative on the boundary term. I suppose I could create such a PDE, but is there any literature where this happens?</p>

答:
Theoretically, you can indeed create such a PDE and for sure you can even design the boundary condition with time derivative in a way that the existence of some kind of solution is easy to prove. However, just because such PDEs exist does not mean they are useful. Time dependent PDEs describe evolution of some quantity according to some laws. Analogously, if you allow the boundary conditions to be so complicated, they should describe evolution of that quantity on the boundary under some local laws. In other words, you should be able to provide some justification for your boundary conditions, other than that they annihilate some boundary integral in the weak formulation (because there are always many ways how to write down the weak formulation).

I can think of two examples from fluid mechanics, where the boundary conditions involve the time derivative of the velocity:

Dynamic wall slip, observed experimentally in some molten polymers, see Hatzikiriakos and also the recent work Abbatiello et. al. for the corresponding analysis.

Outflow boundary conditions, see, e.g., Section 2.1. in Li et. al. and references therein. Note, however, that in this case the boundary conditions are only heuristical and to this day there is no agreement on what outflow boundary condition is the ``right'' one.


问:
Determinant of Vector
Is posible obtain the determinant of any vector?.How I will be able to obtain the determinant of any vector $v=[v_1,v_2,\cdots,v_n]\in \mathbb{R}^n$?</p>

答: Determinants are defined for square matrices, only. If you are trying to use the determinant of a vector (and $n\neq1$) then you are somehow on a very wrong track. </p>

Title: Determinant of Vector


答:
Here's one example --- how $\Gamma$ made my life a bit more pleasant. When I first learned how to integrate, I decided it would be fun to figure out formulas for the volumes of $n$-dimensional spheres, generalizing the $\pi r^2$ and $\frac43\pi r^3$ that I had learned earlier (for $n=2$ and $n=3$). After some calculation (and, if I remember correctly, some use of the integral tables at the end of my calculus book), I had the formulas for $n$ up to about $7$. For even $n$, there was an easily detectable pattern in the results, an $n$-dimensional sphere of radius $r$ has volume $\pi^{n/2}r^n/((n/2)!)$.  But for odd $n$, my formulas were not so nice. They had funny fractions in them (like that $\frac43$ when $n=3$, but worse for bigger $n$), and even the exponent of $\pi$ was not fully cooperative --- it was $(n-1)/2$ rather than $n/2$. ($r$ did have the expected exponent $n$, but that's unavoidable in an $n$-dimensional volume.)

It was many years later that I learned that the nice formula for even $n$ actually works perfectly for odd $n$ also, provided one uses the Gamma function to understand the $(n/2)!$ in the denominator. That factorial, understood as $\Gamma(\frac n2+1)$, provides all the funny fractions and the $\sqrt\pi$ that had bothered me years earlier.

Title: Questions about the Gamma function

Body:
tl;dr What is the import of the Gamma function?

So, I've read everything I can about the Gamma function, including a whole book ("Gamma") on it. I've scoured the internet on the subject. I've watched videos in Hindi, which I don't speak, and no kidding every video on Youtube when you search "Gamma beta function" comes up with a video meant for Indian engineers. Thanks to this site, I found some links to academic articles that discuss the history of the gamma function, too. All cool.

But still, I don't get it.

I do understand the various expressions of it, and even how to solve a problem that involves the Gamma function. What I don't get is the importance of this.

For example, I have learned that the gamma function was developed to deal with the interpolation problem that the Eulers, Bernoullis, Wallis, Stirling and gang were all interested in. I do love the math history. So is it right to say that the main use of the gamma function is to calculate non-integer factorials, which might have use in probability? (It does appear, and I'm not well-tutored in this, that the gamma function is useful to statisticians and coders.)

And why the Gauss/Euler/Weierstrass variations? So, I can express the gamma function in several ways. Gauss uses the product operator; Euler uses the integral; etc. What is the use of all these different forms?

And Weierstrass, it is rarely stated (in my reading) but I found one source who talked about its derivation as coming from using complex numbers in the Euler formula. Is that what makes the Weierstrass version so significant?

In the end, is gamma just an easy way of calculating an integral? That is, most problems I've solved involve rewriting integrals to meet the canonical gamma form. Once done, the integral's value is just gamma(n). Easy. Is that what this is all about? I need a hack to evaluate nasty integrals?

And I could rewrite all of this about the beta function. Hmmmm....... I'd welcome the hivemind to show me what I'm not getting about this.


问:
Stationary Distribution of a Stochastic Processes
A Markov Chain with states 0,1,... has transition probabilities
$$p_{jk}=e^{-a} \sum_{r=0}^k \left( \begin{matrix} j \\ r \end{matrix} \right) p^r (1-p)^{j-r} a^{k-r} / (k-r)!$$
Show that the limiting distribution $\{ v_k\}$ is Poisson with parameter $a/(1-p)$.

My attempt was using:

$$ \sum_{k=0}^\infty p_{jk} = 1 $$ and

$$ v_k = \sum_{j=0}^\infty v_j p_{jk} $$

But I am really out of order on how to simplify

$$ v_k = \sum_{j=0}^\infty v_j e^{-a} \sum_{r=0}^k \left( \begin{matrix} j \\ r \end{matrix} \right) p^r (1-p)^{j-r} a^{k-r} / (k-r)! $$

答:
Let $\mu=a/(1-p)$. Then interchanging order of summation and using $\mu(1-p)=a$,
$$S_k:=\sum_{j\ge 0} \frac{\mu^j}{j!} p_{jk}= \sum_{r=0}^k \left(\frac{p}{1-p}\right)^r \frac{a^{k-r}}{(k-r)!} \, e^{-a}\sum_{j=r}^\infty {j \choose r} \frac{a^j}{j!}$$
$$ =  \sum_{r=0}^k \left(\frac{p}{1-p}\right)^r \frac{a^{k}}{(k-r)! r!} \, e^{-a}\sum_{j=r}^\infty   \frac{a^{j-r}}{(j-r)!}$$
the inner sum yields $e^a$, so by the binomial formula,
$$S_k = \frac{a^k}{k!} \sum_{r=0}^k \left(\frac{p}{1-p}\right)^r {k \choose r}   = \frac{a^k}{k!} \left(1+\frac{p}{1-p}\right)^k  \, =\frac{\mu^k}{k!} \,.$$
Dividing both sides by $e^\mu$ shows that Poisson$(\mu)$ is a stationary distribution for the chain. Thus the chain is positive recurrent and by irreducibility, the stationary measure is unique and also serves as a limiting distribution for any initial distribution.


问:
If condition in Objective function.
I am new to optimization problems. I'm solving an integer programming problem, but I have a problem with the objective function. The question states that if the Quantity allocated in a store (Q) is greater than the required quantity (RQ), the penalty is P1 and if the Quantity allocated is less than RQ, the penalty is P2. How can I frame the objective function for this since it involves conditional logic?

答:
Introduce binary variables $x_1$ and $x_2$, and impose linear big-M constraints
\begin{align}
Q - RQ &\le M_1 x_1 \tag1 \\
RQ - Q &\le M_2 x_2 \tag2 \\
\end{align}
Here, $M_1$ is a small constant upper bound on $Q-RQ$, and $M_2$ is a small constant upper bound on $RQ-Q$.
Constraint $(1)$ enforces $Q>RQ \implies x_1=1$.
Constraint $(2)$ enforces $Q<RQ \implies x_2=1$.
Now minimize $P_1 x_1 +P_2 x_2$.

Title: If condition in Objective function.


问:
How to write that the only integers that have a multiplicative inverse are $1$ and $-1$, in symbolic form
Could it be something like for $(1,-1) \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$, such that $xy = 1$?

答:
The units in $\mathbb Z$ are $1,-1$. (Units are just invertable elements in a group).

More formally from wikipedia, a unit of a ring $\displaystyle R$R is any element $\displaystyle u\in R$$\displaystyle u\in R$ that has a multiplicative inverse in $\displaystyle R$R: an element $\displaystyle v\in R$ $\displaystyle v\in R$ such that

$\displaystyle vu=uv=1$$\displaystyle vu=uv=1$, where 1 is the multiplicative identity.

So using this definition, it is quite obvious that the units in $\mathbb Z$ are $1,-1$.
(I hope this helped? I am known not to explain things the best).


问:
Show that $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$

Show that $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$

Letting $y=e^x-1\implies e^x=y+1\implies x=\log(y+1)$ the evaluation is easy.

But I can't understand how to express given function as a composition of two functions so that the following rule can be used limit operation can be done.

Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$

答:
Use L'Hopital's rule:

$$\frac{d}{dx}(e^x -1 ) = e^x$$

And the denominator is $1$. Filling with the limit, gives $e^0/1 = 1$.


问:
Chain Rule in Leibniz and Lagrange notation: questions
I'm learning multiple applications of the chain rule and the notation surrounding it.  Does the following notation with example correctly reflect the chain rule in both Lagrange and Leibniz notation?

The derivative:
$$
y = f(x)
$$
$$
\frac {dy}{dx} = f'(x) = \frac {dy}{dx} \Big|_{x}
= \lim_{h \to 0} {\frac {f(h+x) - f(x)}{(h+x) -x}}
= \lim_{x_1 \to x} {\frac {f(x_1) - f(x)}{x_1 -x}}
$$
or when:
$$
x=a, \frac {dy}{dx} = f'(a)
$$

The chain rule:
$$
y = f(u), u =g(x)
$$
$$
\frac {dy}{du} \Big|_{g(x)} \cdot \frac {du}{dx} \Big|_{x}
= (f \circ g)'(x) = (f' \circ g)(x) \cdot g'(x)
$$

The chain rule applied two times:
$$
f(x) = (f_1 \circ (f_2 \circ f_3))(x)
$$

$$
\begin{align*}
f'(x) &= (f_1 \circ (f_2 \circ f_3))'(x)
= (f_1' \circ (f_2 \circ f_3))(x) \cdot (f_2 \circ f_3)'(x) \\
&= (f_1' \circ (f_2 \circ f_3))(x) \cdot (f_2' \circ f_3)(x) \cdot f_3'(x)
\end{align*}
$$

Decompose composite into separate functions:
$$
f'(x) = \frac {d f} {dx}
= \frac {dh} {dj} \frac {dj} {dk} \frac {dk} {dx}
$$
$$
f = h = f_1(j), j = f_2(k), k = f_3(x)
$$

$$
f'(x) = \frac {d f} {dx} \Big|_{x}
= \frac {dh} {dj} \Big|_{f_2(f_3(x))}\frac {dj} {dk} \Big|_{f_3(x)} \frac {dk} {dx} \Big|_{x}
$$

which agrees with above.

答:
I think that the source of your confusion arises from blurring the distinction between a function and its value. For example, if we write $f(x) = y$, then $x$ and $y$ are numbers and $f$ is the function that maps the number $x$ to the number $y$. In the two notations for the derivative, we write
$$
f'(x) = \frac{dy}{dx}
$$

Notice that I don't ever write the function name in the Leibniz differential notation. For compositions such as
$$
x \overset{g}{\longmapsto} u \overset{f}{\longmapsto} y,
$$
the derivative in the two notations are
$$
g'(x) = \frac{du}{dx},
\qquad
f'(u) = \frac{dy}{du},
$$
and
$$
(f \circ g)'(x) = f'\bigl( g(x) \bigr) \, g'(x)
= f'(u) \, g'(x)
= \frac{dy}{du} \, \frac{du}{dx}.
$$

This generalizes naturally to chains of functions of any length, hence the name the chain rule.


答: I have seen $U(R)$ for the group of units in a ring R (I mean it is a group iff the ring contains a unit, so it is not a group if your ring is $2\mathbb{Z}$). I think I have also seen $R^{\times}$. I think the latter is probably the more common one.

Title: How to write that the only integers that have a multiplicative inverse are $1$ and $-1$, in symbolic form

Body: Could it be something like for $(1,-1) \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$, such that $xy = 1$?


问:
How many times can quadric kiss cosine at given point?
Let a quadric $ax^2+2bxy+cy^2+dx+ey+f=0$ touches the plot of $y=\cos(x)$ at the point $(0,1)$ with multiplicity $n$. What is the maximum possible value of $n$? Recall that a joint point $P$ of parametrized curves $r_1(t)$ and $r_2(t)$ is said to be a tangent point of multiplicity $n$ if the first $n$ derivatives of $r_1(t)$ and  $r_2(t)$ at the point $P$ are equal.

PS. That was asked as one of the problems of a student mathematical olimpiad in Russia. The author gives general directions and the answer $n=5$.

答:
There are $4$ free parameters when passing a conic through a given point, so fourth-order contact for a general smooth function at a general point, but it is possible to have higher order contact at special points of a general curve, or at many points of a special curve (such as an arc of a conic).

[Edit: this corrects the value of $3$ given earlier.  Increase all values in the comments by $1$ for the order of tangency.

There is a symmetry of the graph of $y=\cos x$ relative to the line $x=0$, so that the local approximating conic should have enhanced contact by order at least $1$, or order $\geq 5$ tangency.  There is no reason to suppose an additional accident occurs, so order $= 5$ stated in the revised question is perfectly sensible and would say that the quality of approximation is the same as for any generic smooth even function at $x=0$.  A computation is needed to check this by finding the approximating conic and verifying that there is no extra tangency more than predicted.]

[More edit: in the power series for $y = \cos x$ we can, up to $O(x^6)$ error, replace the $\frac{x^4}{24}$ by $\frac{(1-y)^2}{6}$ and this gives order $\geq 5$ tangency to a conic: $y = 1 - \frac{x^2}{2} + \frac{(1-y)^2}{6} + O(x^6)$. A computation of the $x^6$ term is needed to show that $\geq 6$ does not happen. As written earlier, there is nothing unique to $\cos x$ here, and a generic even power series would have the same order of tangency at $0$.]


问:
Are limits of integration defined with respect to the pronumeral or the differential?
Consider the indefinite integral $$\int x^2 \,d(x^2).$$ It evaluates fairly easily to $\tfrac{x^4}{2} + C$. My question is about what happens when we start evaluating definite integrals with respect to these functions. In a specific example, how would $\displaystyle \int_0^2 x^2 \,d(x^2)$ be evaluated?
Would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot\left(x^2\right)^2\right]_0^2 =\frac{4}{2}-0=2 $$ with the limits applying to $x^2$, or would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot x^4\right]_0^2 =\frac{16}{2}-0=8 $$ with the limits applying to $x$.

Hope someone can clarify, thanks in advance!

答:
Confusion

If this is a Riemann integral I think I have to disappoint you. I don't think there is something like
$$\int_a^b f(x)\,\mathrm{d}g(x) = \int_a^b f(x)\,\frac{g(x)}{x}\,\mathrm{d}x.$$
It just does not make sense to define something like that. What is the problem with the integral on the right hand side? Why do we need a new notation? If there is, I take everything back. But it is not widely known, you should have defined it.

A similar integral

In fact there is a well-defined integral that looks like yours. It's called the Stieltjes integral, a generic term for the Riemann-Stieltjes integral and the Lebesgue-Stieltjes integral.
$$\int_a^bf(x)\,\mathrm{d}g(x).$$
You may look up it's definition here. I will not go into detail but it has a nice property if $g(x)$ is absolutely continous
$$\int_a^b f(x)\,\mathrm{d}g(x) = \int_a^b f(x)g'(x)\,\mathrm{d}x.$$

Stieltjes integral of your problem

In your example take $f(x)=g(x)=x^2$. Since $g(x) = x^2$ is absolutely continous we can write
$$\int_0^2 x^2\,\mathrm{d}x^2 = 2\int_0^2 x^3\,\mathrm{d}x=\left[\frac{x^4}{2}\right]_{0}^{2} = 8.$$


问:
Computing the parametric equation of the line of intersection of two planes
I'm given the following problem:

Find the parametric equation of the line of intersection of the two given planes: $y-6z = 7$ and $9x - 8y = 5$

My attempted solution follows:

I take the cross product of two vectors normal to the given planes, $<0,1,-6> \times <9,-8,0>$, and obtain the vector $<-48,-54,-9>$. I then set $z=0$, and solve the given simultaneous equations to determine a point along the line of intersection, $(\frac{61}{9}, {7}, 0)$. I can then write the parametric equation of intersection as:

$x = \frac{61}{9} - 48t$, $y = 7 - 54t$, $z = -9t$, using parameter $t$

Unfortunately, however, it seems I've erred in my solution. Is this the proper technique?

答:
The cross product will give you the direction of the line of intersection, yes!

However the line may or may not pass through $z=0$ (e.g. it could be parallel to the XY plane) so this method doesn't guarantee a solution.

The simplest way to get a solution would just be to solve the system of equations, by substitution:

$y-6z = 7$ implies $y = 6z + 7$.

Plugging this into the $9x -8y=5$, you get $9x-42z-56 = 5$ which is the line you are looking for.


问:
I understand the intuition of the Monty hall problem, I do not understand the math
Take the famous Monty Hall problem. There are three doors, one has a prize. If you pick the door with a goat, he reveals the other door with the goat. If you picks the door with the car, he randomizes and reveals one of the goats.

The calculations of the probabilities are what is confusing to me. So say you pick door 1 initially. and I define event $A_i =$ car is behind door $i$, and event $B_i$, Monty Hall reveals door $i$.

I have seen this calculation made:

$\mathbb{P}(A_1 \cap B_2)=$ probability the car is behind door 1 and he opens door 2 $= \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.

$\mathbb{P}(A_1 \cap B_3)= \frac{1}{3}\times\frac{1}{2} = \frac{1}{6}$

$\mathbb{P}(A_2 \cap B_3)= \frac{1}{3}\times 1 = \frac{1}{3}$

$\mathbb{P}(A_3 \cap B_2)= \frac{1}{3}\times 1= \frac{1}{3}$

It can then be shown using conditional probabilities that you have a $\frac{2}{3}$ chance of winning if you switch, and $\frac{1}{3}$ if you stay, and this is because Monty's choice is not random, it is conditional on your choice.

I actually do not understand why we can calculate the above probabilities. For instance, take the first one, $\mathbb{P}(A_1 \cap B_2)$. We are calculating above it seems as $\mathbb{P}(A_1)\times \mathbb{P}(B_2)$. Are we assuming $A_1$ and $B_2$ are independent? How can we do that - isn't the probability that he reveals door 2 dependent on whether or not the prize is behind door 1? So shouldn't this itself be a conditional probability, i.e. something like $\mathbb{P}(A_1)\times \mathbb{P}(B_1\mid A_1)$?

答:
so shouldn't this itself be a conditional probability, i.e. something like pr(A1)*pr(B1|A1)?

It should, and it is.  That is why the second factor in each expression have differing values: they depend on the first event.

$\qquad\begin{align}\Pr(A_1\cap B_2)&=\Pr(A_1)\cdot\Pr(B_2\mid A_1)\\&=(1/3)\cdot(1/2)&&\raise{1ex}{\small\text{Monty has equal chance to reveal door 2 or 3}\\\small\text{ when the car is behind door 1}}\\&=1/6\\[2ex]\Pr(A_2\cap B_2) &=\Pr(A_2)\cdot\Pr(B_2\mid A_2)\\ &= (1/3)\cdot (0)&&\raise{1ex}{\small\text{Monty can not reveal door 2 when}\\\small\text{ the car is behind that door}}\\&=0\\[2ex]\Pr(A_3\cap B_2)&=\Pr(A_3)\cdot\Pr(B_2\mid A_3)\\&=(1/3)\cdot(1)&&\raise{1ex}{\small\text{Monty can only reveal door 2}\\\small\text{ the car is behind door 3}}\\&=1/3\end{align}$

Hence:

$\qquad\begin{align}\Pr(A_3\mid B_2)&=\dfrac{\Pr(A_3\cap B_2)}{\Pr(A_1\cap B_2)+\Pr(A_2\cap B_2)+\Pr(A_3\cap B_2)}\\&=\dfrac{1/3}{1/6+0+1/3}\\&=2/3\end{align}$


问:
Reality check: irreducible over a non-algebraically closed field means replace closed sets with Galois-stable closed sets, right?
I am requesting a reality check. I'm pretty sure of something but my algebraic geometry education is insufficiently comprehensive to give me full conviction, and I can't readily find the answer in texts.

Say $V=\operatorname{Spec}A$ is an affine variety (i.e. reduced, finite-type scheme) over a field $k$ which is not algebraically closed. Then $V$ is irreducible as a topological space if and only if $A$ is an integral domain (so that there is a unique minimal prime ideal, namely zero). I would like to understand irreducibility in terms of the Zariski topology on the $\overline k$-points, where $\overline k$ is $k$'s algebraic closure. It seems to me it's the following, and my question is if this is right:

Let $\Gamma=\operatorname{Gal}(\overline k/ k)$ be the absolute Galois group of $k$. Then $\Gamma$ acts in a natural way on the set $V(\overline k)$ of $\overline k$-points of $V$. The set $V(\overline k)$ gets a topology via its natural bijection with the closed points of $V\otimes_k \overline k$, which have a subspace topology induced from the Zariski topology on $V\otimes_k \overline k$. I think $V$'s irreducibility should be equivalent to the statement that $V(\overline k)$ is not the union of two proper closed subsets each of which is stable under the $\Gamma$-action. Is this right?

I think this because my understanding is that stability under $\Gamma$-action is equivalent to "defined over $k$", and irreducibility of $V$ (since it is a $k$-scheme) ought to mean that it is not reducible into proper closed subsets that are defined over $k$. But I am not immediately seeing how to connect this with the definition in terms of the irreducibility of $V$ itself (i.e. $\operatorname{Spec} A$) as a topological space.

I will be satisfied with an explanation, but if you have recommendations of chapters or sections of texts where this issue is discussed, I would be delighted by that too. Thanks in advance.

答:
Yes, this is true, and your concept of why is basically accurate as well. Here's a rigorous accounting. Your condition on $V(\overline{k})$ can be phrased in terms of the map $V_{\overline{k}}\to V$ by a combination of the following four facts:

$V(\overline{k}) = (V\times_k \overline{k})(\overline{k})$;

the topology on $V_{\overline{k}}(\overline{k})$ is the induced topology from $V_\overline{k}$;

the $\overline{k}$-points of $V_{\overline{k}}$ are dense in every closed subset;

and the Galois action on $V_{\overline{k}}(\overline{k})$ is the restriction of the Galois action on $V_\overline{k}$.

From these we may conclude that irreducible closed subsets of $V(\overline{k})$ are in bijection with irreducible closed subsets of $V_\overline{k}$, and this bijection respects the Galois action.

Next, every irreducible component of $V_\overline{k}$ maps down to an irreducible component of $V$ and this map is surjective (Stacks 04KX). Further, the Galois group acts on the irreducible components of $V_\overline{k}$, and the fibers of the map from the set of irreducible components of $V_\overline{k}$ to the set of irreducible components of $V$ are exactly the Galois orbits (Stacks 04KY).

This proves your assertion. In one direction, if $V=W\cup Z$ is reducible, then $W_\overline{k}$ and $Z_\overline{k}$ are proper closed $G$-invariant subsets of $V_\overline{k}$ having union $V_\overline{k}$. On the other hand, if $V_\overline{k}$ can be written as a union of two proper closed subsets $X,Y$ which are stable under the $G$-action, then we can partition the irreducible components of $V_\overline{k}$ in to two disjoint nonempty collections, both of which are stable under the Galois action: those irreducible components that are only in $X$, and all the others. Since the fibers of the map from irreducible components of $V_\overline{k}$ to irreducible components of $V$ are exactly the Galois orbits, $V$ must have multiple irreducible components.


问:
Solving conservation law IVP when characteristics intersect at $t=0$
This is my first post here so I apologize if there is anything wrong with it. If so, please let me know and I can edit the post.

I just tried to solve the following question:

Problem 2.

Give the explicit solution for all $t\leqslant 1/2$, to the following equation:
$$
\left\lbrace\begin{aligned}
&u_t+f(u)_x=0,\\
&u(x,0)=\begin{cases}
1, & x\leqslant 1\\
1/2, &1<x\leqslant 3\\
3/2, &x>3
\end{cases}
\end{aligned}\right.
$$
where $f(u)=4u(2-u)$.

Represent the solution at times $t=0$, $t=1/4$ and $t=1/2$.

Determine the solution beyond $t=1/2$?

What I have so far:

$f'(u) = 8 - 8u$, and for a solution $u(x,t), u$ is constant along the characteristic curves defined by $x'(t) = 8 - 8u, x(0) = a$ ($a$ real number), ie $u(x(t), t) = u(x(0), 0) = u(a, 0)$.

Then $u$ does not depend on $t$, so the characteristic curves are $x(t) = (8 - 8\cdot u(a,0)) \cdot t + a$.

ie $x(t)$ is given by $$\begin{cases}
   0t + a & \text{if $a<=1$} \\
   4t + a & \text{if $1 <= a <= 3$} \\
   -4t + a & \text{if $a>3$}
  \end{cases}$$

The question asks for the solution before $t = 1/2$, however I'm struggling with the following:

Based on the above, it seems the characteristic curves already intersect at $t = 0$ (and $x = 3$), long before $t = 1/2$. Am I supposed to construct a shock wave solution using the Rankine-Hugoniot jump condition? I actually think there may be 2 shock wave solutions needed, because the characteristic curves do not cover a space around $(1, 0)$, it seems, since the curve $x = -4t + 3$ intersects the t-axis at $t=4/3$... However if I do this, wouldn't it be a solution for some range of t that isn't $t \leq 1/2$?

On the other hand, is my reasoning above (before bullet #1) correct? This is, I think, analogous to what I have seen in class and in some other resources I looked at, however they all used the function $f(u) = u^{2} / 2$ and while I don't think this would change much, maybe I'm missing something vital.

Thank you in advance for any help.

答:
The situation is very similar to this post and related ones, where piecewise constant initial data is considered (i.e, a series of Riemann problems).

At $x=1$, we have the characteristic speeds $f'(1) = 0$ on the left side and $f'(1/2) = 4$ on the right side of the discontinuity. Since the right slope $4>0$ is larger than that on the left, characteristic lines separate: a rarefaction wave forms over $1\leq x\leq 1+4t$, with a linear fan solution.

At $x=3$, we have the characteristic speeds $f'(1/2) = 4$ on the left side and $f'(3/2) = -4$ on the right side of the discontinuity. Since the right slope $-4<4$ is smaller than that on the left, characteristic lines intersect: a shock wave forms, which speed $s=0$ follows from the Rankine-Hugoniot condition (it's a static shock located at $x=3$).

These two waves interact at the time $t^*$ such that $1+4t^* = 3$, i.e $t^*=1/2$. Thus, one may distinguish the case $t<1/2$ from the case $t\geq 1/2$ (cf. linked post).

To link with the Burgers equation, one may consider the change of variable $v=8(1-u)$ such that
$$
v_t + (\tfrac12 v^2)_x = 0\, ,
$$
$$
v(x,0) = \begin{cases}
0, & x\leq 1\\
4, & 1<x\leq 3\\
-4, & x>3
\end{cases}
$$