答:
Let eigenvalues of $A$ be denoted by $2m$ with $m \in \mathbb Z$ and let $u,v$ be the left/right eigenvector respectively and note that $A^2 v = 4m^2 v$. Then,
$$ u^H X v + 2m u^H X v - 4m^2 u^H X v = 0 $$
This implies that $1 + 2m - 4m^2 = 0$ or $u^H X v = 0$. Note that the $m$ satisfying the first equality will not be an integer and so $u^H X v = 0$. Can you finish it from here?

Title: Matrix equation $X+AX-XA^2=0$ has only zero solution

Body: Let $A$ be a $n\times n$ complex matrix, with even numbers as eigenvalues. Show that the matrix equation $X+AX-XA^2=0$ has only zero solution. My attempt is just consider an eigenvector $v$ of $A$, then $Xv+AXv-4m^2Xv=0.$ Then what to do? Or just $v'(X+AX-XA^2)v=0$. But $A$ is not symmetric, how to proceed?


问:
Matrix equation $X+AX-XA^2=0$ has only zero solution
Let $A$ be a $n\times n$ complex matrix, with even numbers as eigenvalues. Show that the matrix equation $X+AX-XA^2=0$ has only zero solution. My attempt is just consider an eigenvector $v$ of $A$, then $Xv+AXv-4m^2Xv=0.$ Then what to do? Or just $v'(X+AX-XA^2)v=0$. But $A$ is not symmetric, how to proceed?

答:
$(I+A)X+X(-A^2)= IX+AX-XA^2=\mathbf 0$

s a Sylvester Equation and is isomorphic to

$\big((I+A)\oplus (-A^2)^T\big)\text{vec}\big(X\big) = \text{vec}\big(\mathbf 0\big)= \mathbf 0_{n^2}$

where $\oplus$ denotes Kronecker Sum.

The eigenvalues of $(I+A)$ are necessarily odd, and the eigenvalues of $(-A^2)^T$ are necessarily even since $A$ has real even eigenvalues.  Thus each eigenvalue of $\big((I+A)\oplus (-A^2)^T\big)$ is an odd number plus an even number $\neq 0$.  Therefore $\det\big((I+A)\oplus (-A^2)^T\big)\neq \mathbf 0$ and $\text{vec}\big(X\big) =\mathbf 0_{n^2}$ is the only solution $\implies X=\mathbf 0$ is the only solution for the original problem.

ContentLicense: CC BY-SA 4.0


问:
How to generate random numbers from density function if it's known that if $X_i$ are iid and $X_i \sim F$, then $F(x_i) \sim U[0,1]$
The given density function:
$$f(x) = \frac{1}{2}\phi(x, 0, 1) + \frac{1}{10}\sum^4_{j=0}\phi(x; (j/2)-1, 1/10) $$
where $\phi$ is normal distribution.

```<code>f <- function(x){
  dnorm(x)/2 +
    dnorm(x,- 1, 1/10)/10 +
    dnorm(x, 1/2 - 1, 1/10)/10 +
    dnorm(x, 2/2 - 1, 1/10)/10 +
    dnorm(x, 3/2 - 1, 1/10)/10 +
    dnorm(x, 4/2 - 1, 1/10)/10
}
</code>```

It is known that if random variables $X_i$ are iid, and $X_i \sim F$, then $F(X_i)\sim U[0,1]$
I tried to define the cumulative function:

```<code>F <- function(x){integrate(f, lower = -Inf, upper = x)}$value
</code>```

And then to find zeroes of function $F(x)-\text{runif}(x)$

```<code>zeros <- function(x){uniroot(F(x)-runif(x), c(-10,10))}
</code>```

But the program R gives me an error:

```<code>Error in uniroot(F(x) - runif(x), c(-10, 10)) :
  f() values at end points not of opposite sign
</code>```

Anyway, I am at loss as to the way, this problem could be possibly solved. Does anybody got any insights?

答:
You didn't show how you are calling the function zeros, so I am not sure how to explain why you get this error.  Probably, when you are calling zeros, it tries to evaluate runif(x) where x is a number between 0 and 1 and that is the source of the problem.

Try this.

```<code>f <- function(x){
  dnorm(x)/2 +
    dnorm(x,- 1, 1/10)/10 +
    dnorm(x, 1/2 - 1, 1/10)/10 +
    dnorm(x, 2/2 - 1, 1/10)/10 +
    dnorm(x, 3/2 - 1, 1/10)/10 +
    dnorm(x, 4/2 - 1, 1/10)/10
}
integrate(f,lower=-Inf,upper = Inf)
F <- function(x){integrate(f, lower = -Inf, upper = x)}$value
Finv <- function(t) uniroot(function(x,t) F(x)-t,c(-10,10),t=t)$root
(u=runif(1))
F(Finv(u))
Finv(F(u))
</code>```

This verifies that Finv is the inverse of F.

Then, you can simulate a random variable with distribution $F$ by using Finv(runf(1)).

Body:
You didn't show how you are calling the function zeros, so I am not sure how to explain why you get this error.  Probably, when you are calling zeros, it tries to evaluate runif(x) where x is a number between 0 and 1 and that is the source of the problem.

Try this.

f <- function(x){
  dnorm(x)/2 +
    dnorm(x,- 1, 1/10)/10 +
    dnorm(x, 1/2 - 1, 1/10)/10 +
    dnorm(x, 2/2 - 1, 1/10)/10 +
    dnorm(x, 3/2 - 1, 1/10)/10 +
    dnorm(x, 4/2 - 1, 1/10)/10
}
integrate(f,lower=-Inf,upper = Inf)
F <- function(x){integrate(f, lower = -Inf, upper = x)}$value
Finv <- function(t) uniroot(function(x,t) F(x)-t,c(-10,10),t=t)$root
(u=runif(1))
F(Finv(u))
Finv(F(u))

This verifies that Finv is the inverse of F.

Then, you can simulate a random variable with distribution $F$ by using Finv(runf(1)).


问:
Why is the ideal $\langle x-2,x+3\rangle$ in $Z[x]$ prime?
Why is the ideal $\langle x-2,x+3\rangle$ in $Z[x]$ prime?

I know it is equivalent to show the quotient is an integral domain, but I am having trouble showing this, or showing it is prime directly.

This question was already asked here: Show the ideal $(x-2,x+3)\subset\mathbb{Z}[x]$ is prime but not principal

But there isn't an accepted answer (or even an answer demonstrating why the ideal is prime)

答:
It's easier to first rewrite $\langle x-2,x+3\rangle$ as $\langle x-2, 5\rangle$ and then "substitute" $y=x-2$ to get the ideal $\langle y,5\rangle$ in $\mathbb Z[y]$. This has an easy description: it's an ideal of all polynomials in $y$ with constant term a multiple of $5$.

Now checking that this ideal is prime in $\mathbb Z[y]$ is equivalent to checking that $\langle 5\rangle$ is prime in $\mathbb Z$, since the constant terms aren't affected by any other terms when we add or multiply.


问:
What is the logical operator for but?
I saw a sentence like,

```<code>I am fine but he has flu.
</code>```

Now I have to convert it into logical sentence using logical operators. I do not have any idea what should but be translated to. Please help me out.

Thanks

答:
But is a set meta-operator. It means "Union NOT".
Everything BUT the kitchen sink. {not K}
Nothing BUT the kitchen sink. {K}
All kitchen sinks BUT the ones with pink plugs. {K Subset not P}
(All)Kitchen sinks with pink plugs BUT in the case of round sinks only the sinks that are Metal {K Subset P Union (M Intersection R)

BUT, BUT ONLY - is an intersection (which can be defined in terms of NOT and AND) of one set with another set (sometimes with an implied set)
BUT NOT is a union of one set with the inverse of another set

ContentLicense: CC BY-SA 3.0

Title: What is the logical operator for but?


问:
"Face-contraction" of Simplicial Complex Preserving Homology
Let $X$ be a simplicial complex on a vertex set $V=\{1,\dots,n\}$. Suppose there is some subset $U\subseteq V$ such that every facet of $X$ contains $U$ or is disjoint with $U$.

Then, define $X'$ on a new vertex set $V'=(V-U)\cup \{v_U\}$ and facets $\sigma'$ where $\sigma'=\sigma$ if $\sigma$ is a facet of $X$ disjoint to $U$, and $\sigma'=(\sigma-U)\cup \{v_U\}$ if $\sigma$ is a facet of $X$ containing $U$. Then $X$ and $X'$ have the same homology.

I was able to prove this by cases (There are a few details I haven't written at the end so it might have some mistakes, but I don't think so): First, if every facet contains $U$, then I used a Mayer-Vietoris sequence to prove there is no homology in this case. Second, if no facet contains $U$ (Thus every facet is disjoint to $U$), then the simplicial complex remains unchanged by the operation, therefore there is also no change in its homology.

Lastly, in the general case, I used Mayer-Vietoris sequences for both $X$ and $X'$ taking $X=A\cup B,X'=A'\cup B$ where $B$ consists in the facets disjoint to $U$ (Or not containing $v_U$ in $X'$) and $A,A'$ consist in the facets containing $U,v_U$ respectively. After using the fact that the homologies of $A,A'$ are zero, and that the intersections $A\cap B=A'\cap B$ remain unchanged by the operation it's easy to define a chain isomorphism between the exact sequences by the five lemma (Where four of the morphisms are identity maps and the fifth one is induced by the inclusion $v_U\mapsto v$ for any fixed $v\in U$).

My question is about the following remarks:

First, the operation done is analogous to the operation of edge contraction. We can interpret it in some ways. It identifies the vertices in $U$, and faces not disjoint with $U$, causing some of them to reduce their dimension (If they contain more than one vertex of $U$). In this case if I'm not mistaken we can also interpret it like removing all the vertices of $U$ but one.

Second: It doesn't always preserve homology, but sometimes it does. In this case the map $v_U\mapsto v$ (for a fixed $v\in U$) even seems to be a good candidate for a homotopy equivalence.

My question is the following: Is there any literature about the "edge contraction" operator applied to simplicial complexes? It seems like a rigid version of Proposition 0.17 in Hatcher's book.

答:
$\newcommand{\cU}{\mathcal{U}}$
This actually is related to the construction of the "Nerve" of a simplicial complex.

Let $\Delta$ be a simplicial complex with facet set $F(\Delta)=\{\sigma_1,\dots,\sigma_r\}$ and vertices $v_1,\dots,v_n$.
For each facet of $\Delta$, consider a small enough open set around it. Now you have an open cover $\cU=\{U_1,\dots,U_r\}$. The nerve $N=N(\cU)$ is defined to be a simplicial complex with vertex set $\cU$ given by $N=\{\tau\subseteq\cU:\cap\tau\neq\varnothing\}$, i.e. the subcollections of open sets with nonempty intersection.

You can make a similar discrete construction, by setting a complex $D=D(\Delta)$ with vertices $V=\{\sigma_1,\dots,\sigma_r\}$ defined by
$D=\{\tau\subseteq V: \cap \tau\neq \varnothing\}$.

You can choose $\cU$ appropiately, so that the open sets are not "too big" so that they only intersect each other if the corresponding facets do:
And in this case, it's clear that $N$ and $D$ are isomorphic by the map $\sigma_i\mapsto U_i$. So we'll call $D$ the nerve of $\Delta$.

Now, let $D^2$ be $D(D(\Delta))$. Naively, the facets of $D$ are given by $F(D)=\{\tau_1,\dots,\tau_n\}$ where $\tau_i=\{\sigma\in F(\Delta):v_i\in\sigma\}$, so the vertices of $D^2$ are $\tau_1,\dots,\tau_n$. The facets of $D^2$ are thus $\rho_1,\dots,\rho_r$ where $\rho_j=\{\tau\in F(D): \sigma_j\in\tau\}$.

Now, $\sigma_j\in \tau_i$ iff $v_i\in \sigma_j$, therefore after renaming $\tau_i$ by $v_i$ you get $D^2=\Delta$.

Is this correct? Actually, it's not. We're missing that $\tau_i$ may equal $\tau_{i'}$ for $i\neq i'$, so $D^2$ will be a little smaller than $\Delta$. When you remove repeated elements, you'll find that $D^2$ is what results of $\Delta$ after contracting all the vertex sets with the condition that every facet of $\Delta$ contains it or is disjoint with it. And then you have the Nerve Theorem of Leray which proves that $D$ is homotopically equivalent to $\Delta$ which means that of course $D^2$ is also homotopically equivalent to it. I expect the proof of that theorem should help to prove that the intermediate steps of contracting those sets one by one also preserve homotopy.


问:
Help finding the sum $\sum_{n=1}^{\infty} \frac{n^2+1}{5^n}$
I have:

$$\sum_{n=1}^{\infty} \frac{n^2+1}{5^n}$$

I have separated the fraction in two, the second one, $\frac{1}{5^n}$ is clearly just a geometric series and simple enough. But what about the first one? Looking through my calculus book and notes I haven't found an example of this.

Any help is appreciated, especially if there's a general method to go around finding these kinds of sums. Thanks.

Body:
Let $f(x) = \sum_{n=1}^{\infty}x^{-n}=\frac{1}{x-1}$. Then, we have
\begin{align}
f'(x)&=-\sum_{n=1}^{\infty} nx^{-n-1}= \frac{-1}{(x-1)^2}\\
f''(x) &= \sum_{n=1}^{\infty} n(n+1)x^{-n-2} = \frac{2}{(x-1)^3}.
\end{align} Therefore, $$\sum_{n=1}^{\infty}\frac{n^2+1}{5^n} = \sum_{n=1}^{\infty}\frac{n(n+1)-n+1}{5^n}=25f''(5)+5f'(5)+f(5)= \frac{23}{32}.$$


问:
Constants in ultraproduct are well-defined
When defining ultraproduct it is defined to be product of domains of models $A_\xi$ modulus the equivalence relation by ultrafilter on index set. The relation on the product are defined by $(a_0/_{\mathcal U}, \dots, a_n/_{\mathcal U}) \in R$ iff $\{\xi : (a_0(\xi), \dots, a_n(\xi))\in R_\xi\}$ where $R_\xi$ is relation on model with domain $A_\xi$ and similar for the function.

I define the constant $C \in \prod_{\xi \in X}A_\xi / \mathcal U$ as $C = (\prod_\xi C_\xi)/\mathcal U$ where $C_\xi$ is the constant in model with the domain $A_\xi$. How to show it is well-defined?

答:
There is no well-definedness issue at all. If $C$ is a constant symbol in the language $L$, then the interpretation $C_{\xi}$ of  $C$ in the model with domain $A_{\xi}$ is uniquely defined. So the "sequence" $(C_{\xi})$, what you call $\prod_{\xi} C_{\xi}$, is uniquely determined, and hence so is $(\prod_{\xi} C_{\xi})/\mathcal{U}$

The situation is different with say a unary function symbol $f$ of the language $L$. For there we define $f$ at $(\prod_{\xi} a_{\xi})/\mathcal{U}$ is defined in terms of the $f(a_{\xi})$.  It is conceivable that different choices of representatives  $\prod_{\xi} a_{\xi}$ and $\prod_{\xi} a_{\xi}'$ yield different results. Easily, they don't, but this has to be checked.


问:
Let $B = (B_t)_{t \geq 0}$ be a standard brownian motion and $I_t = \inf_{0 \leq s \leq t} B_s$, show $F(t, B_t, I_t)$ is a martingale.
Let $B = (B_t)_{t \geq 0}$ be a standard brownian motion and $I_t = \inf_{0 \leq s \leq t} B_s$, show that

$F(t, B_t, I_t) = (B_t−I_t)^6−15t(B_t−I_t)^4+45t^2(B_t−I_t)^2−15t^3$

is a martingale. I'm really not sure where to begin with this, any pointers would be appreciated?

答:
The application of Ito's formula may look intimidating, but it's not too bad. The $dB_t$ terms will combine into a martingale, so it suffices to check that the $dt$ and $dI_t$ terms cancel each other out. In fact, the combined $dI_t$ terms  vanish because $B_t-I_t$ vanishes at points of decrease of $I_t$ and the partial derivative of $F(t,x,y)$ with respect to $y$  vanishes whenever $x=y$. This leaves only the $dt$ terms
$$
F_t(t,B_t,I_t)dt + {1\over 2} F_{xx}(t,B_t,I_t) dt,
$$
which I will leave to you.


问:
Expectation of sum of random variables
I'm having trouble proving the following lemma for my statistics course:

Let $X_1,...,X_n$ be a random sample from $P$ to $\mathbb{R}$, $X$~$P$, $g$ measurable so that $\mathrm{E}g(X)$ and $\mathrm{var}g(X)$ exist. Then

$\mathrm{E}(\sum_{i=1}^ng(X_i))=n\cdot\mathbb{E}g(X)$

$\mathrm{var}(\sum_{i=1}^ng(X_i))=n\cdot\mathrm{var}g(X)$

I have a vague conception of its proof and know that it is directly related to $\sum_{i=1}^{n}{X_i}=n\bar{X}$ and correspondingly the random variables being i.i.d.

答:
The first is simply linearity of expectation applied to $g(X_i)$. So $\mathbb{E}[\sum_{i=1}^ng(X_i)] = \sum_{i=1}^n \mathbb{E}[g(X_i)] = n\cdot\mathbb{E}[g(X)]$

Assuming the $X_i$ are independent then so too are the $g(X_i)$, meaning $\mathbb{E}[g(X_i)g(X_j)]=\mathbb{E}[g(X_i)]\mathbb{E}[g(X_j)]=(\mathbb{E}[g(X)])^2$ when $j \not = i$, so:

$\mathrm{var}(\sum_{i=1}^ng(X_i)) \\= \mathbb{E}[(\sum_{i=1}^ng(X_i))^2] - (\mathbb{E}[\sum_{i=1}^ng(X_i)])^2 \\= \sum_{i=1}^n\mathbb{E}[(g(X_i)^2)]+\sum_{i=1}^n\sum_{j\not=i}\mathbb{E}[g(X_i)g(X_j)]  - (n\cdot\mathbb{E}[g(X)])^2 \\=n\cdot\mathbb{E}[(g(X)^2)]+n(n-1)(\mathbb{E}[g(X)])^2  - n^2\cdot(\mathbb{E}[g(X)])^2 \\=n\cdot\mathbb{E}[(g(X)^2)] - n\cdot(\mathbb{E}[g(X)])^2 \\=n\cdot\mathrm{var}(g(X))$

Title: Expectation of sum of random variables


答:
What is your definition of expectation? For simplicity, let's assume that your random variables are continuous, i.e. $\mathbb{E}[X] = \int x f_X(x) \,\mathrm{d}x$ for some probability density function. (Of course, this works for general random variables that are discrete, continuous, mixed, etc.)

The first property follows from the linearity of expectation, and the fact that each $X_i$ is identically distributed (independence is actually not necessary). That is, we use the fact that for random variables $X$ and $Y$, $\mathbb{E}[X+Y] = \mathbb{E}[X] + \mathbb{E}[Y]$. This follows from the linearity of the integral: assuming the joint density exists for simplicity again,
$
\iint (x + y)f_{X+Y}(x,y) \,\mathrm{d}x \,\mathrm{d}y = \int x f_X(x) \,\mathrm{d} x + \int x f_Y(y) \,\mathrm{d} y .
$

The second property does indeed require independence. That comes into play through the following property: if $X$ and $Y$ are independent random variables, then $\mathrm{Var}(X+Y) = \mathrm{Var}(X) + \mathrm{Var}(y)$. If you can prove this, then your desired result follows. To do so, you can use the definition of the variance, and how independence implies $\mathbb{E}[XY] = \mathbb{E}[X] \mathbb{E}[Y]$.

Title: Expectation of sum of random variables

Body:
I'm having trouble proving the following lemma for my statistics course:

Let $X_1,...,X_n$ be a random sample from $P$ to $\mathbb{R}$, $X$~$P$, $g$ measurable so that $\mathrm{E}g(X)$ and $\mathrm{var}g(X)$ exist. Then

$\mathrm{E}(\sum_{i=1}^ng(X_i))=n\cdot\mathbb{E}g(X)$

$\mathrm{var}(\sum_{i=1}^ng(X_i))=n\cdot\mathrm{var}g(X)$

I have a vague conception of its proof and know that it is directly related to $\sum_{i=1}^{n}{X_i}=n\bar{X}$ and correspondingly the random variables being i.i.d.


问:
Relation R=2K (two dimensions)
I'm looking for a proof of the relation shown above, where $R$ is the scalar curvature and $K$ the gaussian curvature in two dimensions. I've ended up in the following post:

Physics Exchange Post

Looking at @Prahar 's answer, is the following relation valid?

$$K=\frac{R^1_{212}}{g_{22}}$$

If not, what would be a valid proof of $R=2K$?

Thank you in advance,

A.

答: If the curvature tensor of $(M^n,g)$ has the form $$R(X,Y)Z = K(\langle Y,Z\rangle X - \langle X,Z\rangle Y)$$for some function $K\colon M \to \Bbb R$ (which is necessarily constant if $n \geq 3$ -- this is called Schur's Theorem), we trace both sides on the variable $X$ to obtain $${\rm Ric}(Y,Z) = K(n\langle Y,Z\rangle - \langle Y,Z\rangle) = (n-1)K\langle Y,Z\rangle.$$Now take the $g$-trace on the variables $Y$ and $Z$ to obtain $${\rm s} = n(n-1)K,$$where ${\rm s}$ is the scalar curvature. For $n=2$, the curvature tensor of $(M^2,g)$ has the above form, where $K$ is its Gaussian curvature. Thus ${\rm s} = 2K$ in this case.

Title: Relation R=2K (two dimensions)


问:
Kernel of a Linear transformation.
Sorry if this is a repeted question. I did not find it. The question is

Suppose $T:U \rightarrow V$ is a linear transformation between two vector spaces over the same field, such that $dim\, V < dim\, U < \infty$. Show that $\exists u \in U, u\neq 0$ such that $T(u) = 0$.

Thanks!

答:
This is an immediate consequence of rank-nullity:
$$\dim U=\dim\operatorname{im}T+\dim\ker T\leq\dim V+\dim\ker T\implies\dim\ker T>0.$$
So $\ker T$ has a non-zero element.

Title: Kernel of a Linear transformation.


答: Use the kernel-image theorem that says that if you have a linear transformation $T:U\rightarrow V$ between two vector spaces, then: $\textrm{dim}(U)= \textrm{dim ker}(T)+\textrm{dim Im}(T) $. Now, you know that $\textrm{dim Im}(T) \leq \textrm{dim}(V) \leq \textrm{dim}(U)$. Use that to get that $\textrm{dim ker}(T)>0$

Title: Kernel of a Linear transformation.

Body:
Sorry if this is a repeted question. I did not find it. The question is

Suppose $T:U \rightarrow V$ is a linear transformation between two vector spaces over the same field, such that $dim\, V < dim\, U < \infty$. Show that $\exists u \in U, u\neq 0$ such that $T(u) = 0$.

Thanks!


问:
Integral inside a limit?
Problem gives me a review of FTC2, saying $\frac{d}{dx}(\int_{a}^{x}f(t)\,dt) = f(x)$ then tells me:

For $f(t)$ continuous, calculate the limit.

$$\lim_{h\to 0}1/h\int_{h}^{2h}f(t)\,dt$$

I started by putting it in the format of $\lim_{h\to 0}\frac{F(2h)-F(h)}{h}$, but now I am kind of stuck. I'm not sure what to do next. Thank you for any help in advance!

答:
Apply l'Hopital's rule.

$$\lim_{h\to0}\frac{F(2h)-F(h)}{h}$$
$$=\lim_{h\to0}\frac{\frac d{dh}[F(2h)-F(h)]}{\frac d{dh}h}$$
$$=\lim_{h\to0}\frac{\frac d{dh}F(2h)-\frac d{dh}F(h)}1$$
$$=\lim_{h\to0}\frac d{dh}[2h]F'(2h)-F'(h)$$
$$=\lim_{h\to0}2F'(2h)-F'(h)$$

Now recall $F'(x)=f(x)$. Replace $h$ with $0$ and simplify.

Title: Integral inside a limit?


答:
Define a function $F:x \mapsto \int_{a}^{x} f(t) \ dt$, where $f$ is continuous on an interval $[a,b]$ and $x \in [a,b]$. In that case, $f$ is Riemann Integrable and $F$ is differentiable so it is continuous. Now, you correctly deduced that you have to calculate the limit:

$$\lim_{h \to 0} \frac{F(2h)-F(h)}{h}$$

Observe that this is just:

$$\lim_{h \to 0} \frac{F(2h)-F(0)+F(0)-F(h)}{h} = \lim_{h \to 0} \left(\left( 2 \cdot \frac{F(2h)-F(2 \cdot 0)}{2h} \right) - \left( \frac{F(h)-F(0)}{h} \right) \right)$$

The first limit is just $2f(0)$ and the second limit is just $f(0)$. Hence, the entire limit evaluates to $f(0)$ and we are done.

Title: Integral inside a limit?

Body:
Problem gives me a review of FTC2, saying $\frac{d}{dx}(\int_{a}^{x}f(t)\,dt) = f(x)$ then tells me:

For $f(t)$ continuous, calculate the limit.

$$\lim_{h\to 0}1/h\int_{h}^{2h}f(t)\,dt$$

I started by putting it in the format of $\lim_{h\to 0}\frac{F(2h)-F(h)}{h}$, but now I am kind of stuck. I'm not sure what to do next. Thank you for any help in advance!


问:
How do I adjust action outputs when accounting for volume at which said actions are played?
I apologize for the vague and potentially misleading title, I am very new to statistics and do not yet have a handle on the jargon. Essentially, I have the table below:

```<code>             A          B
C       0.925       1.008
D       1.033       0.871
</code>```

This is a table of the payoff when a given player plays option A or B versus what the opponent plays (C or D). For those of you familiar with game theory, this is simply a standard action and output table. Solving for the Nash Equilibrium, I assign option A to be played at probability "p", and thus B is played at 1-p. Multiplying the values by the probability at which they are played and setting row C equal to row D, I have found that it is optimal to play option A 55.95% of the time. However, assuming a self-interested opponent, they would catch on to this pattern, thus forcing an adjustment to volume at which each option is played. How would I adjust the values within the matrix when accounting for the volume at which each is played to find a continuous equilibria? Please let me know if I wasn't clear about anything.

答:
$55.92\%$ would be marginally closer to the equilibrium point but it makes little difference

The point about a mixed strategy with this value is that no matter whether the opponent plays $C$ or $D$ or some mixture of the two, the expectation of the outcome will always be about $0.9616$ since this is $55.95\% \times 0.925 + 44.05\% \times 1.008$ and is also $55.95\% \times 1.033 + 44.05\% \times 0.871$

So it does not matter whether the opponent catches on or not. There is nothing the opponent can do, good or bad, to make the expected outcome worse for the original player.


问:
Confusion about calculating cdf and pdf of random variable
Let $X$ be a random variable with pdf:
$$p_X(x)=\begin{cases}
1-|x| & x\in[-1,1] \\
0 & \text{Otherwise}
\end{cases}$$

Let $Y$ be
$$Y=g(X)=\begin{cases}
-X & x<0 \\
2X & x\geq 0
\end{cases}$$
Find cdf and pdf of $Y$.
This is my attempt:
If $0\leq y\leq 1$, then $\mathbb{P}(Y\leq y)=\mathbb{P}(-X\leq y)=\mathbb{P}(X\geq -y)=1-\mathbb{P}(X\leq -y)$. Furthermore, $\mathbb{P}(X\leq -y)=\int_{-1}^{-y}(1+x)dx=\dfrac{1}{2}-y+\dfrac{y^2}{2}$, hence $\mathbb{P}(Y\leq y)=\dfrac{1}{2}+y-\dfrac{y^2}{2}$. On the other hand, $\mathbb{P}(Y\leq y)=\mathbb{P}(2X\leq y)=\mathbb{P}(X\leq \dfrac{y}{2})=\int_0^{y/2}(1-x)dx=\dfrac{y}{2}-\dfrac{y^2}{8}$. So, that would yield
$$P_y(Y)=\begin{cases}
\dfrac{1}{2}+\dfrac{3y}{2}-\dfrac{5y^2}{8} & y\in[0,1] \\
\dfrac{y}{2}-\dfrac{y^2}{8} & y\in(1,2] \\
0 & \text{Otherwise}
\end{cases}$$

Then, I can take the derivative and thus find the pdf. However, I have several problems, because there is a jump at $y=0$, namely $P_y(0^-)=0$, but $P_y(0^+)=\dfrac{1}{2}$; also $P_y(1)=1.375$, which doesn't make sense. I know I am doing something wrong, but I am not sure where or what. Could someone give me some hint? Thanks!

Edit:
Since I know the pdf of $Y$ and I know that the cdf is just the integral and its value final value should be $1$, I calculated:
$$P_y(Y)=\begin{cases}
\dfrac{3y}{2}-\dfrac{5y^2}{8} & y\in[0,1] \\
\dfrac{1}{2}+\dfrac{y}{2}-\dfrac{y^2}{8} & y\in(1,2] \\
0 & \text{Otherwise}
\end{cases}$$

Everything seems to work, but I still can't "formally" get the cdf.

答:
From the other direction: find the Y-pdf from the X-pdf, then integrate to obtain the Y-CDF

(NOTE: that is  CDF for "Cumulative Distribution Function").

We have $Y= -X\,\mathbf 1_{X<0}+ 2X\,\mathbf 1_{X\geqslant 0}$ and $p_{\small X}(x)=(1-\lvert x\rvert)\,\mathbf 1_{x\in[-1..1]}$ , so using the Jacobian Transformation technique:

$$\begin{align}p_{\small Y}(y) &=p_{\small X}(-y)\left\lvert\dfrac{\mathrm d (-y)}{\mathrm d y}\right\rvert\mathbf 1_{y\in(0..1]}+p_{\small X}(y/2)\left\lvert\dfrac{\mathrm d (y/2)}{\mathrm d y}\right\rvert\mathbf 1_{y\in[0..2]}\\[2ex] &=(1-y)\mathbf 1_{y\in(0..1]}+\tfrac 12(1-y/2)\mathbf 1_{y\in[0..2]}\\[2ex]&=\frac {6-5y}4\mathbf 1_{y\in[0..1)}+\frac{2-y}4\mathbf 1_{x\in[1..2]}\\[4ex]F_{\small Y}(y) &=\int_0^y\dfrac{6-5t}{4}\,\mathrm d t\cdot\mathbf 1_{y\in[0..1)}\,+\left(\int_0^1\dfrac{6-5y}{4}\,\mathrm d t+\int_1^y\dfrac{2-y}{4}\,\mathrm d t\right)\mathbf 1_{y\in[1..2]} \\[2ex]&=\left[\dfrac{12t-5t^2}{8}\right]_{t=0}^{t=y}\mathbf 1_{y\in[0..1)}+\left(\left[\dfrac{12t-5t^2}{8}\right]_{t=0}^{t=1}+\left[\dfrac{4t-t}{8}\right]_{t=1}^{t=y}\right)\mathbf 1_{y\in[1..2)}+\mathbf 1_{y\in[2..\infty)}\\[2ex]&=\frac{12y-5y^2}{8}\mathbf 1_{y\in[0..1)}+\frac{4+4y-y^2}{8}\mathbf 1_{y\in[1..2)}+\mathbf 1_{y\in[2..\infty)}\end{align}$$

Which agrees with your answer.

Of course, when given that CDF you may find the pdf by differentiating.

$$\begin{align}p_{\small Y}(y)&=\dfrac{\mathrm d ~~}{\mathrm d y}\left(\frac{12y-5y^2}{8}\mathbf 1_{y\in[0..1)}+\frac{4+4y-y^2}{8}\mathbf 1_{y\in[1..2)}+\mathbf 1_{y\in[2..\infty)}\right)\\[2ex]&=\dfrac{6-5y}{4}\mathbf 1_{y\in[0..1)}+\dfrac{2-y}{4}\mathbf 1_{y\in[1..2]}\end{align}$$


问:
If a winning strategy does not exist for player 2, does it exist for player 1?
I have been studying some instances of strategy-stealing. The issue is that I am not sure whether proving the non-existence of a winning strategy for player 2 necessarily proves a winning (or at least tying) strategy for player 1. Wikipedia says

In combinatorial game theory, the strategy-stealing argument is a general argument that shows, for many two-player games, that the second player cannot have a guaranteed winning strategy.

The arguments that I have seen make sense, in terms of proving that player 2 cannot have a winning strategy. For example, the Wikipedia article on Chomp makes the modest claim

The second player therefore cannot have a winning strategy.

However, in Engel's Problem-Solving Strategies, Double Chess is analyzed in problem #18 of the games chapter and the following seemingly grander claim is made:

The rules of chess are changed as follows: Black and White make alternately two legal moves. Show that there exists a strategy for white which guarantees him at least a tie.

The proof is

Suppose $B$ can win no matter what $A$ does. On his first move, $A$ moves one of his knights to any one of the two possible squares and then back to its original position. Now all the pieces are in their original position, but $A$ has becmoe the second player and must win. Contradiction!

When does the negation of the existence of a winning strategy for player 2 imply the existence of a winning/tying strategy for player 1?

I might be confused because I am actually not sure of how to formally describe a winning strategy, so a rigorous explanation of that might be helpful as a preface. (I considered using alternating existential and universal quantifiers to describe winning strategies, but didn't know how to terminate the sequence.)

P.S. I have included set theory as a tag because, according to the Wikipedia article on determinacy, the subject is a subfield of set theory. However, I am not familiar with this part of set theory and I did not really understand the article.

答: The positions of a perfect information game with no randomness fall into four categories:  win for the first player, win for the second player, draw in finite time, infinite play.  Not all games will have positions in all categories.  Most games, including chess, have something that prevents the game going on forever.  In those cases, showing there is no second player win is sufficient to show there is either a first player win or both players can force at least a draw.  Chomp has no ties, so once you eliminate a second player win you know there is a first player win.</p>


问:
Finding a minimal polynomial with a modified root
Suppose you have a real number $x$. Given that its minimal polynomial is, say, $P(x)$, is there a simple way that you can find the minimal polynomial when $x$ is modified? For example, what the minimal polynomial of $x+\sqrt{2}$ is?

答:
Say you want to find the minimal polynomial of $y=p(x)/q(x)$, where $p,q$ are polynomials with rational coefficients. Then the minimal polynomial of $y$ is (a factor of) the resultant eliminating $x$ of $P(x)=0$ and $yq(x)-p(x)=0$; getting integers is a matter of multiplication.

In the case of non-rational coefficients, the second argument of the resultant will involve higher degrees of $y$, but the idea is still the same. For $y=x+\sqrt2$, the second polynomial will be $y^2-2xy+x^2-2=0$.


问:
Prove or Disprove that there is an integer $n$ such that $n^4 + 1 < 2n^2$
I am a beginner in writing proofs and want to show that an integer $n$ such that the title is true does not exist. I went about it this way and am wondering if this is even considered a valid proof, and if not, some assistance.

First, let $n = 0$

$0^4 + 1< 2(0)^2$

$1 < 0$. False, $1 > 0$, therefore, at $n = 0$, $ n^2 > 2n^2$

Let us simplify $n^4 + 1 < 2n^2$

$= n^4 < 2n^2 -1$

$= (n^2)(n^2) < 2n^2-1$

$= n^2 < \frac{(2n^2-1)}{n^2}$

We know that the $\lim_{n\to\infty} n^2 = \infty$ and $\lim_{n\to\infty}\frac{(2n^2-1)}{n^2} \approx 2$

So we will use candidate values $n = 1$ and $n = 2$

$ (1)^2 < \frac{(2(1)^2-1)}{(1)^2}$

$=  1 < 1/2$

False, $1 > 1/2$, therefore, at $n = 1$, $ n^2 > \frac{(2(n)^2-1)}{(n)^2}$

Then,

$ (2)^2 < \frac{(2(2)^2-1)}{(2)^2}$

$=  4 < 7/4$

False, $4 > 7/4$, therefore, at $n = 2$, $ n^2 > \frac{(2(n)^2-1)}{(n)^2}$

We have already verified that at $n = 0$ and $n = 1$, $ n^2 > \frac{(2(n)^2-1)}{(n)^2}$. Because at  $n = 2$, the value of $n^2$ is already greater than $\lim_{n\to\infty}\frac{(2n^2-1)}{n^2} = 2$, and $\lim_{n\to\infty} n^2 = \infty$, $n^2$ will continue to grow to $\infty$ while $\frac{(2(n)^2-1)}{(n)^2}$ approches $2$. Thus we have shown that there is no integer $n$ such that $n^4 + 1 < 2n^2$

答:
IMHO, it's a (mostly) valid proof, but it's a bit sloppy. It does more work than it needs to, and neglects mentioning the case of $n<0$.

Also, good proofs are economical: they achieve the task with the minimal mathematical machinery required. Eg, you really don't need that stuff about limits. (BTW, $\lim_{n\to\infty}\frac{(2n^2-1)}{n^2} = 2$. The limit is exact, it's not an approximation).

The comments show a couple of very succinct ways to prove your statement. However, it's not expected that your proof is the smallest possible proof. The history of mathematics is full of examples of proofs by good mathematicians that were later superceded by new proofs that were more compact, elegant, or more general. The main thing is to write a proof in such a way that your readers will be able to follow your train of thought, and verify that your claims are legitimate. You'll get better at that with practice, and by reading good proofs written by others. In that respect, the art of proof writing is similar to that of writing computer programs.


问:
Is The Inverse Laplace Transform of $e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)$ doable using inversion formula?
I'm trying to solve inverse laplace transform using inversion formula and given by this integral:
$$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i \infty} e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)\,\Bbb ds.$$

Here is my contour, since the branch points of $\operatorname{Log} \left(\frac{s+1}{s}\right)$ are $0$ and $-1$

First, i want to show integral on $L_u\cup L_d$ is $0$ by bounding the integral with ML and then take the limit when $R$ goes to $\infty$.

By letting $L_u,\, L_d: s= \xi\pm iR,0\leq \xi\leq \gamma$, where $\gamma$ is the real number that the vertical line of the given contour passed by.

Since the $L$ is $\lvert e^{t(\xi\pm iR)} \rvert$, then i have ML inequality as below:

$$\lvert F(s)e^{st} \rvert \leq M_R \lvert e^{t(\xi\pm iR)} \rvert = M_R e^{\xi t} \leq M_R e^{at}$$

Next, i need to find $M_R$ and take the limit.
$$\begin{align}
\left|F(s)\right| &= \left|\operatorname{Log}\left(\frac{s+1}{s}\right)\right|\\
&= \left|\operatorname{Log}\left(\frac{\xi\pm iR+1}{\xi\pm iR}\right)\right| = M_R
\end{align}$$

And by taking the limit of the last expression when $R$ goes to infinity yields $0$. Meaning the integrals along those lines are $0$.

So, from here, am i doing this right? I'm not sure my work is correct. Maybe there are some mistakes there. Help me please!

Edit:
Working with my $L_u$ with $ML$ inequality, i have $L=\gamma$. Assuming $-\pi<\operatorname{arg}{s}\leq \pi$ and parametrizing $s=-\xi+iR$, $\xi\in [-\gamma,0]$:

$\begin{align}
\left|\int_{L_u}\right|
&\leq \left|e^{st} \log\left(1 + \frac 1s\right)\right|\\
&\leq \left|e^{-\xi t}\right| \left|e^{iRt}\right|\left|\ln\left|1+ \frac{1}{-\xi+iR}\right| + i\pi\right|\\
&\leq 1\cdot 1 \cdot \ln\left(1+\frac 1R\right) + \pi\\
&\approx \frac 1R + \pi
\end{align}$

Combining the $ML$ i have
$$\frac{\gamma}{R} +\gamma\pi$$
Which does NOT approach to $0$. Why? Please spot my mistake. I can't think about how to make it goes to $0$ since yesterday. Hope you kind to help me.

New Edit: Now my main question is not about ML. I managed to set the big and small arc goes to $0$. My main question now is "is it possible to evaluate this by using inversion formula (without differentiate both sides) and use log form instead?"

What i mean by differentiate both sides is:
$$\begin{align}
\mathcal{L}^{-1} &= \operatorname{Log}\left(\frac{s+1}{s}\right)\\
(\mathcal{L}^{-1})' &= \frac{1}{1+s} - \frac{1}{s}
\end{align}$$

I don't want this kind of solution. What i really want is evaluating $\operatorname{Log}\left(\frac{s+1}{s}\right)$ by inversion formula. That's it.
because I keep getting integrals on the contour lines above and below the branch cut cancel each other and make everything 0, which of course I don't want it.

Attempt:

答:
Let's pick a sufficiently large $\gamma$ such that $|s|>1$, so that we can plug in

$$
\log\left(1+\frac1s\right)=\sum_{k=1}^\infty{(-1)^{k+1}\over ks^k}
$$

and get

$$
\begin{aligned}
f(t)
&={1\over2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\sum_{k=1}^\infty{(-1)^{k+1}e^{st}\over s^k}\mathrm ds \\
&=\sum_{k=1}^\infty{(-1)^{k+1}\over k}{1\over2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}{e^{st}\over s^k}\mathrm ds
\end{aligned}
$$

Provided that

$$
\mathcal L\{t^{k-1}\}(s)={(k-1)!\over s^k}
$$

we can deduce

$$
\begin{aligned}
f(t)
&=\sum_{k=1}^\infty{(-t)^{k-1}\over k!} \\
&=-\frac1t\left[\sum_{k=0}^\infty{(-t)^k\over k!}-1\right] \\
&={1-e^{-t}\over t}
\end{aligned}
$$

where the final step follows from the fact that

$$
e^z=\sum_{n=0}^\infty{z^k\over n!}
$$


问:
Complement of a set and inverse image.
I'm currently taking a real analysis class and we are working on measurable functions (the notes can be found here under "Measurable Functions"; the exercise is the first one on the last page).

$$\text{Let }f:E \to \mathbb{R}\text{ and }A\subseteq \mathbb{R}.\\\text{
Show that }f^{-1}(A^{C}) = E - f^{-1}(A).$$

I can easily visualize that this is indeed true (I drew a graph), but I'm unsure how to approach the proof. This chapter is dealing with measurable functions, however we do not have in the hypothesis that $f$ is measurable, and I'm not sure this is even needed in proving the claim.

My first strategy was assuming in two cases that $A$ is closed and open (therefore $A^{C}$ is open and closed respectively) and then going from there using neighborhoods, but this route seems like I'm making the problem harder than it is. I'm also unsure if I could prove the claim this way.

I'm also curious if I can prove this equality by basic set inclusion, i.e.: an element from set $A$ is in set $B$ and vice versa.

Any assistance in approach would be greatly appreciated.

EDIT: Also, I believe this could be a simple property of preimages and such, but still I'm unsure how to approach.

答:
Here's another way to think about it that might help.

Let $g$ be an indicator function defined as follows.

$$ \begin{align} g(x) = 1 & \;\; \text{if and only if} \;\; f(x) \in A \\ g(x) = 0 & \;\; \text{otherwise} \end{align} $$

$f^{-1}(A^C)$ is $g^{-1}(0)$ and $E - f^{-1}(A)$ is $E - g^{-1}(1)$ . Since the image of $g$ is $\{0, 1\}$ everything that isn't sent to $1$ is sent to $0$. Therefore $g^{-1}(0)$ is equal to $E - g^{-1}(1)$. Therefore, as desired

$$ f^{-1}(A^C) = E - f^{-1}(A)$$

Another way of looking at it is, since $f^{-1}(A^C)$ and $f^{-1}(A)$ are disjoint and $f$ is a total function on $E$ (presumably), the following is true because, forall $x$ in $E$, $f$ either sends $x$ to $A$ or it doesn't.

$$ f^{-1}(A^c) \sqcup f^{-1}(A) = E $$

$\sqcup$ denotes the disjoint union.

Title: Complement of a set and inverse image.


问:
Blocking directed paths on a DAG with a linear number of vertex defects.
Let $G=(V,E)$ be a directed acyclic graph.
Define the set of all directed paths in $G$ by $\Gamma$.
Given a subset $W\subseteq V$, let
$\Gamma_W\subseteq \Gamma$ be the set of all paths $\gamma\in\Gamma$ supported on $V\backslash W$ (i.e all vertices in $\gamma$ belong to $V\backslash W$).
Now define $l(W)$ to be:
$$l(W)=\max_{\gamma\in \Gamma_W} |\gamma|$$
Where $|\gamma|$ is the number of vertices in $\gamma$.

I want to prove (or disprove) the following claim:

${\bf Claim:}$ For every $\epsilon>0$ and every $k>0$, there are constants $L$ and $N$  such that  for any directed acyclic graph $G=(V, E)$ satisfying $|V|>N$ with the sum of incoming and outgoing degrees bounded by $k$, there exists a subset $W\subseteq V$ such that $\frac{|W|}{|V|}<\epsilon$  and $l(W)<L$.

The claim is true for directed trees (see edit 1 for a proof) but the same proof idea fails to work in more general DAGs.
Moreover, the statement fails to be true if we remove the constant degree requirement for $G$. Indeed, the maximal topological order on vertices indexed from 1 to n can not be "blocked" for any $\epsilon>0$ by any set $W$ of size linear in $n$.

Any direction or idea would be welcome.

Edit 1:

For trees, a standard proof would go like this: For $0\leq i\leq L-1$, Define  $W_L^i$  to be the set of all vertices reachable from the root with a directed path of length $i \pmod L$. Since the graph is a tree, any such path is uniquely defined for every vertex and therefore, for a given $L$, the set $\{W_L^i\}_{0\leq i\leq L-1}$ gives a partition of $V$. Therefore choosing $L=\frac{1}{\epsilon}$, there is some $i_0$ such that $|W_L^{i_0}|$ is at most $\frac{|V|}{L}=\epsilon |V|$. It is left to show that every $W_L^i$ is indeed $L$-blocking, but this is trivial since any step in a directed path down the tree increases the distance from the root by exactly 1, so the longest path containing no vertices from $W_L^i$ has to be of length at most $L-1$ (Connecting 2 adjacent floors in $W_L^i$)

Edit 2:

In general, the claim is true for any DAG for the special case of $\epsilon = \frac{2k}{2k+1}$ and $L=1$. To see that consider the following algorithm:

1- choose a vertex $v$ in the graph that still has neighbours. Keep $v$, and remove all of its neighbours (in both directions) from graph

2 - if any non isolated vertex is left, go back to 1. Otherwise exit.

The resulting graph is completely disconnected ($L=1$) and we removed at most an $\epsilon=\frac{2k}{2k+1}$ fraction of vertices from the graph.
The claim follows.

Edit 3:

As Misha Lavrov showed, the previous bound can be made tighter and we can prove the claim for $\epsilon=\frac{k}{k+1}$.
I discovered that this bound is not tight when the DAG has total degree bounded by 3. In this case, I will prove the claim for any $\epsilon>\frac{1}{2}$ where the previous bound is only $\epsilon=\frac{2}{3}$.
Define the in-degree and out-degree of a vertex $v$ in $G$ by $in(v)$ and $out(v)$ respectively. From the assumption, for all $v \in V$, $in(v)+out(v)\leq 3$. Define 4 sets: $\{V_i\}_{i=0}^3$ by:
$$V_i=\{v\in V | in(v)=i\}$$

Obviously, $\{V_i\}_{i=0}^3$ forms a partition of $V$. Therefore, either of the sets $V_1$ or $V_2$ has cardinality at most $\frac{n}{2}$. W.l.o.g, assume it is $V_1$.
Let $G'$ be the subgraph of $G$ induced by $V_2$. Obviously, for all $v\in V_2$, $out(v)\leq 1$ and therefore, $G'$ is a disjoint union of directed trees with one vertex sink. Using the proof for trees, for any $\epsilon>0$,  we can find a subset $W\subseteq V_2$ such that $\frac{|W|}{|V_2|}<\epsilon$ and $W$ is $\frac{1}{\epsilon}$-blocking in $G'$.
Finally, define $W'= V_1 \cup W$. On one hand, $|W'|$ is upper bounded by $(\frac{1}{2}+\epsilon)n$ and on the other hand $W'$ is $(\frac{1}{\epsilon}+2)$-blocking since a directed path in $G$ can either stay in $V_2$ and get blocked by $W$ or get outside of $V_2$ and either be blocked by $V_1$ or do one last step from $V_0$, to $V_3$ or both.

This proves the claim.

PS. Crossposted at MO.

答: I know it is a bad style but I have no time for proper typing now (maybe I'll do it later), so here is a set of handwritten notes with a counterexample. Questions are welcome, as usual :)</p>


问:
showing ($\psi - \text{ id}_V)(W) = W$ iff $U \cap W = \{0\}$
Let $V$ be a finite $K$-vector space of dimension $n$. let $\phi$ and $\psi$ be linear transformations from $V \longrightarrow V$ such that $\psi \circ \phi = \phi \circ \psi$. Let
$$ W := \{v \in V | \phi(v)=v \} \text{ and } U := \{v \in V | \psi(v)=v \}.$$

How do I show:  $$(\psi - \text{ id}_V)(W) = W \text{ iff } U \cap W = \{0\}?$$

I've already shown the following from the previous part of the question:

($\psi - \text{ id}_V)(W) \subseteq W$,

$W \neq \lbrace 0 \rbrace  \text{ iff } \det(\text{Id}_n -A)=0$,

where $A :=$ the matrix of $\phi$.

答:
From basic linear algebra, we know that $W$ is an invariant subspace of $\psi$ (hence an invariant subspace of $\psi-id$).

$(\psi -id)(W)=W$

iff $\psi -id$ is invertible on $W$

iff for any $0\neq w\in W$, $(\psi -id)(w)\neq 0$

iff $U\cap W=0$

We are done.


答:
If a strategy stealing argument applies, we can rule out that the second player will win by imagining an (unspecified, hypothetical) strategy that is "optimal" for them, and showing that a sequence of moves from the first player - a sequence of moves which is allowed to incorporate this imagined strategy - is strictly better than the strategy we started out with. Since we assumed that that strategy was optimal to begin with, we show that the first player cannot do worse than the second.

However, we can argue further from this point that, therefore, this stealing is an optimal strategy for the first player. The question is then, in response to this stolen strategy, whether the 2nd player's responses can force a draw or whether they lose. They cannot win, because if they could, the 1st player would have pre-empted them by doing it first.

A game of finite length can only have three outcomes - P1 wins, a draw, or P2 wins. We know P2 cannot win, but we cannot use strategy-stealing alone to show whether P1's stolen strategy can force a win for them or whether P2 can respond in such a way that they draw. But by showing P2 cannot win, we entirely elimate one of the three outcomes and so all possible perfect plays will result in the other two - P1 wins, or draws.

Title: If a winning strategy does not exist for player 2, does it exist for player 1?

Body:
I have been studying some instances of strategy-stealing. The issue is that I am not sure whether proving the non-existence of a winning strategy for player 2 necessarily proves a winning (or at least tying) strategy for player 1. Wikipedia says

In combinatorial game theory, the strategy-stealing argument is a general argument that shows, for many two-player games, that the second player cannot have a guaranteed winning strategy.

The arguments that I have seen make sense, in terms of proving that player 2 cannot have a winning strategy. For example, the Wikipedia article on Chomp makes the modest claim

The second player therefore cannot have a winning strategy.

However, in Engel's Problem-Solving Strategies, Double Chess is analyzed in problem #18 of the games chapter and the following seemingly grander claim is made:

The rules of chess are changed as follows: Black and White make alternately two legal moves. Show that there exists a strategy for white which guarantees him at least a tie.

The proof is

Suppose $B$ can win no matter what $A$ does. On his first move, $A$ moves one of his knights to any one of the two possible squares and then back to its original position. Now all the pieces are in their original position, but $A$ has becmoe the second player and must win. Contradiction!

When does the negation of the existence of a winning strategy for player 2 imply the existence of a winning/tying strategy for player 1?

I might be confused because I am actually not sure of how to formally describe a winning strategy, so a rigorous explanation of that might be helpful as a preface. (I considered using alternating existential and universal quantifiers to describe winning strategies, but didn't know how to terminate the sequence.)

P.S. I have included set theory as a tag because, according to the Wikipedia article on determinacy, the subject is a subfield of set theory. However, I am not familiar with this part of set theory and I did not really understand the article.


问:
Homeomorphism examples
What's a homeomorphism?

a. any countable $f\colon \mathbb {N} \to \mathbb {Q}$

b. $f\colon \mathbb {R} \to \mathbb {R}$, $x \mapsto x^3$

c. $f\colon [0,2\pi ) \to S^1 = \{ z \in \mathbb {C}\,|\,|z|=1\}$, $\varphi \mapsto e^{i \varphi }$

I'm very sure that b doesn't work because it's not bijective and I think that everything is fine with a. But I'm really unsure about c, can someone explain it for me?

答:
Definition Let $X$ and $Y$ be topological spaces ; let $f\colon \Bbb R\to \Bbb R$ be a bijection. If $f$ and $f^{-1}$ are continuous, then $f$ is a homeomorphism. Now, back to your examples

(a) $f\colon \Bbb N \to \Bbb Q$, you must define before saying $f$ is homeomorphism.

(b) the function $f(x)=x^3$ is indeed  bijection and $f$ and $f^{-1}$ are continuous. So, $f$ is homeomorphism.

Try to check $c$. I hope that helps

Title: Homeomorphism examples


问:
Distribution of time that a flashlight can operate
The lifetimes of batteries are independent exponential random variables , each having parameter $\lambda$. A flashlight needs two batteries to work. If one has a flashlight and a stockpile of n batteries, What is the distribution of time that the flashlight can operate?

What I have so far:

Let $Y$ be the lifetime of the flashlight; $Y_1=min(X_1,...,X_n)$ where $X_i$ is the lifetime of a battery ($1\le i\le n$), and $Y_2$ the second smallest of the $X_i$ (so $Y_1\le Y_2$)

I wanted to compute: $P[Y\le t]=P[Y_2\le k+m|Y_1\le m]$ where $k+m=t$ then we have that $$P[Y_2\le k+m|Y_1\le m]={P[Y_2\le k+m, Y_1\le m]\over P[Y_1\le m]}={P[Y_2\le k+m] P[Y_1\le m]\over P[Y_1\le m]}=P[Y_2\le k+m=t]$$ (because of the independence of the random variables)

So $P[Y_2\le t]=P[min(X_1,...,X_{j-1},X_{j+1},...X_n)\le t]$ (assuming $X_j=min(X_1,...,X_n)$) hence:

$$P[min(X_1,...,X_{j-1},X_{j+1},...X_n)\le t]= 1-P[min(X_1,...,X_{j-1},X_{j+1},...X_n)\ge t]=1-P[X_1\ge t,...,X_{j-1}\ge t, X_{j+1}\ge t,... X_n\ge t]=1-e^{(n-1)\lambda t}$$

I would really appreciate if you can tell me if this is the correct approach :)

答:
We are given $n$ batteries, and we put them two at a time in the flashlight. When one of the batteries fails, we swap it out and replace it with an unused one, if possible. The key here is that the battery that we did not swap out now behaves as a completely new battery. That's because the exponential distribution has the special property: given that a battery has survived to time $t$, the distribution of the additional time that it survives is independent of $t$.

Hence, the lifetime of every pair of batteries follows the distribution of the minimum of two exponential random variables. The probability that the minimum is less than $a$ is given by $1 - \exp(-2\lambda a)$, hence the minimum is itself an exponential with parameter $2 \lambda$.

Now since we swap out a battery exactly $n-1$ times, the total lifetime of the flashlight is the sum of $n-1$ independent, identically distributed exponentials of $2 \lambda$, which is a Gamma distribution with parameters $n-1$ and $2 \lambda$. (This result can be seen easily by noting that the sum of two Gamma-distributed variables is also Gamma distributed, and the exponential is a special case of the Gamma).


问:
If $V$ is a vector space over a division ring $K$, and $A=\mathrm{End}_K(V)$, then every quotient ring of $A$ is a prime ring
Let $K$ be a division ring, let $V=V_{K}$ a vector space over $K$, and let $A=\mathrm{End}_{K}(V)$. Could anyone give me an idea of how to prove that every quotient ring of $A$ is a prime ring?</p>

答:
(The question clearly is asking about quotients by proper ideals.)

The ideal structure of this sort of ring is very simple (although it will take some work for you to show that it is true.) It turns out that the proper ideals of this ring form a chain, and they are exactly of the form $I_\alpha=\{f\in A\mid \dim_K(Im(f))\leq\alpha\}$ where $\alpha$ denotes a cardinal, $\omega\leq \alpha<\dim_K(V)$ if $\dim_K(V)$ is infinite, and $0$ if $\dim_K(V)<\omega$ (is finite, naturally because in that case it's just a matrix ring over a division ring, which is simple, hence it trivially has all factor rings prime).

So you can break this task into parts:

Show that each $I_\alpha$ is an ideal of $A$.

Show that each $I_\alpha$ is a prime ideal of $A$.

Show that the $I_\alpha$ are the only ideals.

Then you will be done: the quotient by any of these prime ideals creates a prime ring.

FYI, there is a connection to this question which might furnish an alternative proof: Exercise about prime ideals. The ring you are talking about is an example of a ring with all ideals idempotent and forming a chain.

It is an elementary exercise to show that $A$ is von Neumann regular. Further, it is elementary to show that every ideal of a VNR ring is idempotent. You can use this to replace point #2 above, because with #1 and #3, the linked question gives you that the ideals are prime. However, it's probably just as easy to prove they are prime directly :)


问:
Why do we not define vector multiplication and division?
Consider, two planar vectors:

$$V=  a \hat{x} + b \hat{y}$$

And
$$ U = a' \hat{x} + b' \hat{y}$$

These are analogous to the complex numbers:

$$ v = a + bi$$

and,

$$u= a' + b' i$$

Now, there are clear rules for multiply $ v \cdot u$ and also the expression: $ \frac{v}{u}$ , also there exists a geometric interpretation (if you view in polar form). Then, why is it that we don't bring up these products when speaking vectors and only discuss dot and cross product?

答:
Your proposed product is geometrical in a sense, but there is a crucial difference between it and the dot and cross products. Imagine you have a table, or some other flat horizontal surface, and two arrows drawn on it that represent vectors. Here is a question: What is the "complex number" product of these two vectors?

It is not possible to answer this question. This is perhaps most easily seen by noting that you cannot find the polar form of the vectors, because the angle depends on a coordinate system ("basis"), and we haven't specified one. In other words, the "complex number" product depends not only on the vectors, but on an (entirely arbitrary) choice of coordinates.

By contrast, the dot product and cross product can be understood without reference to any particular coordinate system. The dot product is the same in any orthonormal basis (that is, a coordinate system made from vectors of unit length that are all perpendicular to each other). This is also true of the cross product, provided the two bases have the same handedness (that is, you can rotate one basis into the other).


问:
Example of non-polynomial functions on infinite ring
I would like to finish the following exercise:

R is a infinite commutative ring with $1 \neq 0$.
We know that not every function $f:R \to R$ is a polynomial function (as opposed to when the ring is finite).
Give an example of $f:R\to R$ that is not a polynomial function (and proof that it's not a polynomial function).

My attempt:
For $x\neq 0$ we define $f(x)=\prod_{i=0}^{\infty}(a_i+(-x))$ where $-x$ is inverse element of $x$ with respect to $+$ operation and $a_i \in R, a_i\neq a_j$ for $i\neq j$. From this we know that $f(x)=0$ for infinitely many $x$.

And in this moment I have a question. ,,number of roots of polynomial $\leq$ degree of polynomial" does it always work for $R$ defined like that?

If yes, then $f$ got $\infty$ many roots so its either no-polynomial function or zero function (polynomial deg 0). So by defining $f(0)=1$ we know that it's not polynomial function.

Is this correct? Thanks in advance.

Next attempt, I will try to show that there is no polynomial equal to complex conjugation:
$f(a+bi)=\sum_{n=0}^{n'}(a_n+b_ni)(a+bi)^n$ for some $a_n,b_n\in \mathbb{R}$.
$(a+bi)^n=\sum_{k=0}^n {n\choose k}a^{n-k}(bi)^k=\sum_{k=0}^{\lceil \frac{n-1}{2} \rceil} {n\choose 2k}a^{n-2k}b^{2k}(-1)^k+$ $+i\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} {n\choose 2k+1}a^{n-(2k+1)}b^{2k+1}(-1)^k$

Thus

$Im(f(a+bi))=\sum_{n=0}^{n'}(a_n\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} {n\choose 2k+1}a^{n-(2k+1)}b^{2k+1}(-1)^k)+$ $+(b_n \sum_{k=0}^{\lceil \frac{n-1}{2} \rceil} {n\choose 2k}a^{n-2k}b^{2k}(-1)^k)$

And

$Im(f(a-bi))=\sum_{n=0}^{n'}(-a_n\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} {n\choose 2k+1}a^{n-(2k+1)}b^{2k+1}(-1)^k)+$ $+(b_n \sum_{k=0}^{\lceil \frac{n-1}{2} \rceil} {n\choose 2k}a^{n-2k}b^{2k}(-1)^k)$

And because $Im(f(a+bi))=-Im(f(a-bi))$ if we want it to be complex conjugation then every $b_n=0$, and by very simmular way we can show that every $a_n=0$ from $Re(f(a+bi))=Re(f(a-bi))$, thus $f(x)=0$, thus there is no polynomial equal to complex conjugation.
Is it correct now?

答:
Every function $R\to R$ is represented by a polynomial iff  $R$ is a finite field.

Let $f(0)=1, f(a)=0$ for $a\ne 0$.

If $R$ is a field then it is easy as polynomials have finitely many roots.

Otherwise take a non-unit $a$, if $f$ was a polynomial we'd have $f(a)\equiv f(0)\bmod (a)$.


问:
Solve equations $\sqrt{t +9} - \sqrt{t} = 1$
Solve equation:  $\sqrt{t +9} - \sqrt{t} = 1$

I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$

I squared both sides $(\sqrt{t+9})^2  = (1)^2 (\sqrt{t})^2$

Then I got $t + 9 = 1+ t$

Can't figure it out after that point.

The answer is $16$

答:
From $\sqrt{t +9} - \sqrt{t} = 1$ you don't get $\sqrt{t +9} = 1 -\sqrt{t}$ but $$\sqrt{t +9} = 1 +\sqrt{t}.$$

You can solve as follows, using the algebraic identity $(a+b)^2=a^2+2ab+b^2$:
$$
\begin{eqnarray*}
\sqrt{t+9}-\sqrt{t} &=&1\Leftrightarrow \sqrt{t+9}=1+\sqrt{t}\tag{1} \\
\text{Square both sides of $(1)$} &\Rightarrow &\left( \sqrt{t+9}\right) ^{2}=\left(
1+\sqrt{t}\right) ^{2} \\
\text{Compute and simplify} &\Leftrightarrow &t+9=1+2\sqrt{t}+t\Leftrightarrow 9=1+2\sqrt{t} \\
\text{Simplify} &\Leftrightarrow &9-1=2\sqrt{t}\Leftrightarrow 8=2\sqrt{t} \\
\text{Simplify} &\Leftrightarrow &\frac{8}{2}=\sqrt{t}\Leftrightarrow 4=
\sqrt{t}\tag{2} \\
\text{Square both sides of $(2)$} &\Rightarrow &4^{2}=\left( \sqrt{t}\right) ^{2} \\
&\Leftrightarrow &16=t.\tag{3}
\end{eqnarray*}
$$

Final comment. When we square both sides of an equation we get a new equation which has the same solutions of the original equation, but can have additional solutions. However in this case we got only the solution $t=16$, which is a solution of $(1)$ too.

ADDED. In your recent question solve the equation $\sqrt{3x-2}+2-x=0$, we get two solutions after squaring
$$
\begin{equation*}
\sqrt{3x-2}+2-x=0\Rightarrow 3x-2=x^{2}-4x+4\Leftrightarrow x\in \{1,6\}
\end{equation*}
$$
but only $x=6$ is a solution of $\sqrt{3x-2}+2-x=0$, as explained in
this comment by Glen O.


问:
Uniqueness of endpoints of a closed bounded interval in a partial order.
Let $P$ be a partial order, and let $a,b,c,d$ be elements of $P$ such that $a \leq b$ and $c \leq d$. If the interval $[a,b]=[c,d]$, then does it follow that $a=c$ and $b=d$? In case it needs clarification, the notation $[x,y]$ is shorthand for $\{p \in P| x \leq p \leq y\}$

答: Of course. $b \in [a, b] = [c, d]$ implies $b \leq d$. $d \in [c, d] = [a, b]$ implies $d \leq b$. So $b = d$ by anti-symmetry. Similarly $a = c$.


问:
What is the probability of winning in a dice game
John and Lucy take turns rolling two dice:

If John rolls a pair larger than or equal to $10$ he wins and the game stops.

If Lucy rolls a pair strictly smaller than $10$, she wins and the game stops.

They keep rolling in turn until one of them wins.

Suppose John rolls first. What is the probability that Lucy wins $?$.

My attempt:

First, we have to figure out the probability of John winning on any particular round. That is if he rolls: (5,5), (5,6), (6,5), (4,6), (6,4), (6,6). Thus, his chances of winning are $\frac{6}{36}=\frac{1}{6}$. Then since Lucy wins if she rolls everything else, that means she has a $5/6$ chance of winning on any particular round.

Now to figure out the probability of Lucy winning if John goes first. For this to happen, John has to lose his first roll and Lucy has to win after that. We know that P(John Losing)=$\frac{5}{6}$ and P(Lucy winning)=$\frac{5}{6}$. Thus, we have $(\frac{5}{6})(\frac{5}{6})=\frac{25}{36}$ chance of Lucy winning the game.

However, I don't think that this is right because it does not account for playing additional rounds (i.e. she doesn't win on the second round but on some nth round afterward). Could I get some help? Thanks.

答:
Suppose John's winning probability is $p$. Then by the law of total probability,
$$p=\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}\cdot p\implies\boxed{p=\frac{6}{31}.}$$


答: If $[x,y]\ne\varnothing$, it is still the case that $x=\min[x,y]$ and $y=\max[x,y]$. Therefore $\varnothing\ne [a,b]=[c,d]$ implies $a=c$ and $b=d$.

Title: Uniqueness of endpoints of a closed bounded interval in a partial order.

Body: Let $P$ be a partial order, and let $a,b,c,d$ be elements of $P$ such that $a \leq b$ and $c \leq d$. If the interval $[a,b]=[c,d]$, then does it follow that $a=c$ and $b=d$? In case it needs clarification, the notation $[x,y]$ is shorthand for $\{p \in P| x \leq p \leq y\}$


问:
Is there a function discontinuous on $C = \{\frac{m}{2^{n-1}} \in \Bbb R ; m \in \Bbb Z, n \in \Bbb N \} $?
Given the set $C = \{\frac{m}{2^{k-1}} \in \Bbb R ; m \in \Bbb Z,k \in \Bbb N \} $, is there a function $f: \Bbb R \to \Bbb R$ such that the set of discontinuous points of f are equal to C?

Information that maybe be useful: C is dense on $\Bbb R$.

I think there might be something in the way of $f(x) =
\begin{cases}
0,  & \text{if $x \notin C$} \\
\frac{1}{k}, & \text{if $x = \frac{m}{2^{k-1}} \in C$}
\end{cases}$

In this case:

If $a \in C$, there is a sequence $(x_n) \subset \Bbb R \setminus C$ such that $\lim(x_n)=a$. Then $\lim f(x_n)=0\neq f(a)$. Therefore, f is discontinuous at $a \in C$.

If $a \in \Bbb R \setminus C$, we must show that: $\forall \epsilon \gt 0 (\exists \delta \gt 0(|x-a| \lt \delta \to |f(x)-f(a)| \lt \epsilon))$. For $x \in \Bbb R \setminus C$ it's trivial, because in this case $|f(x)-f(a)|=0 \lt \epsilon$. But now I'm having trouble to show for $x \in C$. As I see it, intuitively, if $x = \frac{m}{2^{k-1}}$ is very close to $a \notin C$, then k must be very large.

答:
You are on the right track. You correctly demonstrated that $f$ is discontinuous at every $a \in C$.

Now let $a \in \Bbb R \setminus C$ and $\epsilon > 0$. The interval $[a-1, a+1]$ contains only finitely many fractions $\frac{m}{2^{k-1}}$ with $m \in \Bbb Z$, $k \in \Bbb N$, and $k \le 1/\epsilon$. Therefore we can define
$$
 \delta = \min \{ | a - \frac{m}{2^{k-1}}| : m \in \Bbb Z, k \in \Bbb N, k \le \frac{1}{\epsilon}\} > 0 \, .
$$
If $|x - a| < \delta$ then either $x \in \Bbb R \setminus C$ or $x = \frac{m}{2^{k-1}}$ with $k > \frac{1}{\epsilon}$. In the first case is $|f(x) - f(a)| = 0$, and in the second case is $|f(x) - f(a)| = \frac 1k < \epsilon$.

This shows that $f$ is continuous at every $a \in \Bbb R \setminus C$.


问:
I need an example of the use of Euler's Theorem for polynomials
I need an example of the use of this theorem for $\mathbb Z / 3$ if possible.

 Let  ∈ [],   a finite field and: ()={  ∈ []: 0 ≤  ≤  and (,)=1} So, for all  ∈ [] with (,)=1, we have: ^(())≡ 1 ( )

For instance, apply this to $x^3 + 2x + 1$

PS: Consider   as the degree of a polynomial

答:
Perhaps this is a bit un-imaginative, but we can perform pseudo-primality tests:

Claim: $m=x^3+x+1$ is not irreducible in $\mathbb{Z}_3[x]$.

Proof: If not, then $\varphi(m)=\left|\mathbb{Z}_3\right|^{\deg m}-1=26$. Thus by Euler's theorem, $x^{26}\equiv 1\pmod{m}$. However by explicit division with remainder, we find that
$$x^{26}\bmod m=x^2\ne 1$$
Thus $m$ is not irreducible. Indeed, $m=(x-1)(x^2+x-1)$.


问:
Checking the measurability of a function
I have to check check the measurability of such a function.
$f(x):\mathbb{R} \rightarrow \mathbb{R}$

$f(x) =  \left\{ \begin{array}{ll}
4x-3 & \textrm{when $x \in V$}\\
0 & \textrm{when $x \in  \mathbb{R}\V$}\\
\end{array} \right. $.

When $V$ is vitaly set. I know how to do it when the whole graph of a function is above or below the OX axis, here the situation seems different and I do not know what to do.

答:
To disprove that the function is measurable, it suffices to find a measurable set $B$, such that $f^{-1}(B)$ is not measurable. For instance we could take $B=\{0\}$, and note that $$f^{-1}(\{ 0 \})= (\mathbb{R}\setminus V) \cup \{\frac34\}$$
which is not a measurable set, since neither the vitali set $V$ nor the complement $\mathbb{R}\setminus V$ are measurable.


问:
Intuition behind closed subsets of a metric space?
Reading for my exam in real analysis, I struggle with the definition of a closed subset of a metric space.

Consider a metric space $$(X,d)$$

Then consider a subset of this space$$F$$

What the book is trying to convince me of is that F is closed iff a convergent sequence in F always has a limit belonging to F.

The intuition that:

$$x_1,\dots,x_n,\dots\in F$$
$${x_n}\rightarrow a $$
where
$$a\in F$$
for all convergent sequences in F implies that F is closed makes sense to me.

What I am struggling with is understanding the converse. What is it about F containing all of its boundary points that makes it impossible to have a sequence in F converging to some limit outside of F (even though the sequence will never get there)?

Edit:
Is the last sentence in the parenthesis actually the reason I am looking for? That it is impossible to get arbitrarily close to some point outside of F, because when you get really close, then you're not necessarily in F at all?

答:
I'm slightly confused about what you think the converse is, and what your definition of a closed set is. You're happy that $$\text{all convergent sequences in $F$ converge to a limit in }F \implies F \text{ is closed}$$ so what you're missing is $$ F \text{ is closed}\implies\text{all convergent sequences in $F$ converge to a limit in }F $$

But "$F$ consisting only of interior points" is equivalent to $F$ being open, which doesn't have anything to do with this statement. Since $F$ closed $\iff$ the complement of $F$ is open, one can argue the following if you want to think about interior points:

Suppose a sequence converges to some point $x\not\in F$. But the complement of $F$ is open, so $x$ is interior to the complement, and there is an open ball $B_x$ which also does not intersect $F$. But then the sequence cannot lie in $F$ since it must eventually lie entirely within $B_x$ since it tends to $x$. Intuitively: there's a neighbourhood of the limit which is stuck outside $F$.

Edit: If instead you want to think in terms of $F$ containing all 'boundary' points or limit points, then the result you want is true by definition! 'Closed' literally means 'closed under taking limits'.


问:
Geometric probabilities solution verification

An unbiased coin is tossed until a head appears and then tossed until a tail appears. If the tosses are independent, what is the probability that a total of exactly n tosses will be required?

My attempt:

P($n$ tosses required to produce one head and one tail) $=$ P($x$ tosses needed for first head) $\times$ P($y$ tosses needed for first tail) where $x+y=n$.

So, the probability becomes
$$\begin{align}
\left(\frac{1}{2}\right)^{x-1} \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^{y-1} \cdot \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{n}
\end{align}$$
This is not the correct answer, however. The correct answer is $\left(\frac{1}{2}\right)^{n} \cdot (n-1)$. Can someone pleas explain what I did incorrectly and where the $n-1$ factor is coming from?

答:
Your thinking is almost there - except in your argument, $x$ could be anything from $1$ to $n-1$ (since getting a head on $x$th toss and then getting a tail on the $y = n - x$th toss gives you a head and tail on exactly the $n$th toss). So if we sum your expression over all possible combinations of $x$ and $y = n-x$, we have $n - 1$ times the expression you got.

More generally, this is an example of (a variant of) the Negative Binomial distribution, which counts the number of successes needed for $r$ events to happen.


问:
How should I visualize mappings in topology?
When I'm trying to visualize mappings from one topological space to another, say from (X,T) to (Y,T1), should I be thinking of the elements X being mapped to elements Y or should I be thinking in terms of open sets in T mapped to open sets in T1?</p>

答:
The map itself goes from $X$ to $Y$, and that's how you should visualize it.

A good example of this that you can draw for yourself is the notion of a path in topology. Intuitively, if I have a topological space $(Y,\sigma)$, a path in $Y$ should be a way of "drawing a line" in $Y$ in a continuous way. This winds up being best phrased as a continuous map from $[0,1]$ with the usual topology. (There's a bit of a subtlety here; what we think of colloquially as a path is really the image of such a map, but the term "path" winds up being used for the map itself because that's ultimately a more useful thing to consider.

That said, what $f$ does to sets is very important ... but you want to look backwards! Specifically, if $f:(X,\tau)\rightarrow (Y,\sigma)$ is a continuous function (= the right kind of map to consider when doing topology), we have $$f^{-1}(U)\in\tau\mbox{ whenever }U\in\sigma.$$ The original $f$ was, set-theoretically, a function from $X$ to $Y$, but it induces a function $\hat{f}:\sigma\rightarrow\tau$.

By contrast, continuous maps need not send open sets to open sets. Consider for example $$f: (\mathbb{R}, \tau_{Euc,1})\rightarrow (\mathbb{R}^2, \tau_{Euc,2}): x\mapsto (x,0).$$ (Here $\tau_{Euc, n}$ is the usual Euclidean topology on $\mathbb{R}^n$.) This is continuous but does not send open sets to open sets.


答:
For planar vectors we can indeed multiply and divide.

First identify $V = a \hat{x} + b \hat{y}$ with $(a, b)$ and

$U = c \hat{x} + d \hat{y} $ with $(c, d)$

Now define $(a, b) \times (c, d) := (ac - bd, ad + bc)$.  You can check that this definition of multiplication is well-defined.

Also, $\frac{(a,b)}{(c,d)} = \frac{1}{c^2 + d^2}(ac + bd, ad - bc)$ is well defined.

These formulas may look foreign at first site but a closer inspection reveals that they are just multiplication and division of complex numbers.

Title: Why do we not define vector multiplication and division?

Body:
Consider, two planar vectors:

$$V=  a \hat{x} + b \hat{y}$$

And
$$ U = a' \hat{x} + b' \hat{y}$$

These are analogous to the complex numbers:

$$ v = a + bi$$

and,

$$u= a' + b' i$$

Now, there are clear rules for multiply $ v \cdot u$ and also the expression: $ \frac{v}{u}$ , also there exists a geometric interpretation (if you view in polar form). Then, why is it that we don't bring up these products when speaking vectors and only discuss dot and cross product?


问:
Solving for $x$: $a \sin(bx + c) = \sin(x)$
Ok, I scoured the internet for more than a few months for this one (whenever I had the time). But just because of this question I've created my first Stack Exchange account.

I have thrown this at all Computer Algebra System software I was accessible to. (not Maple or Mathematica though)

I also tried complex transformation, no luck there (not too much experience there)

I know this could be easily numerically solved because it is guaranteed that there is exactly one solution for every interval when: $a \sin(bx + c) \le 1$. In my case, the general solution is crucial because I'm applying this to an optimization where I can't afford to have an additional dimension to be optimized.

This problem can be visualized as the intersection of two sine waves.

A simplified version of the problem (when a = 1) could be easily solved with the sum-to-product identity:

$\sin(bx + c) = \sin(x)$

$0 = \sin(bx + c) - \sin(x)$

$0 = 2 \cos(\frac{x(b + 1) + c}{2})\sin(\frac{x(b - 1) + c}{2})$

$0 = \cos(\frac{x(b + 1) + c}{2})$ and $0 = \sin(\frac{x(b - 1) + c}{2})$ etc...

But for the time being, the only valid assumption in my case is that b is rational

答:
COMMENT.-I doubt that there is a general closed solution as a function of $a, b, c$. However, I offer you here an equation of which approximate solutions could be found.

We have $$a\sin(bx+c)=a(\sin(c)\cos(bx)+\cos(c)\sin(bx))$$ from which we get the equation
$$a\sin(c)\sum_{k=0}^{\infty}\frac{(-1)^k(bx)^{2k}}{(2k)!}+a\cos(c)\sum_{k=0}^{\infty}\frac{(-1)^k(bx)^{2k+1}}{(2k+1)!}=\sum_{k=0}^{\infty}\frac{(-1)^k(x)^{2k+1}}{(2k+1)!}$$
This is a really difficult problem and what you want is probably impossible, I think.


问:
Notation used in the proof of $^C(^B A) \approx \ ^{C \times B}A$: $\ \ \ \Phi (f)(c,b)$ and $\big(f(c)\big)(b)$
I am trying to step through a proof provided by my book of this statement:

$^C(^B A) \approx \  ^{C \times B}A$

The author writes the following:

Define $\Phi : \ ^C(^B A) \xrightarrow[onto]{1-1}\ ^{C \times B}A $ by saying that $\Phi (f)(c,b) = \big(f(c)\big)(b)$

I have never seen these notations before: $\Phi (f)(c,b)$ or $\big(f(c)\big)(b)$. What do they signify?

$^C(^B A)$ and $^{C \times B}A$ are both sets of functions...but I am not certain how to parse the above notation.

Focusing just on $\ ^C(^B A) $, I can decompose this set into first $^B A$, which is a set of functions from $B$ to $A$. Which I'll illustrate as $\{g_{B \to A_1}, g_{B \to A_2},...,g_{B \to A_n}\}$.

Then $\ ^C(^B A) $ is the set of functions from $C$ to $\{g_{B \to A_1}, g_{B \to A_2},...,g_{B \to A_n}\}$.

So one such function $f$ would look like: $f_1= \{\langle c_1, g_{B \to A_1} \rangle,...,\langle c_m,g_{B \to A_m} \rangle \}$

Given that $g_{B \to A_1}$ would look something like $\{\langle b_1, a_1 \rangle, ..., \langle b_p, a_p \rangle \}$, the first element of $f_1$...i.e. $\langle c_1, g_{B \to A_1} \rangle$...would look like:

$\langle c_1, \{\langle b_1, a_1 \rangle, ..., \langle b_p, a_p \rangle \} \rangle$

Thus, $f_1$ contains different variations of this.

But even after breaking this down, I still do not see what $\Phi (f)(c,b)$ is specifically referencing.

答:
Since the domain of $\Phi$ is ${^C}\left({^B}A\right)$, the $f$ in $\Phi(f)$ must be a function from $C$ to ${^B}A$. The codomain of $\Phi$ is ${^{C\times B}}A$, so $\Phi(f)$ must be a function from $C\times B$ to $A$; for convenience call this function $g$, so that $\Phi(f)=g:C\times B\to A$. Now we can apply this function $g$ to any $\langle c,b\rangle\in C\times B$ to get $g(c,b)\in A$. And $g$ is just a simpler name for the function $\Phi(f)$, so we can equally well write this element of $A$ as $\big(\Phi(f)\big)(c,b)$. Finally, context makes it clear that $\Phi(f)$ is a function from $C\times B$ to $A$, so we can drop one set of parentheses and write simply $\Phi(f)(c,b)$. That is, we know that the function $\Phi$ takes a single argument, so $f$ must be that argument, and we know that $\Phi(f)$ is a function that takes arguments in $C\times B$, so $\langle c,b\rangle$ must be the input to the function $\Phi(f)$.

The rest of the quoted definition tells you which function from $C\times B$ to $A$ the function $\Phi(f)$ actually is. Recall that $f$, the input to $\Phi$, must be a function from $C$ to ${^B}A$. And $c\in C$, so we can apply $f$ to $c$ to get a function $f(c)$ from $B$ to $A$. For convenience I’ll call that function $h$, so that $f(c)=h:B\to A$. Now $b\in B$, the domain of $h$, so $h(b)$ makes sense and is some element of $A$, say $a$. Finally, we can get rid of the abbreviation $h$: $h$ is just handier name for the function $f(c)$, so $h(b)=a$ is just an abbreviation for $\big(f(c)\big)(b)=a$.

To sum up, for each $f:C\to{^B}A$ we define a function $\Phi(f):C\times B\to A$ by letting $\Phi(f)$ take the ordered pair $\langle c,b\rangle\in C\times B$ to the same element of $A$ to which the function $f(c):B\to A$ takes $b$.


问:
Find the 2nd-degree polynomial that approximates with the method of the least squares the:$f(x)=\frac{1}{10}x^2-2x+10$
It is known that a rectangular set of polynomials $\phi_k(x), k=0,1,\cdots,n$ for each $x\in[a,b]$ as to a weight function $w(x)$ can be constructed with the use of the following recursive type (Gram-Schmidt process):

$\phi_0(x)=1$

$\phi_1(x)=x- a_0$, where $a_0=\frac{(x\phi_0, \phi_0)}{(\phi_0,\phi_0)}$

and

$\phi_{k+1}(x)=(x-a_k)\phi_k(x)-\beta_k\phi_{k-1}(x),
 k=1,2,\cdots,n-1$,

where $a_k=\frac{(x\phi_k, \phi_k)}{(\phi_k,\phi_k)}$, and $b_k=\frac{(x\phi_k, \phi_{k-1})}{(\phi_{k-1},\phi_{k-1})}$.

Using the Gram-Schmidt process above to find: the second-degree polynomial $p_2(x)$ that approximates with the method of the least squares the
$f(x)=\frac{1}{10}x^2-2x+10$

in points $(x_i,f_i), i=1,2,\cdots,6$

with $x_i=10+\frac{i-1}{5}$, $f_i=f(x_i)$ and $w(x)=1.$

答:
First make a table of your data. $$\begin {array}{r|r|r} i&x_i&f(x_i)\\ \hline 1&10&0\\2&10.2&0.004\\3&10.4&0.016\\4&10.6&0.036\\5&10.8&0.064\\6&11&0.1 \end {array}$$
Then, as the instructions say, $\phi_0=1$ and $\phi_1=x-\frac {(x\phi_0,\phi_0)}{\phi_0,\phi_0}=x-\frac {(x,1)}{(1,1)}$  The terms in parentheses are sums over the products at the points.  So $(x,1)=10*1+10.2*1+\ldots 11*1=63, (1,1)=6$, which gives $\phi_1(x)=x-10.5$.   Now you need to find $\phi_2$, which will be a quadratic.

Then you can express $f(x)=a_0\phi_0+a_1\phi_1+a_2\phi_2$  You can either find the $a$'s by expanding out the right side and equating like coefficients of $x$, or use the orthogonality to write $a_i=\frac {(f(x_i),\phi_i)}{(\phi_i,\phi_i)}$


问:
How to prove that the intersection of all inductive sets is equal to the intersection of all inductive subsets of an arbitrary inductive set?
A set $y$ is defined as inductive if and only if $\emptyset \in y$ and $x^{+} \in y$ for all $x \in y$, where $x^{+} = x \cup \left\{x\right\}$. The set of all natural numbers $\mathbb{N}$ is then defined as the intersection of all inductive sets. It is also pointed out that $\mathbb{N}$ can be defined as the intersection of all inductive subsets of an arbitrary inductive set (the existence of one inductive set is assumed). I am wondering how to prove that these two definitions are equivalent.

P.S. I have no problem admitting that "the intersection of all inductive subsets of an arbitrary inductive set" is a subset of "the intersection of all inductive sets". However, how should I prove the inverse direction?

答:
Fix some inductive set $I$.

One direction is immediate: intersections of bigger families are smaller (if $A\subseteq B$ then $\bigcap B\subseteq\bigcap A$), so the intersection of all inductive sets is a subset of the intersection of all inductive subsets of $I$.

This is the opposite of the direction you've said is already understood, but I suspect that was a typo.

For the converse inclusion, suppose $a$ is not in every inductive set; we want to find an inductive subset of $I$ not containing $a$. So suppose $J$ is an inductive set with $a\not\in J$. There's a useful lemma which lets us build new inductive sets from old ones: if $X,Y$ are inductive, then so is $X\cap Y$. This means that $I\cap J$ is an inductive set. Now, is $a\in I\cap J$?


问:
Method of characteristics and $\partial_x F(x,\phi(x)) = \partial_y F(x,\phi(x)) = 0$.
Let us assume that we have the following linear PDE.

$$ a(x,y) \partial_x F(x,y) + b(x,y) \partial_y F(x,y) = 0 $$

One way to solve it is to assume that $y = \phi(x)$ is some curve, and consider all points in the plane which satisfy:

$$ F(x,\phi(x)) = c $$

In that case, we have the following relation on partial derivatives.

$$ \partial_x F(x,\phi(x)) + \phi'(x) \partial_y F(x,\phi(x)) = 0 $$

One can now solve for $\partial_x F$ from here and put it into the linear PDE on such curve, which is given by the following.

$$ a(x,\phi(x)) \partial_x F(x, \phi(x)) + b(x,\phi(x)) \partial_y F(x,\phi(x)) = 0 $$

Inserting expression for $\partial_x F$ one gets the following.

$$ \partial_y F(x,\phi(x)) \left( - a(x,\phi(x)) \phi'(x) + b(x,\phi(x)) \right) = 0 $$

The usual solution is given by solving single ODE in parentheses. However, another solution is possible which is the following.

$$ \partial_x F(x,\phi(x))=\partial_y F(x,\phi(x)) = 0$$

This satisfies both equation, so my question is why such solution is never considered? In some sense, I guess it has to be some "trivial" and "not interesting" solution, but I have trouble seeing in which way is that the case. For example, even if $\partial_x F(x,\phi(x)) = 0$, it doesn't mean (I believe) that $\partial_x F(x,y) = 0$ for all $y$, so we cannot conclude that $F(x,y) = g(y)$ for some $g$.

答: The goal of using characteristics is to solve for $\phi$ and then to use that information to then solve for $F$. Therefore when solving for $\phi$ you should not need any information on the specifics of $F$. However were you to try and use $ \partial_y F(x, \phi(x)) = 0 $ to get $\phi$ you would need to already know what $F$ is, which defeats the purpose of introducing $\phi$ in the first place.


答:
It’s actually the other inclusion that is automatic.

Let $S$ be an inductive set, let $\mathscr{S}$ be the set of all inductive subsets of $S$, and let $\mathscr{I}$ be the class of all inductive sets. Then $\mathscr{S}\subseteq\mathscr{I}$, so $\bigcap\mathscr{I}\subseteq\bigcap\mathscr{S}$: when you intersect more sets, you get a smaller intersection.

To show that $\bigcap\mathscr{S}\subseteq\bigcap\mathscr{I}$, we prove the contrapositive. Suppose that $x\notin\bigcap\mathscr{I}$. Then there is an inductive set $J$ such that $x\notin J$. $S\cap J$ is inductive (why?), so $x\notin S\cap J\in\mathscr{S}$, and therefore $x\notin\bigcap\mathscr{S}$.

Title: How to prove that the intersection of all inductive sets is equal to the intersection of all inductive subsets of an arbitrary inductive set?

Body:
A set $y$ is defined as inductive if and only if $\emptyset \in y$ and $x^{+} \in y$ for all $x \in y$, where $x^{+} = x \cup \left\{x\right\}$. The set of all natural numbers $\mathbb{N}$ is then defined as the intersection of all inductive sets. It is also pointed out that $\mathbb{N}$ can be defined as the intersection of all inductive subsets of an arbitrary inductive set (the existence of one inductive set is assumed). I am wondering how to prove that these two definitions are equivalent.

P.S. I have no problem admitting that "the intersection of all inductive subsets of an arbitrary inductive set" is a subset of "the intersection of all inductive sets". However, how should I prove the inverse direction?


问:
Height of minimal primes
Let $R$ be a Noetherian ring and an integral domain. Let $\mathfrak p \subset \mathfrak q$ be prime ideals with $\mathrm{ht}(\mathfrak q) = \mathrm{ht}(\mathfrak p) + n$. I am trying to show that there exist $x_1,\dots,x_n \in R$ such that $\mathfrak q$ is a minimal prime of $\mathfrak p + (x_1,\dots,x_n)$.

I know that $\mathrm{ht}(x_1,\dots,x_n) \leq n$ and by Krull's theorem, any minimal prime of $(x_1,\dots,x_n)$ must have height $\leq n$. The problem is I don't know much about the ideal $\mathfrak p + (x_1,\dots,x_n)$, it would be helpful I think to know its height but I don't believe there is a general result on the height of the sum of two ideals. Any hints on approaching this?

答:
Consider the quotient ring $R/P$. $Q/P$ is a prime in $R/P$ of height $n$, hence there exists $\bar x_1, \dots, \bar x_n \in R/P$ such that $Q/P$ is a minimal prime over the ideal $(\bar x_1, \dots, \bar x_n)$. This is equivalent to saying that $Q$ is a minimal prime over $P+(x_1, \dots, x_n)$.

Let me know if this helps!

Title: Height of minimal primes


问:
Proof verification: An arrow which is monic under a faithful functor is itself monic
Context:

To introduce some symbols and such, what I'm seeking to prove is this:

Let $F$ be a faithful functor. Suppose $F(f)$ is a monic arrow. Show $f$ is monic.

This came up as part of a class assignment recently, tied to an introduction to category theory as we began to introduce free objects. I have an attempt ready, but I have a slight feeling that something's wrong with it. I can't really pin it down on what, it's just a lurking feeling; alternatively, I could be overthinking.

Any thoughts?

My Attempt at a Proof:

Let $F : \scr{C} \to \scr{D}$ be a faithful functor. Let us also have arrows $g,h : A \to B$ and $f : B \to C$, where $A,B,C,f,g,h \in \scr C$. Further, let $F(f)$ be monic and let $f \circ g = f \circ h$.

Since $F$ is faithful, it is injective on the hom-sets of $\scr C$. Thus,

$$F(f \circ g) = F(f \circ h) \tag 1$$

Since $F$ is a functor, it preserves composition: thus,

$$F(f \circ g) = F(f) \circ F(g) = F(f) \circ F(h) = F(f \circ h) \tag 2$$

Since $F(f)$ is monic, $(2)$ yields

$$F(f) \circ F(g) = F(f) \circ F(h) \implies F(g) = F(h) \tag 3$$

Again since $F$ is faithful, $(3)$ yields

$$F(g) = F(h) \implies g = h \tag 4$$

Thus, since our assumptions $F$ is faithful, $F(f)$ is monic, and $f \circ g = f \circ h$ imply $g = h$ we can conclude $f$ is monic.

答:
As noted in the comments, it seems (at this time) the consensus is that this is indeed correct as an approach (though these days I would probably phrase it a little differently).

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Generalisations of the identity $\tan{\frac{3\pi}{11}}+4\sin{\frac{2\pi}{11}}=\sqrt{11}$
I recently came across this curious trigonometric sum:

$$\tan{\frac{3\pi}{11}}+4\sin{\frac{2\pi}{11}}=\sqrt{11}$$

which has a neat proof here: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$

For what values of $k$ does the following general identity have integer solutions for $a,b,x,y$?

$$a\tan{\frac{x\pi}{k}}+b\sin{\frac{y\pi}{k}}=\sqrt{k}$$

答:
Here are a few thoughts :

Let  $\zeta_n = e^{2i\pi/n}$ and $K_n = \Bbb Q(\zeta_n)$ the cyclotomic extension.

It is well-known that this extension is Galois and its Galois group is $\{\sigma_k :
\zeta_n \mapsto \zeta_n^k, k \in (\Bbb Z/n\Bbb Z)^*\}$.

$K_n$ contains $\cos(2\pi/n), i\sin(2\pi/n)$, and $i\tan(\pi/n)$.

Hence, $i(a\tan(x\pi/n)+b\sin(y\pi/n)) \in K_n \cap i \Bbb R$.

So first, it would be good to know when $\sqrt{-n} \in K_n$.

If $n$ is even, then $K_n$ never contains $\sqrt {\pm n}$ (you would need $K_{4n}$ for that).

If $n \equiv 1 \pmod 4$, then $K_n$ contains $\sqrt n$ and not $i$, so it doesn't contain $\sqrt{-n}$, and we can also stop.

If $n \equiv 3 \pmod 4$, then $K_n$ does contain $\sqrt {-n}$. Let $\chi_n : (\Bbb Z/n\Bbb Z)^* \to \{\pm 1\}$ be the character satisfying $\sigma_k(\sqrt{-n}) = \chi_n(k) \sqrt{-n}$, and $H_n = \ker \chi_n$.

We have $\Bbb Q(\sqrt{-n}) = K_n^{H_n}$, and the numbers of the form $a\sqrt{-n} \in K_n^{H_n}$ are the numbers $x \in K$ satisfying $\sigma_k(x) = \chi_n(k)x$. Since $\sigma_k(i\sin(x\pi/n)) = i\sin(kx\pi/n)$ and similarly with $\tan$, we want to find integers $a,b,x,y$ such that forall $k$ prime with $n$, $a\tan(kx\pi/n)+b\sin(2ky\pi/n) = \chi_n(k)(a\tan(x\pi/n)+b\sin(2y\pi/n))$.

It is enough to find one with $x=1$ and to include only a set of generators of $(\Bbb Z/n\Bbb Z)^*$ for $k$. After a bit of rewriting, we have to find an integer $y$ such that forall $k$ in a generating set,
$\frac{\chi_n(k)\tan(\pi/n)-\tan(k\pi/n)}{\chi_n(k)\sin(2y\pi/n)-\sin(2ky\pi/n)} = -b/a \in \Bbb Q$

Forgetting that this rational has to be independant from $k$, checking that this is a rational is the same as checking that it is invariant by the $\sigma_k$, which means we should check that forall $k,k'$ in a generating set,

$(\chi_n(k)\tan(\pi/n) - \tan(k\pi/n))(\chi_n(k)\sin(2k'y\pi/n)-\sin(2kk'y\pi/n)) -
(\chi_n(k)\tan(k'\pi/n) - \tan(k'k\pi/n))(\chi_n(k)\sin(2y\pi/n)-\sin(2ky\pi/n)) = 0$

For example with $n=11$, we can pick $k=k'=2$, we have $\chi_{11}(2) = -1$, and indeed $y=4$ or $y=7$ works.

For $n=7$, $3$ is a primitive root, and $y= 1$ works, which gives $-b/a = 4$, and finally we get $4\sin(2\pi/7) - \tan(\pi/7) = \sqrt 7$.

For $n=19$, $2$ is again a primitive root, and computing this expression for $y= 1 \ldots 18$, we don't find any zero, which shows that there is no such formula (neither for $n$ prime less than $700$, and probably never again)


问:
Monty Hall Problem with Multiple Players?
I understand the common Monty Hall Problem and why switching provides a 2/3 chance of winning, but I'm having trouble wrapping my head around how the probabilities work when multiple players are involved, as their probabilities seem to be contradictory.

Let's say we have two players and four doors.
Each door has a 1/4 probability of hiding a prize.
Let's say contestant 1 chooses door A and contestant 2 chooses door B.
Door D is then revealed to be empty, leaving A, B, and C.
From contestant 1's perspective the odds are 1/4 for door A, 3/8 for door B, and 3/8 for door C.
But from contestant 2's perspective the odds are 3/8 for door A, 1/4 for door B, and 3/8 for door C.

What are the actual odds for each door hiding the prize, does it change depending on if they know each other's choices, and does it differ from the normal example of the problem? If so why, and if not, why not?

答:
In the case where both contestants can see everything, it's more or less the same as a single contestant getting to select two doors. The probabilities under both doors effectively freeze, and the the $1/4$ for door D flows to door C, giving C a $1/2$ chance of having the prize, leaving A and B at their original $1/4$.

In the case where contestants can't see what door the other contestant picks, then you have the situation where their door has $1/4$ chance, and the other two remaining doors both have $3/8$. It's fine that both contestants have different probabilities for the doors; it results from both the contestants having different information about the probability distribution, since only they know which door they selected.


问:
Gaussian Integration, Trouble with Error Function
I'm tasked with evaluating this integral for class:

$\int_{-\infty}^\infty e^{-(x-\mu)^2 / (2\sigma^2)} \, dx$, assume $\mu$ and $\sigma^2$ are real scalars and $\sigma^2 > 0$.

At one point in my solution, I come up against the error function. I've gone over the math a few times and can't seem to figure out what happens. It'll be obvious where my question is in the work below.
Let the Gaussian integral equal $I$ such that

Firstly, take $a = 1/(2\sigma^2)$, $u = x-\mu$, $ du = dx$. Notice that when $x=0$, $u=-\mu$. Now, the integral we must solve is the following:
\begin{equation}
    \int_{0}^\infty
        e^{-(x-\mu)^2 / (2\sigma^2)}
    \, dx
        =
            \int_{-\mu}^\infty
                e^{-a u^2}
            \, du
    \nonumber
\end{equation}
To further simplify this expression, take $t = \sqrt{a} \, u$. Note that when $u = \mu$, $t = -\sqrt{a} \, \mu$. Then,
\begin{align}
    \int_{-\mu}^\infty
        e^{-a u^2}
    \, du
        &=
            \dfrac{1}{\sqrt{a}} \int_{-\mu \sqrt{a}}^\infty
                e^{-t^2}
            \, dt
        \nonumber
        \\[.5em]
        &=
            \dfrac{1}{\sqrt{a}} \int_{-\mu \sqrt{a}}^0
                e^{-t^2}
            \, dt
            +
            \dfrac{1}{\sqrt{a}} \int_{0}^\infty
                e^{-t^2}
            \, dt
        \nonumber
        \\[.5em]
        &=
            -\dfrac{1}{\sqrt{a}} \int_0^{-\mu \sqrt{a}}
                e^{-t^2}
            \, dt
            +
            \dfrac{1}{\sqrt{a}} \int_{0}^\infty
                e^{-t^2}
            \, dt.
        \label{eq:hw1_prob1.2}
\end{align}
We should pause here and take a look at each expression in the sum separately. First, we should note the expression of the Error Function,  \begin{equation}
    Erf(z)
        =
            \dfrac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} \, dt.
    \nonumber
\end{equation}
Applying this to the first expression in our sum, we have

BREAK

At this point in my work, I assume the following expression is true. I know the final answer is correct, but unless $-Erf(-z) = Erf(z)$ (I'm not sure this is true), I'm not sure why this works.

\begin{equation}
    -\dfrac{1}{\sqrt{a}} \int_0^{-\mu \sqrt{a}}
            e^{-t^2}
        \, dt
        =
            \dfrac{1}{\sqrt{a} \,}Erf{\left(\dfrac{\mu}{\sigma \sqrt{2}}\right)}
    \nonumber
\end{equation}

Now, let's turn our focus to the second sum. Using the 'polar' method of solving this expression, we have
\begin{align}
    I
        &=
            \int_0^\infty e^{-t^2} \, dt
        =
            \dfrac{1}{2} \int_{-\infty}^\infty e^{-t^2} \, dt
    \nonumber
    \\[.5em]
    I^2
        &=
            \dfrac{1}{4}  \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)} \, dx dy.
    \nonumber
\end{align}
Take $x = r\cos{\theta}$, $y = r \sin \theta$, $
dx dy = r \, dr d\theta$ such that
\begin{equation}
    \dfrac{1}{4} \int_0^{2\pi} \int_{-\infty}^\infty
        r e^{-r^2} \, dr d\theta.
    \nonumber
\end{equation}
Now, let $u = -r^2$ so that $du = -2 r dr$. So,
\begin{equation}
    I^2
        =
            -\dfrac{1}{8} 2\pi \int_0^\infty e^u \, du
        =
            \dfrac{\pi}{4},
    \nonumber
\end{equation}
and
\begin{equation}
    I
        =
        \dfrac{\sqrt{\pi}}{2}.
    \nonumber
\end{equation}

Putting this all together and skipping over some tedious algebra, we have
\begin{equation}
\boxed{
    \int_{0}^\infty e^{-(x-\mu)^2 / (2\sigma^2)} \, dx
        =
            \sigma \sqrt{\dfrac{\pi}{2}} \left[ 1 + Erf\left(\dfrac{\mu}{\sigma\sqrt{2}}\right) \right]
}
\nonumber
\end{equation}

Any help with this is appreciated. I'm sure it's a simple mistake that I am overlooking.

答:
Your approach is unnecessarily complicated.   $\int_Re^{-(x-\mu)^2/(2\sigma^2)}dx=\int_Re^{(-y^2)/(2\sigma^2)}dy=\sigma \int_Re^{-u^2/2}du=\sigma\sqrt{2\pi}$

The error function is needed when you integrate over part of the real line.


问:
How to understand orbit of size $1$ in this case
I am a self-studying beginner in group theory, so please bear with this question which could have some simple answers. Given a $p$-group $G$ for some prime $p$, let $H$ be a subgroup of $G$. Let $X$ be the set of all conjugates of $H$.

Now, $H$ acts on $X$ by conjugation. I read that there are at least $p$ orbits of size $1$ in $X$.

One example of an orbit with size $1$ is $\{H\} \in X$. This example follows since $aHa^{-1}=H$ for any $a \in H$ since $H$ is a subgroup, and we have $\text{Orb}(H)=H$.

But I read that since $p$ is prime, that there are at least $p-1$ other orbits of size $1$. So there should be another orbit $gHg^{-1} \neq H$ of size $1$ in $X$.

What I don't understand is how $gHg^{-1}$ could be of size $1$ under the action of $H$. Shouldn't this mean that $\text{Orb}(gHg^{-1})=\{agHg^{-1}a^{-1} | a \in H\}$ and $\text{Orb}(gHg^{-1})$ may not necessarily be equal to $gHg^{-1}$. However, it should have size $1$, which means that $\text{Orb}(gHg^{-1})$ should in fact be equal to $gHg^{-1}$.

For reference, this result came from Rotman's Theorem 4.6, where no extra conditions were imposed on $H$ and $G$ except that $H$ is a subgroup of the $p$-group $G$ ... What am I missing here?

答:
The first thing to note is that if $|X| = 1$ then we will not have $p-1$ other orbits so we will also need to assume $|X| \gt 1$.

We will use these two properties of orbits to prove our statement:

Orbits are disjoint and their union is the entire set $X$ (this should be easy to see).

The orbit size divides group order (this is proven in the Orbit-stabilizer theorem)

By property (1) we have that $$|X| = \sum_{Y \in \mathcal{O}} |Y|$$ where $\mathcal{O}$ is the set containing all the orbits of the action. Now we split $\mathcal{O}$ into two disjoint subsets: $\mathcal{O'}$ and $\mathcal{O''}$ where $\mathcal{O'}$ is the set of all orbits of size $1$ and $\mathcal{O''}$ is the set of all orbits of size greater than $1$. This means $$|X| = \sum_{Y' \in \mathcal{O'}} |Y'| + \sum_{Y'' \in \mathcal{O'}} |Y''| = |\mathcal{O'}| + \sum_{Y'' \in \mathcal{O'}} |Y''|$$ since $|Y'| = 1$. By property (2) we know that $|Y''|$ divides $|X| = p^n$ and $|Y''| > 1$ which means that $|Y''| = p^k$ where $k > 1$ which means $p$ divides $|Y''|$. We can view $X$ as an orbit where the group action is conjugation by the group $G$. This means that $|X|$ divides $|G| = p^n$. Since $|X| > 1$ we have that $p$ divides $|X|$. Since $|X| =  |\mathcal{O'}| + \sum_{Y'' \in \mathcal{O'}} |Y''|$, $p$ also has to divide $|\mathcal{O'}|$ which means $|\mathcal{O'}| = pm$ for some $m \gt 1$ which means $|\mathcal{O'}| \geq p$ which is what we were trying to prove.