问:
Less common probabilities and expected values related to the coupon collector problem

A company brings out a new set of $n$ collectible cards. The cards are
made in equal number, so there is the same probability of any given
card being in a random set. They are purchased one-by-one.

Question 1: If I buy $x$ cards, on average how many unique cards $(k)$
will I have? $k$ does not need to be a whole number.

Question 2: How large does x need to be that k/n is over y%?

Question 3: If I buy $x$ cards, what is the probability that I have $j$
unique cards. $j$ is a whole number.

EDIT: Thank you to all who have replied! As a first time poster, it is amazing to get so many helpful responses so quickly. I have learnt that this known as the coupon collector's problem. There is a very straightforward answer to the question, "What is the expected value of x needed to complete the set on n cards"

\begin{align}
E[x]={}&x\sum_{i=1}^n\frac1i\;.
\end{align}

I am struggling to find equations as simple as this to questions 1 and 2, so will post a second edit once I have them. On question 3 "user 4" kindly gave what I think is the right answer:
\begin{align}
\frac{n!}{(n-j)! n^x}{x \brace j}
\end{align}

答:
Q1:

Let us ask first what is the expected value of the binary event $X_i$ which value is $1$ if a certain card was bought at least once and 0 if it was not bought. Clearly the expected value is just the probability to buy the card in at least one of $x$ trials:
$$
\mathsf E(X_i)=1-\left(\frac{n-1}n\right)^x.\tag1
$$
Now due to linearity of the expectation the expected number $k$ of unique cards bought at least once is:
$$
k=\sum_{i=1}^n\mathsf E(X_i)=n\left[1-\left(\frac{n-1}n\right)^x\right].\tag2
$$

Q2:

As I already mentioned in a comment the question is ambiguous. If you mean the ratio $\frac kn$ it is given by (1).

Q3:

The probability can be computed as follows:
$$
p_j(n,x)=\frac1{n^x}\binom nj j!{x \brace j}=\frac{n!}{(n-j)! n^x}{x \brace j},\tag3
$$
where ${x \brace j}$ is the Stirling number of the second kind

In the expression (3)
$n^x$ counts the number of ways to dictribute $x$ labeled cards into $n $ labeled boxes with empty boxes being allowed, $\binom nj$ counts the number of ways to choose $j$ out of $n $ labeled boxes, and $j!{x \brace j}$ counts the number of ways to dictribute $x$ labeled cards into $j $ labeled boxes with empty boxes being not allowed. The cards are labeled by the order of their arrival. The boxes are labeled by the types of unique cards.

Observe that we factually proved the relation:
$$
\sum_{j=1}^n jp_j(n,x)=n\left[1-\left(\frac{n-1}n\right)^x\right].
$$


问:
parametrization of path in contour integral and substitution
I just started studying complex analysis, and we just covered the basics of complex differentiable functions and the technical aspects of performing a contour integral.
I thought the theory made really good sense. But suddenly while trying to perform a contour integral, I started questioning everything I learned and what we are doing. So I hope some of you can help me clarify what I'm allowed to do. I think it's best to go by an example:

Say I have a complex function:
$$
\begin{array}{l}
f: \mathbb{C} \rightarrow \mathbb{C} \\
f(z)=\frac{1}{(1-z)^{2}}
\end{array}
$$
And say I want to perform a integral along a straight line from $z_1$ to $z_2$:

Then I can parameterize the path by:
$$
\begin{array}{l}
\gamma:[0,1] \rightarrow \mathbb{C} \\
\gamma(t)=z_{1}+t\left(z_{2}-z_{1}\right)
\end{array}
$$

And calculate the integral by:
$$
\int_{z_{1}}^{z_{2}} f(z) d z=\int_{0}^{1} f(\gamma(t)) \gamma^{\prime}(t) d t=\int_{0}^{1} \frac{1}{(1-\gamma(t))^{2}} \gamma^{\prime}(t) d t
$$
First question: Can I just differentiate $\gamma(t)$ like a regular real function (so $\gamma^{\prime}(t)=z_2-z_1$) even though it maps to $\mathbb{C}$? Isn't t in some way indirectly dependent on a complex variable and thus a complex function?

Say, that I don't know the primitive/antiderivative to the function $f(\gamma(t)) \gamma^{\prime}(t)=\frac{1}{(1-\gamma(t))^{2}} \gamma^{\prime}(t)$.

Second: Am I then allowed to make a substitution?

Fx:$$u=1-\gamma(t), \frac{d u}{d t}=-\gamma^{\prime}(t) \Rightarrow d t=-\frac{1}{\gamma^{\prime}(t)} d u$$

$$=\int_{u(0)}^{u(1)}-\frac{1}{u^{2}} d u=\left[\frac{1}{u}\right]_{u(0)}^{u(1)}=\frac{1}{u(1)}-\frac{1}{u(0)}$$

But this would be the same as a substitution $u=1-z$ from the beginning. Is the problem that I'm not allowed to have complex numbers in the integration limits? Or why do I even need to parameterize a line to start with?

答:
In this case no parametrization is required, because the function $z\mapsto\frac1{(1-z)^2}$ has a primitive: $z\mapsto\frac1{1-z}$. So, for any differentibale path $\gamma\colon[a,b]\longrightarrow\Bbb C\setminus\{1\}$, you have$$\int_\gamma\frac1{(1-z)^2}\,\mathrm dz=\frac1{1-u(b)}-\frac1{1-u(a)}.$$In general, if $f$ has a primitive $F$, then\begin{align}\int_\gamma f&=\int_a^bf\bigl(\gamma(t)\bigr)\gamma'(t)\,\mathrm dt\\&=\int_a^bF'\bigl(\gamma(t)\bigr)\gamma'(t)\,\mathrm dt\\&=\int_a^b(f\circ\gamma)'(t)\,\mathrm dt\\&=F\bigl(\gamma(b)\bigr)-F\bigl(\gamma(a)\bigr).\end{align}

And, yes, you can differentiate your $\gamma$ the way you did it.

ContentLicense: CC BY-SA 4.0


问:
How to evaluate this integral in terms of a beta function?
I need some help evaluating an integral using beta function.

\begin{equation*}
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4 \phi d\phi \cdot \frac{2\pi}{5}
\end{equation*}

I am not sure how to transform lower bound and cosine to fit the equation for beta function. I have to solve it using beta function because that's what the exercise says.

答:
Hint.

For any even (integrable) function $f$,
$$
\int_{-a}^af(x)dx=2\int_0^{a}f(x)dx
$$

Also, one of the various ways to write the Beta function is:

$$
{\displaystyle {\begin{aligned}\mathrm {B} (x,y)&=2\int _{0}^{\pi /2}(\sin \theta )^{2x-1}(\cos \theta )^{2y-1}\,d\theta \\[6pt]&
\end{aligned}}}
$$


问:
Standard Deviation in Normal Distribution (with only mean and unknown division)
My question comes when I revise the topic.
If the mean and simple information is given, how can I find the standard deviation?
i.e. For normally distributed data, let's say mean = 65, and only 5 % or the samples are >=90. How can I find the standard deviation within so lack of information. Is it possible? Thanks to anyone who can suggest and share your idea.</p>

答:
Hint:

$\Phi^{-1}(0.95)\approx 1.644854$ for a standard normal distribution

so $5\%$ of a normal distribution is more than about $1.644854$ standard deviations above the mean

and you know the mean


问:
Linear Algebra Question: get amount each person tipped from same amount of central tip jars
Essentially, there are six guests and six tip jars (1 between each pair of guests). If we know the amount in each tip jar, and know each guest gives half their money as a tip, can we figure out the starting balances of each guest? Here is the full question text:

```<code>Suppose six guests (represented by green circles) sit around a hexagonal table and there are six jars of
tips (represented by black dots). If we know the amount of tip in each jar, J1 to J6, can we determine
each individual’s tip amount, G1 to G6? If yes, explain why by examining the relationship between the
jar values, J1 to J6, and guest tips, G1 to G6. If not, give two different assignments of G1 to G6 that
will result in the same J1 to J6.
</code>```

I think you want to set this up as:
\begin{pmatrix}0.5&0.5&0&0&0&0&a\\ 0.5&0&0.5&0&0&0&b\\ 0&0.5&0&0.5&0&0&c\\ 0&0&0.5&0&0.5&0&d\\ 0&0&0&0.5&0&0.5&e\\ 0&0&0&0&0.5&0.5&f\end{pmatrix}

Where a,b, etc. are the guest balances. But I'm unsure where to progress from here.

I'm also interested how this changes if the number of guests is 5.

Any help is appreciated
Sincerely,
D

答:
Since the unknowns are the guest balances you should not write them in the last column (the "solution column" of the linear system of equations) since the unknowns of a system of the kind $Ax=b$ are the components of $x$. So in this cas $b= (J1, ..., J6)$ and $x=(G1,..., G6)$ and the matrix you write down to solve the system of equations is $(A |b)$. So the values of the last column should be the jar balances J1 to J6 and each column represents one of the guests and how this guest splits up his/her tip.
The resulting equations are (if J1 ist between Guest 1 and Guest, J2 is between Guest 2 and Guest 3 and so on...):
$$J1 = 0,5 \cdot G1 + 0,5 \cdot G2$$
$$J2 = 0,5 \cdot G2 + 0,5 \cdot G3$$
Putting it all together in a matrix where G1 to G6 are the unknowns we get:
$$\begin{pmatrix} 0,5&0,5&0&0&0&0&J1\\
0&0,5&0,5&0&0&0&J2\\
0&0&0,5&0,5&0&0&J3\\
0&0&0&0,5&0,5&0&J4\\
0&0&0&0&0,5&0,5&J5\\
0,5&0&0&0&0&0,5&J6
\end{pmatrix}$$
The matrix looks like this becaus Guest 1 puts his tip in the Jars 1 and 6

For any other number $n$ of guests we will get a similar looking matrix $A \in M_{n \times (n+1)}$ where the last column is the "solution column" with the jar balances.

Hope this explanation helps you.


问:
Is a function differentiable at the end points of its domain?
Is the function f(x) = x(sqrt(x) + sqrt(x + 9)) differentiable at the point x = 0, which lies at the end point of its domain?
I found differing answers from different sources. Some say it is not even continuous at x=0 while some go on to say that it is continuously differentiable at x=0. What is the right answer?</p>

答: Yes, it is differentiable at $0$. Note that $f(0)=0$ and that\begin{align}f'(0)&=\lim_{x\to0}\frac{f(x)-f(0)}x\\&=\lim_{x\to0}\sqrt x+\sqrt{x+9}\\&=\sqrt0+\sqrt9\\&=3.\end{align}


问:
Norm of the operator $T(x_1,x_2,\cdots , x_n,\cdots )=(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots).$
Let $T$ be a linear operator from  $l_1$ space  to itself defined as  $$T(x_1,x_2,\cdots , x_n,\cdots )=(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots).$$ Then which of the following statements are true ?

$1.$ $\|T\|=1$.

$2.$ $\|T\|>2$.

$3.$ $1<\|T\|\leq 2.$

$4.$ none .

I tired it as $$\|T(x)\|=\|T(x_1,x_2,\cdots , x_n,\cdots )\|=\|(x_2-x_1, x_3-x_2,\cdots, x_{n+1}-x_n, \cdots)\|=\sum |x_{i+1}-x_i|\leq 2\|x\|.$$ So $\|T\|\leq 2$. Now how to process further ? If I choose $x=(1,0,0,\cdots)$ then $$\|T(x)\| =\|x\|$$ Now I am confused. Thank you .

Body:
Well consider $x=(1,-1,0,0,0,\dots)$. Then $Tx=(-2,1,0,0,\dots)$ so $\|Tx\|=3$ while $\|x\|=2$, thus $\|T\|\geq\frac{3}{2}$. Continuing in this fashion, take $x=(1,-1,1,-1,0,0,\dots)$. Then $Tx=(-2,2,-2,1,0,0,0,\dots)$ so $\|Tx\|=7$ while $\|x\|=4$. In general, if $x=(1,-1,1,-1,\dots,1,-1,1,-1,0,0,0,\dots)$ where we have $n$ pairs of $(1,-1)$ appearing, then $Tx=(-2,2,-2,2,\dots,2,-2,1,0,0,\dots)$ so $\|Tx\|=2(2n-1)+1$ and $\|x\|=2n$, so
$$\|T\|=\sup_{y\neq0}\frac{\|Ty\|}{\|y\|}\geq\frac{2(2n-1)+1}{2n}\to2.$$

So $\|T\|=2$.


问:
Real-valued part of complex inner product
$\DeclareMathOperator{\Re}{Re}$Let $\left<f, g\right>$ be the complex inner product defined by
$$\left<f, g\right>=\Re\left(\int_{-\infty}^{\infty}f^{*}(x)g(x)\,dx\right) $$
Could I say that this is symmetric, i.e. does $\left<f, g\right> = \left<g, f\right>$?

I know that if $f, g$ are of the form $f = a+bi$ and $g = c+di$, taking the real part of $f^{*}g$ gives
$$\Re((a-bi)(c+di)=ac+bd$$

Similarly, taking the real part of $g^{*}f$ gives
$$\Re((c-di)(a+bi))=ac+bd$$ so I know that this is symmetric.

As a function $f(x)$ can be written in the form $f(x)=a+bi$, and similarly $g(x)$ can be written in the form $g(x)=c+di$, does this mean that $$\left<f, g\right>=\Re\left(\int_{-\infty}^{\infty}f^{*}(x)g(x)\,_dx\right) = \Re\left(\int_{-\infty}^{\infty}f(x)g^{*}(x)\,dx\right) = \left<g, f\right>$$

Finally, if this is the case, and we assume that $f$ and all of its derivatives vanish at $\pm\infty$, could we say that

$$\left<f, \frac{\partial f}{\partial x}\right>=\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)=\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\, dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)$$

Integrating the first (or the second) by parts gives
$$\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)=-\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}\frac{\partial f^{*}}{\partial x}f(x)\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\, dx\right)$$

As this is symmetric,
$$-\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}\frac{\partial f^{*}}{\partial x}f(x)\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\,dx\right)\\=-\frac{1}{2}\Re\left(\int_{-\infty}^{\infty}\frac{\partial f}{\partial x}f^{*}(x)\,dx\right) + \frac{1}{2}\Re\left(\int_{-\infty}^{\infty}f^{*}(x)\frac{\partial f}{\partial x}\, dx\right)=0$$

I feel like this doesn't work, but I can't for the life of me see why. I feel like the real-valued part of the complex inner product is symmetric, but I don't know if we can apply this to the integral? Any help would be greatly appreciated, thank you!

答:
The definition you give does not define an inner product in the usual sense.  An inner product on a complex vector space $V$ must satisfy three condition for all $\alpha, \beta \in \mathbb C$ and all $u, v, w \in V$,

Linearity.
$\langle u +v, w \rangle = \langle u, w \rangle + \langle v, w \rangle$ and either of $\langle \alpha u, v\rangle = \alpha \langle u,v \rangle$ or $\langle \alpha u, v\rangle = \alpha^* \langle u,v \rangle$.

Symmetry.
$\langle u, v \rangle = \langle v, u \rangle^*$.

Positive definite.
$\langle u, u \rangle > 0$ unless $u=0$.

In the example you give,
$$ \langle u, v \rangle = \Re \Bigg ( \int_{\mathbb R} u^*(x)v (x) dx \Bigg).$$
Here, the product is symmetric as you say, so conditions (2) holds;  moreover, condition (3) holds because $v^*\cdot v = (\Re v)^2 + (\Im v) ^2$.  However,
conditions (1) gives problems.  The first part is alright, but, for instance, if $v(x)=a(x) + ib(x)$ is any non-zero member of $V$, $\langle iv, v \rangle = 0$ (because the real part is zero) but $i\langle v, v \rangle \neq 0$ (product of $i$ and a positive real). Condition (1) requires these to be equal, so it fails with this definition.

In the second part of the question, it is not clear why you want to split the integral into two parts.  In any case, you will have,
\begin{align*}
\int_{\mathbb R} f^*(x) \frac{df}{dx}(x) ~dx = -\int_{\mathbb R} \frac{df^*}{dx}(x) f(x) ~  dx = 0
\end{align*}
which is a general property of functions that vanish at the limits $\pm\infty$.  Recall for a real valued function $g$ with this condition,
\begin{align*}
\int_{\mathbb R} g(x)g'(x) dx = \Big [ \frac{1}{2} g(x)^2 \Big]_{-\infty}^{+\infty} = 0.
\end{align*}


问:
Show that $\langle(1234),(12)\rangle = S_4$
I am trying to show that $\langle(1234),(12)\rangle = S_4$. I can multiply and get all $24$ permutations manually but isn't there a more compact solution?</p>

答: I'm not sure if this a more "compact" solution, but this follows from a more general result that the $\langle (123\ldots n),(12) \rangle = S_{n}$, which I think would be cleaner and more instructive to show. Hint: see if you can show that this subgroup generates all transpositions in $S_{n}$, and therefore generates $S_{n}$. </p>

ContentLicense: CC BY-SA 3.0


问:
Diameter of $\gamma(x,y) = \frac{|x-y|}{1+|x+y|}$
Let $\gamma : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be defined as $\gamma(x,y)=\dfrac{|x-y|}{1+|x-y|}$.

Proof that $\gamma(x,y)$ is indeed a metric in $\mathbb{R}$, and show that $Diam_{\gamma}(\mathbb{R})=1$.

Showing that's is a metric is eazy using the auxiliary function $f(t)=\frac{t}{1+t}$, but I don't know where to start to demonstrate the diameter.

Any leads?

答:
Hint

You have $$|x|<1+|x|,$$
for all $x\in\mathbb R$.


问:
Another pigeonhole principle question
Have another question for you today:

A course has seven elective topics, and students must complete exactly three of them in order to pass the course. If 200 students passed the course, show that at least 6 of them must have completed the same electives as each other.

Now I know this is related to counting and the pigeonhole principle, and there are a couple of other related questions already asked but I couldn't apply them to my quuestion.

I know that the (informal) pigeonhole principle states that if you have $n$ boxes, and you have more than $n$ pigeons to distribute between those boxes, then at least one of the boxes will contain more than one pigeons, but I'm not sure what my boxes and what my pigeons are in this problem

答:
Instead of thinking about pigeons to be put into boxes, you could approach this the way Edsger Dijkstra does in http://www.cs.utexas.edu/users/EWD/transcriptions/EWD10xx/EWD1094.html

Specifically, he restates the principle this way

For a nonempty, finite bag of real numbers, the maximum is at least the average (and the minimum is at most the average).

This is a simple generalization of the familiar principle. In your case, let the bags be the number of possible selections of 3 electives out of 7 = $\binom{7}{3}$ =35.

Each bag is going to get a certain number of students. The number of students is always an integer $\geq 0$.

The average is 200/35 = 5.72 students per bag on average. The maximum out of the set of 35 bags must be greater than this average, i.e., at least 6 of the students must have completed the same set of 3 electives.

Title: Another pigeonhole principle question


答: Write $H$ for the subgroup generated by those two permutations. Then $(1234)(12)=(234)$, so $H$ contains certainly the elements of $\langle(1234)\rangle$, $\langle (234)\rangle$ and $(12)$, hence $\vert H\vert \geq 7$ and therefore $\vert H\vert \geq 8$ ($\vert H\vert$ must divide $24=\vert S_{4}\vert$). Since $(1234)\in H$ but such a permutation is not even, i.e it doesn't belong to $A_{4}$, $H$ is not a subgroup of $A_{4}$. By Lagrange's theorem, $\vert H\vert$ must equal $24$ (the only subgroup of order $12$ in $S_{4}$ is $A_{4}$) and we're done.</p>

Title: Show that $\langle(1234),(12)\rangle = S_4$

Body: I am trying to show that $\langle(1234),(12)\rangle = S_4$. I can multiply and get all $24$ permutations manually but isn't there a more compact solution?</p>


答:
$\gamma(n,0) = \frac{n}{n+1} \to 1$ while $\gamma(x,y) < 1$ for all $x,y$.

So $1$ is an upperbound for $\{\gamma(x,y)\mid x,y \in \Bbb R\}$ and the limit implies there can be no smaller one. So the diameter (i.e. the supremum, least upper bound,  of that set) equals $1$..

Title: Diameter of $\gamma(x,y) = \frac{|x-y|}{1+|x+y|}$

Body:
Let $\gamma : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be defined as $\gamma(x,y)=\dfrac{|x-y|}{1+|x-y|}$.

Proof that $\gamma(x,y)$ is indeed a metric in $\mathbb{R}$, and show that $Diam_{\gamma}(\mathbb{R})=1$.

Showing that's is a metric is eazy using the auxiliary function $f(t)=\frac{t}{1+t}$, but I don't know where to start to demonstrate the diameter.

Any leads?


问:
Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$
I accidentally found out that the two integrals below

$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluated explicitly. For example, the first one can be carried out via double integration,  as sketched below.

\begin{align}
I_1&=\int_0^\infty \left(\int_0^1 \frac{x^2}{1+y^2x^4}dy\right)\frac{1}{1+x^2}dx\\
&= \frac\pi2\int_0^1 \left( \sqrt{\frac y2}+ \frac1{\sqrt{2y} }-1\right)\frac{1}{1+y^2}dy=\frac{\pi^2}8
\end{align}

Similarly, the second one yields $I_2=\frac{\pi^2}8$ as well.
The evaluations are a bit involved, though, and it seems an overreach to prove their equality this way, if only the following needs to be shown

$$\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx=0$$
The question, then, is whether there is a shortcut to show that the above integral vanishes.

Body:
By the known formula for the difference of two inverse tangents, your last integral is
$$
\int_0^{ + \infty } {\frac{{\tan ^{ - 1} \left( {\frac{{x^2  - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}\mathrm{d}x}  = \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2  - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}\mathrm{d}x}  + \int_1^{ + \infty } {\frac{{\tan ^{ - 1} \left( {\frac{{x^2  - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}\mathrm{d}x} .
$$
Taking $y=1/x$ in the last integral gives
\begin{align*}
& \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2  - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}\mathrm{d}x}  + \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{y^{1/2}  - y^2 }}{{1 + y^{5/2} }}} \right)}}{{1 + y^2 }}\mathrm{d}y} \\ & = \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2  - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}\mathrm{d}x}  - \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{y^2  - y^{1/2} }}{{1 + y^{5/2} }}} \right)}}{{1 + y^2 }}\mathrm{d}y}  = 0.
\end{align*}


问:
"Cuboid" not the correct name for 3d rectangle?
I have always thought the best name of the 3d equivalent of a rectangle was "cuboid". I am talking about the 3d shape with 6 rectangular faces shown below.

However, when looking up the name of this shape, I get confusing results. Many sources including most dictionaries and geometry textbooks do list cuboid as the name. However the wikipedia article on cuboid goes out of its way to distinguish cuboid as being actually a hypernym of the target shape I describe:

a cuboid is a convex polyhedron bounded by six quadrilateral faces, whose polyhedral graph is the same as that of a cube. While mathematical literature refers to any such polyhedron as a cuboid,1 other sources use "cuboid" to refer to a shape of this type in which each of the faces is a rectangle

The wikipedia page cites Polytopes and symmetry by Robertson, Stewart Alexander for this fact, even thought it contradicts many other geometry textbooks.

If I need a way to unequivocally refer to the shape in question, do I really have to say "right cuboid" or "rectangular cuboid" every time? Or is the Wikipedia article wrong and simply reflecting some specific unconventional view of one geometer in 1933?

答:
Rectangular Prism, brings forth the image of a triangular prism, then the reader morphs it into a rectangle in the minds eye next.
Cuboid, brings forth the image of a cube, then the reader imagines a cube-like shape.

i feel the former is easier for a reader/listener to mentally picture, which the cuboid is rather ambiguous.
Having said that, cuboid does feel like the more succinct way of writing, so I will opt to use "cuboid" in writing and "rectangular prism" in speech.

It really does seem like a new term should be coined that listeners/readers can directly equate to the shape, rather than a stepping stone building off other known shapes.


问:
Expected number of days
If it is a good day (G) there are 60% chances tomorrow will be G and 40% chances tomorrow will be bad (B). If it is a B day, there 30% chances tomorrow will be G and 70% chances tomorrow will be B. If today is B, what is the expected number of days before seeing another B?

What I think is right: $E(days)=0.7(1)+0.12(2)+0.072(3)+...$

Is this the right way? If it is, how would I sum this? If not, please let me know.

答:
If you read well my previous answer I wrote:

In other words, the sequence, expressed in days, is the following

$1 \xrightarrow{\text{with probability}} p=0.7$

$2\xrightarrow{\text{with probability}} p=0.3\times0.4$

$3\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.4$

$4\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.4$

$5\xrightarrow{\text{with probability}} p=0.3\times0.6\times0.6\times0.6\times0.4$

...

$E(X)=0.7+0.3[2\times0.4+3\times0.4\times0.6+4\times0.4\times0.6^2+5\times0.4\times0.6^3+\dots]$

The probabilities in the [ brackets are the one of a geometric: $p(1-p)^x$ where $p=0.4$ and $(1-p)=0.6$

Is it clear now?

If the support was correctly $Y=1,2,3,4,...$ the average would have been $\frac{1}{0.4}$. Given that the support is $2,3,4,...$ instead...  you can think that your rv is

$$Z=Y+1$$

Thus for linearity of the mean you can have

$$E(Z)=1+E(Y)$$

Title: Expected number of days


问:
Positive integer solutions of equations of the form $ x^4+bx^2y^2+dy^4=z^2$
A method used to answer the post Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution? showed that the only positive integer solutions of the equations $$ x^4-3x^2y^2+3y^4=z^2$$ $$X^4+6X^2Y^2-3Y^4=Z^2,$$ are  $(1,1,1)$ and $(1,1,2)$, respectively. Furthermore these solutions are related in that each can be deduced from the other.

In general, let $b,d,D$ be integers such that $b^2=4d+D$ and consider the pair of equations  $$ x^4+bx^2y^2+dy^4=z^2\tag 1$$ $$X^4-2bX^2Y^2+DY^4=Z^2\tag 2$$ where we can assume that $(x,y)=(X,Y)=1$. Then for the method of the aforementioned post to produce an infinite sequence of solutions, we require a solution $(x,y,z)$ of $(1)$ to generate the solution $(\frac{z}{x},y,|\frac{x^4-dy^4}{x^2}|)$ of $(2)$ and we require a solution $(X,Y,Z)$ of $(2)$ to generate the solution $(\frac{Z}{2X},Y,|\frac{X^4-DY^4}{4X^2}|)$ of $(1)$.

I can prove that for this to occur $y=Y$ is odd. Also, since $x$ must be a factor of $z$, $x^2$ is a factor of $d$. Therefore there are only a finite number of possibilities for $x$ and so we are dealing with a finite loop of solutions. Beyond that, I only know that when solving specific equations these loops are rare and, when they occur, seem to have period $2$.

Questions

What periods are possible for such loops of solutions?

What else can be determined about them?

答:
SOME PROGRESS

Let the loop of solutions have period $2n$. We can suppose that successive values taken by $x$ and $X$ are $a_1,a_2,...a_n$ and $A_1,A_2,...A_n$, respectively, where for each $i$ there are integers $c_i$ and $C_i$ such that $d=a_i^2c_i$ and $D=A_i^2C_i$.

Then the conditions given in the post simplify to:-
$$a_i^2+by^2+c_iy^4=A_i^2 \tag 1$$
$$A_i^2-2by^2+C_iy^4=4a_{i+1}^2 \tag 2$$
$$2a_{i+1}A_i=|a_i^2-c_iy^4| \tag 3$$
$$4a_{i+1}A_{i+1}=|A_i^2-C_iy^4| \tag 4$$
$y^2$ is a factor of $4^n-1$

Consider equations $(1)$ and $(2)$ modulo $y^2$.
$$a_1^2\equiv A_1^2\equiv 4a_2^2\equiv 4A_2^2\equiv 16a_3^2\equiv 16A_3^2\equiv ... $$ Hence $a_1^2\equiv 4^na_1^2$ and so, since $x$ and $y$ are coprime, $y^2$ is a factor of $4^n-1.$

This significantly limits the values of $y$. For periods of less than or equal to $18$ the only possible values for $y$ are $1$ and $3$.

A general formula for loops of period $2$

If $n=1$, then $y=1$ and the equations further simplify to

$$a^2+b+c=A^2 \tag 1$$
$$A^2-2b+C=4a^2 \tag 2$$
$$2aA=|a^2-c| \tag 3$$
$$4aA=|A^2-C| \tag 4$$
From equation $(3)$ we have $c=a^2\pm 2aA$. Then $b=A^2\mp 2aA-2a^2$ and $C=A^2\mp4aA.$

The original equations are then $$ x^4+\left(A^2\mp 2aA-2a^2\right)x^2y^2+\left(a^4\pm2a^3A\right)y^4=a^2A^2$$ $$X^4-2\left(A^2\mp 2aA-2a^2\right)X^2Y^2+\left(A^4\mp4aA^3\right)Y^4=4a^2A^2,$$
with a loop of solutions $(a,1,aA)$ and $(A,1,2aA).$


问:
Divisibility by n for 3- digit number
We have 3 numbers:
$\; \overline{xyz}=100x+10y+z$, $\; \overline{yzx}$ and $\; \overline{zxy}$, all 3 divisible by the same number $n$.

Prove that the number $P=x^3+y^3+z^3-3xyz$ is also divisible by n.

$n \mid 100x+10y+z$

$n \mid 100y+10z+x$

$n \mid 100z+10x+y$

$n \mid 111x+111y+111z \Leftrightarrow n \mid 111(x+y+z)$ but since 111 is not a prime, we can't say for sure if $n \mid (x+y+z)$.

Also $P = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$.

So if I managed to prove that $n \mid (x+y+z)$ I think it would be sufficient, right?

Can you help me?

答:
COMMENT.-We have the linear system $$\begin{pmatrix}100&10&1\\1&100&10\\10&1&100\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}na\\nb\\nc\end{pmatrix}$$ whose determinant is equal to
$\Delta=998001=3^6\cdot37^2$. Calculations give
$$x=\frac{n999(10a-b)}{3^6\cdot37^2}=\frac{n(10a-b)}{3^3\cdot37}(\text{ note that }3^3\cdot37=\sqrt{\Delta})$$ and
$$y=\frac{n(10b-c)}{3^3\cdot37}\\z=\frac{n(10c-a)}{3^3\cdot37}$$
Therefore we have after calculations $$x^3+y^3+z^3-3xyz=\frac{n^3}{\Delta}(a^3+b^3+c^3-3abc)$$ This shows that if the proposition is true then $$\frac{n^2}{\Delta}(a^3+b^3+c^3-3abc)\in \mathbb Z$$ This should not be difficult to prove while holding clearly from the system the equality $n(a+b+c)=3\cdot37(x+y+z)$.

Title: Divisibility by n for 3- digit number


问:
A sum involving Logarithms
I am wondering if a closed form for the following sum exists:

$$S(N,a)=\frac{1}N\sum_{n=1}^N n\, \log\Big(1-\frac{n-a}N\Big),\qquad 0<a<1.$$

If I feed it to Mathematica, it gives me some complex expression (in terms of derivatives of Zeta functions), though the sum is real. I can easily access its asymptotics $N\rightarrow\infty$ (i.e. $S(N\gg1,a)\simeq -\frac{3}4N+\mathcal{O}(\log N)$), but given the fact that this sum does not look that complicated, I can't help but think that a closed form may exist...

答:
The obvious first step is to write it as
$$ \eqalign{\frac{1}{N} \log \left( \prod_{n=1}^N \left(1 - \frac{n-a}{N}\right)^n \right)
&= \frac{1}{N} \log\left(N^{-N(N+1)/2} \prod_{n=1}^N (a - n + N)^n \right)\cr
&= - \frac{N+1}{2} \log(N) + \frac{1}{N} \log\left(\prod_{n=1}^N (a - n + N)^n \right)}$$
The product is a polynomial in $a$ of degree $N(N+1)/2$ with roots of multiplicity $n$ at $n-N$.  I see no reason for that to have a "simpler" closed form.

Title: A sum involving Logarithms


问:
When reducing $A$ to $I$ via elementary matrices $E_k...E_1A=I$, why is it true also that $AE_k...E_1=I$
Say we have a $3 \times 3$ matrix $A$ and we perform the row operation $Z_1:(R_1 \iff R_2)$ of switching the first and second row to get $A'$.

Now, if we perform the same row operation $Z_1:(R_1 \iff R_2)$ on the identity matrix $I_3$ we get the elementary matrix $E_1$ and it is a common fact of linear algebra that $E_1A = A'$. Note that it is important that we multiply $E_1$ on the left of $A$.

Now, say further that $A$ is invertible and that we want to find $A^{-1}$. Then what we do is set up the augmented matrix $[A | I_3]$ and perform row operations $Z_1,...,Z_k$ until we we get an equivalent augmented matrix of the form $[I_3|B]$. It is a common fact of linear algebra that $B = A^{-1}$.

Now, the row operations $Z_1,...,Z_k$ correspond to the elementary matrcies $E_1,...,E_k$ and we can thus see that $A^{-1}=E_k...E_1$.

This is great and makes perfect sense! We reduced $A$ to $I_3$ via row operations corresponding to elementary matrices, so it is clear that $A^{-1}A = E_k,...E_1A=I_3$.

However, it is also true that $AA^{-1}=I_3$ and thus we have that $AE_k,...E_1 = I_3$. In the case when applying a string of elementary matrices to a matrix $A$ such that the string reduces $A$ to the identity, it is no longer important whether we apply the multiplication of the matrices on the right or left.

The bold font was meant to emphasize the phenomonon that I was hoping somebody could clarify. Understand that I am not asking why $A^{-1}A = AA^{-1}$ but rather seeking to clarify the more subtle observation of how the elementary matrices that embody the row operations can be applied to either side of $A$ in the case when they reduce $A$ to $I$.

Thank you.

答:
If $$E_2E_1A=I\tag{1},$$ then $A=E_1^{-1}E_2^{-1}$ and thus
$$
AE_2E_1=(E_1^{-1}E_2^{-1})E_2E_1=I\tag{2}
$$

This simply says that if you can transform $A$ into the identity matrix by pure row operations, you can also do it by pure column operations.

But the orders of the operations you do are different.

In (1), you first do the row operation $E_1$ and then the operation $E_2$.
But in (2), you first do the column operation $E_2$ and then the operation $E_1$.

it is no longer important whether we apply the multiplication of the matrices on the right or left

If this were true, what you should expect is
$$
E_2E_1A=AE_1E_2
$$
which is in general not true.


问:
Matrix ring over a field and its ideals
Let $M_n(R)$ be the matrix ring over a commutative ring $R$ and let $I$ be an ideal of $R$.

1) Show that if $R=F$ is a field then the only nonzero ideal of $M_n(F)$ is $M_n(F)$ itself.

2) Let $M_n(I)$ be the subset of $M_n(R)$ consisting of matrices with entries in the ideal $I$. Show that $M_n(I)$ is an ideal of $M_n(R)$ and describe the quotient ring $M_n(R)/M_n(I)$ by a matrix ring.

3) Show that if $I$ is a maximal ideal then $M_n(I)$ is a maximal ideal of $M_n(R)$.

Any advice on how to do these questions would be greatly appreciated!

答:
1) See Why is the ring of matrices over a field simple? for this.

2) By the way matrix multiplication works, $M_n(R)M_n(I)\subseteq M_n(R)$. One might guess that the quotient is $M_n(R/I)$, so lets look at it backwards: What is the kernel of the canonical homomorphism $M_n(R)\to M_n(R/I)$? (Note that this latter question at the same time shows that $M_n(I)$ is an ideal and what th equotient is)

3) Follows from 1 and 2 as $R/I$ is a field.

Title: Matrix ring over a field and its ideals


问:
network pagerank algorithm -- custom initial weights vs personalization
I am working with a graph and I want to compute the pagerank of its nodes. I want to emphasize some nodes more than others (and I use the networkx python package).

I can think of two ways of doing this:

start with custom initial scores, rather than 1/n for each node,

use personalized pagerank, so that the random walk restarts are biased towards some nodes.
There are parameters for both in networkx.

Which one makes more sense? What are the benefits of one method over the other?

答:
Starting with custom initial scores will not change final scores: the algorithm converges towards the same result independently of its initialization. The initial scores may only have a (minor) impact on convergence speed.

Instead, personalized pagerank iteratively jumps towards the chosen set of nodes, every time a teleportation occurs. This will definitely change the final scores.

Therefore, what you want to use is personalized pagerank.

In addition, you may use the same vector for initialization, in the hope that convergence will be faster.


问:
What is the formal way of constructing limit ordinals larger than $\omega$ (for example, $\omega + \omega$)?
I am trying to figure out how one constructs limit ordinals that are larger than $\omega$. In my book, the Axiom of Infinity (to create a set $X$ that contains the natural numbers) followed by the Separation / Comprehension Schema (with the formula $\varphi(n):= n \text{ is a natural number }$) is used to carve $\omega$ from $X$...the author calls this set $\omega$ the least limit ordinal.

I have read that, starting at $\omega$, you would (informally) apply the successor function an infinite number of times. This yields $S(\omega) = \omega \cup \{\omega\}=\omega+1$, $S(S(\omega))= \omega+1\  \cup \ \{\omega+1 \}=\omega+2$, etc.

Is the process as straightforward as defining the following set:

$T = \{ \omega + n\,|\, n\in \omega\}$

And then taking the union $\bigcup T$? (Knowing that such a union yields the supremum \ least upper bound of the set $T$)

If the above construction is the standard construction, then my once concern is, "From which set am I selecting the $\omega + n$ elements?".

The strategy I am familiar with is invoking the Separation / Comprehension schema to carve away subsets. But I think that must not be what I am doing above...because that would presuppose the existence of $\omega + \omega$ (or at least some ordinal that is larger than $\omega + \omega$).

So what is the method that I am actually using to create $T$ in the first place?

答:
You're right that Separation doesn't do the job. Instead, this is exactly what the axiom (scheme) of Replacement does!

We first write down a formula $\varphi(x,y)$ in two free variables which intuitively says "$y=\omega+x$." We then prove that for each $x\in\omega$ there is a unique $y$ such that $\varphi(x,y)$ holds - at which point Replacement lets us form $$\{y:\exists x\in\omega(\varphi(x,y))\}=\{\omega+n: n\in\omega\}.$$ Applying the Union axiom then finishes the job.

Meanwhile, to see that Separation won't suffice, it's a good exercise to check that $V_{\omega+\omega}\models\mathsf{ZC}$ (= the $\mathsf{ZFC}$ axioms except Replacement). Since $\omega+\omega\not\in V_{\omega+\omega}$, this means that $\mathsf{ZC}$ alone can't prove that $\omega+\omega$ exists. (See here for further discussion of this point.)


问:
Prove $\{0,1\}\cup\{\tfrac{1}{n},\tfrac{n-1}{n}:n=1,2,\cdots\}$ is compact using the open covering definition
I am studying from old exams and have run across this question.  In working with compactness, we only used sequential compactness.  This problem specifically asks for an open covering proof, which I do not know how to work with, as it was not something covered in my class, but is fair game for my comp.  Could someone provide a detailed proof so I can work on understanding the process?

Prove, using the open covering definition of compactness, that the set $$\{0,1\}\cup\left\{\frac{1}{n}\mathrel{}\middle|\mathrel{} n=1,2,\ldots\right\}\cup\left\{\frac{n-1}{n}\mathrel{}\middle|\mathrel{} n=1,2,\ldots\right\}$$ is compact.

答:
Hint: Suppose that you have an open covering $\mathscr{U}$ of the set
$$S=\{0,1\}\cup\{\tfrac{1}{n},\tfrac{n-1}{n}:n=1,2,\cdots\}.$$
Let $U_0,U_1\subseteq\mathbb{R}$ be open sets in $\mathscr{U}$ that cover the points $0$ and $1$ respectively. Because these sets are open and contain the points $0$ and $1$ respectively, they must also contain points "nearby" to $0$ and $1$. Show that there will only be finitely many elements of $S$ not covered by $U_0$ and $U_1$.


答: Since $\int_0^\infty\frac{f(x)dx}{1+x^2}=\int_0^1\frac{f(x)+f(1/x)}{1+x^2}dx$, $\int_0^\infty\tfrac{\arctan x^kdx}{1+x^2}=\tfrac{\pi}{2}\int_0^1\tfrac{dx}{1+x^2}=\tfrac{\pi^2}{8}$ for all $k\in\Bbb R$.

Title: Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$

Body:
I accidentally found out that the two integrals below

$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluated explicitly. For example, the first one can be carried out via double integration,  as sketched below.

\begin{align}
I_1&=\int_0^\infty \left(\int_0^1 \frac{x^2}{1+y^2x^4}dy\right)\frac{1}{1+x^2}dx\\
&= \frac\pi2\int_0^1 \left( \sqrt{\frac y2}+ \frac1{\sqrt{2y} }-1\right)\frac{1}{1+y^2}dy=\frac{\pi^2}8
\end{align}

Similarly, the second one yields $I_2=\frac{\pi^2}8$ as well.
The evaluations are a bit involved, though, and it seems an overreach to prove their equality this way, if only the following needs to be shown

$$\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx=0$$
The question, then, is whether there is a shortcut to show that the above integral vanishes.


答:
For each $n$ you construct by successively applying the successor function ordinals $\omega+n$ for each $n \in \omega$. Then the axiom of replacement tells us that

$$\{\omega+n \mid n \in \omega\}$$ is a set too and we can take its union (which order-theoretically is the sup) of that set to form $\omega+\omega$.

In that case it’s also enough to have one uncountable ordinal (which exists by power set and AC, or use Hartog’s trick) and note that all the above countable ordinals must be elements (and initial segments) of it, and apply separation. The way one goes about it depends on your text’s setup, I suppose.

Title: What is the formal way of constructing limit ordinals larger than $\omega$ (for example, $\omega + \omega$)?

Body:
I am trying to figure out how one constructs limit ordinals that are larger than $\omega$. In my book, the Axiom of Infinity (to create a set $X$ that contains the natural numbers) followed by the Separation / Comprehension Schema (with the formula $\varphi(n):= n \text{ is a natural number }$) is used to carve $\omega$ from $X$...the author calls this set $\omega$ the least limit ordinal.

I have read that, starting at $\omega$, you would (informally) apply the successor function an infinite number of times. This yields $S(\omega) = \omega \cup \{\omega\}=\omega+1$, $S(S(\omega))= \omega+1\  \cup \ \{\omega+1 \}=\omega+2$, etc.

Is the process as straightforward as defining the following set:

$T = \{ \omega + n\,|\, n\in \omega\}$

And then taking the union $\bigcup T$? (Knowing that such a union yields the supremum \ least upper bound of the set $T$)

If the above construction is the standard construction, then my once concern is, "From which set am I selecting the $\omega + n$ elements?".

The strategy I am familiar with is invoking the Separation / Comprehension schema to carve away subsets. But I think that must not be what I am doing above...because that would presuppose the existence of $\omega + \omega$ (or at least some ordinal that is larger than $\omega + \omega$).

So what is the method that I am actually using to create $T$ in the first place?


问:
Projection orthogonal to two vectors
Consider a unit vector $\mathbf{u}$, and the projection operator $\mathbb{P} = I - \mathbf{u}\mathbf{u}^T$. Then $\mathbb{P}$ projects an input vector into the space orthogonal to $\mathbf{u}$.

Now consider two unit vectors $\mathbf{u},\mathbf{v}$. What's the formula for the projection operator that projects an input vector into the subspace orthogonal to both $\mathbf{u}$ and $\mathbf{v}$?

If $\mathbf{u},\mathbf{v}$ are orthogonal, the formula is $\mathbb{P} = I - \mathbf{u}\mathbf{u}^T - \mathbf{v}\mathbf{v}^T$. I am interested in the general case, where $\mathbf{u},\mathbf{v}$ are not necessarily orthogonal.

答:
I will assume that $\mathbf u$ and $\mathbf v$ are linearly independent. Per the original version of this question, I do not assume that $\mathbf u, \mathbf v$ are unit vectors.

There are two standard approaches. One approach is to start by finding an orthonormal basis for the space spanned by $\mathbf u,\mathbf v$. If we follow the Gram-Schmidt process, then we get
$$
\mathbf w_1 = \mathbf u/\|\mathbf u\|, \\
\mathbf v_\perp = \mathbf v - \frac{\mathbf u^T \mathbf v}{\mathbf u^T\mathbf u}\mathbf u, \quad \mathbf w_2 = \mathbf v_\perp/\|\mathbf v_\perp\|.
$$
From there, we have
$$
\mathbb P = I - \mathbf w_1\mathbf w_1^T - \mathbf w_2\mathbf w_2^T = I - \frac{\mathbf u \mathbf  u^T}{\mathbf u^T\mathbf u} - \frac{\mathbf v_\perp\mathbf v_\perp^T}{\mathbf v_\perp^T \mathbf v_\perp}.
$$
Another option is to use a direct formula. For a given matrix $A$ with linearly independent columns, the projection onto the column space of $A$ is given by
$$
\mathbb P_A = A(A^TA)^{-1}A^T.
$$
Thus, if $A$ denotes the matrix whose columns are $\mathbf u,\mathbf v$, then your projection can be obtained as
$$
\mathbb P = I - \mathbb P_A = I - A(A^TA)^{-1}A^T.
$$
Because $A^TA$ is only a $2 \times 2$ matrix, you could expand this into a formula written completely in terms of $\mathbf u,\mathbf v$ if you wanted to. In particular, we find that
$$
A^TA = \pmatrix{\mathbf u^T\mathbf u & \mathbf u^T \mathbf v\\ \mathbf v^T \mathbf u & \mathbf v^T\mathbf v}, \quad
(A^TA)^{-1} =
\frac{1}{\|\mathbf u\|^2\|\mathbf v\|^2 - (\mathbf u^T\mathbf v)^2}\pmatrix{
\mathbf v^T\mathbf v & -\mathbf u^T \mathbf v\\
-\mathbf v^T\mathbf u & \mathbf u^T\mathbf u
}.
$$
If we define $\delta = \frac{1}{\|\mathbf u\|^2\|\mathbf v\|^2 - (\mathbf u^T\mathbf v)^2}$, then we have
$$
\mathbb P_A = \delta\cdot [\|\mathbf v\|^2 \mathbf u\mathbf u^T
+ \|\mathbf u\|^2 \mathbf v \mathbf v^T
- \mathbf u^T\mathbf v(\mathbf u\mathbf v^T + \mathbf v \mathbf u^T)].
$$

Title: Projection orthogonal to two vectors


问:
Convergence of $x_n = \frac{n+\tan^{-1} n}{n^2 + \ln n}$
We know $\max\{\tan^{-1}\}=\frac{\pi}{2}.$ Therefore, for sufficiently large $n$, we have that $n>\tan^{-1}n.$ The limit of this sequence is zero. Therefore $$\frac{n+\tan^{-1} n}{n^2 + \ln n}\leq\frac{2n}{n^2 + \ln n}\leq\frac{2}{n}$$

and we can choose $N\geq \frac{2}{\epsilon}$ for the proof. Correct? I think the initial assumption is justified, since we are looking at the behaviour when $n\to \infty$.

答:
The basic idea behind your proof is solid and would readily convince most people. However, if this is for a class or some assignment, there is some sloppiness that might cause you to lose points. Let me restate your argument in a more air-tight manner:

Let $\epsilon>0$ be given and choose $N=\max\left\{\left\lceil\frac{2}{\epsilon}\right\rceil,2 \right\}$. Then for $n\geq N$ we have

$$\left|\frac{n+\tan^{-1}(n)}{n^2+\ln(n)}-0\right|=\frac{n+\tan^{-1}(n)}{n^2+\ln(n)}<\frac{n+\frac{\pi}{2}}{n^2}<\frac{n+\frac{4}{2}}{n^2}\leq\frac{n+n}{n^2}=\frac{2}{n}\leq \frac{2}{N}\leq \epsilon$$

Note the three things I tightened up:

$1)$ It's $\sup\{\tan^{-1}(n)\}=\frac{\pi}{2}$ (not maximum)

$2)$ I covered the case where $\epsilon$ is large (as it is all $\epsilon>0$) which would imply that $N\geq 2$

$3)$ I made sure that the argument is of the form: For all $\epsilon>0$, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|x_n-L|<\epsilon$

But again, your reasoning is good and the argument is convincing, and the details I mentioned are so minor that they wouldn't effect anything outside of a class or assignment setting.


问:
Taking the second derivative of a parametric curve
I understand that for the parametric equations

$$\begin{align*}x&=f(t)\\
y&=g(t)\end{align*}$$

If $F(x)$ is the function with parameter removed then $\displaystyle F'(x) = \frac{\text{d}y}{\text{d}t}\big/\frac{\text{d}x}{\text{d}t}$

But the procedure for taking the second derivative is just described as " replace $y$ with dy/dx " to get

$$\frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)=\frac{\left[\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}t}\right)\right]}{\left(\frac{\text{d}x}{\text{d}t}\right)}$$

I don't understand the justification for this step. Not at all.

But that's all my book says on the matter then it launches in to plugging things in to this formula, and it seems to work well enough, but I don't know why.

I often find answers about question on differentials are beyond my level, I'd really like to get this, it'd mean a lot to me if someone could break it down.

答:
I've come up with a 'simple enough' proof for the parametric derivative. Begin by consider a curve parameterized in the way :$<x(t), y(t) >$, then I will rewrite derivative operator with $x$ :

$$ \frac{dy(\theta)}{dx} = \frac{d \theta}{dx} \frac{d y}{d \theta}$$
The above equality holds by the chain rule, derivative of inner function times derivative of outer function, now I will remove the $y$:

$$ \frac{d}{dx} = \frac{d \theta}{dx} \frac{d}{ d \theta} \tag{1}$$

Now, the above can be thought of as an operator which eats a function and maps to another function, for example if we multiply both side on right by $y$ , then it gives the $ \frac{dy}{dx}$ with the parametric derivative of $y$.

And, I will introduce one more identity:

$$ \theta( x( \theta) ) = \theta \tag{2}$$

The above can be thought of in the following way: The function $x(\theta)$ maps from $ \theta \to x$ , the function $ \theta(x)$ maps from $ x \to \theta$, simply the inverse mapping and hence above equality holds true for at least a small interval of where we can say that $ \theta $ and $x$ are bijective functions of each other. Taking the derivate of $(2)$ with $ \theta$,:

$$ \frac{ d \theta}{dx} \frac{dx}{d \theta} = 1$$

Or,

$$ \frac{d\theta}{dx} = \frac{1}{ \frac{ d x}{d \theta}} \tag{2}$$

Hence, we can rewrite (1) as:

$$ \frac{d}{dx} = \frac{ \frac{d }{d \theta} }{\frac{dx}{d \theta}} \tag{3}$$

Now, multiply $y$ on right side of (3):
$$ \frac{dy}{dx} = \frac{ \frac{dy }{d \theta} }{\frac{dx}{d \theta}} \tag{4}$$

This is the first parameteric derivative. Now, use (3) on equation (4):

$$ \frac{d^2 y}{dx^2} = \frac{1}{ \frac{dx}{d \theta}} \left[ \frac{ \frac{d^2 y}{d \theta^2} \frac{dx}{d \theta} - \frac{dy}{d \theta} \frac{dx}{ d \theta}}{ ( \frac{dx}{d \theta})^2} \right]=  \left[ \frac{ \frac{d^2 y}{d \theta^2} \frac{dx}{d \theta} - \frac{dy}{d \theta} \frac{dx}{ d \theta}}{ ( \frac{dx}{d \theta})^3} \right]$$

Which is what we sought to prove.


问:
Please Help explain a solution in a Geometric distribution problem
someone please explain to me the solution for this
Assuming that each student visits 5 booths during a typical career fair, find the probability that an A student in 313 will not get an off-campus interview invitation. Similarly,
find the probability that a C student in 313 will get an invitation during a typical career
fair.

There a solution here (question c: https://courses.engr.illinois.edu/ece313/fa2012/homework/sol04.pdf)
for this question but I don't get it. I don't understand why there is a 6 in the sum if we only have 5 trials

答:
I don't understand why there is a 6 in the sum if we only have 5 trials

simply because  you are calculating

$$\mathbb{P}[X>5]=\mathbb{P}[X=6]+\mathbb{P}[X=7]+\mathbb{P}[X=8]+\dots$$

Obvioulsly it is simplier to do

$$(1-p)^5=0.05^5\approx 3\cdot 10^{-7}$$


问:
Difficulty in understanding significance of Grelling's Paradox.
Background: I'm a math rookie, yet to enrol in university. I randomly started reading Mendelson's Introduction to Mathematical Logic, when I stumbled upon this paradox in the introductory section:

Grelling's Paradox: An adjective is called autological if the property denoted by the adjective holds for the adjective itself. An adjective is called heterological if the property denoted by the adjective does not apply to the adjective itself. For example, 'polysyllabic' and 'English' are autological, whereas 'monosyllabic' and 'French' are heterological. Consider the adjective 'heterological'. If 'heterological' is heterological, then it is not heterological. If 'heterological' is not heterological, then it is heterological. In either case, heterological is both heterological and not heterological.

I'd like to understand the following:

What is the source of logical fallacy in this paradox? If I formulate a set $A$ of all adjectives and subsets $A_a$ and $A_h$ corresponding to autological and heterological adjectives, respectively, then it could be the case that $\text{(heterological)}\in A-(A_a\cup A_h)$, i.e., it belongs to neither of the two sets(unless $A_a\cap A_h=\emptyset$ and $A_a\cup A_h=A$).

On a lighter note, I'd like to know about the mathematical significance of this paradox, and how it's dealt with in modern set theories.

Although I understand the answer(s) could be very abstract, please add a simpler analogy along with a necessary technical explanation, if possible.

答:
Logically speaking, Grelling's Paradox is just Russell's Paradox--only the names have been changed.

For better or worse, natural language allows us to define things that cannot logically exist. Logically, just like there can exist no set of all those sets that are not elements of themselves (Russell's Paradox), there can exist no word that describes all those words that do not describe themselves (Grelling's Paradox).

Russell's Paradox:

$\neg \exists r:  [Set(r) \land \forall a: [Set(a) \implies [a\in r
 \iff a\notin a]]]$

We can substitute:

$Set(x) \to Word(x)~~~$ (unary predicate or domain of discourse)

$x \in y \to Describes(y,x)~~~$ (binary predicate)

$r \to h~~~$ ($h$ = "heterological")

Grelling's Paradox:

$\neg \exists h:  [Word(h) \land \forall a: [Word(a) \implies
[Describes(h,a)  \iff \neg Describes(a,a)]]]$


问:
Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

I did this by creating 3 sets:

$S\,\,=\{a,b,c,d\}$

$T\,\,=\{c,d,e,f\}$

$T'=\{a,b,c,d\}$

And then I performed the operations and checked they are the same. Is there something else I should do or only this will suffice?

答:
Let $X$ be a set. For every $A,B,C$ subsets of $X$, you can check that
$$\frac{(A\cap B)\subseteq C}{B\subseteq (A\rightarrow C)}$$
where $A\rightarrow C$ is defined as $(A\rightarrow C):=(\mathscr{C}A)\cup C$ and $\mathscr{C}A$ is the complementary set of $A$ in $X$ and horizontal line means "if and only if", i.e. the line above is true if and only if the line below is true.

More concisely,
$$A\cap (-)\dashv A\rightarrow (-)$$
where $\dashv$ means left adjoint, $A\cap(-)$ is the function sending a subset $B$ of $X$ to $A\cap B$ and $A\rightarrow (-)$ the function sending a subset $C$ of $X$ to $A\rightarrow C$.

Being a left adjoint, $A\cap (-)$ preservs colimits, in particular it preservs unions, hence your desidered distributive law.

$\textrm{EDIT}$: let $f:X\longrightarrow Y$ be a function, $X,Y$ sets, $A$ subset of $X$, $B$ subset of $Y$. Then you have
$$\frac{f(A)\subset B}{A\subset f^{-1}(B)}$$
Proceeding as in the main part of this answer, you have that the direct image of $f$ is left-adjoint to the inverse image $f^{-1}$, thus, being a left adjoint, direct image preserves colimits, for example unions, so that you immediately prove that $f(A\cup C)=f(A)\cup f(C)$, without computing on elements ( pick an element in $A\cup C$, do the image under $f$, check this image is in the image of $A$ or in the image of $C$, conversely, pick an element which is in $f(A)$ or in $f(B)$, check that it comes from an element....), while inverse image $f^{-1}$, being a right-adjoint, preserve limits, for example intersections, so that you have an "automatic" proof that $f^{-1}(A\cap C)=f^{-1}(A)\cap f^{-1}(C)$

So two facts apparently different as your distributive law and these properties of $f$ and $f^{-1}$ appear in some sense as two different manifestations of the same phenomenon


答:
The event "an A student does not get an invitation in $5$ trials" is the same event as "it takes at least $6$ trials for an A student to get an invitation". The corresponding equation with probabilities is

$$P(X\geq 6)=\underbrace{1-P(X\leq 5)}_{\textrm{Student does not get an invitation in 5 trials}}$$

Title: Please Help explain a solution in a Geometric distribution problem

Body:
someone please explain to me the solution for this
Assuming that each student visits 5 booths during a typical career fair, find the probability that an A student in 313 will not get an off-campus interview invitation. Similarly,
find the probability that a C student in 313 will get an invitation during a typical career
fair.

There a solution here (question c: https://courses.engr.illinois.edu/ece313/fa2012/homework/sol04.pdf)
for this question but I don't get it. I don't understand why there is a 6 in the sum if we only have 5 trials


问:
Tool for doing linear algebra with algebra instead of numbers
I have some linear algebra equations, which are a few succinct lines of pytorch code, where the matrices involved have dimensions N and D, and N and D are usually very large. I wanted to really understand what they are doing, so started working through them for N=3, D=2. So I started with this:

```<code>  |a  b|   |p  r|   |ap + bq    ar + bs|
  |c  d| x |q  s| = |cp + dq    cr + ds|
  |e  f|            |ep + fq    er + fs|
</code>```

A couple of long, tedious, calculations later, and after gathering terms, I had this:

```<code> aa(pt+rv) + ab(pu+rw) + ab(qt+sv) + bb(qu+sw)     ac(pt+rv) + ad(pu+rw) + bc(qt+sv) + bd(qu+sw)     ae(pt+rv) + af(pu+rw) + be(qt+sv) + bf(qu+sw)

 ac(pt+rv) + bc(pu+rw) + ad(qt+sv) + bd(qu+sw)     cc(pt+rv) + cd(pu+rw) + cd(qt+sv) + dd(qu+sw)     ce(pt+rv) + cf(pu+rw) + de(qt+sv) + df(qu+sw)

 ae(pt+rv) + be(pu+rw) + af(qt+sv) + bf(qu+sw)     ce(pt+rv) + de(pu+rw) + cf(qt+sv) + df(qu+sw)     ee(pt+rv) + ef(pu+rw) + ef(qt+sv) + ff(qu+sw)
</code>```

Which was very helpful, but I'd cut one of the calculations out, and also made me see I want to know what it looks like when D and N are slightly bigger.

Is there any software or library I can use to have the computer do is this hard work for me? I.e. where I would write this code:

```<code>x = torch.tensor([ [1.0,2], [3,4], [5,6] ])
M = torch.tensor([ [0.1, 0.3], [0.7, 0.5] ])
y = x.matmul(M)
</code>```

I want to write something like:

```<code>x = torch.tensor([ ['a', 'b'], ['c', 'd'], ['e', 'f'] ])
M = torch.tensor([ ['p', 'r'], ['q', 's'] ])
y = x.matmul(M)
</code>```

If it gathered like terms for me that would be a bonus, but it is not essential.
Not tied to a particular computer language, so any of Python, R, Octave/Matlab, C++, etc.

答:
Thanks to the hint in the comments, I took a look at Sage, specifically inside CoCalc.

To do symbolic linear algebra you define your matrices like this:

```<code>x=matrix(3,2, var('a,b,c,d,e,f'))
M=matrix(2,2, var('p,q,r,s'))
V=matrix(2,2, var('あ,い,う,え'))
</code>```

(That last line was just to confirm I can something other than a..z; in fact only valid Python identifiers are allowed. Which means lots of Unicode is available, but things like circled numbers, negative letters and emoji were all disallowed.)

* does matrix multiplication as you'd expect, and there is a function to expand out brackets, and a choice of functions to simplify.

(I look at Sage first, as it is open source, and based on Python; there appear to be a number of other choices.)

Body:
Thanks to the hint in the comments, I took a look at Sage, specifically inside CoCalc.

To do symbolic linear algebra you define your matrices like this:

x=matrix(3,2, var('a,b,c,d,e,f'))
M=matrix(2,2, var('p,q,r,s'))
V=matrix(2,2, var('あ,い,う,え'))

(That last line was just to confirm I can something other than a..z; in fact only valid Python identifiers are allowed. Which means lots of Unicode is available, but things like circled numbers, negative letters and emoji were all disallowed.)

* does matrix multiplication as you'd expect, and there is a function to expand out brackets, and a choice of functions to simplify.

(I look at Sage first, as it is open source, and based on Python; there appear to be a number of other choices.)


问:
Using implicit function theorem without using the inverse function theorem.
Let $f:U\rightarrow \mathbb{R}$ defined on the open set $U \subset \mathbb{R}^m$. It the function $g(x):U\rightarrow \mathbb{R}$, given by the expression $$g(x)= \int_{0}^{f(x)} (t^2 + 1)dt,$$ of class $C^{\infty}$, then f is $C^{\infty}$.Explicitly using the implicit function theorem, knowing that the inverse function is equivalent.</p>

答:
Let $H:U\times\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}$ the function
$$
H(x,y,z)= \int_{\langle x,x\rangle -{\langle z,z\rangle}}^{\mbox{sgn}\big(f(x)\big)\cdot\langle y,y\rangle} \big( t^2+1\big) \; \mathrm d t
$$
By chain rule we have
\begin{array}{rl}
\left.\frac{\partial H(x,y,z)}{\partial y}\right|_{(x,y,z)=(x_0,y_0,x_0)}
=
&
\left. \frac{d}{du}\left( \int_0^{u} \big( t^2+1 \big) \mathrm d t\right) \right|_{u=\mbox{sgn}\big(f(x_0)\big)\cdot\langle y_0,y_0\rangle}
\cdot
\left.\frac{\partial}{\partial x} \Big( \mbox{sgn}\big( f(x)\big)\langle y,y\rangle \Big)\right|_{y=y_0}
\\
&
\\
&
\\
&
\\
&
\\
=
&
\left.  \big( u^2+1 \big) \right|_{u=\mbox{sgn}\big(f(x_0)\big)\cdot\langle y_0,y_0\rangle}
\cdot
\left.2 \cdot\mbox{sgn}\big( f(x)\big)\cdot y\right|_{(x,y)=(x_0,y_0)}
\\
&
\\
&
\\
&
\\
=
&
2 \cdot\mbox{sgn}\big( f(x_0)\big)\cdot\big( \langle y_0,y_0\rangle^2+1 \big)
\cdot y_0
\\
\end{array}
Then for $x_0\in U$ such that $f(x_0)\neq 0$ and $y_0\neq 0$ we have
$$
\left.\frac{\partial H(x,y,z)}{\partial y}\right|_{(x,y,z)=(x_0,y_0,x_0)}\neq 0
$$
By Implicit Function Theorem there is:

a open sets $V=V_{(x_0,z_0)}\subset U\times\mathbb{R}^n$ and $W=W_{y_0}\subset \mathbb{R}^n$ such that $V_{(x_0,z_0)}\times W_{y_0}\subset U\times\mathbb{R}^n\times\mathbb{R}^n$.

a function $\varphi: V\to W$ such that
$$
   H(x,\varphi(x,z),z)=H(x_0,y_0,z_0=x_0) \quad \forall (x,z)\in V \quad \forall y\in W
   $$

The function $\varphi: V\to W$ is unique in the sense that if there is a function $\psi :V\to W$ with the previous property
$$
   H(x,\psi (x,z),z)=H(x_0,y_0,z_0) \quad \forall (x,z)\in V \quad \forall y\in W
   $$
then $\varphi = \psi$.

The function $\varphi$ has the same class of differentiability of $H$. In particular if $H$ is of class $C^\infty$ then $\varphi$ is class $C^\infty$.

Now to finalize the point suffices to observe that if $f(x)=\mbox{sgn}\big( f(x)\big)|f(x)|=\mbox{sgn}\big( f(x)\big)\langle y,y\rangle $
$$
g(x)=H(x,y,x)= \int^{\mbox{sgn}\big( f(x)\big)\langle y,y\rangle}_0 (t^2+1) \mathrm d t
=\int^{\mbox{sgn}\big( f(x)\big)f(x)}_0 (t^2+1) \mathrm d t
$$
On the other hand, for all  neighborhood $V_{(x_0,z_0)}\times W_{y_0}$ of a point $(x_0,y_0,z_0)$, with $ x_0 = z_0 $,  that   implicit function $\varphi: V_{(x_0,z_0)}\to W_{y_0}$ of  Implicit Function Theorem ensures the equality
$$
g(x)=H(x,\varphi(x,x),x)= \int^{\varphi(x,x)}_0 (t^2+1) \mathrm d t
=\int^{\mbox{sgn}\big( f(x)\big)\langle y,y\rangle}_0 (t^2+1) \mathrm d t
$$
By uniqueness of $\varphi : V_{(x_0,z_0)}\to W_{y_0}$ for each point $(x_0,y_0,z_0)=(x_0,x_0,z_0)$ we have
$$
\varphi(x,x)= \mbox{sgn}\big( f(x) \big)\cdot |f(x)|=f(x)
$$
Then we can conclude $f$ is of class $C^\infty$ by cause $\varphi$ is $C^\infty$. How is any point $(x_0,y_0,z_0)$ in $U\times\mathbb{R}^n\times\mathbb{R}^n$  we can extend $\varphi(x,x)$ (again because of the uniqueness) for all $U\times U = \bigcup_{x_0,z_0\in U} V_{x_0,z_0}$.


问:
Multivariate residues in simple cases
I have been looking at residues of multivariate functions and found there are quite a few difficulties (see e.g. Multivariate Residue Theorem? or Multivariate/multidimensional residues). In the literature, this is discussed in the context of manifolds, 1-forms and currents. Unfortunately, I am not an expert on manifolds.

Question: Are there "simple rules" deriving from the general treatment that can be applied in more basic cases.

I am thinking of multivariate functions $f(x,y,z, ...)$ with simple poles at equal points $x=y,~ x=z, ...$, where I would like to evaluate the residue at multiple, coinciding points $x=y=z$ as consecutive residues $\text{Res}_{x=y} \text{Res}_{y=z} \cdots$ in a consistent way.

Example: Consider the function $f(x,y,z) = \frac{1}{(x-y)(x-z)}$ defined on $\mathbb{C}^3$. It has singularities on and the 1-dimensional subspaces $\{(x,y,z) | x=y \}$ and $\{ (x,y,z) | x=z\}$ which intersect at $x=y=z$. Computing the residue on the intersection can be done through consecutive application of residues:
$$
\text{Res}_{y=z} \text{Res}_{x=y} \frac{1}{(x-y)(x-z)} = \text{Res}_{y=z} \frac{1}{(y-z)} = 1.
$$
However, exchanging the residues leads to a wrong result:
$$
\text{Res}_{x=y} \text{Res}_{y=z} \frac{1}{(x-y)(x-z)} = 0.
$$

Is there a procedure that tells me how to correctly take certain residues or at least relate different combinations of residues which give the same result? (For example $\text{Res}_{y=z} \text{Res}_{x=y}$ and $\text{Res}_{z=x} \text{Res}_{y=x}$ in the previous example)

Background: In quantum field theory, amplitudes (vacuum expectation values of time-ordered products of fields) are meromorphic functions in $\mathbb{C}^k$. Poles correspond to the temporary fusion of particles and higher-order poles at the intersection of more than two points (which I would like to evaluate as consecutive residues) appear when there are more complicated composite particles.

I appreciate any help or literature recommendation!

答:
If think I have an idea which works in some cases at least. The idea is to cancel all poles and consider residue as limits of this (now) analytic function. This only works for simple poles of course. Here is what I mean:

Assuming that $f(x,y,z)$ has simple poles at $x=z$ and $y=z$. It follows that $(x-z)(y-z) f(x,y,z)$ is analytic around $x=z$ and $y=z$.

Therefore, it is possible to take limits of this continous function
\begin{align}
&\lim_{x\to z} \lim_{y\to z} (x-z)(y-z) f(x,y,z) = \lim_{y\to z} \lim_{x\to z} (x-z)(y-z) f(x,y,z) \\
\Leftrightarrow& \lim_{x\to z} (x-z) \lim_{y\to z} (y-z) f(x,y,z) = \lim_{y\to z} (y-z) \lim_{x\to z} (x-z) f(x,y,z)
\end{align}
The limits commute of course, because for an analytic function it does not matter from which direction we approach the point $x=y=z$.

These limits yield the residues at $x=z$ and $y=z$, so it follows that:
\begin{align}
&\text{Res}_{x=z} \text{Res}_{y=z} = \lim_{x\to z}(x-z) \lim_{y\to z} (y-z) = \lim_{y\to z} (y-z) \lim_{x\to z}(x-z) = \text{Res}_{y=z} \text{Res}_{x=z}.
\end{align}
This means: If one can exchange the limits consistently, one can exchange the residues.

In the example I gave, the function $f(x,y,z) = \frac{1}{(x-y)(x-z)}$ has poles at $y=x$ and $z=x$. Looking at the residue $\text{Res}_{y=z} \text{Res}_{x=y}$ means to consider $(y-z) (x-z) f(x,y,z)$. This is not analytic because the pole at $x=y$ is not cancelled. We can approach $x=y=z$ only by going first to $x=z$ and then $y=x$, not the other way around.

But $\text{Res}_{z=x} \text{Res}_{y=x} = \text{Res}_{y=x} \text{Res}_{z=x}$ can be exchanged consistently here.

Title: Multivariate residues in simple cases


问:
Show a Subspace of regular space is regular

I use this definition .Proofs of this theorem
I saw are  based on a similar def,  but with functions.
Source:
(https://www.math.tamu.edu/~tomzz/math636/ass7.pdf)

Attempt

Let (X$,\mathcal{T}$) be a regular topology.
Let $\mathcal{T}_S$={O$\cap$ S :O$\in\tau$}

Suppose X is a regular and $S\subseteq X$,S closed and any point  x$\in A$ and V$\subset A$

Let x$\in S\setminus A$, then $S\setminus A$ is open

Then there exists U open set in X s.t U $\cap S$ =V

Clearly $ x\notin$ U else x$\in V$

Since X is regular  (S$\setminus A )\cap S=\emptyset$

答:
I'm not following your proof.  You need to show that a subspace of X is regular.   You need to give it a name, if you are using $A$ that might confuse you with the $A$ in the theorem so I suggest calling it something else, like let $S\subseteq T$ be equipped with the subspace topology (IE a set is open in $S$ if and only if it is equal to the intersection of an open set $U$ in $T$ with $S$).

Now, within $S$ you need to show regularity, so you need to take a closed set in $S$, call it $A\subseteq S$ and a point  $x\in S\setminus A$  and show there exists open sets $U,V$ in the subspace topology that separate $A$ and $x$ in $S$.  Hint: Consider what it means to be closed set in $S$ compared to $T$,  then use the regularity of $T$ to generate $U',V'$ in $T$ that you can then derive your $U,V$ in S from.


问:
Question regarding the formal completion of affine schemes
Let $k$ be any field. We look at the formal completion of $\mathbb{A}^{1}_{k}$ at $0$, i.e. the colimit $X$ of the following diagram in the category of locally ringed spaces,
$$\operatorname{Spec}k[x]/(x) \rightarrow \operatorname{Spec}k[x]/(x^2) \rightarrow \operatorname{Spec}k[x]/(x^3) \rightarrow \cdots  $$

I am trying to understand the underlying topology of $X$. Now each of the affine schemes in the diagram is one point set, so intuitively it seems the colimit should have one point. But rigorously if I need to get any information of the topology of a locally ringed space, I need to apply the forgetful functor $ For : LRS \rightarrow Tops$. And the forgetful functors are usually the right adjoint (though I don't know if there is any adjoint for this one!), so it may not preserve colimit for this diagram.

Maybe I am missing some easy way of understanding this. Thanks in advance!!

答:
The forgetful functor from locally ringed spaces to topological spaces is actually a left adjoint, not a right adjoint; in particular, it preserves colimits.  You can easily see it is not a right adjoint because it does not preserve limits: the terminal locally ringed space is $\operatorname{Spec}(\mathbb{Z})$, but its underlying space is not the terminal topological space (a singleton).

The right adjoint to the forgetful functor is fairly complicated.  Roughly speaking, it takes a topological space $X$, equips it with the constant sheaf $\mathbb{Z}$ (which gives the right adjoint if you were just taking ringed spaces, not locally ringed spaces), and then "expands out" every point of $X$ into new points for each prime ideal in the stalk in order to get a locally ringed space.  For more discussion and references, see this nice answer.

For the specific example you are interested in, you can just easily directly construct the colimit.  Namely, let $X$ be a singleton equipped with the sheaf of rings whose global sections are $k[[x]]$.  This is a locally ringed space, and it has compatible maps from your diagram.  To see that it is the colimit, let $Z$ be any locally ringed space with compatible maps from your diagram.  All these maps must have image that is the same single point $z\in Z$, and we get a compatible system of maps $O_{Z,z}\to k[x]/x^n$ for each $n$.  These induce a map $O_{Z,z}\to k[[x]]$, and this gives a morphism of locally ringed spaces $X\to Z$ that sends the point of $X$ to $z$ and sends a section of $O_Z$ in a neighborhood of $z$ to its image in $k[[x]]$.  Moreover, it is easy to see this is the unique such morphism compatible with the maps from the diagram.

(More generally, you can explicitly construct colimits of locally ringed spaces by taking the colimit of the underlying topological space and then equipping it with essentially the only obvious sheaf of rings you can define.  Of course, the details are much more complicated to check in general than in your example where everything is just a singleton.)


问:
$P^1$ not a regular level surface of a $C^1$ function on $P^2$
I'm working through the first chapter of Morris Hirsch's "Differential Topology". On Chapter 1, section 3 exercise 11, I encountered the following question.

"Regarding $S^1$ as the equator of $S^2$, we obtain $P^1$ as a submanifold of $P^2$  (Hirsch uses $P^n$ to denote real projective n-space). Show that $P^1$ is not a regular level surface for any $C^1%$ map on $P^2$. Hint: no neighbourhood of $P^1$ in $P^2$ is separated by $P^1$."

I attempted to solve this by contradiction, supposing there was a $C^1$ function of $P^2$ such that $P^1=f^{-1}(y)$ and that $T_pf$ is surjective for each $p\in P^1$. Then the inverse function theorem would imply that every open neighbourhood of $P^1$ is diffeomorphic to some open neighbourhood of $y$. I have a feeling that I'm meant to apply the hint here, and arrive at a contradiction by showing a topological property is not preserved under homeomorphism (path connectedness hopefully).

Unfortunately, I am having trouble visualising why the hint it true. I don't think I can go further without understanding that.

答:
Two minor comments:

Say $f\in C^1(P^2, N)$ is an arbitrary map with $P^1=f^{-1}(y)$ for some regular value $y\in N$. Submersivity a priori implies only that the dimension of $N$ is not more than $2$. It would be good form to add a sentence to justify $\dim N = 1$. Then by the local character of the problem one can assume $N$ is an open interval.

The Inverse Function Theorem (here and in general) does not imply what you say it does. Say $\dim(N)=2$, so that IFT kicks in along $P^1$. Taking small open sets and collating their diffeomorphic images gives that small open sets containing $P^1$ (which are homeomorphic to Mobius bands without boundary, if small enough) is open in the target. We don't have a diffeomorphism though, since $P^1$ is collapsed to a point (or alternatively orientability is a diffeomorphism invariant).


问:
Show that $\langle(1234),(12)\rangle = S_4$
I am trying to show that $\langle(1234),(12)\rangle = S_4$. I can multiply and get all $24$ permutations manually but isn't there a more compact solution?</p>

答:
The two methods I would use are either

show all transpositions appear. If $c$ is the cycle and $t$ the transposition, start by considering $c^{-1}tc$.

note that we have an element of order 4 already, and multiplying the ones we have together gives an element of order 3. Hence the group generated has order divisible by 12. But the only subgroup of size 12 is $A_4$ which this clearly is not.


问:
When does a function not have continuity where it's defined?
I.e. a function such as $\tan \sqrt{\log x}$.

答:
The most that can be said about this in general is that one should learn to apply the definition of continuity and the theorems of calculus regarding continuous functions.

Your specific example is a composition of three functions each of which is continuous at every point of its domain, namely $f \circ g \circ h(x)$ where $f(x) = \tan(x)$ is a continuous function with domain $x \in \mathbb R - \{(2k+1)\pi/2 \mid k \in \mathbb Z\}$; and $g(x) = \sqrt{x}$ is a continuous function with domain $x \in [0,\infty)$; and $h(x) = \log(x)$ is a continuous with domain $x \in (0,\infty)$. Continuity of all three of these functions is proved in advanced calculus; here's an outline.

Continuity of $\sqrt{x}$ is pretty easy exercise directly from the $\epsilon,\delta$ definition of continuity.

Continuity of the logarithm function is probably best proved by using the fundamental theorem of calculus.

Continuity of $\tan(x)$ and other trig functions are perhaps best proved using theorems about infinite series expansions, although they can also be proved using a combination of the fundamental theorem of calculus and the inverse function theorem for continuity.

Another theorem of calculus says that a composition of continuous functions is continuous at all points where it is defined; perhaps this theorem gives the most direct answer to your question.

Putting this all together, $\tan \sqrt{\log(x)}$ is continuous at all points where it is defined.


问:
Knowing number of elements in a relation from this information about equivalence classes.
I am trying to understand this question:

Let $A$ be a set with $24$ elements, and let $E$ be an equivalence relation on $A$.

$E$  divides $A$ to $7$ equivalence classes in the following way:

$3$ equivalence classes that has $4$ elements each.

$4$ equivalence classes that has $3$ elements each.

What is $|E|$ or how many elements are in the relation.

My Attempt:

I know that $[a]_E=\{b\in A \mid (a,b)\in E\}$ is the definition of equivalence class. so if I have $3$ equivalence classes with $4$ elements each that means I have $12$ pairs in $E$, but $E$ is a symmetric relation, so if $(a,b)\in E$ automatically we get $(b,a)\in E$ so $24$, and same thing for the other equivalence classes we get more $24$ pairs. but the answer was $84$.

After thinking about it, I thought about the reflexive property and we can add $24$ pairs more since for each element in $A$ we get $(a,a)\in E$, but thinking about it again, in every equivalence class, I have already counted some of these elements, because every element is in it's own equivalence class, so I'm wondering why don't we subtract $7$ from the answer.

All help is appreciated, Thanks in advance.

答: Within an equivalence class,  every element is paired with every element, every possible ordered pair is there.   So for instance, if one equivalence class is $\{a,b,c,d\}$ you have every possible ordered pair $(x,y)$ where $x,y\in\{a,b,c,d\}$,  i.e you have the full cross product and thus 16 ordered pairs relating those 4 elements.   Same thing with the 3 sets,  you have $3^2$ there for each one.  Add them up.


问:
Is there hope for Einstein tensor notation in Quantum Mechanics?

The Core Question

How would one define and justify the tensor notation for vectors and operators on a separable Hilbert space?

Motivation

Whenever I work with finite-dimensional linear algebra, I find it very illuminating to think about vectors, linear maps, bilinear forms etc. collectively as just various types of tensors. I like how the tensor formalism puts emphasis on representation-free formulations and how even a very complicated multi-linear map can be understood in terms of its individual „indices“.

When I started to learn Quantum Mechanics, I immediately tried to use the tensor notation. At a surface level it seemed to work just fine. One can formally rewrite the bra-ket expressions on left with a „tensor-like“ notation on the right:
$$
\begin{aligned}
  \left< \psi \mid \varphi \right>
  \qquad &\leftrightsquigarrow \qquad
  \overline{\psi^{\,\mu}} \; g_{\overline{\mu} \nu} \; \varphi^\nu
  \\
  \left| \psi \right> = \hat A \hat B \left| \varphi \right>
  \qquad &\leftrightsquigarrow \qquad
  \psi^{\,\mu} = A^\mu_\nu \; B^\nu_\kappa \; \varphi^\kappa
  \\
  W = \left| \psi \right>\left< \psi\right|, \;
  \operatorname{Tr} W = 1
  \qquad &\leftrightsquigarrow \qquad
  W^\mu_\nu = \psi^{\,\mu} \; g_{\overline{\kappa}\nu} \; \overline{\psi^\kappa}, \;\;
  W^\mu_\mu = 1
\end{aligned}
$$
Furthermore, the tensor notation feels more natural when dealing with composite systems, where the state is a tensor product of states of the underlying systems. However, when I tried to put the notation on a more rigorous ground, I learned that the topic is much more difficult and nuanced than I thought.

The Problem

In finite-dimensional spaces, the tensor notation is made possible by several circumstances:
\begin{gather}
V^* \otimes V \simeq \operatorname{Hom}(V, V)
\label{homomorphism} \tag{1} \\[5pt]
T(\mathbf{v}) = T( \; \sum_k v^k \mathbf{e}_k \;) = \sum_k v^k \; T(\mathbf{e}_k)
\label{continuity} \tag{2} \\
\big| \operatorname{Tr}(T) \, \big| < \infty
\label{trace} \tag{3}
\end{gather}
While \eqref{homomorphism} essentially makes sure that all $k$-covariant $l$-contravariant tensors are from the same space, \eqref{continuity} lets us describe all tensors using their coefficients wrt. some basis and \eqref{trace} lets us express all operations on tensors using just tensor product and contraction.

In a sense, neither of these is true for the separable Hilbert space $\mathcal H$ where QM is done. By a basis on a Hilbert space one usually means the Schauder basis, which describes vectors with an infinite series of coefficients. This, combined with the fact that most interesting operators in QM aren't continuous, gives us:
$$
  T(\mathbf{v})
  = T( \; \sum_{k=1}^\infty v^k \mathbf{e}_k \;)
  = T( \; \lim_{N\to\infty}\sum_{k=1}^N v^k \mathbf{e}_k \;)
  \neq \lim_{N\to\infty} T( \; \sum_{k=1}^N v^k \mathbf{e}_k \;)
  = \sum_{k=1}^\infty v^k \; T(\mathbf{e}_k)
$$
The sequence on the RHS might either diverge, or even converge to a different value than LHS (although this is pathological and usually not considered in practise). Condition \eqref{trace} only holds for the so-called trace-class operators (not even the identity is trace-class) and the space $\mathcal H^* \widehat{\otimes} \mathcal H$ (tensor product of Hilbert spaces) is isomorphic to finite-rank operators, a proper subset of $\operatorname{Hom}(\mathcal H, \mathcal H)$.

A Possible Solution? (and more questions)

These unfortunate circumstances mean, that if it's even possible to justify the tensor notation in infinite-dimensional spaces, it can't be by a simple generalization of the finite-dimensional case. So, is it possible to make it work?

The condition \eqref{continuity} seems relatively easy to fix – one just needs to replace the Schauder basis with a Hamel basis. A serious downside to this would be that a typical operator would have an uncountable number of coefficients.

For \eqref{homomorphism} it would be great if one could find a space $\mathcal G$, such that the Hilbert space $\mathcal H$ is embedded in it $\mathcal H \subset \mathcal G$, with the property $\mathcal G \otimes \mathcal G \simeq \operatorname{Hom}(\mathcal H^*, \mathcal H)$. Vectors of the bigger space could then be used to construct operators and bilinear forms. Here, one immediately thinks about the Gelfand triple $\Phi \subset \mathcal H \subset \Phi^*$ – could it be that $\mathcal G = \Phi^*$ for an appropriate choice of rigging? Sadly it appears that no, as the Schwartz kernel theorem says that $\Phi^* \otimes \Phi^* \simeq \operatorname{Hom}(\Phi, \Phi^*)$. Is such a space $\mathcal G$ even possible? If yes, would this approach generalize to tensors of higher degree?

Finally, the problem with \eqref{trace} seems to be there to stay. There is no reasonable way to define the trace of identity, for example. What I'm interested in is whether, given a complicated expression, one can tell a priori which indices are contractible and which will inevitably diverge.

Are these proposed “solutions” any good? Is there any literature that would investigate this topic? Or am I on the wrong path and should I stop wasting my time with tensors in the infinite dimension?

答:
In general, this sounds like a bad idea.

First, the notation thing. When your Hilbert space is $\mathbb R^n$ or $\mathbb C^n$, there is a clearly preferred basis, the canonical basis. The same is kind of true in some infinite-dimensional space, like $\ell^2(\mathbb N)$. But what would you do if your Hilbert space is presented as $L^2[0,1]$? In that case the usual inner product is
$$
\langle \psi\,|\,\varphi\rangle=\int_0^1 \varphi(t)\overline{\psi(t)}\,dt,
$$
which you propose to write as $\overline{\psi^{\,\mu}} \; g_{\overline{\mu} \nu} \; \varphi^\nu$; I cannot imagine a single advantage of doing so.

It is also bad when you want to write operators in terms of coordinates; it has been known for a very long time that thinking of operators on a Hilbert space as infinite matrices is not very fruitful.

Writing the trace in coordinates is not nice either, as it is not defined for all operators: for a non-trace-class operator, the infinite sum of its diagonal entries will usually differ with respect to different bases. The expression $W_\mu^\mu$ is defined every time you have a basis, but it will give you different values for different bases when $W$ is not trace-class. So $\operatorname{Tr}(W)$ is a much better notation, as it emphasizes that it can only be used where it can be reasonably defined.

I think that it is much easier to glance that $W=|\psi\rangle\,\langle\psi|$ is a rank-one operator from this notation than from $ W^\mu_\nu = \psi^{\,\mu} \; g_{\overline{\kappa}\nu} \; \overline{\psi^\kappa}$. And this is not me defending the bra-ket notation, which I think is bad and would be better replaced with $\psi^*\varphi$ instead of $\langle \psi\,|\,\varphi\rangle$ and $\psi\varphi^*$ instead of $|\psi\rangle\,\langle\varphi|$.

As for tensors, the Hilbert space tensor product $H^*\otimes H$ is canonically identified with the Hilbert-Schmidt operators. This is far less than the whole algebra $B(H)$ of bounded operators on $H$. If you want the norm of $H^*\otimes H$ to have any relation with that of $H$, you'll get that $H^*\otimes H$ is separable, while $B(H)$ isn't in the norm topology. This doesn't mean that it is impossible to somehow achieve what you want, but it is hard for me to see what could be achieved.

Your "fix" to your condition (2) is worse than a "serious downside". Basically what you want is to write things in such a way that continuity is invisible. In other words, you want to drop all topological considerations; for example, now you cannot say that if $\varphi_n\to\varphi$, then you have in coordinates $(\varphi_n)_\mu\to\varphi_\mu$. The "Schauder" bases that you mention, in a Hilbert space, are way more than that. They are "orthonormal bases". And, ironically since you want to avoid them, they are good precisely because they are the closest you can get to the finite-dimensional case. Orthonormal bases allow you to easily express the norm of a vector in terms of its coefficients. They allow you to easily define isomorphisms of Hilbert spaces: the unitary operators, which incidentally play a crucial role in QM. Dealing with orthonormal bases, unitaries are simply the operators such that their columns are orthonormal bases.

Just a few thoughts off the top of my head.


问:
Formulate the integer linear programming problem to maximize the expected aggregate action for the team of gymnastics
For future Olympics, the Gymnastics Federation can choose
between 6 contestants, and you need 3 of them to participate both in the balance beams
as in parallel bars. The Olympic committee requires that 4
people at each event. The following table shows the average ratings of
each gymnast and their coach assumed that their performance would be similar in the future
competition. Formulate the integer linear programming problem to choose the
gymnasts who participate in each exercise in such a way as to maximize the
expected aggregate action (sum of each gymnast's score) for the team.

Gymnasts
Balance beams
Bars

1
88
79

2
94
83

3
92
85

4
75
87

5
87
81

6
91
86

My try:

Maximize:  $$88x_1+79x_2+94x_3+83x_4+92x_5+85x_6+75x_7+87x_8+87x_8+87x_9+81x_{10}+91x_{11}+86x_{12}$$

restricted to: $$\sum_{i=1}^6x_{2i}=4$$
$$\sum_{i=1}^6x_{2i-1}=4$$

For $x_i\in\{0,1\}$ for all $i\in\{1,2,3,4,5,6\}$

I'm not sure how to simplify the restriction of having 3
gymnasts to participate both in the balance beams as in parallel bars. Any suggestions would be great!

答:
You need to set $x_i \le 1$ to finish off your LP formulation. Strictly you have an IP (integer programming) and depending on how you're solving the LP, you can usually specify that the values must be integers in the software. In any case this type of problem may well give you integer values for the $x_i$. If not, then you need to search for integer points near your optimal solution.

It worries me that you don't seem to have addressed the constraint that three must compete in both events...


问:
how to continue after this substitution?

Given $abc=1$, prove that $${\frac{a}{a^2+2}}+ {\frac{b}{b^2+2}}+{\frac{c}{c^2+2}} \leqslant 1 $$

I tried substitution  $a={\frac{1}{x}},b={\frac{1}{y}},c={\frac{1}{z}}$ but can't finish.

答:
Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ we have to prove

$$\sum_{cyc}\frac{xy}{x^2+2y^2}\le 1$$ Now as $x^2+y^2\ge 2xy$ it suffices to prove

$$\sum_{cyc}\frac{x}{2x+y}\le 1 $$ $$\iff \sum_{cyc} \frac{y}{2x+y}\ge 1$$ which is true as

$$\sum_{cyc} \frac{y}{2x+y}= \sum_{cyc} \frac{y^2}{2xy+y^2}\ge \frac{{(x+y+z)}^2}{2xy+y^2+2zx+x^2+2yz+z^2}=1$$


答: Conjugate $(12)$ by powers of $(1234)$ to get $(12), (23), (34)$. Then, you can conjugate further to get all transpositions.</p>


答:
The set $E$ contains all pairs $(x_1, x_2)$ such that $x_1 \sim x_2$ or in other words it contains all pairs for which the two elements of the pair are in the same equivalence class. An equivalence class with three elements, say $x,y,z$ will make the following 9 pairs appear in $E$:
$$(x,x), (x,y), (x,z), (y,x), (y,y), (y,z), (z,x), (z,y), (z,z)$$
The pattern here is that each combination of two elements of the equvialence class is an element of $E$ resulting in $3 \cdot 3 = 9$ elements in E. Similarly, an equivalence class with 4 elements will result in 16 more elements in the set $E$ (since we have all possible pairs of two elements again). In general, one can conclude from that that an equivalence class with $n$ elements will make $n^2$ elements "join" the set $E$.

Adding everything up in this special case, we will get:
$$|E| = 4\cdot 9 + 3 \cdot 16 = 84$$

I hope this helps you.

Title: Knowing number of elements in a relation from this information about equivalence classes.

Body:
I am trying to understand this question:

Let $A$ be a set with $24$ elements, and let $E$ be an equivalence relation on $A$.

$E$  divides $A$ to $7$ equivalence classes in the following way:

$3$ equivalence classes that has $4$ elements each.

$4$ equivalence classes that has $3$ elements each.

What is $|E|$ or how many elements are in the relation.

My Attempt:

I know that $[a]_E=\{b\in A \mid (a,b)\in E\}$ is the definition of equivalence class. so if I have $3$ equivalence classes with $4$ elements each that means I have $12$ pairs in $E$, but $E$ is a symmetric relation, so if $(a,b)\in E$ automatically we get $(b,a)\in E$ so $24$, and same thing for the other equivalence classes we get more $24$ pairs. but the answer was $84$.

After thinking about it, I thought about the reflexive property and we can add $24$ pairs more since for each element in $A$ we get $(a,a)\in E$, but thinking about it again, in every equivalence class, I have already counted some of these elements, because every element is in it's own equivalence class, so I'm wondering why don't we subtract $7$ from the answer.

All help is appreciated, Thanks in advance.


答:
The easiest way to do it would be to actually have three sets of variables:

$x_1, x_2, \dots, x_6$ to indicate that a contestant is participating in the first event only;

$y_1, y_2, \dots, y_6$ to indicate that a contestant is participating in the second event only;

$z_1, z_2, \dots, z_6$ to indicate that a contestant is participating in both events.

You'll need to modify your existing constraints to use $x_i + z_i$ to test if a contestant is participating in the first event, and $y_i + z_i$ for the second. To make sure that at most one of these is true for a contestant, you'll need to add the constraint $x_i + y_i + z_i \le 1$ for all $i$. Finally, the condition you had trouble with can now be written as $$z_1 + z_2 + z_3 + z_4 + z_5 + z_6 = 3.$$

Title: Formulate the integer linear programming problem to maximize the expected aggregate action for the team of gymnastics

Body:
For future Olympics, the Gymnastics Federation can choose
between 6 contestants, and you need 3 of them to participate both in the balance beams
as in parallel bars. The Olympic committee requires that 4
people at each event. The following table shows the average ratings of
each gymnast and their coach assumed that their performance would be similar in the future
competition. Formulate the integer linear programming problem to choose the
gymnasts who participate in each exercise in such a way as to maximize the
expected aggregate action (sum of each gymnast's score) for the team.

Gymnasts
Balance beams
Bars

1
88
79

2
94
83

3
92
85

4
75
87

5
87
81

6
91
86

My try:

Maximize:  $$88x_1+79x_2+94x_3+83x_4+92x_5+85x_6+75x_7+87x_8+87x_8+87x_9+81x_{10}+91x_{11}+86x_{12}$$

restricted to: $$\sum_{i=1}^6x_{2i}=4$$
$$\sum_{i=1}^6x_{2i-1}=4$$

For $x_i\in\{0,1\}$ for all $i\in\{1,2,3,4,5,6\}$

I'm not sure how to simplify the restriction of having 3
gymnasts to participate both in the balance beams as in parallel bars. Any suggestions would be great!


问:
The way to show equivalence of two objectives in Conjugate Gradient Methods
I am studying Numerical Optimisation and now it comes to Conjugate Gradient Methods

Intuitively, I can accept

"Solving $Ax=b$ is equivalent to solving $min_x~\phi(x)=\frac{1}{2}x^\top Ax-b^\top x$"

How to formally show that these two are equivalent?

Thank you very much

答:
Presumably, it is assumed in this context that $A$ is symmetric and positive definite.

Let $x_*$ denote a solution to $Ax = b$. Note that
$$
\begin{align}
x^TAx &= (x - x_* + x_*)^TA(x - x_* + x_*)
\\ &= (x-x_*)^TA(x-x_*) + x^TAx_* + x_*^TAx + x_*^TAx_*
\\ &= (x-x_*)^TA(x-x_*) + 2x_*^TAx + x_*^TAx_*
\\ &= (x-x_*)^TA(x-x_*) + 2(Ax_*)^Tx + x_*Ax_*
\\ &= (x-x_*)^TA(x-x_*) + 2b^Tx + x_*Ax_*.
\end{align}
$$
It follows that
$$
\begin{align}
\phi(x) &= \frac 12 (x-x_*)^TA(x-x_*) + b^Tx + \frac 12 x_*Ax_* - b^Tx
\\ & = \frac 12 (x - x_*)^TA(x - x_*) + \frac 12 x_*Ax_*.
\end{align}
$$
Note that $x_*Ax_*$ is constant. Because $A$ is positive semidefinite, we can therefore see from the above that $\phi(x)$ is minimized when $(x-x_*)^TA(x - x_*)$ attains its minimal value of $0$, and this minimum occurs when $x = x_*$.

In other words, $x_*$ (the solution to $Ax = b$) is the value at which the minimum is attained.


问:
Estimate the area under $f(x) = \frac{1}{2}x^2 - 1$ from $x = 0$ to $4$ using two right hand rectangles.
Help with beginner college calc plz:

I can't upload pictures on my computer (it's a very old desktop) but I'll try and explain what I did. So obviously I started with graphing f(x);

(2, 1)

(-2, 1)

(0, -1)

(3, 4)

(-3, 4)

Then I drew my rectangles. The first one's width was 0 to 1.5, and the 2nd's width was 1.5 to 3 (this is where my problem came in - I will explain). My height for the 1st rectangle was 2 and the 2nd was 4. To reiterate these are right hand rectangles.

Then I did l . w so the 1st rectangle's area would be 3 and the 2nd's 12 adding up to 15.

My question though is how would you draw your two rectangles? The estimate should be from 0 to 4, but my 2nd rectangle's upper right corner only touched 3,4.

Edit: Or did I graph my parabola completely wrong?

答:
Your graph should look like this:

Also note that $f(4)=7$ and $f(3)=\frac{1}{2}\cdot 9-1$.


问:
Mean and Variance of Normal Distribution with parameters
I see a special pdf function on textbook. I guess that it is the pdf of normal distribution but with a strange parameter t. Is my guessing right? And could someone explain why this pdf has a t? I hope to calculate the mean and variance to see whether it is normal distribution. I can remember that $\int_{-\infty}^{\infty}\frac{1}{\sqrt[]{2\pi}}exp(-\frac{x^2}{2})dx=1$. But I don't know how to work in the case with this t. Could someone give some hints about it? Thank you very much.
$$\frac{1}{\sqrt[]{2\pi\sigma^2t}\ }\exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$$

答: This is just a case of the usual normal distribution in which the mean and variance parameters are linked through a third parameter $t$.  The usual parametrization for a normal density is $$f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp \left(-\frac{(x-\mu)^2}{2\sigma^2}\right).$$  Here the mean is $\mu$ and the variance is $\sigma^2$.  Now in your case, the mean is $\mu t$ and the variance is $\sigma^2 t$.  Note we require $t > 0$.


问:
Another pigeonhole principle question
Have another question for you today:

A course has seven elective topics, and students must complete exactly three of them in order to pass the course. If 200 students passed the course, show that at least 6 of them must have completed the same electives as each other.

Now I know this is related to counting and the pigeonhole principle, and there are a couple of other related questions already asked but I couldn't apply them to my quuestion.

I know that the (informal) pigeonhole principle states that if you have $n$ boxes, and you have more than $n$ pigeons to distribute between those boxes, then at least one of the boxes will contain more than one pigeons, but I'm not sure what my boxes and what my pigeons are in this problem

答:
Note the generalized form of the pigeonhole principle: If more then mr elements are put in r classes then at least one class contains more then m elements. (You can prove it by contradiction, or just think about it intuitively. Also the normal pigeonhole principle follows by taking m=1.)

Now here we have r=35 classes (as $\tbinom{7}{3}=35$ ) and more then 35*5=175 students are being put in them. The result follows.(Let m=5)

Title: Another pigeonhole principle question


问:
A more logical approach to a word problem
I was taught in high school algebra to translate word problems into algebraic expressions. So when I encountered this problem I tried to reason out an algebra formula for it

For every dollar Ben spent on bagels, David spent 25 cents less. Ben
paid $12.50 more than David. How much did they spend in the bagel store
together?

To solve this I imagined a series of comparisons when Ben spends $x$, David spends $.75x$. Loop this relationship until $x - .75x \approx 12.50$. Good. Done. $x = 50$, then add David's for the answer. Coming from computers, I would have set this up in code where a loop (recursion) would increase $x$ until the condition $x - .75x = 12.50$ was met, then the "loop counter/accumulator" would be how much Ben spent, i.e., $50$, etc.

I'm a beginner with math, but it seems like there should be a better approach, something with series and sequences or even calculus derivatives, something better than my brute-force computer algorithm. Can someone enlighten? The "answer" given at the site (see link) is its own brute-force and hardly satisfying. I'm thinking there should be something more formal -- at least for the first part that derives $50$.

Update

I think everyone so far has missed my point. Many of you simply re-did the problem again. I'm wondering if there is a more formal way to do this other than just "figuring it out" (FIO). The whole FIO routine is murky. It looks like a limit problem; it looks like a system of equations, but I'm not experienced enough to know exactly. If there isn't, then let's call it a day....

答: From the step $x-.75x=12.50$ simplify to $.25x=12.5$,  divide by .25  to immediately get $x=50$.  No looping/brute force needed.


问:
Prove this equation involving Landau Notation
I started working through this work about number theory, but I've never really used Landau notation before so I am sometimes struggling to understand the steps used in the proofs. Specifically in Theorem $1.8$ in the last equation I believe it is asserted that
$$
 \int_Y^\infty\frac{u-\lfloor u \rfloor}{u^2}du = \mathcal{O}(\frac{1}{Y})
$$
How can I see this to be the case?

答: It's between $0$ and $\int_Y^\infty\tfrac{1}{u^2}du=\tfrac1Y$.


问:
examples of functions analytic only in strip (0,1) and with natural boundaries Re(z)=0,1?
Looking for examples of functions that are analytic only in the strip $(0,1)$ and with natural boundaries $\Re(z)=0,1.$

I thought of modular forms but they only have one natural boundary I think.

答: Take a known example with natural boundary.  Say $f(z) = \sum_{n=0}^\infty z^{2^n}$ with natural boundary $|z|=1$.  Then use a conformal mapping $\varphi$ that maps your strip onto $|z| < 1$.  Then $f(\varphi(z))$ is your answer.


问:
Linear Function and Newton's Method
I want to show that if a function is linear for example F(x) = 3x + 3, then after one iteration of Newton's method I can find the x-value such that F(x) = 0. In this case, after one iteration x = -1. I see that it works after one iteration, but how do I prove that??</p>

答:
Usual setup for Newton's method:

$$f'\left(x_0\right) = \lim_{x \rightarrow x_0} \frac{f\left(x\right) - f\left(x_0\right)}{x-x_0}$$

Look for intersections, i.e. $f\left(x\right)=0$:

$$x = x_0 - \frac{f\left(x_0\right)}{f'\left(x_0\right)}$$

Now stray from the general case and suppose $f\left(x\right)$ is linear, as in $f\left(x\right) = mx+b$, so $f'\left(x\right) = m$. Then,

$$x = x_0 - \frac{mx_0 + b}{m} = x_0 - x_0 - \frac{b}{m} = - \frac{b}{m}$$

... and you have just $-b/m$ after one iteration.

Title: Linear Function and Newton's Method